Elliptic Partial Differential Equations: Existence and Regularity of Distributional Solutions 9783110315424, 9783110315400

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Lucio Boccardo, Gisella Croce Elliptic Partial Differential Equations

De Gruyter Studies in Mathematics

Editorial Board Carsten Carstensen, Berlin, Germany Nicola Fusco, Napoli, Italy Fritz Gesztesy, Columbia, Missouri, USA Niels Jacob, Swansea, United Kingdom Karl-Hermann Neeb, Erlangen, Germany

Volume 55

Lucio Boccardo, Gisella Croce

Elliptic Partial Differential Equations Existence and Regularity of Distributional Solutions

Mathematics Subject Classification 2010 35J20, 35J25, 35J60, 35J61, 35J62, 35J70, 49J45, 49R05 Authors Prof. Lucio Boccardo Sapienza Università di Roma Dipartimento di Matematica Piazzale Aldo Moro 5 00185 Roma ITALY [email protected] Prof. Gisella Croce Laboratoire de Mathématiques Appliquées du Havre Université du Havre 25, rue Philippe Lebon 76058 Le Havre FRANCE [email protected]

ISBN 978-3-11-031540-0 e-ISBN 978-3-11-031542-4 Set-ISBN 978-3-11-031543-1 ISSN 0179-0986

Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.dnb.de. © 2014 Walter de Gruyter GmbH, Berlin/Boston Typesetting: le-tex publishing services GmbH, Leipzig Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen Printed on acid-free paper Printed in Germany www.degruyter.com

Bernardo m’accennava, e sorridea, perch’io guardassi suso . . . (Paradiso, XXXIII, vv. 49–50) This book is dedicated to Bernard Dacorogna for his 60th birthday.

Contents Notations 1

x

Introduction

1

Part I 2 2.1 2.2 2.3 2.4

Some fixed point theorems 5 5 Introduction Banach–Caccioppoli theorem Brouwer’s theorem 6 10 Schauder’s theorem

3 3.1 3.2 3.3 3.4

13 Preliminaries of real analysis Introduction 13 Nemitski’s composition theorem 18 Marcinkiewicz spaces Appendix 22

4 4.1 4.2 4.3 4.4 4.5 4.6

24 Linear and semilinear elliptic equations 24 Introduction The Lax–Milgram and Stampacchia’s theorems 27 Linear equations 28 Some semilinear monotone equations Sub and supersolutions method 31 34 Appendix

5 5.1 5.2 5.3

37 Nonlinear elliptic equations Introduction 37 37 Surjectivity theorem The Leray–Lions existence theorem

6 6.1 6.2 6.3 6.4 6.5

Summability of the solutions 47 47 Introduction 48 Preliminaries Sources in Lebesgue spaces 51 Sources in Marcinkiewicz spaces 58 Sources in divergence form

5

13

40

56

24

viii

Contents

7 7.1 7.2 7.3

H 2 regularity for linear problems

8 8.1 8.2 8.3 8.4

67 Spectral analysis for linear operators Introduction 67 67 Eigenvalues of linear elliptic operators Applications to some semilinear equations 74 82 Appendix

9 9.1 9.2 9.3 9.4 9.5 9.6

84 Calculus of variations and Euler’s equation 84 Introduction Direct methods in the calculus of variations 84 Euler equation 86 Summability of minimizers of integral functionals 95 The Ekeland variational principle 102 Appendix

61

Introduction 61 Preliminaries 61 2 H (Ω) regularity of the solutions

63

Part II 10 10.1 10.2 10.3

105 Natural growth problems 105 Introduction A problem with bounded solutions A problem with unbounded solutions

11 11.1 11.2 11.3 11.4 11.5 11.6

121 Problems with low summable sources 121 Introduction A priori estimates 123 130 Distributional solutions 130 The linear case: a different proof Entropy solutions 132 A comparison between entropy solutions and 137 distributional solutions Measure sources 138 The regularizing effects of a lower order term T -minima 143 Appendix 148

11.7 11.8 11.9 11.10

106 112

140

90

Contents

12 12.1 12.2 12.3 12.4

Uniqueness 150 Introduction 150 Monotone elliptic operators 150 152 A nonmonotone elliptic operator A uniqueness result for measure sources

13 13.1 13.2

A problem with polynomial growth 157 Introduction Existence results 158

14 14.1 14.2 14.3 14.4

166 A problem with degenerate coercivity Introduction 166 167 The case 0 < θ < 1 The case θ > 1: existence and nonexistence The regularizing effects of a lower order term

Bibliography Index

191

187

154

157

172 175

ix

Notations Ω

open bounded set of RN

∂Ω

boundary of Ω

meas

Lebesgue measure in RN

a.e.

almost everywhere (with respect to the Lebesgue measure)

X

dual of X (i.e., space of linear and continuous functionals on X ) where X is a Banach space

X 

dual of X  where X is a Banach space

L(X)

space of continuous and linear operators from a Banach space X into X

ϕ, v

ϕ(v) if ϕ ∈ X  and v ∈ X

Lp (Ω)

space of functions f such that |f |p is Lebesgue integrable over Ω ⊂ RN

C k (Ω)

space of k times differentiable functions on Ω with continuity

C0k (Ω)

space of k times differentiable functions on Ω, with continuity, 0 on ∂Ω

C 1 (Ω, RN )

space of differentiable maps from Ω onto RN with continuity

C (R , R )

space of differentiable maps from RN onto RN with continuity

W 1,p (Ω)

(Sobolev) space of functions with distributional gradient in (Lp (Ω))N

1,p W0 (Ω)

closure of C 1 (Ω) functions with compact support in Ω with respect to the W 1,p (Ω) norm

H01 (Ω)

W0 (Ω)

H −1 (Ω)

(H01 (Ω))

1

W

N

−1,p 

N

(Ω)

1,2

1,p

(W0 (Ω))

W m,2 (Ω)

space of W m−1,2 (Ω) functions, whose distributional gradient is in W m−1,2 (Ω)

H m (Ω)

W m,2 (Ω)

p

p p−1

p∗ χE (x)

Np N−p

⎧ ⎨1,

if x ∈ E

⎩0,

elsewhere

for a given set E ⊂ RN

{u ≥ 0}

{x ∈ RN : u(x) ≥ 0} for a given function u

for a given function u,

u+ = uχ{u≥0} ; u− = −uχ{u≤0} ⎧ ⎪ ⎪ −k, x ≤ −k ⎪ ⎨ x, |x| ≤ k for k > 0 ⎪ ⎪ ⎪ ⎩k, x≥k

Tk (x) Gk (x)

x − Tk (x) for k > 0

B(x0 , r )

{x ∈ RN : |x − x0 | < r } for a given x0 ∈ RN and r > 0

supp

support of a function

sgn(x)

sign of x for x ≠ 0, that is,

x |x|

1 Introduction This book is the result of undergraduate and Ph.D. courses that the first author has given at La Sapienza University of Rome on elliptic problems. With the aim of giving the background to those who want to do research in this mathematical field, we have divided this book into two parts. In the first one we present some classical results about the existence and the regularity of weak solutions to elliptic problems in divergence form (so that we do not discuss the theory of classical solutions). After the semilinear equations, we study the Leray–Lions problem ⎧ ⎨−div(a(x, u, ∇u)) = f , in Ω , ⎩u = 0 , on ∂Ω , where a is an elliptic operator, that is, a(x, s, ξ) · ξ ≥ α|ξ|2 , and f belongs to H −1 (Ω). We prove the existence and regularity results due to Leray–Lions and Stampacchia. We also treat the spectral theory of linear operators and the H 2 regularity of solutions to linear problems. Even if in this book we focus our attention on partial differential equations, we have presented a chapter about the Calculus of Variations, since some of the differential problems that we treat are motivated by the minimization of integral functionals. The Leray–Lions problem is the origin of an important and still active area of research. In the second part of this book we exhibit some possible directions: the existence of solutions when the source f has a low summability (for instance f is L1 (Ω) function or it is a measure) and the uniqueness of solutions. Moreover we will study three problems where the hypotheses of the Leray–Lions problems are relaxed. Indeed we will present a problem defined by an elliptic operator with a quadratic term in the gradient; we will also study an equation defined by an elliptic operator with a polynomial growth term. Finally we will analyze a problem defined by an elliptic operator with degenerate coercivity. We have made an effort to keep this book self-contained. However, for background in real analysis, functional analysis, and Sobolev spaces, we refer to the book Functional analysis, Sobolev spaces and partial differential equations of Haïm Brezis [22]. For the reader’s convenience we have collected in the Appendices all the main prerequisites. Lucio Boccardo and Gisella Croce

Part I

2 Some fixed point theorems 2.1 Introduction In the study of the existence and uniqueness of solutions to differential equations the fixed point theorems have a very important role. Indeed, it is often possible to look for the solutions to a differential problem among the fixed points of a certain operator related to the differential problem. We will present three fixed point theorems in this chapter. The first result, the Banach–Caccioppoli theorem, states that a function from a complete metric space to the same space which contracts the distances has a unique fixed point. Thereafter, we will prove Brouwer’s theorem, which assures the existence of a fixed point for a continuous map from a convex bounded closed subset of RN to itself. We will finally present Schauder’s theorem, which is a useful tool in the study of certain differential problems, as we will see. In some sense, it is the analog of Brouwer’s theorem for operators defined on any Banach spaces.

2.2 Banach–Caccioppoli theorem Theorem 2.1. Let (X, d) be a complete metric space and let F : X → X be a map with the following property: there exists θ ∈ (0, 1) such that d(F (x), F (y)) ≤ θd(x, y) ,

∀x, y ∈ X .

(2.2.1)

Then there exists a unique x ∈ X such that F (x) = x , i.e., a unique fixed point of F . Remark 2.2. A map F that satisfies condition (2.2.1) is called contraction. Remark 2.3. Among the fixed point theorems that we present in this chapter, only the Banach–Caccioppoli theorem gives the uniqueness. We recall that it is used in the study of Cauchy problems for ordinary differential equations. Proof. The proof is based on an elementary iteration argument. Let us fix any x0 ∈ X and let us define xn := F (xn−1 ), n ≥ 1 . (2.2.2) Using hypothesis (2.2.1), we have d(xn+1 , xn ) = d (F (xn ), F (xn−1 )) ≤ θd(xn , xn−1 ) ≤ · · · ≤ θ n d(x1 , x0 ) (2.2.3)

for every n ≥ 0. The triangle inequality and (2.2.3) give, for every p ∈ N, 

p+1

d(xn+p+1 , xn ) ≤

i=1

d(xn+i , xn+i−1 ) ≤ (θ n+p + · · · + θ n ) d(x1 , x0 ) .

6

Some fixed point theorems

 n Now, applying Cauchy’s criterion for series to ∞ n=1 θ , we deduce that xn is a Cauchy series. Since X is complete, xn converges to some x ∈ X . The continuity of F , given by (2.2.1), implies that F (xn ) converges to F (x). Passing to the limit in (2.2.2), we get the existence of a fixed point. For the uniqueness, let x , y be two fixed points of F . By hypothesis (2.2.1) d(x, y) = d(F (x), F (y)) ≤ θ d(x, y) .

Since θ < 1 we deduce that x = y , that is, there exists one and only one fixed point.

2.3 Brouwer’s theorem Theorem 2.4 (Brouwer). Let K be a convex, closed, and bounded subset of RN and let f : K → K be a continuous function. Then f has a fixed point. Remark 2.5. The hypotheses on f are different from those of Theorem 2.1: indeed, in Brouwer’s theorem f is only assumed to be continuous; however the existence of a convex, closed, and bounded invariant set is required. We will follow the proof given in [35] (note that there exist several different proofs: for example there is one which uses the notion of topological degree and another one uses the notion of the homology group). Observe that we will make use of Banach– Caccioppoli theorem. We will denote with B(0, r ) the set {x ∈ RN : |x| < r }, for r > 0. Theorem 2.6. Let F : B(0, 1) → ∂B(0, 1) be a continuous function. Then there exists x ∈ ∂B(0, 1) such that F (x) = x . Proof. We assume by contradiction that F (x) = x for every x ∈ ∂B(0, 1). Let us define the following continuous extension of F : ⎧ ⎪ ⎨F (x) , if |x| ≤ 1 , f˜(x) = x ⎪ ⎩ , if |x| > 1; |x| we remark that |f˜(x)| = 1. The Weierstrass density theorem implies the existence of a C 1 (RN , RN ) map f1 such that   1   sup f˜(x) − f1 (x) < . (2.3.1) 2 x∈B(0,2) Let us consider any φ ∈ C 1 (R) such that 0 ≤ φ ≤ 1 and ⎧ ⎨1 , if t ≤ 3/2 , φ(t) = ⎩0 , if t ≥ 2 ,

Brouwer’s theorem

7

with φ decreasing for t ∈ (3/2, 2). We define the following combination of f˜ and f1 : fc (x) = [1 − φ(|x|)]f˜(x) + φ(|x|) f1 (x) .

Finally, we set N(x) =

fc (2x) . |fc (2x)|

With these definitions in mind, we prove the theorem in three steps. Step I: Let us prove that N is C 1 (RN , RN ) and Lipschitz continuous. For the regularity it suffices to prove that fc ∈ C 1 (RN , RN ) and that fc (x) ≠ 0 for every x ∈ RN . Now, we observe that if |x| > 1 fc (x) = (1 − φ(|x|))

x + φ(|x|) f1 (x) |x|

and so fc is a C 1 function in the set {x ∈ RN : |x| > 1}. On the other hand, if |x| < 3/2 then fc = f1 , and f1 ∈ C 1 (RN , RN ) by definition; therefore fc ∈ C 1 (RN , RN ). Also fc ≠ 0 for every x ∈ RN . Indeed              fc (x) ≥  f˜(x) − φ(|x|) f˜(x) − f1 (x) ≥ 1 − f˜(x) − f1 (x) > 12 , due to (2.3.1). Thus N ∈ C 1 (RN , RN ). Moreover N is Lipschitz continuous in B(0, 1), since N ∈ C 1 (B(0, 1), RN ). Away from B(0, 1), it holds that x . fc (2x) = [1 − φ(2|x|)]f˜(2x) + φ(2|x|)f1 (2x) = f˜(2x) = |x|

Therefore, away from B(0, 1) N(x) =

x |x|

which is clearly Lipschitz continuous if |x| > 1. Consequently there exists M > 0 such that |N(v) − N(w)| ≤ M |v − w| , ∀ v, w ∈ RN . (2.3.2) Step II: Let us prove that I + t N is a diffeomorphism, for t ∈ (0, M1 ), from B(0, 1) to B(0, t + 1). It is easily seen that the image of B(0, 1) is contained in B(0, t + 1), since, if |x| ≤ 1, then |x + t N(x)| ≤ |x| + t |N(x)| ≤ 1 + t . Moreover, if y ∈ B(0, t + 1), there exists a unique x ∈ B(0, 1) such that y = x + tN(x). Indeed the map T : RN → RN , defined by T (x) = y − t N(x), is a contraction by our choice of t , since |T (v) − T (w)| = t |N(v) − N(w)| ≤ t M |v − w| ,

∀ v, w ∈ RN ,

by (2.3.2). Theorem 2.1 implies the existence of a unique x0 = y − t N(x0 ). Let us prove that |x0 | ≤ 1. Assume by contradiction that |x0 | > 1; we have    x0   = |x0 | + t > 1 + t . |y| =  x + t  0 |x0 | 

8

Some fixed point theorems

This is absurd, since y ∈ B(0, t + 1). It remains to prove that det(D(I + tN)) ≠ 0

where D denotes the gradient of the map. If there were y ≠ 0 such that y = −t(DN)y , we would have |y| ≤ t|DN||y| ≤ t M|y|; this is in contradiction with the choice of t . Thus I + t N is a diffeomorphism between B(0, 1) and B(0, t + 1). Step III: Let us prove the existence of x0 ∈ B(0, 1) such that det DN(x0 ) ≠ 0. Let us distinguish the cases det(I + t DN) > 0 and det(I + t DN) < 0. In the first case, using a change of variables we have (1 + t)N dy = dy = det(I + tDN(x)) dx . B(0,1)

B(0,t+1)

B(0,1)

The last term is a polynomial in t of degree N , whose leading coefficient is det(DN(x))dx . B(0,1)

The identity principle of polynomials implies that dy = det(DN(x))dx . B(0,1)

B(0,1)

From this identity we infer the existence of x0 ∈ B(0, 1) such that det DN(x0 ) ≠ 0. In the case det(I + DN) < 0, with a similar argument, one has N (1 + t) dy = dy = − det(I + tDN(x)) dx. B(0,1)

As before,

B(0,t+1)

B(0,1)



dy = − B(0,1)

det(DN(x))dx B(0,1)

and so there exists x0 ∈ B(0, 1) such that det DN(x0 ) ≠ 0. Now, since DN is an isomorphism, its kernel is composed of only the zero vector. N(x0 ) belongs to the kernel of DN(x0 ): indeed, from the identity (N(x)|N(x)) = 1 ,

we deduce that DN(x) N(x) = 0 for every x ∈ RN and so N(x0 ) = 0. This is a contradiction, as |N(x)| = 1 for every x ∈ RN . We can now prove Brouwer’s theorem.

9

Brouwer’s theorem

Proof. We divide the proof into two steps. Step I: Let us prove the result in the case where K = B(0, 1). By contradiction, assume that f (x) ≠ x , for every x ∈ B(0, 1). Define F (x), for every x ∈ B(0, 1), as the intersection point between the half-line f (x)+λ(x −f (x)), λ ≥ 0, and ∂B(0, 1). We claim that F is continuous. We have, for some t(x) ∈ (0, 1] x = t(x)F (x) + (1 − t(x))f (x) .

(2.3.3)

This yields F (x) = s(x) x + (1 − s(x)) f (x) , 1 where s(x) = t(x) ≥ 1 is such that |F (x)| = 1. We are going to prove that s(x) is continuous; this will imply that F is continuous. One has, since F (x) ∈ ∂B(0, 1)

1 = |F (x)|2 = s 2 (x) |x − f (x)|2 + |f (x)|2 + 2s(x) (x − f (x)|f (x)) ,

that is, s 2 (x) |x − f (x)|2 + 2s(x) (x − f (x)|f (x)) + |f (x)|2 − 1 = 0 .

Now, for a fixed x , ψ(s) = s 2 |x − f (x)|2 + 2s(x − f (x)|f (x)) + |f (x)|2 − 1

is a polynomial of second degree, and so it has at most two zeros. Since ψ(1) ≤ 0 and lims→∞ ψ(s) = +∞, ψ has only one zero in [1, +∞). This implies that s(x) is well defined. Moreover, s is continuous, since it is a zero of a polynomial of second degree with continuous coefficients. Let us prove that F (x) = x , for |x| = 1. From (2.3.3) we have to prove that t(x) = 1. If t(x) ≠ 1, squaring (2.3.3) we obtain t 2 (x) + (1 − t(x))2 |f (x)|2 + 2t(x) (1 − t(x)) (F (x)|f (x)) = 1 ,

that is, (1 − t(x))2 |f (x)|2 + 2t(x) (1 − t(x)) (F (x)|f (x)) = (1 − t(x)) (1 + t(x)) .

Dividing by 1 − t(x) we get t(x) [2(F (x)|f (x)) − 1 − |f (x)|2 ] = 1 − |f (x)|2 ,

that is, t(x) |F (x) − f (x)|2 = −1 + |f (x)|2 .

This is a contradiction, since |f |2 − 1 < 0. Therefore t(x) = 1 and x = F (x). We have thus defined a continuous map F : B(0, 1) → ∂B(0, 1) and all the points of ∂B(0, 1) are fixed points: this is in contradiction with Theorem 2.6.

10

Some fixed point theorems

Step II: Let us study the case where K is any convex compact set. Since K is bounded, there exists R > 0 such that K ⊂ B(0, R). Let PK be the projection over K and f˜ :

B(0, R) → K ⊂ B(0, R) x → f (PK (x)) .

Due to Step I , f˜ has a fixed point x ∈ B(0, R); the image of f˜ is in K , so x ∈ K . Remark 2.7. In dimension 1, Brouwer’s theorem states that if f : [a, b] → [a, b] is continuous, then f has a fixed point. In this case, the existence of a fixed point can be proved in a very simple way: it is sufficient to apply the intermediate value theorem to ψ(t) = t − f (t).

2.4 Schauder’s theorem Schauder’s theorem is a fixed point result for operators over Banach spaces. We will present two different versions (see [46]). Theorem 2.8. Assume that X is a Banach space. Let K be a compact convex subset of X invariant under a continuous map F : K ⊂ X → X . Then F has a fixed point in K . To state the second version of Schauder’s theorem, we need the following definition: Definition 2.9. A map T : X → X is completely continuous if it is continuous and if, for every bounded subset B of X , T (B) is compact. Theorem 2.10. Let F be a completely continuous map and let K be a convex, bounded, closed and invariant subset of X . Then F has a fixed point in K . We are going to prove Theorem 2.8. We observe that we will make use of Brouwer’s theorem. x will denote the norm of an element x ∈ X and B(y, r ) = {x ∈ X : x − y < r }. Proof (of Theorem 2.8). Let ε > 0. Since K is compact, there exist x1 , . . . , xNε ∈ K such that Nε

K⊂ B(xi , ε) . (2.4.1) i=1

Now, let Eε be the vector space generated by {x1 , . . . , xNε } and let bj : K → R, j = 1, ..., Nε , be defined by bj (x) = (ε − x − xj )+ . Since for every x ∈ K , not every bj (x) is zero, we can define Nε j=1 bj (x) xj Gε (x) = Nε . j=1 bj (x)

Schauder’s theorem

11

Gε is a convex combination of x1 , . . . , xnε . This implies that Gε (K) ⊂ Eε ∩ K , as K is convex. We remark that Gε is continuous. We can apply Brouwer’s theorem to the function Gε ◦ F and to the compact convex set Eε ∩ K . Then there exists xε ∈ K ∩ Eε

such that Gε (F (xε )) = xε .

(2.4.2)

Since K ∩ Eε is compact, there exists a subsequence xε converging to some x0 ∈ K . The continuity of F implies that F (xε ) → F (x0 ). On the other hand, observe that for every x ∈ K it holds     Nε     Nε b (x)x   j=1 bj (x)(xj − x)  j=1 j j  − x = Gε (x) − x =  Nε   Nε   j=1 bj (x) j=1 bj (x) (2.4.3) Nε Nε j=1 bj (x)xj − x j=1 bj (x)ε ≤ ≤ Nε = ε, Nε j=1 bj (x) j=1 bj (x) by (2.4.1). From (2.4.2) and the fact that F (xε ) ∈ K one has ε > Gε (F (xε )) − F (xε ) = xε − F (xε ) .

At the limit as ε → 0, one gets x0 = F (x0 ). Proof (of Theorem 2.10). Let us fix ε > 0. Since F (K) is compact, there exist v1 , . . . , vNε ∈ F (K) ⊂ K such that F (K) ⊂

Nε

B(vi , ε) .

(2.4.4)

i=1

Now, let Eε be the vector space generated by {v1 , . . . , vNε } and let bj : K → R be defined by bj (x) = (ε − x − vj )+ . For every u ∈ F (K), we define Nε j=1

Gε (u) = Nε

bj (u)vj

j=1

bj (u)

.

Gε (u) is a convex combination of v1 , . . . , vNε : this implies that Gε ◦ F (K ∩ Eε ) ⊂ K ∩ Eε , as K is convex. We remark that Gε ◦ F is continuous. Brouwer’s theorem gives the existence of xε ∈ K ∩ Eε such that Gε (F (xε )) = xε . We observe that, for x ∈ K ,

one has

   Nε   j=1 bj (F (x))(vj − F (x)) Gε (F (x)) − F (x) = Nε j=1 bj (F (x)) Nε j=1 bj (F (x))vj − F (x) ≤ ≤ε , Nε j=1 bj (F (x))

12

Some fixed point theorems

by (2.4.4). This estimate and the fact that Gε (F (xε )) = xε yield F (xε ) − xε  → 0 ,

ε →0.

On the other hand, F (xε ) ∈ F (K) which is compact; up to a subsequence, F (xε ) → x0 . Now, xε − x0  ≤ xε − F (xε ) + F (xε ) − x0  and the right-hand side tends to 0, as ε → 0. This implies that xε → x0 and F (xε ) → F (x0 ) from the continuity of F . We have already proved that F (xε ) → x0 . The uniqueness of the limit of F (xε ) gives the result. Remark 2.11. In Brouwer’s theorem, the invariant set is closed and bounded, that is, compact, since it is a set of RN . In infinite dimensional spaces, due to a theorem by Riesz, the closed and bounded sets are not necessarily compact in general (see [22]). This is why compactness is required in Schauder’s theorems. The following example, due to Kakutani [31], shows that we can define in l2 a continuous operator which maps the unit ball to itself without admitting a fixed point. Example 2.12. Let T : l2 → l2 be defined by

1 (1 − x2 ), x1 , x2 , . . . , T (x) = 2  2 for every x = (x1 , x2 , . . . ) ∈ l2 , where x2 = ∞ i=1 |xi | . T is continuous, since T (x) − T (y)2 =

2 1 y2 − x2 + x − y2 . 4

Moreover the unit ball is invariant: indeed, if x ≤ 1, one has  T (x)2 =

2 1 (1 − x2 ) + x2 ≤ 1 , 2

since t → 1 − 14 (1 − t 2 )2 − t 2 is positive and decreasing for t ∈ [0, 1]. On the other hand, it is easily seen that T has no fixed point in the unit ball. Indeed, if x = 1, we have T (x) = (0, x1 , x2 , . . . ); if T (x) = x , then xj = 0, for every j , and so 1 x = 0 ≠ 1. If ||x|| = θ < 1, then T (x) = ( 2 (1 − θ 2 ), x1 , x2 , . . . ); if T (x) = x , 1 2 then xj = 2 (1 − θ ), for every j . This is impossible, since x ∈ l2 .

3 Preliminaries of real analysis 3.1 Introduction This chapter, devoted to some important results of real analysis, is divided into two parts. In the first one we will present Nemitski’s composition theorem. This result establishes the continuity of an operator defined between two Lebesgue spaces, through the composition with a real function. In the second part of this chapter we will define the Marcinkiewicz spaces M p , since they will be used later in several regularity results.

3.2 Nemitski’s composition theorem The aim of this section is to study the continuity of Lp (Ω) → Lq (Ω)

Φ:

u(x) → f (x, u(x))

defined between two Lebesgue spaces through the composition with a function f . The proof of the result that we will present is based on several theorems of real analysis. The following ones give some convergence results in Lp . In the sequel Ω will be an open bounded subset of RN . Theorem 3.1. Let fn be a sequence of functions and f be a function in Lp (Ω), p > 1. Assume that (1) fn is bounded in Lp (Ω); (2) fn → f a.e. in Ω. Then fn → f in Lq (Ω), for every q ∈ [1, p) and weakly in Lp (Ω). Proof. Assumption (1) implies the existence of a constant L > 0 such that fn − f Lp (Ω) ≤ L ,

∀n ∈ N.

Let k ∈ R+ . We have

(3.2.1)



kp meas({|fn − f | > k}) ≤

|fn − f |p {|fn −f |>k}



(3.2.2)

|fn − f |p ≤ Lp .

≤ Ω

For every q ∈ [1, p), one has |fn − f |q = Ω



{|fn −f |>k}

|fn − f |q +

|fn − f |q . {|fn −f |≤k}

14

Preliminaries of real analysis

Using Hölder’s inequality with exponent get

⎛ ⎜ |fn − f | ≤ ⎝

Ω

on the right-hand side of this equality, we ⎞q



q

p q

p

p⎟

q

1− p

|fn − f | ⎠ meas({|fn − f | > k})

{|fn −f |>k}



|fn − f |q .

+

{|fn −f |≤k}

Inequalities (3.2.1) and (3.2.2) imply then p−q L |fn − f |q ≤ Lq + k Ω

|fn − f |q . {|fn −f |≤k}

For every fixed k ∈ R+ Lebesgue’s theorem implies that the second term of the righthand side goes to 0, as n → ∞. Moreover, for every fixed ε > 0 there exists kε such that the first term is smaller than ε, for k ≥ kε . For such kε , there exists nε such that the second term is smaller than ε, for n ≥ nε . In conclusion fn → f in Lq (Ω) if q < p. Let us prove that fn → f weakly in Lp (Ω), p ≥ 1. Since fn is bounded in Lp (Ω), we can extract a subsequence which converges weakly in Lp (Ω). The limit is necessarily f , since fn → f in Lq (Ω), q < p . To prove that fn → f weakly in Lp (Ω) one can argue by contradiction. The following theorem gives us some sufficient conditions for the convergence in Lp (Ω) .

Theorem 3.2 (Vitali). Let fn be a sequence of functions and f be a function in Lp (Ω). Assume that (1) fn → f a.e. in Ω;  (2) limmeas(E)→0 E |fn |p = 0, uniformly with respect to n, if E is a measurable subset of Ω. Then fn → f in Lp (Ω). Proof. Let us fix ε > 0. Let E ⊂ Ω be a measurable set; we have |fn − f |p ≤ |fn − f |p + 2p−1 [|fn |p + |f |p ] . Ω

Ω\E

(3.2.3)

E

Using assumption (2), there exists δ1 (ε) > 0 such that, if meas(E) < δ1 (ε), then |fn |p < ε , ∀ n ∈ N . E

Nemitski’s composition theorem

15

Since f ∈ Lp (Ω) there exists δ2 (ε) > 0 such that, if meas(E) < δ2 (ε), then |f |p < ε . E

In conclusion the second term of the right-hand side of (3.2.3) is less than 2p ε. Let us study the first one. Setting δ = min{δ1 (ε), δ2 (ε)} and using Egorov’s theorem (Theorem 3.19), there exist νε ∈ N and a measurable set E0 ⊂ Ω such that meas(E0 ) < δ and |fn − f |p < ε , Ω\E0

for every n > νε . Choosing E = E0 in (3.2.3), we get the result. The following result is a corollary of Vitali’s theorem. Theorem 3.3. Let fn be a sequence of functions and f be a function in Lp (Ω), p ≥ 1. Then fn → f in Lp (Ω) if and only if (1) fn → f in measure;  (2) limmeas(E)→0 E |fn |p = 0 uniformly with respect to n, where E is a measurable subset of Ω. Proof. We divide the proof into two parts. Part I: assume that fn → f in Lp (Ω). Clearly fn → f in measure. Moreover, if E is any measurable subset of Ω, one has |fn |p = |fn − f + f |p ≤ 2p−1 |fn − f |p + 2p−1 |f |p . E

E

E

E

Let us fix ε > 0; since f ∈ Lp (Ω) there exists δ(ε) > 0 such that, if meas(E) < δ(ε), then ⎡ ⎤1 p ⎢ p⎥ (3.2.4) ⎣ |f | ⎦ < ε . E

On the other hand, since fn → f in Lp (Ω), there exists νε in N such that ⎛ ⎞1 ⎛ ⎞1 ⎛ ⎞1 p p p ⎜ ⎜ ⎜ p⎟ p⎟ p⎟ ⎝ |fn | ⎠ − ⎝ |f | ⎠ ≤ ⎝ |fn − f | ⎠ < ε , E

E

E

This implies that ∀ n > νε , if meas(E) < δ(ε) then ⎛ ⎞1 ⎛ ⎞1 p p ⎜ ⎟ ⎜ p p⎟ ⎝ |fn | ⎠ ≤ ε + ⎝ |f | ⎠ ≤ 2ε E

E

∀ n > νε .

(3.2.5)

16

Preliminaries of real analysis

by (3.2.4). Since f1 , . . . , fνε ∈ Lp (Ω), there exists δ1 (ε) such that if meas(E) < δ1 (ε) then |fj |p < ε , ∀ j = 1, . . . , νε . E

This proves the result. Part II: let us prove that assumptions (1) and (2) imply that fn → f in Lp (Ω). Since fn → f in measure, we can extract a subsequence such that fnk → f a.e. in Ω. Vitali’s theorem (Theorem 3.2) implies that fnk → f in Lp (Ω). Indeed, the whole sequence fn converges to f in Lp (Ω). If there exist a subsequence fnj and ε0 > 0 such that fnj − f Lp (Ω) ≥ ε0 , (3.2.6) using the previous argument, we could extract from fnj a subsequence converging to f in Lp (Ω). This is in contradiction with (3.2.6). We now give the definition of a Carathéodory function. Definition 3.4. A function g = g(x, ξ) : Ω × Rm → R is a Carathéodory function if it is continuous with respect to ξ , for almost every x in Ω and measurable with respect to x for every ξ in Rm . Lemma 3.5. Let f (x, t) : Ω×R → R be a Carathéodory function. Let un be a sequence of functions and u0 be a measurable function such that un → u0 in measure. Then f (x, un ) → f (x, u0 ) in measure. Proof. Let ε > 0 and let u be any measurable function. We set, for k > 0  1 Ωk = x ∈ Ω : |u0 (x) − u(x)| < ⇒ |f (x, u0 (x)) − f (x, u(x))| < ε . k ! Since f is continuous with respect to s , one has k∈N Ωk = Ω. Moreover limk→∞ meas(Ωk ) = meas(Ω), as Ωi ⊂ Ωj , for i < j . Therefore, for any fixed η > 0, there exists k0 such that η meas(Ω) − meas(Ωk0 ) < . 2 Let us set  1 An = x ∈ Ω : |un (x) − u0 (x)| < ; k0 since un → u0 in measure, there exists n0 such that meas(Ω) − meas(An )
n0 . We set Dn = {x ∈ Ω : |f (x, un (x)) − f (x, u0 (x))| < ε} .

17

Nemitski’s composition theorem

By definition, one has An ∩ Ωk0 ⊂ Dn and this yields " # meas(Ω) − meas(Dn ) < [meas(Ω) − meas(An )] + meas(Ω) − meas(Ωk0 ) η η < + = η. 2 2

This proves the result. We can now prove Nemitski’s composition theorem. Theorem 3.6 (Nemitski’s composition theorem). Let p, q ≥ 1. Let f (x, t) : Ω × R → R be a Carathéodory function. Assume that there exist a positive function a ∈ Lq (Ω) and a constant b > 0 such that p

|f (x, t)| ≤ a(x) + b|t| q .

(3.2.7)

Then the operator Φ:

Lp (Ω) → Lq (Ω) u(x) → f (x, u(x))

is continuous. Proof. Assume that un → u in Lp (Ω): we have to prove that Φ(un ) → Φ(u) in Lq (Ω). We will prove that Φ(un ) satisfies hypotheses 1 and 2 of Theorem 3.3. Clearly un → u in measure and so, using Lemma 3.5, f (x, un (x)) → f (x, u(x)) in measure, that is, Φ(un ) → Φ(u) in measure. Let us prove that lim |Φ(un )|q = 0 meas(E)→0

E

uniformly with respect to n. If E is a measurable subset of Ω, using (3.2.7), we get |f (x, un (x))|q ≤ 2q−1 a(x)q + 2q−1 b |un (x)|p , E

E



E

and so limmeas(E)→0 E |f (x, un = 0, uniformly with respect to n. Indeed the q 1 first term tends to 0, since a ∈ L (Ω); the second term tends to 0 by Theorem 3.3 applied to the sequence un . Theorem 3.3 applied to the sequence f (x, un (x)) concludes the proof. (x))|q

We end this section with a result that we will use in Chapter 5. Theorem 3.7. Let q > 1 and p ≥ 1. Let f (x, t) : Ω × R → R satisfy the same hypotheses as Theorem 3.6. If un → u weakly in Lp (Ω) and a.e. in Ω, then f (x, un ) → f (x, u) weakly in Lq (Ω). Proof. It is clear that f (x, un ) → f (x, u) a.e. in Ω. Since un Lp (Ω) is bounded, due to (3.2.7), f (x, un )Lq (Ω) is bounded too. Therefore, applying Theorem 3.1 to f (x, un ), we get that f (x, un ) → f (x, u) weakly in Lq (Ω).

18

Preliminaries of real analysis

3.3 Marcinkiewicz spaces In this section, we will define a functional space that will be natural in the study of the regularity of the solutions to some differential problems. Definition 3.8. Let Ω be a bounded open subset of RN . Let p ≥ 0. The Marcinkiewicz space M p (Ω) is the space of all measurable functions f : Ω → R with the following property: there exists a constant γ > 0 such that meas({|f | > λ}) ≤

γ , λp

∀λ > 0.

(3.3.1)

The norm of f ∈ M p (Ω) is defined by p

f M p (Ω) = inf{γ > 0 : (3.3.1) holds} .

It is easy to see that M p (Ω) ⊆ Lp (Ω) for p ≥ 1, as the following proposition shows: Proposition 3.9. Let p ≥ 1. Then Lp (Ω) ⊂ M p (Ω). Proof. Let f be an Lp (Ω) function. Then p |f | ≥ λp = λp meas ({|f | > λ}) , Ω

{|f |>λ}

that is, f belongs to M p (Ω). Remark 3.10. The inclusion Lp (Ω) ⊆ M p (Ω) is strict; indeed it is sufficient to consider the function f (x) = x1 in Ω = (0, 1) ⊂ R. This function does not belong to L1 ((0, 1)), but to M 1 ((0, 1)), because  

1 1  meas  ≤ . x  > λ λ We now want to prove that M p (Ω) is included in some Lebesgue space. We will denote by Ak the set {|f | ≥ k}, and by Bk the set {k ≤ |f | < k + 1}. The following lemma will prove to be useful: Lemma 3.11. Let r ≥ 1 and f : Ω → R be a measurable function. f ∈ Lr (Ω) if and  r −1 only if ∞ meas(Ak ) < +∞. k=0 k Proof. We begin with some simple remarks. Note that ∞  |f |r = |f |r . k=0 B

Ω

Moreover Ak = ∞  k=0

!∞

i=k Bi ,

(3.3.2)

k

and this union is disjoint; therefore

kr −1 meas(Ak ) =

∞  k=0

kr −1

∞  i=k

meas(Bi ) =

∞  i=0

meas(Bi )

i  k=0

kr −1 .

(3.3.3)

Marcinkiewicz spaces

19

Moreover, if g : R+ → R+ is an increasing and continuous function, then n 

n+1

g(k) ≤

g(t) dt ≤

k=0

In particular, we may set g(t) = t m−1 

j r −1 ≤

j=0

m

Step I: Assume that f ∈ Lr (Ω). equality of (3.3.4) in (3.3.3), we get ∞ 

, for r ≥ 1; hence,

t r −1 dt =

0

m−1  mr ≤ (j + 1)r −1 . r j=0

∞

r −1 meas(A ) k k=0 k

kr −1 meas(Ak ) ≤

k=0

g(k + 1) .

k=0

0 r −1

n 

∞ 

meas(Bi )

i=0

(3.3.4)

< +∞. Using the first in-

(i + 1)r . r

Since |f | ≥ i over Bi , one has ∞ 

∞  1 1 (1 + |f |)r = (1 + |f |)r r r i=0 Bi Ω ⎡ ⎤ r −1 2 ⎢ ⎥ ≤ ⎣meas(Ω) + |f |r ⎦ r

kr −1 meas(Ak ) ≤

k=0

Ω

where we have used (3.3.2). Step II: Assume that

∞ 

kr −1 meas(Ak ) < +∞ .

k=0

We are going to prove that f belongs to Lr (Ω). One has, using (3.3.3) and the last inequality of (3.3.4) ∞ 

∞ 

kr −1 meas(Ak ) =

k=0

meas(Bi )

i=0 ∞ 

=

kr −1

k=0

meas(Bi )

i=0

and so

i 

∞ 

i−1 

(h + 1)r −1 ≥

h=0

∞  i=0

meas(Bi )

ir r

meas(Bi ) ir < ∞ .

i=0

By the definition of Bi ∞ 

meas(Bi ) ir ≥

i=0

∞  i=2 B

This implies that f ∈ L (Ω). r

i

(|f | − 1)r ≥

1 2r −1

|f |r − meas(Ω) . Ω\(B0 ∪B1 )

20

Preliminaries of real analysis

Proposition 3.12. Let p > 1 and 0 < ε ≤ p − 1. Then M p (Ω) ⊂ Lp−ε (Ω). Proof. Let f ∈ M p (Ω). Using Lemma 3.11, it suffices to prove that ∞ 

kp−ε−1 meas(Ak ) < ∞ .

k=0

Since meas(Ak ) ≤

γ kp

for some γ > 0, one has ∞ 

kp−ε−1 meas(Ak ) ≤

k=0

∞ 

kp−ε−1

k=0

γ ; kp

the last series is finite since ε > 0. This proves the result. Proposition 3.13. Let f be an M p (Ω) function, p > 1. Then there exists B > 0 such that, for every measurable set E ⊂ Ω 1− 1 |f | ≤ B meas(E) p , (3.3.5) E

where B = B(f M p (Ω) , p). Proof. Once we prove that for every f ∈ L1 (Ω) +∞

|f | = E

meas(At ∩ E) dt ,

(3.3.6)

0

where E is any measurable set of Ω, the statement of the proposition is easy to prove. Indeed, if (3.3.6) holds, one has

+∞

|f | = E

meas(At ∩ E) dt = 0 1 −p

meas (E)

+∞

meas(At ∩ E) dt +

= 0

meas(At ∩ E) dt −1 meas(E) p

1− p1

≤ meas(E)

+∞

+

meas(At ) dt −1 meas(E) p

1− p1

≤ meas(E)

+

p f M p (Ω)

+∞

t −p dt 1 −p

meas(E) 1− p1

≤ B meas(E)

,

where B depends on f M p (Ω) and on p .

Marcinkiewicz spaces

21

 Step I: Assume that f (x) = α χE (x), α > 0. Then E |f (x)| = α meas(E). On the other hand, ⎧ ⎨E , if t ≤ α , At ∩ E = {x ∈ E : |f (x)| > t} = (3.3.7) ⎩∅ , if t > α ,

and so

+∞

α

meas(At ∩ E) dt = 0

meas(E) dt = α meas(E) . 0

Therefore (3.3.6) holds true for f (x) = α χE (x), α > 0.  Step II: Assume that f (x) = M i=1 αi χEi , where Ei are measurable subsets of E such that M

Ei = E , Ei ∩ Ej = ∅ se i ≠ j , i=1 +

αi ∈ R and M ∈ N. From Step I , one has

M 

|f | =

αi χ E i =

i=1 E

E

M +∞ 

meas(Ai,t ) dt

i=1 0

where Ai,t = {x ∈ Ω : αi χEi (x) > t} for i ∈ {1, . . . , M}. Now, ⎫ ⎧ M M ⎨ M M ⎬



 At ∩ E = (At ∩ Ei ) = αi χ E i > t = Ai,t . x ∈ Ei : ⎭ ⎩ i=1

i=1

i=1

i=1

This implies that M +∞ 

meas(Ai,t ) dt =

+∞  M

i=1 0

+∞

meas(Ai,t ) dt =

0 i=1

meas(At ∩ E) dt . 0

Therefore, (3.3.6) holds true for positive step functions. Step III: Let f be any function in L1 (Ω). There exists a sequence of positive step functions such that sn (x)  |f (x)| a.e. in E . From Beppo Levi’s theorem and from Step II , one has

|f | = lim

n→∞

E

+∞

sn = lim

meas{x ∈ E : |sn (x)| > t} dt

n→∞

E

0



+∞

= lim

dt

n→∞ 0

χ{x∈E:|sn (x)|>t} . E

(3.3.8)

22

Preliminaries of real analysis

Lebesgue’s theorem implies that lim χ{x∈E:|sn(x)|>t} = χAt ∩E .

(3.3.9)

n→∞

E

E



We set gn (t) =

χ{x∈E:|sn(x)|>t} . E

Then (3.3.8) is equivalent to

+∞

|f | = lim

gn (t) dt

n→∞

E

0

and (3.3.9) means that gn (t) → meas(At ∩ E) if n → ∞. To prove the result it suffices to prove that +∞ +∞ lim gn (t) dt = meas(At ∩ E) dt . n→+∞

0

0

We have that gn (t) → meas(At ∩ E) a.e. in (0, +∞); moreover |gn (t)| ≤ meas(E); it is sufficient to prove that meas(At ∩ E) belongs to L1 ((0, +∞)), and then apply Lebesgue’s theorem. This is easy, since +∞

1

meas(At ∩ E)dt ≤ 0

+∞

meas(At ∩ E) dt + 0

meas(At ) dt 1

+∞

≤ meas(E) + 1

p

f M p (Ω) tp

dt < +∞ .

Therefore (3.3.6) is proved.

3.4 Appendix We recall here the results on Lebesgue spaces that we use in this book (see [29] for more details). Let E be a Lebesgue measurable subset of RN , N ≥ 1. Let 1 < p < ∞; p  will p denote the number p−1 . 

Theorem 3.14 (Hölder’s inequality). Let f ∈ Lp (E) and g ∈ Lp (E). Then f g ≤ f Lp (E) gLp (E) . E

Appendix

23

Theorem 3.15 (Interpolation inequality). Let p, q, r ∈ [1, +∞) such that p < r < q. Let f ∈ Lq (E). Then f Lr (E) ≤ f θLp (E) f 1−θ Lq (E) , where θ is such that

1 r

=

θ p

+

1−θ q .

Theorem 3.16 (Beppo Levi). Let fn be a sequence of L1 (E) functions such that (1) 0 ≤ fn (x) ≤ fn+1 (x) a.e. in E for every n ∈ N;  (2) E fn < +∞ for every n ∈ N. Then fn → f in L1 (E). Theorem 3.17 (Lebesgue). Let fn be a sequence of L1 (E) functions such that (1) fn → f a.e. in E ; (2) there exists g ∈ L1 (E) such that |fn (x)| ≤ g(x) a.e. in E . Then fn → f in L1 (E). Theorem 3.18 (Fatou). Let fn be a sequence of L1 (E) functions such that (1) fn ≥ 0 a.e. in E ;  (2) E fn < +∞ for every n ∈ N.   Let f (x) = lim infn→∞ fn (x) for a.e. x ∈ E . Then E f ≤ lim infn→∞ E fn . Theorem 3.19 (Egorov). Let fn be a sequence of functions and f be a function defined on E , with meas(E) < +∞. Assume that fn → f a.e. in E . Then for every ε > 0 there exists a measurable subset A of E such that meas(E \ A) < ε and fn → f uniformly on A, as n → ∞. Theorem 3.20. Let 1 < p < ∞. A sequence fn of Lp (E) functions converges weakly   to f in Lp (E) if E (fn − f ) g → 0 for every g ∈ Lp (E). A sequence fn of L1 (E)  functions converges weakly to f in L1 (E) if E (fn − f )g → 0 for every g ∈ L∞ (E). Theorem 3.21. Let 1 < p < ∞. Every bounded sequence fn in Lp (E) has a subsequence weakly converging to some f ∈ Lp (E). Theorem 3.22 (Dunford–Pettis). Let fn be a bounded sequence of L1 (E) functions.  Assume that for every measurable subset A ⊂ E , one has A |fn | → 0, as meas(A) → 0, uniformly with respect to n. Then fn has a subsequence weakly converging to some f ∈ L1 (E). Definition 3.23. We recall that fn converges to f in measure in E if meas({x ∈ E : |fn − f | ≥ ε}) → 0 as ε → 0.

4 Linear and semilinear elliptic equations 4.1 Introduction In this chapter, we will study the existence of solutions to linear and semilinear elliptic problems. More precisely, let Ω be an open bounded subset of RN , N > 3. We assume that M(x) is an N × N symmetric matrix with the following properties: (1) M is elliptic, that is, there exists α > 0 such that M(x)ξ · ξ ≥ α|ξ|2 , ∀ ξ ∈ RN ; (2) M is bounded, meaning there exists β > 0 such that |M(x)| ≤ β , ∀ x ∈ Ω. The first class of problems that we will consider is linear: ⎧ ⎨−div(M(x)∇u) = f , in Ω , ⎩u = 0 , on ∂Ω . Afterward we will add a nonlinearity in u of the form ⎧ ⎨−div(M(x)∇u) + g(u) = f , in Ω , ⎩u = 0 , on ∂Ω . The existence results that we will present are based on Functional Analysis results by Lax, Milgram, and Stampacchia (see [32, 48, 51]).

4.2 The Lax–Milgram and Stampacchia’s theorems In this section, we present the Lax–Milgram theorem and Stampacchia’s theorem. They will be useful to study the elliptic problems of this chapter. We will use the following notations. Let H be a Hilbert space: (u|v) will denote ' the scalar product of two elements u, v ∈ H , and u = (u|u) will denote the norm of u ∈ H . If ϕ ∈ H  , ϕ, v will be the value of ϕ at v ∈ H . Proposition 4.1 (Stampacchia). Let H be a Hilbert space and K ⊆ H a closed convex set. Let a : K × K → R be a bilinear, continuous, coercive form, that is, there exist α, β > 0 such that a(u, v) ≤ βuv and a(u, u) ≥ αuv

for every u, v ∈ H . Then, for every fixed g ∈ H  there exists only one u ∈ K such that a(u, v − u) ≥ g, v − u ,

∀v ∈ K .

The Lax–Milgram and Stampacchia’s theorems

25

Proof. For g ∈ H  fixed, Riesz theorem (Theorem 4.16) posits the existence of a unique f ∈ H such that g, v = (f |v) , ∀ v ∈ H . Moreover, for any fixed u, the map v → a(u, v) is continuous and linear on H . Theorem 4.16 again implies that there exists A(u) ∈ H such that a(u, v) = (A(u)|v). The proposition will follow once we prove that there exists a unique u ∈ K such that (A(u) − f |v − u) ≥ 0 ,

∀v ∈ K ,

that is, (−λ A(u) + λ f |v − u) ≤ 0 ,

∀v ∈ K ,

for some λ > 0. Let C : H → K ⊂ H be the map which associates with z ∈ H the projection over K of z − λA(z) + λ f . Property (4.6.1) of the projection implies that (z − λA(z) + λ f − C(z)|v − C(z)) ≤ 0 ,

∀v ∈ K.

To prove the result, it is sufficient to find a fixed point of C . Using property (4.6.2) of the projection one has C(z1 ) − C(z2 )2 ≤ z1 − z2 − λ(A(z1 ) − A(z2 ))2 .

Since A is continuous and coercive, the following estimates hold: C(z1 ) − C(z2 )2 ≤ z1 − z2 2 + λ2 A(z1 ) − A(z2 )2 − 2λ (z1 − z2 |A(z1 ) − A(z2 )) ≤ z1 − z2 2 + λ2 β2 z1 − z2 2 − 2α λ z1 − z2 2 = (1 + λ2 β2 − 2α λ)z1 − z2 2 .

Therefore C is a contraction if 0 < λ < 2α/β2 . By Theorem 2.1 C has a unique fixed point. Theorem 4.2 (Lax–Milgram). Let H be an Hilbert space and a : H × H → R a bilinear, continuous, coercive form. Then, for every fixed ϕ ∈ H  , there exists a unique u ∈ H such that a(u, v) = ϕ, v , ∀ v ∈ H . Proof. Let ϕ ∈ H  ; by Proposition 4.1 there exists a unique u ∈ H such that a(u, v − u) ≥ ϕ, v − u ,

∀v ∈ H .

In particular a(u, t v − u) ≥ ϕ, t v − u ,

∀v ∈ H ,

∀t ∈ R,

26

Linear and semilinear elliptic equations

so that t [a(u, v) − ϕ, v] ≥ a(u, u) − ϕ, u ,

∀t ∈ R.

Hence a(u, v) − ϕ, v = 0

for every v ∈ H . Theorem 4.3 (Stampacchia). Let H be a Hilbert space. Let a : H × H → R be a continuous and linear form in the second variable such that (1) |a(ψ1 , w) − a(ψ2 , w)| ≤ β ψ1 − ψ2  w , ∀ ψ1 , ψ2 , w ∈ H ; (2) a(ψ1 , ψ1 − ψ2 ) − a(ψ2 , ψ1 − ψ2 ) ≥ C ψ1 − ψ2 2 , ∀ ψ1 , ψ2 ∈ H . Then, for every ϕ ∈ H  there exists a unique u ∈ H such that a(u, w) = ϕ(w) for every w ∈ H . Proof. We divide the proof into two steps. Step I: Let us prove that if A : H → H satisfies for some positive α, γ (1) A(x) − A(y) ≤ γ x − y , ∀ x, y ∈ H (2) (x − y|A(x) − A(y)) ≥ α x − y2 , ∀ x, y ∈ H then, for every fixed f ∈ H , there exists a unique u ∈ H such that A(u) = f . For this it is sufficient to prove that R(v) = v − λ A(v) + λ f

is a contraction for some λ. By definition R(v) − R(w)2 = (R(v) − R(w)|R(v) − R(w)) = v − w2 + λ2 A(v) − A(w)2 − 2λ (v − w|A(v) − A(w)) .

By the hypotheses on A, one has R(v) − R(w)2 ≤ (1 + λ2 γ 2 − 2λ α) v − w2 .

Thus R is a contraction as soon as 0 < λ < 2α γ2 . Step II: By Riesz theorem (Theorem 4.16), for every fixed ϕ ∈ H  there exists a unique f0 ∈ H such that ϕ(w) = (f0 |w) ,

∀w ∈ H .

To prove the theorem we must find u ∈ H such that a(u, w) = (f0 |w)

Linear equations

27

for every w ∈ H . Now, for every v ∈ H we can define the following linear continuous functional on H : Tv : H → R w → a(v, w) .

By Theorem 4.16 again, there exists a unique v0 ∈ H such that Tv (w) = a(v, w) = (v0 |w) for every w ∈ H . The operator H →H

A:

v → v0

satisfies inequalities (1) and (2) of Step I . Consequently, given f ∈ H there exists a unique u such that A(u) = f , that is, a(u, w) = (A(u)|w) = (f |w) for every w ∈ H . Therefore there exists a unique u such that a(u, w) = (f0 |w) = ϕ(w) for every w ∈ H .

4.3 Linear equations In this section we consider the linear problem ⎧ ⎨−div(M(x)∇u) = f , ⎩u = 0 ,

in Ω , on ∂Ω ;

(4.3.1)

we remark that on taking M as the identity matrix, one has the Dirichlet problem for the Laplacian operator: ⎧ ⎨−Δu = f , in Ω , ⎩u = 0 , on ∂Ω . The proof of the existence and uniqueness of a solution to problem (4.3.1) is based on Lax–Milgram theorem. 2N Theorem 4.4. Let f ∈ Lm (Ω), m ≥ N+2 . Then there exists a unique weak solution 1 u ∈ H0 (Ω) to problem (4.3.1). In other words, M(x)∇u · ∇v = f v , ∀ v ∈ H01 (Ω) . Ω

Proof. Define a :

H01 (Ω)

Ω

× H01 (Ω)

→ R by a(u, v) = M(x)∇u · ∇v. Ω

It is easily seen that a is continuous. Indeed, since M is bounded by hypothesis 2, the Cauchy–Schwarz inequality gives |a(u, v)| ≤ |M(x)∇u · ∇v| ≤ β ∇uL2 (Ω) ∇vL2 (Ω) , ∀ u, v ∈ H01 (Ω) . Ω

28

Linear and semilinear elliptic equations

On the other hand, a is coercive, since M(x)ξ · ξ ≥ α|ξ|2 and so M(x)∇u · ∇u ≥ α |∇u|2 , ∀ u ∈ H01 (Ω) . Ω

Ω

The result follows from the Lax–Milgram theorem. Remark 4.5. We will see a different proof of this theorem in Chapter 9.

4.4 Some semilinear monotone equations In this section we study the following semilinear problem ⎧ ⎨−div(M(x)∇u) + g(u) = f , in Ω , ⎩u = 0 , on ∂Ω .

(4.4.1)

Theorem 4.6. Let g : R → R be an increasing function. Suppose that g is Lipschitz continuous, that is, there exists a positive constant C such that |g(s) − g(t)| ≤ C |s − t| ,

∀

s, t ∈ R .

(4.4.2)

2N . Then there exists a unique solution u ∈ H01 (Ω) to probLet f ∈ Lm (Ω), m ≥ N+2 lem (4.4.1) in the following sense: M(x)∇u · ∇v + g(u)v = f v , ∀ v ∈ H01 (Ω) . Ω

Ω

Ω

Proof. Define the following form on H01 (Ω) × H01 (Ω): a(u, w) = M(x)∇u · ∇w + g(u)w . Ω

Ω

By the hypotheses on M and (4.4.2), one has |a(u, w)| ≤ β |∇u| |∇w| + [C |u| + g(0)] |w| , Ω

Ω

that is, a is well defined. The form a is continuous and linear in the second variable. Indeed, if wn → w in H01 (Ω), then M(x)∇u · ∇wn → M(x)∇u · ∇w Ω

and

Ω



g(u)wn → Ω

g(u)w Ω

Some semilinear monotone equations

29

since wn → w weakly in L2 (Ω) and |g(u)| ≤ [C|u| + g(0)] ∈ L2 (Ω). In other words, a(u, wn ) → a(u, w). Moreover, a satisfies hypothesis 1 of Theorem 4.3 because         |a(u1 , w) − a(u2 , w)| =  M(x)∇(u1 − u2 ) · ∇w + [g(u1 ) − g(u2 )]w    Ω  Ω ≤ β ∇(u1 − u2 )L2 (Ω) ∇wL2 (Ω) + C u1 − u2 L2 (Ω) wL2 (Ω)

the last inequality following from hypothesis 2 on M and hypothesis (4.4.2) on g . Finally a satisfies hypothesis 2 of Theorem 4.3: a(u1 , u1 − u2 ) − a(u2 , u1 − u2 ) = M(x)∇(u1 − u2 ) · ∇(u1 − u2 ) Ω



+ [g(u1 ) − g(u2 )](u1 − u2 ) Ω

≥ α∇(u1 − u2 )2L2 (Ω)

since M is elliptic and g is increasing. The result follows from Theorem 4.3. Theorem 4.7. Let g : R → R be an increasing locally Lipschitz continuous function. 2N Let f ∈ Lm (Ω), m ≥ N+2 . Then there exists a unique solution u ∈ H01 (Ω) to problem (4.4.1) in the following sense: M(x)∇u · ∇v + g(u)v = f v , ∀ v ∈ H01 (Ω) ∩ L∞ (Ω) . Ω

Ω

Ω

Moreover, g(u) ∈ L (Ω). 1

In the proof we make use of the following function defined for k > 0: ⎧ ⎪ ⎪ −k , s ≤ −k , ⎪ ⎨ Tk (s) = s , |s| ≤ k , ⎪ ⎪ ⎪ ⎩k , s ≥ k.

(4.4.3)

Proof. Step I: We prove the existence of a solution by approximation. To this end, let gn (t) = Tn (g(t)); let un ∈ H01 (Ω) be the solutions to problems ⎧ ⎨−div(M(x)∇un ) + gn (un ) = f , in Ω , ⎩u = 0 , on ∂Ω .

(4.4.4)

n

Such solutions exist due to Theorem 4.6, since gn is increasing and Lipschitz continuous. Considering un as a test function in (4.4.4) one gets M(x)∇un · ∇un + un gn (un ) = f un . Ω

Ω

Ω

30

Linear and semilinear elliptic equations

Hölder’s inequality on the right-hand side implies M(x)∇un · ∇un + un gn (un ) ≤ f 

2N

L N+2 (Ω)

Ω

un L2∗ (Ω) .

(4.4.5)

Ω

The ellipticity of M and the monotonicity of g used on the left-hand side of the above inequality give α∇un 2L2 (Ω) ≤ f  2N un L2∗ (Ω) . N+2 L

(Ω)

By Sobolev’s inequality on the left-hand side, un H01 (Ω) is uniformly bounded. We

deduce the existence of a H01 (Ω) function u such that un → u weakly in H01 (Ω)  and a.e., up to a subsequence. Moreover, since Ω M(x)∇un · ∇un ≥ 0, we deduce from (4.4.5) that there exists a constant C > 0 such that un gn (un ) ≤ C (4.4.6) Ω

for every n. Let us prove now that gn (un ) → g(u) in L1 (Ω). It is clear that gn (un ) → g(u) a.e. in Ω by the continuity of g . Moreover, if E is any subset of Ω and t ∈ R+ one has |gn (un )| = |gn (un )| + |gn (un )| {x∈E:|un (x)|≤t}

E



≤ E

1 |gn (t)| + t

{x∈E:|un (x)|>t}



un g(un ) {x∈E:|un (x)|>t}

≤ |g(t)| meas(E) +

C t

due to (4.4.6). Consequently |gn (un )| ≤

lim

meas(E)→0 E

C , t

∀t > 0.

By Theorem 3.2, gn (un ) → g(u) in L1 (Ω). Therefore for every ϕ ∈ H01 (Ω) ∩ L∞ (Ω) we can pass to the limit in M(x)∇un · ∇ϕ + gn (un )ϕ = f ϕ Ω

to get

Ω

Ω



M(x)∇u · ∇ϕ + Ω

for every ϕ ∈ H01 (Ω) ∩ L∞ (Ω).

g(u)ϕ =

Ω

fϕ Ω

31

Sub and supersolutions method

Step II: Let us prove that the solution to problem (4.4.1) is unique. By contradiction, if u1 and u2 are two solutions, then M(x)∇u1 · ∇Tk (u1 − u2 ) + g(u1 )Tk (u1 − u2 ) = f Tk (u1 − u2 ); Ω

Ω

Ω





M(x)∇u2 · ∇Tk (u1 − u2 ) + Ω



g(u2 )Tk (u1 − u2 ) = Ω

f Tk (u1 − u2 ) . Ω

This implies that M(x)∇(u1 − u2 ) · ∇Tk (u1 − u2 ) = M(x)∇Tk (u1 − u2 ) · ∇Tk (u1 − u2 ) ≤ 0 Ω

Ω

by the monotonicity of g . Since M is elliptic, Tk (u1 − u2 ) = 0 a.e. for every k; therefore u1 = u2 a.e. in Ω. Example 4.8. Using the previous theorem one sees immediately that there exists a solution u to u ∈ H01 (Ω) ∩ Lp−1 (Ω) : eu − 1 ∈ L1 (Ω) :

−div(M(x)∇u) + |u|p−2 u = f u ∈ H01 (Ω), −div(M(x)∇u) + eu − 1 = f .

Remark 4.9. In Chapters 10 and 11, we shall again study approximating problems to get a priori estimates and then pass to the limit.

4.5 Sub and supersolutions method In this section we will study the semilinear problem ⎧ ⎨−div(M(x)∇u) = g(u) + f , ⎩u = 0 ,

in Ω , on

∂Ω ,

(4.5.1)

under the following hypotheses: 2N (1) f ∈ L N+2 (Ω); (2) g : R → R is increasing and continuous; there exists γ > 0 such that |g(s)| ≤ γ|s|a ,

a≤

N +2 . N −2

We recall that (1) M is elliptic, that is, there exists α > 0 such that M(x)ξ · ξ ≥ α|ξ|2 , ∀ ξ ∈ RN ; (2) M is bounded, meaning there exists β > 0 such that |M(x)| ≤ β , ∀ x ∈ Ω.

32

Linear and semilinear elliptic equations

We remark that as in the previous section g is assumed to be increasing, but here it appears on the right-hand side. We will solve this problem using the sub and supersolutions method (see [45]). Definition 4.10. A function u ∈ H01 (Ω) is a subsolution to problem (4.5.1) if for every positive v ∈ H01 (Ω) M(x)∇u · ∇v ≤ g(u)v + f v . Ω

Ω

Ω

A function u ∈ H01 (Ω) is a supersolution to problem (4.5.1) if for every positive v ∈ H01 (Ω) M(x)∇u · ∇v ≥ Ω

g(u)v + Ω

fv . Ω

Theorem 4.11. Under the previous hypotheses, let u and u be a sub and a supersolution to (4.5.1) such that u ≤ u a.e. in Ω. Then there exists u ∈ H01 (Ω) to problem (4.5.1); moreover u ≤ u ≤ u a.e. in Ω. Theorem 4.12 (Maximum Principle). Let u ∈ H01 (Ω).  (1) Assume that Ω M(x)∇u · ∇v ≤ 0 for every positive v ∈ H01 (Ω). Then u ≤ 0.  (2) Assume that Ω M(x)∇u · ∇v ≥ 0 for every positive v ∈ H01 (Ω). Then u ≥ 0. Proof. (1) Choose v = u+ , that is, the positive part of u as a test function. Then M(x)∇u+ · ∇u+ = M(x)∇(u+ − u− ) · ∇u+ ≤ 0 . Ω

Ω

Using the ellipticity of M , u+ = 0. (2) Choose v = u− , that is, the negative part of u as a test function. Then + − − 0 ≤ M(x)∇(u − u ) · ∇u = − M(x)∇u− · ∇u− . Ω

Ω

The ellipticity of M gives u− = 0. We can now prove Theorem 4.11. Proof. The proof is divided into two steps: in the first one we construct a sequence un in H01 (Ω) by induction and prove that u ≤ u1 ≤ · · · ≤ un ≤ · · · ≤ u; in the second one we prove that this sequence converges to a solution u to problem (4.5.1). Step I: Set u1 =u; let un ∈ H01 (Ω) be the solution to ⎧ ⎨−div(M(x)∇un ) = g(un−1 ) + f , ⎩u = 0 , n

in Ω , on

∂Ω .

Sub and supersolutions method

33

2N

Such a solution exists by Theorem 4.4, since g(un−1 ) + f ∈ L N+2 (Ω). Let us prove by induction that the sequence un is increasing. For every ϕ ≥ 0, by the hypotheses and the definition of u2 , one has M(x)∇u1 · ∇ϕ ≤ [g(u1 ) + f ] ϕ Ω

Ω





M(x)∇u2 · ∇ϕ = Ω

Consequently

[g(u1 ) + f ] ϕ . Ω

M(x)∇(u1 − u2 ) · ∇ϕ ≤ 0 . Ω

Theorem 4.12 implies that u1 − u2 ≤ 0. Let us prove that un ≤ un+1 , assuming that un ≥ un−1 . We have M(x)∇un · ∇ϕ = [g(un−1 ) + f ] ϕ

Ω

Ω



M(x)∇un+1 · ∇ϕ = Ω

[g(un ) + f ] ϕ . Ω

Consequently

M(x)∇(un − un+1 ) · ∇ϕ =

Ω

[g(un−1 ) − g(un )] ϕ. Ω

Since g is increasing, g(un−1 ) ≤ g(un ) and so un ≤ un+1 by Theorem 4.12. Therefore un is increasing. With the same technique one can prove that un ≤ u. Step II: We want to prove that un converges to a solution to problem (4.5.1). Let ∗ us first prove that the un converges to some u ∈ L2 (Ω). Since un is increasing and un ≤ u, un (x) has a limit a.e., say u(x). It follows from Step I that |un | ≤ |u|+|u|; ∗ passing to the limit, one has |u| ≤ |u| + |u| ∈ L2 (Ω). Hence, for a positive constant C(N) ∗

∗

∗

∗

|un − u|2 ≤ C(N)(|un |2 + |u|2 ) ≤ C(N)(|u| + |u|)2 ∈ L1 (Ω) . ∗

Lebesgue’s theorem implies that un → u in L2 (Ω). Using Theorem 3.6 and hypoth2N esis 2 on g , g(un ) → g(u) in L N+2 (Ω). Let us prove that u is a solution to problem (4.5.1). By the definition of un , one has M(x)∇un · ∇un = (g(un−1 ) + f ) un . Ω

Ω

Using Hölder’s inequality on the right-hand side and the ellipticity of M on the lefthand side, one gets α∇un 2L2 (Ω) ≤ g(un−1 ) + f 

2N

L N+2 (Ω)

un L2∗ (Ω) .

34

Linear and semilinear elliptic equations

The right-hand side is uniformly bounded; therefore un is bounded in H01 (Ω) and, up to a subsequence, has a weak limit in H01 (Ω), which necessarily is u. We can now pass to the limit in M(x)∇un+1 · ∇v = [g(un ) + f ] v , ∀ v ∈ H01 (Ω) Ω

to get

Ω



M(x)∇u · ∇v = Ω

[g(u) + f ] v ,

∀ v ∈ H01 (Ω)

Ω

that is, u is a solution to problem (4.5.1). Example 4.13. Let f be a positive L∞ (Ω) function and Ω ⊂ RN , N ≤ 6. Using the previous theorem it is easy to prove that the following problem: ⎧ ⎨−Δu = u2 − f , in Ω , (4.5.2) ⎩u = 0 , on ∂Ω , has a solution. Indeed u = 0 is a supersolution. On the other hand, using Theorem 4.12 the solution ψ to problem ⎧ ⎨−Δψ = −f , in Ω , ⎩ψ = 0 , on ∂Ω , is a negative subsolution to problem (4.5.2).

4.6 Appendix We recall here the results on Hilbert spaces and Sobolev spaces that we have used in this chapter (we refer to [22] for more details).

4.6.1 A brief review of functional analysis

Definition 4.14. Let H be a Hilbert space. Let a : H × H → R be a bilinear form. (1) a is continuous if there exists β > 0 such that |a(u, v)| ≤ βuv ,

∀ u, v ∈ H .

(2) a is coercive if there exists α > 0 such that a(u, u) ≥ αu2 ,

∀u ∈ H .

35

Appendix

Theorem 4.15 (Projection). Let H be a Hilbert space and let K ⊂ H be a closed, not empty, convex set. Then for every g in H there exists a unique point in K , denoted by PK g , such that g − PK g ≤ g − v , ∀ v ∈ K . Moreover, PK g satisfies (g − PK g|v − PK g) ≤ 0 ,

∀v ∈ K

(4.6.1)

∀ g1 , g2 ∈ K .

(4.6.2)

and PK g1 − PK g2  ≤ g1 − g2  ,

Theorem 4.16 (Riesz). Let H be a Hilbert space and let ϕ : H → R be a continuous linear functional. Then there exists a unique g ∈ H such that ϕ, v = (g|v) ,

∀v ∈ H .

4.6.2 A brief review on Sobolev spaces

Let 1 ≤ p < N ; we will denote by p ∗ the real number such that 1 1 1 − . = p∗ p N

Theorem 4.17 (Sobolev embeddings). The following embeddings are continuous: ∗ (1) W 1,p (Ω) ⊂ Lp (Ω) if 1 ≤ p < N ; (2) W 1,p (Ω) ⊂ Lq (Ω) , ∀ q ∈ [p, +∞) if p = N ; (3) W 1,p (Ω) ⊂ L∞ (Ω) if p > N . 1,p

In particular, for every u ∈ W0 (Ω), there exists a positive constant S depending only on N and p such that SuLp∗ (Ω) ≤ ∇uLp (Ω) ;

this inequality is called Sobolev’s inequality. Theorem 4.18 (Rellich–Kondrachov). The following embeddings are compact: (1) W 1,p (Ω) ⊂ Lq (Ω) , ∀q ∈ [1, p ∗ ) if 1 ≤ p < N ; (2) W 1,p (Ω) ⊂ Lq (Ω) , ∀q ∈ [1, +∞) if p = N ; (3) W 1,p (Ω) ⊂ C(Ω) if p > N . In particular, W 1,p (Ω) ⊂ Lp (Ω) for every p and the embedding is compact. Remark 4.19. We recall that if X and Y are Banach spaces, an operator T : X → Y is compact if the image of a bounded subset of X is relatively compact in Y .

36

Linear and semilinear elliptic equations

1,p

Theorem 4.20. Let 1 < p < +∞. Then W0 (Ω) is reflexive, that is, the unit ball is 1,p compact for the weak topology in W0 (Ω). Theorem 4.21 (Poincaré’s inequality). Let 1 ≤ p < +∞. Then there exists a positive constant c = c(Ω, p) such that 1,p

uLp (Ω) ≤ c∇uLp (Ω) , ∀u ∈ W0 (Ω). 1,p

In particular, ∇u(Lp (Ω))N is a norm on W0 (Ω) which is equivalent to the norm uW 1,p (Ω) . 0

1,p

Corollary 4.22. Let 1 < p < +∞. Let un be a bounded sequence of W0 (Ω) func1,p tions. Then there exist a subsequence and a W0 (Ω) function u such that ∇un → ∇u p weakly in L (Ω). 

Theorem 4.23. Let F ∈ W −1,p (Ω), p ∈ (1, ∞). Then there exist f0 , f1 , . . . , fN ∈  Lp (Ω) such that F , v =

f0 v + Ω

N  i=1 Ω

fi

∂v , ∂xi

1,p

∀v ∈ W0 (Ω) .

5 Nonlinear elliptic equations 5.1 Introduction In this chapter, we prove an existence result for solutions to certain nonlinear elliptic problems. More precisely, we study the following boundary value problem: ⎧ ⎨−div(a(x, u, ∇u)) = F (x, u, ∇u) , in Ω , (LL) ⎩u = 0 , on ∂Ω . The existence of solutions u was proved by Leray and Lions in [34]. This is the reason why we call (LL) Leray–Lions problem. Theorem 5.1 (Leray–Lions). Let Ω be an open bounded set of RN and p ∈ (1, ∞). Let a : Ω × R × RN → RN and F : Ω × R × RN → R be two Carathéodory functions, with the following properties: (1) there exists β > 0 such that |a(x, s, ξ)| ≤ β [|s|p−1 + |ξ|p−1 ]; (2) there exists α > 0 such that a(x, s, ξ) · ξ ≥ α|ξ|p , ∀ ξ ∈ RN ; (3) [a(x, s, ξ) − a(x, s, η)] · [ξ − η] > 0 if ξ = η;  (4) there exists f ∈ Lp (Ω) such that |F (x, s, ξ)| ≤ f (x). 1,p

Then there exists a solution u ∈ W0 (Ω) to problem (LL), that is, 1,p  a(x, u, ∇u) · ∇v = F (x, u, ∇u) v , ∀ v ∈ W0 (Ω) . Ω

Ω

We point out that the proof of this theorem is based on an abstract result of surjectivity. Related results proved by Brezis, Browder, and Minty can be found in [21], [25], and [39], respectively. As we will see, the nonlinearity a makes the problem much more difficult than the semilinear problems of Chapter 4.

5.2 Surjectivity theorem We will need the following definitions. If V is a Banach space, ||x|| will denote the norm of an element x ∈ V , ϕV  the norm of an element ϕ ∈ V  and finally ϕ, v will denote ϕ(v), for ϕ ∈ V  and v ∈ V .

38

Nonlinear elliptic equations

Definition 5.2. Let V be a Banach reflexive space. An operator A : V → V  is pseudomonotone if (1) A is bounded, that is, the image of a bounded subset of V is a bounded subset of V ; (2) if uj → u weakly in V and if lim supj→+∞ A(uj ), uj − u ≤ 0, then lim infA(uj ), uj − v ≥ A(u), u − v j→+∞

for every v in V . Definition 5.3. Let V be a Banach reflexive space. An operator A : V → V  is coercive if A(v), v → +∞, as v → +∞ . v In the proof of the surjectivity theorem we will use the following lemma: Lemma 5.4. Let T : Rm → Rm be a continuous map, m ≥ 1. Assume that there exists ρ > 0 such that T (ξ) · ξ ≥ 0, for every ξ with |ξ| = ρ . Then there exists ξ with |ξ| ≤ ρ such that T (ξ) = 0. Proof. By contradiction, assume that T (ξ) ≠ 0 in B(0, ρ) = {ξ ∈ Rm : |ξ| ≤ ρ}. We consider the following continuous map from B(0, ρ) to itself: ξ→

−T (ξ)ρ . |T (ξ)|

Brouwer’s theorem (Theorem 2.4) implies that there exists a fixed point, that is, there exists ξ such that T (ξ)ρ . ξ=− |T (ξ)| We deduce that |ξ| = ρ . On the other hand, T (ξ) · ξ = −ρ|T (ξ)| < 0. This is in contradiction with the hypothesis that T (ξ) · ξ ≥ 0 for every ξ such that |ξ| = ρ . We can now prove the surjectivity theorem. Theorem 5.5 (Surjectivity Theorem). Let V be a Banach reflexive, separable space. Let A : V → V  be a pseudomonotone coercive operator. Then A is surjective, that is, for every f in V  there exists u in V such that A(u) = f . Proof. We will divide the proof into several steps. Step I: Let {w1 , . . . wn , . . . } be a dense and countable subset of V . Let us denote with Vn the subspace of V generated by {w1 , . . . wn }. We define T :

Vn → Vn v →< A(v) − f , v > v .

Surjectivity theorem

39

We are going to prove that T is continuous. It suffices to prove that v → A(v), v is continuous on Vn . Assume that wm → w : we claim that A(wm ) → A(w) weakly in V  . Since A is bounded, A(wm )V  is bounded uniformly. This implies that lim A(wm ), wm − w = 0 .

m→∞

Moreover there exists a subsequence such that A(wmk ) → g weakly in V  , as V  is reflexive. This and the pseudomonotonicity imply that lim infA(wmk ), wmk − v = g, w − v ≥ A(w), w − v mk →∞

for every v ∈ V . Then g = A(w) and therefore A(wmk ) → A(w) weakly in V  . Now, assume that A(wm ) does not converge weakly to A(w) in V  : the previous argument implies a contradiction. Since A(wm ) → A(w) weakly in V  and wm → w in V , then A(wm ), wm  → A(w), w. This implies the continuity of T . Step II: We are going to prove that for every n ∈ N there exists un ∈ Vn such that A(un ), wj  = f , wj  ,

1 ≤ j ≤ n.

(5.2.1)

We claim that T satisfies T (v) · v ≥ 0 for every v with v = ρ , for a certain ρ > 0. Since T is continuous by the previous step, Lemma 5.4 will imply the existence of un . We have that T (v) · v ≥ 0, because A(v), v − f , v ≥ A(v), v − f V  v ≥ 0 ,

if v = ρ , for ρ sufficiently large, due to the coercivity of A. Step III: From (5.2.1), we deduce that A(un ), un  = f , un  ≤ f V  un  .

Using the coercivity of A, we found that un  is uniformly bounded. Since A is bounded, A(un )V  is uniformly bounded too. Consequently there exists a subsequence unk such that ⎧ ⎨un → u k ⎩A(u ) → χ nk

weakly in V , weakly in V  .

Passing to the limit for n → +∞ in (5.2.1), (with j fixed) we have for every j χ, wj  = f , wj  .

Since {w1 , . . . , wn , . . . } is a dense set of V , χ = f . Let us prove that χ = A(u) .

(5.2.2)

40

Nonlinear elliptic equations

We claim that lim supA(unk ), unk − u ≤ 0 . nk →∞

(5.2.3)

For a given ε > 0, there exists u0 ∈ ∪m Vm such that ||u − u0 || ≤ ε. Now, A(unk ), unk − u = f , unk  − A(unk ), u − u0  − A(unk ), u0 .

By (5.2.2) the first term of the right-hand side tends to f , u and the third one tends to χ, u0  = f , u0 ; the second one can be estimated by Cε, for some C > 0, since A is bounded. Using that ||u − u0 || ≤ ε we get (5.2.3). The pseudomonotonicity of A implies that A(u), u − v ≤ lim infA(unk ), unk − v ≤ χ, u − v nk →∞

for every v ∈ V . It follows that χ = A(u), that is, f = A(u).

5.3 The Leray–Lions existence theorem We are going to prove the Leray–Lions theorem. We recall the statement Theorem 5.6. Let Ω be an open bounded set of RN and p ∈ (1, ∞). Let a : Ω × R × RN → RN and F : Ω × R × RN → R be two Carathéodory functions, with the following properties: (1) there exists β > 0 such that |a(x, s, ξ)| ≤ β [|s|p−1 + |ξ|p−1 ]; (2) there exists α > 0 such that a(x, s, ξ) · ξ ≥ α|ξ|p , ∀ ξ ∈ RN ; (3) [a(x, s, ξ) − a(x, s, η)] · [ξ − η] > 0 if ξ = η;  (4) there exists f ∈ Lp (Ω) such that |F (x, s, ξ)| ≤ f (x). 1,p

Then there exists a solution u ∈ W0 (Ω) to problem (LL), that is, 1,p  a(x, u, ∇u) · ∇v = F (x, u, ∇u) v , ∀ v ∈ W0 (Ω) . Ω

Ω

Remark 5.7. An application a satisfying hypothesis 2 will be called elliptic. In the Leray–Lions theorem we will apply the surjectivity theorem to the operator A:

1,p

W0 (Ω) → W −1,p (Ω) v → −div(a(x, v, ∇v)) − F (x, v, ∇v) .

We will use the following lemma to prove that A is pseudomonotone.

41

The Leray–Lions existence theorem

1,p

Lemma 5.8. Assume that un → u weakly in W0 (Ω) and that [a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u) → 0 a.e. in Ω. Then ∇un → ∇u a.e. in Ω. Proof. Since [a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u) → 0

a.e. in Ω

one has |[a(x, unk , ∇unk ) − a(x, u, ∇u)] · ∇(unk − u)| ≤ C(x)

for some function C(x). Up to a Lebesgue measure zero set Z , the above inequality holds pointwise. Let us prove that there exists a function c such that |∇unk (x)| ≤ c(x) .

(5.3.1)

One has, by hypotheses 1 and 2 on a C(x) ≥ [a(x, unk , ∇unk ) − a(x, u, ∇u)] · ∇(unk − u) ≥ α[|∇unk |p + |∇u|p ] − |∇unk |[β(|u|p−1 + |∇u|p−1 )] p−1

− |∇u|[β(|unk |

(5.3.2)

p−1 + |∇unk |)] .

1,p

The W0 (Ω) weak convergence of un to u implies the existence of a subsequence (still denoted by un ) and of a function g in L1 (Ω) such that |un |p−1 |∇u| ≤ g

and un → u a.e. in Ω.

Since in (5.3.2) we have a polynomial in |∇un |, (5.3.1) follows. We are going to prove that ∇un (x) → ∇u(x) in Ω \ Z . (5.3.3) Assume by contradiction that there exists x0 ∈ Ω \ Z such that ∇un (x0 ) does not converge to ∇u(x0 ). The Bolzano–Weierstrass theorem implies that ∇un (x0 ) → b, for some b ∈ RN , up to a subsequence. Passing to the limit in [a(x0 , un (x0 ), ∇un (x0 )) − a(x0 , u(x0 ), ∇u(x0 ))] · ∇(un (x0 ) − u(x0 ))

we get [a(x0 , u(x0 ), b) − a(x0 , u(x0 ), ∇u(x0 ))] · (b − ∇u(x0 )) = 0

which yields b = ∇u(x0 ) by hypothesis 3 on a. 1,p

1,p

Lemma 5.9. Let un , u be W0 (Ω) functions such that un → u weakly in W0 (Ω). If [a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u) → 0 , (5.3.4) Ω 

then a(x, un , ∇un ) → a(x, u, ∇u) weakly in Lp (Ω).

42

Nonlinear elliptic equations

Proof. We claim that [a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u) → 0

in

L1 (Ω) .

(5.3.5)

We can write [a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)

as the sum of αn = [a(x, un , ∇un ) − a(x, un , ∇u)] · ∇(un − u)

and βn = [a(x, un , ∇u) − a(x, u, ∇u)] · ∇(un − u) .

We observe that βn → 0 in L1 (Ω), since by Hölder’s inequality |βn | ≤ a(x, un , ∇u) − a(x, u, ∇u)Lp (Ω) ∇un − ∇uLp (Ω) . Ω 

Now, ∇un − ∇uLp (Ω) is bounded, and a(x, un , ∇u) → a(x, u, ∇u) in Lp (Ω)  due to Theorem 3.6, as un → u in Lp (Ω). By (5.3.4) Ω αn + βn → 0 and βn → 0  in L1 (Ω) as we have just proved. This implies that Ω αn → 0; since αn ≥ 0 for the monotonicity of a, we have that αn + βn → 0 in L1 (Ω), that is, (5.3.5) holds. Up to a subsequence, [a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u) → 0

a.e. in Ω. By Lemma 5.8 ∇un → ∇u a.e. in Ω. Theorem 3.7 implies the result. We can now prove the Leray–Lions theorem. Proof. We will prove that A(v) = −div(a(x, v, ∇v)) − F (x, v, ∇v) is coercive and pseudomonotone; the result will follow from Theorem 5.5. Step I: Hypotheses 2 and 4 on a give A(v), v ≥ α |∇v|p − |f ||v| ≥ α |∇v|p − f Lp (Ω) vLp (Ω) . Ω

Ω

Ω

Poincaré’s inequality implies the coercivity of A. Step II: By hypotheses 1 and 4 on a one has A(v), w = a(x, v, ∇v) · ∇w − F (x, v, ∇v)w Ω

⎡

Ω



⎢ ≤ β ⎣ |v|p−1 |∇w| + Ω

Ω

⎤

⎥ |∇v|p−1 |∇w|⎦ +

|f ||w| Ω

 p−1 p−1 ≤ β vLp (Ω) ∇wLp (Ω) + ∇vLp (Ω) ∇wLp (Ω)  + f Lp (Ω) wLp (Ω)

The Leray–Lions existence theorem

43

and so A(v)W −1,p (Ω) is bounded, if vW 1,p (Ω) is bounded. 0

1,p

Step III: Assume that un → u weakly in W0 (Ω) and 0 ≥ lim supA(un ), un − u n→+∞ (5.3.6) = lim sup a(x, un , ∇un ) · ∇(un − u) + F (x, un , ∇un )(un − u) . n→+∞

Ω

Ω

We are going to prove that lim [a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u) = 0 . n→+∞

(5.3.7)

Ω

This will be used in Step IV to prove the pseudomonotonicity of the operator.  We observe that Ω F (x, un , ∇un )(un −u) → 0, since un → u in Lp (Ω) and the  sequence F (x, un , ∇un ) is bounded in Lp (Ω) due to the hypotheses on F . By (5.3.6) this implies that lim sup [a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u) ≤ 0 . n→+∞

Ω

Using hypothesis 3 on a, we have [a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u) Ω

[a(x, un , ∇u) − a(x, u, ∇u)] · ∇(un − u)

≥ Ω

and this last term goes to 0, since a(x, un , ∇u) → a(x, u, ∇u) by Theorem 3.6. We have thus proved (5.3.7). 1,p Step IV: Assume that un → u weakly in W0 (Ω) and (5.3.6). We will prove that 1,p for every w ∈ W0 (Ω) lim infA(un ), un − w ≥ A(u), u − w . n→+∞

We remark that



A(un ), un − w =

(5.3.8)

a(x, un , ∇un ) · ∇(un − w) +

Ω

F (x, un , ∇un )(un − w) . Ω

We will separately study the two terms of the right-hand side. The limit of (5.3.7) allows us to apply Lemma 5.9 to deduce that a(x, un , ∇un ) →  a(x, u, ∇u) weakly in Lp (Ω), and so a(x, un , ∇un ) · ∇w → a(x, u, ∇u) · ∇w (5.3.9) Ω

Ω

44

Nonlinear elliptic equations

1,p

for every w ∈ W0 (Ω). On the other hand, by Lemma 5.8, a(x, un , ∇un ) → a(x, u, ∇u) and ∇un → ∇u a.e. in Ω. Since a(x, un , ∇un ) · ∇un ≥ 0 by the ellipticity of a, Fatou’s Lemma implies that lim inf a(x, un , ∇un ) · ∇un ≥ a(x, u, ∇u) · ∇u. (5.3.10) n→+∞

Ω

Ω

From (5.3.9) and (5.3.10), we deduce that lim inf a(x, un , ∇un ) · ∇(un − w) ≥ a(x, u, ∇u) · ∇(u − w) . n→+∞

Ω

Let us finally study

Ω

F (x, un , ∇un )(un − w) . Ω

Since un → u and ∇un → ∇u a.e. in Ω and weakly in Lp (Ω), using Theorem 3.7 one  has that F (x, un , ∇un ) → F (x, u, ∇u) weakly in Lp (Ω). This implies that F (x, un , ∇un )(un − w) → F (x, u, ∇u)(u − w) Ω

Ω

since un → u in Lp (Ω). We thus have obtained lim infA(un ), un − w ≥

a(x, u, ∇u) · ∇(u − w) +

n→+∞

Ω

F (x, u, ∇u)(u − w) Ω

= A(u), u − w .

Therefore A is pseudomonotone. We observe that in the following chapters we will essentially use the case p = 2, that is, the following corollary: Corollary 5.10. Let a : Ω×R×RN → RN be a Carathéodory function, with the following properties: (1) there exists β > 0 such that |a(x, s, ξ)| ≤ β [|s| + |ξ|]; (2) there exists α > 0 such that a(x, s, ξ) · ξ ≥ α|ξ|2 ; (3) [a(x, s, ξ) − a(x, s, η)] · [ξ − η] > 0 if ξ = η. Then the operator A : v → −div(a(x, v, ∇v)),

defined on H01 (Ω) into H −1 (Ω), is surjective. In particular, if f ∈ Lm (Ω), with 2N 2N m ≥ N+2 , or f ∈ M m (Ω), with m > N+2 , there exists u ∈ H01 (Ω) solution to −div(a(x, u, ∇u)) = f .

The Leray–Lions existence theorem

45

We will also study elliptic problems with lower order terms, that is, ⎧ ⎨−div(a(x, u, ∇u)) + u = f , in Ω , ⎩u = 0 , on ∂Ω . Theorem 5.11. Let Ω be an open bounded set of RN and p ∈ (1, ∞). Let a : Ω × R × RN → RN be a Carathéodory function, with the following properties: (1) there exists β > 0 such that |a(x, s, ξ)| ≤ β [|s|p−1 + |ξ|p−1 ]; (2) there exists α > 0 such that a(x, s, ξ) · ξ ≥ α|ξ|p , ∀ ξ ∈ RN ; (3) [a(x, s, ξ) − a(x, s, η)] · [ξ − η] > 0 if ξ = η. 2N Let f ∈ Lm (Ω), with m ≥ N+2 . Then there exists a solution u ∈ H01 (Ω) to problem −div (a(x, u, ∇u)) + u = f .

Proof. We are going to prove that A:

H01 (Ω) → H −1 (Ω) v → −div(a(x, v, ∇v)) + v

is pseudomonotone, in order to use Theorem 5.5. In the proof of the Leray–Lions theorem we proved that v → −div(a(x, v, ∇v)) is a pseudomonotone coercive operator from H01 (Ω) to H −1 (Ω). For the first point of the definition of a pseudomonotone operator (see Definition 5.2), we have only to prove that v → v is a bounded operator from H01 (Ω) to H −1 (Ω). Hölder’s and Sobolev’s inequalities imply that 1 v, w = vw ≤ v 2N w 2N ≤ v 2N wH01 (Ω) L N+2 (Ω) L N−2 (Ω) L N+2 (Ω) S Ω

and so A(v)H −1 (Ω) is bounded, if vH01 (Ω) is bounded. Moreover, to prove point 2 of Definition 5.2, we only have to show that if uj → u weakly in H01 (Ω), as j → ∞, and    if lim supj→+∞ Ω uj (uj − u) ≤ 0, then lim infj→+∞ Ω uj (uj − v) ≥ Ω u(u − v) for every v in H01 (Ω). This is clear, since uj (uj − v) − u(u − v) = u2j − u2 − v(uj − u) → 0 , Ω

Ω

as uj → u weakly in H01 (Ω) and uj → u in L2 (Ω). Finally, A is coercive, since v → −div(a(x, v, ∇v)) is coercive, as we have already proved. The Leray–Lions theorem can be proved under less strict hypotheses on a and F : Theorem 5.12. Let Ω be an open bounded set of RN and p ∈ (1, ∞). Let a : Ω × R × RN → RN and F : Ω × R × RN → R be two Carathéodory functions, with the following

46

Nonlinear elliptic equations

properties: (1) (2) (3) (4)

p∗ −ε

there exists β1 > 0 such that |a(x, s, ξ)| ≤ β1 [|s| p + |ξ|p−1 ], for some ε > 0; there exists α > 0 such that a(x, s, ξ) · ξ ≥ α|ξ|p , ∀ ξ ∈ RN ; [a(x, s, ξ) − a(x, s, η)] · [ξ − η] > 0 if ξ = η; ∗ there exist k1 ∈ Lp +ε (Ω) and β2 > 0 such that |F (x, s, ξ)| ≤ k1 (x) + p−ε 

∗ −ε−1

+ |ξ| p∗ ]; (5) there exist k2 ∈ L1 (Ω) and β2 > 0 such that F (x, s, ξ)s ≥ β3 |s|q −k2 (x), q ≤ p . β2 [|s|p

∗

1,p

Then for every g ∈ Lp (Ω) there exists a function u ∈ W0 (Ω) such that 1,p  a(x, u, ∇u) · ∇v + F (x, u, ∇u)v = gv , ∀ v ∈ W0 (Ω) . Ω

Ω

Ω

6 Summability of the solutions 6.1 Introduction In the previous chapter, we have proved the existence of weak solutions to the Leray– Lions problem: ⎧ ⎨−div(a(x, u, ∇u)) = f , in Ω , ⎩u = 0 , on ∂Ω . In this chapter we are going to present some regularity results proved in [17, 49, 50, 52]. We will see that the regularity of the solution depends on the regularity of the source. The starting points are some regularity results by Stampacchia. We will assume that the source is a function f belonging to a Lebesgue space or to a Marcinkiewicz space; we will also treat the case where the source is of divergence type. We illustrate here the results of this chapter in a schematic way. We recall that for an open bounded subset Ω of RN , N ≥ 3, a : Ω × R × RN → RN is a Carathéodory map with the following properties: (1) there exists β > 0 such that |a(x, s, ξ)| ≤ β [|s| + |ξ|]; (2) there exists α > 0 such that a(x, s, ξ) · ξ ≥ α|ξ|2 , ∀ ξ ∈ RN ; (3) [a(x, s, ξ) − a(x, s, η)] · [ξ − η] > 0 if ξ = η. We will show the following results: f ∈ Lm (Ω) , m > N/2 ⇒ u ∈ L∞ (Ω) ∗∗

f ∈ Lm (Ω) , m ∈ [2N/(N + 2), N/2) ⇒ u ∈ Lm (Ω) N

∗

f ∈ L 2 (Ω) ⇒ eλ|u| ∈ L2 (Ω) , ∀λ > 0 f ∈ M m (Ω) , m > N/2 ⇒ u ∈ L∞ (Ω) ∗∗

f ∈ M m (Ω) , m ∈ [2N/(N + 2), N/2) ⇒ u ∈ M m (Ω) N f ∈ M 2 (Ω) ⇒ ∃b > 0 : eb|u| < ∞ . Ω

When the source is the divergence of a vector field F : Ω → RN such that div F ∈ L2 (Ω), if u is a solution, we will prove that F ∈ (Lm (Ω))N , m > N ⇒ u ∈ L∞ (Ω) ∗

F ∈ (Lm (Ω))N , 2 < m < N ⇒ u ∈ Lm (Ω) .

In the case where a satisfies the following coercivity: a(x, s, ξ) · ξ ≥ α|ξ|p ,

∀ ξ ∈ RN

it is possible to prove similar results with the numbers (p ∗ ) , N/p playing the role of 2N/(N + 2), N/2.

48

Summability of the solutions

6.2 Preliminaries In this chapter we will often use the function ⎧ ⎪ ⎪ −k , s ≤ −k , ⎪ ⎨ Tk (s) = s , |s| ≤ k , ⎪ ⎪ ⎪ ⎩k , s ≥ k, for k > 0 and Gk (s) = s − Tk (s). Moreover, if f : Ω → R is a measurable function, we will use the following notation: g(k) = |Gk (f )| Ω

and Ak = {|f | > k} .

Lemma 6.1. Let f ∈ L1 (Ω). Then g(k) is differentiable a.e. and g  (k) = −meas(Ak ). Proof. Let us prove that

(w − k)

˜ g(k) = {w−k>0}

is differentiable with respect to k. To this end, let us set Ak,+ = {w − k > 0} . ˜ The function g(k) is differentiable a.e., since it is monotone. Let us compute its ˜ is derivative. Let h ∈ R+ ; the incremental rate of g ⎛ ⎞ ˜ + h) − g(k) ˜ 1⎜ g(k ⎟ = ⎝ (w − k − h) − (w − k)⎠ h h Ak+h,+

⎛ 1⎜ = ⎝ h =−

Ak,+



{kk+h} − Ω

1 h



We have 0≤

0≤

1 h



(w − k) . {k 1 and C > 0. Then f ∈ L∞ (Ω) and there exists a constant γ = γ(α, Ω) such that f L∞ (Ω) ≤ C γ . Proof. Using Lemma 6.1, one has g(k) ≤ C [−g  (k)]α

that is, 1

g  (k)[g(k)]− α ≤ −

1

.

1

(6.2.1)

Cα

Integrating this inequality on (0, k) we get

1 1 1 1 k 1 1− α 1− α − 1− − g(0)1− α = g(k)1− α − f L1 (Ω) . 1 ≥ g(k) α Cα Consequently, for every k > 0,

1 1 k 1 1− α g(k)1− α ≤ f L1 (Ω) − 1− . α C α1 1− 1

1

In particular this inequality holds true for k0 =

α C α f L1 (Ω) 1 1− α

0, that is,

1− 1

1

|f | ≤ k0 =

. This implies that g(k0 ) =

α C α f L1 (Ω)

1−

1 α

.

By the Hölder inequality 1

|f | ≤

that is,

1− 1

1

α C α f L∞ (Ω) meas(Ω)1− α

1−

1 α

,

1 −α f L∞ (Ω) ≤ 1 − meas(Ω)α−1 C . α

50

Summability of the solutions

Remark 6.3. If the inequality g(k) ≤ C meas(Ak )α

holds for every k ≥ h0 , for some h0 ∈ N, then one can slightly modify the above proof to prove that f L∞ (Ω) ≤ γ, where γ = γ(C, α, h0 , Ω). Indeed it is sufficient to integrate (6.2.1) on (h0 , k) and to follow the same technique as above. We will use the following results in the case where the source f in the Leray– N Lions problem belongs to M 2 (Ω). Proposition 6.4. Let f be a measurable function defined on Ω and let a be a positive   ak constant. Then Ω ea |f | < ∞ if and only if ∞ meas(Ak ) < ∞. k=0 e Proof. We observe that ∞ 

∞ 

ea k meas(Ak ) =

k=0

ea k

k=0

∞ 

meas(Bi ) =

i=k

∞ 

meas(Bi )

i=0

i 

ea k

(6.2.2)

k=0

if Bi = {i < |f | < i + 1}. Since t → eat is increasing and continuous n  k=k0

n+1

e

ak

eat dt ≤

≤

n 

ea (k+1) .

(6.2.3)

k=k0

k0

We will divide the proof into two parts.   ak Step I: Assume that Ω ea|f | < ∞; we are going to prove that ∞ k=0 e meas(Ak ) is finite. Using (6.2.2) and the first inequality of (6.2.3), one has ∞ 

e

ak

meas(Ak ) ≤

k=0

since

 Ω

ea|f | < ∞.

∞ 

i+1

i=0 a

=

e a

≤

ea a

=

ea a

eat dt

meas(Bi ) 0

  1 meas(Bi ) eai − a e i=0   ∞  1 ea |f | − a e i=0 B i   1 ea |f | − a < ∞ , e ∞ 

Ω

51

Sources in Lebesgue spaces



 ak Step II: Assume that ∞ meas(Ak ) < ∞; we are going to prove that k=0 e a|f | e < ∞ . Using (6.2.2) and the second inequality of (6.2.3), we have Ω ∞ 

ea k meas(Ak ) =

k=0

∞ 

meas(Bi )

i=0

=

∞ 

i−1 

meas(Bi )

ea(j+1)

j=−1

i

∞ 

eat dt

meas(Bi )

i=0

=

ea k

k=0

i=0

≥

i 

∞ 

−1

(

meas(Bi )

i=0

ei a − e−a a

)

∞ 1  [ea (|f |−1) − e−a ] a i=0 Bi 1 [ea (|f |−1) − e−a ] , = a

≥

and so

 Ω

Ω

e

a|f |

< ∞.

Proposition 6.5. Let f be a measurable function defined on Ω; if there exist a, c > 0,  c k0 ∈ N such that meas(Ak ) ≤ k ea k for every k ≥ k0 , then Ω eb|f | < ∞ for every b < a.  Proof. Using Proposition 6.4, one has Ω eb|f | < ∞ if and only if ∞ 

ebk meas(Ak ) < ∞ .

k=0

By hypothesis we have ∞ 

∞ 

ebk meas(Ak ) ≤

k=k0

ebk

k=k0

The last series is finite due to the choice of b, and so

c . k eak

∞

k=0 e

bk

meas(Ak ) is finite.

6.3 Sources in Lebesgue spaces We are now going to prove some regularity results of the solutions to ⎧ ⎨−div(a(x, u, ∇u)) = f , in Ω , ⎩u = 0 , on ∂Ω , when the source f belongs Lm (Ω), with m ≥

2N N+2 .

(6.3.1)

52

Summability of the solutions

Theorem 6.6. Let f ∈ Lm (Ω) with m > N/2. Then every solution u ∈ H01 (Ω) to problem (6.3.1) is bounded; moreover the estimate uL∞ (Ω) ≤ C f Lm (Ω)

holds, where C depends on N, α and m. Proof. Let us take v = Gk (u) as a test function in the weak formulation of (6.3.1). By the ellipticity of a we have α |∇Gk (u)|2 ≤ f Gk (u) . (6.3.2) Ω

Ω

Sobolev’s inequality implies that ⎛ ⎞ 2∗ 2 ∗⎟ ⎜ αS2 ⎝ |Gk (u)|2 ⎠ ≤ f Gk (u) . Ω

(6.3.3)

Ω

Let us now study the right-hand side: applying the Hölder inequality twice, first with exponent 2∗ and then with m N+2 2N , we get

⎛ ⎞ 1∗ ⎛ ⎞ N+2 2N 2 2N ⎟ ⎜ ⎜ 2∗ ⎟ N+2 f Gk (u) ≤ ⎝ |Gk (u)| ⎠ ⎝ |f | ⎠

Ω

Ω

Ak

⎛ ⎞ 1∗ 2   2N N+2 ⎜ 1− (N+2)m 2∗ ⎟ 2N . m ≤ f L (Ω) ⎝ |Gk (u)| ⎠ meas(Ak )

(6.3.4)

Ω

Estimates (6.3.3) and (6.3.4) thus give ⎛ ⎞ 2∗ ⎛ ⎞ 1∗ 2 2   2N N+2 ⎜ 1− (N+2)m 2⎜ 2∗ ⎟ 2∗ ⎟ 2N , αS ⎝ |Gk (u)| ⎠ ≤ f Lm (Ω) ⎝ |Gk (u)| ⎠ meas(Ak ) Ω

that is,

Ω

⎛ ⎞ 1∗ 2   2N N+2 ∗ 1− (N+2)m 2N 2⎜ 2 ⎟ αS ⎝ |Gk (u)| ⎠ ≤ f Lm (Ω) meas(Ak ) . Ω

By Hölder’s inequality with exponent 2∗ one has Ω

⎛ ⎞ 1∗ 2 N+2 ⎜ 2∗ ⎟ |Gk (u)| ≤ ⎝ |Gk (u)| ⎠ meas(Ak ) 2N Ω

(6.3.5)

Sources in Lebesgue spaces

53

and so (6.3.5) implies that 2 1 1 g(k) = |Gk (u)| ≤ f Lm (Ω) meas(Ak )1+ N − m . 2 αS Ω

Lemma 6.2 with α = 1 +

2 N

−

1 m

and C =

1 m αS2 f L (Ω)

gives the result.

By considering the same proof and dropping the positive contribute given by the lower order term, one has the following result: Theorem 6.7. Let f ∈ Lm (Ω) with m > N/2. Then every solution u ∈ H01 (Ω) to problem ⎧ ⎨−div(a(x, u, ∇u)) + u = f , in Ω , ⎩u = 0 , on ∂Ω . is bounded. Remark 6.8. Observe that if a function u satisfies inequality (6.3.2), then u is bounded. We can now pass to the regularity of the solutions in the case where f ∈ Lm (Ω), 2N N with N+2 ≤m< 2. 2N ≤ m < N2 . Then every solution u ∈ H01 (Ω) to Theorem 6.9. Let f ∈ Lm (Ω) with N+2 ∗∗ problem (6.3.1) belongs to Lm (Ω); moreover the estimate

uLm∗∗ (Ω) ≤ C f Lm (Ω)

holds, where C depends on N, m and α. Proof. Consider as a test function v=

|Tk (u)|2λ Tk (u) , 2λ + 1

with λ > 0. We will study separately the two sides of the weak formulation of (6.3.1). For the left-hand side we have 1 2λ + 1



⎛ ⎞ 2∗ 2 ∗ 2λ 2 ⎜ (λ+1)2 ⎟ a(x, u, ∇u) · ∇(|Tk (u)| Tk (u)) ≥ αS ⎝ |Tk (u)| ⎠

Ω

Ω

by using the ellipticity of a and Sobolev’s inequality. For the right-hand side, we get 1 2λ + 1

f |Tk (u)|2λ Tk (u) ≤ Ω

1 2λ + 1

⎛ ⎞ 1 m ⎟ ⎜ f Lm (Ω) ⎝ |Tk (u)|(2λ+1)m ⎠ Ω

54

Summability of the solutions

by using Hölder’s inequality with exponent m. We have thus proved that ⎡ ⎤ 2∗ ⎡ ⎤ 1 2 m ⎢ ⎢ 2 (λ+1)2∗ ⎥ (2λ+1)m ⎥ αS (2λ + 1) ⎣ |Tk (u)| ⎦ ≤ ⎣ |Tk (u)| ⎦ f Lm (Ω) . Ω

Ω

Now it is sufficient to choose λ such that (λ + 1)2∗ = m (2λ + 1), that is, −mN + 2N − 2m . 4m − 2N

λ=

Using that

2 2∗

−

1 m

> 0, one deduces that Tk (u)Lm∗∗ (Ω) ≤ C(α, S, m, N) f Lm (Ω) .

Fatou’s Lemma, as k → ∞, implies the result. N

Let us now study the limit case where f ∈ L 2 (Ω). N

Theorem 6.10. Let f ∈ L 2 (Ω). Then every solution u ∈ H01 (Ω) to problem (6.3.1) is ∗ such that eλ|u| belongs L2 (Ω) for every λ > 0. Proof. Let us take v = [e2λ |Gk (u)| − 1]sgn(Gk (u)), k > 0 as a test function in the weak formulation of problem (6.3.1) and study the two sides separately. We can estimate the right-hand side using the following inequality, satisfied by every t ≥ 0 and every Q > 1: 1 . |t 2 − 1| ≤ Q(t − 1)2 + Q−1 We then obtain f [e2λ |Gk (u)| − 1]sgn(Gk (u)) ≤ Q |f |[eλ |Gk (u)| − 1]2 + Ω

Ak

Hölder’s inequality with exponent

N 2

1 Q−1

|f | . Ak

implies

f [e2λ |Gk (u)| − 1]sgn(Gk (u))

Ω

⎡ ≤ Qf 

N L2

(Ak )



⎤

⎢ ⎥ ⎣ [eλ |Gk (u)| − 1] ⎦ Ω

2∗

2 2∗

+

1 Q−1

|f | . Ak

55

Sources in Lebesgue spaces

As a is elliptic, the left-hand side can be estimated from below by + 2 * 1    ∇ eλ |Gk (u)| − 1  . 2λα |∇Gk (u)|2 e2λ |Gk (u)| = 2λα 2 λ Ω

Ω

Sobolev’s inequality yields

⎡ a(x, u, ∇u) · ∇[e2λ |Gk (u)| − 1]sgn(Gk (u)) ≥

Ω

S2 λ

⎤



2 2∗

∗⎥ ⎢ 2α ⎣ [eλ |Gk (u)| − 1]2 ⎦

.

Ω

Therefore by the previous estimates we get ⎡ ⎤ 2∗ 2 S2 ∗ ⎢ λ |Gk (u)| 2 ⎥ 2α ⎣ [e − 1] ⎦ λ Ω

⎡ ≤ Qf 

N

L 2 (Ak )

⎤



2 2∗

∗⎥ ⎢ ⎣ [eλ |Gk (u)| − 1]2 ⎦

Ω

1 + Q−1

|f | , Ak

that is, ,

S λ2 2

2λα −

2

λ Q f  N L 2 (Ak ) S2

⎡ ⎤ 2∗ 2 - ⎢ λ |Gk (u)| 2∗ ⎥ − 1] ⎦ ≤ ⎣ [e Ω

1 Q−1

|f | . Ak

There exists kλ such that 2λα −

since f 

L

N 2

λ2 Qf 

N

L 2 (Ak ) S2

> 0,

∀ k ≥ kλ

→ 0 se k → +∞. Therefore the previous inequality implies that the

(Ak ) eλ |Gk (u)|

∗

− 1, for k ≥ kλ , is bounded in L2 (Ω). It is easy to deduce that sequence λ |u| 2∗ e belongs to L (Ω). Indeed ∗

∗

∗

[eλ |u| − 1]2 = [eλ |Tk (u)+Gk (u)| − 1]2 ≤ [eλk eλ |Gk (u)| − eλk + eλk − 1]2 ∗ −1

≤ 22

∗

∗

∗ −1

eλk2 [eλ |Gk (u)| − 1]2 + 22

∗

[eλk − 1]2 .

Therefore, for every k ≥ kλ ∗ ∗ ∗ ∗ ∗ ∗ [eλ |u| − 1]2 ≤ 22 −1 eλk2 [eλ |Gk (u)| − 1]2 + 22 −1 [eλk − 1]2 meas(Ω) , Ω

Ω 2∗

that is, eλ |u| belongs to L (Ω) for every λ > 0.

56

Summability of the solutions

6.4 Sources in Marcinkiewicz spaces We are now going to study the regularity of the weak solutions to (6.3.1), in the case where f belongs to the Marzinkiewicz space M m (Ω). Theorem 6.11. Let f ∈ M m (Ω), with m > lem (6.3.1) is bounded.

N 2.

Then any solution u ∈ H01 (Ω) to prob-

Proof. The result follows from Theorem 6.6, and from the inclusion M m (Ω) ⊂ Lm−ε (Ω) (0 < ε < m − 1), proved in Proposition 3.12. Theorem 6.12. Let f ∈ M m (Ω) with (2∗ ) < m < N/2. Then any solution u ∈ ∗∗ H01 (Ω) to problem (6.3.1) belongs to M m (Ω). Proof. We take v = Gk (u) as a test function in the weak formulation of problem (6.3.1). We study separately the two sides of the equation. For the first one, we have a(x, u, ∇u) · ∇Gk (u) ≥ α |∇u|2 Ω

Ak

⎛ ⎞ 2∗ 2 2 ⎜ 2∗ ⎟ ≥ α S ⎝ |Gk (u)| ⎠ Ω

by using the ellipticity of a and Sobolev’s inequality. For the right-hand side, we get ⎛ ⎞1/2∗ ⎛ ⎞ N+2 2N 2N ⎟ ⎜ ⎜ 2∗ ⎟ N+2 f Gk (u) ≤ ⎝ |Gk (u)| ⎠ ⎝ |f | ⎠ Ω

Ω

Ak

⎛ ⎞ 1∗ 2   2N N+2 ⎜ 1− (N+2)m 2N 2∗ ⎟ ≤ c ⎝ |Gk (u)| ⎠ meas(Ak ) , Ω

where c is a constant depending on m, f M m (Ω) ), by using Hölder’s inequality with 2N m(N+2) exponent 2∗ and Proposition 3.13 (applied to |f | N+2 ∈ M 2N ). We have thus obtained ⎛ ⎞ 1∗ 2 1 1 1 c ∗ ⎜ 2 ⎟ meas(Ak ) 2 + N − m . ⎝ |Gk (u)| ⎠ ≤ 2 αS Ω

Using Hölder’s inequality on the left-hand side of the above estimate we get 2 1 c |Gk (u)| ≤ meas(Ak )1+ N − m . α S2 Ω

By Lemma 6.1, this is equivalent to -ν , α S2 − ≥ g  (k)g(k)−ν c

57

Sources in Marcinkiewicz spaces

where ν=

mN > 1. mN + 2m − N

Integrating over (0, k) we get -ν , α S2 (ν − 1) k ≥ [g(k)]1−ν − [g(0)]1−ν , c that is, . g(k) ≤

, (ν − 1)

α S2 c

-ν

1 / 1−ν

1−ν

k + [g(0)]

≤

C(α, N, m, Ω) 1

,

k ν−1

since ν > 1. We note that A2k ⊂ Ak , and so g(k) ≥ |Gk (u)| dx ≥ (|u| − k) dx ≥ kmeas(A2k ) ; A2k

(6.4.1)

A2k

therefore meas(A2k ) ≤

C(α, N, m, Ω) ν

k ν−1

,

∗∗

that is, u ∈ M m (Ω). N

Theorem 6.13. Let f ∈ M 2 (Ω). Then there exists a constant b > 0 such that eb |u| < ∞ Ω

for every solution u ∈

H01 (Ω)

to problem (6.3.1).

Proof. Choosing v = Gk (u) as a test function in the weak formulation of (6.3.1), and arguing as in the proof of the previous theorem, we get 1≤−

c g  (k) . α S2 g(k)

Integrating with respect to k we have k 0

c dt ≤ − α S2

k 0

g  (t) dt , g(t)

that is, k≤−

c c [ln g(k) − ln g(0)] = ln α S2 α S2

This implies that ekα S

2 /c

≤

,

uL1 (Ω) g(k)

uL1 (Ω) . g(k)

If k ≥ 1, we get, arguing as in (6.4.1) k meas(A2k ) ≤ g(k) ≤

The theorem follows from Lemma 6.5.

uL1 (Ω) . 2 ekα S /c

.

58

Summability of the solutions

6.5 Sources in divergence form In this section, we focus our attention on ⎧ ⎨−div(a(x, u, ∇u)) = −div F , ⎩u = 0 ,

in Ω , on ∂Ω ,

(6.5.1)

under the hypothesis that F : Ω → RN is a vector field such that divF ∈ L2 (Ω). We note that the existence of a weak solution in H01 (Ω) is guaranteed by the Leray–Lions theorem (Theorem 5.1). Theorem 6.14. Let F ∈ (Lm (Ω))N , m > N . Then any weak solution u ∈ H01 (Ω) to problem (6.5.1) is bounded; moreover, the estimate uL∞ (Ω) ≤ C F Lm (Ω)

holds, where C depends on N, m and α. Proof. Take Gk (u) as a test function in the weak formulation of (6.5.1): a(x, u, ∇u) · ∇Gk (u) = F · ∇Gk (u) . Ω

Ω

Using the ellipticity of a on the left-hand side and the Cauchy–Schwarz inequality on the right one we get ⎛ ⎞1/2 ⎛ ⎞1/2 ⎜ ⎟ ⎜ ⎟ α |∇Gk (u)|2 ≤ ⎝ |F |2 ⎠ ⎝ |∇Gk (u)|2 ⎠ , Ω

Ω

Ak

where Ak = {|u| > k}. Using Hölder’s inequality with exponent m/2 on the righthand side, one has ⎡ ⎤1 ⎛ ⎞1/m 2 1 1 ⎜ ⎟ ⎢ ⎥ α ⎣ |∇Gk (u)|2 ⎦ ≤ ⎝ |F |m ⎠ meas(Ak ) 2 − m ; Ω

Ak

Sobolev’s inequality allows us to say that ⎛ ⎞1/2∗ 1 1 ∗ ⎜ ⎟ αS ⎝ |Gk (u)|2 ⎠ ≤ ||F ||Lm (Ω) meas(Ak ) 2 − m . Ω

We again use Hölder’s inequality on the left-hand side of (6.5.2): ⎛ ⎞1/2∗ ∗ ∗ ⎜ 2 ⎟ |Gk (u)| ≤ ⎝ |Gk (u)| ⎠ meas(Ak )1−1/2 Ω

Ω

⎛ ⎞1/2∗ 1 1 ⎜ 2∗ ⎟ = ⎝ |Gk (u)| ⎠ meas(Ak ) 2 + N . Ω

(6.5.2)

Sources in divergence form

59

Estimate (6.5.2) therefore yields 1 1 |Gk (u)| ≤ C(α, S)||F ||Lm (Ω) meas(Ak )1− m + N . Ω

Using Lemma 6.2 we get the result. In the case m < N , we prove the following result: Theorem 6.15. Let F ∈ (Lm (Ω))N , 2 < m < N . Then any weak solution u ∈ H01 (Ω) ∗ to problem (6.5.1) belongs to Lm (Ω); moreover the estimate uLm∗ (Ω) ≤ C F Lm (Ω)

holds, where C depends on N, m and α. Proof. Let us consider v = |Tk (u)|2γ Tk (u) as a test function in the weak formulation of problem (6.5.1). We have, due to the ellipticity of a 2 2γ α |∇Tk (u)| |Tk (u)| ≤ F · ∇Tk (u)|Tk (u)|2γ . Ω

Ω

Hölder’s inequality on the right-hand side yields |∇Tk (u)|2 |Tk (u)|2γ

α Ω

⎛ ⎞1/2 ⎛ ⎞1/2 ⎜ ⎟ ⎜ ⎟ ≤ ⎝ |F |2 |Tk (u)|2γ ⎠ ⎝ |∇Tk (u)|2 |Tk (u)|2γ ⎠ , Ω

Ω

and consequently

α2

|∇Tk (u)|2 |Tk (u)|2γ ≤ Ω

|F |2 |Tk (u)|2γ . Ω

Using again Hölder’s inequality with exponent m/2 on the right-hand side, one gets ⎛ ⎞ ⎛ ⎞ m−2 m  2 2mγ ⎜ ⎟ 2⎜  γ+1  ⎟ 2  ⎠ ≤ ||F ||Lm (Ω) ⎝ |Tk (u)| m−2 ⎠ α ⎝ ∇|Tk (u)| . Ω

Ω

We use Sobolev’s inequality on the left-hand side of the previous inequality: ⎛ ⎞2/2∗ ⎛ ⎞ m−2 m 2mγ ⎜ ⎟ 2 2⎜ (γ+1)2∗ ⎟ 2 α S ⎝ |Tk (u)| ≤ ||F ||Lm (Ω) ⎝ |Tk (u)| m−2 ⎠ . ⎠ Ω

Ω

Now, let us choose γ in such a way that (γ + 1) 2∗ =

2mγ . m−2

(6.5.3)

60

Summability of the solutions

With this choice, estimate (6.5.3) can now be written as ⎛

⎞



⎜ ⎝ |Tk (u)|

m∗

Ω

We note that

2 2∗

−

m−2 m

⎟ ⎠

2 2∗

− m−2 m

≤

1 α2 S 2

||F ||2Lm (Ω) .

> 0, since m < N . Fatou’s lemma implies ||u||Lm∗ (Ω) ≤ C ||F ||Lm (Ω) ,

with 0 < C = C(α, N, m).

7 H 2 regularity for linear problems 7.1 Introduction In this chapter, we focus our attention on the regularity of the solutions to linear elliptic problems. We study ⎧ ⎨−div(M(x)∇u) = f , in Ω , (7.1.1) ⎩u = 0 , on ∂Ω , where Ω is an open bounded subset of RN , N ≥ 3, and f ∈ L2 (Ω). Moreover M(x) is a RN×N matrix such that M(x)ξ · ξ ≥ α|ξ|2 for every ξ ∈ RN and the entries mij of M are Lipschitz continuous, that is, |mij (x) − mij (y)| ≤ K |x − y| ,

∀ x, y ∈ Ω ,

∀ i, j = 1, ..N .

In Chapter 4, we proved the existence of a unique distributional solution u ∈ H01 (Ω). Now we prove that u belongs to H 2 , following the proof by Nirenberg, given in [40]. Theorem 7.1. Let u ∈ H01 (Ω) be the solution to problem (7.1.1). Then, for every Ω ⊂⊂ Ω, u belongs to H 2 (Ω ) and the following estimate holds: uH 2 (Ω ) ≤ C (uH01 (Ω) + f L2 (Ω) )

(7.1.2)

where C = C(Ω, K, α, d), with d = dist(Ω , ∂Ω).

7.2 Preliminaries We define the incremental rate of a function. We denote by ei ∈ RN the vector having all the coefficients 0, except the ith, which is 1. Definition 7.2. Let f : Ω → R and h ∈ R \ {0}. The incremental rate of f with respect to ei is the function Δh i f : {x ∈ Ω : x + hei ∈ Ω} → R , defined by Δh i f (x) :=

f (x + hei ) − f (x) . h

We remark that Δh i f is defined in Ω|h| := {x ∈ Ω : dist(x, ∂Ω) > |h|} .

From now on, we will write Δh instead of Δh i . We will denote by ∇i f the ith component of the vector ∇f .

62

H 2 regularity for linear problems

Proposition 7.3. The incremental rate of a function has the following properties: (1) If f ∈ W 1,p (Ω), then Δh f ∈ W 1,p (Ω|h| ) and ∇i (Δh f ) = Δh (∇i f ) .

(7.2.1)

(2) If one of the functions f or g has support contained in Ω|h| , then f Δh g = − gΔ−h i i f . Ω

(7.2.2)

Ω

(3) The following equality holds: h h Δh i (f g)(x) = f (x + hei )Δi g(x) + g(x)Δi f (x) .

(7.2.3)

Lemma 7.4. Let v ∈ W 1,p (Ω). Then Δh v ∈ Lp (Ω ) for every Ω ⊂⊂ Ω such that h < dist(Ω , ∂Ω) and the following estimate holds: p  p  Δh i vL (Ω ) ≤ ∇i vL (Ω ) .

Proof. Assume that v ∈ C 1 (Ω) ∩ W 1,p (Ω). Then Δh i v(x)

1 v(x + hei ) − v(x) = = h h

h ∇i v(x1 , . . . , xi−1 , xi + ξ, xi+1 , . . . , xN )dξ . 0

Hölder’s inequality implies p |Δh i v(x)|

and so

1 ≤ h

h |∇i v(x1 , . . . , xi−1 , xi + ξ, xi+1 , . . . , xN |p dξ 0

p |Δh i v| Ω

1 ≤ h



h |∇i v|p dξ ≤ 0 Ω

|∇i v|p . Ω

Using the density of C 1 (Ω) ∩ W 1,p (Ω) in W 1,p (Ω) we can prove the general result. 1,p

We now prove that the previous estimate is sufficient for f to be in W0 (Ω), in some sense. Lemma 7.5. Let v ∈ Lp (Ω), 1 < p < ∞, and suppose that there exists K such that Δh vLp (Ω ) ≤ K for every h > 0 and Ω ⊂⊂ Ω such that h < dist(Ω , ∂Ω). Then the distributional gradient ∇v exists and satisfies ∇vLp (Ω) ≤ K .

H 2 (Ω) regularity of the solutions

Proof. Let hn → 0; let us define, for i = 1, . . . , N , ⎧ ⎨Δhn v in Ω|h | n i gn := ⎩0 in Ω \ Ω

|hn |

63

.

The sequence gn is bounded in Lp (Ω) and so there exists a subsequence, still denot˜ ∈ Lp (Ω), with v˜i Lp (Ω) ≤ K . Therefore for ed by gn , which converges weakly to v 1 every ϕ ∈ C0 (Ω) one has h

ϕΔi n v → Ω

ϕv˜i . Ω

On the other hand, for hn < dist(supp ϕ, ∂Ω), we have, using (7.2.2) and Lebesgue’s theorem hn −hn ϕΔi v = − vΔi ϕ → − v∇i ϕ, n → ∞ . Ω

Ω

We deduce that for every ϕ ∈

Ω

C01 (Ω)

ϕv˜i = − Ω

v∇i ϕ Ω

and so v˜i = ∇i v .

7.3 H 2 (Ω) regularity of the solutions This section is devoted to the proof of Theorem 7.1. Note that this result allows us to say that u solves −div(M(x)∇u) = f a.e. in Ω ⊂⊂ Ω, since u ∈ H 2 (Ω ). Proof of Theorem 7.1. The solution u satisfies N 

mij (x)∇j u∇i v =

i,j=1 Ω

fv ,

∀ v ∈ H01 (Ω) .

Ω

Let ϕ be a function with compact support in Ω and |2h| < dist(supp ϕ, ∂Ω). Let v = Δ−h k ϕ , for 1 ≤ k ≤ N : properties (7.2.1) and (7.2.2) imply N 

Δh k (mij ∇j u)∇i ϕ = −

i,j=1 Ω

N  i,j=1 Ω

mij ∇j u∇i Δ−h k ϕ =−



f Δ−h k ϕ.

Ω

Using property (7.2.3), h h Δh k (mij ∇j u)(x) = mij (x + hek )Δk ∇j u(x) + ∇j u(x)Δk mij (x)

H 2 regularity for linear problems

64

one has N 

mij (x + hek )Δh k ∇j u(x)∇i ϕ

i,j=1 Ω

N    −h m (x)∇ u(x)∇ ϕ + m (x)∇ u(x)∇ Δ ϕ Δh ij j i ij j i k k

=−

i,j=1 Ω N 

=−

−h [Δh k mij (x)∇j u(x)∇i ϕ + f Δk ϕ] .

i,j=1 Ω

Therefore N 

mij (x + hek )Δh ∇ u(x)∇ ϕ = − [g · ∇ϕ + f Δ−h j i k k ϕ] ,

i,j=1 Ω

Ω

where g = (g 1 , . . . , g n ) with g i = Δh k mij ∇j u. Using the Cauchy–Schwarz inequality and Lemma 7.4 we get N 

mij (x + hek )Δh k ∇j u(x)∇i ϕ ≤ (K∇uL2 (Ω) + f L2 (Ω) )∇ϕL2 (Ω) ,

i,j=1 Ω

since mij are Lipschitz continuous functions. Property (7.2.1) implies N 

mij (x + hek )∇j Δh k u(x)∇i ϕ ≤ (KuH01 (Ω) + f L2 (Ω) )∇ϕL2 (Ω) . (7.3.1)

i,j=1 Ω

Let η ∈ C01 (Ω) be such that 0 ≤ η ≤ 1 and define ϕ = η2 Δh k u. Writing ∇i ϕ in an explicit way and using (7.2.1), we get N 

h η2 mij (x + hek )∇i Δh k u∇j Δk u

i,j=1 Ω

=

N 

h mij (x + hek )∇j Δh k u(∇i ϕ − 2Δk uη∇i η) .

i,j=1 Ω

The ellipticity of M gives 2 |η∇Δh k u| ≤

α Ω

N  i,j=1 Ω

h η2 mi,j (x + hek )Δh k ∇i uΔk ∇j u .

(7.3.2)

H 2 (Ω) regularity of the solutions

65

As for the right-hand side of (7.3.2) one has, due to (7.3.1) and the Cauchy–Schwarz inequality N 

h mi,j (x + hek )∇j Δh k u(∇i ϕ − 2Δk uη∇i η)

i,j=1 Ω

≤

N  i,j=1 Ω

        h h h mi,j (x + hek )∇j Δk u∇i ϕ + 2  mi,j (x + hek )∇j Δk uΔk uη∇i η   Ω 

h h ≤ (KuH01 (Ω) + f L2 (Ω) )∇(η2 Δh k u)L2 (Ω) + 2Kη∇Δk uL2 (Ω) Δk u∇ηL2 (Ω) h 2 ≤ (KuH01 (Ω) + f L2 (Ω) )2η∇ηΔh k u + η ∇Δk uL2 (Ω) h + 2Kη∇Δh k uL2 (Ω) Δk u∇ηL2 (Ω) .

(7.3.3) We remark that h h h 2 2η∇ηΔh k u + η ∇Δk uL2 (Ω) ≤ 2∇ηΔk uL2 (Ω) + η∇Δk uL2 (Ω)

since 0 ≤ η ≤ 1. We deduce from the previous estimates and (7.3.3) that N 

h mi,j (x + hek )∇j Δh k u(∇i ϕ − 2Δk uη∇i η)

i,j Ω h ≤ (KuH01 (Ω) + f L2 (Ω) )(2∇ηΔh k uL2 (Ω) + η∇Δk uL2 (Ω) ) h + 2Kη∇Δh k uL2 (Ω) Δk u∇ηL2 (Ω) .

Therefore we have got from (7.3.2) 2 h αη∇Δh k uL2 (Ω) ≤ η∇Δk uL2 (Ω) (KuH01 (Ω) + f L2 (Ω) )

+ 2Δh k u∇ηL2 (Ω) (KuH01 (Ω) h + f L2 (Ω) ) + 2Kη∇Δh k uL2 (Ω) Δk u∇ηL2 (Ω) .

It follows from the Young inequality that 2 αη∇Δh k uL2 (Ω) ≤

1 ε 2 (KuH01 (Ω) + f L2 (Ω) )2 + η∇Δh k uL2 (Ω) 2ε 2 1 2 + (KuH01 (Ω) + f L2 (Ω) )2 + εΔh k u∇ηL2 (Ω) ε 1 h 2 2 + εKη∇Δh k uL2 (Ω) + Δk u∇ηL2 (Ω) , ε

or equivalently,

ε 2 α − − εK η∇Δh k uL2 (Ω) 2

3 1 2 (KuH01 (Ω) + f L2 (Ω) )2 + ε + Δh ≤ k u∇ηL2 (Ω) . 2ε ε

66

H 2 regularity for linear problems

Choosing ε sufficiently small, one gets, if c denotes a constant depending just on k and α h 2 2 2 η∇Δh k uL2 (Ω) ≤ c (KuH01 (Ω) + f L2 (Ω) ) + cΔk u∇ηL2 (Ω) 2 ≤ c (uH01 (Ω) + f L2 (Ω) + Δh k u∇ηL2 (Ω) ) .

By using (7.2.1) h ηΔh k ∇uL2 (Ω) ≤ c (uH01 (Ω) + f L2 (Ω) + sup |∇η|Δk uL2 (spt η) ) Ω

where spt η is the support of η. Lemma 7.4 implies ηΔh k ∇uL2 (Ω) ≤ c (uH01 (Ω) + f L2 (Ω) + sup |∇η|∇uL2 (Ω) ) Ω

and so, for every Ω ⊂⊂ Ω fixed ηΔh k ∇uL2 (Ω ) ≤ c (1 + sup |∇η|)(uH01 (Ω) + f L2 (Ω) ) . Ω

Now, η can be chosen as a cut-off function, such that η = 1 on Ω and |∇η| ≤ 2  1  d , where d = dist(Ω , ∂Ω). From Lemma 7.5 we get that ∇u ∈ H (Ω ) for every  2 Ω ⊂⊂ Ω, and so u ∈ H (Ω). Estimate (7.1.2) is thus proved, using Poincaré’s inequality.

8 Spectral analysis for linear operators 8.1 Introduction In this chapter, we will focus on the eigenvalue problem ⎧ ⎨−div(M(x)∇u) = λu , in Ω , ⎩u = 0 , on ∂Ω ,

(8.1.1)

under the following assumptions. Ω is an open bounded subset of RN , N ≥ 3, and M(x) is a N ×N symmetric matrix, with bounded entries such that there exists α > 0 satisfying M(x)ξ · ξ ≥ α|ξ|2 for every ξ ∈ RN . We will first study the existence and some properties of the eigenvalues of L(v) = −div(M(x)∇v). Later we will see some applications of the spectral theory to some semilinear (noncoercive) equations. We will present some existence and multiplicity results by Dolph [36], Ambrosetti and Prodi [2].

8.2 Eigenvalues of linear elliptic operators Theorem 8.1. There exists an orthonormal basis wm ∈ L2 (Ω) and a sequence of real positive numbers λm such that (1) λm → +∞, as m → +∞; (2) for every m ∈ N, wm is a solution to ⎧ ⎨−div(M(x)∇v) = λm v , in Ω , ⎩v = 0 , on ∂Ω .

The proof of this theorem is based on important properties of T :

L2 (Ω) → L2 (Ω) f →u

(8.2.1)

where u ∈ H01 (Ω) solves −div(M(x)∇u) = f . Lemma 8.2. Let T be defined by (8.2.1). Then T is self-adjoint and compact. Proof. T is well defined by Theorem 4.4 and is linear. We claim that T is self-adjoint (according to Definition 8.22). Let U = T (u) and V = T (v). Then for every ϕ ∈ H01 (Ω) M(x)∇U · ∇ϕ = Ω

uϕ Ω

68

Spectral analysis for linear operators





and

M(x)∇V · ∇ϕ =

v ϕ.

Ω

Ω

Let us choose ϕ = V in the first equation and ϕ = U in the second one. The symmetry of M implies that uV = Ω

vU, Ω

that is, T is self-adjoint. Let us prove that T is compact. If T (f ) = u, then α∇u2L2 (Ω) ≤ M(x)∇u·∇u = f u ≤ f L2 (Ω) uL2 (Ω) ≤ c f L2 (Ω) ∇uL2 (Ω) , Ω

Ω

using the ellipticity of M on the left-hand side and Hölder’s and Poincaré’s inequality on the right one. We deduce that c T (f )H01 (Ω) ≤ f L2 (Ω) , ∀ f ∈ L2 (Ω) . α Since the embedding H01 (Ω) → L2 (Ω) is compact, T is compact. We can now prove Theorem 8.1. Proof. Using Lemma 8.2, we can apply the spectral theorem (Theorem 8.25) to the operator T defined by (8.2.1). Therefore there exists an orthogonal basis wn of L2 (Ω) with |wn |2 = 1 (8.2.2) Ω

and a sequence μn converging to 0, as n → ∞, such that T (wn ) = μn wn , that is, 1 M(x)∇wn · ∇ϕ = wn ϕ , ∀ ϕ ∈ H01 (Ω) . μn Ω

Ω

Observe that wn ∈ H01 (Ω). Moreover μn ≥ 0, since by the ellipticity of M , we have 1 α |∇wn |2 ≤ wn2 . μn Ω

Ω

On the other hand μn ≠ 0, otherwise wn = 0 from T (wn ) = 0. Definition 8.3. According to the notations of the previous theorem, we will say that {λm }m∈N is the set of the eigenvalues of L(v) = −div(M(x)∇v): this means that { λ1m }m∈N is the set of eigenvalues of T :

L2 (Ω) → L2 (Ω) f →u

where u ∈ H01 (Ω) satisfies −div(M(x)∇u) = f . By an eigenfunction of L(v) = −div(M(x)∇v) we shall mean an eigenfunction of T .

69

Eigenvalues of linear elliptic operators

The following theorem concerns λ1 , the first eigenvalue of L(v) = −div(M(x) ∇v). Let

 A(v) =

Ω

M(x)∇v · ∇v  . 2 Ωv

Theorem 8.4. Let λ1 be the smallest eigenvalue of L(v) = −div(M(x)∇v). Then λ1 = min A(v) . 1 (Ω) v∈H0 v≠0

Moreover, each function u that minimizes A is an eigenfunction of L corresponding to λ1 . Remark 8.5. Poincaré’s inequality, for p = 2 reads: there exists a positive constant c = c(Ω) such that 1 uL2 (Ω) ≤ ∇uL2 (Ω) , ∀u ∈ H01 (Ω) . c

Theorem 8.4 allows us to say that the smallest eigenvalue for the operator L(v) = −Δv is equal to the square root of the best constant in Poincaré’s inequality.

We will use the following lemma to prove Theorem 8.4: Lemma 8.6. A(v) has a minimizer over H01 (Ω). Proof. First of all we observe that A is bounded from below, by Poincaré’s inequality. Let vn be a minimizing sequence, that is, A(vn ) → inf A, as n → +∞. Note that zn =

vn vn H01 (Ω)

satisfies zn H01 (Ω) = 1. Up to a subsequence, zn → z in L2 (Ω) and weakly in H01 (Ω). Observe that 0 ≤ M(x)∇(zn −z)·∇(zn −z) = M(x)∇zn ·∇zn −2M(x)∇zn ·∇z+M(x)∇z·∇z .

Therefore,

M(x)∇zn · ∇zn ≥ 2 Ω

M(x)∇zn · ∇z −

Ω

M(x)∇z · ∇z . Ω

If we pass to the limit in the above inequality, we have lim inf M(x)∇zn · ∇zn ≥ M(x)∇z · ∇z . n→∞

Ω

(8.2.3)

Ω

Note that zn is a minimizing sequence for A, as A is homogeneous. Thus   M(x)∇z · ∇z Ω M(x)∇zn · ∇zn   2 inf A = lim A(zn ) = lim inf ≥ Ω = A(z) , 2 n→∞ n→∞ Ωz Ω zn

70

Spectral analysis for linear operators

where we have used (8.2.3) and the fact that zn → z in L2 (Ω). Consequently z is a minimizer for A. We have only to check that z ≠ 0. Since zn is a minimizing sequence, A(zn ) is a bounded sequence, that is, there exists a positive constant C such that 2 M(x)∇zn · ∇zn ≤ C zn . Ω

Ω

The ellipticity of M gives



|∇zn |2 ≤ C

α Ω

2 zn . Ω

This implies that

C zn 2L2 (Ω) . 1 = zn 2L2 (Ω) + ∇zn 2L2 (Ω) ≤ 1 + α

At the limit as n → ∞ we get zL2 (Ω) > 0. We can now prove Theorem 8.4. Proof. Let u ∈ H01 (Ω) be a minimizer for A. Using the previous lemma the function g(t) = A(u + tw), where w ∈ H01 (Ω) attains its minimum at 0. Since g is differentiable, g  (0) = 0, that is,  · ∇u Ω M(x)∇u  M(x)∇u · ∇w = uw = (inf A) uw , ∀ w ∈ H01 (Ω) . 2 Ωu Ω

Ω

Ω

(8.2.4) In other words, if u ∈ H01 (Ω) minimizes A, then u is an eigenfunction of L(v) = −div(M(x)∇v) and  · ∇u Ω M(x)∇u  2 u Ω is the corresponding eigenvalue. Let us prove that inf A = λ1 . Since λ1 is the smallest eigenvalue of −div(M(x)∇v), one has λ1 ≤ inf A . Let us prove the opposite inequality. Let w1 be the eigenfunction corresponding to λ1 ; we have M(x)∇w1 · ∇v = λ1 w1 v , ∀ v ∈ H01 (Ω) . Ω

Ω

Consequently, choosing v = w1 we obtain    2 M(x)∇v · ∇v Ω M(x)∇w1 · ∇w1 Ω λ 1 w1    inf A = inf Ω ≤ = = λ1 . 2 2 2 Ωv Ω w1 Ω w1 Remark 8.7. As we will see in Chapter 9, Example 9.9, (8.2.4) is the Euler equation for the functional A.

Eigenvalues of linear elliptic operators

71

Corollary 8.8. Any eigenfunction w1 of L(v) = −div(M(x)∇v) corresponding to λ1 is of constant sign on Ω. Proof. By the definition of eigenfunction corresponding to λ1 , w1 solves M(x)∇w1 · ∇v = λ1 w1 v , ∀ v ∈ H01 (Ω) . Ω

Ω

Considering v = w1+ one has  λ1 =

Ω

M(x)∇w1+ · ∇w1+  , + 2 Ω (w1 )

that is, w1+ is a minimum for A. By Theorem 8.4, w1+ is an eigenfunction of L(v) = −div(M(x)∇v) corresponding to λ1 . Remark 8.9. One can prove more about λ1 (see [37]). Indeed λ1 is simple and the corresponding eigenfunctions are strictly positive (negative). This is a consequence of Harnack’s inequality (see [50]): if u is a nonnegative solution to L(u) = −div(M(x)∇u) = λu, then for any compact G ⊂ Ω there exists a positive constant K , independent of u, such that maxG u ≤ K minG u. The constant K depends on α, the L∞ (Ω) norm of the matrix M , λ, G, and Ω. More generally, we have the following expression for λm : Theorem 8.10. Let m > 1. Then   · ∇v Ω M(x)∇wm · ∇wm Ω M(x)∇v   λm = = inf 2 2 v∈Pm−1 ,v≠0 Ωv Ω wm 0 1  where Pm = v ∈ H01 (Ω) : Ω wn v = 0 , n = 1, . . . , m . We can now prove the last result of this section: the eigenfunctions of (8.1.1) are bounded. Theorem 8.11. Let u be an eigenfunction of L(v) = −div(M(x)∇v) corresponding to an eigenvalue λ. Then u is bounded and the following estimate holds: N

uL∞ (Ω) ≤ c(α, N)λ 2 uL1 (Ω) .

Proof. We take Gk (u) = u − Tk (u) as a test function in (8.1.1): in this way α |∇Gk (u)|2 ≤ M(x)∇u · ∇u = λ u Gk (u) , Ak

Ω

Ω

(8.2.5)

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Spectral analysis for linear operators

by the ellipticity of M . Let us now estimate the right-hand side. Writing u as u−k+k, we get, by Young’s inequality λ uGk (u) ≤ λ |Gk (u)|2 + λ k |Gk (u)| Ω

Ak

Ak

⎧ ⎫ ⎪ ⎪ ⎨ ⎬ λ ≤ λ |Gk (u)|2 + |Gk (u)|2 + k2 meas(Ak ) ⎪ 2 ⎪ ⎩ ⎭ Ak Ak λ λ =3 |Gk (u)|2 + k2 meas(Ak ) , 2 2

Ak

where Ak = {|u| ≥ k}. We have thus proved that λ λ 2 α |∇Gk (u)| ≤ 3 |Gk (u)|2 + k2 meas(Ak ) . 2 2 Ak

(8.2.6)

Ak

We are going to estimate the first term of the right-hand side by using Hölder’s and Sobolev’s inequalities: ⎡ ⎢ S⎣



⎤1

⎡

2

⎥ ⎢ |Gk (u)|2 ⎦ ≤ S ⎣

Ak



⎤

1 2∗

∗⎥ |Gk (u)|2 ⎦

1

meas(Ak ) N

Ak

⎡ ⎢ ≤⎣



⎤1

(8.2.7)

2

1 ⎥ |∇Gk (u)|2 ⎦ meas(Ak ) N .

Ak

Inequality (8.2.6) implies that 2 3λ λ N α |∇Gk (u)|2 ≤ meas (A ) |∇Gk (u)|2 + k2 meas(Ak ) . k 2S2 2 Ak

Ak

Let us now consider k ≥ k0 , where k0 = k0 (λ) is such that α≥

2 3λ meas(Ak0 ) N . S2

Note that k0 meas(Ak0 ) ≤ uL1 (Ω) . Therefore, in the sequel we will consider  k ≥ k0 =

In this way

3λ S2 α

 N2

uL1 (Ω) .

(8.2.8)

|∇Gk (u)|2 ≤ k2 λ meas(Ak ) .

α Ak

(8.2.9)

73

Eigenvalues of linear elliptic operators

From estimates (8.2.9) and (8.2.7) we deduce that ⎡ ⎢ ⎣

⎤1



2

2⎥

|Gk (u)| ⎦ ≤

Ak

k S



λ meas(Ak ) α

 12

1

meas(Ak ) N .

By Hölder’s inequality on the left-hand side one has

⎡ ⎢ |Gk (u)| ≤ ⎣

Ak



⎤1 2

2⎥

1

1

1

1

|Gk (u)| ⎦ meas(Ak ) 2 ≤ meas(Ak )1+ N k(S2 α)− 2 λ 2 .

Ak

Setting g(k) =

 Ak

|Gk (u)| and using Lemma 6.1, the last estimate is equivalent to 1

1

1

S α 2 g(k) ≤ [−g  (k)]1+ N k λ 2 ,

that is, N

N

N

1

N

g  (k) g(k)− N+1 λ 2(N+1) ≤ −k− N+1 (Sα 2 ) N+1 .

Integrating over (k0 , k) we have 1

1

1

N

N

1

N

N

1

1

g(k) N+1 ≤ g(k0 ) N+1 + (S α 2 ) N+1 λ− 2(N+1) [k0N+1 − k N+1 ] .

Since g(k0 ) ≤ uL1 (Ω) we get 1

1

1

1

− 2 N+1 λ 2(N+1) [k N+1 − k N+1 ] . g(k) N+1 ≤ uLN+1 1 (Ω) + (S α ) 0

˜, For k = k

 N+1 1 1 N 1 N N+1 ˜ = (S α 2 )− N+1 λ 2(N+1) u N+1 k + k , 1 0 L (Ω)

(8.2.10)

˜ ≥ k0 . ˜. Observe that k the right-hand side is zero. Thus g(k) = 0, if k ≥ k Note that, by using (8.2.8), we get N

˜ ≤ c(α, N) λ 2 uL1 (Ω) . k

Therefore N

uL∞ (Ω) ≤ c(α, N) λ 2 uL1 (Ω) .

Remark 8.12. That the eigenfunctions of L(v) = −div(M(x)∇v) are bounded is a consequence of the following bootstrapping technique. By definition, wm ∈ H01 is a solution to ⎧ ⎨−div(M(x)∇wm ) = λm wm , in Ω , ⎩w = 0 , on ∂Ω . m

74

Spectral analysis for linear operators

∗

∗∗∗

Since H01 (Ω) ⊆ L2 (Ω), using Theorem 6.9 we have wm ∈ L2 (Ω). Using Theorem 6.9 in an iterative way we get, after a finite number of steps, the Lp (Ω) summability with p > N/2, for any N ≥ 3. To this end, we define ⎧ ∗ ⎪ ⎪ ⎪q0 = 2 ⎪ ⎨ .. . ⎪ ⎪ ⎪ ⎪ Nqk ∗∗ ⎩q k ≥ 0. k+1 = qk = N−2qk , By contradiction, assume that qk ≤ N/2

for every k. Since qk is strictly monotone, there exists l := lim qk .

One has, necessarily 0 < l ≤ N/2 .

Passing to the limit on k, we get l=

Nl : N − 2l

¯ ≥ 0 this equality implies l = 0 which is a contradiction. Therefore there exists k such that qk¯ > N/2. It is sufficient to use Theorem 6.6 to get the result. We note that the previous theorem gives us an additional information with respect to the above bootstrapping technique: estimate (8.2.5).

Corollary 8.13. Let u be an eigenfunction of L(v) = −div(M(x)∇v) corresponding to an eigenvalue λ. Then N

1

uL∞ (Ω) ≤ c(α, N) λ 2 meas(Ω) 2 .

Proof. It is sufficient to apply Hölder’s inequality on the right-hand side of estimate (8.2.5) and to use (8.2.2).

8.3 Applications to some semilinear equations In this section we will study some semilinear (noncoercive) equations using the results of spectral analysis of the previous section. More precisely we will study ⎧ ⎨−div(M(x)∇u) = g(u) + f , in Ω , (8.3.1) ⎩u = 0 , on ∂Ω ,

Applications to some semilinear equations

75

under different hypotheses on g . Note that we have already studied some semilinear equations in Chapter 4. We will begin by the linear problem ⎧ ⎨−div(M(x)∇u) = μu + f , in Ω , (8.3.2) ⎩u = 0 , on ∂Ω , where μ ∈ R. Theorem 8.14. Let μ ∈ R with μ ≠ λk for every k ∈ N, where {λk }k∈N are the eigenvalues of L(v) = −div(M(x)∇v). Then for every f ∈ L2 (Ω) there exists a unique solution w to problem (8.3.2). Proof. We can suppose that μ ≠ 0, because we have already seen in Theorem 4.4 that there exists a unique H01 (Ω) solution to problem −div(M(x)∇u) = f . Let T be the operator defined by (8.2.1). Then μ −1 is not an eigenvalue of T . Consequently, using Theorem 8.27 (Fredholm alternative), for every f ∈ L2 (Ω) there exists a unique solution to the equation μ −1 u − T u = μ −1 T f , that is, there exists a unique solution to problem (8.3.2). Theorem 8.15. Let μ = λk , where for some k ∈ N, λk is an eigenvalue of L(v) =  −div(M(x)∇v)). Let f ∈ L2 (Ω) be such that Ω f wk = 0 for every eigenfunction wk corresponding to λk . Then there exists a solution to problem (8.3.2). Proof. As in the above theorem we will use the operator T defined by (8.2.1). By Theorem 8.27 (Fredholm alternative), there exists a solution to μ −1 u − T u = μ −1 T f  (i. e., to problem (8.3.2)) provided Ω T f ϕ = 0 for every ϕ such that μ −1 ϕ = T ϕ, that is, for every eigenfunction of L(v) = −div(M(x)∇v)) corresponding to μ . Now, since T is self-adjoint, one has −1 0 = Tfϕ = f Tϕ = μ f ϕ. Ω

Ω

Ω

The existence of the eigenfunctions of L(v) = −div(M(x)∇v) is sometimes useful to find a subsolution (or a supersolution) in view of an application of Theorem 4.11, as the following theorem shows. Theorem 8.16. Let θ ∈ (0, 1). Then there exists a positive solution u ∈ H01 (Ω) to problem ⎧ ⎨−Δu = uθ , in Ω , (8.3.3) ⎩u = 0 , on ∂Ω .

76

Spectral analysis for linear operators

Proof. Let ψ ∈ H01 (Ω) be the positive solution to problem ⎧ ⎨−Δψ = 1 , ⎩ψ = 0 ,

in

Ω,

on

∂Ω .

From Theorem 6.6 we know that ψ is bounded; then for any a > 0 such that a

1−θ θ

≥ ψL∞ (Ω) ,

u = aψ is a supersolution to problem (8.3.3). On the other hand, let ϕ1 be an eigenfunction corresponding to the first eigenvalue λ1 of L(v) = −Δv : ϕ1 can be chosen positive by Theorem 8.8 and is bounded by Theorem 8.11. Choosing t > 0 such that λ1 (t ϕ1 )1−θ ≤ 1 ,

one has that u = t ϕ1 is a subsolution to problem (8.3.3). To prove that u ≤ u, we remark that, by linearity, −Δ(a ψ − t ϕ1 ) = a − λ1 t ϕ1 .

It is then sufficient to consider T such that T ≥ λ1 tϕ1 L∞ (Ω) : in this way u ≥ u from Lemma 4.12. The result follows from Theorem 4.11. In the following two theorems, the hypotheses on the function g in problem (8.3.1) are related to the eigenvalues of L(v) = −div(M(x)∇v). We are now going to state Dolph’s theorem. Theorem 8.17 (Dolph). Let g : R → R be a function with the property that there exists δ > 0 such that 0 < λk + δ
¯ t , there is no solution to problem (8.3.1);  t , there exists a solution to problem (8.3.1); (2) if Ω f ϕ1 = ¯  t , there exist two solutions to problem (8.3.1). (3) if Ω f ϕ1 < ¯ We will use the following result. Lemma 8.21. Under the same hypotheses on g as in Theorem 8.20, let un , u ∈ L2 (Ω) be such that un → u in L2 (Ω), as n → +∞. (1) If tn → +∞, then g(tntnun ) → γ+ u+ − γ− u− in L2 (Ω). (2) If tn → −∞, then

g(tn un ) tn

→ γ− u+ − γ+ u− in L2 (Ω).

Proof. Set ρn (x) =

⎧ ⎨ g(tn un (x)) ,

if un (x) ≠ 0 ,

⎩0 ,

if un (x) = 0 .

tn un (x)

We observe that the sequence ρn is bounded, due to the hypotheses on g . Consequently ρn (un − u) → 0 in L2 (Ω). (1) Assume that tn → +∞. Studying separately − Ωn = {x ∈ Ω : un (x) < 0} 0 Ωn = {x ∈ Ω : un (x) = 0} + Ωn = {x ∈ Ω : un (x) > 0}

one gets that ρn u → γ+ u+ − γ− u− a.e. in Ω. Lebesgue’s theorem implies that ρn u → γ+ u+ − γ− u− in L2 (Ω). Therefore ρn un → γ+ u+ − γ− u− in L2 (Ω). (2) The case where tn → −∞ is similar to the previous case. We can now prove Theorem 8.20. Proof. Let ϕ1 be a positive eigenfunction of L(v) = −div(M(x)∇v) corresponding to the first eigenvalue λ1 , such that ϕ1 L2 (Ω) = 1. We will prove that for every s ∈ R there exists a unique solution z = zs ∈ H01 (Ω) to M(x)∇z · ∇w = g(z + s ϕ1 )w + f w , ∀w ∈ H01 (Ω) : wϕ1 = 0 (8.3.8) Ω

Ω

Ω

Ω

80

and

Spectral analysis for linear operators

 Ω

zϕ1 = 0. We will afterward study the existence of a real s such that λ1 s − g(z + s ϕ1 )ϕ1 = f ϕ1 . Ω

(8.3.9)

Ω

It is easily seen that the existence of a solution to problem (8.3.1) is equivalent to the existence of z and s . Step I: Let us study problem (8.3.8). For every fixed s ∈ R, we will find a unique solution zs to problem (8.3.8) using Theorem 4.3. Indeed, define a(ψ, w) = M(x)∇ψ · ∇w − g(ψ + s ϕ1 )w Ω

Ω

 on the Hilbert space made up of functions w ∈ H01 (Ω) such that Ω wϕ1 = 0. This form is linear in the second variable. Using the Cauchy–Schwarz inequality and the fact that g is Lipschitz continuous, we can find a positive constant C such that |a(ψ1 , w) − a(ψ2 , w)| ≤ C ∇wL2 (Ω) ∇(ψ1 − ψ2 )L2 (Ω) .

Moreover it is easy to see that a(ψ1 , ψ1 − ψ2 ) − a(ψ2 , ψ1 − ψ2 ) ≥ M(x)∇(ψ1 − ψ2 ) · ∇(ψ1 − ψ2 ) − γ+ |ψ1 − ψ2 |2 . Ω

Ω

By Theorem 8.10 and the ellipticity of M we deduce that

γ+ α∇(ψ1 − ψ2 )2L2 (Ω) . a(ψ1 , ψ1 − ψ2 ) − a(ψ2 , ψ1 − ψ2 ) ≥ 1 − λ2 We can then use Theorem 4.3 and state that for every s ∈ R there exists a unique solution zs to problem (8.3.8) . Step II: Let us prove that h(s) = λ1 s − g(zs + sϕ1 )ϕ1 Ω

is a continuous function. Obviously it is sufficient to prove that the second term is a continuous function from R to R. To this end, let sn be a real sequence. We claim that the corresponding sequence zsn of solutions to problem (8.3.8) is uniformly bounded in H01 (Ω). We can write M(x)∇zsn · ∇zsn = g(zsn + sn ϕ1 )zsn + f zsn . Ω

Ω

Ω

Applications to some semilinear equations

81

Multiplying and dividing by zsn + sn ϕ1 in the right-hand side and using the limit hypothesis on g , we have M(x)∇zsn · ∇zsn ≤ γ+ zs2n + f L2 (Ω) zsn L2 (Ω) . Ω

Ω

By Theorem 8.10 and the ellipticity of M we have

γ+ α 1− ∇zsn 2L2 (Ω) ≤ f L2 (Ω) zsn L2 (Ω) . λ2 Poincaré’s inequality implies that the sequence zsn is bounded in H01 (Ω). Now, assume that sn → s0 as n → ∞. Up to a subsequence, zsn converges weakly to a function w in H01 (Ω). Theorem 3.6 implies that g(zsn + sn ϕ1 ) → g(w + s0 ϕ1 )

in L2 (Ω). We have to prove that w = zs0 , that is, w is the solution to problem (8.3.8) corresponding to s0 . Passing to the limit in M(x)∇zsn · ∇ψ = g(zsn + sn ϕ1 )ψ + f ψ , ∀ψ : ψ ϕ1 = 0 Ω

we get

Ω

Ω

Ω



M(x)∇w · ∇ψ = Ω

g(w + s0 ϕ1 )ψ +

Ω

f ψ. Ω

From Step I there exists a unique solution to problem −div(M(x)∇zs0 ) = g(zs0 + s0 ϕ1 ) + f and so necessarily w = zs0 . Consequently, up to a subsequence, g(zsn + sn ϕ1 )ϕ1 → g(zs0 + s0 ϕ1 )ϕ1 . Ω

Ω

Arguing by contradiction, it is easily seen that g(zsn + sn ϕ1 )ϕ1 → g(zs0 + s0 ϕ1 )ϕ1 Ω

Ω

and not only a subsequence. This implies that h is a continuous function. Step III: Let us prove that lim

s→±∞

h(s) = λ 1 − γ± . s

Since h(s) = λ1 − s

 * + g s zs + ϕ 1 s Ω

s

ϕ1 ,

82

Spectral analysis for linear operators

it is sufficient to study the last term. We remark that zs is uniformly bounded in z H01 (Ω) as we have already proved in the previous step. Consequently ss → 0 in zs 1 2 H0 (Ω) as s → ∞. Setting vs = s + ϕ1 we have that vs → ϕ1 in L (Ω). (1) Assume that s → +∞. By Lemma 8.21, we have g(svs ) → γ+ ϕ1+ − γ− ϕ1− s

in L2 (Ω). Since ϕ1 is positive,  * + g s zs + ϕ 1 s Ω

s

ϕ 1 → γ+ .

(2) In a similar way, one can study the case s → −∞. Step IV: Problem (8.3.9) is equivalent to h(s) =

 Ω

f ϕ1 . Since

λ 1 − γ + < 0 < λ 1 − γ− ,

using the result of the previous step, we can say that h has a maximum. Therefore  (1) if Ω f ϕ1 < maxR h, then problem (8.3.1) has at least two solutions;  (2) if Ω f ϕ1 = maxR h, then problem (8.3.1) has at least one solution;  (3) if Ω f ϕ1 > maxR h, then problem (8.3.1) has no solution.

8.4 Appendix We recall here some classical results of spectral theory for linear operators. For the proofs, see [22]. Definition 8.22. Let H be a Hilbert space. Let T : H → H be a linear operator. Let BH = {x ∈ H : x ≤ 1}. We will say that (1) T is self-adjoint if (T u|v) = (u|T v) for every u, v ∈ H ; (2) T is compact if T (BH ) is relatively compact for the strong topology. Definition 8.23. Let E be a Banach space. Let T : E → E be a linear operator. (1) we set ρ(T ) = {λ ∈ R : T − λI : E → E is one-to-one}; (2) the spectrum of T is σ (T ) = R \ ρ(T ); (3) λ is an eigenvalue if Ker (T − λI) ≠ 0. We will denote by AT (T ) the eigenvalues set. Remark that the eigenvalues set of T is contained in σ (T ), but it is not equal in general.

Appendix

83

Theorem 8.24. Let E be Banach space. Let T : E → E be a linear compact operator. Then (1) 0 ∈ σ (T ); (2) σ (T ) \ {0} = AT (T ) \ {0}; (3) either σ (T ) \ {0} is finite or σ (T ) \ {0} is a sequence which goes to 0. Theorem 8.25 (Spectral Theorem). Let H be a Hilbert separable space. Let T be a linear, self-adjoint compact operator. Then H has an orthonormal basis composed on eigenvectors of T . Moreover, the sequence of the corresponding eigenvalues λn is such that |λn | → 0 as n → ∞. Theorem 8.26. Let H be a Hilbert space. Let T : H → H be a linear, compact, selfadjoint operator. Then, if μn are the eigenvalues of T , one has T L(H) = sup |μn | . n

Theorem 8.27 (Fredholm alternative). Let H be a Hilbert space. Let T : H → H be a linear, compact, self-adjoint operator. Let λ ∈ R \ {0}. Then the following alternative holds: either for any ξ ∈ H the equation λx − T x = ξ has a unique solution or else λx = T x has solutions x ≠ 0 and λx − T x = ξ has a solution if ξ ⊥ Ker (λI − T ).

9 Calculus of variations and Euler’s equation 9.1 Introduction Although this book is devoted to partial differential equations, we would like to recall some results of the Calculus of Variations. We will see how the Calculus of Variations can be used to study the existence of solutions to differential problems. We will first present a simple but significant version of De Giorgi’s theorem on weak lower semicontinuity of integral functionals. We will then write their Euler’s equation and study the summability of minimizers. In the last part of this chapter we will study Ekeland’s principle.

9.2 Direct methods in the calculus of variations In this section we will recall some classical results of the Calculus of Variations which will be useful to study the minimization of a functional. We refer to [27] for the proofs. We will often use the following two theorems (in the Appendix we will recall some definitions). Theorem 9.1 (Weierstrass). Let X be a Banach reflexive space. Let J : X → R be coercive, bounded from below and weakly lower semicontinuous. Then J has a minimizer. Proof. Let xn be a minimizing sequence, that is, J(xn ) → inf J . Note that J(xn ) is bounded. Since J is coercive, xn is bounded in X . Therefore, up to a subsequence, xn → x0 weakly in X , for some x0 ∈ X , since X is reflexive. Since J is weakly lower semicontinuous, lim infn→∞ J(xn ) ≥ J(x0 ). Hence J(x0 ) = inf J . We will essentially consider integral functionalsin this chapter: J(v) = j(x, v, ∇v) , Ω 1,p W0 (Ω)

defined on with p ∈ (1, ∞), with Ω bounded open subset of RN . Note that these spaces are reflexive. The following theorem gives a sufficient condition for the 1,p functional to be weakly lower semicontinuous in W0 (Ω) (see [26]). Theorem 9.2 (De Giorgi). Let j : Ω × R × RN → R be a Carathéodory function, convex with respect to the last variable. Let p > 1 and 1 ≤ q < p ∗ if p < N and 1 ≤ q < +∞  if p ≥ N . Assume that there exists α1 ∈ (Lp (Ω))N , α2 ∈ R, α3 ∈ L1 (Ω) such that j(x, s, ξ) ≥ α1 (x) · ξ + α2 |s|q + α3 (x) . 1,p

Let un , u ∈ W0 (Ω) be such that un → u weakly in W 1,p (Ω). Then j(x, u, ∇u) ≤ lim inf j(x, un , ∇un ) . n→+∞

Ω

Ω

Direct methods in the calculus of variations

85

The proof of this theorem is quite complicated. We will present it for a particular class of functionals. The symbol ∇ξ j will denote the gradient of j with respect to ξ . Theorem 9.3. Let j : Ω × R × RN → R be a Carathéodory, convex function with respect to the last variable. Assume that there exist α, β > 0 such that α|ξ|p ≤ j(x, s, ξ) ≤ β|ξ|p . Assume that for a.e. x ∈ Ω and for every s ∈ R j(x, s, ·) is differentiable and  there exists ν > 0 such that |∇ξ j(x, s, ξ)| ≤ ν|ξ|p−1 . Then J(v) = Ω j(x, v, ∇v), 1,p defined on W0 (Ω), has a minimizer. Proof. J is coercive and bounded from below, since j(x, s, ξ) ≥ α|ξ|p . Let us prove that it is weakly lower semicontinuous. Let vn be a sequence weakly converging to v 1,p in W0 (Ω). The hypotheses of convexity and differentiability on j imply j(x, s, ξ) ≥ j(x, s, η) + ∇ξ j(x, s, η) · (ξ − η) ,

and therefore j(x, vn , ∇vn ) ≥ j(x, vn , ∇v) + ∇ξ j(x, vn , ∇v) · (∇vn − ∇v). Ω

Ω

(9.2.1)

Ω

By the continuity of j(x, ·, ξ) and the growth conditions on j , we can apply Theorem 3.6 to have that j(x, vn , ∇v) converges in L1 (Ω) to j(x, v, ∇v). Moreover ∇vn − ∇v converges weakly to 0 in (Lp (Ω))N by hypothesis. By the continuity of j(x, ·, ξ) and the growth conditions on ∇ξ j we can apply again Theorem 3.6 to have  that ∇ξ j(x, vn , ∇v) converges to ∇ξ j(x, v, ∇v) in (Lp (Ω))N . Passing to the lim inf in (9.2.1), we get lim inf j(x, vn , ∇vn ) ≥ j(x, v, ∇v) . n→+∞

Ω

Ω

Example 9.4. Let J be defined on H01 (Ω) by 1 1 J(v) = |∇v|2 − |v|p − f v 2 p Ω

Ω

Ω

with p ∈ [1, 2) and f ∈ L2 (Ω). We claim that this functional has a minimizer. Note that, by Hölder’s and Poincaré’s inequalities 1 1 2 J(v) ≥ ∇vL2 (Ω) − |v|p − c f L2 (Ω) ∇vL2 (Ω) . 2 p Ω

By Rellich–Kondrachov theorem, there exists a positive constant C such that p p vLp (Ω) ≤ C∇vL2 (Ω) . Therefore J(v) ≥

1 C p ∇v2L2 (Ω) − ∇vL2 (Ω) − c f L2 (Ω) ∇vL2 (Ω) . 2 p

86

Calculus of variations and Euler’s equation

The last quantity converges to +∞ if ∇vL2 (Ω) → +∞, since p < 2. This means that J is coercive. Moreover the last inequality implies that J is bounded from below. Finally it is easy to verify that J satisfies the hypotheses of Theorem 9.2. By Theorem 9.1 J has a minimizer.

9.3 Euler equation We will now give the definition of the Euler equation associated with a functional. The Euler equation is related to the Gâteaux differentiability of the functional. We will use the same notations as in Definition 9.29. Definition 9.5. Let X be a Banach space. Let J : X → R be a functional attaining its minimum at u. Suppose that J is Gâteaux differentiable. The equation < J  (u), ϕ >= 0, ϕ ∈ X , is the Euler equation associated with J . We are now going to state a result that allows us to write the Euler equation associated with integral functionals. We will assume a p -growth in the gradient, with p < N . We refer to [27] for the case p ≥ N and for the proofs. Theorem 9.6. Let j : Ω × R × RN → R. Let J(v) = j(x, v, ∇v) Ω 1,p

∂j

be defined on W0 (Ω). Assume that j = j(x, s, ξ), ∂s , ∇ξ j are Carathéodory functions. Let u0 be a minimizer for J . Then the Euler equation ∂j (x, u0 , ∇u0 )ϕ = 0 ∇ξ j(x, u0 , ∇u0 ) · ∇ϕ + ∂s Ω

Ω

Np

1,p

is satisfied for every ϕ ∈ W0 (Ω) if there exist α1 ∈ L1 (Ω), α2 ∈ L Np−N+p (Ω),  α3 ∈ Lp (Ω) and β ≥ 0 such that ∗ (1) |j(x, s, ξ)| ≤ α1 (x) + β (|u|p + |ξ|p ); ∂j (2) | ∂s | ≤ α2 (x) + β (|u|r1 + |ξ|r2 ), |∇ξ j| ≤ α3 (x) + β (|u|q + |ξ|p−1 ) with r1 ≤ Np−N+p Np−N p ∗ − 1, r2 ≤ , q ≤ N−p . N j satisfies condition 1 of the above theorem and the Remark 9.7. In the case  where   ∂j    derivatives of j satisfy ∂s ≤ α1 (x) + β|ξ|p , |∇ξ j| ≤ α2 (x) + β|ξ|p−1 for some 

1,p

α2 ∈ Lp (Ω), the Euler equation is verified for every ϕ ∈ W0 (Ω)∩L∞ (Ω), provided the minimizer u0 is bounded. We will see in the next section sufficient conditions

assuring the boundedness of minimizers. We are going to give some applications of the previous results.

Euler equation

87

Example 9.8. Let J:

be defined by 1 J(v) = 2

H01 (Ω) → R ,

M(x)∇v · ∇v − Ω

fv Ω

where M(x) is a RN×N matrix with bounded entries and such that M(x)ξ ·ξ ≥ α|ξ|2 for some α > 0; moreover f ∈ L2 (Ω). It is easy to see that J attains its minimum at u using Theorems 9.1 and 9.2. Note that the coercivity of J is easily obtained by Hölder’s and Poincaré’s inequalities: J(v) ≥

α α ∇v2L2 (Ω) − f L2 (Ω) vL2 (Ω) ≥ ∇v2L2 (Ω) − c f L2 (Ω) ∇vL2 (Ω) . 2 2

The Euler equation that u satisfies is ⎧ ⎨−div(M(x)∇u) = f , ⎩u = 0 ,

in Ω , on ∂Ω ,

by Theorem 9.6. With the same technique one can prove the existence of a minimizer u ∈ H01 (Ω) of 1 J(v) = M(x)∇v · ∇v − f v + G(v) , 2 Ω

Ω

Ω

where G : R → R is a C function such that 0 ≤ G(s) ≤ γ1 |s|q1 , q1 ≤ 2∗ and |g(s)| = |G (s)| ≤ γ2 |s|q2 , q2 ≤ 2∗ − 1, with γ1 , γ2 ≥ 0. Therefore there exists a solution u ∈ H01 (Ω) to ⎧ ⎨−div(M(x)∇u) + g(u) = f , in Ω , ⎩u = 0 , on ∂Ω . 1

Example 9.9. Let J:

be defined by

J(v) =

v2

M(x)∇v · ∇v − λ1 Ω

where M(x) is a RN×N

H01 (Ω) → R ,

Ω

matrix with bounded entries and such that M(x)ξ ·ξ ≥ α|ξ|2 for some α > 0; moreover λ1 is the smallest eigenvalue of L(v) = −div(M(x)∇v). As a consequence of Theorem 8.4, J attains its minimum at u ∈ H01 (Ω), which satisfies ⎧ ⎨−div(M(x)∇u) = λ1 u , in Ω , ⎩u = 0 , on ∂Ω , by Theorem 9.6.

88

Calculus of variations and Euler’s equation

Remark 9.10. There exist however equations that are not the Euler equation associated with some functional, as we will see in Example 9.16. In the next example we will solve a constrained minimization problem. We will see a different proof in Theorem 9.25. Example 9.11. Let J be defined on H01 (Ω) by 1 1 2 J(v) = |∇v| − |v|p 2 p Ω

Ω

with 2 < p < 2∗ . We claim that J does not have a minimizer in H01 (Ω). Indeed, if ϕ1 ∈ H01 (Ω) is the eigenfunction corresponding to the first eigenvalue of the Laplacian operator (that is, −Δϕ1 = λ1 ϕ1 ), one has, for t > 0, J(tϕ1 ) =



t2 2

|∇ϕ1 |2 − Ω

tp p

|ϕ1 |p = Ω

t2 λ1 2

ϕ12 − Ω

tp p

|ϕ1 |p . Ω

The limit as t → +∞ proves that J is unbounded from below, since p > 2. Nevertheless we are going to prove that J has a minimizer on ⎧ ⎫ ⎪ ⎪ ⎨ ⎬ A = v ∈ H01 (Ω) : |v|p = 1 . ⎪ ⎪ ⎩ ⎭ Ω

It is clear that J is bounded from below on A, by Poincaré’s inequality: J(v) ≥

c2 1 1 p v2L2 (Ω) − vLp (Ω) ≥ − . 2 p p

Moreover if vn is a minimizing sequence, up to a subsequence, vn converges weakly in H01 (Ω) to some function u in A, which is a minimizer, as J is weakly lower semicontinuous. Let us write the equation that u satisfies. We have that , u + tv , J(u) ≤ J u + tvLp (Ω) for every t ∈ R and for every v ∈ H01 (Ω). Let us set , u + tv . g(t) = J u + tvLp (Ω) We note that ⎛



⎞

d ⎜ ⎟ ⎝ |u + tv|p ⎠ = p dt Ω

|u + tv|p−2 (u + tv)v . Ω

Euler equation

89

Using that u ∈ A we obtain ⎡ ⎤ ⎢ ⎥ g  (0) = ⎣ ∇u · ∇v − |∇u|2 |u|p−2 uv ⎦ = 0 . Ω

Ω

Ω

Therefore u satisfies −Δu − u2H 1 (Ω) |u|p−2 u = 0 . 0

Now, let w = tu, for t > 0; then w satisfies −Δw =

Choosing t such that −Δw = |w|p−2 w .

tu2 1

H0 (Ω)

t p−1

tu2H 1 (Ω) 0

t p−1

|w|p−2 w .

= 1 we infer the existence of a H01 (Ω) solution to 1,p

Example 9.12. We prove the existence of a W0 (Ω) solution to the following problem: ⎧ ⎨−div(|∇u|p−2 ∇u) = b(x, u) , in Ω , (9.3.1) ⎩u = 0 , on ∂Ω , where p ≥ 2 and b(x, s) : Ω × R → R is a Carathéodory bounded function. We remark that this problem is a Leray–Lions problem (see Chapter 5). Here we want to show how it can be solved using the tools of the Calculus of Variations and Schauder’s theorem. 1,p 1,p Let us define σ : W0 (Ω) → W0 (Ω) as the map that associates with w ∈ 1,p 1,p W0 (Ω) the solution z ∈ W0 (Ω) of ⎧ ⎨−div(|∇z|p−2 ∇z) = b(x, w) , in Ω , (9.3.2) ⎩z = 0 , on ∂Ω . This map is well defined, since by Theorems 9.1 and 9.2 1 J(v) = |∇v|p − b(x, w)v p Ω

Ω

has a minimizer and the Euler equation is −div(|∇z|p−2 ∇z) = b(x, w) .

Moreover such a minimizer is unique, by the strict convexity of J . Let us prove that there exists a bounded, convex, invariant set for σ and that σ is completely continuous. The existence of a solution to problem (9.3.1) will follow from Theorem 2.10. Let us consider z in the weak formulation of problem (9.3.2) as a test function: by Poincaré’s inequality, we have p p−2 ∇zLp (Ω) = |∇z| ∇z · ∇z = b(x, w)z ≤ C∇zLp (Ω) Ω

Ω

90

Calculus of variations and Euler’s equation

since b is bounded. This implies that there exists R such that if ∇wLp (Ω) ≤ R then ∇zLp (Ω) ≤ R , that is, there exists a convex, bounded, invariant set for σ . To prove that σ is completely continuous, it is sufficient to prove that if wn → w 1,p 1,p weakly in W0 (Ω) then σ (wn ) → σ (w) in W0 (Ω). To this end, by choosing zn −z as a test function, we have |∇zn |p−2 ∇zn · ∇(zn − z) = b(x, wn )(zn − z) Ω

Ω





|∇z|p−2 ∇z · ∇(zn − z) = Ω

b(x, w)(zn − z) . Ω

Subtracting side by side   |∇zn |p−2 ∇zn − |∇z|p−2 ∇z · ∇(zn − z) = [b(x, wn ) − b(x, w)](zn − z) . Ω

Ω

Now, let us use the inequality [|s| s − |t| t](s − t) ≥ C|s − t|p for s, t ∈ R on the left-hand side and Hölder’s inequality on the right-hand side. We obtain C |∇(zn − z)|p ≤ b(x, wn ) − b(x, w)Lp (Ω) zn − zLp (Ω) . p−2

p−2

Ω

Poincaré’s inequality implies p−1

C∇(zn − z)Lp (Ω) ≤ b(x, wn ) − b(x, w)Lp (Ω) . 

By Theorem 3.6 b(x, wn ) → b(x, w) in Lp (Ω), since wn → w in Lp (Ω). Conse1,p 1,p quently zn → z in W0 (Ω), that is, σ (wn ) → σ (w) in W0 (Ω).

9.4 Summability of minimizers of integral functionals In this section we study the regularity of the minimizers of the functionals J(v) = j(x, v, ∇v) − f v , v ∈ H01 (Ω) , Ω

where j : Ω × R × R such that

N

Ω

→ R is a Carathéodory function, convex in the last variable, j(x, s, ξ) ≥ α|ξ|2

(9.4.1)

for some positive α. Moreover we assume that j(x, s, 0) = 0. Observe that if f be2N longs to L N+2 (Ω), the assumptions on j assure the existence of a minimizer u ∈ H01 (Ω) of J , by Theorems 9.1 and 9.2. We are going to present some regularity results on u, according to the summability of f , proving the following results (see [18]): f ∈ Lm (Ω) , f ∈ Lm (Ω) ,

m > N/2 ⇒ u ∈ H01 (Ω) ∩ L∞ (Ω) ∗∗

m ∈ (2N/(N + 2), N/2) ⇒ u ∈ H01 (Ω) ∩ Lm (Ω)

91

Summability of minimizers of integral functionals

Remark 9.13. We observe that we have the same summability of the solutions to Leray–Lions problem (see Theorems 6.6 and 6.9). Remark 9.14. In the case where f has a lower summability, we will introduce the notion of T -minimum in Chapter 11. In the sequel we will use the following sets: Ak = {|u| ≥ k} ,

Bk = {k ≤ |u| < k + 1} .

Theorem 9.15. Let u ∈ H01 (Ω) be a minimizer of J . Let f ∈ Lm (Ω), with m > Then u is bounded.

N 2.

Proof. By the minimality of u, we have J(u) ≤ J(Tk (u)), that is, j(x, u, ∇u) ≤ j(x, Tk (u), ∇Tk (u)) + f Gk (u) . Ω

Ω

This is equivalent to

Ω



j(x, u, ∇u) ≤

f Gk (u) . Ω

Ak

Assumption (9.4.1) on j implies that α |∇Gk (u)|2 ≤ f Gk (u) .

(9.4.2)

Ω

Ak

As observed in Remark 6.8 we deduce from estimate (9.4.2) that u is bounded. Example 9.16. Let J : H01 (Ω) → R be defined by 1 2 J(v) = a(v)|∇v| − f v 2 Ω

(9.4.3)

Ω

under the following hypotheses: (1) 0 < α ≤ a(s) ≤ β, for some α, β > 0; (2) a is differentiable and there exists γ > 0 such that |a (s)| ≤ γ ; 2N (3) f belongs to L N+2 (Ω). By the above result, J attains its minimum at u ∈ H01 (Ω) ∩ L∞ (Ω) satisfying 1 a(u)∇u · ∇v + a (u)|∇u|2 v − f v = 0 , ∀ v ∈ H01 (Ω) ∩ L∞ (Ω) 2 Ω

Ω

Ω

by Remark 9.7. Consequently the equation −div(a(u)∇u) + g(u)|∇u|2 = f

is the Euler equation associated with a functional if and only if g = a /2.

92

Calculus of variations and Euler’s equation

Theorem 9.17. Let u ∈ H01 (Ω) be a minimizer of J . Let f ∈ Lm (Ω), with N m∗∗ (Ω). 2 . Then u ∈ L

2N N+2

1 and let M be a positive integer. Multiplying inequality (9.4.4) by (1 + k)2λ−3 and summing up for k from 0 to M we have M M   α (1 + k)2λ−3 |∇u|2 ≤ (1 + k)2λ−3 |f | |u| . (9.4.5) k=0

k=0

Ak

Ak

We are going to study the right-hand side of (9.4.5). Since Ak = M 

(1 + k)2λ−3

k=0

|f ||u| =

M 

(1 + k)2λ−3

k=0

Ak

=

+∞ 

∞ 

!+∞

j=k Bj ,

we obtain

|f ||u|

h=k B h

|f ||u|

h=0 B h

TM (h) 

(1 + k)2λ−3 ,

k=0

by exchanging the summation order. We now observe that for any h ∈ N it results TM (h) 

(1 + k)2λ−3 ≤

k=0

1 (2 + TM (h))(λ−1)2 . 2(λ − 1)

Moreover on Bh one has (1 + TM (h))(λ−1)2 ≤ (1 + |TM (u)|)(λ−1)2 . Therefore, the right-hand side of (9.4.5) can be estimated by M ∞   1 2λ−3 (1 + k) |f | |u| ≤ |f ||u|[2 + |TM (u)](λ−1)2 2(λ − 1) h=0 k=0 Bh Ak 1 ≤ |f ||u|[2 + |TM (u)](λ−1)2 . 2(λ − 1) Ω

We handle the left-hand side of (9.4.5) in a similar way: M M ∞    (1 + k)2λ−3 |∇u|2 = (1 + k)2λ−3 |∇u|2 k=0

k=0

Ak

=

+∞  h=0 B h

h=k B h 2

|∇u|

TM (h) 

(9.4.6) 2λ−3

(1 + k)

k=0

.

93

Summability of minimizers of integral functionals

We observe that

TM (h)  [1 + TM (h)](λ−1)2 ≤ (1 + k)2λ−3 . 2(λ − 1) k=0

Moreover on Bh one has |TM (u)|(λ−1)2 ≤ (1 + TM (h))(λ−1)2 . Therefore, by (9.4.6), the left-hand side can be estimated from below by α 2(λ − 1)

|∇u|2 |TM (u)|(λ−1)2 ≤ α



+∞ 

|∇u|2

h=0 B

Ω

=α

M 

TM (h) 

(1 + k)2λ−3

k=0

h

2λ−3

|∇u|2 .

(1 + k)

k=0

Ak

In conclusion, if λ is any real greater than 1, it holds α |∇u|2 |TM (u)|(λ−1)2 ≤ |f ||u|(2 + |TM (u)|)(λ−1)2 . Ω

(9.4.7)

Ω

Step II: We are going to prove that |f | |u|(λ−1)2+1 ∈ L1 (Ω) .

(9.4.8)

Let η0 = 2∗ . Since f belongs to Lm (Ω) and u belongs to Lη0 (Ω), by Hölder’s in1 + 2λ−1 = 1, that is, equality, (9.4.8) holds true if λ > 0 is such that m η λ = λ(η) =



1 2

Define λ0 = λ(η0 ) =

η − 1 + 1. m

 1  2∗ − 1 + 1. 2 m

Thus (9.4.8) holds for λ = λ0 . We have λ0 > 1, since m > (2∗ ) . Letting M tend to infinity in (9.4.7), we get α |∇u|2 |u|(λ0 −1)2 ≤ |f | (2 + |u|)(λ0 −1)2+1 . Ω

Ω

The Sobolev inequality on the left-hand side implies ⎡



⎤

∗⎥ ⎢ ⎣ |u|λ0 2 ⎦

Ω

2 2∗

≤C

|f | (2 + |u|)(λ0 −1)2+1 ,

(9.4.9)

Ω

where C denotes a positive constant depending on S, λ0 , α. Define, for η in R, γ(η) = λ(η) 2∗ =

+ 1* η − 1 2∗ + 2∗ . 2 m

(9.4.10)

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Calculus of variations and Euler’s equation

η0 By the definition of λ0 (λ0 − 1)2 + 1 = m  ; by using Hölder’s inequality on (9.4.9), we have ⎧ ⎡ ⎤ 2∗ ⎡ ⎤ 1 ⎫ ⎪ 2 m ⎪ ⎪ ⎪ ⎨ ⎬ ⎢ ⎢ ⎥ γ(η0 ) ⎥ η 0 . ⎣ |u| ⎦ ≤ C f L1 (Ω) + f Lm (Ω) ⎣ |u| ⎦ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ Ω Ω

Let θ =

2∗ 2m ,

and observe that 0 < θ < 1 since m < |u|γ(η0 ) Ω

N 2.

We then have

⎧ ⎡ ⎤θ ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 2∗ 2∗ ⎢ 2 2 η0 ⎥ ≤ C f L1 (Ω) + f Lm (Ω) ⎣ |u| ⎦ . ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ Ω

(9.4.11)

Set η1 = γ(η0 ); we remark that η0 < η1 < m∗∗ . Thus, we deduce from (9.4.11) that u belongs to Lη1 (Ω). Step III: Since η1 > η0 , we have λ(η1 ) > 1; arguing as in Step II we can say that (9.4.8) holds true with λ1 = λ(η1 ). Thus, it is possible to pass to the limit in (9.4.7) as M tends to infinity obtaining (9.4.9) with λ = λ1 , that is, ⎡

⎤ 2∗ 2 ∗ ⎢ λ1 2 ⎥ ⎣ |u| ⎦ ≤ C |f | (2 + |u|)(λ1 −1)2+1 . Ω

Ω

The same passages as above yield the following inequality, by definition of γ(η) (see (9.4.10)): ⎧ ⎡ ⎤θ ⎫ ⎪ ⎪ ⎪ ⎪ ∗ ∗ ⎨ ⎬ 2 2 ⎢ 2 2 γ(η1 ) η1 ⎥ |u| ≤ C f L1 (Ω) + f Lm (Ω) ⎣ |u| ⎦ . ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ Ω Ω By induction we get, setting ηk+1 = γ(ηk ), |u|ηk+1 ≤ C + C Ω

⎧ ⎪ ⎨ ⎪ ⎩

Ω

|u|ηk

⎫θ ⎪ ⎬ .

⎪ ⎭

(9.4.12)

The positive constant C above depends on ηk , f L1 (Ω) , α, S and meas(Ω). Since ηk < m∗∗ , as m > 1, we can assume that C does not depend on k. Step IV: We first observe that ηk is an increasing sequence. Using Hölder’s inequality with exponent Ω

ηk+1 ηk ,

we have ⎡



⎤

⎢ ⎥ |u|ηk ≤ ⎣ |u|ηk+1 ⎦ Ω

ηk ηk+1

1− η

meas(Ω)

ηk k+1

.

(9.4.13)

The Ekeland variational principle

95

By Hölder’s inequality, (9.4.12) and (9.4.13) give: 



θ 1− η

|u|ηk+1 ≤ C + C meas(Ω) Ω

⎡ ⎢ ≤ C + C ⎣1 +



⎢ ≤ C + C ⎣1 +



k+1

⎤θ ⎥ |u|ηk+1 ⎦

Ω

⎡

ηk



⎡



⎤θ

⎢ ⎥ ⎣ |u|ηk+1 ⎦

ηk ηk+1

Ω ηk ηk+1

⎤θ ⎥ |u|ηk+1 ⎦

Ω

⎡

⎤θ ⎢ ⎥ ≤ C + C ⎣ |u|ηk+1 ⎦ . Ω

The last inequality implies that |u|ηk+1 ≤ C ,

(9.4.14)

Ω

as θ < 1. Since {ηk } is an increasing and bounded sequence, it converges to some ρ > 0. By (9.4.10), the limit ρ > 0, as k tends to infinity, is such that ρ=

+ 1* ρ − 1 2∗ + 2∗ ,  2 m

that is, ρ = m∗∗ . Letting k tend to infinity in (9.4.14) we obtain that u belongs to ∗∗ Lm (Ω).

9.5 The Ekeland variational principle In the previous section, we have seen the importance of the minimizing sequences of a functional. In this section, we explain the Ekeland variational principle, which is a useful tool in studying their behavior (see [28]). Indeed it allows us in some sense to have a “good” minimizing sequence, whose elements have some minimizing properties. Theorem 9.18 (Ekeland principle). Let (X, d) be a complete metric space. Let J : X → R ∪ {+∞} be a lower semicontinuous functional which is bounded from below. Let u ∈ X be such that 1 J(u) ≤ inf J + . (9.5.1) X n

96

Calculus of variations and Euler’s equation

Then there exists v ∈ X such that J(v) ≤ J(u)

(9.5.2)

d(v, u) ≤ 1

(9.5.3)

J(v) < J(w) +

1 n d(v, w) ,

∀

w ∈X,

w ≠v.

(9.5.4)

Proof. Let us define by induction a sequence uk ⊂ X . We set u1 = u. Assume we have defined u1 , u2 , . . . uk . Let 0 1 1 Sk = w ∈ X : J(w) ≤ J(uk ) − n d(uk , w) . Sk is not empty, since uk ∈ Sk . By definition of infimum, there exists uk+1 ∈ Sk such

that 1 J(uk+1 ) ≤ 2

.

/ J(uk ) + inf J Sk

.

(9.5.5)

Let us prove that uk is a Cauchy sequence. Since uk+1 ∈ Sk then 1 n d(uk , uk+1 )

≤ J(uk ) − J(uk+1 ) .

(9.5.6)

By the triangle inequality m 1  1 d(uk , uk+m ) ≤ d(uk+j , uk+j−1 ) ≤ J(uk ) − J(uk+m ) . n n j=1

(9.5.7)

Now, from (9.5.6) it follows that J(uk ) is decreasing; since J is bounded from below in X , one has that lim J(uk ) = α k→∞

for some α ∈ R. Inequality (9.5.7) proves that uk is a Cauchy sequence. Consequently there exists v ∈ X such that v = limk→∞ uk . On the other hand, since J is lower semicontinuous, J(v) ≤ lim inf J(uk+m ) = α . m→∞

This inequality and the limit for m → +∞ in (9.5.7) imply that 1 d(uk , v) ≤ J(uk ) − J(v) . n

(9.5.8)

For k = 1 we have 1 1 d(u, v) ≤ J(u) − J(v) ≤ J(u) − inf J ≤ X n n

by hypothesis (9.5.1). Hence d(u, v) ≤ 1 and J(v) ≤ J(u), that is, (9.5.2) and (9.5.3) hold. To prove (9.5.4), assume by contradiction that there exists w ∈ X such that J(w) < J(v) −

1 d(w, v) . n

(9.5.9)

The Ekeland variational principle

97

Using (9.5.8) one has J(w) < J(uk ) −

1 1 1 d(uk , v) − d(w, v) < J(uk ) − d(uk , w), n n n

that is, w ∈ Sk , for every k. Therefore inf J ≤ J(w). Sk

By (9.5.5) and (9.5.9) we obtain 2J(uk+1 ) − J(uk ) ≤ J(w) < J(v) −

1 d(w, v) . n

Passing to the limit for k → ∞ we get J(v) ≤ J(w) < J(v) −

1 d(v, w) n

which is a contradiction.

2 Remark 9.19. Let us introduce in X the distance d1 = n1 d. Then (X, d1 ) is a complete metric space. From Theorem 9.18 it follows that if un is a minimizing sequence, there exists vn ∈ X such that (1) J(vn ) ≤ J(un2) (2) d(un , vn ) ≤ n1 2 (3) J(vn ) ≤ J(w) + n1 d(vn , w) , ∀w ∈ X .

Hence vn is a minimizing sequence and its elements have some minimizing properties. We now apply Ekeland’s principle to study some regular functionals (see the appendix for the definitions). Theorem 9.20. Let (X, ·) be a Banach space and let J : X → R be a semicontinuous functional that is bounded from below. Let J be Gâteaux differentiable in every direction w ∈ X . Then for every n > 0 there exists un ∈ X such that J(un ) ≤ inf J + X

J  (un )X  ≤

1 n

1 . n

Proof. The first inequality follows from the definition of infimum. Let us prove the second one. Theorem 9.18 implies that there exists un ∈ X such that J(un ) ≤ J(v) +

1 v − un , n

for all v ∈ X .

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Calculus of variations and Euler’s equation

Let w ∈ X and t > 0 be fixed. Choosing v = un + tw in the previous inequality, we get 1 J(un ) − J(un + tw) ≤ w . t n Passing to the limit as t → 0 in the previous inequality, one has J  (un ), w ≤ for every w ∈ X . Since this equality is valid for w and −w we get |J  (un ), w| ≤

1 w , n

Therefore J  (un )X  =

sup w∈X , w≠0

1 n w

∀w ∈ X .

1 J  (un ), w ≤ . w n

Definition 9.21. Let (X, ·) be a Banach space and J : X → R be a C 1 functional. We say that J satisfies the Palais–Smale condition if every sequence un in X such that the sequence |J(un )| is bounded and J  (un ) → 0 in X  has a converging subsequence. We are going to prove the following theorem (see [41]): Theorem 9.22 (Minimization with the Palais–Smale condition). Let (X,  · ) be a Banach space and J : X → R a C 1 functional satisfying the Palais–Smale condition. Assume that J is bounded from below. Then J attains its minimum at u0 ∈ X and u0 is a critical point for J , that is, J  (u0 ) = 0. Proof. By Theorem 9.20 for every n there exists un ∈ X such that J(un ) ≤ inf J + X

1 , n

J  (un )X  ≤

1 . n

The Palais–Smale condition implies the existence of a subsequence unj and u0 ∈ X such that unj → u0 . From the continuity of J and J  , passing to the limit as n → ∞ we get J(u0 ) = inf J , J  (u0 ) = 0 . X

We are going to give an application of the previous theorem to the study of the critical points of a functional. Theorem 9.23. Let λ1 be the first eigenvalue of L(v) = −Δv . Let f ∈ Lp (Ω) with 2 < p < 2∗ . Then 1 λ1 1 J(v) = |∇v|2 − v2 + |v|p − f v 2 2 p Ω

Ω

Ω

Ω

defined on H01 (Ω), attains its minimum at u which satisfies u ∈ H01 (Ω) : −Δu + |u|p−2 u = λ1 u + f .

The Ekeland variational principle

99

Proof. We will prove the existence of a minimizer using Theorem 9.22. J is a C 1 functional. Moreover it is bounded from below, because 1 λ1 |∇v|2 − v2 ≥ 0 2 2 Ω

Ω

as we have seen in Theorem 8.4. On the other hand, using Hölder’s inequality we have that 1 1 |v|p − f v ≥ |v|p − f Lp (Ω) vLp (Ω) p p Ω

Ω

Ω

which is bounded from below. Let us prove that J satisfies the Palais–Smale condition. Let un be a sequence such that |J(un )| ≤ R

(9.5.10)

J  (un ) → 0 .

(9.5.11)

for some R > 0 and We want to prove that un converges in H01 (Ω), up to subsequence. Inequality (9.5.10) is equivalent to 1 λ1 1 −R ≤ |∇un |2 − u2n + |un |p − f un ≤ R . 2 2 p Ω

Since

Ω

1 2



Ω

λ1 |∇un | − 2



2

Ω

Ω

u2n ≥ 0 Ω

one has, using Young’s inequality, 1 1  |un |p ≤ R + f un ≤ R + |un |p + c(p) |f |p . p 2p Ω

Ω

Ω

Ω

Consequently the sequence un is bounded in Lp (Ω) and so, up to a subsequence, un → u weakly in Lp (Ω). On the other hand (9.5.11) gives us − Δun + |un |p−2 un − λ1 un − f = yn ,

(9.5.12)

with yn ∈ H −1 (Ω) converging to 0 in H −1 (Ω). Considering un as a test function, one has |∇un |2 ≤ |∇un |2 + |un |p = λ1 u2n + f un + yn , un  . Ω

Ω

Ω

Ω

Ω

The right-hand side is uniformly bounded and so un → u weakly in H01 (Ω) and un → u in Lp (Ω), since p < 2∗ . We claim that un → u in H01 (Ω). Choosing un −u as a test

100

Calculus of variations and Euler’s equation

function, (9.5.12) implies ∇(un − u) · ∇(un − u) = − |un |p−2 un (un − u) + λ1 un (un − u) Ω

Ω

Ω





f (un − u) + yn , un − u −

+ Ω

∇u · ∇(un − u) . Ω

It is easily seen that the right-hand side goes to zero, by the fact that un → u weakly in H01 (Ω) and strongly in Lp (Ω); therefore un → u in H01 (Ω). Due to the previous theorem, J attains its minimum at u ∈ H01 (Ω). The Euler equation is u ∈ H01 (Ω) :

−Δu + |u|p−2 u = λ1 u + f .

For functionals that are unbounded from below, the mountain pass theorem by Ambrosetti and Rabinowitz [1]) can be useful: Theorem 9.24 (Mountain Pass theorem). Let H be a Hilbert space. Let J be a C 1 functional defined on H satisfying the Palais–Smale condition and J(0) = 0. Assume that there exist positive constants r and a such that J(u) ≥ a if u = r , and there exists v ∈ H with v > r such that J(v) ≤ 0. Let Γ = {g ∈ C([0, 1]; H) | g(0) = 0, g(1) = v}

Then c = inf max I[g(t)] , g∈Γ 0≤t≤1

is a critical value of J . Theorem 9.25. The functional 1 1 J(v) = |∇v|2 − |v|p , 2 p Ω

2 < p < 2∗ ,

Ω

defined on H01 (Ω), has a critical point at u which satisfies u ∈ H01 (Ω) :

−Δu = |u|p−2 u .

Remark 9.26. We have already seen in Example 9.11 that J is unbounded from below. Proof. We are going to prove that J satisfies the hypotheses of Theorem 9.24. This will imply the existence of a function u ≠ 0 in H01 (Ω) such that −Δu = |u|p−2 u. It is not difficult to prove that J satisfies the Palais–Smale condition. Indeed let un be a sequence such that |J(un )| ≤ R (9.5.13) for some R > 0 and J  (un ) → 0 .

(9.5.14)

The Ekeland variational principle

101

We want to prove that un converges in H01 (Ω), up to subsequence. Choosing un as a test function in (9.5.14) we get |un |p ≤ un H01 (Ω) + un 2H 1 (Ω) . 0

Ω

On the other hand, (9.5.13) gives 1 1 un H01 (Ω) ≤ R + 2 p

|un |p . Ω

The last two estimates give that un H01 (Ω) is uniformly bounded, since p > 2. There-

fore there exists u ∈ H01 (Ω) such that, up to a subsequence, un → u weakly in H01 (Ω) and un → u in Lp (Ω), as p < 2∗ . We claim that un → u in H01 (Ω). Indeed, (9.5.14) gives, for yn ∈ H −1 (Ω) converging to 0 in H −1 (Ω) ∇(un − u) · ∇(un − u) = |un |p−2 un (un − u) Ω

Ω



+ yn , un − u −

∇u · ∇(un − u) . Ω

It is easily seen that the right-hand side goes to zero, by the fact that un → u weakly in H01 (Ω) and strongly in Lp (Ω); hence un → u in H01 (Ω). Therefore J satisfies the Palais–Smale condition.  2 Now, let u ∈ H01 (Ω), with uH01 (Ω) = r . Then J(u) = r2 − p1 Ω |u|p . By the Hölder and the Sobolev inequalities one has

p

|u|p ≤ meas(Ω)1− 2∗ Ω

⎡ ⎤ p∗ 2 p p meas(Ω)1− 2∗ meas(Ω)1− 2∗ p ⎢ p 2∗ ⎥ ⎣ |u| ⎦ ≤ uH 1 (Ω) ≤ r 0 Sp Sp Ω

2

p 1− ∗ 2

and then J(u) ≥ r2 − meas(Ω) r p = a > 0 if r is sufficiently small, as p > 2. Now, Sp 1 let u ∈ H0 (Ω) be fixed and let v = tu, t > 0. Then J(v) =

t2 2

|∇u|2 − Ω

tp p

|u|p < 0 Ω

if t is sufficiently large, as p > 2. Remark 9.27. For the regularity of minimizing sequences for integral functionals of the Calculus of Variations via the Ekeland’s principle see [11].

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Calculus of variations and Euler’s equation

9.6 Appendix Let X be a Banach space and x denote the norm of x ∈ X . Definition 9.28. Let J : X → R be a functional. (1) J is weakly lower semicontinuous if lim infn→∞ J(xn ) ≥ J(x) for every sequence xn weakly converging to x . (2) J is coercive if limx→+∞ J(x) = +∞. Let us recall the Gâteaux differentiability of a functional J : X → R (see [2] for more details). Definition 9.29. J is Gâteaux differentiable at x ∈ X in the direction h ∈ X if there exists a linear continuous functional J  (x) : X → R such that lim t→0

J(x + th) − J(x) = J  (x), h . t

Let us recall the Fréchet differentiability. Definition 9.30. J is Fréchet differentiable at x ∈ X if there exists Ax ∈ X  such that J(x + h) − J(x) − Ax (h) = 0. h h→0 lim

If the map X → X x → Ax

is continuous, we will say that J is C 1 . We note that if J is Fréchet differentiable, then it is Gâteaux differentiable. Conversely, if the map x → J  (x) is defined in a neighborhood of x0 and is continuous at x0 , then J is Fréchet differentiable at x0 and J  (x0 ) = Ax0 .

Part II

10 Natural growth problems 10.1 Introduction In Chapter 9, we have seen that a minimizer of F (v) = a(v)|∇v|2 − f v Ω

Ω

satisfies the Euler equation −div(a(u)∇u) + a (u)|∇u|2 = f

in which there appears the natural growth term |∇u|2 . In this chapter we are going to study the boundary value problem (not necessarily variational) ⎧ ⎨−div(M(x, u)∇u) + μu = b(x, u, ∇u) + f (x) , in Ω , (10.1.1) ⎩u = 0 , on ∂Ω , under the following assumptions. The set Ω is an open bounded subset of RN , with N ≥ 3. Moreover b : Ω × R × RN → R is a Carathéodory function such that, for some γ>0 (10.1.2) |b(x, s, ξ)| ≤ γ|ξ|2 ; M = M(x, s) is a symmetric matrix with Carathéodory entries satisfying M(x, s)ξ · ξ ≥ α|ξ|2 ,

|M(x, s)| ≤ β .

(10.1.3)

Finally μ > 0. Observe that, due to the quadratic growth of b with respect to the gradient, Leray–Lions theorem cannot be applied, since the composition operator v → b(x, v(x), ∇v(x)) does not map H01 (Ω) to its dual, but only to L1 (Ω). We refer to [20] for more details and for further references. As well, a natural growth term appears in the case where one considers the minimization of integral functionals as 1 I(v) = (1 + |v|r )|∇v|2 − f v , v ∈ H01 (Ω) . 2 Ω

Ω

Indeed, the Euler equation for I is (at least formally) ⎧ ⎨−div((1 + |u|r )∇u) + r u|u|r −2 |∇u|2 = f , 2 ⎩u = 0 ,

in Ω , on ∂Ω .

(10.1.4)

Note that the lower order term depends quadratically on the gradient and satisfies v g(x, v, ∇v) ≥ 0. In the second part of this chapter we will study the boundary

value problem ⎧ ⎨−div([a(x) + |u|q ]∇u) + b(x)u|u|p−1 |∇u|2 = f , ⎩u = 0 ,

in Ω , on ∂Ω ,

(10.1.5)

106

Natural growth problems

under the following assumptions. The functions a and b are measurable and 0 < α ≤ a(x) ≤ β ,

(10.1.6)

0 < μ ≤ b(x) ≤ ν,

(10.1.7)

where α, β, μ , ν are fixed real numbers, and p, q are positive. The case p = q − 1 is related with the minimization problems. We refer to [6] for more details and further references.

10.2 A problem with bounded solutions We will present the following existence and regularity result. Theorem 10.1. Let Ω be an open bounded subset of RN , with N ≥ 3. Let μ > 0. Assume (10.1.2) and (10.1.3). Let f be in Lm (Ω), m > N2 . Then there exists a solution u ∈ H01 (Ω) ∩ L∞ (Ω) to problem (10.1.1) in the following weak sense: M(x, u)∇u·∇ϕ+μ uϕ = b(x, u, ∇u)ϕ+ f ϕ , ∀ ϕ ∈ H01 (Ω)∩L∞ (Ω) . Ω

Ω

Ω

Ω

We will work by approximation on the following problems: ⎧ ⎨−div(M(x, u)∇u) + μu = bn (x, u, ∇u) + fn (x) , in Ω , ⎩u = 0 , on ∂Ω , where bn (x, s, ξ) =

b(x, s, ξ) 1+

and fn (x) =

(10.2.1)

1 n |b(x, s, ξ)|

f (x) 1+

1 n |f (x)|

.

Corollary 5.10 of the Leray–Lions theorem and Stampacchia’s theorem (Theorem 6.7) imply that for every n ∈ N there exists a weak solution un in H01 (Ω) ∩ L∞ (Ω), since |bn | ≤ n. The first step of the proof consists in getting uniform estimates of the H01 (Ω) and L∞ (Ω) norms of the solutions; we therefore may pass to the limit and prove the existence of a solution to problem (10.1.1). The following lemmata will be useful for us. Lemma 10.2. Let un be the sequence of solutions to problems (10.2.1). Then the sequence un is bounded in H01 (Ω).

A problem with bounded solutions

107

γ Proof. Let us consider ϕn = (e2λ|un | − 1)sgn(un ), λ > 2α , as a test function in the weak formulation of problems (10.2.1): this is possible since un ∈ L∞ (Ω). Step I: As for the left-hand side, using the ellipticity of M , one has M(x, un )∇un ·∇ϕn +μ un ϕn ≥ 2α λ |∇un |2 e2λ|un | +μ |un |(e2λ|un | −1) . Ω

Ω

Ω

Ω

(10.2.2) Regarding the right-hand side, the growth hypothesis on b allows us to say that 2 2λ|un | b(x, un , ∇un )ϕn + fn ϕn ≤ γ |∇un | e + |f |(e2λ|un | − 1) . Ω

Ω

It is easy to prove that

Ω

|∇un |2 e2λ|un | =

Ω

Ω

1 λ2



|∇(eλ|un | − 1)|2 .

Ω

Therefore, by (10.2.2), we obtain 2αλ − γ ∇(eλ|un | − 1)2L2 (Ω) + μ λ2

Ω

|un |(e2λ|un | − 1) ≤



|f | (e2λ|un | − 1) .

Ω

Step II: Let us now estimate the last term of the above inequality. Let R > 1. Using 1 that e2t − 1 ≤ R (et − 1)2 + R−1 , t ∈ R+ and Hölder’s inequality we have 1 |f | (e2λ|un | − 1) ≤ |f | + Rf Lm (Ω) eλ|un | − 12L2m (Ω) . R−1 Ω

Ω

2N We note that 2 < 2m < N−2 = 2∗ . The interpolation inequality gives 1 2(1−θ) |f | (e2λ|un | − 1) ≤ |f | + Rf Lm (Ω) eλ|un | − 12θ eλ|un | − 1L2 (Ω) , ∗ L2 (Ω) R−1

Ω

Ω

where θ satisfies

1 2m

=

θ 2∗

+

1−θ 2 .

Young’s inequality on the right-hand side implies

|f | (e2λ|un | − 1)

Ω

1 ≤ R−1

where Cε = ε



1

|f | + ε(Rf Lm (Ω) ) θ eλ|un | − 12L2∗ (Ω) + Cε eλ|un | − 12L2 (Ω) ,

Ω

θ 1−θ

. Summarizing we have obtained 2αλ − γ λ|un | 2 ∇(e − 1)L2 (Ω) + μ |un |(e2λ|un | − 1) λ2 Ω 1 1 |f | + ε(Rf Lm (Ω) ) θ eλ|un | − 12L2∗ (Ω) + Cε eλ|un | − 12L2 (Ω) . ≤ R−1 Ω

108

Natural growth problems

Step III: Let us use Sobolev’s inequality on the right-hand side and afterward choose ε in such a way that 1 1 2λα − γ = 2 ε(Rf Lm (Ω) ) θ . 2 2λ S This implies that 2αλ − γ ∇(eλ|un | − 1)2L2 (Ω) + μ 2λ2



|un |(e2λ|un | − 1)

Ω

≤

1 R−1



|f | + Cε (eλ|un | − 1)2 .

Ω

(10.2.3)

Ω

Since (eλt − 1)2 ≤ e2λt − 1, for t ≥ 0, we have 2αλ − γ ∇(eλ|un | − 1)2L2 (Ω) + μ 2λ2 ≤

1 R−1





{Cε ≤μ|un |}

|f | + Cε Ω

|un |(e2λ|un | − 1)

(e2λ|un | − 1) + Cε

{Cε ≤μ|un |}

(e2λ|un | − 1) .

{Cε >μ|un |}

Consequently Cε f L1 (Ω) 2αλ − γ + Cε (e2λ μ − 1) meas(Ω) . ∇(eλ|un | − 1)2L2 (Ω) ≤ 2 2λ R−1  The last inequality tells us that the sequence Ω e2λ|un | |∇un |2 is bounded. As λ > 0,  the sequence Ω |∇un |2 is bounded.

Lemma 10.3. Let un be the sequence of solutions to problems (10.2.1). Then the sequence un is bounded in L∞ (Ω). Proof. Let us consider vn = (e2λ|Gk (un )| − 1)sgn(un ) as a test functions in problems (10.2.1). Since Gk (un ) = 0 in {|un | ≤ k}, we have M(x, un )∇un ·∇vn +μ un vn = b(x, un , ∇un )vn + fn vn . Ω

{|un |≥k}

{|un |≥k}

{|un |≥k}

Using exactly the same arguments used in the previous lemma to get (10.2.3) we have 2λα − γ ∇(eλ|Gk (un )| − 1)2L2 (Ω) + μ 2λ2 1 ≤ R−1

|un |(e2λ|Gk (un )| − 1) {|un |≥k}



(eλ|Gk (un )| − 1)2 .

|f | + Cε {|un |≥k}

{|un |≥k}

109

A problem with bounded solutions

By Sobolev’s inequality on the left-hand side we deduce that (2λα − γ)S2 λ|Gk (un )| e − 12L2∗ (Ω) + μ 2λ2 1 ≤ R−1

|un |(e2λ|Gk (un )| − 1) {|un |≥k}





(eλ|Gk (un )| − 1)2 .

|f | + Cε {|un |≥k}

{|un |≥k}

The inequality (eλt − 1)2 ≤ e2λt − 1, for t ≥ 0 on the second term of the left-hand side gives 1 (2λα − γ)S2 λ|Gk (un )| e − 12L2∗ (Ω) + μ 2 λ2 1 ≤ R−1

Choosing k ≥

Cε μ ,



k(eλ|Gk (un )| − 1)2 {|un |≥k}

(eλ|Gk (un )| − 1)2 .

|f | + Cε {|un |≥k}

{|un |≥k}

we get

1 1 (2λα − γ)S2 λ|Gk (un )| e − 12L2∗ (Ω) ≤ 2 2 λ R−1

|f |. {|un |≥k}

On the left-hand side we use that et − 1 ≥ t for every t ≥ 0 and Hölder’s inequality with exponent 2∗ . This gives N+2 λ |Gk (un )| ≤ |eλ|Gk (un )| − 1| ≤ meas({|un | > k}) 2N eλ|Gk (un )| − 1L2∗ (Ω) Ω

Ω

⎛ ≤ C1 meas({|un | > k})

N+2 2N

⎜ ⎝

⎞1



2

⎟ |f |⎠

{|un |≥k}

where C1 denotes a constant depending on λ, α, S and R . From Hölder’s inequality with exponent m on the right-hand side we get, for k ≥ Cμε ,

1

|Gk (un )| ≤ C1 f L2m (Ω) meas({|un | > k})

λ Ω

Remark 6.3 and Lemma 6.2 imply that un L∞ (Ω) is bounded.

N+2 1 2N + 2m

.

110

Natural growth problems

We can now prove the existence theorem of this section. Proof. We divide the proof into two steps. Step I: Let un be the solutions to problems (10.2.1). The previous lemmata imply the existence of a function u ∈ H01 (Ω) such that ∇un → ∇u weakly in L2 (Ω), up to a subsequence. We are going to prove that un → u in H01 (Ω). To this end, let us consider as a test function in problems (10.2.1) vn = ψ(un − u), where ψ(t) = (eλ|t| − 1)sgn(t). We can write M(x, un )(∇un − ∇u) · (∇un − ∇u)ψ (un − u) Ω



un vn +

= −μ Ω



bn (un , ∇un )vn Ω

M(x, un )∇u · (∇un − ∇u)ψ (un − u) +

− Ω



≤− Ω

fn vn Ω

M(x, un )∇u · ∇(un − u)ψ (un − u)

Ω

|∇un |2 |vn | +

un vn + γ

−μ

Ω

|f ||vn | . Ω

Using the ellipticity of M , one gets α |∇un − ∇u|2 ψ (un − u) Ω



≤− Ω

M(x, un )∇u · ∇(un − u)ψ (un − u)

Ω

|∇un |2 |vn | +

un vn + γ

−μ

Ω

|f ||vn | . Ω

Now, |∇un | = |(∇un − ∇u) + ∇u| ≤ 2|∇un − ∇u| + 2|∇u|2 : this implies that |∇un − ∇u|2 [αψ (un − u) − 2γ|ψ(un − u)|] 2

Ω



≤ 2γ Ω

2

2

|∇u|2 vn − M(x, un )∇u·(∇un −∇u)ψ (un −u)−μ un vn + |f ||vn | . Ω

Ω

2γ α 

Ω



Observe that if λ > then αψ (un − u) − 2γ|ψ(un − u)| ≥ 1. With this choice, setting C = supn ψ (un − u)L∞ (Ω) , one has ∇un − ∇u2L2 (Ω) ≤ C M(x, un )∇u · (∇un − ∇u) − μ un vn Ω

|∇u|2 |vn | +

+ 2γ Ω

Ω

|f ||vn | . Ω

(10.2.4)

A problem with bounded solutions

111

We now want to prove that the right-hand side of the previous inequality goes to 0. It is clear that vn → 0 a.e. in Ω and that |vn | ≤ C since ψ is continuous and un , u  are bounded in L∞ (Ω); Lebesgue’s theorem implies then Ω |∇u|2 |vn | → 0 and   Ω |f ||vn | → 0. Hölder’s inequality with exponent 2 implies that Ω un vn → 0. Let us now study the first term. One has that M(x, un )∇u → M(x, u)∇u a.e. and consequently in L2 (Ω) for Lebesgue’s theorem. Since ∇un → ∇u weakly in L2 (Ω), the first term goes to 0. Therefore (10.2.4) implies that ∇un − ∇uL2 (Ω) → 0. Step II: We now prove that u is a solution. For every ϕ ∈ H01 (Ω) ∩ L∞ (Ω) one has M(x, un )∇un · ∇ϕ + μ un ϕ = bn (x, un , ∇un )ϕ + fn ϕ . Ω

Ω

Ω

Ω



The first term tends to

M(x, u)∇u · ∇ϕ , Ω

as un → u in H01 (Ω) and M(x, un )∇v → M(x, u)∇v in L2 (Ω) for every     v ∈ H01 (Ω). Clearly Ω fn ϕ → Ω f ϕ and Ω un ϕ → Ω uϕ. We claim that   Ω bn (x, un , ∇un )ϕ → Ω b(x, u, ∇u)ϕ . The sequences un and ∇un converge a.e. in Ω to u and ∇u respectively, and so bn (x, un , ∇un ) → b(x, u, ∇u) a.e.. Moreover |bn (x, un , ∇un )| ≤ γ|∇un |2 : this last sequence converges in L1 (Ω). Due to Lebesgue’s theorem bn (x, un , ∇un )ϕ → b(x, u, ∇u)ϕ

in L1 (Ω), up to a subsequence. We can therefore pass to the limit and get M(x, u)∇u · ∇ϕ + μ uϕ = b(x, u, ∇u)ϕ + f ϕ Ω

Ω

Ω

Ω

for every ϕ ∈ H01 (Ω) ∩ L∞ (Ω). Remark 10.4. The μu term, with μ > 0, has an important role for the existence of solutions. Indeed, let us consider ⎧ ⎨−Δu = |∇u|2 + f , in Ω , ⎩u = 0 , on ∂Ω , where f is bounded and strictly positive. By contradiction, let u ∈ H01 (Ω) ∩ L∞ (Ω) be a solution to the previous problem. By the Maximum Principle (Theorem 4.12) u is positive. Let us set z = eu − 1. Then z belongs to H01 (Ω) ∩ L∞ (Ω) and is positive in Ω. Consequently ∇z = (z + 1)∇u ,

−Δz = −(z + 1)[Δu + | ∇u |2 ] = f (z + 1)

112

Natural growth problems

in the distributional sense. Choosing ϕ1 , the first eigenfunction of L(v) = −Δv (see Chapter 8), as a test function, one gets 0 ≤ λ1 ϕ1 z = ∇ϕ1 · ∇z = − f (z + 1)ϕ1 < − f zϕ1 ≤ 0 , Ω

Ω

Ω

Ω

that is, a contradiction.

10.3 A problem with unbounded solutions In this section we study problem (10.1.5), proving the existence of weak H01 (Ω) solutions. In the first result, we will study the case where the source f belongs to Lm (Ω), with m > N2 . Theorem 10.5. Let Ω be a bounded open subset of RN , N ≥ 3. Assume (10.1.6) and (10.1.7). Let f be a positive function belonging to Lm (Ω), m > N2 . Then there exists a positive solution u ∈ H01 (Ω) ∩ L∞ (Ω) to ⎧ 3" # 4 ⎨−div a(x) + uq ∇u + b(x)up |∇u|2 = f , ⎩u = 0 ,

in

Ω,

(10.3.1)

on ∂Ω .

Proof. Step I: Let un ∈ H01 (Ω) ∩ L∞ (Ω) be a solution of the Dirichlet problem ⎧ 3" # 4 ⎪ − div a(x) + |Tn (un )|q ∇un ⎪ ⎪ ⎪ ⎨ 1 + b(x)Tn (un )|Tn (un )|p−1 |∇un |2 + un = f , ⎪ ⎪ n ⎪ ⎪ ⎩ un = 0 ,

in Ω ,

(10.3.2)

on ∂Ω .

The existence of such a solution is assured by Theorem 10.1, since a is bounded by assumption (10.1.6), |Tn (un )| ≤ n and the second term is bounded by νnp |∇un |2 , due to (10.1.7). Choosing u− n as a test function, it is easy to see that un is positive, since f ≥ 0. Now we use un as a test function. By (10.1.6) and (10.1.7), dropping positive terms, we have |∇un |2 ≤

μ Ω

fn un . Ω

It is easy to deduce that the sequence un is bounded in H01 (Ω). Moreover, the use of Gk (un ) as test function yields, dropping again positive terms, α |∇Gk (un )|2 ≤ f Gk (un ) . Ω

Ω

By Remark 6.8 we infer that the sequence un is bounded in L∞ (Ω).

113

A problem with unbounded solutions

Step II: With the same technique as in the previous section, one can prove that un converges in H01 (Ω) to some function u ∈ H01 (Ω) ∩ L∞ (Ω). Indeed, up to a subsequence, there exists u ∈ H01 (Ω) such that ∇un → ∇u weakly in L2 (Ω). Let us consider as a test function vn = ψ(un − u), where ψ(t) = (eλ|t| − 1)sgn(t). We can

write

[a(x) + |Tn (un )|q ](∇un − ∇u) · (∇un − ∇u)ψ (un − u)

Ω

b(x)Tn (un )|Tn (un )|p−1 |∇un |2 vn

+ Ω

1 =− n



un vn − [a(x) + |Tn (un )|q ]∇u · (∇un − ∇u)ψ (un − u) + f vn .

Ω

Ω

Ω

By assumptions (10.1.6) and (10.1.7), one gets α

|∇un − ∇u|2 ψ (un − u)

Ω

≤ − [a(x) + |Tn (un )|q ]∇u · ∇(un − u)ψ (un − u) Ω

−

1 n



Ω

|∇un |2 vn +

un vn + ν C Ω

f vn , Ω

as un L∞ (Ω) is bounded. Now, |∇un |2 = |(∇un − ∇u) + ∇u|2 ≤ 2|∇un − ∇u|2 + 2|∇u|2 : this implies that

|∇un − ∇u|2 [αψ (un − u) − 2ν C |ψ(un − u)|]

Ω

≤ 2ν C

|∇u|2 vn − [a(x) + |Tn (un )|q ]∇u · (∇un − ∇u)ψ (un − u)

Ω

1 − n





un vn + Ω

Ω

f vn . Ω

 Observe that, if λ > 2γ α , then αψ (un − u) − 2γ|ψ(un − u)| ≥ 1. With this choice,  using that ψ (un − u)L∞ (Ω) and un L∞ (Ω) are bounded, one has

∇un − ∇u2L2 (Ω) ≤ C

∇u · (∇un − ∇u) − Ω



|∇u|2 vn +

+ 2ν C Ω

1 n

un vn Ω

f vn . Ω

(10.3.3)

114

Natural growth problems

We now want to prove that the right-hand side of the previous inequality goes to 0. It is clear that vn → 0 a.e. in Ω and that the sequence vn is bounded since ψ is continuous  and un , u are bounded in L∞ (Ω). Lebesgue’s theorem implies that Ω |∇u|2 vn → 0  and Ω f vn → 0. Using Hölder’s inequality with exponent 2, one can prove that  2 Ω un vn → 0. Since ∇un → ∇u weakly in L (Ω), the first term goes to 0. Therefore (10.3.3) implies that ∇un − ∇uL2 (Ω) → 0. Step III: It is now easy to pass to the limit in (10.3.2) to get a H01 (Ω) ∩ L∞ (Ω) solution to ⎧ 3" # 4 ⎨−div a(x) + uq ∇u + b(x)up |∇u|2 = f , in Ω , ⎩u = 0 , on ∂Ω . In the following result we will get weak positive H01 (Ω) solutions to problem (10.1.5), in the sense thatb(x)up |∇u|2 ∈ L1 (Ω), and q p 2 [a(x) + u ]∇u · ∇ϕ + b(x)u |∇u| ϕ = f ϕ , (10.3.4) Ω

Ω

Ω

for every

ϕ ∈ H01 (Ω) ∩ L∞ (Ω) .

(10.3.5)

Theorem 10.6. Let Ω be a bounded open subset of RN , N ≥ 3. Assume (10.1.6) and (10.1.7). Let f be a positive function belonging to Lm (Ω), m ≤ N2 . Then there exists a solution u in the sense of (10.3.5) such that N (1) if m = 1, p ≥ 2q, then u belongs to L(p+2) N−2 (Ω); ∗∗ 2(q+1)N (2) if 2N+p(N−2)+4q ≤ m ≤ N2 , 2q ≥ p ≥ q − 1, then u belongs to L(p+2)m (Ω); (3) if

2N N+2

≤m≤

N 2,

∗∗

q ≥ 1, 2p ≥ q − 1 ≥ p , then u belongs to L(q+1)m (Ω).

In the case where the summability of f is lower, we will show the existence of a weak positive H01 (Ω) solution u to (10.1.5) in the sense that b(x)up |∇u|2 ∈ L1 (Ω), and [a(x) + uq ]∇u · ∇ϕ + b(x)up |∇u|2 ϕ = f ϕ , (10.3.6) Ω Ω Ω for every

1,∞

ϕ ∈ W0

(Ω) .

Theorem 10.7. Let Ω be a bounded open subset of RN , N ≥ 3. Assume (10.1.6) and (10.1.7), 2q ≥ p . Let f be a positive function belonging to Lm (Ω), with m ≤ N2 . Then there exists a weak solution in the sense of (10.3.6) such that ∗∗ 2(q+1)N (1) if 1 ≤ m ≤ 2N+p(N−2)+4q , p ≥ q − 1, then u ∈ L(p+2)m (Ω); (2q−p)N

(2) if max[1, 2(2q−p)+(q+1)N ] ≤ m ≤

2N N+2 ,

∗∗

q − 1 ≥ p , then u ∈ L(q+1)m (Ω).

115

A problem with unbounded solutions

For the sake of simplicity we will present the above results in the case q = p − 1: Theorem 10.8. Let Ω be a bounded open subset of RN , N ≥ 3. Assume (10.1.6) and (10.1.7). Let f be a positive function belonging to Lm (Ω), with m ≤ N2 . Then there exists 0 ≤ u ∈ H01 (Ω), weak solution of (10.1.5) in the sense of (10.3.5) such that N (1) if m = 1, p ≤ 2, then u belongs to L(p+2) N−2 (Ω); ∗∗ 2pN (2) if 2N+p(N−2)+4q ≤ m ≤ N2 , p ≥ 2, then u belongs to L(p+2)m (Ω). Theorem 10.9. Let Ω be a bounded open subset of RN , N ≥ 3. Assume (10.1.6), (10.1.7) and p ≥ 2. Let f be a positive function belonging to Lm (Ω), with m ≤ N2 . If 1 ≤ m ≤ 2pN 1 (p+2)m∗∗ (Ω) weak solution 2N+p(N−2)+4(p−1) , then there exists 0 ≤ u ∈ H0 (Ω) ∩ L of (10.1.5) in the sense of (10.3.6). We will work by approximation on the following sequence of problems: ⎧ 3" # 4 p ⎨−div a(x) + up−1 ∇un + b(x)un |∇un |2 = fn , in Ω , n (10.3.7) ⎩u = 0 , on ∂Ω , n

where fn = Tn (f ). These problems are well posed due to Theorem 10.5. In the following lemma we prove that the sequence un is bounded in H01 (Ω). Lemma 10.10. Let f ∈ L1 (Ω). (1) There exists R > 0 such that un 2H 1 (Ω) ≤ Rf L1 (Ω) ;

(10.3.8)

0

(2) if Ak = {un ≥ k}, then



p b(x)un |∇un |2

Ak

≤

f.

(10.3.9)

Ak

Moreover, there exists u ∈ H01 (Ω) such that un → u in H01 (Ω) up to a subsequence. Proof. The use of Tj (un ) as a test function in (10.3.7) implies " p−1 # p a(x) + un ∇un · ∇Tj (un ) + b(x)un |∇un |2 Tj (un ) ≤ fn Tj (un ) . Ω

Ω

Ω

By assumptions (10.1.6) and (10.1.7) one has α |∇Tj (un )|2 + μ j p+1 Ω

We deduce that, for every j ∈ N, |∇un |2 = |∇un |2 + Ω

{0≤un h}

By Hölder’s inequality with exponent r2 on the last two terms and estimate (10.3.8) we deduce that |∇(un − u)|r ≤ |∇Th (un − Tk (u))|r Ω

Ω r

+ 2r −1 R r f rL1 (Ω) meas({u > k})1− 2 r

+ 2r −1 R r f rL1 (Ω) meas({|un − u| > h})1− 2 .

A problem with unbounded solutions

117

Thus, for every h > 0 and k > 0, we deduce from (10.3.10) and the L2 (Ω) convergence of un to u, that lim sup n→∞

Ω

⎡ ⎢ 2 |∇(un − u)|r ≤ ⎣h α



⎤r

2

r r ⎥ f ⎦ meas(Ω)1− 2 + 2r −1 R r meas({u > k})1− 2 .

Ω

The limit as h → 0 and then k → +∞ gives |∇(un − u)|r → 0 ,

∀ r < 2.

(10.3.11)

Ω

Then (up to subsequences) ∇un (x) converges a.e. in Ω to ∇u(x). Fatou’s lemma implies the following corollary. Corollary 10.12. Let u be the function found in Lemma 10.10. Then 0 ≤ b(x)up |∇u|2 ≤ f . Ω

Ω

Proof. It is sufficient to pass to the limit in (10.3.9) written for k = 0 using the previous lemma. Remark 10.13. We point out that in Lemma 10.10 and in Lemma 10.11 we only require f to belong to L1 (Ω). " p−1 # Lemma 10.14. Under the assumptions of Theorem 10.8, a(x) + un ∇un converges weakly in (L2 (Ω))N to [a(x) + up−1 ]∇u and a.e. in Ω. Proof. By the previous lemmata it is sufficient to prove that the sequence [a(x) + p−1 un ]∇un is bounded in L2 (Ω). Step I: Assume m = 1 and p ≤ 2. Estimate (10.3.9) and assumption (10.1.7) imply that p μ un |∇un |2 ≤ f . (10.3.12) Ω

Ω

Since p ≤ 2, by (10.1.6), (10.3.8) and (10.3.12) we get " " p−1 #2 p# a(x) + un |∇un |2 ≤ 2 β2 + un |∇un |2 + Ω

{1≤un }

{un 0, define ⎧ ⎪ 1 if s ≤ j , ⎪ ⎪ ⎨ Rj (s) = j + 1 − s if j ≤ s < j + 1 , ⎪ ⎪ ⎪ ⎩0 if s > j + 1 , ν

ν

and choose e− α H(un ) e α H(Tk (u)) Rj (un ) ϕ as a test function in (10.3.7). We conclude the proof, as in the Theorem 10.8 (proof of (10.3.15)), letting first k tend to infinity, and then j tend to infinity, observing that Rj (s) tends to 1.

11 Problems with low summable sources 11.1 Introduction In Chapters 5 and 6, we focused our attention on the existence of solutions to the Leray–Lions problem ⎧ ⎨−div(a(x, u, ∇u)) = f , in Ω (11.1.1) ⎩u = 0 , on ∂Ω , 2N assuming that the source f belongs to Lm (Ω) with m ≥ N+2 and under the following N assumptions on a Carathéodory map a : Ω × R × R → RN : (1) there exists β > 0 such that |a(x, s, ξ)| ≤ β[|s| + |ξ|]; (2) there exists α > 0 such that a(x, s, ξ) · ξ ≥ α|ξ|2 , ∀ ξ ∈ RN ; (3) [a(x, s, ξ) − a(x, s, η)] · [ξ − η] > 0 if ξ = η.

Moreover Ω is an open bounded subset of RN , N ≥ 3. In this chapter we study the existence and the summability of the solutions in the 2N case where f belongs to Lm (Ω) with 1 ≤ m < N+2 . Note that we cannot use the Leray–Lions theorem (Theorem 5.1), since the source does not belong to H −1 (Ω). We will prove the existence of distributional solutions u, that is u will be a function in 1,1 W0 (Ω), at least, such that a(x, u, ∇u) · ∇ϕ = f ϕ , ∀ ϕ ∈ C0∞ (Ω) . Ω

Ω

Although u does not belong to H01 (Ω), we will show that, for every k > 0, |∇Tk (u)|2 ≤ C0 k, Ω

which is fundamental for the uniqueness of solutions, as we will see in Chapter 12. For the reader’s convenience we summarize here the main results of existence of solutions u in function of the summability of the source f : N

N

f ∈ L1 (Ω) or f is a bounded measure ⇒ u ∈ M N−2 (Ω), |∇u| ∈ M N−1 (Ω) ; f ∈ Lm (Ω), 1 < m ≤

2N 1,m∗ ⇒ u ∈ W0 (Ω) . N +2

The last result can be seen as a nonlinear Calderon–Zygmund theory for elliptic problems with infinite energy solutions. We will follow the proofs of [12, 13]. The reader can find further references there; for recent developments see [38].

122

Problems with low summable sources

In the linear case, the existence results (if f is a bounded measure or if f belongs 2N to Lm (Ω), 1 ≤ m ≤ N+2 ) are due to G. Stampacchia. They can be seen as a Calderon– Zygmund theory for linear operators with discontinuous coefficients. We observe that general results can be proved in the case where a satisfies the coercivity assumption a(x, s, ξ) · ξ ≥ α|ξ|p ,

∀ ξ ∈ RN

(see Remark 11.18). Moreover, we will study the regularizing effects of a lower order term. More precisely, we will consider the problem ⎧ ⎨−div(a(x, u, ∇u)) + |u|p−1 u = f , in Ω , ⎩u = 0 , on ∂Ω , for a source f ∈ Lm (Ω). It is not difficult to prove that |u|p ∈ Lm (Ω). We will prove that 1 ⇒ u ∈ H01 (Ω); m−1 1 2pm 1,q ⇒ u ∈ W0 (Ω) , ∀ q < ; f ∈ Lm (Ω), m > 1, p < m−1 1 + pm 2p 1,q , f ∈ L1 (Ω) ⇒ u ∈ W0 (Ω) , ∀ q < 1+p

f ∈ Lm (Ω), m > 1, p ≥

following the proofs in [16]. Note that the study by Brezis and Strauss of semilinear equations like ⎧ ⎨−div(M(x)∇u)) + |u|p−1 u = f , in Ω , ⎩u = 0 , on ∂Ω , was the bridge between the linear case and the nonlinear case. In the last section we will study the regularity of minimizers for integral functionals of the form J(v) = j(x, ∇v) − f v Ω

Ω

where j : Ω × R → R is a Carathéodory function, convex in the last variable, such that j(x, ξ) ≥ α|ξ|2 N

for some positive α and f belongs L1 (Ω). Due to the summability of f , the nonexistence of minima led to the definition of T -minima, in analogy with the definition of entropy solutions to elliptic problems (see [5] and [33]).

123

A priori estimates

11.2 A priori estimates We will work by approximation to prove the existence of distributional solutions and entropy solutions to problem (11.1.1). We consider ⎧ ⎨−div(a(x, un , ∇un )) = fn , in Ω , (11.2.1) ⎩u = 0 , on ∂Ω , n

where fn is a sequence of functions in H −1 (Ω) ∩ L∞ (Ω) such that fn → f in Lm (Ω), fn Lm (Ω) ≤ f Lm (Ω) and |fn (x)| ≤ |f (x)| a.e. in Ω (for example fn = Tn (f )). The existence of solutions un , for every n, follows from Theorem 5.1; moreover un belongs to H01 (Ω) ∩ L∞ (Ω) due to Theorem 6.6. Our strategy is as follows. We get the 1,m∗ (Ω) when m > 1 and in following estimates: the sequence un is bounded in W0 1,q N W0 (Ω), q < N−1 , when m = 1. This allows us to get a subsequence converging to some function u. We prove that ∇un → ∇u a.e. in Ω. In this way we can pass to the limit in (11.2.1) and prove that u is a solution to problem (11.1.1). Lemma 11.1. Let f ∈ L1 (Ω). Then the sequence of solutions un to problems (11.2.1) is 1,q N bounded in W0 (Ω) with q < N−1 . Proof. We consider vn = [(1 + |un |)2λ−1 − 1]sgn(un ) as a test function in (11.2.1). Let λ < 1/2: in this way |vn | ≤ 1. As for the right-hand side we have fn vn ≤ |fn | ≤ f L1 (Ω) . (11.2.2) Ω

Ω

Using the ellipticity of a in the left-hand side we have f L1 (Ω) ≥ a(x, un , ∇un ) · ∇vn ≥ α |∇un |2 (1 − 2λ)(1 + |un |)2λ−2 Ω

Ω

   ∇[(1 + |u |)λ ] 2   n  . = (1 − 2λ)α    λ Ω

The above estimates imply Ω

f L1 (Ω) |∇un |2 . ≤ 2(1−λ) (1 + |un |) α(1 − 2λ)

On the other hand |∇un |q = Ω

Ω

|∇un |q (1 + |un |)

(11.2.3)

q

q

2(1−λ) 2

(1 + |un |)2(1−λ) 2 .

124

Problems with low summable sources

Using Sobolev’s inequality on the left-hand side and Hölder’s inequality with exponent q2 in the right one, we obtain ⎛ ⎞ q∗ q ∗ q⎜ q ⎟ S ⎝ |un | ⎠ ≤ |∇un |q Ω

Ω

⎛ ⎜ ≤⎝

⎞q ⎛ 2



|∇un | ⎟ ⎜ ⎠ ⎝ (1 + |un |) (1 + |un |)2(1−λ) 2

Ω

⎞1− q 2

(1−λ)2q 2−q

⎟ ⎠

.

Ω

Inequality (11.2.3) implies ⎛ ⎞ q∗ ⎛ ⎞1− q 2 q (1−λ)2q ∗ ⎜ ⎟ q⎜ q ⎟ q 2−q S ⎝ |un | ⎠ ≤ |∇un | ≤ C + C ⎝ |un | ⎠ Ω

Ω

(11.2.4)

Ω

where C denotes a constant depending on q, λ, α, f L1 (Ω) , independent on n. Let (1−λ)2q q(N−2) us choose λ such that 2−q = q∗ , that is, λ = 2(N−q) . Since λ < 12 , this implies q
1. Then the sequence of the solutions un to prob1,m∗ (Ω). lems (11.2.1) is bounded in W0 Proof. We divide the proof into two steps. ∗∗ Step I: We prove that the sequence un is bounded in Lm (Ω). Let us consider vn = [(1 + |un |)2λ−1 − 1]sgn(un ) as a test function in (11.2.1); λ > 12 will be defined later. For the right-hand side we have, 2λ−1 fn vn ≤ |f |[(1 + |un |) − 1] ≤ |f |[(1 + |un |)2λ−1 + 1] . Ω

Ω

Ω

By Hölder’s inequality with exponent m we get Ω

⎛ ⎞1 m ⎟ ⎜ (2λ−1)m fn vn ≤ f L1 (Ω) + f Lm (Ω) ⎝ (1 + |un |) ⎠ . Ω

A priori estimates

125

For the left-hand side of (11.2.1), using the ellipticity of a and Sobolev’s inequality, we have a(x, un , ∇un ) · ∇vn ≥ α |∇un |2 (2λ − 1)(1 + |un |)2λ−2 Ω

Ω

   ∇[(1 + |u |)λ ] 2   n  = (2λ − 1)α    λ Ω

⎛ ⎞ 2∗ 2 (2λ − 1)αS ⎜ λ 2∗ ⎟ ≥ ⎝ [(1 + |un |) ] ⎠ . λ2 Ω

Summarizing ⎛ ⎞1 m ⎟ ⎜ (2λ−1)m f L1 (Ω) + f Lm (Ω) ⎝ (1 + |un |) ⎠ Ω

≥ (2λ − 1)α Ω

|∇un |2 (1 + |un |)2(1−λ)

⎛ ⎞ 2∗ 2 (2λ − 1)αS2 ⎜ ∗ λ 2 ⎟ ≥ ⎝ [(1 + |un |) ] ⎠ . λ2

(11.2.5)

Ω

Now, let us fix λ such that λ2∗ = (2λ − 1)m , that is, λ = choice, we deduce from the previous estimate that

m∗∗ 2∗ (>

1/2). With this

⎛ ⎞ 2∗ 2 m∗∗ (2λ − 1)αS2 ⎜ ∗ ⎟ 2 ∗ ⎝ |(1 + |un |) 2 | ⎠ λ2 Ω

⎛ ⎞ 1 m ⎜ m∗∗ ⎟ m ≤ f L1 (Ω) + f L (Ω) ⎝ |1 + un | . ⎠ Ω

Since

2 2∗

>

1 m

this implies |un |m Ω

∗∗

|1 + un |m

≤

∗∗

≤C,

(11.2.6)

Ω

where C denotes a constant depending on S, m, α, f L1 (Ω) , f Lm (Ω) . 1,m∗ (Ω). We observe Step II: Let us prove that the sequence un is bounded in W0  |∇un |2 that estimates (11.2.5) and (11.2.6) imply that the sequence Ω (1+|u 2(1−λ) is bounded. n |)  2N Assume that λ < 1, that is, 1 < m < N+2 . Let 1 ≤ q < 2 and write Ω |∇un |q as q |∇un |q q 2(1−λ) 2 |∇un | = . q (1 + |un |) 2(1−λ) 2 (1 + |un |) Ω

Ω

126

Problems with low summable sources

By Hölder’s inequality with exponent

⎛ ⎜ |∇un |q ≤ ⎝

Ω

Ω

2 q

we get

⎞q ⎛ ⎞1− q 2 2 2q |∇un |2 ⎟ ⎜ (1−λ) 2−q ⎟ . ⎠ ⎝ (1 + |un |) ⎠ (1 + |un |)2(1−λ) Ω

Let q = m∗ ; in this way  m∗ is bounded. Ω |∇un |

(1−λ)2q 2−q

= m∗∗ . From (11.2.6) it follows that the sequence

The following lemma will be useful for us: Lemma 11.3. (1) Let f ∈ L1 (Ω). Then, if un are the H01 (Ω) solutions to (11.2.1), there 1,q N exists u ∈ W0 (Ω), q < N−1 , such that ∇un → ∇u a.e. in Ω, up to a subsequence. 1,m∗ (Ω). (2) If f ∈ Lm (Ω), m > 1, u ∈ W0 Proof. The estimates of the previous lemmata allow us to get a subsequence, still de1,q N noted by un , weakly converging to some u ∈ W0 (Ω), q < N−1 . We note that if 1,m∗

f ∈ Lm (Ω), m > 1, then u belongs to W0

(Ω).* + To prove the result, we will prove that for θ ∈ 0, q4 one has lim {[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)}θ = 0 n→∞

(11.2.7)

Ω

and then we will use Lemma 5.8. In the proof C denotes a constant independent of n (depending on β, θ, meas(Ω)). We write Ω as the union of Ak and Ck , where Ak := {|un | ≥ k} ,

Ck := {|un | ≤ k}.

Step I: Let us estimate θ a(x, un , ∇un ) − a(x, u, ∇u) · ∇(un − u) . Ak

By the growth assumption on a we get |[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)|θ Ak



|a(x, un , ∇un ) · ∇(un − u)|θ +

≤ Ak



≤ 2β

θ

|un | |∇u|θ + 2β Ak



θ

+ 2β

θ

2θ

|u| |∇u| + 2β Ak



θ

|un | |∇un | + 2β Ak

Ak

θ

|a(x, u, ∇u) · ∇(un − u)|θ

|∇un | Ak

|u|θ |∇un |θ Ak



|∇u|2θ + 2β

+ 2β Ak

|∇u|θ |∇un |θ . Ak

A priori estimates

127

The Cauchy–Schwarz inequality and the estimate on un W 1,q (Ω) of the above lem0 mata give |[a(x, un , ∇un ) − a(x, un , ∇u)] · ∇(un − u)|θ ≤ C meas(Ak )1/2 . Ak

Step II: Let us now estimate {[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)}θ . Ck

We observe that |u| ≤ k on Ck (since un → u a.e. in Ω and |un | ≤ k on Ck ). Therefore one has {[a(x, un , ∇un ) − a(x, un , ∇u)] · ∇(un − u)}θ Ck

{[a(x, un , ∇un ) − a(x, u, ∇Tk (u))] · ∇(un − Tk (u))}θ

= Ck



{[a(x, un , ∇un ) − a(x, u, ∇Tk (u))] · ∇(un − Tk (u))}θ .

≤ Ω

The last integral can be estimated as follows, if we set Vj := {|un − Tk (u)| ≤ j} ,



Vj = {|un − Tk (u)| > j} :

{[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)}θ Ck

{[a(x, un , ∇un ) − a(x, u, ∇Tk (u))] · ∇Tj (un − Tk (u))}θ

≤ Ω

{[a(x, un , ∇un ) − a(x, u, ∇Tk (u))] · ∇(un − Tk (u))}θ .

+ Vj

Let us now estimate the first term of the right-hand side of the previous inequality using Hölder’s inequality with exponent 1/θ : in this way {[a(x, un , ∇un ) − a(x, u, ∇Tk (u))] · ∇Tj (un − Tk (u))}θ Ω

⎛ ⎞θ ⎜ ⎟ ≤ ⎝ {[a(x, un , ∇un ) − a(x, u, ∇Tk (u))] · ∇Tj (un − Tk (u))}⎠ meas(Ω)1−θ . Ω

On the other hand, we can estimate the second term of the right-hand side using the same argument used for |[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)|θ : Ak

128

Problems with low summable sources

we get 1 |[a(x, un , ∇un ) − a(x, u, ∇Tk (u))] · ∇(un − Tk (u))|θ ≤ C meas(Vj ) 2 . Vj

Summarizing we have {[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)}θ Ck 1

≤ C meas(Vj ) 2 ⎛ ⎞θ ⎜ ⎟ + ⎝ {[a(x, un , ∇un ) − a(x, u, ∇Tk (u))] · ∇Tj (un − Tk (u))}⎠ meas(Ω)1−θ . Ω

(11.2.8) The use of Tj (un − Tk (u)) as a test function in (11.2.1) yields a(x, un , ∇un ) · ∇Tj (un − Tk (u)) = fn Tj (un − Tk (u)) . Ω

Ω

Estimate (11.2.8) thus implies {[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)}θ Ck

⎛

⎞θ



⎜ ⎟ ≤ C ⎝ {fn Tj (un − Tk (u)) − a(x, u, ∇Tk (u)) · ∇Tj (un − Tk (u))}⎠ Ω

(11.2.9)

1

+ C meas(Vj ) 2 .

Lebesgue’s theorem gives lim fn Tj (un − Tk (u)) = f Tj (u − Tk (u)) . n→∞

Ω

Ω

On the other hand, since un converges to u in measure, one has lim meas(Vj ) = meas({|u − Tk (u)| ≥ j}) .

n→∞

We claim that, as n → ∞, a(x, u, ∇Tk (u)) · ∇Tj (un − Tk (u)) → a(x, u, ∇Tk (u)) · ∇Tj (u − Tk (u)) . Ω

Ω

129

A priori estimates

It suffices to prove that Tj (un − Tk (u)) → Tj (u − Tk (u)) weakly in H01 (Ω). To this end we observe that |∇un − ∇Tk (u)|2 ≤ 2 |∇un |2 + 2 |∇Tk (u)|2 . {|un −Tk (u)| 0 such that M(x)ξ · ξ ≥ α|ξ|2 ,

∀ξ ∈ RN .

The linear case: a different proof

131

1,q

Let f ∈ L1 (Ω). Then there exists a distributional solution u ∈ W0 (Ω) to problem ⎧ ⎨−div(M(x)∇u) = f , ⎩u = 0 ,

satisfying for every q
0. Proof. Let fn = Tn (f ) and un ∈ H01 (Ω) be the weak solution to problem ⎧ ⎨−div(M(x)∇un ) = fn , ⎩u = 0 , n

in Ω ,

(11.4.3)

on ∂Ω .

Theorem 6.6 guarantees that un ∈ L∞ (Ω), for every n ∈ N. Define the following “dual” problem: for every n, let wn be the solution to M(x)∇wn · ∇v = χk |∇un |q−2 ∇un · ∇v , (11.4.4) Ω

Ω

for every v ∈ H01 (Ω), where χk stands for the characteristic function of {|∇un | ≤ k}. We observe that χk |∇un |q−2 ∇un ∈ L∞ (Ω), for every n. Therefore problem (11.4.4) has a solution by Leray–Lions theorem (Theorem 5.1). Moreover we can use Theorem 6.14 to say that wn ∈ L∞ (Ω). Let us now consider ϕ = wn as a test function in problem (11.4.3) and v = un in ((11.4.4)). In this way M(x)∇un · ∇wn = fn wn Ω

Ω





χk |∇un |q−2 ∇un · ∇un .

M(x)∇wn · ∇un = Ω

Ω

The symmetry of M gives χk |∇un |q = fn wn ≤ ||fn ||L1 (Ω) ||wn ||L∞ (Ω) . Ω

Ω

By Theorem 6.14 applied with m =

q q−1

> N , we get ⎛



⎜ ⎟ χk |∇un | ≤ C||fn ||L1 (Ω) ⎝ χk |∇un |q ⎠ q

Ω

⎞ q−1

Ω

q

,

132

Problems with low summable sources

where C depends on N, q and α. Therefore ⎛ ⎞1/q ⎜ ⎟ ≤ C||fn ||L1 (Ω) . ⎝ χk |∇un |q ⎠

(11.4.5)

Ω

Passing to the limit as k → +∞ in (11.4.5), Fatou’s lemma implies ||∇un ||Lq (Ω) ≤ C||f ||L1 (Ω) , 1,q

N for every q ∈ [1, N−1 ). Therefore, un is bounded in W0 (Ω) and there exists a sub1,q

sequence weakly converging to some u in W0 (Ω). It easily seen that u is a weak solution to problem (11.4.1). Moreover estimate (11.4.2) follows from the weak lower semicontinuity of the Lq norm.

11.5 Entropy solutions In this section, we introduce a new notion of solution to problem (11.1.1): the entropy solution, which is a particular distributional solution. We give the definition, some properties and then we prove the existence. We will understand the importance of entropy solutions in Chapter 12 where we will study the uniqueness of solutions. Definition 11.6. We set T (Ω) = {u : Ω → R

measurable and finite a.e. such that Tk (u) ∈ H01 (Ω) ,

∀ k > 0} .

Let us study the gradient of a T (Ω) function. Lemma 11.7. For every u ∈ T (Ω) there exists a unique measurable map v : Ω → RN such that ∇Tk (u) = v χ{|u| 0; then Tk (Tk+ε (u)) = Tk (u);

since Tj (u) belongs H01 (Ω) for every j , one has ∇Tk (Tk+ε (u)) = ∇Tk (u) .

In Ωk = {|u| < k} the previous equality is equivalent to ∇Tk+ε (u) = ∇Tk (u) for ! every ε > 0. Since k>0 Ωk = Ω, setting ∇u(x) = ∇Tk (u(x))

the lemma is proved.

a.e. in Ωk ,

Entropy solutions

133

We are now in position to give the definition of entropy solution to problem (11.1.1). Definition 11.8. Let f be an L1 (Ω) function. A T (Ω) function u is an entropy solution to problem (11.1.1) if a(x, u, ∇u) · ∇Tk (u − ϕ) ≤ f Tk (u − ϕ) (11.5.1) Ω Ω ∀k > 0

and ∀ϕ ∈ H01 (Ω) ∩ L∞ (Ω) .

Remark 11.9. The entropy inequality (11.5.1) is well defined. Indeed, the right-hand side is finite, since Tk (u − ϕ) is bounded. As for the left one, we observe that Tk (u − ϕ) belongs to H01 (Ω) and ∇Tk (u − ϕ) = ∇(u − ϕ) χ{|u−ϕ|≤k} .

Therefore

a(x, u, ∇u) · ∇Tk (u − ϕ) =

Ω

a(x, u, ∇u) · ∇(u − ϕ) . {|u−ϕ|≤k}

In the set {|u − ϕ| ≤ k}, ∇u belongs to L2 (Ω) and consequently a(x, u, ∇u) belongs to L2 (Ω) too. The left-hand side is thus well defined. Remark 11.10. A H01 (Ω) function u satisfying a(x, u, ∇u) · ∇v = f v , Ω

∀ v ∈ H01 (Ω)

Ω

is clearly an entropy solution to problem (11.1.1). This implies that the H01 (Ω) distri2N ), are butional solutions found in Chapter 5 (when the source f ∈ Lm (Ω), m ≥ N+2 entropy solutions to problem (11.1.1). Let us study the main properties of the entropy solutions. Proposition 11.11. Let u ∈ T (Ω) be an entropy solution to problem (11.1.1). Then u N belongs to M N−2 (Ω). Proof. In the whole proof C will denote a constant independent of u and k. Consider ϕ = 0 in (11.5.1): the ellipticity of a implies 2 α∇Tk (u)L2 (Ω) ≤ a(x, u, ∇u) · ∇Tk (u) ≤ k f L1 (Ω) , ∀k > 0 . Ω

Let 0 < h < k. By Sobolev’s inequality on the left-hand side, the above estimate gives ∗ ∗ ∗ h2 meas({|u| > h}) < |u|2 ≤ C k2 /2 . {|u|≤k}

134

Problems with low summable sources

Now, let h = k2 . We deduce that N

meas({|u| > h}) ≤ C h− N−2 , N

meaning, u ∈ M N−2 (Ω). Remark 11.12. The same technique can be used to prove the following statement. Assume that un is a sequence of functions such that for a positive constant C it holds ∇Tk (un )2L2 (Ω) ≤ C k ,

∀k > 0.

Then there exists a positive constant c (independent on n and k) such that meas({|un | > k}) ≤

c N

.

k N−2

Proposition 11.13. Let u be an entropy solution to problem (11.1.1). Then |∇u| ∈ N M N−1 (Ω). Proof. In the proof C will denote a constant independent of u and k. As in the previous proposition, consider ϕ = 0 in (11.5.1). We get |∇Tk (u)|2 ≤ C k . Ω

Therefore

t 2 meas({|u| ≤ k, |∇u| > t}) ≤

|∇u|2 ≤ C k . {|u|≤k,|∇u|>t}

The last estimate and Proposition 11.11 give meas({|u| ≤ k, |∇u| > t}) + meas({|u| > k}) ≤ C Consequently meas({|∇u| > t}) ≤ C

k C + N ; t2 k N−2

minimizing with respect to k the function k→

k 1 + , N t2 N−2 k

one has meas({|∇u| > t}) ≤

C N

t N−1

.

k C + . N t2 k N−2

Entropy solutions

135

Remark 11.14. With the same technique one can prove the following result. Assume that un is a sequence of functions such that for a positive constant C it holds ∇Tk (un )2L2 (Ω) ≤ C k ,

∀k > 0.

Then there exists a positive constant c such that meas({|∇un | > k}) ≤

c k

N N−1

.

Corollary 11.15. Let u ∈ T (Ω) be an entropy solution to problem (11.1.1). Then |a(x, u, ∇u)| ∈ L1 (Ω). Proof. The growth assumption on a implies meas({|a(x, u, ∇u)| > k}) ≤ meas({|u| + |∇u| > k}) .

(11.5.2)

N

The above results give |u| + |∇u| ∈ M N−1 (Ω). We deduce from (11.5.2) that N

|a(x, u, ∇u)| ∈ M N−1 (Ω) ⊆ L1 (Ω) .

We now prove that the entropy solutions are distributional solutions. Proposition 11.16. Let u ∈ T (Ω) be an entropy solution to problem (11.1.1). Then u is a distributional solution, that is, a(x, u, ∇u) · ∇ψ = f ψ , ∀ ψ ∈ C0∞ (Ω) . Ω

Proof. Let ψ ∈

Ω

C0∞ (Ω);

consider ϕ = Th (u) − ψ ∈ H01 (Ω) ∩ L∞ (Ω) in (11.5.1). Then a(x, u, ∇u) · (∇u χ{|u|>h} + ∇ψ) ≤ f Tk (u − Th (u) + ψ) . Ω

{|u−Th (u)+ψ| 0, one can choose Tk (un ) in (11.2.1), as a test function. By the ellipticity of a, it is easy to see that α∇Tk (un )2L2 (Ω) ≤ kf L1 (Ω) .

Since ∇un → ∇u a.e. in Ω, passing to the limit as n → ∞ we have that α∇Tk (u)2L2 (Ω) ≤ kf L1 (Ω) .

Thus Tk (u) belongs to H01 (Ω) for every k > 0. As for the summability of u: (1) if f belongs to L1 (Ω), it is sufficient to use Propositions 11.11 and 11.13; (2) if f belongs to L1 (Ω), m > 1, one can use Lemma 11.2. Remark 11.18. We observe that similar results can be proved (see [12, 13]) in the case where a satisfies the coercivity assumption a(x, s, ξ) · ξ ≥ α|ξ|p ,

∀ ξ ∈ RN .

Indeed, if the source f ∈ L1 (Ω), 1 < p < N , then there exists an entropy solution N(p−1) N(p−1) 1 u ∈ M p1 (Ω) with |∇u| ∈ M p2 (Ω), where p1 = N−p , p2 = N−1 . If p > 2 − N 1,q

then there exists a distributional solution u ∈ W0 (Ω), q < p2 .

A comparison between entropy solutions and distributional solutions

137

11.6 A comparison between entropy solutions and distributional solutions In the previous sections we have proved the existence of entropy solutions and distributional solutions to problem (11.1.1). We now want to compare these two notions, showing that they are not equivalent. Proposition 11.16 establishes that an entropy solution to problem (11.1.1) is a distributional solution. We now present a linear problem having a distributional solution which is not an entropy solution (according to an idea given originally by Serrin in [47] and presented differently in [44]). N×N Let M = (mij )N be the following matrix: if x = (x1 , .., xN ) i,j ∈ R ⎧ 2 ⎨mij = δij + 1−ε2 x2i xj 2 , i, j = 1, 2 ε x1 +x2 (11.6.1) ⎩m = δ , i, j ≠ 1, 2 . ij

ij

M is elliptic and bounded. We claim that u0 (x) =

x1 (x12

1+ε + x22 ) 2

,

ε
j} and Ck = {|un | ≤ k}. Regarding the first term of the right-hand side we observe that ⎞θ ⎛ ⎟ ⎜ ⎝ {fn Tj (un − Tk (u)) − a(x, u, ∇Tk (u)) · ∇Tj (un − Tk (u))}⎠ Ω

 θ  θ             ≤ C  fn Tj (un − Tk (u)) + C  a(x, u, ∇Tk (u)) · ∇Tj (un − Tk (u))     Ω  Ω   θ ⎛ ⎞θ     ⎜ ⎟   ≤ C ⎝ j|fn |⎠ + C  a(x, u, ∇Tk (u)) · ∇Tj (un − Tk (u))   Ω  Ω  θ       θ θ ≤ Cj μM(Ω) + C  a(x, u, ∇Tk (u)) · ∇Tj (un − Tk (u)) .   Ω 

Therefore |[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)|θ Ck

≤ Cj θ μθM(Ω)

 θ     1   + C  a(x, u, ∇Tk (u)) · ∇Tj (un − Tk (u)) + C meas(Vj ) 2 .   Ω 

One can then use the same arguments as in Lemma 11.3 for the second and the third term of the last estimate. In conclusion lim {[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)}θ n→∞

Ω 1

1

≤ Cj θ + C [meas({|u − Tk (u)| ≥ j})] 2 + C [meas({|u| > k})] 2 ⎛ ⎞θ ⎜ ⎟ + C ⎝ |[f Tj (u − Tk (u)) − a(x, u, ∇Tk (u)) · ∇Tj (u − Tk (u))|⎠ . Ω

140

Problems with low summable sources

As k → ∞, the last three terms tend to 0. As j → 0, one has lim {[a(x, un , ∇un ) − a(x, u, ∇u)] · ∇(un − u)}θ = 0 ; n→∞

Ω

this implies that, up to a subsequence, ∇un → ∇u a.e. in Ω. Using the previous lemmata one can prove Theorem 11.19 in the same way as Theorem 11.4.

11.8 The regularizing effects of a lower order term In this section, we are going to study the following problem for p ≥ 1: ⎧ ⎨−div(a(x, u, ∇u)) + |u|p−1 u = f , in Ω , ⎩u = 0 , on ∂Ω ,

(11.8.1)

proving that the lower order term has some regularizing effect on the solutions and on their gradient. This phenomenon was already observed in [23] and [24] for the solutions of ⎧ ⎨−Δu + |u|p−1 u = f , in Ω , ⎩u = 0 , on ∂Ω : indeed if u is a solution, then |u|p belongs to Lm (Ω) under the assumption f ∈ Lm (Ω), m ≥ 1. We will prove the following result (see [16]) on the existence of distributional solutions to problem (11.8.1). Theorem 11.22. Let f ∈ Lm (Ω), m ≥ 1. 1 , there exists a distributional solution u ∈ H01 (Ω). (1) If m > 1 and p ≥ m−1 (2) If m > 1 and p < 2pm every q < 1+pm .

1 m−1 ,

1,q

there exists a distributional solution u ∈ W0 (Ω), for 1,q

(3) If m = 1, there exists a distributional solution u ∈ W0 (Ω), for every q
1 we choose |Tn (un )|p(m−1) sgn(un ) as a test function in (11.8.2). Dropping the operator term of the left-hand side, and using Hölder’s inequality on the right one, we obtain

|Tn (un )|pm ≤ Ω

|f ||Tn (un )|p(m−1)

⎡ ⎤1− 1 m ⎢ ⎥ pm ≤ f Lm (Ω) ⎣ |Tn (un )| ⎦ .

Ω

Ω

n) This implies (11.8.3). In the case m = 1 we choose Tk (u as a test function in (11.8.2). k Again, dropping the operator term from the left-hand side, one has Tk (un ) ≤ |f | . |Tn (un )|p−1 Tn (un ) k

Ω

Ω

It is now sufficient to pass to the limit as k → 0 and use Fatou’s lemma. Lemma 11.24. Let m > 1. Let un be the solutions to problems (11.8.2). 1 , then the sequence Tn (un ) is bounded in H01 (Ω). (1) If p ≥ m−1 (2) If p
1−p

2−q 2q ,

2

Ω

(1−λ)2q 2−q

since q
t}

Let ψi be a sequence of increasing, positive, such that ⎧ ⎪ ⎪ 1, ⎪ ⎨ ψi (s) → 0 , ⎪ ⎪ ⎪ ⎩−1 ,

{|Tn (un )|>t}

uniformly bounded C ∞ (Ω) functions, s ≥ t, |s| < t , s ≤ −t .

T -minima

143

Choosing ψi (Tn (un )) in (11.8.2), we get |Tn (un )|p−1 Tn (un ) ψi (Tn (un )) ≤ Tn (f ) ψi (Tn (un )) . Ω

Ω

The limit on i implies (11.8.5). We are now going to prove that if E is any measurable subset of Ω, then lim |Tn (un )|p = 0 uniformly with respect to n . meas(E)→0

E

By (11.8.5), for any t > 0 we have |Tn (un )|p E





≤ t p meas(E) +

|Tn (un )|p ≤ t p meas(E) +

|f | . {|Tn (un )|>t}

E∩{|Tn (un )|>t}

By Lemma 11.23 the sequence Tn (un ) is bounded in M mp (Ω). This and the fact that f ∈ L1 (Ω) allow us to say that for any given ε > 0, there exists tε such that |f | ≤ ε . {|Tn (un )|>tε }

In this way



p

|Tn (un )|p ≤ tε meas(E) + ε E



and so

|Tn (un )|p ≤ ε ,

lim

meas(E)→0

∀ε > 0 .

E

 We thus proved that limmeas(E)→0 E |Tn (un )|p = 0 uniformly with respect to n. Vitali’s theorem (Theorem 3.2) implies that |Tn (un )|p−1 Tn (un ) → |u|p−1 u in L1 (Ω). Therefore we can pass to the limit in (11.8.2) to get a solution to problem (11.8.1).

11.9 T -minima The last paragraph of this chapter is devoted to the study of integral functionals unbounded from below. More precisely we will be interested to functionals of the form J(v) = j(x, ∇v) − f v Ω

Ω

144

Problems with low summable sources

where j : Ω × RN → R is a Carathéodory function, convex in the last variable, such that α|ξ|2 ≤ j(x, ξ) ≤ β|ξ|2 (11.9.1) for some positive α and f belongs L1 (Ω). Due to the summability of f , J is not bounded from below in H01 (Ω). For example, in Ω = B(0, 1), for β < N , consider v J(v) = |∇v|2 − . |x|β Ω

Ω

Then the sequence vn = Tn ( |x|1 α ), α > N − β, is such that J(vn ) → −∞. The nonexistence of minima led to the definition of T -minima, in analogy with the definition of entropy solutions to elliptic problems (see [5] and [33]). Definition 11.26. Let f ∈ L1 (Ω). A function u ∈ T (Ω) is a T -minimum for the functional J(v) = j(x, ∇v) − f v Ω

H01 (Ω)

Ω

∞

if, for every ϕ in ∩ L (Ω) and every k > 0, we have j(x, ∇(ϕ + Tk (u − ϕ)) ≤ j(x, ∇ϕ) + f Tk (u − ϕ) . Ω

Ω

(11.9.2)

Ω

Observe that if J has a minimum u ∈ H01 (Ω) (see Chapter 9, in the case where 2N f ∈ L N+2 (Ω)), then u satisfies (11.9.2). Lemma 11.27. Let un ∈ T (Ω) be a sequence of functions such that for a positive constant C the following estimate holds: |∇Tk (un )|2 ≤ C k . Ω

Then there exists u ∈ T (Ω) such that, up to a subsequence, Tk (un ) → Tk (u) weakly in H01 (Ω). Proof. Since the sequence Tk (un ) is bounded in H01 (Ω), there exists vk in L2 (Ω) such that, up to a subsequence Tk (un ) → vk weakly in H01 (Ω) and strongly in L2 (Ω). By Remark 11.12, there exists a positive constant c such that meas({|un | > k}) ≤

c N

k N−2

for every n ∈ N. Now, fix ε > 0. The above estimate allows us to choose k such that meas({|un | > k}) ≤

ε , 3

meas({|um | > k}) ≤

ε 3

(11.9.3)

145

T -minima

for every n, m ∈ N and for every k > k. Moreover, since Tk (un ) is a Cauchy sequence in measure (recall that Tk (un ) → vk in measure, as n → ∞), there exists ηε such that for every n, m > ηε meas({|Tk (un ) − Tk (um )| > σ }) ≤

ε 3

(11.9.4)

for every σ > 0. We remark that for every k, σ > 0 and for every n, m ∈ N, one has {|un − um | > σ } ⊆ [{|un | > k} ∩ {|un − um | > σ }] ∪ [{|un | ≤ k} ∩ {|um | > k} ∩ {|un − um | > σ }] ∪ [{|un | ≤ k} ∩ {|um | ≤ k} ∩ {|un − um | > σ }] ⊆ {|un | > k} ∪ {|um | > k} ∪ {|Tk (un ) − Tk (um )| > σ } .

By estimates (11.9.3) and (11.9.4) we deduce that, for k > k and for every n, m > ηε {|un − um | > σ } ≤ ε .

This implies that un is a Cauchy sequence in measure. Therefore there exists a function u such that un converges to u in measure and consequently, for every k > 0, Tk (un ) → Tk (u) in measure. By uniqueness of the limit Tk (u) = vk . Since Tk (un ) → vk weakly in H01 (Ω) we deduce that Tk (u) belongs to H01 (Ω) for every k. Theorem 11.28. Let f ∈ L1 (Ω). Then there exists a T -minimum u of J such that, for every k, h > 0, f L1 (Ω) k; |∇Tk (u)|2 ≤ (11.9.5) α Ω k 2 |∇u| dx ≤ |f | . (11.9.6) α {h≤|u|≤h+k}

{|u|≥h}

Proof. Let fn = Tn (f ). Then the functions v → j(x, v) − fn v Ω

Ω

have minimizers un ∈ H01 (Ω) ∩ L∞ (Ω) by Theorem 9.15. Then un satisfies j(x, ∇un ) − fn un ≤ j(x, ∇(un − Tk (un ))) − fn (un − Tk (un )) Ω

Ω

Ω

Ω

for every k > 0. By (11.9.1), we have 2 α |∇Tk (un )| ≤ fn Tk (un ) ≤ f L1 (Ω) k . Ω

Ω

146

Problems with low summable sources

For any fixed k > 0, the sequence Tk (un ) is bounded in H01 (Ω). Therefore, by Lemma 11.27 there exist a subsequence (not relabeled) and a function u ∈ T (Ω) such that un converges to u a.e. in Ω and, for every k > 0, Tk (un ) converges to Tk (u) weakly in H01 (Ω). Let ϕ ∈ H01 (Ω) ∩ L∞ (Ω); again the minimality of un yields: j(x, ∇un ) ≤ j(x, ∇(un − Tk (un − ϕ))) + fn Tk (un − ϕ) . Ω

Ω

Ω

This is equivalent to j(x, ∇(ϕ + Tk (un − ϕ))) + {|un −ϕ|≤k}

j(x, ∇un ) {|un −ϕ|>k}



j(x, ∇ϕ) +

≤ {|un −ϕ|≤k}



j(x, ∇un ) +

fn Tk (un − ϕ) Ω

{|un −ϕ|>k}

which finally implies j(x, ∇(ϕ + Tk (un − ϕ))) ≤ j(x, ∇ϕ) + fn Tk (un − ϕ) . Ω

Ω

(11.9.7)

Ω

 The weak H01 (Ω) lower semicontinuity of v → Ω j(x, ∇v) (assured by Theorem 9.2) and the weak H01 (Ω) convergence of Tk (un − ϕ) to Tk (u − ϕ) allow us to pass to the limit in (11.9.7) and to obtain the existence of a T -minimum u, according to Definition 11.26. Assumption (11.9.1) and the choice of ϕ = 0 in (11.9.7) give (11.9.5). As well, using ϕ = Th (u) in (11.9.7), we easily get (11.9.6).

We assume in the sequel that j(x, ξ) is differentiable with respect to ξ and a(x, ξ) = ∇ξ j(x, ξ) satisfies (1) there exists β > 0 such that |a(x, ξ)| ≤ β|ξ|; (2) there exists α > 0 such that a(x, ξ) · ξ ≥ α|ξ|2 , ∀ ξ ∈ RN ; (3) [a(x, ξ) − a(x, η)] · [ξ − η] > 0 if ξ = η. We show the strict relationship between T -minima and entropy solutions to the boundary value problem ⎧ ⎨−div(a(x, ∇u)) = f , in Ω , (11.9.8) ⎩u = 0 , on ∂Ω . Proposition 11.29. Let f ∈ L1 (Ω). Under the above assumptions, let u be an entropy solution to (11.9.8), that is, ∇ξ (x, ∇u)∇Tk (u − ϕ) ≤ f Tk (u − ϕ) , Ω

Ω

∀ ϕ ∈ H01 (Ω) ∩ L∞ (Ω). Then u is a T -minimum of J .

T -minima

147

Proof. The convexity of j(x, ξ) with respect to ξ implies that j(x, ∇ϕ) ≥ j(x, ∇u) + ∇ξ j(x, ∇u) · (∇ϕ − ∇u) .

The integration of the previous inequality on the set {|u − ϕ| ≤ k} yields the conclusion. Theorem 11.30. Let f ∈ L1 (Ω). Under the previous assumptions, let u be a T minimum of J . Then u is an entropy solution to the boundary value problem (11.9.8). Proof. It is clear that j(x, ∇u) ≤ {|u−ϕ|≤k} ∩{|u|≤h}







j(x, ∇u) + {|u−ϕ|≤k} ∩{|u|≤h}

j(x, ∇u) =

j(x, ∇u) . {|u−ϕ|≤k}

{|u−ϕ|≤k} ∩{h 0 if ξ = η. Definition 12.1. Let V be a Banach reflexive space. An operator A : V → V  is monotone if A(u) − A(v), u − v ≥ 0 for all u, v ∈ V . Observe that assumption 3 on a implies that −div(a(x, ∇u)) is a monotone operator. We will prove the uniqueness of solutions among the entropy solutions.

Monotone elliptic operators

151

Theorem 12.2. Let f in L1 (Ω). Then there exists a unique entropy solution to problem (12.2.1). Proof. Let u be a solution obtained by approximation as in Theorem 11.4. It is sufficient to prove that any other entropy solution is equal to u. We briefly recall how we obtained u. Let un ∈ H01 (Ω) be the solutions to − div(a(x, ∇un )) = fn

in Ω ,

(12.2.2)

where fn = Tn (f ); un ∈ L∞ (Ω) for every n from Theorem 6.6. Moreover, un (x) and ∇un (x) converge, respectively, to u(x) and ∇u(x), a.e. in Ω (see Lemmata 11.1 and 11.3). We recall that u is an entropy solution. Let z be another entropy solution, that is, Tk (z) ∈ H01 (Ω) for every k and z satisfies a(x, ∇z) · ∇Tk (z − ϕ) ≤ f Tk (z − ϕ) , (12.2.3) Ω

Ω

for every ϕ ∈ H01 (Ω) ∩ L∞ (Ω). Let us choose ϕ = un in (12.2.3). We get a(x, ∇z) · ∇Tk (z − un ) ≤ f Tk (z − un ) . Ω

(12.2.4)

Ω

On the other hand, using Tk (z − un ) as a test function in (12.2.2), one has − a(x, ∇un ) · ∇Tk (z − un ) = − fn Tk (z − un ) . Ω

(12.2.5)

Ω

Adding up (12.2.4) and (12.2.5) gives [a(x, ∇z) − a(x, ∇un )] · ∇Tk (z − un ) ≤ (f − fn )Tk (z − un ). Ω

Ω

The integrand of the left-hand side of the previous inequality is positive from the monotonicity of a. Moreover [a(x, ∇z) − a(x, ∇un )] · ∇Tk (z − un ) converges a.e. in Ω to [a(x, ∇z) − a(x, ∇u)] · ∇Tk (z − u). The right-hand side goes to 0, for every fixed k, by Lebesgue’s theorem. Using Fatou’s lemma, we can pass to the limit for n → +∞ getting [a(x, ∇z) − a(x, ∇u)] · ∇Tk (z − u) ≤ 0 . Ω

The monotonicity of a implies Tk (z − u) = 0 a.e. in Ω for every k > 0. Therefore, z = u a.e. in Ω. 2N

In the case where f ∈ L N+2 (Ω), from Chapters 5 and 11 we know that there exists 1 a H0 (Ω) solution, which is also an entropy solution. One can prove the uniqueness of solutions in a more simple way as the following theorem shows.

152

Uniqueness

2N

Theorem 12.3. Let f ∈ L N+2 (Ω). Then problem ⎧ ⎨−div(a(x, ∇u)) = f , ⎩u = 0 ,

in Ω , on

∂Ω ,

has a unique solution u ∈ H01 (Ω). Proof. Let u1 and u2 ∈ H01 (Ω) be two solutions. Let us use u1 − u2 as a test function: a(x, ∇u1 ) · (∇u1 − ∇u2 ) = f (u1 − u2 ) Ω

Ω





a(x, ∇u2 ) · (∇u1 − ∇u2 ) = Ω

f (u1 − u2 ) . Ω

Subtracting side by side and using assumption 3 on a we get u1 = u2 a.e. in Ω. The following example shows that even a linear problem may have more than one distributional solution. Remark 12.4. Let M , v and f be defined by (11.6.1), (11.6.4), and (11.6.3) respectively. The Dirichlet problem ⎧ ⎨−div(M(x)∇v) = f , in Ω , ⎩v = 0 , on ∂Ω , where Ω = {x ∈ RN : |x| < 1} has two distributional solutions. Indeed v is a solution, as seen in Section 11.6; moreover there exists a solution v1 in H01 (Ω), since f ∈ L∞ (Ω). By linearity one can give the following striking example. Let w = v − v1 : note 1,q N . By linearity, w solves that w ∈ W0 (Ω), q < N−1 ⎧ ⎨−div(M(x)∇w) = 0 , in Ω , ⎩w = 0 , on ∂Ω . It is clear that 0 is a solution to this problem. Therefore, the previous problem has two solutions: the zero solution (entropy solution) and the distributional solution w (which is not an entropy solution).

12.3 A nonmonotone elliptic operator In this section, we will give a uniqueness result for ⎧ ⎨−div(M(x, u)∇u) = f , in Ω , ⎩u = 0 , on ∂Ω .

(12.3.1)

A nonmonotone elliptic operator

153

We observe that A(v) = −div(M(x, v) ∇v) is not monotone (see [9] for more details). We will prove the following theorem: Theorem 12.5. Let M(x, s) be a N × N matrix composed by Carathéodory, bounded and Lipschitz continuous functions in the second variable. We assume that there exists α > 0 such that M(x, s)ξ · ξ ≥ α |ξ|2 . 2N

Then for every f ∈ L N+2 (Ω) there exists a weak solution u ∈ H01 (Ω) to problem (12.3.1). Proof. The existence of solutions follows from Leray–Lions theorem (Theorem 5.1). Now, let u1 and u2 be two solutions, that is, u1 and u2 satisfy M(x, u1 )∇u1 · ∇v = f v Ω

Ω





and

M(x, u2 )∇u2 · ∇v = Ω

fv, Ω

for every v ∈ H01 (Ω). We then get M(x, u1 )∇(u1 − u2 ) · ∇v = [M(x, u2 ) − M(x, u1 )] ∇u2 · ∇v . Ω

(12.3.2)

Ω

Let us fix b and B such that 0 < b < B . Since mij is Lipschitz continuous, there exists L > 0 such that |M(x, u1 ) − M(x, u2 )| ≤ L |u1 − u2 | . The use of v =

u1 −u2 b+|u1 −u2 |

as a test function in (12.3.2) implies

M(x, u1 )∇(u1 − u2 ) · ∇(u1 − u2 ) Ω

1 (b + |u1 − u2 |)2

=

[M(x, u2 ) − M(x, u1 )]∇u2 · ∇(u1 − u2 ) Ω

1 . (b + |u1 − u2 |)2

Consequently, the ellipticity of M and the fact that mij is Lipschitz continuous give α Ω

|∇(u1 − u2 )|2 ≤ [b + |u1 − u2 |]2

Ω

L|∇u2 ||∇(u1 − u2 )| . b + |u1 − u2 |

Hölder’s inequality on the right-hand side implies ⎛

⎞1/2 

2  ⎟ − u | |u ⎜  1 2  ⎠ α⎝  ≤ L u2 H01 (Ω) .  ∇ log 1 + b Ω

154

Uniqueness

By using Poincaré’s inequality on the left-hand side we get ⎛ ⎞1/2 

2  ⎟ − u | |u ⎜  1 2  ⎠ αc ⎝  ≤ L u2 H01 (Ω) .  log 1 + b Ω

Therefore



 1  B   ≤ L u2  1 α c meas({|u1 − u2 | > B}) 2  log 1 +  H0 (Ω) . b 

If b → 0+ , necessarily meas({|u1 −u2 | > B}) = 0 for every B > 0. Therefore u1 = u2 a.e. in Ω.

12.4 A uniqueness result for measure sources In this section, we study the elliptic problem ⎧ ⎨−div(M(x)∇u) = μ , ⎩u = 0 ,

in Ω ,

(12.4.1)

on ∂Ω ,

where μ ∈ M(Ω) and M ∈ RN×N is a symmetric matrix with L∞ (Ω) entries, such that M(x)ξ · ξ ≥ α|ξ|2 . We assume that ∂Ω ∈ C 1 . Theorem 11.4 guarantees the existence of a solution. We recall that we obtained this solution by approximation in Theorem 11.19. In this section we give a uniqueness result of solutions as limit of the solutions to ⎧ ⎨−div(M(x)∇un ) = fn , in Ω , (12.4.2) ⎩u = 0 , on ∂Ω , n

where fn is any sequence of regular functions that converges weakly (in the sense of measures) to μ and which is bounded in L1 (Ω). Theorem 12.6. Under the above hypotheses, let u and v be two solutions to (12.4.1) obtained by approximation from problems (12.4.2). Then u = v a.e. in Ω. The following lemma will be useful to us: Lemma 12.7. Let fn : Ω → R be a sequence of functions such that fn → 0 in L1 (Ω). Proof. We have

Ω

fn2 ≥ 1 + |fn |

For every fixed t > 0 one has fn2 ≥ 1 + |fn | Ω

{|fn |>1}

{t 0 . n→∞

Ω

Since t is arbitrary, fn → 0 in L1 (Ω). The following theorem will be useful too (for the proof use Theorem 8.29 in [30], due to De Giorgi, Theorems 6.6 and 6.14): Theorem 12.8. Under the previous hypotheses on M and Ω, let F ∈ (Lq (Ω))N , q > N , and f ∈ (Ls (Ω))N , s > N2 . Then the solution u ∈ H01 (Ω) to ⎧ ⎨−div(M(x)∇u) = divF + f , in Ω , ⎩u = 0 , on ∂Ω , is Hölder continuous in Ω and satisfies |u(x) − u(y)| ≤ C[F Lq (Ω) + f Ls (Ω) ] |x − y|β x,y∈Ω sup

where C = C(Ω, α, N, q, s) and β = β(Ω, α, N, q, s). We can now prove Theorem 12.6. Proof. Let u and v be two solutions obtained by approximation from problems (12.4.2). Then there exist two sequences of regular functions, fn and gn , converging weakly in the sense of measures to μ and fn and gn are bounded in L1 (Ω). Moreover, if un and vn are the H01 (Ω) solutions to the approximating problems (12.4.2), un and vn 1,q N converge weakly to u and v , respectively, in W0 (Ω), ∀ q < N−1 , by Lemma 11.1. By linearity we have that M(x)∇(un − vn ) · ∇ϕ = (fn − gn )ϕ , ∀ ϕ ∈ H01 (Ω) . (12.4.3) Ω

Ω

On the other hand, let zn ∈

H01 (Ω)

be a solution to

∇(un − vn ) . − div(M(x)∇zn ) = −div 1 + |∇(un − vn )|

(12.4.4)

156

Uniqueness

∇(u −v )

n n Fn = 1+|∇(u belongs to (L∞ (Ω))N and divFn ∈ L2 (Ω). The existence of zn ∈ n −vn )| 1 H0 (Ω) is guaranteed for every n by Theorem 5.1. We observe that zn are uniformly Hölder and uniformly bounded in Ω by Theorem 12.8. The Arzelà–Ascoli theorem implies that, up to subsequence, zn → z uniformly in Ω for some continuous function z in Ω. We have, choosing un − vn as a test function in (12.4.4)

Ω

|∇(un − vn )|2 = 1 + |∇(un − vn )|

M(x)∇zn · ∇(un − vn ) Ω





M(x)∇(un − vn ) · ∇zn =

= Ω

3 4 fn − gn zn

Ω

4 3 by (12.4.3). Writing (fn − gn )zn as (fn − gn )(zn − z) + fn − gn z, it is easy to prove that the last integral goes to 0, since zn → z uniformly in Ω, fn − gn goes to 0 in the sense of measures and is bounded in L1 (Ω). Therefore Ω

|∇(un − vn )|2 → 0, 1 + |∇(un − vn )|

n → ∞.

By Lemma 12.7, this implies that ∇(un −vn ) → 0 in L1 (Ω). Since un and vn converge 1,q N weakly to u and v in W0 (Ω), for every q < N−1 , one has ∇(u − v) = 0 and so u = v a.e. in Ω. Remark 12.9. We are going to prove that it is not possible to find a solution to the 1,p Dirichlet problem u ∈ W0 (Ω): −div(M(x)∇u) = −divF for every F ∈ (Lp (Ω))N , p > N , where M is defined by (11.6.1). Assume, by contradiction that there exists a solution u, that is, 1,p  M(x)∇u · ∇v = F · ∇v, ∀ v ∈ W0 (Ω) . Ω

Ω 1,q

In particular one can choose, as a test function, the not null function w ∈ W0 (Ω), N q < N−1 , defined in Remark 12.4, and F = |∇w|q−2 ∇w , so that



|∇w|q .

M(x)∇u · ∇w = Ω

Ω

On the other hand, one can choose u as a test function in the weak formulation of the problem solved by w , that is, M(x)∇w · ∇u = 0 . Ω

By the symmetry of M , the last two equalities give w = 0, which is a contradiction.

13 A problem with polynomial growth 13.1 Introduction In this chapter, we will study a polynomial growth elliptic equation. More precisely, we will prove existence and regularity results for the solutions to ⎧ ⎨−div(a(x, u, ∇u)) = f , in Ω , (13.1.1) ⎩u = 0 , on ∂Ω , under the following assumptions. The set Ω is an open bounded subset of RN , N > 2, and a : Ω × R × RN → RN is a Carathéodory map, with the following properties: (1) there exists β > 0 such that |a(x, s, ξ)| ≤ β b(|s|)|ξ|; (2) a(x, s, ξ) · ξ ≥ b(|s|)|ξ|2 , ∀ ξ ∈ RN ; (3) [a(x, s, ξ) − a(x, s, η)] · [ξ − η] > 0 if ξ = η where b : [0, +∞) → (0, +∞) is a continuous function such that b(|s|) ≥ γ |s|r ,

∀ |s| ≥ s0 ,

(13.1.2)

γ > 0, r ≥ 0 and s0 > 0. Moreover, there exists α > 0 such that for every s ∈ R b(|s|) ≥ α .

(13.1.3)

The polynomial growth of a will give us more regular solutions than the solutions found in Chapters 6 and 11 for the Leray–Lions problem. We will prove the existence of distributional solutions u, that is, a(x, u, ∇u) · ∇ϕ = f ϕ , ∀ ϕ ∈ C0∞ (Ω) . Ω

Ω

We describe in a schematic way the existence results of distributional solutions u that we are going to prove (following [42]). (1) Let f ∈ L1 (Ω). N(r +1) N−2 . N(r +1) N−1+r .

(a) If r > 1, then u belongs to H01 (Ω) ∩ Ls (Ω), s < 1,q

(b) If 0 ≤ r ≤ 1, then u belongs to W0 (Ω), q < (2) Let f ∈ Lm (Ω), 1 < m < (a) If r ≥ 1 −

2∗ m ,

(b) If 0 ≤ r < 1 −

then u belongs to H01 (Ω) ∩ L 2∗ m ,

(3) Let f ∈ L (Ω), m ≥ m

(a) If m > (b) If

2N N+2

N 2,

2N N+2 . Nm(r +1) N−2m

Nm(r +1) 1, N−m(1−r )

then u belongs to W0

(Ω).

(Ω).

2N N+2 .

then u belongs to H01 (Ω) ∩ L∞ (Ω).

≤m
0, Tk (un )2H 1 (Ω) ≤ k

f L1 (Ω) ; α

t (Ω), where B(t) = 0 b(|s|); (2) |∇B(un )| is bounded in M N . (3) a(x, Tn (un ), ∇un ) is bounded in Lq (Ω) for every q < N−1 0 N N−1

Proof. (1) Taking Tk (un ) as a test function in (13.2.1) and using (13.1.3) we get that Tk (un )2H 1 (Ω) ≤ k 0

f L1 (Ω) . α

t (2) Let B(t) = 0 b(|s|); it is easy to see that Tk (B(un )) belongs to H01 (Ω), so that it can be taken as a test function in (13.2.1). The assumptions on a give b(|un |)2 |∇un |2 {|B(un )|≤k}

≤

a(x, Tn (un ), ∇un ) · ∇un b(|un |) ≤ kf L1 (Ω) , {|B(un )|≤k}

and then the following estimate holds: |∇Tk (B(un ))|2 ≤ k f L1 (Ω) .

(13.2.2)

Ω N

Proposition 11.13 implies that |∇B(un )| is bounded in M N−1 (Ω), that is, b(|un |)|∇un | N is bounded in M N−1 (Ω). (3) Since {|a(x, Tn (un ), ∇un )| > k} is contained in {βb(|un |)|∇un | > k}, we N conclude that a(x, Tn (un ), ∇un ) is bounded in Lq (Ω) for every q < N−1 . In the sequel C denotes a positive constant independent of n ∈ N.

Existence results

159

Lemma 13.2. Let f ∈ L1 (Ω). (1) If r > 1, then the sequence of the solutions un to problem (13.2.1) is bounded in N(r +1) H01 (Ω) ∩ Ls (Ω), s < N−2 . (2) If 0 ≤ r ≤ 1, then the sequence of the solutions un to problem (13.2.1) is bounded 1,q +1) in W0 (Ω), q < N(r N−1+r . Proof. Let un be a solution to problem (13.2.1). We consider T1 (un − Tk (un )), with k ≥ s0 , as a test function in (13.2.1). Since ∇T1 (un − Tk (un )) = ∇un χ{k≤|un | 1; from (13.2.3) we soon obtain the estimate

|∇un |2 ≤ Ω

|∇Ts0 (un )|2 + Ω

∞ f L1 (Ω)  1 . γ kr k=1

This estimate and point 1 of the above lemma imply that un is bounded in H01 (Ω). On N . Due to the other hand, by Lemma 13.1, B(un ) belongs to Lq (Ω) for every q < N−2 the definition of B there exists a positive constant C such that |t|r +1 ≤ C(1 + B(t)). +1) Therefore, un belongs to Lq (Ω) for every q < N(r N−2 . Let 0 < r ≤ 1. For every 1 < q < 2 and λ > 1 − r , we can write, using Hölder’s inequality with exponent q2 and (13.2.3),



|∇un |q

|∇un |q = Ω

λq 2

(1 + |un |)

λq 2

(1 + |un |) ⎡ ⎤q ⎡ ⎤1− q 2 2 2 λq |∇un | ⎢ ⎥ ⎢ ⎥ ≤⎣ ⎦ ⎣ (1 + |un |) 2−q ⎦ (1 + |un |)λ Ω

Ω

Ω

⎞1− q ⎤q ⎛ 2 2 ∞  λq C ⎜ ⎟ ⎦ ⎣ 2−q ≤ C+ ⎝ (1 + |un |) ⎠ kλ+r k=1 ⎡

⎡

Ω

⎤1− q

λq ⎢ ⎥ ≤ C ⎣ (1 + |un |) 2−q ⎦ Ω

2

.

160

A problem with polynomial growth

N(r +1) If we set λ = N−q , the condition λ > 1 − r implies that q < N−(1−r ) . The above inequality then yields, by Sobolev’s inequality, ⎛ ⎛ ⎞ q∗ ⎞1− q 2 q ⎜ ⎜ q∗ ⎟ q q∗ ⎟ . ⎝ |un | ⎠ ≤ |∇un | ≤ C ⎝ (1 + |un |) ⎠ N(2−q)

Ω

Since

q q∗

Ω

Ω 1,q

q

> 1 − 2 , we deduce that un is bounded in W0 (Ω) for every q


1 m ,

Ω

Ω ∗

we deduce that un is bounded in L2 (Ω) and in H01 (Ω).

Existence results

161

We now consider as a test function T1 (|un |r un − Tk (|un |r un )) in (13.2.1) in order to have |un |r a(x, Tn (un ), ∇un ) · ∇un ≤ |fn | . {k≤|un |r +1 0, one has Tk (un )Lm∗∗ (r +1) (Ω) ≤ C f Lm (Ω) .

Fatou’s Lemma, as k → ∞, implies the result.

164

A problem with polynomial growth

Theorem 13.5. Let f ∈ L1 (Ω). There exists a function u such that Tk (u) ∈ H01 (Ω), N a(x, u, ∇u) belongs to Lq (Ω) for every q < N−1 and u solves (13.1.1) in the sense of distributions. +1) (1) If r > 1, then u belongs to H01 (Ω) ∩ Ls (Ω), s < N(r N−2 . 1,q

(2) If 0 ≤ r ≤ 1, then u belongs to W0 (Ω), q < Let f ∈ Lm (Ω), 1 < m < (1) If r ≥ 1 −

∗

2 m

2N N+2 .

then u belongs to H01 (Ω) ∩ L

(2) If 0 ≤ r < 1 −

2∗ m

then u belongs to

N(r +1) N−1+r .

Nm(r +1) N−2m

2N . Let f ∈ Lm (Ω), m ≥ N+2 N (1) If m > 2 , then u belongs to H01 (Ω) ∩ L∞ (Ω).

(2) If

2N N+2

≤m
0, n ≥k > 0, we take Tε (un − Tk (u)) as a test function in (13.2.1); we have a(x, Tn (un ), ∇un ) · ∇Tε (un − Tk (u)) ≤ ε f L1 (Ω) . (13.2.8) Ω

Since ∇Tε (un − Tk (u)) = 0 if |un | > k + ε and a(x, s, ξ) · ξ ≥ 0, we have a(x, Tn (un ), ∇un ) · ∇Tε (un − Tk (u)) Ω

a(x, Tk (un ), ∇Tk (un )) · ∇Tε (Tk (un ) − Tk (u))

≥ Ω

|a(x, Tk+ε (un ), ∇Tk+ε (un ))| |∇Tk (u)| .

− {|un |>k}

Now, |∇Tk (u)|χ{|un |>k} strongly converges to 0 in L2 (Ω), so that last term in the above inequality goes to 0 as n → ∞ for every fixed ε > 0. We will denote by δεn any quantity converging to 0 as n → ∞, for every fixed ε > 0. Thus by (13.2.8) it follows that a(x, Tk (un ), ∇Tk (un )) · ∇Tε (Tk (un ) − Tk (u)) ≤ ε f L1 (Ω) + δεn . (13.2.9) Ω

165

Existence results

Now, let 0 < θ < 1 and set Ekε = { |Tk (un ) − Tk (u)| > ε}. We have

{(a(x, Tk (un ), ∇Tk (un )) − a(x, Tk (un ), ∇Tk (u))) · ∇(Tk (un ) − Tk (u))}θ Ω

{(a(x, Tk (un ), ∇Tk (un )) − a(x, Tk (un ), ∇Tk (u))) · ∇Tε (Tk (un ) − Tk (u))}θ

= Ω

{(a(x, Tk (un ), ∇Tk (un )) − a(x, Tk (un ), ∇Tk (u))) · ∇(Tk (un ) − Tk (u))}θ .

+ Ekε

By Hölder’s inequality with exponent

1 θ

in the second integral we get, from (13.2.9),

{(a(x, Tk (un ), ∇Tk (un )) − a(x, Tk (un ), ∇Tk (u))) · ∇(Tk (un ) − Tk (u))}θ Ω

⎛ ≤ meas(Ω)

1−θ

⎜ ⎝εf L1 (Ω) + δεn −



⎞θ ⎟ a(x, Tk (un ), ∇Tk (u)) · ∇Tε (Tk (un ) − Tk (u))⎠

Ω

+ {(a(x, Tk (un ), ∇Tk (un )) − a(x, Tk (un ), ∇Tk (u))) · ∇(Tk (un ) − Tk (u))}θ . Ekε

(13.2.10)

We are going to study the right-hand side of (13.2.10). Since Tk (un ) is bounded in H01 (Ω) and weakly converges to Tk (u), by the assumptions on a the first integral goes to zero as n goes to infinity. Moreover the sequence {|a(x, Tk (un ), ∇Tk (un )) − a(x, Tk (un ), ∇Tk (u))|} is bounded in L2 (Ω) for every fixed k > 0. Thus, using Hölder’s inequality on the last integral, we get {(a(x, Tk (un ), ∇Tk (un )) − a(x, Tk (un ), ∇Tk (u)))∇(Tk (un ) − Tk (u))}θ Ω

3 4θ ≤ meas(Ω)1−θ ε f L1 (Ω) + 2δεn + C meas({|Tk (un ) − Tk (u)| > ε})1−θ .

Since Tk (un ) converges to Tk (u) in measure, letting first n go to infinity and then letting ε go to zero, we find {(a(x, Tk (un ), ∇Tk (un ))

lim

n→+∞ Ω

− a(x, Tk (un ), ∇Tk (u))) · ∇(Tk (un ) − Tk (u))}θ = 0 .

We deduce that ∇Tk (un ) converges a.e. to ∇Tk (u) for every k > 0. This implies that ∇un converges to ∇u a.e. in Ω, so that a(x, Tn (un ), ∇un ) converges to N a(x, u, ∇u) strongly in Lq (Ω) for every q < N−1 . The above argument implies the existence of a distributional solution u to problem (13.1.1). The regularity of u follows from the a.e. convergence of un to u and Lemmata 13.2, 13.3, and 13.4.

14 A problem with degenerate coercivity 14.1 Introduction In this chapter we will treat some problems defined by an elliptic operator with degenerate coercivity, that is, an elliptic operator which does not satisfy assumption 2 of Leray–Lions theorem (Theorem 5.1). More precisely let Ω be a bounded, open subset of RN , with N > 2. We assume that a(x, s) : Ω × R → R is a Carathéodory function satisfying the following conditions: α ≤ a(x, s) ≤ β , (1 + |s|)θ

(14.1.1)

for almost every x ∈ Ω, for every s ∈ R, where α, β, and θ are positive constants. Note that, because of assumption (14.1.1), the differential operator A(v) = −div(a(x, v)∇v) is not coercive on H01 (Ω), even if it is well defined between H01 (Ω) n(2−N)

and its dual. To see that it is sufficient to consider the sequence un (x) = |x| 2(n+1) −1 defined in Ω = B1 (0). It satisfies un H01 (Ω) → ∞ and, at the same time, 1 ||un ||H01 (Ω)

Ω

|∇un |2 → 0. (1 + |un |)p

We will begin by studying the existence and regularity of solutions to ⎧ ⎨−div(a(x, u)∇u) = f , in Ω , ⎩u = 0 , on ∂Ω ,

(14.1.2)

where the source f belongs to Lm (Ω), with m ≥ 1. In the case where θ < 1 we will prove the following results (following [10]): (1) if m > N2 there exists a solution u in H01 (Ω) ∩ L∞ (Ω); ∗∗ 2N ≤ m < N2 there exists a solution u in H01 (Ω) ∩ Lm (1−θ) (Ω); (2) if N+2−θ(N−2) N < N+1−θ(N−1) Nm(1−θ) N−m(1+θ) < 2.

(3) if

m
1, some nonexistence phenomena appear. In Section 14.3 we will prove that for constant sources sufficiently large, there is no solution. However, we will prove an existence theorem for “small” sources in Theorem 14.10. The existence of solutions to problem (14.1.2) can be recovered for every value of θ by adding a lower order term. More precisely we will study the existence and

167

The case 0 < θ < 1

regularity of solutions to ⎧ ⎨−div(a(x, u)∇u) + u = f , ⎩u = 0 ,

in Ω , on

(14.1.3)

∂Ω ,

proving the following results (see [8] and [19]): (1) if m ≥ θ + 2, θ > 0, there exists a solution u in H01 (Ω) ∩ Lm (Ω); 1,

2m

θ+2 (Ω) ∩ Lm (Ω); (2) if θ+2 2 < m < θ + 2, θ > 0, there exists a solution u in W0 N 1 ∞ (3) if m > θ 2 and θ > 1, there exists a solution u in H0 (Ω) ∩ L (Ω).

1,1

In the case where θ = 2 = m, we will prove the existence of a unique W0 (Ω) solution obtained by approximation and nonexistence in the case of a Dirac mass source.

14.2 The case 0 < θ < 1 In this section, we are going to study problem (14.1.2) under the hypothesis that θ < 1. We will work by approximation. Let fn = Tn (f ). We define the following sequence of problems: ⎧ ⎨−div(a(x, Tn (un ))∇un ) = fn , in Ω , (14.2.1) ⎩u = 0 , on ∂Ω . n

We remark that, due to the fact that we have truncated the second variable of a, problems (14.2.1) are coercive, so that the existence of weak solutions un in H01 (Ω) ∩ L∞ (Ω) to (14.2.1) is assured by Theorems 5.1 and 6.6. Lemma 14.1. Let θ > 0 and let f belongs to Lm (Ω), m > N2 . Assume that un is a solution to problem (14.2.1) with fn = f for every n ∈ N. Then the norms of un in L∞ (Ω) and in H01 (Ω) are bounded. Proof. We define, for s in R and for k > 0 s H(s) = 0

1 dt . (1 + |t|)θ

For k > 0, if we take Gk (H(un )) as a test function in (14.2.1) and use assumption (14.1.1), we obtain α |∇(H(un ))|2 ≤ f Gk (H(un )) , (14.2.2) Ak

Ak

where we have set Ak = {x ∈ Ω : |H(un (x))| > k} .

168

A problem with degenerate coercivity

By Remark (6.8), there exists a constant C > 0 (depending on θ , m, N , α, meas(Ω) and on the norm of f in Lm (Ω)) such that H(un )L∞ (Ω) ≤ C .

The limits lims→+∞ H(s) = +∞, lims→−∞ H(s) = −∞ yield a bound for un in L∞ (Ω): un L∞ (Ω) ≤ C . (14.2.3) The uniform estimate of un in H01 (Ω) is now very easy. Taking un as test function in (14.2.1), one obtains, by assumption (14.1.1) and (14.2.3) α 2 |∇u | ≤ f un ≤ Cf L1 (Ω) . n (1 + C)θ Ω

Ω

Theorem 14.2. Let f be a function in Lm (Ω), with m > N2 . Then there exists a weak solution u ∈ H01 (Ω) ∩ L∞ (Ω) to problem (14.1.2) , that is, a(x, u)∇u · ∇ϕ = f ϕ , ∀ ϕ ∈ H01 (Ω) . Ω

Ω

Proof. In the above lemma we proved the existence of a positive constant C such that un L∞ (Ω) ≤ C for every n ∈ N. It is then sufficient to take ν ∈ N with ν > C , so that Tν (uν ) = uν . Then uν is a weak solution to problem (14.1.2) belonging to H01 (Ω) ∩ L∞ (Ω). Remark 14.3. We observe that in Theorem 14.2 we do not assume that θ < 1. This will be useful to prove Theorem 14.10. We now weaken the summability of the source f . 2N N Lemma 14.4. Let 0 < θ < 1 and N+2−θ(N−2) ≤ m < 2 . Let un be the solutions ∗∗ to (14.2.1). Then the norms of un in Lm (1−θ) (Ω) and in H01 (Ω) are bounded.

Proof. We consider [(1 + |un |)p − 1]sgn(un ), with p = (1−θ)N(m−1) , as a test funcN−2m tion in (14.2.1). We observe that p satisfies pm = (p+1−θ)2∗ /2 = m∗∗ (1−θ) > 1. By assumption (14.1.1), we have p+1−θ 4α |∇un |2 2 2 |∇((1 + |u |) − 1)| = α n 2 (p + 1 − θ) (1 + |un |)θ−p+1 Ω Ω (14.2.4) ≤ C |f ||un |p . Ω

Hölder’s inequality on the right-hand side and Sobolev’s one on the left-hand side imply ⎡ ⎤ 2∗ ⎡ ⎤ 1 2 m 2 |∇un | ⎢ ⎢ (p+1−θ)2∗ /2 ⎥ pm ⎥ m ≤ Cf L (Ω) ⎣ |un | . ⎣ |un | ⎦ ≤C ⎦ (1 + |un |)θ−p+1 Ω

Ω

Ω

The case 0 < θ < 1

169

Since 22∗ > m1  , the above estimate gives that the sequence un is bounded in ∗∗ Lm (1−θ) (Ω). From this uniform estimate we deduce that the sequence  2 p−1−θ is bounded. As p − 1 − θ > 0, by the assumptions Ω |∇un | (1 + |un |) on m, this implies a uniform bound for ∇un in L2 (Ω). 2N N Theorem 14.5. Let f be a function in Lm (Ω), with N+2−θ(N−2) < m < 2 . Then there ∗∗ exists a weak solution u ∈ H01 (Ω) ∩ Lm (1−θ) (Ω) to problem (14.1.2) , that is, a(x, u)∇u · ∇ϕ = f ϕ , ∀ ϕ ∈ H01 (Ω) . Ω

Ω

Proof. The estimate for un in H01 (Ω) implies that, up to a subsequence, un converges to some function u ∈ H01 (Ω) weakly in H01 (Ω) and a.e. in Ω. By Lebesgue’s theorem the coefficient a(x, un ) converges to a(x, u) in Lq (Ω), for any q, due to assumption (14.1.1). Thus it is possible to pass to the limit in (14.2.1) in order to obtain the existence of a weak solution u of problem (14.1.2). The last result gives the existence of distributional solutions in the case where N 2N N+1−θ(N−1) < m < N(1−θ)+2(θ+1) .

f ∈ Lm (Ω), with

2N . Then N(1−θ)+2(θ+1) 1,q Nm(1−θ) bounded in W0 (Ω), with q = N−m(1+θ) .

Lemma 14.6. Assume that f ∈ Lm (Ω), with the sequence of the solutions un to (14.2.1) is

N N+1−θ(N−1)

0. By the Sobolev inequality and the Hölder one with exponent q2 , we have ⎡



⎤

q q∗

∗⎥ ⎢ S ⎣ |un |q ⎦

q

|∇un |q

≤

Ω

Ω



= Ω

|∇un |q (1 + |un |)

⎡ ⎢ ≤⎣ Ω

λq 2

(1 + |un |) ⎤q ⎡

|∇un | ⎥ ⎦ (1 + |un |)λ 2

2



λq 2

⎢ ⎣ (1 + |un |)

(14.2.8) ⎤ 2−q 2

λq 2−q

⎥ ⎦

.

Ω

Estimate (14.2.7) implies ⎧ ⎫ ⎧ ⎡ ⎤ q∗ ⎡ ⎤ q ⎫⎪ ⎤ 2−q ⎪ ⎡ ⎪ ⎪ 2 ⎪ q 2m ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ λq ∗⎥ ⎥ ⎢ ⎢ ⎥ ⎢ 1 + ⎣ |un | 2−q ⎦ . ⎣ |un |q ⎦ ≤ C + C ⎣ |un |(1+θ−λ)m ⎦ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭⎪ ⎪ ⎪ ⎩ ⎭ Ω Ω Ω

171

The case 0 < θ < 1

Since

λq 2−q

= q∗ and (1 + θ − λ)m ≤ q∗ , one obtains that the sequence un is ∗

bounded in Lq (Ω). From (14.2.8) and (14.2.7) we deduce that ⎧ ⎫ ⎧ ⎡ ⎤ q ⎫⎪ ⎤ 2−q ⎪ ⎡ ⎪ ⎪ 2 ⎪ 2m ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬⎨ ⎬ λq ⎥ ⎢ ⎥ ⎢ |∇un |q ≤ C + C ⎣ |un |(1+θ−λ)m ⎦ 1 + ⎣ |un | 2−q ⎦ , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎪ ⎪ ⎩ ⎭ Ω Ω Ω 1,q

that is, the sequence un is bounded in W0 (Ω). Passing to the limit in problems (14.2.1), as in Theorem 14.5, it is easy to prove the following result. Observe that the choice of test functions is different from the above theorems, due to the regularity of the solutions. Theorem 14.7. Assume that f ∈ Lm (Ω), with

N N+1−θ(N−1)

Nm(1−θ) N−m(1+θ) .

Then there exists a distributional solution u ∈ Let q = lem (14.1.2), that is, a(x, u)∇u · ∇ϕ = f ϕ , ∀ ϕ ∈ C0∞ (Ω) . Ω

2N N(1−θ)+2(θ+1) . 1,q W0 (Ω) to prob-

< m ≤

Ω

Remark 14.8. We observe that q∗ = Theorem 14.5.

Nm(1−θ) N−2m ,

which is the summability found in

We end this section by studying under which conditions on the summability of the source f the solution u can be chosen as a test function in problem (14.1.2). N(2−θ) Proposition 14.9. Assume that m ≥ N+2−N θ . Let u be a solution to problem (14.1.2) found by approximation. Then a(x, u) |∇u|2 = f u . Ω

Ω

Proof. Step I: We are going to prove that a(x, u) |∇u|2 ≤ f u < ∞ . Ω

(14.2.9)

Ω

Let un be a solution to (14.2.1). We have proved in Lemmata 14.1, 14.4 and 14.6, that the Nm(1−θ) sequence un is bounded in Lr (Ω), where r = Nm(1−θ) = q∗ , with q = N−m(1+θ) . N−2m Moreover, choosing Tk (un ) as a test function in (14.2.1), we obtain 2 a(x, Tn (un )) |∇Tk (un )| = fn Tk (un ) . (14.2.10) Ω

Ω

By using (14.1.1) on the left-hand side and Hölder’s inequality on the right one we have |∇Tk (un )|2 α ≤ kf Lm (Ω) , (1 + k)θ Ak

172

A problem with degenerate coercivity

that is, for a fixed k > 0, the sequence Tk (un ) is bounded in H01 (Ω). It is easy to see that 2 a(x, Tn (un )) ∇Tk (un ) · ∇Tk (u) − a(x, Tn (un )) |∇Tk (u)|2 Ω

Ω

a(x, Tn (un )) |∇Tk (un )|2 .

≤ Ω

Since a is bounded by assumption (14.1.1), the left-hand side of the above inequality  converges to Ω a(x, u) |∇Tk (u)|2 . Therefore, by (14.2.10), we have a(x, u) |∇Tk (u)|2 ≤ f Tk (u) . Ω

Ω

 At the limit as k → ∞, the right-hand side tends to Ω f u, which is finite by Hölder’s inequality, since q∗ = r ≥ m . By Fatou’s Lemma on the left-hand side we obtain (14.2.9). Step II: Let ϕn be a sequence of C0∞ (Ω) functions, converging to Tk (u) in H01 (Ω) and ∗-weakly in L∞ (Ω). Then a(x, u) ∇u · ∇ϕn = f ϕn . Ω

Ω

The sequence a(x, u) ∇u·∇ϕn converges to a(x, u)|∇Tk (u)|2 a.e. in Ω. Moreover, if E is a measurable subset of Ω, we have, by Hölder’s inequality and (14.1.1)

⎛ ⎞1 ⎛ ⎞1 2 2 ⎜ 2⎟ ⎜ 2⎟ a(x, u) ∇u · ∇ϕn ≤ ⎝ a(x, u) |∇u| ⎠ ⎝ β |∇ϕn | ⎠ .

E

E

E

By Vitali’s theorem (Theorem 3.2) the right-hand side converges to 0, as meas(E) → 0, due to (14.2.9) and the hypotheses on ϕn . At the limit as n → ∞, Vitali’s theorem implies that a(x, u) |∇Tk (u)|2 = Ω

f Tk (u) . Ω

It is now sufficient to pass to the limit as k → ∞.

14.3 The case θ > 1: existence and nonexistence If θ > 1, some nonexistence phenomena occur for problem (14.1.2), even for constant sources. We observe that one would expect these kinds of sources to give bounded solutions, if one thinks to Theorem 6.6.

The case θ > 1: existence and nonexistence

Assume that u is a solution to

⎧ ∇u ⎪ ⎨−div = A > 0, (1 + |u|)θ ⎪ ⎩u = 0 , Then v(x) =

173

in B1 (0) , ∂B1 (0) .

on

1 − (1 + |u|)1−θ θ−1

solves −Δv = A. Observe that v≤

1 θ−1

in B1 (0) .

(14.3.1)

A Now, the H01 (Ω) solution to −Δv = A is v = 2N (1 − |x|2 ). If A is sufficiently large, 1 . This contrathere exists a subset E0 of B1 (0), with meas(E0 ) > 0, where v > θ−1 dicts (14.3.1). However, if A is sufficiently small a solution exists, as stated in the following existence result in the case where θ > 1.

Theorem 14.10. Let Ω be a bounded open subset of RN with ∂Ω of class C 1 . Assume that a satisfies (14.1.1) with θ > 1. Let f belong to Lm (Ω), with m > N2 . Then there exists λ∗ > 0 such that, if |λ| < λ∗ , problem ⎧ ⎨−div(a(x, u)∇u) = λ f , in Ω , (14.3.2) ⎩u = 0 , on ∂Ω , has a bounded H01 (Ω) solution. Proof. We will use Theorem 2.10 (Schauder’s fixed point Theorem) to prove the result. We define S : L∞ (Ω) → L∞ (Ω) as the operator which maps v into z, where z ∈ H01 (Ω) is the weak solution to − div(a(x, v)∇z) = λ f .

(14.3.3)

S is well defined, due to Theorem 14.2 and Remark 14.3. Step I: We prove that S has an invariant convex, closed, bounded set. Let us consider Gk (z) as a test function in (14.3.3). By assumption (14.1.1), one has α Ω

|∇Gk (z)|2 ≤ (1 + |v|)θ



a(x, v)|∇Gk (z)|2 ≤ λ

Ω

f Gk (z) . Ω

Since v is bounded, we deduce that 2 θ α |∇Gk (z)| ≤ λ (1 + vL∞ (Ω) ) f Gk (z) . Ω

Ω

174

A problem with degenerate coercivity

By Remark (6.8), there exists a positive constant C such that zL∞ (Ω) ≤ C |λ| f Lm (Ω) (1 + vL∞ (Ω) )θ .

We define h : R+ → R by h(R) = R − C f Lm (Ω) |λ|(1 + R)θ .

We will prove that there exists R0 such that h(R0 ) > 0. The maximum of h is attained at 1 − 1 = R1 ; 1 (C|λ|θf Lm (Ω) ) θ−1 this value is well defined if 1 1

(C|λ|θf Lm (Ω) ) θ−1

> 1,

that is, |λ| ≤

1 . Cθf Lm (Ω)

Now, ⎞

⎛ h(R1 ) =

1 1

(|λ|θf Lm (Ω) ) θ−1

if and only if

θ−1 θ

θ θ−1

− 1 − Cλ ⎝

1 θ

(C|λ|θf Lm (Ω) ) θ−1

⎠>0

1

> (C|λ|f Lm (Ω) ) θ−1 .

Therefore it is sufficient to choose λ such that / . (θ − 1)θ−1 (θ − 1)θ−1 1 = . |λ| < min , θ θ Cf Lm (Ω) C f Lm (Ω) θ θ θ f Lm (Ω) C In this way the closed L∞ (Ω) ball B of radius R1 centered in 0 is invariant under S . It is clear that B is bounded and convex. Step II: We will show that S is completely continuous. We are going to prove that if vn → v in L∞ (Ω), then S(vn ) = zn → S(v) = z in L∞ (Ω). By subtracting side by side of the equalities a(x, vn )∇zn · ∇w = Ω

Ω





and

a(x, v)∇z · ∇w = Ω

we obtain

λf w , Ω



a(x, vn )∇zn · ∇w = Ω

λf w

a(x, v)∇z · ∇w . Ω

(14.3.4)

The regularizing effects of a lower order term

175

The choice of zn as a test function gives |∇zn |2 α ≤ a(x, vn )|∇zn |2 ≤ β ∇zL2 (Ω) ∇zn L2 (Ω) (1 + |vn |)θ Ω

Ω

by assumption (14.1.1) and Hölder’s inequality on the right-hand side. Since vn belongs to B (see Step I), we obtain that zn H01 (Ω) is bounded, since α ∇zn 2L2 (Ω) ≤ zL2 (Ω) zn L2 (Ω) . (1 + R1 )θ

Therefore there exist a subsequence and a function z0 in H01 (Ω) such that znk → z0 weakly in H01 (Ω) and a.e. in Ω. Fixing w in H01 (Ω), we have that a(x, vnk )∇znk · ∇w → a(x, v)∇z0 · ∇w Ω

Ω

for every w in H01 (Ω), since a(x, vn )∇w → a(x, v)∇w in (L2 (Ω))N by Lebesgue’s theorem. On the other hand, we deduce from (14.3.4) that a(x, vnk )∇znk · ∇w = a(x, v)∇z · ∇w Ω

for every w in

H01 (Ω);

Ω

therefore a(x, v)∇(z0 − z) · ∇w = 0 Ω

for every w in H01 (Ω). Choosing z0 − z as a test function and using that a(x, s) > 0, we deduce that z0 = z. By Theorem 12.8, znk are uniformly Hölder in Ω, since they α are uniformly bounded and (1+R θ ≤ a(x, vnk ) ≤ β (that is, a(x, vn ) are uniform1) ly coercive and bounded). Arzelà-Ascoli theorem applies and gives the existence of a continuous function g in Ω such that znk → g in C(Ω). Since znk → z a.e. in Ω, we infer that znk → z in L∞ (Ω). As the limit is independent of the subsequence, zn → z in L∞ (Ω). Let us show that for every bounded C ⊂ L∞ (Ω), then S(C) is compact. Let vn be a sequence contained in the invariant set B . Then zn L∞ (Ω) ≤ R1 . As above we can say that zn are uniformly Hölder. Arzelà-Ascoli theorem implies that S is completely continuous.

14.4 The regularizing effects of a lower order term In this section we are going to study the Dirichlet problem ⎧ ⎨−div(a(x, u)∇u) + u = f , in Ω , ⎩u = 0 , on ∂Ω .

(14.4.1)

176

A problem with degenerate coercivity

We will show that the presence of the lower order term changes the nature of the existence results. Let fn = Tn (f ) and define the following sequence of problems: ⎧ ⎨−div(a(x, Tn (un ))∇un ) + un = fn , in Ω , (14.4.2) ⎩u = 0 , on ∂Ω . n

It is clear from Theorems 5.11 and 6.7 that there exists a H01 (Ω) ∩ L∞ (Ω) solution to problem (14.4.2). Lemma 14.11. If un is a solution of (14.4.2), then un Lm (Ω) ≤ fn Lm (Ω) . Proof. In the case where f belongs to Lm (Ω) with m > 1, we choose |un |m−2 un as a test function in problems (14.4.2). By dropping the operator term and using the Hölder’s inequality on the right-hand side, one has |un |m ≤ fn |un |m−2 un ≤ fn Lm (Ω) un m−1 Lm (Ω) Ω

which gives the result. n) as a test function In the case where f ∈ Lm (Ω) with m = 1, we choose Tk (u k in (14.4.2). Again, dropping the operator term of the left-hand side, one has Tk (un ) ≤ |f | . un k Ω

Ω

It is now sufficient to pass to the limit as k → 0 and use Fatou’s Lemma. Now we assume that (14.1.1) holds with θ > 1. Let f ∈ Lm (Ω), with m > θ N2 ; we will prove the existence of bounded solutions. Theorem 14.12. If f belongs to Lm (Ω), with m > θ N2 and θ > 1, then there exists a solution u ∈ H01 (Ω) ∩ L∞ (Ω) of the boundary value problem (14.4.1).   θ−1 θ−1 + Proof. The use of (1+|un |) θ−1−(1+k) sgn(un ), k > 0, as a test function in (14.4.2) and Young’s inequality imply that

a(x, Tn (un )) |∇un |2 (1 + |un |)θ−2 + Ak

1 ≤ θ−1 Cε ≤ θ−1

Using the inequality θ−1

[(1 + t)

θ−1

− (1 + k)

|un | Ak



(1 + |un |)θ−1 − (1 + k)θ−1 θ−1

|fn | [(1 + |un |)θ−1 − (1 + k)θ−1 ] Ak



|fn |θ + Ak

⎧ ⎨cθ t θ−1 , ]≤ ⎩c t θ−1 , θ

ε θ−1



θ

[(1 + |un |)θ−1 − (1 + k)θ−1 ] θ−1 . Ak

θ−2

∀ t > k ≥ 2 θ−1 − 1 , cθ = 2θ−2 , if ∀ t > k , cθ = 1 , if

1 h}. Thus the last inequality gives ⎡ ⎢ ⎣

⎤



∗⎥ |vn − h|2 ⎦

2 2∗ θ

≤ Cf θLm (Ω) [meas({vn > h})]1− m .

{vn >h} ∗

θ 2 N Note that (1 − m ) 2 > 1, since m > θ 2 . Then Lemma 6.2 applies and gives that the sequence vn L∞ (Ω) =  log(1 + |un |)L∞ (Ω) is bounded, that is, the sequence un L∞ (Ω) is bounded. As in the proof of Theorem 14.2, un is a solution for n sufficiently large.

Theorem 14.13. Let f ∈ Lm (Ω), with m > 1. (1) Assume that m ≥ θ + 2. Then there exists a distributional solution u ∈ H01 (Ω) ∩ Lm (Ω) to problem (14.4.1). < m < θ + 2. Then there exists a distributional solution u ∈ (2) Assume that θ+2 2 2m 1, θ+2

Lm (Ω) ∩ W0

(Ω) to problem (14.4.1).

178

A problem with degenerate coercivity

Proof. Step I: Assume that m ≥ θ + 2. We use [(1 + |un |)1+θ − 1]sgn(un ) as a test function in (14.4.2). Hölder’s inequality on the right-hand side and assumption (14.1.1) on the left one imply |∇un |2 ≤ C{1 + fn Lm (Ω) }|un |(1+θ) Lm (Ω) , Ω

where C is a positive constant depending on α and θ . By Lemma 14.11, the last term is bounded uniformly if (1 + θ)m ≤ m, that is, m ≥ 2 + θ . Therefore, there exists a function u ∈ H01 (Ω) such that, up to a subsequence, un → u weakly in H01 (Ω), strongly in L1 (Ω) and a.e. in Ω. It is easy to pass to the limit in problems (14.4.2) in order to prove that u is a solution to problem (14.4.1). For the first term of the left-hand side, it is sufficient to observe that ∇un → ∇u weakly in L2 (Ω) and by Lebesgue’s theorem a(x, Tn (un ))∇ϕ → a(x, u)∇ϕ in (Lm (Ω))N for every m, due to assumption (14.1.1). (m−1) − Step II: Assume that θ+2 2 < m < θ + 2. If we choose ϕ = [(1 + |un |) 1]sgn(un ) as a test function in (14.4.2), we get, by the assumptions on a |∇un |2 ≤ C |f | |un |m−1 . (1 + |un |)θ−m+2 Ω

Ω

Now, using Hölder’s inequality on the right-hand side of the previous inequality and Lemma 14.11, we get Ω

⎡

⎤1− 1



m

|∇un | ⎢ ⎥ ≤ C ⎣ |un |m ⎦ (1 + |un |)θ−m+2 2

m 

≤ Cf Lmm (Ω) ,

∀n ∈ N,

(14.4.3)

Ω

 where C depends on m and θ . On the other hand, let σ < 2. Writing Ω |∇un |σ as σ |∇un |σ |∇un |σ = (1 + |un |) 2 [θ−m+2] σ [θ−m+2] (1 + |un |) 2 Ω

Ω

and using Hölder’s inequality with exponent

2 σ

, we get

⎛ ⎞1− σ 2 σ ⎜ ⎟ σ [θ−m+2] |∇un | ≤ C ⎝ (1 + |un |) 2−σ , ⎠

Ω

Ω

σ where we have used (14.4.3). Due to Lemma 14.11, if 2−σ [θ −m+2] = m, that is, σ = 2m θ+2 , the right-hand side is uniformly bounded. Notice that σ < 2, by the assumptions  2m 2m on m and θ . Since θ+2 > 1, the fact that the sequence Ω |∇un | θ+2 is bounded, 2m 1, θ+2

implies the existence of a function u ∈ W0

(Ω) such that, up to a subsequence,

2m

un → u weakly in

1, W0 θ+2 (Ω)

and a.e. in Ω; moreover u ∈ Lm (Ω) by Lemma 14.11. One can prove that u is a distributional solution to problem (14.4.1), as in Step I.

The regularizing effects of a lower order term

179

We are now going to study the case where m = 2, for θ = 2 proving the existence 1,1 of a W0 (Ω) solution to problem ⎧ ⎨−div(a(x, u)∇u) + u = f , ⎩u = 0 ,

in Ω ,

(14.4.4)

∂Ω .

on

We observe that the existence of a solution belonging to the nonreflexive space 1,1

W0 (Ω) is quite unusual for an elliptic problem. However, also the Dirichlet problem ⎧ ⎨−div(|∇u|p−2 ∇u) = f , ⎩u = 0 ,

with 1 < p < 2 − solution u ∈

1 N,

in Ω , on

and f belonging to Lm (Ω), m =

1,1 W0 (Ω),

∂Ω , N (p−1)N+1 ,

has a distributional

as proved in [14].

Theorem 14.14. Let f be a L2 (Ω) function. Then there exists a distributional solution 1,1 u in W0 (Ω) ∩ L2 (Ω) of (14.4.4), that is, 1,∞ a(x, u)∇u · ∇ϕ + u ϕ = f ϕ , ∀ ϕ ∈ W0 (Ω) . (14.4.5) Ω

Ω

Ω

If fn = Tn (f ), by Theorem 14.12, there exists a bounded H01 (Ω) solution to the following problems: ⎧ ⎨−div(a(x, un )∇un ) + un = fn , in Ω , (14.4.6) ⎩u = 0 , on ∂Ω . n

We are going to use the previous approximating problems in order to prove Theorem 14.14. In the following, C will denote a positive constant depending on meas(Ω). Proof. Step I: We prove some a priori estimates on un . Let k ≥ 0, i > 0, and let ψi,k (s) be the function defined by ⎧ ⎪ ⎪ ⎪0 , ⎪ ⎪ ⎪ ⎨i(s − k) , ψi,k (s) = ⎪ ⎪1 , ⎪ ⎪ ⎪ ⎪ ⎩ψ (s) = −ψ (−s) , i,k i,k

Note that

⎧ ⎪ ⎪ 1, ⎪ ⎨ lim ψi,k (s) = 0 , ⎪ i→+∞ ⎪ ⎪ ⎩−1 ,

if 0 ≤ s ≤ k , if k < s ≤ k + if s > k + if s < 0.

if s > k , if |s| ≤ k , if s < −k.

1 i

,

1 i

,

180

A problem with degenerate coercivity

We will choose |un | ψi,k (un ) as a test function in (14.4.6); such a test function is admissible since un belongs to H01 (Ω) ∩ L∞ (Ω) and ψi,k (0) = 0. We obtain, by assumption (14.1.1) |∇un |2 α |ψ (u )| + u |u |ψ (u ) ≤ fn |un |ψi,k (un ) , i,k n n n i,k n (1 + |un |)2 Ω

since

 ψi,k (s)

α Ω

Ω

Ω

≥ 0. Using that |fn | ≤ |f |, we have

|∇un |2 |ψi,k (un )| + (1 + |un |)2



un |un |ψi,k (un ) ≤

Ω

|f ||un ||ψi,k (un )| . Ω

Letting i tend to infinity, we thus get, by Fatou’s lemma on the left-hand side and by Lebesgue’s theorem on the right-hand side, |∇un |2 2 α + |u | ≤ |f | |un | . (14.4.7) n (1 + |un |)2 Ak

Ak

Ak

Dropping the nonnegative first term in (14.4.7), and using Hölder’s inequality on the right-hand side we obtain

⎡ ⎢ |un | ≤ ⎣



2

Ak

⎤1 ⎡ 2

2⎥

⎢ |f | ⎦ ⎣

Ak

⎤1



2

2⎥

|un | ⎦ .

Ak

Simplifying equal terms we thus have |un |2 ≤ |f |2 . Ak

(14.4.8)

Ak

The previous inequality for k = 0 reads |un |2 ≤ |f |2 , Ω

(14.4.9)

Ω

so that un is bounded in L2 (Ω). This fact implies in particular that lim meas({|un | ≥ k}) = 0 ,

k→+∞

uniformly with respect to n.

(14.4.10)

From (14.4.7), written for k = 0, dropping the nonnegative second term, we have |∇un |2 α ≤ |f | |un | . (1 + |un |)2 Ω

Ω

Hölder’s inequality on the right-hand side then gives α Ω

|∇un | (1 + |un |)2 2

⎡ ⎤1 ⎡ ⎤1 2 2 ⎢ 2⎥ ⎢ 2⎥ ≤ ⎣ |f | ⎦ ⎣ |un | ⎦ , Ω

Ω

The regularizing effects of a lower order term

so that, by (14.4.9) we infer that α Ω

|∇un |2 ≤ (1 + |un |)2

181

|f |2 .

(14.4.11)

Ω

Step II: We prove that, up to a subsequence, the sequence un strongly converges 1,1 in L2 (Ω) to some function u belonging to W0 (Ω) ∩ L2 (Ω). Starting from (14.4.11), we deduce that vn = log(1 + |un |)sgn(un ) is bounded in H01 (Ω). Therefore, up to subsequences, it converges to some function v weakly in H01 (Ω), strongly in L2 (Ω), and a.e. in Ω. If we define u = [e|v| − 1]sgn(v), then un converges a.e. to u in Ω. Let now E be a measurable subset of Ω; then 2 2 2 |un | ≤ |un | + |un | ≤ |f |2 + k2 meas(E) , E∩{|un |≥k}

E

E∩{|un | 0 there exists a subset Eδ of Ω, with meas(Eδ ) < δ, such that lim a(x, un )∇ϕ = a(x, u)∇ϕ

n→+∞

uniformly in Ω \ Eδ .

(14.4.12)

We now have          a(x, un )∇un · ∇ϕ − a(x, u)∇u · ∇ϕ    Ω Ω           a(x, un )∇un · ∇ϕ − a(x, u)∇u · ∇ϕ + β |∇ϕ||∇un | . ≤    Ω\E E δ

δ

Using the equiintegrability of |∇un | proved above, and the fact that |∇u| belongs to L1 (Ω), we can choose δ such that the second term of the right-hand side is arbitrarily small, uniformly with respect to n. We then use (14.4.12) to choose n large enough so that the first term is arbitrarily small. We are going to prove the uniqueness of the solution obtained by approximation. Let f ∈ L2 (Ω), let fn be a sequence of L∞ (Ω) functions converging to f in L2 (Ω), with |fn | ≤ |f |, and let un be a solution of (14.4.6). We have just proved the exis1,1 tence of a distributional solution u in W0 (Ω) ∩ L2 (Ω) to (14.4.4), such that, up to a subsequence, lim un − uW 1,1 (Ω)∩L2 (Ω) = 0 . (14.4.13) n→+∞

0

Now, let g ∈ L2 (Ω), let gn be a sequence of L∞ (Ω) functions converging to g in θ+2 L 2 (Ω), with |gn | ≤ |g|, and let zn in H01 (Ω) ∩ L∞ (Ω) be a weak solution of ⎧ ⎨−div(a(x, un )∇un ) + zn = gn , in Ω , (14.4.14) ⎩z = 0 , on ∂Ω. n

Then, up to a subsequence, we can assume that lim zn − zW 1,1 (Ω)∩L2 (Ω) = 0 ,

n→+∞

0

(14.4.15)

1,1

where z in W0 (Ω) ∩ L2 (Ω) is a distributional solution of ⎧ ⎨−div(a(x, z)∇z) + z = f , in Ω , ⎩z = 0 , on ∂Ω.

(14.4.16)

Our result, which will imply the uniqueness of the solution by approximation of (14.1.2), is the following.

The regularizing effects of a lower order term

183

Theorem 14.15. that a = a(x, s) is differentiable with respect to the second  Assume     variable, and  ∂a , for some positive constant C . Assume that un and zn are ≤ C ∂s solutions of (14.4.6) and (14.4.14), respectively, and that (14.4.13) and (14.4.15) hold true, with u and z solutions of (14.1.2) and (14.4.16) respectively. Then |u − z| ≤ |f − g| . (14.4.17) Ω

Ω

Moreover, f ≤ g a.e. in Ω

implies

u ≤ z a.e. in Ω.

(14.4.18)

Proof. Subtracting (14.4.14) from (14.4.6) we obtain −div(a(x, un )∇un − a(x, zn )∇zn ) + un − zn = fn − gn .

Choosing Th (un − zn ) as a test function we have

[a(x, un )∇un − a(x, zn )∇zn ] · ∇Th (un − zn ) + (un − zn )Th (un − zn )

Ω



Ω

(fn − gn )Th (un − zn ) .

= Ω

This equality can be written in an equivalent way as

a(x, un )[∇un − ∇zn ] · ∇Th (un − zn ) + (un − zn )Th (un − zn )

Ω

Ω





(fn − gn )Th (un − zn ) − [a(x, un ) − a(x, zn )]∇zn · ∇Th (un − zn ) .

= Ω

Ω

The first term of the left-hand side is nonnegative, so that it can be dropped; using Lagrange’s theorem on the last term of the right-hand side, we therefore have, since the absolute value of the derivative of a with respect to the second variable is bounded, (un − zn )Th (un − zn ) ≤ (fn − gn )Th (un − zn ) + Ch |∇zn ||∇Th (un − zn )| . Ω

Ω

Ω

Dividing by h we obtain Th (un − zn ) |Th (un − zn )| ≤ |fn −gn | +C |∇zn ||∇Th (un −zn )| . (un −zn ) h h

Ω

Ω

Ω

Since, for every fixed n, un and zn belong to H01 (Ω), the limit as h tends to zero gives |un − zn | ≤ |fn − gn | , Ω

Ω

184

A problem with degenerate coercivity

which then yields (14.4.17) passing to the limit. The use of Th (un − zn )+ as a test function and the same technique as above imply that (un − zn )+ ≤

Ω

(fn − gn ) . {un ≥zn }

Hence, passing to the limit as n tends to infinity, we obtain, if we suppose that f ≤ g a.e. in Ω, (u − z)+ ≤

Ω

(f − g) ≤ 0 , {u≥z}

so that (14.4.18) is proved. Due to (14.4.17), we can prove that problem (14.1.2) has a unique solution obtained by approximation. Corollary 14.16. There exists a unique solution obtained by approximation of (14.1.2), 1,1 in the sense that the solution u in W0 (Ω)∩L2 (Ω) obtained as limit of the sequence un of solutions of (14.4.6) does not depend on the sequence fn chosen to approximate the datum f in L2 (Ω). Our last result is a nonexistence result for solutions of (14.1.2) in the case where the source is a Dirac mass. Theorem 14.17. Let μ denote a Dirac mass concentrated on a point of Ω. Then there is no solution to

⎧ ∇u ⎪ ⎨−div + u = μ , in Ω , (1 + |u|)2 ⎪ ⎩u = 0 , on ∂Ω . More precisely, if fn is a sequence of nonnegative L∞ (Ω) functions which converges to μ in the tight sense of measures, and if un is the sequence of solutions to (14.4.6), then un tends to zero a.e. in Ω and 1,∞ lim un ϕ = ϕ dμ , ∀ ϕ ∈ W0 (Ω) . n→+∞

Ω

Ω

Remark 14.18. More in general the previous result can be proved for bounded Radon measures concentrated on a set of zero harmonic capacity (see [8]). Proof. For every ε > 0 there exists a function ψε in C0∞ (Ω) such that 2 0 ≤ ψε ≤ 1 , |∇ψε | ≤ ε , (1 − ψε )dμ ≤ ε . Ω

Ω

Note that, as a consequence of the estimate on ψε in H01 (Ω), and of the fact that 0 ≤ ψε ≤ 1, ψε tends to zero in the weak∗ topology of L∞ (Ω) as ε tends to zero.

The regularizing effects of a lower order term

185

If fn is a sequence of nonnegative functions which converges to μ in the tight convergence of measures, then 0 ≤ lim fn (1 − ψε ) = (1 − ψε ) dμ ≤ ε . (14.4.19) n→+∞

Ω

Ω

Let un be the nonnegative solution to the approximating problem (14.4.6). If we choose 1 − (1 + un )−1 as a test function in (14.4.6), we have, dropping the nonnegative lower order term,    ∇un 2    (1 + u )2  ≤ fn . n Ω

Ω

Therefore, up to a subsequence, there exist σ in (L2 (Ω))N and ρ in L2 (Ω) such that    ∇un  ∇un   = ρ, lim = σ , lim (14.4.20)   n→+∞ (1 + un )2 n→+∞ (1 + un )2 weakly in (L2 (Ω))N and L2 (Ω) respectively. The choice of [1 − (1 + un )−1 ](1 − ψε ) as a test function in (14.4.6) gives |∇un |2 (1 − ψ ) + un [1 − (1 + un )−1 ](1 − ψε ) ε (1 + un )4 Ω Ω ∇un · ∇ψε [1 − (1 + un )−1 ] = fn [1 − (1 + un )−1 ](1 − ψε ) + (1 + un )2 Ω Ω ∇un · ∇ψε [1 − (1 + un )−1 ] . (14.4.21) ≤ fn (1 − ψε ) + (1 + un )2 Ω

Ω

We study the right-hand side. For the first term, (14.4.19) implies that lim+ lim fn (1 − ψε ) = 0 , ε→0

n→+∞

Ω

while for the second one, we have, using (14.4.20), and the boundedness of [1 − (1 + un )−1 ], ∇un · ∇ψε −1 lim [1 − (1 + u ) ] = σ · ∇ψε [1 − (1 + u)−1 ] . n n→+∞ (1 + un )2 Ω

Ω

Recalling that σ belongs to (L2 (Ω))N , that ψε tends to zero in H01 (Ω), and using the boundedness [1 − (1 + u)−1 ], we have ∇un · ∇ψε lim+ lim [1 − (1 + un )−1 ] = 0 . ε→0 n→+∞ (1 + un )2 Ω

Therefore, since both terms of the left-hand side of (14.4.21) are nonnegative, we obtain |∇un |2 lim+ lim (1 − ψε ) = 0 . ε→0 n→+∞ (1 + un )4 Ω

186

A problem with degenerate coercivity



The functional v ∈ L2 (Ω) →

|v|2 (1 − ψε ) Ω

is weakly lower semicontinuous on L2 (Ω); this implies that

|ρ|2 = lim+

Ω

ε→0

   ∇un 2   (1 − ψε ) = 0 , n→+∞  (1 + u )2 

|ρ|2 (1 − ψε ) ≤ lim+ lim ε→0

Ω

Ω

n

that is, ρ = 0. Thus, since

∇un −1 , = ∇ 1 − (1 + u ) n (1 + un )2

by (14.4.20) the sequence 1 − (1 + un )−1 weakly converges to zero in H01 (Ω), and so, up to a subsequence, it strongly converges to zero in L2 (Ω). Therefore un tends to zero a.e. in Ω. Since the limit does not depend on the subsequence, the whole sequence un tends to zero a.e. in Ω. For Φ in (L2 (Ω))N , by (14.4.20), one has         ∇un  ≤ ρ |Φ| = 0 ,  σ · Φ  = lim  · Φ  n→+∞  (1 + |u |)2   n Ω

Ω

Ω

which implies that σ = 0. Therefore, passing to the limit in (14.4.6), that is, in ∇un · ∇ϕ 1,∞ + u ϕ = fn ϕ , ϕ ∈ W0 (Ω) , n (1 + un )2 Ω

Ω

Ω

we get, since the first term tends to zero, lim un ϕ = ϕ dμ , n→+∞

Ω 1,∞

for every ϕ ∈ W0

(Ω), as desired.

Ω

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Index A approximation 29, 106, 115, 123, 140, 151, 154, 155, 167, 171, 182, 184

first eigenvalue 69, 76, 79, 88 fixed point 5, 6, 9, 10, 12 Fredholm alternative 75, 83

B Banach–Caccioppoli theorem 5, 6 Beppo Levi’s theorem 23 bootstrapping technique 73 Brouwer’s theorem 5, 6, 10, 12, 38

G Gâteaux differentiable 86, 97, 102

C Calculus of Variations 84 Carathéodory function 16 coercive form 34 coercive functional 102 coercive operator 38 compact operator 35, 82 completely continuous map 10 continuous form 34 contraction 5, 25, 26, 77, 78 D De Giorgi’s theorem 84 degenerate coercivity 166 direct methods 84 distributional solution 121, 123, 130, 131, 135–138, 140, 152, 157, 169, 171, 177–179, 182 divergence type source 47, 58 Dunford–Pettis theorem 23 E Egorov’s theorem 23 eigenfunction 68–71, 73–77, 79, 88 eigenvalue 67–71, 74–79, 82, 83, 87 eigenvalue problem 67 Ekeland’s principle 95 elliptic 1, 24, 31, 37, 40, 61, 67, 121, 137, 144, 150, 152, 154, 157, 166, 179 ellipticity 30, 32, 33, 44, 52, 53, 56, 58, 59, 64, 68, 70, 72, 80, 81, 107, 110, 123, 125, 133, 135, 136, 141, 153, 163 entropy solution 122, 123, 132–137, 144, 146, 147, 150–152, 166 Euler equation 70, 84, 86, 87, 89, 91, 100, 105 F Fatou’s theorem 23

H H 2 regularity 61 Hölder’s inequality 22

I incremental rate 61 integral functional 84, 86, 90, 101, 122, 143 interpolation inequality 23 L Laplacian 27, 88 Lax–Milgram theorem 24, 25, 27 Lebesgue space 22 Lebesgue’s theorem 23 Leray–Lions problem 1, 37, 47, 89, 91, 121 Leray–Lions theorem 40, 42, 45, 58, 105, 106, 131, 153, 166 linear equation 27, 61, 130 lower order term 45, 53, 105, 122, 140, 166, 175, 176 M Marcinkiewicz space 13, 18, 47, 56 maximum principle 32, 111 measure 148 measure source 138, 154 monotone operator 150, 152, 153 mountain pass theorem 100 N natural growth 105 Nemitski’s theorem 13, 17 nonlinear equation 37 P Palais–Smale condition 98–101 Poincaré’s inequality 36, 69 polynomial growth 157 projection theorem 35 pseudomonotone operator 38, 40, 42, 44, 45

192

Index

R Rellich–Kondrachov theorem 35 Riesz theorem 35

S Schauder’s theorem 5, 10, 12, 89, 173 self-adjoint operator 82 semilinear equation 31, 67, 74, 122 semilinear monotone equation 28 Sobolev inequality 30, 35 Sobolev space 34, 35 spectral analysis 67 spectral theorem 68, 83 spectral theory 82 spectrum 82 Stampacchia’s theorem 24, 26 subsolution 31, 32, 34, 76 supersolution 31, 32, 34, 76 surjectivity theorem 37, 38, 40

T test function 29, 32, 52–54, 56–59, 71, 77, 89, 90, 99–101, 107, 108, 110, 112, 113, 115, 116, 118–120, 123, 124, 128, 129, 131, 136, 138, 141, 142, 151–153, 156, 158–161, 163, 164, 180, 183–185 T -minimum 91, 122, 143–147 truncated function 29 U unique solution 27–29, 75, 151, 152, 167, 184 uniqueness 5, 150–152, 154, 182 V Vitali’s theorem 14 W weak solution 27, 167–169, 173, 182 weakly lower semicontinuous functional 102 Weierstrass theorem 84

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