Oblique Derivative Problems for Elliptic Equations 9814452327, 9789814452328

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Oblique Derivative Problems for Elliptic Equations

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He Yue – ODPEE – a21/Chp. 12

Oblique Derivative Problems for Elliptic Equations Ω

Lu=f

β•Du=g

β

β

Gary M Lieberman Iowa State University, USA

World Scientific NEW JERSEY

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LONDON

8679hc_9789814452328_tp.indd 2

•

SINGAPORE

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BEIJING

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SHANGHAI

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HONG KONG

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TA I P E I

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CHENNAI

14/2/13 10:23 AM

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

OBLIQUE DERIVATIVE PROBLEMS FOR ELLIPTIC EQUATIONS Copyright © 2013 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

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ISBN 978-981-4452-32-8

Printed in Singapore.

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This book is dedicated to my wife Linda, our four grandchildren (Eliott, Leo, Thisbe, and Edie), and to the many people who helped make this book possible; in particular, Xing-Bin Pan, Chunquan Tang, and Hien Nguyen.

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Preface

Oblique derivative problems are those in which the boundary condition has the form b(x, u, Du) = 0 when b depends on the gradient Du of the unknown function u in a suitable way. The model boundary condition is the normal derivative condition γ · Du = f (x), where γ denotes the unit inner normal to the given domain Ω in which the boundary value problem is studied. When the boundary condition is linear, that is, b(x, u, Du) = β(x)·Du+β 0 (x)u+g(x), then the boundary condition is oblique if β ·γ > 0. For nonlinear boundary conditions, we write p as dummy variable for Du, and then the condition is oblique if β = ∂b/∂p satisfies p · γ > 0. When I first started to study oblique derivative problems over thirty years ago, I had a grand ambition: to put the theory of such problems in the same shape as the theory for the Dirichlet problem. It was clear even then that there was a major difference in attitudes towards the two problems in the eyes of most researchers in partial differential equations. The book [91] of Ladyzhenskaya and Ural0 tseva devotes nine chapters to the Dirichlet problem and then one chapter to “other boundary value problems”, and the book [64] of Gilbarg and Trudinger only devoted a few pages to oblique derivative problems. Fortunately, I was aware of some work on the topic: for example, the papers of Fiorenza [55, 54, 52, 53] suggested that there was some interest in the oblique derivative problem and the work of Concus and Finn [31, 32] indicated that there was even some practical interest in these problems. In writing this book, I started with a similarly grand ambition: to write a complete description of the theory of oblique derivative problems. With the books [91, 64] as models (and my own book [114]), I felt that such an ambition might be within reach. Unfortunately, the theory is not yet in the complete state that I had hoped, and the theory is more technical

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than that for the Dirichlet problem. Nevertheless, what I present here is a first attempt at delineating the ideas for oblique derivative problems in a systematic fashion. An important point at the heart of my choice of topics is to find elements of the theory which are not immediate consequences of the corresponding theory of the Dirichlet problem. Hence, I have made a conscious choice not to duplicate the content of [91, 64] as much as possible. Specifically, the proofs of properties of Sobolev spaces and of much of the measure theory are not given here; I only refer to [64] or [114] for the proofs. Furthermore, I have given alternative proofs of the classical Schauder theory in Chapter 2; my point is not that I consider the original proof using potential theory to be obsolete but rather that the modern proof given there allows us to derive the appropriate Schauder theory for oblique derivative problems in Chapter 4. For the most part, I have included proofs of results from real analysis and functional analysis whenever they are not in [64, 114] or in standard graduate texts on those topics. Hence, the Weierstrass approximation theorem is proved here because the form I use is not the one most familiar to modern audiences. On the other hand, I have chosen not to prove the Fredholm-Schauder-Riesz theorem (Theorem 2.29) because the proof is essentially contained in Section 5.3 of [64]. It is also important to recognize that the oblique derivative problem is best understood not in a vacuum but, rather, in conjunction with the theory of the Dirichlet problem. For this reason, I have abbreviated some of the proofs since they are very close to results from [91, 64]; any reader who has difficulty understanding the proofs here is recommended to peruse these books for additional assistance. Moreover, even though I don’t expect prospective readers to have a detailed knowledge of these books, it is extremely helpful to have some familiarity with their contents. I have provided quite a few detailed references to these books, too, at appropriate points in the text. There are also other books which describe the theory of oblique derivative problem in some detail: [151, 155, 205], but they are really interested in what is more properly called the tangential-oblique derivative problem. When they consider the boundary condition β(x) · Du = g(x) on ∂Ω where the vector β satisfies the condition β·γ ≥ 0, where γ is the unit inner normal to ∂Ω. For the problem to be oblique, we need to have β · γ > 0 everywhere, and the tangential-oblique derivative problem has several features not present in the oblique derivative problem (or the Dirichlet problem); the most important ones are that the solution need not be as smooth as

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the data would suggest and that the difference between the dimensions of the kernel and cokernel of the operator need not be zero (or even finite in more than two space dimensions). To keep this book from becoming even larger, I have not discussed a number of topics that are relevant to the general theory of oblique derivative problems. For example, the theory presented in this book has played a major role in the study of reflected shocks in transonic flow. The first application of the theory was made in [27] (for steady transonic shocks), which led to more papers than we care to mention here; MathSciNet lists 68 papers referring to [27]. Another important example is the capillary problem; although we prove some simple estimates for this problem, there is a vast theory which provides a complete description of the regularity of the solutions under various hypotheses. The book [46] by Finn gives a description of this theory as was known in 1986, and his more recent survey articles [49, 48, 47] present some more recent advances. In addition, Finn suggested in [50] that a slight variation of the historical capillary problem provides a more accurate model of the underlying physical situation, and the work in [51, 4] suggests that the theory in this book can be applied to this problem. I have essentially ignored the two-dimensional theory of oblique derivative problems. The basic theory is quite old and appears in [10, 52], and many of the results in this book can be proved much more easily (or under weaker hypotheses) when there are only two dimensions. For example, a good part of the theory in this book is easily proved in two dimensions in [105], which is the source of the ideas in Chapter 4. I also cannot resist mentioning one application of oblique derivative problems which affects my life outside mathematics: overtone series for musical instruments, specifically, three of the instruments that I play regularly. The overtone series of the clarinet differs from that of most other instruments (such as the flute and oboe) in that the even-numbered overtones are extremely faint, and hence the timbre of the clarinet is very different from those instruments. This difference in overtones comes from oblique boundary conditions, but there is no space here to justify this claim. Here, I thank all the people who have helped with this book. First and foremost are Xing-Bin Pan, who invited me (several times) to present lecture series on oblique derivative problems at East China Normal University, my Ph.D. student Chunquan Tang, who gave me new impetus to examine oblique derivative problem more closely and who helped proofread parts of this book, and Hien Nguyen, who helped with the proofreading. Of course,

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I take full responsibility for any errors in this book, and a website for errata will be available at lieb.public.iastate.edu/ODPBOOK/errata.pdf. I also thank the editors at World Scientific. In addition, I thank Iowa State University, East China Normal University, and the Centre for Mathematics and its Applications at the Australian National University for their support.

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Contents

Preface 1.

Pointwise Estimates 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12

2.

vii 1

The maximum principle . . . . . . . . . . . . . . . . . . The definition of obliqueness . . . . . . . . . . . . . . . The case c < 0, β 0 ≤ 0 . . . . . . . . . . . . . . . . . . . A generalized change of variables formula . . . . . . . . The Aleksandrov-Bakel0man-Pucci maximum principles The interior weak Harnack inequality . . . . . . . . . . . The weak Harnack inequality at the boundary . . . . . . The strong maximum principle and uniqueness . . . . . H¨ older continuity . . . . . . . . . . . . . . . . . . . . . . The local maximum principle . . . . . . . . . . . . . . . Pointwise estimates for solutions of mixed boundary value problems . . . . . . . . . . . . . . . . . . . . . . . Derivative bounds for solutions of elliptic equations . . .

. . . . . . . . . .

3 6 7 11 13 18 21 29 31 34

. .

35 38

Classical Schauder Theory from a Modern Perspective 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Definitions and properties of H¨older spaces . . . An alternative characterization of H¨older spaces An existence result . . . . . . . . . . . . . . . . . Basic interior estimates . . . . . . . . . . . . . . The Perron process for the Dirichlet problem . . A model mixed boundary value problem . . . . . Domains with curved boundary . . . . . . . . . . Fredholm-Riesz-Schauder theory . . . . . . . . . xi

47 . . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

47 58 60 61 65 71 81 83

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xii

3.

The Miller Barrier and Some Supersolutions for Oblique Derivative Problems 3.1 3.2 3.3 3.4 3.5 3.6

4.

Theory of ordinary differential equations . . . . . . . . The Miller barrier construction . . . . . . . . . . . . . Construction of supersolutions for Dirichlet data . . . Construction of a supersolution for oblique derivative problems . . . . . . . . . . . . . . . . . . . . . . . . . . The strong maximum principle, revisited . . . . . . . . A Miller barrier for mixed boundary value problems .

. . 89 . . 95 . . 102 . . 105 . . 109 . . 111 117

1,α

C estimates for continuous β . . . . . . . . . . . . . . Regularized distance . . . . . . . . . . . . . . . . . . . . Existence of solutions for continuous β . . . . . . . . . . H¨ older gradient estimates for the Dirichlet problem . . . C 1,α estimates with discontinuous β in two dimensions . C 1,α estimates for discontinuous β in higher dimensions C 2,α estimates . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

Weak Solutions 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14

6.

89

H¨ older Estimates for First and Second Derivatives 4.1 4.2 4.3 4.4 4.5 4.6 4.7

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Definitions and basic properties of weak derivatives . Sobolev imbedding theorems . . . . . . . . . . . . . Poincar´e’s inequality . . . . . . . . . . . . . . . . . . The weak maximum principle . . . . . . . . . . . . . Trace theorems . . . . . . . . . . . . . . . . . . . . . Existence of weak solutions . . . . . . . . . . . . . . Higher regularity of solutions . . . . . . . . . . . . . Global boundedness of weak solutions . . . . . . . . The local maximum principle . . . . . . . . . . . . . The DeGiorgi class . . . . . . . . . . . . . . . . . . . Membership of supersolutions in the De Giorgi class Consequences of the local estimates . . . . . . . . . . Integral characterizations of H¨older spaces . . . . . . Schauder estimates . . . . . . . . . . . . . . . . . . .

Strong Solutions 6.1 6.2

117 129 131 132 136 159 165 171

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

173 174 177 180 182 187 190 193 201 203 208 210 211 214 227

Pointwise estimates for strong solutions . . . . . . . . . . 227 A sharp trace theorem . . . . . . . . . . . . . . . . . . . . 231

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6.3 6.4 6.5 6.6 6.7 6.8 6.9

7.

7.4 7.5 7.6 7.7

. 235 . 238 . 241 . 246 . 249 . 251 . 258 265

Definitions and notation . . . . . . . . . . . . . . . . . The Theorem of Aleksandrov . . . . . . . . . . . . . . Preliminary results for the comparison theorem for viscosity solutions . . . . . . . . . . . . . . . . . . . . The comparison principle for viscosity sub- and supersolutions . . . . . . . . . . . . . . . . . . . . . . . A test function construction for the oblique derivative problem . . . . . . . . . . . . . . . . . . . . . . . . . . The comparison principle for oblique derivative problems . . . . . . . . . . . . . . . . . . . . . . . . . . Existence and uniqueness of viscosity solutions . . . .

. . 265 . . 266 . . 274 . . 283 . . 283 . . 291 . . 295

Pointwise Bounds for Solutions of Problems with Quasilinear Equations 8.1 8.2 8.3 8.4

9.

Results from harmonic analysis . . . . . . . . . . . . . . Some further estimates for boundary value problems in a spherical cap . . . . . . . . . . . . . . . . . . . . . . . Lp estimates for solutions of constant coefficient problems in a spherical cap . . . . . . . . . . . . . . . . Local estimates for strong solutions of constant coefficient problems . . . . . . . . . . . . . . . . . . . . . Local interior Lp estimates for the second derivatives of strong solutions of differential equations . . . . . . . . Local Lp second derivative estimates near the boundary . . . . . . . . . . . . . . . . . . . . . . . . . . Existence of strong solutions for the oblique derivative problem . . . . . . . . . . . . . . . . . . . . . . . . . . .

Viscosity Solutions of Oblique Derivative Problems 7.1 7.2 7.3

8.

xiii

Maximum estimates H¨ older estimates for Maximum estimates H¨ older estimates for

for nondivergence equations nondivergence equations . . for conormal problems . . . conormal problems . . . . .

Gradient Estimates for General Form Oblique Derivative Problems 9.1 9.2

301 . . . .

. . . .

. . . .

. . . .

. . . .

301 305 307 313

321

Interior gradient bounds . . . . . . . . . . . . . . . . . . . 321 A simple boundary value problem . . . . . . . . . . . . . 329

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xiv

9.3 9.4 9.5 9.6 9.7

Gradient estimates for general boundary conditions. General considerations . . . . . . . . . . . . . . . . . Global gradient estimates for general boundary conditions and false mean curvature equations I . . . Global gradient estimates for general boundary conditions and false mean curvature equations II . . Local gradient estimates . . . . . . . . . . . . . . . . Gradient estimates for capillary-type problems . . .

. . . 330 . . . 333 . . . 340 . . . 346 . . . 352

10. Gradient Estimates for the Conormal Derivative Problems 10.1 10.2 10.3 10.4

The Sobolev inequality of Michael and Simon The interior gradient bound . . . . . . . . . . Preliminaries for estimates . . . . . . . . . . . Gradient bounds for the conormal problem .

. . . .

. . . .

. . . .

365 . . . .

. . . .

11. Higher Order Estimates and Existence of Solutions for Quasilinear Oblique Derivative Problems 11.1 11.2 11.3 11.4 11.5

11.6 11.7 11.8 11.9

The H¨ older gradient estimate for conormal problems . A solvability theorem . . . . . . . . . . . . . . . . . . . Existence results and estimates for linear equations and nonlinear boundary conditions in spherical caps . Estimates and existence results for linear equations and nonlinear boundary conditions in general domains Mixed boundary value problems for simple quasilinear differential equations and nonlinear boundary conditions in spherical caps . . . . . . . . . . . . . . . H¨ older gradient estimates for quasilinear equations . . A basic existence theorem for quasilinear elliptic equations with nonlinear boundary conditions . . . . . Second derivative H¨older estimates . . . . . . . . . . . Existence theorems for our examples . . . . . . . . . .

12. Oblique Derivative Problems for Fully Nonlinear Elliptic Equations 12.1 12.2

. . . .

. . . .

365 368 382 392

407 . . 407 . . 410 . . 412 . . 424

. . 426 . . 437 . . 443 . . 444 . . 447

457

Maximum estimates, comparison principles, and a uniqueness theorem . . . . . . . . . . . . . . . . . . . . . . 459 Second derivative H¨older estimates . . . . . . . . . . . . . 462

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Second derivative H¨older estimates for solutions of oblique derivative problems . . . . . . . . . . . . . . . . . 471 Uniformly elliptic fully nonlinear problems . . . . . . . . . 482

Bibliography

493

Index

507

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Chapter 1

Pointwise Estimates

Introduction This chapter examines some basic estimates for solutions of linear oblique derivative problems, although some of the techniques used are quite sophisticated. The heart of all the estimates in this chapter (and, by extension, most of the estimates in the entire book) is the maximum principle, which is quite straightforward for the Dirichlet problem but not so for the oblique derivative problem. To illustrate the differences between these two problems, we look first at a simple one-dimensional example. If u00 ≥ 0 on the interval (0, 1), then u attains its maximum at one of the endpoints, so prescription of the values for u at these points leads to a simple estimate of the maximum over the interval. If we prescribe values for the first derivative of u at the endpoints, then we get no information about the maximum value for u because we can add an arbitrary constant (positive or negative in this case) to u without changing the differential equation or boundary conditions. The control of the maximum in this case is achieved via one of three modifications, and we shall see instances of the higher-dimensional analog of each one at various points within the text. First, we can modify the boundary conditions to read u0 (0) + au(0) ≥ −A and −u0 (1) + bu(1) ≥ −B for negative constants a and b, and nonnegative constants A and B. (The sign conventions here are designed to match our normalization of the boundary condition in higher dimensions.) It follows that the maximum of u is bounded by max{A/|a|, B/|b|}. Second, we can modify the boundary conditions to u0 (0) ≥ −A and u(1) ≤ B for nonnegative constants A and B. (The case u(0) ≤ B and u0 (1) ≤ A is handled similarly.) In this case, the maximum of u is no greater than B + A. Our third modification is to the differential equation; if we assume that u00 + cu ≥ 0 for some negative 1

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constant c and that u0 (0) ≥ −A and u0 (1) ≤ A for some nonnegative constant A, then it is possible to estimate the maximum of u in terms of A and c. Since the details are not immediately obvious, we defer a discussion of this case to Section 1.3 where the higher-dimensional version is examined. In fact, there is another important point for the oblique derivative problem which does not cause major problems for the Dirichlet problem. The classical definition of the oblique derivative condition is that some directional derivative of the solution is prescribed on the boundary under the assumption that the direction is nowhere tangential. If the boundary of the domain is sufficiently smooth, then this definition makes sense; however, in many applications, the domain is not assumed smooth. Such is the case if we look at the Neumann problem for harmonic functions in a square, and it will appear in many other contexts as well. For this reason, we devote Section 1.2 to a definition of obliqueness appropriate to more general, usually Lipschitz, domains. Although it is relatively simple to develop a number of pointwise estimates for solutions by the classical maximum principle (which estimates the solution in terms of simple properties of the coefficients of the differential equation and boundary condition and the maximum of the inhomogeneous term in the differential equation and boundary condition), there are some significant results which allow us to study a large variety of nonlinear equations but for which the classical maximum principle is inadequate. A particular instance of such a result is a modulus of continuity estimate for the solution in terms of bounds of the coefficients in the differential equation and boundary condition. For oblique derivative problems, the appropriate starting point is a maximum principle, first presented by Bakel0 man [5], Aleksandrov [2], and Pucci [157] for the Dirichlet problem in the 1960’s, which estimates the maximum of the solution of the boundary value in terms of the Ln norm of the right hand side of the differential equation, where n is the number of spatial dimensions in the problem. We present several forms of this estimate which fit the oblique derivative context very well, and we prove several important estimates from this maximum principle. We also prove a simple gradient estimate for solutions of equations with constant coefficients in Section 1.12.

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Pointwise estimates

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3

The maximum principle

The most fundamental estimate for solutions of elliptic equations is an estimate on the maximum of the solution in terms of data. We shall provide several different estimates based on different information. Our first estimate is a simple one which will be generalized very quickly. To apply it to oblique derivative problems, we need to introduce a basic concept that will be used repeatedly in this book. Let Ω be an open subset of Rn with boundary ∂Ω and let β be a vector field defined on ∂Ω. We say that a vector β is inward pointing at x ∈ ∂Ω if there is a positive number h0 such that x + hβ ∈ Ω for any h ∈ (0, h0 ). We also write β · Du(x) ≤ g if lim sup h→0+

u(x + hβ) − u(x) ≤ g. h

If β is a vector field defined on some subset S of ∂Ω such that β(x) is inward pointing at x for each x ∈ S, then we say that β is inward pointing on S and we write β · Du ≤ g on S (for some function g defined on S) if β · Du(x) ≤ g(x) for all x ∈ S. In addition, we define β · Du(x) ≥ g to mean that lim inf h→0+

u(x + hβ) − u(x) ≥ g, h

and we write β · Du(x) = g if β · Du(x) ≤ g and β · Du ≥ g. Note that, if u is differentiable at a point x ∈ ∂Ω, then for any inward pointing vector field β, our definitions of β · Du(x) ≤ g, β · Du(x) ≥ g, and β · Du(x) = g agree with the inequalities (or equality) satisfied by the expressions involving the gradient of u. More precisely, our inequalities on β · Du are equivalent to inequalities on the one-sided derivate of u in the direction parallel to β and any equation involving β · Du provides (at least for nonzero β) a prescription of the directional derivative of u in the direction of β. The most important fact we use is that, if u has a maximum at x ∈ ∂Ω and β is an inward pointing vector at x, then β · Du(x) ≤ 0. In addition, we introduce two operators to simplify our notation. Let [aij ] be a positive definite matrix-valued function defined on some bounded open set Ω, let b = (b1 , . . . , bn ) be a vector-valued function defined on Ω, and let c be a function defined on Ω. We then write L for the operator defined on C 2 (Ω) by Lu = aij Dij u + bi Di u + cu.

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4

If β is a vector field defined on ∂Ω and β 0 is a scalar field defined on ∂Ω, we define the operator M by M u = β · Du + β 0 u.

If u is continuous on Ω, we take the inequality M u ≥ g on some subset S of ∂Ω to mean β(x) · Du(x) ≥ −β 0 (x)u(x) + g(x)

for all x ∈ S and we define M u ≤ g or M u = g similarly. Unless otherwise stated, we always assume that the functions [aij ], b, c, β, and β 0 are given, subject to any explicitly stated hypotheses. Lemma 1.1. Suppose [aij ] is positive semidefinite matrix and β is inward pointing on ∂Ω. Suppose also that c < 0 in Ω and β 0 < 0 on ∂Ω. If u ∈ C 2 (Ω) ∩ C 0 (Ω) satisfies then u ≤ 0 in Ω.

Lu ≥ 0 in Ω,

M u ≥ 0 in ∂Ω,

(1.1)

Proof. Let x0 be a point at which u attains its maximum value. If x0 ∈ Ω, then aij Dij u(x0 ) ≤ 0 and bi Di u(x0 ) = 0, so cu(x0 ) ≥ 0. Hence u(x0 ) ≤ 0. On the other hand, if x0 ∈ ∂Ω, then β · Du ≤ 0, so β 0 u(x0 ) ≥ 0 and hence u(x0 ) ≤ 0 again. Since u(x0 ) ≤ 0, it follows that u ≤ 0 in Ω. 

The restrictions c < 0 and β 0 < 0 can be relaxed somewhat but it is important to recognize that they cannot be eliminated completely. If c ≡ 0 and β 0 ≡ 0, then any constant will satisfy the inequalities in (1.1). Moreover, the case β ≡ 0 (which is an inward pointing vector field) gives a maximum principle for the Dirichlet problem. We shall take advantage of this observation several times. Our next step is a variant on our original maximum principle in which c ≤ 0 and β 0 < 0. In this case, we need to make some assumptions about the relationship between [aij ] and b.

Lemma 1.2. Suppose that there is some nonnegative function λ such that n

aij ξi ξj ≥ λ|ξ|2

(1.2)

for all ξ ∈ R and suppose that, for any compact subset K of Ω, there is a nonnegative constant µ1 (K) such that |b| ≤ µ1 λ on K. Suppose also that c ≤ 0 in Ω, β is inward pointing vector field on ∂Ω, and β 0 < 0. If (1.1) holds, then u ≤ 0 in Ω.

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Pointwise estimates

5

Proof. Fix ε > 0 and let x0 ∈ Ω. Setting M0 = sup∂Ω u, we see that there is a compact set K ⊂ Ω such that x0 is an interior point of K and u ≤ M0 + ε on ∂K. We now set γ = 1 + µ1 (K), and we define the function ϕ by ϕ(x) = exp(−γx1 ). A simple calculation yields aij Dij ϕ + bi Di ϕ = (a11 γ 2 − b1 γ) exp(−γx1 ) ≥ λ(γ 2 − µ1 γ) exp(−γx1 ) > 0.

Now let Ω+ denote the subset of K on which u > 0, and suppose u + εϕ has a positive maximum at some x1 ∈ Ω+ . If x1 ∈ Ω+ , then we have aij Dij (u + εϕ) + bi Di (u + εϕ) = aij Dij (u + εϕ) ≤ 0

at x1 . However, in Ω+ , we have

aij Dij (u + εϕ) + bi Di (u + εϕ) ≥ Lu + ε(aij Dij ϕ + bi Di ϕ) > 0. It follows that u + εϕ can’t have an interior positive maximum, so u(x0 ) + εϕ(x0 ) ≤ sup u+ + εϕ. ∂K

From our choice of ϕ, it follows that u(x0 ) ≤ M0 + 2ε, and hence sup u ≤ sup u+ . Ω

∂Ω

In other words, if u has a positive maximum, it must occur on ∂Ω, say at some x2 ∈ ∂Ω. Then β · Du ≤ 0 at x2 and the boundary condition implies that β · Du ≥ −β 0 u at x2 . Hence u(x2 ) ≤ 0. It follows that u cannot attain a positive maximum and hence u ≤ 0.  An analogous result is true if c < 0 and β 0 ≤ 0 but several new ideas are needed. The most important one is our definition of obliqueness of a vector field, and that definition is the topic of the next section.

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The definition of obliqueness

We now turn to the definition of oblique vector fields, which will be a special case of inward pointing vector fields. Our description of oblique vector fields starts with the same scheme as for inward pointing vector fields. We say that a vector β is oblique at a point x ∈ ∂Ω if there is a cone of finite height, vertex at x and axis parallel to β which lies inside Ω. We also say that a vector field β defined on a subset S of ∂Ω is oblique on S if β(x) is oblique at x for every x ∈ S. Note that every oblique vector field on S is inward pointing on S but an inward pointing vector field need not be oblique. Two special situations are worth noting. First, if Ω has an outward pointing cusp at x ∈ ∂Ω, then the axis of the cusp is inward pointing but not oblique. Second, if β is tangential to ∂Ω, it will be inward pointing but not oblique. In particular, if Ω = {x = (x1 , x2 ) : |x| < 1, x2 > 0}, then the vector (1, 0) is inward pointing but not oblique at (0, 0). For many of our results, a quantitative measure of obliqueness is useful. For ε ∈ (0, 1), we say that a vector field β defined on a subset S of ∂Ω has modulus of obliqueness at least ε on S if there are positive constants R and ω0 , and, for each s ∈ S, a function ω : Bn−1 (0, R) and a coordinate system (y 0 , y n ) = (y 1 , . . . , y n ) centered at s such that Ω ∩ B(s, R) = {(y 0 , y n ) : y n > ω(y 0 ), |y| < R} |ω(y10 )

− 0

ω(y20 )|

≤

ω0 |y10 − n

y20 |,

ω0 |β (x)| ≤ (1 − ε)β (x)

(1.3a) (1.3b) (1.3c)

for all x ∈ S with |x − s| < R. (Here β n is the component of β in the y n direction, and β 0 is the projection β onto the y 0 hyperplane.) If we don’t wish to emphasize the number ε, we say only that β is uniformly oblique on S. We say that ∂Ω ∈ C 1 if there is a continuously differentiable function g : Rn → R such that Ω = {x ∈ Rn : g(x) > 0} and Dg is nowhere zero on ∂Ω. Our definition of obliqueness on S ⊂ ∂Ω in this case is equivalent to the condition that inf S β · γ/|β| > 0, where γ is the unit inner normal to ∂Ω (so γ = Dg/|Dg|). One of the main points in this book is that the more general definition, which allows domains which are not C 1 , still leads to a rich and useful theory.

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1.3

7

The case c < 0, β 0 ≤ 0

We now consider the case c < 0 and β 0 ≤ 0 in proving a maximum principle. An important point for the maximum principle is that, under our general definition, there is a smooth function ρ on Ω such that β · Dρ > 0 on ∂Ω. ( Such a function essentially plays the role of g in the case of ∂Ω ∈ C 1 .) Such a result can be proved once we have two simple results concerning continuous functions. The first result is a simple extension theorem. To state the theorem, we recall that a modulus of continuity is an increasing function ω : [0, ∞) → [0, ∞) such that ω(s + t) ≤ ω(s) + ω(t) for all s, t > 0 and ω(0) = 0. Lemma 1.3. Let S ⊂ Rn and let f : S → R be bounded from below. Suppose that there is a modulus of continuity µf such that |f (s1 ) − f (s2 )| ≤ µf (|s1 − s2 |) (1.4) n ˆ for all s1 and s2 in S. Then there is a function f : R → R such that |fˆ(x1 ) − fˆ(x2 )| ≤ µf (|x1 − x2 |) (1.5a) for all x1 and x2 in Rn , and fˆ(s) = f (s)

(1.5b)

for all s ∈ S. In addition, if there are constants m and M such that m ≤ f ≤ M on S, then fˆ can be chosen so that m ≤ fˆ ≤ M on Rn . Proof.

We define g : Rn → R by

g(x) = inf {f (y) + µf (|x − y|)}. y∈S

Now let x1 and x2 be points in Rn . For any ε > 0, there is a point y ∈ S such that g(x2 ) ≥ f (y) + µf (|x2 − y|) − ε. Then we have g(x1 ) − g(x2 ) ≤ f (y) + µf (|x1 − y|) − f (y) − µf (|x2 − y|) + ε = µf (|x1 − y|) − µf (|x2 − y|) + ε

≤ µf (|x1 − x2 |) + ε.

Since ε is arbitrary, we conclude that g(x1 ) − g(x2 ) ≤ µf (|x1 − x2 |) and, by reversing the roles of x1 and x2 , we conclude that |g(x1 ) − g(x2 )| ≤ µf (|x1 − x2 |).

In addition, if s ∈ S, then it’s easy to see that g(s) ≤ f (s). On the other hand, for any y ∈ S, we have f (s) ≤ f (y) + µf (|s − y|),

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so g(s) = f (s). For the first part of the lemma, we take fˆ = g. For the second part of the lemma, we first observe that, if f ≥ m on S, then g ≥ m on Rn . The proof is completed by taking fˆ(x) = min{M, g(x)}.



Our next step is the Weierstrass Approximation Theorem. Lemma 1.4. Let f be a uniformly continuous function defined on some bounded set K and suppose there is a modulus of continuity µf such that |f (x) − f (y)| ≤ µf (|x − y|) for all x and y in K. Then there is a sequence (Pm ) of polynomials which converges uniformly to f on K and such that |Pm (x) − Pm (y)| ≤ µf (|x − y|) for all x and y in K. Proof. First, we note that there is a positive number R such that |x| ≤ R for all x ∈ K. Then we define f¯ = f − inf f on K and we set f¯(x) = 0 if |x| ≥ 2R. Note that f¯ satisfies the inequality |f¯(x) − f¯(y)| ≤ µf (|x − y|) for all x and y in the domain of f¯. We now extend f¯ to a function fˆ defined on all of Rn by using Lemma 1.3. For our next step, we define the constants am , for m a nonnegative integer, by Z am = [(4R)2 − |x|2 ]m dx, B(0,4R)

and we define ϕm by ϕm (x) =

[(4R)2 − |x|2 ]m . am

Our polynomials Pm are then given by the equation Z Pm (x) = ϕm (y)fˆ(x − y) dy + inf f. Rn

We note that Pm is well-defined because f¯ has compact support. It follows from a change of variables that Z Pm (x) = − ϕm (x − y)fˆ(y) dy + inf f, Rn

and hence Pm is a polynomial.

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Furthermore, if x1 and x2 are in K, then |x1 −y| ≥ 2R and |x2 −y| ≥ 2R if |y| ≥ 4R, so Z |Pm (x1 ) − Pm (x2 )| = | ϕm (y)(fˆ(x1 − y) − fˆ(x2 − y)) dy| Rn Z =| ϕm (y)(fˆ(x1 − y) − fˆ(x2 − y)) dy| B(0,4R) Z ≤ ϕm (y)µf (|x1 − x2 |) dy = µf (|x1 − x2 |). B(0,4R)

To see that (Pm ) converges uniformly to f on K, we examine the difference Z Dm (x) = fˆ(x) − ϕm (y)fˆ(x − y) dy Rn

for |x| ≤ R. We first observe that Z Dm (x) = ϕm (y)[f¯(x) − f¯(x − y)] dy. B(0,4R)

With δ ∈ (0, 4) to be chosen, we now rewrite Dm (x) = I1,m (x) + I2,m (x) with I1,m (x) =

Z

B(0,δR)

I2,m (x) =

ϕm (y)[f¯(x) − f¯(x − y)] dy,

Z

B(0,4R)\B(0,δR)

ϕm (y)[f¯(x) − f¯(x − y)] dy.

Then |I1,m (x)| ≤

Z

B(0,δR)

because δ ≤ 4, and |I2,m (x)| ≤ 2 sup f¯

ϕm (y)µf (δR) dy ≤ µf (δR) Z

ϕm (y) dy.

B(0,4R)\B(0,δR)

We now observe that (1 − s2 )m =0 m→∞ (1 − (s/4)2 )m lim

for any s ∈ (0, 1). It follows that supB(0,4R)\B(0,δR) ϕm lim = 0, m→∞ inf B(0,δR/4) ϕm

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and hence lim

m→∞

so lim

m→∞

R

B(0,4R)\B(0,δR)

R

B(0,δR)

Z

ϕm (y) dy

ϕm (y) dy

= 0,

ϕm (y) dy = 0.

B(0,4R)\B(0,δR)

Hence, if ε > 0 is given, we can find first δ sufficiently small so that |I1,m (x)| ≤ ε/2 for all x with |x| ≤ R and then m sufficiently large so that |I2,m (x)| ≤ ε/2. Hence |Dm (x)| ≤ ε if m is sufficiently large, so (Pm ) converges uniformly to f on K.  We can now prove the existence of the function ρ described earlier in this section. Lemma 1.5. Let ε ∈ (0, 1) be given and let Ω be a domain which possesses a vector field with modulus of obliqueness at least ε. Then there is a function ρ ∈ C ∞ (Ω) such that β · Dρ > |β| for any vector field β with modulus of obliqueness at least ε. Proof. With R as in the definition of the modulus of obliqueness, we cover ∂Ω by finitely many balls B1 , . . . , BN with radius R centered on ∂Ω, and we let (ψj )N j=1 be a partition of unity on ∂Ω subordinate to this cover. In each ball Bj , we use a coordinate system Y (centered at the center of Bj ) so that Ω ∩ Bj = {y : y n > ωj (y 0 ), |y| < R} for some Lipschitz function ωj . Then, with η > 0 to be chosen, the Weierstrass Approximation Theorem provides a function ωj,η ∈ C ∞ (Bn−1 (0, R)) such that |ωj (y 0 ) − ωj,η (y 0 )| < η and |ωj,η (x0 ) − ωj,η (y 0 )| < ω0 |x0 − y 0 | for all x0 and y 0 in Bn−1 (0, R). We define wj on Bj by wj (x) = y n − ωj,η (y 0 ). Next, we set κ = 1/(1 + ω02 )1/2 and note that β n ≥ κ|β|, using the y P coordinates in any ball Bj . It follows that w = ψj wj satisfies X X β · Dw = ψj β · Dwj + wj β · Dψj X ≥ [εκ − η |Dψj |]|β|. Since the cover is finite, we choose εκ P η= 2 sup |Dψj | to infer that β · Dw ≥ 21 εκ|β| on ∂Ω. Hence the desired function is 2 ρ= w.  εκ

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With these preliminaries, the maximum principle follows easily. Lemma 1.6. Suppose that there is a constant C2 such that |aij | + |bi | < −C2 c in Ω. Suppose also that β is uniformly oblique and that β 0 ≤ 0 on ∂Ω. If u satisfies (1.1), then u ≤ 0 in Ω. Proof. Since β is uniformly oblique, there is ε ∈ (0, 1) such that β has modulus of obliqueness less than ε. For any K ∈ R, we set wK = K −ρ with ρ the function from Lemma 1.5. If K ≥ sup ρ, then β · DwK + β 0 wK < 0 on ∂Ω and, if K is sufficiently large (determined only by C2 and sup |D2 ρ| + |Dρ| + |ρ|), we also have aij Dij wK + bi Di wK + cwK = −aij Dij ρ − bi Di ρ + c(K − ρ) < 0.

As before, it follows that u − δwK can’t have a positive maximum for any δ > 0 and hence u ≤ 0.  1.4

A generalized change of variables formula

For our next collection of maximum principles, we need an expression which generalizes the classical change of variables formula. Since the result is not generally known in the United States, we provide a simple proof. In fact, a more precise version of this theorem can be proved by similar methods (see the Notes for details), but we give a result which suffices for our purposes. Theorem 1.7. Let Ω be an open subset of Rn and let F : Ω → Rn be a C 1 map. Then Z Z u ◦ F (x)| det DF (x)| dx ≥ u(y) dy (1.6) Ω

F (Ω)

for any nonnegative measurable function u. Proof. First, we use Sard’s Theorem to observe that the set of y such that F (x) = y for some x ∈ Ω with det DF (x) = 0 has measure zero. Hence, if we write Ω∗ for the subset of Ω on which det DF 6= 0, we have Z Z u ◦ F (x)| det DF (x)| dx = u ◦ F (x)| det DF (x)| dx, (1.7a) ∗ Ω Z ZΩ u(y) dy = u(y) dy. (1.7b) F (Ω)

F (Ω∗ )

Suppose now that u is bounded above. Now let D be a compact subset of Ω∗ . Then, by the Inverse Function Theorem, for each y ∈ D, there is an

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open subset Uy of Ω∗ such that F is a homeomorphism from Uy to F (Uy ) and y ∈ Uy . Since D is compact, we can cover D by finitely many of these open sets, which we label as U1 , . . . , UM for convenience. It follows that there is a positive constant δ such that any subset of D with diameter less than δ is contained in some Uj . We now fix ζ > 0 and take a countable collection {Dj } of open subsets of Ω∗ with diameter δ/2 such that D ⊂ ∪Dj P and |Dj | ≤ |D| + ζ. The classical change of variables formula implies that Z Z u ◦ F (x)| det DF (x)| dx = u(y) dy, Dj

F (Dj )

and hence XZ

Dj

u ◦ F (x)| det DF (x)| dx ≥

Since F (D) ⊂ ∪F (Dj ), it follows that Z XZ u(y) dy ≥ F (Dj )

XZ

u(y) dy.

F (Dj )

u(y) dy.

F (D)

In addition there is a constant K such that u ◦ F | det(DF )| ≤ K on ∪Dj , and this constant can be chosen independent of ζ. It follows that Z Z u ◦ F (x)| det DF (x)| dx + Kζ ≥ u(y) dy. D

F (D)

Since this inequality is true for all ζ > 0, we conclude that Z Z u ◦ F (x)| det DF (x)| dx ≥ u(y) dy. D

F (D)

Next, we note that there are an increasing sequence of bounded measurable functions (uj ) with uj → u a.e. and an increasing sequence of compact sets (Dj ) such that ∪Dj = Ω∗ , so the Monotone Convergence Theorem gives Z Z u ◦ F (x)| det DF (x)| dx ≥ u(y) dy Ω∗

F (Ω∗ )

for any nonnegative measure u. The proof is completed by combining this inequality with (1.7). 

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1.5

13

The Aleksandrov-Bakel0 man-Pucci maximum principles

There are times when the maximum principles described so far are inadequate for important estimates. In this section, we present results in which the maximum of the solution is estimated in terms of some integral norm of data. For these estimates, we introduce an important set and an important function. If u is a continuous function defined on some open set Ω, we write Γ(u) for the upper contact set of u, that is, the set of all x ∈ Ω such that there is a vector p with u(y) ≤ u(x) + p · (x − y)

(1.8)

for all y ∈ Ω. In addition, we write χu for the normal mapping, that is, χu (x) is the set of all vectors p such that (1.8) holds for all y ∈ Ω. In particular, χu (x) is non-empty if and only if x ∈ Γ(u). As a first step, we present a simple version of the main result. Lemma 1.8. Let Ω be a bounded open set, let β be inward pointing on ∂Ω, and let β 0 < 0 on ∂Ω. Suppose that there is a constant B0 such that |β| ≤ −B0 β 0

(1.9)

on ∂Ω. Let h be a positive function in L1loc (Rn ) and define H by Z H(t) = h(p) dp.

(1.10)

{|p| 0, there is a vector p with |p| < R + ε and p∈ / Γ+ (u). Hence (1.13) gives sup u ≤ g + (B0 + diam Ω)(R + ε), and hence sup u ≤ g + (B0 + diam Ω)R. Combining this inequality with (1.14) and (1.15) yields (1.11).



We point out that, if h ∈ L1 (Rn ), then (1.11) only provides a bound for the supremum if Z Z h(Du)| det D2 u| dx < h(p) dp. Γ+ (u)

Rn

To continue, we study the corresponding estimate when u satisfies a linear differential inequality in Ω. Theorem 1.9. Suppose [aij ] is positive definite and set D = det[aij ]. Suppose c ≤ 0 in Ω. Suppose β is inward pointing and β 0 < 0 and suppose that there is a nonnegative constant B0 such that (1.9) holds. Suppose also that there is a nonnegative constant B1 such that Z |b|n dx ≤ B1 . (1.16) Ω D

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If u ∈ C 2 (Ω) ∩ C(Ω) and if

Lu ≥ −f in Ω,

15

M u ≥ gβ 0 on ∂Ω

(1.17)

for some nonnegative function f and some nonnegative constant g, then

f

u ≤ g + C(n, B1 )(B0 + diam Ω) 1/n , (1.18) D n + L (Γ (u))

where Γ+ (u) is the subset of Γ(u) on which u > g.

With µ a constant to be further determined, we define 1 h(p) = n µ + |p|n and note that   tn H(t) = ωn ln 1 + n . µ Hence Lemma 1.8 gives ! Z −1 2 sup u ≤ g + H h(Du)| det D u| dx .

Proof.

Γ+ (u)

Next, we estimate the integral on the right. For simplicity, we abbreviate Γ+ (u) to G. First observe that, for any x ∈ G and all sufficiently small h, we have 2

u(x + h) + u(x − h) − 2u(x) ≤ 0

so [D u] is negative semidefinite on G. det(−D2 u), and the matrix inequality

It follows that | det D2 u| =

tr(AB)n n (valid for all positive semidefinite matrices A and B) allows us to conclude that (−aij Dij u)n . det(−D2 u) ≤ nD Since u > 0 on G, we infer that (f + |b||Du|)n det(−D2 u) ≤ nD on G. Now we observe that f f + |b||Du| ≤ 2 max{ , |b|} max{µ, |Du|} µ fn ≤ 2( n + |b|n )1/n (µn + |Du|n )1/n , µ det A det B ≤

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and therefore h(Du) det(−D2 u) ≤

2n n



|b|n fn + n µ D D



.

Therefore 2n h(Du)| det(D u)| dx ≤ n G

Z

2

Z

G

 fn dx + B1 . µn D

We now take µ to satisfy µ>

Z

G

1/n fn dx D

to conclude that sup u ≤ g + (B0 + diam Ω)H −1 (1 + B1 )µ, Sending µ→

Z

G

1/n fn dx D

completes the proof.



Note that we can replace the integral of |b|n /D over Ω with the integral over G in this result. For our applications to strong solutions in Chapter 6, we point out a further variant. Corollary 1.10. Theorem 1.9 remains valid if we take G to be the set of all x ∈ Γ(u) for which u(x) > g and (1.8) holds with some p satisfying u(x) ≥ B0 |p| + g. Proof. We only need to observe that, if p is chosen not in χu (G) and x ∈ Γ(u) is a point at which u(x) > g and (1.8) holds, then we have u(x) ≤ B0 |p| + g. Hence (1.13) holds. The remainder of the proof is unchanged.  Another variation can be obtained if the vector β is further restricted to a cone. Theorem 1.11. Suppose [aij ] is positive definite with determinant D and that c ≤ 0. Suppose β is inward pointing and β 0 ≤ 0. Suppose further there is a cone K with opening angle θ∗ such that inf

v∈K

β·v = B0 > 0. |β||v|

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If u ∈ C 2 (Ω) ∩ C(Ω) satisfies Lu ≥ −f in Ω for some nonnegative function f and M u ≥ −g|β| for some nonnegative constant g on the subset of ∂Ω on which u > 0, then !

f g ∗

sup u ≤ C(n, θ , B1 ) diam Ω + , B0 D1/n n L (G)

where G is the subset of Γ(u) on which u > 0 and (1.8) holds for some p ∈ −K with |p| > g/B0 . Proof. This time, we take p to be a vector in −K \χu (G) with |p| > g/B0 and we define w by w(x) = u(x) − p · x. Just as before, we take x0 ∈ Ω to be a point at which w attains its maximum. The proof of Lemma 1.8 shows that x0 can’t be in Ω unless u(x0 ) ≤ 0, and u(x0 ) ≤ 0 implies that sup u ≤ diam Ω|p|. (1.19) If x0 ∈ ∂Ω and u(x0 ) > 0, then we have 0 ≥ β · Dw = β · Du − β · p ≥ −g|β| − β · p

≥ |β|(−g + B0 |p|). Hence |p| ≤ g/B0 , which contradicts our assumption on p. Hence x0 can’t be in ∂Ω if u(x0 ) > 0, so we obtain (1.19). We now define ωn (K) to be the measure of the unit ball intersected with K (note that this positive number is completely determined by n and θ∗ ), and we choose R so that Z R Z n−1 ωn (K) h(r)r dt = h(|y|) dy 0

χu (G)

with h from the proof of Theorem 1.8. The proof of that theorem now shows that Z 1/n fn R ≤ C(n, θ∗ , B1 ) dx . G D We now note that, for any ε > 0, p can be chosen so that |p| ≤ max{g/B0, R} + ε. It follows that (after sending ε → 0) that sup u ≤ diam Ω max{g/B0 , R}, and this inequality along with our estimate for R finishes the proof.  Just as before, we can repeat the proof with the integral of |b|n /D over Ω with the integral over G, and we can take G to be an even smaller set. Corollary 1.12. Theorem 1.11 remains valid if we take G to be the set of all x ∈ Γ(u) for which u(x) > 0 and (1.8) holds with some p satisfying the additional conditions u(x) ≥ diam Ω|p| and |p| ≥ g/B0 .

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1.6

He Yue – ODPEE – a21/Chp. 1

The interior weak Harnack inequality

We can prove a number of useful estimates on solutions of oblique derivative problems from the pointwise estimates of the previous section. One of the most important of these is known as the weak Harnack inequality because it is related to the classical inequality for harmonic functions proved by Harnack (see paragraph 19, page 2 from [70]). Although we could prove this result directly for solutions of oblique derivative problems, it will be more helpful to recall the result for solutions of elliptic equations. We follow the basic plan from Section 9.8 of [64] with a slight variation. Our first step is a pointwise estimate for solutions of elliptic equations. Lemma 1.13. Suppose that there are positive constants µ, µ1 and µ2 , and a positive function λ such that λ|ξ|2 ≤ aij ξi ξj ≤ µλ|ξ|2

(1.20a)

|b| ≤ µ1 λ,

(1.20b)

for all ξ ∈ Rn , c ≥ −µ2 λ.

(1.20c)

Let x1 ∈ Rn and let r be a positive constant. Then, for any constants θ1 and θ2 in (0, 1) with θ1 < θ2 , there are positive constants C1 and ζ, determined only by n, θ1 , θ2 , µ, µ1 r, and µ2 r2 , such that, if u ∈ C 2 (B(x1 , r)) ∩ L∞ (B(x1 , r)) satisfies Lu ≤ λf in B(x1 , r)

(1.21)

for some nonnegative function f ∈ Ln (B(x1 , r)), with kf kn = F , then, for any nonnegative constant h such that |{x ∈ B(x1 , θ1 r) : u¯(x) < h}| ≤ ζ n rn ,

(1.22)

C1 u¯ ≥ h in B(x1 , θ2 r),

(1.23)

we have

where u¯ = u + F r. Proof.

We define η by

|x − x1 |2 , r2 and we let q > 2 be a constant at our disposal. η(x) = 1 −

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Because u ≥ 0, we may assume without loss of generality that c ≤ 0 and then c L(η q ) = qη q−2 [ηaij Dij η + (q − 1)aij Di ηDj η + ηbi Di η + η 2 ]. q Since |Dη| ≥ θ1 /r in B(x1 , r) \ B(x1 , θ1 r) and aij Dij η ≥ −µλ/r2 , it follows that there is a number q0 > 2 such that L(η q ) ≥ 0 in B(x1 , r) \ B(x1 , θ1 r) if q ≥ q0 . We now choose q = q0 and we set v = hη q − u. It follows that there are positive constants C2 and C3 such that L(η q ) ≥ −C2 λ/r2 in B(x1 , r) and

(Lv)−

≤ C3 hζ/r + F,

D1/n n L (G) where G is the subset of Γ(v) on which v ≥ 0. Theorem 1.9 now implies that v ≤ C4 (hζ + F r) in B(x1 , r). Next, we set C5 = C5 . Then

1 2

inf B(x1 ,θ2 r) η q and choose ζ so that C3 ζ ≤

u¯ ≥ −C4 rF + C5 . Since u ¯ ≥ rF , this inequality implies (1.23) with C1 = C5 /(1 + C5 ).



Our proof of the weak Harnack inequality rests on two results from harmonic analysis. We refer the reader to Section 9.2 of [64] for proofs; specifically, our Lemma 1.14 is Lemma 9.2 from [64] and our Lemma 1.15 is Lemma 9.23 from [64]. Lemma 1.14. Let f be a measurable function on some open subset Ω of Rn such that |f |p ∈ L1 (Ω) for some p > 0, and set µ(t) = {x ∈ Ω : |f (x)| ≥ t}. Then Z

Ω

p

|f | dx = p

Z

∞

tp−1 µ(t) dt.

0

For our next lemma, we write Kr (z) for the cube {y ∈ Rn : |y i − z i | < r for i = 1, . . . , n}.

(1.24)

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Lemma 1.15. Let K0 be a cube in Rn and let w ∈ L1 (K0 ). For k ∈ R, set Γk = {z ∈ K0 : w(z) < k},

and suppose that there are positive constants δ < 1 and C0 such that sup K0 ∩K3ρ (z)

(w − k) ≤ C0

(1.25)

whenever k, z, and ρ satisfy Kρ (z) ⊂ K0 and

|Γk ∩ Kρ (z)| ≥ δ|Kρ (z)|.

(1.26)

Then, for any k ∈ R, we have

  ln(|Γk |/|K0 |) sup(w − k) ≤ C0 1 + . ln δ K0

Our weak Harnack inequality takes the following form. Theorem 1.16. Let L satisfy the hypotheses from Lemma 1.13 and suppose u ∈ C 2 (Ω) satisfies Lu ≤ λf , for some f ∈ Ln (Ω), and u ≥ 0 in Ω. Then there are positive constants κ and C such that !1/κ   Z −n κ R u dx ≤ C inf u + Rkf kLn(B(2R)) B(R)

B(R)

for any R such that B(2R) ⊂ Ω. Proof.

We set F = kf kLn(B(2R)) ,

u¯ = u + F R,

w = − ln u ¯.

We then let ζ and C1 be the constants from Lemma 1.13 corresponding to √ θ1 = 1/(3n) and θ2 = 1/ n. By using h = e−k , we conclude from Lemma 1.13 that (1.26) (with √ r = 3ρ) implies (1.22) provided K0 = K(x0 , 3R/ n). It follows that (1.25) is satisfied with C0 = − ln C1 . We now use Lemma 1.15 to infer that µ(t) = |{z ∈ K0 : u ¯(z) > t}|

(which is equal to |Γk | for k = e−t ) satisfies the inequality µ(t) ≤ C(inf u/t)κ Rn

(1.27)

for some positive constants C and κ determined by the same quantities as for C0 . If we multiply this equation by t−1+κ/2 , integrate t from inf u ¯ to ∞, and set p = κ/2, we see that Z ∞ tp−1 µ(t) dt ≤ C(inf u ¯)p Rn . inf u ¯

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On the other hand, µ(t) ≤ CRn , so we also have Z inf u¯ tp−1 µ(t) dt ≤ C(inf u ¯)p Rn . 0

Adding these inequalities and using Lemma 1.14 tells us that !1/p Z R−n

u ¯p dx

B(R/3n)

and a simple chaining argument yields !1/p Z R−n

u ¯p dx

B(R/2)

≤ C( inf

B(R/3n)

u ¯),

≤ C( inf u ¯). B(R/2)

The proof is completed by recalling the definition of u ¯.



Although we can derive a number of important results from the weak Harnack inequality, we defer their statements (and proofs) to a later section.

1.7

The weak Harnack inequality at the boundary

In deriving the interior weak Harnack inequality, a crucial role was played by the function η in Lemma 1.13. The corresponding function for the oblique derivative boundary condition is more complicated. To see why this is so, let us first consider a special case. Suppose β = (0, . . . , 0, 1), and set S = {x : |x| < 1, xn = 0} and B + = {x : |x| < 1, xn > 0}. Then w(x) = 1 − |x|2 satisfies β · Dw = 0 on S, w > 0 on B + and w vanishes on ∂B + \ S. (Note that w is exactly the function η from Lemma 1.13.) For more general uniformly oblique β, it’s easy to modify this function to obtain a function satisfying the following conditions: w is a function such that β · Dw ≤ 0 in some neighborhood N 0 of 0 in S and w > 0 in some open set Ω∗ with w = 0 on ∂Ω∗ \ N 0 and N 0 ⊂ ∂Ω∗ . Specifically, set xτ = (0, . . . , 0, −τ ) for τ ∈ (0, 1) and define w(x) = 1 − |x − xτ |2 . By choosing τ sufficiently close to 1 and choosing Ω∗ to be the subset of B + on which w > 0, we obtain the desired function. In fact, if ∂Ω is sufficiently smooth, it is not hard to adapt this construction. Our main point here is to provide such a function for more general domains. We begin by taking α1 , ε, and ω0 to be positive constants with α1 < 1 and ε < 1. We then set A = 21+2ε ε−4ε , and we define G : Rn × (0, ∞) → R

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by G(y, r) =



|y 0 |2 + α21 r2

(1+ε)/2

+

(y n )2 . (Aω0 r)2

(1.28)

We also introduce, for any x1 ∈ Rn , the set

E(x1 , r) = {x : G(x − x1 , r) < 1}.

(1.29)

With these definitions, we can state the following lemma. Lemma 1.17. Let R > 0 and suppose {x : xn > ω0 |x0 |, |x| < R} ⊂ Ω, 0

n

ω0 |β | ≤ (1 − ε)β on ∂Ω[R],

(1.30a) (1.30b)

where ∂Ω[R] is the intersection of ∂Ω with the ball of radius R centered at the origin. Let r ∈ (0, R) and suppose x1 = (0, . . . , 0, xn1 ) satisfies the conditions   ε6 xn1 ≥ A − ω0 r (1.31) 4 and E(x1 , r) ⊂ B(0, R). Then η defined by η(x) = 1 − G(x − x1 , r) satisfies the inequality β · Dη ≥ ε4

βn Aω0 r

on S, the subset of ∂Ω on which η < 1. Proof.

It follows from (1.30a) that   ε6 xn1 − xn ≥ A − ω0 r − ω0 |x0 | 4

for all x ∈ S. Since ε6 /4 < 1 and A > 2, we have A − ε6 /4 > 1. Moreover, because |x0 | < r on S, we have xn1 − xn > 0 on S. We now set N=

xn1 − xn , Aω0 r

P =

|x0 | . r

We also set α = ε6 /4. Our previous inequality then becomes N ≥1−

α P − , A A

(1.32)

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and the definition of G implies that (P 2 + α21 )(1+ε)/2 + N 2 ≤ 1 on S. Combining these inequalities yields  2 α P 1+ε P + 1− − ≤ 1, A A and hence

P 1+ε −

2P 2α ≤ . A A

(1.33)

We now claim that P ≤



2 + 2ε2 A

If this claim is false, then we have

1/ε

.

(1.34)

2P P P 2ε2 P = [AP ε − 2] > [(2 + 2ε2 ) − 2] = . A A A A But P ≥ (2/A)1/ε = ε4 /4, so ε2 P ≥ α, and hence P 1+ε −

2P 2α > . A A It follows that (1.33) does not hold if (1.34) does not hold. Hence (1.34) is valid. Our next step is a suitable lower bound for N . From (1.34), we have P 1+ε −

ε6 (1 + ε2 )1/ε + ε4 . 4 4 A straightforward calculation shows that the function H defined by α+P ≤

H(s) = (1 + s2 )1/s satisfies H 0 (s) =

H(s) g(s2 ) s2

with g(σ) =

2σ − ln(1 + σ). 1+σ

Since g 0 (σ) = (1 − σ)/(1 + σ)2 and g(0) = 0, it follows that g(σ) ≥ 0 for σ ∈ [0, 1], and hence H is increasing on the interval (0, 1). Observing also that H(1) = 2 implies that α+P ≤

ε6 ε4 + ≤ ε4 . 4 2

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Since A ≥ 2, it follows that

1 N ≥ 1 − ε4 . 2 We are now in a position to compute β · Dη on S. We have (xn − xn ) x0 β · Dη = 2β n 1 − (1 + ε)β 0 · 2 (P 2 + α21 )(ε−1)/2 2 (Aω0 r) r 1 βn ≥2 (N − (1 − ε2 )AP ε ) Aω0 r 2 1 4 βn (1 − ε − 1 + ε4 ) ≥2 Aω0 r 2 n β = ε4 . Aω0 r



Although the explicit form of this estimate for β · Dη may be useful, we shall typically just use the form βn β · Dη ≥ C(ε, ω0 ) . (1.35) r We now discuss some geometric aspects of this lemma. Note that, if r is sufficiently small, the set S may be empty. In this situation, the computation of β · Dη is irrelevant and, in our applications, this function η can be replaced by a simpler one, namely η(x) = 1 − |x − x1 |2 /r2 . The reason for introducing the constant α1 is a technical one; we want η to be C 2 and this smoothness is true if and only if α1 ∈ (0, 1). Of course, if α1 ≥ 1, the function G is still defined, but then E(x1 , r) is always empty. In practice, the choice of α1 will not be arbitrary; we shall impose a suitable smallness condition, which will be made explicit at the appropriate time. The main application of this lemma is to prove a more sophisticated minimum principle. Roughly speaking, we show that if u is nonnegative in an open set and greater than some positive constant in some small ball contained in the set, then we can obtain a quantitative lower bound on u. To obtain a form of this bound which leads to useful results, we need to choose our sets rather carefully. We also introduce some notation that will be used throughout this book (although often in a somewhat modified format). Assuming that condition (1.30) holds, we write Ω[R] and ∂Ω[R] for the intersection of Ω and ∂Ω, respectively, with the ball B(0, R). Lemma 1.18. Suppose conditions (1.30) are satisfied and that there are positive constants µ, µ1 , µ2 , and µ3 and a positive function λ such that conditions (1.20) hold and β 0 ≥ −µ3 β n .

(1.36)

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Let α1 < 1 and r be positive constants, and let x1 = (0, xn1 ) be a point in Rn such that E(x1 , r) ⊂ B(0, R) and (1.31) holds. Then, for any constants θ1 , θ2 in (0, 1) with θ1 < θ2 , there are positive constants C1 and ζ, determined only by α1 , n, ε, ω0 , µ, µ1 r, µ2 r2 , and µ3 r, such that, if u ∈ C 2 (Ω[R]) ∩ C(Ω[R]) satisfies Lu ≤ λf in Ω[R],

M u ≤ β n G on ∂Ω[R]

(1.37)

n

for some nonnegative function f ∈ L (Ω[R]), with kf kn = F , and nonnegative constant G, then, for any nonnegative constant h such that we have

|{x ∈ E(θ1 r) : u ¯(x) < h}| ≤ ζ n rn ,

(1.38)

C1 u ¯ ≥ h in E(θ2 r),

(1.39)

where E(s) = E(x1 , s) ∩ Ω and u ¯ = u + (F + G)r. Proof. As in Lemma 1.13, we may assume that c and β 0 are nonpositive. Let η be the function from Lemma 1.17. Then, for any q > 1, we have c L(η q ) = qη q−2 [ηaij Dij η + (q − 1)aij Di ηDj η + ηbi Di η + η 2 ]. q From the explicit form of Dη, it follows that there are positive constants C2 (ω0 , n, ε, θ1 ) and C3 (α1 , ω0 , n, ε, θ1 ) such that |Dη| ≥ C2 /r on E(r) \ E(θ1 r) and |D2 η| ≤ C3 /r2 in E(r), so there is a constant q0 (α1 , n, µ, µ1 r, µ2 r2 , ω0 , ε, θ1 ) such that L(η q ) ≥ 0

on E(r) \ E(θ1 r) if q ≥ q0 . In addition, we have β0 M (η q ) = qη q−1 [β · Dη + η ] ≥ 0 q on E(r) ∩ ∂Ω[R] if q ≥ Aω0 µ3 r/ε4 . We now fix q = max{q0 , Aω0 µ3 r/ε4 }. Then L(η q ) ≥ −C3 λ/r2

on E(θ1 r). Writing B ∗ for the subset of E(r) on which u < h and v = hη q − u, we conclude that

(Lv)−

D1/n n ∗ ≤ C4 hζ/r + F L (B )

and M v ≥ −Gβ n on ∂B ∗ ∩ ∂Ω[R]. In addition v ≤ 0 on ∂B ∗ \ ∂Ω[R], so Theorem 1.11 implies that ∗

v ≤ C5 r(F + G) + C4 hζ

in B . The proof is completed in the same way as for Lemma 1.13.



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Although this result can be used to prove many interesting estimates, it will be convenient to show that a lower bound can be obtained in a much simpler geometry. Given a function ω (as in conditions (1.3)), we introduce the Lipschitz cube K(y0 , r) as the set of all points x such that the corresponding y coordinates satisfy the inequalities: n

|y i − y0i | < r if i < n,

|y −

y0n

+

ω(y00 )

0

− ω(y )| < r.

(1.40a) (1.40b)

Note that the change of variables z 0 = y 0 , z n = y n − ω(y 0 ) converts K(y0 , r) into the ordinary cube {z ∈ Rn : |z i − z0i | < r for i − 1, . . . , n}. We shall take advantage of this fact in the next section. For our preliminary results, we write K ∗ (y0 , r) for the subset of K(y0 , r) on which y n > ω(y 0 ); this set is just the intersection of K(y0 , r) and Ω. Lemma 1.19. Suppose conditions (1.3) are satisfied and that there are positive constants µ, µ1 , µ2 , and µ3 and a positive function λ such that (1.20) and (1.36) hold. Let r ∈ (0, ∞) and x0 satisfy K ∗ (y0 , 3(1 + Aω0 )r) ⊂ Ω[R]. Then there are positive constants C2 and ζ1 , determined only by n, ε, ω0 , µ, µ1 r, µ2 r2 , and µ3 r, such that, if u ∈ C 2 (Ω[R]) ∩ C(Ω[R]) satisfies (1.37) for some nonnegative function f ∈ Ln (Ω[R]), with kf kn = F and some nonnegative constant G, then, for any nonnegative constant h such that |{x ∈ K ∗ (y0 , r) : u ¯(x) < h}| ≤ ζ1n rn ,

(1.41)

C2 u¯ ≥ h

(1.42)

then in K ∗ (y0 , 3r).

Proof. We start by proving a weaker result, namely, that there are constants C3 and δ1 (determined by the same quantities as C2 and η1 ) so that (1.41) implies C3 u ¯ ≥ h in K ∗ (y0 , (1 + δ1 )r).

(1.43)

This result will follow from Lemma 1.18 via a suitable covering of Lipschitz cubes by sets of the form E(y1 , r˜). To create the covering, we take α1 so that 1/2 (1 − α1+ε =1− 1 )

ε6 . 32A

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In addition, we work exclusively in y-coordinates now. For y∗0 ∈ Rn−1 , we set y∗n = ω(y∗0 ) + min{0, y0n − ω(y00 ) − r}.

(1.44)

With r˜ to be chosen, we set y10 = y∗0 ,

  ε6 y1n = y∗n + 1 − Aω0 r˜. 16A

The key observation here is that, if y 0 = y∗0 , then the point (y 0 , y n ) is in 1/2 E(y1 , r˜) if and only if |y n − y∗n | < (1 − α1+ε Aω0 r. This inequality can 1 ) be rewritten as   ε6 3ε6 n n − Aω0 r˜ < y − y∗ < 2 − Aω0 r˜. 32A 32A If y ∈ K ∗ (y0 , r) and y 0 = y∗0 , we have 0 < y n − y∗n < 2r, so we want to choose r˜ such that   3ε6 Aω0 r˜ > 2r. 2− 32A For the purpose of this proof, the exact choice of r˜ is not important, but we shall take r . r˜ = ε6 )Aω0 (1 − 16A

With this choice for r˜, there is a positive constant δ0 , determined only by ε, n, and ω0 , such that {y : y 0 = y∗0 , −δ0 < y n − y∗n < (2 + δ0 )r} ⊂ E(y1 , r˜). Hence, there is a positive constant θ0 , determined only by ε, n, and ω0 , such that {y : |y 0 − y∗0 | < θ0 , − δ20 < y n − y∗n < (2 +

δ0 2 )r}

⊂ E(y1 , r˜).

Therefore there are positive constants δ1 and ε1 , determined only by ε, n, and ω0 , such that, for any y ∈ K ∗ (y0 , (1 + δ1 )r), there is a y∗0 ∈ Rn−1 with |y 0 − y∗0 | < θ0 r,

max |y∗i − y0i | < (1 − ε)r. i 1, and then the constant C1 would also depend on N . The number 3 will be very useful in our proof of the weak Harnack inequality which we now present. Theorem 1.20. Let u ∈ C 2 (Ω[R]) ∩ C(Ω[R]), and suppose u ≥ 0 in Ω[R]. Suppose that conditions (1.3), (1.20a,b), and |c| ≤ µ2 λ, 0

|β | ≤ µ3 β

n

(1.46a) (1.46b)

are satisfied. If there are a nonnegative function f ∈ Ln (Ω[R]) and a nonnegative constant G such that (1.37) holds, then there are positive constants C and p, determined by ε, µ, µ1 R, µ2 R2 , µ3 R, and ω0 such that !1/p Z R−n

up dx

Ω[R/2]

≤ C( inf u + (F + G)R), Ω[R/2]

(1.47)

where F = kf kn . Proof.

We set u ¯ = u + (F + G)R,

we use the change of variables z 0 = y 0 , z n = y n − ω(y 0 ) to define w(z) = − ln u¯(x) for z n ≥ 0, and we extend w as an even function of z n for z n < 0. By using k = − ln h, we conclude that there is a positive δ such that (1.26) implies (1.41) for the constant ζ1 from Lemma 1.19. Converting the result of that lemma into the language of this lemma, we see that (1.25) is satisfied with C0 = − ln C1 . From this point, the argument is the same as for Theorem 1.16. 

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The strong maximum principle and uniqueness

From Lemma 1.18, we can derive several useful consequences about the behavior of solutions of oblique derivative problems. The first assertion is a strong maximum principle which asserts that if a subsolution has a nonnegative maximum, then that subsolution must be a constant. We proceed in several steps. The first step is to prove the strong maximum principle if the nonnegative maximum occurs in the domain. Corollary 1.21. Let u ∈ C 2 (Ω) and suppose u has a nonnegative maximum value at some point in Ω. Suppose also that the coefficients of L satisfy (1.20) for some positive constants λ, µ1 , and µ2 and that c ≤ 0 in Ω. If Lu ≥ 0 in Ω, then u is constant in Ω. Proof. We argue by contradiction. If u is nonconstant and it has an interior maximum value M , then there is a point x1 ∈ Ω on the boundary of the set on which u = M . Since x1 ∈ Ω, there is a positive constant r1 such that the ball B(x1 , r1 ) ⊂ Ω and a point x2 ∈ B(x1 , r1 /4) such that u(x2 ) < M . By continuity, u(x) < M in some ball centered at x2 with radius θ1 r1 /4. We now use Lemma 1.13 with r = r1 /2 and u replaced by M − u to see that M − u > 0 in B(x2 , r/2). But x1 is in this ball, which contradicts our assumption that u(x1 ) = M . Hence u can’t have an interior maximum if it is nonconstant.  Next, we show that, if the nonnegative maximum occurs on the boundary and if a suitable oblique condition is satisfied near that maximum, then u is constant. Corollary 1.22. Let u ∈ C 2 (Ω[R]) ∩ C(Ω[R]) and suppose u has a nonnegative maximum value at 0. Suppose also that conditions (1.20), (1.30) and (1.36) are satisfied. If c and β 0 are nonpositive and if Lu ≥ 0 in Ω[R],

M u ≥ 0 on ∂Ω[R],

then u is constant in Ω[R]. Proof. Again we argue by contradiction. With r sufficiently small, we can find an x1 with x01 = 0 and xn1 = (A − ε6 /4)ω0 r such that η(0) < 1 and a constant θ1 ∈ (0, 1) such that 0 ∈ / E(θ1 ) by choosing α1 sufficiently small. From Corollary 1.21, it follows that u < u(0) in E(θ1 r). Lemma 1.18 then implies that u < u(0) in E(θ2 r) for any θ2 ∈ (θ1 , 1). In particular, we have u(0) < u(0). This contradiction proves the result. 

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The strong maximum principle then produces a simple uniqueness result. Corollary 1.23. Let β be uniformly oblique on ∂Ω and suppose that |aij /λ|, |b/λ|, and |c/λ| are uniformly bounded in Ω. Suppose also that |β 0 |/|β| is uniformly bounded on ∂Ω. If c ≤ 0 in Ω and β 0 ≤ 0 on ∂Ω, then any two solutions of Lu = f in Ω,

M u = g on ∂Ω

(1.48) 0

differ by a constant. Moreover, if c is somewhere nonzero or if β is somewhere nonzero, then (1.48) has at most one solution. Proof. Let u1 and u2 be two solutions of (1.48) and suppose v = u1 − u2 is nonconstant. Since v satisfies Lv = 0 in Ω,

M v = 0 on ∂Ω,

the strong maximum principle implies that v can’t have a nonnegative maximum. In other words, v ≤ 0. On the other hand, −v satisfies the same differential equation and boundary condition as v, so it can’t have a nonnegative maximum, so −v ≤ 0. It follows that v = 0 and hence u1 = u2 . If v is a constant and c(x0 ) 6= 0, then the differential equation implies that c(x0 )v(x0 ) = 0 and hence v ≡ 0. A similar argument applies if β 0 (x0 ) 6= 0.  A related result is known as the boundary point lemma. Although its statement uses a stronger smoothness assumption about ∂Ω, the conclusion is stronger. Lemma 1.24. Suppose |aij |/λ, |b|/λ, and |c|/λ are uniformly bounded in Ω and suppose Lu ≥ 0 in Ω. Suppose also that there is a point x0 ∈ ∂Ω such that u is continuous at x0 , u(x0 ) > u(x) for all x ∈ Ω, and ∂Ω satisfies an interior sphere condition at x0 . If c ≤ 0 in Ω and if u(x0 ) ≥ 0, or if u(x0 ) = 0, then β · Du(x0 ) < 0 for any vector β which is oblique to the interior sphere. Proof. We first suppose that c ≤ 0. Let B(y, R) be a ball which lies in Ω with x0 ∈ ∂B(y, R). Choose ρ ∈ (0, R), set A = B(y, R) \ B(y, ρ) and define v in A by v(x) = exp(−α|x − y|2 ) − exp(−αR2 )

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for a positive constant α at our disposal. It is not hard to see that α can be chosen so that Lv ≥ 0 in A. On ∂B(y, ρ), there is a constant ε so that u − u(x0 ) + εv ≤ 0, and the definition of v guarantees that this inequality is also true on ∂B(y, R). Hence u − u(x0 ) + εv ≤ 0 on ∂A and L(u − u(x0 ) + εv) ≥ 0 in A. The maximum principle implies that u − u(x0 ) + εv ≤ 0 in A and hence β · Du ≤ εβ · Dv < 0.

If u(x0 ) = 0, then we can repeat the preceding argument with L0 , defined by L0 w = aij Dij w + bi Di w, in place of L.  Note that Lemma 1.24 implies Corollary 1.23 under the stronger assumption that Ω satisfies an interior sphere condition at every point of ∂Ω. Hence, for our purposes, this lemma is not very important; however, it has a number of important consequences. We mention here that Lemma 1.24 is the basis for studying many symmetry properties of solutions of elliptic equations. (See for example, [62].)

1.9

H¨ older continuity

We now prove a result that is fundamental in the theory of nonlinear elliptic (and parabolic) equations. Such a result was first proved (although only for weak solutions of equations in divergence form, and only away from the boundary) by De Giorgi [35] and Nash [146]. A key step is the following iteration result. Although the proof is identical to that of, for example, Lemma 4.6 from [114], we include it to emphasize that it is true under slightly different hypotheses from the usual ones. Lemma 1.25. Let w and σ be nonnegative functions defined on some interval [0, R0 ], and suppose that there are positive constants α, δ, and τ with τ < 1 and δ < α such that σ(r) σ(s) ≤ δ (1.49) δ r s if 0 < s ≤ r ≤ R0 and w(τ k R0 ) ≤ τ α w(τ k−1 R0 ) + σ(τ k−1 R0 )

(1.50)

if 0 < r ≤ R0 . If σ is increasing on (0, R0 ] and w is increasing on each interval of the form (τ k+1 R0 , τ k R0 ], then there is a constant C determined only by α, δ, and τ such that  α  r w(r) ≤ C w(R0 ) + σ(r) (1.51) R0

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for all r ∈ (0, R0 ]. Proof.

If r > τ R0 , we have

α r w(r) ≤ w(R0 ) ≤ τ w(R0 ). R0 If r ∈ (τ k+1 R0 , τ k R0 ] for some positive integer k, then induction gives k−1 X w(r) ≤ w(τ k R0 ) ≤ τ kα w(R0 ) + τ −δ σ(τ k R0 ) τ (α−δ)j . −α



j=0

The sum is a finite geometric series with positive terms, so it can be estimated from above by its (finite) limit 1/(1 − τ δ−α ), and, because r > τ k+1 R0 , we have τ k R0 < r/τ , so σ(τ k R0 ) ≤ σ(r/τ ) ≤ τ −δ σ(r).

Combining these three inequalities yields (1.51) with C = max{τ −α , 1/(1 − τ δ−α )}.  We shall use this lemma repeatedly in this book. Typically, w will be assumed increasing (and this is the case for Lemma 4.6 in [114]), but we shall have several occasions to use this stronger result. It is important to point out that (1.51) provides a rate at which w(r) goes to zero with r if σ(r) → 0 as r → 0. In most applications, we shall even assume that σ(r) = Arδ , but sometimes a general σ will be useful. Theorem 1.26. Let L be as in Theorem 1.20. If Lu = f in Ω[R] and M u = g on Σ[R], then there are constants C, α and δ determined only by ε, µ, µ1 R, µ2 R2 , µ3 R, and ω0 such that ! rα δ 1−δ sup u + (F + G)r R , (1.52) osc u ≤ C Rα Ω[R] Ω[r] where F = kf /λkn and G = sup |g/β n |. Proof.

If r ≥ R/4, then the result is immediate. If r < R/4, we write Mi = sup u, Ω[ir]

mi = inf u Ω[ir]

for i = 1, 4. Then L(M4 − u) = f and L(u − m4 ) = f in Ω[4r], and M (M4 − u) = −g and M (u − m4 ) = g on Σ[4r]. We may therefore apply the weak Harnack inequality to these functions to obtain Z (r−n (M4 − u)p dx)1/p ≤ C(M4 − M1 + σ(r)), Ω[2r] Z −n (r (u − m4 )p dx)1/p ≤ C(m1 − m4 + σ(r)), Ω[2r]

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where σ is defined by σ(s) = (F + G)s. We now add these two inequalities and take advantage of the appropriate version of Minkowski’s inequality: Z Z Z ( (f + g)p )1/p ≤ C(p)[(

f p )1/p + (

g p )1/p ]

which is valid for all p > 0; we recall that C(p) = 1 if p ≥ 1. It follows that M4 − m4 ≤ C(M4 − m4 + m1 − M1 + σ(r)). We now set w(s) = oscΩ[s] u to obtain w(r) ≤ γw(4r) + σ(r) with γ = 1 − C −1 , so γ ∈ (0, 1). The desired result now follows from this inequality via Lemma 1.25 with α and δ chosen to satisfy 4−α ≥ γ and δ < α < 1. 

Although this result can be taken without further explanation, it is useful to give a brief description here of the H¨older spaces because they will play a crucial role in the remainder of this book. Let α ∈ (0, 1]. We say that a real-valued function f , defined on some set S is H¨ older continuous with exponent α on S if there is a constant M such that |f (x) − f (y)| ≤ M |x − y|α for all x and y in S. The smallest such M for which this inequality holds is called the H¨ older seminorm of f on S and is written as [f ]α:S . We write Hα (S) for the set of all functions f which are bounded and H¨older continuous on S and we use the norm |f |α;S = [f ]α;S + sup |f |. S

We also frequently abbreviate supS |f | to |f |0;S . Sometimes we write C α (S) in place of Hα (S) when S is closed. A related concept is that of local H¨older continuity. We say that f is locally H¨ older continuous on an open set Ω if f is H¨older continuous on any compact subset S of Ω, and we use C α (Ω) to denote the set of all locally H¨ older continuous functions on an open set Ω. In this language, we can restate Theorem 1.26 to say that solutions of oblique derivative problems in Lipschitz domains are H¨ older continuous. The concept of H¨ older continuity is paramount in our later study of solutions. For that study, we need to define additional function spaces in terms of H¨ older continuity of derivatives, but we save the details for Chapter 4.

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1.10

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The local maximum principle

Occasionally, one wishes to estimate the maximum value of a solution in a neighborhood of a point in terms of local data. Since it is not generally possible to obtain such a bound without some information on the solution (for example, any constant is a solution of the equation Aij Dij u = 0 but the maximum value in a neighborhood of a given point is exactly the unknown constant), we look for weak information on the solution which implies such a bound. In this section, the additional information is a bound on some Lp norm of the solution. Theorem 1.27. Let u ∈ C 2 (Ω[R]) ∩ C(Ω[R]), and suppose u ≥ 0 in Ω[R]. Suppose that conditions (1.3), (1.20a,b), and (1.46) are satisfied. If there are a nonnegative function f ∈ Ln (Ω[R]) and a nonnegative constant g such that Lu ≥ −λf in Ω[R],

M u ≥ −β n g on ∂Ω[R],

(1.53)

then, for any p > 0, there is a constant C, determined only by µ, ω0 , ε, µ1 r, µ2 r2 , µ3 r, and p such that   !1/p Z sup u ≤ C  R−n (u+ )p dx (1.54) + (F + g)R , Ω[R/2]

Ω[R]

where F = kf kn .

Proof. Since the conclusion is clear if supΩ[R/2] u ≤ 0, we shall assume that this supremum is positive. Working in y-coordinates, we let y0 be a point in Ω[R/2] such that u(y0 ) ≥ 12 supΩ[R/2] u. We note that there is a constant A˜ ∈ (0, 12 ), determined only by ω0 and ε, so that, if y10 = y00 , ˜ then y1n = y0n + AR, ˜ ⊂ B((y00 , ω(y00 )), R). y0 ∈ E(y1 , AR)

˜ and q ≥ 2 to be chosen, With η from Lemma 1.17 corresponding to r = AR q we set v = η u and we define the operators L0 and M0 by L0 w = aij Dij w,

M0 w = β i Di w.

Then L0 v = qη q−2 (ηaij Dij η + (q − 1)aij Di ηDj η − cη 2 )u + η q−1 [2qaij Di η − bi ]Dj u − η q f

≥ −CR−2 η q−2 u − CR−1 η q−1 |Du| − f.

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To proceed, we shall invoke Theorem 1.11, and we choose the cone K so that any vector in K with vertex in E(y1 , r) must intersect ∂E(y1 , r) inside Ω[R]. (Such will be the case if every vector ξ ∈ K satisfies ξ0 |v 0 | ≤ ξ n .) It then follows that, on G, we have |Dv| ≤ Cv/η. Since Dv = η q Du + qη q−1 uDη, it follows that L0 v ≥ −CR−2 η q−2 u − f on G. In addition, M0 v = η q−1 (qβ i Di η − ηβ 0 )u − η q g ≥ −β n g if q is sufficiently large, say q ≥ q0 . From Theorem 1.11, we therefore infer that sup v ≤ CR−1 kη q−2 ukn + C(F + g)R. We now observe that 2/q

kη q−2 ukn ≤ (sup v)1−(2/q) kuk2n/q , and hence, for any ζ > 0, there is a constant C(ζ, n, q) such that kη q−2 ukn ≤ ζ sup v + C(ζ, n, q)kuk2n/q . If p ≤ min{n, 2n/q0}, then we take q = 2n/p and then ζ sufficiently small to conclude (1.54). The case p > min{n, 2n/q0 } is then obtained via H¨older’s inequality. 

1.11

Pointwise estimates for solutions of mixed boundary value problems

For our later work, we wish to have estimates for solutions of the mixed boundary value problem Lu = f in Ω,

M u = g on Σ,

u = ψ on σ

when Σ and σ are disjoint subsets of ∂Ω whose union is all of ∂Ω. Without any restrictions on Σ and σ, this problem can be extremely complicated and a complete description is beyond the scope of this book. Here, we make some assumptions on Σ and σ which are fairly general and easy to state. The model configuration is commonly called a wedge or dihedral angle, which means that, in cylindrical coordinates (r, θ, z), we have Σ = {r > 0, θ = θ0 },

σ = {r ≥ 0, θ = θ1 },

Ω = {r > 0, θ0 < θ < θ1 }

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for some numbers θ0 and θ1 satisfying θ0 < θ1 < θ0 +π. (In fact, our results apply if we only assume that θ0 < θ1 < θ0 + 2π, but this extra step requires a more complicated analysis to handle the general setting, and it will not be needed in this book.) More generally, we shall always assume that there is a Lipschitz function ω, defined for all x0 with |x0 | < R (for some positive R), such that ω(0) = 0 and |ω(x0 ) − ω(y 0 )| ≤ ω0 |x0 − y 0 |. Moreover, we write Ω[R] = {x ∈ Ω : |x| < R},

and we assume that ∂Ω[R] consists of two disjoint pieces: Σ[R] which is an open subset of the surface {xn = ω(x0 ), |x0 | < R} and σ[R]. We further assume that y n > ω(y 0 ) on σ[R]. (As already mentioned, this further assumption is not necessary for our analysis but it simplifies the proof of results and it will suffice for our applications.) For simplicity, we refer to this set of hypotheses as condition (ω). Our first step is to obtain a weak Harnack inequality under these hypotheses and the ones previously stated for L and M . Theorem 1.28. Let u ∈ C 2 (Ω[R]) ∩ C(Ω[R]) and suppose u ≥ 0 in Ω[R]. Suppose conditions (ω), (1.3c), (1.20), and (1.36) are satisfied. Suppose further that there are constants G ≥ 0 and m > 0, and a function f ∈ Ln (Ω[R]) such that M u ≥ β n G on Σ[R],

Lu ≥ λf in Ω[R],

u ≥ m on σ[R].

Then there is a constant C such that !1/κ Z −n − κ ≤ C( inf u + R(F + G)), R (um ) dx Ω[R/2]

˜ B(R/2)

(1.55)

(1.56)

where u− m (x)

=

(

min{u(x), m} m

if x ∈ Ω[R] ∩ Σ[R], otherwise,

˜ F = kf kn, and B(R/2) is the subset of B(R/2) on which y n > ω(y 0 ). Proof. We first observe that Lemmata 1.18 and 1.19 are still true in this case (with the same proofs) under the additional assumption that h ≤ m. The key point in Lemma 1.19 is that K ∗ (y0 , r) is defined to be K(y0 , r)∩Ω. It follows that (1.25) holds for w defined by w(z) = − ln(u− m (y) + (F + G)r)

as long as k ≥ − ln m. We then infer (1.27) for t ∈ (0, m), and then (1.56) follows as before because µ(t) = 0 if t > m. 

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This weak Harnack inequality leads to a modulus of continuity estimate if we strengthen condition (ω) just a little. Corollary 1.29. Let u ∈ C 2 (Ω[R]) ∩ C(Ω[R]), and suppose conditions (ω), (1.3c), (1.36a,b), and (1.46) are satisfied. Suppose also that there is a constant A0 ∈ (0, 1) such that ˜ ˜ |B(r) ∩ Ω| < A0 |B(r)|

(1.57)

for all r ∈ (0, R). If there is a concave function σ1 such that oscσ[r] u ≤ σ1 (r), then there are constants C, α and δ (determined by the same quantities as in Theorem 1.26) such that δ ∈ (0, 1) and   α r δ 1−δ δ 1−δ osc u + (F + G)r R + σ (r R ) . osc u ≤ C 1 0 0 R0α Ω[R] Ω[r] Proof. With Mi and mi as in the proof of Theorem 1.26, we apply Theorem 1.28 to M4 − u and obtain !1/p Z  p − −n R (M4 − u)M −M˜ dx ≤ C(M4 − M1 + (F + G)R), 4

˜ B(2R)

4

˜ 4 = supσ[4r] u. The definition of u− where M m implies that r

−n

Z

˜ B(2r)



(M4 −

u)− ˜4 M4 −M

p

!1/p

dx

˜ ˜ 4) ≥ (r−n |B(2r) \ Ω|)1/p (M4 − M ˜ 4) ≥ C(n, ω0 , A0 )1/p (M4 − M

using also condition (1.57), and hence ˜ 4 ≤ C(M4 − M1 + (F + G)r). M4 − M A similar argument with u − m4 yields m ˜ 4 − m4 ≤ C(m1 − m4 + (F + G)r), using the notation m ˜ 4 = inf σ[4r] u. Adding these results and rearranging yields osc u ≤ γ osc u + (F + G)r + σ1 (r),

Ω[r]

Ω[4r]

and the proof is completed by applying Lemma 1.25.



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For our study of problems with discontinuous β, we shall also use the local maximum principle, which takes the following simple form. Since the proof is identical to that of Theorem 1.27, we omit it. Theorem 1.30. Let u ∈ C 2 (Ω[R]) ∩ C(Ω[R]) and suppose u ≥ 0 in Ω[R]. Suppose conditions (ω), (1.3c), (1.20), and (1.36) are satisfied. Suppose further that there are constants G ≥ 0 and m > 0, and a function f ∈ Ln (Ω[R]) such that Lu ≤ λf in Ω[R],

M u ≤ β n G on Σ[R],

u ≤ m on σ[R].

Then, for any p > 0, there is a constant C such that   !1/κ Z sup u ≤ C  R−n (u+ )κ dx + R(F + G) , Ω[R/2]

(1.58)

(1.59)

Ω[R]

where F = kf kn .

1.12

Derivative bounds for solutions of elliptic equations

The maximum principle can also be used to derive bounds on the derivatives of solutions of constant coefficient equations. There are actually several different ways to obtain these bounds, but we present here a straightforward approach that doesn’t make undue assumptions about the solution itself. Most of our estimates are in balls and half-balls, and the estimates in more general domains will derived from those in the simpler domains. We begin with a simple pointwise estimate for solutions of general elliptic equations in a half-ball with zero data on the flat portion of the boundary. Lemma 1.31. Let [aij ] be a positive-definite matrix-valued function defined on B + (R) for some R > 0, and suppose that λ and µ are positive constants such that the eigenvalues of [aij ] are in the interval [λ, µλ]. If v ∈ C 2 (B + (R)) ∩ C(B + (R)) satisfies aij Dij v = f in B + (R),

v = 0 on B 0 (R),

then |v(x)| ≤ 16nµ for all x ∈ B + (7R/8).

sup |v|

B + (R)

!

xn + R

!

sup |f | Rxn

B + (R)

(1.60)

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Proof. Let x ∈ B + (7R/8). If xn ≥ R/8, then (1.60) is immediate. On the other hand, if xn < R/8, we set x0 = (x0 , 0), V = supB + (R) |v|, F = supB + (R) |f |, and r = R − |x0 |, and we define the function w by   n  |y − x0 |2 (y n )2 y w(y) = V + nµ − 2 + F (ry n − (y n )2 ). r2 r r Then aij Dij w ≤ f in B + (x0 , r),

w ≥ v on ∂B + (x0 , r),

so the maximum principle implies that v ≤ w in B + (x0 , r). Similarly v ≥ −w in B + (x0 , r) and hence |v| ≤ w in B + (x0 , r). In particular, this inequality is true at x, so  n   n 2 (x ) x (xn )2 |v(x)| ≤ V + nµ − + F (rxn − (xn )2 ) r2 r r2

The proof is completed by using the inequalities 0 < xn < r and R/8 ≤ r ≤ R. 

Our next step is to obtain an interior gradient bound. We do so by applying our preceding estimate in a suitable way. Lemma 1.32. Let [Aij ] be a constant positive-definite matrix and suppose λ ≤ Λ are positive constants satisfying the inequality λ|ξ|2 ≤ Aij ξi ξj ≤ Λ|ξ|2 .

(1.61)

If u ∈ C 2 (B(0, r)) satisfies the equation Aij Dij u = f in B(0, r) for some bounded function f , then there is a constant C, determined only by n and Λ/λ such that   sup |u| |Du(0)| ≤ C + r sup |f | . (1.62) r Proof. Note that the eigenvectors of [Aij ] are orthogonal. Hence it suffices to prove that this bound is true for the directional derivative of u in the direction of any eigenvector. Let us fix an eigenvector, which we may take to be a unit vector. Since the hypotheses are invariant under rotation of axes, we may assume this vector is (0, . . . , 0, 1). We then define ϕ and ψ in B + (r) by 1 [u(x0 , xn ) − u(x0 , −xn )], 2 1 ψ(x0 , xn ) = [f (x0 , xn ) − f (x0 , −xn )], 2 ϕ(x0 , xn ) =

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so Aij Dij ϕ = ψ in B + (r),

ϕ = 0 on B 0 (r).

It then follows from Lemma 1.31 and the definitions of ϕ and ψ that ! ! xn n |ϕ(0, x )| ≤ 16nµ sup |ϕ| + sup |ψ| rxn r B + (r) B + (r) ! ! xn ≤ 16nµ sup |v| + sup |f | rxn r + B (r) B + (r) we now divide this inequality by xn and take the limit as xn → 0 to infer (1.62), recalling that Dn v(0) = lim

xn →0+

ϕ(0, xn ) . xn



From this estimate, we obtain an interior estimate which will be very useful in several later chapters. Theorem 1.33. Let Ω be an open subset of Rn and let u be a bounded solution of Aij Dij u = f in Ω, where Aij is a positive definite, constant matrix. If λ and Λ are positive constants satisfying (1.61), then there is a constant C(n, Λ/λ) so that d(x)|Du(x)| ≤ C[sup |u| + sup d2 |f |)

(1.63)

for all x ∈ Ω. Proof. Fix x ∈ Ω and set r = d(x). If we define u0 and f 0 by u0 (y) = u(y− x) and f 0 (y) = f (y − x), then u0 is a solution of Aij Dij u0 = f 0 in B(0, r), so we can apply Lemma 1.32 to conclude that |Du0 (0)| ≤ C[sup |u0 |/r + r sup |f 0 |]. Simple rearrangement now gives (1.63).  We shall also take advantage of some higher order derivative estimates. Theorem 1.34. Let Ω be an open subset of Rn and let u be a bounded solution of Aij Dij u = f in Ω, where Aij is a positive definite, constant matrix. If f ∈ C N (Ω) for some nonnegative integer N , then u ∈ C N +1 (Ω). Moreover, if λ and Λ are positive constants satisfying (1.61), then there is a constant C(n, N, Λ/λ) so that d(x)N |DN u(x)| ≤ C[sup |u| + for all x ∈ Ω.

N −1 X j=0

sup dj+2 |Dj f |]

(1.64)

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Proof. We only prove the C 2 estimate since the full strength of the theorem is proved by a simple modification of this argument. We shall estimate the second derivative Dkm u at a point x for k and m integers between 1 and n. We then apply Theorem 1.33 to the function uk,h , defined (for √ h < d(x)/(2 n)) by u(x + hek ) − u(x) , h in the ball B(x, d(x)/2). It follows that uk,h =

|Dm uk,h (x)| ≤ C

sup B(x,d(x)/2)

|uk,h |/d(x) + d(x)

sup B(x,d(x)/2)

!

|fk,h | .

But sup B(x,d(x)/2)

|uk,h | ≤

B(x,3d(x)/4)

|fk,h | ≤

B(x,3d(x)/4)

sup

|Dk u|

and sup B(x,d(x)/2)

sup

|Dk f |

so another application of Theorem 1.33 to u gives   sup |u| sup |Dk u| ≤ C + sup d|f | . d(x) B(x,3d(x)/4) Hence |Dm uk,h (x)| ≤ C



 sup |u| + sup d|f | + sup d2 |Dk f | . d(x)

The proof is completed by recalling that

Dkm u(x) = lim Dm uk,h (x).



h→0

We shall use the corresponding boundary version of this theorem. Lemma 1.35. Let [Aij ] be a positive definite matrix satisfying (1.61) for some positive constants λ and Λ, and let R > 0. If v ∈ C 2 (B + (R)) ∩ C(B + (R)) satisfies Aij Dij v = 0 in B + (R),

v = 0 on B 0 (R),

(1.65)

then there is a constant C, determined only by n and Λ/λ such that sup B + (R/2)

where V = supB + (R) |v|.

|D2 v| ≤ CR−2 V,

(1.66)

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Proof.

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Our first step is to use Lemma 1.31 to infer that xn in B + (7R/8). R

|v(x)| ≤ 16nAV

Our next step is a gradient bound for v. We now fix x1 ∈ B + (3R/4) and use Ω = B(x1 , xn1 /8) in Theorem 1.33 to conclude that xn1 xn |Dv(x1 )| ≤ C sup |v| ≤ CV 1 . 8 R Ω It follows that sup B + (3R/4)

|Dv| ≤ C

V . R

(1.67)

Now let k ∈ {1, . . . , n − 1} and apply the preceding argument to ∆h v defined by v(x + hek ) − v(x) h for 0 < |h| < R/16 to conclude that ∆h v(x) =

sup B + (R/2)

|D∆h v| ≤ CR−1

sup B + (2R/3)

|∆h v|.

Since ∆h v → Dk v as h → 0, we conclude that sup B + (R/2)

|Dik v| ≤ CR−1

sup B + (2R/3)

|Dv| ≤ CR−2 V

by a direct application of (1.67) as long as i ∈ {1, . . . , n} and k ∈ {1, . . . , n− 1}. The proof is completed by using the differential equation to conclude that |Dnn v| ≤ C

n n−1 X X i=1 k=1

|Dik v|.



In fact, these estimates can be expanded considerably. We refer the reader to the Notes of this chapter for detailed information. Notes The maximum principle is one of the most important tools in the theory of elliptic differential equations. The books [156] and [158] give an introduction to this fruitful theory, and the first few results in this chapter (namely, Lemmata 1.1 and 1.2) are minor variations on well-known results. We also

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43

mention Chapter 3 in [64] for a discussion of other, related maximum principles, but most of that chapter is concerned with maximum principles for the Dirichlet problem; we mention that [156] has a thorough bibliography on the topic, current at the time (1967) that book was written. The definition of obliqueness in Section 1.2 come from [116], but it is just a slight restatement of the definition in [142]. The condition (1.3b), universally called Lipschitz continuity of ω, was first used by Lipschitz in [123] to study existence and uniqueness of solutions for ordinary differential equations, a topic we return to in Chapter 3. Lemma 1.3 is just Theorem 2 from [128], and the Weierstrass approximation theorem first appeared in [200] (in the special case of one spatial dimension); our proof is based on the one in Section IV.15 of [36] (which seems to be the original proof), but the idea to use this approximation to control the modulus of continuity comes from the observation in Proposition I.1.4 from [165] that the polynomial approximation also approximates derivatives uniformly. Lemmata 1.5 and 1.6 are taken directly from [116]. The proof of the change of variables formula is a simplified version of the argument in Appendix F of [176]. That source gives a sharper version of the result presented here; it defines, for y ∈ Rn , ν(y) to be the number of points x in Ω such that F (x) = y and then shows that the function ν : R → [0, ∞] is measurable with Z Z u ◦ F (x)| det DF (x)| dx = ν(y)u(y) dy. Rn

Ω

The basic ideas for the estimates in Section 1.5 come from the work of Aleksandrov [2], but the technical details in Lemma 1.8 are based on Theorem 2.1 from [122] and our choice of the function h in the proof of Theorem 1.9 comes from the calculation in the proof of Theorem 9.1 from [64]. (Both [64] and [2] use h(t) = 1/(tn/(n−1) + µn/(n−1) )n−1 . In [2], a more careful estimate is made to evaluate the integral Z R h(r)rn−1 dr, 0

which leads to a more precise description of the way in which the maximum estimate depends on B1 .) Theorem 1.11 comes from [198]. The arguments in Section 1.6 are taken from Section 9.8 in [64], which adapted the discussion in [178]. That work, in turn, took the ideas of [88] for parabolic equations and modified the proofs to handle elliptic equations. The function G in Section 1.7 and the arguments in Sections 1.7 and 1.8 were first introduced in [116]; the present analysis fills in some of the

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details omitted from that paper. The interior strong maximum principle Corollary 1.21 is standard in the theory of elliptic equations; it was first proved, by different techniques, in [72]. The value of the present proof is its application to the oblique derivative problem. Although the strong maximum principle Corollary 1.22 was first proved (under slightly stronger hypotheses, in particular, continuity of the vector β) by Nadirashvili [142] as his Lemma on the Inner Derivative, the method of proof here is quite different. The boundary point lemma (Lemma 1.24) was first proved independently by Hopf [74] and Oleinik [150] at the same time, but, in the West, the result is commonly known as the Hopf boundary point lemma. The material in Sections 1.7, 1.9, and 1.10 has become quite standard in the literature. Our proofs are based very strongly on those in Sections 9.7 and 9.8 from [64]. We point out that the present author’s work [103] is somewhat incautious about its proof of the local maximum principle (Theorem 3.3 of that work) by not taking the definition of Γ+ v into account correctly. Fortunately, the use of Wang’s maximum principle (our Theorem 1.11) avoids that difficulty. Earlier versions of the results were proved by Nadirashvili [142–145] under stronger hypotheses on the coefficients (for example, aij is assumed continuous in [144] in order to prove a H¨older estimate for u in terms of the Ln norm of f /λ). The material in Section 1.11 comes from Section 8 of [116]. The higher order derivative estimates in Section 1.12 are based on work of Brandt [14], [15]. He was able to prove the interior Schauder estimates (see Chapter 2) for elliptic and parabolic differential equations using only maximum principle arguments; in particular, we mention (see also Section 3.4 in [64]) that these arguments can be used to prove a modulus of continuity estimate for the first derivatives of solutions to the equation Aij Dij u = f based only on a pointwise bound for f .

Exercises 1.1 Show that if the hypotheses of Lemma 1.6 are satisfied, then there is a constant C, determined only by Ω, C2 , and the modulus of ellipticity of β such that   Lu− M u− sup u ≤ C sup + sup . |c| |β| 1.2 Prove the following variant of Lemma 1.25.

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45

Lemma 1.36. Let w and σ be increasing functions defined on some interval (0, R0 ], and suppose that there are positive constants α and τ with τ < 1 such that (1.50) holds for all r ∈ (0, R0 ). Then, for any µ ∈ (0, 1), there is a positive constant C(τ, α, µ) such that  α(1−µ) R ω(R0 ) + σ(Rµ R01−µ )]. ω(R) ≤ C[ R0 (Hint: See Lemma 8.23 in [64].) 1.3 Show that Theorem 1.28 remains true if we replace the assumption that y n > ω(y 0 ) on σ[R] by the assumption that there is a number A1 ∈ (0, 1) such that, for any r ∈ (0, R), there is a ball B ∗ (A1 r) centered on the set {y n = ω(y 0 )} which is a subset of Ω. 1.4 Show that the higher order version of Lemma 1.35 is valid. In other words, show that the hypotheses of the lemma imply that, for any positive integer N , there is a constant C, determined only by n, N , and Λ/λ such that sup B + (R/2)

|DN v| ≤ CR−N V.

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Chapter 2

Classical Schauder Theory from a Modern Perspective

Introduction A key element in the theory of the Dirichlet problem for linear elliptic equations is known as Schauder theory. Developed by Schauder [168, 167], this theory has two elements: it provides estimates on the H¨older norm of any solution of the Dirichlet problem and of the first and second derivatives of this solution, and it gives an existence result under appropriate conditions on the coefficients of the operator. Schauder’s method of proof is based on classical potential theory (much of which is an improvement of the results of H¨ older [71]). We refer to Chapter V in [136] for a discussion of the contributions of Cacciopoli [20], Hopf [73], and Giraud [65]. A detailed description of Schauder’s approach (with various minor adjustments for better results) appears in Chapter 6 of [64], and we recommend the interested reader to refer to that work for a better understanding of the theory. In this chapter, however, we use a number of ideas that were developed much later to prove all these classical results. In particular, the works of Safonov [163] and Caffarelli [21] form the basis of the methods used here, and they will be extremely useful in later chapters when we extend the results in this chapter to domains with less regular boundary.

2.1

Definitions and properties of H¨ older spaces

In addition to the H¨ older spaces from Section 1.9, we shall use several other H¨ older spaces, most of them involving the derivatives of the functions in the spaces. From now on, we use Ω to denote a bounded open subset of Rn which may be further restricted in certain circumstances, but we always indicate when this is the case. 47

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48

For a positive real number a = k + α, where k is a nonnegative integer and α ∈ (0, 1], we write Ha (Ω) for the set of all k times continuously differentiable functions with uniformly bounded derivatives of all orders less than or equal to k such that its kth order derivatives are H¨older continuous with exponent α. We then define a norm on Ha by |f |a =

k X j=0

|Dj f |0 + [Dk f ]α .

We use without further comment the fact that Ha is a Banach space under this norm. Sometimes we use C k,α (Ω) to denote Ha . We shall also make use of several weighted H¨older spaces. To describe them, we write (as before) d for the distance from a point to the boundary of Ω. For b ∈ R, we then define the norm (b)

|f |0 = sup | min{db , (diam Ω)b }f |, Ω

and the seminorm [f ](b) α =

sup x6=y∈Ω |x−y| 0, then Ha = Ha . Our next step is to collect some useful facts about all these norms. First, if f and g are in Ha with a > 0, then f g ∈ Ha and |f g|a ≤ |f |a |g|a .

(2.1)

If a ∈ (0, 1], then this inequality follows from the observation that [f g]a ≤ [f ]a |g|0 + |f |0 [g]a , and the general case follows from this one via the product rule for derivatives. More generally, if a, α and β satisfy a + α ≥ 0,

a + β ≥ 0,

a + α + β ≥ 0,

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and if f ∈ Ha

(β)

(α+β)

and g ∈ Ha , then f g ∈ Ha |f g|(α+β) a

≤

49

and

(β) |f |(α) a |g|a .

(2.2)

Next, we have that Ha ⊂ Hb if b ≤ a ≤ 1 and

|f |b ≤ max{1, (diam Ω)a−b }|f |a .

(2.3)

When a > 1, we need a restriction on Ω for Ha ⊂ Hb . For example, we consider the following example from page 53 in [64]. Let Ω be the set of all (x, y) ∈ R2 such that x2 + y 2 < 1 and y < |x|1/2 and let a ∈ (1, 2). Then the function f , defined in Ω by  a  if x, y > 0,  y f (x, y) = −y a if y > 0 > x,   0 if y ≤ 0, is in Ha but not in Hb if b ∈ (a/2, 1). A more careful examination of this issue is made in [56]. In this book, we usually assume that Ω is a bounded Lipschitz domain and then we have Ha ⊂ Hb as long as a ≥ b. Lemma 2.1. Let a be a positive number and let Ω be a bounded Lipschitz domain. If b ∈ (0, a), then Ha ⊂ Hb and |f |b ≤ C(Ω, a)|f |a .

(2.4)

Proof. Suppose a = k + α and b = j + β. If j = k, then (2.4) follows from (2.3), so we just have to prove the result for k > j, and an induction argument shows that we only need to check the case k = 1, j = 0, and β = 1. Let f ∈ H1+α . We now show that [f ]1 ≤ C(Ω)(|Df |0 + |f |0 ).

(2.5)

We first note that, if x and y are points in Ω such that the line segment joining x and y is a subset of Ω, then the mean value theorem gives a point x1 on that line segment such that f (x) − f (y) =

∂f (x1 )(x − y) ∂ξ

where ξ is the unit vector in the direction of x − y. Hence for any two such points, we have |f (x) − f (y)| ≤ |Df |0 |x − y|. In particular, this equation holds if |x − y| ≤ d(x).

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Since Ω is Lipschitz, there are positive constants R and ω0 such that, for each x0 ∈ ∂Ω, there is a function ω and a coordinate system Z = (z 1 , . . . , z n ) centered at x0 such that Ω ∩ B(x0 , R) = {x ∈ B(x0 , R) : z n > ω(z 0 )} and |ω(z10 ) − ω(z20 )| < ω0 |z10 − z20 |

p for z10 and z20 with |z10 |, |z20 | < R. We now set k1 = 2 1 + ω02 and K = 2(1 + k1 ), and we suppose that d(x) ≤ |x − y| ≤ R/K. Using x0 , a nearest point to x on ∂Ω as the center of our Z coordinates and writing ξ for the unit vector parallel to the positive z n axis, simple calculation shows that t min{d(x + tξ), d(y + tξ)} ≥ p 1 + ω02

for any t > 0 such that x+tξ and y+tξ are in B(x0 , R)∩Ω. In particular, t = k1 |x − y| satisfies this condition, so x ¯ = x + k1 |x − y|ξ and y¯ = y + k1 |x − y|ξ are in B(x0 , R) ∩ Ω and |¯ x − y¯| ≤ d(¯ x)/2. In addition, the line segments joining x and x ¯, x ¯ and y¯, and y¯ and y are in Ω and hence |f (x) − f (y)| ≤ |f (x) − f (¯ x)| + |f (¯ x) − f (¯ y )| + |f (¯ y) − f (y)| ≤ |Df |0 (|x − x ¯| + |¯ x − y¯| + |¯ y − y|)

= (2k1 + 1)|Df |0 |x − y|. If |x − y| ≥ R/K, then

2K |f |0 |x − y|. R Combining the three cases yields (2.5), and then we have |f (x) − f (y)| ≤ 2|f |0 ≤

|u|1 ≤ C|u|1+α for any α ∈ (0, 1) because |u|0 + |Du|0 ≤ |u|1+α .



In fact, a similar monotonicity result is true for weighted norms. Lemma 2.2. Let a and α be real numbers with a > 0 and a + α ≥ 0, and let Ω be an open subset of Rn . (α)

(1) If b ∈ (0, a) and b + α ≥ 0, then Ha (α)

for all f ∈ Ha .

(α) |f |b

≤

(α)

⊂ Hb

|f |(α) a

with (2.6)

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(2) If γ ≥ α, then Ha for all f ∈

(α)

⊂ Ha

|f |(α) a

(α) Ha .

51

with ≤ (diam Ω)γ−α |f |(γ) a

(2.7)

Proof. To prove (2.6), we consider several cases. First, if a ∈ (0, 1], then (α) (α) we use the definition of [f ]b and [f ]b to conclude that (α)

[f ]b

≤ [f ](α) a .

(2.8)

Next, the argument in the proof of Lemma 2.1 implies that (α)

[f ]1

(1+α)

≤ |Df |0

,

(2.9)

and hence (2.8) holds for b = 1 and a ∈ (1, 2]. An easy induction argument completes the proof. Similarly, it’s easy to check that γ−α [f ](γ) [f ](α) a ≤ (diam Ω) a

for any a and γ ≥ α ≥ −a, so (2.7) is also valid.



Another useful property of the H¨older norms is an interpolation inequal(α) ity. For the Ha norms with fixed α, the interpolation inequality is valid for any domain. The case α = 0 appears in Proposition 4.2 of [114] and Lemma 6.32 of [64], and we give a quick description for the case of arbitrary α (which is very close to the case α = 0). Lemma 2.3. Let 0 < a < b be real numbers, let α ≥ −a, and let σ ∈ (0, 1). Then there is a constant C(b) such that (α)

(α)

σ 1−σ [u]σa+(1−σ)b ≤ C(b)(|u|(α) . a ) (|u|b )

Proof. define

We consider several cases, and, to simplify some notation, we (α)

[u]0

= sup{d(x)α |u(x) − u(y)| : |x − y|
0. Then (σα+(1−σ)β)

|f |σa+(1−σ)b

(β)

σ 1−σ ≤ C(b)(|f |(α) . a ) (|f |b )

(2.12)

Proof. We first note that the inequality is immediate if a = b or if b ≤ 1. Now suppose that 0 < a < 1 < b ≤ 2 and a + β ≥ 0. Let us fix a point x ∈ Ω and apply Lemma 2.4 in the ball B(x, r) with r = d(x)/5. Then, with σ chosen so that 1 = σa + (1 − σ)b and t = σα + (1 − σ)β), we have r1+t |Df (x)| ≤ C(ra+α [f ]a;B(x,r) )σ (rb+β [f ]b;B(x,r) + ra+β [f ]a;B(x,r) )1−σ .

Now we estimate the factors on the right side of this equation. Let y and z be points in B(x, r). Then d(y) ≥ d(x) − |y − z| ≥

4 d(x) 5

and |y − z| ≤

2 1 d(x) ≤ d(y), 5 2

so −a−α |y − z|a |f (y) − f (z)| ≤ [f ](α) a d(y)

−a−α ≤ C[f ](α) |y − z|a a d(x)

and hence (taking the supremum over all y and z in B(x, r)), we have ra+α [f ]a;B(x,r) ≤ C[f ](α) a . Therefore (σα+(1−σ)β−1)

|Df |0

(β)

σ ≤ C([f ](α) a ) ([f ]b

1−σ + [f ](β) . a )

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Using (2.6), we see that (σα+(1−σ)α)

[f ]1

(β)

σ 1−σ ≤ C([f ](α) . a ) (|f |b )

(2.13)

Next, we remove the restriction a + β ≥ 0 in this inequality by considering two cases. First, if a + σα + (1 − σ)β > 0, then there is a θ ∈ (σ, 1) such that a + θα + (1 − θ)β ≥ 0. We now set c = θa + (1 − θ)b and γ = θα + (1 − θ)β. With δ ∈ (0, 1) taken so that 1 = δa + (1 − δ)c, we infer from (2.13) (with b replaced by c) that (σα+(1−σ)β)

[f ]1

δ (γ) 1−δ . ≤ C([f ](α) a ) (|f |c )

The case b = 1 now implies that (γ−1)

[Df ]c−1

(σα+(1−σ)β−1) λ

(β−1)

) (|Df |b−1 )1−λ ,

≤ ([Df ]0

where λ satisfies λ + (1 − λ)b = c. It follows that (σα+(1−σ)β) λ

≤ ([f ]1 [f ](γ) c

(β)

) (|Df |b )1−λ , (σα+(1−σ)β

and combining this inequality with the one for [f ]1 yields (2.13) in this case. On the other hand if a + σα + (1 − σ)β ≤ 0, then there is a constant θ ∈ (0, σ) such that c = θa + (1 − θ)b is less than 1 and c + β ≥ 0. (Here, we note that 1 + σα + (1 − σ)β > 0.) It follows that (σα+(1−σ)β)

[f ]1

(β)

δ 1−δ ≤ C([f ](γ) , c ) (|f |b )

where γ = θα+(1−θ)β, just as before, but now δ satisfies δγ +(1−δ)β = 1. In addition, we conclude from the b ≤ 1 case that (σα+(1−σ)β) 1−λ

λ [f ](γ) ≤ ([f ](α) c a ) ([f ]1

)

,

with λa + 1 − λ = c. Combining these inequalities yields (2.13) in this case as well. From this estimate, we obtain the full interpolation inequality by arguing as in Lemma 2.3.  Note that the case a + α = b + β = 0 is covered by Lemma 2.4. In this case, Ω must be a bounded Lipschitz domain and the constant C depends also on Ω. We shall also have occasion to consider weighted H¨older spaces where the weighting is with respect to distance to just a part of the boundary. It’s possible to give a thorough treatment under very weak assumptions (for example, the weighting can be with respect to an arbitrary non-empty portion of the boundary and there can be different weightings with respect

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to distance to different parts of the boundary), but we just examine a relatively simple situation. In this way, we avoid some technical points that are not necessary for our program. We refer the reader to the Notes for a discussion of some of these issues. Let Σ be a non-empty subset of ∂Ω and write d∗ for the distance to Σ. (b) We then define Ha (Ω; Σ) to be the set of all functions with finite norm (b) (b) | · |a:Ω,Σ , where the norm is defined by replacing d in the definition of | · |a;Ω by d∗ . These spaces are still Banach spaces but some care must be taken in order to guarantee that the interpolation theorems are still valid. As before, the correct condition is a Lipschitz one. We say that Σ is a C 0,1 subset of ∂Ω ∈ C 0,1 if, for each x0 ∈ Σ, there are two Lipschitz functions ω and ωσ defined on some ball B 0 (0, R) ⊂ Rn−1 and a coordinate system Y centered at x0 such that and

Ω ∩ B(x0 , R) = {x ∈ B(x0 , R) : y n > ω(y 0 )} Σ ∩ B(x0 , R) = {x ∈ B(x0 , R) : y n = ω(y 0 ), ωσ (y 0 ) > 0}.

As a simple representative example, if Ω is the half-ball {(x, y) ∈ R2 : x2 + y 2 < 1, y > 0}, then Σ = {(x, y) : x2 + y 2 < 1, y = 0}

is a C 0,1 subset of ∂Ω. The assumption ∂Ω ∈ C 0,1 can be relaxed if we appropriately modify the definition of C 0,1 subset of ∂Ω but there is no reason to do so here. It’s then easy to check that, if we replace d by d∗ in the proofs of (2.2), Lemma 2.2, and Lemma 2.5, then we obtain the corresponding results with these partially interior norms, so we just state the results. Lemma 2.6. Let a and α be real numbers with a > 0 and a + α ≥ 0, let Ω be a bounded Lipschitz domain, and let Σ be a C 0,1 portion of ∂Ω. (α)

(β)

(1) Let β ≥ max{−a, −a − α}. If f ∈ Ha (Ω, Σ) and g ∈ Ha (Ω, Σ), then (α+β) f g ∈ Ha (Ω, Σ) and (α+β)

(α)

(β)

|f g|a;Ω,Σ ≤ |f |a;Ω,Σ |g|a;Ω,Σ .

(2) Let b ∈ (0, a] and β ≥ max{α, −b}. Then, for any f ∈ have (β)

(α)

|f |b;Ω,Σ ≤ C(Ω, Σ, a)|f |a;Ω,Σ .

(2.14) (α) Ha (Ω, Σ),

we

(2.15)

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(3) Let b ∈ (0, a), β ≥ −b, and σ ∈ (0, 1) be real numbers. Then, for any (α) (β) f ∈ Ha (Ω, Σ) ∩ Hb (Ω, Σ), we have (σα+(1−σ)β)

|f |σa+(1−σ)b

(β)

σ 1−σ ≤ C(b, Ω, Σ)(|f |(α) . a ) (|f |b )

(2.16)

As before, the constant C in (2.16) is independent of Ω and Σ unless a + α = b + β = 0. We close this section with a simple rule which governs the behavior of weighted H¨ older norms under a change of independent variable. Lemma 2.7. Let a and α be real numbers with a > 0 and α ≥ −a. Let Ω and Ω0 be open subsets of Rn and suppose that there is a one-to-one, onto function Ψ : Ω → Ω0 such that Ψ ∈ Hmax{1,a} (Ω) and Ψ−1 ∈ H1 (Ω0 ). If (α)

(α)

f ∈ Ha (Ω0 ), then f ◦ Ψ ∈ Ha (Ω) and there is a constant C, determined only by a, α, |Ψ|max{1,a} , and |Ψ−1 |1 such that (α) |f ◦ Ψ|(α) a ≤ C|f |a .

(2.17)

Proof. Note that there is a constant C1 , determined only by the Lipschitz norms of Ψ and Ψ−1 , such that 1 |x − y| ≤ |Ψ(x) − Ψ(y)| ≤ C1 |x − y| C1 and 1 d(x) ≤ d(Ψ(x)) ≤ C1 d(x) C1 for all x and y in Ω. It immediately follows that (α)

|f ◦ Ψ−1 |0

(α)

≤ C|f |0 .

In addition, there is a positive integer N (determined only by the Lipschitz norms of Ψ and Ψ−1 ) such that, if x and y satisfy |x − y| ≤ 12 d(x), then there are points x0 , . . . , xN with x0 = x and xN = y such that |Ψ−1 (xi ) − Ψ−1 (xi−1 )| ≤

1 d(Ψ−1 (xi )) 2

for i ∈ {1, . . . , N }. This observation implies the result for a ≤ 1, and the case a > 1 follows from this one by the chain rule. 

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An alternative characterization of H¨ older spaces (b)

The definition of Ha will be very useful in understanding the behavior of solutions of oblique derivative problems; however, our analysis of this behavior is greatly facilitated by the use of other means of showing that a function is in some H¨ older space. In this section, we present a characterization that is relevant to the work in this chapter. This time we refer to Lemma 12.12 from [114] for a complete proof (in the parabolic case) and just give an outline of the proof. Lemma 2.8. Let N be a nonnegative integer, let α ∈ (0, 1], let R > 0, let x0 ∈ Rn , and let u ∈ C(B(x0 , R)). Suppose that there is a positive constant H such that, for each y ∈ B(x0 , R/2) and each r ∈ (0, R/2), there is a polynomial PN (y, r) of degree no greater than N with sup |u − PN (y, r)| ≤ HrN +α .

(2.18)

B(y,r)

Then u ∈ HN +α (B(x0 , R/2)) and [u]N +α;B(x0 ,R/2) ≤ C(N, α)H. Proof. We just prove the result for N = 0. Let ϕ be a C ∞ (Rn ) function with support in the unit ball B(0, 1) such that ϕ ≥ 0 and Z ϕ(x) dx = 1. Rn

Set K = sup |Dϕ| and note that ϕ can be chosen so that K is determined only by N . For τ ∈ (−R/2, R/2) and x ∈ B(x0 , R/2), we define Z u ¯(x, τ ) = u(x − τ z)ϕ(z) dz. Rn

By writing this equation in the more usual form   Z x−w −n u ¯(x, τ ) = (−τ ) u(w)ϕ dw, τ Rn we see that u ¯ is differentiable with respect to x and τ as long as τ 6= 0. Moreover, Z ∂u ¯ 1 (x, τ ) = − u(x − τ z)Dϕ(z) dz, ∂x τ Rn Z ∂u ¯ 1 (x, τ ) = − u(x − τ z)Di (z i ϕ(z)) dz. ∂τ τ Rn

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Hence, if we write ∇¯ u for the (n vector  + 1)-dimensional  ∂u ¯ ∂u ¯ , , ∂x ∂τ we see after a quick calculation that |∇¯ u(x, τ )| ≤ C(α, n)Hτ α−1 . We now fix x and y in B(x0 , R/2) and take τ = |x − y|/2. Since u(x) = u ¯(x, 0), it follows that |u(x) − u(y)| ≤ |¯ u(x, 0) − u ¯(w, τ )| + |¯ u(w, τ ) − u¯(y, 0)| for w = (x + y)/2. Since Z 1 u ¯(x, 0) − u ¯(w, τ ) = ∇u(sx + (1 − s)w, sτ ) · (x − y, τ ) dx, 0

it follows by integration that |¯ u(x, 0) − u ¯(w, τ )| ≤ CHτ α , and a similar argument gives |¯ u(w, τ ) − u ¯(y, 0)| ≤ CHτ α . The proof is completed by adding these inequalities and recalling the definitions of τ and [u]α;B(x0 ,R) . 

We shall also need a corresponding result near the boundary, which is proved by modifying the proof of Lemma 12.12 in [114] in the same way that the proof of Lemma 4.3 in [114] was modified to obtain Lemma 4.11 of [114], specifically, we assume that ϕ(x) = 0 for any x with xn < 0. In the statement, we write B + (x0 , r) and B 0 (x0 , r) for the set of all x ∈ B(x0 , r) for which xn > 0 or xn = 0, respectively. Lemma 2.9. Let N be a nonnegative integer, let α ∈ (0, 1], let R > 0, let x0 ∈ Rn with xn0 = 0, and let u ∈ C(B(x0 , R)). Suppose that there is a positive constant H such that, for each y ∈ B 0 (x0 , R/2) and each r ∈ (0, R/2), there is a polynomial PN (y, r) of degree no greater than N with sup |u − PN (y, r)| ≤ HrN +α . (2.19) B + (y,r)

0

Then u ∈ HN +α (B (x0 , R/2)) and [u]N +α;B 0 (x0 ,R/2) ≤ C(N, α)H. Suppose also that, for every y ∈ B + (x0 , R/2) and each r ∈ (0, y n ), there is a polynomial PN (y, r) of degree no greater than n with sup |u − PN (y, r)| ≤ HrN +α . B(y,r)

Then u ∈ HN +α (B + (x0 , R/2)) and [u]N +α;B + (x0 ,R/2) ≤ C(N, α)H.

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An existence result

In this section, we show that the Dirichlet problem for a constant coefficient equation can always be solved in a ball. More precisely, we have the following existence result. Theorem 2.10. Let [Aij ] be a positive definite matrix, let x0 ∈ Rn , let R > 0, and set B = B(x0 , R). Then, for any polynomials f and ψ, there is a unique solution u ∈ C(B) ∩ C 2 (B) of Aij Dij u = f in B,

u = ψ on ∂B.

(2.20)

Proof. First, note that Aij Dij ψ is also a polynomial, so there is no loss of generality in assuming that ψ = 0. For each nonnegative integer N , write PN for the set of all polynomials of degree no greater than N , and define the operator T (which maps PN into PN for any N ) by T w(x) = Aij Dij ((R2 − |x − x0 |2 )w(x)). If T w = 0, then the maximum principle implies that u, defined by u(x) = (R2 − |x − x0 |2 )w(x),

(2.21)

must attain its maximum and minimum values on ∂B. Since u = 0 on ∂B, it follows that u = 0 in B and hence w = 0, so T is one-to-one. By linear algebra, we conclude that T is also onto. Now, let f be a given polynomial. Then there is a unique polynomial w such that T w = f , and hence u given by (2.21) solves the boundary value problem (2.20) (with ψ = 0). The maximum principle implies that this u is the unique solution of this boundary value problem.  In a later section, we shall prove that (2.20) has a unique solution for a very broad choice of functions f and ψ. As a further preliminary to this result, we note that the Weierstrass approximation theorem implies that, for any continuous function ψ defined on ∂B, there is a sequence of polynomials converging uniformly on ∂B to ψ. The maximum principle then implies the following result. Corollary 2.11. Let [Aij ] be a positive definite matrix, let x0 ∈ Rn , let R > 0, and set B = B(x0 , R). Then, for any polynomial f and any continuous function ψ, there is a unique solution u ∈ C(B) ∩ C 2 (B) of (2.20).

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Basic interior estimates

We are now ready to prove some estimates for solutions of constant coefficient equations in a ball. As a first step, we obtain an estimate with constant Aij but variable f . Proposition 2.12. Let Ω be a non-empty, open subset of Rn , let [Aij ] be a positive-definite matrix, let α ∈ (0, 1) and suppose u ∈ C 2 (Ω) ∩ L∞ (Ω) satisfies Aij Dij u = f in Ω (2)

(2.22)

(0)

for some f ∈ Hα . Then u ∈ H2+α and (0)

|u|2+α ≤ C(α, n, µ)(|u|0 + |f /λ|(2) α ),

(2.23)

where λ and µ are constants such that Aij ξi ξj ≥ λ|ξ|2 , max |Aij | ≤ µλ i,j

(2.24)

for all ξ ∈ Rn . Proof. Fix x0 ∈ Ω. We shall estimate [u]2+α;B(x0 ,d(x0 )/2) . (As a byproduct of our estimate, we show that this quantity is finite.) To this end, we set R = d(x0 )/2, and we fix y ∈ B(x0 , R). Because of Lemma 2.8, we need to examine the supremum of the difference between u and some polynomials on balls centered at y. To this end, we obtain the polynomials as the second order Taylor polynomials of appropriate solutions of boundary value problems. We begin by setting u0 = u and defining v0 to be the solution, given by Corollary 2.11, of Aij Dij v0 = f (y) in B(y, R),

v0 = u0 on ∂B(y, R).

Since f (y) is a constant, we can use Theorem 1.34 to infer that there is a constant C0 (n, µ) so that sup B(y,R/2)

|D3 v0 | ≤ C0 (R−3 sup |v0 | + R−1 |f (y)/λ|) B(y,R)

≤ C0 (R

−3

sup |u0 | + R−1 sup |f /λ|).

B(y,R)

B(y,R)

We next define the second degree polynomial Q0 by 1 Q0 (x) = v0 (y) + Di v0 (y)(x − y)i + Dij v0 (y)(x − y)i (x − y)j . 2

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For any r ∈ (0, R/2), it follows that  r 3 r3 sup |v0 − Q0 | ≤ C0 sup |u0 | + C0 sup |f /λ|. R B(y,R) R B(y,R) B(y,r)

(2.25)

It also follows from the maximum principle that

sup |u0 − v0 | ≤ C1 R2 sup |(f − f (y))/λ|.

B(y,R)

B(y,R)

(2)

Setting F = |f /λ|α and combining these two inequalities, we find that  r 3 sup |u| + (C0 + C1 )F. sup |u0 − Q0 | ≤ C0 R B(y,R) B(y,r)

We now set θ = (1 + α)/2 and fix τ ∈ (0, 1/4) so that C0 τ 3 ≤ τ 2+θ to conclude that sup |u0 − Q0 | ≤ τ 2+θ sup |u| + (C0 + C1 )F.

B(y,τ R)

(2.26)

B(y,R)

We obtain a suitable sequence of polynomials by induction. The method can be described briefly by saying that we replace u by uk , and R by Rk = τ k R, but we shall be more specific. For each positive integer m, we set Rm = τ m R. Suppose that we have already determined second degree polynomials Q0 , . . . , Qk−1 , and functions u0 , . . . , uk−1 so that sup B(y,τ Rm )

|um − Qm | ≤ τ 2+α

sup B(y,Rm )

|um | + (C0 + C1 )

2+α Rm F R2+α

(2.27)

and ij

A Dij Qm =

(

f (y)

if m = 0,

0

if m > 0

for m = 0, . . . , k − 1. We then set uk = uk−1 − Qk−1 and take vk to be the solution of Aij Dij vk = 0 in B(y, R),

vk = uk on ∂B(y, R).

We also take Qk to be the second degree Taylor polynomial of vk centered at y. Our preceding argument shows that sup B(y,τ Rk )

|vk − Qk | ≤ τ 2+θ sup |uk |, B(y,R)

and the maximum principle gives sup |uk − vk | ≤ C1 Rk2 sup |(f − f (y))/λ| ≤ C1

B(y,Rk )

B(y,Rk )

Rk2+α F R2+α

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because Aij Dij uk = f − f (y). Combining these two estimates yields (2.27) for m = k as well (because C0 ≥ 0), and clearly Aij Dij Qk = 0. For our next step, we define polynomials P2 (y, r) for any r ∈ (0, R) as Pk−1 follows: if r ∈ (Rk , Rk−1 ], we set P2 (y, r) = m=0 Qm . Then w defined by w(r) = sup |u − P2 (y, r)| B(y,r)

is easily seen to be increasing on each interval (Rk , Rk−1 ], so Lemma 1.25 (with σ(r) = F (r/R)2+α , 2 + θ in place of α, and 2 + α in place of δ) implies that  r 2+α [|u|0 + F ], w(r) ≤ C R

and the proof is completed by invoking Lemma 2.8 and the definition of (0) [u]2+α . 

A slight modification of the preceding argument allows us to consider variable coefficients in the second order terms and nonzero coefficients in lower order terms as well. Proposition 2.13. Let Ω be a non-empty, open subset of Rn , let [aij ] be a positive-definite matrix-valued function defined on Ω, let α ∈ (0, 1) and suppose u ∈ C 2 (Ω) ∩ L∞ (Ω) satisfies aij Dij u + bi Di u + cu = f (2)

(0)

(1)

(2.28) (2)

for some f ∈ Hα . If aij ∈ Hα , bi ∈ Hα , and c ∈ Hα , and if the minimum eigenvalue of [aij ] is bounded away from zero and infinity, then (0) u ∈ H2+α . Moreover, if A, λ, and µ are positive constants such that aij ξi ξj ≥ λ|ξ|2 , ij

a ξi ξj ≤ µλ|ξ|

2

(2.29a) (2.29b)

for all ξ ∈ Rn , |aij |(0) α ≤ λA, |bi |(1) α ≤ λA, |c|(2) α

≤ λA,

(2.29c) (2.29d) (2.29e)

then (0)

|u|2+α ≤ C(A, α, n, µ)(|u|0 + |f /λ|(2) α ).

(2.30)

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Proof. First, we suppose that bi and c are zero and that |u|2 is finite. As in the proof of Proposition 2.12, we fix x0 ∈ Ω, set R = d(x0 )/2, and let y ∈ B(x0 , R). We then set u0 = u and define v0 to be the solution of aij (y)Dij v0 = f (y) in B(y, R), v0 = u0 on ∂B(y, R). Just as before, there are a polynomial Q0 and a constant C0 (n, µ) so that (2.25) is satisfied. We now compute aij (y)Dij (u0 − v0 ) = F0 , where F0 = (aij (y) − aij )Dij u0 + f − f (y). Since (0) −2 , |F0 | ≤ C(n)R−2 |u|2 Aλ + [f ](2) α R we infer from the maximum principle that |u0 − v0 | ≤ C(n, A, µ)F. where now (0) F = |u|2 + |f /λ|(2) (2.31) α . With τ chosen as in the proof of Proposition 2.12, we infer (2.26). A similar argument produces (2.27) (but F given by (2.31)) and hence (0) (0) |u|2+α ≤ C(|u|0 + |u|2 + |f /λ|(2) (2.32) α ). The desired result follows from this inequality via Lemma 2.3. To remove the restriction that bi and c are zero, we set f˜ = f −bi Di u−cu. Then (2.2) and (2.6) imply that (0) (2) |f˜|(2) α ≤ |f |α + C(n, A, α)λ|u|2 . The preceding argument shows that (0) (0) |u|2+α ≤ C(|u|0 + |u|2 + |f˜/λ|(2) α ), and hence (2.32) holds in this case as well. (0) To remove the restriction that |u|2 is finite, we first write Ωk (for k a positive integer) for the set of all x ∈ Ω satisfying |x| < k and d(x) > 1/k. For k sufficiently large, Ωk is nonempty, and the assumption u ∈ C 2 (Ω) (0) implies that u ∈ H2 (Ωk ). Hence (0)

(2)

|u|2+α;Ωk ≤ C(|u|0;Ωk + |f /λ|α;Ωk ). We now observe that (2) |u|0;Ωk + |f /λ|α;Ωk ≤ |u|0 + |f /λ|(2) α and that (0) (0) |u|2+α = lim |u|2+α;Ωk k→∞

to complete the proof.



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For our study of oblique derivative problems, we shall need a similar estimate for a slightly different set of functions on the right hand side. Lemma 2.14. Let Ω be an open subset of Rn and let u ∈ C 2,α (Ω) satisfy the equation Lu = f in Ω for some α ∈ (0, 1). Suppose the coefficients of L satisfy (2.29). Let b ∈ R. Then (b)

|u|2+α ≤ C(n, α, µ, A)(sup |db u| + |f /λ|(2+b) ). α

(2.33)

Ω

Proof. Let x 6= y be points in Ω with |x − y| ≤ 12 d(x). If |x − y| ≤ 14 d(x), we apply Proposition 2.13 in the ball B = B(x, d(x)/2) to conclude that d(x)2+α+b

|D2 u(x) − D2 u(y)| (2) ≤ Cd(x)b (|u|0;B + |f /λ|α;B ) |x − y|α ≤ C(sup |db u| + |f /λ|(2+b) ). α Ω

If |x − y| ∈ d(x)2+α+b It follows that

( 14 d(x), 21 d(x)), 2

we have

2

|D u(x) − D u(y)| ≤ Cd(x)2+b (|D2 u(x)| + |D2 u(y)|). |x − y|α (b)

(2+b)

[u]2+α ≤ C(sup |db u| + |f /λ|(2+b) + |D2 u|0 α (2)

).

(0)

Since |D2 u|0;B ≤ C|u|2+α;B , we can use a similar argument to show that (2+b)

|D2 u|0;B

(2+b) ≤ C(sup |db u| + |f /λ|α ),

and the proof is completed by combining these inequalities.



Of course, when b = 0, this result is identical to Proposition 2.13. If b < 0, it’s clear that sup |db u| can only be finite if u = 0 on ∂Ω. 2.5

The Perron process for the Dirichlet problem

There are many ways in which we can use the estimates from the preceding section to show that the Dirichlet problem has a classical solution under weak conditions on the domain and on the coefficients of the differential equation. This section is devoted to a way that will be used also to prove the existence of solutions of oblique derivative problems via the maximum principle. It is based on Perron’s method of subharmonic functions [152] with some crucial modifications from [129]. We present this method in an

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axiomatic format; a more relaxed description is presented in many sources (for example, Section 2.8 in [64]), but we want to make the transition to the oblique derivative problem as simple as possible. In this section, [Aij ] denotes a constant positive definite matrix. We use λ and µ to denote positive constants so that conditions (2.24) hold. To begin, we say that the differential equation Aij Dij u = f is locally solvable in Ω if, for every x0 ∈ Ω, there is an open set N with x0 ∈ N and N ⊂ Ω such that the Dirichlet problem Aij Dij w = f in N,

w = g on ∂N

(2.34)

has a solution w for every continuous function g. We emphasize that the open set N (for each x0 ) and the function f are considered fixed here. (At present, we only know that the differential equation Aij Dij u = f is locally solvable if f is a polynomial, but this situation will change soon.) We then say that v is a Perron supersolution of Aij Dij u = f in Ω if, for every x0 ∈ Ω and every continuous function g such that v ≥ g on ∂N (the open set from the definition of local solvability), we have that the solution w of (2.34) is no greater than v in N . Perron subsolutions are defined similarly. To solve the Dirichlet problem Aij Dij u = f in Ω,

u = 0 on ∂Ω,

(2.35)

we assume certain properties about f and Ω. In particular, we shall assume that f ∈ C α (Ω) for some α ∈ (0, 1). Solvability for nonzero boundary data will be discussed later. We now assume that Aij Dij u = f in Ω is locally solvable, we suppose that there is a Perron subsolution v − of Aij Dij u = f in Ω such that v − ∈ C(Ω) and v = 0 on ∂Ω, and we assume that S + , the set of all bounded Perron supersolutions v of Lu = f in Ω such that lim v(x) = 0

x→y x∈Ω

for all y ∈ ∂Ω and v ≥ v − in Ω, is nonempty. The solution of (2.35) will be given by u(x) = inf+ v(x) v∈S

for all x ∈ Ω. So we need to verify that u really is the solution of (2.35). First, we verify that u ∈ S + . It’s easy to check that u ≥ v − and that lim u(x) = 0

x→y x∈Ω

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for all y ∈ ∂Ω. In addition, if x0 ∈ Ω and g is a continuous function such that g ≤ u on ∂N , then g ≤ v on ∂N for any v ∈ S + and the definition of Perron supersolution implies that g ≤ v in N for any v ∈ S + , and hence g ≤ u in N . Proving that u actually solves the differential equation requires a little more work. We start with a modification of the harmonic lifting (see (2.33) in [64]). For x0 ∈ Ω and v ∈ S + , we define a function vN in three steps as follows: First, for any continuous function g with g ≤ v on ∂N , we write gN for the solution of (2.34). We then define vN on N by vN (x) = sup{gN (x) : g is continuous and v ≥ g ≥ v − on ∂N }, and we define vN on Ω \ N by vN (x) = v(x). We want to show that vN ∈ S + , that LvN = f in N , and that vN ≤ v in Ω. It’s immediate that lim v (x) x→y N x∈Ω

=0

because vN = v in a neighborhood of ∂Ω. Before completing the verification that vN ∈ S + , we show that Aij Dij vN = f in N . In particular, it will follow that vN ∈ C 2 (N ). Now let x1 ∈ N and let R be a positive number so that B(x1 , R) ⊂ N . Then there is a constant G0 such that every g with v ≥ g ≥ v − satisfies |g| ≤ G0 . By the interior Schauder estimate, Proposition 2.13, we have that |D2 gN |α;B(x1 ,R/2) ≤ C(R, G0 ) so the Arzela-Ascoli theorem implies that we can find a sequence (g (m) ) of continuous functions satisfying v ≥ g (m) ≥ v − (m) and gN converges in C 2 (B(x1 , R/2)) to a limit function which must be vN . Hence Aij Dij vN = f in N . Next, we show that vN ≤ v in Ω. Since the two functions are equal on Ω \ N , we just have to check that vN ≤ v in N , so let g be a continuous function such that v ≥ g ≥ v − on ∂N . Then gN ≤ v in N and the definition of VN implies that vN ≤ v in N . To see that vN is a supersolution, we fix x1 and let g be a continuous function such that vN ≥ g ≥ v − on ∂N1 , where N1 is the open set corresponding to x1 . We write w1 for the solution of Aij Dij w1 = f in N,

w1 = g on ∂N1 .

Then v ≥ w1 and hence vN ≥ w1 in N1 \ N . To proceed, we observe that N ∗ = N1 ∩ N is an open set and, if N ∗ is not empty, then v = vN on ∂N ∗ .

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It follows that vN ≥ w1 on ∂N ∗ and then the maximum principle implies that vN ≥ w1 in N ∗ as well, so vN ≥ w1 in N1 as required. To show that u is a solution of the differential equation, we need to show that u = uN for any open set N corresponding to a point x0 ∈ Ω. But this is easy. We have just shown that uN is in S + and that u ≥ uN , and the definition of u implies that u ≤ uN , so u = uN and hence u satisfies Aij Dij u = f in N for any such N . The only remaining step in our existence program is to show that S + is non-empty. For arbitrary domains, this set may be empty, but we show now that we can solve the Dirichlet problem in a ball under general hypotheses. Theorem 2.15. Let B be a ball in Rn and let α and δ be constants in (2−δ) (0, 1). Then, for any f ∈ Hα , there is a unique C 2 (B) ∩ C(B) solution u of Aij Dij u = f in B,

u = 0 on ∂B.

(2.36)

Moreover, we have (−δ)

|u|2+α ≤ C(α, δ, n, µ)|f /λ|(2−δ) . α

(2.37)

Proof. We begin by verifying local solvability under this hypothesis. For each x0 ∈ B, we write N for a ball centered at x0 , the closure of which is contained in B. It follows from Corollary 2.11 that (2.34) has a unique (2−δ) solution if f is a polynomial. For arbitrary f ∈ Hα , we note that f ∈ Hα (N ), so there is a sequence of polynomials (fm ) which converge in Hα (N ) to f . Let us now fix a continuous function g and write wm for the solution of Aij Dij wm = fm in N,

wm = g on ∂N.

It follows from the maximum principle that (wm ) is uniformly Cauchy and hence it has a limit w. The interior Schauder estimate, Proposition 2.13, implies that (D2 wm ) converges uniformly on compact subsets of N to D2 w and hence Aij Dij w = f in N . Our next step is to show the existence of subsolutions and supersolutions. To this end, we write y for the center of the ball and set h(x) = (R2 − |x − y|2 )δ . Then h ∈ C 2 (B) ∩ C(B) and Aij Dij h = δ(R2 − |x − y|2 )δ−2 ((R2 − |x − y|2 )(−2T) + 4(δ − 1)E),

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where T=

n X

aii ,

E = aij (x − y)i (x − y)j .

i=1

By examining the two cases |x − y| < ij

R 2

and |x − y| ≥

R 2,

we see that

2δ δ−2

A Dij h ≤ −c0 λR d

(2−δ)

for some positive constant c0 determined by n, δ, and µ. For F = |f /λ|0 it follows that F v− = − h c0 R2δ

,

is a subsolution satisfying lim v − (x) = 0

x→y x∈B

for all y ∈ ∂B and that v+ =

F h c0 R2δ

is in S + . The preceding argument shows that (2.36) has a solution u, and the maximum principle implies that it’s unique. Since −v − ≥ u ≥ v − , it also follows that |u| ≤ CF dδ and Lemma 2.14 implies (2.37).  From this result, we obtain an existence theorem and estimate for arbitrary continuous data. Corollary 2.16. With B, α, δ, and f as in Theorem 2.15, for any continuous function ψ, there is a unique C 2 (B) ∩ C(B) solution u of Aij Dij u = f in B,

Moreover, if ψ ∈ Hδ , we have (−δ)

u = ψ on ∂B.

(2.38)

(−δ)

(2.39)

|u|2+α ≤ C(α, δ, n, µ)(|ψ|δ

+ |f /λ|(2−δ) ). α

(−δ)

Proof. Suppose first that ψ ∈ Hδ . Then there is a function G ∈ H2+α with G = ψ on ∂B and (−δ)

(−δ)

|G|2+α ≤ C(α, n, δ)|ψ|δ

.

If v is the unique solution of Aij Dij v = f − Aij Dij G in B,

v = 0 on ∂B

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given by Theorem 2.15, then u = v + G solves (2.38) and (−δ)

(−δ)

(−δ)

|u|2+α ≤ |v|2+α + |G|2+α

(−δ)

≤ C|f − Aij Dij G|(2−δ) + |G|2+α α (−δ)

≤ C(|f |(2−δ) + |G|2+α ) α (−δ)

≤ C(|f |(2−δ) + |g|δ α

).

The existence for continuous g follows from this result because any continuous function is the uniform limit of a sequence of Hδ functions.  For later applications, we also obtain an existence result in any bounded convex domain. Corollary 2.17. Let Ω be a bounded convex open subset of Rn . Then, for any continuous function ψ defined on ∂B + , there is a unique C 2 (Ω) ∩ C(Ω) solution u of Aij Dij u = 0 in Ω,

u = ψ on ∂Ω.

(2.40)

Proof. Let (ψm ) be a sequence of polynomials converging uniformly to ψ, and set fm = Aij Dij ψm and Fm =

1 sup |fm |. λ Ω

Our first step is to show that, for any m, there is a solution um of Aij Dij um = fm in Ω,

um = 0 on ∂Ω.

(2.41)

Local solvability of this problem was established in Theorem 2.15, so we only need to prove the existence of a subsolution and a supersolution. To this end, we recall that at each point y ∈ ∂Ω, there is a supporting hyperplane and, therefore, a unit vector ξ(y) such that ξ(y) · (x − y) > 0 for all x ∈ Ω. If we set R = diam Ω and define hy by hy (x) = R[ξ(y) · (x − y)] − [ξ(y) · (x − y)]2 , then a direct calculation gives Aij Dij hy ≤ −λ in Ω for all y ∈ ∂Ω. We now claim that h, defined by h(x) = inf Fm hy (x) y∈∂Ω

is a supersolution. To verify this statement, let x0 ∈ Ω and let N be the corresponding neighborhood. If g is a continuous function on ∂N with

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g ≥ h on ∂N , then g ≥ hy on ∂N (for any y ∈ ∂Ω), so, if w is the solution of Aij Dij w = fm in N, w = g on N, we have Aij Dij hy ≤ fm in N , so the maximum principle implies that w ≤ hy in N . Since the set of functions {hy } is equicontinuous, it followss that h is continuous with h = 0 on ∂Ω, and hence h is a supersolution. A similar argument shows that −h is a subsolution, and hence (2.41) has a solution. We now set vm = um + ψm and observe that vm solves Aij Dij vm = 0 in Ω,

vm = ψm on ∂Ω.

Since (ψm ) converges uniformly to ψ, it follows from the maximum principle that (vm ) is uniformly Cauchy in Ω, so this sequence converges uniformly to a limit function u which satisfies the boundary condition u = ψ on ∂Ω. The interior Schauder estimate implies that (D2 vm ) is also uniformly Cauchy on compact subsets of Ω, so it converges uniformly on compact subsets of Ω to a limit, which must be D2 u, and hence u also satisfies the differential equation Aij Dij u = 0. 

2.6

A model mixed boundary value problem

The next step in our study of H¨older estimates for oblique derivative problems is an examination of a mixed boundary value problem, that is, one in which Dirichlet data are prescribed on part of the boundary and an oblique derivative condition is prescribed on the remainder of the boundary. For simple operators and domains, there are many ways to examine this problem, and we shall return to mixed boundary value problems several times in this book. The present approach will be useful in laying the groundwork for our estimates in later chapters. We begin by defining a standard set in which to study our problem. For a number κ ∈ (0, 1) to be further specified, we define the spherical cap Γ(R) for R > 0 by Γ(R) = {x ∈ Rn : |x0 |2 + |xn + Rκ|2 < R2 , xn > 0}.

(2.42)

(The introduction of the spherical cap should be viewed in the same light as that for the set E in Section 1.7; it will simplify certain calculations but it leads to more elaborate geometric considerations.) We then write Γ∗ (R)

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and Γ0 (R) for the subsets of ∂Γ(R) on which |x0 |2 + |xn + Rκ|2 = R2 and |x0 |2 + |xn + Rκ|2 < R2 , respectively. (Here, we have to be careful that Γ∗ is closed and Γ0 is open.) We are concerned in this section with solutions of the problem aij Dij u = f in Γ(R), i

0

β Di u = g on Γ (R), ∗

u = ψ on Γ (R)

(2.43a) (2.43b) (2.43c)

ij

under various hypotheses on the coefficients a and β and the functions f , g and ψ. Our goal is to show that this problem has a unique solution and that the solution satisfies appropriate estimates. We shall eventually see that the problem has a solution without any restrictions on κ, but, in the present section, we shall have to make a restriction on κ based on information about β. When we make the assumption, the reason for making it will become clear. In addition, we shall suppress the symbols Γ(R) and Γ∗ (R) from our weighted norms because we always consider weighted norms with respect (b) (b) to distance to Γ∗ (R). So we write Ha for Ha;Γ(R),Γ∗ (R) and similarly we abbreviate the weighted norms. As a first step, we prove everything for aij = δ ij and β i = δ in . This choice is dictated by our simple existence theorem for the Dirichlet problem, which we now extend to a somewhat different set, and by the nature of the mixed boundary value problem. We define ˆ Γ(R) = {x ∈ Rn : |x0 |2 + |xn + Rκ|2 < R2 , |x0 |2 + |xn − Rκ|2 < R2 },

ˆ and then our existence result can be extended to Γ(R).

Lemma 2.18. Let α and δ be constants in (0, 1). then, for any F ∈ (2−δ) ˆ ˆ ˆ ˆ Hα (Γ(R)) and Ψ ∈ C(∂ Γ(R)), there is a unique C 2 (Γ(R)) ∩ C(Γ(R)) solution v of ˆ ˆ ∆v = F in Γ(R), v = Ψ on ∂ Γ(R). Proof. As in Corollary 2.16, we may assume that Ψ = 0, and then it suffices to show that there exist a subsolution v − and a supersolution v + ˆ ˆ (both in C(Ω)) such that v − ≤ v + in Γ(R) and v ± = 0 on ∂ Γ(R). If we set h = min{(R2 − |x0 |2 − |xn − κR|2 )δ , (R2 − |x0 |2 − |xn + κR|2 )δ },

then we can take

v ± = ±Ch

for a suitable constant C, determined by R, n, and δ.



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From this existence result, we can infer an existence theorem for simple mixed boundary value problems. (2−δ)

Lemma 2.19. Let α and δ be constants in (0, 1). If f ∈ Hα , g ∈ (1−δ) H1+α , and ψ ∈ C(Γ∗ (R)), then there is a unique C 2 (Γ(R)) ∩ C(Γ(R)) solution of ∆u = f in Γ(R), ∂u = g on Γ0 (R), ∂xn u = ψ on Γ∗ (R).

(2.44a) (2.44b) (2.44c)

(−δ)

Moreover, if ψ ∈ Hδ , then u ∈ H2+α and (−δ)

(1−δ)

(−δ)

|u|2+α ≤ C(n, α, δ, κ)(|f |(2−δ) + |g|1+α + |ψ|δ α (1−δ)

).

(2.45)

(−δ)

Proof. Since g ∈ H1+α , there is an H2+α function G such that ∂G/∂xn = g on Γ0 (R). We set F = f − ∆G and Ψ = ψ − G, and then we ˆ extend F and Ψ as even functions to Γ(R). (That is, we define F (x0 , xn ) = F (x0 , −xn ),

Ψ(x0 , xn ) = Ψ(x0 , −xn ) ˆ for xn < 0.) It follows that Ψ is continuous on ∂ Γ(R) and F is H¨older ˆ continuous on Γ(R) so there is a unique solution v of the Dirichlet problem ˆ ˆ ∆v = F on Γ(R), v = Ψ on ∂ Γ(R). Since w defined by w(x0 , xn ) = v(x0 , −xn ) is another solution of this problem, it follows that v(x0 , xn ) = v(x0 , −xn ) and hence ∂v/∂xn = 0 on Γ0 (R). It’s now easy to verify that u = v + G is the desired function. If ψ ∈ Hδ , then we infer (2.45) from the argument leading to (2.39). 

Our next step is to prove the corresponding result with ∆ and ∂/∂xn replaced by constant coefficient operators. In this case, it can be shown that the existence result is still valid but the regularity of the solution near points of the form (x0 , 0) ∈ Γ∗ (R) is not as good as in this theorem, that is, even for smooth data, there may not be a solution in Hδ . For this reason and for computational simplicity, we put a restriction on κ which is connected to the vector β in the boundary condition β · Du = g. To be specific, we assume now that [Aij ] is a constant positive-definite matrix and that B is a constant vector with B n > 0. To state our results succinctly, we assume that there are positive constants λ, µ, χ, and µ0 such that Aij ξi ξj ≥ λ|ξ|2 , ij

2

A ξi ξj ≤ µλ|ξ| ,

(2.46a) (2.46b)

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for all ξ ∈ Rn , B n ≥ χ, 0

(2.46c) n

|B | ≤ µ0 B .

(2.46d)

We now fix κ so that κ∈



 µ0 , 1 . (1 + µ20 )1/2

(2.47)

The following preliminary estimate justifies the choice for κ. If we define h by h(x) = (R2 − |x0 |2 − |xn + κR|2 )δ , then a straightforward calculation shows that there are positive constants c1 (determined only by n and µ) and c2 (determined also by κ) such that Aij Dij h ≤ −c1 dδ−2 Rδ in Γ(R),

B i Di h ≤ −c2 dδ−1 Rδ on Γ0 (R).

We emphasize that this estimate on B i Di h is only valid if κ satisfies (2.47). In fact c2 → 0 as κ → µ0 /(1 + µ20 )1/2 . Our next step is a local gradient bound for solutions of the oblique derivative problem with constant right hand sides. Lemma 2.20. Under the preceding hypotheses on Aij and B, if v ∈ C 2 (B + (R)) ∩ C 1 (B + (R) ∪ B 0 (R)) solves Aij Dij v = F in B + (R),

B · Dv = G on B 0 (R)

(2.48)

for constants F and G, then there is a constant C (determined only by n, µ, and µ0 ) such that   G sup |v| RF + + (2.49) |Dv(x)| ≤ C λ χ R for |x| < R/4. Proof. The proof proceeds in several steps. The main step is to estimate the maximum of the gradient in terms of its L1 norm. This estimate proceeds differently for different components of the gradient. Let k ∈ {1, . . . , n − 1}, and set w=

v(x + hek ) − v(x) h

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for h small and nonzero. Then Aij Dij w = 0 in B + (R/2) and B ·Dw = 0 on B 0 (R/2), and it follows from the local maximum principle, Theorem 1.27, that Z sup

B+(R/4)

w ≤ CR−n

w dx.

B + (R/2)

Sending h → 0, we see that sup B + (R/4)

|Dk v| ≤ CR

−n

Z

B + (R/2)

|Dv| dx.

Next, let ε > 0 be given, and, for h small and positive, we define w by v(x + hβ) − v(x) w(x) = − G. h Then Aij Dij w = 0 in B + (R/2) and, if h is sufficiently small depending on ε, |w| ≤ ε on B 0 (R/2). It follows that Z −n sup w ≤ C[R w dx + ε] B + (R/2)

B+(R/4)

as well. Sending ε → 0 (and hence h → 0), we obtain ! Z −n sup |β · Dv| ≤ C R |β · Dv| dx + G . B + (R/4)

B + (R/2)

Combining this estimate with the one for Dk v, we see that ! Z G |Dv| dx + sup |Dv| ≤ C R−n . χ B + (R/4) B + (R/2) To proceed, we use the interior gradient bound, Theorem 1.33, (with u replaced by v − inf v) to infer that 1 |Dv(x)| ≤ C n osc v x B(x,xn /2) for any x ∈ B + (R/2). Then the H¨older estimate, Theorem 1.26, for v implies that osc

B(x,xn /2)

v ≤ CU R−δ (xn )δ

for U=

RG R2 F + + sup |v|. λ χ B + (R)

(2.50)

Therefore sup B + (R/4)

|Dv| ≤ CU R−n−δ

which is just the estimate (2.49).

Z

(xn )δ−1 dx = CU R−1 ,

B + (R/2)



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In fact, we can obtain estimates for higher order derivatives as well. Corollary 2.21. Under the hypotheses of Lemma 2.20, there is a constant C, determined by the same quantities as in that lemma, such that   F G sup |v| 2 sup |D v| ≤ C + + , (2.51a) λ Rχ R2 B + (R/2)   G sup |v| F + 2 + . (2.51b) sup |D3 v| ≤ C Rλ R χ R3 B + (R/2) Proof. We begin by combining the argument in the proof of Theorem 1.33 with (2.49) and Theorem 1.33 itself to infer that sup B + (3R/4)

|Dv| ≤ CU/R,

where U is defined by (2.50). To prove (2.51a), we define vk,h with k ∈ {1, . . . , n − 1} and h small and nonzero by v(x + hek ) − v(x) vk,h (x) = , h and use the preceding estimate for v and vk,h to infer that sup |Dvk,h | ≤ CU/R2 .

B ( R/2)

Sending h → 0 yields

sup B + (R/2)

|Dik v| ≤ CU/R2

for i ∈ {1, . . . , n} and k ∈ {1, . . . , n − 1}, and the differential equation for v gives the same estimate for i = k = n. To prove (2.51b), we define vjk,h with j and k in {1, . . . , n − 1} and h small and nonzero by v(x + hej + hek ) − 2v(x + hej ) + v(x) vjk,h = . h2 As before, we obtain sup B + (R/2)

|Dvjk,h | ≤ CU/R3 ,

and we infer from Theorem 1.34 that v ∈ C 3 (B + (R/2)), so sup

B + (R/2)

|Dijk v| ≤ CU/R3 ,

for i ∈ {1, . . . , n} and j and k in {1, . . . , n − 1}. For each k, we use the differential equation for Dk v to obtain the corresponding result for Dnnk v and the estimate for Dnnn v follows from the differential equation for Dn v. 

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A simple induction argument shows that we can estimate derivatives of v of any order in the corresponding way but with C depending also on the number of derivatives. From this estimate, we can derive a Schauder estimate and an existence result for solutions of the mixed boundary value problem with constant coefficients. Lemma 2.22. Let α and δ be constants in (0, 1) and suppose κ satisfies (2−δ) (1−δ) (2.47). Suppose conditions (2.46) are satisfied. If f ∈ Hα , g ∈ H1+α , and ψ ∈ C(Γ∗ (R)), then there is a unique C 2 (Γ(R)) ∩ C(Γ(R)) solution of Aij Dij u = f in Γ(R),

(2.52a)

0

B · Du = g on Γ (R),

(2.52b)

∗

u = ψ on Γ (R).

(2.52c)

(−δ)

Moreover, if ψ ∈ Hδ , then u ∈ H2+α and there is a constant C = C(n, α, δ, ω0 , κ, µ, µ0 ) such that (−δ)

(1−δ)

(−δ)

+ |g/χ|1+α + |ψ|δ |u|2+α ≤ C(|f /λ|(2−δ) α

).

(2.53)

Proof. For simplicity of notation, we replace Aij by Aij /λ and B by B/χ, so there is no loss of generality in assuming that λ = χ = 1. Moreover, as (−δ) usual, we may assume that ψ = 0. We also write X for the set of all H2+α (2−δ)

(1−δ)

functions which vanish on Γ∗ (R) and Y = Hα × H1+α . With these simplifications, we note that the lemma has already been proved for Aij = δ ij and B i = δ in . We then define i in ij ij i Aij s = sA + (1 − s)δ , Bs = sB + (1 − s)δ

for s = (0, 1), and we define the operator Js : X → Y by i Js = (Aij s Dij u, Bs Di u).

We now imitate the proof of the method of continuity to show that Js is onto if s is sufficiently small. Since J0 is invertible, we see that the equation Js u = y (for y = (f, g) ∈ Y ) is equivalent to J0 u = y + (J0 − Js )u = y + s(J0 − J1 )u, or u = J0−1 y + sJ0−1 (J0 − J1 )u. We now define the map T from X to itself by T u = J0−1 y + sJ0−1 (J0 − J1 )u.

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Then T is a contraction if s is sufficiently small and hence it has a unique fixed point, which is the unique solution of Aij s Dij u = f in Γ(R),

(2.54a)

0

Bs · Du = g on Γ (R),

(2.54b)

∗

us = 0 on Γ (R).

(2.54c)

From this argument, we conclude that (2.54) has a unique solution for any s ∈ [0, 1] if we can show that there is some constant C, independent of s, such that the uniform estimate kukX ≤ CkJs ukY

(2.55)

is valid for all u ∈ X, whenever (2.54) is solvable for all y = (f, g) in Y . To prove this estimate, we proceed in several steps. With h defined (as above) by h(x) = (R2 − |x0 |2 − (xn + κR)2 )δ , it’s easy to check from the maximum principle applied to  2  R F RG −δ CR + h±u λ χ that there is a constant C such that  2  R F RG |u| ≤ C + R−δ dδ . λ χ The proof is completed by using this estimate along with Corollary 2.21 in the proof of Lemma 2.14.  Once we have this general existence result, we infer the appropriate local estimates and existence for the oblique derivative problem. (0)

Theorem 2.23. Let aij ∈ Hα and suppose that there are positive con(0) stants A, λ and µ such that conditions (2.29a,b) are satisfied. Let β ∈ H1+α and suppose that there are positive constants B0 , χ, and µ0 such that β n ≥ χ,

0

(2.56a) n

|β | ≤ µ0 β ,

(2−δ)

(0) |β|1+α

(2.56b)

≤ B0 χ.

(2.56c)

(1−δ)

Then for every f ∈ Hα , g ∈ H1+α and ψ ∈ C 0 , there is a unique (−δ) solution u of (2.43). Moreover, if ψ ∈ Hδ , then u ∈ H2+α and there is a constant C, determined only by A, B0 , n, α, δ, µ, and α such that (−δ)

(−δ)

(−δ)

|u|2+α ≤ C(|f /λ|(2−δ) + |g/χ|1+α + |ψ|δ α

).

(2.57)

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Proof. As usual, we prove the result for ψ ≡ 0. From Lemma 2.22 and the proof of Proposition 2.13, we see that (−δ)

(−δ)

|u|2+α ≤ C(|aij Dij u/λ|(2−δ) + |β i Di u/χ|1+α ) α (−δ)

for all u ∈ H2+α which vanish on Γ∗ (R). We then use the method of continuity with Js defined by Js u = ([saij + (1 − s)δ ij ]Dij u, [sβ i + (1 − s)δ in ]Di u) to infer the existence result for ψ = 0.



For our existence program for the oblique derivative problem, an existence result and estimate for operators with lower order terms will be very useful. Note that an additional condition is placed on b beyond (2.29d). We refer to the Notes for a brief discussion of the significance of this addition, and only point out here that it is used to prove one of the intermediate estimates. For simplicity of notation, we introduce the operators L and M given by Lu = aij Dij u + bi u + cu,

M u = β i Di u + β 0 u.

(0)

(1)

(2.58)

(2)

Theorem 2.24. Let aij ∈ Hα , let bi ∈ Hα , let c ∈ Hα , and suppose that there are positive constants A, λ and µ such that conditions (2.29) are (0) (1) satisfied. Let β ∈ H1+α , let β 0 ∈ H1+α , and suppose that there are positive constants B0 , χ, and µ0 such that (2.56) holds and (1)

|β 0 |1+α ≤ B0 χ.

(2.59)

Suppose also that c ≤ 0, β 0 ≤ 0, and there are nonnegative constants B1 and B2 with B1 < 2(1 − δ)κ/(1 + κ) such that   B1 B2 b(x) · x¯ ≥ − + |¯ x| (2.60) d(x) R (2−δ)

for every x ∈ Γ(R), where x ¯ = (x0 , xn + κR). Then for every f ∈ Hα (1−δ) g ∈ H1+α and ψ ∈ C 0 , there is a unique solution u of Lu = f in Γ(R),

M u = g on Γ0 (R),

u = ψ on Γ∗ (R).

,

(2.61)

(−δ) H2+α

Moreover, if ψ ∈ Hδ , then u ∈ and there is a constant C, determined only by A, B0 , B1 , B2 , n, α, δ, µ, µ0 , and κ such that (2.57) holds. Proof. Again, we may assume that ψ = 0, and we now apply the method of continuity with Js defined by Js u = (aij Dij u + s(bi Di u + cu), β i Di u + sβ 0 u).

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The only step in the proof of Theorem 2.23 that needs modification is the estimate of d−δ u. To this end, we define w by w(x) = R2 − |¯ x |2 , and we set h=

Z

w

sδ−1 exp(−B2 s/(κR2 )) ds.

0

It’s easy to check that M h ≤ −χCRδ dδ−1 on Γ0 (R) and that h = 0 on P Γ∗ (R). Next, we set T = i aii and E = aij x ¯i x ¯j , and 1 B2 L1 = −2T − 2bi x ¯i + 4(δ − 1) E − 4 2 E. w κR

Then Lh ≤ aij Dij h + bi Di h = exp(−B2 |¯ x|/(Rκ))wδ−1 L1 . Because |¯ x| ≥ κR, T ≥ nλ and E ≥ λ|¯ x|2 in Γ(R), we have that Lh ≤ B0 λwδ−2 |¯ x|/d ≤ −C0 λRδ dδ−2 ,

where B0 = [4(δ − 1)κ/(1 + κ) + 2B1 ] exp(−B2 ) and C0 = −B0 κ(1 + κ)δ−1 . (Note that B0 < 0 < C0 ). Applying the maximum principle to (2−δ (1−δ) (2−δ) C0 (|f /λ|0 + |g/χ|0 )R−δ h ± u, we see that |u| ≤ C0 (|f /λ|0 + (1−δ) δ |g/χ|0 )d . From this bound on d−δ |u|, the proof proceeds as before.  The proof of the preceding theorem uses condition (2.60), but a weaker result is still true if we remove this assumption. (0)

(1)

(2)

Corollary 2.25. Let aij ∈ Hα , let bi ∈ Hα , let c ∈ Hα , and suppose that there are positive constants A, λ and µ such that conditions (2.29) are (0) (1) satisfied. Let β ∈ H1+α , let β 0 ∈ H1+α , and suppose that there are positive constants B0 , χ, and µ0 such that (2.56) and (2.59) hold. Then there is a constant C, determined only by A, B0 , n, α, δ, µ, µ0 , and κ, such that, if (0) u ∈ H2+α , then (0)

(1)

|u|2+α ≤ C(|u|0 + |Lu/λ|(2) α + |M u/χ|1+α ).

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2.7

81

Domains with curved boundary

We are now ready to study problems in more general domains. Our first step is an appropriate (and obvious, based on our earlier discussion of Lipschitz domains,) definition of Ha domains. For a ≥ 1, we say that a bounded domain Ω is an Ha domain or that ∂Ω ∈ Ha if there is a number R > 0, and for each x0 ∈ ∂Ω, there are a coordinate system Y centered at x0 and an Ha function ω such that Ω ∩ B(x0 , R) = {x ∈ Rn : |x − x0 | < R, y n ≥ ω(y 0 )}. Since Ω is bounded, we can cover ∂Ω by finitely many balls of radius R, so we may assume that the functions ω have uniformly bounded Ha norm. Again, we refer to [56] for details. Because of the invariance of the norms under appropriate changes of variables, we obtain the following result immediately. To simplify writing, we define the operators L and M by (2.58). Theorem 2.26. Let aij , bi , and c be in Hα for some α ∈ (0, 1) and suppose that β and β i are in H1+α and that ∂Ω ∈ H2+α . Suppose A, B0 , λ, χ, µ, and µ0 are constants such that conditions (2.29a,b) hold and such that β · γ ≥ χ,

ij

(2.62a)

|β| ≤ µ0 χ,

(2.62b)

|a |α + |b|α + |c|α ≤ Aλ, 0

|β|1+α + |β |1+α ≤ B0 χ.

(2.62c) (2.62d)

Then there is a constant C, determined only by A, B0 , n, α, µ, µ0 , Ω such that |u|2+α ≤ C(|Lu/λ|α + |M u/χ|1+α + |u|0 )

(2.63)

for any u ∈ H2+α . Proof. By virtue of the interpolation inequality (2.11), we may assume that bi , c, and β 0 are all zero. Using Theorem 2.23 with ψ = u at each boundary point along with the change of variables theorem for H¨older norms, we find that |u|2+α ≤ C(|u|2+α;Ω0 + |f /λ|α + |g/χ|1+α + |u|δ )

for some open set Ω0 such that d(Ω0 , ∂Ω) > 0. Since (0)

|u|2+α;Ω0 ≤ C(Ω0 , Ω)|u|2+α ,

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It then follows from the interior Schauder estimate that |u|2+α ≤ C(|u|δ + |f /λ|α + |g/χ|1+α ), and another application of (2.11) completes the proof.



In order to obtain further information, we now examine the existence theory. Our approach is a variant of the Perron process from Section 2.5. Theorem 2.27. Let aij , bi , and c be in Hα for some α ∈ (0, 1) and suppose that β and β i are in H1+α and that ∂Ω ∈ H2+α . If c ≤ 0 in Ω and β 0 < 0 on ∂Ω, then Lu = f in Ω,

M u = g on ∂Ω

(2.64)

2

has a unique solution in C (Ω) ∩ C(Ω). Moreover u ∈ H2+α and |u|2+α ≤ C(Ω, α, A, B0 )(|f /λ|α + |g/χ|1+α ).

(2.65)

Proof. The main point is to consider the combination of differential equation and boundary condition in the same light that was used for just the differential equation in Section 2.5. Here we say that (2.64) is locally solvable if for each x0 ∈ Ω, there is a non-empty, relatively open subset N of Ω such that, for any continuous function ψ (defined on ∂N ∩ Ω), there is a unique solution of Lu = f in N ∩ Ω,

M u = g on N ∩ ∂Ω,

u = ψ on ∂N ∩ Ω.

(2.66)

For x0 ∈ Ω, we use the theory of the Dirichlet problem to infer that this problem is solvable provided we take N to be a ball B centered at x0 such that B ⊂ Ω. For x0 ∈ ∂Ω, we need to be more subtle. First, by the definition of ∂Ω ∈ C 2,α , we know that there is a ball B centered at x0 such that, after a suitable rotation, we can write Ω ∩ B = {x ∈ Rn : |x − x0 | < R, xn ≥ ω(x0 )} for some H2+α function ω and some R > 0. We then make the change of variables y 0 = x0 − x00 ,

y n = xn − ω(x0 ).

In the y-coordinates, we see that, for some r > 0, {x ∈ Rn : |y| < r, y n > 0} is a subset of Ω, {x ∈ Rn , |y| < r, y n }

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is a subset of ∂Ω, and y = 0 corresponds to x0 . If we write the differential equation and boundary condition in y-coordinates, we see that local solvability is achieved via Theorem 2.24. The calculations in the proof of Lemma 1.2 show that the function ϕ, defined by ϕ(x) = exp(−(1 + A)x1 ), satisfies Lϕ > 0 in Ω. Hence, there are positive constants K1 and K2 such that L(K1 ϕ − K2 ) ≥ |f | in Ω and M (K1 ϕ − K2 ) ≥ |g| on ∂Ω. It follows that K1 ϕ − K2 is a subsolution and that −K1 ϕ + K2 is a a supersolution. Just as in the case of the Dirichlet problem, it follows that (2.64) has a solution.  Theorem 2.28. Let aij , bi , and c be in Hα for some α ∈ (0, 1) and suppose that β and β i are in H1+α and that ∂Ω ∈ H2+α . If u ∈ C 2 (Ω) ∩ C(Ω) is a solution of (2.64) with f ∈ Hα and g ∈ H1+α is in H2+α and there is a constant C, determined only by Ω, A, B0 , α, µ, and µ0 such that |u|2+α ≤ C(|f /λ|α + |g/χ|1+α + |u|0 ).

(2.67)

Proof. As a first step, we note that Lemma 1.26 implies that u ∈ Hδ for some δ > 0. Then we can use Theorem 2.27 to infer that there is a unique solution v ∈ H2+δ of ¯ = −cu + f in Ω, Lv

¯ v = −(β 0 + 1)u + g on ∂Ω, M

(2.68)

¯ = a Dij v + b Di v and M ¯ v = β Di v − v. It then follows that where Lv u = v so u ∈ H2+δ . Therefore, v ∈ H2+α and hence u ∈ H2+α .  ij

2.8

i

i

Fredholm-Riesz-Schauder theory

Finally, we conclude that (L, M ) is a Fredholm operator of index zero. To explain this terminology and to prove the result, we give a brief account of Fredholm-Riesz-Schauder theory. In what follows, X always denotes a Banach space. The dual X ∗ of X is the set of all bounded, linear, real-valued functions defined on X. Note that X ∗ is a Banach space. In addition, if T is any bounded linear function defined on X with values in X, we define its adjoint T ∗ by defining T ∗ y ∗ , for any y ∗ ∈ X ∗ , to be the element of X ∗ given by (T ∗ y ∗ )x = y ∗ (T x). (Although we shall not need the additional generality, we observe that this expression allows us to define T ∗ y ∗ as a real-valued function on X assuming

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only that T is a function defined on X with values in X.) Also, for any subset W of X, we define its orthogonal complement W ⊥ to be the set of all y ∗ ∈ X ∗ such that y ∗ (w) = 0 for all w ∈ W . Similarly, if W ∗ ⊂ X ∗ , its orthogonal complement (W ∗ )⊥ is the set of x ∈ X such that w∗ (x) = 0 for all w∗ ∈ W ∗ . We say that T is a compact linear operator on X if T is a linear function defined on X with values in X and if T maps bounded sets into compact sets. It can be shown that, for any compact linear operator T on X, T ∗ is a compact linear operator on X ∗ . For a given compact linear operator T on X and µ ∈ R \ {0}, we define the following sets: Nµ = {x ∈ X : x = µT x},

Nµ∗ = {x∗ ∈ X ∗ : x∗ = µT ∗ x∗ }.

We also write Rµ for the set of all y ∈ X such that y = x − µT x for some x ∈ X and Rµ∗ or the set of all y ∗ ∈ X ∗ such that y ∗ = x∗ − µT ∗ x∗ for some x∗ ∈ X ∗ . With these definitions, we can state the basic results of the FredholmRiesz-Schauder theory. We refer to Theorems 5.2.10 and 5.3.2 in [57] for details. Theorem 2.29. Let X be a Banach space and let T be a compact linear operator on X. Then, for any nonzero µ, the spaces Nµ and Nµ∗ have the same finite dimension. In addition Rµ = (Nµ∗ )⊥ and Rµ∗ = (Nµ )⊥ . Finally, the set of all numbers µ for which Nµ has nonzero dimension is countable and it has no finite limit points. Note the numbers µ for which Nµ has nonzero dimension are the reciprocals of the eigenvalues of T ; however, this choice will be convenient for our application. Theorem 2.30. Let aij , bi , and c be in Hα (Ω) and let β 0 and β be in H1+α (∂Ω) for some domain Ω with ∂Ω ∈ H2+α . Set X = Hα (Ω) × H1+α (∂Ω). Suppose also that the minimum eigenvalue of the matrix [aij ] is bounded from below by a positive constant and that β · γ is bounded from below by a positive constant. Then there is a countable set Σ of real numbers with no finite limit points such that the problem (L + σ)u = f in Ω,

(M + σ)u = g on ∂Ω

(2.69)

has a unique solution for all (f, g) ∈ X if and only if µ ∈ / Σ. In addition, for each σ ∈ S, there is a positive integer Iσ such that the dimension of the solution set of the problem (L + σ)v = 0 in Ω,

(M + σ)u = 0 on ∂Ω

(2.70)

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is Iσ , and there is an Iσ -dimensional subspace Wσ∗ of X ∗ such that (2.69) has a solution if and only if (f, g) ∈ (Wσ∗ )⊥ . Proof. Let K be a constant such that K > max{|c|0 , |β 0 |0 }. Then the problem (L − K)u = f in Ω,

(M − K)u = g on ∂Ω

has a unique solution u ∈ H2+α for any (f, g) ∈ X. We now define the operator T : X → X by T (f, g) = (u, u|∂Ω ), where u is the solution of this problem. Since T maps X into H2+α (Ω) × H2+α (∂Ω), it follows that T is a compact linear operator on X. Suppose first that µ is a nonzero number such that Nµ has dimension zero. Then, for any (f, g) ∈ X, there is (f1 , g1 ) ∈ X such that (I − µT )(f1 , g1 ) = (f, g). Let u be the solution of (L − K)u = f1 in Ω,

(M − K)u = g1 on ∂Ω.

Then it’s easy to check that (L − K − µ)u = f in Ω,

(M − K − µ)u = g on ∂Ω,

and u is the only solution of this boundary value problem. Hence if σ = −K − µ for some µ 6= 0 with Nµ having dimension zero (or if σ = −K), then (2.69) has a unique solution for any (f, g) ∈ X. We define Σ to be the set of all numbers σ of the form σ = −K − µ with µ nonzero and with Nµ having nonzero dimension. If σ ∈ Σ, we set µ = −K − σ and note that Nµ has nonzero dimension, which we denote by Iσ . For simplicity of notation, if u ∈ H2+α (Ω), we write u¯ = (u, u|∂Ω ). It’s easy to see that u solves (2.70) if and only if u ¯ = (K + µ)¯ u, so the solution set of (2.70) has dimension Iσ . (In particular, if (f, g) = µT (f, g) for any constant µ, then g must be the restriction of f to ∂Ω.) In addition, u solves (2.69) if and only if u ¯ − µT u ¯ = (f, g), so (2.69) has a solution if and only if (f, g) ∈ (Wσ∗ )⊥ provided Wσ∗ is just Nµ∗ (that is, the set of all x∗ ∈ X ∗ such that x∗ = µT ∗ x∗ ).  In the language of linear operators on vector spaces, the second part of this theorem just says that the codimension of the range of L + λk is equal to the dimension of its kernel.

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Notes As previously mentioned, the H¨older condition was introduced by H¨older [71] to study potentials. The weighted norms are based on those in Chapter 6 from [64]; we have modified the definition there (which comes from Section 2 in [39], and that source suggested that the norms were well-known in the 1950’s). In fact, the norm in [64] is equivalent to the one given here (b) (see Exercise 2.1). The weighted H¨older spaces Ha with b < 0 were used by Michael [129, 130] to study the Dirichlet problem for linear elliptic equations. The results in Section 2.1 are well-known (see, for example, [63]) and the proofs given here are easy modifications of the ones in Section 6.8 in [64]. The alternative characterization of H¨older norms in Section 2.2 was first explicitly noted by Safanov (see Section 2 in [163]); he ascribes the characterization to Campanato [25], who uses the Lp norm (with p finite) in place of the supremum norm. Our proof is different from those in [25, 163, 21] and follows instead the one in Section 3 of [180]. Note that the function u ¯ in the proof of Lemma 2.8 is commonly called a mollification of u with ϕ known as a mollifier; this terminology comes from Friedrich’s paper [58], and the story of the origin of this name is recounted by Peter Lax in his commentary on [58] on page 117 of [59]. The basic existence result (Theorem 2.10) comes from Theorem 2.7.2 in [87]. If we assume that Aij = δ ij , that f ∈ Hα (B), and that ψ ∈ H2+α , then the conclusion of Theorem 2.10 is also true; a proof of this fact without using Theorem 2.10 is given in, for example, Theorem 4.3 of [64]. In fact, that reference proves the result via an integral representation for solutions of the differential equation ∆v = f . Interior Schauder estimates were proved by Schauder in [168] (for constant coefficient equations) and in [167] (for general equations). His primary tools were integral representations of solutions of solutions of differential equations and appropriate functional analysis. We refer to Chapter 6 in [64] for more details on this approach. In particular, our Lemma 2.14 is just Lemma 6.20 from [64] (which, in turn, is essentially Theorem 4.3 of [129]), and its derivation from Proposition 2.13 is taken from [64]. The Perron process described in Section 2.5 is essentially the same as in Section 6.3 of [64] (see also Section 6.5 of [64], which is based on Section 4 of [129]). In particular, Theorem 2.15 is a simple modification of Theorem 4.9 in [64] (which comes from Theorem 4.5 in [129]). Section 2.6 comes from Section 2 of [100] with one important exception:

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the Schauder-type estimates (Lemma 2.20, Corollary 2.21, and (2.53)) were proved in [100] by using integral representations (more precisely, by pointing out that the techniques in [64] could be applied to that situation). We present independent proofs here for the first time; however, the method of proof used here is straightforward. Condition (2.60) in Theorem 2.24 is not surprising for the existence result. The theory of Fichera [44] (see also [118]) shows that the Dirichlet problem for the differential equation 1 aij Dij u + (bi Di u − u) = 0 in Ω, d with aij and b continuously differentiable, has a solution for arbitrary boundary data if and only if b · γ < aij γi γj on ∂Ω, and (2.60) is just a restatement of this condition. The Fredholm-Riesz-Schauder theory is taken from Sections 5.2 and 5.3 in [57]; this theory is an extension by Schauder [166] of earlier work by Riesz [161]. Schauder’s main advance was to introduce the idea of adjoint operators, thus allowing the theory to be applied in general Banach spaces. Exercises (b)

2.1 Define the seminorm Mα (for α ∈ (0, 1] and b ≥ −α) by Mα(b) (f ) = sup min{d(x), d(y)}b+α x6=y∈Ω (b)

and define the norm Na

Na(b) (f ) =

|f (x) − f (y)| , |x − y|α

(for a = k + α) by k X j=0

(b+j)

|Dj f |0

+ Mα(b+k) (f ).

Show that there is a positive constant C such that 1 (b) |f | ≤ Na(b) (f ) ≤ C|f |(b) a . C a (−δ)

2.2 For g ∈ Hδ (B), show that there is a function G ∈ H2+α (B) with G = g on ∂B and (−δ)

(−δ)

|G|2+α ≤ C(n, α, δ)|g|δ

.

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2.3 Suppose that ∂Ω ∈ Hk+2+α with k a positive integer. Suppose also that aij , bi and c are in Hk+α and that β and β 0 are in Hk+1+α . Prove that |u|k+2+α ≤ C(|Lu|k+α + |M u|k+1+α + |u|0 ). 2.4 Prove Corollary 2.25. 2.5 Suppose that ∂Ω ∈ Hk+2+α with k a positive integer. Suppose also that aij , bi and c are in Hk+α , and that there is a function ϕ ∈ Hk+2+α such that u = ϕ on ∂Ω. Prove that |u|k+2+α ≤ C(|Lu|k+α + |ϕ|k+2+α + |u|0 ). (Hint: Use Exercise 1.4.)

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Chapter 3

The Miller Barrier and Some Supersolutions for Oblique Derivative Problems Introduction In this chapter, we construct various supersolutions for both the Dirichlet problem and the oblique derivative problem in Lipschitz domains. This study is based on the construction of a suitable comparison function in a cone, due to Miller, and this construction is based the solution of a singular ordinary differential equation. We start with a discussion of solutions for ordinary differential equations in Section 3.1; some of the results are quite standard but others are not, so we give brief proofs of the results that we use. (The Notes at the end of the chapter give further references for all these results.) Section 3.2 uses the results for ordinary differential equations to construct the barriers.

3.1

Theory of ordinary differential equations

In this section, we present a few basic existence, uniqueness and continuous dependence results for some special second order differential equations. Although many of the results are true under much more general hypotheses, we shall give simple proofs only for the cases of interest in this chapter. Our first step is a basic existence and uniqueness result. Lemma 3.1. Let a < b be real numbers and let H : [a, b] × R × R → R be a continuous function. Suppose H is uniformly Lipschitz with respect to its second and third arguments. Then, for any s0 ∈ [a, b] and any real numbers f0 and f1 , there is a unique solution f ∈ C 2 [a, b] of f 00 + H(s, f (s), f 0 (s)) = 0 for s ∈ (a, b), (3.1a) f 0 (s0 ) = f1 .

f (s0 ) = f0 ,

89

(3.1b)

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Proof. Suppose first that s0 < b. Write K for the Lipschitz constant of H with respect to its second and third arguments and set h = min{b − s0 , 1, 1/(2K)}. We first show that the initial value problem f 00 + H(s, f (s), f 0 (s)) = 0 for s ∈ (s0 , s0 + h), f (s0 ) = f0 ,

0

f (s0 ) = f1

(3.2a) (3.2b)

has a unique solution. To this end, let X denote the set of all C 1 [s0 , s0 + h] functions g with g(s0 ) = f0 and g 0 (s0 ) = f1 with the norm |g|X = sup |g(s)| + sup |g 0 (s)|;

Note that X is a complete metric space. We then define the operator T on X by  Z s Z τ 0 T g(s) = f0 + f1 (s − s0 ) − H(σ, g(σ), g (σ)) dσ dτ. s0

s0

It’s easy to check that T g ∈ X and that

(T g)00 (s) + H(s, g(s), g 0 (s)) = 0.

A simple calculation shows that 1 |T g1 − T g2 |X ≤ K[h + h2 ]|g1 − g2 |X . 2 Our choice of h implies that K[h + 12 h2 ] < 1, so T is a contraction. Hence it has a unique fixed point f , which will be the unique solution of (3.2). We then repeat this process a finite number of times to obtain a solution on [s0 , b]; a similar argument gives a solution on [a, b].  When H is singular, then the existence of a unique solution is no longer guaranteed for arbitrary initial data. As a simple example, consider the equation f 00 + f 0 /s = 0. It’s easy to check that the general solution of this equation is f (s) = c1 + c2 ln s. Hence any solution which is bounded at 0 must have zero derivative there. This is the situation which we wish to analyze more closely for nonlinear problems. For our applications, we now specialize the interval of existence to [0, π). Lemma 3.2. Let G : [0, π) × R × R be continuous, Lipschitz with respect to its second and third arguments, and suppose that there is a constant s0 ∈ (0, π) such that G(s, z, p1 ) ≤ G(s, z, p2 ) for s ∈ [0, s0 ], z ∈ R and p1 ≤ p2 in R. Then, for any real number f0 , there is a unique solution f ∈ C 2 [0, π) of the initial value problem f 00 + G(s, f (s), f 0 (s)/s) = 0 for s ∈ (0, π), f (0) = f0 ,

0

f (0) = 0.

(3.3a)

(3.3b)

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Proof. We first note that, by continuity, if f ∈ C 2 [0, π) is a solution of (3.3), then f 00 (0) = h, the unique fixed point of the equation h = −G(0, f0 , h).

We now define two functions fˆ and θ by

h fˆ(s) = f0 + s2 , θ(s) = h + G(s, fˆ(s), h). 2 For each ε > 0, there is a constant η(ε) ∈ (0, min{ε, π/2}) such that |θ(s)| ≤ ε if 0 ≤ s < η(ε). We now define θˆ by   if 0 ≤ s < η(ε),  θ(s) ˆ ε) = θ(2s − η(ε)) if η(ε) ≤ s < 2η(ε), θ(s;   0 if 2η(ε) ≤ s < π, and we write fˆε for the unique solution of

ˆ ε) = 0 for s ∈ (η(ε), π), fˆε00 + G(s, fˆε , fˆε0 /s) + θ(s; 1 fˆε (η(ε)) = f0 + hη(ε)2 , fˆε0 (η(ε)) = hη(ε) 2 given by Lemma 3.1. If we now define fε by ( fˆ(s) if 0 ≤ s < η(ε), fε (s) = fˆε (s) if s ≥ η(ε), then fε ∈ C 2 [0, π) with

|fε00 + G(s, fε , fε0 /s)| < ε for s ∈ (0, π),

fε (0) = f0 ,

fε0 (0) = 0.

To complete the proof, we must show that the family of functions (fε ) converges in C 2 to a limit function which must be our solution. This convergence is proved by proving a suitable a priori bound for fε − fδ and its derivatives for ε > δ > 0. To prove the estimate, fix ε > δ > 0, set y = fε −fδ , and write K for the Lipschitz constant of G with respect to the second argument. If y 0 (s) > 0, then |y 0 | is differentiable at s with |y 0 |0 (s) = y 00 (s) and hence |y 0 |0 (s) ≤ −G(s, fε (s), fε0 (s)/s) + G(s, fδ (s), fδ0 (s)/s + 2ε) ≤ K|y| + 2ε

0

wherever y (s) > 0 and s ∈ (0, s0 ). A similar argument (but noting that |y 0 |0 (s) = −y 00 (s) wherever y 0 (s) < 0 and s ∈ (0, s0 )) shows that |y 0 |0 (s) ≤ K|y| + 2ε

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wherever y 0 (s) 6= 0. For ζ ∈ (0, s0 ), we now set

m(ζ) = sup{|y 0 |0 (s) : 0 ≤ s < ζ, y 0 (s) 6= 0}.

If s ∈ [0, ζ), then |y 0 (s)| ≤ m(ζ)s and |y(s)| ≤ 21 m(ζ)s2 , so 1 m(ζ) ≤ Kζ 2 m(ζ) + 2ε. 2 It follows that, for ζ sufficiently small (so that Kζ 2 ≤ 1 and ζ ≤ s0 ), we have m(ζ) ≤ 4ε.

(3.4)

|fε (s) − fδ (s)| ≤ 2εs2 , |fε0 (s) − fδ0 (s) ≤ 4εs.

(3.5)

With this ζ fixed, it follows that

Since fε and fδ are C 2 , it follows that the limit f˜ = lim fε ε→0

exists on [0, ζ] and (3.4) and (3.5) imply that the convergence is in C 2 [0, ζ]. To extend f˜ to [0, π), we just take F to be the solution of F 00 + G(s, F, F 0 /s) for s ∈ (ζ, π), F (ζ) = f˜(ζ), F 0 (ζ) = f˜0 (ζ) given by Lemma 3.1 and define f by ( f˜(s) f (s) = F (s)

if 0 ≤ s < ζ,

if ζ ≤ s < π.



In fact, the solution in this lemma depends continuously on the function G. Corollary 3.3. Let G1 and G2 be two continuous functions defined on [0, π) × R × R which are uniformly Lipschitz with respect to their second and third arguments, and suppose that there is a constant s0 ∈ (0, π) such that Gj (s, z, p1 ) ≤ Gj (s, z, p2 ) for j = 1, 2, s ∈ [0, s0 ], z ∈ R and p1 ≤ p2 in R. For j = 1, 2, let fj be the solution of fj00 + Gj (s, fj , fj /s) for s ∈ [0, π),

fj (0) = 1,

fj0 (0) = 0.

Let K be a Lipschitz constant for G1 and G2 with respect to the second argument and suppose Kh2 ≤ 21 for some h ∈ (0, s0 ). Set ε = sup{G1 (s, f1 , f10 /s) − G2 (s, f1 , f10 /s)| : s ∈ [0, h]}.

Then y = f1 − f2 satisfies the inequalities |y(s)| ≤ 2εs2 ,

for all s ∈ [0, h].

|y 0 (s)| ≤ 4εs,

|y 00 (s)| ≤ 4ε

(3.6)

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Proof.

93

Just as in the proof of Lemma 3.2, we set m = sup{|y 0 |0 (s) : y 0 (s) 6= 0, 0 ≤ s ≤ h}.

At a point where y 0 (s) > 0, we have

|y 0 |0 (s) = y 00 (s) = G2 (s, f2 , f20 /s) − G1 (s, f1 , f10 /s)

≤ ε + G2 (s, f2 , f20 /s) − G2 (s, f1 , f10 /s) ≤ ε + K|y(s)|.

0

Arguing similarly where y (s) < 0, we obtain |y 0 |0 (s) ≤ ε + K|y(s)|

for all s at which y 0 (s) 6= 0. Again, |y(s)| ≤ 21 ms2 , so m ≤ 4ε,

and this inequality easily implies (3.6).



We shall also use a result concerning the existence and structure of solutions of linear degenerate ordinary differential equations. Lemma 3.4. Let a and b be continuous functions on [0, π) and suppose that there are positive constants h < π and α such that a(s) > 1 − αs for 0 ≤ s ≤ h. Then there are two functions H ∈ C 2 [0, π) and K ∈ C 2 (0, π) such that a(s) 0 H 00 + H + b(s)H = 0 for 0 < s < π, H(0) = 1, H 0 (0) = 0 (3.7) s and a(s) 0 K 00 + K + b(s)K = 0 for 0 < s < π, lim K(s) = ∞. (3.8) s s→0+ Moreover, for any solution u ∈ C 2 (0, π) of a(s) 0 u + b(s)u = 0 for 0 < s < π, s there are unique constants c1 and c2 such that u = c1 H + c2 K. u00 +

(3.9)

Proof. The existence of H satisfying (3.7) follows immediately from Lemma 3.2. To obtain K, we use the method of reduction of order. Specifically, we seek a nonconstant function w such that K = Hw satisfies the correct differential equation. By differentiating Hw and using the differential equation for H, we see that   a(s) 2H 0 00 + w0 = 0. w + s H

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Of course, this equation only makes sense where H 6= 0, so we now work with this equation only for 0 < s < δ, where δ ∈ (0, h) is so small that H(s) ≥ 12 for 0 ≤ s ≤ δ. Dividing our equation by w0 and integrating yields ln |w0 | =

Z

s

δ

a(σ) dσ − ln(H 2 ) + k σ

for some constant k. Hence ek |w | = 2 exp H 0

k

e ≥ 2 exp H

!

Z

δ

a(σ) dσ σ

Z

δ

1 − ασ dσ σ

s

s

!

k

=

e α(s−δ) 1 e δ . H2 s

It follows that |w0 | ≥ C/s for some positive constant C, so w0 is not integrable at 0, which means (if we choose w to be positive) that w(s) → ∞ as s → 0. Since H ≥ 21 for s < δ, it follows that the K we have constructed tends to infinity at 0 as well and that K(δ) and K 0 (δ) are finite. Just as in Lemma 3.2, we can extend K to a solution of the differential equation on (0, π). Next, we observe that H and K are linearly independent. If x0 ∈ (0, π) and b1 and b2 are constants such that b1 H(x0 ) + b2 K(x0 ) = 0,

b1 H 0 (x0 ) + b2 K 0 (x0 ) = 0,

then, by linearity, the function v = b1 H + b2 K solves the initial value problem v 00 +

a(s) 0 v + b(s)v = 0 for 0 < s < π, s

v(x0 ) = 0,

v 0 (x0 ) = 0,

and hence v ≡ 0 by Lemma 3.1. It follows that b1 = b2 = 0. So, if u is a solution of (3.9), we can choose c1 and c2 so that u(1) = c1 H(1) + c2 K(1),

u0 (1) = c1 H 0 (1) + c2 K 0 (1).

Lemma 3.1 then implies that u = c1 H + c2 K.



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3.2

95

The Miller barrier construction

In this section, we construct an optimal supersolution for a class of linear Dirichlet problems in an infinite cone. We assume that the axis of the cone is the positive xn axis and we introduce polar coordinates r and θ via the equations r = |x|,

r cos θ = xn .

(Strictly speaking, these are two of the n spherical coordinates, but we follow Miller’s terminology here.) Our cones will then be defined by the inequalities r > 0 and θ ∈ [0, θ0 ) for some θ0 ∈ (0, π), and our supersolutions will have the form w(x) = rα f (θ) for some positive constant α and a C 2 [0, π) function f . Of course, f need only be defined for θ ∈ [0, θ0 ) but it will be very convenient to have f defined on the larger interval. For simplicity of notation, we also introduce Lµ , the set of all matrixvalued functions with eigenvalues in the interval [1, µ]. We begin our analysis with a polar representation. Lemma 3.5. Let µ ≥ 1. Then for any [aij ] ∈ Lµ , there are scalar functions a, b, c, and d such that s2 + t2 ≤ as2 + 2bst + ct2 ≤ µ(s2 + t2 )

(3.10a)

for all real numbers s and t, 1 ≤ d ≤ µ,

(3.10b)

and aij Dij w = rα−2 [aα(α − 1)f + 2b(α − 1)f 0 + c(f 00 + αf )

+ d(n − 2)(αf + cot θf 0 )].

(3.11)

Conversely, given a, b, c, and d satisfying (3.10), there is [aij ] ∈ Lµ such that (3.11) holds. Proof.

To begin, we define two vectors ξ and η by

xi xi − csc θδ in , ηi = . r r It is easy to check that ξ and η are orthogonal unit vectors and that ξi = cot θ

∂r = η, ∂x

∂θ ξ = . ∂x r

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Hence, if w = rα f (θ), then Dw = rα−1 (αf η + f 0 ξ) and Dij w = rα−2 (α(α − 1)f ηi ηj + αf 0 ηi ξj + αrf

∂ηi ∂xj

+ (α − 1)f 0 ξi ηj + f 00 ξi ξj + rf 0

∂ξi . ∂xj

Computing the partial derivatives of η and ξ yields 1 ∂ηi = (δij − ηi ηj ) ∂xj r and i 1 1 ∂ξi 2 x = (− csc θ + csc θ cot θ δin )ξj + cot θ (δij − ηi ηj ). ∂xj r2 r r 2 2 Using the trigonometric identity csc θ = cot θ + 1 and the definition of ξ, we see that ∂ξi 1 1 1 = − cot θξi ξj − ξi ηj + cot θ (δij − ηi ηj ). j ∂x r r r From these calculations of partial derivatives, we find that

r2−α Dij w = α(α − 1)f ηi ηj

+ (α − 1)f 0 (ξi ηj + ξj ηj ) + (f 00 + αf )ξi ξj

+ (αf + cot θf )(δij − ξi ξj − ηi ηj ). Our polar representation now follows with a = aij ηi ηj , b = aij ξi ηj , c = aij ξi ξj , 1 d= aij (δij − ξi ξj − ηi ηj ). n−2 Straightforward algebra now gives (3.10). Specifically, if v = sη + tξ, then |v|2 = s2 + t2 and aij vi vj = as2 + 2bst + ct2 .

Since |v|2 ≤ aij vi vj ≤ µ|v|2 , we immediately obtain (3.10a). If v (k) (k = 1, . . . , n − 2) are orthonormal vectors all perpendicular to ξ and η, then (k) (k)

(n − 2)d = aij vi vj ,

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so (3.10b) follows. Conversely, given a, b, c, and d satisfying (3.10), we take v (k) as above, and choose [aij ] to be a symmetric matrix with eigenvalue d corresponding to each eigenvector v (k) and then (because the other two eigenvectors must be perpendicular to all v (k) ) we can take the other two eigenvalues with corresponding eigenvectors in the ξη-plane as follows: We write (si , ti ) (i = 1, 2) for the eigenvectors of the matrix   ab bc with corresponding eigenvalues λi . The other two eigenvectors of [aij ] are v (1) = s1 η + t1 ξ, v (2) = s2 η + t2 ξ, with corresponding eigenvalues of λ1 and λ2 .

 2

For µ ≥ 1, we now define the operator Mµ on C by

Mµ h(x) = sup{aij (x)Dij h(x) : [aij ] ∈ Lµ }.

As we shall see, Mµ is a fully nonlinear operator and a complete existence and uniqueness theory is known for this operator. For now, we make a few simple but crucial observations. First, if we write L∗µ for the set of all constant matrix-valued functions in Lµ , then we have Mµ h(x) = sup{aij (x)Dij h(x) : [aij ] ∈ L∗µ }.

Second, for a given h ∈ C 2 (Ω), there is a matrix-valued function [aij h ] ∈ Lµ such that Mµ h(x) = aij (x)D h(x). To prove the second observation, we ij h ij fix x ∈ Ω and let [ak ] be a sequence of matrices such that lim aij k Dij h(x) = Mµ h(x).

k→∞

Since Lµ is a compact subset of the set of all symmetric matrices, it follows ij that there is a convergent subsequence of [aij k ]; we take [ah (x)] to be the limit of this convergent subsequence. Our next goal is to obtain solutions of Mµ w = 0 when w = rα f (θ), and we do so by solving a suitable initial value problem of the type discussed in the previous section. We first define the function g by a b g(α, a, b, c, d; θ, z, p) = α(α − 1)z + 2 (α − 1)θp + αz c c d + (n − 2)(αz + θ(cot θ)p), c

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and then we define G by G(α; θ, z, p) = sup{g(α, a, b, c, d; θ, z, p) : a, b, c, d satisfy (3.10)}. Although g and G clearly depend on the parameter µ, we suppress this dependence from the notation because it is not as important for our purposes as the dependence on α. It then follows that Mµ w = 0 (and hence aij Dij w ≤ 0 for all [aij ] ∈ Lµ ) if f is a solution of the initial value problem f 00 + G(α; θ, f, f 0 /θ) = 0 for θ ∈ (0, π),

f (0) = 1,

f 0 (0) = 0.

(3.12)

2

We saw in Lemma 3.2 that this initial value problem has a unique C [0, π) solution, which we denote by Fα . In the next lemma, we collect some important properties of the function Fα , especially in terms of its dependence on α. Lemma 3.6. Suppose n ≥ 3. Then Fα has a first zero, which we denote by Ψ(α), in (0, π). Moreover, the function Ψ is continuous and strictly decreasing, Ψ maps (0, ∞) onto (0, π), and Ψ(1) = π/2. Proof.

To show that Ψ is defined, let Φα be the solution of Φ00α

+ (n − 2) cot θΦ0α + [α2 + (n − 2)α]Φα = 0 for θ ∈ (0, π), Φα (0) = 1,

Φ0α (0) = 0.

(In other words, Φα is just Fα corresponding to µ = 1.) Our first step is to show that Φα is zero somewhere in (0, π). As a preliminary to this result, we show that Φα is decreasing as long as it remains positive. To this end, suppose Φα > 0 on [0, h] for some h ∈ (0, π). If Φ0α (θ∗ ) = 0 for some θ∗ ∈ (0, h), then Φ00α (θ) < 0 from the differential equation, so Φ0α can’t change from negative to positive as θ increases across θ∗ . Hence Φ0α is negative in (0, h). It therefore follows that, if Φα > 0 on (0, π), then Φα is bounded on (0, π). Using the change of variables s = π − θ, we see from Lemma 3.4 that Φα ∈ C 2 (0, π] with Φα (π) 6= 0 and Φ0α (π) = 0. Applying the preceding argument, we see that Φ0α must be positive on (0, π). Since we have shown that Φ0α is also negative on this interval, we conclude that Φα must vanish somewhere in the interval (0, π). Since Φα (0) = 1 and Φα is continuous on [0, π), it follows that Φα has a first zero in (0, π). We denote this zero by hα . If Fα > 0 on [0, hα ], then there is a β > α such that Fβ > 0 on [0, hα ] by continuous dependence. We now set v = rα [Φα (θ) − 12 Fα (θ)] and, for any ε > 0, we set vε = v − εrβ Fβ (θ). For ε > 0, there is a positive constant Rε such that vε ≤ 0 if r ≥ Rε . In the cone Kε = {x : 0 < r < Rε , 0 ≤ θ < hα },

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we have that 1 ∆vε = − ∆(rα Fα (θ)) − ∆(rβ Fβ (θ)) 2 1 ≥ − Mµ (rα Fα (θ)) − Mµ (rβ Fβ (θ)) 2 = 0. Since vε ≤ 0 on ∂Kε , it follow from the maximum principle that vε ≤ 0 in Kε . Therefore vε ≤ 0 in K = {x : 0 < r, 0 ≤ θ < hα }. Sending ε → 0, we conclude that v ≤ 0 in K. But v = 21 when r = 1 and θ = 0. This contradiction shows that Fα must be zero somewhere in [0, hα ] and hence Fα has a first zero in (0, π). The continuity of Ψ follows from the inverse function theorem after noting that Fα0 (Ψ(α)) 6= 0. To show that Ψ is decreasing, let α > β and set h = Ψ(β). If Ψ(α) = h, then, by the uniqueness part of Lemma 3.1, we know that Fα0 (h) 6= 0. Hence, there is a positive constant k such that Fβ ≤ kFα on [0, h]. On the other hand, if Ψ(α) > h, then such a constant exists because Fα is positive on [0, h]. Now we write K for the cone with θ < h, and set w1 = rα Fα (θ) and w2 = rβ Fβ (θ). For any ε > 0, there is a positive number Rε such that εw1 − w2 > 0 for r ≥ Rε , and we now write Ω(ε) for the subset of K on which r < Rε . Then εw1 − w2 ≥ 0 on ∂Ω(ε). We now choose [aij ] ∈ Lµ so that aij Dij w2 = 0. Since aij Dij w1 ≤ 0, it follows that aij Dij (εw1 −w2 ) ≤ 0 in Ω(ε), and then the maximum principle implies that εw1 − w2 ≥ 0 in Ω(ε). From our choice of Ω(ε), we therefore conclude that εw1 − w2 ≥ 0 in K. Sending ε → 0, we conclude that −w2 ≥ 0 in K, which contradicts our construction of w2 (which is positive in K). Hence we must have Ψ(α) < Ψ(β) and hence Ψ is strictly decreasing. Since d 2b G(0; θ, z, p) = sup{ θp + (n − 2) θ(cot θ)p}, c c it follows that F0 ≡ 1. By continuity, Fα (θ) → 1 as α → 0 for any θ ∈ [0, π) and hence Ψ(α) → π as α → 0. To show that Ψ(α) → 0 as α → ∞, we show that Φα must have a first zero in the interval (0, π/(2α)). To this end, set ϕα (θ) = cos αθ. Then ∆(rα ϕα (θ)) = (n − 2)α sin αθ[cot αθ − cot θ]

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as long as θ ≤ π/(2α). Since cot is decreasing on the interval (0, π/2), we conclude that ∆(rα ϕα (θ)) ≤ 0 in K = {x : 0 < r, 0 ≤ θ < π/(2α)}. If Φα > 0 in [0, π/(2α)], then our previous argument would show that Φα ≤ ϕα on [0, π/(2α)], which contradicts the assumption Φα > 0. It follows that Φα must have its first zero in the interval [0, π/(2α)] and then the proof of Lemma 3.6 shows that Ψα is in this interval as well. Therefore Ψα → 0 as α → ∞. Since Ψα is a continuous decreasing function of α, we conclude that Ψ maps (0, π) onto (0, ∞). Finally, we have d G(1; θ, z, p) = z + sup{(n − 2) (z + θ(cot θ)p)}, c so F1 (θ) = cos θ and hence Ψ(1) = π/2.



In the case of two space dimensions, the arguments are generally easier because the ordinary differential equation is nondegenerate; however, because θ is allowed to be greater than π, some of the arguments need a little modification. Using the language of Riemann surfaces would allow us to use the same arguments as before, but we shall not do so here. Lemma 3.7. Suppose n = 2. Then Fα has a first zero, which we denote by Ψ(α), in (0, ∞). In addition Ψ is a continuous strictly decreasing function which maps (0, ∞) onto (0, ∞), and Ψ(1) = π/2. Proof. To show that Fα has a first zero in (0, ∞), we note that Fα00 + α2 F ≤ 0, so Fα00 < 0 wherever Fα > 0. It follows that Fα0 is decreasing and, because Fα0 (0) = 0, Fα0 < 0 as long as Fα > 0. Hence Fα can’t be positive for all θ > 0. Continuity of Ψ again follows from the Inverse Function Theorem. To show that Ψ is strictly decreasing, we again let α > β and set h = Ψ(β). If Ψ(α) ≥ h, then the proof of Lemma 3.6 implies that there is ˜ for a positive constant k such that Fβ ≤ kFα on [0, h]. We now write K the set of all (r, θ) such that 0 < r and θ ∈ (−h, h) and we set w ˜1 (r, θ) = rα Fα (θ), w ˜2 (r, θ) = rβ Fβ (θ). For any ε > 0, there is a positive number Rε such that εw1 − w2 > 0 for ˜ ˜ on which r < Rε . r ≥ Rε , and we now write Ω(ε) for the subset of K ˜ Then εw ˜1 − w ˜2 ≥ 0 on ∂ Ω(ε). If εw ˜1 − w ˜2 is ever negative, then it attains

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˜ a negative minimum at some point (r0 , θ0 ) ∈ Ω(ε). We now introduce coordinates x1 = r cos θ,

x2 = r sin θ

in the neighborhood ˜ = {(r, θ) : 0 < r < R − ε, |θ − θ0 | < min{ π , h − |θ|}} N 2 of (r0 , θ0 ) and we abbreviate x = (x1 , x2 ). (The neighborhood is chosen ˜ so small that it is contained in Ω(ε) and so that distinct points in N have distinct x coordinates. Hence it is mapped to an x-neighborhood of the point x0 corresponding to (r0 , θ0 ).) If we define wj (j = 1, 2) by wj (x) = w ˆj (r, θ), then it follows that there is a matrix [aij ] ∈ Lµ such that aij Dij w2 = 0 in N and that aij Dij w1 ≤ 0 in N . The strong maximum principle now implies that εw1 − w2 is equal to its minimum in all of N and hence εw ˜1 − w ˜2 is ˜ . Hence the set on which εw equal to its minimum in all of N ˜1 − w ˜2 attains ˜ its minimum is an open subset of Ω(ε) and this set is also closed by the continuity of w ˜1 and w ˜2 . Hence εw ˜1 − w ˜2 must be a negative constant in ˜ Ω(ε), in contradiction to the assumption that εw ˜1 − w ˜2 is nonnegative on ˜ ˜ Arguing as the boundary. Hence εw ˜1 − w ˜2 ≥ 0 in Ω(ε) and therefore in K. in Lemma 3.6, we conclude that Ψ is strictly decreasing. As in Lemma 3.6, we obtain F0 ≡ 1, so Ψ(α) → ∞ as α → 0. The proof that Ψ(α) → 0 as α → ∞ is similar to that in the lemma with the observation that, now, Φα (θ) = cos αθ. Finally, the equation Ψ(1) = π2 follows from the same argument as in Lemma 3.6.  From these lemmata, we obtain the following result on positive solutions of linear elliptic equations. Theorem 3.8. Let K be an infinite cone with opening angle θ0 and let µ ≥ 1. Then there is a positive constant α0 (θ0 , µ, n) with the following property: For any α ∈ (0, α0 ), there is a function wα ∈ C 2 (K) ∩ C(K) with wα ≥ |x|α and Lwα ≤ 0 in K for all L ∈ Lµ . In addition, there is a constant C(α, µ) such that |Dw| ≤ C(α, µ)|x|α−1 . Proof. Without loss of generality, the axis of the cone is the positive xn axis. Then Lemma 3.6 implies that there is a unique α0 ∈ (0, π) such that Ψ(α0 , µ) = θ0 . In addition, if α ∈ (0, α0 ), then Fα > 0 on [0, θ0 ] and Fα is C 2 on this interval. The proof is completed by taking w = rα Fα (θ)/Fα (θ0 ). 

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It will also be useful to obtain a supersolution with a stronger condition on Lw. Corollary 3.9. Let K be an infinite cone with opening angle θ0 and let µ ≥ 1. Then there is a positive constant α0 (θ0 , µ, n) with the following property: For any α ∈ (0, α0 ), there is a function wα ∈ C 2 (K) ∩ C(K) with wα ≥ |x|α and Lwα ≤ −|x|α−2 in K for all L ∈ Lµ . In addition, there is a constant C(α, µ) such that |Dw| ≤ C(α, µ)|x|α−1 . Proof. With the same α0 as in Theorem 3.8, for ε ∈ R, we define Hε to be the solution of the initial-value problem Hε00 + G(α; θ, Hε , Hε0 /θ) + ε = 0 for θ ∈ (0, π), f (0) = 1,

f 0 (0) = 0.

Then Corollary 3.3 implies that Hε > 0 on [0, θ0 ] for some ε > 0. For this particular ε, we take wα =

3.3

1 rα Hε (θ). min{ε, Hε (θ0 )}



Construction of supersolutions for Dirichlet data

In this section, we are concerned with obtaining supersolutions of the equation Mµ u = −dα−2 which vanish on ∂Ω for suitable positive α. There are several ways to obtain the supersolution, and we include one here that is simple but it requires a rather severe limitation on α. It is relatively elementary to obtain the same result without the strong restriction on α provided we are only interested in obtaining a generalized supersolution (for example, a Perron supersolution), and we discuss this situation in the Notes. Lemma 3.10. and

Let ω : Rn−1 → R be a Lipschitz function with ω(0) = 0 |ω(x0 ) − ω(y 0 )| ≤ ω0 |x0 − y 0 |

(3.13)

for some nonnegative constant ω0 . Set Ω = {x = (x0 , xn ) : xn > ω(x0 )},

(3.14)

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and let µ ≥ 1. Then, for any σ ∈ (0, α0 (arctan(−ω0 ), µ, n)/(n + 1)), there are a constant C0 (µ, σ, n) and a function w : Ω → R such that dσ ≤ w ≤ C0 dσ , σ−1

|Dw| ≤ C0 d

,

σ−2

Mµ w ≤ −d

(3.15a) (3.15b) (3.15c)

in Ω. Proof. From Theorem 3.8 with α = (n+1)σ, there are a positive constant C1 and a function w0 such that |x|α ≤ w0 ≤ C1 |x|α , |Dw0 | ≤ C1 |x|α−1 , Lw0 ≤ −|x|α−2

in the infinite cone K with vertex at the origin, axis along the positive xn axis, and opening angle arctan(−ω0 ). We now set F = Rn \ Ω, and we define W on Ω by Z −σ W (x) = w0 (x − y)−1/σ dy . F

To analyze this function, we also define, for each x ∈ Ω, the set H(x) = {y ∈ F : |y − x| < 2d(x)}.

We also observe that, for any number P > n, we have Z Z ωn |x − y|−P dy ≤ |x − y|−P dy = d(x)n−P , P −n F Rn \B(x,d(x)) Z Z |x − y|−P dy ≥ |x − y|−P dy ≥ c(ω0 , n, P )d(x)n−P F

(3.16a) (3.16b)

H(x)

because |x − y| ≤ 2d(x) in H(x) and there is a ball of radius C(ω0 )d(x) in H(x). It follows that Z c1 c2 ≤ w0 (x − y)−1/σ dy ≤ d(x) d(x) F

for positive constants c1 and c2 , and hence 1 1 d(x) ≤ W (x) ≤ σ d(x). σ c2 c1 Next, we write Di W

−1/σ

1 =− σ

Z

F

−1−1/σ

w0

(x − y)Di w0 (x − y) dy,

(3.17)

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so Di W (x) = W

1+1/σ

(x)

Z

−1−1/σ

F

w0

(x − y)Di w0 (x − y) dy.

(3.18)

Using (3.16) with P = 1 + 1/σ and (3.17), we conclude that |DW (x)| ≤ c3 d(x)σ−1 .

(3.19)

To estimate the second derivatives of W , we first fix a constant matrix [aij ] in Lµ , differentiate (3.18) with respect to xj and then multiply by aij and sum on i and j. It follows that       1 1 aij Dij W (x) = W 1+1/σ W (x)1/σ 1 + I1 − 1 + I2 + I3 , σ σ where Z I1 = w0 (x − y)−1−1/σ w0 (x − z)−1−1/σ aij Di w0 (x − y)Dj w0 (x − z) dy dz, F ×F Z I2 = w0 (x − y)−2−1/σ aij Di w0 (x − y)Dj w0 (x − y) dy, F Z I3 = w0 (x − y)−1−1/σ aij Dij w0 (x − y) dy. F

From Cauchy’s inequality for quadratic forms, we have aij Di w0 (x − y)Dj w0 (x − z) so


1/2 ij
1/2 ≤ aij Di w0 (x − y)Dj w0 (x − y) a Di w0 (x − z)Dj w0 (x − z) ,

I1 ≤

Z

F


1/2 w0 (x − y)−1−1/σ aij Di w0 (x − y)Dj w0 (x − y) dy

2

and another application of Cauchy’s inequality (this time for integrals) gives Z
1/2 1/2 w0 (z−y)−1−1/σ aij Di w0 (x − y)Dj w0 (x − y) dy ≤ I2 W (x)−1/(2σ) . F

It follows that

1/σ

W (x)

    1 1 I1 − 1 + I2 ≤ 0 1+ σ σ

and hence aij Dij W (x) ≤ W 1+1/σ I3 ≤ −c4 d(x)σ−2 .

(3.20)

The proof is completed by taking w = W/ min{c1 , c4 } and using (3.17), (3.19) and (3.20). 

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Construction of a supersolution for oblique derivative problems

Our next step is to obtain a supersolution for the oblique derivative problem. The basic idea of the proof is to reverse the process used to derive the H¨ older gradient estimate for the oblique derivative problem: We obtain our supersolution by integrating a modified form of the supersolution from Lemma 3.10. Lemma 3.11. Under the hypotheses of Lemma 3.10, for each r > 0 and each α ∈ (0, 1), there are a nonnegative function w ˆ defined in Ω[r] = {x : xn > ω(x0 ), |x| < r} and a constant C(n, α, µ) such that w ˆ ≤ Cr1+α ,

(3.21a)

α

|Dw| ˆ ≤ Cr ,

(3.21b)

α−1

Mµ w ˆ ≤ −d

(3.21c)

in Ω[r]. Proof.

Let σ=

1 min{α, α0 (arctan(−ω0 ), µ, n)/(n + 1)}. 2

With w from Lemma 3.10 and κ = (1 + ω02 )1/2 , we set Z 2κr w1 (x) = w(x0 , s) ds. xn

Since 0 ∈ ∂Ω, it follows that d(x0 , s) ≤ (|x0 |2 + s2 )1/2 ≤ (1 + 4κ2 )1/2 r, and hence 0 ≤ w1 (x) ≤ C0 2κ(1 + 4κ2 )σ/2 r1+σ , where C0 is the constant from Lemma 3.10. Next, (R 2κr 0 xn Di w(x , s) ds Di w1 (x) = −w(x)

if i < n, if i = n.

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Since d(x0 , s) ≥ (s − xn )/κ, it follows that Z 2κr Z |D0 w1 | ≤ |D0 w| ds ≤ C0 κ1−σ xn

2κ

xn

= C0 κ1−σ

(s − xn )σ−1 ds

(2κr − xn )σ κ ≤ 2C0 rσ . σ σ

Since |Dn w1 | ≤ C0 rσ , we conclude that |Dw1 | ≤ C0

2κ + σ σ r . σ

To estimate aij Dij w1 for [aij ] ∈ Lµ , we first rewrite Z 2κr Dn w1 (x) = −w(x0 , xn ) = −w(x0 , 2κr) + Dn w(x0 , s) ds, xn

so, for i = 1, . . . , n, we have Di w1 (x) = −δin w(x0 , 2κr) +

Z

2κr

Di w(x0 , s) ds.

xn

We therefore have that

so

 0   Z 2κ Dj w(x , 2κr) Dij w1 (x) = Dij w(x0 , s) ds − Di w(x0 , 2κr)  xn  0 aij Dij w1 (x) =

Z

2κr

xn

if i = n, j < n, if j = n, otherwise,

aij Dij w(x0 , s) ds − 2ain Di w(x0 , 2κr).

Then Z

2κr

xn

aij Dij w(x0 , s) ds ≤ −

Z

2κr

d(x0 , s)σ−2 ds.

xn

Because d(x0 , s) ≤ d(x) + s − xn , it follows that Z 2κr Z 2κr aij Dij w(x0 , s) ds ≤ − (d(x) + s − xn )σ−2 ds xn

xn

1 1 = (d(x) + 2κr − xn )σ−1 − d(x)σ−1 . 1−σ 1−σ

We now observe that 2κr − xn ≥ 2κr − r ≥ r ≥ d(x) to conclude that Z 2κr 2σ−1 − 1 σ−1 1 aij Dij w(x0 , s) ds ≤ d ≤ − dσ−1 . 1−σ 2 xn

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In addition, it’s easy to check that the graph of ω does not intersect the ball B((x0 , 2κr), r), so d(x0 , 2κr) ≥ r and hence there is a constant c1 , determined only by n, α, and µ, such that −2ain Di w(x0 , 2κr) ≤ c1 rσ−1 . Combining these estimates with our expression for aij Dij w1 , we find that 1 Mµ w1 ≤ − dσ−1 + c1 rσ−1 . 2 The proof is completed by taking w ˆ = 2rα−σ w1 + c1 rα−1 (r2 − |x|2 ) and noting that rα−σ dσ−1 ≥ dα−1 because σ ≤ α.



It is interesting to observe that this estimate can be proved for any α < 1 even though σ in the previous supersolution construction had to be much smaller than 1. From this construction, we can prove a weak Harnack inequality when Lu behaves like dα−1 near Σ[r]. More precisely, we have the following estimate. Theorem 3.12. Let R > 0, and let ω : Bn−1 (R) → R be a Lipschitz function with ω(0) = 0 and satisfying (3.13) (for all x and y in Bn−1 (R)) for some nonnegative constant ω0 . Set Ω[R] = {x = (x0 , xn ) : |x| < R, xn > ω(x0 )}, 0

n

n

0

Σ[R] = {x = (x , x ) : |x| < R, x = ω(x )}

(3.22a) (3.22b)

and let µ ≥ 1. Suppose that there are positive constants λ, µ1 , and µ2 such that λ|ξ|2 ≤ aij ξi ξj ≤ µλ|ξ|2

(3.23a)

|b| ≤ µ1 λ,

(3.23b)

for all ξ ∈ Rn , c ≥ −µ2 λ.

(3.23c)

Suppose also that there are positive constants ε and µ3 such that ω0 |β 0 | ≤ (1 − ε)β n , 0

n

β ≥ −µ3 β .

(3.24a) (3.24b)

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If u ∈ C 2 (Ω[R]) ∩ C(Ω[R]) is nonnegative and satisfies aij Dij u + bi Di u + cu ≤ λ(F dα−1 ) in Ω[R], i

0

β Di u + β u ≤ Gβ

n

(3.25a) (3.25b)

for some nonnegative constants F , G and α ∈ (0, 1), then there are positive constants p and C, determined only by ε, µ, µ1 R, µ2 R2 , µ3 R, ω0 , and α such that !1/p Z R−n

up dx

≤ C( inf u + F R1+α + GR). Ω[R/2]

Ω[R/2]

Proof. by

(3.26)

With w ˆ from Lemma 3.11, we set u ˆ = u + F2 w. ˆ With L defined Lh = aij Dij h + bi Di h + cu

and M defined by M h = β i Di h + β 0 h, we see that Lˆ u ≤ λF1 ,

Mu ˆ ≤ β n (G + CF Rα )

for some C determined only by α, µ, µ1 R, µ2 R2 , and µ3 R. It then follows from the weak Harnack inequality Theorem 1.20 that R

−n

Z

p

!1/p

u ˆ dx Ω[R/2]

≤ C( inf u ˆ + F R1+α + GR), Ω[R/2]

and (3.26) follows from this inequality because 0 ≤ w ˆ ≤ CR1+α .



Note that conditions (3.23) are the same as (1.20) from Lemma 1.20, and conditions (3.24) are the same as (1.3c) and (1.36). The argument of Theorem 1.26 gives the corresponding H¨older estimate. Corollary 3.13. Let ω, L, and M be as in Theorem 3.12. Then there are positive constants C and ζ determined only by such that osc u ≤ C

Ω[r]

rζ (sup u + F Rα+1 + GR). Rζ Ω[R]

(3.27)

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The strong maximum principle, revisited

In the next chapter, we study problems with aij ∈ Mµ with b and c possibly growing near ∂Ω like dα−1 for some α ∈ (0, 1). It is possible to use the function from Lemma 3.11 to extend the theory developed in Chapter 1 for equations with bounded coefficients to one for equations with coefficients of this type, but we shall not do so here, referring the reader to [115] for a details. Instead, we just derive a strong maximum principle along the lines of Corollary 1.22. For the reader’s convenience, we recall the notation from Chapter 1. First, let Ω be an open subset of Rn , and suppose that there are positive constants R and ω0 such that {x : |x| < R, xn > ω0 |x0 |} ⊂ Ω. Let ε ∈ (0, 1) and suppose that β is a nonzero vector field defined on ∂Ω ∩ B(0, R) such that ω0 |β 0 | ≤ (1 − ε)β n .

With α1 ∈ (0, 1) to be chosen and A = 21+2ε ε−4ε , we define the function G on Rn × (0, ∞) by  02 (1+ε)/2 |y | (y n )2 2 G(y, r) = + α + , 1 r2 (Aω0 r)2

and for x1 ∈ Rn and r > 0, we write E(x1 , r) for the set of all x ∈ Rn with G(x − x1 , r) < 1. We also abbreviate Ω ∩ B(0, R) to Ω[R] and ∂Ω ∩ B(0, R) to ∂Ω[R]. A major step is the following analog of Lemmata 1.13 and 1.18. Lemma 3.14. With Ω, β, ω0 , ε, and R as above, suppose aij ∈ Mµ (Ω[R]) for some µ ≥ 1, and suppose that b and c are defined on Ω[R] with |b| ≤ µ1 dα−1 ,

c ≥ −µ2 dα−1

(3.28)

for some nonnegative constants µ1 and µ2 . Let β 0 be defined on ∂Ω[R] and suppose that there is a nonnegative constant µ3 such that β 0 ≥ −µ3 β n . Finally, let α1 ∈ (0, 1) and r be positive constants and let x1 = (0, xn1 ) be a point in Rn with   ε6 xn1 ≥ A − ω0 r, 4

E(x1 , r) ⊂ B(0, R), and E(x1 , r) ∩ ∂Ω[R] = 6 ∅. Then, for any constants θ1 < θ2 and α in (0, 1), there are positive constants C1 , determined only

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by α, α1 , n, ε, ω0 , and µ3 R, and r0 , determined only by α, ω0 , n, ε, θ1 , θ2 , µ1 , µ2 , and µ3 R such that, if r ≤ r0 and u ∈ C 2 (Ω[R]) ∩ C(Ω[R]) is nonnegative with Lu ≤ 0 in Ω[R],

M u ≤ 0,

and if h is a nonnegative constant such that u ≥ h in E(x1 , θ1 r) ∩ Ω,

(3.29)

C1 u ≥ h in E(x1 , θ2 r) ∩ Ω.

(3.30)

then

Proof. For simplicity, we write E(s) for E(x1 , s) ∩ Ω and E 0 (s) for E(x1 , s) ∩ ∂Ω. As before, we may assume that c ≤ 0 and β 0 ≤ 0. With η defined by η(x) = 1 − G(x − x1 , r), we conclude (as in Lemma 1.18) that there is a constant q ≥ 2 determined only by µ, n, θ1 , and µ3 R such that L(η q ) ≥ −qη q−1 µ1 dα−1 |Dη| − µ2 η q dα−1

in E(r)\E(θ1 r) and M (η q ) ≥ 0 on E 0 (r). It follows that there is a constant k1 , determined only by ω0 and ε so that µ  1 L(η q ) ≥ −k1 q + µ2 dα−1 . r Next, we note that there is a constant k2 , determined only by ω0 and ε, such that E(r) ⊂ B(0, k2 r). With w ˆ taken from Lemma 3.11 corresponding to R = k2 r and K a positive constant to be determined, we now set w2 = w ˆ + Krα (k2 r − xn ). It follows that there is a constant k3 , determined only by α, µ, ω0 , and ε, such that Lw2 ≤ dα−1 [−1 + (K + k3 µ1 )rα ] in E(r), and M w2 ≤ β n [k3 − K]rα .

We now choose K = k3 and suppose that r0 ≤ (k3 (2µ1 +2))−1/α to conclude that 1 Lw2 ≤ − dα−1 in E(r), M w2 ≤ 0 on E 0 (r). 2 We now apply the maximum principle to µ  1 hη q − u − k1 q + µ2 hw2 r

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in E(r) \ E(θ1 r) to infer that hη q ≤ u + hk1 k4 (µ1 rα + µ2 rα+1 ) for some further constant k4 , determined only by α, µ, n, and k3 . We now choose r0 sufficiently small that k1 k4 (µ1 r0α + µ2 rα+1 ) ≤

1 η0 , 2

where η0 is the (nonzero) minimum of η q on E(θ2 r). Then (3.30) holds with C1 = η0 /2.  Just as in Corollary 1.22, we have a strong maximum principle for oblique derivative problems with appropriate unbounded coefficients as well. Corollary 3.15. Let u ∈ C 2 (Ω[R]) ∩ C(Ω[R]) and suppose u has a nonnegative maximum value at 0. Suppose also that the hypotheses of Lemma 3.14 are satisfied for L, M , and Ω. If c and β 0 are nonpositive and if Lu ≥ 0 in Ω[R],

M u ≥ 0 on ∂Ω[R],

then u is constant in Ω[R].

3.6

A Miller barrier for mixed boundary value problems

In order to study the oblique derivative problem in a domain with an edge when the vector β changes discontinuously across the edge, we need to examine the behavior of a mixed boundary value problem. Although this problem is interesting in its own right, we shall confine ourselves to the points that are applicable to the oblique derivative problem. Our first step is a variant of the Miller barrier in two dimensions. The notation will be slightly modified from what was used earlier in the chapter to conform with standard polar coordinates. In other words, we define r and θ by x1 = r cos θ,

x2 = r sin θ.

A simple calculation shows that, for w = rα f (θ), we have D1 w = rα−1 [α cos θf (θ) − sin θf 0 (θ)], D2 w = rα−1 [α sin θf (θ) + cos θf 0 (θ)].

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It follows that, on the line θ = 0, we have β · Dw = rα−1 [β 1 (r, 0)αf (0) + β 2 (r, 0)f 0 (0)].

Our polar representation is essentially the same as before: If [aij ] ∈ Lµ for some µ ≥ 1, then there are scalar functions a, b, and c satisfying (3.10a) such that aij Dij w = rα−2 (aα(α − 1)f + 2b(α − 1)f 0 + c(f 00 + αf )).

Conversely, given a, b, and c satisfying (3.10a), there is a matrix [aij ] ∈ Lµ such that this representation is valid. We now define the functions g and G by a b g(α, a, b, c; z, p) = α(α − 1)z + 2 (α − 1)p + αz, c c G(α; z, p) = sup{g(α, a, b, c; z, p) : a, b, c satisfy (3.10a)}. Assuming that β 1 ≤ µ0 β 2 for some constant µ0 , we then focus on the initial value problem Fα00 + G(α; Fα , Fα0 ) = 0 on (0, ∞), Fα (0) = 1, Fα0 (0) = −µ0 α.

As before, if Fα is positive on the interval (0, θ0 ), then w satisfies the conditions Mµ w ≤ 0 if 0 < θ < θ0 , β · Dw ≤ 0 if θ = 0, w ≥ 0 if θ = θ0 ,

so we are again interested in determining Ψ(α), the first zero of Fα , as a function of α. The proof of Lemma 3.7 shows immediately that Ψ is continuous and strictly decreasing and that Ψ maps (0, ∞) onto itself. Moreover, because G(1; z, p) = z, we see that F1 (θ) = cos θ − µ0 sin θ and hence Ψ(1) = arccot µ0 . (As before, Ψ(1) is independent of µ.) We therefore obtain (just as for Corollary 3.9) the following supersolution result. Proposition 3.16. Let K be the sector 0 < θ < θ0 for some θ0 > 0, and let µ ≥ 1 and µ0 be constants. Then there is a positive constant α1 (θ0 , n, µ, µ0 ) with the following properties: For any α ∈ (0, α1 ), there is a function wα ∈ C 2 (K) ∩ C(K) with wα ≥ |x|α and Lwα ≤ −|x|α−2 in K for all L ∈ Lµ . In addition, β · Dwα ≤ −|x|α−1 if θ = 0 for all vectors β satisfying β 1 ≤ µ0 β 2 and β 2 > 0. Finally, there is a constant C(α, µ, µ0 ) such that |Dw| ≤ C|x|α−1 in K.

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Our study of nonlinear problems will be based on a slight variation of the barrier constructed in this proposition. We begin by recalling the geometric setting from Section 2.6. With κ ∈ (0, 1) fixed, we define the spherical cap Γ(R) for R > 0 by Γ(R) = {x ∈ Rn : (x0 )2 + (xn + κR)2 < R2 , xn > 0} and we write Γ∗ (R) and Γ0 (R) for the subsets of ∂Γ(R) on which (x0 )2 + (xn + κR)2 = R2 and (x0 )2 + (xn + κR)2 < R2 , respectively. We also set Γ∗∗ (R) = Γ∗ (R) ∩ Γ0 (R) and we write d∗∗ (x) for the distance from x to Γ∗∗ (R). Our barrier result takes the following form. Proposition 3.17. Let µ ≥ 1 and µ0 ≥ 0 be constants, and set α1 = α1 (arctan κ, n, µ0 , µ). Then, for any α ∈ (0, α1 ), there are a positive constant ε(n) and a function w ˜α ∈ C 2 (Γ(R)) ∩ C(Γ(R)) such that w ˜α ≥ (d∗∗ )α ,

Lw ˜α ≤ −(d∗∗ )α−2

(3.31a)

on the subset of Γ(R) where d∗∗ < ε(n)R, β · Dw ˜α ≤ −(d∗∗ )α−1

(3.31b)

on the subset of Γ0 (R) where d∗∗ < ε(n)R. Moreover, there is a constant C1 (α, µ, µ0 , n) such that w ˜α ≤ C1 (d∗∗ )α ,

|Dw ˜α | ≤ C1 (d∗∗ )α−1

on the subset of Γ(R) where d∗∗ < ε(n)R. Proof. First, we note that we can cover Γ∗∗ (R) by a finite number of open sets U1 , . . . , UN such that, for each k ∈ {1, . . . , N }, there is a C 2 function Φk : Uk → Rn which has a C 2 inverse (defined on Φk (Uk )) such that xn = xn−1 = 0 for all x ∈ Φk (Uk ∩ Γ∗∗ (R)), xn = 0 for all x ∈ Φk (Uk ∩Γ0 (R)), xn = κxn−1 for all x ∈ Φ(Uk ∩Γ∗ (R)), and 0 < xn < κxn−1 for all x ∈ Φk (Uk ∩ Γ(R)). Moreover, we can arrange that the number of sets N and an upper bound for C 2 norms of all Φk s and their inverses are completely determined by n. With c˜(n) a constant at our disposal, we now set Wk = c˜(n)Φ−1 k (wα ),

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where wα is the function from Proposition 3.16. By choosing c˜(n) appropriately, we can arrange that Wk ≥ (d∗∗ )α , in Uk ∩ Γ(R), and

Lwk ≤ −(d∗∗ )α−1

β · DWk ≤ −(d∗∗ )α−1

on Uk ∩ Γ0 (R). Moreover there is a constant C, determined by the same quantities as for C1 such that Wk ≤ C(d∗∗ )α ,

|DWk | ≤ C(d∗∗ )α−1

in Uk ∩ Γ(R). If (ηk ) is a partition of unity on Γ∗∗ (R) subordinate to the cover (Uk ), it follows that N X w ˜α = ηk wk k=1

satisfies the conditions

w ˜α ≥ (d∗∗ )α , "  ∗∗ 2 # d∗∗ d ∗∗ α−2 Lw ˜α ≤ (d ) −2 + C2 + C2 R R wherever

P

ηk = 1 in Γ(R) and   d∗∗ β · Dw ˜α ≤ (d∗∗ )α−1 −2 + C2 R

P wherever ηk = 1 on Γ0 (R) for some constant C2 . By choosing ε(n) P sufficiently small, we infer that ηk = 1. Further decreasing ε gives the desired result.  Notes The basic material in Section 3.1 on ordinary differential equations is standard and can be found in any text on the subject; the first proof of the result in this form is in [123]. Lemma 3.2 and Corollary 3.3 are simplified versions of the results in [134]. The polar representation Lemma 3.5 is proved in [133] and the barrier construction comes from [135] (see Section 2 of that work for the case n ≥ 3 and Section 3 for n = 2). Our proofs of these results differ from Miller’s only in some minor technical details. We

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mention here that Oddson [149] derived an explicit formula for α0 in case n = 2. The construction of a supersolution for the Dirichlet problem (Lemma 3.10) is a simplified version of the construction in Theorem 3.7 of [130]. In fact, the existence of a viscosity supersolution is fairly straightforward, given the theory of viscosity solutions, and we shall present that theorem in Chapter 7. The existence of a Perron supersolution was first presented in Theorem 4.2 of [118]; that theorem actually shows that there is a classical supersolution by first showing that a Perron supersolution exists and then using the standard theory of fully nonlinear equations, some of which we present in Chapter 12. The construction of a supersolution for the oblique derivative problem with singular coefficients comes from [115]. Our Lemma 3.11 is essentially Lemma 2.1 of [115], and our Theorem 3.12 and Corollary 3.13 are essentially Theorems 4.2 and 4.3 from [115]. In fact the results in [115] allow more general unbounded lower coefficients; we present a simplified version of some of these results in the Exercises 3.1, 3.2, and 3.3. The material in Section 3.6 is taken from [110] although, as is emphasized here, it is only marginally different from Miller’s work. We point out that most of the results in this chapter can be proved via the arguments in [133] if we are not interested in obtaining the optimal H¨ older exponent of our function w. Our main interest in this sharp result is to show that the H¨ older exponent depends on simple, sharp geometric quantities, and that the exponent can be made arbitrarily close to 1 by taking the opening angle of the exterior cone sufficiently close to π/2. Exercises 3.1 Let B1 be a positive constant. Adapt the proof of Lemma 3.11 to construct a function w ˆ1 such that w ˆ1 ≤ Cr1+α ,

|Dw ˆ1 | ≤ Crα ,

Mµ w ˆ1 + B1 λ|Dw ˆ1 | ≤ −dα−1 in Ω[r]. 3.2 Prove the corresponding maximum principle to Theorem 1.9 with condition (1.16) replaced by the condition |b| ≤ λB1 dα−1 . Specifically, suppose that the eigenvalues of the [aij ] are in the range [λ, µλ] for some positive function λ and a positive constant µ, let β be an inward

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pointing vector and suppose that there are nonnegative constants B0 and B1 such that |β| ≤ −B0 β 0 on ∂Ω,

|b| ≤ λB1 (diam Ω)α dα−1 in Ω. Show that, if u ∈ C 2 (Ω) ∩ C(Ω) satisfies Lu ≥ −f in Ω,

M u ≥ gβ 0 on ∂Ω

for some nonnegative function f and some nonnegative constant g, then sup u ≤ g + C(α, n, B1 )(B0 + diam Ω)(diam Ω)α sup f d1−α . Ω

3.3 Prove the weak Harnack inequality for supersolutions under hypotheses similar to those in the preceding exercise. In other words, show that Theorem 3.12 is still true if we replace conditions (3.23bc) by |b| ≤ µ1 λdα−1 , c ≥ −µ2 λdα−1 and if p and C depend on µ1 Rα and µ2 Rα+1 . 3.4 Prove that solutions of Lu = f in Ω, M u = g on ∂Ω are H¨older continuous under the hypotheses in Exercise 3.3.

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Chapter 4

H¨ older Estimates for First and Second Derivatives

Introduction This chapter provides H¨ older estimates for the first and second derivatives of solutions of linear oblique derivative problems. As we shall see, several different methods are used to cover various situations. In particular, we show that solutions of linear oblique derivative problems are typically smoother than the domains in which they are posed. For example, solutions in Lipschitz domains are C 1,α (for some α > 0 determined by the domain and the differential equation but not the boundary condition) while solutions in C 1,α domains are in C 2,α (with α ∈ (0, 1) arbitrary). Sections 4.1 and 4.5 give C 1,α estimates for linear oblique derivative problems in two special circumstances. The C 2,α estimates are given in Section 4.7 as a straightforward corollary of our C 1,α estimates.

4.1

C 1,α estimates for continuous β

We start with a model problem which will be used to study the general problem. The most important features of the model problem are that the differential equation and boundary condition have constant coefficients (and no lower order terms) and that we do not make any change of variables, so our estimates are derived in the original domain Ω. To avoid later complications, we fix R0 > 0 and a Lipschitz function ω (defined only for |x0 | < R0 ) with Lipschitz constant no greater than a positive constant ω0 . For |x0 | < R0 and R ∈ (0, R0 − |x0 |), we introduce the sets Ω[x0 , R] = {x = (x0 , xn ) : |x − x0 | < R, xn > ω(x0 )}, 0

n

n

0

Σ[x0 , R] = {x = (x , x ) : |x − x0 | < R, x = ω(x )}, 117

(4.1a) (4.1b)

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σ[x0 , R] = {x = (x0 , xn ) : |x − x0 | = R, xn ≥ ω(x0 )}.

(4.1c)

Our first step is to prove a H¨older gradient estimate for solutions of constant coefficient differential equations with constant coefficient oblique boundary condition on Σ[x0 , R]. This estimate naturally comes in two parts, which are most easily described in the special case that the oblique boundary condition is Dn u = 0 on Σ[x0 , R]. In this case, we first obtain a H¨ older estimate on Dn u by observing that it satisfies the same differential equation as u and that it satisfies a Dirichlet boundary condition on Σ[x0 , R]; then we obtain a H¨older estimate on the derivatives D1 u, . . . Dn−1 u. The exact formulation of our estimates is actually somewhat more subtle, as we shall see, but this description is the motivation for our estimates. The H¨ older estimate for Dn u (or more generally β · Du) is proved exactly as described and it requires no restriction on the direction of the vector β. Lemma 4.1. Let u ∈ C 2 (Ω[x0 , R]) ∩ C 1 (Ω[x0 , R] ∪ Σ[x0 , R]) be a solution of Aij Dij u = 0 in Ω[x0 , R], β0i Di u

= 0 on Σ[x0 , R]

(4.2a) (4.2b)

for some positive definite (constant) matrix [Aij ] and some nonzero (constant) vector β0 . Suppose also that |β0i Di u| is bounded in Ω[x0 , R] and that there are positive constants λ and µ such that λ|ξ|2 ≤ Aij ξi ξj ≤ µλ|ξ|2

(4.3)

for all ξ ∈ Rn . Suppose also that there is a cone K with opening θ0 with vertex at x0 which contains Ω[x0 , R]. If α ∈ (0, α0 (θ0 , µ, n)) (the constant from Theorem 3.8), then there is a constant C(n, α, θ0 , µ) such that |β0i Di u(x)| ≤ C sup |β0i Di u|

|x − x0 |α . Rα

(4.4)

Proof. From Lemma 2.1, we know that u ∈ C 3 (Ω[x0 , R]), so v = β0i Di u ∈ C 2 (Ω[x0 , R]) ∩ C 0 (Ω[x0 , R] ∪ Σ[x0 , R]). For simplicity of notation we set H = sup |β0i Di u|. If w is the function from Theorem 3.8 and if we write K 0 for the intersection of K and Ω[x0 , R], then we see that L(Hw ± v) ≤ 0 in K 0 and Hw ± v ≥ 0 on ∂K 0 . It follows from the maximum principle that Hw ± v ≥ 0 in K 0 and this inequality, in conjunction with the upper bound for w in Theorem 3.8, immediately yields (4.4). 

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119

The remainder of the H¨ older gradient estimate is much more delicate. To begin with, we need to make some quantitative hypotheses on the vector β0 which measure its obliqueness. Furthermore (as will be apparent from the proof), we do not estimate the H¨older constant of the gradient directly. Instead, we obtain a pointwise estimate on u which relates to the results in Section 2.2. To state our condition on the H¨older exponent more succinctly in its statement, we define the Miller angle for Ω[x0 , R] to be the largest number θ such that, for each x ∈ Σ[x0 , R], there is a cone K with vertex x and opening angle θ which is exterior to Ω[x0 , R]. We also use α0 to denote the function from Theorem 3.8. Lemma 4.2. Suppose, in addition to the hypotheses of Lemma 4.1, that there are positive constants χ and ε with ε < 1 such that 1−ε n |β00 | ≤ β , β0n ≥ χ. (4.5) ω0 0 If i β0 D i u , H0 = sup χ Ω[x0 ,R]

then there are constants C0 (n, α, α0 (θ0 , n, µ), ε, ω0 ) ≥ 0, where θ0 is the Miller angle for Ω[x0 , R], and κ(ε, ω0 ) ∈ (0, 1) such that, for any r ∈ (0, R), there is a linear function P1 (x0 , r) with β0i Di P1 (x0 , r) = 0, sup |u − P1 (x0 , r)| ≤ C0 (τ 2 sup |u| + H0

Ω[x0 ,τ r]

Ω[x0 ,r]

(4.6a) 1+α

r ) Rα

(4.6b)

for all τ ∈ (0, κ). Proof.

We set ξ = β0 /|β0 | and

1 x ¯0 = x0 + rξ. 2 Then there is a positive constant κ(ε, ω0 ) ∈ (0, 1) such that B(¯ x0 , κr) ⊂ Ω[x0 , r]. With σ ∈ (0, 1) to be further specified, we consider x1 ∈ Ω[x0 , σκr] and set 1 x ¯1 = x1 + rξ. 2

We also set p0 = Du(¯ x0 ) − ξ · Du(¯ x0 )ξ,

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and we define P1 (x0 , r)(x) = p0 · (x − x0 ) + u(x0 )

and

v(y) = Then

β0 · Du(y). |β0 |

u(x1 ) − P1 (x0 , r)(x1 ) = u(¯ x1 ) − Du(¯ x0 ) · (¯ x1 − x ¯0 ) − u(¯ x0 ) + ξ · Du(¯ x0 )ξ · (x1 − x2 ) Z r/2 + v(x1 + sξ) − v(x0 + sξ) ds. 0

The second derivative estimate for u then implies that

|u(¯ x1 ) − Du(¯ x0 ) · (¯ x1 − x¯0 ) − u(¯ x0 )| ≤ C(n, µ)σ 2

sup B(¯ x0 ,κr)

|u|

and Lemma 4.1 gives |ξ · Du(¯ x0 )ξ · (x1 − x2 )| ≤ CH0

r1+α Rα

and Z

0

r/2

v(x1 + sξ) − v(x0 + sξ) ds ≤ CH0

Z

0

r/2

sα ds Rα

r1+α = CH0 α . R The proof is completed by choosing σ = κτ , taking the supremum over all x1 ∈ Ω[x0 , τ r], and combining these three inequalities with the observation that sup B(¯ x0 ,κr)

|u| ≤ sup |u|. Ω[x0 ,r]



We next prove a H¨ older gradient estimate for constant coefficient problems. Although it is possible to prove the corresponding estimate for variable coefficient problems without this intermediate result, we include it for several reasons. First, the result is of independent interest; second, the techniques involved in the variable coefficient case are more complicated versions of those for the constant coefficient case; and, third, we will use the same method to prove the variable coefficient estimate. Proposition 4.3. Under the hypotheses of Lemma 4.2, we have u ∈ H1+α (Ω[x0 , R/2]) and there is a constant C such that |u|1+α;Ω[x0 ;R/2] ≤ CR−1−α |u|0 .

(4.7)

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Proof. Our goal is to obtain an estimate of the form (2.19) for each y ∈ Ω[x0 , R/2] ∪ Σ[x0 , R/2] and appropriate values of r (as in Lemma 2.9). To this end, we consider separately the cases when y ∈ Σ[x0 , R/2] and y ∈ Ω[x0 , R/2]. If y ∈ Σ[x0 , R/2], we take κ and C0 from Lemma 4.2 and set τ = min{κ, (2C0 )1/(α−1) }. Our construction of P1 is obtained via an iteration argument very similar to the one used in proving Proposition 2.12. With u0 = u, r = R/2 and R replaced by R/2, Lemma 4.2 gives a linear function L0 such that sup Ω[y,τ R/2]

|u0 − L0 | ≤ C0 (τ 2

sup |u0 | + H0 R),

Ω[y,R/2]

where here i β Di u . sup 0 χ

H0 =

Ω[y,R/2]

Repeated applications of Lemma 4.2 provide a sequence of linear functions (Lm ) such that the sequence of functions (um ) defined by um = um−1 − Lm−1 satisfies the estimate sup Ω[y,τ m+1 R/2]

|um+1 | ≤ C0 (τ 2

sup Ω[x1 ,τ m R/2]

|um | + H0 Rτ m(1+α) )

(with the same constant C0 ). For r ≤ R/2, we take k to be the integer such that τ k R/2 < r ≤ τ k−1 R/2 and define P1 (y, r) =

k X

Lm .

m=0

It follows from Lemma 1.25 that

sup |u − P1 (y, r)| ≤ Cr1+α (R−1−α |u|0 + H0 R−α ).

(4.8)

Ω[y,r]

The proof of (4.8) when y ∈ Ω[x0 , R/2] is more subtle, but easier. First, recalling that d denotes distance to ∂Ω, we set r0 = 12 d(y) and we choose y1 ∈ ∂Ω so that |y1 − y| = d(y). We also observe that Ω[y, r0 ] ⊂ Ω[y1 , 3r0 ]. It follows that sup |u − P1 (y1 , 3r0 )| ≤

Ω[y,r0 ]

sup Ω[y1 ,3r0 ]

|u − P1 (y1 , 3r0 )|

≤ Cr01+α (R−1−α |u|0 + H0 R−α ). So we take P1 (y, r0 ) = P1 (y1 , 3r0 ), and we abbreviate L0 = P1 (y1 , 3r0 ). We now set u1 = u − L0 and then we define L1 to be the first order

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Taylor polynomial of u1 centered at y. The second derivative estimate for u2 = u1 − L1 then implies that  1+α r sup |u2 | ≤ C sup |u1 |. r0 Ω[y,r] Ω[y,r0 ] We now take P1 (y, r) =

   L 0 + L 1

P1 (y1 , r0 )   P (y , r) 1

1

if 0 < r ≤ r0 ,

if r0 < r ≤ 3r0 , if r > 3r0

to infer (4.8) for r ∈ (0, R], and therefore

|u|1+α;Ω[x0 ;R/2] ≤ C(R−1−α |u|0 + R−α H0 ).

To complete the proof, we need to eliminate H0 from this estimate. This elimination follows from the estimate H0 ≤ C|u|0 /R, which is an immediate consequence of Lemma 2.20.

(4.9) 

Corresponding estimates for variable coefficient problems with inhomogeneous differential equation and boundary condition follow the same basic scheme but several new technical details arise. Most of these details are algebraic in nature (for example, the estimate corresponding to (4.8) must take the inhomogeneity of the problem into account, leading to a more complicated right hand side) but a key point that requires more concern is related to the perturbation argument that we use here. Since the perturbation is based on an existence theorem, we need to take advantage of the rather complicated estimates from Chapter 2. We begin with the existence theorem. Lemma 4.4. Let x0 , R, Aij and β0 be as in Lemma 4.2. Then, for any g ∈ C(σ[x0 , R]), there is a unique solution u ∈ C 2 (Ω[x0 , R]) ∩ C 0 (Ω[x0 , R]) of Aij Dij u = 0 in Ω[x0 , R], β0i Di u

(4.10a)

= 0 on Σ[x0 , R],

(4.10b)

u = g on σ[x0 , R].

(4.10c)

Proof. Let (ωm ) be a sequence of C 2,α functions which converge uniformly to ω and which satisfy |ωm (x0 ) − ωm (y 0 )| ≤ ω0 |x0 − y 0 |

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for all x0 and y 0 in Rn−1 with |x0 |, |y 0 | < R. (The existence of this sequence follows from Lemma 1.4.) Without loss of generality, Ωm [x0 , R] is nonempty for all m. We then obtain a solution um to Aij Dij um = 0 in Ωm [x0 , R], β0i Di um = 0 on Σm [x0 , R], um = g on σm [x0 , R]. The maximum principle implies that the sequence (um ) is uniformly bounded and then Corollary 1.29 implies that the sequence is equicontinuous, so the Arzela-Ascoli theorem provides a uniformly convergent subsequence. Proposition 4.3 shows that (Dum ) is uniformly bounded in Hα (K) for any compact subset K of Ω[x0 , R] ∪ Σ[x0 , R] and the interior second derivative estimates imply that (D2 um ) is uniformly bounded in Hα (K 0 ) for any compact subset K 0 of Ω[x0 , R]. Hence (Dum ) converges uniformly to Du on compact subsets of Ω[x0 , R] ∪ Σ[x0 , R] and (D2 um ) converges uniformly to D2 u on compact subsets of Ω[x0 , R]. It follows that u is a solution of (4.10), and another application of the maximum principle shows that u is unique.  We are now ready to prove our Schauder estimate for variable coefficient equations. Theorem 4.5. Let θ be a continuous, increasing function on [0, 1/2] with θ(0) = 0. Let R0 be a positive number and let ω be a Lipschitz function. Suppose aij satisfies (4.3) and |aij (x) − aij (y)| ≤ θ(|x − y|/d∗ (x)) for all x and y in Ω[R0 ] with |x − y| < (4.5) and

1 ∗ 2 d (x).

(4.11)

Suppose also that β satisfies

|β|(0) α ≤ B0

(4.12)

for some α ∈ (0, α0 (θ0 , µ, n)), where θ0 is the Miller angle for Ω[R0 ]. Let (0) (1) u ∈ C 2 (Ω[R0 ]) ∩ H1 (Ω[R0 ]) with M u ∈ Hα , and suppose that there is a nonnegative constant F such that |Lu| ≤ λF (d∗ )−α−1 dα−1

(4.13) (0) H1+α

in Ω[R0 ], where L = aij Dij and M = β i Di . Then u ∈ and (1) Mu (0) + |u|0 ). |u|1+α ≤ C(θ, µ, B0 , ω0 , ε)(F + χ α

(4.14)

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Proof. It suffices to obtain a H¨older estimate for Du. To this end, we fix a point y ∈ Ω[R0 ] ∪ Σ[R0 ] and prove a suitable estimate for the difference between u and some linear polynomial. To simplify writing, we set g = (1) M u/χ and G = [g]α . We first assume that y ∈ Σ[R0 ]. We set σ = 12 (α + α0 ), and, with δ1 ∈ (0, 1/2] to be further specified, we set R = δ1 d∗ (y). Further we define u0 by u0 (x) = u(x) + and we note that

β(y) · (y − x)M u(y), |β(y)|2

|β(x) · Du0 (x)| ≤ χG

|x − y|α R1+α

(4.15)

for x ∈ Σ[R0 ]. Our next step is to define quantities rk , Uk , Lk , uk , and vk inductively as follows. Specifically, rk and Uk are numbers, uk and vk are functions, and Lk is a linear polynomial with β(y) · DLk = 0. First, with τ ∈ (0, 1/2) to be chosen, we set rk = τ k R/2 and Uk = sup |uk |, Ω[y,rk ]

we write Lk for the first order Taylor polynomial of vk centered at y, and we set uk+1 = uk − Lk . Assuming that we have already uk , we infer from Lemma 4.4 that there is a solution vk of aij (y)Dij vk = 0

in Ω[y, rk ],

(4.16a)

β(y) · Dvk = 0

on Σ[y, rk ],

(4.16b)

on σ[y, rk ].

(4.16c)

vk = uk

This construction guarantees that β(y) · DLk = 0. Proposition 4.3 then implies that there is a constant C0 (determined only by ω0 , σ, n, ε, and µ) such that sup Ω[y,rk+1 ]

|vk − Lk | ≤ C0 τ 1+σ sup |vk |, Ω[y,rk ]

and the maximum principle implies that sup |vk | ≤ Uk .

Ω[y,rk ]

and hence sup Ω[y,rk+1 ]

|vk − Lk | ≤ C0 τ 1+σ Uk .

(4.17)

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To continue, we infer from Lemma 2.20 that |Dvk (y)| ≤ C

sup |vk | , rk

and hence |β · DLk | ≤ Cχ

Uk |x − y|α . rk Rα

It follows that |M uk (x)| ≤ |M uk−1 (x)| + Cδ1α χ

Uk−1 . rk−1

Combining this inequality with (4.15) yields |M uk | ≤ Cχ(G + δ1α

k−1 X j=0

Uj rkα ) τ j R1+α

on Σ[y, rk ]. We now abbreviate Gk = G + δ1α

k−1 X j=0

Uj . τj

It follows from Corollary 3.13 that there are positive constants C and ζ such that  ζ rk1+α |x − x1 | |uk (x) − uk (x1 )| ≤ C(Uk + [F + Gk ] 1+α ) R R for all x and x1 in Ω[y, rk ]. In addition, Corollary 1.29 implies that  ζ r1+α |x − x1 | |vk (x) − vk (x1 )| ≤ C(Uk + [F + Gk ] k1+α ) R R for all x ∈ Ω[y, rk ] and all x1 ∈ σ[y, rk ]. It follows that, for any δ ∈ (0, 1), we have rk1+α R1+α if (1 − δ)rk < |x − y| < rk (since uk (x1 ) = vk (x1 ) where x1 is the closest point on σ[y, rk ] to x). The H¨older gradient estimate (Proposition 4.3) and the second derivative estimate (Theorem 1.34) for vk then imply that |uk (x) − vk (x)| ≤ Cδ ζ U0 + C[F + Gk ]

|D2 vk | ≤ C(δ)rk−1−α dα−1 Uk in Ω[y, (1 − δ)rk ]. It follows that

|Lvk | ≤ C(δ)θ(δ1 )λrk−1−α dα−1 Uk

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and hence |L(uk − vk )| ≤ Aλrk−1−α dα−1 in Ω[y, (1 − δ)rk ] for

rk1+α . R1+α Using the function w ˆ from Lemma 3.11 with rk in place of r, we infer that w± = Ar−1−α w ˆ ± (uk − vk ) satisfy A = C(δ)θ(δ1 )Uk + F

L(w± ) ≤ 0

in Ω[y, (1 − δ)rk ],

  A rkα M (w± ) ≤ χ C + Gk 1+α rk R

on Σ[y, (1 − δ)rk ],

rk1+α R1+α The maximum principle then implies that w± ≥ −Cδ ζ U0 − C[F + Gk ]

w± ≥ −Cδ ζ Uk − C[F + Gk ]

on σ[y, (1 − δ)rk ]. rk1+α − CA R1+α

in Ω[y, (1 − δ)rk ], and it follows from the inequality 0 ≤ w ˆ ≤ Crk1+α that

rk1+α R1+α in Ω[y, (1 − δ)rk ], and hence this inequality is true in Ω[y, rk ]. Combining this inequality with (4.17), we conclude that |uk − vk | ≤ [C(δ)θ(δ1 ) + Cδ ζ ]U0 + C(F + Gk )

Uk+1 ≤ [C(δ)θ(δ1 ) + Cδ ζ + C0 τ 1+σ ]Uk + C(F + Gk )

rk1+α . R1+α 0

We now set σ 0 = (α + σ)/2 and choose τ so that C0 τ 1+σ ≤ 31 τ 1+σ . 0 Then we choose δ so that Cδ ζ ≤ 13 τ 1+σ , and finally δ1 so that C(δ)θ(δ1 ) ≤ 0 1 1+σ . We then see that 3τ 0

Uk+1 ≤ τ 1+σ Uk + C(F + G)τ k(1+α) + Cδ1α τ (1+α)k

k−1 X j=0

Uj τj

for all nonnegative integers k. If we set H = C(F + G), this inequality takes the form Uk+1 ≤ τ

1+σ0

Uk + Hτ

k(1+α)

+

Cδ1α τ (1+α)k

k−1 X j=0

Uj , τj

(4.18)

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and we now claim that Uk ≤ τ (1+α)k U0 + 2Hτ (1+α)(k−1)

(4.19)

0

provided we take τ also small enough that τ σ −α ≤ 1/4 and δ1 small enough that ∞ X Cδ1α τ αj ≤ τ 1+α /2. j=0

To verify this claim, we first note that it is true for k = 0. Assuming that we have verified (4.19) for all k less than or equal to some positive integer m, we infer from (4.18) that Um+1 ≤ U0 τ (1+α)(m+1) [τ σ + Hτ (1+α)m [2τ σ

0

0

−α

−α

+ Cδ1α τ −1−α

m−1 X

τ αj ]

j=0

+ 1 + Cδ1α τ −1−α

m−1 X

τ αj ].

j=0

Our smallness assumptions on τ and δ1 then imply that (4.19) is also valid for k = m + 1, and hence, by induction, this estimate is true for all k. Rewriting (4.19) in terms of the norms of u and g and observing that sup |u0 | ≤ sup |u| + C|g|(1) α

Ω[y,R]

now gives, for each r ∈ (0, 12 d∗ (y)), a linear polynomial P1 (y, r) such that sup |¯ u − P1 (y, r)| ≤ C

Ω[y,r]

r1+α (1) (sup |u| + F + |g|1+α ). R1+α

In particular P1 (y, r) = 0 if r ≥ δ1 d∗ (y). Arguing as in Proposition 4.3 gives the analogous estimate for y ∈ Ω[x0 , R] as well and hence we obtain (4.14).  Once we have this local C 1,α estimate, it is relatively straightforward to obtain estimates for the second derivatives. Unfortunately, an attempt to prove local estimates leads to a consideration of weighted H¨older norms involving two weights, and this is a technical complication that we wish to avoid. For this reason, we proceed to a global estimate. Theorem 4.6. Let Ω be a bounded Lipschitz domain, and let β be a uniformly oblique vector field defined on ∂Ω (with constants R, ω, and ε). Let [aij ] be a matrix-valued function defined on Ω and suppose (4.3) is satisfied. Let α0 be the Miller exponent for µ and Ω, and let α ∈ (0, α0 ). Suppose

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(0)

(1−α)

(1−α)

aij ∈ Hδ (Ω)∩C(Ω), bi ∈ Hδ , c ∈ Hδ . Suppose further that β and (−1−α) β 0 are in Hα (∂Ω). If u ∈ H2+δ (Ω) and if θ is an increasing continuous function with θ(0) = 0 such that |aij (x) − aij (y)| ≤ θ(|x − y|),

(4.20)

and if A and B0 are nonnegative constants with (0)

|aij |δ ≤ λA,

(4.21a)

(1−α) |bi |δ ≤ λA, (1−α) |c|δ ≤ λA,

(4.21b) (4.21c)

|β|α ≤ χB0 ,

(4.21d)

0

|β |α ≤ χB0 ,

(4.21e)

then there is a constant C, determined only by A, B0 , α, ε, ω0 , R, θ, and µ, such that (−1−α)

|u|2+δ Proof.

(1−α)

≤ C(|Lu|δ

+ |M u|α + |u|0 ).

(4.22)

ˆ and M ˆ by We first define L ˆ = aij Dij w, Lw

ˆ w = β i Di w. M

It follows from Theorem 4.5 (and a partition of unity argument) that ˆ (1−α) + |M ˆ u|α + |u|0 ). |u|1+α ≤ C(|Lu| 0 In conjunction with Lemma 2.14, this inequality implies that (−1−α)

|u|2+δ

(1−α)

ˆ ≤ C(|Lu| δ

ˆ u|α + |u|0 ). + |M

ˆ + bi Di u + cu, it follows that, for any ζ > 0, there is a Since Lu = Lu constant C(ζ) such that ˆ (1−α) ≤ C|u|(0) ≤ ζ|u|(−1−α) + C(ζ)|u|0 . |Lu − Lu| δ 1+δ 2+δ Similarly, we have ˆ u| ≤ ζ|u|(−1−α) + C(ζ)|u|0 . |M u − M 2+δ Hence there is a constant C1 such that (−1−α)

|u|2+δ

(1−α)

≤ C(|Lu|δ

(−1−α)

+ |M u|α ) + C1 ζ|u|2+δ

+ C(ζ)|u|0 .

The desired result follows from this inequality with ζ = 1/(2C1 ).



We now recall from Proposition 2.13 that C 2 solutions of Lu = f are in C 2,α if f and the coefficients of L are in C α . Theorem 4.5 shows that, near a Lipschitz boundary portion T , u ∈ C 1,α (Ω ∪ T ).

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129

Regularized distance

To continue our study of boundary value problems in domains with boundaries that are not necesarily C 2 , we introduce a modification of the distance function. We say that ρ is a regularized distance for a domain Ω if ρ ∈ C 2 (Ω) and if there are positive constants C1 and C2 such that C1 ρ ≤ d ≤ C2 ρ in Ω. There are many ways to construct such a function, and the one of most use here is a local construction. Lemma 4.7. Let R > 0, let ω ∈ H1 (Bn−1 (0, R)), and write Ω for the set of all x ∈ Rn with xn > ω(x0 ) and |x| < R. Then there is a function ρ ∈ H1 (Ω) ∩ C 3 (Ω) such that 3 C(sup[ω]1 )ρ ≤ d ≤ ρ, (4.23a) 2 |Dρ| ≤ C(n)[ω]1 , (4.23b)

|D2 ρ| ≤ C(n)[ω]1 d−1 . Furthermore, for any ε > 0, the function ρ can be taken so that Dn ρ ≥ 1 − ε,

|D0 ρ| ≤ [ω]1 (1 + ε),

(4.23c) (4.24)

in which case the constant in (4.23c) is determined also by ε. Proof. First, we observe that ω can be extended to an H1 (Bn−1 (0, 4R)) function ω ˜ with [˜ ω ]1 ≤ [ω]1 . R 3 n−1 Let ϕ be a nonnegative C (R ) function with Rn−1 ϕ(x) dx = 1 such that ϕ(x) = 0 if |x| > 1, and define Z τ ω ¯ (x0 , τ ) = ω ˜ (x0 − y 0 )ϕ(y 0 ) dy 0 L Rn−1 ¯ is three times continuwith L = 2 max{1, [ω]1 }. It’s easy to check that ω ously differentiable with respect to all its arguments wherever it’s defined and τ 6= 0 and that |D2 ω ¯ (x, τ )| ≤ C[ω]1 τ −1 . We then define ρ(x) to be the unique number τ such that xn − ω ¯ (x0 , τ ) = τ. Such a number exists because ∂ω ¯ ≤ 1. ∂τ 2 We leave it to the reader to check (in Exercise 4.1) that this ρ is the desired function satisfying (4.23). If we take L sufficiently large (determined only by ε and [ω]1 ), then (4.24) is also satisfied. 

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In the next section, we shall need a global version of this regularized distance which interacts well with the vector field β. Here we give the result, which follows by a simple partition of unity argument. Lemma 4.8. Let Ω be a bounded Lipschitz domain in Rn and suppose that β is a continuous vector field defined in Ω which is uniformly oblique on ∂Ω. Then, for any α ∈ (0, 1), there is a sequence of H2+α domains (Ωk ) with Ωk ⊂ Ω, and β is uniformly oblique on Ωk (with constants independent of k). Proof. We first proceed locally. Fix a point x0 ∈ ∂Ω and rotate axes so that Ω[R] = Ω ∩ B(x0 , R) can be represented as the set of all x with |x| < R and xn > ω(x0 ) for some Lipschitz function ω and some positive R. To match previous notation, we set ω0 = [ω]1 . Since β is uniformly oblique on ∂Ω, we may assume also that there is a constant ε ∈ (0, 1) such that ω0 |β 0 | ≤ (1 − ε)β n

on ∂Ω[R]. From the continuity of β, it follows that there is a constant r1 such that ε ω0 |β 0 | ≤ (1 − )β n 2 if d(x) < r1 , and hence this inequality holds on the level sets of ρ close enough to ∂Ω[R]; in other words there is a positive constant r2 such that this inequality holds on the set ρ = s if s ∈ (0, r2 ). We then take Ωk [R] to be the level set ρ = r2 /k and observe that this level set is the graph of an H3 function. The proof is completed via a straightforward partition of unity argument.  If we assume that the function ω is more regular, then we can make stronger assertions about ρ. This result will be used in Section 4.4. Lemma 4.9. Let R > 0, let α ∈ (0, 1], let ω ∈ H1+α (Bn−1 (0, R)), and write Ω for the set of all x ∈ Rn with xn > ω(x0 ) and |x| < R. Then there is a function ρ ∈ H1+α (Ω) ∩ C 2 (Ω) such that 3 C(sup |Dω|)ρ ≤ d ≤ ρ, (4.25a) 2 [Dρ]α ≤ C(n)[Dω]α , (4.25b) |D2 ρ| ≤ C(n)[Dω]α dα−1 .

(4.25c)

Proof. We construct ρ just as in Lemma 4.7. We leave the verification of (4.25) to the reader as Exercise 4.2. 

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Existence of solutions for continuous β

We can use the regularized distance from Lemma 4.8 to show that the problem Lu = f in Ω,

M u = g on ∂Ω

(4.26)

has a unique solution under the regularity assumptions of Theorem 4.6 and the additional assumptions c ≤ 0 and β 0 ≤ 0 but these functions are not both identically zero. Theorem 4.10. Let Ω be a bounded Lipschitz domain and let β be a uniformly oblique vector field on ∂Ω (with constants R, ω, and ε). Let [aij ] be a matrix-valued function defined on Ω and suppose (4.3) is satisfied. Let α0 be the Miller exponent for µ and Ω, and let α ∈ (0, α0 ). Suppose (0) (1−α) (1−α) aij ∈ Hδ (Ω) ∩ C(Ω), bi ∈ Hδ , c ∈ Hδ . Suppose further that 0 β and β are in Hα (∂Ω). Suppose also that c ≤ 0 in Ω and β 0 ≤ 0 on ∂Ω but that at least one of these functions is not identically zero. Then, (2−α) for all f ∈ Hδ (Ω) and all g ∈ Hα (∂Ω), there is a unique solution u ∈ C 2 (Ω) ∩ C(Ω) of (4.26). Moreover, there is a constant C, determined only by Ω, and the coefficients of L and M such that (−1−α)

|u|2+δ

(1−α)

≤ C(|Lu|δ

+ |M u|α ).

(4.27)

Proof. First, let f and g be fixed functions, and extend β, β 0 , and g as (−α) H1+δ functions so that β 0 ≤ 0 in Ω. Let Ωk be the sequence of domains from Lemma 4.8 and observe that there is a positive integer k0 such that c < 0 somewhere in Ωk for all k ≥ k0 if c < 0 somewhere in Ω and such that β 0 < 0 somewhere on ∂Ωk if β 0 < 0 somewhere on ∂Ω. Then, for each k ≥ k0 , the strong maximum principle implies that the only solution of Lvk = 0 in Ωk ,

M vk = 0 on ∂Ωk

is vk = 0, so Theorem 2.30 gives a unique solution of Luk = f in Ωk ,

M uk = g on ∂Ωk .

We now claim that the sequence (uk ) is uniformly bounded. If this is not the case, then we may assume (by taking a subsequence, if necessary) that Uk = |uk |0 → ∞. If we set wk = uk /Uk , fk = f /Uk , and gk = g/Uk , then (1−α) (−α) |wk |0 = 1, and |fk |δ and |gk |1+δ are uniformly bounded, so Theorem 4.6 and the Arzela-Ascoli theorem imply that there is a subsequence (wkm ) (−1−α) which converges uniformly to an H2+δ function w. Moreover Lw = 0 in Ω,

M w = 0 on ∂Ω,

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so Corollary 3.15 implies that w ≡ 0, which contradicts the two conditions |wkm |0 = 1 and (wkm ) converges uniformly to w. Since (uk ) is bounded, there is a uniformly convergent subsequence such that its derivatives also converge uniformly on Ω and its second derivatives converge uniformly on compact subsets of Ω. Hence the limit u is a solution of (4.26), and another application of Corollary 3.15 gives the uniqueness of the solution. 

4.4

H¨ older gradient estimates for the Dirichlet problem

In this section, we prove a H¨older estimate for the gradient of the solution to the Dirichlet problem in an H1+α domain. This estimate will be used for several different purposes. In the next section, we use it to obtain H¨ older gradient estimates for problems with discontinuous β; in Section 4.7, we use it to obtain H¨ older estimates on the second derivatives of solutions of suitable oblique derivative problems; in Chapter 11 to obtain both first and second derivative H¨older estimates for nonlinear oblique derivative problems. The model for the proof is that of Theorem 4.5. Our first step is an interior H1+α estimate, which we prove by combining ideas from the proofs of Proposition 2.13 and Theorem 4.5. Proposition 4.11. Let Ω be a non-empty, open subset of Rn , let [aij ] be a positive-definite matrix-valued function defined on Ω, and suppose that there are positive constants λ and µ such that (4.3) is satisfied in Ω. Suppose also that there is a continuous, increasing function θ defined on [0, 1/2] with θ(0) = 0 such that   |x − y| |aij (x) − aij (y)| ≤ θ d(x) for all x and y in Ω with |x − y| ≤ 12 d(x). Suppose finally that bi and c are defined on Ω and that there is a positive constant A such that (1)

(2)

|bi |0 + |c|0 ≤ λA.

Then, for any α ∈ (0, 1), there is a constant C, determined only by A, n, α, µ, and θ such that if u ∈ C 2 (Ω) ∩ L∞ (Ω) satisfies aij Dij uib u + cu = f (2)

(0

in Ω for some function f ∈ H0 , then u ∈ H1+α and (0)

(2)

|u|1+α ≤ C(|u|0 + |f /λ|0 ).

(4.28)

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Proof. Suppose first that bi and c are zero. With δ1 ∈ (0, 1/2) to be chosen, we fix x0 ∈ Ω and set R = δ1 d(x0 ). Then set u0 = u and write v0 for the solution of aij (x0 )Dij v0 = 0 in B(x0 , R),

v0 = u0 on ∂B(x0 , R)

given by Corollary 2.11. From the second derivative bound Theorem 1.34, there are a linear polynomial L0 and a constant C0 , determined only by n and µ such that sup B(x0 ,τ R)

|v0 − L0 | ≤ C0 τ 2 sup |v0 | B(x0 ,R)

for all τ ∈ (0, 1). The argument in the proof of Theorem 4.5 then gives positive constants C1 and ζ and, for any δ ∈ (0, 1), a constant C(δ) such that sup B(x0 ,τ R)

|u0 − L0 | ≤ [C0 τ 2 + C(δ)θ(δ1 ) + C1 δ ζ ] sup |u0 | + C1 F, B(x0 ,R)

(2)

where F = |f /λ|0 . With σ = (1 + α)/2, we can find τ , δ and δ1 so that C0 τ 2 + C(δ)θ(δ1 ) + C1 δ ζ ≤ τ 1+σ , and hence sup B(x0 ,τ R)

|u0 − L0 | ≤ τ 1+σ sup |u0 | + C1 F. B(x0 ,R)

By working inductively, we obtain, for each r ∈ (0, d(x0 )/2), a linear polynomial P1 (x0 , r) such that sup |u − P1 (x0 , r)| ≤ C

B(x0 ,r)

r1+α (|u|0 + F ), R1+α

and then we obtain (4.28) by invoking Lemma 2.8 and the definition of (0) [u]1+α . Now suppose that bi and c are not zero. By arguing as in Proposition (0) 2.13, we may assume that |u|1 is finite. The preceding argument with i f − b Di u − cu implies that (0)

(0)

(2)

|u|1+α ≤ C(|u|0 + |u|1 + |f /λ|0 ), and the proof is completed by using the interpolation inequality Lemma 2.3.  We now provide some estimates in a set that is essentially a half-ball. To be specific, we say that an open subset G of Rn is a generalized half-ball if there is an open subset G0 of Rn−1 such that every point x = (x0 , xn ) in G satisfies the conditions x0 ∈ G0 and xn > 0. We also write G0 = G0 × {0}. Our first estimate is a straightforward modification of Lemma 1.31.

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Lemma 4.12. Let G be a generalized half-ball and suppose aij satisfies (4.3) in Ω. Suppose also that there are constants B0 ≥ 0 and α ∈ (0, 1) such that b satisfies |b(x)| ≤ B0 (xn )α−1 d∗ (x)−α (4.29) ∗ 0 2 for all x ∈ G, where d denotes the distance to ∂G \ G . Let u ∈ C (G) ∩ C(G) with u = 0 on G0 , and suppose that there is a nonnegative constant F such that (4.13) holds in G, where L = aij Dij + bi Di . Then there is a constant C, determined only by n, µ, α, and B0 , such that xn |u(x)| ≤ C(sup |u| + F ) ∗ (4.30) d (x) G for all x ∈ G. Proof. If xn ≤ 14 d∗ (x), we set r = 31 d∗ (x) and x0 = (x0 , 0). We also abbreviate B + (x0 , r) to Ω and we observe that d∗ (y) ≥ r for all r ∈ Ω. Next, we define h by   Z yn B0 + 1 α h(y) = exp − s ds αrα 0 and abbreviate   B0 + 1 n α E = exp − (x ) . αrα It follows that Lh = E(ann (B0 + 1)r−α + bn ) ≤ Eλ[−(B0 + 1)r−α + B0 rα ] where

= −Eλr−α ≤ −E0 r−α λ,

E0 = exp(−(B0 + 1)/α). If we now set U = supG |u| and define w by   |y − x0 |2 2nµh(y) h(y) w(y) = U + +F , r2 E0 r E0 r it follows that L(w ± u) ≤ 0 in Ω and w ± u ≥ 0 on ∂Ω, and the maximum principle implies that w ± u ≥ 0 in Ω. Therefore   n 2 (x ) 2nµh(x) h(x) |u(x)| ≤ w(x) = U + +F . 2 r r E0 r Since h(x) ∈ (0, xn ) and xn /r ∈ (0, 1), it follows that   2nµ F xn |u(x)| ≤ [ 1 + U+ ] , E0 E0 r which is just (4.30) with 8nµ C =4+ . E0 Since C ≥ 4, (4.30) is also true if xn ≥ 14 d∗ (x). 

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From this estimate, we obtain a regularity result with estimate for solutions of the Dirichlet problem. Lemma 4.13. Suppose, in addition to the hypotheses of Lemma 4.12 that there is a continuous increasing function θ on [0, 1/2] with θ(0) = 0 such that aij satisfies (4.11) holds for all x and y in G with |x − y| < 21 d∗ (x). (0) Then u ∈ H1+α and (0)

|u|1+α ≤ C(θ, µ, B0 )(F + |u|0 ).

(4.31)

Proof. We first suppose that bi = 0. Here, we argue as in Proposition 4.11 with Corollary 2.17 in place of Corollary 2.11 and Lemma 1.35 in place of Theorem 1.34. If bi is not necessarily zero, then Lemma 4.12 gives |u(x)| ≤ C(F + sup |u|)

xn d∗ (x)

and then Proposition 4.11 (applied in B(x, min{d∗ (x), d(x)}/2) implies that |Du(x)| ≤ C(F + sup |u|)

1 . d∗ (x)

Then, if we set f ∗ = f − bi Di u, we conclude that |f ∗ | ≤ C(F + sup |u|)(xn )α−1 (d∗ (x))−1−α , so we can apply the case bi = 0 with f ∗ in place of f to infer (4.31) in this case as well.  From Lemmata 4.13 and 4.9, we obtain a local H1+α estimate for solutions of the Dirichlet problem in an H1+α domain. Proposition 4.14. Let ω be an H1+α (Bn−1 (x00 , R)) function for some positive numbers α < 1 and R and some x0 ∈ Rn with ω(x00 ) = xn0 , and write Ω[R] and Σ[R] for the sets of all x ∈ B(x0 , R) with xn > ω(x0 ) or xn = ω(x0 ), respectively. Suppose aij satisfies (4.3) and there is a continuous increasing function θ on [0, 1/2] with θ(0) = 0 such that (4.11) holds for all x and y in Ω[R] with |x − y| < 12 d∗ (x). Suppose also that b satisfies the condition |b| ≤ B0 dα−1 (d∗ )−α

(4.32)

in Ω[R]. If aij Dij u + bi Di u = f in Ω[R],

u = ϕ on Σ[R]

(4.33)

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for some function f satisfying |f | ≤ λF0 dα−1 (d∗ )−1−α and some ϕ ∈ H1+α (Σ[R]), then there is a constant C, determined by n, µ, α, B0 , and |ω|1+α such that (0)

(0)

|u|1+α ≤ C(|u|0 + F0 + |ϕ|1+α ).

(4.34)

Proof. As a first step, we use Exercise 4.3 to conclude that there is a (0) function ϕ∗ ∈ C 2 (Ω[R]) such that |D2 ϕ∗ | ≤ C|ϕ|1+α dα−1 (d∗ )−α−1 in Ω[R] and ϕ∗ = ϕ on Σ[R]. Hence, there is no loss of generality in assuming that ϕ = 0. We then note that the function V : Ω[R] → Rn , defined by V (x) = (x0 , ρ(x)), transforms Ω[R] into a generalized half-ball, and that the hypotheses of Lemma 4.13 are all satisfied in the new variables y = V (x). Changing back to the original variables gives (4.34).  For our application in Section 4.7, we use the following result, which is proved just like Lemma 2.14. Corollary 4.15. Suppose that all the hypothesis of Proposition 4.14 are satisfied except that |f | ≤ λF0 dα−1 (d∗ )s−1−α for some s > 0. Then there is a constant C, determined by n, µ, α, B0 , s, and |ω|1+α such that (s)

(s)

(s)

|u|1+α ≤ C(|u|0 + F0 + |ϕ|1+α ).

4.5

(4.35)

C 1,α estimates with discontinuous β in two dimensions

Our next step is to study solutions of the oblique derivative problem in a two dimensional domain when the direction of the vector β has a jump discontinuity at isolated points on the boundary. As mentioned in the introduction to Chapter 1, the most straightforward example of this situation is the Neumann problem for Laplace’s equation in a square, but there are several important elements in the generalization of this situation that we wish to consider.

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137

We first consider the proper interpretation of the condition β ·Du = g at a point of discontinuity for β. Although we can analyze this discontinuity in great generality, we only study here a very simple case to illustrate the issues. We take Ω to be the set of all x ∈ B(0, 1) with x1 > 0 and x2 > 0, and we work in R2 . We then introduce the sets Σ1 = {x ∈ R2 : |x| < 1, x1 > 0, x2 = 0},

Σ1 = {x ∈ R2 : |x| < 1, x2 > 0, x1 = 0},

and we also choose β so that β = (0, 1) on Σ1 and β = (1, 0) on Σ2 . In this case, neither (0, 1) nor (1, 0) is oblique at (0, 0). Since our theory requires a boundary condition to be prescribed at (0, 0) and we are interested in oblique derivative problems, we note that the condition β · Du = g at (0, 0) will be oblique if and only if the two components of β(0, 0) are both positive. Moreover, if u ∈ C 1 (Ω) and if g is uniformly continuous on the two sets Σ1 and Σ2 , then the gradient of u is prescribed at (0, 0). Specifically, if we write g1 and g2 for the limits of g(x) as x approaches (0, 0) along Σ1 and Σ2 , respectively, then Du(0, 0) = (g2 , g1 ). It follows that the condition β · Du = g at (0, 0) can only be satisfied if g(0, 0) is appropriately determined from the vector β(0, 0) and the numbers g1 , g2 . In what follows, we shall use this observation explicitly when writing the boundary condition as β·Du = g (in which case, β(0, 0) will be given and g(0, 0) is determined from β(0, 0), g1 , and g2 ) or we shall use it implicitly when writing the boundary condition as β · Du = g on Σ1 ∪ Σ2 . These two cases are also distinguished by the explicit regularity assumption that Du is continuous at (0, 0) (or a condition that implies this assumption) when we do not explicitly prescribe a boundary condition at (0, 0). For problems with variable coefficients, we use techniques that should be familiar at this stage, so the difficult part of this section is a study of the dependence of the regularity estimates on the geometry of the domain and on the interaction between the domain and the jump in the direction of the prescribed directional derivative. In fact, the analysis of this problem is relatively simple (see the proof of Lemma 4.16), and the real difficulty is that we must prove a suitable series of existence results based on a simple existence result (Lemma 4.19). As a first step, we assume that the geometry is particularly simple, and we study the regularity estimate only for constant coefficient problems. We also emphasize that we must assume that a boundary condition is prescribed

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at all points of the boundary, and we examine the smoothness of the solution only near a corner point of the domain. Moreover, we shall find it convenient to introduce polar coordinates (r, θ) satisfying x1 = r cos θ,

x2 = r sin θ

again. Lemma 4.16. Let R > 0, θ1 , and θ2 be constants with θ1 < θ2 < θ1 + 2π, and define the sets Ω = {x ∈ R2 : |x| < R, θ1 < θ < θ2 }, 2

(4.36a)

Σk = {x ∈ R : 0 < r < R, θ = θk }

(4.36b)

σ = {x ∈ R2 : r = R, θ1 ≤ θ ≤ θ2 }.

(4.36c)

for k = 1, 2,

Let β1 and β2 be unit vectors such that β1 is oblique on Σ1 and β2 is oblique on Σ2 . Suppose also that β1 and β2 are not parallel. Let [Aij ] be a positive definite matrix. Suppose more specifically that there are positive constants λ and µ such that (4.3) is satisfied and βk · γk ≥ 1/µ

(4.37)

for k = 1, 2, where γk is the unit inner normal to Σk . Then there are constants α ∈ (0, 1) and C, determined only by θ2 − θ1 , µ, and B0 = min{|β1 − β2 |, |β1 + β2 |},

such that, if u ∈ C 1 (Ω) ∩ C 2 (Ω) satisfies Aij Dij u = 0

in Ω,

(4.38a)

β1 · Du = 0

on Σ1 ,

(4.38b)

on Σ2 ,

(4.38c)

β2 · Du = 0 then |Du(x)| ≤ C sup |Du| Ω



|x| R

α

.

(4.39)

Proof. First, set B1 = β11 β22 − β21 β12 , and note that B1 is just the determinant of the matrix with columns β1 and β2 . If we write θ¯ for the angle ¯ When θ¯ ≤ π/2, we (with 0 < θ¯ < π) between β 1 and β 2 , then |B1 | = sin θ. ¯ note that cos θ ≥ 0, and ¯ |β 1 − β 2 |2 = 2(1 − cos θ)

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by the law of cosines. Then sin2 θ¯ 1 − cos θ¯ = ≤ sin2 θ¯ 1 + cos θ¯ √ and hence |B1 | ≥ B0 / 2 in this case. On the other hand, if θ¯ > π/2, ¯ so θ0 ∈ (0, π/2) and then the angle between β 1 and −β 2 is θ0 = π − θ, ¯ Our previous case shows that 1 − cos θ0 ≤ sin2 θ0 and |β 1 + sin θ0 = sin θ. √ β 2 |2 = 2(1 − cos θ0 ), so we infer that |B1 | ≥ B0 / 2 in this case as well. To continue, we note that the statement of the theorem is invariant under a rotation of axes, so we now assume that θ1 = 0. We now set v2 = β2 · Du, and note that Aij Dij v2 = 0 in Ω, v2 = 0 on Σ2 , |v2 | ≤ sup |Du| on σ.

After this rotation, the vector β1 satisfies β12 > 1/µ. Moreover, a simple reflection argument (like the one in Lemma 2.19) shows that u ∈ C 2 (Ω∩Σ1 ). It follows that Dv2 is continuous up to Σ1 and then, with T1 a constant to be chosen, we write β¯2 for the vector (T1 , 1). Then β¯2 is oblique on Σ1 , and β¯2 · Dv2 = T1 [β 1 D11 u + β 2 D12 u] + β 1 D12 u + β 2 D22 u 2

2

2

2

on Σ1 . Also, we can differentiate the boundary condition β1 · Du = 0 in the direction (0, 1), which is tangential to Σ1 , to infer that β11 D11 u + β12 D12 u = 0 which implies that β11 D11 u β12 on Σ1 . Finally, the differential equation implies that
1 D22 u = − 22 A11 D11 u + 2A12 D12 u A  1 2A12 β11 11 = 22 − A D11 u A β12 on Σ1 . It follows that   D11 u 2A12 2 2 A11 2 2 1 1 − β¯ · Dv2 = T B + β β − β β β β 1 1 2 1 β12 A22 2 1 A22 2 1 on Σ1 , so β¯ · Dv2 = 0 on Σ1 if we choose D12 u = −

12

− 2A22 β22 β12 + β21 β11 + T1 = A B1

A11 2 2 A22 β2 β1

.

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Since |T1 | ≤ C(µ, B0 ), Proposition 3.16 gives positive constants α and K, determined only by B0 , θ2 , and µ, along with a nonnegative function wα satisfying Aij Dij wα ≤ 0 in Ω, wα ≤ KRα in Ω, β¯ · Dwα ≤ 0 on Σ1 ,

wα ≥ Rα on σ. The maximum principle implies that ±v2 ≤ sup |Du|wα /Rα in Ω and hence  α |x| |β2 · Du(x)| ≤ K sup |Du| R in Ω. A similar argument (but also using a reflection across the x1 axis) shows that  α |x| |β1 · Du(x)| ≤ K sup |Du| R in Ω, with the same constants K and α, and the combination of these two estimates, along with the positivity of B0 , implies (4.39).  Although this lemma is stated under fairly general geometric conditions, we shall see eventually that the assumed regularity for u is only satisfied under stronger assumptions on β1 , β2 , Σ1 , and Σ2 . In particular, we shall use a stronger form of the estimate in this lemma, which is based on the following H¨ older estimate. Lemma 4.17. Let R > 0, θ0 , θ1 , and θ2 be constants with θ1 < θ0 < θ2 < θ1 + 2π, and define the sets Ω, Σk , and σ by (4.36). Let θ10 and θ20 satisfy (4.40a) θ2 − π < θ20 ≤ θ0 ≤ θ10 ≤ θ1 + π, (4.40b) θ10 < θ0 + π/2, θ20 > θ0 − π/2, 0 0 and set βk = (cos θk , sin θk ) for k = 1, 2. Also, define β on Σ = Σ1 ∪Σ2 ∪{0} by   if x ∈ Σ1 ,  β1 β(x) = β2 (4.41) if x ∈ Σ2 ,   (cos θ , sin θ ) if x = 0. 0

0

ij

Let [a ] be a positive definite matrix-valued function defined on Ω. Suppose more specifically that there are positive constants λ and µ such that (4.3) is satisfied and θ1 + π − θ10 ≥ 1/µ, θ20 + π − θ2 ≥ 1/µ, (4.42a) θ0 + π/2 − θ10 ≥ 1/µ,

θ20 + π/2 − θ0 ≥ 1/µ.

(4.42b)

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Then there are constants α ∈ (0, 1) and C, determined only by θ2 − θ1 and µ such that, if u ∈ C(Ω) ∩ C 2 (Ω) satisfies aij Dij u = 0 in Ω, β · Du = 0 on Σ, (4.43) then  α |x| |u(x) − u(0)| ≤ C sup |u| . (4.44) R Ω Proof. Note that β is a unit vector field which is oblique on Σ. Moreover, after a rotation of axes, we may assume that θ0 = π/2, so our hypotheses imply that β 1 (x)x1 ≤ 0 and β 2 > 0 on Σ. Furthermore, there is a positive constant m such that x2 ≤ m|x1 | on Σ. We now define G by   1 2 2 −2 1 2 (y ) , G(y, ρ) = ρ (y ) + 16m2 we set x1 = (0, 8mρ), and we define E(x1 , ρ) to be the set of all x such that G(x − x1 , ρ) < 1. For all sufficiently small ρ, we see that η, defined by η(x) = 1 − G(x − x1 , ρ) satisfies 7β 2 β · Dη ≥ 8mρ on Σ ∩ E(x1 , ρ). By using this function in place of the function η from Lemma 1.17, we infer the H¨older estimate by following the proofs of Lemmata 1.18 and 1.19 and Theorems 1.20 and 1.26.  We then obtain the following pointwise version of Lemma 4.16. Lemma 4.18. Under the hypotheses of Lemma 4.17, there is a constant C, determined only by µ such that |u(x) − u(0)| ≤ CR−1−α |u|0 |x|1+α (4.45) for |x| ≤ R/2. From Lemma 4.16 and integration, we obtain C |u(x) − u(0)| ≤ α |Du|0,Ω[R/2] |x|1+α . R Then we apply the local maximum principle Theorem 1.30 to β1 · Du and β2 · Du to conclude that Z Proof.

|Du|0,Ω[R/2] ≤ CR−2

Ω[3R/4]

|Du| dx.

From Lemma 4.17 and Z the proof of Lemma 2.20, we conclude that |Du| dx ≤ CR|u|0 , Ω[3R/4]

and combining all these estimates yields (4.45).



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From this estimate, we obtain an estimate for solutions of a particular mixed boundary value problem, which will eventually lead to an existence result for this problem. Our next step is an existence theorem for a very special mixed boundary value problem. Lemma 4.19. Let R > 0 and set Q(R) = {x ∈ R2 : |x| < R, x1 > 0, x2 > 0},

(4.46a)

2

1

2

(4.46b)

2

1

2

(4.46c)

2

1

2

(4.46d)

S1 (R) = {x ∈ R : |x| < R, x > 0, x = 0},

S2 (R) = {x ∈ R : |x| < R, x = 0, x > 0}, s(R) = {x ∈ R : |x| = R, x ≥ 0, x ≥ 0}.

Let α ∈ (0, 1). Then, for any f ∈ Hα (Q(R)), any g1 ∈ Hα (S1 (R)), any g2 ∈ Hα (S2 (R)), and any ϕ ∈ C(s(R)), there is a unique solution w ∈ C 2 (Q(R)) ∩ C 1 (Q(R) \ s(R)) ∩ C(Q(R)) of ∆w = f in Q(R),

(4.47a)

D2 w = g1 on S1 (R),

(4.47b)

D1 w = g2 on S2 (R),

(4.47c)

w = ϕ on s(R).

(4.47d)

Moreover, there is a constant C, determined only by α, such that (0)

(2)

(1) |u|1+α ≤ C(|u|0 + |f |0 + |g1 |(1) α + |g2 |α ),

(4.48)

where the weight in these norms is the distance to s. Proof. Suppose first that g2 ≡ 0 and g1 ∈ H1+α (S1 (R)) with D1 g1 (0) = 0. Then extend f , g1 , and ϕ as even functions of x1 to all of B + (0, R), B 0 (0, R), and B ∗ (0, R), respectively, where B ∗ (0, R) = {x ∈ R2 : |x| = R, x2 ≥ 0}. It follows from the proof of Lemma 2.19 that the problem ∆w ˜ = f in B + (0, R), 0

D2 w ˜ = g1 on B (0, R), ∗

w ˜ = ϕ on B (0, R)

(4.49a) (4.49b) (4.49c)

has a unique solution w ˜ ∈ C 2 (B + (0, R)) ∩ C(B + (0, R)). If we only have g ∈ Hα (S1 (R)), then we first note that there is a sequence (g1,m ) of H1+α (S1 (R)) functions which converges uniformly to g1 with D1 g1,m (0) = 0 and [g1,m ]α ≤ C(α)[g1 ]α . If we write w ˜m for the solution of (4.49) with g1,m in place of g1 , we see that (w˜m ) is uniformly Cauchy by the maximum

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principle, Theorem 1.11, so it converges uniformly to a limit function w. ˜ It then follows from Proposition 2.12 that the sequence (D2 w ˜m ) is uniformly Cauchy on compact subsets of B + so ∆w ˜ = f in Q, and a similar argument using Theorem 4.5 shows that D2 w ˜ = g1 on B 0 . In other words, (4.49) has a solution in this case as well. Now we observe that w, ˆ defined by 1 w(x ˆ 1 , x2 ) = w(−x ˜ , x2 ),

is a solution of the same boundary value problem as w, ˜ so the maximum principle implies that w ˆ = w. ˜ In other words, w ˜ is an even function of x1 , so we have ∆w ˜ = f in Q(R), D2 w ˜ = g1 on S1 (R), D1 w ˜ = 0 on S2 (R), w ˜ = ϕ on s(R). To handle the case of nonzero g2 , we write w1 for the unique solution of ∆w1 = 0 in {|x| < R, x1 > 0},

D1 w1 = g2 on {|x| < R, x1 = 0}, w1 = 0 on {|x| = R, x1 > 0}

given by the previous argument and w2 for the solution of ∆w2 = f in Q(R), D2 w2 = g1 − D2 w1 on S1 (R), D1 w2 = 0 on S2 (R), w2 = ϕ on s(R) that we have just derived. It follows that w = w1 + w2 is the desired solution. Another application of Theorem 4.5 gives (4.48).  For our further analysis, we modify the notation slightly from (4.46). For R > 0 and κ ∈ (0, 1), we set x1 (κ, R) = (0, −κR), and we define the sets Q(κ, R) = {x ∈ R2 : |x − x1 (κ, R)| < R, π/4 < θ < 3π/4},

S1 (κ, R) = {x ∈ R2 : |x − x1 (κ, R)| < R, θ = π/4},

S2 (κ, R) = {x ∈ R2 : |x − x1 (κ, R)| < R, θ = 3π/4},

s(κ, R) = {x ∈ R2 : |x − x1 (κ, R)| = R, π/4 ≤ θ ≤ 3π/4},

Q0 (κ, R) = S1 (κ, R) ∪ S2 (κ, R).

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For α ∈ (0, 1), we write Hα (Q0 ) for the set of all functions g defined on Q0 such that the restriction of g to S1 (κ, R) is in Hα (S1 (κ, R)) and the restriction of g to S2 (κ, R) is in Hα (S2 (κ, R)). Since κ and R will usually be fixed in what follows, we generally omit them from the notation. In what follows, γ denotes the unit inner normal to Q0 . Lemma 4.20. Let α ∈ (0, 1), R > 0, and κ ∈ (0, 1). Then the problem ∆u = f in Q,

(4.50a) 0

γ · Du = g on Q , u = ϕ on s

2

(4.50b) (4.50c)

1

has a unique solution u ∈ C (Q) ∩ C (Q \ s) ∩ C(Q) for any f ∈ Hα (Q), any g ∈ Hα (Q0 ), and any ϕ ∈ C(s). Proof. When ϕ = 0, we modify the Perron process from Section 2.5 only slightly. First, we say that ∆u = f in Q,

γ · Du = g on Q0

(4.51)

is locally solvable if, for every x0 ∈ Q \ s, there is an open set N with x0 ∈ N and N ∩ s = ∅ such that, for any ψ ∈ C(Q ∩ ∂N ), there is a solution ψN C 2 (Q ∩ N ) ∩ C 1 (Q ∩ N ) ∩ C(Q ∩ N ) to the problem ∆ψN = f in Q ∩ N, 0

γ · DψN = g on Q ∩ N, ψN = ψ on Q ∩ ∂N.

(4.52a) (4.52b) (4.52c)

Lemma 4.19 gives local solvability of this problem in a neighborhood of 0, Lemma 2.19 gives local solvability in a neighborhood of any point in Q0 , and Corollary 2.16 gives local solvability in a neighborhood of any point in Q. For future reference, we observe that the set N corresponding to any x0 6= 0 can be taken so that 0 ∈ / N. Next, we say that v is a Perron supersolution of (4.51) if, for every x0 ∈ Q \ s and every continuous function ψ with v ≥ ψ on Q ∩ ∂N , we have v ≥ ψN in N . If we define w by w(x) = (R2 − |x − x1 |2 )1/2 ,

then it’s easy to check that ∆w ≤ −R2 (R2 − |x − x1 |2 )−3/2 in Q and that w = 0 on σ. In addition, a direct calculation gives 1 γ · Dw = − √ (R2 − |x − x1 |2 )−1/2 κR 2

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on Q0 . Hence, ∆w ≤ −C(κ)/R in Q,

γ · Dw ≤ −C(κ) on Q0 , w ≥ 0 on s.

Therefore −Kw is a subsolution and Kw is a supersolution if we choose K sufficiently large in terms of |f |0 , |g|0 , and R. We now write S + for the set of Perron supersolutions v which are greater than −Kw and which satisfy lim v(x) = 0

x→y x∈Q

for all y ∈ s. (Note that S + is nonempty because Kw ∈ S + .) We then define u by u(x) = inf v(x) v∈S +

and note that −Kw ≤ u ≤ Kw, so lim u(x) = 0

x→y x∈Q

for all y ∈ s. Next, for each x0 ∈ Q \ s, we define the lift vN of a supersolution v by vN (x) = sup{ψN (x) : ψ is continuous and v ≥ g ≥ −Kw on Q ∩ ∂N } if x ∈ Q ∩ N , and vN (x) = v(x) if x ∈ Q \ N . It follows from Proposition 2.13 and the argument in Section 2.5 that ∆vN = f in Q ∩ N and Lemma 2.19 implies that γ · DvN = g on Q0 ∩ N . Moreover, if x0 = 0, we use (4.48) from Lemma 4.19 to infer that vN has continuous second derivatives near 0. Just as in Section 2.5, we conclude that u solves (4.50). For nonzero ϕ, we take (ϕm ) to be a sequence of H2+α (Q) functions converging uniformly on s to ϕ. Then our Perron process argument gives a solution u ¯m of ∆¯ um = f − ∆ϕm in Q, ¯ m = g − γ · Dϕm on Q0 , γ · Du u ¯m = 0 on s,

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and then um = u ¯m + ϕm solves ∆um = f in Q, γ · Dum = g on Q0 , um = ϕm on s.

By using the maximum principle along with Proposition 2.12, Theorem 4.5, and (4.48), we conclude that (um ) converges to a solution of (4.50).  Our next step is to obtain estimates involving less regular functions f . To simplify notation, we write g = 0 on Q0 ∩ s as shorthand for the statement lim g(x) = 0

x→x0 x∈Q0

for all x0 ∈ Q0 ∩ s. Lemma 4.21. Let δ ∈ (0, 1). For any α ∈ (0, 1), there is a constant κ1 ∈ (0, 1), determined only by α such that if κ ∈ (κ1 , 1), then any solution u of (4.50) with ϕ = 0 on s and g = 0 on Q0 ∩ s satisfies the estimate (−1−α)

|u|2+δ

(1−α)

≤ C(α, δ, κ)[|f |δ

+ |g|(−α) ]. α (1−α)

Moreover, (4.50) has a unique solution for any f ∈ Hδ with g = 0 on Q0 ∩ s, and ϕ = 0. Proof.

(4.53)

, any g ∈ Hα (Q0 )

We first estimate |u|0 . To this end, we define the function w by

w(x) = R1−α (R2 − |x − x1 |2 )α + R1+α − (dS1 )1+α + R1+α − (dS2 )1+α ,

where dSk denotes the distance to Sk . A simple calculation shows that there is a constant C(α) such that ∆w ≤ −C(α, κ)dα−1 in Q,

γ · Dw ≤ −C(α, κ)Rα on Q0 , w ≥ 0 on s,

and then the maximum principle implies that 1 (1−α) (−α) [|f |0 + |g|0 ]w |u| ≤ C(α, κ) (1−α)

≤ CR1+α [|f |0

(−α)

+ |g|0

].

We now observe that the function G in Section 3.6 has the special form G(α; z, p) = α2 z when µ = 1, which will be the case for the Laplace operator. Moreover the constant µ0 in that section is zero for the normal derivative operator. It follows that the constant α1 (π/4, n, 1, 0) in Proposition

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3.16 is equal to 2, and hence, for κ1 sufficiently close to 1 (and determined just by α), Proposition 3.16 provides a function v1 such that ∆v1 ≤ −|x − y1 |α−1 in Q ∩ N1 ,

γ · Dv1 ≤ −|x − y1 |α on S1 ∩ N1 , v1 ≥ |x − y1 |1+α in Q ∩ N1 ,

where y1 is the point in σ ∩ S1 and N1 is the set of all x such that |x − y1 | < C1 R for some sufficiently small positive constant C1 (determined only by κ) such that N1 ∩ S2 = ∅. Further, C1 can be chosen so that dS2 ≥ |x − y1 | in N1 . It follows that satisfies

w = 2v1 − (dS1 )1+α − (ds )1+α ∆w ≤ −dα−1 in Q ∩ N1 ,

γ · Dw ≤ −|x − y1 |α on S1 ∩ N1 , w ≥ 0 on σ ∩ N1 ,

w ≥ (C1 R)1+α in Q ∩ ∂N1 ,

so an application of the maximum principle shows that (1−α)

|u(x)| ≤ C(d∗∗ )1+α [|f |0

+ |g|(−α) ], α

where d∗∗ denotes distance to σ ∩ (S1 ∪ S2 ), if |x − y1 | ≤ C1 R. A similar argument near S2 ∩ σ shows that (1−α)

|u(x)| ≤ C(d∗∗ )1+α [|f |0

] + |g|(−α) α

if |x − y2 | ≤ C1 R. Next, we replace Lemma 4.2 by Lemma 4.18 in the proof of Theorem 4.5 to conclude that (1−α)

|u(x) − L · x − u(0)| ≤ C|x|1+α [|f |0

where L is the (unique) vector such that

] + |g|(−α) α

γk · L = lim g(x) x→0 x∈Sk

for k = 1, 2, and γk denotes the restriction of γ to Sk . Furthermore, Proposition 4.14 gives a corresponding estimate near points of s, so the argument in Theorem 4.5 then allows us to conclude that (−1−α)

|u|1+α

(1−α)

≤ C[|f |0

+ |g|(−α) ], α

and an application of Lemma 2.14 completes the proof of (4.53). Since (4.50) has a solution for any f ∈ Hδ , the existence result follows by approximation and (4.53). 

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148

From these lemmata, we now derive an existence result and estimate for problems in the standard domain but with arbitrary constant coefficients in the differential equation. Proposition 4.22. Let θ1 = θ0 = π/4 and let δ ∈ (0, 1). Then for any µ ≥ 1, there is a constant α0 ∈ (0, 1), determined only by µ, such that, if α ∈ (0, α0 ) and [Aij ] is a constant matrix in Mµ , then the problem Aij Dij u = f in Q,

(4.54a)

0

γ · Du = 0 on Q ,

(4.54b)

u = 0 on s

(4.54c)

(1−α)

for any f ∈ Hδ . Moreover there is a constant C, determined only by α, δ, κ, and µ, such that (−1−α)

|u|2+δ

(1−α)

≤ C|f |δ

.

(4.55)

As a first step, we observe that
α w(x) = R1−α R2 − |x − x1 |2 + R1+α − (dS1 )1+α + R1+α − (dS2 )1+α

Proof.

satisfies the estimate

Aij Dij w ≤ −Cdα−1 for some positive constant C, determined only by κ, α, and µ, so the maximum principle implies (just as in the proof of Lemma 4.21) that (1−α)

|u|0 ≤ C(κ, α, µ)R1+α |f |0

.

(4.56)

Next, we observe that the constant α1 from Proposition 3.16 must be greater than 1 if κ > 0, and hence the argument in Lemma 4.21 also implies that (1−α)

|u| ≤ C(κ, α, µ)(d∗∗ )1+α |f |0

.

(4.57)

Now, for t ∈ R, we define

ij ij Aij t = tA + (1 − t)δ ,

and we write T for the set of all t ≥ 0 such that the problem Aij t Dij u = f in Q, 0

γ · Du = 0 on Q , u = 0 on s

(1−α)

(4.58a) (4.58b) (4.58c)

has a solution for all f ∈ Hδ . It follows from Lemma 4.21 that 0 ∈ T and we want to show that 1 ∈ T . As a first step towards this end, we show

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that if t ∈ T , then (4.55) holds. If t ∈ T , then, for any ρ ∈ (0, R), there is a unique solution ψ of Aij t Dij ψ = −f in Q(κ, ρ), γ · Dψ = 0 on Q0 (κ, ρ),

ψ = u on s(κ, ρ), and hence v = u + ψ solves Aij t Dij v = 0 in Q(κ, ρ), 0

(4.59a)

γ · Dv = 0 on Q (κ, ρ),

(4.59b)

γ · Du = 0 on Q0 ,

(4.61)

v = u on s(κ, ρ). (4.59c) Using this existence result, we argue as in Lemma 4.21 to conclude (4.55) provided α0 is chosen to be no greater than the constant from Lemma 4.18 in addition to our previous restriction that α0 ≤ α1 − 1. (1−α) New, we fix t ∈ T and f ∈ Hδ . With τ > 0 to be chosen, we write (−1−α) B for set of all u ∈ H2+δ satisfying γ · Du = 0 on Q0 , u = 0 on s, and we define the operator J on B by u = Jv if u is the unique solution of ij ij Aij (4.60) t Dij u = f + [At − At+τ ]Dij v in Q, u = 0 on s. (4.62) If C is the constant from (4.55), then we obtain kJv1 − Jv2 kB ≤ Cτ kv1 − v2 kB , so J will be a contraction if Cτ ≤ 1/2. It follows in this case that J has a fixed point u which solves Aij t+τ Dij u = f in Q, γ · Du = 0 on Q0 ,

u = 0 on s. Therefore t + τ ∈ T , and a finite repetition of this argument shows that 1 ∈ T , which then implies both the existence result and the estimate.  Corollary 4.23. Let θ1 = θ0 = π/4 and let δ ∈ (0, 1). Then for any µ ≥ 1, there is a constant α0 ∈ (0, 1), determined only by µ, such that, if α ∈ (0, α0 ) and [Aij ] is a constant matrix in Mµ , then the problem Aij Dij u = 0 in Q, (4.63a) γ · Du = g on Q0 , u = ϕ on s

(4.63b) (4.63c)

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¯ \ s) ∩ C(Q) ¯ for any g ∈ Hα (Q0 ) has a unique solution u ∈ C 2 (Q) ∩ C 1 (Q and any ϕ ∈ C(s). Proof. First, extend ϕ to all of B(x1 , R) as a continuous function. Then (−1−α) there is a sequence (ϕm ) of H2+δ (B(x1 , R)) functions converging uniformly to ϕ. In addition, there is an increasing sequence (ζm ) of C 2 (R2 ) functions with 0 ≤ ζm ≤ 1 in R2 and ζm ≡ 1 on B(x1 , (1 − 1/m)R). (−1−α) We infer from Lemma 4.21 that there is a unique solution vm ∈ H2+δ of ∆vm = −∆ϕm in Q,

γ · Dvm = (g − γ · Dϕm )ζm on Q0 , vm = ϕm on s,

and Proposition 4.22 gives a unique solution w of Aij Dij wm = −Aij Dij vm − Aij Dij ϕm in Q, γ · Dwm = 0 on Q0 , wm = 0 on s.

Then um = vm + wm + ϕm satisfies Aij Dij um = 0 in Q, γ · Dum = gm on Q0 , um = ϕm on s,

where gm = ζm g + (1 − ζm )γ · Dϕm . Since gm and ϕm are uniformly bounded, the sequence (um ) is bounded in C(Q), and the uniform continuity of (ϕm ) implies (via the arguments in Lemma 4.17 and Corollary 1.29) a uniform modulus of continuity for (um ). The Arzela-Ascoli theorem implies that (um ) converges uniformly to a function u, which is easily verified to be the desired solution of (4.63).  Our analysis of mixed problems for equations with arbitrary (constant) coefficients and the normal derivative condition in this standard domain allows us to study more general boundary conditions in domains with angles other than π/2 at the corner; however, our approach needs two further steps to get to the most general results. In the first step, we fix a constant θ1 ∈ (0, π/2) and we set θ2 = π − θ1 . Then there is a linear transformation T which preserves area and orientation such that T maps the region 0 < θ < π/2 onto the region θ1 < θ
0, and let θ1 < π/2 < θ2 be constants with θ2 < θ1 + π. With m > 0 a constant (to be further specified), we define the sets 1 (x2 + 8mR)2 < R2 , θ1 < θ < θ2 }, 16m2 1 Ek (m, R) = {x ∈ R2 : (x1 )2 + (x2 + 8mR)2 < R2 , θ = θk } 16m2 E(m, R) = {x ∈ R2 : (x1 )2 +

for k = 1, 2, 1 (x2 + 8mR)2 = R2 , θ1 ≤ θ ≤ θ2 }, 16m2 E 0 (m, R) = E1 (m, R) ∪ E2 (m, R), e(m, R) = {x ∈ R2 : (x1 )2 +

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and we define Hα (E 0 (m, R)) to be the set of all functions g defined on E 0 (m, R) such that the restriction of g to Ek (m, R) is in Hα (Ek (m, R)) for k = 1, 2. We now prove the existence of solutions for Aij Dij u = 0 in E(m, R), 0

β · Du = g on E (m, R), u = ϕ on e(m, R)

(4.64a) (4.64b) (4.64c)

under various assumptions on the constant m, and the functions g and ϕ. Furthermore β will be a vector field which is constant on E1 (m, R), constant on E2 (m, R) and oblique on E 0 (m, R), and [Aij ] will be a fixed positive definite matrix satisfying (4.3) for positive constants λ and µ. In this context, our first step is to prove the existence of a solution for all g which are sufficiently smooth on E1 (m, R) and on E2 (m, R), ϕ = 0 and β = β0 . Lemma 4.24. Let [Aij ] ∈ Mµ for some µ ≥ 1. Suppose also that θ1 and θ2 satisfy 1 π 1 ≤ θ1 ≤ − , µ 2 µ

θ 2 = π − θ1 .

(4.65)

Then there are constants m0 > 0 and α0 ∈ (0, 1), determined only by µ, such that the problem Aij Dij u = 0 in E(m, R), 0

β0 · Du = g on E (m, R), u = ϕ on e(m, R)

(4.66a) (4.66b) (4.66c)

has a unique solution u ∈ C 2 (E(m, R)) ∩ C 1 (E(m, R) \ e(m, R)) ∩ C(E(m, R)) for any g ∈ Hα (E 0 (m, R)) and any ϕ ∈ C(e(m, R)) provided m ≥ m0 and α ∈ (0, α0 ). Proof. This time we suppress m and R from the notation. When ϕ ≡ 0, we proceed as in Lemma 4.20 via the Perron process. By invoking Corollary 4.23 (and our change of variables) in place of Lemma 4.19, we infer that Aij Dij u = 0 in E,

β0 · Du = g on E 0

is locally solvable, that is, for every x0 ∈ E \ e, there is an open set N with x0 ∈ N and N ∩ s = ∅ such that, for any continuous function ψ defined on

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E ∩ ∂N , there is a unique C 2 (E ∩ N ) ∩ C 1 (E ∩ N ) ∩ C(E ∩ N ) solution ψN to the problem Aij Dij ψN = 0 in E ∩ N,

β0 · DψN = g on E 0 ∩ N, ψN = ψ on E ∩ ∂N.

Next, we note that w1 , defined by

w1 (x) = (R2 − (x1 )2 −

1 (x2 + 8mR)2 )1/2 , 16m2

satisfies the conditions Aij Dij w1 ≤ 0 in E,

β0 · Dw1 ≤ −C(m) on E 0 , w1 ≥ 0 on e

provided m is sufficiently large. By following the proof of Lemma 4.20 with Aij Dij in place of ∆, E in place of Q, E 0 in place of Q0 , e in place of s, and w1 in place of the function w in that proof, we obtain the existence result for ϕ = 0. For nonzero ϕ, we first extend ϕ as a continuous function to the set 1 e∗ = {x ∈ R2 : (x1 )2 + (x2 + 8mR)2 = R2 } 16m2 and we take (ϕk ) to be a sequence in H1+α (e∗ ) converging uniformly to ϕ on e∗ . If we also define 1 E ∗ = {x ∈ R2 : (x1 )2 + (x2 + 8mR)2 < R2 }, 16m2 it follows from Corollary 2.17 and Proposition 4.14 that there is a solution u ¯k ∈ H1+α (E ∗ ) to Aij Dij u¯k = 0 in E ∗ ,

uk = ϕk on e∗

and our Perron process argument gives a solution vk ∈ H1+α to Aij Dij vk = 0 in E,

β0 · Dvk = g − β0 · Du¯k on E 0 , vk = 0 on e.

Hence uk = u ¯k + vk solves Aij Dij uk = 0 in E, β0 · Duk = g on E 0 , vk = ϕk on e.

As before, (uk ) converges uniformly to the unique solution of (4.66).



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To continue, we introduce one further function space. For α and δ in (1−δ) (0, 1), we denote by Hα (E 0 (m, R)) the set of all functions g defined on (1−δ) 0 E (m, R) such that the restriction of g to Ek (m, R) is in Hα (Ek (m, R)) for k = 1, 2 with the weight being distance to e(m, R). We then have the (1−δ) following existence result and estimate when g ∈ Hα . Corollary 4.25. If Aij , θ1 , m and α satisfy the hypotheses of Lemma 4.24 and if δ ∈ (0, 1), then (4.66) has a unique solution in C 2 (E(m, R)) ∩ (1−δ) C 1 (E(m, R) \ e(m, R)) ∩ C(E(m, R)) for any g ∈ Hα (E 0 (m, R)) and (δ) any ϕ ∈ C(e(m, R)). Moreover, if ϕ ≡ 0, then u ∈ H1+α (E(m, R)) and (−δ)

|u|1+α ≤ C(m, α, µ)|g|(1−δ) . α

(4.67)

We first prove (4.67) by observing that wδ , defined by  δ 1 −δ 2 1 2 2 2 wδ (x) = R R − (x ) − (x + 8mR) , 16m2 satisfies the conditions ∆wδ ≤ 0 in E,

Proof.

β · Dwδ ≤ −C(m)β 2 dδ−1 on E 0 , e

wδ = 0 on e. Since wδ ≤ C(m)dδe , the maximum principle implies that (1−δ) δ |u| ≤ C|g|0 R . Using our local H¨ older gradient estimates then yields (4.67). When ϕ = 0, we approximate g by a sequence of Hα functions and note (−δ) that the corresponding solutions to (4.66) are bounded in H1+α , so they have a limit which is the unique solution of (4.66). When ϕ ∈ Hδ , then we (−δ) can obtain an H1+α extension of ϕ into E with Aij Dij ϕ = 0, and hence our previous case yields the existence result here. For general nonzero ϕ, we take a sequence of Hδ functions which converge uniformly to ϕ and note that the corresponding solutions of (4.66) converge uniformly to the desired solution.  With this existence result in hand, we can modify the proof of Proposition 4.22 to prove an existence result for a more general vector field β. Proposition 4.26. Let [Aij ] ∈ Mµ for some µ ≥ 1 and let θ1 and θ2 satisfy (4.65). Let θ10 and θ20 satisfy the conditions 1 1 π + ≤ θ10 ≤ π − , (4.68a) 2 µ µ 1 π 1 ≤ θ20 ≤ − (4.68b) µ 2 µ

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and define β on E 0 (m, R) by ( (cos θ10 , sin θ10 ) β(x) = (cos θ20 , sin θ20 )

155

if x ∈ E1 (m, R),

(4.69)

if x ∈ E2 (m, R).

Also, let δ ∈ (0, 1). Then there are constants m1 > 0 and α1 ∈ (0, 1), determined only by µ such that, if m ≥ m1 and α ∈ (0, α1 ), then (4.64) has a solution u ∈ C 2 (E(m, R)) ∩ C 1 (E(m, R) \ e(m, R)) ∩ C(E(m, R)) for any (1−δ) g ∈ Hα (E 0 (m, R)) and any ϕ ∈ C(e(m, R)). Proof.

For t ∈ (0, 1), we define βt by

βt (x) = (cos(tθ10 + (1 − t)θ2 ), sin(tθ10 + (1 − t)θ2 ))

if x ∈ E1 (m, R) and

βt (x) = (cos(tθ20 + (1 − t)θ1 ), sin(tθ20 + (1 − t)θ1 ))

if x ∈ E2 (m, R). We then define T to be the set of all t ∈ (0, 1) such that the problem Aij Dij u = 0 in E,

(4.70a)

0

βt · Du = g on E ,

(4.70b)

u = 0 on e

(−δ)

(4.70c) (1−δ)

has a solution u ∈ C 2 (E) ∩ H1+α (E) for any g ∈ Hα by noting that, if t ∈ T , then the problem

(E 0 ). We begin

Aij Dij u = 0 in E,

βt · Du = g on E 0 , u = ϕ on e

(1−δ)

(E 0 ) and has a solution u ∈ C 2 (E) ∩ C 1 (E \ e) ∩ C(E) for any g ∈ Hα any ϕ ∈ C(e). Corollary 4.25 implies that 0 ∈ T , and the proof of (4.67) shows that there is a constant C (independent of t) such that whenever t ∈ T , we have (−δ)

|u|1+α ≤ C|g|(1−δ) α

for any solution of (4.70). This estimate allows us to conclude that T is closed. Now suppose t ∈ T , let τ be a constant at our disposal, and let (1−δ) (−δ) g ∈ Hα . We then define B to be the set of all functions u ∈ H1+α such that Aij Dij u = 0 in E,

u = 0 on e,

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and we define a map J : B → B by defining Jv to be the solution u of the problem Aij Dij u = 0 in E, βt · Du = g − (βt+τ − βt ) · Dv on E 0 , u = 0 on e

which we know to exist because t ∈ T . Since

kJv1 − Jv2 kB ≤ C|βt+τ − βt |0 |D(v1 − v2 )|(1−δ) , α

it follows that J is a contraction mapping if τ is sufficiently small and hence T is also open. Therefore T = [0, 1], which means that (4.64) has a (1−δ) solution for all g ∈ Hα (E 0 ) whenever ϕ ≡ 0, and the case of nonzero ϕ is handled as before.  We can now deal with the most general assumptions of interest by using our preceding results. Theorem 4.27. Let [Aij ] ∈ Mµ for some µ ≥ 1 and let θ0 , θ1 , and θ2 satisfy θ1 ≤ θ0 −

1 , µ

θ0 +

1 ≤ θ2 < θ1 + π. µ

(4.71)

Let θ10 and θ20 satisfy the conditions (4.42), and define β on E 0 (m, R) by (4.69). Also, let δ ∈ (0, 1). Then there are constants m1 > 0 and α1 ∈ (0, 1), determined only by µ such that (4.64) has a solution u ∈ C 2 (E(m, R)) ∩ C 1 (E(m, R) \ e(m, R)) ∩ C(E(m, R)) for any (1−δ) g ∈ Hα (E 0 (m, R)) and any ϕ ∈ C(e(m, R)). Proof. Suppose first that θ2 < θ1 + π. We set θ¯ = 12 (θ1 + θ2 ) and note that Proposition 4.26 gives the result if θ¯ = π/2. Since the cases θ¯ > π/2 and θ¯ < π/2 are symmetric, we only prove the theorem when θ¯ > π/2. ¯ and we define βt for t ∈ [0, 1] by In this case, we set θ00 = 21 (θ2 + θ) βt (x) = (cos(tθ10 + (1 − t)θ00 ), sin(tθ10 + (1 − t)θ00 ))

if x ∈ E1 and βt = β on E2 . We then write T for the set of all t ∈ [0, 1] such (−δ) (1−δ) that (4.70) has a solution in C 2 (E) ∩ H1+α (E) for any g ∈ Hα (E 0 ). It follows from Proposition 4.26 that the problem Aij Dij u = 0 in E,

β0 · Du = g on E 0

is locally solvable at x0 = 0, so the proof of Lemma 4.24 implies that 0 ∈ T . The proof of Proposition 4.26 then shows that T is relatively open

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and relatively closed in [0, 1], so 1 ∈ T and hence (4.64) has a solution if ϕ = 0, and the case of nonzero ϕ is handled in exactly the same way as in that proposition. When θ2 = θ1 + π, we note that there are sequences (θ1k ) and (θ2k ) satisfying all of our hypotheses (with 1/µ replaced by 1/2µ) such that θ1k ≥ θ1 and θ2k ≤ θ2 , so we obtain a sequence of solutions to the problems Aij Dij uk = 0 in E(k), β · Duk = g on E 0 (k), uk = ϕ on e(k),

with E(k), E 0 (k), and e(k) defined in the same way as E, E 0 , and e, respectively, with θjk in place of θj for j = 1, 2. Our estimates are all uniform in k, so there is a subsequence (uk(j) ) which converges to a solution of (4.64).  With this existence result in hand, our usual argument gives local C 1,α estimates for the oblique derivative problem with variable coefficients. We start with regularity in a wedge (that is, a set with straight sides) because of its application in later sections. Proposition 4.28. Let ζ be a continuous increasing function defined on [0, 1/2] with ζ(0), and let R0 > 0, θ1 , and θ2 be constants with π θ1 < < θ2 ≤ θ1 + π, 2 and define the sets Ω, Σk , and σ by (4.36). Let [aij ] satisfy (4.3) for some constants λ and µ and   |x − y| |aij (x) − aij (y)| ≤ λζ d∗ for all x and y in Ω with |x − y| ≤ 12 d∗ (x), where d∗ denotes distance to σ. Let θk0 be an H1+α (Σk ) function for k = 1, 2 and some α ∈ (0, 1) and suppose that conditions (4.42) are satisfied, and define βk for k = 1, 2 by βk (x) = (cos θk0 , sin θk0 ).

(4.72) 2

Also, define β on Σ = Σ1 ∪ Σ2 by setting β = βk on Σk . Let u ∈ C (Ω) ∩ C 1 (Ω ∪ Σ ∪ {0}) ∩ C(Ω) and set Lu = aij Dij u, M u = β · Du. Then there is a constant α ∈ (0, 1), determined only by µ such that if M u ∈ (1) Hα (Σ1 ∪ Σ2 ) and if there is a positive constant F such that (4.13) holds, (0) then u ∈ H1+α (Ω) and (0)

(1) |u|1+α ≤ C(µ, ζ, |β|(0) α )(F + |M u|α + |u|0 ).

(4.73)

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Proof. We follow the argument in the proof of Theorem 4.5 with Theorem 4.27 in place of Lemma 4.4 and Lemma 4.18 in place of Proposition 4.3 when y = 0.  We now extend this result to allow curved boundary. Proposition 4.29. Let Ω, Σk , and β be as in Proposition 4.28 except that θl and θ2 are now H1+α functions with the same α as in that proposition. (0) If u satisfies the hypotheses of Proposition 4.28, then u ∈ H1+α (Ω) and (4.73) holds with C determined also by the H1+α norms of θ1 and θ2 . (−1−α)

Proof. By an H2+α change of coordinates, we can transform the problem into one with Σ1 a straight line segment. If θ2 = θ1 +π, then this change of coordinates also makes Σ2 a straight line segment, and Proposition 4.28 leads easily to the result. If θ2 < θ1 + π, then, after a rotation, we may assume that θ1 = 0 and that Σ2 can be written as x1 = ω(x2 ) for some H1+α function ω. The change of variables y 1 = x1 − ω(x2 ), y 2 = x2 then converts Σ2 to a segment on the positive y 2 axis while Σ1 is still a segment on the positive y 1 axis. A simple argument involving regularized (−1−α) distance shows that this change of coordinates can be made to be H2+α , and a linear change of variables then converts the problem to one with the same angle at the origin as for the original domain. We leave it to the reader to verify that the constant α is unaffected by the change of coordinates.  As in Section 4.1, we can now derive a global estimate which applies to equations with lower order terms and which estimates second derivatives as well. Although the method used here provides estimates in a wide variety of geometric settings, we only consider here a representative setting. Even in this (relatively simple) case, we need a long explanation of the relationship between the geometry, the differential operator and the boundary condition in order to determine the H¨older exponent of the gradient of the solution of the boundary value problem. Theorem 4.30. Let Ω be a bounded Lipschitz domain in R2 , let β be an oblique vector field defined on ∂Ω, and let [aij ] be a matrix-valued function defined on Ω. Let δ ∈ (0, 1) and µ ∈ [1, ∞) be constants, and suppose [aij ] (0) (1−α) (1−δ) satisfies condition (4.3). Let aij ∈ Hδ ∩ C(Ω), bi ∈ Hδ , c ∈ Hδ . Suppose ∂Ω is the union of finitely many H1+α curves Σ1 , . . . , ΣN which

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meet only at their endpoints and which meet at an angle no greater than π when they do meet. Suppose β is discontinuous wherever any of the curves meet and that β and β 0 are in Hα (Σok ) for k = 1, . . . , N , where Σok denotes the interior of Σk relative to the topology of ∂Ω. Suppose further that, if Σi and Σj (with i 6= j) meet at some point xij , and if polar coordinates (r, θ) centered at xij can be introduced in the intersection of Ω with a neighborhood of xij and i and j are relabeled (if necessary) so that θ1 and θ2 , defined by θ1 =

lim

x→xij x∈Σi

θ(x),

θ2 =

lim

θ(x),

(4.74)

lim

θ(β(x)),

(4.75)

x→xij x∈Σj

satisfy θ1 < θ2 , and if θ10 and θ20 are defined by θ10 =

lim

x→xij x∈Σi

θ(β(x)),

θ20 =

x→xij x∈Σj

then there is a constant θ0 such that conditions (4.42) and (4.71) are satisfied. Then there is a constant α0 (µ) ∈ (0, 1) such if α ∈ (0, α0 ), and if (−1−α) u ∈ H2+δ , then   1 1 (−1−α) (1−α) |u|2+δ ≤C |Lu|δ + |M u|α + |u|0 (4.76) λ χ for some constant C, determined only by the modulus of continuity for aij and upper bounds for the quantities 1 ij (0) |a |δ , λ

1 i (1−α) |b | , λ δ

1 i (1−α) 1 |b | , |β|α , λ δ χ

1 0 |β |α , χ

where χ = inf ∂Ω |β|. Proof. We just replace Theorem 4.5 with Proposition 4.29 in the proof of Theorem 4.6.  4.6

C 1,α estimates for discontinuous β in higher dimensions

It is now possible to obtain estimates for oblique derivative problems with discontinuous β in more than two dimensions provided we consider sufficiently simple geometric configurations. The techniques of the preceding section will be expanded to prove the corresponding estimate when the domain is piecewise smooth with smooth portions meeting only in (n − 2)dimensional sets. In three dimensions, this restriction means that we can consider boundary value problems in the cylinder {|x| < 1, |x3 | < 1}

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but not in the cube {max |xi | < 1}.

Such a restriction seems unnatural for the estimates, but the method of proof proposed here is unable to deal with the corners of the cube. Our scheme is essentially identical to that in Section 4.5, so we focus here on the few changes that are made. First, we modify our notation only slightly to consider the multidimensional problem. We abbreviate x00 = (x3 , . . . , xn ), and we introduce coordinates (r, θ) corresponding to (x1 , x2 ) just as before so that x1 = r cos θ,

x2 = r sin θ.

In the higher dimensional setting, we actually need to prove the local H¨older estimate first, and this estimate is essentially the same as Lemma 4.17. Lemma 4.31. Let R > 0, θ0 , θ1 , and θ2 be constants with θ1 < θ0 < θ2 ≤ θ1 + π, and define the sets Ω, Σk , and σ by Ω = {x ∈ Rn : |x| < R, θ1 < θ < θ2 }, n

(4.77a)

Σk = {x ∈ R : 0 < |x| < R, θ = θk }

(4.77b)

σ = {x ∈ Rn : |x| = R, θ1 ≤ θ ≤ θ2 }.

(4.77c)

for k = 1, 2, ij

θ10

θ20

Let [a ] ∈ Mµ for some µ ≥ 1, and Let and satisfy (4.42). Suppose also that there is a positive constant m0 such that |β 00 | ≤ m0 on Σ Then there are constants α ∈ (0, 1) and C, determined only by θ2 − θ1 , µ, and m0 such that, if u ∈ C(Ω) ∩ C 2 (Ω) satisfies (4.43), then (4.44) is valid. Proof. First, as in the proof of Lemma 4.17, we may assume that θ0 = π/2. Moreover, β 1 x1 ≤ 0 and β 2 > 0 on Σ. With m1 = m0 / inf β 2 , we then define G by   1 1 2 2 00 2 G(y, ρ) = ρ−2 (y 1 )2 + (y ) + |y | 16m2 16m2 m21 and we set x1 = (0, 8mρ, 0, . . . , 0). The remainder of the proof is identical to that of Lemma 4.17.  We then have the following higher dimensional analog of Lemma 4.16. Lemma 4.32. Under the hypotheses of Lemma 4.31, there are constants C > 0 and α ∈ (0, 1), both determined by µ and n, such that  α |x| |Du(x) − Du(0)| ≤ C sup |Du| . (4.78) R Ω

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Proof. We write Ω(ρ) (for ρ ∈ (0, R)) to denote the set of all x ∈ Ω with |x| < ρ. For m ≥ 3, we note that Aij Dij (Dm u) = 0 in Ω,

βk · D(Dm u) = 0 on Σk .

It then follows from that Theorem 1.26 that there are constants C and θ such that [Dm u]θ,Ω(7R/8) ≤ CR−α sup |Dm u|.

(4.79a)

Ω

Then the interior gradient estimate Proposition 2.13 implies that (1−θ)

|DDm u|θ,Ω(3R/4) ≤ CR−θ sup |Du|,

(4.79b)

where the norm is weighted with respect to distance to Σ1 ∪ Σ2 . Now fix x0 ∈ B(0, R/2) such that x10 = x20 = 0 and define w by w(x1 , x2 ) = u(x1 , x2 , x000 ).

It follows that Aˆij Dij w = f in Ω∗ , βˆ · Dw = g on Σ∗ ,

where Aˆij = Aij for i, j = 1, 2, f and g are defined by X f (x1 , x2 ) = − Aij Dij u(x1 , x2 , x000 ), g(x1 , x2 ) =

i>2 or j>2 n X − β j Dj u, j=3

and Ω∗ and Σ∗ are defined by

Ω∗ = {x ∈ R2 : |x| < R/4, θ1 < θ < θ2 },

Σ∗ = {x ∈ R2 : |x| < R/4, θ = θ1 or θ = θ2 }.

From (4.79), we conclude that (1−θ)

|f |θ

+ |g|θ ≤ CR−θ sup |Du|,

so Proposition 4.28 implies that

[Dw] ≤ CR−α sup |Du|

for some α ∈ (0, θ). Another application of Proposition 2.13 yields |Dij u| ≤ CR−α sup |Du|dα−1

for i, j = 1, 2, and (4.79) implies that this inequality is also true for all other choices of i and j. Simple integration now gives (4.39). 

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Our next step is the higher-dimensional analog of Lemma 4.18. The main new element here is that, in general, we cannot expect to have Du(0) = 0. Lemma 4.33. Under the hypotheses of Lemma 4.31, there are constants C > 0 and α ∈ (0, 1), both determined only by µ and n, such that |u(x) − u(0) − Du(0) · x| ≤ CR−1−α |u|0 |x|1+α

(4.80)

for |x| ≤ R/2. Proof.

This time, we use Lemma 4.32 and integration to infer that |u(x) − u(0) − Du(0) · x| ≤ CR−α |Du|0,Ω[R/2] |x|1+α .

Then we apply the local maximum principle Theorem 1.30 to β1 · Du and β2 · Du and the local maximum principle Theorem 1.27 to each component of D00 u to conclude that Z −n |Du|0,Ω[R/2] ≤ CR |Du| dx. Ω[3R/4]

From Lemma 4.31 and the proof of Lemma 2.20, we conclude that Z |Du| dx ≤ CRn−1 |u|0 , Ω[3R/4]

and combining all these estimates yields (4.45).



At this point, the differences between the two-dimensional case and the higher-dimensional case become relatively minor for a while. In fact, the statements and proofs of the higher-dimensional versions of Lemmata 4.19 and 4.20 differ from the two-dimensional ones only in notation (and reference to the appropriate, previously proved results), so we just recall the notation and statement of Lemma 4.20. For R > 0 and κ ∈ (0, 1), we set x1 (κ, R) = (0, −κR, 0, . . . , 0), and we define the sets Q(κ, R) = {x ∈ Rn : |x − x1 (κ, R)| < R, π/4 < θ < 3π/4},

S1 (κ, R) = {x ∈ Rn : |x − x1 (κ, R)| < R, θ = π/4},

S2 (κ, R) = {x ∈ Rn : |x − x1 (κ, R)| < R, θ = 3π/4},

s(κ, R) = {x ∈ Rn : |x − x1 (κ, R)| = R, π/4 ≤ θ ≤ 3π/4},

Q0 (κ, R) = S1 (κ, R) ∪ S2 (κ, R).

For α ∈ (0, 1), we write Hα (Q0 ) for the set of all functions g defined on Q0 such that the restriction of g to S1 (κ, R) is in Hα (S1 (κ, R)) and the

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restriction of g to S2 (κ, R) is in Hα (S2 (κ, R)). Since κ and R will usually be fixed in what follows, we generally omit them from the notation. In what follows, γ denotes the unit inner normal to Q. Lemma 4.34. Let α ∈ (0, 1), R > 0, and κ ∈ (0, 1). Then (4.50) has a unique solution u ∈ C 2 (Q) ∩ C 1 (Q \ s) ∩ C(Q) for any f ∈ Hα (Q), any g ∈ Hα (Q0 ), and any ϕ ∈ C(s). In our estimates for problems with less regular data, we need to take into account one additional fact, namely, there are points x ∈ s with x1 = x2 = 0. Fortunately, this situation is easily handled, and Lemma 4.21 remains true in higher dimensions. Lemma 4.35. Let δ ∈ (0, 1). For any α ∈ (0, 1), there is a constant κ1 in (0, 1), determined only by α such that, if κ ∈ (κ1 , 1), then any solution u of (4.50) with ϕ = 0 on s and g = 0 on Q0 ∩ s satisfies the estimate (1−α) (4.53). Moreover, (4.50) has a unique solution for any f ∈ Hδ , any 0 0 g ∈ Hα (Q ) with g = 0 on Q ∩ s, and ϕ = 0. Proof.

The argument in Lemma 4.21 implies that (1−α)

|u|0 ≤ CR1+α [|f |0

(−α)

+ |g|0

].

We now fix y0 ∈ Q0 ∩ s and note that there is a rotation T such which maps the positive x2 axis to the positive z 2 axis and such that the positive z 1 axis points in the direction of the vector −y0 , where we abbreviate z = T (x − y). To continue, we introduce some notation, based on the proof of Proposition 3.16. We write zˆ = (z 1 , z 2 ), and we introduce polar ˆ via the equations coordinates (ˆ r , θ) ˆ z 1 = rˆ cos θ,

ˆ z 2 = rˆ sin θ.

ˆ then we can A simple calculation shows that, if we take W (z) = rˆ1+α f (θ), choose f so that γ1 · DW and γ2 · DW are bounded from above by −ˆ rα α−1 ˆ when θ = 0 and such that M1 W ≤ −ˆ r when θ ≤ θ0 for some positive θ0 . By continuity, there is a constant θ1 ∈ (0, θ0 ) such that 1 γl · DW ≤ − rˆα 2

if θˆ ∈ (0, θ1 ), and hence, by choosing κ1 sufficiently close to 1, we can arrange that θˆ ≤ θ1 in Ω. We then use the comparison function w(x) = A1 W (z(x)) − A2 (dS1 )1+α − A3 (dS2 )1+α − A4 (dσ )1+α

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with suitable constants A1 , . . . , A4 to infer that (1−α)

|u(x) − u(y0 )| ≤ C(|f |0

(−α)

+ |g|0

]

00

for all x ∈ Ω with z (x) = 0. In particular, this inequality is valid if we choose x ∈ Ω and then take y0 to be the closest point in S0 to x. With this estimate in place, we can proceed as in the proof of Lemma 4.21.  At this point, the proof of all the estimates is the same as in the twodimensional case except for some minor notational issues and the use of the function W from the proof of Lemma 4.35 to prove the analog of Proposition 4.22. We leave the details to the reader and jump directly to the statement of the higher-dimensional version of Theorem 4.27. To state this proposition, we introduce some notation, based on that for Lemma 4.24. For positive constants m, m1 , and R, we set N (x) = (x1 )2 +

1 1 |x00 |2 (x2 + 8mR)2 + 2 16m 16m21

and we define the sets E(m, m1 , R) = {x ∈ Rn : N (x) < R2 , θ1 < θ < θ2 }, Ek (m, R) = {x ∈ Rn : N (x) < R2 , θ = θk }

for k = 1, 2, E0 (m, m1 , R) = {x ∈ Rn : x1 = x2 = 0, |x00 | < 4m1 R}, e(m, m1 , R) = {x ∈ Rn : N (x) = R2 , θ1 ≤ θ ≤ θ2 },

E 0 (m, m1 , R) = E1 (m, m1 , R) ∪ E2 (m, m1 , R).

We also define Hα (E 0 (m, m1 , R)) to be the set of all functions g defined on E 0 (m, m1 , R) such that the restriction of g to Ek (m, m1 , R) is in Hα (Ek (m, m1 , R)) for k = 1, 2. Theorem 4.36. Let [Aij ] ∈ Mµ for some µ ≥ 1 and let θ0 , θ1 , and θ2 satisfy (4.71). Let θ10 and θ20 satisfy (4.68), let β100 and β200 be (n−2)-dimensional vectors with |βk00 | ≤ 1/µ (for k = 1, 2), and define β on E 0 (m, m1 , R) by ( (cos θ10 , sin θ10 , β100 ) if x ∈ E1 (m, m1 , R), β(x) = (4.81) (cos θ20 , sin θ20 , β200 ) if x ∈ E2 (m, m1 , R). Also, let δ ∈ (0, 1). Then there are constants m2 > 0, m3 > 0, and α1 ∈ (0, 1), determined only by µ such that if α ∈ (0, α1 ), m ≥ m2 , and

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m1 ≥ m3 m, then

Aij Dij u = 0 in E(m, m1 , R), 0

β · Du = g on E (m, m1 , R), u = ϕ on e(m, m1 , R)

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(4.82a) (4.82b) (4.82c)

has a solution u ∈ C 2 (E(m, m1 , R)) ∩ C 1 (E(m, m1 , R) \ e(m, m1 , R)) ∩ (1−δ) C(E(m, m1 , R)) for any g ∈ Hα (E 0 (m, m1 , R)) and any ϕ ∈ C(e(m, m1 , R)). We also have the higher-dimensional analog of Theorem 4.30, which we state without proof. Theorem 4.37. Let Ω be a bounded Lipschitz domain in Rn , let β be an oblique vector field defined on ∂Ω, and let [aij ] be a matrix-valued function defined on Ω. Let δ ∈ (0, 1) and µ ∈ [1, ∞) be constants, and suppose [aij ] (1−α) (1−δ) (0) , c ∈ Hδ . satisfies condition (4.3). Let aij ∈ Hδ ∩ C(Ω), bi ∈ Hδ Suppose ∂Ω is the union of finitely many H1+α hypersurfaces Σ1 , . . . , ΣN which meet only at their edges and which meet at an angle no greater than π when they do meet. Suppose β is discontinuous wherever any of the hypersurfaces meet and that β and β 0 are in Hα (Σok ) for k = 1, . . . , N , where Σok denotes the interior of Σk relative to the topology of ∂Ω. Suppose further that, if Σi and Σj (with i 6= j) meet at some point xij , and if cylindrical coordinates (r, θ, x00 ) centered at xij with Σi ∩ Σj tangent to the set on which x00 = 0 can be introduced in the intersection of Ω with a neighborhood of xij and i and j are relabeled (if necessary) so that θ1 and θ2 , defined by (4.74) satisfy θ1 < θ2 , and if θ10 and θ20 are defined by (4.75), then there is a constant θ0 such that conditions (4.42) and (4.71) are satisfied. Then there is a constant α0 (µ) ∈ (0, 1) such if α ∈ (0, α0 ), and if (−1−α) u ∈ H2+δ , then (4.76) holds for some constant C, determined only by the modulus of continuity for aij and upper bounds for the quantities 1 ij (0) 1 i (1−α) 1 i (1−α) 1 1 0 |a |δ , |b | , |b | , |β|α , |β |α , λ λ δ λ δ χ χ where χ = inf ∂Ω |β|. 4.7

C 2,α estimates

In this section, we give the final H¨older estimate of the chapter. Note that it improves the classical estimate from Chapter 2 by reducing the regularity hypothesis on ∂Ω.

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As a first step, we obtain an estimate for the solution of a simpler problem in Ω[R]. Lemma 4.38. With ω as in Proposition 4.14, let [aij ] be a constant matrix with eigenvalues in the interval [λ, µλ], let β be an H1+α vector field defined on Σ[R] with |β 0 | ≤ µ0 for some constant µ0 and β n ≡ 1, and let g ∈ H1+α (Σ[R]). Suppose in addition that R ≤ 1. Let δ ∈ (0, 1). If v satisfies the conditions aij Dij v = 0 in Ω[R],

β · Dv = g on Σ[R],

(4.83)

then there is a positive constant C determined only µ, n, |β|1+α , δ, and |ω|1+α such that (−δ)

|v|2+α ≤ C(|g|1+α + |v|δ ).

(4.84)

Proof. To simplify the writing, we set G = |g|1+α , B = |β|1+α , and V = |v|δ . Extend β and g as C 2 (Ω[R]) functions so that |D2 β(x)| ≤ C(n, ω, α)B(xn )α−1 and |D2 g(x)| ≤ C(n, ω, α)G(xn )α−1 in Ω[R]. It follows that w = β · Dv − g satisfies aij Dij w = f in Ω[R],

w = 0 on Σ[R]

for some function f satisfying (−δ)

|f | ≤ C(n, α, ω, B)(G + |v|1

)(xn )α−1 .

From Theorem 4.5 and the argument in Lemma 2.14, we infer that (−δ)

|v|1

≤ C(G + V ),

so |f | ≤ C(G + V ). It then follows that (1−δ)

|w|1+α ≤ C(G + V ).

(4.85)

We now fix k ∈ {1, . . . , n} and set vk = Dk v. From (4.85), we conclude (1−δ) that there is a function gk satisfying |gk |α ≤ C(G + V ) such that aij Dij vk = 0 in Ω[R],

β · Dvk = gk on Σ[R],

and then Theorem 4.5 and the argument in Lemma 2.14 imply that (1−δ)

It follows that

|vk |1+α ≤ C(G + V ). (1−δ)

|Dv|1+α ≤ C(G + V ), and this estimate implies (4.84).



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167

This result then implies a local C 2,α regularity result for solutions of the oblique derivative problem with C 1,α boundary. Lemma 4.39. Let R0 , α and δ be positive constants with R0 ≤ 1 and α, δ < 1, and let ω be an H1+α (Bn−1 (x00 , R0 )) function for some x0 ∈ Rn with xn0 = ω(x00 ). For R ∈ (0, R0 ], write Ω[R] and Σ[R] for the sets of all x ∈ B(x0 , R) with xn > ω(x0 ) and xn = ω(x0 ), respectively. Suppose aij satisfies (4.3) and β satisfies |β 0 | ≤ µ0 and β n ≡ 1. If aij ∈ Hα and β ∈ H1+α , then there is a constant R ∈ (0, R0 ] such that (−δ)

(2−δ)

|u|2+α,Ω[R] ≤ C(|aij Dij u/λ|α,Ω[R] + |u|δ,Ω[R] + |β · Du|1+α,Ω[R] ) (−δ)

for any u ∈ H2+α,Ω[R] . Proof. First, we abbreviate f = aij Dij u and g = β · Du. With R at our (2−δ) disposal, we define f1 = f + [aij (x0 ) − aij ]Dij u and set F1 = |f1 |α,Ω[R] . We (2−δ)

then extend f1 to B(x0 , R) so that f1 ∈ Hα (2−δ) |f1 |α,B(x0 ,R)

(B(x0 , R)) with

≤ CF1 ,

so Corollary 2.16 implies that there is a unique solution w of aij (x0 )Dij w = f1 in B(x0 , R),

w = 0 on ∂B(x0 , R)

with (−δ)

(−δ)

|w|2+α,B(x0 ,R) ≤ C(F1 + |u|δ

).

Setting g1 = g − β · Dw, we conclude that v = u − w satisfies (4.83) with g1 in place of g. Then (−δ)

(1−δ)

(−δ)

|v|2+α,Ω[R] ≤ C(|g1 |1+α,Σ[R] + |v|δ (−δ)

).

(−δ)

Since |w|2+α,Ω[R] ≤ |w|2+α,B(x0 ,R) , it follows that Therefore

(−δ)

(−δ)

+ |g|1+α + F1 ).

(−δ)

(−δ)

+ |g|1+α + F1 ).

|v|2+α,Ω[R] ≤ C(|u|δ

|u|2+α,Ω[R] ≤ C(|u|δ (−δ)

(1−δ)

(1−δ) (−δ)

If we abbreviate U = |u|2+α,Ω[R] and U0 = |u|δ conclude that

(1−δ)

(2−δ)

+ |g|1+α + |f |α

(−δ)

U ≤ C(U0 + |[aij (x0 ) − aij ]Dij u|(2−δ) ) ≤ CU0 + CRα U + C[u]2 α

, we

.

Using the interpolation inequality Lemma 2.3, we conclude that U ≤ CU0 + CRα U

and the proof is concluded by choosing R so small that CRα ≤ 1/2.



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From this local result, we infer the global estimate. We leave the details of the derivation to the reader in Exercise 4.4. Theorem 4.40. Let α ∈ (0, 1) and let Ω be a bounded H1+α domain. Suppose aij satisfies (4.3) and β is oblique on ∂Ω. If aij ,b and c are in Hα and if β and β 0 are in H1+α , and if u ∈ C 2 (Ω) ∩ C(Ω) with Lu ∈ Hα and M u ∈ H1+α , then u ∈ H2+α and |u|2+α ≤ C(|Lu/λ|α + |M u/χ|1+α + |u|0 ).

(4.86)

As usual, this theorem has an associated existence theorem. Corollary 4.41. With Ω, L, and M as in Theorem 4.40, for any f ∈ Hα and g ∈ H1+α , the problem Lu = f in Ω,

M u = g on ∂Ω

(4.87)

has a unique solution if c ≤ 0, β 0 ≤ 0 and one of these two functions is not identically zero.

Notes The basic ideas in this chapter were first presented in [104] and refined in [117] for continuous β and in [109] for discontinuous β. The proof here for discontinuous β differs from the one in [109] in several ways; the most important one is that [109] relies heavily on the theory of weak solutions (which we discuss in Chapter 5) while we avoid that theory altogether. See also [153] for an alternative approach to the H¨older gradient estimate for continuous β in Lipschitz domains. Various estimates have been proved for discontinuous β as well, but most of them are not the same as the ones in this chapter. We mention [68] and [126] in particular, but refer the reader to [85] for a more detailed examination of the problem in a more general setting. Before comparing the results in this chapter to those in [68], we point out that conditions (4.40) and (4.71) are important not just because they allow our proofs to work. By examining functions of the form rα cos(αθ−δ0 ) with suitable constants α and δ0 , we can show that there are solutions of the boundary value problem ∆u = 0 in E(m, R), β · Du = 0 on E 0 (m, R) which do not have H¨ older continuous gradient at the origin.

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169

More specifically, in Chapter 6 of [68], Grisvard studies the problem ∆u = f in Ω, β · Du = 0 on N , u = 0 on D,

where Ω is a polygon, N is the union of some of the edges of ∂Ω, and D is the union of the remaining edges. Assuming that D is not empty, and some technical conditions are satisfied (primarily that β is not a constant vector which is tangential on all of D and that f ∈ Hα ), he shows that there is a solution of this problem which is globally continuous and satisfies all boundary conditions classically. Moreover, as shown in Theorem 6.4.2.5 of [68], any such solution can be written as the sum of an H2+α function and the sum of functions which near each vertex can be written as ra cos(aθ + b) or (in certain exceptional cases) as ra [ln r cos(aθ + b) + θ sin(aθ + b)], where (r, θ) denote polar coordinates centered at the vertex, and the numbers a and b are chosen so the boundary conditions are satisfied near the vertex and so that a ∈ (0, 2 + α). In fact, the exponent a can be computed rather easily, and is equal to 1+

θ10 − θ20 + mπ θ2 − θ1

for suitable integers m, using our notation. In the cases we have studied, there is at most one such number a in the interval (0, 1) and the corresponding functions do not appear in any solution which satisfies the type of boundary condition which we prescribe at the vertex. For this reason, Grisvard can prove the existence of solutions but not their uniqueness. H¨ older gradient estimates for the Neumann problem (with the Laplace operator) in two-dimensional domains with angles were studied by Volkov, too. In [196], he proved that the solution is in Ha for a given non-integer a proved the angle at each corner is greater than π/a, assuming sufficient smoothness of the data. In [195], he proved that, if the angle at a corner is π/m for some integer m ≥ 2, then the solution is in Ha provided suitable compatibility conditions are satisfied at the corner. Although we shall not write these compatibility conditions out for general m, we note here that the conditions corresponding to a ∈ (2, 3) are exactly those that allow the

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solution to have continuous second derivatives at the corner, namely, if the domain is translated and rotated so that the corner is at the origin and the two sides lie locally on the positive x-axis and the positive y-axis, then the condition on the x-axis prescribes uy (x, 0) while the condition on the y-axis prescribes ux (0, y). By differentiating the first boundary condition with respect to x and the second with respect to y, we obtain two values for uxy (0, 0), and the compatibility condition is that these two values must agree. Further results along these lines can be found in [202] and [203]. We also mention that the results for discontinuous β are closely related to issues of reflected Brownian motion. In this connection, the papers [204, 177, 159] are of especial interest. Exercises 4.1 Verify that the function ρ in Lemma 4.7 satisfies all the properties listed in that lemma. (Hint: Look at Theorems 1.3 and 2.1 in [99].) 4.2 Verify that the function ρ in Lemma 4.9 satisfies all the properties listed in that lemma. (Hint: Look at Theorems 1.3 and 2.1 in [99].) 4.3 Construct the function ϕ∗ described in the proof of Proposition 4.14. (Hint: Look at the proof of Lemma 6.38 in [64] and Lemma 4.24 in [114].) 4.4 Derive Theorem 4.40 from Lemma 4.39.

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Chapter 5

Weak Solutions

Introduction In this chapter, we study weak solutions of the conormal problem. The motivation for this problem is to provide a suitable generalization of the Neumann problem for Laplace’s equation. Of course, our general oblique derivative problem is a generalization of this problem, but such solutions are still required to be C 2 inside Ω. Weak solutions are only required to have first derivatives in some Sobolev space, and this weaker regularity leads to several technical difficulties. Fortunately, these difficulties are easily overcome by well-known methods. In fact, the main part of the theory in this chapter is a reworking of the material in Chapter 8 of [64]; we shall write a few results in somewhat different form (in particular, our approach to the Moser iteration scheme is slightly different from the one in [64], and we use a perturbation argument like the one in Chapter 2 to obtain Schauder estimates), but it is not difficult to rewrite all of our proofs to conform with those in [64]. We include the results just for completeness. In particular, we recall the definition and some basic properties of weak derivatives in Section 5.1; we use these results here. The idea is to start with the Neumann problem for Laplace’s equation ∂u ∆u = 0 in Ω, = g on ∂Ω, (5.1) ∂γ multiply the differential equation by a smooth function ϕ, and then integrate by parts. In this way, we obtain the integral equation Z Z Du · Dϕ dx = − gϕ ds, (5.2) Ω

∂Ω

where ds denotes surface measure on ∂Ω. Noting that this integral equation makes sense for u ∈ W 1,2 (Ω), we take this integral identity as the definition 171

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of a weak solution of the boundary value problem. In other words, we say that u ∈ W 1,2 is a weak solution of (5.1) if the identity (5.2) is satisfied for all ϕ ∈ C 1 (Ω). One of the many advantages of this weak formulation is a weaker regularity hypothesis on the function g since (5.1) makes sense even if g is only in L1 (∂Ω). As we shall eventually see, a somewhat stronger assumption must be made on g; this problem is solvable (in the weak sense) for g ∈ L2 (∂Ω) under the obvious Z compatibility condition g ds = 0.

(5.3)

∂Ω

In fact, there is an important comment to be made here about the regularity results in this chapter as compared with the results in the preceding chapter. We were able to show that solutions of the oblique derivative problem are (globally) C 1,α even if the domain is Lipschitz. Here, we prove this regularity under the stronger hypothesis that the boundary is C 1,α (similarly to the situation in Chapter 2), but under weaker assumptions on the coefficients of the operator L. In addition, the results in this chapter require a relationship between the matrix [aij ] and the vector β which was not needed in the preceding chapters. For linear equations, this connection is relatively unimportant; it is possible to write an oblique derivative problem (with suitably smooth data) in the form studied here by adapting the proof of Proposition 6.21 in the next chapter. For nonlinear equations, we shall see in Chapter 11 that this special relationship is a critical part of the theory. We now give the formal definition of the weak solution of the oblique derivative problem. Let Ω be a bounded, Lipschitz domain with unit inner normal γ. Let aij , bi , ci , c0 , and β 0 be bounded functions and suppose that there is a positive constant λ such that aij (x)ξi ξj ≥ λ|ξ|2 (5.4) n 1,2 for almost all x ∈ Ω and all ξ ∈ R . We then say that u ∈ W (Ω) is a weak solution of Di (aij Dj u + bi u) + ci Di u + c0 u = Di f i + g in Ω, (5.5a) aij γi Dj u + b · γu + β 0 u = ψ + f · γ if the integral Z identity [(aij Dj u + bi u − f i )Di ϕ + (−ci Di u − c0 u + g)ϕ] dx Ω Z = [β 0 u − ψ]ϕ ds

(5.5b)

(5.6)

∂Ω

is satisfied for all ϕ ∈ C 2 (Ω). We also abbreviate (5.5) to Lu = Di f i + g in Ω, M u = ψ on ∂Ω.

(5.7)

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173

Definitions and basic properties of weak derivatives

The key point here is the concept of weak derivatives. Let Ω be an open subset of Rn and let i ∈ {1, . . . , n}. We say that v ∈ L1loc (Ω) is the weak derivative Di u of u ∈ L1loc (Ω) if, for any ϕ ∈ Cc1 (Ω) (the set of all continuous functions which have uniformly continuous gradient on Ω and which have compact support in Ω), we have Z Z vϕ dx = − uDi ϕ dx. Ω

Ω

Integration by parts shows that, if u ∈ C 1 (Ω), then the usual partial derivative Di u is also the weak derivative. There are several basic properties of weak derivatives which we now reproduce, referring to [64] for the proofs. Lemma 5.1. Let u and v be locally integrable in Ω. Then v = Di u weakly if and only if there is a sequence (um ) of C ∞ (Ω) functions such that um → u and Di um → v in L1loc . For our next lemma, we write u+ = max{u, 0},

u− = max{−u, 0},

and we observe that u = u+ − u− and |u| = u+ + u− . Lemma 5.2. Let u ∈ L1loc and suppose that the weak derivative Di u exists. Then u+ , u− , and |u| have weak derivatives satisfying ( Di u where u > 0, Di u + = 0 where u ≤ 0, ( Di u where u < 0, Di u − = 0 where u ≥ 0,   where u > 0,  Di u Di |u| =

0   −D u i

where u = 0, where u < 0.

We also write W 1,2 for the set of all u ∈ L2 such that the weak derivatives D1 u, . . . , Dn u are also in L2 , and we write W01,2 for the closure of Cc1 (Ω). It can be shown that any function in W01,2 has a well-defined trace on ∂Ω, at least for Lipschitz domains, which must be zero. Many of the rules of calculus are also valid for weak derivatives. We note here that the product rule D(uv) = uDv + vDu is valid as long as

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u and v have weak derivatives with uDv and vDu being L1 functions. In ˜ in Rn with addition, if ψ is a function that maps Ω onto some domain Ω ˜ then, for every u which has weak derivatives ψ ∈ C 1 (Ω) and ψ −1 ∈ C 1 (Ω), ˜ in Ω, the corresponding function v = u ◦ ψ −1 is weakly differentiable in Ω and ∂y j ∂v Di u(x) = (y) (5.8) ∂xi ∂y j for almost all x ∈ Ω, where y = ψ(x). Furthermore, if f is continuously differentiable on R with bounded derivative, and if u is weakly differentiable in Ω, then f ◦ u is also weakly differentiable in Ω and D(f ◦ u) = f 0 (u)Du. (This version of the chain rule is the basis for proving Lemma 5.2.) We can also define various inequalities on W 1,2 functions at the boundary. If u and v are in W 1,2 , we say that u ≤ v on ∂Ω if (u − v)+ ∈ W01,2 . This concept will be very useful in later chapters as well. We also recall the following results about the relation between derivatives and difference quotients of functions in Sobolev spaces. To state results succinctly, we introduce the notation u(x + hei ) − u(x) ui,h = h for h 6= 0 and i = 1, . . . , n. We then have the following results (see Lemmata 7.23 and 7.24 in [64]). Lemma 5.3. Let u ∈ W 1,p (Ω) for some p ≥ 1. Then ui,h ∈ Lp (Ω0 ) for any Ω0 such that d(x) > |h| for all x ∈ Ω0 . In addition, kui,h kp;Ω0 ≤ kDi ukp;Ω . Lemma 5.4. Let u ∈ Lp (Ω) for some p > 1, and suppose that there is a constant K such that ui,h ∈ Lp (Ω0 ) and kukp;Ω0 ≤ K for any h > 0 and Ω0 ⊂ Ω with d(x) > |h| for all x ∈ Ω0 . Then the weak derivative Di u exists, Di u ∈ Lp (Ω), and kDi ukp ≤ K. 5.2

Sobolev imbedding theorems

Several deeper results for functions in Sobolev spaces will be needed. The first one is a special case of the Sobolev imbedding theorem. We refer the reader to Theorem 7.10 of [64] or Theorem 6.7 of [114] for a proof. Theorem 5.5. For any domain Ω in Rn , W01,1 (Ω) ⊂ Ln/(n−1) (Ω) and there is a constant C, determined only by n, such that kukn/(n−1) ≤ CkDuk1 . (5.9)

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It will also be useful to have a version of the Sobolev imbedding theorem for functions with nonzero boundary values. For our applications to local estimates, we shall find an intermediate form of this inequality very useful, so we present it here with proof. Lemma 5.6. Let R > 0 and ω0 be positive constants, and let ω be a realvalued function defined on Bn−1 (R) with |ω(x0 ) − ω(y 0 )| ≤ ω0 |x0 − y 0 | 0

(5.10)

0

for all x and y in Bn−1 (R) and ω(0) ∈ (0, R). Define

Ω[R] = {x ∈ Rn : xn > ω(x0 ), |x| < R}, n

n

0

σ[R] = {x ∈ R : x ≥ ω(x ), |x| = R}.

(5.11a) (5.11b)

If u ∈ W 1,1 (Ω[R]) and u = 0 on σ[R], then there is a constant C, determined only by n and ω0 , such that (5.9) holds. Proof. By virtue of Lemma 1.3, we may assume that ω is defined on Rn and that (5.10) is satisfied for all x0 and y 0 . We extend u to the set where xn > ω(x0 ) by defining u(x) = 0 if xn > ω(x0 ) and |x| > R, and we define new coordinates z by z 0 = x0 and z n = xn − ω(z 0 ). Then u ¯ defined by u ¯(z) = u(x) is in W 1,1 (Rn+ ) and the even extension of u¯ is in W 1,1 (Rn ). It follows that k¯ ukn/(n−1) ≤ C(n)kDu¯k1 . But kukn/(n−1) ≤ C(ω0 , n)k¯ ukn/(n−1) and kDu¯k1 = 2kDu¯k1,Rn+ ≤ C(ω0 , n)kDuk1 . Combining these inequalities completes the proof.



From this lemma, a simple partition of unity argument gives the general Sobolev imbedding theorem with nonzero boundary values. Theorem 5.7. Let Ω be a bounded Lipschitz domain. Then W 1,1 (Ω) ⊂ Ln/(n−1) (Ω) and there is a constant C, determined only by Ω and n, such that kukn/(n−1) ≤ C(kuk1 + kDuk1 ) for all u ∈ W 1,1 .

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We also use a multiplicative version of the Sobolev imbedding theorem, which is a special case of the Gagliardo-Nirenberg inequality. We present three versions of this inequality, but first we point out a useful consequence of H¨ older’s inequality: For any measure space (X, µ) and any numbers p, r, and α such that 1 ≤ p < r < ∞ and α ∈ (0, 1), we have 1−α kukαp+(1−α)r ≤ kukα p kukr

(5.12)

for all u ∈ Lp (µ) ∩ Lr (µ). Theorem 5.8. Let p and N be constants such that N > p ≥ 1 and N ≥ n. (a) Then there is a constant C, determined only by n, N , and p such that Z (N −p)/N N p/(N −p) |u| dx Ω (5.13) Z (N −n)/N Z n/N ≤C

Ω

|u|p dx

Ω

|Du|p dx

for any u ∈ W 1,p (Rn ). (b) If R, ω0 , Ω[R], and σ[R] are as in Lemma 5.6, then (5.13) holds with Ω[R] in place of Ω and C determined also by ω0 for any u ∈ W 1,p (Ω[R]) which vanishes on σ[R]. (c) For any bounded Lipschitz domain Ω there is a constant C, determined only by Ω, N , n, and p, such that Z (N −p)/N |u|pN/(N −p) dx Ω Z (N −n)/N Z n/N (5.14) ≤C

Ω

|u|p dx

Ω

[|Du|p + |u|p ] dx

.

for all u ∈ W 1,p (Ω). Proof. Setting r = np(N − 1)/(N − p)(n − 1), we begin by using (5.12) to see that Z α Z 1−α Z N p/(N −p) r p |u| dx ≤ |u| dx |u| dx , where α satisfies the equation αr + (1 − α)p = N p/(N − p). We next infer from Theorem 5.5 that (n−1)/n Z Z r |u| dx ≤ C(n)s |u|s−1 |Du| dx,

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where s = r(n − 1)/n. If p > 1, H¨older’s inequality shows that Z (p−1)/p Z 1/p Z s−1 N p/(N −p) p u |Du| dx ≤ |u| dx |Du| dx . since (s − 1)p/(p − 1) = N p/(N − p), while p = 1 implies s = 1, in which case this inequality is immediate. Combining these three inequalities and using the explicit expression α=

p(n − 1) N + np − n − p

yields (5.13). By invoking Lemma 5.6 in place of Theorem 5.5, we obtain (5.13) with Ω[R] in place of Ω, and we obtain (5.14) by using Theorem 5.7.  Note that the case p < n = N in Theorem 5.8 corresponds to the classical Sobolev imbedding theorem. If p ≥ n, it follows that W 1,p (Ω) ⊂ Lq (Ω) for any finite q, and it is known that functions in W 1,p are H¨older continuous with exponent 1 − n/p for p > n. We refer the interested reader to Theorem 7.17 in [64] or Theorem 6.8 in [114] for details. We also have the following compactness theorem, which is proved as Theorem 7.26 in [64]. Theorem 5.9. Let Ω be a bounded Lipschitz domain in Rn and let p ∈ [1, n). Then the closure of any bounded subset of W 1,p (Ω) is a compact subset of Lq (Ω) for any q ∈ [1, np/(n − p)). 5.3

Poincar´ e’s inequality

A related result is Poincar´e’s inequality. We include a direct proof because the usual one (found, for example, in Lemma 4.1.3 and Theorem 4.2.1 of [206]) is indirect. The first step is the appropriate inequality for a ball. We refer to (7.45) in [64] or Proposition 6.14 in [114] for a proof. Lemma 5.10. LetR B be a ball of radius R > 0 in Rn , and let f ∈ L∞ (B) be nonnegative with B f (x) dx = 1. Then there is a constant C, determined only by n and p such that Z Z inf |u − L|p dx ≤ CRp sup f |Du|p dx (5.15) L∈R

B

for any u ∈ W 1,p (B).

B

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We also use the following result relating the left hand side of (5.15) to other similar integrals. This result is just Lemma 6.12 in [114]. n ∞ RLemma 5.11. Let Ω be a domain in R pand let f ∈ L (Ω) with Ω f (x) dx = 1. Let p ≥ 1. Then, for all u ∈ L (Ω), we have Z Z p p |u(x) − uf | dx ≤ 2 sup |f | inf |u − L|p dx, (5.16) L∈R

Ω

Ω

where

uf =

Z

u(y)f (y) dy.

Ω

From these two lemmata, the general Poincar´e inequality follows by a change of variables and a chaining argument. Theorem 5.12. Let Ω be a bounded Lipschitz domain. Then, for any p ≥ 1, there is a constant C, determined only by Ω and p such that Z Z inf |u − L|p dx ≤ C |Du|p dx (5.17) L∈R

for all u ∈ W

1,p

Ω

Ω

(Ω).

Proof. We observe first that Lemma 5.10 gives (5.17) when Ω is a ball. Now suppose that there is a one-to-one Lipschitz function ψ : Ω → B (for some ball B) such that ψ −1 : B → Ω is also Lipschitz. For later reference, we say that Ω is biLipschitz equivalent to a ball if this condition is satisfied. In this case, there is a constant C0 , determined only by Ω, such that the Jacobian determinant J(ψ) satisfies 1/C0 ≤ |J(ψ)| ≤ C0 , and (5.17) follows from Lemma 5.10 and the change of variables formula for integrals. Next suppose that Ω = S1 ∪ S2 with S1 and S2 both biLipschitz equivalent to the same ball B, and suppose that there is a ball Bδ with radius δ > 0 such that Bδ ⊂ S1 ∩ S2 . Setting Z 1 L0 = u dx, |Bδ | Bδ we infer from Lemma 5.11 and our previous case that Z Z inf |u − L|p dx ≤ C |Du|p dx L∈R

Ωj

Ωj

for j = 1, 2. Combining these two inequalities yields Z Z p |u − L0 | dx ≤ C |Du|p dx. Ω

Ω

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Finally, any bounded Lipschitz domain Ω can be covered by finitely many sets S1 , . . . , SN , each of which is biLipschitz equivalent to a ball and there is a positive constant δ such that, if Si ∩ Sj is non-empty, then this intersection contains a ball of radius δ. The desired result then follows from the preceding argument via induction.  We also have the following variation on the Gagliardo-Nirenberg inequality. Proposition 5.13. Let Ω be a bounded Lipschitz domain. Let p ≥ 1, let N be a positive number such that N ≥ n and N > p, and let θ ∈ (0, 1). Then there is a constant C, determined only by Ω, n, N , p, and θ such that (5.13) holds for any u ∈ W 1,p (Ω) such that u = 0 on a set with measure greater than θ|Ω|. Proof. that

By using f = χS /|S| in Lemma 5.11, we see from Theorem 5.12 kukp ≤ CkDukp .

The proof is completed by using this inequality along with (5.14).



We close this section with a result that will be useful in our analysis of local properties of weak solutions. To state our result, for k ∈ R and u a real-valued function defined on a set S, we define Ak to be the subset of S on which u > k. Lemma 5.14. Let R, ω0 , and Ω[R] be as in Lemma 5.6. Let u ∈ W 1,1 (Ω[R]) and let m and k be real numbers with m > k such that Ak has positive measure. Then there is a constant C, determined only by ω0 , n, and R−n |Ak | such that Z (n−1)/n (m − k)|Am | ≤C |Du| dx. Ak \Am

Proof. We first observe that, when Ω = Ω[R] in Proposition 5.13, then C depends on Ω only through ω0 . Applying this proposition with p = 1 and N = n, and u replaced by w, defined by   if u < k,  0 w(x) = u − k if k ≤ u < m,   m − k if m ≤ u

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we have Z

w

n/(n−1)

!(n−1)/n

dx

Ω[R]

≤C

Z

Ω[R]

|Dw| dx,

with C determined by n, ω0 , and R−n |Ak | since |Ω[R]| ≥ C(ω0 , n)Rn . Moreover Z Z n/(n−1) w dx ≥ wn/(n−1) dx = (m − k)n/(n−1) |Am | Ω[R]

Am

and Z

Ω[R]

|Dw| dx =

Z

Ak \Am

|Du| dx

because Dw = 0 except on Ak \ Am , and Dw = Du on Ak \ Am . 5.4



The weak maximum principle

Just as the maximum principles of Chapter 1 play a critical role in the analysis of classical solutions, a corresponding result is valid for weak solutions, and it plays a similar role. A crucial element of the maximum principles is the inequalities on c and β 0 . For weak solutions, the corresponding inequality is more subtle but is easily guessed due to the nature of our proof below. Theorem 5.15. Let u ∈ W 1,2 satisfy

Di (aij Dj u + bi u) + ci Di u + c0 u ≥ 0 in Ω, ij

0

a γi Dj u + b · γu + β u ≥ 0 on ∂Ω.

(In other other words, Z [(aij Dj u + bi u)Di ϕ + (−ci Di u − c0 u)ϕ] dx Ω Z ≤ β 0 uϕ ds

(5.18a) (5.18b)

(5.19)

∂Ω

1

for all nonnegative ϕ ∈ C (Ω).) Suppose also that Z Z 0 i c w − b Di w dx + β 0 w ds ≤ 0 Ω

(5.20)

∂Ω

for all nonnegative w ∈ C 1 (Ω). Then either u is a constant or else u ≤ 0 in Ω.

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Proof. Let us assume that sup u >Z0. We write (5.18) in weak Z Z form as ij i i 0 i a Dj uDi v − (b + c )vDi u dx ≤ c uv − b Di (uv) dx + β 0 uv ds Ω

Ω

∂Ω

for all nonnegative v ∈ C 1 (Ω). Because we have assumed that all coefficients of L and M are bounded, it follows that this inequality holds for all nonnegative v ∈ W 1,2 such that uv ≥ 0 in Ω. Since c0 and bi are bounded, it follows that (5.20) is valid for all nonnegative w ∈ W 1,1 and, in particular, for w = uv. Therefore Z Z ij a Dj uDi v dx ≤ (bi + ci )vDi u dx. Ω

Ω

From our bounds, Z we conclude that Z aij Di vDj u dx ≤ C v|Du| dx. Ω

(5.21)

Ω

We now let k be a positive constant such that k < sup u+ . We take v = (u − k)+ , and we write Γk for the support of Dv. It follows that Du = Dv in Γk and Dv = 0 inZ Ω \ Γk , and hence Z Ω

aij Di vDj v dx ≤ C

v|Dv| dx.

Γk

ij Using H¨ older’s inequality Z and the ellipticity Z of [a ], we find that |Dv|2 dx ≤ C v 2 dx. Ω

Γk

If we combine this estimate with the Gagliardo-Nirenberg inequality, Proposition 5.13, taking v in place of u and N = n + 1, we infer that 1/q Z (N −n)/N Z n/N Z q 2 2 v dx v dx ≤C v dx Ω

Ω

Γk

for q = 2N/(N − 2). H¨ older’s inequality implies that Z 2/q Z 2 q v dx ≤ v dx |S|(q−2)/q S

S

for any measurable set S, and hence kvkq;Ω ≤ C|Γk |(q−2)n/(2N ) kvkq;Ω . It follows that |Γk | is bounded from below by a positive constant, independent of k, and hence Γ0 = ∩0 n are left to the reader.  More generally, we have the following versions of the GagliardoNirenberg inequality concerning boundary integrals, obtained by modifying the proof of Lemma 5.20 along the lines of Theorem 5.8. Theorem 5.21. Let p > 1, let N be a positive constant with N > p and N ≥ n, and set q = (N − 1)p/(N − p), α=

(N − n)(p − 1) , (N − p)p

β=

N − p + np − n . (N − p)p

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(a) Let Ω be a bounded Lipschitz domain in Rn . Then there is a constant C, determined only by N , n, p, and Ω such that Z α Z β Z q p p p |u| ds ≤ C |u| dx [|Du| + |u| ] dx (5.27) ∂Ω

Ω

Ω

1,p

for all u ∈ W (Ω). (b) Let Ω be a bounded Lipschitz domain in Rn and let θ ∈ (0, 1). Then there is a constant C, determined only by N , n, p, θ, and Ω such that Z α Z β Z |u|q ds ≤ C |u|p dx |Du|p dx (5.28) ∂Ω

Ω

Ω

for all u ∈ W 1,p (Ω) which vanish on a set with measure greater than θ|Ω|. (c) Let R, ω0 , Ω[R], and σ[R] be as in Lemma 5.6, and define Σ[R] = {x ∈ Rn : |x| < R, xn = ω(x0 )}.

(5.29)

Then there is a constant C, determined only by N , n, p, and ω0 such that !α Z !β Z Z q p p |u| ds ≤ C |u| dx |Du| dx (5.30) Σ[R]

Ω[R]

Ω[R]

for all u ∈ W 1,p (Ω[R]) which vanish on σ[R]. We can also prove an interior Lp estimate in terms of the LP norm of the gradient and the the Lp norm of the trace. Lemma 5.22. Let Ω be a bounded Lipschitz domain and let p ≥ 1. Then there is a constant K2 , determined only by Ω and p, such that Z  Z Z |u|p dx ≤ K2 |Du|p dx + |u|p ds (5.31) Ω

Ω

∂Ω

for any u ∈ W 1,p (Ω). Proof.

We imitate the proof of Poincar´e’s inequality. Set Z Z 1 1 L0 = u ds, L1 = u dx. |∂Ω| ∂Ω |Ω| Ω

Then the triangle inequality implies that Z  Z Z p p p |u − L0 | dx ≤ p |u − L1 | dx + |L1 − L0 | dx . Ω

Ω

Ω

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Because L1 − L0 is a constant, Z |L1 − L0 |p dx = |Ω||L1 − L0 |p , Ω

and

p Z 1 (u − L1 ) ds ≤ |u − L1 |p ds |∂Ω| ∂Ω ∂Ω

Z 1 |L1 − L0 | = |∂Ω| p

by H¨ older’s inequality. It follows that Z Z Z |u − L0 |p dx ≤ C |u − L1 |p dx + Ω

Ω

 |u − L1 |p ds ,

∂Ω

and (5.26) gives a constant C, determined only by Ω and p such that Z  Z Z p p p |u − L1 | ds ≤ C |u − L1 | dx + |Du| dx . ∂Ω

Ω

Ω

Applying Poincar´e’s inequality gives Z Z |u − L0 |p dx ≤ C |Du|p dx, Ω

Ω

and the proof is completed by observing that  Z Z Z |Ω| |u|p ds . |u|p dx ≤ p |u − L0 |p dx + |∂Ω| ∂Ω Ω Ω



We also obtain a slightly improved version of the preceding estimate, which will be used later. Corollary 5.23. Let Ω be a bounded Lipschitz domain and let p > 1. Then there is a constant C, determined only by Ω and p, such that Z Z Z |u|p dx ≤ ε |Du|p dx + Cε−1/(p−1) |u|p ds (5.32) Ω

Ω

∂Ω

for any u ∈ W 1,p (Ω) and all ε ∈ (0, 1). Proof.

By using Young’s inequality in the sharper form ab ≤

εap (p − 1)ε−1/(p−1) bp/(p−1) + , p p

we obtain Z

∂Ω

p

|u| ds ≤ ε

Z

Ω

p

|Du| dx + Cε

−1/(1−p)

Z

Ω

|u|p dx.

The proof now follows that of Lemma 5.22 with this inequality replacing (5.26). 

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5.6

187

Existence of weak solutions

To prove the existence of weak solutions to the oblique derivative problem, we use an extension of the Riesz representation theorem for Hilbert spaces, due to Lax and Milgram. We refer to Theorem 5.8 in [64] for a proof of this result. Theorem 5.24. Let H be a Hilbert space and suppose B is a real-valued function defined on H × H such that B(αx1 + βx2 , y) = αB(x1 , y) + βB(x2 , y)

(5.33a)

B(x, αy1 + βy2 ) = αB(x, y1 ) + βB(x, y2 )

(5.33b)

for all real α and β and all x, y, x1 , x2 , y1 , y2 in H and there is a positive constant K such that |B(x, y)| ≤ Kkxkkyk, (5.34a) 1 B(x, x) ≥ kxk2 (5.34b) K for all x and y in H. Then, for any bounded linear functional F defined on H, there is a unique f ∈ H such that B(x, f ) = F (x) for all x ∈ H. We recall that a function B : H × H → R satisfying (5.33) is called a bilinear form. If B also satisfies (5.34a), it is called bounded ; if B also satisfies (5.34b), it is called coercive. To apply this result to the oblique derivative problem, we define the bilinear form L on H = W 1,2 by Z Z L(u, v) = [(aij Dj u + bi u)Di v − (ci Di u + c0 u)v dx − β 0 uv ds, Ω

∂Ω

and we define the bounded linear functional F on H by Z Z F (v) = f i Di v − gv dx − ψv ds. Ω

∂Ω

Then u is a weak solution of the boundary value problem (5.5) if and only if L(u, v) = F (v) for all v ∈ H. Note that L is not coercive if, for example, bi , ci , c0 and β 0 are all zero, so it will not be possible to apply the Lax-Milgram theorem directly to our existence question; however, a coercivity-type estimate is true. Lemma 5.25. Suppose there is a positive constant λ such that (5.4) is satisfied. Suppose further that there are constants µ1 ≥ 0 and q > n such that kb − ckq + kc0 kq/2 + kβ 0 kq−1 ≤ µ1 λ.

(5.35)

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Then there is a constant C, determined only by µ1 and Ω such that Z Z λ 2 L(u, u) ≥ |Du| dx − Cλ u2 dx. (5.36) 2 Ω Ω Proof.

We write L(u, u) =

Z

aij Di uDj u dx + I1 + I2 + I3 ,

Ω

where Z

(bi − ci )Di uu dx, Ω Z I2 = − c0 u2 dx, Ω Z I3 = − β 0 u2 ds. I1 =

∂Ω

We then estimate these three integrals. First, by H¨older’s inequality and (5.14) with N = q, Z (n+q)/2q Z (q−n)/2q 2 2 2 I1 ≥ −Cλ [|Du| + u ] dx u dx Ω

Ω

(with C determined by n, q, µ1 and Ω) and then Young’s inequality yields Z Z 1 2 I1 ≥ − λ |Du| dx − Cλ u2 dx. 6 Ω Ω Similarly, we have 1 I2 ≥ − λ 6

Z

Ω

2

|Du| dx − Cλ

Z

u2 dx. Ω

If we use (5.28) in place of (5.14), we also obtain Z Z 1 2 I3 ≥ − λ |Du| dx − Cλ u2 dx. 6 Ω Ω The proof is completed by using (5.4) to estimate Z aij Di uDj u dx. Ω



From this lemma, we can prove a general existence result. Theorem 5.26. Let Ω be a bounded Lipschitz domain, and suppose that there are positive constants λ and µ1 such that conditions (5.4) and (5.35)

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are satisfied. Then there is a countable set Σ of real numbers with no finite limit points such that the problem (L + σ)u = Di f i + g in Ω,

(M + σ)u = ψ on ∂Ω

(5.37)

has a unique solution for all (f, g, ψ) ∈ X = L2 (Ω)n+1 × L2 (∂Ω) if and only if σ ∈ / Σ. Moreover, for each σ ∈ Σ, the dimension of the solution set of the problem (L + σ)u = 0 in Ω,

(M + σ)u = 0 on ∂Ω

(5.38)

is a positive integer Iσ , and there is an Iσ -dimensional subspace Wσ∗ of X ∗ such that (5.37) has a solution if and only if (f i , g, ψ) ∈ (Wσ∗ )⊥ . Proof. We begin by setting H = W 1,2 (Ω) and noting that any x∗ ∈ H∗ can be represented by some (f, g, ψ) ∈ X so that Z Z x∗ (v) = −f i Di v + gv dx + ψv ds. Ω

∂Ω

Choosing σ0 to be a positive number such that σ0 > Cλ, where C is the constant from Lemma 5.25, we define the bilinear operator B by Z Z B(u, v) = L(u, v) + σ0 uv dx + σ0 uv ds. Ω

∂Ω

It follows from Lemma 5.25 that B is a bounded, coercive bilinear form on H and hence, for any (f, g, ψ) ∈ X, there is a unique function u ∈ W 1,2 satisfying (5.37) with σ0 in place of σ. We now define L0 : H∗ → H by L0 (x∗ ) = u, the solution of (5.37) with σ0 in place of σ and (f, g, ψ) representing x∗ , and observe that L0 is a well-defined continuous, one-toone mapping of H∗ onto H. Next, we define the imbedding I : H → H∗ by Z Z Iu(v) = uv dx + uv ds. Ω

∂Ω

It’s straightforward to check that I is compact (by virtue of Theorem 5.9) and that the equations (5.7) are equivalent to u + σ0 L0 Iu = L0 (f, g, ψ). Then σ0 L0 I is also compact and hence the Fredholm alternative (Theorem 2.29) implies the desired result, just as in the proof of Theorem 2.30. 

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From our general theory, we obtain one simple set of hypotheses which guarantees the unique solvability of (5.5). Corollary 5.27. Let Ω be a bounded Lipschitz domain, and suppose that there are positive constants λ and µ1 such that conditions (5.4) and (5.35) are satisfied. Suppose also that (5.20) holds for all nonnegative w ∈ C 1 (Ω) and that there is some w ∈ C 1 (Ω) such that Z Z i 0 b Di w − c w dx 6= β 0 w ds. (5.39) Ω

∂Ω

Then (5.5) has a unique solution for all f i and g in L2 (Ω) and all ψ ∈ L2 (Ω).

Proof. The proof of the weak maximum principle implies that the only solutions of Lu = 0 in Ω, M u = 0 on ∂Ω are constants, and condition (5.39) implies that zero is the only constant solution of this boundary value problem.  We also note that the case when bi , ci , c0 and β 0 all vanish is particularly simple because then the weak maximum principle implies that the only solutions of problem (5.38) with σ = 0 are constants and it’s immediate to check that any constant is a solution of this problem. Hence I0 = 1, and, by using the test function v = 1 in the weak form of the boundary value problem (5.5), we find that this problem is solvable only if Z Z g dx + ψ ds = 0. Ω

∂Ω

It follows that (5.5) is solvable if and only if this compatibility condition is satisfied. In particular, as mentioned in the introduction to this chapter, problem (5.1) is solvable if and only if (5.3) holds. 5.7

Higher regularity of solutions

We start with an interior regularity result, which is just Theorem 8.8 [64]. Because the method of proof will be important for our boundary estimate, we sketch the proof here. Theorem 5.28. Let Ω be a domain in Rn , and suppose that there is a positive constant λ such that (5.4) holds. Suppose also that aij and bi are Lipschitz and that ci and c0 are bounded. Let u ∈ W 1,2 be a solution of Lu = g in Ω with g ∈ L2 (Ω). Then, for any Ω0 ⊂ Ω with dist(Ω0 , ∂Ω) > 0,

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u ∈ W 2,2 (Ω0 ) and there is a constant C, determined only by n, dist(Ω0 , ∂Ω) and K, where K = |aij /λ|1 + |bi /λ|1 + kci /λk∞ + kc0 /λk∞ ,

(5.40)

kukW 2,2 (Ω0 ) ≤ C(kukW 1,2 (Ω) + kg/λk2 ).

(5.41)

aij Dij u + (bi + ci )Di u + (Di bi + c0 )u = g

(5.42)

such that

Moreover

a.e. in Ω. Proof. First we set g0 = (bi + ci )Di u + (Di bi + c0 )u − g and observe that u is a weak solution of the equation Di (aij Dj u) = g0 in Ω.

(5.43)

Now, let Ω0 be given, set d0 = dist(Ω0 , ∂Ω), and let Ω00 be a subdomain of Ω such that dist(Ω00 , ∂Ω) ≥ d0 /2 and Ω0 ⊂ Ω00 . Also we fix k ∈ {1, . . . , n} and h ∈ (−d0 /4, d0 /4). Then for any function v ∈ C01 (Ω00 ), we take ϕ = vk,h in the weak form of (5.43) to conclude that Z Z ij a (x + hek )Dj uk,h Di v dx = − (f i Di v + g0 v) dx, Ω

Ω

aij k,h Dj u.

where fi = From the definitions of f i and g0 , we infer via Lemma 5.4 that Z aij (x + hek )Dj uk,h Di v dx ≤ Cλ(kukW 1,2 + kg/λk2 )kDvk2 . Ω

In particular, if η ∈ C01 (Ω00 ) satisfies the conditions 0 ≤ η ≤ 1 and |Dη| ≤ 4/d0 in Ω00 and η = 1 on Ω0 , we have Z Z 2 λ |Duk,h | dx ≤ λ η 2 |Duk,h |2 dx Ω0 Ω Z ≤ η 2 aij (x + hek )Di uk,h Dj uk,h dx. Ω

2

With v = η uk,h , we also have η 2 Dj uk,h = Dj v − 2ηuk,h Dj η and hence Z λ |Duk,h |2 dx ≤ Cλ(kukW 1,2 + kg/λk2 )kDvk2 Ω0 Z −2 ηuk,h aij (x + hek )Di uh,k Dj η dx. Ω

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The Schwarz inequality implies that Z −2 ηuk,h aij (x + hek )Di uh,k Di η dx Ω

≤ µλ

Z

Ω

1/2 Z 1/2 2 2 η |Di uh,k | dx |Dη| |uh,k | dx , 2

2

Ω

and it follows from simple rearrangement and Lemma 5.4 that Z |Duk,h |2 dx ≤ C(kuk2W 1,2 + kf k2 ). Ω0

Sending h → 0 completes the proof of (5.41), and (5.42) follows from (5.6).  A slight modification of the proceeding proof gives the corresponding regularity up to the boundary. Theorem 5.29. Let Ω be a domain in Rn with ∂Ω ∈ C 2 , and suppose that there is a positive constant λ such that (5.4) holds. Suppose also that aij , bi and β 0 are Lipschitz and that ci and c0 are bounded. Let u ∈ W 1,2 be a solution of Lu = g in Ω and M u = ψ with g ∈ L2 (Ω) and ψ ∈ W 1,2 (∂Ω). Then u ∈ W 2,2 (Ω) and there is a constant C, determined only by n, K (the constant from (5.40)), and |β 0 /λ|1 , such that kukW 2,2 (Ω) ≤ C(kukW 1,2 (Ω) + kg/λk2 + kψ/λkW 1,2 (∂Ω) ). Proof. From Theorem 5.28, it suffices to prove a local W 2,2 estimate for u near the boundary. Moreover, by a change of independent variable, we only need to prove a bound for D2 u in the half-ball B + (R), assuming that Lu = g in B + (2R) and M u = ψ on B 0 (2R). Using the notation from the proof of Theorem 5.28 and setting Ψ = (β 0 + bn )u − ψ, we find that Z Z ij a (x + hek )Dj uk,h Di v dx = − (f i Di v + g0 vk,−h ) dx B + (2R) B + (2R) Z + Ψk,h v dx0 B 0 (2R)

if k < n and |h| < R. This time, we use the divergence theorem to conclude that Z Z Ψk,h v dx0 = Ψk,n Dn v − Dn ΨDn v dx, B 0 (2R)

B + (2R)

and then the argument of Theorem 5.28 implies that kDik uk2,B + (R) ≤ C(kukW 1,2 + kg/λk2 + kΨkW 1,2 )

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for i ≤ n and k < n. The proof is completed by noting that   X 1  Dnn u = nn g − (bi + ci )Di u + (Di bi + c0 )u + aij Dij u . a (i,j)6=(n,n)



We shall see in Chapter 6 that the hypotheses of these theorems can be relaxed considerably, but the techniques involved are quite different from those of the current chapter. 5.8

Global boundedness of weak solutions

We now show that weak solutions of (5.5) are bounded under appropriate conditions on the data of the problem. Although the arguments we shall use are based on a nonlinear structure (see Section 8.5 in [64] for details), we prefer to write everything in linear form here; the corresponding result for nonlinear problems will be presented in a later chapter. As was the case in Chapter 1, we are concerned with subsolutions and supersolutions of (5.5). We say that u is a weak subsolution of (5.5) if u ∈ W 1,2 and if Z [(aij Dj u + bi u − f i )Di ϕ + (−ci Di u − c0 u + g)ϕ] dx Ω Z (5.44) ≤ [β 0 u − ψ]ϕ ds ∂Ω

1

for all ϕ ∈ C (Ω). For brevity, we also write Lu ≥ Di f i + g in Ω,

M u ≥ ψ in ∂Ω.

Weak supersolutions are defined by reversing all these inequalities. Our first result is that subsolutions are bounded above. Before proving this result, we demonstrate an iteration argument, which is usually called Moser iteration. Lemma 5.30. Let (X, µ) be a measure space, let w ∈ Lp (µ) for all p ≥ 1 be nonnegative, and suppose that there are positive constants K, κ > 1, m ≥ 1, and s0 ≥ 1 such that Z 1/κ Z κs m w dµ ≤ Ks ws dµ (5.45) X

X

∞

for all s ≥ s0 . Then w ∈ L (µ) and there is a constant C, determined only by K, κ, m, and s0 , such that kwk∞ ≤ Ckwk1 .

(5.46)

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Proof.

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For brevity, we set I(q) = kwkqs0 , so (5.45) says that 1/s

I(κs) ≤ K1 sm1 /s I(s) m/s0

for s ≥ 1, where K1 = K 1/s0 s0 that

and m1 = m/s0 . By induction, it follows a(j) b(j)

I(κj ) ≤ K1

κ

I(1)

for any positive integer j, where a(j) =

j X

κ1−i ,

i=1

b(j) =

j X i=1

κ1−i (i − 1)m1 .

Since these are partial sums of convergent series, it follows that there is a constant σ, determined only by κ, so that a(j) ≤ σ and b(j) ≤ σm1 . It follows that w ∈ L∞ with kwk∞ ≤ (K1 κm1 )σ kwks0 .

The desired result follows from this inequality because 1/s0

0 −1)/s0 kwks0 ≤ kwk(s kwk1 ∞

.



Theorem 5.31. Let Ω be a bounded Lipschitz domain, let L and M satisfy (5.4) and (5.35), and suppose there is q > n such that f i ∈ Lq (Ω), g ∈ Lq/2 (Ω), and ψ ∈ Lq−1 (∂Ω). If u ∈ W 1,2 (Ω) is a weak subsolution of (5.5), then u+ ∈ L∞ (Ω). Moreover, if we set k = (kf kq + kgkq/2 + kψkq−1 )/λ, then there is a constant C, determined only by n, q, µ1 , and Ω such that sup u ≤ C(ku+ k2 + k).

(5.47)

Ω

Proof. We begin by proving a basic energy inequality. Let χ be a Lipschitz function defined on [k, ∞), and suppose that there is a nonnegative constant cχ such that 0 ≤ χ0 (s)(s − k) ≤ cχ χ(s)

(5.48)

for all s ≥ k. For notational simplicity, we define χ(s) = 0 if s < k. We then take ϕ = χ(u)(u − k)+ as test function and observe that ( (χ0 (u)(u − k) + χ(u))Du if u > k, Dϕ = 0 if u ≤ k. Hence aij Dj uDi ϕ ≥ χ|Du|2 wherever u > k. It follows that I0 ≤ I1 + I2 + I3 + I4 ,

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195

Z

χ|Du|2 dx,  Z  i 1 f i I1 = − b (χ0 (u − k) + χ)Di uu dx, λ Ω u Z 1 I2 = χci Di uu(u − k) dx, λ Ω Z  g 1 I3 = c0 − χu(u − k) dx, λ Ω u   Z ψ 1 0 β − χu(u − k) ds. I4 = λ ∂Ω u We first use (5.48) and the observation that q Z f − b dx ≤ λq (1 + µ1 )q u I0 =

Ω

{u>k}

to infer that

1/2

I1 ≤ (1 + µ1 )(1 + cχ )I0

Z

Ω

q−2)/2q [χu2 ]q/(q−2) dx .

To estimate the integral in this inequality, we adapt the proof of Theorem 5.8 slightly. We set r = 2n(q − 1)/(q − 2)(n − 1), A = 2(n − 1)/(q + n − 2), and s = r(n − 1)/n. (These correspond to N = q in the proof of Theorem 5.8.) We also set B = −rA/(1 − A). Then H¨older’s inequality implies that !1−A Z A Z  B Z k 2 q/(q−2) 2 r/2 2 [χu ] dx ≤ [χ(u − k) ] dx χu 1 − dx . u Ω Ω Ω Moreover, using the Sobolev imbedding theorem and another application of H¨ older’s inequality, we find that Z (n−1)/n 2 r/2 [χ(u − k) ] dx Ω

≤ C(n, Ω) ×

Z

Z

Ω

Ω

2 q/(q−2)

[χ(u − k) ]

1/2 dx

|D(χ1/2 (u − k))|2 dx +

Z

Ω

1/2 χ(u − k)2 dx .

Because of (5.48), we have |D(χ1/2 (u − k))| ≤ (1 + cχ )χ1/2 |Du| and hence (5.13) implies that Z (q−2)/q [χ(u − k)2 ]q/(q−2) dx Ω

Z (q−n)/q n/q Z 2 2 2 ≤ C(n, Ω, q) χ(u − k) dx (1 + cχ ) I0 + χ(u − k) dx . Ω

Ω

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It then follows, via Young’s inequality and the simple inequality (u − k)2 ≤ u2 (1 − k/u)B that  B Z 1 k (2q+n)/(q−n) 2 I1 ≤ I0 + C(1 + cχ ) χu 1 − dx. 4 u Ω Similar reasoning shows that

1 I2 + I3 + I4 ≤ I0 + C(1 + cχ )(2q+n)/(q−n) 4

 B k χu 1 − dx, u Ω

Z

2

and hence B  Z Z k dx. χ|Du|2 dx ≤ C(1 + cχ )(2q+n)/(q−n) χu2 1 − u Ω Ω

(5.49)

Now choose N ∈ (2, q) so that n ≥ N , set

BN B(N − 2) , β= −N 2 2 and, with s ≥ 1 and Z > k parameters at our disposal, set αs+β  k χ = 1− min{u, Z}2s−2. u + α=N−

It follows that (5.48) is satisfied with cχ = C(n, q)s. We now observe that  (αs+β+2)/2 k h= 1− min{u, Z}s−1 u u + satisfies the inequalities |Dh|2 ≤ Cs2 χ|Du|2 and  B k h ≤ χu 1 − , u 2

2

so we may apply the Gagliardo-Nirenberg inequality (5.14) along with (5.49) to obtain Z 1/κ  B Z k h2κ dx ≤ Cs(2q+n)/(q−n) χu2 1 − dx, u Ω Ω

where κ = N/(N − 2). Since κ(β + 2) = β + B and min{u, Z} ≤ u, it follows that !1/κ Z ακs+β+B 1/κ Z  k u2 2κs 2κ 1− min{u, Z} dx ≤ h dx u + min{u, Z}2 Ω Ω

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and Z

Ω

χu

2



k 1− u

B

197

αs+β+B Z  k u2 dx = 1− min{u, Z}2s dx. u + min{u, Z}2 Ω

We now set m = (2q + n)/(q − n) and α  k min{u, Z}2, w = 1− u + and we define the measure µ by  B−β k u2 dµ = 1 − dx. u + min{u, Z}2 In this way, we conclude that Z 1/κ Z wκs dµ ≤ Csm ws dµ. Ω

Ω

It then follows from Lemma 5.30 that Z Z sup w ≤ C w dµ = C Ω

u2 dx,

Ω(k)

where Ω(k) is the subset of Ω on which u ≥ k. Since this inequality is true for all Z > k, it follows that  α Z k sup 1 − u2 ≤ C u2 dx, u + Ω(k) and the desired result is an immediate consequence of this inequality.



Replacing u by −u gives the corresponding result for supersolutions. Corollary 5.32. Suppose all the hypotheses of Theorem 5.31 are satisfied except that u is a supersolution of (5.5) rather than a subsolution. Then u− ∈ L∞ and sup(−u) ≤ C(ku− k2 + k).

(5.50)

Ω

In general, we can’t expect a global bound for u (independent of the L2 norm) without some restrictions on bi , c0 , and β 0 . For example, if bi , c0 , and β 0 are all zero, then any constant is a solution of Lu = 0 in Ω, M u = 0 on ∂Ω. We now provide some sufficient conditions for a global bound. Our first set of conditions is based on those in Theorem 5.15. Theorem 5.33. Let Ω be a bounded Lipschitz domain, let L and M satisfy (5.4) and (5.35), and suppose there is q > n such that f i ∈ Lq (Ω), g ∈

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Lq/2 (Ω), and ψ ∈ Lq−1 (∂Ω). Suppose also that there is a positive constant ε such that Z Z (c0 + ε)w − bi Di w dx + β 0 w ds ≤ 0 (5.51) Ω

∂Ω

1

for all nonnegative w ∈ C (Ω). Then there is a constant C, determined only by µ1 , ε, Ω, and q such that any weak subsolution u ∈ W 1,2 (Ω) of (5.5) satisfies the estimate sup u ≤ C(kf kq + kgkq/2 + kψkq−1 )/λ.

(5.52)

First, we infer from the proof of Theorem 5.15 that Z aij Dj uDi v dx + λε uv dx Ω Ω Z Z Z Z i i i ≤ (b + c )Di uv dx + f Di v dx + gv dx +

Proof. Z

Ω

Ω

Ω

ψv ds

∂Ω

for all nonnegative v ∈ W 1,2 (Ω) with uv ≥ 0 in Ω. In particular, we can take v = us− with s > 1 at our disposal. We now take k as in Theorem 5.31 and use that theorem to conclude that U = sup u+ is finite. The integrals in this inequality are now easily estimated. First, Z Z aij Dj uDi v dx ≥ λs |Du|2 us−1 + dx. Ω

Ω

Then

Z s s+1 (bi + ci )Di uv dx ≤ λ|Du|2 us−1 , + dx + CλU 2 Ω Ω Z Z s 2 s−1 λ|Du|2 us−1 , f i Di v dx ≤ + dx + Csk λU 2 Ω Ω Z gv dx ≤ CkU s , Ω Z ψv ds ≤ CkU s .

Z

∂Ω

It follows that

C s+1 U + CkU s + Ck 2 U s−1 . s Ω We now use Theorem 5.31 to infer that ε

Z

us+1 + dx ≤

U ≤ C(ku+ k2 + k), and then (5.12) implies that U

s+1

Z s+1 ≤ C( us+1 ). + +k Ω

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Therefore εU s+1 ≤

C s+1 U + CkU s + Ck 2 U s−1 + Ck s+1 . s

We now choose s so that C/s ≤ ε/2 and then use Young’s inequality to complete the proof.  We can also obtain a global bound if we assume that c0 is sufficiently negative in place of (5.51). Theorem 5.34. Let Ω be a bounded Lipschitz domain, let L and M satisfy (5.4) and (5.35), and suppose there is q > n such that f i ∈ Lq (Ω), g ∈ Lq/2 (Ω), and ψ ∈ Lq−1 (∂Ω). Suppose also that there is a positive constant µ ¯ such that kb − ck2 ≤ µ ¯λ,

kβ 0 k1 ≤ µ ¯2 λ.

(5.53)

Then there are positive constants C0 determined only by µ1 , Ω, and q, and C1 , determined also by µ ¯, such that, if c0 ≤ −C0 µ ¯2 λ,

(5.54)

then any weak subsolution u ∈ W 1,2 (Ω) of (5.5) satisfies the estimate sup u ≤ C1 (kf kq + kgkq/2 + kψkq−1 )/λ.

(5.55)

Now we use the test function v = u+ to see that Z ij 2 a Di vDj v dx + C0 µ ¯ λ v 2 dx Ω Ω Z ≤ (bi − ci )Di vv dx ZΩ Z Z + f i Di v dx + gv dx + β 0 v 2 − ψv ds.

Proof. Z

Ω

Ω

∂Ω

With U = sup v, we find from H¨older’s inequality that Z 1 kψk1 + kg 1 C0 µ ¯2 v 2 dx ≤ 2¯ µ2 U 2 + kf k22 /λ2 + U, 2 λ Ω and Theorem 5.31 gives a constant C2 such that Z  2 2 2 U ≤ C2 v dx + k . Ω

The proof is completed by taking C0 = 1/2C2 .



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Finally, we present a result if β 0 is large and negative, rather than c0 . Theorem 5.35. Let Ω be a bounded Lipschitz domain, let L and M satisfy (5.4) and (5.35), and suppose there is q > n such that f i ∈ Lq (Ω), g ∈ Lq/2 (Ω), and ψ ∈ Lq−1 (∂Ω). Suppose also that there is a positive constant µ ¯ such that kb − ck2 ≤ µ ¯λ,

kc0 k1 ≤ µ ¯2 λ.

(5.56)

Then there are positive constants C0 determined only by µ1 , Ω, and q, and C1 , determined also by µ ¯, such that, if β 0 ≤ −C0 max{¯ µ2 , µ ¯4 }λ,

(5.57)

then any weak subsolution u ∈ W 1,2 (Ω) of (5.5) satisfies the estimate (5.55). Proof.

We again use the test function v = u+ , and this time we obtain Z Z ij 2 4 a Di vDj v dx + C0 max{¯ µ ,µ ¯ }λ v 2 ds Ω ∂Ω Z Z ≤ (bi − ci )Di vv dx + c0 v 2 dx Ω ZΩ Z Z i + f Di v dx + gv dx − ψv ds, Ω

Ω

∂Ω

and hence Z Z 1 2 4 2 |Dv| dx + C0 max{¯ µ ,µ ¯ } v 2 ds ≤ 2¯ µ2 U 2 + Ck 2 + CkU. 2 Ω ∂Ω Now Corollary 5.23 gives a constant C3 > 1, determined only by Ω, such that Z Z Z C3 v 2 dx ≤ θ |Dv|2 dx + v 2 ds θ Ω Ω ∂Ω ¯ −2 } to infer for any θ ∈ (0, 1). We now take C0 ≥ 4C3 and use θ = min{1, µ that Z C0 2 µ ¯ v 2 dx ≤ 2¯ µ2 U 2 + Ck 2 + CkU 2C3 Ω and the proof is completed as before.



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5.9

201

The local maximum principle

The preceding maximum estimate also has a local analog similar to Theorem 1.27. We state our results under the hypotheses of Lemma 5.6, that is, given a real-valued Lipschitz function ω defined on Bn−1 (R) with Lipschitz constant ω0 , we define Ω[R] and σ[R] by (5.11) and Σ[R] by (5.29). We say that u ∈ W 1,2 (Ω[R]) is a weak subsolution of Lu = Di f i + g in Ω[R],

M u = ψ on Σ[R]

(5.58)

if (5.44) is satisfied (with Ω replaced by Ω[R] and ∂Ω replaced by Σ[R]) for all ϕ ∈ C 1 (Ω[R]) which vanish on σ[R]. We rewrite condition (5.35) in a way that will be convenient for the form of our estimates. We assume that there are nonnegative constants q > n, µ, µ2 , and µ3 such that aij ξi ζj ≤ µλ|ξ||ζ|

(5.59a)

kbi kq + kci kq ≤ µ2 λ,

(5.59b)

µ22 λ,

(5.59c)

kβ0 kq−1 ≤ µ3 λ.

(5.59d)

for all ξ and ζ in Rn ,

0

kc kq/2 ≤

We also set k(R) =

 1  1−n/q R kf kq + R2(1−n/q) kgkq/2 + R(q−n)/(q−1) kψkq−1 . λ

Theorem 5.36. Let Ω be a bounded Lipschitz domain, let L and M satisfy (5.4) and (5.59), and suppose that f i ∈ Lq (Ω), g ∈ Lq/2 (Ω), and ψ ∈ Lq−1 (∂Ω). Let u be a weak subsolution of (5.58). Then, for any p > 0, there is a constant C, determined only by µ, µ2 R1−n/q , µ3 R(q−n)/(q−1) , n, p, q, and ω0 , such that sup u ≤ C(R−n/p ku+ kp,Ω[2R] + k(R)).

(5.60)

Ω[R]

Proof. We imitate the proof of Theorem 5.31 with one important change. With χ as in the proof of that theorem and ζ a nonnegative C 1 function which vanishes on σ[R], we use the test function ϕ = χ(u)(u − k(R))+ ζ 2 . It follows that I0 ≤ I1 + I2 + I3 + I4 + I5 ,

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where I0 =

Z

χ|Du|2 ζ 2 dx,  Z  i 1 f i − b (χ0 (u − k) + χ)ζ 2 Di uu dx, λ Ω u Z 1 χci Di uu(u − k)ζ 2 dx, λ Ω Z  g 1 c0 − χu(u − k)ζ 2 dx, λ Ω u  Z  1 ψ β0 − χu(u − k)ζ 2 ds, λ ∂Ω u  Z  i 2 f − bi χ(u − k)+ ζDi ζ dx, λ Ω u Z 2 aij Dj uDi ζζχ(u − k) dx. λ Ω Ω

I1 = I2 = I3 = I4 = I5 = I6 =

These integrals are estimated in a similar fashion to that used in Theorem 5.31 except for the use of Lemma 5.6 in place of Theorem 5.5. For example, we have !(q−2)/q Z Ω[R]

[χ(u − k)2 ζ 2 ]q/(q−2) dx

≤C

Z

Ω[R]

2 2

!(q−n)/q

Z

χ(u − k) ζ dx

Ω[R]

|D(χ

1/2

2

!n/q

(u − k)ζ)| dx

,

and straightforward calculation shows that |D(χ1/2 )(u − k)ζ| ≤ |χ(u − k)||Dζ| + (1 + cχ )χ1/2 ζ|Du|. Our energy inequality now becomes Z χ|Du|2 ζ 2 dx Ω[R]

≤ C(1 + cχ )(2q+n)/(q−n) R−2 Z + C(1 + cχ )(2q+n)/(q−n)

 B k χu2 1 − ζ 2 dx u Ω[R]

Z

Ω[R]

χ(u − k)2 |Dζ|2 dx

with B=−

4n(q − 1) (q − 2)(q − n)

(5.61)

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as before. To continue, we fix a cut-off function η with |Dη| ≤ C/R (for example, η = (1 − |x|2 /R2 )+ ), we set (2 − B)N B(N − 2) α= , β= − N, p 2 and, with s ≥ 1 and Z > k constants at our disposal, we take αs+β  k min{u, Z}2s−2 , ζ = η (αs+β)/2 . χ= 1− u + Using

 (αs+β+2)/2 k h= 1− min{u, Z}s−1 η (αs+β+2)/2 u +

in the Gagliardo-Nirenberg inequality, we find that  αp k w = 1− η αp min{u, Z}p u +

satisfies the inequality (5.45) with m = (4q − n)/(q − n) and µ defined by  β+B k u2 −n dµ = R 1− dx u + min{u, Z}2

and s0 sufficiently large, depending on p. Again, Moser iteration and sending Z → ∞ completes the proof.  5.10

The DeGiorgi class

In this section, we discuss a special subset of W 1,p (Ω[R]) that will be very useful in the next few sections and also when we study local regularity of weak solutions of nonlinear problems. Let S be a measurable subset of Rn with positive measure, let u ∈ W 1,p (S) for some p ≥ 1, and set A(k, S) = {x ∈ S : u(x) < k}

for k ∈ R. For nonnegative constants χ and K and ε ∈ (max{0, (p − n)/np}, 1/n], we then define DG− (p, S, χ, ε, K) to be the set of all nonnegative u ∈ W 1,p (S) such that Z (u − k)− |Du|p dx ≤ K p sup S∩B(r) (1 − σ)r A(k,S∩B(σr)) (5.62) !p + kr−nε |A(k))|ε−1/n

|A(k)|

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for all balls B(r) with center in S and all k ≥ χrnε , and we have abbreviated A(k, S ∩ B(r)) to A(k). Our first step is to estimate the size of the set on which u is larger than some constant in some ball if we know that u is larger than a given constant on some fraction of a given, smaller ball. Lemma 5.37. Let p ≥ 1. For any constants θ1 and θ2 in (0, 1), there is a positive integer s, determined also by K, p, and ε such that, for any u ∈ DG− (p, Ω[2R], χ, ε, K) with µ = inf u ≤ 2−s h,

(5.63a)

χ(2R)nε ≤ 2−s h,

(5.63b)

|A(µ + χ(2R)nε + 2−s h, Ω[R])| ≤ θ2 |Ω[R]|.

(5.64)

Ω[2R]

nε

|A(µ + χ(2R)

+ h, Ω[2R])| ≤ (1 − θ1 )|Ω[2R]|

(5.63c)

for some h > 0, we have

Proof.

With s to be determined, we define kj = µ + χ(2R)nε + 2−j h

for j ∈ {1, . . . , s}. From (5.62) with σ = 12 , r = 2R, and k replaced by kj , we infer that Z |Du|p dx A(kj ,Ω[R])

≤K

p

(u − kj )− sup + kj (2R)−nε |A(kj )|ε−1/n R Ω[2R]

!p

|A(kj )|

≤ (c(n, ε)K)p kjp Rn−p

for j = 1, . . . , s since |A(kj )| ≤ 2n ωn Rn and (u − kj )− ≤ kj . For ease of notation, we now set Aj = A(kj , Ω[R]) and ∆j = A(kj , Ω[R])\A(kj+1 , Ω[R]) and conclude that !1/p Z Z (p−1)/p p |Du| dx ≤ |∆j | |Du| dx ∆j

Aj

by H¨ older’s inequality. Combining these inequalities gives Z |Du| dx ≤ |∆j |(p−1)/p c(n, ε)R(n−p)/p Kkj , ∆j

and then Lemma 5.14 implies that (kj − kj+1 )|Aj+1 |(n−1)/n ≤ c(n, ε, K, θ1 )|∆j |(p−1)/p kj R(n−p)/p .

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We now rewrite this inequality. First, it follows from (5.63a) and (5.63b) that kj ≤ 2−j+1 h, while kj − kj+1 = 2−j−1 h, so |Aj+1 |(n−1)/n ≤ c(n, ε, K, θ1 )|∆j |(p−1)/p R(n−p)/p . Next, we set Xj = |Aj |/|Ω[R]|, Yj = |∆j |/|Ω[R]|, and η=

np − p np − n

to rewrite our estimate as η Xj+1 ≤ CYj .

Summing this inequality on j from 1 to s − 1 then yields (s − 1)Xsη ≤ C

s−1 X j=1

Yy ≤ C

because the ∆j ’s are disjoint subsets of Ω[R] and Aj+1 ⊂ Aj . Noting that η > 0, we conclude that  1/η C Xs ≤ s−1 and the proof is completed by choosing s so that the right hand side of this inequality is no greater than θ2 .  To proceed, we need a simple iteration lemma. Lemma 5.38. Let (Yj )∞ 0 be a sequence of nonnegative numbers and suppose that there are positive constants C, η, and b with b > 1 such that Yj+1 ≤ Cbj Yj1+η

(5.65)

for all j. If, in addition, 2

Y0 ≤ C −1/η b−1/η ,

(5.66)

then limj→∞ Yj = 0. Proof.

It’s easy to check by induction that 2

Yj ≤ C −1/η b−j/η b−1/η , so Yj → 0 as j → ∞.



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Using this iteration lemma, we now show that, if u is larger than a suitable positive constant on most of Ω[2R], then it is larger than a corresponding positive constant on all of Ω[R]. Lemma 5.39. Let p ≥ 1. Then there is a constant θ∗ ∈ (0, 1) such that, if u ∈ DG− (p, Ω[2R], χ, ε, K) and h is some positive constant such that |A(µ + χ(2R)nε + h, Ω[2R])| ≤ θ∗ |Ω[2R]|, nε

µ + χ(2R)

then u ≥ µ + χ(2R)nε + 12 h on Ω[R]. Proof.

≤ h,

(5.67a) (5.67b)

For each nonpositive integer j, we set ρj = (1 + 2−j )R,

kj = µ + χ(2R)nε +

h (1 + 2−j ), 2

and we define the sets Aj = A(kj , Ω[ρj ]),

∆j = A(kj , ω[ρj+1 ]) \ Aj+1 .

We then apply (5.62) for each j with σj = ρj+1 /ρj in place of σ, ρj in place of r, and kj in place of k. Setting Yj = |Aj |/|Ω[2R]| and noting that 1 − σj ≥ 2−j−1 , Yj ≤ 1, and sup (u − kj )− ≤ kj ,

Ω[ρj ]

we conclude that Z

1+pε−p/n

∆j

|Du|p dx ≤ Ckjp 2jp Yj

Rn−p .

and hence, from H¨ older’s inequality and the inequality |∆j | ≤ CYj Rn , we have Z 1+ε−1/n n−1 |Du| dx ≤ Ckj 2j Yj R . ∆j

Now we use Lemma 5.14 along with the inequality kj ≤ 2h and the equality k + j − kj+1 = 2−j−1 to infer that Z (n−1)/n n−1 |Du| dx. 2j+2 kj Yj+1 R ≤C ∆j

(n−1)/n

ε+(n−1)/n

It follows that Yj+1 ≤ C4j Yj . We now invoke Lemma 5.38 n/(n−1) with b = 4 and η = εn/(n − 1) to see that there is a constant θ ∈ (0, 1) such that Yj → 0 as j → ∞ if Y0 ≤ θ. We take θ∗ = min{1/2, θ}, so (5.67a) implies that Y0 ≤ θ. The proof is completed by observing that Yj → ∞ implies that u ≥ µ + χ(2R)nε + h/2 in Ω[R]. 

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Our final step is to prove an inequality which is the analog of Lemma 1.19. We begin by recalling some notation from Section 1.7. The Lipschitz cube K(y0 , r) is the set of all points x such that n

|xi − y0i | < r if i < n,

|x −

y0n

+

ω(y00 )

0

− ω(x )| < r.

(5.68a) (5.68b)

∗

For our preliminary results, we write K (y0 , r) for the subset of K(y0 , r) on which xn > ω(x0 ). Lemma 5.40. Let p > 1, and let K, χ and ε be nonnegative constants with 0 < ε ≤ 1/n. There are constants C2 (determined only by K, n, ω0 , ε) and k1 (determined only by ω0 and n) such that if y0 ∈ Ω[R] and K(y0 , 3r) ⊂ B(x0 , k1 R), u ∈ DG− (p, Ω[R], χ, ε, K), and 1 |{x ∈ K ∗ (y0 , r) : u¯(x) < h}| ≤ |K ∗ (y0 , r)|, (5.69) 2 then C2 u ¯ ≥ h in K ∗ (y0 , 3r). Proof. As a first step, we observe that there is a constant k2 determined only by ω0 and n such that K(y0 , r) ⊂ B(y0 , k2 r). We then take k1 = k2 /12. Under our hypotheses, we then have that Ω[y0 , 4k1 r] ⊂ Ω[R]. It follows that there is a constant θ1 ∈ (0, 1), determined only by ω0 and n such that 1 ∗ |K (y0 , r)| ≥ θ1 |Ω[y0 , 4k1 r]| 2 and hence |{x ∈ Ω[y0 , 4k1 r] : u ¯(x) ≤ h}| ≤ (1 − θ1 )|Ω[y0 , 4k1 r]|.

We use s to denote the constant from Lemma 5.37 corresponding to θ2 = θ∗ , where θ∗ is the constant from Lemma 5.39. Also we set µ=

inf

Ω[y0 ,4k1 r]

u.

If h ≤ 2s+1 µ, then it follows that u ¯ ≥ u ≥ 2s+1 h in Ω[y0 k1 r]. If s+1 nε nε h ≤ 2 ((4k1 ) + 1)χr , we conclude that 1 u ¯≥ h (4k1 )nε + 1 in Ω[y0 , k1 r]. Hence we may assume that h > 2s+1 max{µ, χ((4k1 )nε + 10rnε }.

It then follows from Lemma 5.37 with h replaced by (1 − 2−s−1 )h that |{x ∈ Ω[y0 , 2k1 r] : u ¯ < 2−s−1 h}| ≤ θ∗ |Ω[y0 , 2k1 r]|.

and the proof is completed by invoking Lemma 5.39 with h replaced by 2−s−1 and R replaced by k1 r. 

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Although this lemma, along with the arguments in Chapter 1, immediately yields results for functions in the De Giorgi class, we shall not state the results here. Instead, we apply them in the next several sections to weak solutions of the differential equations. Of course, the applications in this chapter only need the case p = 2, but the more general case will be very useful in studying nonlinear problems, and it requires no additional effort.

5.11

Membership of supersolutions in the De Giorgi class

We now show that positive weak supersolutions of (5.1) are in a suitable De Giorgi class provided the coefficients of L and M belong to certain Lebesgue spaces. Specifically, we have the following result. Theorem 5.41. Suppose that there are positive constants q > n, λ, µ, µ2 , and µ3 such that conditions (5.4) and (5.59) are satisfied. Suppose also that f ∈ Lq , g ∈ Lq/2 and ψ ∈ Lq−1 . Then any positive supersolution of (5.58) is in DG− (p, Ω[R/2], χ, ε, K) with χ = kf kq + kgkq/2 R1−n/q + kψkq−1 R(q−n)/q(q−1) ,

ε = (q − n)/nq and K determined only by n, q, ω0 , µ2 R1−n/q , and µ3 R(q−n)/(q−1) . Proof. Let r > 0, σ ∈ (0, 1), x0 ∈ Ω. We then use ϕ = (u − k)− η 2 as test function in the weak form of the inequalities for u, where η is Lipschitz, with 0 ≤ η ≤ 1 in B(x0 , r) and η = 0. Writing w = (u − k)− , we obtain I1 ≤ where

I3 I4 I5

Im ,

m=2

aij Di wDj wη 2 dx, Z = − 2aij Di wDj ηηw dx, Z = (bi u − f i )Di wη 2 dx, Z = 2(bi u − f i )Di ηηw dx, Z = ci Di wwη 2 dx,

I1 = I2

Z

7 X

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Z I6 = − (c0 u − g)wη 2 dx, Z I7 = (β 0 u − ψ)wη 2 ds. ∂Ω

We now proceed to estimate these integrals. First is the easy one: Z I1 ≥ λ |Dw|2 η 2 dx.

Next, from the Cauchy-Schwarz inequality, we have Z 1 2 w2 |Dη|2 dx. I2 ≤ I1 + 8µ 8 The Cauchy-Schwarz inequality along with H¨older’s inequality then yields 1 I3 ≤ I1 + 2λA1−2/q [k 2 µ21 + χ2 ], 8Z I4 ≤ λ

w2 |Dη|2 dx + λA1−2/q [k 2 µ21 + χ2 ],

1 I0 + 2λk 2 A1−2/q µ21 , 8 where we use A to denote the measure of the support of ϕ, and we have used the inequality 0 ≤ u ≤ k wherever ϕ > 0. In a similar vein (but only using H¨ older’s inequality), we also find that I5 ≤

I6 ≤ λA1−2/q [k 2 µ22 + kχR(n−q)/q ]. To estimate I7 , we imitate the proof of Lemma 5.20. First, H¨older’s inequality implies that !(q−2)/(q−1) Z (wη 2 )(q−1)/(1−2) ds I7 ≤ [kµ3 + χR(n−q)/q(q−1) ] . Σ[R]

and (5.30) (with q = p = 1) implies that Z Z 2 (q−1)/(1−2) (wη ) ds ≤ C(ω0 , q) Σ[R]

Ω[R]

(wη 2 )1/(q−2) |D(wη 2 )| dx.

We now apply the Cauchy-Schwarz inequality and note that |D(wη 2 )|2 ≤ 4η 2 |Dw|2 + 4w2 |Dη|2 to infer that Z Z 2 1/(q−2) 2 (wη ) |D(wη )| dx ≤ 2(I1 + w2 |Dη|2 dx)1/2 k 1/(q−2) A1/2 , Ω[R]

Ω[R]

and hence Z  1 2 2 (n−q)/q(q−1) 2−2/q 2/q I7 ≤ I1 + C(ω0 , q)λ w |Dη| dx + (µ3 k + χR ) k . 8

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Combining all these inequalities gives us Z  Z |Dw|2 η 2 dx ≤ C w2 |Dη|2 dx + k 2 r−nε A1−2/q with C determined only by ω0 , q, µ2 R1−n/q , µ3 R(q−n)/(q−1) , and the proof is completed by taking   if |x − x0 | ≤ σr,  1 1−|x−x0 |/r η(x) = if σr < |x − x0 | ≤ r, 1−σ   0 if |x − x0 | > r.  5.12

Consequences of the local estimates

Lemma 5.40 leads to several important consequences. We begin with a weak Harnack inequality, which follows from that lemma via the argument of Theorem 1.20. Theorem 5.42. Let Ω be a bounded Lipschitz domain, let L and M satisfy (5.4) and (5.59), and suppose that f i ∈ Lq (Ω), g ∈ Lq/2 (Ω), and ψ ∈ Lq−1 (∂Ω). Let u be a nonnegative weak supersolution of Lu = Di f i + g in Ω[4R],

M u = ψ on Σ[4R].

Then there are positive constants C and p, determined only by µ, µ2 R1−n/q , µ3 R(q−n)/(q−1) , n, and Ω, such that !1/p Z R−n

Ω[2R]

|u|p dx

≤ C( inf u + k(R)). Ω[R]

(5.70)

In fact, (5.70) is true for any p ∈ (0, n/(n − 2)) if n > 2 and any p > 0 if n = 2. Since this stronger result will not be used in this book, we defer the proof to Exercise 5.1. Next, by arguing exactly as in Corollary 1.22, we obtain the following strong maximum principle, which may also be deduced in several other ways. We sketch some of these methods in Exercise 5.2. Theorem 5.43. Let Ω be a bounded Lipschitz domain, and let L and M satisfy (5.4), (5.59), and (5.20). Suppose u ∈ W 1,2 (Ω) satisfies Lu ≥ 0 in Ω and M u ≥ 0 on ∂Ω. If sup u ≥ 0, then u is constant.

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Combining Theorems 5.36 and 5.42 gives a full Harnack inequality. Theorem 5.44. Let Ω be a bounded Lipschitz domain, let L and M satisfy (5.4) and (5.59), and let q > n. Then there is a constant C, determined only by µ, µ2 R, µ3 R2 , n, p, and Ω, such that sup u ≤ C( inf u + k(R))

Ω[R]

Ω[R]

for any nonnegative weak solution u of (5.5) with f i ∈ Lq (Ω), g ∈ Lq/2 (Ω), and ψ ∈ W 1−1/q,q (∂Ω). Also, the proof of Theorem 1.26 (with Theorem 5.42 in place of Theorem 1.20) gives the following H¨ older estimate for weak solutions of (5.5). Theorem 5.45. Let Ω be a bounded Lipschitz domain, let L and M satisfy (5.4) and (5.59), and suppose there is q > n such that f i ∈ Lq (Ω), g ∈ Lq/2 (Ω), and ψ ∈ W 1−1/q,q (∂Ω). If Lu = Di f i + g in Ω[R] and M u = ψ on Σ[R], then there are constants C, α and δ determined only by µ, µ2 R, µ3 R2 , n, p, and Ω, such that ! rα osc u ≤ C sup u + k(R) . (5.71) Rα Ω[R] Ω[r] For solutions of the Dirichlet problem, a similar result is true. Theorem 5.46. Let Ω be a bounded Lipschitz domain, let L satisfy (5.4), and suppose there is q > n such that f i ∈ Lq (Ω) and g ∈ Lq/2 (Ω). If Lu = Di f i + g in Ω[R] and u = ψ on Σ[R], then there are constants C, α and δ determined only by µ, µ2 R, µ3 R2 , n, p, and Ω, such that ! rα osc u ≤ C sup u + k(R) + osc ψ . (5.72) Rα Ω[R] Σ[R] Ω[r]

5.13

Integral characterizations of H¨ older spaces

We now show that certain integral inequalities for functions imply that these functions are in certain H¨older spaces. To this end, we first recall the definition of the Campanato spaces L1,q for q ≥ 0. We say that u ∈ L1,q (Ω) for some open set Ω if u ∈ L1 (Ω) and if there is a constant H such that Z inf |u(x) − L| dx ≤ Hrq L∈R

B(y,r)∩Ω

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for all r ∈ (0, diam Ω) and all y ∈ Ω. The smallest such constant H is called the L1,q seminorm of u and is written [u]L1,q . Note that adding a constant to u gives another function with the same L1,q seminorm, so we define a norm on L1,q by kuk1,q = [u]L1,q + kuk1 .

It’s easy to check that u must be a constant if u ∈ L1,q with q > n + 1 and that C α (Ω) ⊂ L1,n+α if α ∈ (0, 1]. In fact, for most domains of interest (and, in particular, all bounded Lipschitz domains), these spaces are the same. For our purposes, we shall prove a slightly weaker result first. Lemma 5.47. Let f ∈ L1 (B(x0 , 2R)) for some x0 ∈ R and R > 0. If there are positive constants α ≤ 1 and H and a function g(y, r) defined on B(x0 , R) × (0, R) such that Z |f (x) − g(y, r)| dx ≤ Hrn+α (5.73) B(y,r)

for all r ∈ (0, R) and all y ∈ B(x0 , R), then f ∈ C α (B(x0 , R)) and [f ]α;B(x0 ,R) ≤ c(n, α)H.

Proof. The proof is essentially the same as for Lemma 2.8. As in that lemma, we let ϕ be a nonnegative C 1 (Rn ) function with support in B(0, 1) such that Z ϕ(x) dx = 1, Rn

and set K = sup |Dϕ|. For (x, τ ) ∈ B(x0 , R)×(0, R), we define the function f¯ by Z ¯ f (x, τ ) = f (x − τ )ϕ(z) dz. (5.74) Rn

We then have that   Z ∂ f¯ x−w −n−1 (x, τ ) = τ [f (w) − g(x, τ )]Dϕ dw, ∂x τ B(x,τ ) and a similar expression for ∂ f¯/∂τ . Writing ∇f¯ for the (n + 1)-dimensional vector

we conclude that

 ¯  ∂ f ∂ f¯ , , ∂x ∂τ |∇f¯(x, τ )| ≤ (n + 1)HKτ α−1 ,

and the proof is completed exactly as for Lemma 2.8.



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A similar result is true near the boundary of the domain provided a suitable assumption is made about the regularity of the boundary. Here we only consider Lipschitz boundaries, but the result is actually true under a weaker assumption on the regularity of the boundary. Lemma 5.48. Let Ω be a bounded domain in Rn and, for any x0 ∈ Rn and R > 0, set Ω[x0 , R] = Ω ∩ B(x0 , R). Let x0 ∈ ∂Ω and R > 0, and suppose there is a function ω such that Ω[x0 , 2R] = {(x0 , xn ) : xn > ω(x0 ), |x| < 2R}

x01

x02

|ω(x01 )

−

n−1

ω(x02 )|

≤

|x00

ω0 |x01

x0i |

−

x02 |

(5.75a) (5.75b)

for any and in R with − < R for i = 1, 2. Let f ∈ L1 (Ω[x0 , 2R]) and suppose that there are positive constants α ≤ 1 and H and a function g defined on Σ[x0 , R] × (0, R), where Σ[x0 , R] = ∂Ω ∩ B(x0 , R), so that Z |f (x) − g(y, r)| dx ≤ Hrn+α . (5.76) Ω[y,r]

Then [f ]α;Σ[x0 ,R] ≤ c(n, α, ω0 )H. Proof. The proof is essentially the same as for Lemma 5.47 but more care must be taken on some technical points. First, we let ϕ0 be a nonnegative C 1 (Rn ) function supported in B(0, 1) R and Rn ϕ0 (x) dx = 1, we set κ = (1 + ω02 )1/2 , and we define ϕ by ϕ(x) = κn ϕ0 (κ(2x − en )),

where en = (0, . . . , 0, 1). ItRfollows that ϕ is a nonnegative C 1 (Rn ) function supported in B(0, 1) and Rn ϕ(x) dx = 1. Moreover if ϕ(z) 6= 0, if y ∈ Σ[x0 , R], and if r ∈ (0, R), then y + rz ∈ Ω[x0 , R]. Because ∂Ω is only Lipschitz, it no longer makes sense to talk about f¯ (defined by (5.74)) being continuously differentiable. Instead, we use a slightly different expression, which will be continuously differentiable with respect to r but only Lipschitz with respect to y. To write everything conveniently, we first write B 0 (x00 , R) for the ball in Rn−1 with center x00 and radius R. We define f¯ on B 0 (x00 , R) × (0, R) by Z 0 ¯ f (x , τ ) = f ((x0 , ω(x0 )) − τ z)ϕ(z) dz, Rn

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so f¯ is Lipschitz with respect to x0 and continuously differentiable with respect to τ (for τ 6= 0. It follows now that Z ∂ f¯ 0 1 (x , τ ) = − f ((x0 , ω(x0 )) − τ z)[Di ϕ(z) + Di ω(x0 )Dn ϕ(z)] dz ∂xi τ Rn (for i = 1, . . . , n − 1) almost everywhere with respect to x because ω is almost everywhere differentiable. It follows that ∂ f¯ 0 α−1 , ∂xi (y , τ ) ≤ c(n, ω0 )HKτ and then the proof is completed as before.

5.14



Schauder estimates

In this section, we prove some H¨older estimates for the derivatives of weak solutions of conormal problems. Clearly, we shall need stronger hypotheses on the coefficients than just boundedness. Conveniently, H¨older continuity of (some) coefficients implies H¨older continuity of the gradient of the solution with the same exponent. We begin with an improved version of the regularity results from Section 5.7 for problems with constant coefficients. Lemma 5.49. Let [Aij ] be a constant, positive definite matrix, let F be a constant vector, and suppose v is a solution of Di (Aij Dj v) = 0 in B + (0, R),

Anj Dj v = 0 on B 0 (0, R).

(5.77)

Then v ∈ C 2 (B + (0, R/2)). Proof. We begin by showing that D1 v, . . . , Dn−1 v are uniformly H¨older continuous in B + (0, 3R/4). To this end, we fix k ∈ {1, . . . , n − 1} and let h ∈ (0, R/8) be a parameter at our disposal. We then define vh in B + (7R/8) by vh (x) =

v(x + hek ) − v(x) , h

where ek is the vector with kth component equal to one and all other components equal to zero. Then vh satisfies the conditions Di (Aij Dj vh ) = 0 in B + (0, 7R/8),

Anj Dj vh = 0 on B 0 (0, 7R/8),

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so there are positive constants γ (determined only by n and the ratio of the maximum and minimum eigenvalues of [Aij ]) and C (determined also by R) so that [vh ]γ;B + (0,3R/4) ≤ Ckvh k2;B + (0,7R/8) ≤ CkDk vk2;B + (0,R) .

Since vh → Dk v in L2 as h → 0+ , it follows that Dk u is H¨older continuous in B + (0, 3R/4) with [Dk v]γ;B + (0,3R/4) ≤ C. The second derivative estimate for v then implies that |Dik v(x)| ≤ C(xn )γ−1

for i = 1, . . . , n, k = 1, . . . , n − 1 and any x ∈ B + (0, 5R/8). The differential equation then implies that this inequality is also valid for i = k = n. It then follows from simple integration that [Dn v]γ;B + (0,9R/16) ≤ C.

Hence v ∈ C 1,γ (B + (0, 9R/16)). We now repeat this argument with vh in place of v to conclude that [Dvh ]γ;B + (0,R/2) ≤ C,

and hence we get a uniform H¨older estimate for Dik v in B + (0, R/2) with i = 1, . . . , n, k = 1, . . . , n − 1. The differential equation then gives a H¨older estimate for Dnn v and hence v ∈ C 2 (B + (0, R/2)).  From this regularity result, we easily obtain some useful integral estimates for v. Lemma 5.50. Let [Aij ] be a constant, positive definite matrix, and suppose v is a solution of (5.77). Suppose λ and µ are positive constants such that ij

λ|ξ|2 ≤ Aij ξi ξj , 2 1/2

A ξi ηj ≤ (µλ|ξ| )

n

ij

(5.78a) 1/2

(A ηi ηj )

(5.78b)

for all ξ and η in R . Then there is a constant C, determined only by n and µ so that Z Z |Dv|2 dx ≤ CR−2 v 2 dx, (5.79a) B + (R/2) B + (R) Z  r n Z v 2 dx ≤ C v 2 dx, (5.79b) R + + B (r) B (R) Z  r n+2 Z |v − {v}r |2 dx ≤ C |v − {v}R |2 dx (5.79c) R + + B (r) B (R)

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for r ∈ (0, R), where

Proof.

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1 {v}r = + |B (r)|

Z

v dx.

(5.80)

B + (r)

Set ζ = 1 − |x|2 /R2 . By using ζ 2 v as test function, we see that Z Z 2 ij ζ A Di vDj v dx = −2 ζvAij Dj vDi ζ dx. B + (R)

B + (R)

We then estimate the right hand side through Cauchy’s inequality and (5.78b): Z −2 ζvAij Dj vDi ζ dx B + (R)

Z

≤2

!1/2

Z

ζ 2 Aij Di vDj v dx

B + (R)

It follows that Z

2

B + (R)

µλ|Dζ|2 v 2 dx

B + (R)

Z

ij

ζ A Di vDj v dx ≤ 4µλ

!1/2

B + (R)

.

|Dζ|2 v 2 dx.

Using (5.78a) and dividing the resulting inequality by λ yields Z Z 2 2 ζ |Dv| dx ≤ 4µ |Dζ|2 v 2 dx. B + (R)

B + (r)

We now infer (5.79a) by observing that |Dζ| ≤ 2/R in B + (R) while ζ ≥ 1/2 in B + (R/2). To prove (5.79b), we note that this inequality is clear for r ≥ R/2, so we assume that r < R/2. It then follows that !2 Z B + (r)

v 2 dx ≤ C(n)rn

sup

B + (R/2)

|v|

.

We now use the local maximum principle to infer that sup B + (R/2)

|v| ≤ C(n, µ) R−n

Z

!1/2

v 2 dx

B + (R)

.

Combining these two inequalities gives (5.79b). For our final inequality, we first use Poincar´e’s inequality to infer that Z Z |v − {v}r |2 dx ≤ C(n)r−2 |Dv|2 dx. (5.81) B + (r)

B + (r)

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If r < R/2, we then apply (5.79b) with Dk v (k = 1, . . . , n − 1) in place of v to obtain Z  r n Z |Dk v|2 dx ≤ C(n, µ) |Dk v|2 dx. R B + (r) B + (R/2) A similar result is true with n in place of k but the proof requires a little more care. If we set w = Anj Dj v, then we have Di (Aij Dj w) = 0 in B + (R),

w = 0 on B 0 (R).

A straightforward modification of the proof of the local maximum principle as given here shows that w satisfies the estimate !1/2 Z sup

B + (R/4)

w≤C

R−n

w2 dx

B + (R/2)

and hence Z

w2 dx ≤ C(n, µ)

B + (r)

We now use the inequalities n−1 X

|w| ≤ Cλ( |Dn v| ≤ C(

k=1

 r n Z R

|Dk v| + |Dn v|), n−1

X |w| +µ |Dk v|) λ k=1

to conclude that Z  r n |Dn v|2 dx ≤ C(n, µ) R B + (r) and hence Z

B + (r)

2

|Dv| dx ≤ C(n, µ)

w2 dx.

B + (R/2)

 r n R

Z

2

B + (R/2)

Z

|Dv| dx ,

2

B + (R/2)

!

!

|Dv| dx .

(5.82)

We now repeat the proof of (5.79a) with v − {v}R in place of v to infer that Z Z |Dv|2 dx ≤ CR−2 |v − {v}R |2 dx. B + (R/2)

B + (R)

In combination with (5.81) and (5.82), this inequality implies (5.79c) provided r < R/4. For r ≥ R/4, we observe that Z Z |v − {v}r |2 dx = inf |v − L|2 dx, (5.83) B + (r)

L∈R

B + (r)

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so Z

B + (r)

|v − {v}r |2 dx ≤

Z

B + (r)

|v − {v}R |2 dx

Z

|v − {v}R |2 dx  r n+2 Z n+2 ≤4 |v − {v}R |2 dx. R B + (R) ≤

B + (R)



The Schauder estimates are now proved by a simple idea: Consider the solutions of general problems as perturbations of solutions of constant coefficients. Estimates for the constant coefficient problems are already presented, so we just need to prove the appropriate perturbation estimates. These are relatively straightforward but they involve a large amount of calculation. Rather than provide one, extremely long calculation, we break the procedure down into several steps, all of which refer to the problem Di (aij Dj u + f i ) = g in B + (x0 , R0 ), nj

(5.84a)

0

a Dj u + ψ = 0 on B (x0 , R0 ).

(5.84b)

In order to describe our results in a convenient form, we make some definitions. First, for any y ∈ R with y n = 0 and any r > 0, we set B ∗ (y, r) = {x ∈ Rn , |x − y| = r, xn > 0} and we write d∗ for distance to B ∗ (x0 , R0 ). For q ∈ (0, n) and δ ≥ 0, we write M q,δ for the set of all functions g such that the norm !1/2 Z kgkq,δ =

sup

y∈B + (x0 ,R0 ) r∈(0,d∗ (y))

r−q d∗ (y)−δ

B(y,r)∩B + (x0 ,R0 )

|g|2 dx

is finite. Note that M q,δ is a weighted Morrey space. In [114], the (b) corresponding space for parabolic equations was denoted M2,q with b = (δ + n + 2 − q)/2. We assume that there are positive constants λ and µ so that (5.78) holds (0) and we assume that aij ∈ Hα (B + (x0 , R0 )). To quantify our regularity assumption on aij , we take A to be a constant such that [aij ](0) α ≤ Aλ. Assumptions on the regularity of f i , g, and ψ will be made below.

(5.85)

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Lemma 5.51. Suppose that conditions (5.78) and (5.85) are satisfied. If u satisfies (5.84), then there is a constant C, determined only by n, µ, α, and A such that Z rn−α |Du|2 dx ≤ C ∗ 2−α (sup |u| + F0 )2 (5.86) d (y) B + (y,r) for all y ∈ B 0 (x0 , R0 ) and all r ∈ (0, d∗ (y)/2), where (1)

(1)

F0 = |f /λ|0 + kg/λkn−2+2α,2+2α + |ψ/λ|0 .

(5.87)

Proof. Fix y ∈ B 0 (x0 , R0 ). As a first step, we obtain an estimate for the L2 norm of Du over + B (y, d∗ (y)/2). Define ζ by   4|x − y|2 ζ = 1− ∗ 2 d (y) + and note that 0 ≤ ζ ≤ 1 and |Dζ| ≤ 8/d∗ (y) in Rn . Using ζ 2 u as test function in the weak form of the (5.84) yields Z Z ζ 2 aij Di uDj u dx = − 2ζaij Di ζDj uu dx + + B ZB − ζ 2 f i Di u + 2ζf i Di ζu dx B+ Z Z 2 gζ u dx + ψζ 2 u ds, − B+

B0

where we have abbreviated B + (y, d∗ (y)/2) to B + and B 0 (y, d∗ (y)/2) to B 0 . From (5.78) and the Cauchy-Schwarz inequality, we infer that Z ζ 2 aij Di uDj u dx ≤ C(µ, n)λd∗ (y)n−2 [sup |u|2 + F02 ] B+

and hence (after noting that ζ ≥ 3/4 on B + (y, d∗ (y)/4)) Z |Du|2 dx ≤ C(µ, n)d∗ (y)n−2 [sup |u|2 + F02 ].

(5.88)

B + (y,d∗ (y)/4)

With τ ∈ (0, 1) and θ ∈ (0, 1/4) to be further specified, we set r0 = θd∗ (y) and we define rk = τ k r0 for positive integers k. We also take vk to be the solution of Di (aij (y)Dj vk ) = 0 in B + (y, rk ), anj (y)Dj vk = 0 on B 0 (y, rk ), vk = u on B ∗ (y, rk ).

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It then follows from (5.82) that Z Z |Dvk |2 dx ≤ Cτ n B + (y,r

B + (y,r

k+1 )

k)

|Dvk |2 dx.

(5.89)

Moreover, if we use wk = vk − u as test function in the weak forms of the equations for u and vk , we find that Z Z aij (y)Di wk Dj wk dx = [aij − aij (y)]Di uDj wk dx B + (y,rk ) B + (y,rk ) Z + f i Di wk + gwk dx + B (y,rk ) Z − ψwk dx. B 0 (y,rk )

We now observe that |aij (x) − aij (y)| ≤ λA|x − y|α d∗ (y)−α ≤ Cλθα

for y ∈ B + (y, rk ) and hence, after some rearrangement along with the Cauchy-Schwarz inequality and the Sobolev inequality, we obtain Z Z |Dw|2 dx ≤ Cθ2α |Du|2 dx + CF02 rkn−α d∗ (y)α−2 . B + (y,rk )

B + (y,rk )

In combination with (5.89), this inequality implies that Z Z 2 n 2α |Du| dx ≤ C[τ + θ ] |Du|2 dx + CF02 rkn−α d∗ (y)α−2 . B + (y,rk+1 )

B + (y,rk )

n

We now choose τ so that Cτ ≤ τ n−α/2 /2 and then θ so that Cθ2α ≤ τ n−α/2 /2. Applying Lemma 1.25 and (5.88) completes the proof.  Essentially the same idea then provides a mean oscillation estimate for (1) Du if f and ψ are in Hα . To simplify the notation, we define (similarly to (5.80)) Z 1 {g}y,r = + g(x) dx |B (y, r)| B + (y,r)

for any y ∈ B + (x0 , R0 ), any r ∈ (0, d∗ (y)) and g ∈ L1 (B + (y, r)).

Lemma 5.52. Assume the hypotheses of Lemma 5.51. Also, for y ∈ B 0 (x0 , R0 ) and r ∈ (0, d∗ (y)), define {Du}∗y,r to be the vector with components L1 , . . . , Ln defined by ( {Di u}y,r if i < n, Li = Pn−1 nj nn [f (y) − ψ(y) − j=1 a (y)Lj ]/a (y) if i = n.

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Then there is a constant C, determined only by A, α, n, and µ such that Z rα |Du − {Du}∗y,r |2 dx ≤ C(sup |u| + F )2 ∗ δ (5.90) d (y) B + (y,r) for any y ∈ B 0 (x0 , R0 ) and r ∈ (0, d∗ (y)), where

(0) F = |f /λ|(1) α + kg/λkn−2+2α,2+2α + |ψ/λ|α .

(5.91)

Proof. With y fixed and τ ∈ (0, 1) to be chosen, we define r0 = d∗ (y) and rk = τ k r0 , and we write vk for the weak solution of Di (aij (y)Dj vk + f i (y)) = 0 in B + (y, rk ), anj (y)Dj vk + f n (y) = ψ(y) on B 0 (y, rk ), vk = u on B ∗ (y, rk ). Then (5.79c) applied to Di vk with i < n yields Z

2

B + (y,rk+1 )

|Di vk − {Di vk }y,rk+1 | dx ≤ Cτ

n+2

Z

B + (y,rk )

|Di vk − {Di vk }y,rk |2 dx,

and a similar argument applied to vk,n = anj (y)Dj vk + f n (y) − ψ(y) yields Z Z |vk,n |2 dx ≤ Cτ n+2 |vk,n |2 dx. B + (y,rk+1 )

B + (y,rk )

Combining these two inequalities, we obtain Z Z ∗ 2 n+2 |Dvk −{Dvk }y,rk+1 | dx ≤ Cτ B + (y,rk+1 )

B + (y,rk )

|Dvk −{Dvk }∗y,rk |2 dx.

Again, we set wk = vk − u and use this function as test function in the weak forms of the boundary value problems for vk and u to obtain Z Z aij (y)Di wk Dj wk dx = [aij − aij (y)]Di uDj wk dx B + (y,rk ) B + (y,rk ) Z + [f i − f i (y)]Di wk + gwk dx + B (y,rk ) Z − [ψ − ψ(y)]wk dx. B 0 (y,rk )

This time, we observe that |aij (x) − aij (y)| ≤ λ

rkα d∗ (y)α

and use (5.86) to conclude that Z |Dwk |2 dx ≤ C(sup |u| + F )2 B + (y,rk )

rkn+α . ∗ d (y)2+α

(5.92)

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If we set I(r) =

Z

B + (y,r)

|Du − {Du}∗y,r |2 dx,

we conclude that I(rk+1 ) ≤ Cτ n+2 I(rk ) + C(sup |u| + F )2

rkn+α , d∗ (y)2+α

and (5.90) follows from this inequality via Lemmata 1.25 and 5.51.



(0)

Of course, this mean oscillation estimate only provides an H1+α/2 esti(0)

mate for u, but the full H1+α estimate follows by a easy extra step. Proposition 5.53. Let α ∈ (0, 1), let x0 ∈ Rn with xn0 = 0, and let R0 > 0. Suppose that there are positive constants A, λ, and µ such that conditions (5.78) and (5.85) are satisfied. If u is a weak solution of (5.84) (1) (1) with f ∈ Hα , g ∈ M n−2+2α,2+2α , and ψ ∈ Hα , then there is a constant C, determined only by A, µ, n such that (0)

|u|1+α ≤ C(|u|0 + F )

(5.93)

with F given by (5.91). Proof. By repeating the argument of Lemma 5.52 with y ∈ B + (x0 , R0 ) and using Lemma 5.48, we conclude that (0)

[u]1+α/2 ≤ C(|u|0 + F ), and the interpolation inequality (Lemma 2.3) implies that (0)

[u]1 ≤ C(|u|0 + F ). With the notation from the proof of Lemma 5.52, this inequality implies that Z rn+2α |Dwk |2 dx ≤ C(sup |u| + F )2 ∗ k 2+2α . d (y) B + (y,rk ) We then argue as before but with this inequality in place of (5.92) to infer (5.93).  Via a simple change of variables, we obtain a global H¨older gradient estimate. To state this result, for q ∈ (0, n), we write M q for the set of all g ∈ L2 (Ω) such that the norm !1/2 Z −q 2 kgkM q = sup r |g| dx y∈Ω r∈(0,diam Ω)

B(y,r)∩Ω

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is finite. Although our notation here is somewhat nonstandard, we point out that our M q is the usual Morrey space (written as M 2,q in [114]). Theorem 5.54. Let α ∈ (0, 1), and let ∂Ω ∈ H1+α . Suppose aij ∈ Hα (Ω) satisfies (5.78). Suppose b ∈ Hα , and c and c0 are in L2 (Ω) and that there is a constant A such that [aij ]α ≤ Aλ,

(5.94a)

0

|b|α + kckn−2+2α + kc kn−2+2α ≤ Aλ.

(5.94b)

If u is a weak solution of (5.7) with f and ψ in Hα and g ∈ M n−2+2α , then u ∈ H1+α and there is a constant C determined only by A, n, µ,and Ω, such that |u|1+α ≤ C(|u|0 + |f /λ|α + kg/λkM q + |ψ/λ|α ). Proof.

(5.95)

Assuming initially that u ∈ H1+α , we see that

Di (aij Dj u) = gˆ − Di fˆi in Ω,

aij Dj uγi = ψˆ + fˆ · γ on ∂Ω

with fˆi = −bi u + f i ,

gˆ = −ci Di u + −c0 u + g,

ψˆ = −β 0 u + ψ.

By combining Proposition 5.53 with an H1+α change of variable and the corresponding interior estimate, we find that |u|1+α ≤ C(|u|1 + |f /λ|α + kg/λkM q + |ψ/λ|α ), and the interpolation inequality (2.11) completes the proof in this case. The general case follows by a simple approximation argument which we omit.  We note here that g ∈ Ln/(1−α) implies that g ∈ M n−2+2α . Notes In many ways, this chapter is just a rehash of Chapters 7 and 8 of [64] and Chapter 2 of [91], and we refer to those chapters for a history of the theory of Sobolev spaces and weak solutions of elliptic equations although we mention the pioneering work [172] of Sobolev for an early look at the theory of what are now called Sobolev spaces and [173] for Sobolev’s more thorough description of the theory. Further information can be found in [206, 127]. We also mention [124] as a source for reading about the ideas

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that led to Sobolev’s theory of weak solutions as well as the deep theory of distributions. Some of our results on Sobolev spaces do not appear in [91] or [64]. Our proof of Poincar`e’s inequality is based on that in [11]; a more general result using this argument was proved by Hurri-Syrj¨anen [75]. In addition, the various Sobolev inequalities and trace theorems on Ω[R] are not stated in [91] or [64], but the proofs are easy modifications of the ones in those sources. Specifically, Corollary 5.23 seems to be new, but it most likely appeared much earlier in a reference unknown to this author because versions of Lemma 5.22 are present in [172] and [76]. In fact, our approach combines certain elements from [91] with others from [64]. Like [91], we have used De Giorgi’s method to prove H¨older continuity of weak solutions, but, in order to keep the proof close to that in Chapter 1, we have followed the argument in [37] to prove a weak Harnack inequality as an intermediate result towards proving H¨older continuity. On the other hand, we used Moser’s method (see [64] for a discussion of this method which is based on [140]) to prove the maximum and local maximum principles. Our proof of the Schauder estimates differs from that in [91] and [64]. We have followed Campanato’s proof (from [26]; see also [102]) of the Schauder estimates in Section 5.14. The boundary estimates in [26] are only presented for the Dirichlet problem, but the extension to conormal problems given here is a minor variant of the techniques in [26]. Exercises 5.1 Modify the proof of the local maximum principle to show that, if u is a nonnegative supersolution of (5.58), then, for any two numbers p1 and p2 with 0 < p1 < p2 (and p2 < n/(n − 2) if n > 2), there is a constant C, determined only by p1 , p2 , µ, µ2 R1−n/q , µ3 R(q−n)/(q−1) , n, and ω0 , such that   !1/p2 !1/p1 Z Z R−n |u|p2 dx ≤ C  R−n |u|p1 dx + k(R) . Ω[R/2]

Ω[R]

Use this estimate to show that (5.70) is valid for any p > 0 with p < n/(n − 2) if n > 2 provided the constant C depends also on p. (See also the proof of Theorem 8.18 in [64].) 5.2 In this exercise, we provide two alternative proofs for the strong maximum principle Theorem 5.43.

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(a) Imitate the proof of Theorem 5.15 to prove the strong maximum principle. (b) Use the weak Harnack inequality (5.70) with p = 1 to prove the strong maximum principle. (This is essentially the proof of Theorem 8.19 in [64].) R 5.3 Show that (5.83) holds. (Hint: Note that B + (r) |v − L|2 dx is a quadratic function of L.)

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Chapter 6

Strong Solutions

Introduction In this chapter, we study strong solutions of the oblique derivative problem. We define a strong solution of the problem Lu = f in Ω,

M u = g on ∂Ω

(6.1)

2,n Wloc (Ω)

to be a function u ∈ ∩ C(Ω) which satisfies the equation Lu = f a.e. in Ω and which satisfies the boundary condition M u = g is the usual sense. 6.1

Pointwise estimates for strong solutions

We begin by showing that strong solutions satisfy the AleksandrovBakel0 man-Pucci maximum principle. To this end, we recall some definitions from Section 1.5. For a continuous function u defined on Ω, we write Γ(u) for the set of all x ∈ Ω for which there is a vector p ∈ Rn such that u(y) ≤ u(x) + p · (x − y)

(6.2)

for all y ∈ Ω. In addition, we write χu for the normal mapping, that is, χu (x) is the set of all vectors p such that (1.8) holds for all y ∈ Ω. In particular, χu (x) is non-empty if and only if x ∈ Γ(u). We also recall that a vector field β is inward pointing if, for every x0 ∈ ∂Ω, there is a positive constant h such that x0 + tβ(x0 ) ∈ Ω for all t ∈ (0, h). Our first result is that Theorem 1.9 is valid for strong solutions. Theorem 6.1. Suppose [aij ] is positive definite and set D = det[aij ]. Suppose c ≤ 0 in Ω. Suppose β is inward pointing and β 0 < 0 and suppose that 227

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there are nonnegative constants B0 and B1 such that |β| ≤ −B0 β 0 on ∂Ω, Z |b|n dx ≤ B1 . Ω D

2,n If u ∈ Wloc (Ω) ∩ C(Ω) and if

Lu ≥ −f in Ω,

(6.3a) (6.3b)

M u ≥ gβ 0 on ∂Ω

(6.4)

for some nonnegative function f and some nonnegative constant g, then

f

u ≤ g + C(n, B1 )(B0 + diam Ω) 1/n , (6.5) D n + L (Γ (u))

+

where Γ (u) is the subset of Γ(u) on which u > g.

Proof. We proceed via approximation from the proof of Theorem 1.9. As a first step, we let ε > 0 be given and we write Gε for the subset of all x ∈ Γ(u) on which u(x) ≥ g + ε + B0 |p| for some p satisfying (6.2). We want to show that there is a δ > 0 such that d(x) > 2δ for all x ∈ Gε , and we argue by contradiction. If there were no such δ, then there would be a sequence (xm ) in Gε with a corresponding sequence (pm ) in Rn such that u(y) ≤ u(xm ) + pm · (y − xm )

for all y ∈ Ω and u(xm ) ≥ g+ε+B0|pm |, and xm → x0 , a point in ∂Ω. Since (pm ) is bounded (assuming without loss of generality that B0 > 0), we can take subsequences (also labeled (xm ) and (pm )) so that (pm ) converges to some vector p0 with u(y) ≤ u(x0 ) + p0 · (y − x0 )

(6.6a)

u(x0 ) ≥ g + ε + B0 |p0 |.

(6.6b)

for all y ∈ Ω and By taking y = x0 + hβ(x0 ) for all sufficiently small h in (6.6a) and then dividing the resultant inequality by h and sending h → 0, we conclude that β(x0 ) · Du(x0 ) ≤ β(x0 ) · p0 .

Combining this inequality with the boundary condition for u, we conclude that β 0 g ≤ β(x0 ) · p0 + β 0 (x0 )u(x0 ) and then the inequality (6.6b) implies that β(x0 ) · p0 ≤ −β 0 B0 |p0 | ≤ −β 0 (x0 )u(x0 ) + β 0 (x0 )[g + ε].

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Combining these inequalities gives β 0 (x0 )g ≤ β 0 (x0 )[g + ε], which contradicts β 0 < 0 and ε > 0. Hence there is some δ > 0 such that d(x) > 2δ if x ∈ Gε . Now let η be a C 2 (Ω) function vanishing in a neighborhood of ∂Ω such that η(x) = 1 if d(x) > δ, let (˜ um ) be a sequence of C 2 (Ω) functions which converge on the support of η to u, and set um = η˜ um + (1 − η)u. We also write Gε,m for the set of all x ∈ Ω such that there is a vector p ∈ Rn such that um (y) ≤ um (x) + p · (y − x) for all y ∈ Ω and um (x) ≥ g + ε + B0 |p|. Our next step is to show that, for any open set U such that Gε ⊂ U , there is a positive integer m0 such that Gε,m ⊂ U if m ≥ m0 . To this end, we let (xm ) be a convergent sequence with xm ∈ Gε,m \ U and write pm for the corresponding vector. If x0 = limm→∞ xm , then, without loss of generality, we may assume that (pm ) converges to some p0 . It follows that u(y) ≤ u(x0 ) + p0 · (y − x0 ) for all y ∈ Ω and u(x0 ) ≥ g + ε + B0 |p0 |. Hence x0 ∈ Gε , and hence there is a positive integer m0 such that xm ∈ U for m ≥ m0 . Now let h be a positive L1loc (Rn ) function, and define the function H by Z H(r) = h(p) dp. B(0,r)

For any open set U with Gε ⊂ U , and m ≥ m0 , we then have β · Dum + β 0 um ≥ β 0 g on ∂Ω, so Corollary 1.10 implies that Z  sup um ≤ g + ε + (B0 + diam Ω)H −1 h(Dum ) det D2 um dx U

because u ∈ C 2 (Gε,m ). Let us now assume further that h is bounded and then write Z h(Dum ) det D2 um − h(Du) det D2 um dx = I1,m + I2,m , U

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where I1,m = I2,m =

Z

Z

U

h(Dum )(det D2 um − det D2 u) dx,

U

[h(Dum ) − h(Du)] det D2 u dx.

Since h is bounded, we have Z |I1,m | ≤ sup h | det D2 um − det D2 u| dx → 0 U

as m → ∞. Also, [h(Dum ) − h(Du)] det D2 u converges in measure to 0 as m → ∞ and |[h(Dum ) − h(Du)] det D2 u| ≤ 2 sup h| det D2 u|, so the dominated convergence theorem implies that I2,m → 0 as m → ∞. It follows that Z Z h(Dum ) det D2 um dx → h(Du) det D2 u dx U

U

and hence sup u ≤ g + ε + (B0 + diam Ω)H

−1

Z

U

 h(Du) det D u dx . 2

Since this inequality is true for all open U with Gε ⊂ U , it follows that Z  sup u ≤ g + ε + (B0 + diam Ω)H −1 h(Du) det D2 u dx G Z ε  ≤ g + ε + (B0 + diam Ω)H −1 h(Du) det D2 u dx . G

Sending ε → 0, we infer that sup u ≤ g + (B0 + diam Ω)H −1

Z

G

 h(Du) det D2 u dx .

The proof is completed by noting that the argument giving Theorem 1.9 2,n from Lemma 1.8 is still valid for Wloc (Ω) ∩ C(Ω) functions.  It follows immediately that all of the results in Chapter 1 are valid for strong solutions. We leave the exact statements of these results to the reader.

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231

A sharp trace theorem

In Section 5.5, we showed that the trace map T maps W 1,p into Lq (∂Ω) for q = (n − 1)p/(n − p) if p < n. A natural question to raise is: What is the range of T ? The purpose of this section is to answer that question. We begin with a definition. For q > 1 and α ∈ (0, 1), we define the W α,q (∂Ω) seminorm h·iα,q;∂Ω by Z Z 1/q |u(x) − u(y)|q huiα,q;∂Ω = dsx dsy . n−2+qα ∂Ω ∂Ω |x − y|

We then use W α.q (∂Ω) to denote the set of elements of Lq (∂Ω) with huiα,q;∂Ω finite. It is a Banach space under the norm k · kα,q;∂Ω defined by kukα,q;∂Ω = huiα,q;∂Ω + kukq,∂Ω . Theorem 6.2. Let Ω be a bounded Lipschitz domain and let p > 1. Then there is a bounded linear function T : W 1,p (Ω) → W 1−1/p,p (∂Ω) such that T (u) is the restriction of u to ∂Ω for any u ∈ C 1 (Ω). Moreover, T maps W 1,p (Ω) onto W 1−1/p,p (∂Ω). The proof of this theorem is based on several basic calculus results. The first one is known as Hardy’s inequality.

Lemma 6.3. Let p > 1, let a < b be real numbers, and let f ∈ Lp (a, b). Then p  p Z b Z b Z x 1 p (6.7) f (t) dt dx ≤ |f (x)|p dx. p−1 a x−a a a

Proof.

Set

1 F (x) = x−a

Z

a

x

|f (t)| dt.

Integration by parts yields Z ∞ Z ∞ F p (x) dx = (b − a)F p (b) − p (x − a)F p−1 (x)F 0 (x) dx a Za∞ Z ∞ p = (b − a)F (b) − p |f (x)|F p−1 (x) dx + p F p (x) dx a

because

F 0 (x) =

|f (x)| − F (x) x−a

a

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on (a, b). By rearrangement, H¨older’s inequality and the nonnegativity of F , we obtain Z ∞ Z ∞ p b−a p F p (x) dx = |f (x)|F p−1 (x) dx − F (b) p−1 a p−1 a Z ∞ 1/p Z ∞ (p−1)/p p p p ≤ |f (x)| dx F (x) dx . p−1 a a One more rearrangement gives (6.7).  Related to Hardy’s inequality is an inequality due to Morrey. Lemma 6.4. Let h ∈ (0, 1) and let f ∈ Lp (0, 1) for some p ≥ 1. Then p Z 1−h Z x+h Z 1 1 f (t) dt dx ≤ |f (x)|p dx. h x 0 0

Proof. that

We begin by setting g = |f |p . Then H¨older’s inequality implies Z

0

1−h

Z p Z 1−h Z x+h 1 x+h 1 f (t) dt dx ≤ g(t) dt dx. h x h 0 x

(So there is really no loss of generality in assuming that p = 1 and f is nonnegative.) We then then write Z x+h Z h g(t) dt = g(x + t) dt x

0

and then apply Fubini’s theorem to obtain Z 1−h Z h Z h Z 1−h 1 1 g(x + t) dt dx = g(x + t) dx dt h h 0 0 0 0 Z h Z 1−h+t 1 = g(ξ) dξ dt. 0 h t The proof is completed by noting that g is nonnegative, so Z 1−h+t Z 1 g(ξ) dξ ≤ g(x) dx. t

0



Our next lemma is a simplified form of the general trace theorem for a two-dimensional region. Lemma 6.5. Let T be the triangle with vertices (0, 0), (1, 0), and (1, 1), let p > 1, and set cp = 2p+1 (p/(p − 1))p . Then Z 1Z 1 Z V (t, t) − V (τ, τ ) p dt dτ ≤ cp |DV |p dx (6.8) t−τ 0 0 T for any V ∈ W 1,p (T ).

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Proof.

where

233

By symmetry we have Z 1Z 1 V (t, t) − V (τ, τ ) p dt dτ = 2I, t−τ 0 0

Z t V (t, t) − V (τ, τ ) p dt dτ. t−τ 0 0 Hence we only need to estimate I. To this end, we first observe that, for any t and τ with 0 ≤ τ < t ≤ 1, we have I=

Z

1

V (t, t) − V (τ, τ ) = V (t, t) − V (t, τ ) + V (t, τ ) − V (τ, τ ) Z t Z t = D2 V (t, s) ds + D1 V (σ, τ ) dσ, τ

and hence

τ

p Z t V (t, t) − V (τ, τ ) p ≤ 2p−1 1 D1 V (σ, τ ) dσ t−τ t−τ τ p Z t p−1 1 +2 D2 V (t, s) ds t − τ τ

We now use Hardy’s inequality to estimate the integrals of the terms on the right hand side of this inequality. Specifically, p  p Z 1 Z 1 Z t 1 p D1 V (σ, τ ) dσ dt ≤ |D1 V (σ, τ )|p dσ t − τ p−1 τ τ τ so p  p Z Z 1Z 1 Z t 1 p dt dτ ≤ D V (σ, τ ) dσ |D1 V |p dx, 1 t − τ p−1 0 τ τ T and a similar argument shows that p  p Z Z 1Z t Z t 1 p D2 V (t, s) ds dτ dt ≤ |D2 V |p dx. t − τ p−1 0

0

τ

It follows that I ≤2

p−1



p p−1

T

p Z

T

1 |D1 V | + |D2 V | dx ≤ cp 2 p

p

Z

T

|DV |p dx.



By a simple change of variables, we obtain an actual trace theorem. Lemma 6.6. Let S be the square with vertices (0, 0), (1, 0), (1, 1) and (0, 1). Then Z 1 Z 1−h Z U (0, y + h) − U (0, y) p dy dh ≤ cp |DU |p dx (6.9) h 0 0 S for all U ∈ W 1,p (S).

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Proof. We define V in T by V (x, y) = U ((x − y)/2, (x + y)/2). It’s easy to check that Z Z |DU |p dx ≥ |DV |p dx S

I

and that Z 1Z 1 Z 1 Z 1−h V (t, t) − V (τ, τ ) p U (0, y + h) − U (0, y) p dy dh ≤ dt dτ. h t−τ 0 0 0 0  With these preliminaries, we are ready to prove our theorem.

Proof. [Proof of Theorem 6.2] We use the map T from Section 5.5. By virtue of a partition of unity argument and the invariance of the norm with respect to a uniformly Lipschitz change of independent variable, there is no loss of generality in assuming that ∂Ω is on the hyperplane xn = 0 and that xn > 0 in Ω. To prove that T maps into W 1−1/p,p (∂Ω), we may assume that ∈ W 1,p (Ω) vanishes for any x such that xi is outside of the interval (0, 1/2) for some i. It is elementary to verify that, in this setting, the seminorm hui1−1/p,p;∂Ω is equivalent to one given by !1/p n−1 X Z 1Z u(x0 + hej ) − u(x0 ) p 0 dx dh Λp (u) = , h j=1

0

Sj (h)

where

Sj (h) = {x0 : 0 ≤ xi ≤ 1 if i 6= j, 0 ≤ xj ≤ 1 − h} and ej is the (n − 1)-dimensional vector with ith component eij = δji . It follows from Lemma 5.20 that T maps into Lp (∂Ω), so we just need to show that Λp (u) ≤ CkDukp , and this inequality follows from Lemma 6.6. (If n = 2, the inequality is exactly (6.9), and the case n > 2 follows by simple integration.) To show that T is onto, we write Π for the pyramid defined by Π = {x ∈ Rn : 0 ≤ xi ≤ 1 − xn if i < n, 0 ≤ xn ≤ 1}, and we write S for the intersection of Π with the hyperplane xn = 0. For any element ϕ of W 1−1/p,p (S), we define u on Π by the expression Z x1 +xn Z xn−1 +xn 1 u(x) = n n−1 ... ϕ(ξ) dξ. (x ) x1 xn−1

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(Note that, here, ξ = (ξ 1 , . . . , ξ n−1 ).) Because of Lemma 5.22, we just have to show that kDukp ≤ CΛp (ϕ). If i < n, we use the simple differentiation expression Z 1 ¯ dξ 0 , Di u(x) = n n−1 [ϕ(ξ¯ + xn ei ) − ϕ(ξ)] i (x ) Σi (xn ) where ξ¯ is the vector with ξ¯j = ξ if j 6= i and ξ¯i = xi , ξi0 is the (n − 2)dimensional vector consisting of all ξ j with j < n and j 6= i, and Σi (xn ) is the set on which ξ j is between xj and xj + xn . Integrating with respect to x1 , . . . , xn and using Lemma 6.4, we conclude that Z Z 1Z ϕ(x0 + hei ) − ϕ(x0 ) p 0 p dx dh |Di u| dx ≤ C h Π 0 Si (h)

for i = 1, . . . , n − 1. To estimate the Lp norm of Dn u, we first observe that Z xn−1 +xn n−1 X 1 Z x1 +xn Dn u = ... (ξ j − xj )Dj ϕ(ξ) dξ n )n (x 1 n−1 x x j=1 =

n−1 X j=1

1 (xn )n

Z

x1 +xn

...

x1

Z

xn−1 +xn

xn−1

[ϕ(ξ¯j ) − ϕ(ξ)] dξ,

where ξ¯j has ith component ξ i if i 6= j and jth component xj + xn . Since xj + xn − ξ j ≥ xn , we can imitate the proof of the estimate for Di u to conclude that Z |Dn u|p dx ≤ CΛp (ϕ). Π

It follows that u ∈ W 6.3

1,p

as required.



Results from harmonic analysis

In this section, we recall some results from harmonic analysis that will be useful in our further study of strong solutions. The first result is a covering lemma of Wang. Lemma 6.7. Let A be a bounded subset of Rn and let C be a covering of A by balls of bounded radius. Then there is a sequence (B(xk , Rk )) of disjoint elements of C such that [ A⊂ B(xk , 5Rk ). (6.10)

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Proof. We define the sequence (B(xk , Rk )) as follows, writing Bk = B(xk , Rk ). First, we choose B1 ∈ C so that R1 ≥

1 sup{R : R is the radius of some B ∈ C}. 2

Once we have B1 , . . . , Bk−1 , we write Ck for the set of all B ∈ C such that B ∩ Bi = ∅ for all i ≤ k − 1, and we take Bk ∈ Ck so that Rk ≥

1 sup{R : R is the radius of some B ∈ Ck }. 2

To prove that (6.10) holds, we fix x ∈ A and choose a ball B(y, r) ∈ C so that x ∈ B(y, r), and we consider two cases. If the sequence (Bk ) is finite, then there is a last integer k such that Rk ≥ 21 r. If the sequence is infinite, then, without loss of generality, we may assume that B ∩ A is nonempty P for all B ∈ C and hence |Bk | = | ∪ Bk | is finite. It follows that |Bk | → 0 as k → ∞ and hence there is a last integer k such that Rk ≥ 12 r in this case as well. Letting k be this last integer, we note that the definition of Ck implies that Bi ∩ B(y, r) 6= ∅ for some i ≤ k. For this i, we write yi for any point in Bi ∩ B(y, r) and note that |x − xi | ≤ |x − y| + |y − yi | + |yi − xi | ≤ r + r + Ri < 5Ri .

In other words, x ∈ B(xi , 5Ri ), as required.



We now introduce the Hardy-Littlewood maximal operator M, defined on L1 (Rn ) by Z 1 Mv(x) = sup |v(y)| dy, r>0 |B(x, r)| B(x,r) and we recall the following properties of this operator, proved, for example, as Lemma 7.9 of [114]. Lemma 6.8. If v ∈ L1 (Rn ), then sup α|{x : Mv(x) > α}| ≤ 5n kvk1 .

α>0

If v ∈ Lp (Rn ) for some p > 1, then kMvkp ≤ C(n, p)kvkp .

(6.11)

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As a consequence, we have the Lebesgue differentiation theorem and a key inequality between v and Mv. Corollary 6.9. If v ∈ L1 (Rn ), then 1 r→0 |B(x, r)|

v(x) = lim

Z

v(y) dy

B(x,r)

for almost all x ∈ Rn . Hence |v| ≤ Mv a.e. We also have a further variant of the Vitali covering lemma, which is proved as Lemma 7.11 of [114]. Lemma 6.10. Let ε < 1 and ρ be positive constants, let x0 ∈ Rn , and suppose that A and C are measurable subsets of B(x0 , ρ) such that A ⊂ C and |A| < ε|B(x0 , r)|. Suppose in addition that, for any x ∈ B(0, ρ) and any r ∈ (0, ρ) such that |A ∩ B(x, r)| ≥ ε|B(x, r)|, we have B(x, r) ∩ B(x0 , ρ) ⊂ C. Then |A| ≤ 20n ε|C|. Finally, we have two simple geometric covering results. Lemma 6.11. For every positive integer n, there is a constant K, determined only by n, and, for any r > 0, there is a countable set of points (xi ) n in Rn with Rn = ∪∞ i=1 B(xi , r) and no point of R is in more than K balls of the form B(xi , 2r). Proof. We take (xi ) to be the collection of all points with each coordinate √ an integer multiple of r/ n.  Lemma 6.12. For every positive integer n, there is a constant K, determined only by n, and, for any r > 0, there is a countable set of points (xi ) n in Rn+ with xni ≥ r/2 if xni > 0, Rn+ ⊂ ∪∞ i=1 B(xi , ri ) and no point of R is n in more than K balls of the form B(xi , 2ri ), where ri = r if xi = 0 and ri = r/8 if xni > 0. Proof. Now we take (xi ) to be the collection of all points such that each coordinate is an integer multiple of r/n with two additional restrictions: First, xni ≥ 0 and xni ≥ r/2 if xni > 0. 

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Some further estimates for boundary value problems in a spherical cap

We recall the definitions from Section 2.6, but with a special choice of the parameter κ. Throughout this section, we consider R to be a fixed constant, so we do not include it in the notation, and we assume that [Aij ] is a positive definite matrix (not necessarily symmetric) such that there are positive constants λ and µ with λ|ξ|2 ≤ Aij ξi ξj ≤ µλ|ξ|2

(6.12)

for all ξ ∈ Rn . We also assume that β0 is a nonzero vector and that there is a positive constant µ0 such that |β00 | ≤ µ0 β0n .

(6.13)

With κ ∈ (µ0 /(1 + µ20 )1/2 , 1), we then write Γ for the set of all x ∈ Rn with |x0 |2 +|xn +κR|2 < R2 and xn > 0. We write Γ0 and Γ∗ for the subsets of ∂Γ on which |x0 |2 + |xn + κR|2 < R2 and |x0 |2 + |xn + κR|2 = R2 , respectively, and we write Γ∗∗ for the set of all x with xn = 0 and |x0 | = (1 − κ)1/2 R. In other words Γ∗∗ is the intersection of the closures of Γ0 and Γ∗ . We shall also find it convenient to look at the distance from these sets to a point in Γ, so we write d for the distance to ∂Γ, d0 for the distance to Γ0 , d1 for the distance to Γ∗ , and d2 for the distance to Γ∗∗ . Our first result is a pointwise gradient bound for solutions of the oblique derivative problem. Lemma 6.13. Let x0 ∈ Γ and let r ∈ (0, R). If u satisfies the conditions Aij Dij u = 0 in B(x0 , r) ∩ Γ,

(6.14a)

0

β0 · Du = 0 on B(x0 , r) ∩ Γ ,

(6.14b)

∗

u = 0 on B(x0 , r) ∩ Γ ,

(6.14c)

then there is a constant C, determined only by n, κ, µ, and µ0 such that !1/2 Z sup

B(x0 ,r/2)∩Γ

|Du| ≤ C

r−n

B(x0 ,r)∩Γ

|Du|2 dx

.

Proof. To simplify the writing, we note that there is a constant δ ∈ (0, 1/2], determined only by κ, such that min{d0 , d1 } > δd2 . Let y ∈ B(x0 , r/2) ∩ Γ. We also set b = β0 /|β0 | and note that b is a unit vector with bn ≥ 1/(1 + µ20 )1/2 .

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Suppose first that d2 (y) ≥ r/8. If d1 (y) < δr/32, we choose x1 ∈ Γ∗ so that d1 (y) = |x1 − y| and observe that B(x1 , δr/16) doesn’t intersect Γ0 . By arguing as in Lemma 1.35, we find that |Du(y)| ≤

|Du| ≤

sup B(x1 ,δr/32)∩Γ

C r

sup B(x1 ,δr/24)∩Γ

|u|.

A simple variation of the local maximum principle (Theorem 1.27) then shows that !1/2 Z −n 2 sup |u| ≤ C r |u| dx B(x1 ,δr/24)∩Γ

B(x1 ,δr/16)∩Γ

and Poincar`e’s inequality implies that Z Z 2 2 |u| dx ≤ Cr B(x1 ,δr/16)∩Γ

B(x1 ,δr/16)∩Γ

Since B(x1 , δr/16) ⊂ B(x0 , r), we conclude that |Du(y)| ≤ C

r

−n

Z

2

B(x0 ,r)∩Γ

|Du|2 dx.

!1/2

|Du| dx

.

(6.15)

Next, if d1 (y) ≥ δr/32 and d0 (y) < δr/256, we choose x1 ∈ Γ0 so that d0 (y) = |x1 − y| and note that B(x1 , δr/128) doesn’t intersect Γ∗ . If we now set w = |D0 u|2 + |b · Du|2 , then it’s easy to check that ij

a Dij w = 2

n−1 X

aij Dis uDjs u + 2aij Dik ubk Djm ubm

s=1

≥0

in B(x1 , δr/128) ∩ Γ and β0 · Dw = 0 on B(x1 , δr/128), and hence the local maximum principle Theorem 1.27 implies that Z −n sup w ≤ Cr w dx. B(x1 ,δr/256)∩Γ

B(x1 ,δr/128)∩Γ

2

It’s easy to check that |Du| ≥ 2w and |Du|2 ≤ C(µ0 )w, so Z sup |Du|2 ≤ Cr−n |Du|2 dx. B(x1 ,δr/256)∩Γ

B(x1 ,δr/128)∩Γ

Since y ∈ B(x1 , δr/256) ∩ Γ and B(x1 , δr/128) ⊂ B(x0 , r), we again infer (6.15). Further, if d1 (y) ≥ δr/32 and d0 (y) ≥ δr/256, then B(y, δr/256) ⊂ B(x0 , r) ∩ Γ, so we can apply the local maximum principle directly to each component of Du in B(y, δr/256) to infer (6.15).

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Now suppose that d2 (y) < r/8 and choose x1 ∈ Γ∗∗ so that |x1 − y| = d2 (y). To simplify writing, we set ρ = d2 (y) and !1/2 Z −n 2 U= r |Du| dx . B(x0 ,r)∩Γ

As a first step here, we note that B(x1 , r/4) ⊂ B(x0 , r) and use Poincar`e’s inequality along with the local maximum principle to infer that sup B(x1 ,r/6)∩Γ

|u| ≤ CrU.

Now, we note that the constant α1 in Proposition 3.16 corresponding to θ0 = arctan κ is greater than 1 and hence (after using a suitable rotation of axes) there is a nonnegative function w1 such that Aij Dij w1 ≤ −|x − x1 |−1 in B(x1 , r/6) ∩ Γ, b · Dw1 ≤ −1 on B(x1 , r/6) ∩ Γ0 , w ≤ C|x − x1 | in B(x1 , r/6) ∩ Γ.

It follows that there is a constant K1 such that w = |x − x1 |2 /r + K1 w1 satisfies Aij Dij w ≤ 0 in B(x1 , r/6) ∩ Γ, b · Dw ≤ 0 on B(x1 , r/6) ∩ Γ0 , w ≥ 0 on B(x1 , r/6) ∩ Γ∗ ,

w ≥ r on ∂B(x1 , r/6) ∩ Γ.

It follows from the maximum principle in B(x1 , r/6) ∩ Γ that |u| ≤ CU w in B(x1 , r/6) and hence that sup B(y,ρ/4)∩Γ

|u| ≤ CU ρ.

Arguing as in the previous cases (with ρ in place of r), we find that (6.15) holds here as well.  We also need a result which shows that the mixed boundary value problem with a conormal boundary condition has a unique solution which satisfies a corresponding estimate. Lemma 6.14. For any f ∈ L2 (Γ0 ), there is a unique weak solution of Di (Aij Dj u) = Di f i in Γ,

nj

A Dj u = f

n

0

on Γ ,

(6.16a)

∗

u = 0 on Γ .

(6.16b)

Moreover, kDuk2 ≤ kf k2 /λ.

(6.17)

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Proof. We begin by proving (6.17). To this end, we use u as test function in the weak version of the boundary value problem to obtain Z Z Aij Di uDj u dx = Du · f dx. Γ

Γ

ij

2

Applying the lower bound A ξi ξj ≥ λ|ξ| for any vector ξ and the CauchySchwarz inequality gives (6.17). The existence proof is essentially that of Theorem 5.26 along with the observation that σ = 0 is not in the set Σ, and we leave the details of the proof to the reader. 

6.5

Lp estimates for solutions of constant coefficient problems in a spherical cap

In this section, we continue the notation of the previous section with one additional condition: we write β0 for the vector with components β0j = Anj . Since [Aij ] is not assumed to be symmetric, the number µ0 is not controlled by µ, so we continue to use both quantities (and κ). The main part of the proof of our Lp estimates is a measure-theoretic estimate for the maximal function of the solutions of a constant coefficient problem. Specifically, we study the problem Di (Aij Dj u) = Di f i in Γ,

Anj Dj u = f n on Γ0 ,

u = 0 on Γ∗ .

(6.18)

n

In addition, if S is any measurable subset of R , we define the operator MS by Z 1 MS v(x) = sup |v(y)| dy. r>0 |B(x, r)| B(x,r)∩S (In other words, if v is defined on S and w is defined by ( v(x) if x ∈ S, w(x) = 0 if x ∈ / S,

then MS v = Mw.)

Proposition 6.15. Let u be a weak solution of (6.18) for some f ∈ L2 (Γ). Let x1 ∈ Ω and r ∈ (0, R/6). Then, for any ε > 0, there are constants M (µ, µ0 , κ, n) > 0 and δ(ε, µ, µ0 , κ, n) ∈ (0, 1] such that, if there is a point y ∈ B(x1 , r) ∩ Γ with MΓ |f |2 (y) ≤ δ 2 λ2 θ2 and MΓ (|Du|2 )(y) ≤ θ2 for some θ > 0, then |{x ∈ B(x1 , r) ∩ Γ : MΓ (|Du|2 )(x) > θ2 M 2 }| ≤ ε|B(x1 , r)|.

(6.19)

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Proof.

and

For k > 0, we abbreviate B(x1 , kr) ∩ Γ to Bk . We have Z 1 |Du|2 dx ≤ θ2 ωn (6r)n B(y,6r)∩Γ 1 ωn (6r)n

Z

B(y,6r)∩Γ

|f |2 dx ≤ δ 2 λ2 θ2 .

Since B4 ⊂ B(y, 6r) ∩ Γ, it follows that Z |Du|2 dx ≤ 2n θ2 ωn (4r)n . B4

We now let w be the solution of

Di (Aij Dj w) = Di f i in B + (x1 , 4r), Anj Dj w = f n on B 0 (x1 , 4r), w = 0 on B ∗ (x1 , 4r) given by Lemma 6.14 and set h = u − w. Then (6.17) implies that Z Z 1 2 |Dw| dx ≤ 2 |f |2 dx ≤ 2n δ 2 θ2 ωn (4r)n , λ B4 B4

and hence (recalling that δ ∈ (0, 1]) Z Z
2 |Dh| dx ≤ 2 |Dw|2 + |Du|2 dx ≤ 2n+1 θ2 ωn (4r)n . B4

B4

Moreover, Lemma 6.13 implies that there is a constant M0 , determined only by µ and n such that sup |Dh|2 ≤ M0 θ2 . B3

It follows that |Du|2 ≤ 2|Dw|2 + 2M02 θ2 in B3 ,

(6.20)

2

and we infer from (6.11) (with α = θ ) that Z −n 2 2 2 5 θ |{x ∈ B1 : MB3 |Dw| > θ }| ≤ |Dw|2 dx ≤ 2n δ 2 θ2 ωn (4r)n . B3

It follows that

|{x ∈ B1 : MB3 |Dw|2 > θ}| ≤ 40n δ 2 |B(x1 , r)|.

(6.21)

Now let z ∈ B1 be a point at which MB3 |Dw|2 (z) ≤ θ2 . For ρ ∈ (0, 2r], we have B(z, ρ) ∩ Γ ⊂ B3 , and therefore Z 1 |Du|2 ≤ 2MB3 |Dw|2 (z) + 2M0 θ2 ≤ (2M0 + 2)θ2 . |B(z, ρ)| B(z,ρ)∩Γ

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On the other hand, for ρ > 2r, we have B(z, ρ) ⊂ B(y, 2ρ) and therefore Z Z 1 1 2 |Du| dx ≤ |Du|2 dx ≤ 2n θ2 . |B(z, ρ)| B(z,ρ)∩Γ |B(z, ρ)| B(y,2ρ)∩Γ

With M = max{2n /2, (2M0 + 2)1/2 }, we conclude that MΓ |Du|2 (z) ≤ M 2 θ2 and hence {x ∈ B1 : MΩ |Du|2 > M 2 θ2 } ⊂ {x ∈ B(x1 , r); MB3 |Dw|2 > θ2 }. It follows from this inequality and (6.21) that |{x ∈ B1 : MΩ |Du|2 > M 2 θ2 }| ≤ 40n M0 δ 2 |B(x1 , r)|, and (6.19) follows if we take min{1, ε} δ= .  (1 + 40n M0 )1/2 Our next step is to prove some conclusions about the measure of the set on which MΓ |Du|2 is large than a certain number. Corollary 6.16. Let u be a weak solution of (6.18) for some f ∈ L2 (Γ). Let ε > 0, set ε1 = 20n ε, and let M and δ be the constants from Proposition 6.15. If g ∈ L1 (Γ) and h ≥ 0, write {MΓ g > h} for the subset of Γ on which MΓ g > h. If θ is a positive constant such that |{MΓ |Du|2 > θ2 M 2 }| ≤ ε|B(x0 , R)|, (6.22) then |{MΓ |Du|2 > θ2 M 2j }| ≤ εj1 |{MΓ |Du|2 > θ2 }| +

j X i=1

for any positive integer j. Proof.

εi1 |{MΓ |f |2 > θ2 M 2(j−i) δ 2 }|

(6.23)

We set A = {MΓ |Du|2 > θ2 M 2 },

B = {MΓ |Du|2 > θ2 } ∪ {MΓ |f |2 > θ2 δ 2 }. From the contrapositive of Proposition 6.15, we conclude that |A ∩ B(x, r)| ≥ ε|B(x, r)| implies that B(x, r)∩B(x0 , R) ⊂ B, and hence Lemma 6.10 implies (6.23) for j = 1. For any α ≥ 1, we can replace θ by αθ in (6.23) with j = 1 to conclude that |{MΓ |Du|2 > α2 θ2 M 2 }| ≤ εj1 |{MΓ |Du|2 > α2 θ2 }| + ε1 |{MΓ |f |2 > α2 θ2 M 2 δ 2 }|, and (6.23) then follows from this inequality by induction, using α = 1, M, . . . , M j . 

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From this corollary, we can now prove our Lp estimate for p > 2. Proposition 6.17. Let p > 2 and let u be a weak solution of (6.16) for some f ∈ Lp (Γ). Then there is a constant C, determined only by µ, µ0 , κ, p and n such that kDukp ≤ Proof.

C kf kp . λ

(6.24)

With M the constant from Proposition 6.15, set

1 1 min{1, n p }, 2 20 M and let δ be the corresponding constant from Proposition 6.15. We also set ε=

F =

1 kf kp λ

and observe that 1 kf k2 ≤ F |B(x0 , R)|(p−2)/(2p) . λ It then follows from (6.17) that kDuk2 ≤ CF |B(x0 , R)|(p−2)/(2p) , and hence (6.11) gives a constant c1 such that |{x ∈ Γ : MΓ |Du|2 > θ2 M 2 } ≤ c1

1 F 2 |B(x0 , R)| R2n/p θ2 M 2

for any θ > 0. The choice 1/2

θ=

c1 F M Rn/p ε

then gives (6.22). Now, we use the inequality |Du|2 ≤ MΓ |Du|2 from Corollary 6.9 and Lemma 1.14 to infer that Z Z |Du|p dx ≤ (MΓ |Du|2 )p/2 dx Γ Γ Z ∞ =p σ p−1 |{MΓ |Du|2 > σ 2 }| dσ. 0

We then rewrite the right hand side of this inequality as Z θM ∞ Z θM j+1 X p−1 2 2 p σ |{MΓ |Du| > σ }| dσ+p σ p−1 |{MΓ |Du|2 > σ 2 }| dσ. 0

j=1

θM j

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Then p

Z

θM

0

and Z p

θM j+1

θM j

σ p−1 |{MΓ |Du|2 > σ 2 }| dσ ≤ θp M p |B(x0 , R)|,

σ p−1 |{MΓ |Du|2 > σ 2 }| dσ ≤ θp M (j+1)p |{MΓ |Du|2 > θ2 M 2j }|.

From Corollary Z 6.16, we then infer that |Du|p dx ≤ θp M p (|B(x0 , R)| + I1 + I2 ), Γ

where

I1 =

∞ X j=1

I2 =

∞ X j=1

p

M j εj1 |{MΓ |Du|2 | > θ2 M 2 }|, M jp

j X i=1

εi1 |{MΓ |f |2 > M 2(j−i) δ 2 θ2 }.

Since M ε1 ≤ 1/2, it follows that

I1 ≤ |B(x0 , R)|, and simple algebra shows that ∞ ∞ X X I2 = εi1 M ip M (j−i)p |{M|f |2 > M 2(j−i) δ 2 θ2 }| =

i=1 ∞ X i=1

≤

∞ X j=0

εi1 Mip

j=i ∞ X j=0

M jp |{MΓ |f |2 > M 2j δ 2 θ2 }|

M jp |{MΓ |f |2 > M 2j δ 2 θ2 }|

≤ |{MΓ |f |2 > δ 2 θ2 }| + M p Since

∞ X j=1

M (j−1)p |{MΓ |f |2 > M 2j δ 2 θ2 }|.

|{MΓ |f |2 > δ 2 θ2 }| ≤ |B(x0 , R)| and a simple calculation shows that Z ∞ ∞ X M (j−1)p |{MΓ |f |2 > M 2j δ 2 θ2 }| ≤ M p σ p−1 |{MΓ |f |2 > σ 2 δ 2 θ2 }| dσ 0

j=1

= Mp

Z

B + (x0 ,6R)

≤ c(n, p)M p

Z

MΓ |f |p dx (δθ)p

B + (x0 ,6R)

|f |p dx, (δθ)p

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it follows that I2 ≤ C(n, p)|B(x0 , R)|. Combining all these inequalities gives Z |Du|p dx ≤ C(n, µ, p)F p Γ

which is just (6.24).



p

In order to prove the L estimates when p ∈ (1, 2), we argue via duality. Proposition 6.18. Proposition 6.17 is true for any p > 1. Proof. When p > 2, this proposition is exactly Proposition 6.17, and when p = 2, it’s just Lemma 6.14. When p < 2, we set q = p/(p − 1), and g = |Du|p−2 Du. If v is the weak solution of Dj (Aij Di v) = Dj g j in Γ,

Ain Di v = g n on Γ0 ,

v = 0 on Γ∗ ,

then Proposition 6.17 implies that kDvkq ≤ Ckgkq = CkDukp−1 . p By using u as test function in the weak version of the boundary value problem for v and then using v as test function in the weak version of the boundary value problem for u, it follows that Z Z Z Di ug i dx = Aij Dj uDi v dx = Di vf i dx. Γ

i

Γ

Γ

p

But Di ug = |Du| , and hence

kDukpp ≤ kDvkq kf kp ≤ CkDukp−1 kf kp . p

Dividing this inequality by kDukp−1 gives (6.24). p 6.6



Local estimates for strong solutions of constant coefficient problems

We now prove some estimates for strong solutions of constant coefficient problems which will be used, in the next section, to prove our Lp estimates for variable coefficient problems. We first prove an estimate for the solution of a differential equation in a ball and then an estimate for the solution of a boundary value problem in a half-ball. To simplify notation, we use x1 to denote an arbitrary point in Rn and r to denote a fixed positive number.

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We also assume that [Aij 0 ] is a symmetric matrix and that there are positive constants λ and µ such that 2 λ|ξ|2 ≤ Aij 0 ξi ξj ≤ µλ|ξ|

(6.25)

for all ξ ∈ Rn , and we define the operator L0 by L0 h = Aij 0 Dij h.

Our first proposition gives the estimate for strong solutions of constant coefficient equations in a ball. Proposition 6.19. For any p ∈ (1, ∞), there is a positive constant C, determined only by n, p, and µ such that any u ∈ W 2,p (B(x0 , 2r)) satisfies the estimate   1 1 1 2 kD ukp,r ≤ C kL0 ukp,2r + kDukp,2r + 2 kukp,2r , (6.26) λ r r where we have abbreviated kgkp,B(x1 ,ρ) to kgkp,ρ for any function g and any ρ ∈ (0, 2r). Proof. We begin by noting that there is a nonnegative function η ∈ C 2 (Rn ) with support in B(x1 , 2r) which is identically 1 in B(x1 , r) such that r|Dη| + r2 |D2 η| ≤ C(n). For k = 1, . . . , n, we set vk = Dk (ηu) and define fk to be the vector-valued function with components fki = δki [η(L0 u) + 2Amj 0 Dm ηDj u + L0 ηu]. Without loss of generality, we may assume that xn1 = 2r and then we set x0 = (x01 , 0). There is a constant µ0 , determined only by n and µ such that nn Anj 0 ξj ≤ µ0 A0 |ξ|

for any vector ξ ∈ Rn with ξn = 0. With

1/2

κ=

µ0 , (1 + µ20 )1/4

p we have κ ∈ (µ0 / 1 + µ20 , 1) and we can choose R = C(n, µ)r so that B(x1 , 2R) ⊂ Γ. Hence vk is a weak solution of i Di (Aij 0 Dj vk ) = Di fk in Γ,

Anj 0 Dj vk

=

fkn

0

on Γ , ∗

vk = 0 on Γ .

(6.27a) (6.27b) (6.27c)

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It follows from Proposition 6.18 that C kfk kp,Γ , λ and then (6.26) follows from this inequality because X kD2 ukp,r ≤ kDvk kp,Γ kDvk kp,Γ ≤

k

and

kfk kp,Γ ≤ CkLukp,2r + Cλ



 1 1 kDukp,2r + 2 kuk . r r



in order to prove our second derivative estimate for solutions of the oblique derivative problem, we use a special cut-off function. Lemma 6.20. Let µ be a nonnegative constant and let β0 be a constant vector such that |β00 | ≤ µβ0n . Then, there is a nonnegative function η ∈ C 2 (B(0, 2r)) with support in B(0, 3r/2) such that η ≤ 1 and r|Dη| + r2 |D2 η| ≤ C(n, µ) in B(0, 2r), η ≡ 1 in B(0, r), and β0 · Dη ≡ 0 on B 0 (0, 2r). Proof. Let g ∈ C 2 ([0, 1]) with g(0) = 0, g 0 (0) = 1, and 0 ≤ 18µg ≤ 1 on [0, 1]. Also, let h ∈ C 2 ([0, ∞)) be decreasing with h ≡ 1 in [0, 1/3) and h ≡ 0 in [1/2, ∞). We then define G on B(0, 2r) by  n 0   n 2 |x|2 x β0 · x x G(x) = 2 + 9µ2 g −g , 4r 2r 2r β0n r and we take η = h ◦ G. Cauchy’s inequality implies that 2 |x|2 ≤ G(x), 9 r2 and hence η(x) = 0 if |x| ≥ 3r/2, the choice of h implies that 0 ≤ η ≤ 1 and the estimates on the derivatives of η, and Cauchy’s inequality also implies that   n 2 5 |x|2 x 2 G(x) ≤ + 18µ g , 18 r2 2r so η ≡ 1 on B(0, r). A simple calculation (using g(0) = 0 and g 0 (0) = 1) leads to β0 · Dη ≡ 0 on Q0 .  We are now ready for our local second derivative estimate for strong solutions of the oblique derivative problem. Proposition 6.21. Let x0 ∈ Rn0 , let β0 be a constant nonzero vector satisfying (6.13), and let p ∈ (1, ∞). Then there is a constant C, determined only by n, µ, and µ0 , such that any u ∈ W 2,p (B(x0 , 2r)) with β0 · Du = 0 on B(x0 , 2r) satisfies (6.26) with kgkp,ρ now standing for kgkp,B +(x0 ,ρ) .

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Proof. With κ as in the proof of Proposition 6.19 (but with µ0 from the hypotheses of this proposition), we again can choose R = C(n, µ, µ0 )r so that B + (x0 , 2R) ⊂ Γ. With η the function from Lemma 6.20, we set vk = Dk (ηu) and we define fk to be the vector-valued function with components fki = δki [ηL0 u + 2Amj 0 Dm ηDj u + L0 ηu]. If we define ij

A =

 ij   A0

for i, j < n,

Ann β j /β n  0 0 0  Ain + Ani − 0 0

i n Ann 0 β /β0

for i = n, j ≤ n,

for i < n, j = n,

then we find that vk is a weak solution of (6.27) if k = 1, . . . , n − 1. Hence we have C kDvk kp,Γ ≤ kfk kp,Γ λ and   1 1 kfk kp,Γ ≤ CkLukp,2r + Cλ kDukp,2r + 2 kuk r r for this range of k. The proof is completed by using the differential equation Aij 0 Dij u = f to infer that ! n−1 X 1 2 kD ukp,r ≤ C kDvk kp,Γ + kf kp,2r . λ  k=1 6.7

Local interior Lp estimates for the second derivatives of strong solutions of differential equations

Our next step is to prove estimates for strong solutions of equations with variable coefficients. We argue via a standard perturbation argument which takes advantage of the following interpolation lemma, which is proved (as part of Theorem 9.11) on page 237 of [64]. Lemma 6.22. For k = 0, 1, 2, define the weighted seminorms: [v]k = sup (1 − s)k rk kDvkk,B(x1 ,sr) . 0 0 and any p > 1, there is a constant C, determined only by n, p, µ, θ, and ζ such that   1 2 0 kD ukp,Ω ≤ C kLukp,Ω + kukp,Ω (6.32) λ 2,p for any u ∈ Wloc (Ω) and any Ω0 ⊂ Ω with d(Ω0 , ∂Ω) < θ.

Proof. Fix x0 ∈ Ω, define L0 by L0 h = aij (x0 )Dij h, and set f = Lu. For r ∈ (0, d(x0 )/2) to be chosen, let η ∈ C 2 (Rn ) satisfy η = 1 in B(x0 , r), η = 0 outside B(x0 , 2r) and |Dη| ≤

2 , r

|D2 η| ≤

C(n) . r2

Then v = ηu satisfies L0 v = (aij (x0 ) − aij )Dij v + (2aij Di η − bi η)Di u + (aij Dij η − cη)u + ηf. It then follows from Proposition 6.19 that there are constants C1 (µ, n) and C(µ, n, B) such that kD2 vkp ≤ C1 ζ(r/d(x0 ))kD2 vkp   1 1 1 +C kf kp,2r + kDukp,2r + 2 kukp,2r . λ r r

We now choose δ so small that C1 ζ(δ) ≤ 1/2 and assume that r ≤ δd(x0 )/2 to infer that   1 1 1 2 kf kp,2r + kDukp,2r + 2 Ckukp,2r . kD vkp ≤ C λ r r

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Since kD2 ukp,r ≤ kD2 vkp ,

we infer that 

 1 1 1 kf kp,2r + kDukp,2r + 2 Ckukp,2r . λ r r We now fix y ∈ Ω with d(y) > θ/2 and set R = δθ/2. For each s ∈ (0, 1), we use the previous estimate with r = (1 − s)R/4 and x0 replaced by any xi in the covering given by Lemma 6.11 which is in B(y, sR). Modifying notation slightly, we now write kgkp,ρ for kgkp,B(y,ρ) and kgkp,ρ,i for kgkp,B(xi ,ρ) . Then we have  X1 1 1 2 kD ukp,sR ≤ C kf kp,r,i + kDukp,2r,i + kukp,2r,i λ (1 − s)R (1 − s)2 r2 i   1 1 1 ≤ CK kf kp,R + kDukp,(1+s)R/2 + kukp,R λ (1 − s)R (1 − s)2 r2 by Lemma 6.11. Using the notation from Lemma 6.22, we infer that  2  R [u]2 ≤ C [f ]0 + [u]1 + [u]0 . λ so Lemma 6.22 then gives us  2  R [u]2 ≤ C [f ]0 + [u]0 , λ and, therefore, there is a constant C2 (n, µ, B) such that   1 2 2 kD ukp,R/2 ≤ C2 kf kp,R + R kukp,R . λ Arguing again via Lemma 6.11 yields (6.32).  kD2 ukp,r ≤ C

6.8

Local Lp second derivative estimates near the boundary

We now prove the analog of Theorem 6.23 for the oblique derivative problem. In addition to ideas already presented, we shall take advantage of two basic results about functions in W 1,p and W 2,p . Our first result is a variant of the trace theorem. Lemma 6.24. Let p > 1, let w ∈ W 1,p (Rn+ ) and suppose v(x) = 0 if xn > R for some positive constant R. Then, for any α ∈ (1 − 1/p, 1), there is a constant C, determined only by p and α, such that Z Z (xn )p(α−1) |v|p dx ≤ CRpα |Dv|p dx. (6.33) Rn +

Rn +

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Proof. Suppose first that v ∈ C 1 . For fixed x0 ∈ Rn−1 , define w by w(s) = v(x0 , s). Then Z ∞ Z R sp(α−1) |w(s)| ds = sp(α−1) |w(s)| ds 0 0 Z p Z R R (pα−1) 0 = s w (t) dt ds s 0 !p Z R Z R ≤ s(pα−1) |w0 (t)| dt ds. 0

0

We now use H¨ older’s inequality to conclude that !p Z R Z R 0 |w (t)| dt ≤ Rp−1 |w0 (t)|p dt, 0

and hence Z

∞

0

Z R |w0 (t)|p dtRp−1 sp(α−1) ds 0 0 0 Z ∞ 1 pα = R |w0 (t)|p dt. pα − p + 1 0 We now rewrite this inequality in terms of v and x and then integrate over Rn−1 to infer (6.33). The general case follows by approximation.  sp(α−1) |w(s)| ds ≤

Z

R

Our next result concerns the representation of the trace of various functions in terms of other, simpler functions. Lemma 6.25. Let p > 1 and α ∈ (1−1/p, 1), let v ∈ W 1,p (Rn+ ) and suppose that there are a number r > 0 and a point x0 ∈ Rn0 such that v(x) = 0 if |x − x0 | > r. Then there is a constant C, determined only by n, p, and α, and, for any h ∈ Hα (Bn−1 (x00 , r)), there is a function z ∈ W 2,p (Rn+ ) such that z = 0, Dn z = hv on Rn0 ,

(6.34a)

z(x) = 0 if |x − x0 | > r,

(6.34b)

2

α

kD zkp ≤ C(|h|0 + [h]α r )kvkp .

(6.34c)

Proof. First, by virtue of Lemma 1.3, we may assume that h has been extended to all of Rn0 without increasing |h|0 or [h]α . Then we extend h to all of Rn+ so that (xn )1−α |Dh| + (xn )2−α |D2 h| ≤ C(n, α)[h]α .

(6.35)

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Finally, we define z by z(x) = xn h(x)

Z

  xn y v x− ϕ(y) dy, 2 Rn

where, as usual, ϕ ∈ C 2 (Rn ) is nonnegative with kϕkL1 = 1 and ϕ(y) = 0 if |y| ≥ 1. Straightforward algebra verifies (6.34a) and (6.34b). In addition, we have Dij z(x) = I1 + I2 + I3 for   Z xn y ϕ(y) dy, I1 = (xn Dij h(x) + δin Dj h(x) + δjn Di h(x)) v x− 2 Rn    Z xn y ∂ ∂ k I2 = Di h(x) v x− ϕ(y) + δ (y ϕ(y)) dy, jn 2 ∂y j ∂y k Rn    Z xn y ∂ ∂ k I3 = h(x) Di v x − ϕ(y) + δ (y ϕ(y)) dy. jn 2 ∂y j ∂y k Rn

We now use (6.35) along with Tonelli’s theorem and the observation that 1 n xn y 3 x ≤ xn − ≤ xn 2 2 2 if ϕ(y) 6= 0 to conclude that kI1 kp + kI2 kp ≤ C[h]α

Z

n p(α−1)

(x )

Rn +

p

!1/p

|v| dx

,

and then Lemma 6.24 gives kI1 kp + kI2 kp ≤ C[h]α rα kDvkp . A similar argument (but without Lemma 6.24) yields kI3 kp ≤ C|h|0 kDvkp . The combination of these last two inequalities yields (6.34c).



From these results, we can prove estimates for the oblique derivative problem near the boundary. Our first estimate is one near a flat boundary portion. Theorem 6.26. Let [aij ], b, and c be defined on B(x0 , R) for some x0 ∈ Rn0 and some R > 0. Define the operator L by (6.29), and suppose that there are positive constants α < 1, λ, µ, and B such that λ|ξ|2 ≤ aij ξi ξj ≤ µλ|ξ|2

(6.36a)

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for all ξ ∈ Rn , Rα (xn )1−α |b| + R2 |c| ≤ B.

(6.36b)

Suppose further that there is a positive, continuous increasing function ζ defined on [0, 1/2] with ζ(0) = 0 such that   |x − y| ij ij |a (x) − a (y)| ≤ ζ (6.37) R for all x and y in Ω with |x − y| ≤ R/2. Let β and β 0 be defined on B 0 (x0 , R) and suppose that |β| ≤ µβ n ,

R|β 0 | ≤ Bβ n

(6.38a)

on B 0 (x0 , R), Rα |β(x) − β(y)| + R1+α |β 0 (x) − β 0 (y)| ≤ Bβ n (x)|x − y|α

(6.38b)

for all x and y in B0 (x0 , R) with |x − y| ≤ R/2. Then, for any p ∈ (1, 1/(1 − α)), there is a constant C, determined only by n, p, µ, B, and ζ such that !

Mu 1 2

kLukp;R + kukp;R + (6.39) kD ukp;R/2 ≤ C

βn λ 1−1/p,p;R

for any u ∈ W 2,p (B(x0 , R)), where M u = β · Du + β 0 u and kgkp;ρ = kgkp,B +(x0 ,ρ) ,

kgk1−1/p,p;ρ = kgk1−1/p,p,B 0 (x0 ,ρ) .

Proof. We begin by observing that there is no loss of generality in assuming that β n ≡ 1. Fixing x1 ∈ B 0 (x0 , R/2) and r ∈ (0, R/4), we take η from Lemma 6.20 with β0 = β(x1 ), we find that  r  r α  kL0 vkp ≤ C1 ζ +B kD2 vkp R R   1 1 1 kf kp,2r + kDukp,2r + 2 kukp,2r . +C λ r r The only new element is that we estimate the quantity w = bi Di uη in three steps. First, we use the structure condition to see that kbi Di uηkp ≤ C(n)BR−α k(xn )α−1 η|Du|kp . Then Lemma 6.24 implies that k(xn )α−1 η|Du|kp ≤ C(n, p, α)rα kD(η|Du|)kp .

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Finally, C C |Du| + 2 |u|. r r We then introduce three functions from Lemma 6.25. The most straightforward function is z1 , the function (labeled z in the lemma) corresponding to v and h = −β 0 . Next, for each k ∈ {1, . . . , n}, we write zk,2 for the function P corresponding to v = ηDk u and h = (β k (x0 )−β k ), and we set z2 = k zk,2 . For the last function, we write g˜ for the function in W 1,p (Rn+ ) with trace equal to ηM u on Rn0 , and we take z3 to be the function corresponding to v = g˜ and h ≡ 1. With z = z1 + z2 + z3 , it follows that  r α   r +B kD2 vkp kL0 (v − z)kp ≤ C1 λ ζ R R   λ λ + C kf kp,2r + kDukp,2r + 2 kukp,2r + λkηgk1−1/p,p r r and that  r  r α  kD2 zkp ≤ C1 ζ +B kD2 vkp R R  1 1 1 +C kf kp,2r + kDukp,2r + 2 kukp,2r + kηgk1−1/p,p . λ r r |D(η|Du|)| ≤ C|D2 v| +

Moreover, β 0 · D(v − z) = 0 on Rn0 and (v − z)(x) is zero if |x − x1 | ≥ 2r, so we can use Proposition 6.21 to conclude that C kD2 (v − z)kp ≤ kLv0 kp . λ Combining all these inequalities, we see that  r  r α  kD2 vkp ≤ C1 ζ +B kD2 vkp R R   1 1 1 +C kf kp,2r + kDukp,2r + 2 kukp,2r + kηgk1−1/p,p . λ r r

α
We now choose r small enough that C1 ζ Rr + B Rr ≤ 1/2 and then argue as in the proof of Theorem 6.23 to obtain (6.39), except that we use Lemma 6.12 in place of Lemma 6.11. 

For domains with curved boundary, the local estimate is easy to prove if we don’t try to obtain quite the same simple form for the estimate. Instead, we introduce some notation that is used consistently in [64], so we first give the following definition. We say that a domain Ω has a C k,α boundary portion T if T ⊂ ∂Ω and if, for any x0 ∈ T , there is a open set S = S(x0 ) with x0 ∈ S(x0 ) and a one-to-one mapping ψ ∈ Hk+α (S) such that ψ(S) = B(0, 1),

ψ(S ∩ Ω) = B + (0, 1),

ψ(S) ∩ ∂Ω) = B 0 (0, 1)

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and ψ −1 ∈ Hk+α (B(0, 1))). For our purposes, we also note that the function ψ can be taken so that |Dm ψ| ≤ Cdk+α−m in S ∩ Ω and |Dm ψ −1 | ≤ C(xn )k+α−m in B + (0, 1) for any given integer m > k. Our local boundary estimate for domains with curved boundary then takes the following form. Theorem 6.27. Let Ω be a domain in Rn with a C 1,α boundary portion T for some α ∈ (0, 1). Let [aij ], b,and c be defined on Ω and suppose they satisfy condition (6.36a) and d1−α |b| + |c| ≤ Bλ. (6.40) Suppose further that there is a positive, continuous increasing function ζ defined on [0, ∞) with ζ(0) = 0 such that |aij (x) − aij (y)| ≤ ζ(|x − y|) (6.41) for all x and y in Ω. Let β and β 0 be defined on T and suppose that |β| ≤ µβ · γ, |β 0 | ≤ Bβ · γ (6.42a) on T , |β(x) − β(y)| + |β 0 (x) − β 0 (y)| ≤ Bβ · γ(x) (6.42b) for all x and y in T . Then for any open subset Ω0 of Ω such that Ω0 ⊂ Ω∩T and any p ∈ (1, 1/(1 − α)), there is a constant C, determined only by n, p, µ, B, Ω, and Ω0 such that !

Mu 1 2

kD ukp,Ω0 ≤ C kLukp,Ω + kukp,Ω + . (6.43)

β · γ λ 1−1/p,p,T

Of course, we can take T = ∂Ω and Ω0 = Ω to obtain a global W 2,p estimate for u, but, rather than state such an estimate directly, we now look at a related global estimate that will be critical to our existence theory.

Theorem 6.28. Let Ω be a bounded domain in Rn with ∂Ω ∈ C 1,α for some α ∈ (0, 1). Let [aij ], b, and c be defined on Ω and suppose they satisfy condition (6.36a) and (6.40). Suppose further that there is a positive, continuous increasing function ζ defined on [0, ∞) with ζ(0) = 0 such that (6.41) holds for all x and y in Ω. Let β and β 0 be defined on ∂Ω and suppose that they satisfy (6.42) (with T replaced by ∂Ω). Then for any p ∈ (1, 1/(1 − α)), there are positive constants C and σ0 , determined only by n, p, µ, B, and Ω, such that !

Mu 1 2

kD ukp ≤ C kLu − σλukp + (6.44) λ β · γ 1−1/p,p

for any σ ≥ σ0 .

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Proof.

257

Define Σ = Ω × (−1, 1) and define the operator L on Σ by Lh = Lh + λDn+1,n+1 h.

Then v, defined by v(x, xn+1 ) = u(x) cos(σ 1/2 xn+1 ), satisfies Lv = cos(σ 1/2 xn+1 )(Lu − σλu) in Σ and M v = cos(σ 1/2 xn+1 )M u on T = ∂Ω × (−1, 1). It then follows from Theorem 6.27 with Ω0 = Ω × (−ε, ε) (for ε ∈ (0, 1/2) that there is a constant C1 , determined only by n, p, µ, B, and Ω, such that !

Mv 1

kDn+1,n+1 vkp,Ω0 ≤ C1 kLvkp,Σ + kvkp,Σ +

β · γ λ 1−1/p,p !

Mu 1

≤ C1 kLu − σλukp + kukp + . λ β · γ 1−1/p,p We now compute

kDn+1,n+1 vkp,Ω0 = σkvkp,Ω0 = σ

Z

ε

−ε

| cos(σ

1/2

p

t)| dt

1/p

kukp .

If we now choose ε = π/(3σ 1/2 ), then cos(σ 1/2 t) ≥ 1/2 for t ∈ (−ε, ε) and hence kDn+1,n+1 vkp,Ω0 ≥ σ



2π 3σ 1/2

1/p

1 1 kukp = 2 2



2π 3

1/p

σ 1−1/(2p) kukp .

By choosing σ0 sufficiently large, we find that kDn+1,n+1 vkp,Ω0 ≥ 2C1 kukp and hence

Mu 1

. kukp ≤ kLu − σλukp + λ β · γ 1−1/p,p

The proof is completed by combining this inequality with Theorem 6.27 using T = ∂Ω and Ω0 = Ω. 

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Existence of strong solutions for the oblique derivative problem

We are now ready to show that the oblique derivative problem is solvable under suitable hypotheses. There are many ways to use our estimates for this purpose, and we provide here a simple one, namely, the method of continuity. Other methods are suggested in the exercises. Throughout this section, we assume that the hypotheses of Theorem 6.28 are satisfied and that σ is a constant with σ ≥ σ0 , the constant of that theorem. We shall show that the problem Lu − σλu = f in Ω,

M u = g on ∂Ω

L0 h = λ∆h − σλh,

M0 h = β0 · Dh.

2,p

(6.45)

p

has a unique solution u ∈ W (Ω) for any f ∈ L (Ω) and any g ∈ W 1−1/p,p (∂Ω). To simplify the proof, we shall assume that β · γ ≡ 1 on ∂Ω. Also, to apply the method of continuity, we write B for W 2,p (Ω) and V = Lp (Ω) × W 1−1/p,p (∂Ω). We begin by setting β0 = γ, and we define the operators L0 and M0 by Then, for t ∈ [0, 1], we define the operators Lt and Mt by Lt = tL + (1 − t)L0 ,

Mt = tM + (1 − t)M0 .

Since σ > 0, it follows that the problem L0 u = f in Ω,

M0 u = g on ∂Ω

(6.46)

(−1−α) H2+δ

has a unique solution in for any f ∈ H1 (Ω) and g ∈ Hα (Ω), where δ ∈ (0, 1) is arbitrary. Since p(1 − α) < 1, it follows that this solution is in W 2,p . Since H1 (Ω) is dense in Lp (Ω) and Hα (∂Ω) is dense in W 1−1/p,p (∂Ω), a simple approximation argument (using Theorem 6.28) then implies that (6.46) has a unique solution u ∈ W 2,p for any f ∈ Lp and g ∈ W 1−1/p,p . Hence (L0 , M0 ) maps B onto V. In addition, Theorem 6.28 implies that kukB ≤ Ck(Lt u, Mt u)kV

for all t ∈ (0, 1), so the method of continuity implies that (L1 , M1 ) maps B onto V. In other words, (6.45) is solvable for any f and g. The uniqueness of the solution follows from another application of Theorem 6.28. We therefore have the following basic existence and uniqueness theorem. Theorem 6.29. Let Ω be a bounded domain in Rn with ∂Ω ∈ C 1,α for some α ∈ (0, 1). Let [aij ], b,and c be defined on Ω and suppose they satisfy condition (6.36a) and (6.40). Suppose further that there is a positive,

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continuous increasing function ζ defined on [0, ∞) with ζ(0) = 0 such that (6.41) holds for all x and y in Ω. Let β and β 0 be defined on ∂Ω and suppose that they satisfy (6.42) (with T replaced by ∂Ω). Then for any p ∈ (1, 1/(1 − α)), there is a positive constant σ0 , determined only by n, p, µ, B, ζ, and Ω, such that (6.45) has a unique solution for any σ ≥ σ0 . In addition there is a constant C, determined by the same quantities as for σ0 , such that this solution satisfies the estimate !

g 1

kuk2,p ≤ C kf kp + . (6.47) λ β ·γ 1−1/p,p

Note that the condition σ ≥ σ0 is quite strong and we do not have a simple way of estimating the constant σ0 . If we use suitable results from Chapter 4, we can prove a uniqueness and existence result for (6.1) assuming that c and β 0 are nonpositive but not both zero. Theorem 6.30. Let Ω be a bounded domain in Rn with ∂Ω ∈ C 1,α for some α ∈ (0, 1). Let [aij ], b,and c be defined on Ω and suppose they satisfy condition (6.36a) and (6.40). Suppose further that there is a positive, continuous increasing function ζ defined on [0, ∞) with ζ(0) = 0 such that (6.41) holds for all x and y in Ω. Let β and β 0 be defined on ∂Ω and suppose that they satisfy (6.42) (with T replaced by ∂Ω). Suppose finally that c ≤ 0 in Ω and β 0 ≤ 0 on ∂Ω but at least one of these functions is not identically zero. Then for any p ∈ (1, 1/(1−α)), (6.1) has a unique solution u ∈ W 2,p for any f ∈ Lp (Ω) and any g ∈ W 1−1/p,p (∂Ω). In addition there is a constant C, determined only by n, c0 , p, µ, B, ζ, and Ω, such that this solution satisfies (6.47).

Proof. Suppose first that g ∈ H1 and that the functions aij , b, c, f are all in Hα (Ω). In this case, Theorem 4.10 gives a unique solution of (6.1) in (−1−α) H2+α . Next, suppose aij , b, and c satisfy the hypotheses of the theorem, and let ij (ak ), (bk ), and (ck ) be sequences of Hα functions such that (aij k ) converges ij uniformly to a , (bk ) converges a.e. to b, and (ck ) converges to c. We further take the sequences so that each aij k satisfies the hypotheses of this theorem, |bk | ≤ 2Bλdα−1 , and ck ≤ 0 with ck nonzero in some open set if c 6≡ 0. We still assume that f ∈ Hα and that g ∈ H1 , and we write uk for the solution of Lk uk = f in Ω, M uk = g on ∂Ω. We now claim that the sequence (uk ) is uniformly bounded. Suppose to the contrary that this sequence is not uniformly bounded, in which case we may

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assume that kuk k∞ → ∞ as k → ∞. Setting Uk = kuk k∞ , vk = uk /Uk , fk = f /Uk and gk = g/Uk , we conclude that Lk vk = fk in Ω,

vk = gk on ∂Ω

for each k and that (vk ) is bounded in L∞ . Then Theorem 4.5 gives a uniform H1+α bound for vk and Lemma 6.31 gives a uniform W 2,p bound for vk . It follows that some subsequence converges in W 2,q (for any q < p) to a limit function v, which is therefore 2,n a C 1 (Ω) ∩ Wloc (Ω) solution of Lv = 0 in Ω,

M v = 0 on ∂Ω.

Hence v ≡ 0, which contradicts the pair of conditions kvk k∞ = 1 for all k and (vk ) converges uniformly to v. Therefore the sequence (uk ) is uniformly bounded, so Lemma 6.31 gives a subsequence which converges in W 2,q (for any q < p) to a limit function u, which is therefore a W 2,p solution of (6.1). Since Hα (Ω) × H1 (∂Ω) is dense in Lp (Ω) × W 1−1/p,p (∂Ω), we conclude that the range of the map T : W 2,p → Lp (Ω) × W 1−1/p,p (∂Ω) defined by T u = (Lu, M u) is dense in Lp (Ω) × W 1−1/p,p (∂Ω). But Theorem 2.29 and the proof of Theorem 2.30 imply that this range is also closed and hence T is onto, so that (6.1) has a solution for any f and g. Another application of Theorem 2.29 implies that the solution is unique.  We close with a useful local regularity result. Lemma 6.31. Let Ω be a domain in Rn with a C 1,α boundary portion T for some α ∈ (0, 1). Let [aij ], b,and c be defined on Ω and suppose they satisfy condition (6.36a) and (6.40). Suppose further that there is a positive, continuous increasing function ζ defined on [0, ∞) with ζ(0) = 0 such that (6.41) holds for all x and y in Ω. Let β and β 0 be defined on T and 2,q suppose that conditions (6.42) are satisfied. Let u ∈ Wloc (Ω ∪ T ) for some p q ∈ (1, 1/(1 − α)) and suppose that Lu ∈ L (Ω) and M u ∈ W 1−1/p,p (T ) for 2,p some p ∈ (q, 1/(1 − α)). Then u ∈ Wloc (Ω ∪ T ). Proof. Let us fix a point x0 ∈ T and choose R so that B(x0 , R)∩∂Ω ⊂ T . Then there is a set Ω∗ with ∂Ω∗ ∈ C 1,α such that B(x0 , R/2) ∩ Ω ⊂ Ω∗ ⊂ B(x0 , R)∩Ω. Let η ∈ C 2 (Rn ) be nonnegative with η ≡ 1 in B(x0 , R/2) and η ≡ 0 in Rn \ B(x0 , 5R/8). Then there is a vector field β¯ ∈ Hα (∂Ω∗ ) which is oblique on ∂Ω∗ with β¯ = β on ∂Ω ∩ B(x0 , 5R/8), so v = η 2 u satisfies the conditions aij Dij v + bi Di v − σv = f1 in Ω∗ , β¯ · v = g1 on ∂Ω∗

(6.48)

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for any σ ∈ R, with

f1 = 2ηaij Di ηDj u + [aij Dij (η 2 ) − cη 2 − σ]u + 2η(bi Di η)u

and g1 = ηu[2β¯ · Dη − β 0 η].

It follows from the Sobolev imbedding theorem that Du ∈ W 1,r (Ω∗ ) with ( p if q ≥ n, r= min{p, nq/(n − q)} if q < n.

By using Lemma 6.24 and observing that D(ηu|Dη|) ∈ Lr , we see that f1 ∈ Lr (Ω∗ ) and g1 ∈ W 1−1/r,r (∂Ω∗ ). If we choose σ sufficiently large, then Theorem 6.29 gives a unique solution v ∈ W 2,r (Ω∗ ) of (6.48), and the same theorem gives a unique W 2,q (Ω∗ ) solution of this problem, so these two solutions must be equal. It follows that v ∈ W 2,r (Ω∗ ) and hence u ∈ W 2,r (B(x0 , R/2) ∩ Ω). After repeating this argument (with r in place of q) finitely many times, we conclude that u ∈ W 2,p (B(x0 , R/2) ∩ Ω). A similar (but simpler) argument shows that u ∈ W 2,p (B(x0 , ρ)) for any x0 ∈ Ω and ρ ≤ d(x0 )/2, and a covering argument gives the general result.  Notes Section 6.2 is based on the results in [60], and we refer the interested reader to that paper for a more detailed account of the proof of Theorem 6.2. The main differences between [60] and Section 6.2 are that we have only sketched the proof that T is onto and that we have proved the preliminary results Lemmata 6.3 and 6.4. Lemma 6.3 was first proved by Hardy in [69]; we use a simplified version of the proof in that paper. A more detailed account of the history of this standard result appears in [89]. Lemma 6.4 was proved by Morrey as Lemma 1 in [138] and we have used his proof here. Gagliardo cites [139] as a source for this result, but that source doesn’t contain the lemma. Instead, it is stated as Theorem 2.3 in [24], which is part one of [139], and [24] refers to [138] for a proof. The covering lemmata, labeled as Lemmata 6.7 and 6.10 here, were first proved by Wang [199], and the proof of Lemma 6.7 given here is identical to that in [199]. On the other hand, Lemma 6.8 and Corollary 6.9 are standard results and we refer the reader to page 12 of [175] for further

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discussion. Lemmata 6.11 and 6.12 are apparently new here, but they are really just minor modifications of results used by Whitney [201] to prove his general extension theorem (Lemma 1.3 is a very special case of this result); however, nowhere in Whitney’s paper are the results stated in a form immediately recognizable as related to these lemmata. Instead, we mention Propositions 2 and 3 on page 169 of [175] which have an obvious connection, both in statement and in proof, to our lemmata. The method of proof for the Lp estimates for constant coefficient problems in Section 6.5 is due to Wang [199], who developed it as an alternative to earlier proofs; the proof in [199] is just presented for interior estimates, but the generalization to boundary value problems is very straightforward. In particular, this generalization consists primarily of proving the estimates in Section 6.4. Applications of Wang’s method to the oblique derivative problem were first given in Chapter 7 of [114], and the presentation here streamlines that approach by proving all the local pointwise and L2 estimates at once (in Lemmata 6.13 and 6.14). The corresponding boundary estimates for the Dirichlet problem, under various weak regularity assumptions on the boundary, appear in numerous papers which we shall not list here since additional care must be taken with the boundary condition; however, it would be interesting to know if the Dirichlet problem results can be used to prove estimates for the oblique derivative problem along the lines of the proof of Schauder estimates in Chapter 4. To the best of my knowledge, the only other application of Wang’s method to oblique derivative problems is a work by Byun and Wang [19] which gives W 1,p estimates for solutions of the conormal problem in Reifenberg flat domains. We shall not give the precise definition of a Reifenberg flat domain, but we point out that they are not directly comparable to Lipschitz domains; all Lipschitz domains with sufficiently small Lipschitz constant are Reifenberg flat, but Lipschitz domains with large Lipschitz constant need not be Reifenberg flat and Reifenberg flat domains need not be Lipschitz. Although the method of proof for the Lp estimates is relatively recent, the results themselves are not. The basic interior estimate for Poisson’s equation is due to Calder´ on and Zygmund [23], and our Theorem 6.23 was first proved by Greco [67]. Our Theorem 6.26 is a special case of a much more general result, which was first proved by Browder (see Theorem 1 in [16]) and Agmon, Douglis, and Nirenberg (see Theorem 15.1 of [1]). In fact, these results apply to domains with C 2 boundaries, and the extension to H1+α boundaries (with p(1 − α) < 1) seems to have first stated in Theorem 7.35 of [114]. Moreover, the continuity assumption on [aij ] can be relaxed

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to a smallness condition on the BM O modulus of this matrix. In other words, one needs to know only that Z −n lim inf sup inf R |u − L| dx R→0 x∈Ω L∈R

Ω∩B(x,R)

is sufficiently small. This fact was apparently first noted explicitly in [21] for nonlinear equations and for p > n. A proof for linear equations and all p > 1 was first given in [29], and the method developed here was used to prove the same results in [18]. In fact, this smallness condition can be further weakened; see, for example, [79]. Various Lp estimates for derivatives have also been derived in Lipschitz domains by Calder´ on [22] and by Kenig and Pipher [78]. To state their results, we first fix a positive number θ and, for each y ∈ ∂Ω, we write Γ(y) for the set of all x ∈ Ω with |x − y| < (1 + θ)d(x). We also define the nontangential maximal operator N associated with this number θ by setting, for any g defined on Ω and any y ∈ ∂Ω, N [g](y) = sup |g|. Γ(y)

Then, for any continuous oblique unit vector field β, there are finitely many linear conditions and a positive number ε (determined only by β and Ω) such that if g ∈ Lp (∂Ω) for some p > 2 − ε satisfies the finitely many linear conditions, then there is a solution of the problem ∆u = 0 in Ω,

β · Du = g on ∂Ω

such that kN [Du]kp,∂Ω ≤ C(p, Ω, β)kgkp,∂Ω . Specifically, Calder´ on proved this result for p ∈ (2 − ε, 2 + ε) and Kenig and Pipher proved it for p > 2. In fact, uniqueness for this problem is generally not true for p < 2; Verchota [194] has shown that there are continuous oblique vector fields β such that the homogeneous problem has an arbitrarily large number of linearly independent solutions with p < 2 but the homogeneous problem has only the constant solution for p = 2. Techniques for these two cases are quite different from each other and from the ones used in this book.

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Chapter 7

Viscosity Solutions of Oblique Derivative Problems

In this chapter, we provide an extremely brief introduction to viscosity solutions of fully nonlinear elliptic equations with fully nonlinear oblique boundary conditions. 7.1

Definitions and notation

Throughout this chapter, we assume that the function F , defined on Ω × R × Rn × Sn , satisfies the following monotonicity condition: F (x, z1 , p, r1 ) ≤ F (x, z2 , p, r2 )

(7.1)

for all (x, p) ∈ Ω × Rn , all z1 ≥ z2 in R, and all r1 ≤ r2 in Sn . (As before, we use the inequality r1 ≤ r2 to mean that r1 − r2 is a negative semidefinite matrix.) In addition, for any subset S of Rn , we write U SC(S) for the set of all functions which are upper semicontinuous on S and LSC(S) for the set of all functions which are lower semicontinuous on S. If u is defined on S, we write JS2,+ u(x) for the set of all (p, r) ∈ Rn × Sn such that

1 u(y) ≤ u(x) + p · (y − x) + rij (x − y)i (x − y)j + o(|x − y|2 ) 2 for y ∈ S. In other words, for any ε > 0, there is δ > 0 such that 1 u(y) ≤ u(x) + p · (y − x) + rij (x − y)i (x − y)j + ε|x − y|2 2

for all y ∈ S with |x − y| < δ. Similarly, we write JS2,− u(x) for the set of all (p, r) ∈ Rn × Sn such that 1 u(y) ≥ u(x) + p · (y − x) + rij (x − y)i (x − y)j − o(|x − y|2 ) 2 265

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for y ∈ S. A viscosity subsolution of F = 0 in Ω is a function u ∈ U SC(Ω) such that F (x, u(x), p, r) ≥ 0 for all x ∈ Ω and all (p, r) ∈ JΩ2,+ u(x). A viscosity supersolution of F = 0 is defined to be a function u ∈ LSC(Ω) such that F (x, u(x), p, r) ≤ 0 for all x ∈ Ω and all (p, r) ∈ JΩ2,− u(x). (Note that the definition of subsolution puts no restrictions on the function u at a point where JΩ2,+ u is empty.) Although this formulation of viscosity subsolutions and supersolutions is very convenient for our purposes, we point out that it is related to a more usual one involving comparison functions. Specifically, one can show that JΩ2,+ u(x) is equal to the set of all (p, r) ∈ Rn × Sn such that p = Dϕ(x) and r = D2 ϕ(x) for some function ϕ ∈ C 2 (Ω) such that u − ϕ has a local maximum at x. Hence u is a viscosity subsolution if and only if F (x, u(x), Dϕ(x), D2 ϕ(x)) ≥ 0 for any x and ϕ ∈ C 2 (Ω) such that u−ϕ has a local maximum at x. Similar assertions are true for JΩ2,− and viscosity supersolutions. We don’t spell out the details because they won’t be important in our brief development of the theory. We shall discuss the appropriate definition of viscosity subsolutions and viscosity supersolutions for boundary value problems in Section 7.6.

7.2

The Theorem of Aleksandrov

In this section, we present a proof of one of the deepest result in the entire book. We show that every convex function is twice differentiable almost everywhere; this result will be used as one step in the analysis of viscosity solutions. This result was first proved in the case of two dimensions by Busemann and Feller [17] and in an arbitrary number of dimensions by Aleksandrov [3]. We proceed via a sequence of results on convex functions and measures. The first result consists of some elementary estimates on convex functions. Lemma 7.1. Let f : Rn → R be a C 2 convex function. Then D2 f ≥ 0 in

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Rn and there is a constant C, determined only by n, such that Z sup |f | ≤ CR−n |f (y)| dy, B(x,R/2)

sup B(x,R/2)

267

(7.2a)

B(x,R)

|Df | ≤ CR−n−1

for any x ∈ Rn and any R > 0.

Z

B(x,R)

|f (y)| dy,

(7.2b)

Proof. First, we note that, for any h > 0 and any x and y in Rn with |y| = 1, we have f (x + hy) + f (x − hy) − 2f (x) ≥ 0.

Dividing this inequality by h2 and letting h → 0, we see that Dij f (x)y i y j ≥ 0,

and hence D2 f ≥ 0. Next, we use Taylor’s theorem to conclude that, for any y and z in Rn , there is a point x∗ ∈ Rn such that 1 f (y) = f (z) + Df (z) · (y − z) + Dij f (x∗ )(y − z)i (y − z)j , 2 and hence f (y) ≥ f (z) + Df (z) · (y − z).

(7.3)

Now suppose z ∈ B(x, R/2). Integrating (7.3) with respect to y over B(z, R/2) then gives Z Z 1 1 f (z) ≤ |f (y)| dy ≤ 2n |f (y)| dy |B(z, R/2)| B(z,R/2) |B(x, R)| B(x,R) because Z

B(z,R/2)

Df (z) · (y − z) dy = Df (z) ·

Z

B(z,R/2)

(y − z) dy = 0.

On the other hand, by reversing the roles of y and z in (7.3), we conclude that f (z) ≥ f (y) + Df (y) · (z − y). 1

n

(7.4)

There is a function ϕ ∈ C (R ) such that 0 ≤ ϕ ≤ 1 and |Dϕ| ≤ 4/R in Rn with ϕ ≡ 1 on B(x, R/2) and ϕ ≡ 0 on Rn \ B(x, R). Multiplying (7.4) by ϕ(y) and integrating with respect to y over B(x, R) yields Z Z Z f (z) ϕ(y) dy ≥ f (y)ϕ(y) dy + ϕ(y)Df (y) · (z − y) dy. B(x,R)

B(x,R)

B(x,R)

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The divergence theorem implies that Z Z ϕ(y)Df (y) · (z − y) dy = − B(x,R)

B(x,R)

f (y) div[ϕ(y)(z − y)] dy,

and so

f (z)

Z

B(x,R)

ϕ(y) dy ≥

Z

f (y)(ϕ(y) − div[ϕ(y)(z − y)]) dy Z ≥ −(5 + n) |f (y)| dy. B(x,R)

B(x,R)

Since Z

B(x,R)

ϕ(y) dy ≥

it follows that

Z

B(x,R/2)

ϕ(y) dy = |B(x, R/2)| = 2−n |B(x, R)|,

f (z) ≥ −(5 + n)2n

1 |B(x, R)|

Z

B(x,R)

|f (y)| dy.

In combination with our upper bound for f (z), we infer that Z |f (z)| ≤ C(n)R−n |f (y)| dy, B(x,R)

which implies (7.2a). Finally, for z ∈ B(x, R/2), then (7.2b) is immediate if |Df (z)| = 0, so suppose |Df (z)| 6= 0 and set ξ=

Df (z) |Df (z)|

and y = z − (R/4)ξ. Then (7.3) implies that R R f (y) ≥ f (z) + Df (x) · ξ = f (z) + |Df (z)|. 4 4 Hence 4 8 |Df (z)| ≤ |f (y) − f (z)| ≤ sup |f |. R R B(z,R/4) From (7.2a), we infer that sup B(z,R/4)

|f | ≤ C(n)R−n

Z

B(z,R/2)

|f (y)| dy

and hence sup B(z,R/4)

|f | ≤ C(n)R

−n

Z

B(x,R)

|f (y)| dy

Combining this inequality with our estimate for |Df (z)| then yields (7.2b). 

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Our next step is to identify the second derivatives of an arbitrary convex function with appropriate signed Radon measures. Lemma 7.2. Let f : Rn → R be convex. Then, for all i and j in {1, . . . , n}, there is a signed Radon measure µij such that µji = µij and Z Z f (y)Dij ϕ(y) dy = ϕ(y) dµij (7.5) Rn

2

Rn

n

for all ϕ ∈ C (R ) with compact support. Moreover, µii is a measure for i = 1, . . . , n. Proof.

For each vector ξ ∈ Rn , define the linear functional Lξ on Cc2 by Z Lξ (ϕ) = f (y)Dij ϕ(y)ξ i ξ j dy. Rn

Now let ζ be a mollifier, that is, ζ ∈ C 2 (Rn ) with support in B(0, 1), ζ ≥ 0 in Rn and Z ζ(y) dy = 1. Rn

ε

For ε > 0, define f by

f ε (x) =

Z

Rn n

f (x − εy)ζ(y) dy.

Then f ε is convex and C 2 on R , so Z Z f ε (y)Dij ϕ(y)ξ i ξ j dy = Rn

Rn

ϕ(y)Dij f ε (y)ξ i ξ j dy ≥ 0

by Lemma 7.1 if ϕ ≥ 0. Since f ε converges to f in L1 (S), where S is the support of ϕ, it follows that Lξ (ϕ) ≥ 0 for any nonnegative ϕ ∈ Cc2 , and then the Riesz representation theorem (in the form of Corollary 1 in Section 1.8 of [43]) provides a Radon measure µξ such that Z Lξ (ϕ) = ϕ(y) dµξ Rn

Cc2 .

for any ϕ ∈ To define µij , we proceed in two steps. First, for i = 1, . . . , n, we define µii to be µei . Then for i 6= j, we define 1 1 µij = µ(ei +ej )/√2 − µei − µej .  2 2

Our next step is a regularity result for the gradient of convex functions. Corollary 7.3. If f : Rn → R is convex, then Df ∈ BVloc (Rn ).

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Proof. By using the function f ε from the proof of Lemma 7.2, we see that |Df ε | is uniformly bounded on compact subsets of Rn and hence Df ∈ n L∞ loc (R ). Now let ϕ be a vector-valued Cc2 function with |ϕ| ≤ 1 and let V be a bounded open set containing the support of ϕ. Then, for k = 1, . . . , n, we have Z Z Dk f (y) div ϕ dy = − f (y)Dik ϕi (y) dy. Rn

Rn

But the definition of µij implies that Z Z n X f (y)Dik ϕi (y) dy = ϕ(y) dµik ≤ |µik |(V ) < ∞. Rn

Rn

i=1

n

Hence Dk f ∈ BVloc (R ).



Now the Lebesgue decomposition theorem implies that each µij is the sum of a signed measure µij,ac , which is absolutely continuous with respect to Lebesgue measure, and a signed measure µij,s , which is singular with respect to Lebesgue measure. We then write Dij f for the Radon-Nikodym derivative of µij,ac with respect to Lebesgue measure. The final step in the proof of Aleksandrov’s theorem is to show that this definition of Dij f agrees with the second derivative of f almost everywhere. Theorem 7.4. If f is convex, then f is twice differentiable almost everywhere. Specifically, if Dij f is the Radon-Nikodym derivative of µij,ac with respect to Lebesgue measure, then there is a measurable set A ⊂ Rn with |Rn \ A| = 0 such that 1 |f (x)−f (y)−Df (x)·(y −x)− Dij f (x)(y −x)i (y −x)j | = o(|x−y|2 ) (7.6) 2 as y → x for all x ∈ A. Proof. From Lemma 7.2, Corollary 7.3, and the continuity of the L1 norm with respect to translation, it follows that there is a measurable set A ⊂ Rn norm with |Rn \ A| = 0 such that Df (x) exists, lim R−n

R→0

Z

B(x,R)

|Df (y) − Df (x)| dy = 0,

Dij f (x) exists for all i, j, Z lim R−n |Dij f (y) − Dij f (x)| dy = 0

R→0

B(x,R)

(7.7a) (7.7b) (7.7c) (7.7d)

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for all x ∈ A. Moreover, because µij,s is singular with respect to Lebesgue measure, we have lim R−n |µij,s |(B(x, R)) = 0.

R→0

(7.8)

We now fix a point x ∈ A and assume without loss of generality that x = 0. Then we choose R > 0 and y ∈ B(0, R). Finally, we write f ε for the mollification of f described in the proof of Lemma 7.2. Taylor’s theorem then implies that Z 1 f ε (y) = f ε (0) + Df ε (0) · y + (1 − s)Dij f ε (sy)y i y j dy. 0

For ease of notation, we now define F ε and F by

1 F ε (y) = f ε (y) − f ε (0) − Df ε (0) · y − Dij f (0)y i y j , 2 1 F (y) = f (y) − f (0) − Df (0) · y − Dij f (0)y i y j . 2 Then our Taylor’s theorem equation can be rewritten as Z 1 F ε (y) = (1 − s)[Dij f ε (sy) − Dij f (0)]y i y j ds. 0

We now multiply this equation by a C 2 (Rn ) function ϕ with support in B(0, R) satisfying |ϕ| ≤ 1 in Rn and integrate with respect to y over B(0, R) to infer that Z Z Z 1 ϕ(y)F ε (y) dy = ϕ(y) (1−s)[Dij f ε (sy)−Dij f (0)]y i y j ds dy. B(0,R)

B(0,R)

0

Applying Fubini’s theorem and then using the change of variables z = sy, we conclude that Z Z 1Z   z [Dij f ε (z) − Dij f (0)]z i z j dz ds. ϕ(y)F ε (y) dy = ϕ s B(0,R) 0 B(0,sR)

Our next step is to take the limit as ε → 0, and the analysis of this limit procedure is simplified by defining the function gε by Z z 1 gε (s) = ϕ(z) Dij f ε (z)z i z j dz. |B(0, Rs)| B(0,Rs) s Then

so

1 gε (s) = |B(0, Rs)|

  z Dij ϕ z i z j f ε (z) dz, s B(0,Rs)

Z

lim gε (s) = g(s),

ε→0

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with Z   z 1 g(s) = f (z)Dij ϕ z i z j dz |B(0, Rs)| B(0,Rs) s Z   1 z i j = ϕ z z dµij |B(0, Rs)| B(0,Rs) s Z z 1 ϕ z i z j Dij f (z) dz = |B(0, Rs)| B(0,Rs) s Z z 1 + ϕ z i z j dµij,s . |B(0, Rs)| B(0,Rs) s

We wish to show that gε → g in L1 (0, 1). To this end, we prove a uniform upper bound for gε , and this upper bound is proved by noting first that Z C(n) Dij f ε (z)z i z j dz gε (s) ≤ (Rs)n B(0,Rs) because |ϕ| ≤ 1 and Dij f ε (z)z i z j ≥ 0. Then, we recall that Z f ε (z) = ζε (w − z)f (w) dw, B(z,ε)

where ζε is defined by ζε (y) = ε−n ζ(y/ε). Hence, if we set I=

Z

Dij f ε (z)z i z j dz,

B(0,Rs)

−n

then gε (s) ≤ C(n, R)s I and Z Z I= zizj D2 ζε (w − z)f (w) dw dz B(0,Rs) B(z,ε) Z Z = zizj ζε (w − z) dµij dz B(0,Rs)

B(z,ε)

Fubini’s theorem then implies that Z Z I≤ B(0,Rs+ε)

B(0,Rs)∩B(w,ε)

ζε (w − z)z i z j dz, dµij

because the set {(w, z) : |z − w| < ε, z ∈ B(0, Rs)} is a subset of {(w, z) : |z − w| < ε, |w| < Rs + ε}.

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Recalling also that 0 ≤ ζ ≤ C(n), we have I ≤ |Rs|2 C(n)ε−n min{|Rs|n , εn } sup µξ (B(0, Rs + ε)). |ξ|=1

By virtue of (7.7d) and (7.8), it follows that sup µξ (B(0, Rs + ε)) ≤ C(Rs + ε)n

|ξ|=1

and hence there is a constant C, determined only by n and R such that gε (s) ≤ C for all ε and s in (0, 1). The dominated convergence theorem then implies that Z Z 1 −n lim R ϕ(y) (1 − s)[Dij f ε (sy) − Dij f (0)]y i y j ds dy ε→0

B(0,R)

= R−n

0

Z

ϕ(y)

B(0,R)

1

Z

0

(1 − s)[Dij f (sy) − Dij f (0)]y i y j ds dy,

and therefore lim R

ε→0

−n

Z

ϕ(y)F ε (y) dy B(0,R)

is equal to the same quantity. In other words, Z R−n ϕ(y)F (y) dy B(0,R)

=R

−n

Z

ϕ(y)

B(0,R)

Z

1

0

(1 − s)[Dij f (sy) − Dij f (0)]y i y j ds dy.

From (7.7d) and (7.8), we therefore infer that there is a continuous, increasing function ω with ω(0) = 0 such that Z 1 ϕ(y)F (y) dy ≤ ω(R) |B(0, R)| B(0,R) for any function ϕ ∈ C 2 with compact support in B(0, R) such that |ϕ| ≤ 1 and hence Z 1 |F (y)| dy ≤ ω(R). (7.9) |B(0, R)| B(0,R) Our final step in the proof is to show that this integral estimate can be converted into a pointwise estimate. For this step, we first note that h, defined by h(y) = f (y) − f (0) − Df (0) · y,

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is convex, so Lemma 7.1 gives a constant C1 (n) such that Z sup |Dh| ≤ C1 R−n−1 |h| dy B(0,R/2)

B(0,R)

for all R > 0. Writing Λ for the maximum eigenvalue of the matrix [Dij f (0)] and using (7.9), we conclude that there are constants C2 (Λ, n) and R1 such that sup B(0,R/2)

|DF | ≤ C2 R

if 0 < R < R1 . Now, let ε ∈ (0, 1) and infer from (7.9) that there is a constant R0 such that Z |F | dy ≤ 2−n−1 εn+1 |B(0, R)| B(0,R)

if 0 < R < R0 . Hence, if 0 < R < min{R0 , R1 } and y ∈ B(0, R/4), we have Z 1 inf |F | ≤ |F | dy B(y,εR/2)| |B(y, εR/2)| B(y,εR/2) Z 2 n εn 1 ≤ |F | dy ≤ εR2 . |B(y, εR)| B(0,R) 2 In particular, there is a point z ∈ B(y, εR/(1 + 2C2 )) such that |F (z)| ≤ 1 2 2 εR , so ! 1 ε |F (y)| ≤ |F (z)| + sup |DF | |y − z| ≤ εR2 + C2 R R < εR2 . 2 1 + 2C2 B(0,R) Since ε is arbitrary, we infer (7.6).

7.3



Preliminary results for the comparison theorem for viscosity solutions

We wish to prove a comparison theorem for viscosity subsolutions and viscosity supersolutions. This theorem is a consequence of a number of technical results which we collect in this section, and the comparison theorem itself will be proved in the next section. Our first preliminary result concerns the behavior of a certain combination of upper semicontinuous functions.

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Lemma 7.5. Let Ω be a subset of Rn and let u1 and u2 be two upper semicontinuous functions on Ω. For α > 0, set   α Mα = sup u1 (x) + u2 (y) − |x − y|2 , 2 x∈Ω y∈Ω

and suppose that there is a nonnegative constant α0 such that Mα0 is finite. For α ≥ α0 , suppose xα and yα satisfy  h i α lim Mα − u1 (xα ) + u2 (yα ) − |xα − yα |2 = 0. α→∞ 2 Then lim α|xα − yα |2 = 0,

(7.10)

α→∞

and, for any increasing sequence (αk ) which diverges to ∞ such that (xαk ) converges, say to x ˆ, we have lim Mα = u1 (ˆ x) + u2 (ˆ x) = sup(u1 (x) + u2 (y)).

α→∞

Proof.

(7.11)

x∈Ω y∈Ω

We set

and note that

h i α δα = Mα − u1 (xα ) + u2 (yα ) − |xα − yα |2 , 2 lim δα = 0.

α→∞

Since |xα − yα |2 ≥ 0, it follows that Mα is a decreasing function of α and hence lim Mα < ∞.

α→∞

For α ≥ 2α0 , we have

Mα/2 ≥ u1 (xα ) + u2 (yα ) −

α |xα − yα |2 2

= u1 (xα ) + u2 (yα ) − α|xα − yα |2 +

α |xα − yα |2 . 2

It follows that Since

α|xα − yα |2 ≤ 2(Mα/2 − Mα + δα ). lim Mα/2 − Mα = 0

α→∞

and lim δα = 0,

α→∞

we infer (7.10), and the upper semicontinuity of u1 (x) + u2 (y) in Ω × Ω implies (7.11). 

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Our next result is a slight variation of the theorem of Aleksandrov proved in the previous section. To state this variation, we recall that a real-valued function ϕ defined on some subset S of Rn is semiconvex if there is a constant λ ≥ 0 such that the function ϕλ , defined by λ ϕλ (x) = ϕ(x) + |x|2 2 is convex. Any such constant λ is called a semiconvexity constant for ϕ. Theorem 7.6. Let ϕ : Rn → R be semiconvex. Then ϕ is twice differentiable a.e. in Rn . Proof. Since ϕλ is convex for some λ ≥ 0, it follows from Theorem 7.4 that ϕλ is twice differentiable a.e., and hence so is ϕ.  The Aleksandrov theorem is used to prove an estimate similar to the Aleksandrov-Bakel0man-Pucci estimate. Lemma 7.7. Let ϕ be a real-valued semiconvex function defined on Rn and suppose ϕ has a strict local maximum at some x ˆ. For p ∈ Rn , define ϕp by ϕp (x) = ϕ(x) + p · x. Then, for any positive constants ρ and δ, the set, K, of all x ∈ B(ˆ x, ρ) for which there is a p ∈ B(0, δ) such that ϕ has a local maximum at x has positive measure. Proof. There is no loss of generality in assuming that ρ is small enough that ϕ has a unique maximum in B(ˆ x, ρ) at xˆ. We suppose now that ϕ ∈ C 2 . Then there is a positive number δ0 such that |p| < δ0 implies that the maximum value for ϕp in B(ˆ x, ρ) can only be attained somewhere in B(ˆ x, ρ). Since Dϕ(x) + p = 0 at such a point, it follows that B(0, δ) ⊂ Dϕ(K) if δ ≤ δ0 . We may therefore assume that δ ≤ δ0 . Now let λ be a constant of semiconvexity for ϕ. Then −λI ≤ D2 ϕ in B(ˆ x, ρ) and, if x ∈ K, it follows that D2 ϕp (x) ≤ 0 for some p ∈ Rn (and, in fact |p| < δ0 ). Therefore −λI ≤ D2 ϕ ≤ 0 in K, and hence | det D2 ϕ| ≤ λn

in K. We now use the change of variables theorem (Theorem 1.7) to infer that Z |B(0, δ)| ≤ |Dϕ(K)| ≤ | det D2 ϕ| dx ≤ |K|λn . K

Hence

 n δ . |K| ≥ c(n) λ

(7.12)

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If ϕ is an arbitrary semiconvex function, we write ϕ(m) for the mollification of ϕ with parameter 1/m, that is, Z  y (m) ϕ (x) = ϕ(y)ψ x − dy m Rn

for some (fixed) nonnegative C 2 (Rn ) function ψ with support in B(0, 1) and kψkL1 = 1. If λ is a constant of semiconvexity for ϕ, it follows that λ is also a constant of semiconvexity for each ϕ(m) and hence  n δ |Km | ≥ c(n) λ by (7.12), where Km is the set corresponding to ϕ(m) . Next, suppose that ∞ [ ∞ \ Km . x∈ k=1 m=k

Then x ∈ B(ˆ x, ρ), so, for each positive integer m, there is a point pm ∈ Rn (m) m with |p | < δ such that ϕpm has a local maximum at x. In other words, ϕ(m) (x) + pm · x ≥ ϕ(m) (y) + pm · y

for all y ∈ B(ˆ x, ρ). It follows that (pm ) has a convergent subsequence m(j) (p ) with limit p. Since ϕ(m(j)) (x) + pm(j) · x ≥ ϕ(m(j)) (y) + pm(j) · y

for all y ∈ B(ˆ x, ρ), and ϕ(m) converges pointwise to ϕ, we can take the limit as j → ∞ to conclude that ϕ(x) + p · x ≥ ϕ(y) + p · y

for all y ∈ B(ˆ x, ρ); in other words, x ∈ K. Therefore ∞ [ ∞ \ Km ⊂ K, k=1 m=k

which implies that

∞ [ ∞ \ Km ≤ |K|. k=1 m=k

The Borel-Cantelli lemma implies that ∞ ∞ ∞ \ [ [ Km = lim Km . k→∞ k=1 m=k

m=k

In conjunction with our lower bound for |Km |, we therefore conclude that (7.12) is valid for all semiconvex ϕ, and hence |K| has positive measure. 

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To continue, we need to introduce a generalization of JS2,+ . Specifically, if u is defined on a set S and x ∈ S, we define J¯S2,+ u(x) to be the set of all (p, r) such that there is a sequence (xm , pm , rm ) in S × Rn × Sn such that (pm , rm ) ∈ JS2,+ u(xm ) and lim (xm , u(xm ), pm , rm ) = (x, u(x), p, r).

m→∞

We also define J¯S2,− u(x) by replacing JS2,+ with JS2,− . Finally, if S = Rn , we suppress S from the notation. Lemma 7.8. Let f ∈ C(Rn ) and suppose that f is semiconvex with constant of semiconvexity λ. If there is a matrix B ∈ Sn such that 1 max f (x) − Bij xi xj = f (0), 2

x∈Rn

then there is a matrix r ∈ Sn such that (0, r) ∈ J¯2 f (0) and −λI ≤ r ≤ B. Proof.

Define f˜ by 1 f˜(x) = f (x) − Bij xi xj − |x|4 . 2

Then f has second derivatives a.e. by Theorem 7.6 and Lemma 7.7 implies that, for any δ > 0, there is q δ ∈ Rn such that f˜qδ has a maximum at some ξδ with |ξδ | < δ and f has second derivatives at ξδ . Then the differentiability of fqδ at ξδ implies that Dfqδ (ξδ ) = 0 and hence there is a constant C1 , determined only by n and the matrix B, such that |Df (ξδ )| ≤ Cδ.

(7.13)

Using the explicit description for f˜ and the fact that D2 fqδ (ξδ ) ≤ 0, we see that there is a constant C2 , determined only by n and the matrix B, such that D2 f (ξδ ) ≤ B + Cδ 2 I.

(7.14)

Moreover, the semiconvexity of f implies that D2 f (ξδ ) ≥ −λI.

(7.15)

From (7.14) and (7.15), we conclude that there is a sequence (δm ) with δm → 0 such that D2 f (ξδ(m) ) converges to a matrix r. Conditions (7.13), (7.14), and (7.15) then imply the desired result. 

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Our final result is a property of the sup convolution, defined by (7.16) below. Lemma 7.9. Let λ > 0, let v ∈ U SC(Rn ) be bounded above, and define the function vˆ by   λ (7.16) vˆ(ξ) = sup v(x) − |x − ξ|2 . 2 x∈Rm If η and q are in Rn and r ∈ Sn with (q, r) ∈ J 2,+ vˆ(η), then  q , (q, r) ∈ J 2,+ v η + λ  1 2 q vˆ(η) + |q| = v η + . 2λ λ In particular, if (0, r) ∈ J¯2,+ vˆ(0), then (0, r) ∈ J¯2,+ v(0).

(7.17a) (7.17b)

Proof. Fix η, q and r so that (q, r) ∈ J 2,+ vˆ(η), and let y be a point in Rn such that λ vˆ(η) = v(y) − |y − η|2 . 2 Such a point exists because v is upper semicontinuous. Then for all x and ξ in Rn , we have v(x) −

λ |ξ − x|2 ≤ vˆ(ξ) 2 1 ≤ vˆ(η) + q · (ξ − η) + rij (ξ − η)i (ξ − η)j + o(|ξ − η|2 ) 2 λ 1 = v(y) − |y − η|2 + q · (ξ − η) + rij (ξ − η)i (ξ − η)j 2 2 + o(|ξ − η|2 ).

In particular, if ξ = x − y + η, then ξ − x = η − y and ξ − η = x − y, so 1 v(x) ≤ q · (x − y) + rij (x − y)i (x − y)j + o(|x − y|2 ), 2 2,+ and hence (q, r) ∈ J v(y). Now, let α be a real parameter at our disposal and take x = y and ξ = η + α(λ(η − y) + q). It follows that v(y) −

λ |(η − y) + α(λ(η − y) + q)|2 2 λ ≤ v(y) − |y − η|2 + αq · (λ(η − y) + q) 2 + O(|α(λ(η − y) + q)|2 ).

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Simple rearrangement of this inequality shows that 0≤

λ (|ξ − y|2 − |y − η|2 ) + q · (ξ − η) + O(α2 ). 2

Then |ξ − y|2 − |y − η|2 = |ξ − η|2 − 2(y − η) · (ξ − η), and |ξ − η|2 = α2 |λ(η − y) + q|2 = O(α2 ). Hence 0 ≤ α|λ(η − y) + q|2 + O(α2 ). If α < 0, we conclude that |λ(η − y) + q|2 = O(|α|) and sending α → 0 yields λ(η − y) + q = 0, which is the same as q y=η+ . λ Hence  1 2 q vˆ(η) + |q| = v η + . 2λ λ Finally, if (0, r) ∈ J¯2,+ vˆ(0), then for any sequence ((ξm , q m , rm )) in Ω × Rn × S n such that (q m , rm ) ∈ J 2,+ vˆ(ξm )) and (ξm , q m , rm ) → (0, 0, r), we have that   qm (q m , rm ) ∈ J 2,+ v ξm + λ and   qm 1 m2 v ξm + = ˆ(ξm ) + |q | . λ 2λ Moreover, we have   qm v(0) ≥ lim v ξm + m→∞ λ   1 m2 = lim vˆ(ξm ) + |q | m→∞ 2λ = vˆ(0) ≥ v(0),

so vˆ(0) = v(0) and therefore (0, r) ∈ J¯2,+ v(0).



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Our next result concerns a function of 2n variables. In this lemma, we use the matrix norm kAk = sup Aij ξ i ξ j . |ξ|=1

Lemma 7.10. Let u1 and u2 be upper semicontinuous, real-valued functions defined on Rn with u1 (0) = u2 (0) = 0 and define w on R2n by w(x, y) = u1 (x) + u2 (y), and abbreviate (x, y) to X. If there is a matrix A ∈ S2n such that 1 w(X) ≤ Aij X i X j (7.18) 2

for all X ∈ R2n , then, for all ε > 0, there are matrices r1 and r2 in Sn such that (0, ri ) ∈ J¯2,+ ui (0) for i = 1, 2 and   1   1 r 0 ≤ A + εA2 . (7.19) + kAk I ≤ − 0 r2 ε Proof.

Fix ε > 0. Then the Cauchy-Schwarz inequality implies that   1 Aij X i X j ≤ (A + εA2 )ij Ξi Ξj + + kAk |X − Ξ|2 (7.20) ε

for all X and Ξ in R2n . If we now set λ = (7.18) and (7.20) that

1 ε

+ kAk, then it follows form

λ 1 |X − Ξ|2 ≤ (A + εA2 )ij Ξi Ξj . 2 2 Next, we define u ˆi and w ˆ by   λ uˆi (ξ) = sup ui (x) − |x − ξ|2 2 x∈Rn w(X) −

for i = 1, 2, w(Ξ) ˆ = sup X∈R2n



 λ 2 w(X) − |X − Ξ| , 2

and notice that w(ξ, ˆ η) = uˆ1 (ξ) + uˆ2 (η). In addition, we have u ˆi (0) ≥ ui (0) = 0 and u ˆ1 (0) + u ˆ2 (0) = w(0) ˆ ≤0 by virtue of (7.21), so u ˆi (0) = 0.

(7.21)

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Moreover, for each X ∈ R2n , the function WX , defined by λ λ WX (Ξ) = w(X) − |X − Ξ|2 + |Ξ|2 , 2 2 satisfies λ WX (Ξ) = w(X) − |X|2 + λX · Ξ, 2 and hence WX is convex. It follows that w ˆ and u ˆi are semiconvex with constant λ. We can therefore apply Lemma 7.8 to w ˆ with B = A + εA2 to conclude 2n that there is a matrix R ∈ S such that (0, R) ∈ J¯2,+ w(0) and −λI ≤ R ≤ A + εA2 . From the form of w, it follows that there are matrices r1 and r2 in Sn such that  1  r 0 R= 0 r2 and ri ∈ J¯2,+ uˆi (0). Since u ˆi (0) = 0, it follows from Lemma 7.9 that ri ∈ J¯2,+ ui (0).  Finally, we prove a version of the preceding result for functions defined on more general sets. Lemma 7.11. Let S1 and S2 be locally compact subsets of Rn , let ui be upper semicontinuous on Si , let α ≥ 0, and define ϕ on R2n by ϕ(x, y) = α 2 2 |x − y| . Define w on S = S1 × S2 by w(x, y) = u1 (x) + u2 (y) ˆ = (ˆ and suppose that w − ϕ attains a local maximum at some X x, yˆ) ∈ S. 1 2 n Then there are matrices r and r in S such that (α(ˆ x − yˆ), r1 ) ∈ J¯2,+ u1 (ˆ x), (7.22a) 2 2,+ ¯ (−α(ˆ x − yˆ), r ) ∈ J u2 (ˆ y ), (7.22b)  1    r 0 I −I −3αI ≤ ≤ 3α . (7.22c) 0 r2 −I I Proof. We first observe that a simple translation reduces to the case that x ˆ = yˆ = 0. If Si is not all of Rn , we extend ui to all of Rn by setting ui (x) = −∞ if x ∈ / Si . For such an extension, we have J¯S2,+ ui (x) = J¯2,+ ui (x) i for x ∈ Si . Since S1 and S2 are locally compact, it follows that ui ∈ U SC(Rn ), so we can apply Lemma 7.10 with   I −I A=α −I I and ε = 1/α. We just have to note that A2 = 2αI and kAk = 2α. 

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283

The comparison principle for viscosity sub- and supersolutions

A comparison principle follows rather easily from the considerations of the previous section provided F satisfies suitable structure conditions. Theorem 7.12. Let Ω be a bounded open subset of Rn and let F ∈ C(Ω × R × Rn × Sn ). Suppose that there is a positive constant c0 such that c0 (z1 − z2 ) ≤ F (x, z2 , p, r2 ) − F (x, z1 , p, r1 )

(7.23)

for all (x, zi , p, ri ) ∈ Ω × R × Rn × Sn with z1 ≥ z2 and r1 ≤ r2 . Suppose also that there is a continuous increasing function ω defined on [0, ∞) with ω(0) = 0 such that F (y, z, α(x − y), r2 ) − F (x, z, α(x − y), r1 ) ≤ ω(α|x − y|2 + |x − y|) (7.24) for all x and y in Ω, all z ∈ R, and all α > 0 and r1 and r2 in Sn such that    1    I0 r 0 I −I −3α ≤ ≤ 3α . (7.25) 0I 0 −r2 −I I If u ∈ U SC(Ω) is a viscosity subsolution of F = 0 in Ω, and if v ∈ LSC(Ω) is a viscosity supersolution of F = 0 in Ω with u ≤ v on ∂Ω, then u ≤ v in Ω. ˜ ∈ Ω such that u − v attains Proof. If u 6≤ v in Ω, then there is a point x its positive maximum value over Ω at x ˜. We then use Lemma 7.11 with u1 = u and u2 = −v, observing that J¯2,− v(x) = −J¯2,+ (−v(x)) for any v ∈ Ω and that Ω is locally compact. 

7.5

A test function construction for the oblique derivative problem

In order to apply the arguments of the preceding section to the oblique derivative problem, we first construct a suitable test function. This construction takes several steps. Our first step is to solve a Hamilton-Jacobi equation. In what follows, we shall look at functions of two multi-dimensional variables, one in Rn and one in Rm , with n and m positive integers. For ease of notation, we use x as dummy variable in Rn , and we use q as a dummy variable in Rm .

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Lemma 7.13. Let U be a bounded open set in Rn , let m ≥ 1, and let H be a C 2 (U × Rm ) function. Suppose that there is a nonnegative constant A such that |H(x, q)| ≤ A(|q|2 + 1)1/2 ,

|Hx (x, q)| ≤ A(|q|2 + 1)1/2 , A |Hq (x, q)| ≤ A, |Hqq (x, q)| ≤ . 2 (|q| + 1)1/2

(7.26a) (7.26b)

Suppose also that there is a nonnegative constant C0 such that |Hxx (x, q)| ≤ C0 (|q| + 1),

|Hxq (x, q)| ≤ C0 .

(7.27)

If δ ∈ (0, 4/A), then there is a function ψ ∈ C 2 (U × Vδ ), where Vδ = {(y, t) ∈ Rm+1 : |t| ≤ δ(|y|2 + 1)1/2 }

(7.28)

ψt (x, y, t) + H(x, ψy (x, y, t)) = 0

(7.29)

such that

for all (x, y, t) ∈ U × Vδ . Moreover ψ(x, 0, 0) ≤ 1

(7.30)

for all x ∈ U ,

1 (|y|2 + 1) (7.31) 4 for all (x, y, t) ∈ U × Vδ , and there is a constant C1 , determined only by A and C0 such that ψ(x, y, t) ≥

|ψ(x, y, t)| + |ψx (x, y, t)| + |ψxx (x, y, t)| ≤ C1 (|y|2 + 1),

(7.32a)

|ψyy (x, y, t)| + |ψyt (x, y, t)| + |ψtt (x, y, t)| ≤ C1 .

(7.32c)

|ψy (x, y, t)| + |ψt (x, y, t)| + |ψxy (x, y, t)| + |ψxt (x, y, t)| ≤ C1 (|y| + 1), (7.32b)

Proof.

First, we fix (x, y, t) ∈ U × Vδ and define J on Rm by

1 |q − y|2 + tH(x, q). 2 Since J is continuous, and H grows linearly as |q| → ∞, it follows that there is a p ∈ Rm at which J attains its minimum. In addition Jq (p) = 0, so J(q) =

y = p + tHq (x, p).

(7.33)

It follows from (7.26) and the definition of Vδ that |y − p| = |t||Hq (x, p)| ≤ A|t| ≤ Aδ(|y|2 + 1)1/2 .

(7.34)

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In addition, 2 (|p| + 1)1/2 − (|y|2 + 1)1/2 ≤ |y − p|,

so

(1 − Aδ)(|y|2 + 1)1/2 ≤ (|p|2 + 1)1/2 ≤ (1 + Aδ)(|y|2 + 1)1/2 .

(7.35)

Our restriction on δ then implies that 3 5 (|y|2 + 1)1/2 ≤ (|p|2 + 1)1/2 ≤ (|y|2 + 1)1/2 . 4 4 From (7.26), we infer that |tHqq (x, p)| ≤

(7.36)

δ(|y|2 + 1)1/2 A , (|p|2 + 1)1/2

and then from (7.36) and our restriction on δ, we have 1 , 2 (I + Hqq (x, p))−1 ≤ 2. |tHqq (x, p)| ≤

(7.37a) (7.37b)

Before using these calculations to construct ψ, we observe that, if r is another element of Rm that minimizes J, then r + tHq (x, r) = p + tHq (x, p). It follows that 1 |p − r| 2 by virtue of (7.37a), and hence p is unique. In other words, there is a function p(x, y, t) which solves (7.33), and the implicit function theorem implies that p ∈ C 1 (U × Vδ ). We now define ψ by |p − r| = |t||Hq (x, p) − Hq (x, r)| ≤

1 1 (|y|2 + 1) − |p(x, y, t) − y|2 − tH(x, p(x, y, t)). 2 2 Direct calculation of the partial derivatives of ψ along with (7.33) gives us (7.29) and ψ(x, y, t) =

ψx (x, y, t) = −tHx (x, y, t), ψy (x, y, t) = p(x, y, t).

Moreover, we have ψ(x, 0, 0) =

1 1 − |p(x, 0, 0)|2 2 2

(7.38a) (7.38b)

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which easily implies (7.30). In addition, J(p) ≤ J(y) = tH(x, y), so (7.26) implies that   1 1 ψ(x, y, t) ≥ (|y|2 + 1) − tH(x, y) ≥ − Aδ (|y|2 + 1), 2 2 which implies (7.31) because of our restriction on δ. Since 1 ψ(x, y, t) ≤ (|y|2 + 1) − tH(x, p(x, y, t)), 2 the estimates on ψ, ψx , ψy , and ψt in (7.32) now follow from (7.29), (7.38), and (7.35). Finally, we differentiate (7.33) with respect to x, y, and t to deduce that px (x, y, t) = −t(I + tHqq (x, p(x, y, t))−1 Hqx (x, p(x, y, t)), py (x, y, t) = (I + tHqq (x, y, t))−1 ,

It follows that

pt (x, y, t) = −(I + tHqq (x, y, t))−1 Hq . |px | ≤ 2C0 (|y| + 1),

|py | ≤ 2,

|px | ≤ 2A,

and the second derivative estimates for ψ are obtained by differentiating the equations for the first derivatives of ψ and combining the first derivative estimates for p with (7.27), (7.26), and our restriction on δ.  Our next step is to construct a function like ψ but defined for all (y, t) ∈ Rm+1 . To simplify writing, we abbreviate (y, t) to ξ. Lemma 7.14. With U , H, δ, and m as in Lemma 7.13, there is a function ϕ ∈ C 2 (U × Rm+1 ) such that for all x ∈ U ,

ϕ(x, 0) ≤ 1

(7.39)

1 2 |ξ| (7.40) 8 for all (x, ξ) ∈ U × Rm+1 , and there are constants δ˜ and C2 , determined only by A and C0 , such that ˜ ϕt (x, ξ) + H(x, ϕξ (x, ξ)) ≥ 0 if t ≥ −δ|y|, (7.41a) ˜ ϕt (x, ξ) + H(x, ϕξ (x, ξ)) ≤ 0 if t ≤ δ|y|, (7.41b) ϕ(x, ξ) ≥

and

|ϕ(x, ξ)| + |ϕx (x, y, t)| + |ϕxx (x, y, t)| ≤ C2 (|ξ|2 + 1), |ϕξ (x, ξ)| + |ϕxξ (x, ξ)| ≤ C2 (|y| + 1), |ϕξξ (x, y, t)| ≤ C2 .

(7.42a) (7.42b) (7.42c)

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Proof. Let f ∈ C ∞ (R) satisfy 0 ≤ f (r) ≤ 1 and rf 0 (r) ≤ 0 for all r ∈ R, f (r) = 1 if |r| ≤ 1, and f (r) = 0 if |r| ≥ 2. With g defined by g(r) = 1 − f (2r) and α = min{1, δ/3}, we define fα and gα by     t t fα (y, t) = f , gα (y, t) = g . α(|y|2 + 1)1/2 α(|y|2 + 1)1/2 Note in particular that fα has support in Vδ . With β > 0 to be chosen, we then set ϕ(x, y, t) = fα (y, t)ψ(x, y, t) + gα (y, t)βt2 . Since gα (0, 0) = 0 and fα (0, 0) = 1, (7.39) follows from (7.30). Next, if t ≤ α(|y|2 + 1)1/2 , then fα (y, t) = 1, so 1 1 1 1 ϕ(x, y, t) ≥ ψ(x, y, t) ≥ (|y|2 + 1) ≥ |y|2 + 2 t2 ≥ |ξ|2 4 8 8α 8 because α ≤ 1. On the other hand, if t > α(|y|2 + 1)1/2 , then gα (y, t) = 1, so β βα2 2 1 ϕ(x, y, t) ≥ βt2 ≥ t2 + |y| ≥ |ξ|2 2 2 8 provided β ≥ 1/(4α2 ). With this restriction on β, (7.40) holds. To verify the remaining inequalities, we first compute ∂fα 1 ∂gα 1 (y, t) = f 0, (y, t) = g0, ∂t ∂t α(|y|2 + 1)1/2 α(|y|2 + 1)1/2 ∂fα ty ∂gα ty (y, t) = − f 0, (y, t) = − g0, ∂y ∂y α(|y|2 + 1)3/2 α(|y|2 + 1)3/2 where, here and below, f 0 and g 0 are always evaluated at t/(α(|y|2 + 1)1/2 ). When |t| ≤ α(|y|2 + 1)1/2 /2, then ϕ(x, y, t) = ψ(x, y, t) so ψt (x, y, t) + H(x, ψy (x, y, t)) = 0, so (7.41a) and (7.41b) hold in this case. When α(|y|2 + 1)1/2 /2 ≤ t ≤ α(|y|2 + 1)1/2 , we have βt2 ϕt (x, y, t) + H(x, ϕy (x, y, t)) = ψt (x, y, t) + 2βtgα (y, t) + α(|y|2 + 1)1/2 ty + H(x, ψy (x, y, t) − βt2 g0) α(|y|2 + 1)3/2 ≥ ψt (x, y, t) + H(x, ψy (x, y, t))

t|y| βt2 g 0 [1 − A 2 ] 2 1/2 |y| + 1 α(|y| + 1) because |Hq | ≤ A and gα and g 0 are nonnegative. By construction, ψt (x, y, t) + H(x, ψy (x, y, t)) = 0, and we have t|y| A 2 ≤ Aα ≤ Aδ ≤ 1, |y| + 1 +

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so (7.41a) holds in this case as well. A similar argument shows that (7.41b) holds for α(|y|2 + 1)1/2 /2 ≤ −t ≤ α(|y|2 + 1)1/2 . If t ≥ α(|y|2 + 1)1/2 , then gα (y, t) = 1 and g 0 = 0, so ϕt (x, y, t) + H(x, ϕy (x, y, t)) = ψt (x, y, t) +

α(|y|2

1 ψf 0 + 2βt + 1)1/2

+ H(x, fα (y, t)ψy (x, y, t) + f 0 ψ(x, y, t)) F ≥ (|y|2 + 1)1/2 [2βα − C1 − − 8C1 F ] α by virtue of (7.32). (Here F = sup |f 0 |; note that we can take F ≤ 2.) If β is sufficiently large (determined only by α, C1 and sup |f 0 |), we obtain (7.41a) in this case, and a similar argument gives (7.41b) when t ≤ −α(|y|2 + 1)1/2 . We therefore have (7.41) with δ˜ = α/2. Finally, since the derivatives of fα and gα vanish if |t| ≤ 12 α(|y|2 + 1)1/2 or |t| ≥ 2α(|y|2 + 1)1/2 , it follows that ∂gα ∂fα C ∂ξ (x, y, t) + ∂ξ (x, y, t) ≤ |y| + |t| + 1 with a corresponding estimate for the second derivatives of fα and gα . Combining these estimates with (7.32) gives (7.42). 

From this construction, we now produce our test function. At this stage, it will be convenient to work with a function more like our function b in the boundary condition. Theorem 7.15. Let Ω be a bounded open subset of Rn with ∂Ω ∈ C 1 , let x0 ∈ ∂Ω, let U be a neighborhood of x0 , and suppose that ˆb is a real-valued function defined on U × Rn such that γ(x0 ) · ˆbp (x, p) ≥ 1, |ˆbp (x, p)| ≤ B0 ,

|ˆbx (x, p)| + |ˆbxx (x, p)| ≤ B0 (|p| + 1), B0 |ˆbxp (x, p)| ≤ B0 , |ˆbpp (x, p)| ≤ , |p| + 1

(7.43a) (7.43b) (7.43c) (7.43d)

for some positive constant B0 and all (x, p) ∈ U × Rn . Then there are ˜ determined only by Ω and B0 such that, if positive constants C and δ, ˜ |γ(x) − γ(x0 )| ≤ δ for all x ∈ U ∩ ∂Ω, then, for any ε > 0, there is a function wε ∈ C 2 (U × U ) such that wε (x, x) ≤ ε

(7.44)

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for all x ∈ U , wε (x, y) ≥

|x − y|2 8ε

for all (x, y) ∈ U × U , ˆb(x, wε,x (x, y)) ≥ −C

ˆb(x, −wε,y (x, y)) ≤ C for all x ∈ Ω ∩ U and y ∈ ∂Ω ∩ U ,

(7.45)



|x − y|2 +ε ε



|x − y|2 +ε ε

for all x ∈ ∂Ω ∩ U and y ∈ Ω ∩ U ,

289



(7.46)



(7.47)



 |x − y| |wε,y (x, y)| ≤ C +1 , (7.48a) ε   |x − y|2 |wε,x (x, y) + wε,x (x, y)| ≤ C +ε , (7.48b) ε        C I −I |x − y|2 I0 Dwε (x, y), +C +ε ∈ J 2,+ wε (x, y) 0I ε −I I ε (7.48c)

for all (x, y) ∈ U × U . Proof. First, by rotation, we may assume that γ(x0 ) = en . Then, conditions (7.43a) and (7.43b) imply that we can define a function H implicitly by the equation ˆb(x, p0 , −H(x, p0 )) = 0, where p0 = (p1 , . . . , pn−1 ), noting that conditions (7.43) imply that H satisfies the hypotheses of Lemma 7.13 with A and C0 determined only by B0 and m = n − 1. We now take δ˜ to be the constant given by Lemma 7.14, ψ to be the function from Lemma 7.14, and define wε by   x−y . wε (x, y) = εϕ x, ε (Here, y denotes a point in Rn . In this proof, ξ will be the same ndimensional quantity as in Lemma 7.14, but y is not the first n − 1dimensional part of ξ.) With this choice, (7.44) follows from (7.39) and (7.45) follows form (7.40).

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Next, we observe that ˆb(x, p) ≥ 0 if and only if pn + H(x, p0 ) ≥ 0. It then follows from (7.43b) that     ˆb(x, wε,x (x, y)) = ˆb(x, εϕx x, x − y + ϕξ x, x − y ) ε ε     x−y x − y ) − B0 ϕx x, . ≥ ˆb(x, ϕξ x, ε ε ˜ it follows that Since |γ(x) − γ(x0 )| ≤ δ, 0 x − y0 xn − y n ≥ −δ˜ ε ε

if x ∈ ∂Ω ∩ U and y ∈ Ω ∩ U , so (7.41a) implies that   ˆb(x, wε,x (x, y)) ≥ −B0 ϕx x, x − y , ε and (7.46) is immediate. A similar argument gives (7.47). We also infer (7.48a) and (7.48b) from (7.42) along with the observation that   x−y wε,x (x, y) + wε,y (x, y) = εϕx x, . ε For our final estimate, we first infer from Taylor’s theorem and (7.42) that ϕ(x + h, ξ + k) ≤ p · h + q · k

+ C[(|ξ|2 + 1)|h|2 + (|ξ| + 1)|h||k| + |k|2 ]

+ o(|h|2 + |k|2 )

for

p = ϕx (x, ξ),

q = ϕξ (x, ξ).

Taking ξ = (x − y)/ε and k = (h − K)/ε in this inequality yields

wε (x + h, y + K) − wε (x, y) ≤ p · h + q · (h − K) + C[s|h|2 +

|h − K|2 ] + o(|h|2 + |h − K|2 ), ε

for s=

|x − y|2 + ε. ε

We then observe that     |h − K|2 1 I −I h h = s(|h|2 + |K|2 ) + s|h|2 + K K ε ε −I I to infer (7.48c).



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The comparison principle for oblique derivative problems

We are now almost ready to prove the comparison principle for viscosity sub- and supersolutions of oblique derivative problems. We just need two more steps. The first step is a modified version of Lemma 7.11. Lemma 7.16. Let S1 and S2 be locally compact subsets of Rn , let ui ∈ U SC(Si ) (i = 1, 2), and define w on S = S1 × S2 by w(x, y) = u1 (x) + u2 (y). Suppose that there are vectors p1 and p2 in Rn , points xˆ1 and x ˆ2 in Rn and positive constants θ1 and θ2 such that      I −I I0 1 2 p , p , θ1 + θ2 ∈ J 2,+ w(ˆ x1 , x ˆ2 ). −I I 0I Then there are matrices r1 and r2 in Sn such that       1 I0 r − θ1 I 0 I −I ≤ ≤ 3θ −3θ1 1 0I 0 r 2 − θ2 I −I I

(7.49)

and (pi , ri ) ∈ J

2,+

ui (ˆ xi )

for i = 1, 2. Proof. As in the proof of Lemma 7.11, we may assume that x ˆ1 = x ˆ2 = 0. We then define u ˆi by u ˆi (xi ) = ui (xi ) − pi · xi and, for each η > 0, we define vi,η for η > 0 by η vi,η (x) = u ˆi (x) − |x|2 . 2 It’s easy to see that vi,η (0) = 0 for all i and η and that w, defined by 1 w(x1 , x2 ) = v1 (x1 ) + v(x2 ) − Aij xi xj 2 has a local minimum at 0. It follows from Lemma 7.10 that, for each η > 0, there are matrices ri,η such that (7.49) holds with ri,η in place of 2,+ ri and (0, ri,η ) ∈ J vi (0). Using (7.49) again, we infer that there is a sequence (η(j)) with η(j) → 0 as j → ∞ such that (ri,η(j) ) converges to some matrix ri , which must also satisfy (7.49). It is easily verified that 2,+ (pi , ri ) ∈ J ui (ˆ xi ). 

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Our next step is define viscosity subsolutions and supersolutions. We say that u ∈ U SC(Ω) is a viscosity subsolution of F = 0 in Ω, b = 0 on ∂Ω if F (x, u(x), p, r) ≥ 0

for all x ∈ Ω and all (p, r) ∈ JΩ2,+ u(x), and

max{F (x, u(x), p, r), b(x, u(x), p)} ≥ 0

for all x ∈ ∂Ω and all (p, r) ∈ JΩ2,+ u(x). (Note that for x ∈ Ω, we have

JΩ2,+ u(x) = JΩ2,+ u(x).) Similarly, we say that u ∈ LSC(Ω) is a viscosity supersolution of F = 0 in Ω, b = 0 on ∂Ω if F (x, u(x), p, r) ≤ 0

for all x ∈ Ω and all (p, r) ∈ JΩ2,− u(x), and

min{F (x, u(x), p, r), b(x, u(x), p)} ≤ 0

for all x ∈ ∂Ω and all (p, r) ∈ JΩ2,− u(x). The important feature of this definition is that the boundary condition need not be satisfied on the boundary. Our comparison principle now follows via a suitable modification of the argument used to prove Theorem 7.12. Theorem 7.17. Let Ω be a bounded open subset of Rn with ∂Ω ∈ C 1 , let b ∈ C(∂Ω × R × Rn ) and let F ∈ C(Ω × R × Rn × Sn ). Suppose that b is twice continuously differentiable with respect to x and p. Suppose also that, for each (x0 , z) ∈ ∂Ω × R, there are a neighborhood U of x0 and a positive constant χ such that γ(x0 ) · bp (x, z, p) ≥ χ,

(7.50)

and that there is a nonnegative constant B0 such that |bx (x, z, p)| + |bxx (x, z, p)| ≤ B0 χ[|p| + 1], |bp (x, z, p)| + |bpx (x, z, p)| ≤ B0 , B0 χ |bpp (x, z, p)| ≤ |p| + 1

(7.51a) (7.51b) (7.51c)

for all (x, p) ∈ U × Rn such that b(x, z, p) = 0. Suppose further that there is a positive constant such that condition (7.23) holds for all (x, zi , p, ri ) ∈ Ω × R × Rn × Sn with z1 ≥ z2 and r1 ≤ r2 . Suppose finally that there are a continuous increasing function ω, defined on [0, ∞) with ω(0) = 0 and a positive constant δ0 such that F (y, z, p, r2 ) − F (x, z, p, r1 ) ≤ ω(|x − y|[|p| + 1] + α|x − y|2 )

(7.52)

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for all x and y in Ω, all z ∈ R, and all α > 0 and r1 and r2 satisfying (7.25), and |F (x, z, p1 , r1 ) − F (x, z, p2 , r2 )| ≤ ω(|p1 − p2 | + |r1 − r2 |) 1

2

(7.53)

n

for all x ∈ Ω with d(x) < δ0 , all z ∈ R, all p and p in R , and all r1 and r2 in Sn . If u ∈ U SC(Ω) is a viscosity subsolution of F = 0 in Ω, b = 0 on ∂Ω and if v ∈ LSC(Ω) is a viscosity supersolution of F = 0 in Ω, b = 0 on ∂Ω, then u ≤ v in Ω. Proof. By virtue of Theorem 7.12, if u − v has a positive maximum, it must occur at some x0 ∈ ∂Ω. We use σ to denote the positive maximum value. To proceed, we fix z ∈ (v(x0 ), u(x0 )), rotate axes so that γ(x0 ) = en , and define H implicitly by b(x, z, p0 , −H(x, p0 )) = 0.

We then decrease U so that it satisfies the conditions of Theorem 7.15. With θ and ε positive constants at our disposal, we define functions u ¯, v¯, and Φ by u ¯(x) = u(x) + θ2 γ(x0 ) · (x − x0 ) − θ|x − x0 |2 , v¯(x) = v(x) − θ2 γ(x0 ) · (x − x0 ),

Φ(x, y) = u ¯(x) − v¯(y) − wε (x, y).

If (x1 , y1 ) is a point in (U ∩ Ω) × (U ∩ Ω) at which Φ attains its maximum, then Since

σ − ε ≤ Φ(x0 , x0 ) ≤ Φ(x1 , y1 ).

−θ|x1 − x0 |2 = −θ|x1 − x0 + θγ(x0 )|2 − θ|θγ(x0 )|2 − 2θ2 (x1 − x0 ) · γ(x0 )

and θ|θγ(x0 )|2 = θ3 , it follows that

Φ(x1 , y1 ) = u(x1 ) − v(y1 ) − θ|x1 − x0 + θγ(x0 )|2 − θ3 + θ(y1 − x1 ) · γ(x0 ) − wε (x1 , y1 ).

Moreover, because u ˆ − vˆ is bounded from above on (U ∩ Ω) × (U ∩ Ω), it follows that, for fixed θ, we have lim |x1 − y1 | = 0

ε→0+

and |x1 − y1 |2 ≤ θ3 , ε

(7.54a)

lim x1 = θγ(x0 ) + x0 .

(7.54b)

lim sup ε→0+ ε→0+

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Simple algebra now shows that, if x ∈ Ω satisfies |x − x0 − θγ(x0 )| ≤ θ, then |x − x0 | = o(θ) as θ → 0+ . Therefore, there is a positive ρ such that, for x and y in Ω ∩ B(x0 , ρ), the inequality u(x) − v(y) > σ − ρ implies that u(x) ≥ z ≥ v(y). We now take ε sufficiently small that |x1 − x0 | = o(θ) if x1 ∈ ∂Ω and |y1 − x0 | = o(θ) if y1 ∈ ∂Ω. We also take θ sufficiently small that x1 and y1 are in the interior of U and u(x1 ) ≤ z ≤ v(y1 ). Since Φ(x, y) ≤ Φ(x1 , y1 ) for x and y in U ∩ Ω, we conclude that u ¯(x) − v¯(y) ≤ u ¯(x1 ) − v¯(y1 ) + wε (x, y) − wε (x1 , y1 ) for x and y in U ∩ Ω. Setting p = wε,x (x1 , y1 ), s=

q = −wε,y (x1 , y1 ),

|x1 − y1 |2 + ε, ε

we then infer from (7.48c) that u ¯(x) − v¯(y) ≤ u ¯(x1 ) − v¯(y1 ) + p · (x − x1 ) + q · (y − y1 )   C |(x − x1 ) − (y − y1 )|2 + + s[|x − x1 |2 + |y − y1 |2 ] 2 ε + o(|x − x1 |2 + |y − y1 |2 ).

¯ by w(x, ¯ y) = u¯(x) − v¯(y), it follows that If we define w      C I −I I0 p, q, ∈ J 2,+ w(x + Cs ¯ 1 , y1 ), 0I ε −I I

and hence Lemma 7.16 gives matrices r1 and r2 such that    1    C I0 C I −I r − CsI 0 − ≤ ≤ 0 r2 − CsI ε 0I ε −I I

(7.55)

and (p, r1 ) ∈ J

2,+

u ¯(x1 ),

(−q, −r2 ) ∈ J

2,−

v¯(y1 ).

It follows that (p + θ2 γ(x0 ) + 2θ(x1 − x0 ), r1 + 2θI) ∈ J

2,+

u(x1 )

(7.56)

and (−q − θ2 γ(x0 ), −r2 ) ∈ J

2,−

v(y1 ).

By using (7.51), (7.50), (7.46), and (7.54), we infer that b(x1 , u(x1 ), p + θ2 γ(x0 ) + 2θ(x1 − x0 )) ≥ θ2 − Cθo(θ) − Cε

(7.57)

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if x1 ∈ ∂Ω, and hence, if we choose θ sufficiently small and then ε sufficiently small, we conclude that b(x1 , u(x1 ), p + θ2 γ(x0 ) + 2θ(x1 − x0 )) > 0. A similar argument shows that, if y1 ∈ ∂Ω and if we chose first θ and then ε sufficiently small, then b(y1 , u(y1 ), −q − θ2 γ(x0 )) < 0. The definitions of viscosity subsolution and viscosity supersolution then imply that F (x1 , u(x1 ), p + θ2 γ(x0 ) + 2θ(x1 − x0 ), r1 ) ≤ 0

(7.58)

F (y1 , u(y1 ), −q − θ2 γ(x0 ), −r2 ) ≥ 0

(7.59)

and

whether or not x1 or y1 is in ∂Ω. We now use (7.53) and (7.58), along with the observation that (7.48b) implies that |p + q| ≤ Cs, to infer that ω(Cs + Cθ) ≥ F (x1 , u(x1 ), −q − θ2 γ(x0 ), r1 − CsI).

In addition, we infer from (7.53) and (7.59) that F (y1 , u(y1 ), −q − θ2 γ(x0 ), −r2 + CsI) ≤ ω(Cs + Cθ). Combining these two inequalities with (7.23), (7.52), and the inequality r1 − CsI ≤ r2 + CsI (which follow from (7.55)), we conclude that c0 [u(x1 ) − v(y1 )] ≤ 2ω(Cs + Cθ). Once ε and θ are sufficiently small, the left side of this inequality is greater than c0 σ/2 (a fixed positive number) while the right side can be made arbitrarily small. This contradiction shows that u−v cannot have a positive maximum and hence u ≤ v.  7.7

Existence and uniqueness of viscosity solutions

In this section, we prove that a viscosity solution of F = 0 in Ω,

b = 0 on ∂Ω

(7.60)

exists and is unique. The method of proof is a simple variant of the Perron process described in previous sections for linear problems.

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To begin, if u is a function defined on Ω with values in [−∞, ∞], we define functions u∗ and u∗ by u∗ (x) = lim sup{u(y) : y ∈ Ω ∩ B(x, δ)}, δ→0+

u∗ (x) = lim inf{u(y) : y ∈ Ω ∩ B(x, δ)}. δ→0+

Since the idea of the Perron process is to generate a solution as a supremum of subsolutions, we now show that the supremum of a family of viscosity subsolutions generates a viscosity subsolution. Lemma 7.18. Let Ω be a bounded, Lipschitz domain, let F ∈ U SC(Ω×R× Rn × Sn ), let b ∈ U SC(∂Ω × R × Rn ), and let F be a family of subsolutions of (7.60). Define w on Ω by w(x) = sup{u(x) : u ∈ F},

and suppose that w is everywhere finite. Then w∗ is a subsolution of (7.60). Proof. Fix x ∈ Ω and let (p, r) ∈ JΩ2,+ w∗ (x). It follows that there is a sequence (xm , um ) in Ω × F such that lim (xm , um (xm )) = (x, w∗ (x)),

m→∞

and hence, for any ε > 0, there is a δ > 0 such that 1 w∗ (y) ≤ w∗ (x) + p · (y − x) + rij (x − y)i (x − y)j + ε|x − y|2 (7.61) 2 for all y ∈ Ω[x, δ]. We now define vm by 1 vm (y) = um (y) − p · (y − x) + rij (x − y)i (x − y)j + 2ε|x − y|2 2 and, for each m, there is a point ym ∈ Ω[x, δ] at which vm attains its maximum. Since vm (y) ≤ vm (ym ) for all y ∈ Ω[x, δ], it follows that um (y) ≤ um (ym ) + p · (y − ym ) 1 + [rij (y − x)i (y − x)j − rij (ym − x)i (ym − x)j ] (7.62) 2 + 2ε(|y − x|2 − |x − ym |2 ). Since Ω[x, δ] is compact, it follows that, after possibly extracting a subsequence, the sequence (ym ) has a limit point y0 ∈ Ω[x, δ]. If we now take y = xm in (7.62) and then send m → ∞, we infer that 1 w∗ (x) ≤ lim inf (um (xm )) − p · (y0 − x) − rij (y0 − x)i (y0 − x)j − 2ε|x − y0 |2 . m→∞ 2

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Since lim inf (um (xm )) ≤ w∗ (y0 ), m→∞

we infer, after using (7.61), that w∗ (x) ≤ w∗ (x) − ε|x − y0 |2 , and hence y0 = x. Furthermore, um (ym ) → w∗ (x). Since (7.62) implies that (pm , r + 4εI) ∈ JΩ2,+ um (ym ), where pm is the vector with components i j pm i = pi + 4ε(ym − x) + rij (ym − x) ,

we conclude that S, the set of all (q, s) ∈ Rn × Sn such that there are sequences (xm ) in Ω and (pm , rm ) ∈ JΩ2,+ um (xm ) with lim (xm , um (xm ), pm , rm ) = (x, w∗ (x), q, s),

m→∞

contains (p, r + 4εI) for any ε > 0. Since S is also closed, it follows that S also contains (p, r). We now consider two possibilities determined by the sequences (xm ) and (pm , rm ) such that lim (xm , um (xm ), pm , rm ) = (x, w∗ (x), p, r).

m→∞

If there are infinitely many values of m for which F (xm , um (xm ), pm , rm ) ≥ 0, it follows that F (x, w∗ (x), p, r) ≥ 0. On the other hand, if there are only finitely many values of m for which F (xm , um (xm ), pm , rm ) ≥ 0, then only finitely many xm s are in Ω, and hence b(xm , um (xm ), pm ) ≥ 0 for infinitely many values of m, and hence b(x, w∗ (x), p) ≥ 0. It follows that w∗ is a subsolution of (7.60).  We are now ready to prove our existence and uniqueness theorem. Theorem 7.19. Let Ω be a bounded open subset of Rn with ∂Ω ∈ C 1 , let b ∈ C(∂Ω × R × Rn ) and let F ∈ C(Ω × R × Rn × Sn ). Suppose that b is twice continuously differentiable with respect to x and p. Suppose also that, for each (x0 , z) ∈ ∂Ω × R, there are a neighborhood U of x0 and a positive constant χ such that (7.50) holds and that there is a nonnegative constant B0 such that (7.51) holds for all (x, p) ∈ U × Rn such that b(x, z, p) = 0. Suppose further that there is a positive constant such that condition (7.23) holds for all (x, zi , p, ri ) ∈ Ω × R × Rn × Sn with z1 ≥ z2 and r1 ≤ r2 .

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Suppose finally that there are a continuous increasing function ω, defined on [0, ∞) with ω(0) = 0 and a positive constant δ0 such that (7.52) holds for all x and y in Ω, all z ∈ R, and all α > 0 and r1 and r2 satisfying (7.25), and (7.53) holds for all x ∈ Ω with d(x) < δ0 , all z ∈ R, all p1 and p2 in Rn , and all r1 and r2 in Sn . Then there is a unique viscosity solution of (7.60). Proof. From (7.23) and the continuity of F , we infer that there is a positive constant M such that F (x, M, 0, 0) < 0 < F (x, −M, 0, 0) for all x ∈ Ω. Hence M is a supersolution and −M is a subsolution, so F, the set of all viscosity subsolutions of (7.60) w with w ≥ −M in Ω is nonempty, and W , given by W (x) = sup{w(x) : w ∈ F}

satisfies −M ≤ W ≤ M on Ω. It then follows from Lemma 7.18 that W ∗ is a subsolution of (7.60). Since W ∗ ∈ F, it also follows that W = W ∗ . Our next step is to show that W∗ is a viscosity supersolution of (7.60). Suppose to the contrary that W∗ is not a viscosity supersolution of (7.60). Then there is (x, p, r) ∈ Ω × Rn × Sn with (p, r) ∈ JΩ2,− W∗ (x) such that F (x, W∗ (x), p, r) > 0

(and b(x, W∗ (x), p) > 0 if x ∈ ∂Ω). By continuity, there is a positive constant δ0 such that, if κ, δ, and ε are in (0, δ0 ), then uδ,ε , defined by 1 uδ,ε (y) = W∗ (x) + δ + p · (y − x) + rij (y − x)i (y − x)j − ε|y − x|2 , 2 is a viscosity subsolution of F = 0 in Ω[x, κ], b = 0 on ∂Ω[x, κ]. But W (y) ≥ W∗ (y)

1 ≥ W∗ (x) + p · (y − x) + rij (y − x)i (y − x)j + o(|y − x|2 ), 2 so if ε is sufficiently small, if κ taken sufficiently small (determined by ε) and if δ = εκ2 /8, then W (y) > uδ,ε (y) for |y − x| ∈ [κ/2, κ]. With ε, κ, and δ now fixed this way, we infer from Lemma 7.18 that U , defined by ( max{W (y), uδ,ε (y)} if y ∈ Ω[x, κ], U (y) = W (y) if y ∈ Ω[x, κ],

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is in F. Moreover, if (xm ) is a sequence in Ω such that then

xm → x,

W (xm ) → W∗ (x),

lim U (xm ) − W (xm ) = δ > 0,

m→∞

so there is a point in Ω at which U > W . This inequality contradicts the inference that U ∈ F, and hence W∗ is a viscosity supersolution of (7.60). Theorem 7.17 implies that W∗ ≥ W and the definition of W∗ implies that W∗ ≤ W , so W = W∗ , and hence W is a viscosity solution of (7.60). Another application of Theorem 7.17 shows that this viscosity solution is unique. 

Notes The main part of this chapter was taken from Section 3 of [34], and we refer the reader to that work for a more complete description of the theory and for a good bibliography (up to 1992). Two places where we stray from [34] are in our proof of Aleksandrov’s Theorem and the comparison principle for oblique derivative problems. Our proof of Aleksandrov’s Theorem is essentially taken from Sections 6.3 and 6.4 of [43] (which, in turn, is based on Reˇsetnjak’s work [160]); we have expanded the proof slightly and made some inconsequential changes in the arguments. Our proof of the comparison principle is a slight reworking of the argument in [77], although we have taken the proof of Lemma 7.16 from [33]. We also mention several other works related to viscosity solutions of oblique derivative problems. First are two works of Dupuis and Ishii: In [40], the authors study oblique derivative problems in general bounded Lipschitz domains, while, in [41], they study oblique derivative problems in domains which can be represented as the intersection of finitely many smooth domains. Both papers assume that b has the special form b(x, z, p) = β(x) · p + f (x, z).

with β a suitable oblique vector field and f a scalar function satisfying appropriate structure conditions. In [6] and [7], Barles studies oblique derivative problems with b in the same form as in this chapter. He only makes assumptions about the first derivatives of b with respect to (x, z, p) but, as compensation, his domains must have H3 boundary. The primary difference between these two papers is that [6] studies general (fully nonlinear)

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elliptic equations and oblique boundary conditions while [7] mostly studies problems with quasilinear equations and, hence, condition (7.53) is relaxed. In addition, the comparison principles in this chapter (and related ones proved in other sources) can be used to prove a priori estimates. H¨older estimates for viscosity solutions of uniformly elliptic equations with general oblique boundary conditions have been proved in [8]. Higher regularity (but only for solutions of the Neumann problem) has been proved in [132].

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Chapter 8

Pointwise Bounds for Solutions of Problems with Quasilinear Equations

Introduction In this chapter, we study various pointwise estimates for solutions of the boundary value problem Qu = 0 in Ω,

N u = 0 on ∂Ω

(8.1)

under two different basic structures. In Sections 8.1 and 8.2, we assume that Q and N are given by Qu = aij (x, u, Du)Dij u + a(x, u, Du),

(8.2a)

N u = b(x, u, Du),

(8.2b)

ij

assuming that [a ] is a positive definite matrix with minimum eigenvalue λ and b is oblique. For our present purposes, we say that b is oblique if ∂Ω is Lipschitz and if, for all (x, z) ∈ ∂Ω × R), there is an inward pointing vector p0 (x, z) such that b(x, z, p + tp0 (x, z)) ≤ b(x, z, p + sp0 (x, z)) for all p ∈ R and all t ≤ s in R. In Sections 8.3 and 8.4, we assume that n

Qu = div A(x, u, Du) + B(x, u, Du),

(8.3a)

N u = A(x, u, Du) · γ + ψ(x, u)

(8.3b)

for some vector function A and scalar functions B and ψ.

8.1

Maximum estimates for nondivergence equations

A crucial role will be played in the theory by the Bernstein E function, defined by E(x, z, p) = aij (x, z, p)pi pj . 301

(8.4)

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Our first estimate is a quasilinear version of Lemma 1.2. Lemma 8.1. Let u ∈ C 1 (Ω) ∩ C 2 (Ω) be a solution of (8.1) with Q and N given by (8.2). Suppose there are nonnegative constants µ1 , µ2 , M1 , and R such that diam Ω ≤ R and (sgn z)a(x, z, p) µ1 |p| + µ2 ≤ (8.5a) E(x, z, p) |p|2 for all (x, z, p) ∈ Ω× R× Rn with |z| ≥ M1 and 0 < |p| ≤ µ2 exp((1 + µ1 )R), (sgn z)b(x, z, p) < 0

(8.5b)

n

for all (x, z, p) ∈ ∂Ω × R × R with |z| ≥ M1 and |p| ≤ µ2 exp((1 + µ1 )R). Then µ2 exp((1 + µ1 )R). sup |u| ≤ M1 + 1 + µ1 Ω Proof. We use the function ϕ from Lemma 1.2 with a slight change. Setting K = µ1 + 1, we define µ2 ϕ(x) = − exp(K(R − x1 )) K and we assume, without loss of generality, that 0 < x1 < R in Ω. We then let x0 be a point in Ω at which u − ϕ attains its maximum. Our first step is to show that u(x0 ) ≤ M1 . ¯ by Suppose first that x0 ∈ Ω, and define the operator Q ¯ = aij (x, u, Dv)Dij v + a(x, u, Dv). Qv

Since Dϕ(x0 ) = Du(x0 ), it follows that ij ¯ Qϕ(x 0 ) = a (x0 , u, Du)Dij ϕ(x0 ) + a(x0 , u, Du)

≤ aij (x0 , u, Du))Dij u(x0 ) + a(x0 , u, Du)

=0

because D2 (u − ϕ)(x0 ) is negative semidefinite. But E(x0 , u, Dϕ) = a11 (x0 , u, Dϕ))|Dϕ(x0 )|2 , so, if u(x0 ) > M1 , then we have 11 1 ¯ Qϕ(x 0 ) = −a (x0 , u, Dϕ)µ2 K exp(K(R − x0 )) + a(x0 , u, Dϕ)

< a11 (x0 , u, Dϕ) exp(K(R − x10 ))[−µ2 K + µ2 µ1 + µ2 ]

because exp(K(R − x1 )) > 1 in Ω. Since −µ2 K + µ2 µ1 + µ2 = 0, it follows that u(x0 ) ≤ M1 if x0 ∈ Ω. If x0 ∈ ∂Ω, we use the obliqueness condition to conclude that b(x0 , u, Dϕ) ≥ b(x, u, Du) = 0.

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Since |Dϕ(x0 )| ≤ µ2 exp(KR), it follows that u(x0 ) ≤ M1 in this case as well. Hence, u(x) ≤ u(x) − ϕ(x) ≤ u(x0 ) − ϕ(x0 ) ≤ M1 − ϕ(x0 ) µ2 ≤ M1 + exp(KR) K for any x ∈ Ω. This gives the desired upper bound for u and a lower bound is proved similarly.  Note that Lemma 8.1 applies, for example, to the problem ∂u = u + g(x) on ∂Ω, ∂γ

aij (x, Du)Dij u = f (x) in Ω,

(8.6)

provided the minimum eigenvalue λ(x, p) satisfies the lower bound λ ≥ 1/(1 + |p|) and if f and g are bounded. We can modify the proof to obtain a slightly better result along the lines of Theorem 1.9. To proceed, we recall the definition D = det[aij ]. Lemma 8.2. Suppose that there are positive, measurable functions g ∈ Lnloc (Rn ) and h ∈ Ln (Ω) such that Z Z n h(x) dx < g(p)n dx (8.7) Ω

Rn

Suppose also that h(x) (sgn z)a(x, z, p) ≤ 1/n g(p) (nD)

(8.8a)

for all (x, z, p) ∈ Ω × R × Rn and that there is a continuous, increasing function M1 such that (sgn z)b(x, z, p) < 0

(8.8b)

for all (x, z, p) ∈ ∂Ω × R × Rn with |z| ≥ M1 (|p|). Then there is a number t > 0, determined only by h and g such that Z Z n h(x) dx = g(p)n dx. (8.9) Ω

{|p|≤t}

Moreover, sup |u| ≤ M1 (t) + t diam Ω.

(8.10)

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Proof. We recall from the proof of Theorem 1.9 that the contact set Γ(u) is defined to be the set of all x ∈ Ω such that there is a p ∈ Rn with u(y) ≤ u(x) + p · (y − x)

(8.11)

n

for all y ∈ Ω. We also write χ for the set of all p ∈ R such that, for some x ∈ Ω and all y ∈ Ω, (8.11) is satisfied. It then follows that h(x)n (−aij Dij u)n ≤ det(−D2 u) ≤ nD g(Du)n on Γ(u). Hence Z Z Z g(p)n dp = g(Du)n det(−D2 u) dx ≤ h(x)n dx. χ

Γ(u)

Γ(u)

n

Hence, for any T > t, there is a p ∈ R with p ∈ / χ, which means that u−p·x must take its maximum at a point x0 ∈ Ω. If x0 ∈ Ω, then u(x0 ) ≤ M1 (T ) by the definition of χ. If x0 ∈ ∂Ω, then b(x0 , u, p) ≥ b(x0 , u, Du) = 0

so condition (8.8b) implies that u(x0 ) ≤ M1 in this case as well. Then we argue as in Theorem 1.9 to infer (8.10).  Note that Lemma 8.2 applies to the problem (8.6) if λ(x, p) ≥ Ψ(p) for some positive function Ψ, if f and g are bounded provided sup |f | is sufficiently small, determined only by Ψ and Ω. If we wish to consider the slightly different problem ∂u aij (x, Du)Dij u = u + f (x) in Ω, = g(x) on ∂Ω, ∂γ we need to make slightly different hypotheses. To state them more easily, we suppose that ∂Ω ∈ C 2 , and we take advantage of Lemma 5.18. Specifically, we recall that, if ∂Ω ∈ C 2 , then there is a C 2 (Ω) function ρ such that ρ = 0 and Dρ = γ on ∂Ω. Moreover, |Dρ| ≤ 1 in Ω, and there are constants R0 and ρ2 such that |D2 ρ| ≤ ρ2 and 0 < ρ ≤ R0 in Ω. Lemma 8.3. Suppose ∂Ω ∈ C 2 . Suppose also that there are positive constant µ1 and M1 such that µ1 ρ2 Λ(x, z, p) + (sgn z)a(x, z, p) < 0

(8.12a)

for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M1 and |p| ≤ µ1 , (sgn z)b(x, z, −(sgn z)µ1 γ) < 0 n

for all (x, z, p) ∈ ∂Ω × R × R with |z| ≥ M1 and |p| ≤ µ1 . Then sup |u| ≤ M1 + R0 µ1 . Ω

(8.12b)

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Proof. Set w = −µ1 ρ, v0 = sup(u − w), and v = w + v0 . Then sup(u − v) = 0, so there is a point x0 ∈ Ω with u(x0 ) = v(x0 ). If x0 ∈ Ω, then Du(x0 ) = Dv(x0 ) and the matrix D2 (u − v)(x0 ) is negative semidefinite. It follows that Qv(x0 ) = aij (x0 , v, Dv)Dij v(x0 ) + a(x0 , v, Dv) = aij (x0 , u, Du)Dij v(x0 ) + a(x0 , u, Du) ≥ aij (x0 , u, Du)Dij u(x0 ) + a(x0 , u, Du) = 0.

Since |Dρ| ≤ 1, it follows that |Dv| ≤ µ1 , so (8.12a) implies that v(x0 ) ≤ M1 . If x0 ∈ ∂Ω, then Du(x0 ) − Dv(x0 ) is some nonpositive multiple of γ, so b(x0 , v, Dv) = b(x0 , u, Dv) ≥ b(x0 , u, Du) = 0. Since Dv = −µ1 γ on ∂Ω, it follows from (8.12b) that v(x0 ) ≤ M1 in this case as well. We therefore have −µ1 ρ(x0 ) + v0 ≤ M1 and hence v0 ≤ M1 + sup µ1 ρ = M1 + µ1 sup ρ = M1 + µ1 R0 /2. In addition, u(x) ≤ −µ1 ρ(x) + v0 ≤ v0 for any x ∈ Ω, so sup u ≤ M1 + µ1 R0 /2. The lower bound is proved similarly. 

8.2

H¨ older estimates for nondivergence equations

In later chapters, it will be useful to have a H¨older estimate for solutions of nonlinear oblique derivative problems. We begin by recalling the notation from Chapter 1. Let ω be a Lipschitz function defined on the (n − 1)– dimensional ball Bn−1 (0, R) (for some R > 0), and set Ω[R] = {x ∈ Rn : |x| < R, xn > ω(x0 )},

Σ[R] = {x ∈ Rn : |x| < R, xn = ω(x0 )}. Let ω0 be a nonnegative constant such that |ω(x0 ) − ω(y 0 )| ≤ ω0 |x0 − y 0 |

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for all x0 and y 0 in Bn−1 (0, R), and let ε ∈ (0, 1). We say that a vector field β defined on Σ[R] has modulus of obliqueness at least ε if ω0 |β 0 (x)| ≤ (1 − ε)β n (x)

(8.13)

for all x ∈ Σ[R]. Our H¨ older continuity result is then stated in the following form. Lemma 8.4. Let [aij ] be a positive-definite matrix-valued function defined on Ω[R] with minimum eigenvalue λ, and suppose that there is a positive constant µ such that |aij (x)| ≤ µλ(x)

for all x ∈ Ω[R]. Define Lu = aij Dij u. Let β have modulus of obliqueness at least ε, and let u ∈ C 2 (Ω[R]) ∩ C(Ω[R]) satisfy |Lu| ≤ λ(b0 |Du|2 + b1 ) in Ω[R], n

|β · Du| ≤ Gβ on Σ[R]

(8.14a) (8.14b)

for some nonnegative constants b0 , b1 and G. Then there are constants C, α, and δ determined only by ε, µ, ω0 , and b0 sup |u| such that  α  r δ 1−δ osc u ≤ C osc u + (b1 R + G)r R (8.15) Ω[r] Rα Ω[R] for all r ∈ (0, R). Proof. We start with a suitable weak Harnack inequality. Suppose that w is nonnegative and satisfies Lw ≤ λ(b0 |Dw|2 + b1 ) in Ω[R],

β · Dw ≤ Gβ n on Σ[R],

and set v=

1 − e−b0 w . b0

It’s easy to check that Lv ≤ b1 λ in Ω[R] and β · Dv ≤ Gβ n on Σ[R] (recalling that e−b0 w ∈ (0, 1]), so Theorem 1.20 implies that there are positive constants C and p, determined by n, ε, µ, and ω0 such that !1/p Z R−n

v p dx

Ω[R/2]

≤ C( inf v + b1 R2 + GR). Ω[R/2]

(8.16)

Since v ≤ w ≤ exp(b0 sup w)v, this estimate is also valid with w in place of v (and C depending also on b0 sup w). The proof of Theorem 1.26 immediately gives (8.15). 

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We observe here that the inequality |Lu| ≤ λ(b0 |Du|2 + b1 ) is satisfied if Qu = 0 with the structure conditions |aij (x, z, p)| ≤ µ(|z|)λ(x, z, p),

|a(x, z, p)| ≤ (b0 (|z|)|p|2 + b1 (|z|))λ(x, z, p)

for increasing, nonnegative functions µ, b0 , and b1 . The connection of the linear boundary condition in this theorem with the nonlinear structure of current interest will be clarified in later chapters. 8.3

Maximum estimates for conormal problems

In this section, we prove a maximum estimate for solutions of the conormal problem under quite general hypotheses. All of our structure conditions are based on a positive, increasing function g defined on [0, ∞) and its integral G, which is defined by Z s G(s) = g(t) dt. (8.17) 0

There are two inequalities we shall use repeatedly. First, since g is increasing, we have ag(b) ≤ ag(a) + bg(b) for any nonnegative a and b. Also Z s G(s) ≤ g(s) dt = sg(s).

(8.18)

(8.19)

0

We also take G1 to be an increasing C 1 ([t0 , ∞)) function such that G1 (t)tg(t) ≤ G(t)2 ,

tG01 (t) ≤ G(t)

(8.20)

for all t ≥ t0 , where t0 is some positive constant. (If there is a positive constant g0 such that G satisfies the inequality G(s) ≥ g0 sg(s) for all s > 0, then we can take G1 to be a suitable multiple of G. We shall return to this inequality in the next section.) Lemma 8.5. Suppose that there are nonnegative constants a1 and b1 and positive constants c0 < 1, M , and R such that p · A(x, z, p) ≥ |p|g(|p|) − a1

RG(|z|/R)2 , |z|g(|z|/R)

zB(x, z, p) ≤ b0 p · A(x, z, p) + b1 G(|z|/R)

(8.21a) (8.21b)

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for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M , zψ(x, z) ≤ Rc0 G1 (|z|/R)

(8.21c)

n

for all (x, z, p) ∈ ∂Ω × R × R with |z| ≥ M . Suppose also that   M M/R ≥ t0 , G ≥ 1. (8.22) R Then there is a constant c1 , determined only by n, Ω, a1 , b0 , b1 , and c0 , such that  2n     Z M |u| |u| sup 1 − G ≤ c1 G dx. (8.23) |u| R R Ω Ω + Proof.

Since

With q ≥ 1 + b0 a parameter, we define the function ϕ by 2nq−2n+1  q−1  |u| M G u. ϕ(u) = 1 − |u| + R

 2nq−2n  q−1 M |u| M ϕ (u) = (2nq − 2n + 1) 1 − G |u| + R |u| 0

+

ϕ(u) u

 2nq−2n+1  q−2   M |u| |u| |u| + (q − 1) 1 − G g , |u| + R R R it follows that ϕ0 (u) ≥ qϕ(u)/u, and hence ϕ(u) ϕ0 (u) ≥ b0 . (8.24) u Now, we use ϕ(u) as test function in the weak form of the boundary value problem to infer that Z Z Z 0 ϕ (u)Du · A dx = ϕ(u)B dx + ϕ(u)ψ ds. Ω

Ω

∂Ω

From (8.21) and (8.24), we infer (after some rearrangement) that  Z  ϕ(u) 0 g(|Du|)|Du| dx ≤ I1 + I2 + I3 , ϕ (u) − b0 u Ω where  Z  ϕ(u) RG(|u|/R)2 I1 = a1 ϕ0 (u) − b0 dx u |u|g(|u|/R) Ω 2nq−2n+1  q Z  M M I2 = b1 1− G dx, |u| |u| Ω + 2nq−2n+1  q−1   Z  M |u| |u| I3 = Rc0 1− G G1 ds. |u| R R ∂Ω +

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By direct computation, we also obtain  2nq−2n  q−2   M |u| |u| |u| 0 ϕ (u) ≤ 2(n + 1)q 1 − G g , |u| + R R R so, after noting that (1 − M/|u|)+ ≤ 1, we have 2nq−2n  q Z  M |u| I1 + I2 ≤ [2a1 (n + 1)q + b1 ] 1− G dx. |u| R Ω + Next, we define the function ϕ1 by 2nq−2n+1  q−1    M t t G G1 ϕ1 (t) = 1 − t + R R and observe that ϕ01 (t) = Φ1 (t) + Φ2 (t) + Φ3 (t), where  2nq−2n  q−1   M t t M Φ1 (t) = (2nq − 2n + 1) 1 − G G1 , t + R R t2  2nq−2n+1  q−2   M t t g(t/R) Φ2 (t) = (q − 1) 1 − G G1 , t + R R R  2nq−2n+1  q−1   M t t 1 G Φ3 (t) = 1 − G01 . t + R R R Then Lemma 5.17 implies that there is a constant K0 , determined only by Ω, such that Z Z I3 ≤ Rc0 |Du|ϕ01 (|u|) dx + K0 c0 R ϕ1 (|u|) dx. Ω

Ω

We now use (8.18) to conclude that Z |Du|ϕ01 (|u|) dx ≤ I31 + I32 , Ω

with

I31 = I32 =

Z

ZΩ Ω

|Du|g(|Du|)

ϕ01 (|u|) dx, g(|u|/R)

|u| 0 ϕ (|u|), dx. R 1

Next, we infer from the first inequality of (8.20) and the first inequality of (8.22) that G1 (t) ≤ G(t) for t ≥ M/R. In addition, the second inequality of (8.20) and (8.19) imply that G01 (t) ≤ g(t) for t ≥ M/R. Comparing the

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explicit expressions for the derivatives of ϕ and ϕ1 and using (8.19) and the second inequality of (8.22), we see that   1 g(|u|/R) 0 ϕ1 (|u|) ≤ 1 + ϕ0 (|u|) . q−1 R If q is large enough (determined only by c0 ) that   1 1 + c0 c0 1 + ≤ , q−1 2 we infer that

I31

1 + c0 ≤ 2

Z

Ω

|Du|g(|Du|)ϕ0 (|u|) dx.

Using (8.20) and the inequality G1 ≤ G, we conclude that  2nq−2n  q−2 |u| 1 M 0 ϕ1 (|u|) ≤ 2(n + 1)q 1 − G . |u| + R R

It therefore follows that Z 1 + c0 I3 ≤ g(|Du|)|Du|ϕ0 (u) dx 2 Ω 2nq−2n  q Z  M |u| + C(n)c0 q 1− G dx. |u| R Ω +

Combining this inequality with our estimates for I1 and I2 , we find that there is a constant C, determined only by a1 , b0 , b1 , and c0 , such that    Z  ϕ(u) |u| 1 − c0 0 ϕ (u) − b0 |Du|g dx 2 u R Ω 2q−2n  q Z  M |u| ≤ Cq 1− G dx. |u| R Ω To proceed, we note that

ϕ(u) 1 − c0 ϕ(u) − b0 ≥ 2 u

 2nq−2n+1  q−1 M |u| 1− G |u| + R

if q is sufficiently large (now determined by b0 and c0 ), and we define the function H by  2q−2n+2  q M |u| H(x) = 1 − G . |u| + R Then

|DH| ≤

3q R

 2nq−2n+1  q−1   M |u| |u| 1− G g , |u| + R R

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so 2nq−2n  q Z  M |u| |DH| + |H| dx ≤ Cq 1− G dx. |u| R Ω Ω

Z

2

Applying the Sobolev imbedding theorem, we find that !κ 2nκq−2n  κq 2q−2n  q Z  Z  |u| M M |u| 2 G dx ≤ Cq 1− G dx, 1− |u| + R |u| + R Ω Ω where κ = n/(n − 1). If we set 2n    M |u| G , w = 1− |u| + R

dµ =

 −2n M 1− dx, |u| +

then this inequality takes the standard form Z 1/κ Z κq 2 w dµ ≤ Cq wq dµ, Ω

Ω

and then the Moser iteration scheme, Lemma 5.30, implies that Z sup w ≤ C w dµ. Ω

Ω

Rewriting this inequality in terms of u and dx yields (8.23).



Estimates of the right hand side of (8.23) require somewhat stronger conditions on the coefficients. The most important condition is that zB − b0 p · A is negative for large z. More precisely, we have the following result. Lemma 8.6. Suppose there are nonnegative constants a1 , b0 , and c0 , along with positive constants b2 , M , and R (with M/R ≥ t0 ) such that condition (8.21a) holds for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M , condition (8.21c) holds for all (x, z) ∈ ∂Ω × R with |z| ≥ M , and   |z| zB ≤ b0 p · A − b2 G (8.25) R for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M . Suppose also that there is a q ≥ max{1, b0 − 1} and ε ∈ (0, 1) such that c0 (q + 1) ≤ min{q, q + 1 − b0 },

a1 max{q, q + 1 − b0 } + c0 (q + 1 + K0 R) ≤ (1 − ε)b2 .

(8.26a) (8.26b)

Then Z

Ω

G



|u| R



dx ≤ ε

−1/q

G



M R



|Ω|.

(8.27)

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Proof. To simplify the notation, we set τ = G(M/R), we write Ω0 for the subset of Ω on which |u| ≥ M , and we suppress the argument |u|/R from G, G1 , and G01 . If we use the test function ϕ = (Gq − τ q )+ u and apply (8.25), we can write the resulting inequality as I0 + I1 ≤ I2 where Z

[(1 − b0 )(Gq − τ q ) + qGq−1 Z I1 = b2 (Gq − τ q )G dx, Ω0 Z I2 = c0 R G(Gq − τ q )+ ds.

I0 =

Ω0

|u|g(|u|/R) ]Du · A dx, R

∂Ω

From Lemma 5.17, we conclude that Z I2 ≤ K0 c0 R (Gq − τ q )G1 dx Ω0 Z + c0 |Du|((Gq − τ q )G01 + qGq−1 G1 g(|u|/R)) dx. Ω0

We infer from (8.20) that G1 (t) ≤ G(t) and G01 (t) ≤ g(t) for all t > 0, and hence, because τ ≥ 0, we have Z I2 ≤ K0 c0 R Gq+1 dx 0 ZΩ + c0 q |Du|g(|u|/R)Gq−1 G1 dx Ω0 Z + c0 |Du|Gq G01 dx. Ω0

Applying (8.20), (8.18), and (8.19), we find that Z I2 ≤ c0 (q + 1) |Du|g(|Du|)Gq dx 0 Ω Z + c0 (q + 1 + K0 R) Gq+1 dx. Ω0

Our next step is to obtain a lower bound for I0 . We set q1 = min{q, q + 1 − b0 },

q2 = max{q, q + 1 − b0 },

and use (8.19) along with the assumption q ≥ b0 − 1 to conclude that q1 Gq ≤ (1 − b0 )(Gq − τ q )+ + qGq−1

|u|g(|u|/R) |u|g(|u|/R) ≤ q2 Gq−1 . R R

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It follows that I0 ≥ q1

Z

Ω0

Gq |Du|g(|Du|) dx − a1 q2

Z

Gq+1 dx.

Ω0

Taking (8.26a) into account, we conclude that Z Z (Gq − τ q )G dx ≤ Gq+1 (a1 q2 + c0 (q + 1 + K0 R)) dx b2 Ω0 Ω0 Z ≤ (1 − ε)b2 Gq+1 dx. Ω0

Simple rearrangement yields Z Z ε Gq+1 ≤ Ω0

τ q G dx,

Ω0

and hence Z

Ω0

G dx ≤

 Z ε−1 τ q

Ω0

1/(q+1) G dx |Ω0 |q/(q+1)

by H¨ older’s inequality. One more rearrangement then yields (8.27).



It is possible to prove an estimate like that in Lemma 8.6 with the negativity assumption on zB − b0 p · A replaced by one on zψ, but we shall not do so here. 8.4

H¨ older estimates for conormal problems

In this section, we make a slightly stronger assumption on g, namely, we assume that there are positive constants δ and m, with 1 + δ ≤ m such that (1 + δ)G(t) ≤ tg(t) ≤ mG(t)

(8.28)

for all t > 0. As a simple consequence, we observe that G(t) G(s) ≥ 1+δ , 1+δ t s G(t) G(s) ≤ m tm s

(8.29a) (8.29b)

for all s ∈ (0, t]. To see the significance of condition (8.28), we note first that it is satisfied for g(t) = tm with δ = 1 − m for any m > 1. Moreover, if we take g(t) = rtr−1 (ln(t + e))s +

str (ln(t + e))s−1 t+e

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for constants r > 1 and s ≥ 0, then

G(t) = tr (ln(t + e))s ,

so (8.28) is satisfied with δ = r − 1 and m = r + s. On the other hand, tg(t) lim = r, t→∞ G(t) so this example can be considered only a small perturbation of a power function. To see the full strength of our structure condition, we let δ > 0 and m > 1 + δ to be arbitrary and set m−1−δ 1+δ+m √ , b= a= . 2 2 2 Then, for 1 g(t) = [a + b(cos(ln(ln t)) + sin(ln(ln t))] exp(ln t(a + b sin(ln(ln t))), t we have G(t) = exp(ln t(a + b sin(ln(ln t))) = t(a+b sin(ln(ln t))) and (8.28) is satisfied. Moreover, we have tg(t) tg(t) lim inf = 1 + δ, lim sup = m. t→∞ G(t) t→∞ G(t) Hence condition (8.28) is a genuine improvement on the assumption that g is a power function. We next prove a weak Harnack inequality under the structure conditions: p · A ≥ |p|g(|p|) − a1 g(a1 ),

(8.30a)

|A| ≤ a2 g(|p|) + g(a3 ),

(8.30b)

ψ ≥ −g(c0 ).

(8.30d)

B ≥ −b0 |p|g(|p|) − g(b1 )/R,

(8.30c)

which are assumed to hold for all z ≥ 0, all x ∈ Ω[R], and all p ∈ Rn . Lemma 8.7. Set χ = a1 + a3 + b1 + c0 and M = sup u. Then there are positive constants p and C, determined only by n, a2 , b0 M , and K1 , such that !1/p Z −n p R u dx ≤ C( inf u + χR) (8.31) Ω[R/2]

Ω[R]

for any nonnegative supersolution of

div A(x, u, Du) + B(x, u, Du) = 0 in Ω[R],

(8.32a)

A(x, u, Du) · γ + ψ(x, u) = 0 on Σ[R].

(8.32b)

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Proof. Our proof consists of three steps. The main part of the proof is to show that u lies in a space which is almost a De Giorgi class, but based on G rather than a power function. Then, we show that u is in a De Giorgi class, and then the proof is easily finished. For the first step, we let η be a cut-off function (so 0 ≤ η ≤ 1 in B(R) and η = 0 on ∂B(R)). With k at our disposal, we use the test function ϕ = η m exp(−b0 u)(u − k)− .

From our structure conditions, we have that

I1 ≤ I2 + I3 + I4 + I5 + I6 ,

where I1 =

Z

e−b0 u η m g(|Dw|)|Dw| dx,

Ω[R]

I2 =

Z

e−b0 u η m g(b1 )

Ω[R]

I3 =

Z

w dx, R

e−b0 u η m a1 g(a1 )(1 + b0 w) dx,

Ak

I4 =

Z

Ω[R]

I5 =

Z

e−b0 u mη m−1 wa2 |Dη|g(|Dw|) dx, e−b0 u mη m−1 g(a3 )w|Dη| dx,

Ω[R]

I6 =

Z

e−b0 u η m wg(c0 ) ds.

Σ[R]

Here, we have set w = (u − k)− and we have written Ak for the subset of Ω[R] on which w is nonzero. Let us suppose first that k ≤ M , and we write C for any constant determined only by n, a2 , ω0 , b0 M . Then (8.18) implies that   Z Z k I2 ≤ m e−b0 u η m G dx + m e−b0 u η m G(b1 ) dx. R Ω[R] Ω[R] Also Z I3 ≤

e−b0 u η m CG(a1 ) dx

Ω[R]

because w ≤ k ≤ M . Two more applications of (8.18) (along with the monotonicity of g −1 and conditions (8.28) and (8.29)) give Z Z 1 1 I4 ≤ e−b0 u η m |Dw|g( |Dw|) dx + e−b0 u η m G(4a2 mw|Dη|η −1 )) dx 4 4 Ω[R] Ω[R] Z Z 1 −b0 u m ≤ e η |Dw|g(|Dw|) dx + e−b0 u G(Ca2 w|Dη|) dx 4 Ω[R] Ω[R]

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and I5 ≤

Z

e−b0 u mG(w|Dη|) dx +

Ω[R]

Z

e−b0 u M G(η m−1 a3 m) dx. Ω[R]

By repeating the proof of Lemma 5.17 with γ¯ a unit vector in the xn direction, we find that I6 ≤ I7 + I8 , where Z I7 = Cc0 e−b0 u η m [1 + b0 w]|Du| dx, Ω[R] Z I8 = Cc0 e−b0 u mη m−1 w|Dη| dx. Ω[R]

As we have already noted, b0 w ≤ C, so (8.18) implies that Z Z −b0 u m 1 I7 ≤ e η |Dw|g(|Dw|) dx + e−b0 u η m G(Cc0 ) dx, 2 Ω[R] Ω[R] and I8 ≤

Z

Ω[R]

e−b0 u η m−1 G(mw|Dη|) dx +

Z

e−b0 u η m−1 G(mCc0 ), dx.

Ω[R]

Combining all these estimates (and using (8.29)) leads us to the inequality    Z Z k m η G(|Dw|) dx ≤ G C w|Dη| + + χ dx. R Ω[R] A(k) In fact, this inequality is valid for k > M as well. To see why this is so, we note that (u − k)− = k − u if k ≥ M so A(k) = Ω[R] and Dw = −Du. Therefore the left hand side of this inequality is independent of k for k ≥ M while the right hand side is an increasing function of k, and hence we may assume that k ≥ χR to obtain    Z Z k η m G(|Dw|) dx ≤ G C w|Dη| + dx R Ω[R] A(k) for all k ≥ χR. If we choose η so that η ≡ 1 in B(σR) and |Dη| ≤ 2/(1 − σ)R, we obtain ! Z k (u − k)− G(|Du|) dx ≤ G sup + |A(k)|. R A(k) (1 − σ)R A(k,Ω[σR]) Note that this inequality is the analog of (5.62) with G(t) in place of tp . To proceed, we define Z t G(t) h(t) = , H(t) = h(s) ds. t 0

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Then (8.28) implies that g(t)/m ≤ h(t) ≤ g(t)/(1 + δ) for all t > 0 and hence G(t)/m ≤ H(t) ≤ G(t)/(1 + δ) as well. Moreover, if we define Ψ by Ψ(t) = H(t1/(1+δ) ), then  !1+δ  Z (u − k) k −  |A(k)|. Ψ(|Du|1+δ ) dx ≤ Ψ  sup + R A(k) (1 − σ)R A(k,Ω[σR])

(8.33)

We now set A0 (k) = A(k, Ω[σR]), v = |Du|1+δ , and !1+δ k (u − k)− + . J = sup R A(k) (1 − σ)R With these abbreviations, (8.33) becomes Z Ψ(v) dx ≤ Ψ(J)|A(k)|. A0 (k)

We now observe that Ψ is increasing and Ψ00 (t) =

t−(2+δ)/(1+δ) (−δH 0 (t1/(1+δ) ) + t1/(1+δ) H 00 (t1/(1+δ) ). (1 + δ)2

But sH 00 (s) − δH 0 (s) = sh0 (s) − δh(s) = g(s) − (1 + δ)h(s) ≥ 0, so Ψ is convex. It therefore follows from Jensen’s inequality that     Z 1 |A(k)| |A(k)| Ψ v dx ≤ Ψ(J) ≤ Ψ J . |A0 (k)| |A0 (k)| |A0 (k)|

In the second inequality, we have used the estimate KΨ(J) ≤ Ψ(KJ) for K ≥ 1, which follows from the observation that Ψ(t)/t is an increasing function of t. Since Ψ is strictly increasing, we conclude that Z v dx ≤ J|A(k)|, A0 (k)

and this inequality tells us that u ∈ DG− (1+δ, Ω[R], χ, 1/n, K) for some K determined only by m, δ, b0 M , a2 , and ω0 . The weak Harnack inequality is then an immediate consequence of Lemma 5.40 via the argument in Theorem 1.20. 

Finally, we obtain a H¨ older estimate from this weak Harnack inequality by following the argument of Theorem 1.26. Theorem 8.8. Let g satisfy (8.28) for some positive constants δ and m with 1 + δ ≤ m. Let Ω be a bounded Lipschitz domain and let u be a

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bounded weak solution of (8.1) with Q and N defined by (8.3). Suppose also that there are nonnegative constants a1 , a2 , a3 , b0 , b1 , and c0 such that conditions (8.30a), (8.30b), |B(x, z, p)| ≤ b0 |p|g(|p|) +

g(b1 ) diam Ω

(8.34a)

are satisfied for all x ∈ Ω, z = u(x), and all p ∈ Rn , and |ψ(x, z)| ≤ g(c0 )

(8.34b)

n

for all x ∈ ∂Ω, z = u(x), and all p ∈ R . Then there are positive constants α and C, determined only by a2 , b0 M , δ, m, and Ω, such that  α  r osc u ≤ C + (a1 + a3 + b1 + c0 )r (8.35) Rα Ω[r]

for all R ∈ (0, diam Ω) and all r ∈ (0, R).

Notes The results in this chapter are based on well-known results for the Dirichlet problem although certain elements of the oblique derivative problem require a more careful analysis of the maximum estimates. For example, Lemma 8.1 is essentially the same as Theorem 10.3 of [64], and Lemma 8.2 is essentially the same as Theorem 10.5 of [64]. Both results (for the Dirichlet problem) are described in [64] as “basically variants of the Hopf maximum principle”. Lemma 8.3 is a slight rewording of Lemma 3.2 from [98], which, in turn, is based on Theorem 1 in [31]; in this case, all results are for oblique derivative problems. The H¨ older estimate Lemma 8.4 first appears as an interior H¨older estimate in [162] (under the additional assumption that b0 = 0) and in Theorem 1 from [93] (with arbitrary b0 ). For C 2 domains, the estimate first appears as Theorem 2.3 in [120], and, for Lipschitz domains, it was first proved as Corollary 3.5 in [116]. The maximum estimates for conormal problems, Lemmata 8.5 and 8.6, first appear (in slightly different form) as Lemmata 2.4 and 2.5 in [112]. For the Dirichlet problem and with g being a power function, these results appear in Theorem 7.1 from Chapter 4 of [91]; see also Lemma 2.2 from Chapter 10 of [91] for a version of our Lemma 8.5 with g a power function. Finally, the H¨ older estimate Lemma 8.7 is stated here for the first time, in this exact form. As demonstrated in Section 1.9, it is a straightforward

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consequence of the weak Harnack inequality, and various other versions have been stated. For example, when g is a power function, it is proved on page 467 of [91] as a simple modification of the corresponding interior estimate for solutions of such equations, and the interior H¨older estimate for equations of the general type considered here is just Corollary 1.5 of [112]. The proof given here is based on the results of Klimov [81], specifically, Lemma 2 from [81], which asserts that any function satisfying (8.33) is in the De Giorgi class DG− (1 + δ, Ω[R], χ, 1/n, K). A further alternative approach (under slightly different assumptions) is given in [66].

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Chapter 9

Gradient Estimates for General Form Oblique Derivative Problems

In this chapter, we study the gradient estimate for the boundary value problem Qu = 0 in Ω,

M u = 0 on ∂Ω,

(9.1)

with Q and M defined by Qu = aij (x, u, Du)Dij u + a(x, u, Du),

(9.2a)

M u = b(x, u, Du).

(9.2b)

We study a somewhat general differential equation, based on the structure in Chapter 15 of [64]. It is important for the purposes of this chapter that boundary condition (9.2) NOT be restricted to the conormal condition for the differential equation. The main assumption (except in Section 9.7) is that the principal coefficients of Q can be decomposed as follows: aij (x, z, p) = aij ∗ (x, z, p) + τ (x, z, p)pi pj [aij ∗ ] n

1

(9.3)

n

for some positive-definite matrix ∈ C (Ω × R × R ) and some nonneg1 ative function τ ∈ C (Ω × R × R ). The model problem here is the false mean curvature equation: ∆u + Di uDj uDij u = 0, aij ∗

which corresponds to = δ ij and τ ≡ 1. We begin by providing an interior gradient estimate of a very special form, which will then be used to prove gradient estimates near the boundary. 9.1

Interior gradient bounds

Our first estimate is an interior gradient bound. Some notation will prove very useful in this chapter. First, we use numerical superscripts to denote 321

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differentiation with respect to the corresponding component of p, numerical subscripts to denote differentiation with respect to the corresponding component of x, and the subscript z to denote differentiation with respect to z. (In other words, ∂aij , ∂pk

∂a , ∂z and so on.) We also define the operators δ and δ¯ by p ¯ (x, z, p) = p · fp (x, z, p). δf (x, z, p) = fz (x, z, p) + 2 · fx (x, z, p), δf |p| aij,k =

az =

To prove our gradient estimate, we differentiate the equation Qu = 0 with respect to xk for k = 1, . . . , n, multiply the resultant equation by Dk u, and then sum over k. After performing this calculation, we find that w0 = |Du|2 satisfies the differential equation i aij Dij w0 − 2aij ∗ Dik uDjk u + B0 Di w0 + C0 w0 = 0,

(9.4)

B0 = ajk p Djk u + ap

(9.5)

where

and C0 = 2[δaij ∗ Dij u + δa] + δτ Du · Dw0 . For future reference, we note that 1 B0 = ajk ∗,p Djk u + ap + Du · Dw0 τp + τ Dw0 . 2 It will be convenient, however, to perform one more algebraic step here. We write our equation as i aij Dij w − 2aij ∗ Dik uDjk u + B1 Di w0 + C1 w0 = 0

(9.6)

with B1 = B0 + δτ w0 Du and C1 = 2[δaij ∗ Dij u + δa]. Although we can prove a global gradient estimate (under suitable hypotheses on the coefficients of Q) by manipulating this equation appropriately, we refer the reader to Section 15.2 of [64] for a complete description of the procedure, and, instead, we turn to interior estimates.

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Let B be a subset of the ball B(y, R) for some y ∈ Rn and some R > 0, and suppose that u ∈ C 3 (B) satisfies Qu = 0 in B. If we define η by

|x − y|2 , R2 then w1 = η 2 w0 satisfies the differential equation η(x) = 1 −

i i aij Dij w1 − 2η 2 aij ∗ Dik uDjk u + B2 Di w1 + C2 w1 − 2ηw0 B1 Di η = 0,

where

2 B2i = B1i − aij Dj η, η and 2 ij 2 a Dij η + 2 aij Di ηDj η. 2 η η To continue, we rewrite this equation to separate terms involving second derivatives of u. Specifically, we have 1 i (τp · DηDu + 2τ Dη) · Dw1 B1i Di η = Dk ηaij,k ∗ Dij u + a Di η + 2η 2 1 − (τp · DηDu · Dη + 2τ w0 |Dη|2 ), η and 2 ij ij C2 w1 = 2δaij ∗ Dij u w1 + 2δaw1 + 2 [a Di ηDj η − a Di ηDj η]. η and hence i aij Dij w1 − 2η 2 aij ∗ Dik uDjk u + B3 Di w1   (9.7) 2 ij ij,r ˜ 1, = w1 2δa∗ − Dr ηa∗ Dij u + 2δaw1 + Iw η where 1 2 2 B3i = B2i − |Du|2 Dη · τp Du − |Du|2 τ Dη − aij Dj η η η η and 2 2 I˜ = 2 [ηaij Dij η − aij Di ηDj η] + (δτ |Du|2 Du · Dη) η η 2 2 + Dη · ap − 2 [τp · DuDu · Dη + τ |Dη|2 ]. η η For further computation, we set M1 = supB w1 and we introduce a C 2 function h (defined on some interval containing the range of u) such that 0 ≤ h ≤ 1, and we set 1 w = w1 + M1 h(u). 2 C2 = C1 −

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We observe that w1 ≤ w ≤ w1 + M1 /2, so M1 /2 ≤ supB w ≤ 3M1 /2. These comparisons on the maxima of w and w1 will be very useful. To take advantage of the function w, we first recall E(x, z, p) = aij (x, z, p)pi pj from (8.4), and we write aij Dij h = h0 aij Dij u + h00 aij Di uDj u = −h0 a + h00 E and 1 1¯ ¯ − 2 aij Di uDj η ( δτ + τ )Dk uDk w + δa η2 2 η 1 ¯ 1 ¯ + 2τ )Du · Dη (δτ + 2τ )w0 M1 h0 − (δτ 2η η w0 2 + δτ w02 − aij Di uDj η − (τp · Dη|Du|2 + τ Du · Dη). η η It then follows that ¯ ij Dij u + B2i Di u = δa ∗

i ij −2η 2 aij ∗ Dik uDjk u + a Dij w + B4 Di w = S 1 ˜ 0 ] + (K0 + K)w ˜ 1, + M1 [Eh00 + H0 (h0 )2 + (J0 + J)h 2 where M 1 h0 ¯ 1 B4i = B3i − [δτ − τ ]Di u, (9.8a) 2 2η 2 2w1 1 0 ¯ ij S = −2w1 δaij Dk ηaij,k (9.8b) ∗ Dij u − ∗ Dij u + M1 h δa∗ Dij u, η 2 M1 ¯ + 2τ ), H0 = − 2 |Du|2 (δτ (9.8c) 4η 1 J0 = (δ¯ − 1)a + δτ |Du|4 , (9.8d) 2 4 1 ¯ J˜ = − aij Di uDj η − (δτ + 2τ )Du · Dη, (9.8e) η η

K0 = −2δa, (9.8f) w 0 ˜ 0 = I˜ − K (Dη · τp + τ Du · Dη). (9.8g) η (Note that S contains all the terms in which second derivatives of u appear linearly. The terms H0 , J0 , and K0 will be important in determining the nature of the gradient estimate and they do not contain derivatives of η. ˜ 0 contain derivatives of η; they will be treated as perThe terms J˜ and K turbation terms, so our eventual structure conditions will guarantee that they are small, in an appropriate sense, wherever η 2 |Du|2 is large.)

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Our next step is to use the Cauchy-Schwarz inequality to estimate the terms in S: 2 X (δaij 1 2 ij ∗ ) 2 2w1 δaij , ∗ Dij u ≤ η a∗ Dik uDjk u + 4|Du| w1 4 λ∗ i,j 1 2 ij |Dη|2 X (aij,r )2 2w1 2 Dr ηaij,r D u ≤ η a D uD u + 4|Du| w , ij ik jk 1 ∗ ∗ η 4 η 2 i,j,r λ∗

2 2 X (δa ¯ ij 1 ∗ ) ¯ ij Dij u ≤ 1 η 2 aij Dik uDjk u + 1 M0 (h0 )2 − M0 h0 δa ∗ ∗ 2 4 4 η2 λ∗ i,j

and hence aij Dij w + B4i Di w ≥

1 1 ˜ 0 + Kw ˜ 1 M1 [Eh00 + H1 (h0 )2 + J0 h0 ] + K1 w1 + M1 Jh 2 2 (9.9)

where H1 = H0 −

2 ¯ ij 1 M1 X (δa ∗ ) , 2 2 η i,j λ∗

K1 = K0 − 4|Du|2

2 X (δaij ∗ )

λ∗

i,j

˜ =K ˜ 0 − 4|Du|2 |Dη| K η2

2

(9.10a) ,

2 X (aij,r ∗ ) . λ∗ i,j,r

(9.10b) (9.10c)

Next, we suppose that δa ≤ O(E), ¯ (δ − 1)a ≤ O(E), p ij ¯ ij | = O( λ∗ E), |δa∗ | + |δa ∗ ¯ |p|4 ≤ O(E) δτ

(9.11a) (9.11b) (9.11c) (9.11d)

as |p| → ∞, uniformly with respect to x and z. It then follows that there ¯ , H, J, and K (all determined only by the are nonnegative constants M limit behavior in (9.11)) such that H1 ≥ −HE, J0 ≤ JE,

K1 ≥ −KE

(9.12a) (9.12b) (9.12c)

¯ and w1 ≥ M1 /2. The constants H, J, and K will be wherever |Du| ≥ M used to determine the nature of the gradient estimate.

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We are now ready to state our interior gradient estimate. In order to obtain a precise form of the estimate, we consider two possibilities. First, if the oscillation of u is sufficiently small, then we have the following result. Theorem 9.1. Let u ∈ C 3 (B), where B = B(y, R) for some y ∈ Rn and R > 0, and suppose that Qu = 0 in B. Suppose aij can be decomposed as ¯ , H, J, K be the in (9.3) and that conditions (9.11) are satisfied. Let M constants from (9.12) and suppose that there is a nonnegative constant µ1 such that Λ|p|2 ≤ µ1 E, p |p|2 |aij ∗,p | ≤ µ1 λ∗ E,

(9.13b)

|p||τp | ≤ µ1 τ

(9.13d)

|p||ap | ≤ µ1 E,

¯ . If whenever |p| ≥ M osc u ≤ min{ then

1 1 ,p }. 1 + 2J e(1 + 2K)(1 + H)

¯ + C(n, µ1 , H, K) osc u . |Du| ≤ 2M R B(y,R/2) sup

(9.13a) (9.13c)

(9.14)

(9.15)

¯ 2 , then the conclusion is immediate, so we assume that Proof. If M1 ≤ M 2 ¯ M1 > M . We set σ = osc u and choose 1 h(z) = exp(−(z − inf u)2 /σ 2 ). B 1+H Note that 1 1 ≤h≤ , (9.16) (1 + H)e 1+H and hence 2 h00 E + H1 (h0 )2 + J0 h0 ≥ E(h00 − H(h0 )2 + Jh0 ) = 2 hEH ∗ σ for (sup u − u)2 H∗ = 1 + 4 (1 − Hh) − J(sup u − u) σ2 wherever w1 ≥ 12 M1 . From the first inequality of (9.14), we have osc u ≤ 1/(2J), and the the second inequality of (9.16) gives Hh ≤ 1. Hence H ∗ ≥ 1/2, and therefore h h00 E + H1 (h0 )2 + J0 h0 ≥ M1 2 E. σ

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then the second inequality of (9.14) and the first inequality of (9.16) imply that   1 h00 E + H1 (h0 )2 + J0 h0 ≥ M1 E +K . 2 Moreover, from (9.12c), we have K1 w1 ≥ −KM1 E, and the second inequality of (9.14) along with the first inequality of (9.16) imply that −K h M1 E. 1 + 2K σ 2 Since h ≥ 1/(e(1 + H)), we infer that K1 w1 ≥

1 1 M1 [Eh00 + H1 (h0 )2 + J0 h0 ] + K1 w1 ≥ M1 E 2 4e(1 + H)(1 + 2K)σ 2

wherever w1 ≥ M1 /2. Our next step is to estimate 1 ˜ 0 + Kw ˜ 1. I˜ = M1 Jh 2 We first observe that there is a positive constant C1 , determined only by n and µ1 , such that 1 J˜ ≥ −C1 E ηR|Du| and ˜ ≥ −C1 K



1 1 + ηR|Du| η 2 R2 |Du|2



E

¯ . With C2 a constant at our disposal, we assume that whenever w1 ≥ M M1 ≥ C2 σ 2 /R2 . Then √ 1 σ M1 h 2C1 h 0 ˜ M1 Jh ≥ −C1 E ≥ −√ M1 E 1/2 σ 2 2 C 2(1 + H)σ 2 2 Rw1 and ˜ 1 ≥ −C1 Kw

! √ 2 2 M0 h + E. C2 σ C2 σ 2 σ2

By choosing C2 sufficiently small (determined only by H, K, and C1 ) and noting that σ ≤ 1, which follows from (9.14), we obtain aij Dij w + B3i Di w > 0

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on S0 , the set on which w1 ≥ M1 /2. It follows from the maximum principle that w cannot attain its maximum in S and hence it attains its maximum on ∂S0 . On ∂B, we have w1 = 0 and hence sup w ≤ M1 /2 which implies that sup w1 ≤ M1 /2, which can’t happen. On ∂S0 ∩ B, we have w1 = M1 /2, so w ≤ M1 (2 + H)/(2(1 + H)) < M1 , which contradicts our inequality sup w ≥ M1 . Therefore we must have M1 ≤ C2 and then (9.15) follows since η ≥

3 4

σ2 R2

in B(y, R/2).



If the oscillation of u is not small, then a gradient estimate is true provided we make an additional hypothesis on the coefficients H, J, and K. In our next theorem, we assume that K is sufficiently small and we refer the reader to the Notes for a discussion of other possibilities. Theorem 9.2. Let u ∈ C 3 (B), where B = B(y, R) for some y ∈ Rn and R > 0, and suppose that Qu = 0 in B. Suppose aij can be decomposed as ¯ , H, J, K be the in (9.3) and that conditions (9.11) are satisfied. Let M constants from (9.12) and suppose that there is a nonnegative constant µ1 ¯ . If such that condition (9.13) holds whenever |p| ≥ M osc u ≥ min{ B

and K≤

1 1 ,p } 1 + 2J e(1 + 2K)(1 + H)

1 (J + 1) exp(−(J + 1)(H + 1) osc u), B 2

(9.17)

(9.18)

then ¯ + C(n, µ1 , H, J, K) osc u . |Du| ≤ 2M R B(y,R/2) sup

(9.19)

¯ 2 , and we set σ = oscB u. With Proof. Again, we assume that M1 ≥ M A = (1 + J)(1 + H), we choose h(z) =

1 exp(A(inf u − z)). B 1+H

It then follows that 1 1 M1 (h00 E + H1 (h0 )2 + J0 h0 ) + K1 w1 ≥ AhE 2 4 wherever w1 ≥ M1 /2.

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The estimate of I˜ in the proof of Theorem 9.1 is easily modified to show that there is a constant C2 (n, µ1 , H, J, K) such that 1 I˜ ≥ − AhM0 E 8 whenever w1 ≥ M1 /2 provided M1 ≥ C2 σ 2 /R2 . Again, we conclude that aij Dij w + B3i Di w > 0 whenever w1 ≥ M1 /2 provided M1 ≥ C2 σ 2 /R2 , and hence (9.19) follows.  For future reference, we note that a gradient bound can also be proved if we have a bound for the gradient only on the boundary of Ω. Corollary 9.3. Let Ω be an open set and let B = B(y, R) ∩ Ω for some y ∈ Rn and some R > 0. Let u ∈ C 3 (B) ∩ C 1 (B(y, R) ∩ Ω) and suppose that Qu = 0 in B. Under the hypotheses of Theorem 9.15, we have oscB u ¯ + sup |Du| ≤ 2(M sup |Du|) + C(n, µ1 , H, K) . (9.20) R B(y,R/2)∩Ω ∂Ω∩B(y,R) Under the hypotheses of Theorem 9.19, we have sup B(y,R/2)∩Ω

¯+ |Du| ≤ 2(M

sup ∂Ω∩B(y,R)

|Du|)+C(n, µ1 , H, J, K)

oscB u . (9.21) R

Proof. Just note that w can’t have a maximum on ∂Ω ∩ ∂B if M1 ≥ sup∂Ω∩B(y,R) |Du|2 .  9.2

A simple boundary value problem

We now observe that the arguments of the preceding section yield a gradient bound when the boundary condition is Du · γ = 0 on ∂Ω

(9.22)

and Ω is convex. This boundary condition implies that Du is a tangential vector field on ∂Ω and hence we can differentiate the boundary condition in the direction of Du to obtain Du · D(Du · γ) = 0 or Di uDij uγ j = −Dj uDi uDi γ j .

If we set w0 = |Du|2

(9.23)

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(as before), then Dj w = 2Di uDij u, and hence Dw · γ = 2Di uDij uγ j = −2Dj uDi uDi γ j by virtue of (9.23). Since Dγ is negative semidefinite, we infer that Dw0 · γ ≥ 0 on ∂Ω.

(9.24)

If we repeat the proof of Theorem 9.2 with η ≡ 1, then we infer under the hypotheses (9.11), osc u ≥ min{ Ω

and (9.18) that

1 1 ,p }, 1 + 2J e(1 + 2K)(1 + H)

(9.25)

aij Dij w + B4i Di w > 0 wherever w0 ≥ M0 /2, where

1 w = w0 + M0 h(u) 2 and M0 = sup w0 . It’s also easy to check that γ · Dw ≥ 0

¯ , then we have on ∂Ω. If w has a maximum at a point where |Du| ≤ M a bound for Du everywhere. Otherwise, the strong maximum principle implies that w0 must be a constant. Now let x0 be a point on ∂Ω at which u attains its maximum over ∂Ω. It follows from (9.24) that Du(x0 ) = 0 because the tangential derivatives of u must be zero at x0 , which implies that |Du| ≡ 0, which also implies a gradient bound. 9.3

Gradient estimates for general boundary conditions. General considerations

If we want to study oblique boundary value problems under more general hypotheses, then the considerations of the previous section are no longer adequate for obtaining gradient bounds. It is not hard to see that the choice of estimating w = |Du|2 in Section 9.2 is closely tied to the specific boundary condition Du · γ = 0. Hence, we want to estimate a function w with two basic properties: (1) w satisfies a suitable differential inequality in the domain and a suitable boundary inequality which allow the application of the maximum principle.

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(2) Estimates on w imply estimates on Du. There are several considerations that are implied by the general description of these two properties. First, the function w must take the boundary condition into account, and, second, w should be a convex function of Du in order to control the differential inequality for w. (The convexity of w = |Du|2 was used in the previous two sections to obtain a lower bound for aij ∗ Dik uDjk u.) Ideally, w should even be strictly convex in order to estimate |Du| in terms of w. In fact, we shall now use a strictly convex w, and we shall not estimate |Du| directly. For the simple boundary condition Du · γ = 0, any estimate for the tangential gradient of u on ∂Ω implies a corresponding estimate for |Du| on ∂Ω; then we can the argument in Corollary 9.3 to obtain an estimate for |Du| in all of Ω. This is the basis for the argument in the next section. We make some preliminary observations in this section. As a first step, we recall from Lemma 5.18 that, if ∂Ω ∈ C 3 , then there is a positive constant R0 , determined only by Ω such that d ∈ C 3 (Ω∗ ), where Ω∗ is the subset of Ω on which d < R0 . It follows that γ = Dd is a C 2 (Ω∗ ) unit vector field. In fact, many of the results in this chapter can be proved (even for ∂Ω ∈ H2+α ) by using other unit vector fields. We also define and we set

cij = δ ij − γ i γ j , wT = cij Di uDj u.

Under some reasonable assumptions, we can estimate |Du · γ| in terms of v on ∂Ω just by using the boundary condition. Lemma 9.4. Let b ∈ C 0 (∂Ω × I × Rn ) for some set I of real numbers, and suppose that b is differentiable with respect to the third variable with bp · γ > 0 and lim ±b(x, z, p ± tγ) > 0.

(9.26)

t→∞

Suppose also that there are positive constants µ2 , M3 and M4 such that if |p − (p · γ(x))γ(x)| ≤ M3 and b(x, z, p) = 0, then |p · γ| ≤ M4 and |bp (x, z, p)| ≤ µ2 bp (x, z, p) · γ(x) n

(9.27)

1

for all (x, z, p) ∈ ∂Ω × R × R with b(x, z, p) = 0. If u ∈ C (Ω) satisfies b(x, u, Du) = 0 on ∂Ω and u(x) ∈ I for all x ∈ ∂Ω, then 1/2

on ∂Ω.

|Du · γ| ≤ M4 + µ2 wT

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Proof. We set p0 = p − (p · γ)γ and D0 u = Du − (Du · γ)γ (so Di0 u = cij Dj u), we fix x ∈ ∂Ω, and we consider two cases. If |D0 u| ≤ M3 , then we have |Du · γ| ≤ M4 . If |D0 u| > M3 , then we note that the equation b(x, u, p0 + pγ γ) = 0 gives a function F such that pγ = F (p0 ). In other words, if p is any vector such that |p0 | ≥ M2 , then there is a unique pγ ∈ R such that b(x, u, p0 +pγ γ) = 0. Noting that p0 lies in an (n − 1)-dimensional hyperplane, we see that ∂F |b0p | ∂p0 = bp · γ ≤ µ2 .

Since |pγ | ≤ M4 if |p0 | = M3 , it follows from the mean value theorem that |pγ | ≤ M4 +µ2 |p0 | and hence |Du·γ| ≤ M4 +µ2 |D0 u| in this case. Combining 1/2 these inequalities and observing that |D0 u| = wT completes the proof.  A number of other functions will also be useful in our gradient bounds. We write c˜ for the vector with components c˜i defined by c˜i (x, p) = Di (ckm )pk pm .

Since there will be no danger of confusion, we abbreviate c˜i (x, Du) to c˜i . Similarly, we define the vector-valued functions T u and T 0 u with components given by Ti u = cij Dj u,

Ti0 u = Di γ j Dj u,

and we set Γu = Du · γ. Because γ is a unit vector, it follows that γ j Di γ j = 0 and that c˜i = −2ΓuTi0u. We also define the matrix [¯ cij ] with entries given by c¯ij = |T u|−2Dij (ckm )Dk uDm u.

We also define two operators δT and δN , acting on functions of the variables (x, z, p), by 1 δT g = gz + |p0 |−2 p0 · gx − |p0 |−2 c˜ · gp , δN g = γ · gp . 2 Another structure functions that will be very useful is C2T = aij ckm Dik uDjm u. For the estimates, we also use the abbreviations p Du v = 1 + |Du|2 , ν = , g ij = δ ij − ν i ν j . v

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In addition, if [f ij ] is a matrix, we write F in place of f ij γi γj . The only exception to this abbreviation is that we reserve the letter C to denote a constant. Since cij γi γj = 0, such an exception is natural. Although it is possible to prove a number of global gradient estimates directly, we prove our global gradient estimates via local gradient estimates because the additional complications introduced by this step are relatively minor compared to those generated by the general boundary condition we consider and any smooth extension of γ cannot be a unit vector in all of Ω. In fact, the second difficulty is not significant in the proof since the extension can be taken to be a vector with length no greater than one.

9.4

Global gradient estimates for general boundary conditions and false mean curvature equations I

We again assume that the decomposition (9.3) is valid for some positive definite matrix [aij ∗ ] with minimum eigenvalue λ∗ and some nonnegative 1 scalar function τ . We also assume that aij ∗ and τ are C with respect to (x, z, p). To illustrate the use of our decomposition, we begin by deriving a gradient estimate under fairly strong assumptions on the coefficients of Q and M . Because the calculations, even in this case, are delicate, we first derive a global estimate for wT under strong hypotheses. To derive a differential inequality for wT , we first differentiate the equation Qu = 0 in the direction xm and then multiply the resulting equation by ckm Dk u. In this way, we obtain aij Dij wT − 2C2T + B0r Dr wT = S + I1 ,

(9.28)

with S = −2δT aij Dij uwT + 2aij Dim uDj (ckm )Dk u,

I1 = −2δT awT + aij c¯ij wT .

Our goal is now easy to state. As before, the second derivatives of u appear only linearly in S, so we want to use Cauchy’s inequality to estimate S in terms of C2T (which is quadratic in the second derivatives) and then use appropriate structure conditions to show that we obtain a differential inequality for w which yields a maximum principle. Unfortunately, C2T is only quadratic with respect to some of the second derivatives. For example,

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if B 0 lies in the hyperplane xn = 0 and if aij = δ ij , then C2T

n n−1 X X = (Dij u)2 . i=1 j=1

In general, we expect the terms in S to include Dnn u in this special situation, so we must estimate this second derivative, and this estimate is accomplished because the differential equation for u can be solved for Dnn u. We use the decomposition (9.3) of aij to achieve this subgoal by rewriting some of the terms of S using DwT . We begin by showing how to estimate various linear combinations of second derivatives of u. Our first estimate will be used to estimate terms involving derivatives of aij ∗ , so we obtain bounds in terms of λ∗ . To simplify notation, if F = [f ij ] is a matrix, we define  1/2 n X kFk =  (f ij )2  . (9.29) i,j=1

Lemma 9.5. If Qu = 0, then for any matrix [f ij ] and any ε > 0, we have 2F 2 τ wT εAλ∗ 4kFk2 F 2 Λ∗ Fa + + (n + 1) − . λ∗ ε εAλ∗ A

f ij Dij u ≤ εC2T +

Proof.

(9.30)

We set f1 = f ij cjk Dik u, f2 = f im γm cij Djk uγ k , F f3 = − aij cjk Dik u, A F im f4 = − a γm cij Djk uγ k . A

It follows that f ij Dij u = f1 + f2 + f3 + f4 +

F ij a Dij u, A

so the equation Qu = 0 implies that f ij Dij u = f1 + f2 + f3 + f4 − Since 2 aij ξi ξj ≥ aij ∗ ξi ξj ≥ λ∗ |ξ|

F a. A

(9.31)

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for any vector ξ, we conclude directly from the Cauchy-Schwarz inequality that 1 2kFk2 f1 ≤ εC2T + , 8 λ∗ ε 1 2kFk2 f2 ≤ εC2T + . 8 λ∗ ε Next, we use the Cauchy-Schwarz inequality another time to infer that f3 ≤

1 F 2 ij ij 1 2 εCT + a c , 4 ε A2

and we also have ij ij ij ij aij cij = aij ∗ c + τ Di uc Dj u = a∗ c + τ wT ≤ nΛ∗ + τ wT .

(9.32)

Hence, 1 2 1 F2 εCT + [nΛ∗ + τ wT ]. 4 ε A2 Using the decomposition (9.3), we see that f3 ≤

F F ij a γm cij Djk uγk − τ ΓuDi ucij Djk uγ k . A ∗ A Schwarz’s inequality gives X F |F | ij k ij − aim (a∗ γi γj )1/2 (Λ∗ (cij Djk u)2 )1/2 ∗ γm c Djk uγ ≤ A A i,k  1/2 |F | Λ∗ ij km ≤ 1/2 a c Dik uDjm u λ∗ ∗ A f4 = −

because aij ∗ γj γj = A∗ ≤ A, and hence Cauchy’s inequality implies that F im 1 F 2 Λ∗ a∗ γm cij Djk uγ k ≤ εC2T + . A 4 ελ∗ A A final application of the Cauchy-Schwarz inequality yields −

−

F 1 F 2 τ 2 (Γu)2 τ Di uDm uγm cij Djk uγ k ≤ εC2T + wT A 4 εA2 λ∗

and hence, since τ (Γu)2 ≤ A,

1 2 Λ∗ F 2 F 2τ εCT + + wT . 2 εAλ∗ εAλ∗ The proof is completed by combining the estimates on f1 , f2 , f3 , and f4 and noting that A ≥ λ∗ .  f4 ≤

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The estimate for terms involving τ and its derivatives is similar. Lemma 9.6. Let ξ be a vector and set ζ = ξ · γ. If Qu = 0, then for any ε > 0, we have 1 (n + 1)Λ∗ 2 ξ r Dir uDi u ≤ εC2T + cij ξi ξj + ζ ετ ετ A (9.33) ζ 2 A2∗ wT ζΓu + a. − εA2 λ∗ A Proof.

We begin by noting that ξ r Dir uDi u = ξ k ckr Dir uDi u + ζγ r Dir uDi u,

and the Cauchy-Schwarz inequality implies that 1 1 ξ k ckr Dir uDi u ≤ εC2T + cij ξi ξj . 4 ετ To continue, we set ζΓu ij ir Z1 = − [a c Drj u], A ζΓu ij Z2 = − [a γi γr cjm Drm u], A ∗ ζΓu Z3 = − [τ ΓuDj uγr cjm Drm u], A Z4 = ζγ i cjr Dij uDr u, and we observe that ζΓu a. A We then use the Cauchy-Schwarz inequality and (9.32) to infer that ε ζ 2 (Γu)2 Z1 ≤ C2T + (nΛ∗ + τ wT ) 4 εA2 ε ζ2 ≤ C2T + (nΛ∗ + τ wT ) 4 ετ A because τ (Γu)2 ≤ A. By applying the Cauchy-Schwarz inequality and the inequality τ (Γu)2 ≤ A again, we see that ε ζ2 Z2 ≤ C2T + Λ∗ . 4 ετ A Next, we observe that   τ (Γu)2 Z3 + Z4 = ζ 1 − γ i cjr Dij uDr u A A∗ i jr =ζ γ c Dij uDr u. A ζγ r Dir uDi u = Z1 + Z2 + Z3 + Z4 −

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Another application of the Cauchy-Schwarz inequality leads to Z3 + Z4 ≤

ε 2 ζ 2 A2∗ wT CT + . 4 εA2 λ∗

The proof is completed by combining all these estimates.



We are now ready to estimate S. First, we write wT (δT aij )Dij u = wT (δT aij ˜i )Dj uDij u. ∗ )Dij u + (wT δT τ Di u − τ c ij We estimate (δT aij = wT δT aij ∗ )Dij uwT by using f ∗ in Lemma 9.5 with i ε = 1/8, and we estimate (wT δT τ Di u − τ c˜ )Dj uDrj u by using ξ = wT δT τ Du − τ c˜ and ε = 1/8 in Lemma 9.6. Combining these estimates with the inequalities A ≥ A∗ ≥ λ∗ yields

1 −(δT aij )Dij uwT ≥ − C2T + A1 wT , 4

(9.34)

32kδT A∗ k2 16(δT A∗ )2 τ wT2 16(δT τ )2 wT2 − − 2 λ∗ λ∗ τ   2 2 (δT τ ) (δT A∗ ) 16τ |˜ c |2 + − 8(n + 1)Λ∗ wT − 2 2 λ∗ τ wT   |δT A∗ | |δT τ | − |a| + . λ∗ τ

(9.35)

with A1 = −

To obtain a useful lower bound for the second term, 2aij Dim uDj (ckm )Dk u, in S, we note that Dj (ckm ) = −Dj γ m γ k − Dj γ k γ m and therefore (using also (9.3)) −aij Dim uDj (ckm )Dk u = I2 + I3 + I4 , where I2 = aij Dim uDj γ m Γu, m k I3 = aij ∗ Dim uγ Dj γ Dk u,

I4 = τ Di uDim uγ m Dj γ k Dk uDj u. We begin by using the Cauchy-Schwarz inequality to infer that I2 ≤

1 2 C + 2(Γu)2 aij Di γ m Dj γ m . 8 T

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k j Next, we use Lemma 9.5 with f ij = aim ∗ Dm γ Dk uγ and ε = 1/8 to infer that k i 2 k ir j 1 2 32(aim 64aim ∗ Dm γ Dk uγ ) ∗ Dm γ Dk ua∗ Dr γ Dj u I3 ≤ CT + τ wT + 16 Aλ∗ λ∗ im im k i 2 (a Dm γ Dk uγ ) Λ∗ a∗ Dm γ k Dk uγ i − a. + 16(n + 1) ∗ λ2∗ A

To estimate I4 , we first note that Dj γ k Dk uDj u = T 0 u · Du. It then follows from Lemma 9.6 with ξ = τ T 0 u · Duγ and ε = 1/16 that I4 ≤

1 2 16(n + 1)τ Λ∗ (T 0 u · Du)2 τ (T 0 u · Du)Γu CT + − a 16 λ∗ A 16τ 2 (T 0 u · Du)2 + . λ∗

Combining these estimates with the inequality τ (Γu)2 ≤ A yields 1 aij Dim uDj (ckm )Dk u ≥ − C2T + A2 wT 4

with k i 2 2 32(aim ∗ Dm γ Dk uγ ) τ (Γu)2 aij Di γ m Dj γ m − 2 wT λ∗ im k ir j 64a∗ Dm γ Dk ua∗ Dr γ Dj u − λ∗ wT 16(n + 1)Λ∗ im − ((a∗ Dm γ k Dk uγ i )2 + τ (T 0 u · Du)2 ) λ∗ wT 16τ 2 (T 0 u · Du)2 − λ∗ wT  im  a∗ Dm γ k Dk uγ i + τ (Γu)T 0 u · Du +a . AwT

A2 = −

We now collect these estimates to obtain

−C2T + aij Dij wT + B0i Di wT ≥ A3 wT ,

(9.36)

A3 = 2A1 + 2A2 − 2δT a + aij c¯ij .

(9.37)

aij Dij wT + B0i DwT ≥ A3 wT .

(9.38)

where

For later use, the C2T term will be useful, but, for now, we note that To apply the maximum principle, we need to introduce hypotheses that imply A3 > 0 and the complicated form of this quantity make its analysis

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somewhat unwieldy. We start by taking advantage of a few simple inequalities. Specifically, if we assume that there is a positive constant µ3 such that |δT aij ∗ | ≤ µ3 λ∗ ,

|δT τ | ≤ µ3 τ

(9.39a)

Λ∗ ≤ µ3 λ∗

(9.39b)

wherever |Du| ≥ 1, then there is a constant C4 (n, Ω, µ3 ) such that A1 ≥ −C4 (λ∗ |Du|2 + τ |Du|4 + |a|)

(9.40)

wherever wT ≥ 1. Similarly, if there is a positive constant µ3 such that condition (9.39b) holds whenever |Du| ≥ 1, then there is a constant C5 (n, Ω, µ3 ) such that    τ 1/2 A2 ≥ −C5 λ∗ |Du|2 + τ |Du|4 + |a| 1 + 1/2 (9.41) A whenever wT ≥ 1. At this point, we can state a gradient bound under fairly strong hypotheses. The most stringent one is that δT a must be large and negative. Theorem 9.7. Let u ∈ C 3 (Ω) ∩ C 2 (Ω) and suppose that there are a posiij tive definite matrix [aij ∗ ] and a nonnegative scalar τ such that [a ] can be decomposed as in (9.3). Suppose conditions (9.11) are satisfied and let M1 , H, J, and K be the constants from (9.12). Suppose that there is a nonnegative constant µ1 such that (9.13) holds. Suppose also that bp · γ > 0, (9.26) is satisfied, δT b < 0, and that there are positive constants µ2 , M3 and M4 such that (9.27) holds for all (x, z, p) with b(x, z, p) = 0 and that if b(x, z, p) = 0 and |p−(p·γ)γ| ≤ M3 , then |p·γ| ≤ M4 . Set R = min{1, R0 }. Suppose further that, for any y ∈ Ω with d(y) ≥ R, either condition (9.14) or conditions (9.17) and (9.18) are satisfied. Suppose finally that there are positive constants µ3 , µ4 , and µ ¯1 such that conditions (9.39) and τ ≤ µ4 λ∗ ,

|a| ≤ µ ¯1 δT a,

¯ . If are satisfied for |Du| ≥ M

(9.42a)

E≤µ ¯ 1 δT a

(9.42b)

1/2

µ ¯1 (C4 + C5 (1 + µ4 ) + 1) ≤ 2 (for the constants C4 and C5 from (9.40) and (9.41)), then ¯ , M3 , M4 , H, J, K, osc u). sup |Du| ≤ C(n, Ω, µ1 , . . . , µ4 , M Ω

Ω

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Proof. We first fix y ∈ Ω with d(y) ≥ R. If condition (9.14) holds, then Theorem 9.1 gives a constant C6 (determined only by n, µ1 , osc u, H, M1 , M2 , and Ω) such that |Du(y)| ≤ C6 . On the other hand, if conditions (9.17) and (9.18) hold, then Theorem 9.2 gives a constant C7 (determined by J, K, and the same quantities as for C6 ) such that |Du(y)| ≤ C7 . It then follows from Corollary 9.3 and Lemma 9.4 that sup |Du| ≤ C(C6 , C7 , µ2 , M3 , sup wT ), ΓR

and hence we only need to estimate wT in ΓR . Moreover, if we set M6 = max{C6 , C7 }, then wT ≤ M6 on ∂ΓR ∩ Ω. Next, because δT b < 0 on ∂Ω and br Dr wT + (δT b)wT = 0 on ∂Ω, it follows that wT can’t have a maximum on ∂Ω. If wT has a maximum at some x0 ∈ ΓR , then we use (9.38) to infer that A3 (x0 ) ≤ 0. If wT (x0 ) ≥ max{1, M1 , . . . , M5 }, then we have 1/2

−1/2

A3 (x0 ) ≥ −δT a[2 − (2C4 + 2C5 (1 + µ4 ) + C(Ω)wT

)¯ µ1 ].

But this inequality implies that A3 (x0 ) > 0 if wT (x0 ) > C(Ω)2 . It follows that ¯ 2M4 , C(Ω)2 }, sup wT ≤ max{2, 2M, ΓR

and the desired estimate follows. 9.5



Global gradient estimates for general boundary conditions and false mean curvature equations II

If we want to relax the hypotheses δT a < 0 and δT b < 0, we need to apply the maximum principle to a function similar to w from Section 9.1. We set MT = supΓR wT and w = wT +

1 1 MT vd + MT h(u) 4k0 4

with k0 is a constant at our disposal such that dv ≤ k0 ,

(9.43)

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and h a function to be further specified but satisfying 0 ≤ h ≤ 1. (Hypotheses on the operator Q which guarantee such an inequality will be made later.) Using the notation from Section 9.4, we infer from (9.36) that −C2T + aij Dij w + B0i Di w ≥ A3 wT + (A4 + A5 )MT , where 1 [v(aij Dij d + B0i Di d) + d(aij Dij v + B0i Di v) + 2aij Di vDj d], 4k0 1 A5 = [h00 E − h0 a + h0 B0i Di u]. 4 We rewrite the terms in A4 in two steps. The first step is to differentiate the differential equation Qu = 0 to infer that A4 =

|Du|2 (δaij Dij u + δa). v For the second step, we recall that Dd = γ and then define the operator δN by ∂g δN g = γ · , ∂p so aij Dij v + B0i Di v = aij g km Dik uDjm u −

B0i Di d = δN aij Dij u + δN a. For A5 , we note that ¯ ij Dij u + δa, ¯ B0i Di u = δa and hence (since aij g km Dik uDjm u ≥ 0)

1 aij Dij w + B0i Di w ≥ C2T + A3 wT + MT [h00 E + J1 h0 + K2 ] + S2 , 4 where J1 = (δ¯ − 1)a,

1 d|Du|2 (vaij Dij d + vδN a − δa), k0 v MT d|Du|2 ij S2 = [− δa Dij u − 2aij Di vDj d + vδN aij Dij u] 4k0 v MT 0 ¯ ij + h δa Dij u. 4 The terms in S2 are now estimated using Lemmata 9.5 and 9.6. First, we note that MT d|Du|2 ij MT d|Du|2 ij MT d|Du|2 δa Dij u = δa∗ Dij u + δτ Di uDij uDj u. 4k0 v 4k0 v 4k0 v K2 =

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We estimate the first term in this expression via Lemma 9.5 with ε = 1/16 and f ij = MT dδaij ∗ /(4k0 v), and we estimate the second term via Lemma 9.6 with ε = 1/16 and ξ = MT dδτ Du/(4k0 vτ ). From these inequalities, (9.43), and τ (Γu)2 ≤ A,

λ∗ ≤ A,

|Du| ≤ v

(9.44)

we infer that −

MT d|Du|2 ij 1 1 δa Dij u ≥ − C2T + MT K3 , 4k0 v 8 4

where 8MT (δA∗ )2 kδA∗ k2 τ w − 16M T T λ2∗ λ∗   (δτ )2 (δA∗ )2 + − 4(n + 1)MT Λ∗ λ2 τ2   ∗ |δA∗ | |δτ | − |a| + . λ∗ τ

K3 = −

Next, we write Di v = v1 Dik uDk u and hence MT ij MT ik MT 2a Di vDj d = a γk Dj uDij u + τ (Γu)Di uDij uDj u. 2k0 2k0 v ∗ 2k0 v Using ε = 1/16, f ij = MT aik ∗ γk Dj u/(2k0 v), and ξ = MT (Γu)Du/(2k0 v) (along with (9.43) and (9.44) and the definition of Λ∗ ), we find that −

MT ij 1 1 2a Di vDj d ≥ − C2T + MT K4 , 4k0 8 4

24 k02

  2 τ Λ∗ 24n + 32 Λ3∗ 2|a| 1+ τ MT wT − − . 2 λ∗ λ∗ k02 λ2∗ k0

with K4 = −

For the next term, we have MT v MT v MT v δN aij Dij u = δN aij [δN τ Di u + 2τ γ i ]Dij uDj u ∗ Dij u + 4k0 4k0 4k0 Using ε = 1/16, f ij = −MT vδN aij ∗ /(4k0 ), and ξ = −MT v[δN τ Du + 2τ γ]/(4k0 ), we then have MT v 1 1 δN aij Dij u ≥ − C2T + MT K5 4k0 8 4

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with

8MT v 2 (δN A∗ )2 τ wT 16MT v 2 kδN A∗ k2 − k02 λ2∗ k02 λ∗   2 2 2(δN τ )2 8τ 4(n + 1)Λ∗ MT v kδN A∗ k + + − k02 λ2∗ τ2 λ∗ 2Λ∗ MT v 2 (δN τ )2 8τ 2 v 2 MT − − k02 τ λ∗ k02 λ∗   |a| v|δN A∗ | v|δN τ | 2τ 1/2 v − + 1/2 . + k0 λ∗ τ A Next, we write MT 0 ¯ ij MT 0 ¯ ij MT 0 ¯ h δa Dij u = h δa∗ Dij u + h (δ + 2)τ Di uDij uDj u. 4 4 4 ij 0 ¯ ij Using ε = 1/16, f = MT h δa∗ /4 and ξ = MT h0 (δ¯ + 2)τ Du/4, we find that MT 0 ¯ ij 1 1 h δa Dij u ≥ − C2T + MT [H2 (h0 )2 + J2 h0 ] 4 8 4 with ¯ ∗ k2 ¯ ∗ )2 τ MT wT 16MT kδA 2(δA − H2 = − 2 λ∗ λ∗  ¯  2 (δA∗ ) [(δ¯ + 2)τ ]2 − 4(n + 1)Λ∗ MT + λ2∗ τ2 and a ¯ 2 ¯ J2 = − (δA ∗ + (δ + 2)τ (Γu) ). A Combining all these estimates, we see that 1 aij Dij w + B0i Di w ≥ A3 wT + MT [h00 E + H2 (h0 )2 + J3 h0 + K6 ], (9.45) 4 where J3 = J1 + J2 , K 6 = K 2 + K 3 + K 4 + K 5 . We are now ready to state our next gradient bound. K5 = −

Theorem 9.8. Suppose that Λ∗ = O(λ∗ ),

(9.46a)

= o(λ∗ ), |δτ | = o(τ ), ∂aij ∂τ ∗ |p| = O(λ∗ ), |p| = O(τ ), ∂p ∂p ∂a |p| = O(E), ∂p

(9.46b)

|δaij ∗ |

δa ≤ O(E), ¯ (δ − 1)a ≤ O(E).

(9.46c) (9.46d) (9.46e) (9.46f)

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Suppose further that |δT aij ∗ | = o(λ∗ ),

|δT τ | = o(τ ),

|δT a| = o(E),

τ = O(λ∗ ),   1/2 τ |p| 1 + 1/2 |a| = O(E). λ∗

(9.47a) (9.47b) (9.47c) (9.47d)

Suppose also that bp · γ > 0, (9.26) is satisfied and that there are positive constants µ2 , M3 , and M4 such that (9.27) holds for all (x, z, p) with b(x, z, p) = 0 and that if b(x, z, p) = 0 and |p − (p · γ)γ| ≤ M3 , then |p · γ| ≤ M4 . Suppose finally that ¯ ≤ o(|p|bp · γ) δT b, δb

(9.48)

wherever b(x, z, p) = 0. Then there is a constant C determined only by osc u, Ω, µ2 , M3 , M4 , and the limit behavior in (9.46), (9.47), and (9.48) such that |Du| ≤ C in Ω. Proof. We first observe from Theorems 9.1 and 9.2 that there is a constant C1 , determined only by osc u, Ω, and the limit behavior in (9.46) such that |Du(x)| ≤ C1 (1 + 1/d(x)). We now fix R = min{R0 , 1}. It follows that |Du| ≤ C1 /R if d ≥ R and that |Du| ≤ 2C1 /d if d < R. Our next step is to estimate wT where d < R. To this end, we observe ¯ J, ¯ and M6 (determined by the limit that there are positive constants H, behavior in (9.46)) such that ¯ H2 ≥ −HE,

¯ J3 ≤ JE

(9.49)

whenever wT ≥ M6 and wT ≥ 21 MT . Our next step is to proceed as in the proof of Theorem 9.2 and set ¯ + H), ¯ A¯ = (1 + J)(1 1 ¯ h(z) = exp(A(inf u − z)), ¯ Ω 1+H ¯ = exp(−A¯ osc u). K

(9.50a) (9.50b) (9.50c)

We then conclude that ¯ h00 E + H2 (h0 )2 + J3 (h0 ) ≥ KE as long as wT ≥ 21 MT and MT ≥ M62 . Using Theorems 9.1 and 9.2 again, we also see that there are constants C2 and M7 determined only by osc u,

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Ω, and the limit behavior in (9.46) and (9.47) such that k0 ≥ C2 implies that 1 ¯ A3 wT + MT K6 > −KE 4 provided (9.43) holds, wT ≥ 21 MT , and wT ≥ M7 . In particular, this inequality holds if k0 = 1 + C1 + C2 . Hence wT can’t have an interior maximum greater than M6 + M7 + C1 /R. Moreover, on ∂Ω, we have   1 1 0¯ bp · Dw + δT bwT − MT vbp · γ + h δb = 0, 4 k0 and there is a constant M8 (determined only by osc u, Ω, and the limit behavior in (9.46), (9.47), and (9.48)) such that   1 1 0¯ δT bwT − MT vbp · γ + h δb < 0 4 k0 wherever wT ≥ 12 MT and wT ≥ M8 . As before, we infer a bound on wT where d ≤ R, and the proof is completed by using Lemma 9.4 and Corollary 9.3.  The hypothesis (9.47d) is stronger than the usual assumption |a| = O(E) unless τ |p|2 = O(λ∗ ), but it is not clear how to obtain a gradient bound in this generality (specifically with O(E) growth for a); see Theorem 5.1 in [101] for an attempt at weakening (9.47d) by strengthening the assumptions on b. We now look at our hypotheses on b in a representative case. Suppose that b(x, z, p) = F (v)Du · γ + ψ(x, z) for some C 1 functions F and ψ; we assume that sF 0 (s) ≥ −F (s) for all s ≥ 1. Then

(Du · γ)2 ≥ [1 − (ν · γ)2 ]F (v) ≥ c0 F (v) v for some positive constant c0 by virtue of Lemma 9.4, and     2 0 2 ¯ = Du · γ F (v) + F 0 (v) |Du| = −ψ(x, z) 1 + F (v) |Du| . δb v F (v) v bp · γ = F (v) + F 0 (v)

When F (v) = v q−1 for some nonnegative constant q, we conclude that   |Du|2 ¯ δb = ψ 1 + (q − 1) 2 . v

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For q = 0, we have |Du|2 1 = 2 v2 v ¯ = O(bp · γ/|p|). For q > 0, we see that δb ¯ = O(|p|1−q bp · γ). and hence δb ¯ It follows that δb ≤ o(|p|bp · γ) for all q ≥ 0. In addition,   (Du · γ)2 F 0 (v) p0 · ψx 1 δT b = ψz + + ≤ ψ + O + O(F ) z 0 2 0 2 |p | |p | v |p| (using γ · T 0 u = 0 and Lemma 9.4). It follows that δT b = o(|p|bp · γ) if q > 0 or if ψz ≤ 0. Hence, we obtain a gradient bound via Theorem 9.8 if q > 1 or if q = 0 and ψz ≤ 0. If q = 0 and ψz is bounded from above by a sufficiently small positive constant, then the argument of Theorem 9.8 is easily modified to obtain a gradient bound. (See Exercise 9.2 for details.) 1 + (q − 1)

9.6

Local gradient estimates

To proceed, we fix a point y ∈ ∂Ω and a number R ∈ (0, R0 ) (to be further specified), and we write B = B(y, R), B + = B ∩ Ω, B 0 = B ∩ ∂Ω. (Note that this is somewhat different notation from what we have used, but it should not cause any confusion.) We wish to combine the argument of the previous section with the one used to derive an interior gradient bound. The interior gradient bound was proved by using a simple cut-off function η, but, here, we need a more carefully chosen function. In particular, we introduce an important restriction on R. First, we assume that condition (9.27) holds, and we assume that ∂Ω is locally C 1 near y. (Of course, for our gradient estimate, we shall assume more regularity of ∂Ω.) Then there is a positive constant R1 such that 1 |γ(x) − γ(y)| ≤ (4µ2 + 1)(µ2 + 1) if x ∈ B 0 and R ≤ R1 . For any such x, we therefore have bp (x, z, p) · γ(y) = bp (x, z, p) · γ(x) + bp (x, z, p) · [γ(y) − γ(x)] 1 ≥ bp (z, x, p) · γ(x) − |bp (x, z, p)| (4µ2 + 1)(µ2 + 1) µ2 ] ≥ bp (x, z, p) · γ(x)[1 − (4µ2 + 1)(µ2 + 1) 4µ2 ≥ bp (x, z, p) · γ(x) . 4µ2 + 1

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It therefore follows that 1 |bp (x, z, p)| ≤ (µ2 + )bp (x, z, p) · γ(y). 4 To simplify notation, we now rotate and translate variables so that γ(y) points along the positive xn -axis and y = 0, and we define 2|x0 |2 (R − 4(2µ2 + 1)xn )2 − . (9.51) R2 4R2 √ √ |xn | ≥ R/ 2. Also, It’s easy to check that η(x) ≤ 0 if |x0 | ≥ R/ 2 or if √ √ 0 simple algebra shows that |x| ≥ R implies |x | ≥ R/ 2 or |xn | ≥ R/ 2, and hence η is nonpositive outside B(y, R) with η(y) = 3/4. We now write E and E 0 for the subsets of Ω and ∂Ω, respectively on which η > 0, and we set E + = ∂E \ E 0 . It follows that η ∈ C 2 (E) with |Dη| ≤ C(µ2 )/R and |D2 η| ≤ C(µ2 )/R2 . Moreover, on E 0 , we have η(x) = 1 −

x0 xn 2 [−2bp · + (2µ2 + 1)bp · γ(y) − 2(2µ2 + 1)2 bp · γ(y)] R R R   xn 2bp · γ(y) 1 |x0 | ≥ −(2µ2 + ) + 2µ2 + 1 − 8(2µ2 + 1)2 . R 2 R R

bp · Dη =

Next, we note that there is a positive constant R2 such that xn ≤ |x0 |/16(2µ2 + 1)2 if x ∈ E 0 and R ≤ R2 . With this additional restriction on R, we have   bp · γ(y) |x0 | bp · Dη ≥ −(2µ2 + 1) + 2µ2 + 1 . r R Since |x0 | ≤ R wherever η > 0, it follows that bp · Dη ≥ 0 on E 0 . Next, we set MT = sup η 2 wT , B+

and, with a nonnegative function h and a constant k0 also to be further determined, we set w = η 2 wT +

1 1 MT vd + MT h(u). 4k0 4

As before, we assume that condition (9.43) is satisfied and h ≤ 1. Although most of our effort will be directed toward estimating terms in the differential equation for w, we note here that our choice of η gives the boundary condition 1 ¯ ≥ 0. bp · Dw + δT bη 2 wT − MT [vbp · γ + h0 δb] (9.52) 4

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From our previous calculations and the identity B0i Di η = Dk ηaij,k Dij u + Di ηai , we see that 1 aij Dij w + B0i Di w ≥ 2C2T + A6 η 2 wT + MT [h00 E + J1 h0 + K2 ] + S4 , 4 where 2aij Dij η 2ai Di η A6 = A3 + + , η η S4 = S2 + 2ηwT Dk ηaij,k Dij u + 4ηaij Di wT Dj η. By arguing as in the previous section (except that ε = η 2 /16), we can estimate the terms in S2 just as before. In this way, we find that MT d|Du|2 1 ≥ − η 2 C2T 4k0 v 8 1 MT 2aij Di vDj d ≥ − η 2 C2T − 4k0 v 8 MT v 1 δN aij Dij u ≥ − η 2 C2T 4k0 8 MT 0 ¯ ij 1 h δa Dij u ≥ − η 2 C2T 4 8 wherever ηwT ≥ 12 MT , with −

16(δA∗ )2 2 kδA∗ k2 τ wT − 32wT 2 λ∗ λ∗   (δA∗ )2 (δτ )2 − 8(n + 1)wT Λ∗ + λ2 τ2 ∗  |δA∗ | |δτ | + , − |a| λ∗ τ   2 48 τ Λ∗ 2 2|a| 48n + 64 Λ3∗ =− 2 1+ τ wT − − , 2 k0 λ∗ λ∗ k02 λ2∗ k0 8MT v 2 (δN A∗ )2 τ wT 16MT v 2 kδN A∗ k2 =− − 2 2 k0 λ∗ k02 λ∗   4(n + 1)Λ∗ MT v 2 kδN A∗ k2 2(δN τ )2 8τ − + + k02 λ2∗ τ2 λ∗ 2 2 2 2 2Λ∗ MT v (δN τ ) 8τ v MT − − k02 τ λ∗ k02 λ∗   |a| v|δN A∗ | v|δN τ | 2τ 1/2 v + + , − k0 λ∗ τ A1/2

K10 = −

K11 K12

1 + MT K10 , 4 1 + MT K11 , 4 1 + MT K12 , 4 1 + MT [H3 (h0 )2 + J2 h0 ] 4

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¯ ∗ )2 τ w2 ¯ ∗ k2 4(δA 32wT kδA T − λ2∗ λ∗  ¯  (δA∗ )2 [(δ¯ + 2)τ ]2 − 8(n + 1)Λ∗ wT + . λ2∗ τ2

H3 = −

f

ij

Similarly, we can estimate the remaining terms in S4 . Using ε = η 2 /16, k = −2wT ηDk ηaij,k ∗ , and ξ = −2wT η[Dk ητ p + 2τ Dη], we find that 1 2wT ηDk ηaij,k Dij u ≥ − η 2 C2T + A˜1 η 2 wT , 8

where P ij,k 2 (Di ηAi∗ )2 2 i,j (a∗ Dk η) A˜1 = − 128 τ w − 256w T T η 2 λ2∗ η 2 λ∗   2(Dk ητ k )2 (Di ηAi∗ )2 8τ |Dη|2 + − 64(n + 1)Λ∗ wT + η 2 λ2∗ η2 τ 2 η 2 λ∗   k 2 2 (Dk ητ ) wT |Dη| Λ∗(Dk ητ k )2 wT τ 2 |Dη|2 − 128 + 4τ − 128 − 512 η2 τ η2 η 2 λ∗ τ η 2 λ∗ ! |Di ηAi∗ | |Dk ητ k | |Dη|τ 1/2 − 2|a| + + . 1/2 ηλ∗ ητ ηλ∗ Finally, we use the Cauchy-Schwarz inequality to see that 4ηaij Di wT Dj η ≥ −

η2 + A˜2 η 2 wT 8

with aij Di ηDj η A˜2 = −512 . η2 Combining all these estimates, we find that 1 aij Dij w + B0i Di w ≥ [A3 + A˜3 ]η 2 wT + MT [h00 E + H3 (h0 )2 + J3 h0 + K13 ] 4 where 2aij Dij η 2ai Di η A˜3 = + + A˜1 + A˜2 , η η K13 = K10 + K11 + K12 , From these estimates, we can now prove our basic local gradient estimate. Theorem 9.9. Let y ∈ R, let r > 0, and let u be a solution of Qu = 0 in Ω ∩ B(y, r), N u = 0 on ∂Ω ∩ B(y, r),

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with ∂Ω ∩ B(y, r) ∈ C 3 . Suppose that conditions (9.46) and (9.47) are satisfied. Suppose also that bp · γ > 0, (9.26) is satisfied and that there are positive constants µ2 , M3 , and M4 such that (9.27) holds for all (x, z, p) with b(x, z, p) = 0 and that if b(x, z, p) = 0 and |p − (p · γ)γ| ≤ M3 , then |p · γ| ≤ M4 . Suppose finally that (9.48) is satisfied wherever b(x, z, p) = 0. Then there are positive constants R (determined only by r, µ2 and Ω) and C (determined only by Ω, µ2 , M3 , M4 , and the limit behavior in (9.46), (9.47), and (9.48)) such that  oscΩ∩B(y,R) u  |Du(y)| ≤ C 1 + . R

Proof. We choose R = min{r, R1 , R2 , 1, R0 }. It follows (as in Theorem 9.8) that there is a constant C1 such that d|Du| ≤ C1 in B(y, R). In ˆ J, ˆ and M8 so that addition, there are positive constants H, ˆ H3 ≥ −HE,

ˆ J3 ≤ JE,

K13 ≥ −E

if η 2 wT ≥ M8 and η 2 wT ≥ 12 MT . We now distinguish two cases, using the same definition σ = oscE u, as before. First, if 1 1 σ ≤ min{ , √ }, ˆ 1 + 2Jˆ 16e(1 + H) we take 1

exp(−(z − inf u)2 /σ 2 ). ˆ E 1+H In this case, we argue as in Theorem 9.1 to infer that there is a further σ constant M9 so that w can’t have an interior maximum if MT ≥ M9 (1+ R ). On the other hand, if 1 1 σ > min{ , √ }, ˆ ˆ 1 + 2J 16e(1 + H) h(z) =

we take 1 ˆ + H)(inf ˆ exp((1 + J)(1 u − z)) ˆ E 1+H and argue as in Theorem 9.2 to infer that there is a constant M9 so that w σ can’t have an interior maximum if MT ≥ M9 (1 + R ). 2 + Since η wT = 0 on E , it follows that w can’t attain its maximum there. Finally, by virtue of (9.52), there is a constant M10 so that w can’t have a maximum on E 0 if MT ≥ M10 . The theorem is now proved with C = max{M9 , M10 }.  h(z) =

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When the oscillation of u is sufficiently small, we can combine the proof of the preceding theorem with that of Theorem 9.1 to obtain a local gradient bound. In view of Lemma 8.4, such a result will be useful in studying uniformly elliptic problems. Theorem 9.10. Let y ∈ R, let r > 0, and let u be a solution of Qu = 0 in Ω ∩ B(y, r), N u = 0 on ∂Ω ∩ B(y, r),

with ∂Ω ∩ B(y, r) ∈ C 3 . Suppose that

Λ∗ = O(λ∗ ),

(9.53a)

= O(λ∗ ), |δτ | = O(τ ), ∂aij ∂τ ∗ |p| = O(λ∗ ), |p| = O(τ ), ∂p ∂p ∂a |p| = O(E), ∂p

(9.53b)

|δaij ∗ |

δa ≤ O(E),

|δT aij ∗ |

(δ¯ − 1)a ≤ O(E),

= O(λ∗ ),

|δT τ | = O(τ ),

|δT a| = O(E),

(9.53c) (9.53d) (9.53e) (9.53f) (9.53g) (9.53h)

τ = O(λ∗ ), (9.53i)   τ 1/2 |p| 1 + 1/2 |a| = O(E). (9.53j) λ∗ Suppose also that bp · γ > 0, (9.26) is satisfied and that there are positive constants µ2 , M3 , and M4 such that (9.27) holds for all (x, z, p) with b(x, z, p) = 0 and that if b(x, z, p) = 0 and |p − (p · γ)γ| ≤ M3 , then |p · γ| ≤ M4 . Suppose finally that ¯ ≤ O(|p|bp · γ) δT b, δb (9.54) wherever b(x, z, p) = 0. Then there are positive constants σ (determined only by the limit behavior in (9.53) and (9.54)), R (determined only by r, µ2 and Ω) and C (determined only by Ω, µ2 , M3 , M4 , and the limit behavior in (9.53), and (9.54)) such that, if osc

Ω∩B(y,R)

then

u ≤ σ,

 oscΩ∩B(y,R) u  . |Du(y)| ≤ C 1 + R

(9.55)

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Proof.

We now take w = η 2 wT +

1 MT h(u) vd + 4k0 4

with k0 ∈ (0, 1) to be chosen and h defined by h(s) =

1 exp(α1 ( inf u − s)) α1 Ω∩B(y,R)

for R and α1 also to be chosen. By following the calculations in the proof ¯ J, ¯ K, ¯ µ1 of Theorem 9.9, we find that there are positive constants M1 , H, (determined by the limit behavior in (9.53)) such that aij Dij w + B0i Di w ≥

1 ¯ ¯ 0 )2 − Jh ¯ 0 − (k0 )−2 K] MT E[h00 − H(h 4

wherever w ≥ M1 and ηwT ≥ 12 MT . Using the specific form of h and setting m = inf Ω∩B(y,R) u, we infer that ¯ 0 )2 − Jh ¯ 0 − (k0 )−2 K ¯ ≥ exp(α1 (m − u))[α1 − H ¯ − J] ¯ − (k0 )−2 K. ¯ h00 − H(h

¯ + J¯ and Assuming that α1 ≥ H α1

osc

Ω∩B(y,R)

u ≤ ln 2,

(9.56)

we conclude that ¯ 0 )2 − Jh ¯ 0 − (k0 )−2 K ¯ ≥ 1 [α1 − H ¯ − J] ¯ − (k0 )−2 K. ¯ h00 − H(h 2 Moreover, by virtue of (9.52), we also have   1 bp · Dw ≥ vbp · γ − µ1 M T 4k0 for some constant µ1 determined only by the limit behavior in (9.54). We ¯ + J¯ + (k0 )−2 K. ¯ now choose k0 = 4/(1 + µ1 ) and then choose α1 ≥ H Finally, we choose R (according to Theorem 9.1 and our assumed modulus of continuity) so small that (9.56) holds and vd ≤ k0 in Ω ∩ B(y, R). With these choices for R, k0 , and α1 , we infer (9.55). 

9.7

Gradient estimates for capillary-type problems

In this section, we assume that there are a positive function Λ and a positive constant µ1 such that Λg ij ξi ξj ≤ aij ξi ξj ,

|aij | ≤ µ1 Λ.

(9.57)

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(Of course, the two inequalities can’t both be true if µ1 ≤ 0.) As before, it is crucial to estimate terms which are linear in the second derivatives as a first step in obtaining gradient bounds. This time, we must use the identities δ ij = cij + γ i γ j ,

δ ij = g ij + ν i ν j .

Furthermore, we take advantage of a change of independent variable in place of the function h used in the preceding sections. Specifically, we write I = [inf Ω u, supΩ u] and let Ψ ∈ C 3 (I) be a strictly increasing function with Ψ0 never zero. It will be convenient to set ψ = Ψ−1 , ω = ψ 00 /(ψ 0 )2 , and u ˜ = Ψ(u). We also suppress the argument u ˜ from ψ and ω, and we suppress the argument (x, u, Du) from aij , a, E, and their derivatives. For notational simplicity, we then set C˜T2 = aij ckm Dik u˜Djm u˜,

c˜i = Di (ckm )Dk u˜Dm u,

w ˜T = cij Di u ˜ Dj u ˜. We also observe that aij Dij u ˜+

1 (a + ωE) = 0. ψ0

(9.58)

Our bound on terms that are linear in second derivatives will actually be a bound on terms of the form f ij Dij u ˜, which we now state. Lemma 9.11. Suppose F = [f ij ] is a matrix. Then, for any ε > 0, we have   2 2 ˜ 2 + 1 5|F| + 9nµ1 ΛF f ij Dij u˜ ≤ εC T ε ΛG 2A2   1 ij ν·γ (9.59) ˜T + ψ0 f νj + εΛ γr Dmr u ˜cim Di w 2v v 2εΛ 2 |F ||a| ω|F |E 1 ij + |˜ c| + + − f νj c˜i . Gv 2 Aψ 0 Aψ 0 v Proof.

We start by setting F1 = f ij cjk Dik u˜, F2 = f ij γj γ k Dik u ˜,

so f ij Dij u˜ = F1 + F2 .

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To estimate F1 , we write F1 = F11 + F12 + F13 with F11 = f ij g jk ckr Drm u˜cim , F12 = f ij γi g jk ckr Drm u ˜γ m , F13 = f ij ν j ν k ckr Dir u ˜. Also, to simplify some later calculations, we set G0 = 2g jk γk γr Dmr u˜cim cjs Dis u ˜, G1 = g jk ckr Dmr u ˜cim cjs Dis u˜, G2 = Gγ m Dim u ˜cij Djk u ˜γ k , ˜ 2 = g ij ckm Dik u˜Djm u˜. G T Note that ˜ 2 = G0 + G1 + G2 . G T

(9.60)

We then have that F11 ≤ (g ij f ik f jk )1/2 (G1 )1/2 and kr ij

jk ms

F12 ≤ (c f γi g f by Cauchy’s inequality, while

is 1/2

γm g )



G2 G

1/2

ψ 0 ij 1 f νj D i w ˜T − f ij νj c˜i . 2v v Since 0 < G ≤ 1, it follows that, for any ε1 > 0, we have F13 =

ε1 3 ψ0 1 G2 + |F|2 + f ij νj Di w ˜T − f ij νj c˜i . (9.61) 2 4ε1 G 2v v The estimate of F2 is slightly more elaborate in that we must use the differential equation. First, we write F2 = F21 + F22 with F1 ≤ ε1 G1 +

F21 = f ij γj cir Drk u ˜γ k , F F22 = (ADij u ˜γ i γ j ). A As before, Cauchy’s inequality gives  1/2 G2 F21 ≤ (f ij γj cir f rk γk )1/2 . G

We now define

A1 = aij cjk Dik u ˜,

A2 = aij γi γk Dkm u ˜cim ,

A3 = ADij u˜γi γj .

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It’s easy to check that aij Dij u ˜ = A1 + A2 + A3 . Moreover, if we set A4 = aij cjk cim Dkm u˜, then we have A1 = A4 + A2 , and hence aij Dij u ˜ = 2A1 − A4 + A3 . From the equation (9.58) and this identity, we infer that   F a + ωE ij jk ij jk im F22 = − − 2a c Dik u˜ + a c c Dim u˜ . A ψ0 Then Cauchy’s inequality yields −2

F ij jk |F | ˜ 2 )1/2 a c Dik u ˜≤2 (nµ1 Λ)1/2 (C T A A

and F ij jk im |F | ˜ 2 )1/2 . a c c Dim u ˜≤ (nµ1 Λ)1/2 (C T A A It follows that, for any ε1 > 0 (which will be equal to the ε1 above), we have 1 1 ˜2 9 F2 1 |F ||a| |F |ωE |F|2 + nµ1 Λ 2 + F2 ≤ ε1 G2 + εC + (9.62) T + 2 2 4ε1 G 2ε A Aψ 0 Aψ 0 Combining (9.61) and (9.62) (and noting that 0 < G ≤ 1) gives

5 1 ˜2 F2 9 |F|2 + εC nµ1 Λ 2 T + 4ε1 G 2 2ε A (9.63) |F ||a| |F |ωE ψ 0 ij 1 ij + + + f νj D i w ˜T − f νj c˜i . Aψ 0 Aψ 0 2v v ˜ 2 . (We refer the Our next step is to estimate G1 + G2 in terms of G T reader to Exercise 9.3 for an explanation of why this estimate is needed.) To this end, we note that g jk γk = γj − ν j ν · γ and hence f ij Dij u ˜ ≤ ε1 (G1 + G2 ) +

2 G0 = − ν · γγr Drm u ˜cim Dis u ˜cks Dk u v ψ0 ν · γ ν·γ γr Dmr u ˜cim Di w ˜T + γr Dmr ucim c˜i . =− v v From Cauchy’s inequality, we infer that  2 1/2 ν·γ |ν · γ| |˜ c| γr Dmr u ˜cim c˜i ≥ − (G2 )1/2 , v v G

and hence, since G1 ≥ 0, we have G0 ≥ −

ψ0 ν · γ (ν · γ)2 2 1 γr Dmr u˜cim Di w ˜T − |˜ c| − (G1 + G2 ). v 2v 2 G 2

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If we add G1 + G2 to both sides of this inequality and then multiply the resulting inequality by 2, we conclude that 2 ψ0 ν · γ 2 ˜ 2 + (ν · γ) |˜ G1 + G2 ≤ 2G c | + 2 γr Dmr u ˜cim Di w ˜T . T Gv 2 v the proof is completed by using this inequality in (9.63) with choosing ˜2 ≤ C ˜2 . ε1 = εΛ/4 and noting that ΛG  T T From this second derivative estimate, we can derive a gradient bound for solutions of (9.1) if aij satisfies (9.57) along with other appropriate structure conditions (also on a and b), which we describe in detail in the next theorem. Although the basic method of proof is identical to that for Theorem 9.8, there are several important differences in the details. We have already mentioned the change of dependent variable, but another important difference is a much stronger use of an estimate like that in Lemma 9.4 (but valid over Ω instead of just ∂Ω); we state this estimate as structure condition (9.66a) below. Theorem 9.12. Suppose, in addition to (9.57), that (δ¯ − 1)a + δT E ≤ O(E), δT a ≤ O(E),

Λ = O(E), ij ¯ |δa | = o((ΛE)/|p|), ij

(9.64b) (9.64c) (9.64d)

|δT a | = O((ΛE)/|p|),

(9.64e)

δT b ≤ O(bp · γ), ¯ ≤ o(bp · γ). δb

(9.64g)

ij

|δN a | = O(ΛE)/|p|),

Suppose also that

(9.64a)

lim sup sup |p|→∞ x,z

¯ δE < 1, E

(9.64f) (9.64h)

(9.65)

Suppose finally that there are positive constants µ0 and µ1 , and a decreasing function M1 such that sup |Du| ≤ µ0 (sup |D0 u| + 1),

(9.66a)

sup |Du| ≤ M1 (s)

(9.66b)

d≤s

d≤s

d≥s

for all s ∈ (0, µ1 ). Then there is a constant C, determined only by µ0 , µ1 , M1 , and the limit behavior in (9.64) and (9.65) such that, if u solves (9.1), then |Du| ≤ C in Ω.

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Proof. By differentiating (9.58) with respect to xk , multiplying by ckm Dm u˜, and summing on k, we find that 0=

1 ij ˜2 + C ∗w a Dij w ˜T + B4i Di w ˜T − C ˜T + S, T 2

(9.67)

where B4 = ωEp + ψ 0 aij ˜ + ap , p Dij u C∗ =

ω0 E + ω 2 H10 + ωJ10 + K20 , ψ0 S = S11 + S12 + S13 , H10 = (δ¯ − 1)E, J10 = (δ¯ − 1)a + δT E,

1 aij Dij (ckm )Dk uDm u , 2 wT ˜ Dm u ˜, = −2aij Di (ckm )Djk u 0 ¯ ij S12 = ψ ω δa Dij u ˜w ˜T ,

K20 = δT a − S11

S13 = δT aij Dij u ˜w ˜T . By using Lemma 9.11 with f ij = 2ajk Dk (cim )Dm u ˜, ij 0 ij ¯ f = −ψ ω w ˜T δa , f ij = −ψ 0 w ˜T δT aij

to estimate S11 , S12 , and S13 , respectively, we obtain 0≤

1 ij 1 ˜2 a Dij w ˜T + B5i Di w ˜T + C1∗ w ˜T − C , 2 4 T

where 1 ¯ ij νj + ψ 0 w [−ajk Dk (crm )Dm u ˜ νj + ψ 0 ω w ˜T δa ˜T δT aij νj ] 2v 3 + Λν · γk Dkm u ˜cmi , 4v ¯ 2 ¯ 2 ¯ 20wT |δA| 18nµ1 ΛwT |δA| |δA|E H11 = H10 + + + , 2 ΛG A |A| ¯ |ajm Dk γ m Dm uγ j |E |a||δA| ψ 0 ¯ ij |δT A|E = J10 + + + δa νj c˜i + AwT A v A

B5i = B4i +

J11

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and 20

Pn

|ajk Dk (cim )Dm u|2

18nu1 Λ(ajk Dk γ m Dm uγj )2 ΛGwT A2 wT 2 jk m jk c| 24Λ|˜ |a||a Dk γ Dm uγj | a Dm (cim )Dm uci + + − Gv 2 w ˜T AwT vwT 18nµ1 Λ|δT A|2 20wT |δT A|2 + + 20ΛG A2 |a||δT A| δT aij νj ci + + . A v ˜T = To proceed, with s ∈ (0, µ1 ) and k2 to be determined and M supd≤s w ˜T , we set K21 = K20 +

i,j=1

+

˜ T d. w=w ˜T + k2 M It then follows (after noting that cim γm = 0) that 0≤

1 ij 1 ˜2 ˜ T + S14 , a Dij w + B5i Di w + C1∗ w ˜T − C + k2 C2∗ M 2 4 T

where C2∗ = J12 ω + K22 , with J12 = −δN E − K22 = −δN a −

wT ¯ ij δa νj γi , 2v

1 jk a Dk γ m Dm uνj = wT δT aij νj γi , 2v

and ˜ T ψ 0 δN aij Dij u S14 = −k2 M ˜. Applying Lemma (9.11) one last time, we find that 1 ij a Dij w + B6i Di w + C1∗ w ˜T + C3∗ MT 2 with B6i = B5i + k2 ψ 0



 ψ0 ˜ Λν · γ r MT δN aij νj + γ Drm u ˜cim , 2v 4v

C3∗ = k2 C2∗ + ωJ14 + K23 , J14 = k2

|δN A|E , A

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and ˜T k22 (ψ 0 )2 M Λ|˜ c|2 δN aij νj γi + 2v 2Gv 2 w ˜T ˜ T k2 20(ψ 0 )2 M k2 |δN A||a| 2 − k2 δN aij νj ci + |δN A|2 + A ΛG ˜T 18nµ1 Λk22 M + |δN A|2 . 2 A Our next step is to determine k2 . Since δT b = O(bp · γ), there are constants M2 and β1 such that δT b ≤ β1 bp · γ wherever |p0 | ≥ M2 . From the boundary condition, we infer that K23 =

¯ w ˜ T bp · γ − 2(δT b + ω δb) bp · Dw = k2 M ˜T , so choosing k2 = 1 + β1 gives us ¯w ˜ T bp · γ − 2ω δb bp · Dw ≥ M ˜T

wherever |D0 u| ≥ M2 . We then choose s so that k2 s ≤ 21 . We now write Ωs (M2 ) for the subset of Ω where d ≤ s and |D0 u| ≥ M2 . ˆ such Taking (9.66) into account, we infer that there are constants Jˆ and K that 1 ij a Dij w + B5i Di w + C4∗ Ew˜T + C5∗ EMT ≥ 0 2 on Ωs (M2 ), where ω0 H11 ˆ + ω2 + ω Jˆ + K, 0 ψ E ˆ C5∗ = ω Jˆ + K.

C4∗ =

Then conditions (9.65) and (9.64d) imply that there are positive constants M3 ≥ M2 and e1 such that H11 ≤ −e1 E on Ωs (M3 ). Next, we take Ω0 to be the subset of Ω on which d ≤ s, |D0 u| ≥ M3 and ˜ T . then w ˜T ≥ 21 M 1 ij a Dij w + B5i Di w + C6∗ Ew˜T ≥ 0 2

on Ω0 , where C6∗ =

ω0 ˆ − e1 ω 2 + 3ω Jˆ + 3ω K. ψ0

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It follows that we can take ω to be a constant in such a way that C6∗ > 0 on Ω0 . With this choice for ω (and the corresponding function ψ), we finally ¯ < bp · γ wherever |D0 u| ≥ M4 . Then we choose M4 ≥ M3 so that 2ω δb 00 ˜T . write Ω for the subset of Ω on which d ≤ s, |D0 u| ≥ M4 and w ˜T ≥ 21 M All these choices imply that 1 ij a Dij w + B5i Di w > 0 in Ω00 , 2

bp · Dw > 0 on ∂Ω00 ∩ ∂Ω.

Since we have an upper bound for w on ∂Ω00 \ ∂Ω (in terms of M4 and M1 (s)), it follows from the maximum principle that we have an upper bound for w in Ω00 and hence a bound for |Du| in Ω.  Note that the strongest condition in our list of hypotheses is probably ¯ ≤ o(bp · γ) since, in our example δb b(x, z, p) = v q−1 Du · γ + ψ(x, z), this hypothesis is only satisfied if q = 0 or q > 1. Presumably this condition can be relaxed to δb ≤ O(bp · γ) by using a function of the form ˜ T f (v)d w=w ˜T + M with f a suitable function which goes to infinity sufficiently slowly as its argument goes to infinity but the verification of this assertion has not yet been attempted. Moreover, unlike the situation considered in previous sections, there is no reason to expect M1 (s) to behave nicely. For example, Finn has shown in Section V of [45] that the gradient of a solution of the minimal surface equation grows exponentially with respect to distance to the boundary. Our next step is to obtain hypotheses under which conditions (9.66) are satisfied. In the next chapter, we shall derive a number of gradient bounds for equations similar to the prescribed mean curvature equation under the additional assumption that the coefficients aij are given by aij (x, z, p) = ∂Ai (x, z, p)/∂pj for some vector-valued function A. Here, we show, via the maximum principle, that such an estimate is valid if aij = g ij . Lemma 9.13. Suppose a is differentiable with respect to (x, z, p) with δa ≤ O(1), ¯ (δ − 1)a ≤ O(1/|p|), |ap | = O(1/|p|).

(9.68a) (9.68b) (9.68c)

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If u ∈ C 3 (B(0, R)) ∩ L∞ (B(0, R)) for some R > 0 with

g ij Dij u + a(x, u, Du) = 0 in B(0, R),

(9.69)

then there is a constant C0 , determined only by the maximum of |u|, R, and the limit behavior in (9.68) such that |Du(0)| ≤ C0 .

(9.70)

Proof. First, to simplify notation, we assume without loss of generality that u < 0 in B(0, R). With this assumption and positive constants u0 and C1 to be chosen, we define functions ϕ, f , and η by   z |x|2 ϕ(x, z) = +1− 2 , 2u0 R + f (s) = exp(C1 s) − 1, η(x) = f (ϕ(x, u(x)).

Our first step is to derive a suitable differential equation for w = ηv. It follows from our previous arguments that g ij Dij v + (g ij,r Dij u + ar )Dr v + Cv − C2 = 0, with C = v −2 (az |Du|2 + ax · Du) and 1 ij km g g Dik uDjm u. v But a direct calculation shows that 2 2 g ij,r Dij u = − 2 Drj uDj u + 4 Di uDj uDij uDr u = −2g rj Dj v, v v and hence C2 =

g ij Dij v + ar Dr v − 2g ij Di vDj v + Cv − C2 = 0. If we multiply this equation by η, we find that η g ij Dij w + (ar + arj Dj v)Dr w + (C − g ij Dij η − aj Dj η)v − C2 = 0. v From the definition of η, we see that |p|2 1 ¯ x )− (δ − 1)a − 2ap · ) 2 v 2u0 R i j 2 Du · x |Du| x x + f 00 [g ij 2 + + ]. R u0 v 2 R 4u20 v 2

g ij Dij η + aj Dj η = f 0 (−2(n −

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From the limit behavior in (9.68), we conclude that there are constants M0 and µ0 such that −2(n −

|p|2 1 ¯ x )− ≥ −µ0 (δ − 1)a − 2ap · v2 2u0 R

wherever |Du| ≥ M0 . In addition, because g ij xi xj ≥ 0, we have g ij

Du · x |Du|2 |Du|2 xi xj + + ≥ R2 u0 v 2 R 4u20 v 2 8u20 v 2

wherever |Du| ≥ 8u0 /R. Therefore g ij Dij η + aj Dj η ≥ C1 exp(C1 ϕ(x, u))[−µ0 + C1

|Du|2 ]. 8u20 v 2

We now choose u0 = −u(0) and observe that |Du|2 /v 2 ≥ 1/10 if |Du| ≥ 3. Hence, g ij Dij η + aj Dj η ≥ C1 wherever |Du| ≥ max{M0 , 3, 8u0 /R} if C1 is sufficiently large. Moreover, there are constants M1 and µ1 such that C ≤ µ1 wherever |Du| ≥ M1 . We now set M2 = max{M0 , M1 , 3, 8u0/R}. By further increasing C1 so that C1 > µ1 , we conclude that g ij Dij w + (ar +

η rj g Dj v)Dr w ≥ 0 v

wherever |Du| ≥ M2 and w > 0. It follows that w cannot have a maximum at such a point, and therefore, because 0 ≤ η ≤ C1 exp(C1 ), w ≤ exp(C1 )(1 + M22 )1/2 . Since ϕ(0, u(0)) = 1/2, it follows that η(0) = exp(C1 /2) − 1 > 0 and hence |Du(0)| ≤

exp(C1 ) (1 + M22 )1/2 . exp(C1 /2) − 1



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363

Notes The underlying idea in the gradient estimates presented here goes back to Bernstein [9], and we refer the reader to [64] (specifically the Notes to Chapter 15) for further discussion of this approach to gradient estimates which are purely interior (like those in Section 9.1) or which estimate the gradient inside the domain from a gradient estimate on the boundary (like our Corollary 9.3). However, we do point out that the results quoted here are based on corresponding, but more general, results in [92] and [169]. One notable exception to this general rule is our Lemma 9.13, which is a slight reworking of the gradient estimate in Section 1 of [82]. A simple version of the gradient for problem (9.1) appears in Section 4 of [98], and the results in Sections 9.3 through 9.6 come from [101] with one exception: Theorem 9.10 is a combination of Theorem 9.9 with the gradient bound Theorem 3.3 from [120]. The approach for capillary-type problems in Section 9.7 comes from [107] (except for Lemma 9.13 as already pointed out). An alternative maximum principle approach, but restricted to the capillary boundary condition, was presented in [84]. In many ways, the current state of the gradient bound for quasilinear elliptic equations and nonlinear boundary conditions is still in an unsatisfactory form. The bounds in this chapter are the most general ones known to the author and the assumptions in these theorems, especially the decomposition (9.3), seem very restrictive compared to the corresponding ones in, for example, Chapter 15 of [64]. In addition, the computations are rather intensive. An alternative approach for uniformly parabolic equations was presented by Dong in [38] and we discuss this approach more fully in the notes to Chapter 11. It seems that Dong’s approach could be used to simplify some of the proofs in this chapter, but no one has attempted to do so yet. On the other hand, the gradient bounds require very little geometric structure, unlike the corresponding gradient bounds for the Dirichlet problem. In particular, our gradient bound Theorem 9.12 provides a gradient bound for the differential equation g ij Dij u = 0 and the boundary condition Du · γ + ψ(x) = 0 v for any smooth domain Ω as long as |ψ| < 1. For the same differential equation with Dirichlet data u = ϕ on ∂Ω, a gradient bound is only possible (for arbitrary ϕ) if the mean curvature of ∂Ω is everywhere nonnegative. (See Theorem 14.14 of [64] for more information on this situation.)

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Exercises 9.1 Suppose that aij can be decomposed as in (9.3). Suppose also that there are functions r and s and positive constants p k2 and k2 such that ij ij ¯ |(δ + r + 1)a∗ | + |(δ + s)a∗ | ≤ k1 λ∗ E/|p|, ¯ δE, (δ¯ + r)a, (δ + s)a} ≤ k1 E, max{δE, and set

|p|2 X ¯ 2 A1 = (δ¯ + r)E + |(δ + r + 1)aij ∗ | , 4λ∗ i,j

|p|2 X ¯ ij [(δ + r + 1)aij B1 = (δ + s)E + (δ¯ + r)a + ∗ ][(δ + s)a∗ ], 2λ∗ i,j C1 = (δ + s)a + and

|p|2 X ¯ 2 |(δ + r + 1)aij ∗ | , 4λ∗ i,j

a∞ = lim sup

A1 , (x,z)∈Ω×[−M0 ,M0 ] E

b∞ = lim sup

B1 , (x,z)∈Ω×[−M0 ,M0 ] E

|p|→∞

|p|→∞

sup

sup

C1 . |p|→∞ (x,z)∈Ω×[−M0 ,M0 ] E √ Suppose also that a∞ ≤ 0, c∞ ≤ 0, or b∞ ≤ −2 a∞ c∞ . Prove a gradient bound for solutions of Qu = 0 in Ω, Du · γ = 0 on ∂Ω provided Ω is convex and ∂Ω ∈ C 2 . (Hint: Use the calculations from the proof of Theorem 15.2 in [64] along with the argument in Section 9.2.) 9.2 Verify the comment at the end of Section 9.5 by showing that Theorem 9.8 remains valid if we assume that there is a sufficiently small positive constant µ ¯ such that ¯ ≤µ δT b, δb ¯|p|bp · γ for large |p|. 9.3 Define the matrices F and U by f ij = 1 for i, j = 1, 2 and uij = (−1)i+j for i, j = 1, 2. Observe that these matrices are symmetric. Show that they are positive semi-definite with f 11 u11 = f 22 u22 = 1 but f ij uij = ˜2 0. Use this set of matrices to show that we can’t have G1 + G2 ≤ cG c∞ = lim sup

sup

T

for any positive constant c.

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Chapter 10

Gradient Estimates for the Conormal Derivative Problems

In this chapter, we prove gradient bounds for the conormal derivative problem div A(x, u, Du) + B(x, u, Du) = 0 in Ω,

(10.1a)

A(x, u, Du) · γ + ψ(x, u) = 0 on ∂Ω

(10.1b)

under varying hypotheses on A, B, and ψ provided ∂Ω ∈ C 2 . The p model problems here are the capillary problem, in which A(x, z, p) = p/ 1 + |p|2 and B(x, z, p) = −κz for some positive constant κ, and the Neumann problem in which A(x, z, p) = p and B(x, z, p) = f (x) for some function f . In both cases, we have obtained gradient bounds in Chapter 9, provided ∂Ω ∈ C 3 (and f ∈ C 1 (Ω) for the Neumann problem). Here, we weaken the regularity hypotheses on the data of the problem when the problem has this special conormal form. The basic techniques here were presented in Chapter 5, and we shall look at various ideas to extend the applicability of those techniques.

10.1

The Sobolev inequality of Michael and Simon

For the reader’s convenience, we begin by stating the Sobolev inequality of Michael and Simon in its general form. A proof can be found in Theorem 11.8 of [114]. Lemma 10.1. Let m and n be positive integers with 2 ≤ n ≤ m, let U be an open subset of Rm and let M be a subset of U . Let µ be a measure defined on all sets of the form M ∩ B, where B is a Borel subset of Rm , which is finite on any set of the form M ∩ K with K compact. Let γ be a 365

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continuous, symmetric-matrix-valued function with m X

γ ii (x) = n,

(10.2a)

i=1 ij

0 ≤ γ (x)ξi ξj ≤ |ξ|2

(10.2b)

for all x ∈ M and all ξ ∈ Rm , and define the operator δ on C 1 (U ) by δh = (δ 1 h, . . . , δ m h) and δ i h = γ ij Dj h. Suppose also that lim sup ρ→0+

µ(M ∩ B(x, ρ)) ≥1 ω n ρn

for all x ∈ M . If there is a vector-valued function H such that Z (δh + hH) dµ = 0

(10.3)

(10.4)

M

for all h ∈ C 1 (U ) with compact support, then Z (n−1)/n Z n/(n−1) |h| dµ ≤ C(n) (|δh| + |h||H|) dµ M

(10.5)

M

for all h ∈ C 0,1 (U ) with compact support. For our applications, we let n ≥ 2 be a positive integer, and let Ω be an open subset of Rn . We then set v = (1 + |Du|2 )1/2 , g

ij

ij

ν = Du/v,

i j

i

= δ − ν ν , H = Di (ν ).

(10.6a) (10.6b)

The Sobolev inequality we then use is as follows. Lemma 10.2. Let h ∈ C 1 (Ω) have compact support. Then Z (n−1)/n Z |h|n/(n−1) v dx ≤ C(n) (|δh| + |h||H|)v dx. Ω

(10.7)

Ω

If N > 2 is a constant with N ≥ n and if κ = N/(N − 2), then Z 1/κ Z (N −n)/N 2κ 2 h v dx ≤ C(n, N ) h v dx Ω

Ω

×

Z

Ω

n/N (g ij Di hDj h + h2 H 2 )v dx .

(10.8)

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Proof. The main step is to show that the hypotheses of Lemma 10.1 are satisfied with m = n + 1, U = Ω × R, M equal to the graph of u, and a suitable choice of γ, H, and µ. We begin by defining µ. For any Borel set B, there is a Lebesgue measurable subset SB of Ω such that M ∩ B = {(x, u(x)) : x ∈ SB }, and we define µ(M ∩ B) =

Z

v dx.

SB

Next, we set ν n+1 (x) = −1/v, and, for i, j = 1, . . . , n + 1, we define γ ij (x, u(x)) = δ ij − ν i (x)µj (x). Then   g ij (x) if i, j ≤ n,     νi if i ≤ n, j = n + 1, γ ij (x, u(x)) = νvj  if j ≤ n, i = n + 1,  v    |Du|2 if i = j = n + 1. v2

Finally, we set H(x, u(x)) = ν(x)H(x). It’s easy to check (10.2) and (10.3). To verify (10.4), we define ϕ by ϕ(x) = h(x, u(x)), and we use k to denote an integer parameter that only takes values in {1, . . . , n}. It follows that γ ik Dk ϕ = γ ik Dk h + γ ik Dk uDn+1 h.

From the expression for γ ij , we also have that γ ik Dk u = γ i,n+1 , so γ ik Dk ϕ = δi h. Therefore Z

δi h dµ =

M

Z

ik

ik

Ω

γ Dk ϕv dx = − i

Z

Dk (vγ ik )ϕ dx.

Ω

We now evaluate Dk (vγ ) = −vHν , so (10.4) is satisfied. Next, we observe that, if ξ ∈ Rn and Ξ ∈ Rn+1 is defined by Ξi = γ ik ξk , then |Ξ|2 = g ij ξi ξj . 2

ij

0,1

(10.9)

Hence |δh| = g Di hDj h if h ∈ C (Ω), and (10.7) follows from Lemma 10.1. The proof of (10.8) from (10.7) is the same as the proof of Theorem 5.8. 

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For our gradient estimate, one further inequality will be useful, which estimates H 2 by a specific combination of second derivatives of u. To prove this estimate, we define the functions g˜ij = δ ij −

v − 1 pi pj . v |p|2

It is easy to check that [˜ g ij ] is the square root of the matrix [g ij ], that is, ij ik kj g = g˜ g˜ . Hence, if we define bij = g˜ik g˜jm Dkm u then vH = δij bij , so 1/2

|vH| ≤ (δij δik δjk )

δij bik bjk

it follows that H2 ≤


1/2

= n1/2 g ij g km Dik uDjm u

n ij km g g Dik uDjm u. v2


1/2

.

(10.10)

Alternatively, this inequality can be proved by noting that H is the sum of the principal curvatures of the graph of u while v −2 g ij g km Dik uDjm u is the sum of the squares of these principal curvatures.

10.2

The interior gradient bound

In this section, we prove an interior gradient bound for a large class of divergence structure equations. A key assumption is related to the derivatives of A and B. We assume that A is differentiable with respect to x, z, and p and that, for all i and k in {1, . . . , n}, there are functions Cki and Dki such that Dki is differentiable with respect to x, z, and p and ∂Ai ∂Ai pk + + Bδki = Cki + Dki . ∂z ∂xk

(10.11)

For notational convenience, we define Dij = ν k

∂Dki , ∂pj

F = ν k (pi

∂Dki ∂Dki + ). ∂z ∂xi

(10.12)

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We also assume that there are functions Λ0 and µ ¯ and nonnegative constants β1 and τ0 such that 1/2

(10.13a)

1/2

(10.13b)

Cki ν k ξi ≤ β1 Λ0 (aij ξi ξj )1/2 ,

Cki g jk ηij ≤ β1 Λ0 (aij g km ηik ηjm )1/2 , ij

vD ηij ≤

1/2 β1 Λ0 (aij δ km ηik ηjm )1/2 , β12 Λ0 ,

vF ≤ ˆ 2 )1/2 vaij ξˆi ξj ≤ (aij ξi ξj )1/2 (¯ µ|ξ|

(10.13c) (10.13d) (10.13e)

for all vectors ξ and ξˆ and all matrices η wherever v ≥ τ0 . (Note that there is no loss of generality in assuming that τ0 ≥ 1.) For our estimates, we shall also use the following two quantities: C20 = aij g km Dik uDjm u, ij

E0 = a Di vDj v.

(10.14a) (10.14b)

With these definitions and assumptions, we have our first integral inequality, which is the basis for our estimates. Lemma 10.3. Suppose conditions (10.13) are satisfied, and let χ be an increasing function on [τ0 , ∞). Suppose that there are constants τ ≥ τ0 and cχ such that 0 ≤ (s − τ )

χ0 (s) ≤ cχ χ(s)

(10.15)

for all s ≥ τ . If ζ ∈ C 1 (Ω) vanishes on ∂Ω, and if Qu = 0 in Ω, then Z Z τ [(1− )C20 +E0 ]χζ 2 dx ≤ 18(1+cχ ) (β12 Λ0 ζ 2 + µ ¯|Dζ|2 )χ dx, (10.16) v Ωτ Ωτ

where Ωτ is the subset of Ω on which v ≥ τ .

Proof. Let θ be a C 2 (Ω) vector-valued function with compact support in Ω and use Dk θk as test function in the weak form of the differential equation Qu = 0. It follows that Z Z i k A Dik θ dx = BDk θk , Ω

Ω

and an integration by parts yields Z (Dk Ai + Bδki )Di θk dx = 0. Ω

We now observe that

Dk Ai + Bδki = aij Djk u + Cki + Dki ,

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so another integration by parts yields Z (aij Djk uCki )Di θk − Di (Dki )θk dx = 0. Ω

By approximation, this equality is true for any θ ∈ C 1 (Ω) that vanishes on ∂Ω. In particular, we can use θ = ζ 2 (v − τ )+ χν. Since Di θk = [χ + (v − τ )χ0 ]ζ 2 ν k Di v + ζ 2 (v − τ )χg jk Dij u + 2ζ(v − τ )χν k Di ζ on Ωτ , it follows that Z τ 0 = ([χ + (v − τ )χ0 ]E0 + (1 − )χC20 )ζ 2 dx v Z + 2(v − τ )χaij Dj vDi ζζ dx Z τ + [χ + (v − τ )χ0 ]Cki ν k Di vζ 2 + (1 − )Cki g jk Dij uζ 2 dx v Z + 2(v − τ )χCki ν k Di ζζ dx Z − Dij Dij u(v − τ )χζ 2 + F(v − τ )χζ 2 dx,

where here and in the remainder of this proof, all integrals are integrated over Ωτ . We now estimate the integrals in this identity, using the abbreviations Z I0 = χE0 ζ 2 dx, Z I1 = [χ + (v − τ )χ0 ]E0 ζ 2 dx, Z τ I2 = (1 − )χC20 ζ 2 dx, v Z I3 = β12 Λ0 χζ 2 dx, Z I4 = χ¯ µ|Dζ|2 dx. Using (10.13e), we have that Z 1 2(v − τ )χaij Dj vDi ζζ dx ≥ − I0 − 8I4 . 8

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Then (10.13a) and (10.15) imply that Z 1 [χ + (v − τ )χ0 ]Cki ν k Di vζ 2 dx ≥ − I1 − 2(1 + cχ )I3 8 and (10.13b) implies that Z τ 1 (1 − )χζ 2 Cki g jk Dij u dx ≥ − I2 − 4I3 . v 4 Next, (10.13a) gives Z 2(v − τ )χCki ν k Di ζζ dx ≥ −I3 − I4 .

Finally, (10.13c) gives Z 1 1 − Dij Dij u(v − τ )χζ 2 dx ≥ − I1 − I2 − I3 4 4 and (10.13d) implies that Z − F(v − τ )χζ 2 dx ≥ −I3 .

The proof is completed by combining these inequalities and noting that I0 ≤ I1 .  Note that conditions (10.13) are exactly the hypotheses which give (10.16). For this form of the estimate, the use of β1 is not necessary, but its utility will be seen when we present our gradient estimates. Our proof of the gradient bound is effected through suitable choices of the function χ. It will be useful to introduce an auxiliary positive, realvalued function w ∈ C 1 ([τ0 , ∞)). We assume that there is a positive number β such that 0 ≤ sw0 (s) ≤ βw(s)

(10.17)

Λ0 ≤ Λv,

(10.18a)

−β

for all s ≥ τ0 . (Equivalently, s w(s) is a decreasing function of s.) Instead of estimating v directly, we shall estimate w(v). It will also be useful to introduce three more positive, real-valued struc˜ in C 1 ([τ0 , ∞)). We suppose that they ture functions Λ ∈ C(Ω) and λ and λ are related to the operator Q (and to each other) by the following relations

λ(v(x)) ≤ Λ(x),

λ(v) 1 +



µ ¯ ≤ Λv, 2 ! vλ (v) g ij ξi ξj ≤ vaij ξi ξj λ(v) 0

(10.18b) (10.18c) (10.18d)

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for all ξ ∈ Rn , ˜ (Λ(x)/λ(v(x)))N/2 /Λ(x) ≤ λ(v),

(10.18e)

˜ satisfies the where N > 2, N ≥ n, and n ≥ 2. We further suppose that λ monotonicity conditions ˜ βw0 (s)λ(s) + λ0 (s)w(s) ≥ 0, (10.19a) 0 ˜ (s) ≤ β λ(s) ˜ sλ (10.19b)

for all s ≥ τ0 . Under these hypotheses, we can estimate sup w in terms of an integral.

Lemma 10.4. Suppose conditions (10.13), (10.17), (10.18), and (10.19) are satisfied. If Qu = 0 in some ball B(y, R) and τ ≥ τ0 , then Z  τ N ˜ 1− sup w(v) ≤ c1 R−n w(v)Λ(x)λ(v)v dx, (10.20) v + B(y,R/2) Bτ (y,R)

where Bτ (y, R) is the subset of B(y, R) on which v ≥ τ , and c1 is determined only by β, β1 R, n, and N . Proof.

Then

With q > 1 + β a parameter, we set  τ N q−N q ˜ χ(s) = λ(s)w (s) 1 − . s +

 τ N q−N ˜ 0 0 ˜ χ0 (s) = w(s)q−1 1 − [λ (s)w(s) + q λ(s)w (s)] s +  τ N q−N −1 τ q ˜ + N (q − 1)λ(s)w (s) 1 − . s + s2 ˜ 0 w + q λw ˜ 0 ≥ 0, and Since q ≥ β and w0 ≥ 0, it follows from (10.19a) that λ 0 hence χ ≥ 0. Moreover, (10.19b) and (10.17) imply that 0 0 ˜ 0 (s)w(s) + q λ(s)w ˜ ˜ 0 (s)w(s) + q λ(s)sw ˜ (s − τ )[λ (s)] ≤ sλ (s) ˜ ≤ β(1 + q)λ(s)w(s)

and   τ N q−N τ τ N q−N −1 τ (s − τ ) 1 − = 1 − . s + s2 s + s Our choice of q now implies that χ satisfies (10.15) with cχ = C(n, N, β)q, and hence we can use χ in Lemma 10.3. If we also replace ζ 2 by ζ N q−N +2 in Lemma 10.3 and set Z τ I1 = [E0 + (1 − )C20 ]χ(v)ζ N q−N +2 dx, v B(y,R)

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we obtain I1 ≤ C(n, β)q

Z

B(y,R)

[β12 Λ0 ζ 2 + µ ¯|Dζ|2 ]χ(v)ζ N q−N dx.

To proceed, we choose ζ(x) =



1−

|x − y|2 R2



+

and observe that 0 ≤ ζ ≤ 1 and |Dζ| ≤ 2/R. Setting also Z τ ˜ I2 = wq λΛ[(1 − )ζ]N q−N v dx, v Bτ (y,R) we then obtain from (10.18a) and (10.18c) that I1 ≤ C(n, β, β1 R)qR−2 I2 .

(10.21)

We next define the function Ψ by Ψ(v) = χ(v)λ(v)(1 −

τ 2 ) v

and the function h by h(x) = Ψ(v)1/2 ζ(x)[N q−N ]/2 , and note that ˜ q [ζ(1 − τ )]N q−N +2 h2 = λλw v and ˜κ λκ ≥ λΛ ˜ λ by virtue of (10.18e). Then the Sobolev inequality (10.8) yields 1/κ Z (N −n)/N Z τ κN q−N n/N 2 κq ˜ λΛw [ζ(1 − )] v dx ≤ C(n, N ) h v dx I3 , v where I3 =

Z

[g ij Di hDj h + h2 H 2 ]v dx.

Our next step is to estimate the integrals on the right of this inequality. First, we have Z Z ˜ q [ζ(1 − τ )]N q−N v dx h2 v dx ≤ λΛw v

because λ ≤ Λ and ζ(1 − τv ) ≤ 1.

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Next, 2hδh = [(χ0 λ + χλ0 )(1 −

τ 2 τ τ ) + 2χλ(1 − ) 2 ]ζ N q−N +2 δv v v v

τ 2 N q−N +1 ) ζ δζ, v and hence   1 |vλ0 | τ 2h|δh| ≤ C(n, N, β)q 1+ [χλ(1 − )]|δv|ζ N q−N +2 v λ v q τ N q−N +1 + C(n, N ) χλ(1 − )ζ . R v We now square this inequality and divide by h to conclude that (vλ0 )2 λ )χ|δv|2 ζ N q−N +2 |δh|2 ≤ C(n, N, β)q 2 2 (1 + v λ2 q2 + C(n, N ) 2 χλζ N q−N R τ 1 q2 ˜ q ≤ C(n, N, β)q 2 E0 χζ (n+2)q−n + C(n, N ) 2 λΛw [ζ(1 − )]N q−N v R v by virtue of (10.18d). Using (10.21), we conclude that Z Z τ ˜ g ij Di hDj hv dx ≤ C(n, N, β, β1 R)q 2 R−2 wq λΛ[(1 − )ζ]N q−N v dx. v Bτ (y,R) It follows from (10.10) that Z h2 H 2 v dx ≤ C(n)I1 . + χλ(1 −

Using our estimate for I1 once more, we find that Z

wκq [ζ(1 −

1/κ Z τ κN q−N ˜ τ ˜ dx, )] λΛv dx ≤ Cq 3 R−2n/N wq [ζ(1 − )]N q−N λΛv v v

where here, and in the remainder of the proof, we use C to denote a constant determined only by n, N , β, and β1 R. This inequality is now easily rewritten in a form suitable for Moser iteration. We set τ w ¯ = w[ζ(1 − )]N v and we define the measure µ by τ ˜ dµ = R−n λΛ[ζ(1 − )+ ]−N v dx. v It follows that Z 1/κ Z w ¯κq dµ ≤ C(n, β, β1 R)q 3 w ¯q dµ. Our usual iteration scheme then yields

sup w ¯ ≤ C(n, β, β1 R)

Z

w ¯ dµ,

and rewriting this inequality in terms of the original structure functions gives (10.20) because ζ ≥ 12 in B(y, R/2). 

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˜ to the operator Q via the To continue, we relate the functions Λ and λ assumption that there are positive constants β2 and β3 such that ˜ Λ(x)λ(v)v ≤ β2 wβ3 Du · A.

(10.22)

With this assumption, our gradient estimate is reduced to an estimate of Z wq Du · A dx

for q = 1 + β3 . This integral can be estimated if there are positive constants β4 , β5 and β6 and a decreasing function ε such that w0 A · ξ ≤ β4 (aij ξi ξj )1/2 (Du · A)1/2

(10.23a)

w|A| ≤ β4 Du · A,

(10.23b)

µ ¯ ≤ β5 Du · A,

(10.23d)

for all ξ ∈ Rn , 2

Λ0 ≤ ε(v)w Du · A, B ≤ β6 Du · A

(10.23c) (10.23e)

wherever v ≥ τ0 . In fact, we also need a condition connecting the function ε to the oscillation of u. Lemma 10.5. Suppose conditions (10.13) and (10.23) are satisfied, and set σ = oscB(y,R) u and E = exp(β6 σ), and let q ≥ 2 and τ ≥ τ0 . Finally, suppose w satisfies (10.17). If there is a constant τ1 ≥ τ such that 72(1 + βq)β12 β42 σ 2 q 2 Eε(τ1 ) ≤ 1, then we have Z wq Du · A dx Ωτ (y,R/2)

 q Z β4 σ ≤ C(q, β, β5 )E q w(τ1 ) + Du · A dx. R Ωτ (y,R)

(10.24)

(10.25)

Proof. We begin by setting m = minΩ(y,R) u and defining functions f and f1 by f (s) = (s − m) exp(β6 (s − m)),

f1 (s) = exp(β6 (s − m)).

We now set I1 =

Z

Ω

(wq − w(τ )q )+ ζ q f 0 (u)Du · A dx.

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Since f 0 (u)Du = D(f (u)), an integration by parts shows that I1 = I2 + I3 + I4 , where I2 = − I3 = − I4 = −

Z

Ω

Z

Ω

Z

Ω

qf (u)wq−1 ζ q w0 Dv · A dx, qf (u)(wq − w(τ )q )+ ζ q−1 Dζ · A dx, (wq − w(τ )q )+ ζ q div A dx.

On the other hand, if we set Z I5 = (wq − w(τ )q )+ ζ q f1 (u)Du · A dx, ZΩ I6 = (wq − w(τ )q )+ ζ q β6 (u − m)f1 (u)Du · A dx Ω

then I1 = I5 + I6 , so I5 = I2 + I3 + (I4 − I6 ). Moreover, we have − div A = B and hence Z I4 = (wq − w(τ )q )+ ζ q f (u)B dx ≤ I6 Ω

by virtue of (10.23e). We conclude that I5 ≤ I2 + I3 . For notational convenience, we set Z I6 = wq ζ q f1 (u)Du · A dx, ZΩτ I0 = f1 (u)Du · A dx, Bτ (y,R)

and note that I5 = I6 − w(τ )q I0 . It follows from (10.23a) that  1/2 Z f (u) q−1 q 2 1/2 I2 ≤ qβ4 w ζ (σ EE0 ) Du · A dx σ Ω Z 1 ≤ I6 + q 2 β42 σ 2 E wq−2 ζ q E0 dx. 4 Ωτ

(10.26)

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Since χ = wq−2 satisfies (10.15) with cχ = βq, it follows from Lemma 10.3 that Z Z wq−2 ζ q E0 dx ≤18(1 + βq) β12 Λ0 wq−2 ζ q dx Ωτ Ωτ Z 72 + 2 (1 + βq) µ ¯wq−2 ζ q−2 dx R Ωτ and then (10.23c) and (10.23d) imply that 1 I2 ≤ ( + 18(1 + βq)β12 β42 σ 2 Eε(τ ))I6 4 Z β42 σ 2 + C(β, β5 , q) 2 wq−2 ζ q−2 Du · A dx. R Ωτ Next (10.23b) implies that Z I3 ≤ β4 qσ wq−1 ζ q−1 |Dζ| exp(β6 (u − m))Du · A dx Ωτ

≤

1 β 2 σ2 I6 + C(q) 4 2 E 4 R

Z

Bτ (y,R)

wq−2 ζ q−2 Du · A dx.

If τ ≥ τ1 , then (after using (10.24)) we have Z 3 β 2 σ2 I5 ≤ I6 C(q) 4 2 wq−2 ζ q−2 Du · A dx. 4 R Ωτ

A slight rearrangement then yields Z β42 σ 2 I6 ≤ C 2 wq−2 ζ q−2 Du · A dx + 2w(τ )2 I0 , R Ωτ

where here and, in the remainder of the proof, we use C to denote a constant determined only by n, q, β, and β5 . From Young’s inequality along with the inequality f1 (u) ≥ 1, we then infer that Z Z β4 σ q wq ζ q Du · A dx ≤ CE q (w(τ ) + ) Du · A dx. (10.27) R Bτ (y,R) Bτ (y,R) Note that this inequality is true with τ1 in place of τ as well. If τ < τ1 , we use the preceding inequality with τ1 in place of τ and note that Z Z wq ζ q Du · A dx ≤ w(τ1 )q Du · A dx Bτ (y,R)\Bτ1 (y,R)

Bτ (y,R)

to see that (10.27) is valid in the form Z Z β4 σ q wq ζ q Du · A dx ≤ CE q (w(τ1 ) + ) Du · A dx. R Bτ (y,R) Bτ1 (y,R)

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The proof is completed by observing that Z Z wq Du · A dx ≤ 2q Bτ (y,R/2)

Bτ (y,R)

wq ζ q Du · A dx

and Z

Bτ1 (y,R)

Du · A dx ≤

Z

Bτ (y,R)

Du · A dx.



Our final step in the proving an interior gradient bound is to estimate Du · A dx. Such an estimate is possible under very weak conditions on A and B. R

Lemma 10.6. Let u be a bounded weak solution of Qu = 0 in B(y, R) and suppose that there are a decreasing function ε1 and a nonnegative constant β6 such that |A| ≤ ε1 (v)Du · A,

(10.28a)

B ≤ β6 Du · A

(10.28b)

for v ≥ τ0 . Set σ = oscB(y,R) u and E = exp(β6 σ) and let τ ≥ τ0 . If there is a constant τ2 ≥ τ such that σ 4ε1 (τ2 ) ≤ 1, (10.29) R then Z Du · A dx ≤ C(n)Rn E∆1 , Bτ (y,R/2)

where 

σ|A| ∆1 = sup σ(B − β6 Du · A)+ + (Du · A)− + R S0



and S0 is the subset of B(y, R) on which v < τ2 . Proof. For this proof, we need a function ζ which is identically one on B(y, R/2). This can be achieved with, for example,   if |x − y| ≤ R/2,  1 2|x−y| ζ(x) = 1 − R if R/2 < |x − y| ≤ R,   0 if |x − y| ≥ R, and this is the function we shall use in the proof. We now set M = supB(y,R) u and define functions η1 and η by η1 (x) = (M − u(x) − σ(1 − ζ(x)))+ ,

η(x) = exp(β6 (M − u(x)))η1 (x).

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Using η as test function in the weak form of the differential equation for u, we Z find that Z exp(β6 (M − u))Du · A dx = (B − β6 Du · A)η dx supp η Ω Z + exp(β6 (M − u))σDζ · A dx. supp η

To simplify notation, we introduce two sets: Σ, which is the subset of supp η on which v ≥ τ2 , and S, which is the subset of supp η on which v < τ2 . Then Z Z Ω

(B − β6 Du · A)η dx ≤

S

(B − β6 Du · A)η dx

≤ ωn Rn Eσ sup σ(B − β6 Du · A)+ . S0

Since |Dζ| ≤ 2/R and Du · A ≥ 0 if v ≥ τ , we conclude that Z Z 1 exp(β6 (M − u))σDζ · A dx ≤ exp(β6 (M − u))Du · A, dx 2 Σ Σ Z 1 ≤ exp(β6 (M − u))Du · A dx, 2 Σ0

where Σ0 is the subset of supp η on which v ≥ τ , while Z σ exp(β6 (M − u))σDζ · A dx ≤ 2ωn Rn E sup |A|. R S0 S Finally, Z

supp η

exp(β6 (M − u))Du · A dx ≥ +

Z

Z

Σ0

S

and −

Z

S

exp(β6 (M − u))Du · A dx

exp(β6 (M − u))Du · A dx

exp(β6 (M − u))Du · A dx ≤ ωn Rn E sup(Du · A)− . S0

Combining these estimates with the observations that Bτ (y, R/2) ⊂ Σ0 and that exp(β6 (M − u)) ≥ 1 gives the desired result.  To see the applicability of these results, we consider three examples. First, we look at prescribed mean curvature equations. We suppose that A = ν and that B(x, z, p) = B ∗ (x, z, p) + B0 (x, z, p), where B ∗ is differentiable with respect to (x, z, p). We also suppose that there are nonnegative constants θ1 and θ2 such that v|B0 | ≤ θ1 ,

v(|Bp∗ | + |p · Bp∗ |) ≤ θ1 ,

vBz∗ + |Bx∗ | ≤ θ12 ,

B ≤ θ2 .

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If we choose Cki = B0 δki and Dki = B ∗ δki , then conditions (10.13) are satisfied with β1 = θ1 , Λ0 = µ ¯ = v, and conditions (10.17) are satisfied 2 with w = (ln v) and β = 1 provided we take τ0 = 3. In addition (10.18) ˜ = 1. It follows from Lemma 10.4 (with and (10.19) hold with Λ = λ = λ R/4 in place of R) that, for any τ ≥ 3, we have Z  τ N 2 −n sup 1− (ln v) ≤ C(n, N, θ1 R)R (ln v)2 v dx. v B(y,R/8) Bτ (y,R/4)

To continue, √ we observe that (10.22) holds with β2 = 2 and β3 = 0. We now set c1 = 1728eθ1 R and c2 = 2θ2 R, and we chose τ1 so that (c1 + c2 )θ/R ≤ ln τ1 . Then conditions (10.23) and (10.24) are satisfied with β4 = 1, β5 = 2, β6 = 2θ2 / ln τ1 , w = ln v, and ε(t) = 2/(ln t)2 . Hence we can apply Lemma 10.5 (with R/2 in place of R) to conclude that Z Z  σ 2 Du · A dx. (ln v)2 v dx ≤ C(n, θ1 R, θ2 R) ln τ1 + R Bτ (y,R/2) Bτ (y,R/4) This final integral is estimated via Lemma 10.6; conditions (10.28) and (10.29) are satisfied with ε1 (v) = 1/(2v), τ2 = 3 + 8σ/R, β6 = 2θ2 /τ2 . It follows that E ≤ C(θ2 R) and ∆2 ≤ C(θ2 R)σ/R and hence Z  σ Du · A dx ≤ C(n, θ2 R)Rn τ + . R Bτ (y,R/2) Combining our three estimates and taking τ = 3, we conclude that    σ 3/2 v(y) ≤ exp C(n, θ1 R, θ2 R) 1 + . R

Next, we look at uniformly elliptic equations, which we define in terms of a positive C 1 (1, ∞) function Ψ such that

Ψ(s) ≤ sΨ0 (s) ≤ θ0 Ψ(s) (10.30) θ0 for some positive constant θ0 ≥ 1. (In particular, we can take Ψ to be any power function.) We assume that there are positive constants θ1 , . . . θ4 such that Du · A ≥ vΨ(v) − θ1 ,

|A| ≤ θ2 Ψ,

(10.31a)

Ψ(v) 2 Ψ(v) 2 |ξ| ≤ aij ξi ξj ≤ θ3 |ξ| v v

(10.31b)

v|Az | + |Ax | + |B| ≤ θ4 vΨ(v).

(10.31c)

for all ξ ∈ Rn ,

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For future reference, we observe that Ψ(s) → ∞ as s → ∞, so we may assume that τ0 is large enough that τ0 Ψ(τ0 ) ≥ 1 + 2θ1 . Then conditions (10.13) are satisfied with Dki = 0, Λ0 = v 3 Ψ(v), µ ¯ = θ3 vΨ(v), and β1 = θ4 . 2 Furthermore, if we take w = v, Λ = (1 + θ3 )v Ψ(v), λ = Ψ(v)/(1 + θ02 ), and ˜ = (1 + θ3 )(N −2)/2 (1 + θ2 )N/2 v N −1 , then conditions (10.17), (10.18), and λ 0 (10.19) hold with β = N + θ0 . Hence Lemma 10.4 implies that Z  τ n+2 sup 1− v ≤ C(n, θ0 , θ1 , θ2 θ3 , θ4 R)R−n v N +3 Ψ(v) dx v Bτ (y,R/8) Bτ (y,R/4) Z ≤ CR−n v N +2 Du · A dx. Bτ (y,R/4)

Next, conditions (10.23) are satisfied with β4 = 2θ2 , β5 = 2θ3 , β6 = 2θ4 , and ε(v) = 2. Since u is H¨ older continuous (by virtue of Theorem 8.8), it follows that there is a constant R0 (determined only by θ0 , θ1 , θ2 , θ4 R, and n) such that (10.24) holds and β6 σ ≤ 1 as long as R ≤ R0 . Hence Z Z  σ N +2 v N +2 Du · A dx ≤ C τ0 + Du · A dx. R Bτ (y/R/4) Bτ (y,R/2)

Finally, conditions (10.28) hold with ε1 (v) = 2θ2 /v and β6 = 2θ4 , so conσ dition (10.29) holds with τ2 = τ0 + 8θ2 R . It follows that  σ σ  Ψ 1+ ∆1 ≤ C(θ0 , θ1 , θ2 , τ0 ) 1 + R R and hence Z  σ  σ Du · A dx ≤ CRn 1 + Ψ 1+ . R R Bτ (y,R/2) Combining our inequalities, we obtain

 σ N +3  σ v(y) ≤ C(n, θ0 , . . . , θ3 , θ4 R) 1 + Ψ 1+ . R R Finally, we look at a problem with exponential growth, the false mean curvature equation. Specifically, we assume that A = exp(θv 2 )p for some positive constant θ. We also assume that there is a positive constant θ1 such that |B(x, z, p)| ≤ θ1 exp(θv 2 )v 2 . As in the case of uniformly elliptic equations, we choose Dki = 0 and note that Cki = Bδki . Then conditions (10.13) are satisfied with Λ0 = exp(θv 2 )v 3 , µ ¯ = exp(θv 2 )v 2 , and β1 = θ1 . Furthermore, if we take w = v, 2 3 ˜ = v −2 , then conditions (10.17), Λ = exp(θv )v , λ = exp(θv 2 )v −3 , and λ

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(10.18), and (10.19) hold with β = 1 provided θ0 is sufficiently large. We therefore obtain Z  τ n+2 −n sup 1− ≤ C(n, θ1 R)R exp(θv 2 )v 2 dx. v Bτ (y,R/4) Bτ (y/R/2) Since exp(θv 2 )v 2 ≤ 2Du · A for v ≥ 2, we can use Lemma 10.6 to estimate this integral. Conditions (10.28) are satisfied with ε1 (v) = 1/(2v) and σ β6 = 2θ1 , so (10.29) holds with τ2 = τ0 + 2 R . We therefore infer that   σ 2 σ 2 v(y) ≤ C(n, θ, θ1 R) 1 + exp(θ 1 + ). R R 10.3

Preliminaries for estimates

To prove a gradient bound for conormal problems, we must modify the preceding argument in a suitable fashion. In particular, we can no longer estimate w(v) because this function does not satisfy a suitable boundary condition. Instead, we use a construction originally due to Dong to create an analog of the function v. As a preliminary step, we recall from Lemma 5.18 that the distance function in a C 2 domain is C 2 near the boundary. From this lemma, we can generate a function v1 (x, z, p) which behaves like v and which behaves appropriately on ∂Ω. To simplify some writing, we introduce some notation. First, we extend γ into Γ by taking γ(x) = Dd(x). Then, we define cij (x) = δ ij − γi (x)γj (x), and for a vector p, we define p0 to be the vector with components p0i = cij (x)pj . (Although p0 is really a function of x and p,
1/2 we suppress this fact from the notation.) We also set v 0 = 1 + |p0 |2 . Finally, for τ ≥ 1, we write Σ(τ ) and Σ∗ (τ ) for the subsets of ∂Ω × R × Rn and Γ × R × Rn , respectively, on which v ≥ τ . As a first step, we show how to extend a function defined on Στ to one defined on Γ × R × Rn with important properties preserved. The extended function (denoted by g in this lemma) will be used to define a suitable v1 in the next lemma. Lemma 10.7. Let ∂Ω ∈ C 2 , and let R0 and Γ be as in Lemma 5.18. Let τ0 ≥ 1 be given and let G ∈ C 1 (Σ(τ0 ) with lim G(x, z, p − tγ(x)) < 0 < lim G(x, z, p + tγ(x))

t→∞

t→∞

(10.32)

for all (x, z, p) ∈ Σ(τ0 ). Write Σ0 (τ0 ) for the subset of Σ(τ0 ) on which

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G = 0, and suppose that there are positive constants β0 and c0 such that |Gp | ≤ β0 Gp · γ, |p · γ| ≤ c0 v

(10.33a)

0

(10.33b)

on Σ0 (τ0 ). Suppose further that there are decreasing functions εz and εx (defined on [τ0 , ∞)) such that for τ0 ≤ s ≤ t and

sεz (s) ≤ tεz (t),

s2 εx (s) ≤ t2 εx (t)

|Gz | ≤ εz (v)vGp · γ,

(10.34) (10.35a)

2

|Gx | ≤ εx (v)v Gp · γ,

(10.35b)

g(x, z, p1 ) = g(x, z, p2 )

(10.36a)

on Σ0 (τ0 ). Then there is a function g ∈ C 1 (Γ × R × Rn ) such that if (x, z) ∈ Γ × R and p1 · γ(x) = p2 · γ(x), p · γ(x) = g(x, z, p) if G(x, z, p) = 0

(10.36b)

for all (x, z, p) ∈ ∂Ω × R × Rn with v > (1 + c20 )1/2 v, |gz | ≤ εz (v)v,

|gx | ≤ εx (v)v 2 +

v , R0

|gp | ≤ β0

(10.36c)

on Γ × R × Rn , where εz and εx are defined for s ∈ [1, τ0 ) by εz (s) =

s εz (τ0 ), τ0

εx (s) =

s2 εx (τ0 ). τ02

Proof. We begin by observing that the implicit function theorem gives a function g0 , defined for all (x, z, p) ∈ ∂Ω × R × Rn with v 0 ≥ τ0 such that G(x, z, p) = 0 if and only if p · γ = g0 (x, z, p0 ). Moreover g0 is continuously differentiable with respect to its arguments, provided we only differentiate x in tangential directions. We then define g, for v 0 > τ0 by g(x, z, p) = g0 (x, z, p0 ). Implicitly differentiating the equation G(x, z, p + γ[g(x, z, p) − p · γ]) = 0

shows that the conditions in (10.36) are satisfied for v 0 > τ0 and x ∈ ∂Ω. It then follows that we can extend g so that these conditions are satisfied in all of ∂Ω × R × Rn . The important observation is that G(x, z, p) is never zero if |v 0 | ≤ τ0 and v ≥ (1 + c20 )1/2 τ0 . Finally, we extend g to Γ × R × Rn by setting g(x, z, p) = g(y(x), z, p), where y(x) is the closest point in ∂Ω to x. Since γ(y(x)) = γ(x), all the conditions are easily verified. 

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Lemma 10.8. Under the hypotheses of Lemma 10.7, there is a C 1 (Γ × R × Rn ) function N such that 3 1 ≤ Np · γ ≤ , 2 2 |Np | ≤ 4β0 ,

(10.37a) (10.37b)

|Nz | ≤ 8εz (v)v,   2 |Nx | ≤ 24εx (v)v + v R0

(10.37d)

N = 0 on Σ0 ((1 + c20 )1/2 τ0 ).

(10.37e)

v1 = (1 + |p|2 − (p · γ)2 + εN 2 )1/2

(10.38)

(10.37c)

on Σ∗ and

Moreover, if we set

for some ε ∈ (0, 1/(c0 + 1)2 ), then √ ε v ≤ v1 ≤ 2v. 2

(10.39)

If we also define ν1 =

∂v1 , ∂p

(10.40)

then |ν1 | ≤ 2

(10.41a)

ν1 · Du ≥ v1 /8

(10.41b)

in Ω and

wherever v1 ≥ 3. Moreover, if we define bkm = v1

∂ 2 v1 + ν1k ν1m , ∂pk ∂pm

(10.42)

then there are positive constants ε0 (β0 , c0 ) and c1 (ε, β0 , n) such that ε < ε0 implies that c1 |ξ|2 ≤ bkm ξk ξm for all ξ ∈ Rn .

(10.43)

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Proof. The basic idea is to obtain N as a regularized distance to the surface G(x, z, p) = 0 once we have extended G to a function defined on Γ × R × Rn . For the extension, we first recall that γ = Dd in Γ and take g from Lemma 10.7. Then we define g¯ as follows: Let ϕ be a nonnegative C ∞ (Rn ) function having support in B(0, 1) and Z ϕ(x) dx = 1, Rn

and set

g¯(x, z, p, s) =

Z

g(x, z, p +

Rn

sy )ϕ(y) dy. 2β0 + 2c0

Note that |¯ gp | ≤ β0 , g¯p · γ = 0, and |¯ gs | ≤ function ρ(x, z, p) such that

1 2.

Since |¯ gs | ≤

1 2,

there is a

p · γ − g¯(x, z, p, ρ) = ρ.

Since g¯(x, z, p, 0) = g(x, z, p) and |¯ g (x, z, p, ρ) − g¯(x, z, p, 0) ≤ 12 |ρ|, we conclude that 1 3 |ρ| ≤ |p · γ − g(x, z, p)| ≤ |ρ|. 2 2 In particular, we have Hence, if |y| ≤ 1, we have

|ρ| ≤ (1 + c0 )v.

ρ(x, z, p)y 1 2β0 + 2c0 ≤ 2 v because β0 ≥ 1. Moreover, it’s easy to check that |ρp | ≤ 4β0

because β0 ≥ 1, |ρz | ≤ 8εz (v)v,

v . R0 For ε0 a positive constant to be further specified, we then define N as |ρx | ≤ 24εx (v)v 2 +

N (x, z, p) = p · γ − g¯(x, z, p, ε0 ρ(x, z, p)).

To simplify notation, we write X = (x, z, p, ε0 ρ(x, z, p)). From our previous estimates, we then infer that v |¯ gx (X)| ≤ 24εx(v)v 2 + , R0 |¯ gz (X)| ≤ 4εz (v)v.

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Since g¯p · γ = 0, conditions (10.37) are then easily verified if ε0 < 1/(2β0 ). To prove (10.39), we write |N (x, z, p)| = |p · γ − g¯(X)| ≤ |p · γ − g(x, z, p)| + |g(x, z, p) − g¯(X)|

3 ε0 |ρ| + |ρ| ≤ 2(1 + c0 )v 2 2 if ε < 1. It follows that ≤

v1 ≤ (1 + 4ε(1 + c0 )2 )1/2 v ≤ 2v if ε ≤ 3(1 + c0 )2 /4. Also, |N (x, z, p) − p · γ| = |¯ g (X)| ≤ |g(x, z, p)| + |g(x, z, p) − g¯(X)| ≤ c0 v 0 +

ε0 1 (1 + c0 )v ≤ (c0 + 1)v 0 + |p · γ| 2 2

if ε < 1/(c0 + 1), so |N (x, z, p)| ≥

1 |p · γ| − (c0 + 1)v 0 , 2

and hence 1 v12 = (v 0 )2 + εN 2 ≥ (v 0 )2 (1 − 2ε(1 + c0 )2 ) + ε(p · γ)2 . 8 If ε ≤ 1/(4(c0 + 1)2 ), it follows that v12 ≥

ε 2 v , 8

√ and hence v1 ≥ εv/3. Now, we write p0 = p − (p · γ)γ and then ν1 =

p0 + εN Np . v1

It follows that |ν1 |2 ≤

1 1 [2|p0 |2 + 2ε2 N 2 |Np |2 ] ≤ 2 [2|p0 |2 + 32β02 ε2 N 2 ]. v12 v1

If 16β02 ε ≤ 1, we conclude that |ν1 |2 ≤ 2 and hence (10.41a) is verified. Next, we have ν1 · p =

1 0 1 N (p · p + εN Np · p) = (|p0 |2 + εN 2 ) + ε (Np · p − N ). v1 v1 v1

From the definition of N and our observation that g¯p · γ = 0, we conclude that Np · p − N = −¯ g (X) + g¯p (X) · p0 + ε0 g¯s (X)p · ρp ,

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and hence |Np · p − N | ≤ (c0 + β0 )v 0 +

ε0 (c0 + β0 + 1)v. 2

From these inequalities, we conclude that 1 ν1 · p ≥ (|p0 |2 + εN 2 − εk1 |N |(v 0 + ε0 v)) v1 1 = ((v 0 )2 + εN 2 − 1 − εk1 |N |(v 0 + ε0 v)) v1 with k1 = 2(1 + c0 + β0 ). From Cauchy’s inequality, we have 1 εk1 |N |(v 0 + ε0 v) ≤ εN 2 + k12 ε(v 0 )2 + k12 ε(ε0 )2 v 2 , 2 and (10.39) implies that εv 2 ≤ 3v12 . If ε ≤ 1/(2k12 ) and ε0 ≤ 1/(4k1 ), we conclude that 3 εk1 |N |(v 0 + ε0 v) ≤ v12 , 4 and hence we infer (10.41b). To prove (10.43), we write N i = ∂N/∂pi , N ij = ∂ 2 N/∂pi ∂pj and note that ∂ 2 g¯ ∂ 2 g¯ ∂ 2ρ ∂ 2 g¯ N ij ξi ξj = − ξi ξj −2ε0 ξρp ·ξ−(ε0 )2 2 (ρp ·ξ)2 −ε0 g¯s ξi ξj , ∂pi ∂pj ∂pi ∂s ∂s ∂pi ∂pj where g¯ and its derivatives are evaluated at X and ρ and its derivatives are evaluated at (x, z, p). Since |¯ gpp (x, z, p, s)| + |¯ gps (x, z, p, s)| + |¯ gss (x, z, p, s)| ≤ C(n, β0 , c0 )/s and ∂ 2 g¯ i ∂ 2 g¯ i γ = γ = 0, ∂pi ∂pj ∂pi ∂s it follows that there is a constant c3 determined only by n, β0 , and c0 such that   1 0 2 ε0 ij 2 |N ξi ξj | ≤ c3 |ξ | + |ξ · γ| . ε0 N N

In addition,

bkm ξk ξm = |ξ 0 |2 + ε(Np · ξ)2 + εN N km ξk ξm , and hence     c3  1 bkm ξk ξm ≥ |ξ 0 |2 1 − ε β02 + 0 + ε|ξ · γ|2 − c3 ε 0 . ε 8

If ε0 ≤ 1/8(c3 + 1) and then ε ≤ 1/2(β02 + c3 /ε0 ), we thus infer (10.43). 

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So far, we have presented results that apply to any oblique boundary condition. We now turn to results that are specific to the cornormal boundary condition. Our first step extends our previous Sobolev inequalities. Lemma 10.9. Let ∂Ω ∈ C 2 , let R0 be the constant from Lemma 5.18 and suppose u ∈ C 2 (Ω). Suppose also that there are constants c0 ∈ (0, 1) and τ > 1 such that |ν · γ| ≤ c0 on Σ(τ ).

(10.44)

Let N > 2 be a constant such that N ≥ n and set κ = N/(N − 2). Then there is a constant C, determined only by c0 , n, and N , such that Z Z hv ds ≤ C (|δh| + h|H| + h/R0 )v dx, (10.45a) ∂Ω

Z

Ω

Ω

1/κ h2κ v dx ≤C

Z

Ω

(N −n)/N Z n/N h2 v dx (|δh|2 + h2 H 2 + h2 /R02 )v dx Ω

(10.45b)

for all nonnegative Lipschitz functions h which vanish whenever v ≤ τ and d > R0 . Proof.

From (10.44), we see that Z Z 1 h(1 − (ν · γ)2 )v ds, hv ds ≤ 1 − c20 ∂Ω ∂Ω

and then Z

∂Ω

h(1 − (ν · γ)2 )v ds =

Z

∂Ω

h(γ − (ν · γ)ν) · γv ds.

If we define γ = Dd, then γ is C 1 on the support of h, and the divergence theorem gives Z Z 2 h(1 − (ν · γ) )v ds = div(hv(γ − (ν · γ)ν)) dx. ∂Ω

Ω

Moreover, we have div(hv(γ − (ν · γ)ν)) = (γ i − (ν · γ)ν i )Di hv + h div(v(γ − (ν · γ)ν)). Since γ i − (ν · γ)ν i = g ij γj , it follows that (γ i − (ν · γ)ν i )Di h ≤ |δh| and div(v(γ−(ν·γ)ν)) = Di (vg ij γj ) = γj Di (vg ij )+vg ij Di γj = −ν i vH+vg ij Di γj .

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Combining all these estimates with the observation that g ij Di γj ≤ n/R0 yields div(h(γ − (ν · γ)ν)) ≤ (|δh| + h|H| + nh/R0 )v, and then (10.45a) follows. To prove (10.45b), we first define ηk = min{1, kd} for k a positive integer and then replace h by hηk in (10.7) to obtain Z

Ω

1/κ Z (hηk )n/(n−1) v dx ≤ C(n) (|δ(hηk )| + hηk |H|) dx. Ω

From the monotone convergence theorem we conclude that Z Z (hηk )n/(n−1) v dx → hn/(n−1) v dx Ω

Ω

and Z

Ω

hηk |H|v dx →

Z

h|H|v dx

Ω

as k → ∞. It is easy to check that Z Z Z |δ(hηk )|v dx ≤ ηk |δh|v dx + h|δηk |v dx Ω

Ω

Ω

and that |δηk | ≤ k if kd ≤ 1 while |δηk | = 0 if kd > 1. The monotone convergence theorem implies that Z Z ηk |δh|v dx → |δh|v dx, Ω

Ω

and, because ∂Ω ∈ C 2 , we have Z Z k hv dx → Ωk

hv ds

∂Ω

as k → ∞, where Ωk is the subset of Ω on which d ≤ 1/k. We therefore obtain Z 1/κ Z Z hn/(n−1) v dx ≤ C(n) (|δh| + h|H|)v dx + Ω

Ω

∂Ω

 hv ds .

(10.46) The proof is completed by combining this inequality with (10.45a) and the proof of Theorem 5.8. 

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Next, we show that, if u satisfies the boundary condition (10.1b) and A and ψ satisfy appropriate structure conditions, then condition (10.44) is satisfied. Lemma 10.10. Suppose u ∈ C 1 (Ω) and ∂Ω ∈ C 1 . Suppose also that u satisfies (10.1b). If there are constants β1 , β2 , and τ0 with 0 ≤ β1 , β2 < 1 and τ0 ≥ 1 such that β1 |A| ≤ ν · A,

(10.47a)

|ψ| ≤ β2 ν · A

on Σ(τ0 ), then

(10.47b)

 −1/2 (1 − β2 )2 |ν · γ| ≤ 1 + β12

on Σ(τ0 ).

(10.48)

Set

Proof.

F0 = (γ − (γ · ν)ν) · (A + ψγ).

From the Schwarz inequality and condition (10.47a), we conclude that |F0 | ≤ |γ − (γ · ν)ν||A + ψγ| ≤ β1 (1 − (γ · ν)2 )1/2 ν · A. Since A · γ + ψ = 0, we conclude that F0 = −(γ · ν)(A · ν + ψ(γ · ν)),

so

|F0 | ≥ |γ · ν|(A · ν)(1 − β2 )

by virtue of (10.47b). Combining these two inequalities and dividing by A · ν, we obtain (1 − β2 )|γ · ν| ≤ β1 (1 − (γ · ν)2 )1/2 ,

and simple algebra yields (10.48).



Our final step is to derive a differentiated weak form of (10.1). Lemma 10.11. Let u ∈ C 2 (Ω) be a weak solution of (10.1) with ∂Ω ∈ C 2 . Suppose that A is differentiable with respect to x, z, and p. Suppose also that there are functions Cki and Dki satisfying (10.11) and Dki is differentiable with respect to x, z, and p. If η ∈ C 1 (Ω) is a vector field with η · γ = 0 on ∂Ω, then Z Z Ω

[(aij Djk u + Cki )Di η k − Di (Dki )η k ] dx =

where

g k = pk

gk η k ds,

(10.49)

∂Ω

∂ψ ∂ψ + k + Ai cij Dj γk − γi Dki . ∂z ∂x

(10.50)

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Proof. We first take a C 2 (Ω) vector field θ (with no restrictions on its normal component on ∂Ω). With Dk θk as our test function, we see that Z Z i k k (A Dik θ − BDk θ ) dx = ψDk θk ds, Ω

∂Ω

and then the divergence theorem gives Z Z Z Ai Dik θk dx = − Dk Ai Di η k dx − Ω

Ω

Ai Di θk γk ds. ∂Ω

We now rewrite the boundary integrands. Since

Ai Di θk γk = Ai cij Dj θk γk − ψγi Di θk γk

= Ai cij Dj (θ · γ) − Ai cij θk Dj γk − ψγi Di θk γk

and ψDk θk = ψcij Di θj + ψγi Di θk γk , we have Z

Ω

(Dk Ai + δki B)Di θk dx = +

Z

Z

∂Ω

(−ψcij Di θj − Ai cij Dj (θ · γ)) ds Ai cij Dj γk θk ds.

∂Ω

Now, we replace θ by a sequence of C 2 (Ω) vectors which converge in C (Ω) to η and note that cij Dj (η · γ) = 0 on ∂Ω to infer that Z Z (Dk Ai + δki B)Di η k dx = (−ψcij Di η j + Ai cij Dj γk η k ) ds. 1

Ω

∂Ω

We then rewrite

Dk Ai + δki B = aij Djk u + Cki + Dki and integrate by parts to obtain Z [(aij Djk u + Cki )Di η k − Di (Dki )η k ] dx = I, Ω

where

I=

Z

∂Ω

(−ψcij Di η j + Ai cij Dj γk η k − γi Dki η k ) ds.

The proof is completed by observing that we can integrate by parts on ∂Ω to obtain Z Z ij j −ψc Di η ds = cij Di ψη j ds ∂Ω

and using the definition of gk .

∂Ω



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Gradient bounds for the conormal problem

We are now ready to prove some gradient bounds in C 2 domains. In outline, our proof is a simple modification of the method from Section 10.2, but we shall present the proof in several steps so that the changes are easier to digest. Most of the changes (other than the obvious one of using v1 in place of v) appear in the proof of the energy estimate, and we break that proof down into three steps. The first step is to estimate the terms which come from aij Djk uDj η k with a suitable choice of η. (In our previous case, this estimate was simple because Di ν k = g ik /v and Di v = Dij uν j .) We begin by noting that, if there are positive decreasing functions εx and εz satisfying (10.34) and a positive constant β0 such that |Az (x, u, p) · γ| + |ψz (x, u)| ≤ εz (v)vaij (x, u, p)γi γj , (10.51a) 1 |Ax (x, u, p) · γ| + |A(x, u, p)| + |ψx (x, u)| ≤ εx (v)v 2 aij (x, u, p)γi γj , R0 (10.51b) X ij ij |a (x, u, p)γj | ≤ β0 a (x, u, p)γi γj (10.51c) i

for all (x, p) ∈ ∂Ω × Rn with A(x, u, p) · γ + ψ(x, u) = 0 and v ≥ τ0 , then G(x, z, p) = A(x, z, p) · γ + ψ(x, z) satisfies the hypotheses of Lemma 10.7, and hence there is a function v1 as in Lemma 10.8; in what follows, we define ν1 and bkm from that lemma. We also define E1 = aij Di v1 Dj v1 , C21 = aij Dik uDjk u. (10.52)

Lemma 10.12. Let u ∈ C 2 (Ω) and let A satisfy (10.47). Suppose also that u satisfies (10.1b). Suppose further that there are positive decreasing functions εx and εz satisfying (10.34) and a positive constant β0 such that (10.51) holds for all (x, p) ∈ ∂Ω × Rn with A(x, u, p) · γ + ψ(x, u) = 0 and v ≥ τ0 . Suppose finally that there are functions µ ¯ and Λ0 along with a nonnegative constant β3 such that ¯ 2 )1/2 , vaij ξi ξ¯j ≤ (aij ξi ξj )1/2 (¯ µ|ξ| (10.53a) ¯ 2 )1/2 , vaij ξ¯i ξj ≤ (aij ξi ξj )1/2 (¯ µ|ξ| (10.53b) εz (v)vaij ξi Dj u ≤ β3 (aij ξi ξj )1/2 (Λ0 )1/2 , ij

and

ij

1/2

εz (v)va Di uξj ≤ β3 (a ξi ξj ) 

1 + εx (v)v R0

2

1/2

(Λ0 )

µ ¯ ≤ β32 Λ0

(10.53c) (10.53d) (10.54)

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for all ξ and ξ¯ in Rn , as long as v ≥ τ0 and aij is evaluated at (x, u, Du). Let χ be an increasing function defined on [τ, ∞) with τ sufficiently large that v1 ≥ τ implies v ≥ τ0 , and set η k = χ(v1 )(v1 − τ )+ ζ 2 ν1k .

(10.55)

Then η · γ = 0 on ∂Ω, and there are constants c2 and c3 , determined only by n, β0 , β1 , and β2 such that   τ aij Djk uDi η k ≥ c2 [(v1 − τ )χ0 + χ]E1 + (1 − )χC21 ζ 2 v1 (10.56) − c3 [(v1 − τ )χ0 + χ]β32 Λ0 ζ 2 − c3 µ ¯|Dζ|2 χ

on Ωτ , the subset of Ω on which v1 ≥ τ .

Proof. First, because N = 0 on Σ(τ0 ), we immediately conclude that η · γ = 0 on ∂Ω. Next, a direct calculation gives aij Djk uDi η k = H1 + H2 + H3 , where H1 = aij Djk uν1k Di v1 [(v1 − τ ) + χ]ζ 2 , H2 = aij Djk uDi ν1k (v1 − τ )χζ 2 ,

H3 = aij Djk uν1k Di ζ2ζχ(v1 − τ ). Each one of these terms needs special care in its estimation. First, we set ai =

1 1 ( Di (ckm )pk pm + εN Ni ), v1 2

so Di v1 = Dij uν1j + ai +

ε N Nz Di u. v1

It follows that ε N Nz aij Di v1 Dj u). v1 By applying (10.53a) (10.53c), and (10.54), we then infer that H1 = [(v1 − τ )χ0 + χ]ζ 2 (E1 − aij Di v1 aj −

1/2

H1 ≥ [(v1 − τ )χ0 + χ]ζ 2 (E1 − Cβ3 (E1 )1/2 Λ0 ) 1 ≥ [(v1 − τ )χ0 + χ]ζ 2 ( E1 − Cβ32 Λ0 ) 2 In addition, if we set ˜ = aij Dik uν k Djm uν m , E 1 1

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then ˜ + aij Dik uν k aj + H1 = [(v1 − τ )χ0 + χ]ζ 2 (E 1

ε N Nz aij Dik uν1k Dj u). v1

and hence ˜ − Cβ3 (E) ˜ 1/2 Λ1/2 ) H1 ≥ [(v1 − τ )χ0 + χ]ζ 2 (E 0 by virtue of (10.53b) (10.53d), and (10.54). Cauchy’s inequality gives

Another application of

C 2 β Λ0 ) θ 3 for any θ ∈ (0, 1) and hence (by choosing θ so that (1 − θ)(1 − c1 /4) = 1 − c1 /2), c1 c1 ˜ H1 ≥ [(v1 − τ )χ0 + χ]ζ 2 ( θE1 + (1 − )E − Cβ32 Λ0 ). 8 2 Then we have ˜− H1 ≥ [(v1 − τ )χ0 + χ]ζ 2 ((1 − θ)E

H 2 = J1 + J2 + J3 + J4 , where   τ 1− χζ 2 aij bkm Dik uDjm u, v1   τ ˜ J2 = − 1 − χζ 2 E, v1

J1 =

∂ν1k , ∂xi ∂ν k J4 = (v1 − τ )χζ 2 aij Djk u 1 Di u. ∂z

J3 = (v1 − τ )χζ 2 aij Djk u

(Of course, here, we consider ν1k as a function of the variables x, z, p.) From (10.43), it follows that   τ J1 ≥ c 1 1 − χζ 2 C21 , v1 ˜ ≤ C2 implies that and the estimate E 1     1 τ 1 ˜ 2. J2 ≥ − c 1 1 − χζ 2 C21 + c1 − 1 χEζ 2 v1 2 It’s easy to check that   ∂ν ≤ C 1 + εx (v)v ∂x R0

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and ∂ν ≤ Cεz (v), ∂z

so

  τ 1/2 J3 + J4 ≥ −Cβ3 1 − C1 Λ 0 . v1 Finally,   τ H3 ≥ −C 1 − C1 ζ(¯ µ|Dζ|2 )1/2 , v1 and the proof is completed by combining all these inequalities with Cauchy’s inequality. Note that c2 = c1 /8.  We are now ready to present our energy estimate. Lemma 10.13. Let u ∈ C 2 (Ω) be a solution of (10.1) and suppose that ∂Ω ∈ C 2 with conditions (10.47) and (10.51) satisfied on Σ(τ0 ) and conditions (10.53) satisfied wherever v ≥ τ0 . Suppose also that 1/2

Cki ν1k ξi ≤ (aij ξi ξj )1/2 β3 Λ0

(10.57a)

for all ξ ∈ Rn , 1/2

Cki ηik ≤ (aij ηik ηjk )1/2 β3 Λ0 ,

∂Dki vν1k ηij ∂pj

(10.57b) 1/2

≤ (aij ηik ηjk )1/2 β3 Λ0

(10.57c)

for all matrices η, ∂Dki ∂Dki + ν1k ≤ β32 Λ0 (10.57d) ∂z ∂xi wherever v ≥ τ0 . Suppose further that there is a function λ0 such that ν1k pi

λ0 g ij ξi ξj ≤ aij ξi ξj

(10.58)

wherever v ≥ τ0 . Now let τ be a positive constant such that v1 ≥ τ implies v ≥ τ0 and suppose that there is a C 1 ([τ, ∞)) function Λ1 such that gk ν1k ≤ Λ1 (v1 )

(10.59)

on the portion of ∂Ω where v1 ≥ τ and that

(v1 Λ01 (v1 ))2 + Λ1 (v1 )2 ≤ β32 λ0 Λ0 , Λ1 (v1 )v1 ≤ β3 Λ0

(10.60a) (10.60b)

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wherever v1 ≥ τ . If χ is a C 1 ([τ, ∞)) function satisfying (10.15) for all s > τ , then there is a constant c4 , determined only by n, β0 , β1 , and β2 such that  Z  τ [ 1− C21 + E1 ]χζ 2 dx v1 Ωτ Z ≤ c4 (1 + cχ )2 ((β3 + 1/R0 )2 Λ0 ζ 2 + µ ¯|Dζ|2 )χ dx Ωτ

(10.61)

for any C 1 function ζ which vanishes wherever d ≥ R0 . Proof. Using η from Lemma 10.12 and arguing as in that lemma, we find that   Z Z c2 τ Cki Di η k − Di (Dki )η k dx ≤ [(v1 − τ )χ0 + χ]E1 + 1 − χC21 )ζ 2 dx 4 Ωτ v1 Z β32 Λ0 ζ 2 + µ ¯|Dζ|2 dx. + C(1 + cχ ) Ωτ

Using this η in (10.49) and recalling also the result of Lemma 10.12, we find that   Z τ ([(v1 − τ )χ0 + χ]E1 + 1 − χC21 )ζ 2 dx v 1 Ωτ Z Z β32 Λ0 ζ 2 + µ ¯|Dζ|2 dx + C gk η k ds. ≤ C(1 + cχ ) Ωτ

∂Ω

To proceed, we estimate this boundary integral. From (10.59) and (10.39), we have   Z Z τ k gk η ds ≤ C Λ1 (v1 ) 1 − χζ 2 v ds, v1 + ∂Ω ∂Ω and then (10.45a) implies that Z Z τ gk η k ds ≤ C |δ(Λ1 (v1 )(1 − )χ)|ζ 2 v dx v 1 ∂Ω Z Ωτ τ +C (Λ1 χζ|δζ| + λ1 (1 − )χζ 2 |H| + (1/R0 )Λ1 χζ 2 )v dx. v 1 Ωτ These integrals are estimated straightforwardly. First, we differentiate to see that τ τ τ τ |δ(Λ1 (v1 )(1 − )χ)| = |Λ01 (v1 )(1 − )χ + Λ1 2 χ + Λ1 (1 − )χ0 ||δv1 | v1 v1 v1 v1

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and

τ τ 1 + cχ τ 1/2 1/2 )χ + Λ1 2 χ + Λ1 (1 − )χ0 | ≤ β3 λ0 Λ0 v1 v1 v1 v1 by (10.15) and (10.60a). It then follows from (10.39) Z Z and (10.58) that τ 1/2 1/2 |δ(Λ1 (v1 )(1 − )χ)|ζ 2 v dx ≤ C(1 + cχ ) E1 β3 Λ0 dx. v1 Ωτ Ωτ Similarly, Zby invoking (10.10), (10.53a), and (10.39), we see that τ (Λ1 χζ|δζ| + λ1 (1 − )χζ 2 |H| + (1/R0 )Λ1 χζ 2 )v dx v 1 Ωτ Z τ 1/2 ≤C ((1 − )C21 )1/2 β3 Λ0 ζ 2 dx v 1 Ω Z τ 2 2 +C (β3 Λ0 ζ + µ ¯|Dζ|2 ) dx. |Λ01 (v1 )(1 −

Ωτ

The proof is completed by combining these inequalities and applying Cauchy’s inequality. 

We are now in a position to prove our gradient estimates. The first estimate (for the maximum of the gradient in terms of an integral) is proved by making three simple replacements in the proof of Lemma 10.4: we use v1 in place of v, we use our energy inequality Lemma 10.13 in place of Lemma 10.3 and we use the Sobolev inequality Lemma 10.9 in place of Lemma 10.2. Lemma 10.14. Let u ∈ C 2 (Ω) be a solution of (10.1) and suppose that ∂Ω ∈ C 2 with conditions (10.47) and (10.51) satisfied on Σ(τ0 ) and conditions (10.53) satisfied wherever v ≥ τ0 . Suppose also that conditions (10.57) and (10.58) hold wherever v ≥ τ0 , that condition (10.59) holds on the portion of ∂Ω on which v1 ≥ τ , and that conditions (10.60) hold wherever v1 ≥ τ . Suppose further that there are functions Λ ∈ C(Ω) and w, λ, ˜ in C 1 ([τ, ∞)) such that and λ Λ0 ≤ Λv, (10.62a) λ(v1 ) 1 +



λ(v1 ) ≤ Λ(x), 2 ! v1 λ (v1 ) g ij ξi ξj ≤ vaij ξi ξj , λ 0

(10.62b)

(10.62c)

˜ 1) (Λ(x)/λ(v1 ))N/2 /Λ(x) ≤ λ(v (10.62d) wherever v1 ≥ τ . Suppose finally that conditions (10.17) and (10.19) hold. Then there is a constant c5 , determined only by n, N , β, β1 , β2 , and β3 R, such that  N Z τ ˜ dx sup 1− w(v1 ) ≤ c5 R−n w(v1 )Λλv (10.63) v1 Ωτ (y,R/2) Ωτ (y,R)

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for any y ∈ Ω and R > 0 such that d ≤ R0 in Ω[y, R]. The integral in (10.63) is estimated by essentially the same technique that was used to estimate the integral in (10.20). We assume that there are positive constants β4 and β5 such that ˜ 1 )v ≤ β4 wβ5 Du · A, Λ(x)λ(v

(10.64)

so Lemma 10.14 reduces the gradient estimate to one on Z wq Du · A dx

for q = 1 + β5 . To estimate this integral, we set ψ0 = sup |ψ(x, u)| and we assume that there are positive constants β6 , β7 and β8 such that w0 A · ξ ≤ β6 (aij ξi ξj )1/2 (Du · A)1/2 ,

(10.65a)

w0 ψ0 γ · ξ ≤ β6 (aij ξi ξj )1/2 (Du · A)1/2

(10.65b)

w|A| ≤ β6 Du · A,

(10.65c)

for all ξ ∈ Rn , 2

Λ0 ≤ ε(v1 )w Du · A,

(10.65d)

B ≤ β8 Du · A

(10.65f)

µ ¯ ≤ β7 Du · A,

(10.65e)

wherever v ≥ τ0 . (These conditions are essentially the same as (10.23).) We also assume that there is a constant β9 such that ψ0 ≤ β9 |A|

(10.66)

wherever v ≥ τ0 . Note that this inequality is almost the same as (10.47b); in practice, this inequality with β9 < 1 is considered the basic assumption even though it is somewhat stronger than (10.47b). Lemma 10.15. Let u ∈ C 2 (Ω) be a solution of (10.1) and suppose that ∂Ω ∈ C 2 with conditions (10.47) and (10.51) satisfied on Σ(τ0 ). Suppose also that conditions (10.53), (10.57), (10.58), (10.65), and (10.66) are satisfied wherever v ≥ τ0 . Let τ be a positive constant such that v1 ≥ τ implies v ≥ τ0 and suppose conditions (10.17), (10.59), and (10.60) are also satisfied. Let q ≥ 2, set σ = osc u, and E = exp(β8 σ). If there is a τ1 ≥ τ such that 2  1 2 β62 σ 2 q 2 E 2 ε(τ1 ) ≤ 1, (10.67) 16c4 (1 + βq) β3 + R

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then there is a constant C, determined only by β, β0 , β1 , β2 , β7 , β9 , and q, such that Z wq Du · A dx Ωτ (y,R/2)

 q Z β6 σ ≤ CE q w(τ1 ) + Du · A dx R Ωτ (y,R)

(10.68)

for any y ∈ Ω and R > 0 such that d ≤ R0 in Ω[y, R].

Proof. The proof is essentially the same as for Lemma 10.5. With m = inf Ω(R) u, we define f1 and f by and we set

f1 (s) = exp(β8 (s − m)),

f (s) = (s − m)f1 (s),

Z

(w(v1 )q − w(τ )q )+ ζ q f 0 (u)Du · A dx, Z =− qf (u)wq−1 ζ q w0 Dv1 · A dx, Ωτ Z =− qf (u)(wq − w(τ )q )+ ζ q−1 Dζ · A dx, Ω Z = − (wq − w(τ )q ))+ ζ q f (u) div A dx, Z Ω = (wq − w(τ )q ))+ ζ q f1 (u)Du · A dx, ZΩ = (wq − w(τ )q ))+ ζ q β8 f (u)Du · A dx, ZΩ = (wq − w(τ )q )+ ζ q f (u)A · γ ds.

I1 =

Ω

I2 I3 I4 I5 I6 I7

∂Ω

Integration by parts gives

I1 = I2 + I3 + I4 + I7 , the explicit form of f implies that I1 = I5 +I6 , and the argument in Lemma 10.5 shows that I4 ≤ I6 . It follows that I5 ≤ I2 + I3 + I7 .

We now estimate I7 . First, Z I7 ≤ Eψ0 σ (wq − w(τ )q )+ ζ q ds ∂Ω Z = Eψ0 σ (wq − w(τ )q )+ ζ q γ · γ ds ∂Ω

= Eψ0 σ[J1 + J2 + J3 ],

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where J1 = − J2 = − J3 = −

Z

Ωτ

Z

Ω

Z

Ω

qwq−1 w0 γ · Dv1 ζ q dx,

q(wq − w(τ )q )+ ζ q−1 γ · Dζ dx, (wq − w(τ )q )+ ζ q div γ dx.

By invoking (10.65a), (10.65b), (10.65c), and (10.66), we infer that Z 1/2 I5 ≤ 2Eσqβ6 wq−1 ζ q E1 (Du · A) dx Ωτ Z + (1 + β9 )β6 Eσ qwq−1 ζ q−1 |Dζ|Du · A dx Ωτ Z σ wq + β9 E Du · A ζ q dx. R Ωτ v We now choose ζ as in Lemma 10.5 so |Dζ| ≤ 2/R. The Cauchy-Schwarz inequality along with Lemma 10.13 and conditions (10.65d) and (10.65e) imply that Z 1/2 2Eσqβ6 wq−1 ζ q E1 (Du · A) dx Ωτ

≤

1 2

Z

Ωτ

wq ζ q Du · A dx + CE 2 β62

σ2 R2

Z

Ωτ

wq−2 ζ q−2 Du · A dx

provided τ ≥ τ1 . From (10.65c) and the inequality Du · A ≤ v|A|, we conclude that w ≤ β6 v, so Z Z σ wq q−1 q−1 (1 + β9 )β6 Eσ qw ζ |Dζ|Du · A dx + β9 E Du · A ζ q dx R Ωτ v Ωτ Z σ ≤ CEβ6 wq−1 ζ q−1 Du · A dx. R Ωτ Combining these inequalities as in the proof of Lemma 10.5 with Young’s inequality completes the proof.  R Finally, we estimate Du · A dx. To simplify notation, we note that, if Ω is a bounded Lipschitz domain, then there is a constant K ∗ (Ω) such that |∂Ω ∩ B(y, R)| ≤ K ∗ Rn−1 for any y ∈ Ω and any R > 0, where |∂Ω ∩ B(y, R)| denotes (n − 1)dimensional measure.

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Lemma 10.16. Let u be a bounded weak solution of (10.1) and suppose that there are a decreasing function ε1 and a nonnegative constant β6 such that |A| ≤ ε1 (v)Du · A,

(10.69a)

B ≤ β8 Du · A

(10.69b)

for v ≥ τ0 . Set σ = oscΩ[y,R] u and E = exp(β6 σ) and let τ ≥ τ0 . If there is a constant τ2 ≥ τ such that σ 4ε1 (τ2 ) ≤ 1, (10.70) R then Z Du · A dx ≤ C(n)Rn E∆1 + 2K ∗ ψ0 EσRn−1 , (10.71) Ωτ (y,R/2)

where

  σ|A| ∆1 = sup σ(B − β8 Du · A)+ + (Du · A)− + R S0 and S0 is the subset of Ω[y, R] on which v < τ2 . Proof. By following the proof of Lemma 10.6 with β8 replacing β6 and Ω[y, R] replacing B(y, R), we find that Z Z n Du · A dx ≤ C(n)R E∆1 + 2 η(x)ψ ds, Ωτ (y,R/2)

∂Ω

where η is a function satisfying the conditions 0 ≤ ζ ≤ Eσ in Ω and ζ = 0 outside B(y, R). (The proof of Lemma 10.6 requires more of η but these are the only properties of relevance to the estimate of the boundary integral.) It follows that Z η(x)ψ ds ≤ EσK ∗ Rn−1 , ∂Ω

which then implies (10.71).



It is a simple matter to verify that our examples are covered by these results. In our first example, we have A = ν and B(x, z, p) = B ∗ (x, z, p) + B0 (x, z, p) satisfies the conditions |B ∗ |+v|B0 | ≤ θ1 ,

v(|Bp∗ |+|Bp∗ ·p|) ≤ θ1 ,

vBz∗ ≤ θ12 ,

|Bx∗ | ≤ θ12 ,

B ≤ θ2

for some nonnegative constants θ1 and θ2 . If we assume that there are nonnegative constants ψ0 , ψ1 , and ψ2 with ψ0 < 1 such that |ψ| ≤ ψ0 ,

|ψx | ≤ ψ1 ,

−ψ2 ≤ ψz ≤ 0,

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then the hypotheses of Lemmata 10.14, 10.15, and 10.16 are satisfied with 1 1 1 + ψ0 β0 = , β1 = , β2 = , 2 1 − ψ0 2 2 1 β3 = C(n, N, ψ0 )(θ1 + ψ1 + ψ2 + ), R0 β6 = C(n, ψ0 ), β7 = 2, β8 = β10 = θ2 , q = 1, ψ2 ψ1 + 1/R0 , εx (v) = , 1 − ψ02 v(1 − ψ02 ) 1 2n Λ0 = v, λ0 = , Λ1 = 2ψ1 + 2θ1 + , µ ¯ = v, v R0 σ  2 , τ = C + 1 , ε(v) = 1 (ln v)2 R 0 ˜ = 1, ε1 (v) = 2 , λ=Λ=λ v 2 w = (ln v) in Lemma 10.14 and w = ln v in Lemma 10.15. Hence    σ 3/2 v(y) ≤ exp C(n, ψ0 , [θ1 + θ2 + ψ1 + ψ2 ]R) 1 + . R εz (v) =

For uniformly elliptic equations, we assume that Ψ satisfies (10.30) and that A and B satisfy (10.31). In addition, we assume that there are nonnegative constants ψ0 , ψ1 , and ψ2 such that |ψ| ≤ ψ0 ,

|ψx | ≤ ψ1 ,

|ψz | ≤ ψ2 .

(10.72)

(Note that, this time, we do not assume that ψz ≤ 0.) If we take τ0 so large that τ0 Ψ(τ0 ) ≥ 1 + 2θ1 and Ψ(τ0 ) ≥ 2ψ0 , then the hypotheses of Lemmata 10.14, 10.15, and 10.16 are satisfied with 1 1 β 0 = θ3 , β 1 = , β2 = 2θ2 2 1 β3 = C(θ0 , θ1 , θ2 , θ3 )(θ4 + ψ1 + ψ2 + ), R0 β6 = 4θ2 , β7 = 4θ3 , β8 = 2θ4 , q = N + 2, εz (v) = 2θ4 + ψ2 , Λ0 = v 3 Ψ(v),

λ0 =

Ψ(v) , v

εx (v) = θ4 + ψ2 +

1 , R0

Λ1 = C(θ0 , θ1 , θ2 , θ3 )(ψ1 + Ψ2 + µ ¯ = θ3 vΨ(v),

1 )v1 Ψ(v1 ), R0

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˜ = C(θ0 , θ1 , θ2 , θ3 )v N −1 , Λ = (1 + θ3 )v 2 Ψ(v), λ = C(θ0 , θ1 , θ2 , θ3 )Ψ(v1 ), λ 1 2θ2 ε1 (v) = , w=v v provided R is sufficiently small. Finally, we examine the exponential case. We consider (10.1) when A = exp(θv 2 )p for some positive constant θ and B satisfies the condition |B(x, z, p)| ≤ θ1 exp(θv 2 )v 2 for some nonnegative constant θ1 , and we assume that ψ0 , ψ1 , And ψ2 are nonnegative constants such that (10.72) holds. By observing that aij γj = 2θ exp(v 2 )γ i − 2θψpi

wherever A · γ + ψ = 0, we see that the hypotheses of Lemma 10.14 are satisfied with 1 β0 = 2, β1 = β2 = , 2 2θψ1 2θ 1 εz = , εx = [ψ2 + ], v v R0 1 β3 = e(ψ1 + ψ2 + θ1 + ), β8 = 2θ1 , R0 β = 1, Λ = Λ0 = exp(θv 2 )v 3 , µ ¯ = exp(θv 2 )v 2 , λ = exp(θv12 )v1−3 , 1 ˜ = v −2 , ) exp(θv12 )v1 , λ0 = exp(θv 2 ), λ Λ1 = (ψ1 + ψ2 + 1 R0 w = v. We then obtain sup Ωτ (y,R/4)



τ N +2 1− v ≤ CR−n v1

Z

exp(θv 2 )v 2 dx.

Ωτ (y,R/2)

The hypotheses of Lemma 10.16 are satisfied with ε1 (v) = 1/(2v) and β8 = 2θ1 , so we obtain   σ 2 σ 2 v(y) ≤ C 1 + exp(θ 1 + ). R R Notes As mentioned in Chapter 9, the basic idea for deriving gradient estimates is due to Bernstein [9]. The corresponding result for uniformly elliptic equations in divergence form was developed by Ladyzhenskaya and Ural0 tseva

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[91] and extended to the minimal surface equation by Bombieri, De Giorgi, and Miranda [12]. Further extensions can be found in [92] and (for general prescribed mean curvature equations) in [13]. In this chapter, we follow the arguments in [170], which combines arguments from the uniformly elliptic case and the prescribed mean curvature case. In particular, the Sobolev-type inequality (Lemma 10.1), which is taken directly from [131], is a generalized version of Teorema 2 from [137] although these works use very different methods of proof. The interior gradient estimates in Section 10.2 are taken from [170] with a few slight modifications: our structure conditions are written without direct reference to the minimum eigenvalue of [aij ], we incorporate the decomposition in terms of Cki and Dki whereas [170] assumes that one of these two matrices is zero, and our aij differs from that in [170] by a factor of 1/v. (In addition, we use slightly different structure functions; in particular [170] does not distinguish between Λ0 and what would be Λv in our notation or between λ and vλ0 .) Specifically, our Lemma 10.3 is the same as (2.11) of [170], Lemma 10.4 is the same as Lemma 1 of [170], Lemma 10.5 is the same as Lemma 2 of [170], and Lemma 10.6 is the same as Lemma 3 of [170]. As a further refinement, [12] gives the gradient estimate for the prescribed mean curvature equation in the form v(y) ≤ c1 exp(c2 sup(u − u(y))/d(y)) with c1 and c2 determined only by n and d(y)θ1 . (As mentioned in the notes to Chapter 9, such an estimate was first proved by Finn [45] but only for n = 2; moreover, Finn showed there that this form cannot be improved.) Theorem 2 of [170] shows that the method in this chapter can be further modified to obtain an interior estimate of this form, and it is possible (see Section 4 in [106]) to prove corresponding estimates for the conormal problems in this chapter but, due to the length of this chapter, we omit even the statements of such estimates. Moreover, results similar to those in [170] were proved later by Marcellini [125]. The primary difference between the estimates in [170] and those in [125] is a technical one; Marcellini shows directly that weak solutions (which are not assumed to have second derivatives) have bounded gradient while Simon assumes initially more smoothness of the solutions (although a proof that weak solutions of uniformly elliptic equations have appropriate derivatives is given on pages 845 and 846 of [170]), and Simon’s structure conditions are slightly more general. A gradient estimate for oblique derivative problems was announced by Ural0 tseva in [182] and proved by Ladyzhenskaya and Ural0 tseva in Section

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X.2 of [91] under slightly stronger assumptions than given here; it was briefly pointed out in Section V.7 of [90] that the assumptions on the first derivatives of B and the second derivatives of ψ can be eliminated, and it is easy to modify the proof in [91] to handle theses weaker hypotheses. The proof given there was an integral version of the method from Chapter 9: after a suitable change of variable so that ∂Ω is replaced by B 0 (R) and Ω is replaced by B + (R), the authors estimate |D0 u|, and the remaining derivative is estimated as in Chapter 9 (but using integral methods). A major advance was made by Ural0 tseva in [183]: she introduced the function v2 , defined by ¯ u)Du · γ¯ , v2 (x) = F (x, u, Du) + ψ(x, when A behaves in appropriate sense like ν and F depends only on p. In that work, ψ was assumed to be identically zero and Ω was assumed convex. These two restrictions were removed in two subsequent works [184] and [185] with ψ assumed constant but nonzero and Ω assumed convex in [184]; the general case of nonconstant ψ and not necessarily convex Ω (but ∂Ω ∈ C 2 ) was studied in [185] specifically for the case A = ν. A further generalization, allowing F to depend also on x and z, was given in [186]. The results of [185] were also proved in [61] using very similar techniques; as Gerhardt points out in [61], the main justification of that work is that Ural0 tseva’s work was not readily available in the West at the time. (Although [184] was published in 1973, the English translation did not appear until 1979, and [185] was published in 1975 with the English version not appearing until 1980. Gerhardt’s work appeared in 1976.) All of these results were extended to general variational conormal problems in smooth domains by the present author in [97] (using a straightforward combination of ideas in [170], [184], and [61]); we mention in particular that our (10.45a) and its proof come from Lemma 1.4 of [61] and that our (10.46) and its proof come from Lemma 1.1 of [61]. In addition, the proof of our Lemma 10.10 is based on the proof of Lemma 1 in [184]. Corresponding results in nonsmooth domains were proved in [113] by combining the ideas developed in [97] and [84], and we refer the reader to those works for more details. We also mention [83], in which the authors prove that the graph of the solution of the capillary problem is a Lipschitz surface even when the contact angle is zero; their results also give gradient bound for the capillary problem when the contact angle is bounded away from zero. The gradient estimates in Sections 10.3 and 10.4 were first presented in [106] which was primarily concerned with parabolic conormal problems.

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The main new ingredient in that work is the function v1 , which was suggested by [38]; this work studies oblique derivative problems for fully nonlinear parabolic equations and it shows that, under suitable hypotheses, the function v¯1 = (D0 u)2 + ε(Du · γ − g(x, u, Du))2 satisfies a nice boundary value problem. However, this function is only useful under stronger hypotheses than seem appropriate. First, in order to derive the differential equation, one must assume that g is twice differentiable with respect to its arguments (whereas our v1 only uses first derivatives of g), and, second, some of these structure conditions seem to be too strong for even the simple case of linear boundary conditions. The second objection is overcome with a slightly different function: vˆ1 = max{¯ v1 , ε1 |Du|2 } by Ural0 tseva in [187] and [188], and we refer to those works for further discussion of the use of vˆ1 in place of v¯1 . The more elaborate construction of v1 in Lemmata 10.7 and 10.8 eliminates the need for any structure conditions on the second derivatives of the boundary condition, even for problems in nondivergence form. Details of this statement have, so far, only been carried out for uniformly parabolic problems in Theorem 13.13 of [114]. We point out in addition, that Nezhinskaya [147, 148] has derived gradient bounds for conormal problems under a slightly different set of hypotheses than allowed here by obtaining suitable integral estimates for the quantity denoted by wT in Chapter 9; this approach is based on that in [91], and it would be interesting to know if either this approach can be generalized to include the results given here or if the approach given here can be generalized to include the results in [147, 148].

Exercises 10.1 Show that condition (10.23e) can be replaced by ∂Ai ∂Ai Di u + ≤ β6 Du · A ∂z ∂xi in Lemma 10.5 if (10.24) is suitably adjusted. Also prove the corresponding modification of Lemma 10.15. 10.2 Show that Lemma 10.6 remains valid if B is replaced by −B in (10.28b).

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Higher Order Estimates and Existence of Solutions for Quasilinear Oblique Derivative Problems Introduction Our next step in studying the oblique derivative problem is to obtain estimates on higher derivatives. Specifically, we prove a H¨older estimate for the first and second derivatives of u assuming that we have an a priori bound for the maximum modulus of the gradient of u. We also use these estimates to prove a number of existence results. Our first existence results will apply to a number of problems with sufficiently smooth data, and we shall use these results along with suitable estimates to prove the existence of solutions to oblique derivative problems with nonsmooth data. For the nonsmooth data, we shall prove some existence results in which the solutions are known to have some global smoothness (typically C 1,α (Ω)) and some results when the solutions are not known to be so smooth.

11.1

The H¨ older gradient estimate for conormal problems

The easiest of our estimates is the one for the H¨older norm of the gradient of the solution of a conormal problem. This estimate is proved using the observation that the tangential derivatives of the solution are themselves solutions of suitable linear conormal problems. In fact, for the interior H¨ older gradient estimate, all the derivatives of the solution are solutions of suitable linear equations in divergence form. Lemma 11.1. Let u ∈ C 2 (B(R)) for some ball B(R) and suppose u is a solution of the equation div A(x, u, Du) + B(x, u, Du) = 0 in B(R). 407

(11.1)

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If there are positive constants K, λK , ΛK , and µK such that ij

|Du| ≤ K,

a ξi ξj ≥ λK |ξ|

(11.2a) 2

(11.2b)

for all ξ ∈ Rn , |aij | ≤ ΛK ,

|Ax | + K|Az | + |B| ≤ µK

(11.2c) (11.2d)

in B(R), where aij = ∂Ai /∂pj , and A, Ap , Ax , Az , and B are evaluated at (x, u, Du), then there are positive constants α and C, determined only by n and ΛK /λK such that  r α µK osc Du ≤ C ( osc Du + R) (11.3) B(r) R B(R) λK for all r ∈ (0, R). Proof.

For k = 1, . . . , n, set wk = Dk u and

∂Ai ∂Ai + D u + δki B. k ∂xk ∂z The weak form of (11.1) says that Z [Ai Di ζ − Bζ] dx = 0 fki =

B(R)

1

for any C (B(R)) function ζ with compact support. If we let ϕ be an arbitrary C 2 (B(R)) function with compact support, set ζ = Dk ϕ, and integrate by parts, we obtain Z [aij Dj wk + fki ]Di ϕ dx = 0. B(R)

The desired estimate then follows from Theorem 5.45.



A similar argument provides a H¨older gradient estimate up to the boundary. Theorem 11.2. Let ∂Ω ∈ C 2 and suppose u ∈ C 2 (Ω) satisfies div A(x, u, Du) + B(x, u, Du) = 0 in Ω ∩ B(R),

A(x, u, Du) · γ + ψ(x, u) = 0 on ∂Ω ∩ B(R).

(11.4a) (11.4b)

Suppose also that there are positive constants K, λK , ΛK , and µK such that (11.2) holds and that |ψx (x, z)| + |ψz (x, z)|K ≤ µK

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for x ∈ ∂Ω and |z| ≤ sup |u|. Then there are positive constants α < 1 and C, both determined by ΛK /λK and n, such that osc Du ≤ C

Ω∩B(r)

 r α R

( osc Du + Ω∩B(R)

µK R) λK

(11.5)

for all r ∈ (0, R). Proof. Since the hypotheses are invariant under a C 2 change of variables, there is no loss of generality in assuming that Ω ∩ B(R) = B + (R) and ∂Ω ∩ B(R) = B 0 (R). With wk = Dk u as before, we see that Di (aij Dj wk + fki ) = 0 in B + (R). Moreover, for k < n, we also have anj Dj wk + fkn + ψk = 0 on B 0 (R) with ψk =

∂ψ(x, u) ∂ψ(x, u) + Dk u. ∂xk ∂z

It follows from Theorem 5.45 that  r α µK osc D u ≤ C ( osc Dk u + R). k R λK B + (r) B + (R) In addition, for any ρ ∈ (0, R), we have osc Dn u ≤ C

B 0 (ρ)

≤C

X

k K, and hence kvk ≤ I + ε. Since this inequality is true for any ε, it follows that kvk ≤ I, and the definition of I implies that kvk ≥ I. Therefore kvk = I. If I > 0, then we could find a v1 ∈ JU ∩ S with kv1 k < kvk, so I = 0. There is a v ∈ JU ∩ S with kvk = 0, and hence 0 ∈ JU ∩ S, as desired.  Condition (11.6) is easily connected to the concept of Gˆateaux variation, which we now define. Let B1 and B be Banach spaces, and let V be a subset of B1 . For a function J : V → B, we define the Gˆ ateaux variation of J at u ∈ V in the direction ψ ∈ B1 to be J(u + εψ) − Ju . ε→0 ε Note that this limit is undefined unless u + εψ ∈ V for all sufficiently small ε, but it does not require that u be an interior point of V . Since the choice of V depends on the specific problem that we study, we shall only give a particularly simple example of the Gˆateaux variation here. Let Ω be a bounded domain in Rn with ∂Ω ∈ H1+α for some α ∈ (0, 1), and define Q and N by Ju (ψ) = lim

Qu = aij (x, u, Du)Dij u + a(x, u, Du) in Ω,

(11.7a)

N u = b(x, u, Du) on ∂Ω.

(11.7b)

We suppose that the functions aij , a, and b are differentiable with respect to z and p and that these functions, along with their derivatives with respect to z and p, are in Hα (S) for any compact subset of Ω× R× Rn (or ∂Ω× R× Rn (−1−δ) for b and its derivatives). If we take B1 = H2+δ for any δ ∈ (0, α) and (1−δ)

define J by Ju = (Qu, N u) (with B = Hδ ij

Ju ψ = (a Dij ψ +

(aij p Dij u

+ ap ) · Dψ +

× Hα (∂Ω)), then

(aij z Dij u

+ az )ψ, bp · Dψ + bz ψ),

where aij , a, b and their derivatives are all evaluated at (x, u, Du).

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The topology T will always be chosen in the same way, which we now describe. We start by taking Ω to be a bounded subset of Rn with ∂Ω ∈ Hk+α for some positive integer k and some α ∈ (0, 1), and we take B to be one of the following spaces (1) B = Ha (Ω) for some non-integer a ∈ (0, k + α), (2) B = Hb (∂Ω) for some non-integer b ∈ (0, k + α), (3) B = Ha (Ω) × Hb (∂Ω) for some non-integers a and b in (0, k + α). We then take the topology T to be the one induced by the L∞ (S) norm, with S = Ω in the first and third cases and S = ∂Ω in the second case. Now suppose that (vm ) is a sequence in B with a uniform bound M on kvm k which converges in T to some v ∈ B. If B = Ha (Ω), then the Arzela-Ascoli theorem implies that Dj vm converges uniformly to Dj v for any integer j < a, so kvk ≤ M . Similar arguments show that kvk ≤ M in the other two cases. 11.3

Existence results and estimates for linear equations and nonlinear boundary conditions in spherical caps

In our study of the oblique derivative problem for quasilinear elliptic equations, we shall use some results on the existence and regularity of solutions to related, simpler problems. It will be convenient to study problems in a special geometry first, namely, the spherical cap from Section 2.6. With µ0 a fixed positive constant and κ a fixed constant in (µ0 /(1 + µ20 )1/2 , 1), we define, for any r > 0, Γ(r) = {x ∈ Rn : |x0 |2 + |xn + κr|2 < r2 , xn > 0}, and we write Γ0 (r) and Γ∗ (r) for the subsets of ∂Γ(r) on which |x0 |2 + |xn + κr|2 < r2 and |x0 |2 + |xn + κr|2 = r2 , respectively. We also take [aij ] to be a continuous positive-definite matrix-valued functions with eigenvalues in the interval [λ, µλ] for some positive numbers λ and µ, and we take g to be a continuous function defined on Γ0 (r) × Rn−1 such that |g(x, p0 ) − g(x, q 0 )| ≤ µ0 |p0 − q 0 | for all x ∈ Γ0 (r) and all p0 and q 0 in Rn−1 . We now estimate the H¨ older norm of the gradient for solutions of various simpler problems. To write our estimates more simply, we abbreviate [·]α,ρ = [·]α,Γ(ρ)

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for any ρ ∈ (0, r) and α ∈ (0, 1). Lemma 11.4. Suppose that aij and g are independent of x. If v satisfies aij Dij v = 0 in Γ(r), Dn v = g(D0 v) on Γ0 (r), (11.8) then there are positive constants C and σ0 , both determined only by µ0 , n, and µ, such that [Dv]α,r/2 ≤ Cr−1−α |v|0 (11.9) for any α ∈ (0, σ0 ]. First, for k ∈ {1, . . . , n − 1} and h ∈ (0, r/8), we set 1 vk,h (x) = [v(x + ek ) − v(x)]. h n 0 Then there is a vector field βk,h with βk,h ≡ 1 and |βk,h | ≤ µ0 such that ij a Dij vk,h = 0 in Γ(7r/8), βk,h · Dvk,h = 0 on Γ0 (7r/8). The local H¨ older estimate Theorem 1.26 then gives a positive constant σ1 determined only by µ0 , n, Λ/λ such that [vk,h ]α,3r/4 ≤ Cr−α |vk,h |0,7r/8 if α ∈ (0, σ1 ). Sending h → 0, we obtain [Dk v]α,3r/4 ≤ Cr−α |Dk v|0,7r/8 . To continue, we observe that aij Dij Dn v = 0 in Γ(3r/4) and n−1 X [Dn v]α,Γ0 (3r/4) ≤ C [Dk v]α,3r/4 . Proof.

k=1

It follows from Corollary 1.29 that there is a constant σ2 , determined only by Λ/λ and n, such that [Dn v]α,r/2 ≤ C[Dn v]α,Γ0 (3r/4) if α ∈ (0, σ2 ). With this further restriction on α, we have [Dv]α,r/2 ≤ Cr−α |Dv|0 . To continue, we use weighted norms with respect to d∗ , the distance to Γ∗ (r). It follows that (0) (0) [v]1+α ≤ C[v]1 , and the interpolation inequality Lemma 2.3 implies that (0) (0) [v]1 ≤ C([v]1+α + |v|0 )1/(1+α) (|v|0 )α/(1+α) . From these two inequalities, we conclude that (0) [v]1+α ≤ C|v|0 , and (11.9) follows easily from this estimate with σ0 = min{σ1 , σ2 }. 

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To proceed, we now present a key extension lemma. Lemma 11.5. Let α ∈ (0, 1), and suppose there is an nonnegative increasing function µ1 such that |g(x, p0 ) − g(y, p0 )| ≤ µ1 (M )|x − y|α

for all x and y in Γ0 (r) and all p0 ∈ Rn−1 with |p0 | ≤ M . Then, for all (−1−α) ϕ ∈ H1+α (Γ∗ (r)), there is a function ϕ0 ∈ H2 (Γ(r)) ∩ C 2 (Γ(r)) such that ϕ0 = ϕ,

(11.10a)

Dϕ0 · τ = Dϕ · τ

(11.10b)

for any vector field τ tangential to Γ∗ (r), Dn ϕ0 = g(D0 ϕ0 )

(11.10c)

∗∗

on Γ (r). Moreover, there is a constant C, determined only by µ0 , κ, α, n, µ1 (|ϕ|1 ), and r such that (−1−α)

|ϕ0 |2

≤ C|ϕ|1+α .

Proof. We begin by defining K to be the set of all points x with |x0 | = √ 2 1 − κ r. Then, for any function f ∗ ∈ Hα (Γ∗∗ (r)), there is a function ∗ ϕ ∈ H1+α (K) such that ϕ∗ = ϕ and Dn ϕ∗ = f ∗ on Γ∗∗ (r). Furthermore, we may take ϕ∗ so that |ϕ∗ |1 ≤ |ϕ|1 and |ϕ∗ |1+α ≤ C|ϕ|1+α . To define a suitable function f ∗ , we fix a point √ x0 ∈ Γ∗∗ (r) and rotate 2 axes so that, in the new √ coordinates, x0 = (0, . . . ∗, 0, 1 − κ r, 0). Then the 2 vector (0, . . . , 0, −κ, 1 − κ ) is tangential to Γ (r) at x0 , and there is a Lipschitz function G with Lipschitz constant no greater than µ0 such that the equation vn = g(x0 , v 0 ) in the old coordinates is equivalent to vn = G(v 0 ) in the new coordinates. (For simplicity, we suppress the dependence of G on x0 here.) Write h for the directional derivative of ϕ in the direction of τ at x0 , and, for k = 1, . . . , n − 2, write vk for the directional derivative of ϕ in the direction of the positive xk axis. We then define the function H by ! √ s 1 − κ2 00 H(s) = s − G v , h + . κ A simple calculation shows that H 0 has a positive lower bound and hence there is a unique solution t of the equation H(t) = 0; we define f ∗ (x0 ) to be this number t. It is straightforward to verify that |f ∗ |α ≤ C|ϕ|1+α . We now observe that, at x√ 0 , the vector (0, . . . , −1, 0) is normal to K, and the vector v = (v 00 , h + t 1 − κ2 /κ, t) (where t is as above) satisfies

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the conditions v · τ = Dϕ · τ at x0 for any τ tangential to Γ∗ (r) at x0 and vn = G(v 0 ). In other words, v is the appropriate gradient of our extension function ϕ0 . We now define f ∗∗ on K by √ f ∗ (x0 , 0) 1 − κ2 ∗∗ 0 n 0 f (x , x ) = Dτ ϕ(x , 0) + , κ 0 where τ (which depends on x ) is the vector tangent to Γ∗ (r) at (x0 , 0) which lies in the plane containing the normal to K at (x0 , 0) and the point (0, −κr). It follows that f ∗∗ ∈ Hα (K) with |f ∗∗ |α ≤ C|ϕ|1+α √. Finally, we write J for the set of all points x with |x0 | < 1 − κ2 r. The (−1−α) proof is completed by taking ϕ0 to be an H2 (J) ∩ C 2 (J) extension of ϕ∗ with normal derivative equal to f ∗∗ .  Our next step is a conditional H¨older gradient estimate, that is, we prove such an estimate under the additional assumption that a certain class of boundary value problems has a solution. This conditional estimate will be used later to show that the associated problems really do have a solution. Lemma 11.6. Let r > 0 be given, and suppose that there are positive constants µ1 and µ2 such that |g(x, p0 ) − g(y, p0 )| ≤ [µ1 + µ2 |p0 |]|x − y|α

(11.11)

aij (x0 )Dij w ˜ = 0 in Γ(r),

(11.12a)

0

0

n−1

for all x and y in Γ (r) and all p ∈ R . Suppose also that g(x, 0) = 0 for all x ∈ Γ0 (r). Then there is a positive constant σ1 , determined only by µ0 , τ , n, and Λ/λ such that if the problem ∗

w ˜ = ϕ˜ on Γ (r),

(11.12b)

0

0

Dn w ˜ = g(x0 , D w) ˜ on Γ (r)

(11.12c)

0

is solvable for all x0 ∈ Γ (r), all ϕ˜ ∈ H1+α and some α ∈ (0, σ1 ), then there is a constant C, determined only by µ, µ0 , κ, α, n, r, and the modulus of continuity of aij such that any solution v of the problem aij Dij v = f in Γ(r),

(11.13a)

∗

v = ϕ on Γ (r),

(11.13b)

0

0

Dn v = g(D v) + ψ on Γ (r)

(11.13c)

satisfies the estimate (1−α)

|v|1+α ≤ C(|f /λ|0

(1+α)/α

+ |ϕ|1+α + µ1 + µ2

+ |ψ|α ).

(11.14)

Moreover, for any θ ∈ (0, 1), there is a constant C, determined also by θ such that (1+α)/α

|v|1+α,θr ≤ C(|f /λ|1−α + |v|0 + µ1 + µ2 0

+ |ψ|α ).

(11.15)

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Proof.

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We first suppose that µ2 = 0 and we define dˆ by

0 2 n 2 ˆ = r − |x | − (x + κr) . d(x) (11.16) r r There are several main estimates to be proved. The first is an estimate of the maximum of v, and we introduce the function

w0 = (2 + 2A0 )r1+α − dˆ1+α − A0 (xn )1+α − A0 rα xn with A0 a constant at our disposal to effect this estimate. Elementary calculations show that, by choosing A0 sufficiently large (depending only on µ, µ0 , and α), we have aij Dij w0 ≤ −λα(1 + α)dα−1 in Γ(r), w0 ≥ 0 on Γ∗ (r),

Dn w0 ≤ −µ0 (1 + α)rα , (1−α)

If we set F = |f /λ|0

|D0 w0 | ≤ (1 + α)rα on Γ0 (r)

, Φ0 = |ϕ|0 , and Ψ0 = |ψ|0 , and define

w = F w0 + Ψ0 [r − xn ] + Φ0 ,

it follows that aij Dij (w ± v) ≤ −λα(1 + α)dα−1 in Γ(r), w ± v ≥ 0 on Γ∗ (r),

β · D(w ± v) ≤ 0 on Γ0 (r) for some vector β satisfying β n = 1 and |β 0 | ≤ µ0 , so the maximum principle implies that |v| ≤ w, and the definition of w gives us |v|0 ≤ C[F + Ψ0 + Φ0 ].

(11.17)

Our next estimate is on the rate at which v grows near Γ∗∗ , the set of all x ∈ Rn with xn = 0 and |x0 |2 = (1 − κ2 )r2 . As in Section 3.6, we write d∗∗ for the distance to Γ∗∗ . For convenience, we abbreviate |ϕ|1+α to Φ and |ψ|α to Ψ. Then, we use Lemma 11.5 to infer that there is a function ϕ0 ∈ C 2 (Γ(r)) ∩ C 1 (Γ(r)) such that ϕ0 = ϕ on Γ∗∗ ,

Dϕ0 = Dv on Γ∗∗ , |D2 ϕ0 | ≤ C0 Φ(d∗∗ )α−1 .

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for some constant C0 . Further, we note that the constant α1 = α1 (arctan κ, n, µ0 , µ) from Proposition 3.17 is greater than 1. For α ∈ (0, α1 − 1) (to be further specified), we use w ˜1+α , the function from Proposition 3.17, and ε, the constant from that proposition. With A1 , A2 , and A3 positive constants at our disposal, we set w1 = A1 w ˜1+α − A2 (dˆ)1+α − A3 (xn )1+α . Since aij Dij w ˜1+α ≤ 0, we can choose A2 and A3 so that aij Dij w1 ≤ −C0 dα−1

in Γ(r). On Γ∗ (r), we have d∗ = 0 and 0 ≤ xn ≤ Cd∗∗ , so we have w1 ≥ C0 (d∗∗ )1+α

on Γ∗ (r) for all sufficiently large A1 . In addition, where d∗∗ = εr, we have d∗ ≤ Cd∗∗ and xn ≤ Cd∗∗ , so we also have w1 ≥ C0 (d∗∗ )1+α

where d∗∗ < εr for A1 sufficiently large. Finally, we have |D[(d∗ )1+α ]| ≤ C(d∗∗ )α on Γ0 (r), so we also have β · Dw1 ≤ −C0 (d∗∗ )α on the subset of Γ0 (r) where d∗∗ ≤ εr. It follows from the maximum principle that |v − ϕ0 | ≤ C(F + Φ + Ψ)w1 ≤ C(F + Φ + Ψ)(d∗∗ )1+α as long as d∗∗ ≤ εr, and hence, for any x0 ∈ Γ∗∗ (r), we have

|v(x) − v(x0 ) − Dv(x0 ) · (x − x0 )| ≤ C(F + Φ + Ψ)|x − x0 |1+α .

Now we fix x0 ∈ Γ0 (r) and set R0 = d∗∗ (x0 ); we also set Γ(ρ, x0 ) = {x ∈ Rn : x − x0 ∈ Γ(ρ)}

for ρ > 0, with corresponding definitions for Γ∗ (ρ, x0 ) and Γ0 (ρ, x0 ). Further, we note that there is a positive constant ε1 (κ) such that Γ(ε1 R0 , x0 ) ⊂ Γ(r). Choosing x1 ∈ Γ∗ (r) so that |x1 − x0 | = R0 , we infer that |v(x) − v(x0 ) − Dv(x1 ) · (x − x0 )| ≤ CΦ|x − x0 |1+α

if |x − x0 | ≥ ε1 R0 . If |x − x0 | < ε1 R0 , we define v0 by v0 (x) = v(x) − Dv(x1 ) · (x − x0 ), so v0 satisfies the conditions aij Dij v0 = f in Γ(r),

Dn v0 = g0 (x0 , D0 v) + ψ0 on Γ0 (r),

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with g0 defined by g0 (p0 ) = g(x, p0 + D0 v(x1 )) − g(x, D0 v(x1 )), and ψ0 defined by ψ0 (x) = ψ(x) − ψ(x1 ).

With δ1 ∈ (0, 1) and τ ∈ (0, 1) to be chosen, we set ρk = τ k δ1 R0 for any nonnegative integer k, and we define three sequences of functions (vk ), (gk ), and (ψk ), and a sequence of linear polynomials (Lk ) inductively as follows. Given vk and gk , we write wk to be the solution of aij (x0 )Dij wk = 0 in Γ(ρk , x0 ), Dn wk = gk (x0 , D0 wk ) on Γ0 (ρk , x0 ), wk = vk on Γ∗ (ρk , x0 ). Such a function exists by virtue of our assumptions because we can take wk = w ˜ − gk−1 (x0 , D0 Lk−1 )xn + D0 Lk−1 · x, where w ˜ is the solution of aij (x0 )Dij w ˜ = 0 in Γ(ρk , x0 ), Dn w ˜ = gk−1 (x0 , D0 w) ˜ on Γ0 (ρk , x0 ), w ˜ = vk + g(x0 , D0 Lk−1 )xn − D0 Lk−1 · x on Γ∗ (ρk , x0 ) given by our assumption. We then take Lk to be the first degree Taylor polynomial for wk centered at x0 and we set ψk (x) = ψk−1 (x) − ψk−1 (x0 ), noting that ψk (x0 ) = 0 if k ≥ 1. Using Lemma 11.4 to estimate wk , we can imitate the proof of Theorem 4.5 to find constants τ ∈ (0, 1) and σ ∈ (0, 1) such that sup Γ(τ ρk ,x0 )

|vk − Lk | ≤ τ 1+σ

sup |vk | + Cρ1+α (F + Φ + Ψ + µ1 ) 0

Γ(ρk ,x0 )

for Lk = Dwk (x0 ) if δ1 is sufficiently small. Proceeding inductively and arguing as in Theorem 4.5, we obtain that, for every ρ ∈ (0, r), there is a linear polynomial P1 (ρ, x0 ) such that sup Γ(r)∩B(x0 ,ρ)

|v − P1 (ρ, x0 )| ≤ C(|v0 | + F )

ρ1+α . r1+α

(11.18)

Similar arguments show that this estimate is valid for any x0 ∈ Γ∗ (r) and then for any x0 ∈ Γ(r). The estimate (11.14) then follows from Lemma 2.9 in this case. By noting that (11.11) implies that |g(x, p0 ) − g(y, p0 )| ≤ (µ1 + µ2 M )|x − y|α

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for all x and y in Γ0 (r) and all p0 with |p0 | ≤ M , we infer from the case µ2 = 0 that |v|1+α ≤ C(|v|0 + F + Ψ + Φ + µ1 + µ2 |v|1 ). The interpolation inequality Lemma 2.4 then implies (11.14). To complete the proof, we observe that the process used to prove (11.18) implies that, if there is a constant µ3 such that |g(x, p0 ) − g(y, p0 )| ≤ µ3

|x − y|α d∗ (x)

for all x and y in Γ0 (r) with |x − y| ≤ 21 d∗ (x) and all p0 ∈ Rn−1 in place of (11.11), then (0)

(1)

|v|1+α ≤ C(|v|0 + µ3 + F + |ψ|1+α ). It follows that (0)

(0)

(1)

|v|1+α ≤ C(|v|0 + µ1 + µ2 |v|1 + F + |ψ|1+α ), and the proof is completed by using Lemma 2.5.



We are now ready to prove an existence result for linear equations with nonlinear boundary conditions, based on a variation of the method of continuity. Although its primary application is to show that (11.12) is solvable, we state a more general result because that result is critical to the proof of the solvability of (11.12). Lemma 11.7. Let σ1 be the constant from Lemma 11.6, let α ∈ (0, σ1 ), and suppose that there are constants µ1 and µ2 such that (11.11) holds. (0) (1−α) If aij ∈ Hδ for some δ ∈ (0, 1), then, for any f ∈ Hδ (Γ(r)), h ∈ Hα (Γ0 (r)) and continuous ϕ, there is a solution of aij Dij w = f in Γ(r), 0

(11.19a) 0

Dn w = g(x, D w) + h on Γ (r), ∗

w = ϕ on Γ (r).

(11.19b) (11.19c)

Proof. Suppose first that g ∈ C 2 (Γ0 (r)×Rn−1 ) and that g is independent of x. We write T for the set of all t ∈ [0, 1] such that the problem aij Dij w = f in Γ(r), Dn w = tg(D0 w) + h on Γ0 (r), w = ϕ on Γ∗ (r).

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has a solution for every f , h, and ϕ as in the hypotheses of this lemma. (−1−α) From the linear theory, we know that 0 ∈ T , and that w is in H2+δ if ϕ is in H1+α . With ε a positive constant to be further determined, let t ∈ T , let (1−α) s ∈ [0, 1], and suppose that |s− t| ≤ ε. For fixed f ∈ Hδ and ϕ ∈ H1+α , (−1−α) we set B1 = H2+δ and B = Hα (Γ0 (r)), and we write U for the subset of all u ∈ B1 such that aij Dij u = f in Γ(r),

u = ϕ on Γ∗ (r).

We then define the operator J : U → B by setting Ju = Dn u − sg(D0 u) − h.

Our goal is to show that Ju = 0 for some u ∈ U , and we shall do so by appropriate application of Theorem 11.3. As a first step, we note that, if (−1−α) ψ ∈ H2+δ with aij Dij ψ = 0 in Γ(r),

u = 0 on Γ0 (r),

then u + εψ ∈ U for any ε and that, in this case, we have ∂g Ju ψ = Dn ψ − s 0 (D0 u) · D0 ψ. ∂p From the linear theory, it follows that there is a function ψ ∈ B1 such that aij Dij ψ = 0 in Γ(r), ψ = 0 on Γ∗ (r), ∂g (D0 u) · D0 ψ = −Ju on Γ0 (r), ∂p0 and hence, for every u ∈ U , there are u1 ∈ U and ε ∈ (0, 1) such that (11.6) is satisfied with η = 1/2. Our next step is to observe that the definition of J implies that Dn ψ − s

Dn u − tg(D0 u) = (s − t)g(D0 u) + Ju,

so Lemma 11.6 implies that

(1−α)

kukB1 ≤ C[|f |δ

It follows that

+ |ϕ|1+α + |(s − t)g(D0 u) + Ju|α ).

kukB1 ≤ C1 εkukB1 + C2 (0)

with C1 determined only by µ, n, r, α, |aij |δ , and the modulus of continuity of aij , and C2 , determined also by f , h, and ϕ. If C1 ε ≤ 1/2, we conclude that kukB1 ≤ 2C2 .

(11.20)

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As mentioned in the previous section, we now take T to be the topology induced by the L∞ (Γ0 (r)) norm, and we take u0 to be the solution of aij Dij u0 = f in Γ(r), u0 = ϕ on Γ∗ (r), Dn u0 = 0 on Γ0 (r). If (vm ) is a sequence in JU ∩S(Ju0 , 1/2), then it’s bounded in B, and hence it contains a subsequence which converges in T to some function v ∈ B. For each m, there is a um ∈ U such that Jum = vm and (11.20) implies that (um ) is bounded in B1 . Hence, by the Arzela-Ascoli theorem, this sequence contains a subsequence which converges in H1+α/2 (Γ(r)) to some u ∈ B1 , and hence Jum → Ju uniformly on Γ0 (r). It’s easy to check that u ∈ U and that Ju ∈ S(u0 , 1/2), so JU ∩ S(u0 , 1/2) is sequentially compact in T . Theorem 11.3 then implies that there is u ∈ U such that Ju = 0, which means that u solves aij Dij u = f in Γ(r), 0

(11.21a) 0

Dn u = sg(D u) + h on Γ (r), ∗

u = ϕ on Γ (r).

(11.21b) (11.21c)

We now observe that, if ϕ is only assumed continuous, then there is a sequence of functions (ϕm ) in H1+α converging uniformly to ϕ. If we write um for the solution of (11.21) with ϕm in place of ϕ, then the maximum principle shows that (um ) is uniformly Cauchy, so it has a limit v. From (11.15) and the interior Schauder estimate, we infer that Dvm → Dv uniformly on compact subsets of Γ0 (r) and D2 vm → D2 v uniformly on compact subsets of Γ(r), so s ∈ T if |s − t| ≤ ε. Since ε is a constant independent of t, it follows that T = [0, 1] and therefore (11.12) has a solution. When g depends on x, we repeat the above argument, but we have to use the case of g independent of x to infer (11.15). To remove the assumption that g ∈ C 2 , we observe that there is a sequence of C 2 functions (gm ) which converges uniformly on compact subsets of Γ0 (r) × Rn−1 to g such that |gm (x, p0 ) − gm (x, q 0 )| ≤ µ0 |p0 − q 0 |,

|gm (x, p0 ) − gm (y, p0 )| ≤ 2[µ1 + µ2 |p0 |]|x − y|α

for all x and y in Γ0 (r) and all p0 and q 0 in Rn−1 . If ϕ ∈ H1+α , then the solutions of aij Dij wm = f in Γ(r), Dn wm = gm (x, D0 wm ) + h on Γ0 (r), wm = ϕ on Γ∗ (r)

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are uniformly bounded in H2+δ , so there is a subsequence that converges to the unique solution of (11.12). If ϕ is only continuous, then we take a sequence of H1+α functions (ϕm ) converging uniformly to ϕ and argue as before to conclude that (11.12) has a unique solution, which is the limit of the solutions to aij Dij vm = f in Γ(r), Dn vm = g(x, D0 vm ) + h in Γ0 (r), vm = ϕm on Γ∗ (r).



In fact, the same result is true if the equation has lower order terms. To prove this more general result, we use the following version of the LeraySchauder fixed point, which reduces the existence question to a verification of an a priori estimate; we refer to Theorem 11.6 of [64] for a proof. Theorem 11.8. Let B be a Banach space, and let T : B × [0, 1] → B be a compact mapping with T (x, 0) = 0 for all x ∈ B. If there is a constant M such that kxkB < M for any x ∈ B satisfying x = T (x, t) for some s ∈ [0, 1], then there is a fixed point of T (·, 1). The following generalization of Lemma 11.7 to linear equations with lower order terms will be used several times in the rest of this book. Proposition 11.9. Let σ1 be the constant from Lemma 11.6, let α ∈ (0, σ1 ), and suppose that there are constants µ1 and µ2 such that (11.11) holds. Suppose also that there is a nonnegative constant B0 such that |b| ≤ (0) (1−α) (1−α) λB0 (xn )α−1 . Let aij ∈ Hδ , bi ∈ Hδ , and c ∈ Hδ for some (1−α) δ ∈ (0, 1). If c ≤ 0 in Γ(r), then, for any f ∈ Hδ (Γ(r)), h ∈ Hα (Γ0 (r)) and continuous ϕ, there is a solution of aij Dij w + bi Di w + cw = f in Γ(r), (11.22a) Dn w = g(x, D0 w) on Γ0 (r), ∗

(11.22b)

w = ϕ on Γ (r), (11.22c) and this solution satisfies the estimate (1−α) (1+α)/α |w|1+α ≤ C(|f /λ|0 + |ϕ|1+α + |g(·, 0)|0 + µ1 + µ2 ). (11.23) Moreover (even without the assumption c ≤ 0), there is a constant C, (1−α) determined only by n, r, α, µ, µ0 , κ, B0 , |c|0 /λ, and the modulus of ij continuity of a such that (1−α) (1+α)/α |w|1+α,r/2 ≤ C(|w|0 + |f /λ|0 + µ1 + µ2 ). (11.24)

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Proof. Suppose first that ϕ ∈ H1+α . We introduce the Banach space B = H1 (Ω), and we define the function T : B × [0, 1] → B by saying u = T (v, t) if u solves the boundary value problem aij Dij u + tbi Di v + tcv = tf in Γ(r), Dn u = tg(x, D0 u) on Γ0 (r), u = tϕ on Γ∗ (r). It follows from Lemma 11.7 that this problem has a solution in H1+α for any (v, s) ∈ B × [0, 1] and hence T is a compact mapping. The linear theory implies that T (v, 0) = 0 for all v ∈ B, so we just have to provide a uniform bound on kuk1 for all solutions of aij Dij u + tbi Di u + tcu = tf in Γ(r), Dn u = tg(x, D0 u) on Γ0 (r), u = tϕ on Γ∗ (r). Our first step is an estimate on the maximum of u. To obtain this estimate, we define the function h by Z 1 1 + B0 s h(s) = s− exp(B0 σ α /α) dσ. B0 B0 0

Since h00 (s) − B0 sα−1 h0 (s) = −sα−1 for all s > 0 and h0 (0) = −1, it follows that w0 = h(xn ) satisfies aij Dij w0 + tbi Di w0 ≤ −(xn )α−1 in Γ(r), Dn w0 ≤ −1, D0 w0 = 0 on Γ0 (r), 1 + B0 w0 ≥ − r exp(B0 rα /α) on Γ∗ (r). B0 (1−α)

We now set F = |f /λ|0 + |g(·, 0)| and conclude that there are constants A1 and A2 such that w1 = F [A1 + A2 w0 − dˆ1+α ] satisfies aij Dij w1 + tbi Di w1 + tcw1 ≤ f in Γ(r), Dn w1 < g(x, D0 w) on Γ0 (r), w1 ≥ tϕ on Γ∗ (r). It follows that u ≤ w1 , and a similar argument gives a lower bound for u, so |u|0 ≤ CF.

(11.25)

Then Lemmata 11.6 and 11.7 imply that (1+α)/α

|u|1+α ≤ C(|f /λ|0 + µ1 + µ2

(1−α)

+ |b|0

|u|1 ),

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and interpolation, in this case Lemma 2.4, along with (11.25), gives (11.23). It then follows from Lemma 2.14 that (−1−α)

|u|2+δ

(1−α)

≤ C(|f /λ|δ

(1+α)/α

+ µ1 + µ2

).

In particular, this implies an upper bound on |u|1 , so Theorem 11.8 implies that (11.22) has a solution. When ϕ is only assumed continuous, we argue as in Lemma 11.7 to infer (11.24) and solvability of (11.22). 

11.4

Estimates and existence results for linear equations and nonlinear boundary conditions in general domains

From these preliminaries, we can prove a local H¨older gradient estimate for solutions of such problems in arbitrary H1+α domains. Theorem 11.10. Let ω ∈ H1+α (Bn−1 (0, R)) for some α ∈ (0, 1) and some R > 0, and suppose ω(0) = 0. Write Ω[R] and Σ[R] for the subsets of Rn with |x| < R such that xn > ω(x0 ) or xn = ω(x0 ), respectively, and suppose that g is a continuous function defined on Σ[R] × Rn−1 . Let aij ∈ (0) (1−α) (1−α) Hδ , b i ∈ Hδ , c ∈ Hδ with aij continuous on Ω with eigenvalues in the interval [λ, µλ]. (Here the weighted norms are respect to distance from Σ[R].) Suppose finally that there are nonnegative constants µ0 , µ1 , and µ2 such that |g(x, p0 ) − g(x, q 0 )| ≤ µ0 |p0 − q 0 |,

g(x, p0 ) − g(y, p0 )| ≤ [µ1 + µ2 |p0 |]|x − y|α

for all x and y in ∂Ω and all p and q in Rn−1 . Then there are constants α1 , determined only by µ, µ0 , n, and max |Dω|, and C, determined only by (1−α) (1−α) |ω|1+α , R, |b|0 , and |c|0 such that any solution of aij Dij v + bi Di v + cv = f in Ω[R], Dn v = g(x, D0 v) on Σ[R] satisfies the estimate (1−α)

|v|1+α,Ω[R/2] ≤ C(|v|0 + |f /λ|0

(1+α)/α

+ µ1 + µ2

)

provided α ≤ α1 .

(−1−α)

Proof. We note that, after a suitable H2+δ change of variable, we may assume that ω ≡ 0 (with appropriate change in the constants) and then this estimate follows from (11.23). 

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We also obtain an existence result in H1+α domains. Theorem 11.11. Let ∂Ω ∈ H1+α for some α ∈ (0, 1),and suppose that g is a continuous function defined on ∂Ω × Rn such that g(x, p) = g(x, q) (0) (1−α) (1−α) whenever p − q is parallel to γ(x). Let aij ∈ Hδ , bi ∈ Hδ , c ∈ Hδ with aij continuous on Ω with eigenvalues in the interval [λ, µλ]. Suppose finally that there are nonnegative constants µ0 , µ1 , and µ2 such that |g(x, p) − g(x, q)| ≤ µ0 |p0 − q 0 |,

|g(x, p) − g(y, p)| ≤ [µ1 + µ2 |p0 |]|x − y|α for all x and y in ∂Ω and all p and q in Rn . (Here p0 = p − γ(x)p · γ(x).) (1−α) Suppose finally that c ≤ −λ. Then, for any f ∈ Hδ , there is a solution of aij Dij v + bi Di v + cv = f in Ω,

(11.26a)

γ · Dv = g(x, D0 v) on ∂Ω.

(11.26b)

Then there are constants α1 , determined only by µ, µ0 , n, and Ω, and (1−α) (1−α) C, determined also by Ω, |b|0 , and |c|0 , such that any solution of (11.26) satisfies the estimate (1−α)

|v|1+α ≤ C(|f /λ|0

(1+α)/α

+ µ1 + µ2

)

(11.27)

provided α ≤ α1 . Proof. For ease of notation, we write L for the operator defined by Lw = aij Dij w + bi Di w + cw. As in Lemma 11.7, we first assume that g ∈ C 2 (∂Ω × R). Instead of the method of continuity argument used in that lemma, we directly apply (−1−α) (Ω) and B = Hα (∂Ω), and, for a Theorem 11.3, so we set B1 = H2+δ (1−α)

fixed f ∈ Hδ , we write U for the set of all u ∈ B1 such that Lu = f in Ω. Also, we define the operator J : U → B by Ju = γ · Du − g(x, D0 u).

(−1−α)

Now, for any ψ ∈ H2+δ

with Lψ = 0 in Ω, we find that

Ju ψ = γ · Dψ −

∂g · D0 ψ, ∂p0

so, just as in Lemma 11.7, for any u ∈ U , there is u1 ∈ U such that (11.6) holds with η = 1/2. Therefore, Theorem 11.3 reduces the proof of the existence of solutions to the establishment of (11.27).

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To verify this estimate,√ we fix v ∈ U , and let ρ be a regularized distance for Ω such that |Dρ| ≥ 1/ 2 in some neighborhood of ∂Ω, say the set on which ρ < δ0 . With B0 = sup (aij Dij ρ + bi Di ρ)ρ1−α /λ, {ρ 0, there are positive constants C and σ, determined only by µ0 , n, µ, K, and A0 , such that  ρ σ osc Dv ≤ C osc Dv (11.29) R B + (R) B + (ρ)

for all ρ ∈ (0, R).

Proof. We use the function N from Lemma 10.8, noting that N is a function of p only, and fixing a positive constant ε which is less than the number ε0 from that lemma. Moreover, it follows from the proof of that lemma that N (p) = 0 whenever pn = g(p0 ). We also define V = |p0 |2 + εN 2 . (In the notation of Lemma 10.8, we have V = v12 − 1.) Now, with θ a positive constant at our disposal, we define (for k = 1, . . . , n) wk± = ±Dk v + θV, and we define b=

∂aij (Dv) Dij v. ∂p

A simple calculation shows that aij Dij wk± + bi Di wk± = 2θaij bmr Dim vDjr v, where [bkm ] is the matrix from Lemma 10.8, and hence there is a positive constant c1 such that aij Dij wk± + bi Di wk± ≥ c1 θλ|D2 v|2 .

Since |b| ≤ A0 λ|D2 v|, we conclude that c2 aij Dij wk± ≥ − λ|Dwk± |2 θ 2 for c2 = A0 /c1 . The quadratic gradient term is then eliminated by a standard exponential substitution. First, we define Wk± = sup wk± . B + (R)

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Then we define h± k = It follows that

c   θ  2 exp (Wk± − wk± ) − 1 . c2 θ

aij Dij h± k ≤ 0,

+ h± k ≥ 0 in B (R).

(11.30)

Our next step is to prove a suitable weak Harnack inequality for h± k. The case k < n is quite straightforward, so we investigate this one first. To this end, we set   ∂g β = − (D0 v), 1 ∂p and then we have β · DDk v = 0. Since V = |D0 v|2 and Dn V = Dn (|D0 v|2 ) on B 0 (R), it follows that β · Dwk± = 0 on B 0 (R), and therefore 0 β · Dh± k = 0 on B (R).

We then infer from Theorem 1.20 that there are positive constants P and C such that !1/P Z P R−n (h± ≤ C +inf h± (11.31) k ) dx k. B (R/2)

B + (R/2)

For k = n, we define H ± = inf B 0 (R) h± n and ( ± min{h± if x ∈ B + , n (x), H } H± (x) = H± if x ∈ / B+. We now apply Theorem 1.28 with Ω = Rn , ω(x0 ) = |x0 |, Σ[R] = ∅, and G = 0 to infer that !1/P Z R−n (H± )P ≤ C inf H± , B(0,R/2)

B ∗ (0,R/2)

where B ∗ (R/2) is the set of all x ∈ Rn with |x| < R and xn > −|x0 |. Since |B ∗ (R/2) \ B + (R/2)| = C(n)Rn and H± = H ± on B ∗ (R/2) \ B + (R/2), it follows that H± ≤ C

inf

B(0,R/2)

H± .

Since H± ≤ h± n , it follows that inf h± n ≤C

B 0 (R)

inf

B(0.R/2)

h± n.

(11.32)

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We next observe that Wk± − wk± ≤ h± k ≤ exp

c

2

θ

 (Wk± − wk± ) (Wk± − wk± ).

The assumed gradient bound implies that there is a constant Cθ (determined by the known quantities and θ) such that c  2 exp (Wk± − wk± ) ≤ Cθ . θ Hence ± ± Wk± − wk± ≤ h± k ≤ Cθ (Wk − wk ).

This inequality allows us to rewrite our estimates for h± k as estimates for wk± . Specifically, we infer that !1/P Z R−n

B+ (R/2)

(Wk± − wk± )P

≤ Cθ

inf

B + (R/2)

(Wk± − wk± )

if k < n and inf

B 0 (R/2)

(Wn± − wn± ) ≤ Cθ

inf

B + (R/2)

(Wn± − wn± ).

We now introduce some notation to connect these inequalities to oscillations. Specifically, we define ωk (ρ) = osc wk+ + osc wk− + + B (ρ)

B (ρ)

(for k = 1, . . . , n) and V. ω0 (ρ) = osc + B (ρ)

We now note that inf B + (R/2) w ≥ inf B + (R) w for any function w, and, hence, if we set W = supB + (R) w, then inf

B + (R/2)

(W − w) = sup w − B + (R)

sup

w

B + (R/2)

≤ sup w − inf w+ + B + (R)

= osc w− + B (R)

B (R)

osc

B + (R/2)

inf

B + (R/2)

w−

sup

w

B + (R/2)

w.

It follows that inf

B + (R/2)

(Wk+ − wk+ ) +

inf

B + (R/2)

(Wk− − wk− ) ≤ ωk (R) − ωk (R/2).

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In addition, we have aP + bP ≥ C(P )(a + b)P for any nonnegative a and b, so Z 1/P + + − − P (Wk − wk + Wk − wk ) dx S

≤

Z

S

(Wk+

−

wk+ )P

1/P Z 1/P − − P dx + (Wk − wk ) dx . S

Moreover, Wk+ − wk+ + Wk− − wk− = sup wk + sup wk− − 2θV B + (R)

B + (R)

≥ osc Dk v − 2θω0 (R), + B (R)

so our previous estimates (in particular, (11.31)) give osc Dk v ≤ Cθ [ωk (R) − ωk (R/2)] + 2θω0 (R).

B + (R)

(11.33)

Next, we observe that inf

B 0 (R/2)

(Wn+ − wn+ ) +

inf

B 0 (R/2)

(Wn− − wn− )

≥ osc Dn v − +

osc

Dn v − 2θω0 (R).

osc

Dk v,

B 0 (R/2)

B (R)

From the boundary condition, we have that osc 0

B (R/2)

D n v ≤ µ1 n

n−1 X k=1

B 0 (R/2)

and we know that oscB 0 (R/2) Dk v ≤ oscB + (R) Dk v. Hence inf

B 0 (R/2)

(Wn+ − wn+ ) +

inf (Wn− − wn− )

B 0 (R/2)

≥ osc Dn v − nµ0 + B (R)

n−1 X k=1

osc

B + (R/2)

Dk v − 2θω0 (R).

Using this inequality along with (11.32), we conclude that osc D n v ≤ µ0 n +

B (R)

n−1 X k=1

We now set ω ¯=

osc Dk v+2θω0 (R)+Cθ (ωn (R)−ωn (R/2)). (11.34)

B + (R)

Pn

j=1

ωj and we define ω(ρ) =

n X j=1

osc Dj v.

B + (ρ)

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Summing our inequalities (11.33) on k and adding the resultant inequality to (11.34), we find that ω(R) − 2n(1 + µ0 )θω ¯ 0 (R) ≤ Cθ [¯ ω(R) − ω ¯ (R/2)].

Now there is a positive constant K1 such that ω0 (R) ≤ K1 ω ¯ (R) and hence, if we take θ = 1/(2n(1 + µ0 )K1 ), we have and also

ω(R) ≤ C[¯ ω (R) − ω ¯ (R/2)] 1 3 ω ¯ (ρ) ≤ ω(ρ) ≤ ω ¯ (ρ) 2 2

for any ρ ∈ (0, R). Hence, which implies that

ω ¯ (R) ≤ C[¯ ω (R) − ω ¯ (R/2)], ω ¯ (R/2) ≤ δ ω ¯ (R)

with δ = (C − 1)/C ∈ (0, 1). We can repeat this argument with R replaced by any ρ ∈ (0, R) and hence we obtain ω ¯ (τ ρ) ≤ τ α ω ¯ (ρ)

with τ = 1/2 and α = log1/2 δ > 0. Lemma 1.25 then implies that  ρ α ω ¯ (R), ω ¯ (ρ) ≤ C R and therefore  ρ α ω(ρ) ≤ C ω(R). R Simple algebra allows us to infer (11.29).



Note that the constant C in (11.29) depends exponentially on the gradient bound for v, and hence gradient bounds cannot be obtained from this one via interpolation. For this reason, we need to modify our approach to various estimates somewhat compared to the methods in the previous sections. In addition, we need to use the following result of Krylov, related to the H¨ older gradient estimate for solutions of Dirichlet problems for linear equations up to the boundary. To state this lemma, we fix a point x0 ∈ Γ∗ (r) and we define (for s > 0 and dˆ defined by (11.16)) ˆ < s }, G(s) = {x ∈ Rn : |x0 − x00 | < s, 0 < d(x) 32nµ s s 0 n 0 0 ˆ G (s) = {x ∈ R : |x − x0 | < 2s, < d(x) < }. 32nµ 16nµ

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Lemma 11.13. Let u be a nonnegative C 2 (G(2s)) function for some s > 0 so small that G(2s) ⊂ Γ(r), and suppose that there are positive constants α < 1 and F such that aij Dij u ≤ λF dˆα−1 in G(2s).

(11.35)

u u 16 ≤ 2 inf + F sα . ˆ G(s) d α dˆ

(11.36)

Then inf

G0 (s)

We set A = inf G0 (s) u/dˆ and introduce functions ! dˆ |x0 − x00 |2 ˆ d, w1 = −3 − 8nµ + s s2  α 1 16s ˆ w2 = [ − dˆα ]d, α nµ A w = u + w1 + F w2 . 4 Direct calculation shows that ! 16nµdˆ |x0 − x00 |2 ij a Dij w1 = 1 − + aij Dij dˆ − 4aij Dij dˆ s s2 Proof.

+ − where T 0 =

n−1 n−1 X 2 0 ˆ 4 X ij i i ˆ+ 2 T d + a (x − x )D d ain (xi − xi0 )Dn dˆ j 0 2 2 2 s s i,j=1 s i=1

16 ij ˆ ˆ µa Di dDj d, s

Pn−1 i=1

aii . Since aij Dij dˆ = −2T /r and 1−

16nµdˆ |x0 − x00 |2 + ≥0 s s2

in G(2s), we conclude that λ 1 ˆ − 16µ|Dd| ˆ 2 ) + 8nµλ/r ( + 6µ|Dd| s 8 ˆ 2 = 4 − 4d/r, ˆ so |Dd| ˆ ≥ 1 in G(2s) and hence in G(2s). In addition, |Dd| ij a Dij w1 ≤ 0 in G(2s). Next,   α 1 s aij Dij w2 = − dˆα aij Dij dˆ α 16nµ α−1 ˆ ij ˆ ˆ j d). ˆ −d (da Dij dˆ + (1 + α)aij Di dD aij Dij w1 ≤

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ˆ 2 allow us to infer that The explicit expressions for aij Dij dˆ and |Dd|

dˆ aij Dij w2 ≤ λdˆα−1 [−4 − 4α + (2nµ + 4 + 4α)] ≤ −λdˆα−1 . r ij Therefore a Dij w ≤ 0 in G(2s) and w ≥ 0 on ∂G(2s), so the maximum principle implies that w ≥ 0 in G(2s). In particular w ≥ 0 in G(s). Since ˆ in G(s), we have u ≥ ((A/2) − (8F/α))dˆ there. w1 ≤ −2dˆ and w2 ≤ 8sα d/α Simple rearrangement gives (11.36).  ˆ This lemma leads in a standard way to an oscillation estimate for u/d. Lemma 11.14. With x0 and α as in Lemma 11.13, let u ∈ C 2 (G(2R)) ˆ is bounded for some R > 0 such that G(2R) ⊂ Γ(r), and suppose that |u/d| in G(2R). Then there is a constant θ0 ∈ (0, 1] determined by n and µ, and, for any θ and α satisfying 0 < α < θ < θ0 , there is a positive constant C, determined also by α and θ, such that, if there is a nonnegative constant F with |aij Dij u| ≤ λF dˆα−1 in G(2R),

(11.37)

then  s θ u u ≤ C[ sup + F sα ] R G(2R) dˆ G(s) dˆ osc

for any s ∈ (0, 2R).

(11.38)

Proof. Note from Lemma 1.25 that it suffices to show that there is a constant δ(n, µ) ∈ (0, 1) such that u u u s α (11.39) osc ≤ δ osc + Fs + sup ˆ ˆ G(s) d G(2s) d R dˆ G(2R)

for all s ∈ (0, R/2). To prove (11.39), we set

mi = inf

G(is)

u , dˆ

u G(is) dˆ

Mi = sup

for i = 1, 2, and u M = sup . G(2R) dˆ

We then apply the weak Harnack inequality (Theorem 1.16) to u − m2 dˆ in Σ = {x ∈ Rn : |x0 − x00 |
0. We also write G0 = G0 × {0}. Theorem 11.17. Let G be a generalized half-ball, let [aij ] be a matrixvalued function defined on G×Rn , and let g be a real-valued function defined on G0 × Rn−1 . Suppose that there are positive constants α < 1, µ, λ, µ0 , µ1 , A1 , K, such that λ|ξ|2 ≤ aij (x, p)ξi ξj ≤ µλ|ξ|2 , ij ∂a (x, p) ≤ A1 λ ∂p

for all (x, p) ∈ G × Rn with |p| ≤ K and all ξ ∈ Rb , and |g(x, p0 ) − g(y, q 0 )| ≤ µ0 |p0 − q 0 |, 0

0

|g(x, p ) − g(y, p )| ≤ µ1 |x − y|

α

(11.44a) (11.44b)

(11.45a) (11.45b)

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for all x and y in G0 and all p0 and q 0 in Rn−1 with |p0 |, |q 0 | ≤ K. Suppose also that there is a nonnegative, increasing function θ, defined on [0, 1/2] with θ(0) = 0, such that   |x − y| ij ij |a (x, p) − a (y, p)| ≤ θ (11.46) d∗ (x) for all x and y in G with |x − y| ≤ 12 d∗ (x) and all p ∈ Rn with |p| ≤ K. Then is a constant α0 ∈ (0, 1), determined only by K, α, A1 , n, µ, and µ0 , and, for any A2 ≥ 0, there is a constant C, determined also by A2 , µ1 , and G0 = supx∈G |g(x, 0)| such that any function u ∈ C 2 (G) ∩ C 1 (G) satisfying the conditions |Du| ≤ K in G and |aij (x, Du)Dij u| ≤ A2 d∗ (x)−α−1 (xn )α−1 in G, 0

0

Dn u = g(x, D u) on G

(11.47a) (11.47b)

obeys the estimate (0)

|Du|α0 ≤ C. Proof. As a first step, we create extensions of aij and g which satisfy the appropriate conditions for all p, p0 , and q 0 . The argument in Lemma 1.3 gives us a matrix-valued function [˜ aij ] defined on G × Rn such that ij ij a ˜ (x, p) = a (x, p) for all (x, p) ∈ G × Rn with |p| ≤ K, |˜ aij (x, p) − a ˜ij (x, q)| ≤ A1 |p − q|,   |x − y| |˜ aij (x, p) − a ˜ij (y, p)| ≤ θ d∗ (x) for all x ∈ G, all p and q in Rn and all y ∈ G with |x − y| ≤ 12 d∗ (x). It follows that the eigenvalues of [˜ aij ] are in the interval [λ/2, 2µλ] if |p| ≤ −1 K + (2A1 ) . We then let η : [0, ∞] → [0, 1] be a fixed Lipschitz function which is 1 on [0, K] and 0 on [K + (2A1 )−1 , ∞) and set Aij = η˜ aij + (1 − η)λδ ij . It follows that Aij has eigenvalues in the range [λ/2, 2µλ] and satisfies the conditions ij ∂A (x, p) ≤ C(A1 , K)λ, ∂p for all (x, p) ∈ G × Rn and ij

ij

|A (x, p) − A (y, p)| ≤ θ



|x − y| d∗ (x)



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for all x and y in G with |x − y| ≤ 12 d∗ (x) and all p ∈ Rn . Moreover, we have that Aij is independent of p if |p| ≥ K +(2A1 )−1 . For ease of notation, we denote this extension also by aij . Similarly, if we define g for |p0 | ≥ K by g(x, p0 ) = η(|p0 |)g(x, Kp0 /|p0 |),

we find that g satisfies (11.45) for all p0 and q 0 . We need to obtain the usual iterative inequality involving the oscillation of the difference between u and suitable linear functions. Specifically, we fix x0 ∈ G0 and with δ1 ∈ (0, 1/2) to be chosen, we set R = δ1 d∗ (x0 ). With τ ∈ (0, 1/2) also to be chosen, we define Rk = τ k R and we wish to create a sequence of functions (uk ) and a sequence of linear polynomials (Lk ) inductively. As a preliminary step, we set L−1 (x) = g(x0 , 0)(xn − xn0 ), ij 0 0 u0 = u−L−1 , aij 0 (x, p) = a (x, p+DL−1 ), and g0 (x, p ) = g(x, p )−g(x0 , 0). Then u0 satisfies the conditions n α−1 |aij in Γ(x0 , R), 0 (x, Du0 )Dij u0 | ≤ A2 (x )

Dn u0 = g0 (x, D0 u0 ) on Γ0 (x0 , R).

Note that conditions (11.44), (11.45), and (11.46) hold if we we replace aij ij by aij 0 and g by g0 . In addition, a0 is independent of p if |p| ≥ K + G0 and g0 (x0 , 0) = 0. We then define v˜k to be the solution of aij vk )Dij v˜k = 0 in Γ(x0 , Rk ), 0 (x0 , D˜ v˜k = uk on Γ∗ (x0 , Rk ), Dn v˜k = g0 (x0 , D0 v˜k ) on Γ0 (x0 , Rk ) ˜ k for the first order Taylor polynomial given by Lemma 11.16, and we write L ˜ k and we write Lk for the for v˜k centered at x0 . We also define vk = v˜k − L first order Taylor polynomial for vk centered at x0 . Finally, we define uk inductively by uk+1 = uk − Lk . To proceed, we note from Theorem 9.9 (and the assumed bound |Du0 | ≤ K + G0 ) that, for any δ ∈ (0, 1), there is a constant C(δ) such that |D˜ vk | ≤ C(δ) in Γ(x0 , (1 − δ)Rk ). From Lemma 11.12 applied to v˜k in Γ(x0 , Rk /2), we infer that there is a constant σ such that sup Γ(x0 ,Rk )

|vk − Lk | ≤ Cτ 1+σ

sup Γ(x0 ,Rk )

|vk |.

(11.48)

Then, Corollary 3.13 gives a constant ζ such that sup Γ(x0 ,Rk )\Γ(x0 ,(1−δ)Rk )

|uk − vk | ≤ Cδ ζ [ sup

Γ(x0 ,Rk )

|uk | + (A2 + µ1 )Rk1+α ]

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for any δ ∈ (0, 1). (Note that the gradient estimate for u is not used here.) To estimate the supremum of |uk − vk | over Γ(x0 , (1 − δ)Rk ), we need to estimate the second derivatives of vk , and this estimate is more delicate than before. From the gradient bound for v˜k and Lemma 11.12, we conclude that (0)

[aij vk )]σ,S ≤ C(δ) 0 (x0 , D˜

for any open subset S of Γ(x0 , (1 − δ/2)Rk ). By invoking the interior Schauder estimate, Proposition 2.13, we conclude that |D2 v˜k (x)| ≤ C(δ)Rk−σ (xn )σ−1

and

|D2 v˜k | ≤ C(δ)Rk−1−σ sup |vk |(xn )σ−1

for all x ∈ Γ(x0 , (1 − δ)Rk ). We now write

aij ˜k = bik Di wk + f0 , 0 (x, Du0 )Dij v

with

0

bk (x) =

0

wk = v˜k − u0 ,

ij vk (x)) 0 (x))−a0 (x,D˜

( aij (x,Du

|Dwk (x)|

f0 =

[aij vk ) 0 (x, D˜

Then

−

Dwk (x) Dij v˜k (x) |Dw k (x)|

if Dwk (x) 6= 0,

if Dwk (x) = 0,

aij vk )]Dij v˜k . 0 (x0 , D˜

|b| ≤ C(δ)Rk−σ (xn )σ−1

and We now set

|f0 | ≤ C(δ)θ(δ1 )Rk−1−σ sup |vk |(xn )σ−1 . Mk = Cδ ζ [ sup Γ(x0 ,Rk )

|uk | + (A2 + µ1 )Rk1+α ]

and Bk = C(δ)θ(δ1 )Rk−1−σ sup |vk | + A2 .

It follows that there is a vector βk such satisfying βkn = 1 and |β 0 | ≤ µ0 such that |aij Dij wk + bik Di wk | ≤ λBk (xn )σ−1 in Γ(x0 , (1 − δ)Rk ), |wk | ≤ Mk on Γ∗ (x0 , (1 − δ)Rk ),

|β · Dwk | ≤ µ1 Rkσ on Γ0 (x0 , (1 − δ)Rk ).

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A simple calculation shows that hk , defined by   Z Rk [C(δ) + 1]sσ σ hk (x) = [µ1 + Bk ]Rk exp ds + Mk , σRkσ xn satisfies aij Dij hk + bik Di hk ≤ −λBk (xn )σ−1 in Γ(x0 , (1 − δ)Rk ), hk ≥ Mk on Γ∗ (x0 , (1 − δ)Rk ),

β · Dhk ≤ −µ1 Rkσ on Γ0 (x0 , (1 − δ)Rk ), and then the maximum principle implies that ±wk ≤ hk in Γ(x0 , (1−δ)Rk ). Since     Z Rk Z Rk [C(δ) + 1]sσ [C(δ) + 1]sσ exp exp ds ≤ ds σRkσ σRkσ 0 xn   C(δ) + 1 ≤ Rk exp σ and uk − vk = wk , it follows that |uk − vk | ≤ C(δ)Rk1+α + (C(δ)θ(δ1 ) + Cδ ζ )

sup Γ(x0 ,Rk )

|uk |

in Γ(x0 , Rk ). It follows that sup Γ(x0 ,Rk+1 )

|uk − Lk | ≤ C(δ)Rk1+α + (C(δ)θ(δ1 ) + Cδ ζ + Cτ 1+α )

sup Γ(x0 ,Rk )

|uk |.

Just as in the proof of Theorem 4.5, we can choose first τ , then δ, and finally δ1 in a suitable way and then conclude that, for each s ∈ (0, 21 d∗ (x0 )), there is a linear polynomial P (x0 , s) such that sup |u − P (x0 , s)| ≤ C

Γ(x0 ,s)

s1+σ sup |u| + Cs1+α . r1+σ

A similar estimate is true if x0 ∈ G, so the desired estimate follows by virtue of Lemma 2.9.  We then infer a H¨ older gradient estimate near a C 1,α portion of ∂Ω. Proposition 11.18. Let ω ∈ H1+α (Bn−1 (x0 , R)) for some α ∈ (0, 1), x0 ∈ Rn with xn0 = ω(x00 ) and some R > 0, let [aij ] be a matrix-valued function defined on Ω[R] × Rn , and let b be a real-valued function defined on Σ[R] × Rn . Suppose that there are positive constants µ, λ, µ0 , µ1 , A1 ,

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K, and χ such that (11.44) holds for all (x, p) ∈ Ω[R] × Rn with |p| ≤ K and all ξ ∈ Rn , and ∂b(x, p) ∂b(x, p) (11.49a) ∂p ≤ µ0 ∂p · γ, ∂b(x, p) · γ ≥ χ, ∂p

(11.49b)

|b(x, p) − b(y, p)| ≤ µ1 χ|x − y|α

(11.49c)

for all (x, p) ∈ Ω[R] × Rn with |p| ≤ K and all y ∈ Ω[R]. Suppose also that there is a nonnegative, increasing function θ, defined on [0, 1/2] with θ(0) = 0 such that (11.46) for all x and y in Ω[R] with |x − y| ≤ 21 d∗ (x) and all p ∈ Rn with |p| ≤ K. Then is a constant α0 ∈ (0, 1), determined only by K, α, A1 , n, µ, and µ0 , and, for any A2 ≥ 0, there is a constant C, determined also by A2 , µ1 , and G0 = supx∈Σ[R] |b(x, 0)| such that any function u ∈ C 2 (Ω[R]) ∩ C 1 (Ω[R]) satisfying the conditions |Du| ≤ K in Ω[R] and |aij (x, Du)Dij u| ≤ A2 d∗ (x)−α−1 d(x)α−1 in Ω[R], b(x, Du) = 0 on Σ[R]

(11.50a) (11.50b)

obeys the estimate (0)

|Du|α0 ≤ C. Proof.

Note that the boundary condition can be rewritten as Du · γ = g(x, D0 u)

with D0 u = Du − (Du · γ)γ and the hypotheses (11.49) imply that |g(x, p0 ) − g(x, q 0 )| ≤ C(µ0 , n)|p0 − q 0 |, 0

0

|g(x, p ) − g(y, p )| ≤ C(µ0 , µ1 , n, α)|x − y|

(11.51a) α

(11.51b)

for all x and y in Σ[R] and all p0 and q 0 with |p0 |, |q 0 | ≤ K. Since all of the hypotheses of this proposition (with (11.51) in place of (−1−α) (11.49)) are invariant under an H2 change of variables, the conclusion follows from Theorem 11.17.  This local estimate leads to a global H¨older gradient estimate, which we state in a form for applications to existence results. Theorem 11.19. Let ∂Ω ∈ H1+α for some α ∈ (0, 1), let [aij ] be a matrixvalued function defined on Ω×Rn , and let b be a real-valued function defined on ∂Ω × Rn . Suppose that there are positive constants µ, λ, µ0 , µ1 , A1 ,

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K, and χ such that (11.44) holds for all (x, p) ∈ Ω × Rn with |p| ≤ K and all ξ ∈ Rn , and (11.49) holds for all (x, p) ∈ Ω × Rn with |p| ≤ K and all y ∈ Ω. Suppose also that there is a nonnegative, increasing function θ, defined on [0, 1/2] with θ(0) = 0 such that |aij (x, p) − aij (y, p)| ≤ θ(|x − y|) for all x and y in Ω with |x − y| ≤ 21 and all p ∈ Rn with |p| ≤ K. Then is a constant α0 ∈ (0, 1), determined only by K, α, A1 , n, µ, µ0 , and Ω, and, for any A2 ≥ 0, there is a constant C, determined also by A2 , µ1 , and G0 = supx∈∂Ω |b(x, 0)| such that any function u ∈ C 2 (Ω) ∩ C 1 (Ω) satisfying the conditions |Du| ≤ K in Ω and |aij (x, Du)Dij u| ≤ A2 d(x)α−1 in Ω, b(x, Du) = 0 on ∂Ω

(11.52a) (11.52b)

obeys the estimate |Du|α0 ≤ C. 11.7

A basic existence theorem for quasilinear elliptic equations with nonlinear boundary conditions

Our global H¨ older gradient estimate leads to a relatively simple but useful existence theorem. Theorem 11.20. Let ∂Ω ∈ H1+α for some α ∈ (0, 1), and let aij and a be H¨ older continuous functions defined on Ω × R × Rn × [0, 1] such that ij a is Lipschitz with respect to p. Let b be a H¨ older continuous function n defined on ∂Ω × R × R × [0, 1] such that b is Lipschitz with respect to p. Suppose [aij ] is positive definite and that b is oblique. Suppose finally that a(x, z, p; 0) = 0 for all (x, z, p) ∈ Ω × R × Rn and b(x, z, p; 0) = b(x, w, p; 0) for all (x, p) ∈ ∂Ω × Rn and all z and w in R with b(x, 0, 0; 0) = 0 for all x ∈ ∂Ω. If there is a constant M such that any function u ∈ C 2 (Ω) ∩ C 1 (Ω) satisfying aij (x, u, Du; s)Dij u + (s − 1)u + a(x, u, Du; s) = 0 in Ω, b(x, u, Du; s) = 0 on ∂Ω

(11.53a) (11.53b)

for some s ∈ [0, 1] obeys the estimate |u|1 ≤ M,

(11.54)

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then there is a solution u ∈ C 2 (Ω) ∩ C 1 (Ω) of

aij (x, u, Du; 1)Dij u + a(x, u, Du; 1) = 0 in Ω, b(x, u, Du; 1) = 0 on ∂Ω.

Moreover this solution is in

(−1−δ) H2+δ

(11.55a) (11.55b)

for some δ ∈ (0, 1).

Proof. With δ ∈ (0, α) to be chosen, we take B = H1+δ , and we define the operator T : B × [0, 1] → B by writing u = T (v, s) if u is the solution, given by Theorem 11.11, of the problem aij (x, v, Dv; s)Dij u + sv − u + a(x, v, Dv; s) = 0 in Ω, b(x, v, Du; s) = 0 on ∂Ω.

This solution is in H1+α0 for some α0 ∈ (0, 1), so T will be compact if δ < α0 . It is also easy to see that T (v, 0) = 0 for all v ∈ B and that u = T (u, s) for some s if and only if u satisfies (11.53). Finally, Theorem 11.19 gives a constant α00 such that |u|1+α00 ≤ C.

00

If we choose δ so that δ ≤ α and δ ∈ (0, α0 ), it follows from Theorem 11.8 that the operator T (·, 1) has a fixed point, which will be a solution of (11.55). The H¨ older gradient estimate just quoted shows that u ∈ H1+δ , and then the functions a ˜ij and a ˜, defined by a ˜ij (x) = aij (x, u, Du; 1),

a ˜(x) = a(x, u, Du; 1),

(11.56)

are in Hδ and hence the interior Schauder estimate implies that u ∈ (−1−δ) H2+δ .  As stated, our existence theorem has an important limitation which prevents us from applying it directly to the examples from the previous chapters. The gradient estimate from Chapters 9 and 10 always assume the solution of our boundary value problem to be in C 2 (Ω) but the existence theorem refers to solutions that are only assumed to be H1+δ for some δ < 1. The purpose of the next section is to prove that, under simple assumptions on the data, any H1+δ solution is actually in C 2 . 11.8

Second derivative H¨ older estimates

We now show that solutions of the oblique derivative problem have globally H¨ older continuous second derivatives if the data are smooth enough. By

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adapting the proof of the first part of the proof of Lemma 11.6, we can show that u ∈ H2+α if a ˜ij and a ˜ (defined by (11.56)) are H¨older continuous, b has H¨ older continuous first derivatives and ∂Ω ∈ H2+α , but we wish to obtain this regularity even when ∂Ω ∈ H1+α . Our proof is a variant of that for Lemma 4.39, and, in this way, we can make the quasilinear regularity theory closer to that of the linear theory. As before, we need only consider solutions of linear differential equations with nonlinear boundary conditions. Proposition 11.21. Let ω ∈ H1+α (Bn−1 (x00 , R)) for some α ∈ (0, 1), (0) some x0 ∈ Rn with xn0 = ω(x00 ) and some R ∈ (0, 1). Let aij ∈ Hα (Ω[R]), (2) (0) let a ∈ Hα (Ω[R]), and b ∈ H1+α (Σ[R] × B(0, K)) for some positive constant K. Suppose also that aij has its eigenvalues in the interval [λ, µλ] for some positive constants λ and µ, and b is uniformly oblique on ∂Ω. Specifically, suppose that there are nonnegative constants χ > 0, A1 , and B1 such that (2) |aij |(0) α + |a|α ≤ A1 λ, ∂b (x, p) · γ ≥ χ, ∂p

(11.57a) (11.57b)

|Db(x, p) − Db(y, q)| ≤ B1 χ[|p − q|α + d∗ (x)−1−α |x − y|α ] for all (x, p) and (y, q) in Σ[R] × R with |x − y| ≤ Here Db denotes the vector   ∂b ∂b , . ∂x ∂p

1 ∗ 2 d (x)

(11.57c)

and |p|, |q| ≤ K.

Then, there is a constant C, determined only by α, A1 , B1 , µ, n, and |ω|1+α such that (0)

|u|2+α ≤ C

(11.58)

for any solution u ∈ H1 (Ω[R]) ∩ C 2 (Ω[R]) of

aij Dij u + a = 0 in Ω[R],

(11.59a)

b(x, Du) = 0 on Σ[R]

(11.59b)

with |Du| ≤ K in Ω[R]. Proof.

From Proposition 11.18, there is a constant α0 such that (0)

|u|1+α0 ≤ C. (0)

We then take δ = min{α, α0 }, and suppose that u ∈ H2+δ .

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We then fix y0 ∈ Σ[R] and set R0 = θd∗ (y0 ) with θ ∈ (0, 1) to be further specified. We now set E = Ω[y0 , R0 ], E 0 = Σ[y0 , R0 ] and E ∗ = σ[y0 , R0 ], and we note that (0)

(2)

|aij |δ;E + |a|δ;E ≤ CA1 λ.

Moreover, we have β · Du = g on E 0 with β = bp (y0 , Du(y0 )) and g(x) = β · Du − b(x, Du).

A quick calculation shows that

Di g(x) − Di g(y) = [β k − bk (x, Du(x))][Dik u(x) − Dik u(y)] + [bk (y, Du(y)) − bk (x, Du(x))]Dik u

and hence

+ [bi (x, Du(x)) − bi (y, Du(y))],

(1/2)

(−1/2)

|g|1+δ;E 0 ≤ C + Cθδ |u|2+δ;E .

It follows from Lemma 4.39 that (−1/2)

(−1/2)

|u|2+δ:E ≤ C + Cθδ |u|2+δ;E ,

and hence, if Cθδ ≤ 1, we infer that

(−1/2)

|u|2+δ;E ≤ C.

Since this estimate is valid for any y0 ∈ Σ[R] and a corresponding inequality holds for any ball in Ω[R], we have (0)

|u|2+δ,Ω[R] ≤ C.

(11.60)

To see that, in fact, u ∈ H2+δ , we first fix y ∈ Σ[R] and imitate the proof of Lemma 11.7 to conclude that there are a point y ∗ ∈ Rn and a positive number r such that |y − y ∗ | < r, d∗ (y) < R, and the problem aij Dij v = a in Ω ∩ B(y ∗ , r),

b(x, Dv) = 0 on ∂Ω ∩ B(y ∗ , r), v = u on Ω ∩ ∂B(y ∗ , r)

(−1−α0 )

has a solution in H2+δ (Ω ∩ B(y ∗ , r)). Here the weighted norms are with respect to distance to ∂B(y ∗ , r), and we have used the preceding ar(−1−α0 ) gument to estimate |v|2+δ . By the maximum principle v = u and hence (0)

u ∈ H2+δ (Ω[R0 ]) for any R0 ∈ (0, R). Since the constant C in (11.60) with R0 replaced by R is independent of R0 , it follows that (11.60) holds. The proof is completing by noting that, by interpolation, we may assume that α0 = 1. 

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Existence theorems for our examples

In this section, we collect existence theorems for our examples from previous chapters. p For our conormal problems, we recall the abbreviation v = 1 + |p|2 Our first example is the capillary ! problem: div

Du p 1 + |Du|2

+ B(x, u, Du) = 0 in Ω,

(11.61a)

Du · γ p + ψ(x, u) = 0 on ∂Ω. (11.61b) 1 + |Du|2 We prove the existence of solutions for this problem under suitable, standard hypotheses on B, ψ, and Ω. Theorem 11.22. Suppose that lim sup

|z|→∞ (x,p)∈(Ω×Rn )

(sgn z)B(x, z, p) = −∞

(11.62a)

and that there are functions B ∗ and B0 , along with an increasing nonnegative function θ1 such that B = B ∗ + B0 and |B ∗ (x, z, p)| + v|B0 (x, z, p)| ≤ θ1 (|z|),

v(|Bp∗ (x, z, p)| + vBz∗ (x, z, p) ≤ θ12 (|z|),

|p · Bp∗ (x, z, p)|) ≤ θ1 (|z|), |Bx∗ (x, z, p)| ≤ θ12 (|z|).

(11.62b) (11.62c)

(11.62d) Suppose also that there are a constant c0 ∈ (0, 1) and an increasing, nonnegative function ψ0 with ψ0 (t) < 1 for all t > 0, such that sgn zψ(x, z) ≤ c0 , (11.63a) |ψ(x, z)| ≤ ψ0 (|z|),

(11.63b)

ψz (x, z) ≤ 0 (11.63c) for all (x, z) ∈ ∂Ω × R. In addition, suppose that there is a constant α ∈ (0, 1) such that B ∈ Hα (K) for any bounded subset K of Ω × R × Rn , ψ ∈ H1+α (K 0 ) for any bounded subset K 0 of ∂Ω×R, and ∂Ω ∈ H2+α . Then there is a solution of (11.61) in C 0 (Ω) ∩ C 2 (Ω). Moreover u ∈ H2+α0 (Ω) for any α0 ∈ (0, α). Proof.

In this case, we define the coefficients in (11.53) by setting δ ij − ν i ν j aij (x, z, p; s) = , v a(x, z, p; s) = sB(x, z, p), p·γ b(x, z, p; s) = + sψ(x, z), v

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where ν = p/v, so conditions (11.62) and (11.63) are satisfied with B ∗ (x, z, p; s) = (s − 1)z + sB ∗ (x, z, p), B0 (x, z, p; s) = sB0 (x, z, p). In particular, the limit behavior in (11.62a) is independent of s. Hence we may assume for simplicity that s = 1 in obtaining our estimates. We first observe that conditions (8.21) are satisfied with a1 = 1, b0 = 0, R = 1, g ≡ 1, G1 (t) = t, and any M ≥ 1, so Z sup |u| ≤ C[ |u| dx + M ]. Ω

We now choose q ≥ 1 so that c0 ≤ q/(q + 1) and, with K0 from Lemma 5.17, we set b2 = 4(q + 1 + K0 ).

Then there is a positive constant M (determined by the limit behavior in (11.62a)) such that conditions (8.25) are satisfied with a1 = 1, b0 = 0, and ε = 1/2. In this way we obtain from Lemma 8.6 that Z |u| dx ≤ 21/q qM |Ω|. Ω

In conjunction with our previous estimate, this inequality implies a uniform estimate on sup |u|, and the combination of Lemmata 10.4, 10.5, and 10.6 provides a uniform estimate on sup |Du|. Existence of a solution of (11.61) then follows from Theorem 11.20 and Proposition 11.21. An additional application of Proposition 11.21 gives the desired regularity.  Our next example is a uniformly elliptic conormal problem, that is, a problem of the form (10.1): div A(x, u, Du) + B(x, u, Du) = 0 in Ω,

(11.64a)

A(x, u, Du) · γ + ψ(x, u) = 0 on ∂Ω,

(11.64b)

with A, B and ψ satisfying suitable conditions, which we define in term of a positive C 1 (0, ∞) function Ψ, satisfying the inequality Ψ(s) ≤ sΨ0 (s) ≤ θ0 Ψ(s) (11.65) θ0 for some positive constant θ0 and all s > 0, and an increasing function g, satisfying g(t) ≤ Ψ(t),

G(t) ≥ δtg(t)

for all sufficiently large t, where G is defined by Z s G(s) = Ψ(t) dt. 0

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(Note that an easy example of g is g = Ψ, but g = Ψµ for µ ∈ (δ, 1) is also acceptable.) Theorem 11.23. Suppose that there are nonnegative constants a ˆ1 , cˆ0 , and M0 such that p · A(x, z, p) ≥ vΨ(v) − a ˆ1 G(|z|)

(11.66)

zψ(x, z) ≤ cˆ0 G(|z|)

(11.67)

n

for all (x, z, p) ∈ Ω × R × R with |z| ≥ M0 and

for all (x, z) ∈ Ω×R with |z| ≥ M0 . Suppose also that there is a nonnegative constant b0 such that, for any positive constant ˆb2 , there is a constant M1 with zB(x, z, p) ≤ b0 p · A(x, z, p) − ˆb2 G(|z|) (11.68) for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M1 . Suppose further that there are increasing functions θ2 , θ3 , θ4 , ψ0 , and ψ1 such that |A(x, z, p)| ≤ θ2 (|z|)Ψ(v),

(11.69a)

i

Ψ(v) 2 ∂A Ψ(v) 2 |ξ| ≤ (x, z, p)ξi ξj ≤ θ3 (|z|) |ξ| , v ∂pj v

(11.69b)

v|Az (x, z, p)| + |Ax (x, z, p)| + |B(x, z, p)| ≤ θ4 (|z|)vΨ(v)

(11.69c)

for all (x, z, p) ∈ Ω × R × Rn . Suppose finally that there is a constant α ∈ (0, 1) such that B ∈ Hα (K) for any bounded subset K of Ω × R × Rn , ∂Ω ∈ H2+α , and ψ ∈ H1+α (K 0 ) for any bounded subset K 0 of ∂Ω×R. Then there is a solution u ∈ C 0 (Ω) ∩ C 2 (Ω) of (11.64). In fact , u ∈ H2+α0 (Ω) for any α0 ∈ (0, α). This time, we define the coefficients in (11.53) by   Ψ0 (v) ∂Ai (x, z, p) + (1 − s) Ψ(v)δ ij + pi pj , aij (x, z, p; s) = s ∂pj v

Proof.

a(x, z, p; s) = s[Aiz (x, z, p)pi + Aii (x, z, p) + B(x, z, p)],

b(x, z, p; s) = s[A(x, z, p) · γ + ψ(x, z)] + (1 − s)Ψ(v)p · γ.

For our maximum estimate on u, we invoke Lemmata 8.5 and 8.6 with g replaced by the function gs , defined by gs (t) = sg(t) + (1 − s),

θ = 0, and the following prescription for a1 , c0 , b1 , b2 , R, and M . First, we choose R ≥ 1 so that 1 cˆ0 ≤ R−1/δ . 2

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Since G(t) G(τ ) ≤ 1/δ 1/δ t τ for any τ ≤ t, we have (8.21) satisfied with a1 ≥ a ˆ1 R1/δ /δ, b1 = 0, and c0 = 1/2 provided M ≥ M0 . Finally, set q = max{1, b0 − 1} and choose b2 so that 1 1 a1 max{q, q + 1 − b0 } + (q + 1 + K0 )R ≤ b2 . 2 2 Then there is a constant ˆb2 such that ˆb2 G(|z|) ≥ b2 G(|z|/R); with M2 the constant corresponding to this ˆb2 , we set M = max{M1 , M2 , R}, and then infer a maximum estimate for |u|. (Note that, although the function gs depends on s, the constants in the estimates are independent of s.) Then Lemma 10.14, Theorem 8.8, Lemma 10.15, Lemma 10.16 provide the gradient bound, and the proof is completed as before.  Our third and final conormal problem is for the false mean curvature equation: div(exp(θv 2 )Du) + B(x, u, Du) = 0 in Ω, (11.70a) exp(θv 2 )Du · γ + ψ(x, u) = 0 on ∂Ω (11.70b) 2 2 for some θ > 0 with B = O(|p| exp(θ|p| )). We also require suitable conditions on B and ψ to derive a bound on u. Theorem 11.24. Let θ > 0 and suppose there are nonnegative constants b0 , c0 < 1, and M0 , along with an increasing function b2 such that zB(x, z, p) ≤ b0 exp(θv 2 )|p|2 − b2 (|z|) exp(θz 2 ) (11.71a)

for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M0 , c0 zψ(x, z) ≤ 2 2 exp(θz 2 ) 4θ z for all (x, z) ∈ ∂Ω × R with |z| ≥ M0 . Suppose also that lim b2 (s) = ∞. s→∞

(11.71b) (11.72)

Suppose further that there is an increasing function θ1 such that |B(x, z, p)| ≤ θ1 (|z|) exp(θv 2 )v 2 (11.73) n for all (x, z, p) ∈ Ω × R × R . If there is a constant α ∈ (0, 1) such that B ∈ Hα (K) for any bounded subset K of Ω × R × Rn , ∂Ω ∈ H2+α , and ψ ∈ H1+α (K 0 ) for any bounded subset K 0 of ∂Ω × R. Then there is a solution u ∈ C 0 (Ω) ∩ C 2 (Ω) of (11.70). In fact, u ∈ H2+α0 (Ω) for any α0 ∈ (0, α).

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Proof.

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Now Lemmata 8.5 with a1 = 0,

b1 = 0,

R = 1, M = M0 , 1 g(s) = s exp(θs2 ), G1 (s) = 2 2 exp(θs2 ), 4θ s R provides a bound on u in terms of G(|u|) dx, and 8.6 (with the same a1 , g, and G1 ) bounds this integral if we take q so large that c0 ≤ q/ max{q, q + 1 − b0 } and M so large that

1 c0 (q + 1 + K0 ) ≤ ˆb2 (M ). 2 ˆ (Of course, we also take b2 = b2 (M ).) The proof is completed as before, taking into account the gradient bound estimate at the end of Section 10.4.  We also observe that, by taking g to be a power function (or, more generally, a function satisfying G(s) ≥ δsg(s) for some δ > 0), we have a similar result. For the reader’s convenience, we state the theorem here. Theorem 11.25. Let θ > 0 and suppose there are nonnegative constants b0 , c0 , m, and M0 , along with an increasing function b2 such that zB(x, z, p) ≤ b0 exp(θv 2 )|p|2 − b2 (|z|)|z|m

(11.74a)

for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M0 , zψ(x, z) ≤ c0 |z|m

(11.74b)

for all (x, z) ∈ ∂Ω × R with |z| ≥ M0 . Suppose also that (11.72) holds and that there is an increasing function θ1 such that (11.73) holds. If there is a constant α ∈ (0, 1) such that B ∈ Hα (K) for any bounded subset K of Ω × R × Rn , ∂Ω ∈ H2+α , and ψ ∈ H1+α (K 0 ) for any bounded subset K 0 of ∂Ω × R. Then there is a solution u ∈ C 0 (Ω) ∩ C 2 (Ω) of (11.70). In fact, u ∈ H2+α0 (Ω) for any α0 ∈ (0, α). For problems which are not of conormal form, our existence theorems are still straightforward. For capillary-type problems, we recall that g ij = δ ij − ν i ν j , where p 2 ν = Du/ 1 + |Du| , and our boundary value problem can be written as g ij Dij u + a(x, u, Du) = 0 in Ω,

(11.75a)

b(x, u, Du) = 0 on ∂ω.

(11.75b)

Our existence result then takes the following form.

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Theorem 11.26. Suppose that ∂Ω ∈ C 3 and that there are positive constants µ1 and M1 such that µ1 ρ2 + (sgn z)a(x, z, p) < 0

(11.76a)

for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M1 and |p| ≤ µ1 , b(x, z, −(sgn z)µ1 γ) < 0

(11.76b)

for all (x, z, p) ∈ ∂Ω × R × Rn with |z| ≥ M1 and |p| ≤ µ1 . Suppose also that δa, δT a, (δ¯ − 1)a ≤ O(1),

(11.77a)

|ap | = O(1),

(11.77b)

|bp | = O(bp · γ), ¯ ≤ o(bp · γ), δb

(11.77d)

δT b ≤ O(bp · γ),

(11.77c) (11.77e)

where the limit behavior is uniform on bounded subsets of Ω × R. If there is a constant α ∈ (0, 1) such that b ∈ H1+α (K 0 ) for any bounded subset K 0 of ∂Ω × R × Rn , then there is a solution u ∈ C 0 (Ω) ∩ C 2 (Ω) of (11.75). In fact, u ∈ H2+α0 for any α0 ∈ (0, α). Proof. We first note that conditions (11.76a) are just the hypotheses of Lemma 8.3, so we have a bound on u. A quick calculation shows that (9.57) is satisfied with Λ ≡ 1 and µ1 = 1. In addition, E = |p|2 /v 2 , so δT E ≤ O(v −3 ), ¯ ij |, δE ¯ = O(v −2 ), |δg

|δT g ij |, |δN g ij | = O(v −1 ). Hence the hypotheses of Theorem 9.12 and Lemma 9.13 are satisfied. The existence and regularity results then follow from Theorem 11.20 and Proposition 11.21.  For uniformly elliptic problems, we have the most general class of possible boundary conditions. Theorem 11.27. Suppose that ∂Ω ∈ C 3 and that there are positive constants µ1 and M1 such that µ1 ρ2 Λ(x, z, p) + (sgn z)a(x, z, p) < 0

(11.78)

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for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M1 and |p| ≤ µ1 while (11.76b) holds for all (x, z, p) ∈ ∂Ω × R × Rn with |z| ≥ M1 and |p| ≤ µ1 . Suppose also that Λ = O(λ), |a| = O(λ|p|2 ), (11.79a) ij ij |p||aij p | + |az | + |ax | = O(λ),

(11.79b)

2

|p||ap | + |az | + |ax | = O(λ|p| ), (11.79c) with the limit behavior uniform on bounded subsets of Ω × R as |p| → ∞, and |p||bp | + |bz | + |bx | = O(|p|bp · γ) (11.80) with the limit behavior uniform on bounded subsets of ∂Ω × R as |p| → ∞. If there is a constant α ∈ (0, 1) such that b ∈ H1+α (K 0 ) for any bounded subset K 0 of ∂Ω × R × Rn , then there is a solution u ∈ C 0 (Ω) ∩ C 2 (Ω) of (11.55). In fact, u ∈ H2+α0 for any α0 ∈ (0, α). Proof. Now our bounds follow from Lemmata 8.3, 8.4 and Theorems 9.1 and 9.10.  Finally, we write the false mean curvature equation in the form ∆u + θDi uDj uDij u + a(x, u, Du) = 0 for some positive constant θ. Theorem 11.28. Suppose that ∂Ω ∈ C 3 and that there are positive constants µ1 and M1 such that (11.78) for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M1 and |p| ≤ µ1 while (11.76b) holds for all (x, z, p) ∈ ∂Ω × R × Rn with |z| ≥ M1 and |p| ≤ µ1 . Suppose also that |a| = o(|p|3 ), (11.81a)

and

|p||ap | + |δT a| = O(|p|4 ), δa, (δ¯ − 1)a ≤ O(|p|4 )

(11.81b)

(11.81c)

¯ ≤ o(|p|bp · γ). δT b, δb (11.82) 0 If there is a constant α ∈ (0, 1) such that b ∈ H1+α (K ) for any bounded subset K 0 of ∂Ω × R × Rn , then there is a solution u ∈ C 0 (Ω) ∩ C 2 (Ω) of ∆u + θDi uDj uDij u + a(x, u, Du) = 0 in Ω, (11.83a) In fact, u ∈ H2+α0

b(x, u, Du) = 0 on ∂Ω. for any α0 ∈ (0, α).

(11.83b)

Proof. As before, conditions (11.78) and (11.76b) provide a bound on u. ij Then conditions (9.46) (9.47), and (9.48) are satisfied with aij ∗ = δ and τ ≡ θ, so the gradient bound follows from Theorem 9.8. 

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Notes The results in Section 11.1 are due to Ladyzhenskaya and Ural0 tseva (see Chapter X of [91]). Theorem 11.3 and the discussion in Section 11.2 come from [119]. The form of Theorem 11.3 given here is a slight restatement of the version of Theorem 1 in [119], which is based on [96]. In turn, [96] is based on ideas of Kirk and Caristi (see [28, 80]) and a result from Miranda [136]. We refer to the notes for Chapter XII of [114] for a more detailed discussion of this situation, but we point out here that the application of the Gˆateaux variation is taken directly from [80]. We also point out that our self-contained proof is considerably shorter (and more elementary) than that of its earlier form, Lemma 8.5 of [114]. Furthermore, the nonlinear method of continuity from Section 17.2 of [64] can be used in place of Theorem 11.3 everywhere in this book via appropriate modification of the various proofs. The approach to the H¨older gradient estimate given here is a more careful working of the method described in Sections XIII.5 and XIII.6 of [114]. As mentioned in the notes to Chapter XIII of [114], a simpler form of the H¨ older gradient estimate for solutions of equations in divergence form, but with general oblique boundary conditions, was first given as Theorem 5.2 of [54]. This estimate was extended to nonlinear oblique derivative problems with equations in general form (but with more regularity assumed on the coefficients aij , a, and b) in Lemma 2.4 of [98]. A variant of the method used in these papers was presented in Theorem 4.1 of [120], but the current method is based on ideas in [111]. Alternative approaches (which also influenced [111]) are given in Theorem 4.5 of [38] and Theorem 2 of [187]. The existence results in Section 11.9 come from several sources. Existence results for the capillary problem have been proved many times with various methods under varying hypotheses; we mention Chapter 7 of [46] as a more complete bibliographic reference for this topic, but list some of the more relevant theorems here. In [183], Ural0 tseva proved the existence of a solution for the problem div A(Du) = ψ(x, u) in Ω,

A(Du) · γ = 0 on ∂Ω

if A behaves (in a suitable way) like ν, if ψz has a positive lower bound, and if Ω is convex. She allowed the boundary condition to have the form A(Du) · γ = cos θ (for some constant θ ∈ (0, π)) in [184], and, in [185], she considered nonconstant θ and she removed the convexity hypothesis on ∂Ω.

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As previously mentioned, Gerhardt [61] reproduced (via slightly different methods) the results in [185] although he only considered explicitly the case that A = ν; in addition, he studied the obstacle problem. Theorem 11.22 first appears (although in slightly different form) as a brief remark at the end of [97]; see also Theorem 7.6 in [106]. Existence of a solution for the capillary problem under slightly different hypotheses was proved by Spruck [174] in two dimensions and by Simon and Spruck [171] in higher dimensions; these papers allow the contact angle θ to be zero or π on some portions of the boundary, in which case the solution will have an unbounded gradient. Existence of a solution for uniformly elliptic conormal problems is discussed briefly in Section 10.2 of [91] under more restrictive hypotheses than those of Theorem 11.23 although the method of proof there can be adapted to prove our theorem. A more complete version of Theorem 11.23 is given in [97] (for variational problems) and in [106] (for non-variational problems) although the exact statement here is new; it requires the H¨older estimate Theorem 8.8, which wasn’t available at the time of [97, 106]. An existence theorem for the conormal problem with exponential growth first appears in [106] (see p. 62 of that work). Problems for nondivergence problems (or, as a special case, problems with the equation in divergence form but with boundary conditions that are not conormal) were first discussed by Fiorenza ([53–55]) but he did not provide the estimates necessary to prove an existence theorem. Theorem 11.26 seems not to have been stated before, but it is an easy corollary of the gradient estimate in [107]. A restricted version of Theorem 11.27 first appears in Section 5 of [98], and the exact form given here was first proved as Theorem 7.6 in [120]. Theorem 11.28 is new, but it is an easy corollary of the gradient estimate in [101]. We also mention that a number of existence and regularity results are available under hypotheses not covered by the methods in this chapter. In particular, the method used for proving H¨older gradient estimates for discontinuous β in Sections 4.5 and 4.6 has a nonlinear variant, described in [108] for equations in divergence form (focused on the capillary problem). Estimates for nonlinear problems in corners in other types of function spaces have also been extensively studied. We mention here [94] which provides a complete description of the regularity of solutions of the capillary problem near a corner, depending on the data of the problem near that corner. In addition, by using the refined gradient bounds from [113], it was shown (see page 62 of [113]) that Lipschitz solutions exist for conormal

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problems in bounded Lipschitz domains satisfying a uniform exterior sphere condition. A variant of this result was proved using different techniques in [30] for convex domains and uniformly elliptic conormal problems; the most important improvement here is that some of the assumptions on the vector field A are relaxed. In addition, [30] proves some gradient bounds for a subclass of C 1 domains. When the boundary condition is linear, then a H¨older gradient estimate was proved, even for fully nonlinear differential equations, in [117]. The proof is essentially that given for linear boundary conditions with linear differential equations in Section 4.1; it would of great interest to prove a corresponding result for nonlinear boundary conditions, but I know of no such results other than Theorem 1.1 from [108] (and its precursor Theorem 1 from [105]), which is only valid when n = 2.

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Chapter 12

Oblique Derivative Problems for Fully Nonlinear Elliptic Equations

Introduction We now examine the oblique derivative problem for elliptic equations which are fully nonlinear, that is, equations of the form F (x, u, Du, D2 u) = 0

(12.1)

for some function F defined on Γ = Ω × R × Rn × Sn , where Sn denotes the set of all real-valued symmetric matrices. We write (x, z, p, r) for a typical point in Γ and we define the operator F by F [u](x) = F (x, u, Du, D2 u).

(12.2)

We say that the function F is elliptic at a point (x, z, p, r) ∈ Γ if F (x, z, p, r + η) > F (x, z, p, r)

for any positive definite matrix η, and we say that the operator F is elliptic at a function u if the function F is elliptic at (x, u(x), Du(x), D2 u(x)) for every x ∈ Ω. If, in addition there are positive constants Λ and λ such that λ tr η ≤ F (x, z, p, r + η) − F (x, z, p, r) ≤ Λ tr η

(12.3)

for all (x, z, p, r) in some subset Σ of Γ and all positive definite matrices η, we say that F is uniformly elliptic on Σ. In this chapter, we shall be concerned primarily with uniformly elliptic equations. A brief discussion of various nonuniformly elliptic equations will be given in the Notes to this chapter. Note that, if F is continuously differentiable with respect to r, then F is elliptic if and only if Fr is positive definite and F is uniformly elliptic if and only if the eigenvalues of Fr are between λ and Λ. For notational convenience, we write ∂F F ij = . ∂rij 457

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An important subclass of uniformly elliptic equations is given via the set of Bellman operators; we define a Bellman operator from a indexed family of linear operators (Lν )ν∈V and an indexed family of functions (fν )ν∈V for some index set V . In this case, we define F [u](x) = inf (Lν u(x) − fν (x)). ν∈V

Such an operator will be uniformly elliptic if we can write each operator in the form i Lν w = aij ν Dij w + aν Di w + aν w

and if there are positive constants λ and Λ such that 2 λ|ξ|2 ≤ aij ν ξi ξj ≤ Λ|ξ|

for all ν ∈ V and all ξ ∈ Rn . This class of operators includes as a special case the Pucci minimal operator mµ for µ ≥ 1, in which the indexed family of linear operators is the set of all linear operators of the form Lw = aij Dij u with [aij ] a constant matrix with eigenvalues in the interval [1, µ]. This operator is related to the operator Mµ defined in Chapter 3 via the equation Mµ (−u) = −mµ (u). We also write our boundary condition in the form b(x, u, Du) = 0,

(12.4)

and we define the operator N by N [u] = b(x, u, Du).

(12.5)

As before, the most general definition of obliqueness is that the function N ∗ , defined by N ∗ (τ ) = b(x, z, p + τ β) is increasing for some oblique vector field β. For the most part, we shall assume here that N is C 1 with respect to p and that ∂Ω ∈ C 1 , so that obliqueness is equivalent to the condition ∂N (x, z, p) · γ > 0. ∂p

(12.6)

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459

Maximum estimates, comparison principles, and a uniqueness theorem

It is easy to extend the maximum estimates from Chapter 8 so that they apply to fully nonlinear problems, assuming that F is differentiable with respect to p. First, we write F [u] = aij (x, u, Du)Dij u + a(x, u, Du), with aij and a defined by ij

a (x, z, p) =

Z

1

F ij (x, z, p, tD2 u(x)) dt,

0

a(x, z, p) = F (x, z, p, 0). We also define E by

E(x, z, p, r) = F ij (x, z, p, r)pi pj .

From Lemma 8.1, we have the following result. Lemma 12.1. Let u ∈ C 1 (Ω) ∩ C 2 (Ω) be a solution of F [u] = 0 in Ω,

N [u] = 0 on ∂Ω,

(12.7)

with F and N given by (12.2) and (12.5). Suppose there are nonnegative constants µ1 , µ2 , M1 , and R such that diam Ω ≤ R and (sgn z)F (x, z, p, 0) µ1 |p| + µ2 ≤ E(x, z, p, r) |p|2

(12.8a)

for all (x, z, p, r) ∈ Ω× R× Rn × Sn with |z| ≥ M1 and 0 < |p| ≤ µ2 exp((1 + µ1 )R), (sgn z)b(x, z, p) < 0

(12.8b)

n

for all (x, z, p) ∈ ∂Ω × R × R with |z| ≥ M1 and |p| ≤ µ2 exp((1 + µ1 )R). Then µ2 sup |u| ≤ M1 + exp((1 + µ2 )R). 1 + µ1 Ω In a similar vein, Lemma 8.3 gives the following result if we write Λ(x, z, p, r) for the maximum eigenvalue of the matrix Fr (x, z, p, r). Lemma 12.2. Suppose ∂Ω ∈ C 2 . Suppose also that there are positive constant µ1 and M1 such that µ1 ρ2 Λ(x, z, p, r) + (sgn z)a(x, z, p) < 0

(12.9a)

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for all (x, z, p) ∈ Ω × R × Rn with |z| ≥ M1 and |p| ≤ µ1 , b(x, z, −(sgn z)µ1 γ) < 0 for all (x, z, p) ∈ ∂Ω × R × Rn with |z| ≥ M1 and |p| ≤ µ1 . Then

(12.9b)

sup |u| ≤ M1 + R0 µ1 . Ω

In a slightly different vein, we infer from Theorem 1.9 the following estimate for equations of Monge-Amp`ere type, which is essentially the same as Lemma 8.2. Lemma 12.3. Let u ∈ C 2 (Ω) ∩ C 1 (Ω) be a solution of (12.7) with F given by (12.64). Suppose that there are positive, measurable functions g ∈ Lnloc (Rn ) and h ∈ Ln (Ω) such that Z Z h(x)n dx < g(p)n dx (12.10) Rn

Ω

Suppose also that

−(sgn z)ψ(x, z, p) ≤

h(x) g(p)

(12.11a)

for all (x, z, p) ∈ Ω × R × Rn and that there is a continuous, increasing function M1 such that (sgn z)b(x, z, p) < 0

(12.11b)

n

for all (x, z, p) ∈ ∂Ω × R × R with |z| ≥ M1 (|p|). Then there is a number t > 0, determined onlyZ by h and g such Z that h(x)n dx =

Ω

Moreover,

g(p)n dx.

(12.12)

{|p|≤t}

sup |u| ≤ M1 (t) + t diam Ω.

(12.13)

It is also easy to prove a fully nonlinear comparison theorem, which we shall use to derive a uniqueness result. In fact, we present two slightly different comparison theorems, analogous to the two maximum principles, Lemmata 1.2 and 1.6. Theorem 12.4. Let ∂Ω ∈ C 1 , and let u and v be in C 2 (Ω) ∩ C 1 (Ω) with F [u] ≥ F [v] in Ω and N [u] ≥ N [v] on ∂Ω. Suppose that F is elliptic at tu + (1 − t)v for all t ∈ [0, 1] and decreasing in z for each (x, p, r) ∈ Ω × Rn × Sn . Suppose also that F is locally uniformly Lipschitz with respect to z, p, and r with |Fp |/λ locally bounded in Ω×R×Rn ×Sn . Suppose further that N is oblique at u and strictly decreasing in z for each (x, p) ∈ ∂Ω× Rn . Then u ≤ v in Ω.

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461

Proof. We argue as in Lemma 1.2. Let w = u − v, fix ε > 0 and x0 ∈ Ω, and set M0 = sup∂Ω w. Then there is a compact set K ⊂ Ω such that x0 is an interior point of K and w ≤ M0 + ε on ∂K. Since |Fp |/λ is locally uniformly bounded, there is a constant K1 such that |Fp | ≤ K1 λ in K and, with γ = 1 + K1 , we define the function ϕ by ϕ(x) = exp(−γx1 ). We now write Ω+ for the subset of K on which w > 0, and suppose that w + εϕ has a positive maximum at some x1 ∈ Ω+ . If x1 is an interior point of Ω+ , then we have Du + εDϕ = Dv and D2 u + εD2 ϕ ≥ D2 v at x1 , so F (x1 , u(x1 ), Du(x1 ) + εDϕ(x1 ), D2 u(x1 ) + εϕ(x1 )) ≤ F [v](x1 ).

Using the conditions on F , we find that F (x1 , u(x1 ), Du(x1 ) + εDϕ(x1 ), D2 u(x1 ) + εϕ(x1 )) ≥ F [u](x1 ) + ελ[∆ϕ(x1 ) − |Dϕ(x1 )], and direct computation gives ∆ϕ(x1 ) − |Dϕ(x1 )| > 0,

so F [u](x1 ) < F [v](x1 ). Hence w must attain its maximum over Ω+ at a boundary point. Now the proof of Lemma 1.2 implies that, if w has a positive maximum, then it must occur at some x2 ∈ ∂Ω. But then, Du(x2 ) = Dv(x2 ) + τ γ(x2 ) for some nonpositive constant τ , so N [u](x2 ) ≤ b(x2 , u(x2 ), Dv(x2 )) because N is oblique and b(x2 , u(x2 ), Dv(x2 )) < N [v](x2 ) because N is strictly decreasing. These inequalities contradict the assumption that N [u] ≥ N [v], and hence w can’t have a positive maximum so u ≤ v in Ω.  Theorem 12.5. Let Ω be a bounded Lipschitz domain, and let u and v be in C 2 (Ω) ∩ C 1 (Ω) with F [u] ≥ F [v] in Ω and N [u] ≥ N [v] on ∂Ω. Suppose that F is elliptic at tu + (1 − t)v for all t ∈ [0, 1] and decreasing in z for each (x, p, r) ∈ Ω × Rn × Sn . Suppose also that F is locally uniformly Lipschitz with respect to z, p, and r with |Fp |/|Fz | and |Fr |/|Fz | bounded in Ω × R × Rn × Sn and Fz < 0. Suppose further that N is oblique at u and decreasing in z for each (x, p) ∈ ∂Ω × Rn and that there is a constant ε ∈ (0, 1) such that the vector field ∂b(x, z, p)/∂p has modulus of obliqueness at least ε for any (z, p) ∈ R × Rn . Then u ≤ v in Ω.

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From Lemma 1.5, there is a function ρ such that ∂N (x, z, p) · Dρ(x) > 0 ∂p for all x ∈ ∂Ω. With δ > 0 at our disposal, we now set

Proof.

w = u − v − δ exp(−δρ), and we suppose that w has a positive maximum at x0 ∈ Ω. If x0 ∈ Ω, then, at x0 , we have

F [u](x0 ) ≤ F (x0 , v + δψ, Dv − δ 2 ψDρ, D2 v − δ 2 ψD2 ρ + δ 3 ψDρ ⊗ Dρ),

where we have written ψ for exp(−δρ) and Dρ⊗Dρ is the matrix [Di ρDj ρ]. The hypotheses on F imply that F (x0 , v + δψ, Dv − δ 2 ψDρ, D2 v − δ 2 ψD2 ρ + δ 3 ψDρ ⊗ Dρ) ≤ F [v](x0 ) + δψFz [x0 , Ξ](1 − C1 δ)

for Ξ some point on the line segment in R × Rn × Sn joining (v, Dv, D2 v) and (v + δψ, Dv − δ 2 ψDρ, D2 v − δ 2 ψD2 ρ + δ 3 ψDρ ⊗ Dρ) and C1 some positive constant. It follows that x0 can’t be in Ω if C1 δ < 1. Similarly, if x0 ∈ ∂Ω, then N [u](x0 ) ≤ b(x0 , v(x0 ), Du) ≤ N [v](x0 ) + β · D(u − v) for some vector field β with modulus of obliqueness at least ε, and hence N [u](x0 ) ≤ N [v](x0 ) − δψβ · Dρ < N [v](x0 ). (Here, we only use the inequality δ > 0.) Therefore w can’t have a positive maximum if C1 δ < 1. Sending δ → 0 completes the proof.  From the comparison principle, we infer an immediate uniqueness statement. Corollary 12.6. If F , N and Ω satisfy the hypotheses of Theorem 12.4 or Theorem 12.5, then the problem (12.7) has at most one solution in C 2 (Ω) ∩ C 1 (Ω). 12.2

Second derivative H¨ older estimates

Our proof of second derivative H¨older estimates is a combination of standard ideas which we identify more precisely in the Notes. The key to these estimates is an algebraic lemma due to Motzkin and Wasow [141]; we refer the reader to Lemma 17.13 of [64] (or Lemma 14.5 of [114]) for a proof.

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Lemma 12.7. Let λ ≤ Λ be positive constants and let n be a positive integer. Then there is a positive constant λ∗ , determined only by n, λ, and Λ, and a finite set {ξ 1 , . . . , ξ K } of unit vectors such that any matrix [Aij ] satisfying λ|ξ|2 ≤ Aij ξi ξj ≤ Λ|ξ|2 for all ξ ∈ Rn can be written as Aij =

K X

βk ξik ξjk

(12.14)

(12.15)

k=1 1 K for constants β1 , . . . , βk in [λ∗ , Λ]. Moreover, the set √ {ξ , . . . , ξ } can be taken to include the vectors e1 , . . . , en and (ei ± ej )/ 2 for all i 6= j.

From this lemma, we first obtain an estimate in a very special circumstance but under very weak assumptions about the regularity of the functions involved. Lemma 12.8. Let F : Sn → R satisfy λ tr η ≤ F (r + η) − F (r) ≤ Λ tr η

(12.16)

for some positive constants λ ≤ Λ and all positive definite matrices η. Suppose also that F is concave. Then there are constants C and α determined only by n and Λ/λ such that any solution u ∈ C 2 (B(R)) of F (D2 u) = 0 in B(R) satisfies the estimate  ρ α osc D2 u ≤ C osc D2 u (12.17) B(ρ) R B(R) for any ρ ∈ (0, R). Proof.

For h ∈ (0, R/4) and ξ ∈ Rn a unit vector, we define uh by

1 [u(x + hξ) + u(x − hξ)] − u(x). 2 Since F is concave, we have   1 1 F [u(x + hξ) + u(x − hξ)] ≥ [F (u(x + hξ)) + F (x − hξ)] = F u. 2 2 uh (x, ξ) =

It then follows from (12.16) that there is a matrix-valued function [aij ] (determined by F , h, and ξ) with eigenvalues in the interval [λ, Λ] such that aij Dij uh ≥ 0 in B(3R/4).

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Setting Hs,h = supB(sR/4) uh for s = 1, 2, we infer from the interior weak Harnack inequality Theorem 1.16 that there are positive constants C and κ such that !1/κ Z −n κ R (H2,h − uh ) dx ≤ C(H2,h − H1,h ). B(R/4)

We now divide this inequality by h2 /2, send h → 0, and define w(x, ξ) = Dij u(x)ξi ξj ,

Ms =

sup w B(sR/4)

to infer that R

−n

Z

κ

B(R/4)

!1/κ

(M2 − w) dx

≤ C(M2 − M1 ).

(12.18)

Now, for every x and y in B(R/2), there is a matrix [Aij ] satisfying (12.14) such that Aij [Dij u(x) − Dij u(y)] = F [u](x) − F [u](y) = 0,

and hence Lemma 12.7 gives unit vectors ξ 1 , . . . , ξ K , a constant λ∗ and functions β1 , . . . , βK with values in [λ∗ , Λ] such that K X

k=1

βk (x, y)[w(x, ξ k ) − w(y, ξ k )] = 0.

(12.19)

We further abbreviate

w(k) = w(·, ξ k ),

Ms(k) =

sup w(k) , B(sR/4)

m(k) s =

inf

B(sR/4)

w(k)

for s = 1, 2 and set ω(sR/4) =

K X

k=1

Ms(k) − m(k) s .

With j fixed, we now sum (12.18), with ξ replaced by ξk , for k 6= j. Since (k) (k) m2 ≤ m1 , we conclude that  1/κ Z X (k) R−n (M2 − w(k) )κ dx ≤ C(ω(R/2) − ω(R/4)). B(R/4) k6=j

From (12.19), we conclude that

βj (x, y)[w(j) (y) − w(j) (x)] =

X k6=j

βk [w(k) (x) − w(k) (y)],

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and hence (by taking the infimum over all x ∈ B(R/2)) X (k) (j) λ∗ (w(j) (y) − m2 ) ≤ Λ [M2 − w(k) (y)]. k6=j

We now raise this inequality to the power κ and sum on j; similarly, we sum (12.18) with ξ replaced by ξ (k) on k. Adding the resulting inequalities yields ω(R/2) ≤ C[ω(R/2) − ω(R/4)], so ω(R/4) ≤ δω(R/2) for some constant δ ∈ (0, 1). Iteration yields  ρ α ω(ρ) ≤ C ω(R) R

for constants C and α as in the statement of this lemma, and (12.17) follows from this estimate because ω(ρ) ≥ osc D2 u B(ρ)

and ω(R) ≤ K osc D2 u. B(R)



For our further analysis of problems with fully nonlinear equations, we now study the interior H¨ older estimate for second derivatives of solutions of (12.1) when F has a particular form. A key step is the following weighted estimate in a ball. Lemma 12.9. Assume that F is as in Lemma 12.8 and that F ∈ C 2 . Let η be a C 2 function defined on Rn with support in B = B(0, R0 ) for some R0 > 0 with 0 ≤ η ≤ 1 in Rn , let g ∈ C 2 (B), and define F˜ by F˜ (x, r) = ηF (r) + (1 − η)λ tr r − λg(x). If v ∈ C 4 (B) ∩ L∞ (B) satisfies F˜ (x, D2 v) = 0 in B, then there are positive constants θ < 1 and C, determined only by Λ/λ, n, and η, such that (0)

(2)

|u|2+θ ≤ C(|u|0 + |g|2 ).

(12.20)

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Proof. To simplify notation, we assume without loss of generality that λ = 1. If we differentiate the equation F˜ (x, D2 v) = 0 twice in the direction of any unit vector ξ and use the concavity of F , we conclude that h(x, ξ), defined by h(x, ξ) = Dij u(x)ξ i ξ j , satisfies the inequality F˜ ij Dij h(x, ξ) + 2ξ · DηAij Dijk vξ k + A ≥ 0, where

Aij = 2ξ · Dη[F ij (D2 v) − δ ij ]

∂2η i j ξ ξ − λDij gξ i ξ j . ∂xi ∂xj Hence, if we fix a point y ∈ B and set M2 = supB(y,d(y)/2) |D2 v| and G2 = supB(y,d(y)/2) |D2 g|, it follows that there are constants A0 and B0 , determined only by Λ/λ, n, and η such that F˜ ij Dij h(x, ξ) ≥ −λ[A0 |D3 v| + B0 M2 + G2 ] A = (F (D2 v) − λ∆v)

in B(y, d(y)/2). With ξ1 , . . . , ξK the vectors from Lemma 12.7, h(k) = h(·, ξk ), and K 1 X (k) 2 v˜ = (h ) , M2 k=1

it’s easy to check that

4

K X

k=1

and hence

|Dh(k) |2 ≥ |D3 u|2 ,

F˜ ij Dij v˜ ≥ −C(B0 + A20 )λM2 − CG2 .

For k = 1, . . . , K and ε ∈ (0, 1) a constant to be chosen, we now set w(k) = h(k) + ε˜ v , and conclude that F˜ ij Dij w(k) ≥ −C µ ¯M2 − CG2 , with

A20 . ε We now fix R ∈ (0, d(y)/4) and we further define µ ¯ = B0 +

(k)

Wi

(k)

Mi

=

sup

=

w(k) ,

sup

B(y,iR/4)

h(k) ,

B(y,iR/4)

ω(s) =

K X

k=1

(k)

mi

=

osc h(k)

B(y,s)

inf

B(y,iR/4)

h(k) ,

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for i = 1, 2, and s ∈ (0, R). Then the weak Harnack inequality implies that !1/κ Z (k)

R−n

B(y,R/4)

(W2

(k)

− w(k) )κ dx

(k)

≤ C[W2 −W1 + µ ¯M2 R2 +G2 R2 ).

Since (k)

W2

(k)

− w(k) ≥ M2

− h(k) − 2εω(R/2)

and (k)

W2

(k)

− W1

(k)

≤ M2

(k)

− M1

+ 2εω(R/2),

we conclude that, for any j ∈ {1, . . . , K},  1/κ Z X (k) R−n (M2 − h(k) )κ dx B(y,R/4) k6=j

≤ C[(1 + ε)ω(R/2) − ω(R/4) + µ ¯M2 R2 + G2 R2 ). (k)

To obtain a corresponding inequality for h(k) − m2 , we note that, for any x1 and x2 in B(y, R), there are a matrix [aij ] and a constant D0 such that λ|ξ|2 ≤ aij ξi ξj ≤ Λ|ξ|2 for any ξ ∈ Rn and aij [Dij u(x1 ) − Dij u(x2 )] ≤ M2 λD0 |x1 − x2 | + G1 |x1 − x2 |, where G1 = supB(y,d(y)/2) |Dg|. From Lemma 12.7 and the argument in the proof of Lemma 12.8, we infer that X (k) (j) h(j) (x) − m2 ≤ C[D0 M2 R + G1 R + (M2 − h(k) (x))], k6=j

and hence R

−n

Z

B(y,R/4)

(h

(k)

−

(k) m2 )κ

!1/κ

dx

≤ C[(1 + ε)ω(R/2) − ω(R/4) + M2 (¯ µR2 + D0 R) + G1 R + G2 R2 ]. Adding these inequalities then yields ω(R/2) ≤ C[(1 + ε)ω(R/2) − ω(R/4) + µ ¯M2 (R2 + D0 R) + G1 R + G2 R2 ].

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By rearranging this inequality and setting δ = 1 − 1/C, we see that ω(R/4) ≤ (δ + ε)ω(R/2) + M2 (D0 R + µ ¯R2 ) + G1 R + G2 R2 . We now fix ε so that δ + ε < 1 to conclude that there are positive constants θ, determined only by n and Λ/λ, and K, determined also by η such that  θ R ω(R) ≤ C(M2 + G1 R + G2 R2 ) d(y) for any R ∈ (0, d(y)). It follows that osc D2 v ≤ C(M2 + G1 R + G2 R2 )

B(y,R)

so (0)

(0)



R d(y)

θ

,

(2)

|u|2+θ ≤ C(|u|2 + |g|2 ). The desired estimate (12.20) now follows from this one via Lemma 2.3.  In fact, this second derivative H¨older estimate can be extended to study solutions of (12.1) assuming suitable conditions on the second derivatives of F with respect to all arguments, but we shall only use the simpler version described here. Our next step is to prove a corresponding existence result. Lemma 12.10. Suppose F satisfies the hypotheses of Lemma 12.9, and fix R0 > 0. Then, for any ϕ ∈ C(B(0, R0 )), there is a unique solution of F (D2 v) = 0 in B(0, R0 ),

v = ϕ on ∂B(0, R0 ).

(12.21)

Proof. Suppose first that ϕ ∈ Hδ for some δ ∈ (0, 1), fix a function η as in Lemma 12.9, and define F˜ by F˜ (x, r) = ηF (r) + (1 − η)λ tr r. (−δ)

(2−δ)

With θ ∈ (0, 1) arbitrary, we set B1 = H4+θ and B = H2+θ , and we define the map J : B1 → B by J(u)(x) = F˜ (x, D2 u(x)). Recalling the definition of the Gˆ ateaux variation Ju (ψ), from Section 11.2, it is easy to check that Ju (ψ) = F˜ ij (x, D2 u)Dij ψ, and hence, if U denotes the subset of all u ∈ B1 such that u = ϕ on ∂B, it follows that, for any u ∈ U , there are u1 ∈ U and ε ∈ (0, 1) such that kJu1 − (1 − ε)Juk ≤

1 εkJuk, 2

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which is just (11.6) with η = 1/2. By combining the linear theory with Lemma 12.9, we infer that there is a constant C0 (depending on R, θ, δ, and η) such that (−δ) (2−δ) kuk4+θ ≤ C0 (kF˜ (x, D2 u)k2+α + kϕkδ ) for all u ∈ U . It then follows from Theorem 11.3 that there is a solution (−δ) u ∈ H4+θ of F˜ (x, Du) = 0 in B, u = ϕ on ∂B, and the proof of Corollary 12.6 is easily modified to verify that this solution is unique. Now let (ηk ) be an increasing sequence of functions as above such that lim ηk (x) = 1 k→∞

for all x ∈ B, let (ϕk ) be a sequence of Hδ functions converging uniformly to ϕ, write F˜k for the function (denoted above by F˜ ) with ηk in place of η, and write uk for the solution of F˜k (x, Duk ) = 0 in B, uk = ϕk on ∂B that we have just proved to exist. Lemma 12.9 gives a locally uniformly second derivative H¨ older estimate for the sequence (uk ) while Theorem 1.26 and Corollary 1.29 give a uniform global modulus of continuity estimate. In addition, the maximum principle shows that the sequence (uk ) is uniformly bounded, so we can extract a uniformly convergent subsequence which, by virtue of the second derivative H¨older estimate, converges to a solution of the differential equation F (D2 u) = 0 in B.  We can also obtain H¨ older estimates for solutions of (12.1) under fairly weak conditions on the function F . As usual, we prove these estimates via our existence result. Theorem 12.11. Let B = B(x0 , R) for some x0 ∈ Rn and some R > 0, let F : B × Sn → R and suppose that there are positive constants λ ≤ Λ, σ, A1 , and A2 such that λ tr η ≤ F (x, r + η) − F (x, r) ≤ Λ tr η (12.22a) for all x ∈ B and all positive definite matrices η, and

|F (x, r) − F (y, r)| ≤ (A1 d(x)−2 + A2 |r|)λ|x − y|σ d(x)−σ (12.22b) 1 for all x and y in B with |x − y| ≤ 2 d(x). Suppose also that F is concave with respect to r. If σ < α, the constant from Lemma 12.8, then there is a constant C, determined by n, Λ/λ, σ, and A2 such that any solution u ∈ C 2 (B) ∩ L∞ (B) of F (x, D2 u) = 0 in B satisfies the estimate (0) (0) [u]2+σ ≤ C([u]2 + A1 ).

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Proof. Without loss of generality, we may assume that F is C 2 with respect to r. We then follow a familiar path. We fix a point y ∈ B, we set ρ0 = 21 d(y), and we take τ ∈ (0, 1) to be further determined. We then define ρk , for any positive integer k, to be τ −k ρ0 , and we abbreviate B(y, ρk ) to B(k). We take v˜k to be the solution of F (y, D2 v˜k ) = 0 in B(k), v˜k = u on ∂B(k) ˜ k to be the second order Taylor given by Lemma 12.10, and we write Q polynomial of v˜k centered at y. Next, we define uk and Qk inductively by setting u0 = u, Q0 = 0, k−1 X ˜k + Qk = Q Qj , j=0

uk+1 = uk − Qk . For notational convenience, we also set k−1 X vk = v˜k + Qj . j=0

We infer from Lemma 12.8 that sup |vk − Qk | ≤ osc v˜k ≤ Cτ 2+α sup v˜k ≤ Cτ 2+α sup |vk |. B(k+1)

B(k+1)

B(k)

B(k)

In addition, there is a matrix [aij ] with eigenvalues in the interval [λ, Λ] such that aij Dij (u − v˜k ) = F (y, D2 u) − F (y, D2 v˜k ) (which will depend on k but we suppress this dependence from the notation). Using the differential equations for u and v˜k along with (12.22b), we infer that |F (y, D2 u) − F (y, D2 v˜k )| ≤ ρσk A3 in B(k), where (0) A3 = ρ−2 0 (A1 + A2 [u]2 ). In addition, since uk − vk = u − v˜k , the maximum principle implies that |uk − vk | ≤ CA3 ρ2+σ . k Combining all these estimates gives sup |uk+1 | ≤ Cτ 2+α sup |uk | + Cρ2+α A3 . k B(k+1)

If we now set

B(k)

(0)

A4 = ρ−2 0 (A1 + (A2 + 1)[u]2 ), then our iteration scheme implies that, for every y ∈ B and every ρ ∈ (0, d(y)), there is a second degree polynomial P2 (y, ρ) such that sup |u − P2 (y, ρ)| ≤ Cρ2+σ A4 , B(y,ρ)

and the proof is completed by invoking Lemma 2.8.



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Second derivative H¨ older estimates for solutions of oblique derivative problems

We now show how to prove estimates for solutions of oblique derivative problems. Our first step is an estimate near the boundary with a simple differential equation and a constant coefficient linear boundary condition. Although this estimate is not, by itself, a H¨older estimate for the second derivatives of the solutions, it contains the important analytical aspect of the proof of our desired estimate. Lemma 12.12. Let ω ∈ H1+α (Bn−1 (0, R0 )) for some α ∈ (0, 1) and R0 ∈ (0, 1] with ω(0) = 0 and Dω(0) = 0, and, for s ∈ (0, R0 ], set Ω[s] = {x ∈ Rn : |x| < s, xn > ω(x0 )}, n

n

0

Σ[s] = {x ∈ R : |x| < s, x = ω(x )}.

(12.23a) (12.23b)

n

Suppose F : Ω[R0 ]×S → R is differentiable with respect to x and r, concave with respect to r and that there are positive constants λ ≤ Λ and A0 such that λ tr η ≤ F (x, r + η) − F (x, r) ≤ Λ tr η,

(12.24a)

for all positive definite matrices η and all (x, r) ∈ Ω[R0 ] × Sn , |F (x, r) − F (y, r)| ≤ A0 d(x)α−1 |x − y|

(12.24b)

for all x and y in Ω[R0 ] with |x − y| ≤ 21 d(x) and all r ∈ Sn . Let β be a unit vector such that 1−ε n |β 0 | ≤ β (12.25) ω0 for some positive constants ε < 1 and ω0 ≥ sup |Dω|. Then there are constants τ0 ∈ (0, 1), determined only by ε and ω0 , θ ∈ (0, 1), determined only by n, ε, Λ/λ, and ω0 and C, determined also by [Dω]α such that, if u ∈ H1 (Ω[R0 ]) ∩ C 2 (Ω[R0 ]) satisfies F (x, D2 u) = 0 in Ω[R0 ],

β · Du = 0 on Σ[R0 ]

(12.26)

and if α < θ, then, for any R ∈ (0, R0 /2), there is a quadratic polynomial QR such that osc |u−QR | ≤ C(τ 2+θ osc |u−Q|+A0 R2+α +

Ω[τ R]

Ω[R]

R2+α sup |β ·Du|) (12.27) R01+α Ω[R]

for any quadratic polynomial Q and any τ ∈ (0, τ0 ).

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Proof. Suppose first that β = en , and set H = Dnn u(0) and M = supΩ[R0 ] |Dn u|. Let ρ be a regularized distance from Lemma 4.9, and note from (4.24) with ε = 1/2, that we may assume that |Dρ| ≥ 1/2 in Ω[R0 ]. Moreover, the construction shows that Dρ(0) = en , and there is a constant A1 , determined only by n, [Dω]α , and α such that |D2 ρ| ≤ A1 ρα−1 . We now observe that   Z ρ 4A1 Λ α ρˆ = exp − dt t αλ 0

satisfies aij Dij ρˆ ≤ 0 and Cρ ≤ ρˆ ≤ ρ in Ω[R0 ] for any matrix [aij ] with eigenvalues in the interval [λ, Λ]. ˆ It then follows from the proof of Proposition 11.15, with ρˆ replacing d, that there is a constant θ > 0 such that Dn u Dn u(x) |x|θ α ρˆ(x) − H ≤ C[A0 |x| + Rθ sup ρˆ ] 0 Ω[3R0 /4]

for any x ∈ Ω[R0 /2] and that Dn u ≤ C(A0 R0α + M ). sup ρ ˆ R0 Ω[3R0 /4]

From this inequality, we immediately conclude that M H ≤ C(A0 R0α + ). R0 In addition, Dρˆ(0) = en , and hence

|Dn u(x) − H[xn − ω(x0 )]| ≤ C(A0 |x|1+α + M

(12.28)

|x|1+θ ) R01+θ

(12.29)

for all x ∈ Ω[R0 /2]. We next observe that there is a constant τ1 ∈ (0, 1/4) such that B((R/2)en , τ1 R) ⊂ Ω[R] for any R ∈ (0, R0 /2). In what follows, we always assume that τ ∈ (0, τ1 /2), or equivalently, τ0 ≤ τ1 /2. We then define x ¯1 = (R/2)en and we define the polynomial P by 1 P (x) = Du(¯ x1 ) · (x − x ¯1 ) + Dij u(¯ x1 )(x − x ¯1 )i (x − x ¯1 )j 2 and infer from Theorem 12.11 that osc

B(¯ x1 ,τ R)

|u − P | ≤ C[τ 2+θ osc |u − Q| + A0 (τ R)2+α ] Ω[R]

for any quadratic polynomial Q. We now define ( Dij u(¯ x1 ) if i, j < n, bij = Dij u(0) if i = n or j = n

(12.30)

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and we set

Since bin

1 QR (x) = Du(¯ x1 ) · (x − x ¯1 ) + bij (x − x ¯1 )i (x − x ¯1 )j . 2 = bni = 0 if i < n, we have

QR (x) − P (x) =

n−1 1X [Din u(¯ x1 ) − bin ](x − x ¯1 )i (x − x ¯1 )n 2 i=1 n

+

1X [Dnj u(¯ x1 ) − bnj ](x − x¯1 )j (x − x ¯1 )n . 2 j=1

We now define, for i = 1, . . . , n, fi = When i < n, we have bin

Dn u(¯ x1 + τ Rei ) − Dn u(¯ x1 ) . τR = 0, and hence

Din u(¯ x1 ) − bin = (Din u(¯ x1 ) − fi ) + fi . The mean value theorem then gives a point yi on the line segment joining x¯1 and x ¯1 + τ Rei (and hence in B(¯ x1 , τ R)) such that fi = Din u(yi ). Theorem 12.4 then implies that   oscΩ[R] (u − Q) α |Din u(¯ x1 ) − fi | ≤ osc Din u ≤ C τ σ + A τ R . 0 B(¯ x1 ,τ R) R2

Furthermore, (12.29) implies that |fi | ≤

C [(τ R)1+α H + A0 (τ R)1+α + M τR



R R0

1+θ

]

x1 + τ Rei )n = x¯n1 , ω(¯ x01 ) = ω(0) = 0, and |ω(¯ x1 + τ Rei )| ≤ because (¯ 1+α C(τ R) . Using the estimate (12.28) and the inequalities 0 ≤ R0 ≤ 1, we find that  1+θ R 1+α |fi | ≤ C[A0 (τ R) +M ]. R0

To estimate the final term in the expansion of QR − P , we observe that Dnn u(¯ x1 ) − Dnn u(0) = Dnn u(¯ x1 ) − fn + fn − H.

It follows as before that   oscΩ[R] (u − Q) α |Dnn u(¯ x1 ) − fn | ≤ C τ σ , + A τ R 0 R2

and simple algebra yields fn − H =

Dn (¯ x1 + τ Ren ) − (1 + τ R)H H − Dn u(¯ x1 ) + . τR τR

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Since x ¯01 = (¯ x1 + τ Ren )0 = 0 and ω(0) = 0, we conclude from (12.29) that |Dnn u(¯ x1 ) − Dnn u(0)| ≤

C R1+θ [A0 τ R1+α + M 1+θ ]. τR R0

Combining these inequalities, we find that |QR (x) − P (x)| ≤ C[τ 2+θ osc (u − Q) + A0 R2+α + M Ω[R]

R2+α ]. R01+α

(12.31)

In combination with (12.30), this inequality implies that   R2+α 2+θ 2+α osc (u − QR ) ≤ C τ osc (u − Q) + A0 R + M 1+α . (12.32) B(¯ x1 ,τ R) Ω[R] R0 Now let x ∈ Ω[τ R] and set x ¯=x+x ¯1 , w = u − QR . It follows that w(x) − w(0) = w(¯ x) − w(¯ x1 ) + w(x) − w(¯ x) + w(¯ x1 ) − w(0), and |w(¯ x) − w(¯ x1 )| ≤

osc

(u − Q2 ).

B(¯ x1 ,τ R)

In addition, w(x)−w(¯ x )+w(¯ x1 )−w(0) = and Dn w



R 2

Z 1 0

Dn w



   sR sR en − Dn w x + en ds 2 2

   sR sR en − Dn w x + en 2 2       sR sR sR sR n = Dn u en − H − Dn u x + en + H x + . 2 2 2 2

Using (12.29) again, we find that     1+θ Dn w sR en − Dn w x + sR en ≤ H|ω(x0 )| + C(A0 R1+α + M R ). 1+θ 2 2 R0 Since

M R1+α ], R0 it follows (after recalling that R0 ≤ 1 and α ≤ θ) that H|ω(x0 )| ≤ C[A0 R0α R1+α +

|w(x) − w(¯ x) + w(¯ x1 ) − w(0)| ≤ C[A0 R2+α +

M R2+α ], R01+α

and hence osc (u − Q2 ) ≤ C

Ω[τ R]



osc

(u − Q2 ) + A0 R

B(¯ x1 ,τ R)

2+α

 M R2+α + . R01+α

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Combining this inequality with (12.32) and (12.31) yields (12.27) in the special case that β = en . In the case of general β, we begin by defining the linear change of variables T by saying that y = T x if ( xi − (β i /β n )xn if i < n, i y = xn if i = n. We also define u ˜ = u ◦ T −1 and we define the function ω ˜ implicitly in a neighborhood of 0 to satisfy the equation   β0 ω ˜ (y 0 ) = ω y 0 + n ω ˜ (y 0 ) ; β n 0 n ˜ hence x = ω(x ) if and only if x = ω ˜ ((T x)0 ). We then write Ω[s] (for suitable s > 0) for the set of all y ∈ Rn with |y| < s and y n > ω ˜ (y 0 ), and we n n ˜ write Σ[s] for the set of all y ∈ R with |y| < s and y = ω ˜ (y 0 ). It follows that ∂u ˜ =0 ∂y n on T (Σ[R0 ]) and that there are positive constants c1 and c2 , determined only by ω0 and ε, such that |Dω ˜ | ≤ c1 , [Dω ˜ ]α ≤ c1 [Dω]α , ˜ 1 s] T (Ω[s]) ⊂ Ω[c ˜ 1 R0 ] ⊂ T (Ω[R0 ]); we can for all s ∈ (0, R1 ) (where R1 is chosen so that Ω[c take R1 = c2 R0 ), ˜ 2 s] ⊂ T (Ω[s]) Ω[c

for all s ∈ (0, R0 ), and Dω ˜ (0) = 0. Further, there is a function F˜ : T (Ω[R0 ]) × Sn → R such that c2 λ tr η ≤ F˜ (y, r + η) − F˜ (y, r) ≤ c1 Λ tr η for any positive definite matrix η and all (y, r) ∈ T (Ω[R0 ]) × Sn , ˜ α−1 |x − y| |F˜ (x, r) − F˜ (y, r)| ≤ c1 A0 d(x)

for all x and y in T (Ω[R0 ]) and all r ∈ Sn , where d˜ denotes distance to ˜ 0 ], and Σ[R F˜ (y, D2 u ˜) = 0 in T (Ω[R0 ]). Our previous step shows that u ˜ satisfies an estimate like ˜ (12.27) but with Ω[s] in place of Ω[s] for s = τ R, R, and the proof is completed by pointing out that the class of quadratic polynomials is unchanged under the mapping T . 

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At this point, we jump to the relevant existence result with an attendant estimate. Lemma 12.13. Let ω ∈ H1+α (Bn−1 (0, R)) for some α ∈ (0, 1) and R ∈ (0, 1] with ω(0) = 0, for s ∈ (0, R], define Ω[s] and Σ[s] by (12.23), and define σ[s] = {x ∈ Rn : |x| = s, xn ≥ ω(x0 )}.

(12.33)

Suppose F is as in Lemma 12.8 and let β be a unit vector satisfying (12.25) for some positive constants ε < 1 and ω0 ≥ sup |Dω|. Then, for any function ϕ ∈ C(σ[R]), there is a unique solution v ∈ C 2 (Σ[R]) ∩ C(Σ[R]) of F (D2 v) = 0 in Ω[R],

β · Dv = 0 on Σ[R],

v = ϕ on σ[R].

(12.34)

Moreover, there are positive constants θ, determined only by Λ/λ, n, ε, and ω0 , and C, determined also by [Dω]α , such that if α < θ, then ¯ ≤ Cτ 2+α sup |v| sup |v − Q|

Ω[τ R]

(12.35)

Ω[R]

¯ is the second degree Taylor polynomial for v for any τ ∈ (0, 1/2), where Q centered at 0. Proof. First, we note that Lemma 12.12 remains valid even without the assumption that Dω(0) = 0 via a simple rotation of axes. Then, as in Section 12.2, we fix a nonnegative C ∞ function η : Rn → [0, 1] with compact support in B(0, R), define F˜ by F˜ (x, r) = ηF (r) + (1 − η) tr r, and we assume that ϕ ∈ Hδ for some δ ∈ (0, 1). The combination of that lemma with the argument of Theorem 12.11 shows that any solution of the problem F˜ (x, D2 u) = g(x) in Ω[R],

(12.36a)

β · Du = 0 on Σ[R],

(12.36b)

u = ϕ on σ[R]

with g ∈

(2−δ) H2+α

(12.36c)

satisfies the estimate (−δ)

(−δ)

+ |g|2+α + |ϕ|δ

(−δ)

+ |g|2+α + |ϕ|δ

|u|4+α ≤ C(|u|2

(2−δ)

(−δ)

).

(2−δ)

(−δ)

),

Lemma 2.3 shows that (−δ)

|u|4+α ≤ C(|u|0

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and a simple application of the maximum principle yields (−δ)

(2−δ)

(−δ)

|u|4+α ≤ C(|g|2+α + |ϕ|δ

+ F (0)). (−δ)

It then follows from Theorem 11.3 that (12.36) is solvable in H4+α for any g, and the approximation arguments used in the proof of Theorem 12.11 yield our existence result. From Lemmata 12.12, 2.8, and 2.3, we have (0)

|v|2+α ≤ C|v|0

and then integration yields (12.35).



A straightforward modification of our prior perturbation arguments leads to a H¨ older estimate for second derivatives of solutions of fully nonlinear oblique derivative problems. Theorem 12.14. Let α and α0 be positive numbers with α < α0 ≤ 1 and set δ = α/(2α0 − α). Let Ω be a domain in Rn with a C 1,α boundary portion T , and write d∗ for the distance to ∂Ω\T . Suppose F is a real-valued function defined on Ω × Sn which is concave with respect to r and which satisfies (12.24a) for all positive definite matrices η and all (x, r) ∈ Ω × Sn , and |x − y|α |F (x, r) − F (y, r)| ≤ [A1 d∗ (x)−2 + A2 |r|]λ ∗ (d (x))α for all x and y in Ω with |x − y| ≤ d∗ (x)/2 and all r ∈ Sn . Let b be a real-valued function defined on T × Rn and set χ = ∂b/∂p. Suppose that there are positive constants b0 and G1 such that |bp (x, p)| ≤ b0 χ(x, p),

∗

∗

|bx (x, p)| ≤ G1 χ(x, p)/d (x)

(12.37a) (12.37b)

n

for all (x, p) ∈ Σ[R ] × R . Suppose further that there is a nonnegative constant G2 such that   α |x − y|α 1/(1+α) |p − q| + G2 G1 χ(x, p), |bx (x, p) − bx (y, q)| ≤ G1 ∗ (d (x))2+α d∗ (x)2−α (12.38a)   α 0 |x − y| α/(1+α) |bp (x, p) − bp (y, q)| ≤ G1 + G2 |p − q|α χ(x, p) (12.38b) d∗ (x)α (0)

for all (x, p) and (y, q) in T × Rn with |x − y| ≤ d∗ (x)/2. If u ∈ H2 (Ω) ∩ (0) C 2 (Ω) ∩ H1+δ (Ω) (with the weight in the H2 norm being with respect to d∗ ) satisfies F (x, D2 u) = 0 in Ω,

b(x, Du) = 0 on T,

(12.39)

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then there are positive constants θ, determined only by n, Λ/λ, b0 , and Ω, and C, determined also by A2 and G2 , such that 2(1+α)(2α0 −α)/α

(0)

|u|2+α ≤ C(|u|0 + A1 + G1 + [Du]δ

(0)

+ ([u]1 )2+α ). (12.40)

Proof. By arguing as in Theorem 12.11, we may assume that A2 is zero. We begin by fixing a point x0 ∈ T . We then rotate and translate so that γ(x0 ) = en and x0 = 0. Now we choose R0 ∈ (0, 1] so that d∗ (x0 ) ≥ R0 . By further decreasing R0 , we may assume that there is a function ω ∈ H1+α (Bn−1 (0, R0 )) such that ω(0) = 0 and Ω ∩ B(x0 , R0 ) = Ω[R0 ],

T ∩ B(x0 , R0 ) = Σ[R0 ],

using the definitions (12.23). Moreover, there are constants ω0 > 0 and ε ∈ (0, 1) such that |Dω| ≤ ω0 and β = bp (0, Du(0)) satisfies |β 0 | ≤

1−ε n β . ω0

Using (12.37) and (12.38), we infer that there is a constant C such that |β · (Du − Du(0)) + bx (0, Du(0)) · x| ≤ C(I1 + I2 + I3 + I4 ), where |x|1+α , R02+α α 1/(1+α) |Du − Du(0)| |x| I2 = G1 , R02−α α α/(1+α) |x| |Du − Du(0)| I3 = G1 , R0α

I1 = G1

0

I4 = |Du − Du(0)|1+α . (Strictly speaking, bx (0, Du(0)) denotes a tangential derivative along ∂Ω but we can take it to be a vector with nth component equal to zero.) Now (0) we set Uδ = [Du]δ , U2 = [u]2 , and ζ = (2 + α)/(2 + 2α0 ), and we note that |Du − Du(0)| ≤ R0−2 U2 |x|, |Du − Du(0)| ≤

Uδ1−ζ U2ζ R0−2ζ |x

− y|

ζ+(1−ζ)δ

Using (12.41a), we conclude that 1/(1+α)

I2 ≤ G1

U2α

|x|1+α R02+α

(12.41a) .

(12.41b)

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and |x|1+α . R02+α To estimate I4 , we use (12.41b) and the easily checked identities α ζ(1 + α0 ) = 1 + , (ζ + (1 − ζ)δ)(1 + α0 ) = 1 + α 2 to infer that 1+α (1−ζ)(1−α0 ) 1+α/2 |x| I4 ≤ Uδ U2 2+α . R0 Combining these inequalities with Young’s inequality and setting α/(1+α)

I3 ≤ G1

U2

2(1+α)(2α0 −α)/α

U = G1 + Uδ

1+α/2

+ U2

,

we conclude that I1 + I2 + I3 + I4 ≤ CU

|x|1+α . R02+α

We now write βˆ for the vector perpendicular to β with βˆ0 = bx (0, Du(0)) and define u0 by 1 u0 (x) = u(x) − Du(0) · x + (β · x)(βˆ · x). |β|2 Since β · D((β · x)(βˆ · x)) = |β|2 βˆ · x, |βˆ · x − bx (0, Du(0)) · x| = |βˆn ||xn |,

and |xn | = ω(x0 ) ≤ C|x0 |1+α on Σ[R0 ], it follows that  1+α  |x| |Du − Du(0)|1+α n |β · Du0 | ≤ Cβ + . R0α R02+α We therefore infer that |x|1+α |β · Du0 | ≤ β n C4 U 2+α . (12.42) R0 0 Next, we write r0 for the matrix with entries rij = β i βˆj , and we define 0 F0 by F0 (x, r) = F (x, r − r ). With τ ∈ (0, 1) a constant at our disposal, we define Rk = τ k for any nonnegative integer k. We also define Fk , vk , rk , and Qk inductively as follows. First, vk is the solution of Fk (0, D2 vk ) = 0 in Ω[Rk ], β · Dvk = 0 on Σ[Rk ], vk = uk on σ[Rk ].

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We then write Qk for the second order Taylor polynomial of vk centered at 0, and rk = D2 vk (0). Finally, we set uk+1 = uk − Qk and we define Fk+1 by Fk+1 (x, r) = Fk (x, r − rk ). It follows from Lemma 12.13 that there are constants θ (as in the statement of this theorem) and C1 such that sup |vk − Qk | ≤ C1 τ 2+α sup |vk |

Ω[Rk+1 ]

(12.43)

Ω[Rk ]

for α ≤ θ. In addition, the maximum principle implies that ! Rk2+α |β · Duk | |vk − uk | ≤ C2 A1 2+α + sup Rk . βn R0 Σ[Rk ]

(12.44)

On Σ[Rk ], we have β · Duk = β · Duk−1 + β · DQk−1 and it’s easy to check that β · DQk = β i Din vk (0)xn . By arguing as in Proposition 11.15, we find that |β i Din vk (0)| ≤ Cβ n sup |vk |Rk−2 ,

and we have |xn | ≤ C|x|1+α on Σ[Rk ], so

|β · DQk | ≤ C3 βn Rk−2 |x|1+α sup |vk |. Ω[Rk ]

We now set Bk (x) =

|β · Duk (x)| , βn

Vk = sup |vk |, Ω[Rk ]

and infer that −2 Bk (x) ≤ Bk−1 (x) + C3 Vk−1 Rk−1 |x|1+α .

A straightforward induction yields Bk (x) ≤ B0 (x) + C3 |x|1+α and hence, by virtue of (12.42), we have

k−1 X

Rj−2 Vj

j=0

Rk Bk (x) ≤ C4 U τ (2+α)k + C3 R0α τ (2+α)k

k−1 X j=0

Vj τ 2j

for x ∈ Σ[Rk ]. Inserting this inequality into (12.44) gives |vk − uk | ≤ (C2 A1 + C2 C4 U )τ

(2+α)k

+

C3 R0α τ (2+α)k

(12.45)

k−1 X j=0

Vj , τ 2j

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so (12.43) implies that Vk+1 ≤ (C1 + C2 C3 R0α )τ 2+α Vk + Hτ (2+α)k + 2C2 C3 R0α τ (2+α)k

k−1 X j=0

Vj . τ 2j

with H = 2(C2 A1 + C2 C4 U ). We now choose θ ∈ (0, α) and set σ = (θ + α)/2. In addition, we take R0 so small that C2 C3 R0α ≤ 1 and choose τ so small that (C1 + 1)τ 2+α ≤ τ 2+σ . Then our induction inequality becomes Vk+1 ≤ τ 2+σ Vk + Hτ (2+θ)k + 2C2 C3 R0α τ (2+θ)k

k−1 X j=0

Vj . τ 2j

For suitable choices of τ and R0 , it follows from the argument deriving (4.19) from (4.18) that Vk ≤ τ (2+θ)k V0 + 2Hτ (2+α)(k−1) .

(12.46)

We now use this inequality, along with the summability of the infinite series ∞ X τ θj j=0

and (12.45) to infer that

Rk Bk (x) ≤ C[U + V0 + H]τ (2+θ)k on Σ[Rk ]. It then follows from (12.44) and (12.46) that sup |uk | ≤ [U + U0 + H]τ (2+θ)k

Ω[Rk

with U0 = sup |u0 |. Ω[R0 ]

Hence, for any ρ ∈ (0, R), there is a second order polynomial P (ρ) such that sup |u − P (ρ)| ≤ C(U + U0 + H)ρ2+θ . Ω[ρ]

Combining this inequality with the corresponding interior estimates, we find that (0)

which implies that

[u]2+α ≤ C(|u|0 + U + U0 + H), 2(1+α)(2α0 −α)/α

[u]2+α ≤ C(|u|0 + A1 + G1 + [Du]δ

(0)

) + C([u]2 )1+α/2 .

The estimate (12.40) follows from this inequality via Lemma 2.3.



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Our next step is to derive suitable a priori estimates for solutions of uniformly elliptic problems. The H¨ older estimate for the solution is particularly simple. Theorem 12.15. Let Ω be a bounded C 1 domain in Rn and let u ∈ C 2 (Ω) ∩ C(Ω) be a solution of (12.7). Suppose that there are positive constants λ ≤ Λ and an increasing function A0 such that λ tr η ≤ F (x, z, p, r + η) − F (x, z, p, r) ≤ Λ tr η

(12.47a)

for all (x, z, p, r) ∈ Ω × R × Rn × Sn and all positive definite matrices η, |F (x, z, p, 0)| ≤ A0 (|z|)λ(1 + |p|2 )

(12.47b)

for all (x, z, p) ∈ Ω × R × Rn . Suppose also that |b(x, z, p0 )| ≤ A0 (|z|)χ(1 + |p0 |)

(12.48)

for all (x, z, p) ∈ ∂Ω × R × Rn with b(x, z, p) = 0, where p0 = p − (p · γ)p. Then there are constants α ∈ (0, 1) and C > 0, determined only by n, Λ/λ, sup |u| and A0 (sup |u|) such that |u|α ≤ C.

(12.49)

Proof. With aij and a as in the remarks before Lemma 12.1, we see that u satisfies the differential equation aij Dij u + a = 0 in Ω with aij having eigenvalues in the range [λ, Λ] and |a| ≤ A0 (sup |u|)λ(1 + |p|2 ). Similarly, u satisfies the boundary condition β · Du + g = 0 for some vector field β having a modulus of obliqueness determined by A0 (sup |u|) and Ω and g/|β| is bounded by a constant determined by the same quantities. The H¨older estimate then follows from Lemma 8.4.  If we impose suitable assumptions on the derivatives of F and b with respect to (x, z, p, r), then we can imitate the arguments in Chapter 9 to obtain a bound on the maximum of the gradient. As in that chapter, an important part in the derivation of this bound is played by a suitable interior gradient bound, which we now present. It will be convenient to introduce some additional notation here. First, we set E2 = F ij rik rjk and we use the operators δ and δ from Chapter 9. In other words, δg = gz + |p|−2 p · gp ,

¯ = p · gp . δg

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Lemma 12.16. Let u ∈ C 3 (Ω) be a solution of F [u] = 0 in Ω, and suppose that there are nonnegative constants µ, µ1 , and M such that µ1 ε Λ|p|2 ≤ µE, |p||Fp |, δF ≤ + 2 E2 (12.50) ε |p| for all |p| ≥ M and ε ∈ (0, 1). Then there are constants C and θ, determined only by n, µ, and µ1 , such that, for any y ∈ Ω and R ∈ (0, d(y)) such that osc u ≤ θ,

B(y,R)

we have |Du(y)| ≤ C

oscB(y,R) u + M. R

(12.51)

Proof. We set α = oscB(y,R) u and assume first that α ≤ 1/(2µ1 ). We then define η by η(x) = (1 − |x − y|2 )/R2 , and h by h(z) = exp((z − sup u)/α). B(y,R)

With β ∈ (0, 1/2) to be further determined, we set M1 = sup η 2 |Du|2 , B(y,R)

w = η 2 |Du|2 + βM1 h(u).

If we apply the operator Dk uDk to the equation F [u] = 0, then a straightforward calculation yields β F ij Dij w + B i Di w = |Du|2 H0 + M1 hJ0 + 2η 2 E2 , α where 4 B i = F i − F ij Dj η, η H0 = 2ηF ij Dij η − 6F ij Di ηDj η − 2ηF i Di η − η 2 δF, 4 1 J0 = F i Di u − F ij Di ηDj u + F ij Dij u + E. η α Since |Dη| ≤ 2/R and F ij Dij η ≥ −4nΛR−2, we infer from (12.50) with ε = 2αµ1 that F ij Dij w + B i Di w ≥ H1 E + J1 E2 , where H1 = − 36µR−2 − 2η|Du|/(αR) − η 2 |Du|2 /(2α) − βM1 h/(2α2 ) β − 8βM1 hµ/(αη|Du|R) − β 2 M12 h/(2α2 η 2 |Du|2 ) + 2 M1 h α

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and J1 = −8µ1 αη/(R|Du|) − 2µ1 αη 2 − 2µ1 M1 hβ/|Du|2 + 3η/2. Now let x0 be a point at which w attains its maximum. Since β ∈ (0, 1/2] and h(u) ∈ (0, 1), it follows that

M1 , 2 and hence η 2 (x0 )|Du(x0 )|2 ≥ M1 /2. It follows that, at x0 , we have M1 ≤ w(x0 ) ≤ η 2 (x0 )|Du(x0 )|2 +

F ij Dij w ≥ (H2 + H3 hβ/α2 )E + J2 η 2 E2

with 1/2

H2 = −(4n + 32)µ/R2 − 2M1 /(αR) − M1 /α, 1/2

H3 = −8µαM1 /R − βnµM1 + M1 /2, 1/2

J2 = −8µ1 α/(RM1 ) − 2αµ1 − 4βµ1 + 3/2. We now take β = min{1/2, 1/(8µ1), 1/(8nµ)} and note that J2 > 0 if (16µ1 α/R)2 < M1 ,

α ≤ 1/(4µ1 ).

Moreover, if we have ((16(µ+ µ1 )α/R)2 < M1 , it follows that 1/2

F ij Dij w ≥ E[βM1 /(24α)2 − (4n + 32)µ/R2 − 2M1 /(αR) − M1 /α]

at x0 because h(u) ≥ e−1 ≥ 1/3. It follows that there is a constant Cn determined only by n such that F ij Dij w(x0 ) > 0 if α≤ M1 >



Cn , µ + µ1

(µ + µ1 )α Cn R

(12.52) 2

.

(12.53)

Since we must have F ij Dij w(x0 ) ≤ 0, it follows that (  2 ) (µ + µ1 )α 2 M1 ≤ max M , Cn R if α ≤ θ with θ = Cn /(µ + µ1 ), and this inequality implies (12.51).



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Note that (12.50) follows from the assumptions (12.47a) and (1 + |p|)|Fp |, |Fz |, |Fx |/(1 + |p|) ≤ λA1 (|z|)(1 + |p|2 + |r|)

(12.54)

for some increasing function A1 . Our next step is to prove a gradient bound near a flat boundary. Lemma 12.17. Let R ∈ (0, 1), suppose F ∈ C 1 (B + (0, R) × R × Rn × Sn ), and suppose that b ∈ C 1 (B 0 (0, R) × R × Rn) with bn > 0. Suppose also that there are positive constants Λ, λ, and an increasing function A1 such that conditions (12.47a) and (12.54) are satisfied for all (x, z, p, r) ∈ B + (0, R)× R × Rn × Sn . Suppose also that there is an increasing function B1 such that (1 + |p|2 )|bp | + (1 + |p|)|bz | + |bx | ≤ B1 (|z|)(1 + |p|2 )bn 0

n

(12.55) 3

+

for all (x, z, p) ∈ B (0, R)×R×R with b(x, z, p) = 0 If u ∈ C (B (0, R))∩ C 2 (B + (0, R)) satisfies F [u] = 0 in B + (0, R),

N [u] = 0 on B 0 (0, R),

(12.56)

then there are positive constants θ and C, determined only by Λ/λ, A1 (sup |u|), B1 (sup |u|), and n such that, if osc

B + (0,R)

u ≤ θ,

then sup B + (0,R/4)

|Du| ≤ C

 osc

B + (0,R)

R

u

 +1 .

(12.57)

Proof. Set α = oscB + (0,R) u, ρ = min{R/4, α}, and write S for the set of all x ∈ B + (0, R/2) with xn < ρ. Similarly to the proof of Lemma 12.16, we define functions η and h by   z − supB + (0,R) u 4|x|2 η(x) = 1 − , h(z) = exp . R2 α Our next step is to note that a simple modification of the proof of Lemma 10.10 gives a function N ∈ C 2 (B + (0, R) × R × Rn ) and a constant C, determined only by B1 (sup |u|) and n such that 1 3 ≤ N n ≤ , |Np | ≤ 4B1 , 2 2 |p||Nz | + |Nx | ≤ C|p|2 ,

|p|2 |N Npp | + |p||N Npz | + |N Npx | ≤ C|p|2 ,

|p|2 |N Nzz | + |p||N Nzx | + |N Nxx | ≤ C|p|2 ,

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for |p| ≥ C and b(x, u, Du) = 0 on B 0 (0, R) wherever |Du| ≥ C. (Note that the expressions N Npp , etc., are not typographical errors.) By taking ε sufficiently small and setting v1 = (1 + |p0 |2 + εN 2 )1/2 √ (where p0 = (p1 , . . . , pn−1 , 0)), we have that v ε/2 ≤ v1 ≤ 2v, and ν1 = ∂v1 /∂p satisfies |ν1 | ≤ 2,

ν1 · Du ≥ v1 /8

for v1 ≥ 3. Finally, there is a constant C1 such that   ∂ 2 v1 ν1 + ν1i ν1j ξi ξj ≥ C1 |ξ|2 ∂pi ∂pj

for all x ∈ Rn . We now set M1 = supB + (0,R/2) η 2 v12 and, with α1 and α2 positive constants to be further specified, we set w = η 2 v12 + α1 M1 h(u) + α2 M1 vxn ,

and we write x0 for a point in S at which w attains its maximum. By using Lemma 12.16, we find that vxn ≤ C2 α in B + (0, R/2) with C2 determined only by n, Λ/λ and A1 (sup |u|) if α is sufficiently small. Hence, if α1 ≤ 1/4 and C2 αα2 ≤ 1/4,

(12.58)

we conclude that η 2 v12 ≥ M1 /2 and η 2 v 2 ≥ M1 /8 at x0 . If x0 ∈ B + (0, R/2) and |Du(x0 )| ≥ C, then we have α1 bi Di w = 2v12 ηbi Di η − 2η 2 (bz |D0 u|2 + bx · D0 u) + M1 hDu · bp + α2 M1 vbn α   2B0 B0 α1 ≥ bn M1 v α2 − − − 4B1 . ηvR α If M1 ≥ (8α/(α1 R))2 and

3B0 α1 , α then we conclude that bi Di w(x0 ) > 0 which contradicts the assumption that w has a maximum at x0 . We also observe that our choice of α2 implies that (12.58) follows from the inequality C1 M (4B1 α + 3B0 α1 ) ≤ 1/4, and this inequality holds if α and α1 are sufficiently small. If x0 ∈ S and |Du(x0 )| ≥ C, then we define the operator L by 4 Lw = F ij Dij w + (F i − F ij Dj η)Di w. η α2 = 4B1 +

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It follows that there are positive constants C2 and C3 such that Lw(x0 ) ≥ M1 α−2 h(u)E(α1 − C2 α21 − C3 α2 ) so we can’t have x0 ∈ S if |Du(x0 )| ≥ C provided α1 ≤

1 , 2C2

α≤

1 . (2C3 )1/2

It follows that the maximum of w occurs either where |Du| ≤ C or where xn ≥ ρ. Since Lemma 12.16 gives a bound on Du(x0 ) if xn ≥ ρ, we obtain (12.57) by straightforward algebra.  By combining this estimate with Theorem 12.15, Lemma 12.16, and a C 3 change of variables, we infer the following global gradient bound. Theorem 12.18. Let Ω be a bounded domain in Rn with ∂Ω ∈ C 3 . Suppose conditions (12.47), (12.48), (12.54) are satisfied. Suppose also that there is an increasing function B1 such that (1 + |p|2 )|bp | + (1 + |p|)|bz | + |bx | ≤ B1 (|z|)(1 + |p|2 )bp · γ

(12.59)

for all (x, z, p) ∈ ∂Ω × R × Rn with b(x, z, p) = 0. Then, for any solution u ∈ C 3 (Ω) ∩ C 2 (Ω) of (12.7), there is a constant C, determined only by Λ/λ, A0 (sup |u|), A1 (sup |u|), B0 (sup |u|), B1 (sup |u|), and Ω such that sup |Du| ≤ C.

(12.60)

Ω

A H¨ older gradient estimate then follows by straightforward modification of the proof of Lemma 11.12. Theorem 12.19. Let ∂Ω ∈ C 3 and suppose that there are positive constants λ, Λ such that (12.47) is satisfied. Suppose also that there is an increasing function µ1 such that |Fp | + |Fz | + |Fx | ≤ µ1 (|z| + |p|)(1 + |r|)

(12.61)

for all (x, z, p, r) ∈ Ω × R × Rn × Sn and |bp | + |bz | + |bx | ≤ µ1 (|z| + |p|)bp · γ

(12.62)

for all (x, z, p) ∈ Ω × R × Rn with b(x, z, p) = 0. If u ∈ C 3 (Ω) ∩ C 2 (Ω) solves (12.7), then there is are positive constants C and δ, determined only by n, Ω, Λ/λ, and µ1 (|u|1 ) such that [Du]δ ≤ C.

(12.63)

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Proof. We only prove a local H¨older gradient estimate near a point in ∂Ω since the corresponding interior estimate is proved similarly. In addition, since the hypotheses are invariant under a C 3 change of independent variable, we only may assume that u satisfies (12.56). With N as in the proof of Lemma 12.17 and ε > 0 sufficiently small, we set V = |D0 u|2 + εN 2 . Then with θ a positive constant to be further determined, we define, for k = 1, . . . , n, wk± = ±Dk u + θV. Using the given structure conditions, we find that there are constants c2 and c3 (determined only by n, Λ/λ, and µ1 (|u|1 ) such that c2 F ij Dij wk± ≥ − λ|Dwk± |2 − c3 λ. θ The proof of the H¨ older gradient estimate at 0 is now almost identical to that of Lemma 11.12; the only change is that the functions h± k in that proof satisfy inhomogeneous differential equations.  We note that second derivative H¨older estimates for u follow immediately from Theorem 12.14, so we obtain the following existence result very easily. Theorem 12.20. Let ∂Ω ∈ C 3 , let F ∈ C 1 (Ω × R × Rn × Sn ), and let b ∈ C 1,α (∂Ω × R × Rn ) for some α ∈ (0, 1). Suppose F and b satisfy conditions (12.9), (12.47), (12.48), (12.54), and (12.59). Suppose also that Fz (x, z, p, r) < 0 for all (x, z, p, r) ∈ Ω × R × Rn × Sn and bz (x, z, p) ≤ 0 for all (x, z, p) ∈ ∂Ω × R × Rn . Then there is a unique solution of (12.7). Notes The basic material in Sections 12.1 and 12.2 comes from Chapter 17 of [64], and we refer the reader there for detailed references. We only mention that Lemma 12.7 is the lemma on page 255 of [141], and that Lemma 12.8 was first proved simultaneously (using essentially the method described here) by Evans [42] and Krylov [86]. The first study of oblique derivative problems for fully nonlinear elliptic equations was [121]; the authors looked at equations of Bellman type and linear boundary conditions. Shortly thereafter, Trudinger and I, in [120] extended the results of [121] to general uniformly elliptic equations and nonlinear boundary conditions. Unfortunately, our second derivative

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H¨ older estimates required an estimate on the H¨older norm of the derivatives of F with respect to the variables x, z, p, r, although (as pointed out in the Supplementary remarks to [120]) it was shown in [179] that the second derivative H¨ older estimate can be proved without this condition. Further variations on the second derivative H¨older estimate are given in [164] and [117]. The proof in [164] is stated for nonlinear equations with linear boundary conditions, but it can be modified to include nonlinear boundary conditions like the ones studied here. In broad outline, the goal in [164] is to reduce the estimate to one with the boundary condition Dn u = 0 on ∂Ω and then to show that the estimate in this case follows from an estimate for the same boundary condition but with ∂Ω being locally the hyperplane xn = 0. Our proof is based on the one in [117], but we have included more details. We have followed [120] here in most other respects except that we use the H¨ older gradient estimate method described in Chapter 11, and we have used the function N (from Lemma 10.8) to deduce the gradient bound. We refer the reader to [120] for further examination of uniformly elliptic oblique derivative problems; in particular, we mention Corollary 7.10 in [120] which removes the monotonicity assumptions (with respect to z) on F and b in our Theorem 12.20. For nonuniformly elliptic problems, the study of oblique derivative problems is much more complicated and not yet in a final form. An important class of nonuniformly elliptic equations is based on Monge-Amp`ere operators, which have the form F [u] = det D2 u − ψ(x, u, Du).

(12.64)

Since F ij (x, z, p, r) = rij , where rij is the cofactor of rij , it follows that these operators are elliptic only when r is positive definite, which means that solutions must be strictly convex. In this case, we also know that det D2 u > 0, so we must restrict attention to positive functions ψ. This class of operators is important because it is elliptic only for some functions, and the theory presented so far has been concerned with operators that are elliptic for all functions. The Neumann problem for Monge-Amp`ere equations, that is, the problem det D2 u = ψ(x, u, Du) in Ω,

γ · Du = ϕ(x, u) on ∂Ω,

was studied in detail by Lions, Trudinger, and Urbas in [122]. They showed proved the existence of a solution to this problem under fairly simple hypothesis. They assumed that ψ ∈ H2 (K) for any bounded subset K of Ω × R × Rn , ϕ ∈ H3 (K 0 ) for any compact subset K 0 of ∂Ω × R, and that

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∂Ω ∈ H4 . For structure conditions, they assumed ϕz to be bounded from below by a positive constant, and that there are functions g ∈ L1 (Ω) and h ∈ L1loc (Rn ), and a constant N , such that g(x) ψ(x, N, p) ≤ h(p) for all (x, p) ∈ Ω × Rn with Z Z g(x) dx < h(p) dp. Ω

Rn

They also proved solvability of some related boundary value problems, in which the differential equation is slightly modified. The estimates on u and Du in [122] are also valid, with the proofs given there, for a more general class of boundary conditions, but the second derivative bounds rely heavily on the exact form of the boundary condition. The significance of the normal derivative condition for the MongeAmp`ere equation was examined in great detail in [192] and [95]. Although the two papers use slightly different methods and obtain slightly different results, the most important point we wish to make here is that both papers prove second derivative estimates and, hence, existence theorems when b(x, z, p) is sufficiently close to γ · p − ϕ(x, z); specifically they assume that |bp −γ| is sufficiently small in an appropriate norm. Moreover, when |bp −γ| is large, then the estimates fail. The extension of these results to more general differential equations is still not well studied. Urbas [189, 190] studied (12.7) with more general differential equations when b is close to γ · p − ϕ but only when n = 2. An alternative boundary condition, first studied by Pogorelov (see pages 41–44 of [154]), is called the second boundary value problem. In the present situation, this problem is to solve the differential equation subject to the constraint Du(Ω) = Ω∗ for a given domain Ω∗ . For the Monge-Amp`ere equation, the convexity of the solution implies that this constraint is equivalent to the pointwise condition h(Du) = 0 on ∂Ω, where h is a defining function for Ω∗ . In other words h ∈ C 1 (Rn ), Ω is the set on which h is positive, and |Dh| + |h| never vanishes in Rn . When Ω and Ω∗ are convex, Urbas [191] showed that such a function h must be oblique on ∂Ω. Hence the second boundary value problem is an oblique boundary value problem in this case, and the estimates in [122] were adapted in [191] to prove second derivative bounds for solutions of the second boundary value problem for Monge-Amp`ere equations. In addition, the existence of solutions was proved there. Extensions of this result to a slightly larger class of differential equations was given in [193]. We mention also [181] for a discussion

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of the problem of optimal transportation in this context and [197] for an extension of the results in [181] to some more general differential equations. Both papers examine the second boundary problem Du(Ω) = Ω∗ for different differential equations using the idea of conversion to an oblique boundary value problem.

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dients for solutions of capillarity problems, Zap. Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 115, pp. 274–284, [Russian]. English transl. in J. Soviet Math. 28 (1985), 806–813. Ural0 tseva, N. N. (1991). A nonlinear problem with an oblique derivative for parabolic equations, Zap. Nauchn. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 188, pp. 143–158, [Russian]. English transl. in J. Math. Sci. 70 (1994), 1817–1827. Ural0 tseva, N. N. (1992). Gradient estimates for solutions of nonlinear parabolic oblique derivative problem, in Geometry and nonlinear partial differential equations (Fayetteville, AR, 1990), Contemp. Math., Vol. 127 (Amer. Math. Soc., Providence, RI), pp. 119–130. Urbas, J. (1995). Nonlinear oblique boundary problems for Hessian equations in two dimensions, Ann. Inst. H. Poincar´e: Anal. Non Lin´eaire 12, pp. 507–575. Urbas, J. (1996). Nonlinear oblique derivative problems for two dimensional curvature equations, Adv. Differential Equations 1, pp. 301–336. Urbas, J. (1997). On the second boundary value problem for equations of Monge-Amp`ere type, J. Reine Angew. Math. 487, pp. 115–124. Urbas, J. (1998). Oblique boundary value problems for equations of MongeAmp`ere type, Calc. Var. Partial Differential Equations 7, pp. 19–39. Urbas, J. (2001). The second boundary value problem for a class of Hessian equations, Comm. Partial Differential Equations 26, pp. 859–882. Verchota, G. C. (2007). Counterexamples and uniqueness for Lp (∂Ω) oblique derivative problems, J. Funct. Anal. 245, pp. 413–437. Volkov, E. A. (1965a). Differential properties of solutions of boundary value problems for the Laplace equation on polygons, Trudy Mat. Inst. Steklov 77, pp. 113–142, [Russian]. English transl. in Proc. Steklov Math. Inst. 77 (1968), 127–159. Volkov, E. A. (1965b). On differential properties of solutions of boundary value problems for the Laplace and Poisson equations on a rectangle, Trudy Mat. Inst. Steklov 77, pp. 89–112, [Russian]. English transl. in Proc. Steklov Math. Inst. 77 (1968), 101–126. von Nessi, G. T. (2010). On the second boundary value problem for a class of modified-hessian equations, Comm. Partial Differential Equations 35, pp. 745–785. Wang, L. (1992). A maximum principle for elliptic and parabolic equations with oblique derivative boundary problems, J. Partial Differential Equations 5, pp. 23–27. Wang, L. (2003). A geometric approach to the Calder´ on-Zygmund estimates, Acta Math. Sin. (Eng. Ser.) 19, pp. 381–396. ¨ Weierstrass, K. (1885). Uber die analytische Darstellbarkeit sogennater willk¨ urzer Funktionen einer reellen Ver¨ anderlichen, Berliner Berichte 1885, 2, pp. 633–639, 789–805, reprinted in Mathematische Werke, Band 3, Abhandlungen III, pp. 1-37. Whitney, H. (1934). Analytic extensions of differentiable functions defined on closed sets, Trans. Amer. Math. Soc. 36, pp. 63–89.

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[202] Wigley, N. M. (1970). Mixed boundary value problems in plane domains with corners, Math. Z. 115, pp. 33–52. [203] Wigley, N. M. (1988). Schauder estimates in domains with corners, Arch. Rational Mech. Anal. 104, pp. 271–276. [204] Williams, R. J. (1985). Recurrence classification and invariant measure for reflected Brownian motion in a wedge, Ann. Probab. 13, pp. 758–778. [205] Yanushauskas, A. I. (1989). The oblique derivative problem of potential theory, Contemporary Soviet Mathematics (Consultants Bureau, New York), translated from the Russian by Norman Stein. [206] Ziemer, W. P. (1989). Weakly Differentiable Functions, Graduate Texts in Mathematics, Vol. 120 (Springer-Verlag, New York).

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Index

Fredholm-Riesz-Schauder theory, 83–85

Aleksandrov’s theorem, 266–282 Bellman operators, 458

Gˆ ateaux variation, 411 generalized half-ball, 133, 136

C k,α boundary portion, 255 Campanato spaces, 211–214 capillary problem, 352–362, 365, 379–380, 401, 447, 451 change of variables formula, 11, 12, 14 comparison principle for fully nonlinear problems, 460–462 for viscosity solutions, 283, 291–295 conormal derivative problems, 190–225 existence, 187–190, 240–241, 447–451 covering lemma, 235, 237 curved boundaries, 81–83

Ha domains, 81 H¨ older continuity, 31–33, 108, 211, 305–307, 313–318, 482 H¨ older gradient estimate conormal problem, 408–409 fully nonlinear equation nonlinear boundary condition, 487–488 linear equation Dirichlet boundary condition, 433–434 linear boundary condition, 117–128, 136–165, 222–223 nonlinear boundary condition, 413–419, 422–426 quasilinear equation, 407–409 nonlinear boundary condition, 427–431, 437–443 H¨ older second derivative estimate fully nonlinear equation, 462–470 linear boundary condition, 476–477 nonlinear boundary condition, 477–481 linear equation linear boundary condition,

DeGiorgi class, 203–211 derivative bounds, 38–42, 74–77, 190–193, 238–240, 321–406, 483–487 interior, 321–329, 483–485 Dirichlet problem, 102 existence, 60 extension theorem, 7, 414 false mean curvature equation, 321, 333–352, 381, 403, 450, 453 507

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508

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Oblique Derivative Problems for Elliptic Equations

165–168 nonlinear boundary condition, 444–446 H¨ older spaces, 47–59, 211–214 interpolation inequality, 51–55 inward pointing vector, 3 vector field, 3, 4, 6, 13, 14, 16 Kondrachov compactness theorem, 177 Lebesgue differentiation theorem, 237 Leray-Schauder theorem, 422 Lp estimates, 241–251, 253–257 maximal operator, 236, 241 maximum principle, 3–5, 180–181 Aleksandrov-Bakel0 man-Pucci, 13–17, 227–230, 460 conormal problem, 193–200, 307–313 for fully nonlinear equations, 459–460 for quasilinear equations, 301–305 local, 34–35 conormal problem, 201–203 mixed boundary value problem, 38 strong, 29–31, 109–111 conormal problem, 210 Miller barrier, 95–102 mixed boundary value problems, 111–114 mixed boundary value problems, 35–38, 71–80 existence, 79, 419–424, 435–437, 476–477 modulus of continuity, 7, 8, 37 modulus of obliqueness, 6, 10, 11 Morrey space, 222 weighted, 218–222 Moser iteration, 311, 374 oblique

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vector field, 6 oblique derivative problems existence, 131–132, 168, 295–299, 425–426, 443–444, 447–453, 488 obliqueness, 6 ordinary differential equations, 89–94 Perron process Dirichlet problem, 65–71 oblique derivative problem, 82–83, 295–299 Perron subsolution, 66 Perron supersolution, 66, 67 Poincar´e’s inequality, 177–180 regularized distance, 129–130 Schauder estimates boundary conormal problem, 214–223 Dirichlet condition, 132–136 mixed boundary conditions, 73, 77–79 oblique boundary conditions, 81–83, 136–168 interior, 61–65 Schauder theory, 47–83 Sobolev imbedding theorem, 174–177 Sobolev spaces, 173–225 strong solution definition, 227 trace theorems, 182–186, 231–235, 251–253 uniformly oblique, 6, 11, 21, 30 uniqueness, 30, 79, 462 viscosity solutions, 265–300 weak derivatives, 173–225 weak Harnack inequality boundary conormal problem, 210

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Index

mixed boundary value problem, 36 oblique derivative problem, 21–28, 107–108 interior, 18–21 Weierstrass approximation theorem, 8, 10

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509