Fractional Partial Differential Equations [1 ed.] 9789811290404, 9789811290411, 9789811290428

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Preface

The fractional calculus started more than three centuries ago. In the last years, the fractional calculus is playing a very important role in various scientific fields. In fact, it has been recognized as one of the best tools to describe long-memory processes. Fractional-order models are interesting not only for engineers and physicists, but also for mathematicians. Among such models those described by partial differential equations (PDEs) containing fractional derivatives are of utmost importance. Their evolution was more complex than for the classical integer-order counterpart. Nonetheless, classical PDEs’ methods are hardly applicable directly to fractional PDEs. Therefore, new theories and methods are required, with concepts and algorithms specifically developed for fractional PDEs. In the recent years, the theory of fractional PDEs has been highly developed and constitutes an important branch of differential equations. This monograph gives a presentation of the theory for time-fractional partial differential equations. Many of the basic techniques and results recently developed about this theory are presented, including the well-posedness, regularity, approximation and optimal control. The contents of the book cover fractional Navier-Stokes equations, fractional Rayleigh-Stokes equations, fractional Fokker-Planck equations and fractional Schr¨ odinger equations. Several examples of applications relating to these equations are also discussed in detail. The materials in this monograph are based on the research work carried out by the author and some experts during the past five years. It provides the necessary background material required to go further into the subject and explore the rich research literature. It is useful for researchers, graduate or PhD students dealing with differential equations, applied analysis, and related areas of research. I would like to thank Professors B. Ahmad, M. Feˇckan, P. G´orka, M. Kirane, V. Kiryakova, K.N. Le, W. McLean, S.K. Ntouyas, G.M. N’Guerekata, H. Prado, M. Stynes, J.J. Trujillo and N.H. Tuan for their support. I also wish to express my appreciation to doctoral students J.W. He, L. Peng and J.N. Wang for their help. Finally, I thank the editorial assistance of World Scientific Publishing Co., especially Ms. L.F. Kwong.

v

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Fractional Partial Differential Equations

I acknowledge with gratitude the support of National Natural Science Foundation of China (12071396) and the Macau Science and Technology Development Fund (0092/2022/A). Yong Zhou November 2023, Macau, China

About the Author

Dr.Yong Zhou is a Full Professor at Xiangtan University and a Distinguished Guest Professor at Macau University of Science and Technology. His research fields include fractional differential equations, functional differential equations, evolution equations and control theory. Zhou has published seven monographs in Springer, Elsevier, De Gruyter, World Scientific and Science Press respectively, and more than three hundred research papers in international journals including Mathematische Annalen, Journal of Functional Analysis, Nonlinearity, Inverse Problems, Proceedings of the Royal Society of Edinburgh, Bulletin des Sciences Math´ematiques, Comptes rendus Mathematique, International Journal of Bifurcation and Chaos, Zeitschrift f¨ ur Angewandte Mathematik und Physik, and so on. He was included in Highly Cited Researchers list from Thompson Reuters (2014) and Clarivate Analytics (2015–2021). Zhou has undertaken five projects from National Natural Science Foundation of China, and two projects from the Macau Science and Technology Development Fund. He won the second prize of Chinese University Natural Science Award in 2000, and the second prize of Natural Science Award of Hunan Province, China in 2017 and 2021. Zhou had worked as an Associate Editor for IEEE Transactions on Fuzzy Systems, Journal of Applied Mathematics & Computing, Mathematical Inequalities & Applications, and an Editorial Board Member of Fractional Calculus and Applied Analysis.

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Introduction

Fractional calculus has been attracting the attention of mathematicians and engineers from long time ago. The concept of fractional (or, more precisely, noninteger) differentiation appeared for the first time in a famous correspondence between L’Hospital and Leibniz, in 1695. Many mathematicians have further developed this area and we can mention the studies of Euler, Laplace, Abel, Liouville and Riemann. However, the fractional calculus remained for centuries a purely theoretical topic, with little if any connections to practical problems of physics and engineering. In the past thirty years, the fractional calculus has been recognized as an effective modelling methodology for researchers. Fractional differential equations are generalizations of classical differential equations to an arbitrary (noninteger) order. Based on the wide applications in engineering and sciences such as physics, mechanics, electricity, chemistry, biology, economics, and many others, research on fractional differential equations is active and extensive around the world. In the recent years, there has been a significant development in ordinary and partial differential equations involving fractional derivatives, see the monographs of Miller et al. [268], Podlubny [303], Hilfer [171], Kilbas et al. [197], Diethelm [101], Zhou [386, 387] and the references therein. A strong motivation for investigating this class of equations comes mainly from a compelling reasons: the fractional order models of real systems are often more adequate than the classical integer order models, since the description of some systems is more accurate when the fractional derivative is used. As an example, considering anomalous diffusion phenomena, the classical diffusion equation fails to describe diffusion phenomenon in heterogeneous porous media that exhibits fractal characteristics. How is the classical diffusion equation modified to make it appropriate to depict anomalous diffusion phenomena? This problem is interesting for researchers. Fractional calculus has been found effective in modelling anomalous diffusion processes since it has been recognized as one of the best tools to characterize the long memory processes. Consequently, it is reasonable and significative to propose the generalized diffusion equations with fractional derivative operators, which can be used to simulate anomalous diffusion in fractal media. Its evolutionary behave in a much more complex way than in classical integer order case and the corresponding investigation becomes more challenging. ix

x

Fractional Partial Differential Equations

Such models are interesting not only for engineers and physicists but also for pure mathematicians. This monograph gives a presentation of the theory for time-fractional partial differential equations. The contents of the book cover fractional Navier-Stokes equations, fractional Rayleigh-Stokes equations, fractional Fokker-Planck equations and fractional Schr¨ odinger equations. Many of the basic techniques and results recently developed about this theory are presented, including well-posedness, regularity, approximation and optimal control for fractional partial differential equations. Several examples of applications relating to these equations are also discussed in detail. Such basic theory should be the starting point for further research concerning the dynamics, numerical analysis and applications of fractional partial differential equations. This monograph is arranged and organized as follows. In order to make the book self-contained, we devote the first chapter to a description of general information on fractional calculus, Mittag-Leffler functions, integral transforms and semigroups. The second chapter deals with the Navier-Stokes equations with time-fractional derivative of order α ∈ (0, 1). The second section is concerned with the existence and uniqueness of local and global mild solutions. Meanwhile, we also give local mild solutions in Cσ∞ (Ω)

|·|q

. Moreover, we prove the existence and regularity of |·|q

classical solutions for such equations in Cσ∞ (Ω) . The third section is devoted to investigation of existence, uniqueness and H¨older continuity of the local mild solutions. In the fourth section, we obtain the existence and uniqueness of the solutions to approximate equations, as well as the convergence of the approximate solutions. Furthermore, we present some convergence results for the Faedo-Galerkin approximations of the given problems. In the fifth section, we firstly give the concept of the weak solutions and establish the existence criterion of weak solutions by means of the Galerkin approximations in the case that the dimension n ≤ 4. Moreover, a complete proof of the uniqueness is given when n = 2. At last we give a sufficient condition of optimal control pairs. In the sixth section, we make use of energy methods to study the time-fractional Navier-Stokes equations. In the first part, we construct a regularized equation by using a smoothing process to transform unbounded differential operators into bounded operators and then obtain the approximate solutions. The second part describes a procedure to take a limit in the approximation program to present a global solution to the objective equation. In the final section of this chapter, we use the tools from harmonic analysis to study the Cauchy problem for time-fractional Navier-Stokes equations. Two results concerning the local existence of solutions for the given problem in Sobolev spaces are addressed. In the third chapter, we are interested in discussing the nonlinear RayleighStokes problem for a generalized second grade fluid with the Riemann-Liouville fractional derivative. In the first section, we firstly show that the solution operator

Introduction

xi

of the problem is compact, and continuous in the uniform operator topology. Furthermore, we give an existence result of mild solutions for the nonlinear problem. In the second section, we are devoted to the global/local well-posedness of mild solutions for a semilinear time-fractional Rayleigh-Stokes problem on RN . We are concerned with, the approaches rely on the Gagliardo-Nirenberg’s inequalities, operator theory, standard fixed point technique and harmonic analysis methods. We also present several results on the continuation, a blow-up alternative with the blow-up rate and the integrability in Lebesgue spaces. In the third section, we consider the fractional Rayleigh-Stokes problem with the nonlinearity term satisfies certain critical conditions. The local existence, uniqueness and continuous dependence upon the initial data of -regular mild solutions are obtained. Furthermore, a unique continuation result and a blow-up alternative result of -regular mild solutions are given. In the fourth section, we are devoted to the study of the existence, uniqueness and regT ularity of weak solutions in L∞ (0, b; L2 (Ω)) L2 (0, b; H01 (Ω)) of the Rayleigh-Stokes problem. Furthermore, we prove an improved regularity result of weak solutions. In the fifth section, we prove a long time existence result for fractional Rayleigh-Stokes equations. More precisely, we discuss the existence, uniqueness, continuous dependence on initial value and asymptotic behavior of global solutions in Besov-Morrey spaces. The results are formulated that allows for a larger class in initial value than the previous works and the approach is also suitable for fractional diffusion cases. In the final section, we study a fractional nonlinear Rayleigh-Stokes problem with final value condition. By means of the finite dimensional approximation, we first obtain the compactness of solution operators. Moreover, we handle the problem in weighted continuous function spaces, and then the existence result of solutions is established. Finally, because of the ill-posedness of backward problem in the sense of Hadamard, the quasi-boundary value method is utilized to get the regularized solutions, and the corresponding convergence rate is obtained. In the fourth chapter, we study the time-fractional Fokker-Planck equations which can be used to describe the subdiffusion in an external time and spacedependent force field F (t, x). In the first section, we present some results on existence and uniqueness of mild solutions allowing the “working space” that may have low regularity. Secondly, we analyze the relationship between “working space” and the value range of α when investigating the problem of classical solutions. Finally, by constructing a suitable weighted H¨ older continuous function space, the existence of classical solutions is derived without the restriction on α ∈ ( 21 , 1). In the second section, a time-fractional Fokker–Planck initial-boundary value problem is considered, the spatial domain Ω ⊂ Rd , where d ≥ 1, has a smooth boundary. Existence, uniqueness and regularity of a mild solution u are proved under the hypothesis that the initial data u0 lies in L2 (Ω). For 1/2 < α < 1 and u0 ∈ H 2 (Ω) ∩ H01 (Ω), it is shown that u becomes a classical solution of the problem. Estimates of time derivatives of the classical solution are derived. The final chapter is concerned with time-fractional Schr¨odinger equation. In the first section of this chapter, we study the linear fractional Schr¨odinger equation

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Fractional Partial Differential Equations

on a Hilbert space, with a time-fractional derivative of order 0 < α < 1, and a self-adjoint generator A. We show that A generates a family of bounded operators {Uα (t)}t≥0 which are defined by the functional calculus of A via the Mittag-Leffler function when evaluated at A. Using the spectral theorem we prove existence and uniqueness of strong solutions. Moreover, we prove that the solution family Uα (t) converges strongly to the family of unitary operators e−itA , as α approaches to 1. In the second section of this chapter, we apply the tools of harmonic analysis to study the Cauchy problem for nonlinear fractional Schr¨odinger equation. Some fundamental properties of two solution operators are estimated. The existence and a sharp decay estimate for solutions of the given problem in two different spaces are addressed. The materials in this monograph are based on the research work carried out by the author and some experts during the past five years. The contents in Section 2.3 are taken from Zhou, Peng and Huang [404]. The material in Section 2.4 is taken from Peng, Debbouche and Zhou [293]. The results in Section 2.5 are adopted from Zhou and Peng [403]. The contents of Section 2.6 due to Zhou and Peng [406]. Section 2.7 is taken from Zhou, Peng, Ahmad et al. [298]. The results in Section 3.1 due to Zhou and Wang [409]. The main results in Section 3.2 are adopted from He, Zhou and Peng [163]. Section 3.3 is taken from Wang, Alsaedi, Ahmad and Zhou [346]. The contents in Section 3.4 are taken from Wang, Zhou, Alsaedi and Ahmad [348]. The results in Section 3.5 due to Peng and Zhou [296]. Section 3.6 is from Wang, Zhou and He [347]. The contents in Section 4.1 are taken from Peng and Zhou [294]. The results of Section 4.2 are due to Le, McLean and Stynes [219]. The results in Section 5.1 due to G´ orka, Prado and Trujillo [152]. The contents in Section 5.2 are taken from Peng, Zhou and Ahmad [299].

Contents

Preface

v

About the Author

vii

Introduction

ix

1.

Preliminaries 1.1

1.2

1.3

2.

1

Fractional Calculus . . . . . . . . . . . . 1.1.1 Definitions . . . . . . . . . . . . 1.1.2 Properties . . . . . . . . . . . . Some Results from Analysis . . . . . . . 1.2.1 Mittag-Leffler Function . . . . . 1.2.2 Laplace and Fourier Transforms Semigroups . . . . . . . . . . . . . . . . 1.3.1 C0 -Semigroup . . . . . . . . . . 1.3.2 Analytic Semigroup . . . . . . .

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Fractional Navier-Stokes Equations 2.1 2.2

2.3

2.4

1 1 7 9 9 11 12 12 14 15

Introduction . . . . . . . . . . . . Cauchy Problem in Rn . . . . . . 2.2.1 Definitions and Lemmas . 2.2.2 Global/Local Existence . 2.2.3 Local Existence . . . . . 2.2.4 Regularity . . . . . . . . Existence and H¨ older Continuity 2.3.1 Notations and Lemmas . 2.3.2 Existence and Uniqueness 2.3.3 H¨ older Continuous . . . . Approximations of Solutions . . . 2.4.1 Preliminaries . . . . . . . xiii

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15 17 18 21 29 32 40 41 42 47 48 49

xiv

Fractional Partial Differential Equations

2.5

2.6

2.7

2.8 3.

2.4.2 Existence and Convergence of Approximate 2.4.3 Faedo-Galerkin Approximation . . . . . . . Weak Solution and Optimal Control . . . . . . . . 2.5.1 Notations and Definitions . . . . . . . . . . 2.5.2 Existence and Uniqueness . . . . . . . . . . 2.5.3 Optimal Control . . . . . . . . . . . . . . . Energy Methods . . . . . . . . . . . . . . . . . . . 2.6.1 Notations and Lemmas . . . . . . . . . . . 2.6.2 Local Existence . . . . . . . . . . . . . . . 2.6.3 Global Existence . . . . . . . . . . . . . . . Cauchy Problem in Sobolev Space . . . . . . . . . 2.7.1 Definitions and Lemmas . . . . . . . . . . . 2.7.2 Local Existence I . . . . . . . . . . . . . . 2.7.3 Local Existence II . . . . . . . . . . . . . . Notes and Remarks . . . . . . . . . . . . . . . . . .

Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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. 51 . 60 . 62 . 62 . 66 . 73 . 76 . 76 . 78 . 80 . 91 . 91 . 96 . 103 . 113

Fractional Rayleigh-Stokes Equations 3.1

3.2

3.3

3.4

3.5

3.6

Initial Value Problem . . . . . . . . . . . . . . . 3.1.1 Introduction . . . . . . . . . . . . . . . 3.1.2 Existence . . . . . . . . . . . . . . . . . Well-posedness on RN . . . . . . . . . . . . . . 3.2.1 Introduction . . . . . . . . . . . . . . . 3.2.2 Preliminaries . . . . . . . . . . . . . . . 3.2.3 Well-posedness . . . . . . . . . . . . . . 3.2.4 Integrability of Solution . . . . . . . . . Well-posedness and Blow-up . . . . . . . . . . . 3.3.1 Introduction . . . . . . . . . . . . . . . 3.3.2 Preliminaries . . . . . . . . . . . . . . . 3.3.3 Existence of Mild Solutions . . . . . . . 3.3.4 Continuation and Blow-up Alternative . Weak Solutions . . . . . . . . . . . . . . . . . . 3.4.1 Introduction . . . . . . . . . . . . . . . 3.4.2 Preliminaries . . . . . . . . . . . . . . . 3.4.3 Existence, Uniqueness and Regularity . Global Solutions in Besov-Morrey Spaces . . . . 3.5.1 Introduction . . . . . . . . . . . . . . . 3.5.2 Notations and Function Spaces . . . . . 3.5.3 Properties of Solution Operators . . . . 3.5.4 Well-posedness of Global Solutions . . . Final Value Problem . . . . . . . . . . . . . . . 3.6.1 Introduction . . . . . . . . . . . . . . . 3.6.2 Preliminaries . . . . . . . . . . . . . . .

115 . . . . . . . . . . . . . . . . . . . . . . . . .

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116 116 118 124 124 126 131 140 142 142 143 145 154 160 160 161 163 173 173 174 176 182 188 188 189

Contents

3.7 4.

3.6.3 Existence of Solutions . . . . . . . . . . . . . . . . . . . . . 190 3.6.4 Quasi-boundary Value Method . . . . . . . . . . . . . . . . 199 Notes and Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

Fractional Fokker-Planck Equations 4.1

4.2

4.3 5.

xv

205

Operator Method . . . . . . . . . . . . . . . . . . . . . 4.1.1 Introduction . . . . . . . . . . . . . . . . . . . 4.1.2 Preliminaries . . . . . . . . . . . . . . . . . . . 4.1.3 The Solution Operators . . . . . . . . . . . . . 4.1.4 Mild Solutions . . . . . . . . . . . . . . . . . . 4.1.5 Classical Solutions in Case of α ∈ ( 21 , 1) . . . . 4.1.6 Classical Solutions in Case of α ∈ (0, 1) . . . . Galerkin Approximation Method . . . . . . . . . . . . 4.2.1 Introduction . . . . . . . . . . . . . . . . . . . 4.2.2 Notation and Definitions . . . . . . . . . . . . 4.2.3 Technical Preliminaries . . . . . . . . . . . . . 4.2.4 Galerkin Approximation of Solution . . . . . . 4.2.5 Existence and Uniqueness of Mild Solution . . 4.2.6 Existence and Uniqueness of Classical Solution 4.2.7 Regularity of Classical Solution . . . . . . . . Notes and Remarks . . . . . . . . . . . . . . . . . . . .

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Fractional Schr¨ odinger Equations 5.1

5.2

5.3

Linear Equations on Hilbert Space . . . . . 5.1.1 Introduction . . . . . . . . . . . . . 5.1.2 Preliminaries . . . . . . . . . . . . . 5.1.3 Existence of Dynamics . . . . . . . 5.1.4 Properties of the Solution Operator 5.1.5 An Example . . . . . . . . . . . . . Nonlinear Schr¨ odinger Equation . . . . . . . 5.2.1 Introduction . . . . . . . . . . . . . 5.2.2 Solution Operators . . . . . . . . . . 5.2.3 Existence . . . . . . . . . . . . . . . Notes and Remarks . . . . . . . . . . . . . .

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Bibliography

279

Index

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Chapter 1

Preliminaries

In this chapter, we introduce some notations and basic facts on fractional calculus, Mittag-Leffler functions, the Wright-type function, integral transforms and semigroups, which are needed throughout this monograph.

1.1 1.1.1

Fractional Calculus Definitions

A number of definitions for the fractional derivative has emerged over the years, we refer the reader to Diethelm [101], Hilfer [171], Kilbas, Srivastava and Trujillo [197], Miller and Ross [268], Podlubny [303]. In this book, we restrict our attention to the use of the Riemann-Liouville and Caputo fractional derivatives. In this section, we introduce some basic definitions and properties of the fractional integrals and fractional derivatives which are used further in this book. The materials in this section are taken from [197]. As usual Z denotes the set of integer numbers, N+ denotes the set of positive integer numbers and N0 denotes the set of nonnegative integer numbers. R denotes the set of real numbers, R+ denotes the set of nonnegative reals and R+ the set of positive reals, R− denotes the set of nonpositive reals. Let C be the set of complex numbers. Let J = [a, b] (−∞ < a < b < ∞) be a finite interval of R. We assume that X is a Banach space with the norm | · |. Denote C(J, X) be the Banach space of all continuous functions from J into X with the norm kxk = supt∈J |x(t)|, where x ∈ C(J, X). C n (J, X) (n ∈ N0 ) denotes the set of mappings having n times continuously differentiable on J, AC(J, X) is the space of functions which are absolutely continuous on J and AC n (J, X) (n ∈ N+ ) is the space of functions f such that f ∈ C n−1 (J, X) and f (n−1) ∈ AC(J, X). In particular, AC 1 (J, X) = AC(J, X).

1

2

Fractional Partial Differential Equations

Let 1 ≤ p ≤ ∞. Lp (J, X) denotes the Banach space of all measurable functions f : J → X. Lp (J, X) is normed by Z  p1   p  |f (t)| dt , 1 ≤ p < ∞,  J kf kLp J = n o    sup |f (t)| , p = ∞.  µ(inf ¯ J)=0 t∈J\J¯

1

In particular, L (J, X) R is the Banach space of measurable functions f : J → X with the norm kf kLJ = J |f (t)|dt, and L∞ (J, X) is the Banach space of measurable functions f : J → X which are bounded, equipped with the norm kf kL∞ J = inf{c > 0| |f (t)| ≤ c, a.e. t ∈ J}. The Gamma function Γ(z) is defined by Z ∞ Γ(z) = tz−1 e−t dt (Re(z) > 0), 0

where tz−1 = e(z−1) ln(t) . This integral is convergent for all complex z ∈ C (Re(z) > 0). For this function the reduction formula Γ(z + 1) = zΓ(z)

(Re(z) > 0)

holds. In particular, if z = n ∈ N0 , then Γ(n + 1) = n!

(n ∈ N0 )

with (as usual) 0! = 1. Let us consider some of the starting points for a discussion of fractional calculus. One development begins with a generalization of repeated integration. Thus if f is locally integrable on (c, ∞), then the n-fold iterated integral is given by Z t Z s1 Z sn−1 −n D f (t) = ds ds · · · f (sn )dsn c t 1 2 c c c Z t 1 (t − s)n−1 f (s)ds = (n − 1)! c for almost all t with −∞ ≤ c < t < ∞ and n ∈ N+ . Writing (n − 1)! = Γ(n), an immediate generalization is the integral of f of fractional order α > 0, Z t 1 −α (t − s)α−1 f (s)ds (left hand) D f (t) = c t Γ(α) c and similarly for −∞ < t < d ≤ ∞ Z d 1 −α (s − t)α−1 f (s)ds (right hand) t Dd f (t) = Γ(α) t both being defined for suitable f . Definition 1.1. (Left and right Riemann-Liouville fractional integrals) Let J = [a, b] (−∞ < a < b < ∞) be a finite interval of R. The left and right RiemannLiouville fractional integrals a Dt−α f (t) and t Db−α f (t) of order α ∈ R+ , are defined by Z t 1 −α D f (t) = (t − s)α−1 f (s)ds, t > a, α > 0 (1.1) a t Γ(α) a

Preliminaries

3

and −α t Db f (t)

=

1 Γ(α)

b

Z

(s − t)α−1 f (s)ds,

t < b, α > 0,

(1.2)

t

respectively, provided that the right-hand sides are pointwise defined on [a, b]. When α = n ∈ N+ , the definitions (1.1) and (1.2) coincide with the n-th integrals of the form Z t 1 −n (t − s)n−1 f (s)ds a Dt f (t) = (n − 1)! a and −n t Db f (t)

=

1 (n − 1)!

Z

b

(s − t)n−1 f (s)ds.

t

Definition 1.2. (Left and right Riemann-Liouville fractional derivatives ) The left and right Riemann-Liouville fractional derivatives a Dtα f (t) and t Dbα f (t) of order α ∈ R+ , are defined by α a Dt f (t)

dn −(n−α) f (t) aD dtn t Z t  dn 1 n−α−1 (t − s) f (s)ds , = Γ(n − α) dtn a =

t>a

and α t Db f (t)

dn −(n−α) f (t) tD dtn b   Z b 1 dn = (−1)n n (s − t)n−α−1 f (s)ds , Γ(n − α) dt t =(−1)n

t < b,

respectively, where n = [α] + 1, [α] means the integer part of α. In particular, when α = n ∈ N0 , then 0 a Dt f (t) n a Dt f (t)

= f (n) (t)

= t Db0 f (t) = f (t), n t Db f (t)

and

= (−1)n f (n) (t),

where f (n) (t) is the usual derivative of f (t) of order n. If 0 < α < 1, then Z t  1 d α −α (t − s) f (s)ds , t > a a Dt f (t) = Γ(1 − α) dt a and α t Db f (t)

d 1 =− Γ(1 − α) dt

Z

b −α

(s − t)

 f (s)ds ,

t < b.

t

Remark 1.1. If f ∈ C([a, b]; RN ), it is obvious that the Riemann-Liouville fractional integral of order α > 0 exists on [a, b]. On the other hand, following Lemma 2.2 in [197], we know that the Riemann-Liouville fractional derivative of order α ∈ [n − 1, n) exists almost everywhere on [a, b] if f ∈ AC n ([a, b]; RN ).

4

Fractional Partial Differential Equations

The left and right Caputo fractional derivatives are defined via above the Riemann-Liouville fractional derivatives . Definition 1.3. (Left and right Caputo fractional derivatives) The left and right α C α Caputo fractional derivatives C a Dt f (t) and t Db f (t) of order α ∈ R+ are defined by   n−1 X f (k) (a) C α α k D f (t) = D f (t) − (t − a) a t a t k! k=0

and C α t Db f (t)

=

α t Db

 f (t) −

n−1 X k=0

 f (k) (b) k (b − t) , k!

respectively, where n = [α] + 1 for α 6∈ N0 ; n = α for α ∈ N0 .

(1.3)

In particular, when 0 < α < 1, then C α a Dt f (t)

= a Dtα (f (t) − f (a))

C α t Db f (t)

= t Dbα (f (t) − f (b)).

and

The Riemann-Liouville fractional derivative and the Caputo fractional derivative are connected with each other by the following relations. Proposition 1.1. (i) If α 6∈ N0 and f (t) is a function for which the Caputo fractional derivatives C α C α + exist together with the Riemanna Dt f (t) and t Db f (t) of order α ∈ R α Liouville fractional derivatives a Dt f (t) and t Dbα f (t), then C α a Dt f (t)

= a Dtα f (t) −

n−1 X

f (k) (a) (t − a)k−α Γ(k − α + 1)

k=0

and C α t Db f (t)

= t Dbα f (t) −

n−1 X k=0

f (k) (b) (b − t)k−α , Γ(k − α + 1)

where n = [α] + 1. In particular, when 0 < α < 1, we have C α a Dt f (t)

= a Dtα f (t) −

f (a) (t − a)−α Γ(1 − α)

C α t Db f (t)

= t Dbα f (t) −

f (b) (b − t)−α . Γ(1 − α)

and

Preliminaries

5

n (ii) If α = n ∈ N0 and the usual derivative f (n) (t) of order n exists, then C a Dt f (t) C n and t Db f (t) are represented by C n a Dt f (t)

n n (n) = f (n) (t) and C (t). t Db f (t) = (−1) f

(1.4)

Proposition 1.2. Let α ∈ R+ and let n be given by (1.3). If f ∈ AC n ([a, b]; RN ), α C α then the Caputo fractional derivatives C a Dt f (t) and t Db f (t) exist almost everywhere on [a, b]. α C α (i) If α 6∈ N0 , C a Dt f (t) and t Db f (t) are represented by  Z t 1 C α n−α−1 (n) (t − s) f (s)ds a Dt f (t) = Γ(n − α) a

and C α t Db f (t)

(−1)n Γ(n − α)

=

=

 (s − t)n−α−1 f (n) (s)ds ,

t

In particular, when 0 < α < 1, f ∈

respectively, where n = [α] + 1. AC([a, b]; RN ), C α a Dt f (t)

b

Z

1 Γ(1 − α)

t

Z

 (t − s)−α f 0 (s)ds

(1.5)

a

and C α t Db f (t)

1 =− Γ(1 − α)

Z

b

 f (s)ds .

−α 0

(s − t)

(1.6)

t

C α α (ii) If α = n ∈ N0 , then C a Dt f (t) and t Db f (t) are represented by (1.4). In particular, C 0 a Dt f (t)

0 =C t Db f (t) = f (t).

Remark 1.2. If f is an abstract function with values in Banach space X, then integrals which appear in above definitions are taken in Bochner’s sense. The fractional integrals and derivatives, defined on a finite interval [a, b] of R, are naturally extended to whole axis R. Definition 1.4. (Left and right Liouville-Weyl fractional integrals on the real axis) −α The left and right Liouville-Weyl fractional integrals −∞ Dt−α f (t) and t D+∞ f (t) of order α > 0 on the whole axis R are defined by Z t 1 −α (t − s)α−1 f (s)ds (1.7) −∞ Dt f (t) = Γ(α) −∞ and −α t D+∞ f (t)

1 = Γ(α)

respectively, where t ∈ R and α > 0.

Z t

∞

(s − t)α−1 f (s)ds,

6

Fractional Partial Differential Equations

Definition 1.5. (Left and right Liouville-Weyl fractional derivatives on the real axis) The left and right Liouville-Weyl fractional derivatives −∞ Dtα f (t) and α t D+∞ f (t) of order α on the whole axis R are defined by dn −(n−α) α f (t)) −∞ Dt f (t) = n (−∞ Dt dt   Z t dn 1 n−α−1 (t − s) f (s)ds = Γ(n − α) dtn −∞ and n −(n−α) α n d (t D+∞ f (t)) t D+∞ f (t) =(−1) dtn  Z ∞ n 1 n−α−1 n d (s − t) f (s)ds , (−1) = Γ(n − α) dtn t respectively, where n = [α] + 1, α ≥ 0 and t ∈ R. In particular, when α = n ∈ N0 , then 0 0 −∞ Dt f (t) = t D+∞ f (t) = f (t), n (n) n (t) and t D+∞ f (t) = (−1)n f (n) (t), −∞ Dt f (t) = f (n) where f (t) is the usual derivative of f (t) of order n. If 0 < α < 1 and t ∈ R, then Z t  1 d α (t − s)−α f (s)ds −∞ Dt f (t) = Γ(1 − α) dt −∞ Z ∞ α f (t) − f (t − s) = ds Γ(1 − α) 0 sα+1 and Z ∞  1 d α −α (s − t) f (s)ds t D+∞ f (t) = − Γ(1 − α) dt t Z ∞ α f (t) − f (t + s) = ds. Γ(1 − α) 0 sα+1 Formulas (1.5) and (1.6) can be used for the definition of the Caputo fractional derivatives on the whole axis R. Definition 1.6. (Left and right Caputo fractional derivatives on the real axis) The C α left and right Caputo fractional derivatives −∞ Dtα f (t) and C t D+∞ f (t) of order α + (with α > 0 and α 6∈ N ) on the whole axis R are defined by Z t  1 C α n−α−1 (n) (t − s) f (s)ds (1.8) −∞ Dt f (t) = Γ(n − α) −∞ and Z ∞  (−1)n C α n−α−1 (n) D f (t) = (s − t) f (s)ds , (1.9) t +∞ Γ(n − α) t respectively. When 0 < α < 1, the relations (1.8) and (1.9) take the following forms Z t  1 C α −α 0 (t − s) f (s)ds −∞ Dt f (t) = Γ(1 − α) −∞ and Z ∞  1 C α −α 0 (s − t) f (s)ds . t D+∞ f (t) = − Γ(1 − α) t

Preliminaries

1.1.2

7

Properties

We present here some properties of the fractional integral and fractional derivative operators that will be useful throughout this book. Proposition 1.3. If β > 0, then −α a Dt (t

Γ(β) (t − a)β+α−1 (α > 0), Γ(β + α) Γ(β) = (t − a)β−α−1 (α ≥ 0) Γ(β − α)

− a)β−1 =

α a Dt (t

− a)β−1

and Γ(β) (b − t)β+α−1 (α > 0), Γ(β + α) Γ(β) α β−1 (b − t)β−α−1 (α ≥ 0). = t Db (b − t) Γ(β − α) In particular, if β = 1 and α ≥ 0, then the Riemann-Liouville fractional derivatives of a constant are, in general, not equal to zero: (b − t)−α (t − a)−α α , t Dbα 1 = . a Dt 1 = Γ(1 − α) Γ(1 − α) On the other hand, for j = 1, 2, ..., [α] + 1, −α t Db (b

− t)β−1 =

α a Dt (t

− a)α−j = 0,

α t Db (b

− t)α−j = 0.

The semigroup property of the fractional integral operators a Dt−α and t Db−α are given by the following results. Proposition 1.4. If α > 0 and β > 0, then the equations     −β −α−β −α f (t) and t Db−α t Db−β f (t) = t Db−α−β f (t) (1.10) a Dt f (t) = a Dt a Dt are satisfied at almost every point t ∈ [a, b] for f ∈ Lp (a, b; RN ) (1 ≤ p < ∞). If α + β > 1, then the relations in (1.10) hold at any point of [a, b]. Proposition 1.5. (i) If α > 0 and f ∈ Lp (a, b; RN ) (1 ≤ p ≤ ∞), then the following equalities     −α −α α α a Dt a Dt f (t) = f (t) and t Db t Db f (t) = f (t) hold almost everywhere on [a, b]. (ii) If α > β > 0, then, for f ∈ Lp (a, b; RN ) (1 ≤ p ≤ ∞), the relations     β −α+β −α f (t) and t Dbβ t Db−α f (t) = t Db−α+β f (t) a Dt a Dt f (t) = a Dt hold almost everywhere on [a, b]. In particular, when β = k ∈ N+ and α > k, then     −α −α+k k f (t) and t Dbk t Db−α f (t) = (−1)k t Db−α+k f (t). a Dt a Dt f (t) = a Dt

8

Fractional Partial Differential Equations

To present the next property, we use the spaces of functions a Dt−α (Lp ) and defined for α > 0 and 1 ≤ p ≤ ∞ by

−α p t Db (L )

−α p a Dt (L )

= {f : f = a Dt−α ϕ, ϕ ∈ Lp (a, b; RN )}

−α p t Db (L )

= {f : f = t Db−α φ, φ ∈ Lp (a, b; RN )},

and respectively. The composition of the fractional integral operator a Dt−α with the fractional derivative operator a Dtα is given by the following results. −(n−α)

Proposition 1.6. Let α > 0, n = [α] + 1 and let fn−α (t) = a Dt fractional integral (1.1) of order n − α.

f (t) be the

(i) If 1 ≤ p ≤ ∞ and f ∈ a Dt−α (Lp ), then   −α α a Dt a Dt f (t) = f (t). (ii) If f ∈ L1 (a, b; RN ) and fn−α ∈ AC n ([a, b]; RN ), then the equality n (n−j)   X fn−α (a) −α α (t − a)α−j a Dt a Dt f (t) = f (t) − Γ(α − j + 1) j=1 holds almost everywhere on [a, b]. −(n−α)

Proposition 1.7. Let α > 0 and n = [α] + 1. Also let gn−α (t) = t Db the fractional integral (1.2) of order n − α.

g(t) be

(i) If 1 ≤ p ≤ ∞ and g ∈ t Db−α (Lp ), then   −α α t Db g(t) = g(t). t Db (ii) If g ∈ L1 (a, b; RN ) and gn−α ∈ AC n ([a, b]; RN ), then the equality n (n−j)   X (−1)n−j gn−α (a) −α α (b − t)α−j t Db g(t) = g(t) − t Db Γ(α − j + 1) j=1 holds almost everywhere on [a, b]. In particular, if 0 < α < 1, then   g1−α (a) −α α D g(t) = g(t) − D (b − t)α−1 , t b t b Γ(α) where g1−α (t) = t Dbα−1 g(t) while for α = n ∈ N+ , the following equality holds: 



−n n t Db t Db g(t)

= g(t) −

n−1 X k=0

(−1)k g (k) (a) (b − t)k . k!

Proposition 1.8. Let α > 0 and let y ∈ L∞ (a, b; RN ) or y ∈ C([a, b]; RN ). Then     −α −α C α C α a Dt a Dt y(t) = y(t) and t Db t Db y(t) = y(t).

Preliminaries

9

Proposition 1.9. Let α > 0 and let n be given by (1.3). If y ∈ AC n ([a, b]; RN ) or y ∈ C n ([a, b]; RN ), then 

−α C α a Dt a Dt y(t)



= y(t) −

n−1 X k=0

y (k) (a) (t − a)k k!

and 



−α C α t Db t Db y(t)

= y(t) −

n−1 X k=0

(−1)k y (k) (b) (b − t)k . k!

In particular, if 0 < α ≤ 1 and y ∈ AC([a, b]; RN ) or y ∈ C([a, b]; RN ), then     −α C α −α C α D y(t) = y(t) − y(a) and D D y(t) = y(t) − y(b). (1.11) a Dt t a t t b b 1.2 1.2.1

Some Results from Analysis Mittag-Leffler Function

Definition 1.7. [268, 303] The Mittag-Leffler function Eα,β is defined by Z ∞ X λα−β eλ 1 zk = dλ, α > 0, β ∈ R, z ∈ C, Eα,β (z) := Γ(αk + β) 2πi Υ λα − z k=0

where Υ is a contour which starts and ends as −∞ and encircles the disc |λ| ≤ |z|1/α counter-clockwise. The function Eα,β (z) is an entire function, and so it is real analytic when restricted to the real line. Moreover, the approximation form of Mittag-Leffler function is given by   N X 1 1 1 Eα,β (z) = − + O , Γ(β − αk) z k z N +1 k=1

with |z| → ∞, µ ≤ |arg(z)| ≤ π for µ > 0, and N ∈ N+ . In particular,   1 1 1 Eα,1 (z) = − +O , Γ(1 − α) z z2 with |z| → ∞, µ ≤ |arg(z)| ≤ π for µ > 0. For short, set Eα (z) := Eα,1 (z),

eα (z) := Eα,α (z).

Then Mittag-Leffler functions have the following properties. Proposition 1.10. [268, 303] For α ∈ (0, 1) and t ∈ R, (i) Eα (t), eα (t) > 0;

(1.12)

10

Fractional Partial Differential Equations

(ii) (Eα (t))0 = (iii) (iv)

1 α eα (t);

lim Eα (t) = lim eα (t) = 0;

t→−∞

t→−∞

C α α 0 Dt Eα (ωt )

= ωEα (ωtα ), 0 Dtα−1 (tα−1 eα (ωtα )) = Eα (ωtα ), ω ∈ C.

Proposition 1.11. [303] Let 0 < α < 1 and λ > 0. Then (i) (ii) (iii)

d α dt Eα (−λt ) d dt


tα−1 eα (−λtα ) = tα−2 Eα,α−1 (−λtα ), for t > 0;

R∞ 0

= −λtα−1 eα (−λtα ), for t > 0;

e−st Eα (−λtα )dt =

sα−1 sα +λ ,

for Re(s) > λ1/α .

It is well known that Eα,1 (−t) is a positive and completely monotonic function for α ∈ (0, 1), t > 0, that is, for all t > 0, k ∈ N0 , we have d
k (−1)k Eα,1 (−t) ≥ 0. dt Additionally, one can find that ω(t) := Eα,1 (λtα ) is a solution of equation C α 0Dt ω(t) = λω(t), λ ∈ R, α ∈ (0, 2). We use the notation a . b that stands for a ≤ Cb, with a positive constant C that does not depend on a, b. The following propositions will be frequently used and can be found in [303]. Proposition 1.12. For λ > 0, α > 0, β ∈ R and any arbitrary positive number m, we have  m
d tβ−1 Eα,β (−λtα ) = tβ−m−1 Eα,β−m (−λtα ), t > 0. dt In particular, d Eα,1 (−λtα ) = −λtα−1 Eα,α (−λtα ), t > 0. dt Proposition 1.13. If 0 < α < 2, β ∈ R, πα/2 < θ < min{π, πα}, then 1 , z ∈ C, θ ≤ |argz| ≤ π. |Eα,β (z)| . 1 + |z| Proposition 1.14. If 0 < α < 2, β ∈ R, θ is such that πα/2 < θ < min{π, πα}, then
1 |Eα,β (z)| . (1 + |z|)(1−β)/α exp Re(z 1/α ) + , z ∈ C, |argz| ≤ θ. 1 + |z| By the fractional order term-by-term integration of the series, there is a more general relationship obtained as follows Z t 1 (t − s)ϑ−1 sβ−1 Eα,β (λsβ )ds = tβ+ϑ−1 Eα,β+ϑ (λtβ ), ϑ > 0, β > 0, t > 0. Γ(ϑ) 0 (1.13) Proposition 1.15. [14] Let 1 < β < 2, β 0 ∈ R and λ > 0. Then the following estimates hold.

Preliminaries

11

0 0 (i) Let 0 ≤ α ≤ 1, 0 < β 0 < β. Then λα tβ Eβ,β 0 (−λtβ ) . tβ −βα , 0 0 (ii) Let 0 ≤ β 0 ≤ 1. Then λ1−β tβ−2 Eβ,β 0 (−λtβ ) . tββ −2 ,

t > 0.

t > 0.

In what follows, let us state the definition and some properties of a function Mα (·) which is also called the Wright-type function. This function is a special case of the Wright function that plays an important role in different areas of fractional calculus and it is introduced by Mainardi to characterize the solution of initial value problem for fractional diffusion-wave equations. Definition 1.8. [254] The Wright-type function Mα is defined by Mα (z) := =

∞ X

(−z)n n!Γ(−αn + 1 − α) n=0 ∞ 1 X (−z)n Γ(nα) sin(nπα), π n=1 (n − 1)!

for

0 < α < 1, z ∈ C.

For −1 < r < ∞, λ > 0, the Wright-type function has the properties. Proposition 1.16. (W1) Mα (t) ≥ 0, t > 0; Z ∞ α 1 α Mα ( α )e−λt dt = e−λ ; (W2) α+1 t t 0 Z ∞ Γ(1 + r) (W3) Mα (t)tr dt = ; Γ(1 + αr) 0 Z ∞ (W4) Mα (t)e−zt dt = Eα (−z), z ∈ C; 0 Z ∞ (W5) αtMα (t)e−zt dt = eα (−z), z ∈ C. 0

1.2.2

Laplace and Fourier Transforms

In this subsection we present definitions and some properties of Laplace and Fourier transforms. Definition 1.9. The Laplace transform of a function f (t) of a real variable t ∈ R+ is defined by Z ∞ (1.14) (Lf )(s) = L[f (t)](s) = f¯(s) := e−st f (t)dt (s ∈ C). 0

The inverse Laplace transform is given for x ∈ R+ by the formula Z γ+i∞ 1 esx f (s)ds (γ = Re(s)). (L−1 f )(x) = L−1 [f (s)](x) := 2πi γ−i∞

(1.15)

12

Fractional Partial Differential Equations

Proposition 1.17. Let f (t) be defined on (0, ∞) and 0 < α < 1. Then the Laplace transform of fractional integral and fractional differential operators satisfy (i) 0Dt−α f (s) = s−α f¯(s); (ii) 0Dtα f (s) = sα f¯(s) − (0Dtα−1 f )(0); (iii)

CD −α f (s) 0 t

= sα f¯(s) − sα−1 f (0).

Definition 1.10. The Fourier transform of a function f (t) of a real variable t ∈ R is defined by Z ∞ (Ff )(w) = F[f (t)](w) = fˆ(w) := e−it·w f (t)dt (w ∈ R). (1.16) −∞

The inverse Fourier transform is given by the formula Z ∞ 1 1 gˆ(−w) := eit·w g(t)dt (F −1 g)(w) = F −1 [g(t)](w) = 2π 2π −∞

(w ∈ R). (1.17)

The integrals in (1.16) and (1.17) converge absolutely for functions f, g ∈ L1 (R) and in the norm of the space L2 (R) for f, g ∈ L2 (R). Proposition 1.18. Let f (t) be defined on (−∞, ∞) and 0 < α < 1. Then Fourier transform of Liouville-Weyl fractional integral and fractional differential operators satisfy (i)

−α \ −∞ Dt f (w)

= (iw)−α fˆ(w);

−α (ii) t \ D∞ f (w) = (−iw)−α fˆ(w);

(iii)

α \ −∞ Dt f (w)

= (iw)α fˆ(w);

α f (w) = (−iw)α fˆ(w). (iv) t\ D∞

1.3 1.3.1

Semigroups C0 -Semigroup

Let us recall the definitions and properties of operator semigroups, for details see Pazy [291]. Let X be a Banach space and L(X) be the Banach space of linear bounded operators with the norm k · k. Definition 1.11. A one parameter family {T (t)}t≥0 ⊂ L(X) is a semigroup of bounded linear operators on X if (i) T (t)T (s) = T (t + s), for t, s ≥ 0; (ii) T (0) = I; here, I denotes the identity operator in X.

Preliminaries

13

Definition 1.12. A semigroup of bounded linear operators {T (t)}t≥0 is uniformly continuous if lim kT (t) − Ik = 0.

t→0+

From the definition it is clear that if {T (t)}t≥0 is a uniformly continuous semigroup of bounded linear operators, then lim kT (s) − T (t)k = 0.

s→t

Definition 1.13. We say that the semigroup {T (t)}t≥0 is strongly continuous (or a C0 -semigroup) if the mapping t → T (t)u is strongly continuous, for each u ∈ X, i.e., lim T (t)u = u, ∀ u ∈ X.

t→0+

Definition 1.14. Let {T (t)}t≥0 be a C0 -semigroup defined on X. The linear operator A is the infinitesimal generator of {T (t)}t≥0 defined by Au = lim+ t→0

n where D(A) = u ∈ X : limt→0+

T (t)u − u , for u ∈ D(A), t o T (t)u−u exists in X . t

Definition 1.15. The family R(λ, A) = (λI − A)−1 , λ ∈ ρ(A) of bound linear operator is called of the resolvent of A, where ρ(A) is the set of the all complex number λ for which λI − A is invertible. If there are M ≥ 0 and ν ∈ R such that kT (t)k ≤ M eνt , then Z ∞ (λI − A)−1 u = e−λt T (t)udt, Re(λ) > ν, u ∈ X.

(1.18)

0

A C0 -semigroup {T (t)}t≥0 is called exponentially stable if there exist constants M > 0 and δ > 0 such that kT (t)k ≤ M e−δt ,

t ≥ 0.

(1.19)

The growth bound ν0 of {T (t)}t≥0 is defined by ν0 = inf{δ ∈ R : there exists Mδ > 0 such that kT (t)k ≤ Mδ eδt , ∀ t ≥ 0}. (1.20) Furthermore, ν0 can also be obtained by the following formula: ν0 = lim sup t→+∞

ln kT (t)k . t

(1.21)

Definition 1.16. A C0 -semigroup {T (t)}t≥0 is called uniformly bounded if there exists a constant M > 0 such that kT (t)k ≤ M, t ≥ 0.

(1.22)

14

Fractional Partial Differential Equations

Definition 1.17. A C0 -semigroup {T (t)}t≥0 is called compact if T (t) is compact for t > 0. Proposition 1.19. If {T (t)}t>0 is compact, then {T (t)}t>0 is equicontinuous for t > 0. Definition 1.18. A C0 -semigroup {T (t)}t≥0 is called positive if T (t)u ≥ θ for all u ≥ θ and t ≥ 0, where θ is the zero element. 1.3.2

Analytic Semigroup

Definition 1.19. Let ∆ := {z : ϕ1 < arg z < ϕ2 , ϕ1 < 0 < ϕ2 }. The family {T (z)}z∈∆ ⊂ L(X) is called an analytic semigroup in ∆ if (i) z 7→ T (z) is analytic in ∆; (ii) T (0) = I and limz∈∆,z→0 T (z)x = x for every x ∈ X; (iii) T (z1 + z2 ) = T (z1 )T (z2 ) for z1 , z2 ∈ ∆. A semigroup T (t) is called analytic if it is analytic in some sector ∆ containing the nonnegative real axis. Theorem 1.1. Let {T (t)}t≥0 be a uniformly bounded C0 -semigroup. Let A be the infinitesimal generator of {T (t)}t≥0 and assume 0 ∈ ρ(A). The following statements are equivalent: (i) T (t) can be extended to an analytic semigroup in a sector Σδ := {z ∈ C : | arg z| < δ} and kT (z)k is uniformly bounded in every closed subsector Σδ0 , δ 0 < δ, of Σδ ; (ii) there exists a positive constant C such that for every σ > 0, τ 6= 0, kR(σ + iτ, A)k ≤ (iii) there exist 0 < δ
0 such that

ρ(A) ⊃ Σ := {λ ∈ C : | arg λ|
0 and there is a constant C such that kAT (t)k ≤

C , for t > 0. t

Chapter 2

Fractional Navier-Stokes Equations

2.1

Introduction

Navier-Stokes equations have been investigated by many researchers in view of their crucial role in fluid mechanics and turbulence problems. For more details, we refer the reader to the monographs by Ben-Artzi et al. [43], Cannone [59], Lemari´eRieusset [221], Raugel and Sell [308], Temam [330] and Varnhorn [340]. There are extensive literatures on the well-posedness of initial value problems for these equations, see, e.g., Dubois [109], Giga [138], Hieber and Sawada [169], Hieber and Shibata [170], Ibrahim and Keraani [178], Koch et al. [202], Koch and Tataru [203], Lemari´e-Rieusset [221], Miura [269], Von Wahl [343], Yamazaki [370] and references therein. It is worth mentioning that Leray carried out an initial study that a boundaryvalue problem for the time-dependent Navier-Stokes equations possesses a unique smooth solution on some intervals of time provided the data are sufficiently smooth. Since then many results on the existence for weak, mild and strong solutions for the Navier-Stokes equations have been investigated intensively by many authors, see, e.g., Almeida and Ferreira [12], Heck et al. [164], Iwabuchi and Takada [182], Koch et al. [202], Masmoudi and Wong [249], and Weissler [362]. Moreover, one can find results on regularity of weak and strong solutions from Amrouche and Rejaiba [16], Chemin and Gallagher [71], Chemin et al. [72], Choe [80], Danchin [93], Giga [140], Kozono [206], Raugel and Sell [308] and the references therein. In the last few years considerable process has been made in the existence, uniqueness and smoothness properties of weak solutions related to the Navier-Stokes equations, see, ˘ ak [184], e.g., Barbu [33], Duchon and Robert [110], Feireisl et al. [120], Jia and Sver´ Jungel [185], Vasseur and Yu [341] and the references therein. The topic of the global existence of weak, mild and strong solutions supplemented with small initial data received considerable attention. For example, Leray [223] carried out a pioneering study on the existence of global weak solutions in the energy space and the uniqueness of such solutions in R2 . Lemari´e-Rieusset [222] established the existence of global mild solutions in different types of frameworks in Morrey-Campanato spaces. Later, Iwabuchi and Takada [183] discussed the same 15

16

Fractional Partial Differential Equations

problem in function spaces of Besov type. A similar result was established by Lei and Lin [220] in the space X−1 . Similar results were studied by Kato [186] in Ln (Rn ), Giga and Miyakawa [139], and Taylor [328] in Morrey spaces, Cannone [60] −1+n/p and Planchon [302] in the Besov spaces Bp,∞ (Rn ), 1 < p < ∞. It is worth mentioning that Lions [231] was the first to carry out the study that the Navier-Stokes equations have a weak solution with time-fractional derivative of order less than 41 provided the space dimension is not further than four. After that, only a few research results on this subject have been achieved; for example, Zhang [382] proved that this type of equations has a weak solution whose timefractional derivative of order is no more than 12 . A strong motivation for investigating the time-fractional Navier-Stokes equations arises from the fact that the phenomena of anomalous diffusion in fractal media can be simulated by Navier-Stokes equations with the time-fractional derivative. Also, fractional partial differential equations play a significant role in modelling many practical situations such as fluid flow, diffusive transport akin to diffusion and so on. Comparing with theory of the classical Navier-Stokes equations, the researches on time-fractional Navier-Stokes equations are only on their initial stage of development. The main effort on time-fractional Navier-Stokes equations has been put into attempts to derive numerical solutions and analytical solutions, see El-Shahed et al. [113], Ganji et al. [134], and Moman and Zaid [270]. However, to the best of our knowledge, there are few results on time-fractional Navier-Stokes equations. For the first time, Carvalho-Neto [64] dealt with the existence and uniqueness of global and local mild solutions for the time-fractional Navier-Stokes equations. This chapter deals with the Navier-Stokes equations with time-fractional derivative of order α ∈ (0, 1). Section 2.2 is concerned with the existence and uniqueness of local and global mild solutions in H β,q . Meanwhile, we also give local mild solutions in Jq . Moreover, we prove the existence and regularity of classical solutions for such equations in Jq . Section 2.3 is devoted to investigation of existence, uniqueness and H¨ older continuity of the local mild solutions. In Section 2.4, we obtain the existence and uniqueness of the solutions to approximate equations, as well as the convergence of the approximate solutions. Furthermore, we present some convergence results for the Faedo-Galerkin approximations of the given problems. In Section 2.5, we firstly give the concept of the weak solutions and establish the existence criterion of weak solutions by means of the Galerkin approximations in the case that the dimension n ≤ 4. Moreover, a complete proof of the uniqueness is given when n = 2. At last we give a sufficient condition of optimal control pairs. In Section 2.6, we make use of energy methods to study the time-fractional NavierStokes equations. In the first step, we construct a regularized equation by using a smoothing process to transform unbounded differential operators into bounded operators and then obtain the approximate solutions. The second part describes a procedure to take a limit in the approximation program to present a global solution to the objective equation. In Section 2.7, we use the tools from harmonic analysis to

Fractional Navier-Stokes Equations

17

study the Cauchy problem for time-fractional Navier-Stokes equations. Two results concerning the local existence of solutions for the given problem in Sobolev spaces are addressed. 2.2

Cauchy Problem in Rn

In this section, we study the following time-fractional Navier-Stokes equations in an open set Ω ⊂ Rn (n ≥ 3):  α ∂t u − ν∆u + (u · ∇)u = −∇p + f, t > 0,      ∇ · u = 0, (2.1)   u|∂Ω = 0,    u(0, x) = a, where ∂tα is the Caputo fractional derivative of order α ∈ (0, 1), u = (u1 (t, x), u2 (t, x), . . . , un (t, x)) represents the velocity field at a point x ∈ Ω and time t > 0, p = p(t, x) is the pressure, ν the viscosity, f = f (t, x) is the external force and a = a(x) is the initial velocity. From now on, we assume that Ω has a smooth boundary. Firstly, we get rid of the pressure term by applying Helmholtz projector P to equation (2.1), which converts equation (2.1) to  α   ∂t u − νP ∆u + P (u · ∇)u = P f, t > 0,    ∇ · u = 0,  u|∂Ω = 0,     u(0, x) = a. The operator −νP ∆ with Dirichlet’s boundary conditions is, basically, the Stokes operator A in the divergence-free function space under consideration. Then we rewrite (2.1) as the following abstract form ( C α 0 Dt u = −Au + F (u, u) + P f, t > 0, (2.2) u(0) = a, where F (u, v) = −P (u · ∇)v. If one can give sense to the Helmholtz projection P and the Stokes operator A, then the solution of equation (2.2) is also the solution of equation (2.1). In this section, we establish the existence and uniqueness of global and local mild solutions of problem (2.2) in H β,q . Further, we prove the regularity results which state essentially that if P f is H¨ older continuous then there is a unique classical α solution u(t) such that Au and C D older continuous in Jq . 0 t u(t) are H¨ The section is organized as follows. In Subsection 2.2.1 we recall some notations, definitions, and preliminary facts. Subsection 2.2.2 is devoted to the existence and uniqueness of global mild solution in H β,q of problem (2.2), then proceed to study

18

Fractional Partial Differential Equations

the local mild solution in H β,q . In Subsection 2.2.3, we use the iteration method to obtain the existence and uniqueness of local mild solution in Jq of problem (2.2). Finally, Subsection 2.2.4 is concerned with the existence and regularity of classical solution in Jq of problem (2.2). 2.2.1

Definitions and Lemmas

In this subsection, we introduce notations, definitions, and preliminary facts which are used throughout this section. Let Ω = {(x1 , . . . , xn ) : xn > 0} be an open subset of Rn , where n ≥ 3. Let 1 < q < ∞. Then there is a bounded projection P called the Hodge projection on (Lq (Ω))n , whose range is the closure of Cσ∞ (Ω) := {u ∈ (C ∞ (Ω))n : ∇ · u = 0, u has compact support in Ω}, and whose null space is the closure of {u ∈ (C ∞ (Ω))n : u = ∇φ, φ ∈ C ∞ (Ω)}. |·|q

For notational convenience, let Jq := Cσ∞ (Ω) , which is a closed subspace of (Lq (Ω))n . (W m,q (Ω))n is a Sobolev space with the norm | · |m,q . A = −νP ∆ denotes the Stokes operator in Jq whose domain is D(A) = D(∆) ∩ Jq , here, D(∆) = {u ∈ (W 2,q (Ω))n : u|∂Ω = 0}. It is known that −A is a closed linear operator and generates the bounded analytic semigroup {e−tA } on Jq , and we have for θ ∈ (π/2, π), k(z + A)−1 kL(X) ≤ M/|z| for z ∈ Σθ = {z ∈ C : z 6= 0, |argz| < θ}, where L(X) be the Banach space of linear bounded operators with the norm k·kL(X) . So as to state our results, we need to introduce the definitions of the fractional power spaces associated with −A. For β > 0 and u ∈ Jq , define Z ∞ 1 −β tβ−1 e−tA udt. A u= Γ(β) 0 Then A−β is a bounded, one-to-one operator on Jq . Let Aβ be the inverse of A−β . For β > 0, we denote the space H β,q by the range of A−β with the norm |u|H β,q = |Aβ u|q . It is easy to check that e−tA extends (or restricts) to a bounded analytic semigroup on H β,q . For more details, we refer to Von Wahl [343]. Let X be a Banach space and J be an interval of R. C(J, X) denotes the set of all continuous X-valued functions. For 0 < ϑ < 1, C ϑ (J, X) stands for the set of all functions which are H¨ older continuous with the exponent ϑ. Let X be a Banach space, α ∈ (0, 1] and v : [0, T ] → X. The fractional integral of order α with the lower limit zero for the function v is defined as −α 0 Dt v(t)

= (gα ∗ v)(t), v ∈ L1 (0, T ; X), t ∈ [0, T ],

Fractional Navier-Stokes Equations

19

α−1

where gα = tΓ(α) , t > 0 and ∗ denotes the convolution. Further, the Caputo fractional derivative operator of order α for the function v is defined by
 d  −(1−α) C α v(t) − v(0) . 0 Dt 0 Dt v(t) = dt In general, for u : [0, T ] × Rn → Rn , the Caputo fractional derivative with respect to time of the function u can be written as  −(1−α)
 ∂tα u(t, x) = ∂t 0 Dt u(t, x) − u(0, x) . Let us introduce the generalized Mittag-Leffler special functions: Z ∞ Z ∞ α α Eα (−tα A) = Mα (s)e−st A ds, Eα,α (−tα A) = αsMα (s)e−st A ds, 0

0

where Mα (θ) is the Wright-type function defined in Definition 1.8. Proposition 2.1. (i) Eα,α (−tα A) =

1 2πi

Z

Eα,α (−µtα )(µI + A)−1 dµ, where

Γθ

Γθ = {re−iθ : r ≥ δ} ∪ {δeiψ : |ψ| ≤ θ} ∪ {reiθ : r ≥ δ} for δ > 0. Z 1 µγ Eα,α (−µtα )(µI + A)−1 dµ. (ii) Aγ Eα,α (−tα A) = 2πi Γθ R∞ Proof. (i) In view of 0 αsMα (s)e−st ds = Eα,α (−t) and the Fubini theorem, we get Z ∞ α Eα,α (−tα A) = αsMα (s)e−st A ds 0 Z ∞ Z α 1 = αsMα (s) e−µst (µI + A)−1 dµds 2πi 0 Γθ Z 1 Eα,α (−µtα )(µI + A)−1 dµ, = 2πi Γθ where Γθ is a suitable integral path. (ii) A similar argument shows that Z ∞ α γ α A Eα,α (−t A) = αsMα (s)Aγ e−st A ds 0 Z ∞ Z α 1 = αsMα (s) µγ e−µst (µI + A)−1 dµds 2πi 0 Γθ Z 1 γ µ Eα,α (−µtα )(µI + A)−1 dµ. = 2πi Γθ Moreover, we have the following results. Lemma 2.1. [352] For t > 0, Eα (−tα A) and Eα,α (−tα A) are continuous in the uniform operator topology. Moreover, for every r > 0, the continuity is uniform on [r, ∞).

20

Fractional Partial Differential Equations

Lemma 2.2. [352] Let 0 < α < 1. Then (i) for all u ∈ X, limt→0+ Eα (−tα A)u = u; α α α (ii) for all u ∈ D(A) and t > 0, C 0 Dt Eα (−t A)u = −AEα (−t A)u;

(iii) for all u ∈ X, Eα0 (−tα A)u = −tα−1 AEα,α (−tα A)u; −(1−α)

(iv) for all u ∈ X and t > 0, Eα (−tα A)u = 0 Dt


tα−1 Eα,α (−tα A)u .

Before presenting the definition of mild solution of problem (2.2), we give the following lemma for a given function h : [0, ∞) → X. For more details we refer to Zhou [386, 387]. Lemma 2.3. If 1 u(t) = a + Γ(α)

Z

t


(t − s)α−1 − Au(s) + h(s) ds,

for t ≥ 0

(2.3)

0

holds, then we have u(t) = Eα (−tα A)a +

Z

t

(t − s)α−1 Eα,α (−(t − s)α A)h(s)ds.

0

We rewrite (2.2) as Z t
1 u(t) = a + (t − s)α−1 − Au(s) + F (u(s), u(s)) + P f (s) ds, Γ(α) 0

for t ≥ 0.

Inspired by above discussion, we adopt the following concepts of mild solution to problem (2.2). Definition 2.1. A function u : [0, ∞) → H β,q is called a global mild solution of the problem (2.2) in H β,q , if u ∈ C([0, ∞); H β,q ) and for t ∈ [0, ∞), Z t u(t) =Eα (−tα A)a + (t − s)α−1 Eα,α (−(t − s)α A)F (u(s), u(s))ds 0 (2.4) Z t α−1 + (t − s) Eα,α (−(t − s)α A)P f (s)ds. 0

Definition 2.2. Let 0 < T < ∞. A function u : [0, T ] → H β,q (or Jq ) is called a local mild solution of the problem (2.2) in H β,q (or Jq ), if u ∈ C([0, T ]; H β,q ) (or C([0, T ]; Jq )) and u satisfies (2.4) for t ∈ [0, T ]. For convenience, we define two operators Φ and G as follows: Z t Φ(t) = (t − s)α−1 Eα,α (−(t − s)α A)P f (s)ds, 0 Z t G(u, v)(t) = (t − s)α−1 Eα,α (−(t − s)α A)F (u(s), v(s))ds. 0

Fractional Navier-Stokes Equations

21

In subsequent proof we use the following fixed point result. Lemma 2.4. [59] Let (X, | · |) be a Banach space, G : X × X → X a bilinear operator and L a positive real number such that |G(u, v)| ≤ L|u||v|, ∀ u, v ∈ X. Then for any u0 ∈ X with |u0 | < solution u ∈ X. 2.2.2

1 4L ,

the equation u = u0 + G(u, u) has a unique

Global/Local Existence

Our main purpose in this subsection is to establish sufficient conditions for existence and uniqueness of mild solution to the problem (2.2) in H β,q . To this end we assume that: (f) P f is continuous for t > 0 and |P f (t)|q = o(t−α(1−β) ) as t → 0 for 0 < β < 1. Lemma 2.5. [133, 362] Let 1 < q < ∞ and β1 ≤ β2 . Then there is a constant C = C(β1 , β2 ) such that |e−tA v|H β2 ,q ≤ Ct−(β2 −β1 ) |v|H β1 ,q , t > 0, for v ∈ H β1 ,q . Furthermore, limt→0 t(β2 −β1 ) |e−tA v|H β2 ,q = 0. Now, we study an important technical lemma, which helps us to prove the final main theorems of this subsection. Lemma 2.6. Let 1 < q < ∞ and β1 ≤ β2 < β1 + 1. Then for any T > 0, there exists a constant C1 = C1 (α, β1 , β2 ) > 0 such that |Eα (−tα A)v|H β2 ,q ≤C1 t−α(β2 −β1 ) |v|H β1 ,q , |Eα,α (−tα A)v|H β2 ,q ≤C1 t−α(β2 −β1 ) |v|H β1 ,q , for all v ∈ H β1 ,q and t ∈ (0, T ]. Furthermore, lim tα(β2 −β1 ) |Eα (−tα A)v|H β2 ,q = 0.

t→0

Proof. Let v ∈ H β1 ,q . By Lemma 2.5, we estimate Z ∞ α α |Eα (−t A)v|H β2 ,q ≤ Mα (s)|e−st A v|H β2 ,q ds 0 Z ∞  ≤ C Mα (s)s−(β2 −β1 ) ds t−α(β2 −β1 ) |v|H β1 ,q 0

≤C1 t−α(β2 −β1 ) |v|H β1 ,q , for β1 − β2 > −1. More precisely, the Lebesgue’s dominated convergence theorem shows Z ∞ α α(β2 −β1 ) α lim t |Eα (−t A)v|H β2 ,q ≤ Mα (s) lim tα(β2 −β1 ) |e−st A v|H β2 ,q ds = 0. t→0

0

t→0

22

Fractional Partial Differential Equations

Similarly, Z

∞

α

αsMα (s)|e−st A v|H β2 ,q ds 0 Z ∞  ≤ αC Mα (s)s1−(β2 −β1 ) ds t−α(β2 −β1 ) |v|H β1 ,q

|Eα,α (−tα A)v|H β2 ,q ≤

≤C1 t

0 −α(β2 −β1 )

|v|H β1 ,q ,

where the constant C1 = C1 (α, β1 , β2 ) is such that   Γ(1 − β2 + β1 ) αΓ(2 − β2 + β1 ) , . C1 ≥ C max Γ(1 + α(β1 − β2 )) Γ(1 + α(1 + β1 − β2 ))

In the following, we study the global mild solution of problem (2.2) in H β,q . For convenience, we denote M (t) = sup {sα(1−β) |P f (s)|q }, s∈(0,t]

B1 = C1 max{B(α(1 − β), 1 − α(1 − β)), B(α(1 − γ), 1 − α(1 − β))}, L ≥ M C1 max{B(α(1 − β), 1 − 2α(γ − β)), B(α(1 − γ), 1 − 2α(γ − β))}, where M is given later. Theorem 2.1. Let 1 < q < ∞, 0 < β < 1 and (f ) hold. For every a ∈ H β,q , suppose that 1 , 4L

C1 |a|H β,q + B1 M∞
max{β, 12 } and a unique function u : [0, ∞) → H β,q satisfying:

(a) u : [0, ∞) → H β,q is continuous and u(0) = a; (b) u : (0, ∞) → H γ,q is continuous and limt→0 tα(γ−β) |u(t)|H γ,q = 0; (c) u satisfies (2.4) for t ∈ [0, ∞). Proof. Let γ = H β,q such that:

(1+β) 2 .

Define X∞ = X[∞] as the space of all curves u : (0, ∞) →

(i) u : [0, ∞) → H β,q is bounded and continuous; (ii) u : (0, ∞) → H γ,q is bounded and continuous, moreover, limt→0 tα(γ−β) |u(t)|H γ,q = 0 with its natural norm 

α(γ−β)

kukX∞ = max sup |u(t)|H β,q , sup t t≥0

 |u(t)|H γ,q

t≥0

It is obvious that X∞ is a non-empty complete metric space.

.

Fractional Navier-Stokes Equations

23

From an argument of Weissler [362], we know that F : H γ,q × H γ,q → Jq is a bounded bilinear map, then there exists M such that for u, v ∈ H γ,q , |F (u, v)|q ≤M |u|H γ,q |v|H γ,q , |F (u, u) − F (v, v)|q ≤M (|u|H γ,q + |v|H γ,q )|u − v|H γ,q .

(2.6)

Step 1. Let u, v ∈ X∞ . We show that the operator G(u(t), v(t)) belongs to C([0, ∞); H β,q ) as well as C((0, ∞); H γ,q ). For arbitrary t0 ≥ 0 fixed and ε > 0 enough small, consider t > t0 (the case t < t0 follows analogously), we have |G(u(t), v(t)) − G(u(t0 ), v(t0 ))|H β,q Z t (t − s)α−1 |Eα,α (−(t − s)α A)F (u(s), v(s))|H β,q ds ≤ t0

Z + 0

Z

t0


(t − s)α−1 − (t0 − s)α−1 Eα,α (−(t − s)α A)F (u(s), v(s)) β,q ds H t0 −ε

(t0 − s)α−1 Eα,α (−(t − s)α A) 0
− Eα,α (−(t0 − s)α A) F (u(s), v(s)) H β,q ds Z t0 + (t0 − s)α−1 Eα,α (−(t − s)α A) t0 −ε
− Eα,α (−(t0 − s)α A) F (u(s), v(s)) H β,q ds +

= : I11 (t) + I12 (t) + I13 (t) + I14 (t). We estimate each of the four terms separately. For I11 (t), in view of Lemma 2.6, we obtain Z t I11 (t) ≤C1 (t − s)α(1−β)−1 |F (u(s), v(s))|q ds t0

Z

t

(t − s)α(1−β)−1 |u(s)|H γ,q |v(s)|H γ,q ds

≤M C1 t0 Z t

≤M C1

(t − s)α(1−β)−1 s−2α(γ−β) ds sup {s2α(γ−β) |u(s)|H γ,q |v(s)|H γ,q } s∈[0,t]

t0

Z

1

=M C1

(1 − s)α(1−β)−1 s−2α(γ−β) ds sup {s2α(γ−β) |u(s)|H γ,q |v(s)|H γ,q }. s∈[0,t]

t0 /t

By the properties of the Beta function, there exists δ > 0 small enough such that for 0 < t − t0 < δ, Z 1 (1 − s)α(1−β)−1 s−2α(γ−β) ds → 0, t0 /t

which follows that I11 (t) tends to 0 as t − t0 → 0. For I12 (t), since Z t0
I12 (t) ≤C1 (t0 − s)α−1 − (t − s)α−1 (t − s)−αβ |F (u(s), v(s))|q ds 0

24

Fractional Partial Differential Equations t0

Z


(t0 − s)α−1 − (t − s)α−1 (t − s)−αβ s−2α(γ−β) ds

≤M C1 0

× sup {s2α(γ−β) |u(s)|H γ,q |v(s)|H γ,q }. s∈[0,t0 ]

Noting that Z t0 |(t0 − s)α−1 − (t − s)α−1 |(t − s)−αβ s−2α(γ−β) ds 0 Z t0 Z t0 (t0 − s)α−1 (t − s)−αβ s−2α(γ−β) ds (t − s)α−1 (t − s)−αβ s−2α(γ−β) ds + ≤ 0 0 Z t0 ≤2 (t0 − s)α(1−β)−1 s−2α(γ−β) ds 0

=2B(α(1 − β), 1 − 2α(γ − β)), then by the Lebesgue’s dominated convergence theorem, we have Z t0
(t0 − s)α−1 − (t − s)α−1 (t − s)−αβ s−2α(γ−β) ds → 0, as t → t0 , 0

one deduces that limt→t0 I12 (t) = 0. For I13 (t), since Z t0 −ε I13 (t) ≤ (t0 − s)α−1 Eα,α (−(t − s)α A) 0
+ Eα,α (−(t0 − s)α A) F (u(s), v(s)) H β,q ds Z t0 −ε
≤C1 (t0 − s)α−1 (t − s)−αβ + (t0 − s)−αβ |F (u(s), v(s))|q ds 0 Z t0 −ε ≤2M C1 (t0 − s)α(1−β)−1 s−2α(γ−β) ds 0

× sup {s2α(γ−β) |u(s)|H γ,q |v(s)|H γ,q }, s∈[0,t0 ]

using the Lebesgue’s dominated convergence theorem again, the fact from the uniform continuity of the operator Eα,α (−tα A) due to Lemma 2.1 shows Z t0 −ε lim I13 (t) = (t0 − s)α−1 lim Eα,α (−(t − s)α A) t→t0 t→t 0 0
− Eα,α (−(t0 − s)α A) F (u(s), v(s)) H β,q ds =0. For I14 (t), by immediate calculation, we estimate Z t0
I14 (t) ≤C1 (t0 − s)α−1 (t − s)−αβ + (t0 − s)−αβ |F (u(s), v(s))|q ds t0 −ε Z t0 ≤2M C1 (t0 − s)α(1−β)−1 s−2α(γ−β) ds t0 −ε

×

sup

{s2α(γ−β) |u(s)|H γ,q |v(s)|H γ,q }

s∈[t0 −ε,t0 ]

→ 0,

as ε → 0

Fractional Navier-Stokes Equations

25

according to the properties of the Beta function. Thenceforth, it follows |G(u(t), v(t)) − G(u(t0 ), v(t0 ))|H β,q → 0,

as t → t0 .

The continuity of the operator G(u, v) evaluated in C((0, ∞); H γ,q ) follows by the similar discussion as above. So, we omit the details. Step 2. We show that the operator G : X∞ × X∞ → X∞ is a continuous bilinear operator. By Lemma 2.6, we have Z t (t − s)α−1 Eα,α (−(t − s)α A)F (u(s), v(s))ds |G(u(t), v(t))|H β,q ≤ 0 H β,q Z t ≤C1 (t − s)α(1−β)−1 |F (u(s), v(s))|q ds 0 Z t ≤M C1 (t − s)α(1−β)−1 s−2α(γ−β) ds 0

× sup {s2α(γ−β) |u(s)|H γ,q |v(s)|H γ,q } s∈[0,t]

≤M C1 B(α(1 − β), 1 − 2α(γ − β))kukX∞ kvkX∞ and |G(u(t), v(t))|H γ,q

Z t α−1 α (t − s) Eα,α (−(t − s) A)F (u(s), v(s))ds ≤ 0 H γ,q Z t ≤C1 (t − s)α(1−γ)−1 |F (u(s), v(s))|q ds 0 Z t ≤M C1 (t − s)α(1−γ)−1 s−2α(γ−β) ds 0

× sup {s2α(γ−β) |u(s)|H γ,q |v(s)|H γ,q } s∈[0,t]

≤M C1 t−α(γ−β) B(α(1 − γ), 1 − 2α(γ − β))kukX∞ kvkX∞ , it follows that sup tα(γ−β) |G(u(t), v(t))|H γ,q ≤ M C1 B(α(1 − γ), 1 − 2α(γ − β))kukX∞ kvkX∞ . t∈[0,∞)

More precisely, lim tα(γ−β) |G(u(t), v(t))|H γ,q = 0.

t→0

Hence, G(u, v) ∈ X∞ and kG(u, v)kX∞ ≤ LkukX∞ kvkX∞ . Step 3. We verify that (c) holds.

26

Fractional Partial Differential Equations

Let 0 < t0 < t. Since |Φ(t) − Φ(t0 )|H β,q Z t (t − s)α−1 |Eα,α (−(t − s)α A)P f (s)|H β,q ds ≤ t0 t0

Z


(t0 − s)α−1 − (t − s)α−1 |Eα,α (−(t − s)α A)P f (s)|H β,q ds

+ 0

t0 −ε

Z


(t0 − s)α−1 Eα,α (−(t − s)α A) − Eα,α (−(t0 − s)α A) P f (s) H β,q ds

+ 0 t0

Z +

t0 −ε t

Z ≤C1


(t0 − s)α−1 Eα,α (−(t − s)α A) − Eα,α (−(t0 − s)α A) P f (s) H β,q ds

(t − s)α(1−β)−1 |P f (s)|q ds

t0

Z

t0

+ C1


(t0 − s)α−1 − (t − s)α−1 (t − s)−αβ |P f (s)|q ds

0

Z

t0 −ε


(t0 − s)α−1 Eα,α (−(t − s)α A) − Eα,α (−(t0 − s)α A) P f (s) H β,q ds

+ C1 0

Z

t0

(t0 − s)α(1−β)−1 |P f (s)|q ds

+ 2C1 t0 −ε Z t

(t − s)α(1−β)−1 s−α(1−β) ds

≤C1 M (t) t0

Z + C1 M (t)

t0


(t0 − s)α−1 − (t − s)α−1 (t − s)−αβ s−α(1−β) ds

0

Z

t0 −ε

(t0 − s)α−1 Eα,α (−(t − s)α A) 0
− Eα,α (−(t0 − s)α A) s−α(1−β) H β,q ds Z t0 + 2C1 M (t) (t0 − s)α(1−β)−1 s−α(1−β) ds.

+ C1 M (t)

t0 −ε

By the properties of the Beta function, the first two integrals and the last integral tend to 0 as t → t0 as well as ε → 0. In view of Lemma 2.1, the third integral also goes to 0 as t → t0 , which implies |Φ(t) − Φ(t0 )|H β,q → 0, as t → t0 . The continuity of Φ(t) evaluated in H γ,q follows by the similar argument as above. On the other hand, we have Z t |Φ(t)|H β,q ≤ (t − s)α−1 Eα,α (−(t − s)α A)P f (s)ds 0 H β,q Z t ≤C1 (t − s)α(1−β)−1 |P f (s)|q ds (2.7) 0 Z t ≤C1 M (t) (t − s)α(1−β)−1 s−α(1−β) ds 0

=C1 M (t)B(α(1 − β), 1 − α(1 − β))

Fractional Navier-Stokes Equations

27

and |Φ(t)|H γ,q

Z t α−1 α (t − s) Eα,α (−(t − s) A)P f (s)ds ≤ 0 H γ,q Z t ≤C1 (t − s)α(1−γ)−1 |P f (s)|q ds 0 Z t ≤C1 M (t) (t − s)α(1−γ)−1 s−α(1−β) ds 0

=t−α(γ−β) C1 M (t)B(α(1 − γ), 1 − α(1 − β)). More precisely, tα(γ−β) |Φ(t)|H γ,q ≤ C1 M (t)B(α(1 − γ), 1 − α(1 − β)) → 0, as t → 0, since M (t) → 0 as t → 0 due to assumption (f). This ensures that Φ(t) ∈ X∞ and kΦkX∞ ≤ B1 M∞ . For a ∈ H β,q . By Lemma 2.1, it is easy to see that Eα (−tα A)a ∈ C([0, ∞); H β,q ) and Eα (−tα A)a ∈ C((0, ∞); H γ,q ). This, together with Lemma 2.6, implies that for all t ∈ (0, T ], Eα (−tα A)a ∈ X∞ , tα(γ−β) Eα (−tα A)a ∈ C([0, ∞); H γ,q ), kEα (−tα A)akX∞ ≤ C1 |a|H β,q . According to (2.5), the inequality kEα (−tα A)a + ΦkX∞ ≤ kEα (−tα A)akX∞ + kΦkX∞
max{β, 12 } such that for every a ∈ H β,q there exist T∗ > 0 and a unique function u : [0, T∗ ] → H β,q satisfying: (a) u : [0, T∗ ] → H β,q is continuous and u(0) = a; (b) u : (0, T∗ ] → H γ,q is continuous and limt→0 tα(γ−β) |u(t)|H γ,q = 0; (c) u satisfies (2.4) for t ∈ [0, T∗ ]. β,q . Let X = X[T ] be the space of all curves Proof. Let γ = (1+β) 2 . Fix a ∈ H β,q u : (0, T ] → H such that:

(i) u : [0, T ] → H β,q is continuous; (ii) u : (0, T ] → H γ,q is continuous and limt→0 tα(γ−β) |u(t)|H γ,q = 0 with its natural norm kukX = sup {tα(γ−β) |u(t)|H γ,q }. t∈[0,T ]

Similar to the proof of Theorem 2.1, it is easy to claim that G : X × X → X is continuous linear map and Φ(t) ∈ X. By Lemma 2.1, it is easy to see that for all t ∈ (0, T ], Eα (−tα A)a ∈ C([0, T ]; H β,q ), Eα (−tα A)a ∈ C((0, T ]; H γ,q ). From Lemma 2.6, it follows that Eα (−tα A)a ∈ X, tα(γ−β) Eα (−tα A)a ∈ C([0, T ]; H γ,q ). Hence, let T∗ > 0 be sufficiently small such that kEα (−tα A)a + Φ(t)kX[T∗ ] ≤ kEα (−tα A)akX[T∗ ] + kΦ(t)kX[T∗ ] < From Lemma 2.4, there exists a u : [0, T∗ ] → H β,q satisfies (2.4).

1 . 4L

Fractional Navier-Stokes Equations

2.2.3

29

Local Existence

This subsection is devoted to consideration of local mild solution to problem (2.2) in Jq by means of the iteration method. Let γ = (1+β) 2 . Theorem 2.3. Let 1 < q < ∞, 0 < β < 1 and (f ) hold. Suppose that a ∈ H β,q with

n 1 − < β. 2q 2

Then problem (2.2) has a unique mild solution u in Jq for a ∈ H β,q . Moreover, u is continuous on [0, T ], Aγ u is continuous in (0, T ] and tα(γ−β) Aγ u(t) is bounded as t → 0. Proof. Step 1. Set K(t) := sup sα(γ−β) |Aγ u(s)|q s∈(0,t]

and Z Ψ(t) := G(u, u)(t) =

t

(t − s)α−1 Eα,α (−(t − s)α A)F (u(s), u(s))ds.

0

As an immediate consequences of Step 2 in Theorem 2.1, then Ψ(t) is continuous in [0, T ], Aγ Ψ(t) exists and is continuous in (0, T ] with |Aγ Ψ(t)|q ≤ M C1 B(α(1 − γ), 1 − 2α(γ − β))K 2 (t)t−α(γ−β) .

(2.9)

We also consider the integral Φ(t). Since (f) holds, the inequality |P f (s)|q ≤ M (t)s−α(1−β) is satisfied with a continuous function M (t). From Step 3 in Theorem 2.1, we derive that Aγ Φ(t) is continuous in (0, T ] with |Aγ Φ(t)|q ≤ C1 M (t)B(α(1 − γ), 1 − α(1 − β))t−α(γ−β) .

(2.10)

For |P f (t)|q = o(t−α(1−β) ) as t → 0, we have M (t) = 0. Here (2.10) means |Aγ Φ(t)|q = o(t−α(γ−β) ) as t → 0. We prove that Φ is continuous in Jq . In fact, take 0 ≤ t0 < t < T , we have |Φ(t) − Φ(t0 )|q Z t Z α−1 ≤C1 (t − s) |P f (s)|q ds + C1 t0

t0


(t0 − s)α−1 − (t − s)α−1 |P f (s)|q ds

0

Z

t0 −ε

(t0 − s)α−1 kEα,α (−(t − s)α A) − Eα,α (−(t0 − s)α A)k|P f (s)|q ds

+ C1 0

Z

t0

(t0 − s)α−1 |P f (s)|q ds

+ 2C1 t0 −ε Z t

(t − s)α−1 s−α(1−β) ds

≤C1 M (t) t0

30

Fractional Partial Differential Equations

Z

t0


(t0 − s)α−1 − (t − s)α−1 s−α(1−β) ds

+ C1 M (t) 0

Z

t0 −ε

+ C1 M (t)

(t0 − s)α−1 s−α(1−β) ds

0

×

kEα,α (−(t − s)α A) − Eα,α (−(t0 − s)α A)k

sup s∈[0,t−ε]

Z

t0

+ 2C1 M (t)

(t0 − s)α−1 s−α(1−β) ds → 0,

as t → t0 , ε → 0

t0 −ε

by previous discussion. Further, we consider the function Eα (−tα A)a. It is obvious by Lemma 2.6 that |Aγ Eα (−tα A)a|q ≤ C1 t−α(γ−β) |Aβ a|q = C1 t−α(γ−β) |a|H β,q , lim tα(γ−β) |Aγ Eα (−tα A)a|q = lim tα(γ−β) |Eα (−tα A)a|H γ,q = 0.

t→0

t→0

Step 2. Now we construct the solution by the successive approximation: u0 (t) = Eα (−tα A)a + Φ(t), un+1 (t) = u0 (t) + G(un , un )(t), n = 0, 1, 2... .

(2.11)

Making use of above results, we know that Kn (t) := sup sα(γ−β) |Aγ un (s)|q s∈(0,t]

are continuous and increasing functions on [0, T ] with Kn (0) = 0. Furthermore, in virtue of (2.9) and (2.11), Kn (t) fulfills the following inequality Kn+1 (t) ≤ K0 (t) + M C1 B(α(1 − γ), 1 − 2α(γ − β))Kn2 (t).

(2.12)

For K0 (0) = 0, we choose a T > 0 such that 4M C1 B(α(1 − γ), 1 − 2α(γ − β))K0 (T ) < 1.

(2.13)

Then a fundamental consideration of (2.12) ensures that the sequence {Kn (T )} is bounded, i.e., Kn (T ) ≤ ρ(T ), n = 0, 1, 2, ..., where ρ(t) =

1−

p 1 − 4M C1 B(α(1 − γ), 1 − 2α(γ − β))K0 (t) . 2M C1 B(α(1 − γ), 1 − 2α(γ − β))

Analogously, for any t ∈ (0, T ], Kn (t) ≤ ρ(t) holds. In the same way we note that ρ(t) ≤ 2K0 (t). Let us consider the equality Z t wn+1 (t) = (t − s)α−1 Eα,α (−(t − s)α A) 0

× [F (un+1 (s), un+1 (s)) − F (un (s), un (s))]ds,

Fractional Navier-Stokes Equations

31

where wn = un+1 − un , n = 0, 1, ..., and t ∈ (0, T ]. Writing Wn (t) := sup sα(γ−β) |Aγ wn (s)|q . s∈(0,t]

On account of (2.6), we have |F (un+1 (s), un+1 (s)) − F (un (s), un (s))|q ≤ M (Kn+1 (s) + Kn (s))Wn (s)s−2α(γ−β) , which follows from Step 2 in Theorem 2.1 that tα(γ−β) |Aγ wn+1 (t)|q ≤ 2M C1 B(α(1 − γ), 1 − α(1 − β))ρ(t)Wn (t). This inequality gives Wn+1 (T ) ≤2M C1 B(α(1 − γ), 1 − 2α(γ − β))ρ(T )Wn (T ) ≤4M C1 B(α(1 − γ), 1 − 2α(γ − β))K0 (T )Wn (T ).

(2.14)

According to (2.13) and (2.14), it is easy to see that Wn+1 (T ) < 4M C1 K0 (T )B(α(1 − γ), 1 − 2α(γ − β)) < 1, Wn (T ) P∞ thus the series converges. It shows that the series n=0 Wn (T ) P∞ α(γ−β) γ A wn (t) converges uniformly for t ∈ (0, T ], therefore, the sequence n=0 t {tα(γ−β) Aγ un (t)} converges uniformly in (0, T ]. This implies that lim

n→∞

lim un (t) = u(t) ∈ D(Aγ )

n→∞

and lim tα(γ−β) Aγ un (t) = tα(γ−β) Aγ u(t) uniformly,

n→∞

since A−γ is bounded and Aγ is closed. Accordingly, the function K(t) = sups∈(0,t] sα(γ−β) |Aγ u(s)|q also satisfies K(t) ≤ ρ(t) ≤ 2K0 (t), t ∈ (0, T ]

(2.15)

and ςn := sup s2α(γ−β) |F (un (s), un (s)) − F (u(s), u(s))|q s∈(0,T ]

≤M (Kn (T ) + K(T )) sup sα(γ−β) |Aγ (un (s) − u(s))|q s∈(0,T ]

→ 0,

as n → ∞.

Finally, it remains to verify that u is a mild solution of problem (2.2) in [0, T ]. Since Z t |G(un , un )(t) − G(u, u)(t)|q ≤C1 (t − s)α−1 ςn s−2α(γ−β) ds 0

=C1 B(α, 1 − 2α(γ − β))tαβ ςn →0,

as n → ∞,

32

Fractional Partial Differential Equations

we have G(un , un )(t) → G(u, u)(t). Take the limit on both sides of (2.11), we derive u(t) = u0 (t) + G(u, u)(t).

(2.16)

Let u(0) = a, we find that (2.16) holds for t ∈ [0, T ] and u ∈ C([0, T ]; Jq ). Moreover, the uniform convergence of tα(γ−β) Aγ un (t) to tα(γ−β) Aγ u(t) derives the continuity of Aγ u(t) on (0, T ]. From (2.15) and K0 (0) = 0, we get that |Aγ u(t)|q = o(t−α(γ−β) ) obviously. Step 3. We prove that the mild solution is unique. Suppose that u and v are mild solutions of problem (2.2). Let w = u − v, we consider the equality Z t (t − s)α−1 Eα,α (−(t − s)α A)[F (u(s), u(s)) − F (v(s), v(s))]ds. w(t) = 0

Introducing the function e K(t) := max{ sup sα(γ−β) |Aγ u(s)|q , sup sα(γ−β) |Aγ v(s)|q }. s∈(0,t]

s∈(0,t]

By (2.6) and Lemma 2.6, we get Z

γ

e |A w(t)|q ≤ 2M C1 K(t)

t

(t − s)α(1−γ)−1 s−α(γ−β)) |Aγ w(s)|q ds.

0

The Gronwall’s inequality shows that Aγ w(t) = 0 for t ∈ (0, T ]. This implies that w(t) = u(t) − v(t) ≡ 0 for t ∈ [0, T ]. Therefore the mild solution is unique. 2.2.4

Regularity

In this subsection, we consider the regularity of a solution u which satisfies problem (2.2). Throughout this part we assume that: (f1 ) P f (t) is H¨ older continuous with an exponent ϑ ∈ (0, α(1 − γ)], that is, there exists a positive constant L > 0 such that |P f (t) − P f (s)|q ≤ L|t − s|ϑ , for all 0 < t, s ≤ T. Definition 2.3. A function u : [0, T ] → Jq is called a classical solution of problem α (2.2), if u ∈ C([0, T ]; Jq ) with C 0 Dt u(t) ∈ C((0, T ]; Jq ), which takes values in D(A) and satisfies (2.2) for all t ∈ (0, T ]. Lemma 2.7. Let (f1 ) be satisfied. If Z t
Φ1 (t) := (t − s)α−1 Eα,α (−(t − s)α A) P f (s) − P f (t) ds, 0

then Φ1 (t) ∈ D(A) and AΦ1 (t) ∈ C ϑ ([0, T ]; Jq ).

for t ∈ (0, T ],

Fractional Navier-Stokes Equations

33

Proof. For fixed t ∈ (0, T ], tα−1 AEα,α (−tα A) Z 1 = tα−1 AEα,α (−µtα )(µI + A)−1 dµ 2πi Γθ Z Z 1 1 α−1 α = t tα−1 µEα,α (−µtα )(µI + A)−1 dµ Eα,α (−µt ) − 2πi Γθ 2πi Γθ Z Z 1 1 1 1 ξ α−1 = −t Eα,α (ξ) α − tα−1 Eα,α (ξ)(− α I + A)−1 α dξ. 2πi Γθ t 2πi Γθ t t In view of kµI + Ak−1 ≤ such that

C |µ| ,

we derive that there exists a positive constant Cα

ktα−1 AEα,α (−tα A)k ≤ Cα t−1 .

(2.17)

From (2.17) and (f1 ), we have
(t − s)α−1 |AEα,α (−(t − s)α A) P f (s) − P f (t) |q ≤Cα (t − s)−1 |P f (s) − P f (t)|q ≤Cα L(t − s)

ϑ−1

(2.18)

1

∈ L (0, T ; Jq ),

then Z

t


(t − s)α−1 |AEα,α (−(t − s)α A) P f (s) − P f (t) |q ds 0 Z t Cα L ϑ ≤Cα L (t − s)ϑ−1 ds = t < ∞. ϑ 0

|AΦ1 (t)|q ≤

By the closeness of A, we obtain Φ1 (t) ∈ D(A). We need to show that AΦ1 (t) is H¨ older continuous. Since
d α−1 t Eα,α (−µtα ) = tα−2 Eα,α−1 (−µtα ), dt then
d α−1 t AEα,α (−tα A) dt Z 1 = tα−2 Eα,α−1 (−µtα )A(µI + A)−1 dµ 2πi Γθ Z Z 1 1 α−2 α = t Eα,α−1 (−µt )dµ − tα−2 µEα,α−1 (−µtα )(µI + A)−1 dµ 2πi Γθ 2πi Γθ Z Z 1 1 1 ξ ξ 1 α−2 = −t Eα,α−1 (ξ) α dξ − tα−2 Eα,α−1 (ξ) α (− α I + A)−1 α dξ. 2πi Γ0θ t 2πi Γ0θ t t t In view of k(µI + A)−1 k ≤

C |µ| ,

we derive

d


tα−1 AEα,α (−tα A) ≤ Cα t−2 , 0 < t ≤ T.

dt

34

Fractional Partial Differential Equations

By the mean value theorem, for every 0 < s < t ≤ T , we have ktα−1 AEα,α (−tα A) − sα−1 AEα,α (−sα A)k

Z t


d α−1 α

= τ AE (−τ A) dτ α,α

s dτ

Z t

d α−1


AEα,α (−τ α A) ≤

dτ

dτ τ s Z t
≤Cα τ −2 dτ = Cα s−1 − t−1 .

(2.19)

s

Let h > 0 be such that 0 < t < t + h ≤ T , then AΦ1 (t + h) − AΦ1 (t) Z t (t + h − s)α−1 AEα,α (−(t + h − s)α A) = 0

− (t − s)α−1 AEα,α (−(t − s)α A) P f (s) − P f (t) ds Z t
+ (t + h − s)α−1 AEα,α (−(t + h − s)α A) P f (t) − P f (t + h) ds

(2.20)

0

Z +

t+h


(t + h − s)α−1 AEα,α (−(t + h − s)α A) P f (s) − P f (t + h) ds

t

=:h1 (t) + h2 (t) + h3 (t). We estimate each of the three terms separately. For h1 (t), from (2.19) and (f1 ), we have Z t

k(t + h − s)α−1 AEα,α (−(t + h − s)α A)

|h1 (t)|q ≤ 0

− (t − s)α−1 AEα,α (−(t − s)α A)k|P f (s) − P f (t)|q ds Z t ≤Cα Lh (t + h − s)−1 (t − s)ϑ−1 ds 0 Z t =Cα Lh (s + h)−1 sϑ−1 ds Z ≤Cα L 0

0 h

h ϑ−1 s ds + Cα Lh s+h

Z

∞

h

(2.21)

s ϑ−2 s ds s+h

1 Cα Lhϑ . ≤ ϑ(1 − ϑ) For h2 (t), we use Lemma 2.6 and (f1 ), Z t
|h2 (t)|q ≤ (t + h − s)α−1 |AEα,α (−(t + h − s)α A) P f (t) − P f (t + h) |q ds 0 Z t ≤C1 (t + h − s)−1 |P f (t) − P f (t + h)|q ds 0 Z t ≤C1 Lhϑ (t + h − s)−1 ds 0

=C1 L(ln(t + h) − ln h)hϑ . (2.22)

Fractional Navier-Stokes Equations

35

Furthermore, for h3 (t), by Lemma 2.6 and (f1 ), we have Z t+h
(t + h − s)α−1 |AEα,α (−(t + h − s)α A) P f (s) − P f (t + h) |q ds |h3 (t)|q ≤ t t+h

Z

(t + h − s)−1 |P f (s) − P f (t + h)|q ds

≤C1 t

Z ≤C1 L

t+h

(t + h − s)ϑ−1 ds = C1 L

t

hϑ . ϑ (2.23)

Combining (2.21), (2.22) with (2.23), we deduce that AΦ1 (t) is H¨older continuous. Theorem 2.4. Let the assumptions of Theorem 2.3 be satisfied. If (f1 ) holds, then for every a ∈ D(A), the mild solution of (2.2) is a classical one. Proof. For a ∈ D(A). Then Lemma 2.2(ii) ensures that u(t) = Eα (−tα A)a (t > 0) is a classical solution to the following problem ( C α 0 Dt u = −Au, t > 0, u(0) = a. Step 1. We verify that Z t Φ(t) = (t − s)α−1 Eα,α (−(t − s)α A)P f (s)ds 0

is a classical solution to the problem ( C α 0 Dt u = −Au + P f (t),

t > 0,

u(0) = 0. It follows from Theorem 2.3 that Φ ∈ C([0, T ]; Jq ). We rewrite Φ(t) = Φ1 (t) + Φ2 (t), where Z t
Φ1 (t) = (t − s)α−1 Eα,α (−(t − s)α A) P f (s) − P f (t) ds, 0 Z t Φ2 (t) = (t − s)α−1 Eα,α (−(t − s)α A)P f (t)ds. 0

According to Lemma 2.7, we know that Φ1 (t) ∈ D(A). To prove the same conclusion for Φ2 (t). By Lemma 2.2(iii), we notice that AΦ2 (t) = P f (t) − Eα (−tα A)P f (t). Since (f1 ) holds, it follows that |AΦ2 (t)|q ≤ (1 + C1 )|P f (t)|q , thus Φ2 (t) ∈ D(A) for t ∈ (0, T ] and AΦ2 (t) ∈ C ϑ ((0, T ]; Jq ).

(2.24)

36

Fractional Partial Differential Equations

α Next, we prove C 0 Dt Φ ∈ C((0, T ]; Jq ). In view of Lemma 2.2(iv) and Φ(0) = 0, we have
d d α−1 C α Φ(t) = (Eα (−tα A) ∗ P f ). 0 Dt 0 Dt Φ(t) = dt dt It remains to prove that Eα (−tα A) ∗ P f is continuously differentiable in Jq . Let 0 < h ≤ T − t, one derives the following:
1 Eα (−(t + h)α A) ∗ P f − Eα (−tα A) ∗ P f h Z t
1 Eα (−(t + h − s)α A)P f (s) − Eα (−(t − s)α A)P f (s) ds = 0 h Z 1 t+h Eα (−(t + h − s)α A)P f (s)ds. + h t

Notice that for 0 < θ < 1, Z t 1 Eα (−(t + h − s)α A)P f (s) − Eα (−(t − s)α A)P f (s)ds q 0 h Z t = (t − s + θh)α−1 AEα,α (−(t + θh − s)α A)P f (s)ds q

0

Z ≤

t+θh

(t − s + θh)α−1 AEα,α (−(t + θh − s)α A)P f (s)ds . q

0

For any t ∈ (0, T ], due to Φ(t) ∈ D(A), one know that Z t (t − s)α−1 AEα,α (−(t − s)α A)P f (s)ds < ∞, 0

q

then using the dominated convergence theorem, we find Z t
1 Eα (−(t + h − s)α A)P f (s) − Eα (−(t − s)α A)P f (s) ds lim h→0 0 h Z t =− (t − s)α−1 AEα,α (−(t − s)α A)P f (s)ds 0

= − AΦ(t). On the other hand, Z 1 t+h Eα (−(t + h − s)α A)P f (s)ds h t Z 1 h = Eα (−sα A)P f (t + h − s)ds h 0 Z
1 h = Eα (−sα A) P f (t + h − s) − P f (t − s) ds h 0 Z Z
1 h 1 h α + Eα (−s A) P f (t − s) − P f (t) ds + Eα (−sα A)P f (t)ds. h 0 h 0

Fractional Navier-Stokes Equations

37

From Lemma 2.1 and (f1 ), we have Z h 1
α ≤ C1 Lhϑ , E (−s A) P f (t + h − s) − P f (t − s) ds α h 0 q Z h ϑ 1
α ≤ C1 L h . E (−s A) P f (t − s) − P f (t) ds α h ϑ+1 0 q R 1 h α Also Lemma 2.2(i) gives that limh→0 h 0 Eα (s A)P f (t)ds = P f (t). Hence Z 1 t+h lim Eα ((t + h − s)α A)P f (s)ds = P f (t). h→0 h t
d Eα (−tα A) ∗ P f + = We deduce that Eα (tα A) ∗ P f is differentiable at t+ and dt d −AΦ(t) + P f (t). Similarly, Eα (tα A) ∗ P f is differentiable at t− and dt Eα (−tα A) ∗
P f − = −AΦ(t) + P f (t). We show that AΦ = AΦ1 +AΦ2 ∈ C((0, T ]; Jq ). In fact, it is clear that AΦ2 (t) = P f (t)−Eα (tα A)P f (t) due to Lemma 2.2(iii), which is continuous in view of Lemma 2.1. Furthermore, according to Lemma 2.7, we know that AΦ1 (t) is also continuous. α Consequently, C 0 Dt Φ ∈ C((0, T ]; Jq ). Step 2. Let u be the mild solution of (2.2). To prove that F (u, u) ∈ C ϑ ((0, T ]; Jq ), in view of (2.6), we have to verify that Aγ u is H¨older continuous in Jq . Take h > 0 such that 0 < t < t + h. Denote ϕ(t) := Eα (−tα A)a, by Lemmas 2.2(iii) and 2.6, then Z t+h γ γ α−1 γ+1 α |A ϕ(t + h) − A ϕ(t)|q = −s A Eα,α (−s A)ads t q Z t+h ≤ sα−1 |Aγ Eα,α (−sα A)Aa|q ds t

Z ≤C1

t+h

sα(1−γ)−1 ds|Aa|q

t


C1 |a|D(A) (t + h)α(1−γ) − tα(1−γ) α(1 − γ) C1 |a|D(A) α(1−γ) h . ≤ α(1 − γ) =

Thus, Aγ ϕ ∈ C ϑ ((0, T ]; Jq ). For every small ε > 0, take h such that ε ≤ t < t + h ≤ T , since |Aγ Φ(t + h) − Aγ Φ(t)|q Z t+h ≤ (t + h − s)α−1 Aγ Eα,α (−(t + h − s)α A)P f (s)ds t q Z t + Aγ (t + h − s)α−1 Eα,α (−(t + h − s)α A) 0
α−1 α − (t − s) Eα,α (−(t − s) A) P f (s)ds q

=:φ1 (t) + φ2 (t).

38

Fractional Partial Differential Equations

Applying Lemma 2.6 and (f), we get Z t+h φ1 (t) ≤C1 (t + h − s)α(1−γ)−1 |P f (s)|q ds t

Z ≤C1 M (t)

t+h

(t + h − s)α(1−γ)−1 s−α(1−β) ds

t

C1 hα(1−γ) t−α(1−β) α(1 − γ) C1 hα(1−γ) ε−α(1−β) . ≤M (t) α(1 − γ)

≤M (t)

To estimate φ2 , we give the equations Z
d α−1 γ 1 α t A Eα,α (−t A) = µγ tα−2 Eα,α−1 (−µtα )(µI + A)−1 dµ dt 2πi Γ Z ξ ξ 1 1 −(− α )γ tα−2 Eα,α−1 (ξ)(− α I + A)−1 α dξ, = 2πi Γ0 t t t
d this yields that k dt tα−1 Aγ Eα,α (−tα A) k ≤ Cα tα(1−γ)−2 . The mean value theorem shows ktα−1 Aγ Eα,α (−tα A) − sα−1 Aγ Eα,α (−sα A)k

Z t

d α−1 γ
α

dτ ≤ τ A E (−τ A) α,α

dτ

s Z t
Cα sα(1−γ)−1 − tα(1−γ)−1 , ≤Cα τ α(1−γ)−2 dτ = 1 − α(1 − γ) s

(2.25)

thus Z

t

|Aγ (t + h − s)α−1 Eα,α (−(t + h − s)α A)
− (t − s)α−1 Eα,α (−(t − s)α A) P f (s)|q ds Z t
Cα (t − s)α(1−γ)−1 − (t + h − s)α(1−γ)−1 |P f (s)|q ds ≤ 1 − α(1 − γ) 0 Z t Cα M (t) ≤ (t − s)α(1−γ)−1 s−α(1−β) ds 1 − α(1 − γ) 0  Z t+h α(1−γ)−1 −α(1−β) − (t − s + h) s ds

φ2 (t) ≤

0

0

Z t+h Cα M (t) (t − s + h)α(1−γ)−1 s−α(1−β) ds 1 − α(1 − γ) t
Cα M (t) tα(β−γ) − (t + h)α(β−γ) B(α(1 − γ), 1 − α(1 − β)) ≤ 1 − α(1 − γ) Cα + M (t)hα(1−γ) t−α(1−β) α(1 − γ)(1 − α(1 − γ)) +

Fractional Navier-Stokes Equations

≤

39

Cα M (t)hα(γ−β) [ε(ε + h)]α(β−γ) B(α(1 − γ), 1 − α(1 − β)) 1 − α(1 − γ) Cα M (t)hα(1−γ) ε−α(1−β) , + α(1 − γ)(1 − α(1 − γ))

which ensures that Aγ Φ ∈ C ϑ ([ε, T ]; Jq ). Therefore Aγ Φ ∈ C ϑ ((0, T ]; Jq ) due to arbitrary ε. Recall Z t Ψ(t) = (t − s)α−1 Eα,α (−(t − s)α A)F (u(s), u(s))ds. 0

Since |F (u(s), u(s))|q ≤ M K 2 (t)s−2α(γ−β) , where K(t) := sups∈[0,t] sα(γ−β) |u(s)|H γ,q is continuous and bounded in (0, T ]. A similar argument enable us to give the H¨ older continuity of Aγ Ψ in C ϑ ((0, T ]; Jq ). γ γ γ Therefore, we have A u(t) = A ϕ(t) + A Φ(t) + Aγ Ψ(t) ∈ C ϑ ((0, T ]; Jq ). Since F (u, u) ∈ C ϑ ((0, T ]; Jq ) is proved, according to Step 2, this yields that C α C α 0 Dt Ψ ∈ C((0, T ]; Jq ), AΨ ∈ C((0, T ]; Jq ) and 0 Dt Ψ = −AΨ + F (u, u). In this α C α way we obtain that 0 Dt u ∈ C((0, T ]; Jq ), Au ∈ C((0, T ]; Jq ) and C 0 Dt u = −Au + F (u, u) + P f , we conclude that u is a classical solution. Theorem 2.5. Assume that (f1 ) holds and a ∈ H β,q . If u is a classical solution of the equation (2.2), then for every ε > 0, Au ∈ C min{αβ,α(1−β),ϑ} ([ε, T ]; Jq ) and C α min{αβ,α(1−β),ϑ} ([ε, T ]; Jq ). 0 Dt u ∈ C Proof. If u is a classical solution of (2.2), then u(t) = ϕ(t)+Φ(t)+Ψ(t). It remains to show that Aϕ ∈ C min{αβ,α(1−β)} ([ε, T ]; Jq ) for every ε > 0, it suffices to prove that Aϕ ∈ C min{αβ,α(1−β)} ([ε, T ]; Jq ). In fact, take h such that ε ≤ t < t + h ≤ T , by Lemma 2.2(iv) and (2.25), we have |Aϕ(t + h) − Aϕ(t)|q Z t+h 1 −α α−1 1−β α β s (t + h − s) A Eα,α (−(t + h − s) A)A ads ≤ Γ(1 − α) t q Z t 1 + s−α [(t + h − s)α−1 A1−β Eα,α (−(t + h − s)α A) Γ(1 − α) 0 α−1 1−β α β − (t − s) A Eα,α (−(t − s) A)]A ads q

≤

C1 |a|H β,q Γ(1 − α)

Z

t+h

s−α (t + h − s)αβ−1 ds

t

Cα |a|H β,q + (1 − αβ)Γ(1 − α)

Z 0

t

s−α [(t − s)αβ−1 − (t + h − s)αβ−1 ]ds.

40

Fractional Partial Differential Equations

Using the similar arguments as we estimated the two functions φ1 (t) and φ2 (t), one can show that Z t+h 1 αβ −α s−α (t + h − s)αβ−1 ds ≤ h ε , αβ t Z t s−α [(t − s)αβ−1 − (t + h − s)αβ−1 ]ds ≤hα(1−β) [ε(ε + h)]−α(1−β) B(1 − α, αβ) 0

+

1 αβ −α h ε . αβ

Similar to Lemma 2.7, we write Φ(t) as Z t
(t − s)α−1 Eα,α (−(t − s)α A) P f (s) − P f (t) ds Φ(t) = Φ1 (t) + Φ2 (t) = 0 Z t + (t − s)α−1 Eα,α (−(t − s)α A)P f (t)ds, 0

for t ∈ (0, T ]. It follows from Lemma 2.7 and (2.24) that AΦ1 (t) ∈ C ϑ ([0, T ]; Jq ) and AΦ2 (t) ∈ C ϑ ([ε, T ]; Jq ), respectively. Since F (u, u) ∈ C ϑ ([ε, T ]; Jq ), the result related to the function Ψ(t) is proved by similar argument, which means that AΨ ∈ C ϑ ([ε, T ]; Jq ). Therefore Au ∈ ϑ α C ϑ ([ε, T ]; Jq ) and C 0 Dt u = Au + F (u, u) + P f ∈ C ([ε, T ]; Jq ). The proof is completed. 2.3

Existence and H¨ older Continuity

In this section, we discuss the following time-fractional Navier-Stokes equations:  α ∂ u(t, x) − ∆u(t, x) + u(t, x) · ∇u(t, x)   t    (t, x) ∈ (0, T ] × Ω,  = −∇p(t, x) + f (t, x),  (2.26) ∇ · u(t, x) = 0, (t, x) ∈ (0, T ] × Ω,     u(t, x) = 0, (t, x) ∈ (0, T ] × ∂Ω,    u(0, x) = a(x), x ∈ Ω, in a bounded set Ω ⊂ RN (N ≥ 2) with smooth boundary ∂Ω, where ∂tα is the Caputo fractional derivative of order α ∈ (0, 1), u(t, x) represents the velocity field at (t, x), p(t, x) is the pressure, f (t, x) is the external force and a(x) is the initial velocity. Applying the classic approach, the problem of solutions to (2.26) is equivalent to that of the following equations ( C α 0 Dt u = −Au + F (u, u) + P f, t ∈ (0, T ], (2.27) u(0) = a, where P is the Helmholtz projector, F (u, v) = −P (u · ∇)v. The purpose of this section is to investigates the problem of solutions to (2.26). The existence, uniqueness and H¨ older continuity of the local mild solutions are

Fractional Navier-Stokes Equations

41

established. The section is organized as follows. In Subsection 2.3.1 we introduce some notations, definitions, and preliminary results. Subsection 2.3.2 is concerned with the existence and uniqueness of local mild solutions of problem (2.27) by using the iteration method, then proceed to study H¨older continuity of mild solutions. 2.3.1

Notations and Lemmas

In this subsection, we present notations, definitions, and preliminary facts which are used throughout this section. Let H = the closure in (Lr (Ω))N of {u ∈ (C0∞ (Ω))N : ∇ · u = 0}, (1 < r < ∞). We denote by P the orthogonal projection P : (Lr (Ω))N → H. The Stokes operator A is defined by A : D(A) ⊂ H → H, A = −P ∆, D(A) = H ∩ {v ∈ (Wr2 (Ω))N : v|∂Ω = 0}. Here, (Wr2 (Ω))N is a Sobolev space. As such −A generates the bounded analytic semigroup {e−tA }, see Constantin and Foicas [85]. This allows us to define the fractional powers Aβ (β ∈ R) in the usual way. Lemma 2.8. [138] (i) kAβ e−tA k ≤ t−β for 0 < β ≤ e = 2.718..., t > 0. (ii) |(e−hA − I)v| ≤ β1 hβ |Aβ v| for 0 < β < 1, h > 0 and v ∈ D(Aβ ). (iii) Let 0 ≤ δ < 21 + N2 (1 − 1r ). Then |A−δ P (u · ∇)v| ≤ M |Aρ u||Aκ v| with some N + 21 , ρ > 0, κ > 0, κ+δ > constant M = M (δ, r, ρ, κ), provided that δ +ρ+κ ≥ 2r 1 2. Inspired by Zhou [386, 387], we introduce the following definition of mild solutions to problem (2.27). Definition 2.4. Let 0 < T ≤ ∞. A function u : [0, T ] → H is called a mild solution of problem (2.27), if u ∈ C([0, T ]; H) and Z t α u(t) =Eα (−t A)a + (t − s)α−1 (2.28) 0
α × Eα,α (−(t − s) A) F (u(s), u(s)) + P f (s) ds, t ∈ [0, T ], where Z Z ∞

α

Mα (s)e−st

Eα (−tα A) =

∞

A

0

α

αsMα (s)e−st

ds, Eα,α (−tα A) =

A

ds,

0

Mα (θ) =

∞ X

(−θ)n . n!Γ(1 − α(1 + n)) n=0

We state some important technical lemmas that contributes to the main results. Lemma 2.9. [352] Let 0 < α < 1. Then for t > 0, Eα (−tα A) and Eα,α (−tα A) are continuous in the uniform operator topology. Moreover, for every r > 0, the continuity is uniform on [r, ∞).

42

Fractional Partial Differential Equations

Lemma 2.10. [402] Let 0 < α < 1 and γ ∈ [0, 1). Then there exists a constant Cγ = C(α, γ) > 0 such that |Aγ Eα (−tα A)v| ≤ Cγ t−αγ |v| and |Aγ Eα,α (−tα A)v| ≤ Cγ t−αγ |v|, for all v ∈ H and t ∈ (0, T ]. Furthermore, limt→0+ tαγ |Aγ Eα (−tα A)v| = 0. For convenience, let t

Z

(t − s)α−1 Eα,α (−(t − s)α A)P f (s)ds,

Φ(t) = 0 t

Z

(t − s)α−1 Eα,α (−(t − s)α A)F (u(s), v(s))ds.

G(u, v)(t) = 0

2.3.2

Existence and Uniqueness

This subsection is to show the existence and uniqueness of mild solutions to problem (2.27), this is achieved by iteration methods. Set Bγ = B(α(1 − γ − δ), 1 − α(1 − β − δ)). Theorem 2.6. Fix β < 1 and choose δ > 0 such that N 1 − ≤ β, −β < δ < 1 − |β|. 2r 2 Suppose that |A−δ P f (t)| is continuous for t ∈ (0, T ] and |A−δ P f (t)| = o(tα(β+δ−1) ) as t → 0. Then for a ∈ D(Aβ ) there exist T ∗ ∈ (0, T ) and a unique function u(t) satisfying: (a) u ∈ C([0, T ∗ ]; D(Aβ )) and u(0) = a; (b) u ∈ C((0, T ∗ ]; D(Aγ )) and limt→0+ tα(γ−β) |Aγ u(t)| = 0 for all γ with γ ∈ (β, 1 − δ);
Rt (c) u(t) = Eα (−tα A)a + 0 (t − s)α−1 Eα,α (−(t − s)α A) F (u(s), u(s)) + P f (s) ds. Proof. We construct the solution by means of the successive approximation: u0 (t) = Eα (−tα A)a + Φ(t), um+1 (t) = u0 (t) + G(um , um )(t), m = 0, 1, ... .

(2.29)

Step 1. We estimate |Aγ um (t)| by making use of mathematical induction. For u0 (t), in view of Lemma 2.10, we obtain Z t |Aγ u0 (t)| ≤|Aγ Eα (−tα A)a| + (t − s)α−1 |Aγ Eα,α (−(t − s)α A)P f (s)|ds 0 Z t ≤|Aγ Eα (−tα A)a| + Cγ+δ (t − s)α(1−γ−δ)−1 |A−δ P f (s)|ds 0 Z t γ α ≤|A Eα (−t A)a| + Cγ+δ L (t − s)α(1−γ−δ)−1 s−α(1−β−δ) ds 0

=Kγ,0 t−α(γ−β) , for β ≤ γ < 1 − δ,

Fractional Navier-Stokes Equations

43

where L = sup {sα(1−β−δ) |A−δ P f (s)|}, s∈(0,T ]

Kγ,0 = sup tα(γ−β) |Aγ Eα (−tα A)a| + Bγ Cγ+δ L. t∈(0,T ]

Suppose that for some m ≥ 0, |Aγ um (t)| ≤ Kγ,m t−α(γ−β) .

(2.30)

We estimate |Aγ um+1 (t)|. On the one hand, the inequality δ < 1 − |β| implies the existence of ρ and κ such that ρ + κ + δ = 1 + β, β < ρ, κ < 1 − δ, ρ > 0, κ > 0, ρ + κ >

1 . 2

(2.31)

N − 12 ≤ β, N ≥ 2 and δ < 1 − |β|, it is easy to know that δ, ρ, κ satisfy the Since 2r conditions of Lemma 2.8(iii), which together with (2.30) follows that

|A−δ F (um (t), um (t))| ≤ M |Aρ um (t)||Aκ um (t)| ≤ M Kρ,m Kκ,m t−α(1−β−δ) . (2.32) Therefore, (2.29) and Lemma 2.10 yield that |Aγ um+1 (t)| Z t ≤|Aγ u0 | + (t − s)α−1 kAγ+δ Eα,α (−(t − s)α A)k|A−δ F (um (s), um (s))|ds 0 Z t ≤Kγ,0 t−α(γ−β) + Cγ+δ M Kρ,m Kκ,m (t − s)α(1−γ−δ)−1 s−α(1−β−δ) ds 0

=Kγ,m+1 t

−α(γ−β)

,

where Kγ,m+1 = Kγ,0 + Bγ Cγ+δ M Kρ,m Kκ,m .

(2.33)

Thus um (t) satisfies (2.30) with Kγ,m defined recursively as above. Step 2. To prove that um ∈ C((0, T ]; D(Aγ )) and um ∈ C([0, T ]; D(Aβ )) by using mathematical induction. We need to prove u0 ∈ C((0, T ]; D(Aγ )). For ε > 0 enough small, choose h > 0 such that t + h > t > 0 (the case t − h < t follows analogously), we have |Aγ u0 (t + h) − Aγ u0 (t)|
≤ Aγ−β Eα (−(t + h)α A) − Aγ−β Eα (−tα A) Aβ a Z t+h + (t + h − s)α−1 |Aγ+δ Eα,α (−(t + h − s)α A)A−δ P f (s)|ds t Z t
+ (t − s)α−1 − (t + h − s)α−1 |Aγ+δ Eα,α (−(t + h − s)α A)A−δ P f (s)|ds 0

44

Fractional Partial Differential Equations

Z

t−ε

(t − s)α−1 Aγ+δ Eα,α (−(t + h − s)α A) 0
− Eα,α (−(t − s)α A) A−δ P f (s) ds Z t (t − s)α−1 Aγ+δ Eα,α (−(t + h − s)α A) + t−ε
− Eα,α (−(t − s)α A) A−δ P f (s) ds

+

= : I11 (t) + I12 (t) + I13 (t) + I14 (t) + I15 (t). We estimate each of the five terms separately. For I11 (t), it is clear that limh→0 I11 (t) = 0 due to Lemma 2.9. For I12 (t), in view of Lemma 2.10, one finds Z t+h I12 (t) ≤Cγ+δ (t + h − s)α(1−γ−δ)−1 |A−δ P f (s)|ds t

Z

t+h

(t + h − s)α(1−γ−δ)−1 s−α(1−β−δ) ds

≤Cγ+δ L t

=Cγ+δ L(t + h)−α(γ−β)

Z

1

(1 − s)α(1−γ−δ)−1 s−α(1−β−δ) ds.

t/(t+h)

On account of the properties of the Beta function, there exists  > 0 small enough such that for 0 < h < , Z 1 (1 − s)α(1−γ−δ)−1 s−α(1−β−δ) ds → 0, t/(t+h)

which implies that I12 (t) tends to 0 as h → 0. For I13 (t) and I14 (t), since Z t
I13 (t) ≤Cγ+δ L (t − s)α−1 − (t + h − s)α−1 (t + h − s)−α(γ+δ) s−α(1−β−δ) ds 0 Z t ≤2Cγ+δ L (t − s)α(1−γ−δ)−1 s−α(1−β−δ) ds = 2Cγ+δ LBγ t−α(γ−β) 0

and t−ε

Z

(t − s)α−1 |Aγ+δ Eα,α (−(t + h − s)α A)A−δ P f (s)|ds

I14 (t) ≤ 0

Z

t−ε

(t − s)α−1 |Aγ+δ Eα,α (−(t − s)α A)A−δ P f (s)|ds Z t−ε
≤Cγ+δ L (t − s)α−1 (t + h − s)−α(γ+δ) + (t − s)−α(γ+δ) s−α(1−β−δ) ds 0 Z t−ε ≤2Cγ+δ L (t − s)α(1−γ−δ)−1 s−α(1−β−δ) ds, +

0

0

then we know from the Lebesgue’s dominated convergence theorem and Lemma 2.9 that limh→0 I13 (t) = 0 and Z t−ε lim I14 (t) = (t − s)α−1 lim Aγ+δ Eα,α (−(t + h − s)α A) h→0

0

h→0

Fractional Navier-Stokes Equations

45


− Eα,α (−(t − s)α A) A−δ P f (s)|ds =0. For I15 (t), we calculate immediately, Z t
I15 (t) ≤Cγ+δ (t − s)α−1 (t + h − s)−α(γ+δ) + (t − s)−α(γ+δ) |A−δ P f (s) ds t−ε

Z

t

(t − s)α(1−γ−δ)−1 s−α(1−β−δ) ds

≤2Cγ+δ L t−ε

→ 0,

as ε → 0.

Therefore, we show |Aγ u0 (t + h) − Aγ u0 (t)| → 0,

as h → 0.

From (2.29) and (2.32), we derive um ∈ C((0, T ]; D(Aγ )) recursively by the similar discussion as above. So we omit the details. Similarly, we can show um ∈ C([0, T ); D(Aβ )). Step 3. Set Km = max{Kρ,m , Kκ,m }, Then (2.33) yields that Km+1 = that if

C1 = max{Cρ+δ , Cκ+δ },

An elementary calculation shows

1 , 4B1 C1 M

(2.34)

K0 < then for each m ≥ 1, 1−

√

1 − 4B1 C1 M 1 < , 2B1 C1 M 2B1 C1 M ≤ Kγ,0 + Bγ C1 M K 2 =: Kγ0 ,

Km < K := Kγ,m+1

B1 = max{Bρ , Bκ }.

2 . K0 +B1 C1 M Km

(2.35)

|Aγ um+1 (t)| ≤ Kγ0 t−α(γ−β) . Let us consider the equality Z t wm+1 (t) = (t − s)α−1 0

× Eα,α (−(t − s)α A)[F (um+1 (s), um+1 (s)) − F (um (s), um (s))]ds, where wm = um+1 − um , m = 0, 1, ..., and t ∈ (0, T ]. By Lemma 2.8(iii), we have −δ
A F (um+1 (s), um+1 (s)) − F (um (s), um (s))
≤M |Aρ wm ||Aκ um+1 | + |Aρ um ||Aκ wm | , which follows from with an induction on m that |Aγ wm+1 (t)| ≤ 2KBγ C1 (2KM B1 C1 )m−1 t−α(γ−β) , for each γ such that γ < 1 − δ. Since 2B1 C1 M K < 1 by (2.35), this means that there exists u ∈ C((0, T ); D(Aγ )) ∩ C([0, T ); D(Aβ )) such that β
A um (t) − u(t) → 0 uniformly on [0, T ],

46

Fractional Partial Differential Equations

γ
A um (t) − u(t) → 0 uniformly on [ε, T ] (0 < ε < T ) as m → ∞. γ 0 −α(γ−β) Moreover, . This in turn gives −δ |A u(t)| ≤ Kγ t
A F (um (s), um (s)) − F (u(s), u(s))
≤M |Aρ (um (t) − u(t))||Aκ um | + |Aρ um ||Aκ (um (t) − u(t))| → 0, as m → ∞, and |A−δ F (um (t), um (t))| ≤ C2 tα(β+δ−1) with a constant C2 > 0 independent of m. Take the limit on both sides of (2.29), using the dominated convergence theorem, we derive u(t) = u0 (t) + G(u, u)(t). (2.36) Since a ∈ D(Aβ ), Lemma 2.10 implies that tα(γ−β) |Aγ Eα (−tα A)a| → 0 as t → 0+ for γ > β. If T > 0 is chosen sufficiently small then Kγ,0 (β < γ < 1 − δ) becomes small and K0 satisfies (2.34). This shows the existence of T ∗ > 0 with the desired property (b). Step 4. Assume that u and v are mild solutions of problem (2.27). By the above estimates, we suppose without loss of generality that there exists χ, |β| < χ < 1 − δ, such that |Aγ u(t)| = o(t−α(γ−β) ), |Aγ v(t)| = o(t−α(γ−β) ), for all γ withZβ < γ < χ. Let w = u − v, we consider the equality t

(t − s)α−1 Eα,α (−(t − s)α A)[F (u(s), u(s)) − F (v(s), v(s))]ds.

w(t) = 0

Set δ 0 = 1 − χ, then |β| < χ implies δ 0 < 1 − |β|. Note that β + δ 0 > β + δ > 0. As Step 1, we can choose ρ and κ , |β| < ρ, κ < χ such that |Aγ w(t)| Z t (2.37)
0 ≤M Cγ+δ0 (t − s)α(1−γ−δ )−1 |Aρ w(s)||Aκ u(s)| + |Aρ v(s)||Aκ w(s)| ds 0

for β < γ < χ. For γ = ρ and γ = κ, let K(t0 ) (to t0 ∈ (0, T ∗ ]) be a constant such that |Aγ u(t)| ≤ K(t0 )t−α(γ−β) and |Aγ v(t)| ≤ K(t0 )t−α(γ−β) , where K(t0 ) → 0 as t0 → 0. In view of (2.37), we obtain Z t 0 |Aρ w(t)| ≤2K 2 (t0 )M Cρ+δ0 (t − s)α(1−ρ−δ )−1 s−α(ρ+κ−2β) ds 0 Z t 0 + M Cρ+δ0 (t − s)α(1−ρ−δ )−1 |Aρ w(s)||Aκ u(s)|ds 0

≤2K 2 (t0 )M B2 Cρ+δ0 t−α(ρ−β) Z t 0 + K(t0 )M Cρ+δ0 (t − s)α(1−ρ−δ )−1 s−α(κ−β) |Aρ w(s)|ds, 0

where B2 = B(α(1 − ρ − δ 0 ), 1 − α(ρ + κ − 2β)). By induction, we see that |Aρ w(t)| ≤ 2K(t0 )(2K(t0 )M B2 Cρ+δ0 )m t−α(ρ−β) for each m ≥ 1. Similarly, |Aκ w(t)| ≤ 2K(t0 )(2K(t0 )M B2 Cρ+δ0 )m t−α(κ−β) for each m ≥ 1. Choose t0 > 0 sufficiently small such that 2K(t0 )M B2 Cρ+δ0 < 1. Hence w(t) = 0 on [0, t0 ]. Repeating this argument on [t0 , T ∗ ], one finds that w(t) = 0 on [0, T ∗ ]. Therefore, the mild solution is unique.

Fractional Navier-Stokes Equations

2.3.3

47

H¨ older Continuous

Theorem 2.7. Assume that u is the solution given by Theorem 2.6. Then for β < γ < 1 − δ and arbitrary ε ∈ (0, T ∗ ), Aγ u is uniform H¨ older continuous on [ε, T ∗ ]. Proof. Denote ϕ(t) := Eα (−tα A)a. For arbitrary s ≤ t and 0 < ϑ < α, by Lemma 2.8, we have γ −tα θA α α α α ϑ ϑ A [e − e−s θA ]a ≤ [e(s θ−t θ)A − I]A− α · Aγ+ α −β e−s θA Aβ a Z tα θ−sα θ α ϑ ϑ ≤ kA1− α e−τ A kdτ |Aγ+ α −β e−s θA Aβ a| 0 tα θ−sα θ

Z

ϑ

ϑ

τ α −1 dτ (sα θ)β− α −γ |Aβ a|

≤ 0

α ≤ (t − s)ϑ sα(β−γ)−ϑ θβ−γ |Aβ a|. ϑ Choose h satisfying ε ≤ t < t + h ≤ T , by the definition of ϕ, we have Z ∞ α α |Aγ ϕ(t + h) − Aϕ(t)| ≤ Mα (θ)|Aγ [e−(t+h) θA − e−t θA ]a|dθ 0 Z ∞ β ϑ α(β−γ)−ϑ ≤|A a|h t θβ−γ Mα (θ)dθ 0

Γ(β + 1 − γ) . Γ(1 + α(β − γ)) It suffices to prove the H¨ older continuity of Aγ v, where Z t
v(t) = (t − s)α−1 Eα,α (−(t − s)α A) F (u(s), u(s)) + P f (s) ds. ≤|Aβ a|hϑ εα(β−γ)−ϑ

0

Let t1 , t2 satisfy 0 < t1 < t2 ≤ T ∗ . From Lemma 2.10, we find α−1 γ+δ ktα−1 Aγ+δ Eα,α (−tα A Eα,α (−tα 2 A) − t1 1 A)k

2 α−1


α−1 γ+δ α ≤ t 2 − t 1 A Eα,α (−t2 A)


α−1 γ+δ α

A Eα,α (−tα + t1 2 A) − Eα,α (−t1 A)
−α(γ+δ) ≤Cγ+δ tα−1 − tα−1 t2 1 2

Z t2

Z ∞

α−1 α−1 1+γ+δ −θτ α A

+ t1 αθMα (θ) αθτ A e dτ

dθ t1 0
−α(γ+δ) ≤Cγ+δ tα−1 − tα−1 t2 1 2 Z ∞ −α(γ+δ) −α(γ+δ)
− t α2 θ1−γ−δ Mα (θ)dθ + C3 tα−1 t 1 1 2 0 α(1−γ−δ)−1

≤Cα t1

α(1−γ−δ)−1


− t2

,

where C3 is a positive constant depending on α, γ, δ. Take h such that ε ≤ t < t + h ≤ T ∗ , then |Aγ v(t + h) − Aγ v(t)|

48

Fractional Partial Differential Equations

Z ≤

t+h α−1


A Eα,α (−(t + h − s) A) F (u(s), u(s)) + P f (s) ds γ

α

(t + h − s) Z t Aγ (t + h − s)α−1 Eα,α (−(t + h − s)α A) + t

0

α−1

− (t − s)


Eα,α (−(t − s) A) F (u(s), u(s)) + P f (s) ds α




Z t+h
≤Cγ+δ (t + h − s)α(1−γ−δ)−1 |A−δ F (u(s), u(s))| + |A−δ P f (s)| ds t Z t

γ+δ

A (t + h − s)α−1 Eα,α (−(t + h − s)α A) + 0

− (t − s)α−1 Eα,α (−(t − s)α A) |A−δ F (u(s), u(s))| + |A−δ P f (s)| ds Z t+h ≤Cε Cγ+δ (t + h − s)α(1−γ−δ)−1 ds t Z t
+ Cγ+δ (t − s)α(1−γ−δ)−1 − (t + h − s)α(1−γ−δ)−1 s−α(1−β−δ) ds, 0  where Cε = supε≤s≤T ∗ |A−δ F (u(s), u(s))| + |A−δ P f (s)| . By immediate calculation, the first integral of the last inequality is less than Cε C3 hα(1−γ−δ) . Furthermore, since (1 + ht )1−α(1−γ−δ) − 1  h 1−α(1−γ−δ) h 1 − (1 + )α(1−γ−δ)−1 = , ≤ t t (1 + ht )1−α(1−γ−δ) then Z t
(t − s)α(1−γ−δ)−1 − (t + h − s)α(1−γ−δ)−1 s−α(1−β−δ) ds 0

=tα(β−γ) ≤εα(β−γ)

Z

1


h + s)α(1−γ−δ)−1 (1 − s)−α(1−β−δ) ds t 0 Z
1 α(1−γ−δ)−1 h 1 − (1 + )α(1−γ−δ)−1 s (1 − s)−α(1−β−δ) ds t 0 sα(1−γ−δ)−1 − (

≤Bγ εα(1+β−2γ−δ)−1 h1−α(1−γ−δ) . Therefore, Aγ u is uniform H¨ older continuous on every interval [ε, T ∗ ]. 2.4

Approximations of Solutions

In this section, we mainly discuss the approximation of solutions to the following time-fractional Navier-Stokes equations:  α  ∂t u(t, x) − ν∆u(t, x) + u(t, x) · ∇u(t, x)     = −∇p(t, x) + f (t, x), (t, x) ∈ (0, T ] × Ω,   (2.38) ∇ · u(t, x) = 0, (t, x) ∈ (0, T ] × Ω,     u(t, x) = 0, (t, x) ∈ (0, T ] × ∂Ω,    u(0, x) = a(x), x ∈ Ω,

Fractional Navier-Stokes Equations

49

in an open set Ω ⊂ RN (N ≥ 2) with smooth boundary ∂Ω, where ∂tα is the Caputo fractional derivative of order α ∈ (0, 1), u(t, x) represents the velocity field at (t, x), p(t, x) is the pressure, ν the viscosity, f (t, x) is the external force and a(x) is the initial velocity. Firstly, we get rid of the pressure term by applying Helmholtz projector P to equations (2.38), which converts equation (2.38) to  α ∂t u − νP ∆u + P (u · ∇)u = P f, t > 0,      ∇ · u = 0,  u|∂Ω = 0,     u(0, x) = a. The operator −νP ∆ with Dirichlet’s boundary conditions is, basically, the Stokes operator A in the divergence-free function space under consideration. Then the problem of solutions to (2.38) is equivalent to that of the following equations ( C α 0 Dt u = −Au + F (u, u) + P f, t ∈ (0, T ], (2.39) u(0) = a, where F (u, v) = −P (u · ∇)v. Our approach used to prove these results is that numerical and analytic approximations. The section is organized as follows. In Subsection 2.4.1, we introduce some notations, definitions, and preliminary results. In Subsection 2.4.2, we use fixed point method to obtain the existence and uniqueness of mild solutions to each approximate equation, as well as the convergence of the approximate solutions. Subsection 2.4.3 is devoted to some convergence results for the Faedo-Galerkin approximations of problem (2.39). 2.4.1

Preliminaries

Let V = {u ∈ (C0∞ (Ω))N : ∇ · u = 0},

H = the closure of V in (L2 (Ω))N ,

V = the closure of V in (H01 (Ω))N . We denote by P the orthogonal projection P : (L2 (Ω))N → H. Here, (H01 (Ω))N and H 2 (Ω) are Sobolev spaces. The Stokes operator A is defined by A : D(A) ⊂ H → H, A = −P ∆, D(A) = H 2 (Ω) ∩ V. It is known that the Stokes operator A is symmetric, self-adjoint and A−1 is compact operator in H. Therefore, A−1 is self-adjoint, injective and compact, and then, there exists a sequence of positive number µj > 0, µj+1 ≤ µj and an orthogonal basis of −1 H, {ϕj } such that A−1 ϕj = µj ϕj . We denote λj = µ−1 has range in j . Since A D(A), then Aϕj = λj ϕj , for ϕj ∈ D(A); 0 < λ1 ≤ ...λj ≤ λj+1 ≤ ...;

lim λj = ∞.

j→∞

50

Fractional Partial Differential Equations

As such −A generates the bounded analytic semigroup {e−tA }, see Constantin and Foicas [85]. Let X be a Banach space, α ∈ (0, 1] and v : [0, T ] → X. We define the RiemannLiouville integrals of v: −α 0 Dt v(t)

= (gα ∗ v)(t), v ∈ L1 (0, T ; X), t ∈ [0, T ],

α−1

where gα = tΓ(α) , t > 0 and ∗ denotes the convolution. Further, the Caputo fractional derivative operator of order α for the function v is defined by
 d  α−1 C α v(t) − v(0) . 0 Dt 0 Dt v(t) = dt In general, for u : [0, T ] × RN → RN , 
 ∂tα u(t, x) = ∂t 0 Dtα−1 u(t, x) − u(0, x) . Definition 2.5. Let 0 < T ≤ ∞. A function u : [0, T ] → H is called a mild solution of problem (2.39), if u ∈ C([0, T ]; H) and Z t u(t) =Eα (−tα A)a + (t − s)α−1 Eα,α (−(t − s)α A)P f (s)ds 0 (2.40) Z t α−1 α + (t − s) Eα,α (−(t − s) A)F (u(s), u(s))ds, t ∈ [0, T ], 0

where α

Z

∞

Mα (s)e

Eα (−t A) =

−stα A

α

Z

ds, Eα,α (−t A) =

0

∞

α

αsMα (s)e−st

A

ds,

0

where Mα (θ) is the Wright-type function defined in Definition 1.8. We state some important technical lemmas that contribute to the main results. Lemma 2.11. [352] Let 0 < α < 1. Then for t > 0, Eα (−tα A) and Eα,α (−tα A) are continuous in the uniform operator topology. Moreover, for every r > 0, the continuity is uniform on [r, ∞). Lemma 2.12. [138] Let 0 ≤ δ
12 .

N 4

+ 21 , ρ > 0, κ > 0,

Lemma 2.13. [403] Let 1 < q < ∞ and γ ≥ 0. Then there exists a constant Cγ = C(α, γ) > 0 such that |Aγ Eα (−tα A)v| ≤ Cγ t−αγ |v| and |Aγ Eα,α (−tα A)v| ≤ Cγ t−αγ |v|, for all v ∈ H and t ∈ (0, T ]. Furthermore, limt→0+ tαγ |Aγ Eα (−tα A)v| = 0.

Fractional Navier-Stokes Equations

Throughout this section, we fix β < 1 and choose δ > 0 such that N 1 − < β, −β < δ < 1 − |β|, 4 2 and suppose that

51

(2.41)

(H) |A−δ P f (t)| is continuous for t ∈ (0, T ∗ ] and |A−δ P f (t)| = o(tα(β+δ−1) ) as t → 0. For convenience, set L=

sup {tα(1−β−δ) |A−δ P f (t)|}, Bβ = B(α(1 − β − δ), 1 − α(1 − β − δ)) t∈(0,T ∗ ]

and for all γ with γ ∈ (β, 1 − δ), we let Bγ = B(α(1 − γ − δ), 1 − α(1 − β − δ)). Choose 1

µ≤

. 4M (Bβ Cβ+δ + Bγ Cγ+δ ) From Lemma 2.13 and the assumption of f , we know that there exist T ∗ ∈ (0, T ] and  > 0 such that µ µ tα(γ−β) |Aγ Eα (−tα A)a| ≤ , tα(1−β−δ) |A−δ P f (t)| ≤ 4 4Bγ Cγ+δ for all t ∈ (0, T ∗ ] satisfying |t| < . Let  XT ∗ :=

u ∈ C([0, T ∗ ]; D(Aβ )) : tα(γ−β) u ∈ C((0, T ∗ ]; D(Aγ )) and

 sup {tα(γ−β) |u(t)|D(Aγ ) } ≤ µ

t∈(0,T ∗ ]

with the norm kukXT ∗ = sup {|u(t)|D(Aβ ) } + sup {tα(γ−β) |u(t)|D(Aγ ) }. t∈[0,T ∗ ]

2.4.2

t∈(0,T ∗ ]

Existence and Convergence of Approximate Solutions

Denote by Hn ⊂ H the finite dimensional subspace spanned by {ϕ1 , ϕ2 , . . . , ϕn } for n = 1, 2, . . . and Qn : H → Hn the corresponding projection operators. Then kQn k ≤ 1. Set Fn (·, ·) : H × H → Hn ,

Fn (u(t), u(t)) = F (Qn u(t), Qn u(t)).

Define a map Fn : XT ∗ → XT ∗ : (Fn u)(t) =Eα (−tα A)a Z t
+ (t − s)α−1 Eα,α (−(t − s)α A) Fn (u(s), u(s)) + P f (s) ds, t ∈ [0, T ∗ ]. 0

Theorem 2.8. Let (2.41) and the condition (H) be satisfied. Then for a ∈ D(Aβ ) there exists a unique function un ∈ XT ∗ satisfying Fn un = un and limt→0+ tα(γ−β) |Aγ un (t)| = 0 for all γ with γ ∈ (β, 1 − δ).

52

Fractional Partial Differential Equations

Proof. We first prove that Fn u ∈ C((0, T ∗ ]; D(Aγ )) and Fn u ∈ C([0, T ∗ ]; D(Aβ )) for any u ∈ XT ∗ . For ε > 0 enough small, choosing h > 0 such that t + h > t > 0 (the case t − h < t follows analogously), we have |Aγ Fn u(t + h) − Aγ Fn u(t)|
≤ Aγ−β Eα (−(t + h)α A) − Aγ−β Eα (−tα A) Aβ a Z t+h
+ (t + h − s)α−1 Aγ+δ Eα,α (−(t + h − s)α A)A−δ Fn (u(s), u(s)) + P f (s) ds t Z t
+ (t − s)α−1 − (t + h − s)α−1 0
× Aγ+δ Eα,α (−(t + h − s)α A)A−δ Fn (u(s), u(s)) + P f (s) ds Z t−ε + (t − s)α−1 Aγ+δ Eα,α (−(t + h − s)α A) 0

− Eα,α (−(t − s)α A) A−δ Fn (u(s), u(s)) + P f (s) ds Z t + (t − s)α−1 Aγ+δ Eα,α (−(t + h − s)α A) t−ε

− Eα,α (−(t − s)α A) A−δ Fn (u(s), u(s)) + P f (s) ds =:I11 (t) + I12 (t) + I13 (t) + I14 (t) + I15 (t). We estimate each of the five terms separately. For I11 (t), it is clear that limh→0 I11 (t) = 0 due to Lemma 2.11. On the other hand, the inequality δ < 1 − |β| implies the existence of ρ and κ such that ρ + κ + δ = 1 + β, β < ρ, κ < 1 − δ, ρ > 0, κ > 0, ρ + κ >

1 . 2

(2.42)

Since N4 − 12 < β, N ≥ 2 and δ < 1 − |β|, it is easy to know that δ, ρ, κ satisfy the conditions of Lemma 2.12, which together with u ∈ XT∗ follows that |A−δ Fn (u(t), u(t)))| =|A−δ F (Qn u(t), Qn u(t))| ≤M |Aρ Qn u(t)||Aκ Qn u(t)| ≤M kQn k2 µ2 t−α(1−β−δ)

(2.43)

≤M µ2 t−α(1−β−δ) . For I12 (t), in view of Lemma 2.13, one finds Z t+h
I12 (t) ≤Cγ+δ (t + h − s)α(1−γ−δ)−1 |A−δ P f (s)| + |A−δ Fn (u(s), u(s))| ds t

≤Cγ+δ (L + M µ2 )

Z

t+h

(t + h − s)α(1−γ−δ)−1 s−α(1−β−δ) ds

t

=Cγ+δ (L + M µ2 )(t + h)−α(γ−β)

Z

1

t/(t+h)

(1 − s)α(1−γ−δ)−1 s−α(1−β−δ) ds.

Fractional Navier-Stokes Equations

53

On account of the properties of the Beta function, there exists  > 0 small enough such that for 0 < h < , Z 1 (1 − s)α(1−γ−δ)−1 s−α(1−β−δ) ds → 0, t/(t+h)

which implies that I12 (t) tends to 0 as h → 0. For I13 (t) and I14 (t), since I13 (t) ≤Cγ+δ (L + M µ2 ) Z t
(t − s)α−1 − (t + h − s)α−1 (t + h − s)−α(γ+δ) s−α(1−β−δ) ds × 0 Z t (t − s)α(1−γ−δ)−1 s−α(1−β−δ) ds ≤2Cγ+δ (L + M µ2 ) 0

=2Cγ+δ (L + M µ )Bγ t−α(γ−β) 2

and Z

t−ε


(t − s)α−1 |Aγ+δ Eα,α (−(t + h − s)α A)A−δ P f (s) + Fn (u(s), u(s)) |ds

I14 (t) ≤ 0

Z +

t−ε


(t − s)α−1 |Aγ+δ Eα,α (−(t − s)α A)A−δ P f (s) + Fn (u(s), u(s)) |ds

0

≤Cγ+δ (L + M µ2 ) Z t−ε
× (t − s)α−1 (t + h − s)−α(γ+δ) + (t − s)−α(γ+δ) s−α(1−β−δ) ds 0 Z t−ε ≤2Cγ+δ (L + M µ2 ) (t − s)α(1−γ−δ)−1 s−α(1−β−δ) ds, 0

then we know from the Lebesgue’s dominated convergence theorem and Lemma 2.11 that limh→0 I13 (t) = 0 and Z t−ε lim I14 (t) = (t − s)α−1 lim Aγ+δ Eα,α (−(t + h − s)α A) h→0 h→0 0

− Eα,α (−(t − s)α A) A−δ P f (s) + Fn (u(s), u(s)) |ds =0. For I15 (t), we calculate immediately Z t
I15 (t) ≤Cγ+δ (t − s)α−1 (t + h − s)−α(γ+δ) + (t − s)−α(γ+δ) t−ε
× |A−δ P f (s) + Fn (u(s), u(s)) ds Z t ≤2Cγ+δ (L + M µ2 ) (t − s)α(1−γ−δ)−1 s−α(1−β−δ) ds t−ε

→ 0,

as ε → 0.

Thenceforth, we show |Aγ Fn u(t + h) − Aγ Fn u(t)| → 0,

as h → 0.

54

Fractional Partial Differential Equations

Similarly, we can show Fn u ∈ C([0, T ∗ ]; D(Aβ )). To prove that Fn u ∈ XT ∗ , it remains to verify that sup {tα(γ−β) |Fn u(t)|D(Aγ ) } ≤ µ, for any u ∈ XT ∗ . t∈(0,T ∗ ]

Indeed, we have tα(γ−β) |(Fn u)(t)|D(Aγ ) ≤tα(γ−β) |Aγ Eα (−tα A)a| Z t (t − s)α−1 kAγ+δ Eα,α (−(t − s)α A)k|A−δ P f (s)|ds + tα(γ−β) 0 Z t + tα(γ−β) (t − s)α−1 kAγ+δ Eα,α (−(t − s)α A)k|A−δ Fn (u(s), u(s))|ds 0 Z t µ ≤ + Cγ+δ sup {sα(1−β−δ) |A−δ P f (s)|}tα(γ−β) (t − s)α(1−γ−δ)−1 s−α(1−β−δ) ds 4 s∈(0,t] 0 Z t + M Cγ+δ kQn k2 tα(γ−β) (t − s)α(1−γ−δ)−1 |Aρ u(s)||Aκ u(s)|ds 0

µ ≤ + Bγ Cγ+δ sup {sα(1−β−δ) |A−δ P f (s)|} + M Bγ Cγ+δ µ2 4 s∈(0,t] 0 is the viscosity, p = p(t, x) is the pressure, f = (f1 (t, x), f2 (t, x), ..., fn (t, x)) is the external force and a = (a1 (x), a2 (x), ..., an (x)) is the initial velocity. Then we proceed to consider the systems with control in Ω ⊂ R2 :  α ∂t u − ν∆u + (u · ∇)u = −∇p + C0 w + f, (t, x) ∈ [0, T ] × Ω,      ∇ · u = 0, (t, x) ∈ [0, T ] × Ω, (2.52)  u(t, x) = 0, (t, x) ∈ [0, T ] × ∂Ω,     u(0, x) = a, x ∈ Ω, where U is a real Hilbert space, w : [0, T ] → U and the operator C0 : U → (L2 (Ω))2 is linear and continuous. Since many good properties satisfied in integer-order differential equations aren’t generalized directly to fractional-order case, we found it more challenging in dealing with weak solutions of equation (2.51). As we all know, the main difficulty to study weak solutions is how to give an appropriate definition of weak solutions, introduce a suitable “work space” and establish the estimates of inequality by using the Galerkin approximations. We begin in Subsection 2.5.1 with some notations and definitions. Subsection 2.5.2 is devoted to the proof of the existence for weak solutions to equation (2.51), then proceed to study the uniqueness of the weak solution for such equations in R2 . Finally, Subsection 2.5.3 is concerned with an existence result of the optimal control of systems (2.52). 2.5.1

Notations and Definitions

In this subsection, we introduce notations, definitions, and preliminary facts which are used throughout this section. Assume that X is a Banach space. Let α ∈ (0, 1] and v : [0, T ] → X. We recall the left and right Riemann-Liouville integrals of v: Z t Z T −α −α gα (t − s)v(s)ds, gα (s − t)v(s)ds, t > 0, 0 Dt v(t) = t DT v(t) = 0

t

Fractional Navier-Stokes Equations

63

provided the integrals are point-wise defined on [0, ∞), where gα denotes the α−1 α α Riemann-Liouville kernel gα (t) = tΓ(α) , t > 0. Further, C 0 Dt and tDT stand the left Caputo and right Riemann-Liouville fractional derivative operators of order α, respectively; they are defined by Z t d C α D v(t) = g1−α (t − s) v(s)ds, t > 0, 0 t ds 0  Z T d α g1−α (s − t)v(s)ds , t > 0. tDT v(t) = − dt t More generally, for u : [0, ∞) × Rn → Rn , the left Caputo fractional derivative with respect to time of the function u can be written as Z t ∂tα u(t, x) = g1−α (t − s)∂s u(s, x)ds, t > 0. 0

Let v : R → X. We define the Liouville-Weyl fractional integral and the Caputo fractional derivative on the real axis: Z t Z t d −α C α gα (t − s)v(s)ds, D v(t) = g1−α (t − s) v(s)ds, −∞ Dt v(t) = −∞ t ds −∞ −∞ respectively. We also need for our purposes the fractional integration by parts in the formula (see [7]): Z T Z T (∂tα u(t), ψ(t))dt = (u(t), t DTα ψ(t))dt + (u(t), t DTα−1 ψ(t))|T0 , 0

since

0

limt→T t DTα−1 ψ(t)

= 0 for ψ ∈ C0∞ ([0, T ]; X), then Z T Z T (∂tα u(t), ψ(t))dt = (u(t), t DTα ψ(t))dt − (u(0), 0 DTα−1 ψ(t)). 0

(2.53)

0

We can refer to Kilbas et al. [197] and Zhou [386, 387] for more details. Next we declare a compactness theorem in Hilbert spaces. Let X0 , X, X1 be Hilbert spaces with X0 ,→ X ,→ X1 being continuous and X0 ,→ X is compact. Assume that v : R → X1 , vˆ denotes its Fourier transform Z +∞ vˆ(τ ) = e−2iπtτ v(t)dt. −∞

We have C\ α −∞Dt v(τ )

= (2iπτ )α vˆ(τ ).

We introduce a space for given 0 < γ ≤ 1, n W γ (R, X0 , X1 ) = v ∈ L2 (R, X0 ) :

C γ −∞Dt v

o ∈ L2 (R, X1 ) .

(2.54)

64

Fractional Partial Differential Equations

Clearly, it is a Hilbert space for the norm n o 12

2 kvkW γ = kvk2L2 (R,X0 ) + |τ |γ vˆ L2 (R,X1 ) . For any set J ⊂ R, we associate with the subspace WJγ ⊂ W γ defined as:  WJγ (R, X0 , X1 ) = v ∈ W γ (R, X0 , X1 ) : support v ⊂ J . By similar discussion as the proof of Theorem 2.2 in Temam [330], we state the following compactness result. Theorem 2.14. [330] Let X0 , X, X1 be Hilbert spaces and (2.54) hold. Then WJγ (R, X0 , X1 ) ,→ L2 (R, X) is compact for any bounded set J and γ > 0. In order to get the desired results, we introduce some necessary spaces. V = {u ∈ (C0∞ (Ω))n : ∇ · u = 0}. Let us denote by H and V the following sets: H = the closure of V in (L2 (Ω))n ,

V = the closure of V in (H01 (Ω))n .

Here (·, ·) is duality in H, h·, ·i is a duality pairing of V 0 and V . Also, we have hf, ui = (f, u), ∀ f ∈ H, ∀ u ∈ V.

(2.55)

L(U, H) stands for the set of all bounded linear operators from U to H. Denote by P the orthogonal projection P : (L2 (Ω))n → H. Here, the space H01 (Ω) is also a Hilbert space with the associated scalar product n X ((u, v)) = (Di u, Di v). i=1

The Stokes operator A is defined by A : D(A) ⊂ H → H, A = −P ∆, D(A) = H 2 (Ω) ∩ V. Clearly, hAu, vi = ((u, v)), ∀ u, v ∈ D(A)

(2.56)

and Aϕj = λj ϕj , for ϕj ∈ D(A); 0 < λ1 ≤ ...λj ≤ λj+1 ≤ ...; limj→∞ λj = ∞. Setting n Z X b(u, v, w) = ui (Di vj )wj dx, i,j=1

Ω

it is easy to know that b is trilinear continuous. If u ∈ V , then b(u, v, v) = 0, for any v ∈ H01 (Ω). For more details, we can refer the reader to Constantin and Foicas [85] and Temam [330]. We introduce an important space W α ([0, T ], V, V 0 ) := {u ∈ L2 (0, T ; V ) : ∂tα u ∈ L2 (0, T ; V 0 )}. Lemma 2.14. ∂tα (u(t), v) = h∂tα u(t), vi in L(C0∞ ([0, T ]; Ω)) for u ∈ W α ([0, T ], V, V 0 ), v ∈ V .

Fractional Navier-Stokes Equations

65

Proof. Let ψ ∈ C0∞ ([0, T ]; Ω). Notice that u(t), v ∈ V ⊂ H, we know that h·, ·i is consistent with (·, ·) by (2.55). It follows hv, u(t)i = (v, u(t)) = (u(t), v). From (2.53) we obtain Z T Z h∂tα u(t), viψ(t)dt = 0

T

hu(t), vit DTα ψ(t)dt − hu(0), vi0 DTα−1 ψ(t)

0

Z

T

(u(t), v)t DTα ψ(t)dt − (u(0), v)0 DTα−1 ψ(t)

= 0

Z =

T

∂tα (u(t), v)ψ(t)dt.

0

Hence, ∂tα (u(t), v) = h∂tα u(t), vi in L(C0∞ ([0, T ]; Ω)). We assume that u is a classical solution of (2.51) associated with p, it means that u ∈ C 2 ([0, T ] × Ω), p ∈ C 1 ([0, T ] × Ω). Obviously u ∈ L2 (0, T ; V ) and if v ∈ V, it is easy to check ∂tα (u, v) + ν((u, v)) + b(u, u, v) = hf, vi.

(2.57)

This presents the following weak formulation of equation (2.51). Let α1 ∈ (0, α) and Cw ([0, T ]; H) := {u ∈ L∞ (0, T ; H) : u is weakly continuous}.

2

Definition 2.6. Let a ∈ H, f ∈ L α1 (0, T ; V 0 ). A function u ∈ L2 (0, T ; V ) ∩ Cw ([0, T ]; H) is said a weak solution of equation (2.51) if ( ∂tα (u, v) + ν((u, v)) + b(u, u, v) = hf, vi, ∀ v ∈ V, (2.58) u(0) = a. Lemma 2.15. [330] Let u ∈ L2 (0, T ; V ). Then the function Bu(t) defined by (B(u(t), u(t)), v) = b(u(t), u(t), v), for any v ∈ V, a.e. in t ∈ [0, T ]

(2.59)

belongs to V 0 . Now if u satisfies the problem (2.58), then based on (2.55), (2.56) and Lemma 2.15, we rewrite the first equality of (2.58) as ∂tα hu, vi = hf − νAu − B(u, u), vi, ∀ v ∈ V. Since Au ∈ L2 (0, T ; V 0 ), then f (t) − νAu(t) − Bu(t) ∈ V 0 . Thus Lemma 2.14 shows that ∂tα u = f − νAu − B(u, u).

66

Fractional Partial Differential Equations

An equivalent form of problem (2.58) is ( C α 0 Dt u + νAu + B(u, u) = f, u(0) = a.

(2.60)

Lemma 2.16. Suppose that X is a real Hilbert space and v : [0, T ] → X has a derivative, then there holds α (v(t), C 0 Dt v(t)) ≥

1C α D |v(t)|2 . 20 t

Note the above lemma is a generalization of the inequality proved in [10]. Lemma 2.17. Suppose that a nonnegative function v(t) satisfies C α 0 Dt v(t)

+ c1 v(t) ≤ c2 (t)

for almost all t ∈ [0, T ], where c1 > 0 and the function c2 (t) is nonnegative and integrable for t ∈ [0, T ]. Then Z t 1 (t − s)α−1 c2 (s)ds. v(t) ≤ v(0) + Γ(α) 0 α Proof. Let C 0 Dt v(t) + c1 v(t) = g(t), then by [197] we have the following expression: Z t v(t) = v(0)Eα (−c1 tα ) + (t − s)α−1 Eα,α (−c1 (t − s)α )g(s)ds, 0

where Eα (t) =

∞ X k=0

∞

X tk tk and Eα,α (t) = Γ(αk + 1) Γ(αk + α) k=0

α

are Mittag-Leffler functions. Since 0 < Eα (−c1 t ) ≤ 1, 0 < Eα,α (−c1 tα ) ≤ and g(t) ≤ c2 (t), then Z t 1 v(t) ≤v(0)Eα (−c1 tα ) + (t − s)α−1 g(s)ds Γ(α) 0 Z t 1 ≤v(0) + (t − s)α−1 c2 (s)ds. Γ(α) 0

2.5.2

1 Γ(α)

Existence and Uniqueness 2

Theorem 2.15. Let a ∈ H, f ∈ L α1 (0, T ; V 0 ). Then there exists at least one function u ∈ L2 (0, T ; V ) ∩ Cw ([0, T ]; H) which satisfies (2.60).

Fractional Navier-Stokes Equations

67

Proof. Step 1. Let Hm ⊂ H denote the finite dimensional subspace spanned by {ϕ1 , ϕ2 , ..., ϕm } and let Pm : H → Hm be the corresponding projection operators for m = 1, 2, ... . Applying Pm to (2.60) yields the equation C α 0 Dt (Pm u)

+ νA(Pm u) + Pm B(u, u) = Pm f.

The Galerkin system of order m is the system ( C α 0 Dt um + νAum + Pm B(um , um ) = fm , um (0) = am ,

(2.61)

where fm = Pm f and am = Pm a. The function um (t) belongs to Pm H. Taking the scalar product of (2.61) with ϕj we obtain ( α (C 0 Dt um , ϕj ) + ν(Aum , ϕj ) + (Pm B(um , um ), ϕj ) = hfm , ϕj i, (2.62) (um (0), ϕj ) = (am , ϕj ). More precisely, let ξj = ξj (t) denote the j th component of um (t): ξj (t) = (um (t), ϕj ). Also, let ηj (t) = (fm (t), ϕj ) be component of fm (t). Then (2.62) is equivalent to a nonlinear differential system for the functions  m X   α C b(ϕk , ϕl , ϕj )ξk ξl = ηj , j = 1, ..., m, 0 Dt ξj + νλj ξj + (2.63) k,l=1    0 ξj (0) = ξj , where ξj0 = (am , ϕj ). The nonlinear differential system (2.63) has a maximal solution on the interval [0, tm ]. If tm < T , then |ξ(t)| must go to +∞ as t → tm , the estimates that we show indicate that this does not happen. Indeed, taking the scalar product of (2.63) with ξ in R, we obtain α (C 0 Dt ξ(t), ξ(t)) + ν

m X

λj ξj2 (t) = hη(t), ξ(t)i,

j=1

because of X

b(ϕk , ϕl , ϕj )ξk ξl ξj = 0.

k,l,j

From Lemma 2.16 and the Young’s inequality, we know 1C α D |ξ(t)|2 + νλ1 |ξ(t)|2 ≤hη(t), ξ(t)i 20 t νλ1 1 |η(t)|2 + |ξ(t)|2 . ≤ 2νλ1 2 Using Lemma 2.17 and the fact that λ1 > 0, we have the estimation Z t 1 2 2 |ξ(t)| ≤|ξ(0)| + (t − s)α−1 |η(s)|2 ds νλ1 Γ(α) 0

(2.64)

68

Fractional Partial Differential Equations

1 ≤|ξ(0)| + νλ1 Γ(α)

Z

1 ≤|ξ(0)|2 + νλ1 Γ(α)

Z

2

t 2/α1

|η(s)| 0

0

T

1 ds + νλ1 Γ(α)

Z

t

α−1

(t − s) 1−α1 ds 0

T 1+b |η(s)|2/α1 ds + (1 + b)νλ1 Γ(α)

=:M, α−1 where α1 ∈ (0, α), b = 1−α . Therefore tm = T . 1 Step 2. We also estimate the right-hand side of (2.64) differently. Since X  21  X  21 m m 2 |hη(t), ξ(t)i| ≤ λj ξj2 , λ−1 η j j j=1

j=1

and we return to the notation um (t) =

m X

ξj ϕj , fm (t) =

j=1

m X

ηj ϕj ,

j=1

then from (2.64) we get 1C α D |um (t)|2 + νkum (t)k2V ≤hfm (t), um (t)i 20 t 1 ≤|A− 2 fm (t)|kum (t)kV 1 1 ν ≤ |A− 2 fm (t)|2 + kum (t)k2V . 2ν 2

(2.65)

1

We use the notation |h|V 0 = |A− 2 h|, meaning that we identify the dual V 0 1 of V with D(A− 2 ). Integrating (2.65) (with order α) and applying the Young’s inequality, we get Z t ν |um (t)|2 + (t − s)α−1 kum (s)k2V ds Γ(α) 0 Z t 1 (t − s)α−1 |fm (s)|2V 0 ds ≤|um (0)|2 + νΓ(α) 0 Z t Z t α−1 1 1 2/α1 2 ≤|um (0)| + |fm (s)|V 0 ds + (t − s) 1−α1 ds νΓ(α) 0 νΓ(α) 0 Z T 1 T 1+b 2/α ≤|um (0)|2 + . |fm (s)|V 0 1 ds + νΓ(α) 0 (1 + b)νΓ(α) It follows that for a.e. t ∈ [0, T ], |um (t)|2 ≤ |um (0)|2 + ν Γ(α)

1 νΓ(α)

Z 0

T

2/α

|f (s)|V 0 1 ds +

T 1+b , (1 + b)νΓ(α)

t

Z

(t − s)α−1 kum (s)k2V ds

(2.66)

0

1 ≤|um (0)| + νΓ(α) 2

Z 0

T

2/α

|f (s)|V 0 1 ds +

T 1+b . (1 + b)νΓ(α)

Fractional Navier-Stokes Equations

69

This ensures that the sequence {um } is a bounded set of L∞ (0, T ; H).

(2.67)

Moreover, ν T α−1 Γ(α)

Z

t

kum (s)k2V

0

ν ds ≤ Γ(α)

Z

t

(t − s)α−1 kum (s)k2V ds

0

≤|um (0)|2 +

1 νΓ(α)

Z 0

T

2/α

|f (s)|V 0 1 ds +

T 1+b . (1 + b)νΓ(α)

Hence the sequence {um } is a bounded set of L2 (0, T ; V ).

(2.68)

Step 3. Let u ˜m : R → V denote the function defined as ( um (t), t ∈ [0, T ], u ˜m (t) = 0, t ∈ R \ [0, T ], and u ˆm denote the Fourier transform of u ˜m . To show that the sequence {˜ um } remains to be a bounded set of W γ (R, V, H). Applying Theorem 2.14 along with (2.68), we need to verify that Z +∞ |τ |2γ |ˆ um (τ )|2 dτ ≤ const., for some γ > 0.

(2.69)

(2.70)

−∞

In order to prove (2.70), we observe that (2.62) can be written as α (C ˜m , ϕj ) = h˜ gm , ϕj i + (um (0), ϕj )−∞ Dtα−1 δ0 − (um (T ), ϕj )−∞ Dtα−1 δT , (2.71) 0 Dt u

where ( g˜m (t) =

gm (t), t ∈ [0, T ], 0,

t ∈ R \ [0, T ],

here gm = fm − νAum − Bum , δ0 and δT are the Dirac distributions at 0 and T . Indeed, it is classical that since u ˜m has two discontinuities at 0 and T , the Caputo derivative of u ˜m is given by   d α−1 C α ˜m =−∞ Dt u ˜m −∞Dt u dt   d α−1 =−∞ Dt um + um (0)δ0 − um (T )δT dt
α−1 α =C um (0)δ0 − um (T )δT . 0 Dt um + −∞ Dt By the Fourier transform, (2.71) yields (2iπτ )α (ˆ um , ϕj ) =hˆ gm , ϕj i + (um (0), ϕj )(2iπτ )α−1 − (um (T ), ϕj )(2iπτ )α−1 e−2iπT τ , here u ˆm and gˆm denote the Fourier transforms of u ˜m and g˜m , respectively.

(2.72)

70

Fractional Partial Differential Equations

We multiply (2.72) by ξˆj (τ ) (i.e., Fourier transform of ξj ) and plus the resulting equations, it follows (2iπτ )α |ˆ um (τ )|2 =hˆ gm , u ˆm (τ )i + (um (0), u ˆm (τ ))(2iπτ )α−1 − (um (T ), u ˆm (τ ))(2iπτ )α−1 e−2iπT τ .

(2.73)

In view of the inequality kBukV 0 ≤ ckuk2V , for all u ∈ V , Z T Z T
kfm (t)kV 0 + νkum (t)kV + ckum (t)k2V dt, kgm (t)kV 0 dt ≤ 0

0

this remains bounded based on (2.68). Therefore sup kˆ gm (τ )kV 0 ≤ const., ∀ m. τ ∈R

On account of (2.67), |um (0)| ≤ const., |um (T )| ≤ const., and we conclude from (2.73) that |τ |α |ˆ um (τ )|2 ≤ c2 kˆ um (τ )kV + c3 |τ |α−1 |ˆ um (τ )| or |τ |α |ˆ um (τ )|2 ≤ c4 max{1, |τ |α−1 }kˆ um (τ )kV . For γ fixed, γ < α/4, we see that |τ |2γ ≤ c5 (γ) Accordingly, Z +∞ Z |τ |2γ |ˆ um (τ )|2 dτ ≤c5 (γ) −∞

1 + |τ |α . 1 + |τ |α−2γ

+∞

1 + |τ |α |ˆ um (τ )|2 dτ α−2γ −∞ 1 + |τ | Z +∞ Z ≤c6 (γ) kˆ um (τ )k2V dτ + c7 (γ)

+∞

|τ |α−1 kˆ um (τ )kV dτ. 1 + |τ |α−2γ −∞ −∞ Applying the Parseval’s equality, the first integral is bounded as m → +∞, thus (2.70) is showed if one proves that Z +∞ α−1 |τ | kˆ um (τ )kV dτ ≤ const. (2.74) 1 + |τ |α−2γ −∞ Making use of the Schwarz’s inequality we estimate above integral by Z +∞ α−1 |τ | kˆ um (τ )kV dτ 1 + |τ |α−2γ −∞  Z +∞  12  Z +∞  21 |τ |α−1 dτ α−1 2 |τ | kˆ um (τ )kV dτ . ≤ α−2γ )2 −∞ −∞ (1 + |τ | Since Z 1  Z +∞ Z +∞ |τ |α−1 dτ |τ |α−1 dτ |τ |α−1 dτ = 2 + , α−2γ )2 α−2γ )2 (1 + |τ |α−2γ )2 −∞ (1 + |τ | 0 (1 + |τ | 1 it is clear that the first integral is finite and the second integral is also finite due to γ < α/4. On the other hand, it follows from the Parseval’s equality that

Fractional Navier-Stokes Equations

Z

+∞

|τ |

α−1

kˆ um (τ )k2V

+∞

Z

α

k−∞ Dt2

dτ =

−∞

−∞ Z T

α

k0 Dt2

=

− 21

71

− 12

u ˜m (t)kV


2

dt

um (t)k2V dt

0

 ≤

1

α

T 2− 2 Γ( 32 − α2 )

2 Z

T

kum (t)k2V dt,

0

which implies that (2.74) holds by (2.68). In a word, (2.70) and (2.69) hold. Step 4. (2.67) and (2.68) enable us to say the existence of an element in L2 (0, T ; V ) ∩ L∞ (0, T ; H) and a subsequence {um0 } such that um0 * u weakly in L2 (0, T ; V ) and weak − star in L∞ (0, T ; H), as m → ∞. Owing to (2.69) and Theorem 2.14, we get um0 → u strongly in L2 (0, T ; H). Let v ∈ V . Take the scalar product of (2.61) with v and integrate (um0 (t), v) − (um0 (t0 ), v) Z t0
1 = (t0 − s)α−1 − (t − s)α−1 Γ(α) 0 × [ν((um0 (s), v)) + b(um0 (s), um0 (s), Pm0 v) − hfm0 (s), vi]ds Z t 1 − (t − s)α−1 [ν((um0 (s), v)) + b(um0 (s), um0 (s), Pm0 v) − hfm0 (s), vi]ds. Γ(α) t0 (2.75) Since um0 * u weakly in L2 (0, T ; V ), by extracting a subsequence, relabeled um0 , we assume that um0 (t0 ) * u(t0 ) in V for all t0 ∈ [0, T ]\K, for some K satisfying mes(K) = 0. Thus limm→∞ um0 (t0 ) = u(t0 ) in H for t0 ∈ / K. Now by (2.66) and using the Lebesgue’s dominated convergence theorem, we have Z t Z t lim (t − s)α−1 ((um0 (s), v))ds = (t − s)α−1 ((u(s), v))ds, m→∞

t0 t0

Z lim Z t0

m→∞

=

t0

(t0 − s)α−1 − (t − s)


α−1

((um0 (s), v))ds

0


(t0 − s)α−1 − (t − s)α−1 ((u(s), v))ds.

0

A simple argument shows that Z t Z t lim (t − s)α−1 b(um0 (s), um0 (s), Pm0 v)ds = (t − s)α−1 b(u(s), u(s), v)ds, m→∞

t0 t0

Z lim Z t0

m→∞

= 0

t0


(t0 − s)α−1 − (t − s)α−1 b(um0 (s), um0 (s), Pm0 v)ds

0


(t0 − s)α−1 − (t − s)α−1 b(u(s), u(s), v)ds.

72

Fractional Partial Differential Equations

Passing to the limit of (2.75) with the sequence m0 , one finds that the equality for t ≥ t0 and t, t0 ∈ / K, (u(t), v) − (u(t0 ), v) Z t0
1 = (t0 − s)α−1 − (t − s)α−1 [ν((u(s), v)) + b(u(s), u(s), v) − hf (s), vi]ds Γ(α) 0 Z t 1 − (t − s)α−1 [ν((u(s), v)) + b(u(s), u(s), v) − hf (s), vi]ds. Γ(α) t0 (2.76) It is clear that (u(t), v)−(u(t0 ), v) → 0 as t−t0 → 0 after a fundamental calculation. This follows the weak continuity of u(t) in H since V is dense in H. Applying the Caputo fractional derivative of order α on the sides of (2.76) for t0 = 0, we obtain (2.58). We address the uniqueness of weak solutions in the case that n = 2. Theorem 2.16. The solution u of the problem (2.60) given by Theorem 2.15 is unique. Proof. Let us assume that u1 and u2 are two solutions of (2.60) and let u = u1 −u2 . We obtain for u the equation ( C α 0 Dt u + νAu + B(u1 , u) + B(u, u2 ) = 0, (2.77) u(0) = 0. Taking the scalar product of (2.77) with u we obtain α 2 hC 0 Dt u, ui + νkukV + b(u, u2 , u) = 0. We proceed similarly as the proof of Theorem 2.15 to show that C α 2 2 (2.78) 0 Dt |u| + 2νkukV ≤ 2b(u, u, u2 ). By the H¨ older’s inequality, |b(u, u, u2 )| ≤c0 |u|kukV ku2 kV . Then we bound the right-hand side of (2.78) by |b(u, u, u2 )| ≤ c0 |u|kukV ku2 kV ≤ νkuk2V + c1 |u|2 ku2 k2V , we get C α 2 2 2 0 Dt |u(t)| ≤ 2c1 |u(t)| ku2 (t)kV or Z t c1 (t − s)α−1 |u(s)|2 ku2 (s)k2V ds. |u(t)|2 ≤ |u(0)|2 + Γ(α) 0 By the Gronwall’s inequality, (2.66) and (2.67),   Z t c1 |u(t)|2 ≤|u(0)|2 exp (t − s)α−1 ku2 (s)k2V ds Γ(α) 0    Z T c1 1 T 1+b 2/α ≤|u(0)|2 exp |u2 (0)|2 + |f (s)|V 0 1 ds + . ν νΓ(α) 0 (1 + b)νΓ(α) Since |u(0)| = 0, it follows that |u| ≡ 0. The proof is completed.

Fractional Navier-Stokes Equations

2.5.3

73

Optimal Control

In this subsection we study the optimal control problem: Z Z Z T 1 T minimize (u(t, x) − z(t, x))2 dxdt + h(w(t))dt 2 0 Ω 0

(2.79)

2

over w ∈ L α1 (0, T ; U ) and u ∈ L2 ([0, T ] × Ω; R2 ) subject to systems (2.52). Let C ∈ L(U, H) be given by C = P C0 . By the same way as we derived (2.58), then we give the concept of weak solutions of systems (2.52). 2

2

Definition 2.7. Let a ∈ H, f ∈ L α1 (0, T ; V 0 ) and w ∈ L α1 (0, T ; U ). A function u ∈ L2 (0, T ; V ) ∩ Cw ([0, T ]; H) is said a weak solution of systems (2.52) if ( ∂tα (u, v) + ν((u, v)) + b(u, u, v) = hC0 w + f, vi, ∀ v ∈ V, u(0) = a. Similarly we give the equivalent form as follows: ( C α 0 Dt u + νAu + B(u, u) = Cw + f, u(0) = a.

(2.80)

If z ∈ L2 (0, T ; H), rewriting the problem (2.79) in the abstract form, we have Z Z T 1 T 2 (P) minimize J(u, w) = (u(t) − z(t)) dt + h(w(t))dt 2 0 0 2

over (u, w) ∈ (L2 (0, T ; V ) ∩ Cw ([0, T ]; H)) × L α1 (0, T ; U ) subject to (2.80). We assume that: (i) z ∈ L2 (0, T ; H); (ii) the function h : U → R is convex, lower semi-continuous and satisfies 2

|h(w)| ≥ b1 |w|Uα1 + b2 , for some b1 > 0, b2 ∈ R. Theorem 2.17. If hypotheses (i) and (ii) hold, then the problem (P) has at least 2 one solution (u∗ , w∗ ) ∈ (L2 (0, T ; V ) ∩ Cw ([0, T ]; H)) × L α1 (0, T ; U ). Proof. According to Theorems 2.15 and 2.16, by replacing f with Cw + f we show that equation (2.80) has a weak solution corresponding to w. Suppose that (um , wm ) are a minimizing sequence of problem (P), that is, J(um , wm ) → inf J(u, w), C α 0 Dt um

+ νAum + B(um , um ) = Cwm + f, t ∈ [0, T ],

(2.81)

um (0) = a. 2

Hypothesis (ii) ensures that {wm } is bounded in L α1 (0, T ; U ), then there exists a 2 subsequence, relabeled as {wm } and w∗ ∈ L α1 (0, T ; U ) such that 2

wm * w∗ weakly in L α1 (0, T ; U ).

74

Fractional Partial Differential Equations

Similar to (2.65), we get 1C α D |um (t)|2 + νkum (t)k2V ≤hCwm + f (t), um (t)i. 20 t This yields ν Γ(α)

|um (t)|2 + ≤|um (0)|2 +

t

Z

1 νΓ(α)

(t − s)α−1 kum (s)k2V ds

0

Z 0

T

2/α

|Cwm (s) + f (s)|V 0 1 ds +

T 1+b . (1 + b)νΓ(α)

Hence, Z t ν α−1 kum (s)k2V ds T Γ(α) 0 Z t ν (t − s)α−1 kum (s)k2V ds ≤ Γ(α) 0 Z T 1 T 1+b 2/α ≤|um (0)|2 + . |Cwm (s) + f (s)|V 0 1 ds + νΓ(α) 0 (1 + b)νΓ(α) Therefore the sequence {um } is bounded in L∞ (0, T ; H), the sequence {um } is bounded in L2 (0, T ; V ), thus there exists an element u∗ ∈ L2 (0, T ; V ) ∩ L∞ (0, T ; H) such that um *u∗

weakly in L2 (0, T ; V ) and weak − star in L∞ (0, T ; H), as m → ∞.

Meanwhile, we derive from similar argument as given in the Step 3 of Theorem 2.15 that um → u∗ strongly in L2 (0, T ; H). We need to give some uniform bounds on Aum and B(um , um ). Since {um } ⊆ L2 (0, T ; V ) is bounded, then {Aum } ⊆ L2 (0, T ; V 0 ) is bounded. Actually, Z t Z t |Aum (s)|2V 0 ds = kum (s)k2V ds. 0

0

Therefore, choosing a subsequence, if necessary, we have Aum *Au∗ weakly in L2 (0, T ; V 0 ). It remains to investigate the term Pm B(um , um ). Now, consider 1

|(A− 2 B(um , um ), v)| ≤ c˜|um |kum kV kvkV , c˜ > 0, then 1

|B(um , um )|V 0 = |A− 2 B(um , um )| ≤ c˜|um |kum kV .

Fractional Navier-Stokes Equations

75

Since |um (s)| is uniformly bounded and kum (s)k2V is uniformly integrable, we obtain that Z t Z t |B(um (s), um (s))|2V 0 ds ≤˜ c |um (s)|2 kum (s)k2V ds 0 0 Z t ≤˜ c sup |um |2 kum (s)k2V ds < ∞. 0

This implies that there exists η ∗ such that B(um , um ) * η ∗

weakly in L2 (0, T ; V 0 ).

Now by virtue of (2.59) we have |(B(um (t), um (t)) − B(u∗ (t), u∗ (t)), v)| ≤|b(um (t) − u∗ (t), um (t), v)| + |b(u∗ (t), um (t) − u∗ (t), v)| 1

1

1

1

≤C|um (t) − u∗ (t)| 2 kum (t) − u∗ (t)kV2 |um (t)| 2 kum (t)kV2 kvkV 1

1

1

1

+ C|u∗ (t)| 2 ku∗ (t)kV2 |um (t) − u∗ (t)| 2 kum (t) − u∗ (t)kV2 kvkV , ∀ v ∈ V, we infer that B(um , um ) * B(u∗ , u∗ ) weakly in L2 (0, T ; V 0 ). Thus η ∗ (t) = B(u∗ (t), u∗ (t)) a.e. t ∈ [0, T ]. Let v ∈ V . Take the scalar product of (2.81) with v and integrate, it follows (um (t), v) − (um (t0 ), v) Z t0
1 = (t0 − s)α−1 − (t − s)α−1 [ν((um (s), v)) Γ(α) 0 + b(um (s), um (s), Pm v) − hCwm (s) + f (s), vi]ds Z t 1 (t − s)α−1 − Γ(α) t0 × [ν((um (s), v)) + b(um (s), um (s), Pm v) − hCwm (s) + f (s), vi]ds. By similar arguments with Step 4 in Theorem 2.15 and taking the limits as m → ∞, we know that (u∗ (t), v) − (u∗ (t0 ), v) Z t0
1 (t0 − s)α−1 − (t − s)α−1 [ν((u∗ (s), v)) = Γ(α) 0 + b(u∗ (s), u∗ (s), v) − hCw∗ (s) + f (s), vi]ds Z t 1 (t − s)α−1 [ν((u∗ (s), v)) + b(u∗ (s), u∗ (s), v) − hCw∗ (s) + f (s), vi]ds. − Γ(α) t0 This similarly yields the weak continuity of u∗ . By applying the Caputo fractional derivative of order α on the sides of the above equality for t0 = 0, we obtain ( α ∗ ∗ ∗ ∗ ∗ hC 0 Dt u , vi + ν((u , v)) + b(u , u , v) = hCw + f, vi, t ∈ [0, T ], for all v ∈ V, u∗ (0) = a.

76

Fractional Partial Differential Equations

Clearly, an equivalent form of the last equality is ( C α ∗ ∗ ∗ ∗ ∗ 0 Dt u + νAu + B(u , u ) = Cw + f, u∗ (0) = a. Hence, we see that (u∗ , w∗ ) satisfies system (2.81). From weak lower semi-continuity of the function (u, w) 7→ J(u, w) we deduce J(u∗ , w∗ ) = inf J(u, w). Consequently, (u∗ , w∗ ) is an optimal pair for problem (P). This completes the proof. 2.6

Energy Methods

In this section we consider the following Navier-Stokes equations with timefractional derivatives in R3 :  α   ∂t u + (u · ∇)u = −∇p + ν∆u, t > 0, (2.82) ∇ · u = 0,   u(0, x) = u0 , where ∂tα is the Caputo fractional derivative of order α ∈ (0, 1), u = (u1 (t, x), u2 (t, x), u3 (t, x)) represents the velocity field at a point x ∈ R3 and time t > 0, p = p(t, x) is the pressure, ν the viscosity and u0 = u0 (x) is the initial velocity. It is worth-mentioning that the purpose of this section is to develop a new global result when we relax the smallness condition on the initial data. The section is organized as follows. In Subsection 2.6.1 we recall some notations, definitions, and preliminary facts. In Subsection 2.6.2 we construct a regularized equation, and show the local existence of its solutions. Subsection 2.6.3 deals with the global existence and continuation of solutions for a kind of fractional differential equations in a Banach space. Via energy methods, we also obtain the convergence of approximate solutions and the global existence and uniqueness of mild solutions of problem (2.83) in different spaces. 2.6.1

Notations and Lemmas

Here we recall some notations, definitions, and preliminary facts. Denote by k · k the L2 norm on R3 , where Z  12 2 kuk = |u(x)| dx . R3

By H m (R3 ), m ∈ N0 , we denote the Sobolev space consisting of functions u ∈ L2 (R3 ) such that Dβ u ∈ L2 (R3 ), 0 ≤ |β| ≤ m with the norm k · km defined by ! 21 X β 2 kukm = kD uk . 0≤|β|≤m

Fractional Navier-Stokes Equations

77

For s ∈ R, the Sobolev space H s (R3 ) is the completion of S(R3 ) with respect to the norm Z  12 2 s 2 kuks = (1 + |ξ| ) |b u(ξ)| dξ . R3

Here u b represents the Fourier transform of u and S(R3 ) denotes the Schwarz space of rapidly decreasing smooth functions. It is clear that the two norms are equivalent for s = m. Let % be the standard mollifier satisfying Z %(|x|) ∈ C0∞ (R3 ), 0 ≤ % ≤ 1, %(x)dx = 1. R3

We define the mollification Jε u of the function u ∈ Lq (R3 ) (1 ≤ q ≤ ∞) as Z x−y −3 )u(y)dy. (Jε u)(x) = ε %( ε R3 We introduce the space V s := {u ∈ H s (R3 ) : ∇ · u = 0} and call the operator P : H s (R3 ) → V s the Leray projector. Clearly, P commutes with Jε . For more details, we refer the reader to Majda and Bertozzi [256]. Let us denote an important space: for m ∈ Z,   Z Xm = u ∈ D0 (R3 ) : |ξ|m |b u|dξ < ∞ . R3 0

3

Here D (R ) stands for the space of distributions with the norm k · kXm given by Z kukXm = |ξ|m |b u|dξ. R3

It is not difficult to show that if u ∈ H 1 , then u ∈ X−1 . For u : [0, ∞) × R3 → R3 , Caputo time-fractional derivative of the function u can generally be written as
 ∂  α−1 ∂tα u(t, x) = u(t, x) − u(0, x) , t > 0. 0 Dt ∂t Before proceeding further, we present two important results which play a key role in proving the main results. Lemma 2.18. [188] Let T > 0. Then, for any v ∈ L2 ((0, T ] × R3 ; R3 ), there holds Z v(t, x)∂tα v(t, x)dx ≥ kvk∂tα kvk. R3

The following result is a generalization of Lemma 6.1 in Kemppainen et al. [188]. Lemma 2.19. Let the function v : (0, T ] × R3 → C3 and ∂tα v exist. Then v(t)∂tα v(t) + v(t)∂tα v(t) ≥ 2|v|∂tα |v|, where v(t) denotes the dual of v(t).

78

Fractional Partial Differential Equations

Proof. Let v(t) = a(t) + b(t)i. Then v(t)∂tα v(t) + v(t)∂tα v(t)
=2 a(t)∂tα a(t) + b(t)∂tα b(t) 
=2 ∂tα a2 (t) + ∂tα b2 (t) + t−α a2 (t) + b2 (t) Z

t 2

−α−1

|a(t) − a(t − s)| [αs + 0  Z ≥2 ∂tα |v(t)|2 + t−α |v(t)|2 +

Z

t 2

−α−1

|b(t) − b(t − s)| [αs

]ds +

 ]ds

0 t


2 |v(t)| − |v(t − s)| [αs−α−1 ]ds

0

=2|v(t)|∂tα |v(t)|.

2.6.2

Local Existence

This subsection is concerned with the construction of an approximate (regularized) equation for the time-fractional Navier-Stokes equations and obtain the existence and some properties of its solution. To do this, we use the mollifier Jε to regularize equation (2.82):  α ε ε ε ε ε   ∂t u + Jε [(Jε u ) · ∇(Jε u )] = −∇p + νJε ∆(Jε u ), (2.83) ∇ · uε = 0,   ε u (0) = Jε u0 . By applying Leray projector P to equation (2.83), we get rid of the pressure term in equation (2.83), which reduces to an ordinary differential equation (ODE) in V s : ( ∂tα uε + P Jε [(Jε uε ) · ∇(Jε uε )] = νJε2 ∆uε , (2.84) uε (0) = Jε u0 . We set Fε (uε ) = νJε2 ∆uε − P Jε [(Jε uε ) · ∇(Jε uε )]. From the argument of Proposition 3.6 in Majda and Bertozzi [256], we know that kFε (u1 ) − Fε (u2 )km ≤ L(kuj k, ε, M )ku1 − u2 km , for uj ∈ BM , j = 1, 2, (2.85) where m ∈ N0 and BM = {u ∈ V m : kukm < M }. It means that Fε is locally Lipschitz continuous on any open set. Let  C α ([0, T ]; V m ) = u ∈ C([0, T ]; V m ) : ∂tα u ∈ C([0, T ]; V m ) . Theorem 2.18. Assume the initial data u0 ∈ V 0 . Then, for any ε > 0,

Fractional Navier-Stokes Equations

79

(i) there exists Tε = T (kJε u0 km , ε) such that equation (2.84) has a unique solution uε ∈ C α ([0, Tε ); V m ); (ii) uε ∈ C α ([0, T ]; V 0 ) on any interval [0, T ] with sup kuε k ≤ ku0 k.

(2.86)

t∈[0,T ]

Proof. (i) If u0 ∈ V 0 , then Jε u0 ∈ V m , m ∈ N0 . For given r > 0, define   ε m ε B(r, Tε ) = u ∈ C([0, Tε ); V ) : sup ku (t) − Jε u0 km ≤ r . t∈[0,Tε )

Notice that M=

kFε (uε )km < +∞.

sup

(2.87)

(t,uε )∈[0,Tε )×B(1,Tε )

Indeed, kFε (uε )km ≤kFε (uε ) − Fε (Jε u0 )km + kFε (Jε u0 )km ≤Lkuε − Jε u0 km + kFε (Jε u0 )km ≤Lr + kFε (Jε u0 )km < +∞. Consider the operator T : 1 T u (t) = Jε u0 + Γ(α) ε

Z

t

(t − s)α−1 Fε (uε (s))ds.

0

Obviously T (B(r, Tε )) ⊂ B(r, Tε ). Moreover, for u1 , u2 ∈ B(r, Tε ), Ltα kT u1 (t) − T u2 (t)km ≤ sup ku1 (s) − u2 (s)km , for t ∈ [0, Tε ). Γ(α + 1) s∈[0,t]  1  1 Γ(α+1)r α Γ(α+1) α Fixing Tε < min , , it is easy to show that T is a strict M L contraction mapping on Br (r, Tε ). In consequence, we deduce that T has a fixed point. On account of (2.85) and uε ∈ C([0, Tε ); V m ), it is clear that ∂tα uε ∈ C([0, Tε ); V m ). (ii) Take the L2 inner product of equation (2.84) with uε and apply Lemma 2.18 to obtain Z Z kuε k∂tα kuε k ≤ ν

R3

uε Jε2 ∆uε dx −

uε P Jε [(Jε uε ) · ∇(Jε uε )]dx.

R3

Integrating by parts and using the fact that ∇ · uε = 0, we get kuε k∂tα kuε k + νkJε ∇uε k2 ≤ 0. Since ν ≥ 0, we have ∂tα kuε k ≤ 0. Then the definition of ∂tα u implies that sup kuε k ≤ kJε u0 k ≤ ku0 k. t∈[0,T ]

80

Fractional Partial Differential Equations

In the next lemma, we derive a key estimate which plays an important role in proving the main results. Lemma 2.20. The regularized solution uε of equation (2.84) satisfies the inequality: ∂tα kuε km ≤ cm |Jε ∇uε |L∞ kuε km . Proof. We take the derivative Dβ of equation (2.84) and L2 inner product with Dβ uε to obtain

β α ε β ε D ∂t u , D u



 =ν Dβ Jε2 ∆uε , Dβ uε − Dβ P Jε [(Jε uε ) · ∇(Jε uε )], Dβ uε

 = − νkJε Dβ ∇uε k2 − P Jε [(Jε uε ) · ∇(Dβ Jε uε )], Dβ uε

 − Dβ P Jε [(Jε uε ) · ∇(Jε uε )] − P Jε [(Jε uε ) · ∇(Dβ Jε uε )], Dβ uε . From the discussion of Majda and Bertozzi [256] and Lemma 2.18, we deduce that kuε km ∂tα kuε km + νkJε ∇uε k2m ≤cm |Jε ∇uε |L∞ kuε k2m , which takes the following form for ν > 0, ∂tα kuε km ≤ cm |Jε ∇uε |L∞ kuε km .

2.6.3

Global Existence

Here we discuss the global existence of solutions by means of mathematical analysis and previous estimate when the initial data being less than the viscosity. Firstly, we consider the following autonomous equation C α (2.88) 0 Dt z(t) = F (z), z(0) = z0 . Lemma 2.21. Let X be a Banach space and U ⊂ X be an open set. Suppose that the mapping F : U → X satisfies the locally Lipschitz condition, that is, for any z ∈ U there is a constant L > 0 and an open neighborhood Uz ⊂ U of z such that |F (z1 ) − F (z2 )|X ≤ L|z1 − z2 |X , for all z1 , z2 ∈ Uz . Then for any z0 ∈ U , there exists a time T such that equation (2.88) has a unique (local) solution z ∈ C([0, T ); U ). Proof. The method for showing the local existence of solutions for equation (2.88) is similar to that of Theorem 2.18 (i), so we omit it. Now we establish the global existence of solutions for a kind of fractional differential equations in Banach space, which furnishes as the key instrument in the proof of the global existence of solutions for the time-fractional Navier-Stokes equations. Lemma 2.22. Under the conditions of Lemma 2.21, the unique solution z = z(t) (t ∈ [0, T )) of the fractional differential equations (2.88) either exists globally in time, or T < ∞ and z(t) preserves the open set U as t → T − .

Fractional Navier-Stokes Equations

81

Proof. Suppose that the maximum existing interval for the solution z(t) is [0, T ), that is,  T = sup Tz : z(t) is defined on [0, Tz ] and z(t) is a solution of (2.88) on [0, Tz ] . Then T = ∞ or T < ∞. If T = ∞, the conclusion holds. If T < ∞, we show that z(t) preserves the open set U as t → T − . Observe that there exists a sequence {tk } and a positive constant K > 0 such that tn ≤ tn+1 for n ∈ N+ ,

lim tn = T, |z(tn )| ≤ K.

n→∞

(2.89)

According to the equicontinuity of z(t) and Lemma 2.21, {z(tn )} has a convergent subsequence. Without loss of generality, let limn→∞ z(tn ) = z ∗ . This together with (2.89) implies that for sufficiently small τ > 0, there exists n0 such that T − τ < tn0 < T and for n ≥ n0 , we have |z(tn ) − z ∗ | ≤ 2ε . We show that limt→T − z(t) = z ∗ . On the contrary, for n ≥ n0 , there exists ηn ∈ (tn , T ) such that |z(ηn ) − z ∗ | ≥ ε and |z(t) − z ∗ | < ε, for t ∈ (tn , ηn ). By the continuity of F on [0, T ), we denote M = sups∈[0,ηn ] |F (z(s))|. Thus, ε ≤ |z(ηn ) − z ∗ | ≤|z(tn ) − z ∗ | + |z(ηn ) − z(tn )| Z tn
1 ε (tn − s)α−1 − (ηn − s)α−1 |F (z(s))|ds ≤ + 2 Γ(α) 0 Z ηn 1 + (ηn − s)α−1 |F (z(s))|ds Γ(α) tn
ε 1 α ≤ + sup |F (z(s))| (ηn − tn )α + tα n − ηn 2 Γ(1 + α) s∈[0,tn ] +

1 sup |F (z(s))|(ηn − tn )α Γ(1 + α) s∈[tn ,ηn ]

≤ε, which is a contradiction in view of
1 α sup |F (z(s))| 2(ηn − tn )α + tα n − ηn Γ(1 + α) s∈[0,ηn ]
M α 2(ηn − tn )α + tα n − ηn Γ(1 + α) ε ≤ , for sufficiently large n ≥ n0 . 2 Thus limt→T − z(t) exists. Next we show that z ∗ ∈ ∂U . On the contrary, we assume that z ∗ ∈ / ∂U . Since z(t) ∈ U (t ∈ [0, T )), therefore, z ∗ ∈ U . Let z(t) be equal to z ∗ for t = T and itself for 0 ≤ t < T . Denote Z T 1 ϕ(t) ¯ = z0 + (t − s)α−1 F (z(s))ds, t ∈ [T, T1 ], Γ(α) 0 ≤

82

Fractional Partial Differential Equations

with ϕ¯ ∈ C([T, T1 ]; U ), and define the operator S as follow: Z t 1 (t − s)α−1 F (y(s))ds, t ∈ [T, T1 ], Sy(t) = ϕ(t) ¯ + Γ(α) T where y ∈ C([T, T1 ]; U ). Let   Br = (t, y) : t ∈ [T, T1 ], |y(t)| ≤ sup |ϕ(t)| ¯ +r . t∈[T,T1 ]

On account of the continuity of F on Br , let M = supy∈Br |F (y(s))|. Consider   Br (ϕ, ¯ T + h) = y ∈ C([T, T + h]; U ) : sup |y(t) − ϕ(t)| ¯ ≤ r, y(T ) = ϕ(T ¯ ) , t∈[T,T +h]

 
α1  where h ∈ 0, min T1 − T, Γ(1+α)r . M As argued in Lemma 2.21, we can derive that S is a strict contraction mapping on Br (ϕ, ¯ T + h). Therefore, S has a fixed point z˜(t) ∈ Br (ϕ, ¯ T + h), that is, Z t 1 z˜(t) =ϕ(t) ¯ + (t − s)α−1 F (˜ z (s))ds Γ(α) T Z t 1 (t − s)α−1 F (¯ z (s))ds, t ∈ [T, T + h], =z0 + Γ(α) 0 with ( z¯(t) =

z(t), t ∈ [0, T ), z˜(t), t ∈ [T, T + h].

Evidently z¯ ∈ C([0, T + h]; U ). Thus we find that z¯(t) is the solution of (2.88) on the interval [0, T + h], which contradicts the assumption that [0, T ) is the maximum existing interval. Let us introduce 1 L∞ α (R+ , X )

 =

1

Z

u : R+ → X :

t

(t − s)

α−1

ku(s)k

X1

 ds ∈ L (R+ , X ) . ∞

1

0

Theorem 2.19. For u0 ∈ X−1 satisfying ku0 kX−1 < ν,

(2.90)

the problem (2.82) has a unique solution u existing globally in time. Moreover, 1 u ∈ C([0, ∞); X−1 ) ∩ L∞ α (R+ , X ) and the following estimate holds:   Z
t sup ku(t)kX−1 + ν − ku0 kX−1 (t − s)α−1 k∇u(s)kL∞ ds ≤ ku0 kX−1 . 0≤t 0 is the distribution function of f and f ∗ (t) = inf{a : |µf (a)| ≤ t}, t ≥ 0. Let 0 < p < ∞ and 0 < q ≤ ∞. The Lorentz space Lp,q (Ω) collects all measurable functions f in Ω satisfying  Z |Ω|  q dt  q1 1 ∗ p p,q kf kL = t f (t) < ∞, t 0

92

Fractional Partial Differential Equations

when 0 < q < ∞ and kf kLp,q =

1

sup t p f ∗ (t) < ∞ t∈(0,|Ω|]

when q = ∞. For an introduction to Lorentz spaces, see e.g., Bergh and Lofstrom [45]. Throughout this section, the notation U . V means that there exists a constant M such that U ≤ M V . We will need the following useful result of convolution in Lorentz spaces. Lemma 2.24. [221] Assume that 1 < p, p1 , p2 < ∞, 1 ≤ q, q1 , q2 ≤ ∞ satisfy 1 1 1 1 1 1 +1= + and = + . p2 p p1 q2 q q1 Then for any f ∈ Lp,q and g ∈ Lp1 ,q1 , we have kf ∗ gkLp2 ,q2 . kf kLp,q kgkLp1 ,q1 . Here ∗ denotes the convolution. Denote Rj by the Riesz transforms defined as iξj ∂j , i.e., F(Rj g)(ξ) = F(g)(ξ). Rj = √ |ξ| −∆ Then the operator P is defined by X (Pf )j = fj + Rj Rk fk , 1≤k≤d

which is called the Helmholtz-Leray projection onto the divergence-free fields. Let v : [0, ∞) × Rd → Rd . The fractional integral of order α ∈ (0, 1] for the function v is defined as Z t 1 −α (t − τ )α−1 v(τ, x)dτ, t > 0. D v(t, x) = 0 t Γ(α) 0 Further, ∂tα v represents the Caputo fractional derivative of order α for the function v; it is defined by Z t 1 ∂ ∂tα v(t, x) = (t − τ )−α v(τ, x)dτ, t > 0. Γ(1 − α) 0 ∂τ For more details, we refer the reader to Kilbas et al. [197]. Next we study the mild formulation of solutions to equation (2.102). Let Eα (t), Eα,α (t) be the Mittag-Leffler functions as in Definition 1.7 and Mα (θ) be the Wright-type function as in Definition 1.8. Lemma 2.25. If u satisfies equation (2.102), then we have Z t u(t, x) = Eα,1 (tα ∆)u0 (x) − (t − τ )α−1 Eα,α ((t − τ )α ∆)P∇ · (u(τ, x) ⊗ u(τ, x))dτ, 0

(2.103) where Eα,1 (tα ∆) and Eα,α (tα ∆) are expressed as:   Z ∞ | · |2 −d α α −d 2 2 θ Mα (θ) exp(− α )dθ ∗ v (x), Eα,1 (t ∆)v(x) = (4πt ) 4θt 0   Z ∞ | · |2 α α −d 1− d 2 2 Eα,α (t ∆)v(x) = (4πt ) αθ Mα (θ) exp(− α )dθ ∗ v (x). 4θt 0

Fractional Navier-Stokes Equations

93

Proof. We apply Helmholtz-Leray projector P to equations (2.102), which removes the pressure term. Then by using the space Fourier transform, the equations formally reduce to the form ( ∂tα F(u)(t, ξ) = −|ξ|2 F(u)(t, ξ) − F[P∇ · (u ⊗ u)](t, ξ), t > 0, F(u)(0, ξ) = F(u0 ). Making use of [197], we have the following expression F(u)(t, ξ) =Eα,1 (−tα |ξ|2 )F(u0 ) Z t − (t − τ )α−1 Eα,α (−(t − τ )α |ξ|2 )F[P∇ · (u ⊗ u)](t, ξ)dτ. 0

Afterwards, using the Fourier inverse transform allows us to derive that u(t, x) =F −1 [Eα,1 (−tα |ξ|2 )F(u0 )] Z t   − (t − τ )α−1 F −1 Eα,α (−(t − τ )α |ξ|2 )F[P∇ · (u ⊗ u)](τ, ξ) dτ. 0

According to the relationship between convolutions and Fourier transforms, we obtain F −1 [Eα,1 (−tα |ξ|2 )F(u0 )] = {F −1 [Eα,1 (−tα |ξ|2 )]} ∗ u0 (x),   F −1 Eα,α (−tα |ξ|2 )F[P∇ · (u ⊗ u)](t, ξ) ={F −1 [Eα,α (−tα |ξ|2 )]} ∗ [P∇ · (u ⊗ u)](t, x), R∞ it follows from the equality Eα,1 (−z) = 0 Mα (θ) exp(−zθ)dθ and the Fubini’s theorem that Z ∞  1 d Z F −1 [Eα,1 (−tα |ξ|2 )] = exp(ix · ξ) Mα (θ) exp(−θtα |ξ|2 )dθdξ 2π Rd 0 Z  1 d Z ∞ Mα (θ)dθ exp(ix · ξ − θtα |ξ|2 )dξ = 2π 0 Rd Z ∞  d d |x|2  θ− 2 Mα (θ) exp − =(4πtα )− 2 dθ. 4θtα 0 R∞ Notice that Eα,α (−z) = 0 αθMα (θ) exp(−zθ)dθ, a similar argument indicates that Z ∞  d d |x|2  αθ1− 2 Mα (θ) exp − dθ. F −1 [Eα,α (−tα |ξ|2 )] = (4πtα )− 2 4θtα 0 Consequently, u(t, x) =Eα,1 (tα ∆)u0 (x) −

Z 0

t

(t − τ )α−1 Eα,α ((t − τ )α ∆)P∇ · (u(τ, x) ⊗ u(τ, x))dτ.

94

Fractional Partial Differential Equations

Motivated by above discussion, we can define the concept of mild solutions to the objective equations by means of the above representation formula. Definition 2.8. For T > 0, by a mild solution of the equation (2.102) on [0, T ] corresponding to a divergence-free initial value u0 , we mean that u satisfies integral equation (2.103). Before going further, we present the representation and basic estimates for the composition of the operator Λ˙ s and the solution operator Eα,1 (tα ∆), which will be used to show that the initial value part belongs to a particular space given in the later. The main tool we use comes from harmonic analysis. Lemma 2.26. Let s > −1. Then the kernel function of Λ˙ s Eα,1 (tα ∆) is the function Z ∞  x  α(s+d) s+d Kα,s (t, x) = t− 2 dθ. θ− 2 Mα (θ)K √ θtα 0 Here the function K denotes the kernel function of Λ˙ s exp(∆) and the following estimate holds |K(x)| .

1 . 1 + |x|s+d

Proof. By the definition of Λ˙ s , we have Λ˙ s Eα,1 (tα ∆)v =F −1 {|ξ|s F[Eα,1 (tα ∆)v]} =F −1 [|ξ|s Eα,1 (−tα |ξ|2 )F(v)] = {F −1 [|ξ|s Eα,1 (−tα |ξ|2 )]} ∗ v. Then we proceed to deduce F −1 [|ξ|s Eα,1 (−tα |ξ|2 )] as follows: F −1 [|ξ|s Eα,1 (−tα |ξ|2 )] Z ∞  1 d Z s exp(ix · ξ)|ξ| Mα (θ) exp(−θtα |ξ|2 )dθdξ = 2π Rd 0 Z  1 d Z ∞ Mα (θ)dθ |ξ|s exp(ix · ξ − θtα |ξ|2 )dξ = 2π d 0 R Z  1 d Z ∞ s+d  x  α(s+d) − − 2 θ 2 Mα (θ)dθ exp i √ =t · ξ |ξ|s exp(−|ξ|2 )dξ 2π θtα 0 Rd Z ∞   α(s+d) s+d x dθ, θ− 2 Mα (θ)K √ =t− 2 θtα 0 where the change ξ 7→

√ξ θtα

is used.

Next we also state the similar properties for the composition of the operator Λ˙ s and the solution operator Eα,α (tα ∆)P∇·, it will play an important role in obtaining the bound for nonlinear part of the integral equations. Lemma 2.27. Let s > −1. Then the kernel function of Λ˙ s Eα,α (tα ∆)P∇· is the function Z ∞  x  α(s+d+1) 1−s−d − 2 αθ 2 Mα (θ)P √ dθ. Pα,s (t, x) = t θtα 0

Fractional Navier-Stokes Equations

95

Here the function P denotes the kernel function of Λ˙ s exp(∆)P∇· and the following estimate holds C |P (x)| . . 1 + |x|s+d+1 Moreover, Pα,s (t, x) satisfies the inequality |Pα,s (t, x)| .

1 , for λ1 > 0, 0 < λ2 < 2 and λ1 + 2λ2 = d + s + 1. tαλ2 |x|λ1

Proof. Arguing as in the proof of Lemma 2.26, we infer Λ˙ s Eα,α (tα ∆)P∇ · v =F −1 {|ξ|s F[Eα,α (tα ∆)P∇ · v]} =F −1 [|ξ|s Eα,α (−tα |ξ|2 )F(F (x)v)] ={F −1 [|ξ|s Eα,α (−tα |ξ|2 )]F[F ](ξ)} ∗ v, where F (x) is the tensor kernel associated with the operator P∇·, that is   ξj ξk ξl , j, k, l = 1, ..., d. F[Fl,k,j ](ξ) = i δj,k − |ξ|2 Proceeding as before, one has F −1 [|ξ|s Eα,α (−tα |ξ|2 )F[F ](ξ)] Z ∞  1 d Z = exp(ix · ξ)F[F ](ξ)|ξ|s αθMα (θ) exp(−θtα |ξ|2 )dθdξ 2π Rd 0 Z  1 d Z ∞ αθMα (θ)dθ F[F ](ξ)|ξ|s exp(ix · ξ − θtα |ξ|2 )dξ = 2π 0 Rd  1 d Z ∞ 1−s−d α(s+d+1) − 2 αθ 2 Mα (θ)dθ =t 2π 0 Z  x  × exp i √ · ξ F[F ](ξ)|ξ|s exp(−|ξ|2 )dξ θtα Rd Z ∞  x  α(s+d+1) 1−s−d =t− 2 dθ. αθ 2 Mα (θ)P √ θtα 0 On the other hand, by the Young’s inequality, we deduce Z ∞ α(s+d+1) 1−s−d 1 dθ αθ 2 Mα (θ) |Pα,s (t, x)| ≤t− 2 x d+s+1 0 1 + | √θt α Z ∞ 1 = αθMα (θ) dθ (s+d+1) α 0 (θt ) 2 + |x|d+s+1 Z ∞ 1 αθ1−λ2 Mα (θ)dθ ≤ αλ2 λ1 t |x| 0 1 . αλ2 λ1 , t |x| for λ1 > 0, 0 < λ2 < 2 and λ1 + 2λ2 = d + s + 1.

96

2.7.2

Fractional Partial Differential Equations

Local Existence I

The purpose of this subsection is to study the existence of solutions for equation (2.102). In order to prove the version of Khai and Duong [193] in the fractional framework, a fixed point technique is applied. In this way, we define the bilinear operator B as Z t (t − τ )α−1 Eα,α ((t − τ )α ∆)P∇ · (u(τ, x) ⊗ v(τ, x))dτ. B(u, v)(t, x) = 0   Before starting our theorems, let q˜ ≥ q ≥ d, β = d 1q − 1q˜ with 0 ≤ β < 1. We q˜ introduce a suitable spaces Hq,T consisting of the functions u(t, x) such that



αβ αβ



kukHq˜ := sup t 2 |u(t, x)| < ∞ and lim sup τ 2 |u(τ, x)| = 0. q,T

q˜

0 0 such that

Z t



S(t − s)f (u(s))ds ≤ CkukσXpr kukLµ (0,∞;Lr (RN )) ≤ Ckukσ+1 Zpr .

µ r N 0

L (0,∞;L (R ))

Together with above arguments, we conclude that there is a unique global solution which belongs to Lµ (0, ∞; Lr (RN )). The proof of (ii) is analogous, so we omit it. The proof is completed. 3.3 3.3.1

Well-posedness and Blow-up Introduction

In this section, we introduce the fractional Rayleigh-Stokes problem  α   ∂t u − (1 + γ∂t )∆u = f (t, u), t > 0, x ∈ Ω, u(t, x) = 0, t > 0, x ∈ ∂Ω,   u(0, x) = u0 (x), x ∈ Ω,

(3.17)

where γ > 0 is a given constant, ∂tα is the Riemann-Liouville fractional partial derivative of order α ∈ (0, 1). ∆ is the Laplace operator, Ω ⊂ Rd (d ≥ 1) is a bounded domain with smooth boundary ∂Ω, u0 (x) is the initial data for u in L2 (Ω), f : [0, ∞] × R → R is an -regular map defined in the later subsection. The section is organized as follows. In Subsection 3.3.2, the interpolationextrapolation scales and -regular map are briefly introduced, some concepts and lemmas which will be used in this section are given. Subsection 3.3.3 begins with the

Fractional Rayleigh-Stokes Equations

143

definition of -regular mild solutions of the problem (3.17), and then, the properties of solution operators are discussed and the well-posedness results are obtained of the problem (3.17) in the case that the nonlinearity term is an -regular map. Furthermore, the continuation and blow-up alternative results are given in Subsection 3.3.4. 3.3.2

Preliminaries

In this subsection, we recall some concepts and lemmas which are useful in the next. R+ is the set of all non-negative real numbers, N+ denotes the set of all positive integer numbers, L2 (Ω) denotes the Banach space of all measurable functions on 1 Ω ⊂ Rd (d ≥ 1) with the inner product (·, ·) and the norm kvkL2 (Ω) = |(v, v)| 2 . For any given Banach space Y and a set J ⊂ R+ , we denote by C(J, Y ) the space of all continuous functions from J into Y equipped with the norm kukC(J,Y ) = sup ku(t)kY . t∈J

H01 (Ω) is the closure of C0∞ (Ω) in H 1 (Ω), and equipped with the norm kvkH01 (Ω) := k∇vkL2 (Ω) , here C0∞ (Ω) is the space of all infinitely differentiable functions with compact support in Ω, H 1 (Ω) is a Hilbert space. The left Riemann-Liouville fractional partial derivative of order α ∈ (0, 1) is defined by Z t 1 ∂ (∂tα u)(t, x) = ∂t (0 Dtα−1 u(t, x)) = (t − s)−α u(s, x)ds, t > 0. Γ(1 − α) ∂t 0 In the following, we introduce the spectral problem ( − ∆ϕj (x) = λj ϕj (x), x ∈ Ω, x ∈ ∂Ω,

ϕj (x) = 0,

(3.18)

1 where {ϕj (x)}∞ j=1 is an orthogonal basis of H0 (Ω) and an orthonormal basis of 2 L (Ω), λj is the eigenvalue of −∆ corresponding to ϕj which satisfies 0 < λ1 ≤ λ2 ≤ . . . λj ≤ . . . and has the property that λj → ∞ as j → ∞. In the following, denote A = −∆ for the convenience of writing, we define the fractional power operator Aβ for β ≥ 0 as

Aβ v :=

∞ X

λβj (v, ϕj )ϕj (x),

v ∈ D(Aβ ).

j=1

Obviously, D(Aβ ) is a Banach space equipped with the norm   21 ∞ X 2 . kvkD(Aβ ) =  λ2β j |(v, ϕj )| j=1

144

Fractional Partial Differential Equations

Now we introduce the construction of abstract interpolation-extrapolation scales, for more information, we refer readers to the monograph [15]. For Banach spaces E 0 and E 1 , if E 1 ,→ E 0 densely, we say the pair (E 0 , E 1 ) is a densely injected Banach couple. Denote by (·, ·)θ the admissible interpolation functor of exponent θ ∈ (0, 1), we mean that an interpolation functor of exponent θ for the category of densely injected Banach couples such that E 1 ,→ (E 0 , E 1 )θ densely. Let E 1 = D(A) with the graph norm k · kE 1 = kA · kE 0 and A1 : D(A1 ) ⊂ E 1 → E 1 be the realization of A in E 1 , we can define E 2 = (D(A1 ), kA1 · kE 1 ). Similarly, for k ∈ N+ , we can define E k+1 = (D(Ak ), kAk · kE k ) and Ak+1 = E k+1 − the realization of Ak . On the other side, there is no loss of assuming 0 ∈ ρ(A). Consider the space (E 0 , kA−1 · k) and define E −1 to be the completion of E 0 with the norm kA−1 · k, we know that E 0 ,→ E −1 . Then, for k ∈ N+ ∪ {−1} and θ ∈ (0, 1), we denote E k+θ = (E k , E k+1 )θ and Ak+θ = E k+θ −the realization of Ak . We say {(E β , Aβ ) : −1 ≤ β < ∞} the interpolation-extrapolation scale over [−1, ∞) associated to A and (·, ·)θ . In this section, for the reason that the scale of fractional power spaces may not suitable to treat critical problems, we consider the interpolation space X β := β E 2 (β ≥ 0), thus the norm of X β is given as follows   21 ∞ X λβj |(v, ϕj )|2  . kvkX β :=  j=1

It is easy to see that X 0 = L2 (Ω). In the following, we introduce a definition given in [64]. Definition 3.5. For  ≥ 0, we say a map g is an -regular map relative to the pair (X 0 , X 1 ) if there exist ρ > 1, δ() ∈ [ρ, 1) and a positive constant M such that g : X 1+ → X δ() satisfies   ρ−1 kg(u) − g(v)kX δ() ≤ M ku − vkX 1+ kukρ−1 + kvk + 1 , ∀u, v ∈ X 1+ . X 1+ X 1+ Denote by F the family of all -regular maps relative to (X 0 , X 1 ) which satisfy the following conditions: (i) for u, v ∈ X 1+ , ρ > 1 and ρ ≤ δ() < 1, f (t, ·) satisfies ρ−1 ς kf (t, u) − f (t, v)kX δ() ≤ Cku − vkX 1+ (kukρ−1 X 1+ + kvkX 1+ + w(t)t ), (3.19)

where C is a positive constant, w(t) is a non-decreasing function satisfies 0 ≤ w(t) ≤ a, −δ() +  ≤ ς ≤ 0, t > 0; (ii) for v ∈ X 1+ , −δ() ≤ ς˜ ≤ 0, f (t, ·) satisfies kf (t, v)kX δ() ≤ C(kvkρX 1+ + w(t)tς˜).

(3.20)

It is easy to verify that there exist functions belong to F, which implies that F is not empty, for example, f (t, u) = u|u|ρ−1 , f (t, u) = P (u·∇)u, where P is the orthogonal

Fractional Rayleigh-Stokes Equations

145

projection, for more details, we refer readers to [18, 64] and other references cited therein. For any b > 0 and η > 0, we denote by C η ([0, b]; X 1+ ) the space of all functions v ∈ C((0, b]; X 1+ ) with the property tη v(t) ∈ C([0, b]; X 1+ ). It is easy to see that C η ([0, b]; X 1+ ) is a Banach space under the norm kvkC η ([0,b];X 1+ ) := sup tη kv(t)kX 1+ . t∈[0,b]

R zt 1 Let Sα,j (t) = L−1 ((z + (1 + γz α )λj )−1 )(t) = 2πi e (z + (1 + γz α )λj )−1 dz, B −1 where t ≥ 0, j = 1, 2, . . ., L (·) denotes the inverse Laplace transform and B = {z : Re(z) = $, $ > 0} is the Brownwich path. From [39, Theorem 2.2], we know that Z ∞ Sα,j (t) = e−ξt Kj (ξ)dξ, (3.21) 0

where Kj (ξ) =

λj ξ α sin (απ) γ . π (−ξ + λj γξ α cos (απ) + λj )2 + (λj γξ α sin (απ))2

We introduce some properties of Sα,j (t) in the following. Lemma 3.12. [39, Theorem 2.2] The functions Sα,j (t), j = 1, 2, . . . have the following properties: (i) Sα,j (0) = 1, 0 < Sα,j (t) ≤ 1 for t ≥ 0; (ii) for any j ∈ N+ , Sα,j (t) is completely monotone for t ≥ 0. Lemma 3.13. [278, Lemma 2.2] For α ∈ (0, 1) and t ≥ 0, Sα,j (t) satisfies Sα,j (t) ≤

M1 , 1 + λj t1−α

where j ∈ N+ and M1 = 3.3.3

Γ(1 − α) + 1. γπ sin(απ)

Existence of Mild Solutions

At the beginning, we give the definition of -regular mild solutions of problem (3.17). In fact, multiplying both sides of the first equation in (3.17) by ϕj (x) and taking the Laplace transform, we obtain that zuj (z) − uj (0) + (1 + γz α )λj uj (z) = fj (z), that is, uj (z) = (z + (1 + γz α )λj )−1 (uj (0) + fj (z)),

146

Fractional Partial Differential Equations

where uj = (u, ϕj ), fj = (f, ϕj ), uj and fj stand for the Laplace transform of uj and fj , respectively. And then, from the uniqueness of the inverse Laplace transform, we get Z t Sα,j (t − s)fj (s, u(s))ds, uj (t) = Sα,j (t)u0,j + 0

where u0,j = (u0 , ϕj ). Denote Sα (t)v =

∞ X

Sα,j (t)(v, ϕj )ϕj (x),

for v ∈ L2 (Ω),

j=1

then we have t

Z

Sα (t − s)f (s, u(s))ds.

u(t, x) = Sα (t)u0 +

(3.22)

0

Now we can give the definition of -regular mild solutions. Definition 3.6. By an -regular mild solution of the problem (3.17), we mean a function u ∈ C η ([0, b]; X 1+ ) ∩ C([0, b]; X 1 ) and satisfies (3.22). In the following, we give some properties of Sα (t), which will be useful for us to get the main results. Lemma 3.14. Assume 0 ≤ p < ∞ and 0 ≤ q − p ≤ 2, for t > 0 and any v ∈ X p , the following inequality holds t(1−α)(q−p)/2 kSα (t)vkX q ≤ M1 kvkX p ,

(3.23)

where M1 is a positive constant given in Lemma 3.13. Proof. From Lemma 3.13, for t > 0 and any v ∈ X p , we have ∞ X 2 2 kSα (t)vkX q = λqj Sα,j (t)|(v, ϕj )|2 j=1

≤ M12

∞ X j=1

≤ M12

∞ X j=1

≤

λqj |(v, ϕj )|2 (1 + λj t1−α )2 λqj |(v, ϕj )|2 (1 + λj t1−α )q−p

M12 t−(1−α)(q−p) kvk2X p ,

which implies t(1−α)(q−p)/2 kSα (t)vkX q ≤ M1 kvkX p . This concludes the result. Remark 3.5. From the proof process of Lemma 3.14, we can see that (3.23) holds at t = 0 if and only if p = q. When p 6= q, the property λj → ∞ as j → ∞ makes the conclusion (3.23) invalid at t = 0.

Fractional Rayleigh-Stokes Equations

147

Lemma 3.15. For 0 ≤ p < ∞ and 0 ≤ q − p ≤ 2, Sα (t) is strongly continuous in terms of t > 0 on X q . Moreover, for t2 > t1 > 0 and any v ∈ X p , the following inequality holds (q−p−2)/2 α−1 kSα (t2 )v − Sα (t1 )vkX q ≤ M2 λ1 (t1 − t2α−1 )kvkX p , (3.24) where M2 =

Γ(1−α) πγ sin (απ) .

Proof. For any t2 > t1 > 0 and v ∈ X p , by virtue of (3.21), we have ∞ X kSα (t2 )v − Sα (t1 )vk2X q = λqj |Sα,j (t2 ) − Sα,j (t1 )|2 |(v, ϕj )|2 j=1

=

∞ X

λqj

=

−τ t1

(e

−e

−τ t2

)Kj (τ )dτ

|(v, ϕj )|2 (3.25)

0

j=1 ∞ X

2

∞

Z

λqj

∞

Z

Z

0

j=1

t2

e−τ t τ Kj (τ )dtdτ

2

|(v, ϕj )|2 ,

t1

and then, from the expression of Kj (τ ), we can easily obtain that τ −α 0 < Kj (τ ) ≤ , πγ sin (απ)λj which implies Z ∞ Z t2 Z t2 Z ∞ 1 −τ t ≤ e τ K (τ )dtdτ e−τ t τ 1−α dτ dt j πγ sin (απ)λj 0 t1 t1 0 Z t2 Γ(2 − α) (3.26) = tα−2 dt πγ sin (απ)λj t1 Γ(1 − α) (tα−1 − t2α−1 ). = πγ sin (απ)λj 1 Substituting (3.26) into (3.25), we get that 2  Γ(1 − α)(tα−1 − t2α−1 ) 1 λq−p−2 kvk2X p . kSα (t2 )v − Sα (t1 )vk2X q ≤ 1 πγ sin (απ) It shows that in the case t2 → t1 , kSα (t2 )v − Sα (t1 )vkX q → 0, which implies that Sα (t)v is strongly continuous for t > 0. This completes the proof. Remark 3.6. In the case p = q, the strong continuity of Sα (t) still holds at t = 0 on X p . In fact, for any v ∈ X p , taking t1 = 0, t2 > 0, from the fact that Sα,j (t) = 1 at t = 0 in Lemma 3.12, we have Sα (t1 )v = v. By virtue of Lemma 3.13, we get ∞ X kSα (t2 )v − vk2X p = λpj (1 − Sα,j (t2 ))2 |(v, ϕj )|2 ≤ 2(1 + M12 )kvk2X p . j=1

The Weierstrass discriminance implies that kSα (t2 )v−vkX p is uniformly convergent. Then we have   12 ∞ X lim+ kSα (t2 )v − vkX p =  lim+ λpj (1 − Sα,j (t2 ))2 |(v, ϕj )|2  = 0, t2 →0

j=1

t2 →0

which implies that Sα (t) is strongly continuous for all t ≥ 0 on X p .

148

Fractional Partial Differential Equations

In the following, we give some estimations when the source term is F type, which will be useful for verifying the main results. For the convenience in our writing, we use the notation σ as σ = (1 − α)(1 +  − δ()). Obviously, 0 < σ < 1. Lemma 3.16. Assume f ∈ F and 0 < η < ρ1 , then for t ∈ (0, b] and any u ∈ C η ([0, b]; X 1+ ), we have

Z t  

ρ 1−σ/2 −ρη ς˜

≤ M CB t t , kuk + w(t)t S (t − s)f (s, u(s))ds η 1+ 1 1 α C ([0,b];X )

1+

0

X

(3.27) where B1 = max{B(1 − σ/2, 1 − ρη), B(1 − σ/2, 1 + ς˜)}. Proof. For t > 0 and any u(t, x) ∈ C η ([0, b]; X 1+ ), from Lemma 3.14, we have

Z t Z t



kSα (t − s)f (s, u(s))kX 1+ ds ≤ Sα (t − s)f (s, u(s))ds

0 0 X 1+ Z t ≤M1 (t − s)−σ/2 kf (s, u(s))kX δ() ds. 0

Substituting the condition (3.20) into above, we have

Z t



S (t − s)f (s, u(s))ds α

1+ 0 X Z t
≤M1 (t − s)−σ/2 C ku(s)kρX 1+ + w(s)sς˜ ds 0 Z t Z t ρ −σ/2 −ρη ≤M1 C (t − s) s dskukC η ([0,b];X 1+ ) + M1 Cw(t) (t − s)−σ/2 sς˜ds 0

0

=M1 CB(1 − σ/2, 1 − ρη)t1−σ/2−ρη kukρC η ([0,b];X 1+ ) + M1 Cw(t)B(1 − σ/2, 1 + ς˜)t1−σ/2+˜ς , where B(·, ·) denotes the Beta function. From the definition of σ, we know that 1 − σ/2 > 0, and then, combined with η < ρ1 and ς˜ ≥ −δ() > −1, we can easily check that the Beta functions in the last inequality of above are meaningful. The proof is completed. Lemma 3.17. Assume f ∈ F and 0 < η < min{ ρ1 , 1 + ς}, for any u, v ∈ C η ([0, b]; X 1+ ), denote U = max{kukC η ([0,b];X 1+ ) , kvkC η ([0,b];X 1+ ) }, then we have

Z t



S (t − s)(f (s, u(s)) − f (s, v(s)))ds α

0

X 1+

1−σ/2

≤M1 B2 Ct

2U

ρ−1 −ηρ

t

−η+ς

+ w(t)t




ku − vkC η ([0,b];X 1+ ) ,

where B2 = max{B(1 − σ/2, 1 − ηρ), B(1 − σ/2, 1 − η + ς)}. Proof. For any u, v ∈ C η ([0, b]; X 1+ ), from Lemma 3.14, we have

Z t



S (t − s)(f (s, u(s)) − f (s, v(s)))ds α

1+ 0

X

0 < t ≤ b, (3.28)

Fractional Rayleigh-Stokes Equations

Z

149

t

kSα (t − s)(f (s, u(s)) − f (s, v(s)))kX 1+ ds Z t ≤M1 (t − s)−σ/2 kf (s, u(s)) − f (s, v(s))kX δ() ds.

≤

0

0

And then, the condition (3.19) implies that

Z t

Sα (t − s)(f (s, u(s)) − f (s, v(s)))ds

0

X 1+

t

Z

ρ−1 ς (t − s)−σ/2 Cku(s) − v(s)kX 1+ (ku(s)kρ−1 X 1+ + kv(s)kX 1+ + w(s)s )ds Z t ≤2M1 CU ρ−1 (t − s)−σ/2 s−ηρ dsku − vkC η ([0,b];X 1+ ) 0 Z t (t − s)−σ/2 s−η+ς dsku − vkC η ([0,b];X 1+ ) + M1 Cw(t)

≤M1

0

0

=2M1 CU

ρ−1

B(1 − σ/2, 1 − ηρ))t1−σ/2−ηρ ku − vkC η ([0,b];X 1+ )

+ M1 Cw(t)B(1 − σ/2, 1 − η + ς)t1−σ/2−η+ς ku − vkC η ([0,b];X 1+ ) , where B(·, ·) denotes the Beta function. Similarly, combined with the fact σ ∈ (0, 1) and η < min{ ρ1 , 1 + ς}, we know that 1 − σ/2 > 0, ρη < 1 and 1 − η + ς > 0, which imply that the Beta functions in the last inequality of above are meaningful. The proof is completed. In the following, we present an existence result and investigate the behaviors of -regular mild solutions of problem (3.17) at t = 0. Furthermore, the dependence of -regular mild solutions on initial conditions is also obtained. Before giving our main results, let’s verify the relationships between the parameters at first. From the constraint on ς and the relation that δ() < 1, we have that 1 + ς − (1 − α)/2 ≥ 1 − δ() + (1 + α)/2 > 0 and 1 − σ/2 + ς ≥ 1 − σ/2 − δ() +  = (1 − δ() + )(1 + α)/2 > 0. By virtue of the fact that δ() ≥ ρ, we get that 1 − (1 − α)(1 − δ())/2 1+α 1−α − (1 − α)/2 = + (δ() − ρ) > 0. ρ 2ρ 2ρ Theorem 3.6. Assume f ∈ F and u0 ∈ X 1 , let  1 − (1 − α)(1 − δ())/2 (1 − α)/2 < η < min 1 + ς, , ρ then there exists b > 0 such that the problem (3.17) has a unique -regular mild solution in [0, b], and the solution u(t, x) satisfies tη ku(t)kX 1+ → 0,

as

t → 0+ .

150

Fractional Partial Differential Equations

Moreover, if u(t, x) and z(t, x) are two -regular mild solutions of the problem (3.17) corresponding to the initial value conditions u0 , z0 ∈ X 1 , respectively, then for any t ∈ [0, b], we have tη ku(t) − z(t)kX 1+ ≤ C1 ku0 − z0 kX 1 , where C1 = 4M1 bη−(1−α)/2 . Proof. From the condition η < 1−(1−α)(1−δ())/2 , we have η < ρ1 , which implies ρ that B1 and B2 exist and are well-defined, we denote Bmax = max{B1 , B2 }. For u0 ∈ X 1 , we take 0 < r ≤ 1 and choose b > 0 such that for any t ∈ [0, b], the following inequalities hold: r M1 tη−(1−α)/2 ku0 kX 1 ≤ , 2 (3.29) 1 M1 CBmax t1+η−σ/2−ρη rρ−1 ≤ 4 and r (3.30) max{M1 CB1 t1+η−σ/2+˜ς w(t), M1 CB2 t1−σ/2+ς w(t)} ≤ . 4 For any u ∈ C η ([0, b]; X 1+ ), we define an operator T as follows: Z t (T u)(t) = Sα (t)u0 + Sα (t − s)f (s, u(s))ds, t ≥ 0. (3.31) 0

The main idea of our proof here is to prove that T exists a unique fixed point in C η ([0, b]; X 1+ ), and then to check that this fixed point belongs to C([0, b]; X 1 ). Let ( ) B(r, b) =

u ∈ C η ([0, b]; X 1+ ) : sup tη ku(t)kX 1+ ≤ r . t∈[0,b]

Now we shall prove that T maps B(r, b) into itself and T is a contraction mapping. Step I. T maps B(r, b) into B(r, b). We start by showing that for any u ∈ B(r, b), T u ∈ C((0, b]; X 1+ ). In fact, for 0 < t1 < t2 ≤ b, we have k(T u)(t2 ) − (T u)(t1 )kX 1+ ≤kSα (t2 )u0 − Sα (t1 )u0 kX 1+

Z

+

t2

t1

Z

+

0

t1

Sα (t2 − s)f (s, u(s))ds

X 1+

(Sα (t2 − s) − Sα (t1 − s))f (s, u(s))ds

X 1+

=:I1 + I2 + I3 . Since t1 > 0, according to the strong continuity of Sα (t)v for v ∈ X 1 and combining with the inequality (3.24), we have (−2)/2

I1 ≤ M2 λ 1

(tα−1 − t2α−1 )ku0 kX 1 , 1

which implies that I1 goes to zero as t2 → t1 .

Fractional Rayleigh-Stokes Equations

151

And then, from the process of proof of Lemma 3.16, we obtain that Z t2 I2 ≤M1 C (t2 − s)−σ/2 (ku(s)kρX 1+ + w(s)sς˜)ds t1

≤M1 Crρ

Z

t2

(t2 − s)−σ/2 s−ρη ds + M1 Cw(t2 )

t1

t2

(t2 − s)−σ/2 sς˜ds

t1

1−σ/2−ρη

=M1 Crρ t2 +

Z

Z

(3.32)

1 t1 t2

(1 − s)−σ/2 s−ρη ds

1−σ/2+˜ ς M1 Cw(t2 )t2

Z

1 t1 t2

(1 − s)−σ/2 sς˜ds.

The ranges of σ, η, ρ and ς˜ show that −σ/2 > −1, −ρη > −1 and ς˜ ≥ −δ() > −1, which imply that (1 − s)−σ/2 s−ρη and (1 − s)−σ/2 sς˜ are integrable for s ∈ ( tt21 , 1), thus the last inequality of (3.32) converges to zero as t2 → t1 , which implies that I2 converges to zero as t2 → t1 . For any 0 < ξ < t1 , we can estimate I3 as follows Z t1 I3 ≤ k(Sα (t2 − s) − Sα (t1 − s))f (s, u(s))kX 1+ ds 0 t1 −ξ

Z

k(Sα (t2 − s) − Sα (t1 − s))f (s, u(s))kX 1+ ds

= 0

Z

t1

k(Sα (t2 − s) − Sα (t1 − s))f (s, u(s))kX 1+ ds

+ t1 −ξ

=:I31 + I32 . From Lemma 3.14 and (3.20), we have k(Sα (t2 − s) − Sα (t1 − s))f (s, u(s))kX 1+ ≤2M1 (t1 − s)−σ/2 kf (s, u(s))kX δ() ≤2M1 C(t1 − s)−σ/2 (s−ηρ rρ + w(s)sς˜), and then, from −σ/2 > −1, −ρη > −1 and ς˜ ≥ −δ() > −1, we know that 2M1 C(t1 − s)−σ/2 (s−ηρ rρ + w(s)sς˜) is integrable for s ∈ (0, t1 ). Consequently, in the case s ∈ (0, t1 − ξ), by virtue of the Lebesgue’s dominated convergence theorem and Lemma 3.15, we get I31 → 0 as t2 → t1 . When s ∈ (t1 − ξ, t1 ), the above inequality implies that Z t1 Z t1 ρ −σ/2 −ρη I32 ≤ 2M1 Cr (t1 − s) s ds + 2M1 Cw(t1 ) (t1 − s)−σ/2 sς˜ds, t1 −ξ

t1 −ξ

in light of the integrability of the integrand terms, we obtain that I32 → 0 as ξ → 0. Thus we obtain that I3 → 0 as t2 → t1 . Therefore, combined with above discussion, we get that k(T u)(t2 ) − (T u)(t1 )kX 1+ → 0 as t2 → t1 for 0 < t1 < t2 ≤ b, which implies that (T u)(t) is continuous for t ∈ (0, b]. Now we show that tη (T u)(t) ∈ C([0, b]; X 1+ ).

152

Fractional Partial Differential Equations

From the above discussion, we can easily get that tη (T u)(t) ∈ C((0, b]; X 1+ ), there only left to prove the continuity of tη (T u)(t) at t = 0. For any u ∈ B(r, b) and t ∈ (0, b], from Lemma 3.14 and Lemma 3.16, we can see that tη k(T u)(t)kX 1+

Z t

S (t − s)f (s, u(s))ds ≤tη kSα (t)u0 kX 1+ + tη α

1+

0

X

≤M1 tη−(1−α)/2 ku0 kX 1 + M1 CB1 t1+η−σ/2−ρη rρ + M1 CB1 w(t)t1+η−σ/2+˜ς , (3.33) where B1 is given in Lemma 3.16. According to the facts that (1 − α)/2 < η < 1−(1−α)(1−δ())/2 and ς˜ ≥ −δ(), we get 1 + η − σ/2 − ρη > 1 + η − σ/2 − (1 − (1 − ρ α)(1 − δ())/2) = η − (1 − α)/2 > 0 and 1 + η − σ/2 + ς˜ ≥ 1 + η − σ/2 − δ() = (1 − δ())(1 + α)/2 + η − (1 − α)/2 > 0, which imply that tη k(T u)(t)kX 1+ → 0 as t → 0+ , thus we get that tη (T u)(t) ∈ C([0, b]; X 1+ ). Now we show that for any u ∈ B(r, b) and all t ∈ [0, b], tη k(T u)(t)kX 1+ ≤ r. In fact, substituting (3.29) and (3.30) into (3.33), we can get that
tη k(T u)(t)kX 1+ ≤M1 tη−(1−α)/2 ku0 kX 1 + M1 CB1 t1+η−σ/2 t−ρη rρ + w(t)tς˜ r r r ≤ + + 2 4 4 =r. Therefore, we obtain that T maps B(r, b) into itself. Step II. T is a contraction mapping. For any u, v ∈ B(r, b), from Lemma 3.17, (3.29) and (3.30), we have tη k(T u)(t) − (T v)(t)kX 1+

Z t

η Sα (t − s)(f (s, u(s)) − f (s, v(s)))ds =t

1+ 0

X

≤M1 B2 C(2rρ−1 t1+η−σ/2−ηρ + w(t)t1−σ/2+ς )ku − vkC η ([0,b];X 1+ )   1 r ≤ + ku − vkC η ([0,b];X 1+ ) 2 4 3 ≤ ku − vkC η ([0,b];X 1+ ) . 4 Thus we get that T is a contraction mapping. The Banach fixed point theorem implies that T exists a fixed point u(t, x) in B(r, b). Step III. The fixed point u(t, x) ∈ C([0, b]; X 1 ). From Step I, we know that the fixed point u ∈ C((0, b]; X 1+ ), and then, combined with the fact that X 1+ ,→ X 1 , analogous to the process in proving the continuity of (T u)(t) in Step I, we can get that u ∈ C((0, b]; X 1 ), here we omit the proof process. There only left to prove the continuity of u(t) at t = 0 in X 1 . Actually, for t > 0, from the fact that u is the fixed point of T in B(r, b), we have

Z t



ku(t) − u0 kX 1 ≤ kSα (t)u0 − u0 kX 1 + Sα (t − s)f (s, u(s))ds

1 0

=: O1 + O2 .

X

Fractional Rayleigh-Stokes Equations

153

From Remark 3.6, we can easily get that O1 → 0 as t → 0+ . Now we estimate O2 . In fact, by using an analogous argument presented in the process of proof of Lemma 3.16, we have Z t O2 ≤ kSα (t − s)f (s, u(s))kX 1 ds 0 Z t ≤M1 C (t − s)−(1−α)(1−δ())/2 (ku(s)kρX 1+ + w(s)sς˜)ds 0 Z t ≤M1 C (t − s)−(1−α)(1−δ())/2 s−ρη ds( sup sη ku(s)kX 1+ )ρ s∈[0,t]

0 t

Z

(t − s)−(1−α)(1−δ())/2 w(s)sς˜ds

+ M1 C 0

≤M1 CB(1 − (1 − α)(1 − δ())/2, 1 − ηρ) × t1−(1−α)(1−δ())/2−ηρ ( sup (sη ku(s)kX 1+ ))ρ s∈[0,t]

+ M1 Cw(t)B(1 − (1 − α)(1 − δ())/2, 1 + ς˜)t1−(1−α)(1−δ())/2+˜ς , where the facts 0 < α < 1 and −˜ ς ≤ δ() < 1 are used, which show that the Beta functions in the last inequality of above are well-defined, and 1−(1−α)(1−δ())/2+ ς˜ ≥ (1 − δ())(1 + α)/2 > 0 can be checked as well. According to the constraint of η, we can get that the exponents of t in the last inequality of above are positive. Thus, the right-hand side of above inequalities converges to zero as t → 0+ , which implies u(t, x) ∈ C([0, b]; X 1 ). Based on the discussion of above steps, we conclude that u(t, x) is an -regular mild solution of the problem (3.17). Now we show that tη ku(t)kX 1+ → 0 as t → 0+ . From Lemma 3.14, Lemma 3.16 and (3.29), we have tη ku(t)kX 1+ ≤M1 tη−(1−α)/2 ku0 kX 1 + M1 CB(1 − σ/2, 1 − ρη)t1+η−σ/2−ρη sup (sη ku(s)kX 1+ )ρ s∈[0,t] 1+η−σ/2+˜ ς

+ M1 Cw(t)B(1 − σ/2, 1 + ς˜)t

≤M1 tη−(1−α)/2 ku0 kX 1 + M1 CB1 t1+η−σ/2−ρη rρ−1 sup (sη ku(s)kX 1+ ) s∈[0,t] 1+η−σ/2+˜ ς

+ M1 Cw(t)B1 t

≤M1 tη−(1−α)/2 ku0 kX 1 +

1 sup (sη ku(s)kX 1+ ) + M1 Cw(t)B1 t1+η−σ/2+˜ς , 4 s∈[0,t]

which implies sup (sη ku(s)kX 1+ ) ≤ s∈[0,t]

 4 M1 tη−(1−α)/2 ku0 kX 1 + M1 Cw(t)B1 t1+η−σ/2+˜ς 3

→ 0,

as t → 0+ .

154

Fractional Partial Differential Equations

Step IV. The continuous dependence of -regular mild solutions on the initial value conditions. Assume that u(t, x), z(t, x) ∈ C η ([0, b]; X 1+ ) ∩ C([0, b]; X 1 ) are two -regular mild solutions of the problem (3.17) obtained from above under the given initial value conditions u0 and z0 , respectively, thus they have the following forms Z

t

Sα (t − s)f (s, u(s, x))ds,

u(t, x) = Sα (t)u0 + Z

0 t

Sα (t − s)f (s, z(s, x))ds.

z(t, x) = Sα (t)z0 + 0

For u0 , z0 ∈ X 1 , from Lemma 3.14, Lemma 3.17, (3.29) and (3.30), we can estimate tη ku(t) − z(t)kX 1+ as follows tη ku(t) − z(t)kX 1+ η

≤t kSα (t)(u0 − z0 )kX 1+

Z t

Sα (t − s)(f (s, u(s)) − f (s, z(s)))ds +t

η

0

≤M1 t

η−(1−α)/2

ku0 − z0 kX 1

η+1−σ/2−ηρ ρ−1

+ 2M1 B2 Ct

X 1+

sup (sη ku(s) − z(s)kX 1+ )

r

s∈[0,t]

+ M1 B2 Ct1−σ/2+ς w(t) sup (sη ku(s) − z(s)kX 1+ ) s∈[0,t]

≤M1 tη−(1−α)/2 ku0 − z0 kX 1 +

3 sup (sη ku(s) − z(s)kX 1+ ), 4 s∈[0,t]

which implies tη ku(t) − z(t)kX 1+ ≤4M1 tη−(1−α)/2 ku0 − z0 kX 1 ≤4M1 bη−(1−α)/2 ku0 − z0 kX 1 . Thus the continuous dependence of -regular mild solutions on the initial value conditions is proved. The proof is completed. 3.3.4

Continuation and Blow-up Alternative

In this subsection, we consider the continuation and blow-up alternative for the -regular mild solution obtained by Theorem 3.6. We give the definition of continuation of -regular mild solutions at the beginning. Definition 3.7. For the -regular mild solution u ∈ C η ([0, b]; X 1+ ) ∩ C([0, b]; X 1 ) of the problem (3.17), we say that v is a continuation of u in [0, ˜b] if v ∈ C η ([0, ˜b]; X 1+ ) ∩ C([0, ˜b]; X 1 ) is an -regular mild solution for ˜b > b and v(t) = u(t) whenever t ∈ [0, b].

Fractional Rayleigh-Stokes Equations

155

Theorem 3.7. Assume the conditions in Theorem 3.6 hold and let u be an -regular mild solution of the problem (3.17) in [0, b], then there exists a unique continuation u ˜ of u in [0, ˜b] for some ˜b > b. Proof. Take ˜b > b and choose r˜ > 0 such that for u0 ∈ X 1 and any t ∈ [b, ˜b], the following inequalities hold: r˜ t (bα−1 − tα−1 )ku0 kX 1 ≤ , 3 Z 1 r˜ M1 Cw(t)t1+η−σ/2+˜ς (1 − s)−σ/2 sς˜ds ≤ , b 6 t  (−δ()−1)/2 ρ η M2 λ 1 Cm t B(α, 1 − ρη)(bα−ρη − tα−ρη ) Z 1  r˜ α−ρη +t (1 − s)α−1 s−ρη ds ≤ , b 6 t  (−δ()−1)/2 M2 λ 1 Cw(b)tη B(α, 1 + ς˜)(bα+˜ς − tα+˜ς ) Z 1  r˜ (1 − s)α−1 sς˜ds ≤ + tα+˜ς b 6 t (−2)/2 η

M2 λ 1

(3.34)

and M1 Cmρ t1+η−σ/2−ηρ

1

Z

(1 − s)−σ/2 s−ηρ ds ≤ b t

1−σ/2+ς

1

Z

−σ/2 −η+ς

(1 − s)

M1 Cw(t)t

b t

s

r˜ , 6

1 ds ≤ , 3

(3.35)

where m = max{r, r˜+ ˜bη ku(b)kX 1+ }. From the condition (1−α)/2 < η < min{1+ ς, 1−(1−α)(1−δ())/2 }, we can easily check that 1 + η − σ/2 − ρη > 0, 1 + η − σ/2 + ς˜ ≥ ρ 1 + η − σ/2 − δ() > 0. The fact ρη < 1 and 1 + ς˜ ≥ 1 − δ() > 0 imply that the Beta functions in (3.34) are well-defined. We can also verify functions (1−s)−σ/2 sς˜, (1 − s)−σ/2 s−ηρ and (1 − s)−σ/2 s−η+ς are integrable for s ∈ ( bt , 1) and t ∈ (b, ˜b]. We define the set B(˜ r, ˜b) of all functions v ∈ C η ([0, ˜b]; X 1+ ) ∩ C([0, ˜b]; X 1 ) such that v(t) = u(t) for t ∈ [0, b] and sups∈[b,t] sη kv(s) − u(b)kX 1+ ≤ r˜ for t ∈ [b, ˜b]. For any v ∈ B(˜ r, ˜b) and t ∈ [0, ˜b], we still use the definition of T in (3.31). Now we prove that T maps B(˜ r, ˜b) into B(˜ r, ˜b). ˜ For any v ∈ B(˜ r, b), in the case t ∈ [0, b], we know that v(t) = u(t), then from Theorem 3.6, we get that T maps B(˜ r, ˜b) into itself and (T v)(t) = (T u)(t) = u(t). Now we prove that (T v)(t) ∈ C η ([0, ˜b]; X 1+ ) ∩ C([0, ˜b]; X 1 ). In fact, using a similar process of proof provided in Theorem 3.6, we can show that for any v ∈ B(˜ r, ˜b), η 1+ 1 ˜ ˜ (T v)(t) ∈ C ([0, b]; X ) ∩ C([0, b]; X ), here we omit the details. There left to prove that sups∈[b,t] sη k(T v)(s) − u(b)kX 1+ ≤ r˜ holds for t ∈ [b, ˜b]. In fact, for any v ∈RB(˜ r, ˜b), from Theorem 3.6 and the Rdefinition of B(˜ r, ˜b), we b b have u(b) = Sα (b)u0 + 0 Sα (b − s)f (s, u(s))ds = Sα (b)u0 + 0 Sα (b − s)f (s, v(s))ds,

156

Fractional Partial Differential Equations

then for any t ∈ [b, ˜b], we obtain

Z t

Sα (t − s)f (s, v(s))ds ≤k(Sα (t) − Sα (b))u0 kX 1+ +

1+ b X

Z

b

+ (Sα (t − s) − Sα (b − s))f (s, v(s))ds

1+

0

k(T v)(t) − u(b)kX 1+

X

=:Q1 + Q2 + Q3 . From Lemma 3.15, we get that (−2)/2

Q1 ≤ M2 λ1

(bα−1 − tα−1 )ku0 kX 1 .

And then, similar as (3.32), we use Lemma 3.14 to estimate the second term, that is Z t Q2 ≤ kSα (t − s)f (s, v(s))kX 1+ ds b Z t ≤M1 C (t − s)−σ/2 s−ρη ds( sup sη kv(s)kX 1+ )ρ s∈[b,t]

b t

Z

(t − s)−σ/2 sς˜ds

+ M1 Cw(t) b 1−σ/2−ρη

1

Z

(1 − s)−σ/2 s−ρη ds( sup sη kv(s)kX 1+ )ρ

≤M1 Ct

b t

s∈[b,t]

+ M1 Cw(t)t1−σ/2+˜ς

1

Z

(1 − s)−σ/2 sς˜ds. b t

From Lemma 3.15 and (3.20), we can estimate the third term as follows Z b Q3 ≤ k(Sα (t − s) − Sα (b − s))f (s, v(s))kX 1+ ds 0 (−δ()−1)/2

b

Z


(b − s)α−1 − (t − s)α−1 kf (s, v(s))kX δ() ds

≤M2 λ1

0 (−δ()−1)/2 ≤M2 λ1 C

b

Z 0

(−δ()−1)/2 ≤M2 λ1 C(

(−δ()−1)/2

+ M 2 λ1


(b − s)α−1 − (t − s)α−1 (kv(s)kρX 1+ + w(s)sς˜)ds η

ρ

b

Z


(b − s)α−1 − (t − s)α−1 s−ρη ds

sup s kv(s)kX 1+ ) s∈[0,b]

0

Z

b


(b − s)α−1 − (t − s)α−1 sς˜ds

Cw(b) 0

(−δ()−1)/2 =M2 λ1 C(

Z ×

sup s kv(s)kX 1+ )ρ s∈[0,b]

b

(b − s) 0

η

α−1 −ρη

s

Z ds −

!

t α−1 −ρη

(t − s) 0

s

ds

Fractional Rayleigh-Stokes Equations

+ +

(−δ()−1)/2 M2 λ1 C( sup sη kv(s)kX 1+ )ρ s∈[0,b]

α−1 ς˜

(b − s)

+

(t − s)α−1 s−ρη ds

b

Z

s ds −

0 (−δ()−1)/2 M2 λ1 Cw(b)

t

b

Z

(−δ()−1)/2 M2 λ1 Cw(b)

Z

157

!

t α−1 ς˜

(t − s)

s ds

0

t

Z

(t − s)α−1 sς˜ds.

b

From the definition of Beta function, and by virtue of the facts ρη < 1, ς˜ ≥ −δ() > −1, we can compute the integrals of above inequality as follows Z b (b − s)α−1 s−ρη ds = bα−ρη B(α, 1 − ρη), 0 b

Z

(b − s)α−1 sς˜ds = bα+˜ς B(α, 1 + ς˜),

0

similarly, t

Z

(t − s)α−1 s−ρη ds = tα−ρη B(α, 1 − ρη),

0 t

Z

(t − s)α−1 sς˜ds = tα+˜ς B(α, 1 + ς˜).

0

Substituting the above results into the original inequality, we obtain that (−δ()−1)/2

Q3 ≤M2 λ1

CB(α, 1 − ρη)( sup sη kv(s)kX 1+ )ρ (bα−ρη − tα−ρη ) s∈[0,b]

(−δ()−1)/2

+ M2 λ1 +

C( sup sη kv(s)kX 1+ )ρ tα−ρη s∈[0,b]

(−δ()−1)/2 M2 λ1 Cw(b)

+t

α+˜ ς

1

Z



1

Z

(1 − s)α−1 s−ρη ds b t

B(α, 1 + ς˜)(bα+˜ς − tα+˜ς )

 (1 − s)α−1 sς˜ds .

b t

Observing that sup sη kv(s)kX 1+ ≤ sup sη kv(s) − u(b)kX 1+ + tη ku(b)kX 1+ ≤ m, s∈[0,t]

s∈[0,t]

and in virtue of the conditions (3.34) and (3.35), we obtain that tη kT v(t) − u(b)kX 1+ (−2)/2

≤tη M2 λ1

(bα−1 − tα−1 )ku0 kX 1 Z 1 1+η−σ/2−ρη ρ (1 − s)−σ/2 s−ρη ds + M1 Ct m b t

+ M1 Cw(t)t1+η−σ/2+˜ς

1

Z

(1 − s)−σ/2 sς˜ds b t

(−δ()−1)/2

+ M2 λ1

 Cmρ tη B(α, 1 − ρη)(bα−ρη − tα−ρη )

158

Fractional Partial Differential Equations

+t

α−ρη

1

Z

 (1 − s)α−1 s−ρη ds

b t

 (−δ()−1)/2 + M2 λ1 Cw(b)tη B(α, 1 + ς˜)(bα+˜ς − tα+˜ς ) Z 1  α+˜ ς (1 − s)α−1 sς˜ds +t b t

r˜ r˜ r˜ r˜ r˜ ≤ + + + + = r˜. 3 6 6 6 6 Thus we conclude that T maps B(˜ r, ˜b) into B(˜ r, ˜b). In the following, we prove that T is a contraction mapping on B(˜ r, ˜b). For any v1 , v2 ∈ B(˜ r, ˜b), it is easy to see that (T v1 )(t)−(T v2 )(t) = 0 for t ∈ [0, b], we just have to discuss the case when t ∈ [b, ˜b], in this case, we have Z t Sα (t − s)(f (s, v1 (s)) − f (s, v2 (s)))ds. (T v1 )(t) − (T v2 )(t) = b

It follows from the proof process of Lemma 3.17 that tη k(T v1 )(t) − (T v2 )(t)kX 1+ Z t ρ−1 η ≤2M1 C(Vmax ) t (t − s)−σ/2 s−ηρ dskv1 − v2 kC η ([0,˜b];X 1+ ) b Z t η + M1 Cw(t)t (t − s)−σ/2 s−η+ς dskv1 − v2 kC η ([0,˜b];X 1+ ) b

≤2M1 Cmρ−1 t1+η−σ/2−ηρ

1

Z b t

+ M1 Cw(t)t

1−σ/2+ς

(1 − s)−σ/2 s−ηρ dskv1 − v2 kC η ([0,˜b];X 1+ )

1

Z b t

(1 − s)−σ/2 s−η+ς dskv1 − v2 kC η ([0,˜b];X 1+ ) ,

 where Vmax = max kv1 kC η ([0,˜b];X 1+ ) , kv2 kC η ([0,˜b];X 1+ ) . r˜ ≤ 1 into the first condition of (3.35), we get that Applying the fact m Z 1 1 1 r˜ ≤ . M1 Cmρ−1 t1+η−σ/2−ηρ (1 − s)−σ/2 s−ηρ ds ≤ b 6 m 6 t

(3.36)

Combined with the above discussion, by virtue of (3.36) and the second condition of (3.35), we obtain that 2 kv1 − v2 kC η ([0,˜b];X 1+ ) , 3 which implies that T is a strictly contraction mapping. By virtue of the Banach fixed point theorem, we obtain that there exists a unique fixed point u ˜ ∈ B(˜ r, ˜b), which implies that u ˜ is a continuation of u in [0, ˜b]. Now we prove the continuation u ˜ is unique in [0, ˜b]. Assume that u1 and u2 are two continuations of the -regular mild solution u obtained by Theorem 3.6 in [0, ˜b]. From the definition of continuation, it is tη k(T v1 )(t) − (T v2 )(t)kX 1+ ≤

Fractional Rayleigh-Stokes Equations

159

obviously that u1 (t) = u2 (t) = u(t) for t ∈ [0, b]. Thus, we only need to check that u1 (t) = u2 (t) for t ∈ [b, ˜b] in the sequel. From the proof process of Lemma 3.17, we have ku1 (t) − u2 (t)kX 1+

Z t

S (t − s)(f (s, u (s)) − f (s, u (s)))ds = α 1 2

1+

0

X

Z t

Sα (t − s)(f (s, u1 (s)) − f (s, u2 (s)))ds =

b

X 1+

Z

t

ρ−1 ρ−1 ς (t − s)−σ/2 Cku1 (s) − u2 (s)kX 1+ (ku1 (s)kX 1+ + ku2 (s)kX 1+ + w(s)s )ds Z t ≤M1 CM3 (t − s)−σ/2 ku1 (s) − u2 (s)kX 1+ ds,

≤M1

b

b

where M3 = (sups∈[b,˜b] ku1 (s)kX 1+ )ρ−1 + (sups∈[b,˜b] ku2 (s)kX 1+ )ρ−1 + w(˜b)bς . For the reason that (t − s)−σ/2 is continuous and nonnegative for s ∈ (b, t), the singular Gronwall’s inequality implies that the relation u1 (t) = u2 (t) also holds for t ∈ [b, ˜b]. Thus we conclude that u ˜ is the unique continuation of u in [0, ˜b]. This completes the proof. Remark 3.7. We note that the conclusion in Theorem 3.7 still holds if we change the existence time b of u(t) into the maximal time we can get from Theorem 3.6. Theorem 3.8. Assume the conditions in Theorem 3.6 hold and u is an -regular mild solution of the problem (3.17) with maximal time of existence bmax . Then ku(t)kX 1+ = ∞. either bmax = ∞ or lim supt→b− max Proof. Assume that the maximal time of existence bmax < ∞ and there exists a positive constant R such that the solution u(t) satisfies supt∈(0,bmax ) tη ku(t)kX 1+ ≤ R. In the following, we shall prove that this assumption is a contradiction. Taking a sequence {tn }n∈N+ (tn < bmax ) which satisfies tn → b− max as n → ∞. 1+ Now we show that {u(tn )}n∈N+ is a Cauchy sequence in X . In fact, for 0 < tm < tn < bmax , we have Z tn u(tn ) − u(tm ) =(Sα (tn ) − Sα (tm ))u0 + Sα (tn − s)f (s, u(s))ds 0 Z tm − Sα (tm − s)f (s, u(s))ds 0 Z tn =(Sα (tn ) − Sα (tm ))u0 + Sα (tn − s)f (s, u(s))ds tm

Z

tm

(Sα (tn − s) − Sα (tm − s))f (s, u(s))ds.

+ 0

By using an analogous argument presented in the proof process of Theorem 3.6 (Step I) and Theorem 3.7 (Q3 ), we can easily get that

160

Fractional Partial Differential Equations

ku(tn ) − u(tm )kX 1+ ≤k(Sα (tn ) − Sα (tm ))u0 kX 1+

Z

+

tn

tm

Z

+

tm

Sα (tn − s)f (s, u(s))ds

0 (−2)/2 α−1 ≤M2 λ1 (tm

+

X 1+

(Sα (tn − s) − Sα (tm − s))f (s, u(s))ds

X 1+

− tα−1 )ku0 kX 1 n Z 1 M1 CRρ t1−σ/2−ρη (1 − s)−σ/2 s−ρη ds n t m tn

ς + M1 Cw(tn )t1−σ/2+˜ n

(−δ()−1)/2

+ M2 λ 1

Z

Z C

1

(1 − s)−σ/2 sς˜ds tm tn

tm


(tm − s)α−1 − (tn − s)α−1 (Rρ s−ρη + w(tm )sς˜)ds,

0 −σ/2 −ρη


where (1 − s) s , (1 − s)−σ/2 sς˜ and (tm − s)α−1 − (tn − s)α−1 (Rρ s−ρη + w(tm )sς˜) are integrable according to our earlier discussion. Moreover, for given θ > 0, since {tn }n∈N+ (tn < bmax ) is a sequence which converges to b− max , then there exists an N ∈ N+ such that for any n, m ≥ N , |tn − tm | can be small enough so that θ (−2)/2 α−1 M2 λ1 (tm − tα−1 )ku0 kX 1 < , n 3 Z 1 Z 1 −σ/2 −ρη 1−σ/2+˜ ς (1 − s) s ds + M Cw(t )t (1 − s)−σ/2 sς˜ds M1 CRρ t1−σ/2−ρη 1 n n n tm tn


0 is a given time, γ > 0 is a constant, h(x) is the state of u at the initial time, f : [0, b] × Ω → R is the source term of our problem. The outline of this section is as follows. In Subsection 3.4.2, we give some notations, concepts and lemmas which will be used in this section. In Subsection 3.4.3, firstly, we give the definition of weak solutions of the fractional Rayleigh-Stokes problem, and then, by using the Galerkin method, we obtain the existence, uniqueness and the energy estimate of approximate solutions, afterwards, the existence, uniqueness and regularity results of weak solutions are obtained with the initial data h(x) in H01 (Ω) and the source term f (t, x) in H −1 (Ω). Finally, under conditions that h ∈ H 2 (Ω) and f ∈ L2 (0, b; L2 (Ω)), the spacial regularity of weak solutions of the fractional Rayleigh-Stokes problem is improved. 3.4.2

Preliminaries

In this subsection, we recall some notations, concepts and lemmas which are useful in this section. Denote Rd (d ≥ 1) is the d-dimensional Euclidean space, and Ω ⊂ Rd is a bounded domain with smooth boundary ∂Ω. In order to obtain our main results, we introduce the spectral problem: −∆ϕk = 1 λk ϕk in Ω and ϕk = 0 on ∂Ω, where {ϕk }∞ k=1 is both an orthogonal basis of H0 (Ω) 2 and an orthonormal basis of L (Ω), λk is the eigenvalue of −∆ corresponding to ϕk and satisfies 0 < λ1 ≤ λ2 ≤ . . . λk ≤ . . . with the property that λk → ∞ as k → ∞. Some functional spaces are introduced in the following. We denote by L2 (Ω) the Banach space of all measurable functions on Ω endowed with the inner product (·, ·) and the norm ! 21 ∞ X 1 2 kvkL2 (Ω) := (v, v) 2 = |(v, ϕk )| . k=1

H01 (Ω)

denotes the closure of

C0∞ (Ω)

1

in H (Ω), and equipped with the norm

kvkH01 (Ω) := k∇vkL2 (Ω) ,

162

Fractional Partial Differential Equations

here C0∞ (Ω) is the space of all infinitely differentiable functions with compact supP∞ port, and it is easy to see that kvk2H 1 (Ω) = k=1 λk |(v, ϕk )|2 . 0 Denote by H −1 (Ω) the dual space of H01 (Ω) with the norm kvkH −1 (Ω) :=

∞ X

! 21 2 λ−1 k |(v, ϕk )|

.

k=1

We denote by Lp (0, b; X)(p ≥ 1) the space of all strongly measurable functions v with the norm ! p1 Z b

kvkLp (0,b;X) := 0

kv(t)kpX dt

< ∞,

for

1≤p 0 Γ(1 − α) ∂t 0 and C α ∂t u(t, x)

= ∂tα (u(t, x) − u(0, x)),

t > 0,

respectively. In the following, we introduce some inequalities which will be useful in proving our main results. Lemma 3.18. [233] Let a(t) ∈ R be a continuous and differentiable function. Then the following inequality holds 1 α 2 α 0 D a (t) ≤ a(t)0 Dt a(t), 2 t

∀α ∈ (0, 1),

∀t ≥ 0.

Lemma 3.19. [273] If α ∈ (0, 1) and v(·, x) ∈ L2 (0, b) for each x ∈ Ω, then Z t  απ  Z t α

−α

Ds− 2 v(s) 2 2 ds, f or t ∈ [0, b]. (0 Ds v(s), v(s))ds ≥ cos 0 L (Ω) 2 0 0

Fractional Rayleigh-Stokes Equations

163

Now we give an existence result under a special case of Proposition 5.5 and Theorem 5.5 in [197]. Lemma 3.20. Let 0 < α < 1, λ, µ ∈ R and let g(t) be a given real function defined on R+ . Then the Cauchy type problem ( y 0 (t) − λ0 Dtα y(t) − µy(t) = g(t), t > 0, y(0) = w is solvable, and its solution has the form Z t y(t) = Pα,λ,µ (t)w + Pα,λ,µ (t − s)g(s)ds,

(3.38)

0

provided that the right-hand side in (3.38) is convergent, where   1 Pα,λ,µ (t) = L−1 (t), s − λsα − µ L−1 (v(s))(t) stands for the inverse Laplace transform of v. It is easy to see that in the case λ = −λk γ and µ = −λk (k = 1, 2, . . .), the function Pα,λ,µ (t) is equivalent to the function uk (t) in [39], in this section, for the convenience of writing, we denote   1 −1 (t), (3.39) Pα,k (t) = L s + γλk sα + λk and it has the following properties. Lemma 3.21. [39] The functions Pα,k (t), k = 1, 2, . . . , have the following properties: (i) Pα,k (0) = 1, 0 < Pα,k (t) ≤ 1, t ≥ 0; (ii) Pα,k (t) are completely monotone for t ≥ 0. 3.4.3

Existence, Uniqueness and Regularity

In this subsection, we give the definition of weak solutions of problem (3.37) and estimate the existence, uniqueness and regularity of weak solutions by virtue of Galerkin method. Furthermore, an improved regularity result of weak solutions is obtained. T Definition 3.8. We say a function u ∈ L2 (0, b; H01 (Ω)) L∞ (0, b; L2 (Ω)) with ∂tα u ∈ L2 (0, b; H01 (Ω)), ∂t u ∈ L2 (0, b; H −1 (Ω)) is a weak solution of problem (3.37) provided (i) h∂t u, vi + ((1 + γ∂tα )∇u, ∇v) = (f, v), for each v ∈ H01 (Ω) and a.e. t ∈ [0, b]; (ii) u(0) = h.

164

Fractional Partial Differential Equations

For a fixed positive integer m, the work on finding the existence of weak solutions of problem (3.37) is turned to looking for a function um of the following form m X um (t) = dkm (t)ϕk , (3.40) k=1

where um is required to satisfy ( (∂t um (t), ϕk ) + ((1 + γ∂tα )∇um (t), ∇ϕk ) = (f (t), ϕk ), (um (0), ϕk ) = (h, ϕk ),

(3.41)

here, for the convenience of writing, we omit the spatial variable in um and f . Lemma 3.22. Assume h ∈ H01 (Ω) and f ∈ L2 (0, b; H −1 (Ω)). Then for every fixed integer m = 1, 2, . . . and any fixed integer k = 1, . . . , m, the problem (3.41) exists a unique solution um of the form Z t m X um (t) = dkm (t)ϕk = Pαm (t)h + Pαm (t − s)f (s)ds, (3.42) 0

k=1

where (3.39).

Pαm (t)v

=

Pm

k=1

Pα,k (t)(v, ϕk ) for any v ∈ L2 (Ω) and Pα,k (t) is defined in

Proof. Assume that um (t) is given by (3.40). From the assumptions of {ϕk }∞ k=1 , we obtain that 0

(∂t um (t), ϕk ) = dkm (t) and ((1 + γ∂tα )∇um (t), ∇ϕk ) = (1 + γ 0 Dtα )λk dkm (t). Then the problem (3.41) is equivalent to the following fractional ordinary differential equations ( 0 dkm (t) + (1 + γ 0 Dtα )λk dkm (t) = f k (t), (3.43) dkm (0) = hk , where f k (t) := (f (t), ϕk ) and hk := (h, ϕk ). From Lemma 3.20, we can easy to check that the problem (3.43) is solvable, which implies that there exists a unique solution dkm (t) for a.e. t ∈ [0, b] of the following form Z t k k dm (t) = Pα,k (t)h + Pα,k (t − s)f k (s)ds, (3.44) 0

where Pα,k (t) is defined in (3.39). Lemma 3.21 indicates that the solution dkm (t) is continuous for t ∈ [0, b]. Thus we conclude that the problem (3.41) exists a unique continuous solution um of the form Z t m X um (t) = dkm (t)ϕk = Pαm (t)h + Pαm (t − s)f (s)ds, 0

k=1

where the operator

Pαm (t)v

=

Pm

k=1

Pα,k (t)(v, ϕk ) for any v ∈ L2 (Ω).

Fractional Rayleigh-Stokes Equations

165

Before giving the main results, we propose to use the energy estimate method to show the priori estimate of um in the following. Lemma 3.23. Let 0 < α < 21 , and assume h ∈ H01 (Ω), f ∈ L2 (0, b; H −1 (Ω)). Then there exists a constant C > 0 depending only on α, γ and b such that kum kL∞ (0,b;L2 (Ω)) + kum kL2 (0,b;H01 (Ω)) + k∂t um kL2 (0,b;H −1 (Ω)) + k∂tα um kL2 (0,b;H01 (Ω))   ≤C khkL2 (Ω) + khkH01 (Ω) + kf kL2 (0,b;H −1 (Ω)) .

(3.45)

Proof. Multiplying both sides of the first equation in (3.41) by dkm (t) and summing it from k = 1 to m, we obtain the following equation (∂t um (t), um (t)) + ((1 + γ∂tα )∇um (t), ∇um (t)) = (f (t), um (t)). From Lemma 3.18 and the relation (∂t v(t), v(t)) = the left side of (3.46) as follows

1 d 2 dt (v(t), v(t)),

(3.46)

we can estimate

(∂t um (t), um (t)) + ((1 + γ∂tα )∇um (t), ∇um (t)) 1 d γ ≥ kum (t)k2L2 (Ω) + 0 Dtα k∇um (t)k2L2 (Ω) + k∇um (t)k2L2 (Ω) . 2 dt 2 Substituting the above result into (3.46), and then, applying the Young’s inequality and the inequality |hg, vi| ≤ kgkH −1 (Ω) kvkH01 (Ω) ,

for g ∈ H −1 (Ω), v ∈ H01 (Ω),

(3.47)

we get that 1 d γ kum (t)k2L2 (Ω) + 0 Dtα k∇um (t)k2L2 (Ω) + k∇um (t)k2L2 (Ω) 2 dt 2 1 1 ≤|(f (t), um (t))| ≤ kf (t)kH −1 (Ω) kum (t)kH01 (Ω) ≤ kf (t)k2H −1 (Ω) + kum (t)k2H 1 (Ω) , 0 2 2 which implies d kum (t)k2L2 (Ω) + γ 0 Dtα k∇um (t)k2L2 (Ω) + k∇um (t)k2L2 (Ω) ≤ kf (t)k2H −1 (Ω) . dt Integrating both sides of the above inequality from 0 to t, from kum (0)kL2 (Ω) = Pm 1 ( k=1 |(h, ϕk )|2 ) 2 ≤ khkL2 (Ω) , we obtain Z t Z t γ −α 2 2 (t − s) k∇um (s)kL2 (Ω) ds + k∇um (s)k2L2 (Ω) ds kum (t)kL2 (Ω) + Γ(1 − α) 0 0 Z t ≤ kf (s)k2H −1 (Ω) ds + khk2L2 (Ω) , 0

and then, the fact that (t − s)−α ≥ t−α ≥ b−α for 0 ≤ s < t ≤ b implies  Z t γb−α kum (t)k2L2 (Ω) + +1 k∇um (s)k2L2 (Ω) ds Γ(1 − α) 0 Z t ≤ kf (s)k2H −1 (Ω) ds + khk2L2 (Ω) . 0

166

Fractional Partial Differential Equations

Taking the maximum for t ∈ [0, b] of both sides of the above inequality yields that   γb−α 2 kum kL∞ (0,b;L2 (Ω)) + + 1 kum k2L2 (0,b;H 1 (Ω)) 0 Γ(1 − α) (3.48) 2 2 ≤kf kL2 (0,b;H −1 (Ω)) + khkL2 (Ω) . Thus we obtain um ∈ L∞ (0, b; L2 (Ω)) ∩ L2 (0, b; H01 (Ω)). In the following, we will estimate that ∂t um ∈ L2 (0, b; H −1 (Ω)). At the beginning, we multiply both sides of the first equation of (3.43) by −1 k 0 λk dm (t), it follows that 0

0

0

0

−1 k k 2 k α k k k k λ−1 k (dm (t)) + γdm (t)0 Dt dm (t) + dm (t)dm (t) = λk f (t)dm (t), P∞ −1 k and then, let g(t, x) = k=1 λk f (t)ϕk (x) and sum the above equation from k = 1 to m, we have k∂t um (t)k2H −1 (Ω) + γ(∂tα um (t), ∂t um (t)) + (um (t), ∂t um (t)) = (g(t), ∂t um (t)). α−1

More precisely, denote ωα (t) = tΓ(α) , from the facts that ∂tα um (t) = α−1 d ∂t um (t) +ω1−α (t)um (0), (∂t um (t), um (t)) = 12 dt (um (t), um (t)), and on ac0 Dt count of the Young’s inequality and (3.47), we get that 1 d k∂t um (t)k2H −1 (Ω) + γ(0 Dtα−1 ∂t um (t), ∂t um (t)) + (um (t), um (t)) 2 dt =(g(t), ∂t um (t)) − γ(ω1−α (t)um (0), ∂t um (t)) ≤kg(t)kH01 (Ω) k∂t um (t)kH −1 (Ω) + γkω1−α (t)um (0)kH01 (Ω) k∂t um (t)kH −1 (Ω) =kf (t)kH −1 (Ω) k∂t um (t)kH −1 (Ω) + γkω1−α (t)um (0)kH01 (Ω) k∂t um (t)kH −1 (Ω) 1 2 ≤kf (t)k2H −1 (Ω) + γ 2 ω1−α (t)kum (0)k2H 1 (Ω) + k∂t um (t)k2H −1 (Ω) , 0 2 that is, d k∂t um (t)k2H −1 (Ω) + 2γ(0 Dtα−1 ∂t um (t), ∂t um (t)) + (um (t), um (t)) dt 2 ≤2kf (t)k2H −1 (Ω) + 2γ 2 ω1−α (t)kum (0)k2H 1 (Ω) . 0 Integrating the above inequality from 0 to t, then we conclude from Lemma 3.19 that Z t

k∂s um (s)k2H −1 (Ω) ds  Z t α−1 1−α + 2γ cos π k0 Ds 2 ∂s um (s)k2L2 (Ω) ds + kum (t)k2L2 (Ω) 2 0 Z t 2γ 2 t1−2α ≤2 kf (s)k2H −1 (Ω) ds + khk2H 1 (Ω) + khk2L2 (Ω) . 2 (1 − α) 0 (1 − 2α)Γ 0 Taking the maximum of both sides of the above inequality for t ∈ [0, b], we get   1+α 1−α k∂t um k2L2 (0,b;H −1 (Ω)) + 2γ cos π kC∂t 2 um k2L2 (0,b;L2 (Ω)) 2 0

+ kum k2L∞ (0,b;L2 (Ω))

(3.49) 2 1−2α

≤2kf k2L2 (0,b;H −1 (Ω)) +

2γ b khk2H 1 (Ω) + khk2L2 (Ω) . 0 (1 − 2α)Γ2 (1 − α)

Fractional Rayleigh-Stokes Equations

167

Thus we conclude that ∂t um ∈ L2 (0, b; H −1 (Ω)). In the sequel, we shall prove that ∂tα um ∈ L2 (0, b; H01 (Ω)). We multiply both sides of the first equation in (3.41) by 0 Dtα dkm (t) and sum it from k = 1 to m, that is, γ(∂tα ∇um (t), ∂tα ∇um (t)) + (∇um (t), ∂tα ∇um (t)) =(f (t), ∂tα um (t)) − (∂t um (t), ∂tα um (t)), and then, on account of (3.47), the Young’s inequality and Lemma 3.18, we have 1 α 2 0 D k∇um (t)kL2 (Ω) 2 t ≤kf (t)kH −1 (Ω) k∂tα um (t)kH01 (Ω) + k∂t um (t)kH −1 (Ω) k∂tα um (t)kH01 (Ω) 1 1 γ ≤ kf (t)k2H −1 (Ω) + k∂t um (t)k2H −1 (Ω) + k∂tα um (t)k2H 1 (Ω) , 0 γ γ 2 γk∂tα ∇um (t)k2L2 (Ω) +

which shows that γk∂tα ∇um (t)k2L2 (Ω) + 0 Dtα k∇um (t)k2L2 (Ω) ≤

2 2 kf (t)k2H −1 (Ω) + k∂t um (t)k2H −1 (Ω) . γ γ

Integrating the above inequality from 0 to t, we get that Z t Z t 1 (t − s)−α k∇um (s)k2L2 (Ω) ds γ k∂sα ∇um (s)k2L2 (Ω) ds + Γ(1 − α) 0 0 Z Z t 2 2 t kf (s)k2H −1 (Ω) ds + k∂s um (s)k2H −1 (Ω) ds. ≤ γ 0 γ 0 Taking the maximum for t ∈ [0, b] of both sides of the above inequality and using (t − s)−α ≥ t−α ≥ b−α for 0 ≤ s < t ≤ b, we conclude from (3.49) that γk∂tα um k2L2 (0,b;H 1 (Ω)) + 0

b−α kum k2L2 (0,b;H 1 (Ω)) 0 Γ(1 − α)

2 2 ≤ kf k2L2 (0,b;H −1 (Ω)) + k∂t um k2L2 (0,b;H −1 (Ω)) γ γ 6 4γb1−2α 2 ≤ kf k2L2 (0,b;H −1 (Ω)) + khk2H 1 (Ω) + khk2L2 (Ω) , 2 0 γ (1 − 2α)Γ (1 − α) γ which implies that ∂tα um ∈ L2 (0, b; H01 (Ω)). Therefore, there exists a positive constant C such that (3.45) holds. This completes the proof. Theorem 3.9. Assume h ∈ H01 (Ω) and f ∈ L2 (0, b; H −1 (Ω)). Then for 0 < α < 12 , the problem (3.37) exists at least one weak solution u ∈ L2 (0, b; H01 (Ω)) ∩L∞ (0, b; L2 (Ω)) with ∂tα u ∈ L2 (0, b; H01 (Ω)) and ∂t u ∈ L2 (0, b; H −1 (Ω)). Proof. From Lemma 3.23, we know that the sequence {um }∞ m=1 is bounded in 2 1 ∞ 2 ∞ 2 L (0, b; H0 (Ω)) ∩L (0, b; L (Ω)), {∂t um }m=1 is bounded in L (0, b; H −1 (Ω)), and 2 1 {∂tα um }∞ m=1 is bounded in L (0, b; H0 (Ω)), and then, from the weak compactness ∞ theorem in Hilbert space, there exist a subsequence {um ⊂ {um }∞ ˆ }m=1 m=1 and a ˆ

168

Fractional Partial Differential Equations

function u ∈ L2 (0, b; H01 (Ω)) ∩ L∞ (0, b; L2 (Ω)) with ∂t u ∈ L2 (0, b; H −1 (Ω)), ∂tα u ∈ L2 (0, b; H01 (Ω)) such that 2 1 ∗ ∞ 2 um ˆ * u in L (0, b; H0 (Ω)), um ˆ * u in L (0, b; L (Ω)),

(3.50)

2 −1 ∂t um (Ω)) ˆ * ∂t u in L (0, b; H

(3.51)

α 2 1 ∂tα um ˆ * ∂t u in L (0, b; H0 (Ω)).

(3.52)

and

We choose a function v ∈ C 1 ([0, b]; H01 (Ω)) with v(b) = 0, and substitute ϕk in the first equation of (3.41) by v, which implies (∂t um (t), v(t)) + ((1 + γ∂tα )∇um (t), ∇v(t)) = (f (t), v(t)). Integrating the above equation with respect to t from t = 0 to b, we have Z b Z b Z b (∂t um (t), v(t))dt + ((1 + γ∂tα )∇um (t), ∇v(t))dt = (f (t), v(t))dt. (3.53) 0

0

0

Choose m = m ˆ and on account of (3.50)-(3.52), in the limiting case m ˆ → ∞ we get the following equation Z b Z b Z b (∂t u(t), v(t))dt + ((1 + γ∂tα )∇u(t), ∇v(t))dt = (f (t), v(t))dt. (3.54) 0

0

0 1

([0, b]; H01 (Ω)) 2

And then, from the fact that C is dense in L2 (0, b; H01 (Ω)), we conclude that (3.54) holds for all functions in L (0, b; H01 (Ω)). Particularly, for each v ∈ H01 (Ω), the following equation holds: (∂t u(t), v) + ((1 + γ∂tα )∇u(t), ∇v) = (f (t), v),

a.e. 0 ≤ t ≤ b.

(3.55)

It remains to prove that u(0) = h. From the facts that u ∈ L2 (0, b; H01 (Ω)) and ∂t u ∈ L2 (0, b; H −1 (Ω)), we have u ∈ C([0, b]; L2 (Ω)), which implies that the function u(t) is well-defined at t = 0. Integrating (3.53) by parts gives Z b Z b −(um (t), ∂t v(t))dt + ((1 + γ∂tα )∇um (t), ∇v(t))dt 0

Z =

0 b

(f (t), v(t))dt + (um (0), v(0)). 0

Let m = m, ˆ from (3.50)-(3.52), in the case that m ˆ → ∞, we conclude from the fact 2 um ˆ (0) → h in L (Ω) that Z b Z b Z b −(u(t), ∂t v(t))dt + ((1 + γ∂tα )∇u(t), ∇v(t))dt = (f (t), v(t))dt + (h, v(0)). 0

0

0

(3.56)

Fractional Rayleigh-Stokes Equations

169

Similarly, (3.54) gives that Z b Z b ((1 + γ∂tα )∇u(t), ∇v(t))dt −(u(t), ∂t v(t))dt + 0

0

Z =

b

(3.57)

(f (t), v(t))dt + (u(0), v(0)). 0

Comparing (3.56) with (3.57), we can easy to get that (u(0), v(0)) = (h, v(0)), for the reason that v(0) is arbitrarily selected, we conclude u(0) = h. Therefore, u satisfies Definition 3.8, which implies that u is a weak solution of the problem (3.37). The proof is completed. In the following, we prove the uniqueness of weak solutions of the problem (3.37). Theorem 3.10. Assume the conditions in Theorem 3.9 hold. Then the weak solutions of the problem (3.37) is unique. Proof. Suppose that u1 , u2 are two weak solutions of (3.37). If we denote by u = u1 − u2 , then u satisfies ( (∂t u, v) + ((1 + γ∂tα )∇u, ∇v) = 0, (3.58) u(0) = 0. Thus the purpose of getting the uniqueness of weak solutions of the problem (3.37) is converted to check that u ≡ 0. We choose v = u in (3.58), that is, (∂t u(t), u(t)) + (∇u(t), ∇u(t)) + γ(∂tα ∇u(t), ∇u(t)) = 0. d According to the fact 2(∂t u(t), u(t)) = dt (u(t), u(t)) and by virtue of Lemma 3.18, we get 1 d (u(t), u(t)) + (∇u(t), ∇u(t)) = − γ(∂tα ∇u(t), ∇u(t)) 2 dt γ ≤ − 0 Dtα (∇u(t), ∇u(t)), 2 which implies that d (u(t), u(t)) ≤ −γ 0 Dtα (∇u(t), ∇u(t)). dt Integrating both sides of the above inequality from 0 to t, we have (u(t), u(t)) ≤ −γ 0 Dtα−1 (∇u(t), ∇u(t)). It is easy to see that the right-hand side of the above inequality is less than 0, that is, (u(t), u(t)) ≤ 0, which implies that u ≡ 0. Therefore, the uniqueness of weak solutions of the problem (3.37) is concluded. The proof is completed.

170

Fractional Partial Differential Equations

In the following, under the improved regularity conditions, we shall discuss the spacial regularity of weak solutions of problem (3.37). Theorem 3.11. Assume the conditions in Theorem 3.9 hold, and u is the weak solution of the problem (3.37). If h ∈ H 2 (Ω) and f ∈ L2 (0, b; L2 (Ω)), then we have T u ∈ L∞ (0, b; H01 (Ω)) L2 (0, b; H 2 (Ω)), ∂tα u ∈ L2 (0, b; H 2 (Ω)), ∂t u ∈ L2 (0, b; L2 (Ω)), and the following estimation holds kukL∞ (0,b;H01 (Ω)) + kukL2 (0,b;H 2 (Ω)) + k∂t ukL2 (0,b;L2 (Ω)) + k∂tα ukL2 (0,b;H 2 (Ω)) ˜ ≤C(kf kL2 (0,b;L2 (Ω)) + khkH 2 (Ω) ),

(3.59)

where C˜ is a positive constant depending only on γ, b and α. Proof. From the fact that u is the weak solution of the problem (3.37), we know that um satisfies (3.41) under the form (3.40). Now we estimate that u ∈ L2 (0, b; H 2 (Ω)) ∩ L∞ (0, b; H01 (Ω)). Multiplying the first equation in (3.41) by λk dkm (t) and summing it from k = 1 to m, we have (∂t um (t), ∆um (t)) + (∇um (t), ∇∆um (t)) + γ(∂tα ∇um (t), ∇∆um (t)) =(f (t), ∆um (t)), which implies (∂t ∇um (t), ∇um (t)) + (∆um (t), ∆um (t)) + γ(∂tα ∆um (t), ∆um (t)) = − (f (t), ∆um (t)). From Lemma 3.18 and the fact (∂t v(t), v(t)) = the left-hand side of (3.60) as following

1 d 2 dt (v(t), v(t)),

(3.60)

we can estimate

(∂t ∇um (t), ∇um (t)) + (∆um (t), ∆um (t)) + γ(∂tα ∆um (t), ∆um (t)) γ 1 d k∇um (t)k2L2 (Ω) + k∆um (t)k2L2 (Ω) + 0 Dtα k∆um (t)k2L2 (Ω) . ≥ 2 dt 2 Applying the H¨ older’s inequality and the Young’s inequality to the right-hand side of (3.60), we can get that | − (f (t), ∆um (t))| ≤kf (t)kL2 (Ω) k∆um (t)kL2 (Ω) 1 1 ≤ kf (t)k2L2 (Ω) + k∆um (t)k2L2 (Ω) . 2 2 Indeed, the above discussion and (3.60) imply that d k∇um (t)k2L2 (Ω) + k∆um (t)k2L2 (Ω) + γ 0 Dtα k∆um (t)k2L2 (Ω) ≤ kf (t)k2L2 (Ω) , dt

Fractional Rayleigh-Stokes Equations

171

and then, integrating the above inequality with respect to the time variable from 0 to t, we obtain that Z t Z t γ k∇um (t)k2L2 (Ω) + k∆um (s)k2L2 (Ω) ds + (t − s)−α k∆um (s)k2L2 (Ω) ds Γ(1 − α) 0 0 Z t kf (s)k2L2 (Ω) ds + k∇hk2L2 (Ω) . ≤ 0

Taking the maximum of t ∈ [0, b] and applying the fact that (t − s)−α ≥ t−α ≥ b−α for 0 ≤ s < t ≤ b, we have   γb−α 2 k∆um k2L2 (0,b;L2 (Ω)) k∇um kL∞ (0,b;L2 (Ω)) + 1 + Γ(1 − α) ≤kf k2L2 (0,b;L2 (Ω)) + khk2H 1 (Ω) . 0

Taking m = m ˆ in the above inequality, and passing m ˆ → ∞, we can easy to get the same bounds of u and the conclusion T u ∈ L∞ (0, b; H01 (Ω)) L2 (0, b; H 2 (Ω)). We estimate that ∂t u ∈ L2 (0, b; L2 (Ω)) in the sequel. Multiplying both sides of the first equation of (3.41) by (dkm )0 and summing it from k = 1 to m show that (∂t um (t), ∂t um (t)) + γ(∂tα ∇um (t), ∇∂t um (t)) + (∇um (t), ∇∂t um (t)) =(f (t), ∂t um (t)). In view of the H¨ older’s inequality, the Young’s inequality and the fact that d 2(∂t v(t), v(t)) = dt (v(t), v(t)), we have 1 d k∇um (t)k2L2 (Ω) 2 dt ≤kf (t)kL2 (Ω) k∂t um (t)kL2 (Ω) − γ(ω1−α (t)∇um (0), ∇∂t um (t)) 1 ≤kf (t)k2L2 (Ω) + k∂t um (t)k2L2 (Ω) + γω1−α (t)k∆um (0)kL2 (Ω) k∂t um (t)kL2 (Ω) 4 1 2 (t)k∆um (0)k2L2 (Ω) , ≤kf (t)k2L2 (Ω) + k∂t um (t)k2L2 (Ω) + γ 2 ω1−α 2 which implies k∂t um (t)k2L2 (Ω) + γ(0 Dtα−1 ∂t ∇um (t), ∇∂t um (t)) +

k∂t um (t)k2L2 (Ω) + 2γ(0 Dtα−1 ∂t ∇um (t), ∇∂t um (t)) +

d k∇um (t)k2L2 (Ω) dt

(3.61)

2 ≤2kf (t)k2L2 (Ω) + 2γ 2 ω1−α (t)k∆um (0)k2L2 (Ω) .

Integrating (3.61) for both sides from 0 to t, we conclude from Lemma 3.19 that   Z t α−1 1−α k∂s um (s)k2L2 (Ω) ds + 2γ cos π k0 Dt 2 ∂t ∇um (t)k2L2 (Ω) 2 0 + k∇um (t)k2L2 (Ω)

172

Fractional Partial Differential Equations t

Z

kf (s)k2L2 (Ω) ds +

≤2 0

2γ 2 t1−2α k∆um (0)k2L2 (Ω) + k∇um (0)k2L2 (Ω) . (1 − 2α)Γ2 (1 − α)

Then taking the maximum of t ∈ [0, b] of the above inequality implies that   1+α 1−α 2 π kC∂t 2 ∇um k2L2 (0,b;L2 (Ω)) k∂t um kL2 (0,b;L2 (Ω)) + 2γ cos 2 + k∇um k2L∞ (0,b;L2 (Ω)) 2γ 2 b1−2α k∆um (0)k2L2 (Ω) + k∇um (0)k2L2 (Ω) (1 − 2α)Γ2 (1 − α) 2γ 2 b1−2α khk2H 2 (Ω) + khk2H 1 (Ω) . ≤2kf k2L2 (0,b;L2 (Ω)) + 0 (1 − 2α)Γ2 (1 − α)

≤2kf k2L2 (0,b;L2 (Ω)) +

From the above discussion, choose m = m ˆ and let m ˆ → ∞, we conclude that 2 2 ∂t u ∈ L (0, b; L (Ω)). In the following, we estimate ∂tα u ∈ L2 (0, b; H 2 (Ω)). Multiplying the first equation in (3.41) by λk 0 Dtα dkm (t) and summing it from k = 1 to m, we have γ(∂tα ∆um (t), ∂tα ∆um (t)) + (∆um (t), ∂tα ∆um (t)) =(∂t um (t), ∂tα ∆um (t)) − (f (t), ∂tα ∆um (t)). By virtue of Lemma 3.18, the H¨ older’s inequality and the Young’s inequality, we have 1 α 2 0 D k∆um (t)kL2 (Ω) 2 t ≤k∂t um (t)kL2 (Ω) k∂tα ∆um (t)kL2 (Ω) + kf (t)kL2 (Ω) k∂tα ∆um (t)kL2 (Ω) 1 1 γ ≤ k∂t um (t)k2L2 (Ω) + kf (t)k2L2 (Ω) + k∂tα ∆um (t)k2L2 (Ω) , γ γ 2 γk∂tα ∆um (t)k2L2 (Ω) +

that is, γk∂tα ∆um (t)k2L2 (Ω) + 0 Dtα k∆um (t)k2L2 (Ω) ≤

2 2 k∂t um (t)k2L2 (Ω) + kf (t)k2L2 (Ω) . γ γ

Integrating both sides of the above inequality for s ∈ [0, t], and taking the maximum of t ∈ [0, b], we have Z b 1 α 2 γk∂t ∆um kL2 (0,b;L2 (Ω)) + (b − s)−α k∆um (s)k2L2 (Ω) ds Γ(1 − α) 0 2 2 ≤ k∂t um k2L2 (0,b;L2 (Ω)) + kf k2L2 (0,b;L2 (Ω)) , γ γ let m = m, ˆ in the limiting case m ˆ → ∞, we deduce that ∂tα u ∈ L2 (0, b; H 2 (Ω)). Combined with the above discussion, we conclude that (3.59) holds. This completes the proof of the theorem.

Fractional Rayleigh-Stokes Equations

3.5 3.5.1

173

Global Solutions in Besov-Morrey Spaces Introduction

The two of the most important mathematical and physical models among nonNewtonian fluids are the Stokes’s first problem for a flat plate as well as the Rayleigh-Stokes problem for an edge. We refer to [122–124, 275, 314] for details about velocity and temperature of the Rayleigh-Stokes problems. In this section, we study the nonlinear Rayleigh-Stokes equations which involve the Riemann-Liouville fractional derivative operator (in time) as follows: ( ∂t u − (1 + γ0 ∂tα )∆u = f (u), t > 0, x ∈ RN , (3.62) u(0, x) = u0 (x), x ∈ RN , where γ0 > 0 is a constant, ∆ is the Laplacian operator, ∂tα is the Riemann-Liouville fractional derivative of order α ∈ (0, 1) which is defined by Z t 1 ∂ α ∂t u(t, x) = (t − s)−α u(s, x)ds, t > 0, x ∈ RN , Γ(1 − α) ∂t 0 provided the right-handed side of the equality is pointwise defined. In order to gain insights into the behavior of solutions of the model (3.62), there has been substantial interest in a form solution and its existence (see e.g., [39, 40, 248,280,335,338] and references therein). For instance, some recent results on local existence in time in a bounded domain Ω ⊂ RN have been obtained in [210, 409], whose solutions obtained in these studies involve infinite series and special functions, and thus are inconvenient for deriving the properties of solutions in unbounded domains. To the best of the authors’ knowledge, there are limited articles treating with solvability of fractional Rayleigh-Stokes equations in the whole space RN . We would like to mention that He et al. [163] studied the qualitative properties relied on the Gagliardo-Nirenberg’s inequalities, like global/local well-posedness, blow-up results and integrability of mild solutions for small initial value in Lq . Very recently, our proceeding paper [297] obtained two global existence results in the framework N/p−2/(ρ(1−α)) . However, it remains to of Besov type with the initial value u0 ∈ B˙ p,∞ take more efforts to find a solution space as large as possible. Theoretical studies on well-posedness analyses for differential equations involving fractional derivatives have received considerable attention in the last decade, see [155, 298, 322, 360, 383]. Recent results on global solvability of fractional diffusion cases [64, 284, 334] tell us that solutions consisting of C0 -semigroup and the Wright-type function is a key point in deriving the properties of solutions operators in different spaces such as Lq , Besov spaces and Besov-Morrey spaces. Though the solution operator of the considered equation (3.62) can be established in the form: Z ∞ S(t) = φ(t, τ )T (τ )dτ, t > 0, 0

where T (t) is a C0 -semigroup and φ(t, τ ) is Ra probability density, it is a difficulty for ∞ discriminating convergence of the integral 0 φ(t, τ )τ −% dτ for % ∈ (0, 1) and t > 0.

174

Fractional Partial Differential Equations

Therefore, it seems to be infeasible for the equation (3.62) along the way as we studied fractional diffusion case based on C0 -semigroup and thus it is imperative to develop several suitable approaches deriving the properties of its solution operator. 2p 2 For ρ > 0, λ = N − ρ(1−α) and µ = N − ρ(1−α) one has the continuous inclusions Lq provided N r

N q

2 ρ(1−α)

=

N −λ p

Mp,λ

= −s0 +

s0 k0 s0 Np,µ,∞ and B˙ r,∞ ⊂ Np,µ,∞

N −µ p

= −k0 +

N r ,

where s0 =

N −µ p

−

2 ρ(1−α) ,

k0 =

ρN (1−α) . 2

− and 1 ≤ p < r ≤ q ≤ Moreover, there are no results on the global solvability of fractional Rayleigh-Stokes equations in a class such that s0 Z ) Np,µ,∞ . In this sense, we provide a long time existence of solutions for the equation (3.62). The aims of this section is to establish the global well-posedness and asymptotic behavior of solutions for the equation (3.62) in the framework of Besov-Morrey spaces. The main features of our work are three aspects. The first is that our results are improved compared with the existence results obtained in [297]. Furthermore, the approach is also different which depends on real interpolation, a multiplier and some inequalities such as the Gagliardo-Nirenberg’s inequalities and Sobolev-type embedding. The last but not the least is the selection of the initial value that allows for a larger range than the previous works. The rest of the section is organized as follows. In the next subsection, we introduce some notations, function spaces that will be used throughout the section. In Subsection 3.5.3, the several estimates of resolvent operators and solution operators in Besov-Morrey spaces are presented, respectively. In Subsection 3.5.4, the existence, uniqueness and continuous dependence on initial value of global solutions for nonlinear fractional Rayleigh-Stokes equations are investigated in Besov-Morrey spaces. Further, we also establish the asymptotic behavior of solutions at infinity. 3.5.2

Notations and Function Spaces

Here, we recall definitions of function spaces which are main tools for estimates in the following sections. Throughout this section we use the following notations. S and S 0 denote the Schwartz spaces of rapidly decreasing smooth functions and tempered distributions, respectively. Let (X, | · |) be a Banach space. We denote by Cb (I, X) the space of bounded continuous operators from I ⊂ R+ to X, equipped with the norm supt∈I |·| and C∗ (I, X) ={v ∈ C(I, X) : kvkC∗ (I,X) = sup t

1−α ϑ

|v(t)| < ∞}

t∈I

for ϑ > 1. Now we recall the definition of Sobolev-Morrey spaces. Let Uε (x0 ) be the open ball in RN centered at x0 and with radius ε > 0 and let 1 ≤ p < ∞ and 0 ≤ µ < N , the Morrey spaces Mp,µ = Mp,µ (RN ) is defined to be the set of functions v ∈ Lp (Uε (x0 )) such that µ

kvkMp,µ = sup sup ε− p kvkLp (Uε (x0 )) < ∞. x0 ∈RN ε>0

Fractional Rayleigh-Stokes Equations

175

For s ∈ R and 1 ≤ p < ∞, the homogeneous Sobolev-Morrey space Msp,µ = s s (−∆)− 2 Mp,µ with norm kvkMsp,µ = k(−∆) 2 vkMp,µ . Especially, we have Mp,0 = Lp and Msp,0 = H˙ ps is the well known Sobolev space for p > 1. Moreover, l (−∆) 2 Msp,µ = Ms−l p,µ for 0 ≤ l ≤ s. Let {ϕj }j∈Z be the Littlewood-Paley decomposition. We take a function φ ∈ P C0∞ (RN ) with supp(φ) = {ξ ∈ RN : 21 ≤ |ξ| ≤ 2} such that j∈Z φ(2−j ξ) = 1 for all ξ 6= 0. The functions ϕj and ψj are defined by ∞ X cj (ξ) = 1 − ϕ cj (ξ) = φ(2−j ξ), ψ φ(2−j ξ). j=1

For s ∈ R, 1 ≤ p < ∞, 1 ≤ r ≤ ∞ and 0 ≤ µ < N , the homogeneous Besov-Morrey s space Np,µ,r is defined by s s Np,µ,r = {v ∈ S 0 /P : kvkNp,µ,r < ∞}, 0 where the set S /P consists in equivalence classes in S 0 modulo polynomials, and X 1/r   (2sj kϕj ∗ vkMp,µ )r , 1 ≤ r < ∞,  j∈Z s kvkNp,µ,r =    sup(2sj kϕj ∗ vkMp,µ ), r = ∞. j∈Z

s is the homogeneous Besov space when µ = 0. In particular, = B˙ p,r Let 0 ≤ µ < N and 1 ≤ p < ∞. We recall inclusion relations between Morrey spaces and Besov-Morrey spaces (see [207, 328]) 0 0 Np,µ,1 ⊂ Mp,µ ⊂ Np,µ,∞ . (3.63) N −µ = s − , we obtain Let pj , sj , j = 1, 2 such that p2 ≤ p1 , s1 ≤ s2 and s2 − Np−µ 1 p1 2 the Sobolev-type embedding as follows (3.64) Msp22 ,µ ⊂ Msp11 ,µ and Nps22,µ,∞ ⊂ Nps11,µ,∞ . For 1 < p, q < ∞, we define the space based on Besov-Morrey type spaces as s0 ) ∩ C∗ ((0, ∞); Mq,µ ), Zp,q = Cb ((0, ∞); Np,µ,∞ which is a Banach space endowed with the norm 1−α s0 kvkZp,q = sup kv(t)kNp,µ,∞ + sup t ϑ kv(t)kMq,µ . s Np,0,r

t∈(0,∞)

t∈(0,∞)

Here ϑ > 1 and s0 are given by 1 1 N −µ N −µ 2 = − and s0 = − . ϑ ρ(1 − α) 2q p ρ(1 − α) We present an estimate for a multiplier operator on Msp,µ (see [207]). Lemma 3.24. Let m, s ∈ R and 0 ≤ µ < N and P (ξ) ∈ C [N/2]+1 (RN \ {0}). Assume that there is C > 0 such that ∂k k P (ξ) ≤ C|ξ|m−|k| , ∂ξ for all k ∈ (N0 )N with |k| ≤ [N/2] + 1 and for all ξ 6= 0. Then the multiplier operator P (D)v = F −1 [P (ξ)b v ] on S 0 /P satisfies the estimate kP (D)vkMs−m ≤ M CkvkMsp,µ , p,µ for 1 < p < ∞, where M > 0 is a constant independent of v.

176

Fractional Partial Differential Equations

We finish this subsection recalling an elementary fixed point lemma whose proof can be found in [121]. Lemma 3.25. Let (X, | · |) be a Banach space and 1 < ρ < ∞. Suppose that W : X → X satisfies W (0) = 0 and |W (u) − W (v)| ≤ L|u − v|(|u|ρ + |v|ρ ). Let R > 0 be the unique positive root of 2ρ+1 LRρ − 1 = 0. Given 0 < ε < R and ϕ ∈ X such that |ϕ| ≤ ε, there exists a solution u ∈ X for the equation u = ϕ+W (u) which is the unique one in the closed ball {v ∈ X : |v| ≤ 2ε}. Moreover, if |ϕ| ¯ ≤ε and |¯ u| ≤ 2ε satisfies the equation u ¯ = ϕ¯ + W (¯ u) then 1 |u − u ¯| ≤ (|ϕ − ϕ|). ¯ ρ+1 1 − 2 Lερ 3.5.3

Properties of Solution Operators

The goal of this subsection is to derive several estimates of solution operators for the fractional Rayleigh-Stokes equations on Sobolev-Morrey, Besov-Morrey and Besov spaces, respectively. Let 1 < p < ∞. We consider the operator A in Lp (RN ) defined by A = −∆ with the domain D(A) = {u ∈ Lp (RN ) : ∆u ∈ Lp (RN )}. It follows from [63, Theorem 2.3.2] that −A generates a bounded analytic semigroup T (t) and the spectrum is equal to (−∞, 0). Let Σπ−σ = {z ∈ C : z 6= 0, |argz| < cz (ξ) := 1 2 . Then we have that for π − σ} for σ ∈ (0, π/2). Define Gz (x) by G z+|ξ| p N v ∈ L (R ), Z N cz (ξ)b (z + A)−1 v = (4π)− 2 eix·ξ G v (ξ)dξ, for x ∈ RN and z ∈ Σπ−σ . RN s We specify several estimates of the operator (z + A)−1 on Msp,µ and Np,µ,r , which are playing an important role in proving properties of the solution operators. Here and below the letter M1 will denote constants which can change from line to line.

Lemma 3.26. Let s, β ∈ R with s ≤ β, 1 < p ≤ q < ∞ and 0 ≤ µ < N . Assume z ∈ Σπ−σ . (i) Let β − s + ( N p−µ −

N −µ q )

≤ 2. There exists a constant M1 > 0 such that

k(z + A)−1 vkMβq,µ ≤ M1 |z|

β−s N −µ 1 N −µ 2 + 2 ( p − q )−1

kvkMsp,µ ,

(3.65)

Msp,µ .

for v ∈ (ii) Let 1 ≤ r ≤ ∞ and β − s + ( N p−µ − such that k(z + A)−1 vkNq,µ,r ≤ M1 |z| β 0

for v ∈ S /P.

N −µ q )

≤ 2. There exists a constant M1 > 0

β−s N −µ 1 N −µ 2 + 2 ( p − q )−1

s kvkNp,µ,r ,

(3.66)

Fractional Rayleigh-Stokes Equations

(iii) Let 1 ≤ r ≤ ∞ and s < β be such that β − s + ( N p−µ − a constant M1 > 0 such that k(z + A)−1 vkN β

q,µ,1

≤ M1 |z|

177 N −µ q )

β−s N −µ 1 N −µ 2 + 2 ( p − q )−1

< 2. There exists

s kvkNp,µ,r

for v ∈ S 0 /P. Proof. (i) Let 1 < q < ∞ and θ ∈ [0, 2]. Since ∂ k |ξ|θ ≤ C|ξ|−k , ξ 6= 0, ∂ξ k 1 + |ξ|2 θ

for k ∈ (N0 )N with |k| ≤ [N/2] + 1. Using Lemma 3.24, it follows that A 2 G1 v = |ξ|θ b(ξ)] satisfies the estimate F −1 [ 1+|ξ| 2v θ

kA 2 G1 vkMsq,µ ≤ M CkvkMsq,µ , where M may depend on q. θ \ |ξ|θ √ξ Let h(ξ, z) := A 2 Gz = z+|ξ| 2 , we have by changing variable ξ → η =

|z|

θ

h(ξ, |z|) = |z| 2 −1

that

θ ξ |η|θ = |z| 2 −1 h( p , 1). 2 1 + |η| |z|

We can calculate immediately to derive the following relation θ

v (ξ)](x) A 2 (z + A)−1 v =F −1 [h(ξ, |z|)ˆ p p θ N =|z| 2 + 2 −1 F −1 [h(η, 1)ˆ v ( |z|η)]( |z|x) p θ =|z| 2 −1 F −1 [h(·, 1)v\ |z|−1/2 (·)]( |z|x) θ

θ

=|z| 2 −1 [A 2 (1 + A)−1 v|z|1/2 ]|z|−1/2 (x), −N where we have used vc vb( λξ ) for vλ (x) := v(λx). λ (ξ) = λ

kv(λ·)kMsq,µ = λ

s− N −µ p

kv(·)kMsq,µ , it follows that 1

θ

θ

Recalling that

kA 2 (z + A)−1 vkMsq,µ =|z| 2 −1− 2 (s−

N −µ p )

1

θ

≤M1 |z| 2 −1− 2 (s− =M1 |z|

θ 2 −1

β

kA 2 (1 + A)−1 v|z|1/2 kMsq,µ

N −µ p )

kv|z|1/2 kMsq,µ

kvkMsq,µ ,

where M1 = M C. From the Sobolev-type embedding for l = for θ = β − s + l one can see that

N −µ N −µ p − q

β

k(z + A)−1 vkMβq,µ =kA 2 (z + A)−1 vkMq,µ β

≤kA 2 (z + A)−1 vkMlp,µ =|A

β−s l 2 +2

≤M1 |z| Thus (3.65) holds.

(3.67)

(z + A)−1 vkMsp,µ

β−s N −µ 1 N −µ 2 + 2 ( p − q )−1

kvkMsp,µ .

and (3.67)

178

Fractional Partial Differential Equations

s 1 2 (ii) By real interpolation, we see that Np,µ,r = (Msp,µ , Msp,µ )θ1 ,r for 0 ≤ s0 < s < s1 and 0 < θ1 < 1 with s = (1 − θ1 )s0 + θ1 s1 . This together with the interpolation theorem on operators shows that N

1

1

θ

1 s k(z + A)−1 vkNq,µ,r ≤m1−θ mθ11 |z| 2 ( p − q )+ 2 −1 kvkNp,µ,r , β 0

where mi = k(z +A)−1 kMsi →Mβi for i = 0, 1. Using the relation (3.65), we obtain p,µ p,µ the inequality (3.66). (iii) We choose ε > 0 so small that β + ε − s + 12 ( N p−µ − N −µ q ) ≤ 2. Thus we  . It ensures that θ2 ∈ (0, 1) and β = (β + ε)(1 − θ2 ) + sθ2 . can take θ2 = +β−s β β+ s = (Nq,µ,∞ , Nq,µ,∞ )θ2 ,1 , it follows In virtue of the real interpolation, we have Nq,µ,1 from (3.66) that

k(z + A)−1 vkNq,µ,∞ ≤M1 |z| β+

β+−s N −µ + 12 ( N −µ 2 p − q )−1

s ≤M1 |z| k(z + A)−1 vkNq,µ,∞

N −µ 1 N −µ 2 ( p − q )−1

s kvkNp,µ,∞ ,

s kvkNp,µ,∞ .

This yields that k(z + A)−1 vkN β

q,µ,1

≤M1 |z|

N −µ β−s 1 N −µ 2 + 2 ( p − q )−1

s kvkNp,µ,∞ .

The proof is completed. s s , the restriction β − s + ( N p−µ − N −µ = B˙ p,r Despite Msp,0 = H˙ ps and Np,0,∞ q ) 0 is a constant. Therefore, one can see that for v ∈ Lp (RN ), 1

N

1

γ

k(z + A)−1 vkH˙ qγ ≤ M1 |z| 2 ( p − q )+ 2 −1 kvkLp , for γ ∈ [0, 2] and 1 < p ≤ q < +∞. This also implies that N

1

1

k(z + A)−1 vkH˙ qβ ≤ M1 |z| 2 ( p − q )+

β−s 2 −1

kvkH˙ s , p

H˙ ps ,

for v ∈ γ = β − s ∈ [0, 2] and 1 < p ≤ q < +∞. (ii) Let N ≥ 1, s, β ∈ R with β − s ∈ [0, 2], 1 ≤ r ≤ +∞ and 1 < p ≤ q < +∞. s By real interpolation, we see that B˙ p,r = (H˙ ps0 , H˙ ps1 )ν1 ,r for 0 ≤ s0 < s < s1 and 0 < ν1 < 1 with s = (1 − ν1 )s0 + ν1 s1 . This together with the interpolation theorem on operators shows that N

1

1

θ

k(z + A)−1 vkB˙ q,r ≤M1 |z| 2 ( p − q )+ 2 −1 kvkB˙ s , β p,r

Fractional Rayleigh-Stokes Equations

179

for v ∈ B˙ ps , β − s ∈ [0, 2] and 1 < p ≤ q < +∞. s (iii) Let 1 < p < ∞ and β − s ∈ (0, 2). If v ∈ B˙ p,∞ , then there exists a constant M1 > 0 such that θ

k(z + A)−1 vkB˙ β ≤ M1 |z| 2 −1 kvkB˙ s

p,∞

p,1

.

For the proof of (iii), we can see [208]. For δ > 0 and σ ∈ (0, π/2) we introduce the contour Γδ,σ defined by Γδ,σ = {re−iσ : r ≥ δ} ∪ {δeiψ : |ψ| ≤ σ} ∪ {reiσ : r ≥ δ}, where the circular arc is oriented counterclockwise, and the two rays are oriented with an increasing imaginary part. In the sequel, we introduce an operator S(t) as Z 1 ezt H(z)dz, for t > 0, S(t) = (3.68) 2πi Γδ,π−σ where H(z) =

ω(z) (ω(z) + A)−1 , z

ω(z) =

z . 1 + γ0 z α

Note that if γ0 = 0, then we recover the classical heat equation ∂t u − ∆u = f (u), t > 0. Thus following the notation above we have ω(z) = z, H(z) = (z − ∆)−1 . From the Cauchy’s theorem, we have the identity Z 1 S(t)φ(x) = ezt (z − ∆)−1 dzφ(x) = e∆t φ(x) = gt ∗ φ(x), 2πi Γδ,π−σ where gt is the classical heat (Gaussian) kernel. From [163], we give a concept of mild solutions of the problem (3.62). Definition 3.9. Let 1 < p < ∞, 0 ≤ µ < N and s ∈ R. A continuous function s satisfies u : (0, ∞) → Np,µ,∞ Z t u(t) = S(t)u0 + S(t − τ )f (u(τ ))dτ, t > 0, 0 s . we say that u is a global mild solution to the problem (3.62) on Np,µ,∞

Our next lemma describes several estimates of the solution operator S(t)v for β t > 0 on Mβq,µ and Nq,µ,r . Lemma 3.27. Let s, β ∈ R with s ≤ β, 1 < p ≤ q < ∞, 0 ≤ µ < N be such that β − s + ( N p−µ − N −µ q ) < 2. (i) Then there exists a constant M1 > 0 such that kS(t)vkMβq,µ ≤ M1 t−(1−α) for t > 0 and v ∈ Msp,µ .

(β−s) N −µ N −µ − 1−α 2 2 ( p − q )

kvkMsp,µ ,

180

Fractional Partial Differential Equations

(ii) Let 1 ≤ r ≤ ∞. There exists a constant M1 > 0 such that kS(t)vkNq,µ,r ≤ M1 t−(1−α) β

(β−s) N −µ N −µ − 1−α 2 2 ( p − q )

s kvkNp,µ,r ,

for t > 0 and v ∈ S 0 /P. (iii) Let 1 ≤ r ≤ ∞ and s < β. There exists a constant M1 > 0 such that ≤ M1 t−(1−α)

kS(t)vkN β

q,µ,1

(β−s) N −µ N −µ − 1−α 2 2 ( p − q )

s kvkNp,µ,r ,

(3.69)

for t > 0 and v ∈ S 0 /P. Proof. (i) Let σ ∈ (0, π/2). From Lemma 2.1 in [39], we have that w(z) ∈ Σπ−σ and |w(z)| ≤ C0 |z|1−α for any z ∈ Σπ−σ . Then we deduce from (3.65) that k(ω(z) + A)−1 vkMβq,µ ≤ M1 |ω(z)| for β − s + ( N p−µ −

N −µ q )

N −µ β−s 1 N −µ 2 + 2 ( p − q )−1

kvkMsp,µ ,

< 2, this also shows

kH(z)vkMβq,µ ≤

N −µ β−s 1 N −µ M1 |ω(z)| 2 + 2 ( p − q ) kvkMsp,µ |z|

≤M1 C0 |z|(1−α)[

N −µ β−s 1 N −µ 2 + 2 ( p − q )]−1

Since 1 < p ≤ q < ∞ satisfy β − s + ( N p−µ −

N −µ q )

kvkMsp,µ .

< 2, it implies

(1−α) 2 [β

−s+

( N p−µ − N −µ q )] < 1 for α ∈ (0, 1). Let t > 0 and δ > 0. We choose δ = 1/t, it follows from the property of the Gamma function that Z 1 eRe(z)t kH(z)vkMβq,µ |dz| kH(z)vkMβq,µ ≤ 2π Γ1/t,π−σ  Z ∞ (1−α) N −µ N −µ M1 C0 ≤ kvkMsp,µ 2 τ 2 [β−s+( p − q )]−1 e−τ t cos(σ) dτ 2π −1 t Z π−θ  (1−α) N −µ N −µ + t− 2 [β−s+( p − q )] ecos(ψ) dψ −π+θ

≤M1 t−

(1−α) N −µ [β−s+( N −µ 2 p − q )]

kvkMsp,µ ,

for v ∈ Msp,µ , where M1 = M1 (C0 , N, p, q, σ, β, s, µ). (ii) By the same arguments as we derived the result of (i), we can obtain from (3.66) that Z 1 eRe(z)t kH(z)vkNq,µ,r |dz| kS(t)vkNq,µ,r ≤ β β 2π Γ1/t,π−σ ≤M1 t−

(1−α) N −µ [β−s+( N −µ 2 p − q )]

s kvkNp,µ,r .

(iii) Similar arguments follow the inequality (3.69). Therefore, we omit the proof.

Fractional Rayleigh-Stokes Equations

181

β Now we present the continuity of S(t)v for t > 0 in Nq,µ,r and Mβq,µ as follows.

Remark 3.9. Let β, s ∈ R, 1 < p ≤ q < ∞ and 0 ≤ µ < N be such that s ≤ β and β − s + ( N p−µ − N −µ q ) < 2. By differentiating S(t)v we have Z 1 0 ezt zH(z)vdz, for t > 0. S (t)v = 2πi Γδ,π−σ Using the similar arguments as we derived Lemma 3.27, we arrive at Z 1 0 kS (t)vkNq,µ,r ≤ eRe(z)t kzH(z)vkNq,µ,r |dz| β β 2π Γ1/t,π−σ ≤M1 t−

(1−α) N −µ [β−s+( N −µ 2 p − q )]−1

Therefore for any 0 < t1 < t2 < ∞ and β − s + ( N p−µ −

s kvkNp,µ,r .

N −µ q )

6= 0, we have

kS(t2 )v − S(t1 )vkNq,µ,r β Z t2 ≤ kS 0 (t)vkNq,µ,r dt β t1

Z s ≤M1 kvkNp,µ,r

t2

t−

(1−α) N −µ [β−s+( N −µ 2 p − q )]−1

dt

t1

 − (1−α) [β−s+( N −µ − N −µ )] (1−α) N −µ  [β−s+( N −µ − p q p − q )] f1 kvkN s − t2 2 , =M t1 2 p,µ,r f1 = where M

M1 (1−α) N −µ [β−s+( N −µ 2 p − q )]

. In fact, from the above arguments, we can also

β derive the continuity of S(t)v for t > 0 in Nq,µ,r when β − s + ( N p−µ − N −µ q ) = 0. β Therefore, S(·)v ∈ C((0, ∞); Nq,µ,r ). On the other hand, by the similar arguments as above, the continuity of the operator S(t)v in Mβq,µ also holds for t > 0.

In virtue of Remark 3.8, the restriction β − s + ( N p−µ − N −µ q ) < 2 that is indispensable for Lemma 3.2 in [297] can be weaken to the restriction (1 − α)(β − s + Np − Nq ) < 2 when we study the estimates about the operator S(t) for t > 0 on Besov spaces. Remark 3.10. Let 1 ≤ r ≤ +∞ and s, β ∈ R. It holds that + N (1−α) ( p1 − 1q ) < (i) Let β − s ∈ [0, 2] and 1 < p ≤ q < ∞ be such that (1−α)(β−s) 2 2 s 1. If v ∈ B˙ p,r (RN ), then there exists a constant M1 > 0 such that kS(t)vkB˙ q,r ≤ M1 t− β

N (1−α) 1 (1−α)(β−s) ( p − q1 )− 2 2

kvkB˙ s , p,r

for t > 0.

s (ii) Let 1 < p < ∞ and β −s ∈ (0, 2). If v ∈ B˙ p,∞ (RN ), then there exists a constant M1 > 0 such that

kS(t)vkB˙ β ≤ M1 t− p,1

(1−α)(β−s) 2

kvkB˙ s

p,∞

,

for t > 0.

182

Fractional Partial Differential Equations

3.5.4

Well-posedness of Global Solutions

In this subsection we shall study the well-posedness and asymptotic behavior of global solutions for the semi-linear fractional Rayleigh-Stokes equations. Let N ≥ 1 and 0 ≤ µ < N , we recall that a triplet (ϑ, q, q0 ) is admissible if 1 N 1 1 = ( − ), ϑ 2 q0 q where   N q0 , 1 < q0 ≤ q < N − 2  ∞,

N > 2, N ≤ 2.

We denote (ϑ, q, q0 ) as a generalized admissible triplet if N −µ 1 1 1 = ( − ), ϑ 2 q0 q where    q0 (N − µ) , N − µ > 2q0 , 1 < q0 ≤ q < N − µ − 2q0   ∞, N − µ ≤ 2q0 . Clearly, q < ϑ ≤ ∞ if (ϑ, q, q0 ) is an admissible triplet, and 1 < ϑ ≤ ∞ if (ϑ, q, q0 ) is a generalized admissible triplet. Next, the hypothesis of the semilinear term is introduced. (Hf) Let f ∈ C(R, R). We suppose that f (0) = 0 and that there exist ρ > 0 and K > 0 such that |f (u) − f (v)| ≤ K(|u|ρ + |v|ρ )|u − v|, for u, v ∈ R. For 1 < q < ∞ and 0 ≤ µ < N , from the hypothesis (Hf) we deduce that kf (u) − f (v)kM

q ,µ 1+ρ

≤(kukρMq,µ + kvkρMq,µ )ku − vkMq,µ .

(3.70)

The following lemma implies strong continuity of the operator S(t)u0 for t ∈ s0 s0 (0, ∞) with values in Mq,µ ∩ Np,µ,∞ for the initial data u0 ∈ Np,µ,∞ . Lemma 3.28. Let 0 ≤ µ < N , 1 < q0 := (N −µ)ρ(1−α) < p ≤ q and (ϑ, q, q0 ) 2 s0 be a generalized admissible triplet. Assume that u0 ∈ Np,µ,∞ . Then it holds that 1−α s0 S(t)u0 ∈ Cb ((0, ∞); Np,µ,∞ ) and t ϑ S(t)u0 ∈ Cb ((0, ∞); Mq,µ ) with the estimate s0 sup kS(t)u0 kNp,µ,∞ + sup t

t∈(0,∞)

where L0 is a constant.

t∈(0,∞)

1−α ϑ

s0 kS(t)u0 kMq,µ ≤L0 ku0 kNp,µ,∞ ,

(3.71)

Fractional Rayleigh-Stokes Equations

183

2 Proof. Consider p > (N −µ)ρ(1−α) > 1, we have s0 = N p−µ − ρ(1−α) < 0. Notice 2 1 1 2 that N ( p − q ) − s0 = ϑ < 2 for a generalized admissible triplet (µ, q, q0 ), and which gives by Lemma 3.27(iii) and (3.63) that s0 sup kS(t)u0 kNp,µ,∞ + sup t

t∈(0,∞)

1−α ϑ

kS(t)u0 kMq,µ

t∈(0,∞)

s0 ≤M1 ku0 kNp,µ,∞ + sup t

1−α ϑ

t∈(0,∞)

s0 ≤M1 ku0 kNp,µ,∞ + M1 sup t

0 kS(t)u0 kNq,µ,1 (N −µ)(1−α) 1−α 1 −ρ ϑ + 2q

t∈(0,∞)

s0 ku0 kNp,µ,∞

s0 =L0 |u0 kNp,µ,∞

s0 for all u0 ∈ Np,µ,∞ and t > 0, where L0 = 2M1 . This yields (3.71). s0 Next we prove that S(t)u0 is strongly continuous with values in Mq,µ ∩ Np,µ,∞ s0 for all t > 0. Indeed, Remark 3.9 implies the continuity of S(t)u0 in Np,µ,∞ for t > 0. It remains to prove the continuity of S(t)u0 in Mq,µ . For any 0 < t1 < t2 < ∞, we can obtain from (3.63) that 1−α

1−α

t ϑ S(t2 )u0 −t ϑ S(t1 )u0 2 1 Mq,µ 1−α
1−α

1−α

ϑ ϑ 0 ≤ t2 − t1 S(t2 )u0 N 0 + t1 ϑ kS(t2 )u0 − S(t1 )u0 kNq,µ,1 . q,µ,1

The first term for the right-hand side of above inequality is bounded by 1−α s0 M1 ku0 kNp,µ,∞ [1 − (t1 /t2 ) ϑ ] due to Lemma 3.27(iii). Using Remark 3.9 again, the second term tends to 0 as t2 → t1 . Cb ((0, ∞); Mq,µ ). The proof is completed.

So one can see that t

1−α ϑ

S(t)u0 ∈

We state the well-posedness results of global solutions for the equation (3.62). Theorem 3.12. Let 0 ≤ µ < N , 1 < q0 := generalized admissible triplet satisfying

(N −µ)ρ(1−α) 2

< p ≤ q and (ϑ, q, q0 ) be a

max{q0 , 1 + ρ} < q < q0 (1 + ρ).

(3.72)

s0 and (Hf ) holds. If there exist ε > 0 and κ := κ(ε) such that Let u0 ∈ Np,µ,∞ s0 ku0 kNp,µ,∞ ≤ κ, then the equation (3.62) has a unique solution u ∈ Zp,q with kukZp,q ≤ 2ε. Moreover, the map u0 ∈ Uκ 7→ u ∈ Zp,q is Lipschitz continuous, where Uκ is denoted by s0 s0 Uκ = {u0 ∈ Np,µ,∞ : ku0 kNp,µ,∞ ≤ κ}.

Proof. Let u ∈ Zp,q and t > 0. Define the operator Φ in Zp,q as Φ(u)(t) = S(t)u0 + W (u)(t), where Z

t

S(t − τ )f (u(τ ))dτ.

W (u)(t) = 0

(3.73)

184

Fractional Partial Differential Equations

Now we estimate the nonlinear part. For any generalized admissible triplet (ϑ, q, q0 ), we get ρ(N − µ)(1 − α) ρ(1 − α) + = 1, 2q ϑ ρ(N −µ)(1−α) 2q

which implies that

(3.74)

< 1. Let u, v ∈ Zp,q . Notice that Mq0 ,µ ⊂

= s0 − N p−µ , we obtain from ⊂ due to (3.63) and (3.64) for 0 − Nq−µ 0 Lemma 3.27(i), (3.70) and (3.74) that

Nq00 ,µ,∞

s0 Np,µ,∞

s0 kW (u)(t) − W (v)(t)kNp,µ,∞

≤kW (u)(t) − W (v)(t)kMq0 ,µ Z t (N −µ)(1−α) 1+ρ ( q − q1 ) 2 0 kf (u(τ ) − f (v(τ )))k ≤M1 (t − τ )− Mq/(1+ρ),µ dτ 0 Z t (N −µ)(1−α) 1+ρ ( q − q1 ) ρ 2 0 ku(τ ) − v(τ )k ≤M1 K (t − τ )− Mq,µ (ku(τ )kMq,µ 0

+ kv(τ )kρMq,µ )dτ ≤M1 Kku − vkC∗ ((0,∞);Mq,µ ) kukρC∗ ((0,∞);Mq,µ ) + kvkρC∗ ((0,∞);Mq,µ ) Z t ρ(N −µ)(1−α) (ρ+1) 1−α 2q τ − ϑ (1−α) dτ × (t − τ ) ϑ −




0
≤M1 KB(χ, 1 − χ)ku − vkC∗ ((0,∞);Mq,µ ) kukρC∗ ((0,∞);Mq,µ ) + kvkρC∗ ((0,∞);Mq,µ ) ,

(3.75) where B(·, ·) stands for the Beta function, χ = (ρ+1) ϑ (1 − α) ∈ (0, 1), here we have used (3.72). Further, let β = s = 0, it follows from Lemma 3.27(i), (3.70) and (3.74) again that kW (u)(t) − W (v)(t)kMq,µ Z t ρ(N −µ)(1−α) 2q kf (u(τ ) − f (v(τ )))kMq/(1+ρ),µ dτ ≤M1 (t − τ )− 0

≤M1 Kku − vkC∗ ((0,∞);Mq,µ ) kukρC∗ ((0,∞);Mq,µ ) + kvkρC∗ ((0,∞);Mq,µ ) Z t ρ(N −µ)(1−α) (ρ+1) 2q × τ − ϑ (1−α) dτ (t − τ )−




0

≤M1 KB(


χρ , 1 − χ)ku − vkC∗ ((0,∞);Mq,µ ) kukρC∗ ((0,∞);Mq,µ ) + kvkρC∗ ((0,∞);Mq,µ ) . 1+ρ (3.76)

We observe that B(

χρ , 1 − χ) = 1+ρ

Z

1

χρ

τ 1+ρ −1 (1 − τ )χ−1 dτ ≥

0

Z

1

τ χ−1 (1 − τ )χ−1 dτ = B(χ, 1 − χ).

0

Thus we deduce the relation kW (u) − W (v)kZp,q ≤ L1 (kukρZp,q + kvkρZp,q )ku − vkZp,q ,

(3.77)

Fractional Rayleigh-Stokes Equations

185

χρ where L1 = 2M1 KB( 1+ρ , 1 − χ). It remains to prove that W (u) ∈ Zp,q for u ∈ Zp,q . Now we show the continuity s0 of W (u)(t) in Mq,µ ∩ Np,µ,∞ for all t ∈ (0, ∞) and u ∈ Zp,q . For 0 < t1 < t2 < ∞, it holds that

W (u)(t2 ) − W (u)(t1 ) Z t2 Z = S(t2 − τ )f (u(τ ))dτ + t1

t1

[S(t2 − τ ) − S(t1 − τ )]f (u(τ ))dτ

0

=:I1 (t1 , t2 ) + I2 (t1 , t2 ). s0 Similar to the arguments on (3.75), we can calculate kI1 (t1 , t2 )kNp,µ,∞ as follows s0 kI1 (t1 , t2 )kNp,µ,∞

≤M1 Kkukρ+1 C∗ ((0,∞);Mq,µ ) ≤M1 Kkukρ+1 C∗ ((0,∞);Mq,µ )

t2

Z

(t2 − τ )−

t1 Z 1

(1 − τ )

(N −µ)(1−α) 1+ρ ( q − q1 2 0

ρ(N −µ)(1−α) 1−α ϑ − 2q

) − (ρ+1) (1−α) ϑ

τ

dτ

τ −χ dτ

t1 /t2

→0, as t2 → t1 , where we have used the properties of the Beta function. Using the continuity of q S(t)v on Mq0 ,µ in Remark 3.9 for v ∈ M 1+ρ ,µ , it holds that kS(t2 − τ ) − S(t1 − τ )]f (u(τ ))kMq0 ,µ ≤[(t2 − τ )−

(N −µ)(1−α) 1+ρ ( q − q1 2 0

)

− (t1 − τ )−

(N −µ)(1−α) 1+ρ ( q − q1 2 0

)

]kf (u(τ ))kM

q ,µ 1+ρ

.

It yields that s0 kI2 (t1 , t2 )kNp,µ,∞

h Z t1 (N −µ)(1−α) 1+ρ ( q − q1 ) − (ρ+1) (1−α) 2 ϑ 0 τ dτ ≤M1 Kkukρ+1 (t1 − τ )− C∗ ((0,∞);Mq,µ ) 0 Z t1 i (N −µ)(1−α) 1+ρ ( q − q1 ) − (ρ+1) (1−α) 2 ϑ 0 τ − (t2 − τ )− dτ 0

≤M1 Kkukρ+1 C∗ ((0,∞);Mq,µ ) Z −

t1 /t2

(1 − τ )

hZ

1

(1 − τ )

ρ(N −µ)(1−α) 1−α ϑ − 2q

τ −χ dτ

0

ρ(N −µ)(1−α) 1−α ϑ − 2q

τ −χ dτ

i

0

→0, as t2 → t1 . s0 This shows W (u) ∈ C((0, ∞); Np,µ,∞ ). Moreover, in the same way as above we have that

kI1 (t1 , t2 )kMq,µ − 1−α ϑ

≤M1 Kkukρ+1 C∗ ((0,∞);Mq,µ ) t2

Z

1

t1 /t2

(1 − τ )−

ρ(N −µ)(1−α) 2q

τ−

ρ+1 µ (1−α)

dτ,

186

Fractional Partial Differential Equations

kI2 (t1 , t2 )kMq,µ ≤M1 Kkukρ+1 C∗ ((0,∞);Mq,µ ) − 1−α ϑ

h 1−α Z 1 ρ(N −µ)(1−α) ρ+1 − 2q (1 − τ )− t1 ϑ τ − µ (1−α) dτ

t1 /t2

Z

(1 − τ )

− t2

0 ρ(N −µ)(1−α) − 2q

τ−

ρ+1 µ (1−α)

i dτ .

0

The above inequalities also tend to zero as t2 → t1 , and then W (u) ∈ C((0, ∞); Mq,µ ). Consequently, this together with the relation (3.77) for v = 0 shows that W (u) ∈ Zp,q for u ∈ Zp,q . 1 ρ We choose 0 < ε < R = ( 2ρ+1 L1 ) , where L1 > 0 is the constant obtained in ε (3.77). Let κ = L0 , where L0 > 0 is the constant obtained in Lemma 3.28. Lemma 3.25 with X = Zp,q and ϕ = S(t)u0 yields the existence of a unique global mild solution u ∈ Zp,q such that kukZp,q ≤ 2ε. Finally, we show the continuous dependence on the initial value. Let u0 ∈ Uκ and u ¯ be the solution of the equation u ¯(t) = S(t)u0 + W (¯ u)(t). From Lemma 3.25 s0 , we obtain that and Lemma 3.28 for u0 ∈ Np,µ,∞ ku − u ¯kZp,q ≤

1 2ρ+1 L1 ερ

1− The proof is completed.

kS(t)(u0 − u0 )kZp,q ≤

1−

L0 ku0 ρ+1 2 L1 ερ

s0 − u0 kNp,µ,∞ .

The following corollary means that the equation (3.62) is solvable without the restrictions q < q0 (1 + ρ). Corollary 3.2. Let the conditions of Theorem 3.12 hold except for q < q0 (1 + ρ). Then the equation (3.62) has a unique solution u ∈ Zp,q satisfying 1

ku(t)kMq,µ ≤ M1 t − ρ + ˜

(N −µ)(1−α) 2q

.

(3.78)

Proof. Let q ≥ q0 (1 + ρ). Then we can choose ζ > 0 so small that   N −µ 1+ρ 1 − < 1, 2 q˜ q where we denote q˜ = q0 (1 + ρ) − ζ. Observe that the above inequality can be guaranteed by the fact that q < (N − µ)q0 /(N − µ − 2q0 ) if N − µ > 2q0 and q < ∞ if N − µ ≤ 2q0 . Let   1 N −µ 1 1 = − . 2 q0 q˜ ϑ˜ ˜ q˜, q0 ) is a generalized admissible triplet and ϑ˜ > (1 + ρ)(1 − α). UsThen (ϑ, ing Theorem 3.12 we know that the equation (3.62) has a unique solution u ∈ s0 Cb ((0, ∞); Np,µ,∞ ) ∩ C∗ ((0, ∞); Mq˜,µ ). A direct calculation yields that kW (u)(t)kMq,µ Z t (N −µ)(1−α) 1+ρ ( q˜ − q1 ) 2 ≤M1 (t − τ )− kf (u(τ ))kMq/(1+ρ),µ dτ ˜ 0

Fractional Rayleigh-Stokes Equations

≤M1 Kkukρ+1 C∗ ((0,∞);Mq,µ ˜ ) ≤M1 KB1 t−

1−α ϑ

t

Z

(t − τ )−

(N −µ)(1−α) 1+ρ ( q˜ − q1 ) 2

187

τ−

(ρ+1) (1−α) ˜ ϑ

dτ

0

kukρ+1 , C∗ ((0,∞);Mq,µ ˜ )

ρ+1 1 ( 1+ρ where B1 = B(1 − N (1−α) ˜ (1 − α)). Furthermore, by the similar 2 q˜ − q ), 1 − ϑ arguments, the continuity of W (u)(t) on Mq,µ is obvious for t > 0. This together with Lemma 3.28 implies that u ∈ C∗ ((0, ∞); Mq,µ ). This implies u ∈ Zp,q and (3.78). The proof is completed.

Finally we state an asymptotic behavior result of the solution of the equation equation (3.62) as t → ∞. Theorem 3.13. Let the conditions of Theorem 3.12 hold. Assume that u and v be two global mild solutions for the equation (3.62) obtained in Theorem 3.12, with the initial values u0 and v0 , respectively. Then s0 lim ku(t) − v(t)kNp,µ,∞ + lim t

t→∞

1−α ϑ

t→∞

ku(t) − v(t)kMq,µ = 0

if and only if s0 lim kS(t)(u0 − v0 )kNp,µ,∞ + lim t

t→∞

1−α ϑ

t→∞

kS(t)(u0 − v0 )kMq,µ = 0.

Proof. (⇐) We denote J(u, v)(t) = W (u)(t) − W (v)(t). Using (3.75), kukZp,q ≤ 2ε and kvkZp,q ≤ 2ε, we obtain that s0 kJ(u, v)(t)kNp,µ,∞ Z t (N −µ)(1−α) 1+ρ ( q − q1 ) ρ 2 0 ku(τ ) − v(τ )k ≤M1 K (t − τ )− Mq,µ (ku(τ )kMq,µ

0

+ kv(τ )kρMq,µ )dτ Z t (N −µ)(1−α) 1+ρ ( q − q1 ) − ρ(1−α) ρ 2 ϑ 0 τ ≤2M1 K(2ε) (t − τ )− ku(τ ) − v(τ )kMq,µ dτ. 0

For kJ(u, v)(t)kMq,µ , in view of (3.76), one can see that Z t ρ(N −µ)(1−α) ρ(1−α) ρ 2q τ − ϑ ku(τ ) − v(τ )kMq,µ dτ. kJ(u, v)(t)kMq,µ ≤2M1 K(2ε) (t − τ )− 0 1−α

s0 Let Π(t) = ku(t) − v(t)kNp,µ,∞ + t ϑ ku(t) − v(t)kMq,µ . By performing a change of s0 variables in kJ(u, v)(t)kNp,µ,∞ and kJ(u, v)(t)kMq,µ , we get that 1−α

s0 kJ(u, v)(t)kNp,µ,∞ + t ϑ kJ(u, v)(t)kMq,µ Z 1 (N −µ)(1−α) 1+ρ ( q − q1 ) − (ρ+1) (1−α) 2 ϑ 0 τ Π(tτ )dτ ≤2M1 K(2ε)ρ (1 − τ )−

0

+ 2M1 K(2ε)

ρ

Z 0

1

(1 − τ )−

ρ(N −µ)(1−α) 2q

τ−

(ρ+1) (1−α) ϑ

Π(tτ )dτ.

188

Fractional Partial Differential Equations

Now we claim that lim supt→∞ Π(t) = 0. Indeed, we take lim supt→∞ in above inequality in order to obtain ρ(1−α)

s0 lim sup[kJ(u, v)(t)kNp,µ,∞ + t ϑ kJ(u, v)(t)kMq,µ ] t→∞ Z 1 (N −µ)(1−α) 1+ρ ( q − q1 ) − (ρ+1) (1−α) 2 ϑ 0 τ lim sup Π(tτ )dτ (1 − τ )− ≤2M1 K(2ε)ρ

t→∞

0 ρ

1

Z

(1 − τ )

+ 2M1 K(2ε)

ρ(N −µ)(1−α) − 2q

τ

(ρ+1) − ϑ (1−α)

(3.79)

lim sup Π(tτ )dτ t→∞

0 ρ+1 ρ

≤L1 2

ε lim sup Π(t). t→∞

It follows that s0 lim sup Π(t) ≤ lim sup[kS(t)(u0 − v0 )kNp,µ,∞ +t

t→∞

1−α ϑ

t→∞

s0 + lim sup[kJ(u, v)(t)kNp,µ,∞ +t

1−α ϑ

t→∞ ρ+1 ρ

≤L1 2

kS(t)(u0 − v0 )kMq,µ ]

kJ(u, v)(t)kMq,µ ]

ε lim sup Π(t). t→∞

ρ+1 ρ

Since L1 2 ε < 1, we get lim supt→∞ Π(t) = 0, as required. (⇒) According to the relation S(t)(u0 − v0 ) = u(t) − v(t) − (J(u, v)(t)) and (3.79), we have that s0 lim sup[kS(t)(u0 − v0 )kNp,µ,∞ +t

t→∞

1−α ϑ

kS(t)(u0 − v0 )kMq,µ ]

s0 ≤ lim sup Π(t) + lim sup[kJ(u, v)(t)kNp,µ,∞ +t

t→∞

t→∞

1−α ϑ

kJ(u, v)(t)kMq,µ ]

≤(1 + L1 2ρ+1 ερ ) lim sup Π(t) = 0. t→∞

The proof is completed. 3.6 3.6.1

Final Value Problem Introduction

The aim of this section is to obtain the existence result of mild solutions for the nonlinear fractional Rayleigh-Stokes problem under a weaker condition on the nonlinear source term than Lipschitz condition. And the quasi-boundary value method will be used to regularize the problem, which is mainly by perturbing the final condition or boundary condition to a new regularization condition depending on a small parameter. Consider the nonlinear fractional Rayleigh-Stokes problem with final condition  α   ∂t u − (1 + γ∂t )∆u = f (t, x, u), t ∈ (0, b), x ∈ Ω, (3.80) u(t, x) = 0, t ∈ (0, b), x ∈ ∂Ω,   u(b, x) = g(x), x ∈ Ω,

Fractional Rayleigh-Stokes Equations

189

where Ω ⊂ Rd (d ≥ 1) is a bounded domain with smooth boundary ∂Ω, ∆ is the Laplacian operator, b > 0 is a given time, γ > 0 is a fixed constant, g ∈ L2 (Ω) is the final value, f : [0, b] × Ω × R → R is a given function satisfying some assumptions, and ∂tα is the Riemann-Liouville fractional partial derivative of order α ∈ (0, 1) which is defined as follows Z t 1 ∂ ∂tα v(t, x) = (t − s)−α v(s, x)ds, t > 0, Γ(1 − α) ∂t 0 where Γ(·) is the Gamma function. In Subsection 3.6.2, some concepts and lemmas which will be used in this section are given. In Subsection 3.6.3, the compactness and continuity in the uniform operator topology of solution operator for problem (3.80) are discussed at first, then we obtain the existence result of mild solutions for problem (3.80). Furthermore, a regularization method is used in Subsection 3.6.4, the well-posedness and the convergence rate for regularized solutions are obtained. 3.6.2

Preliminaries

Let b > 0 be a given number, N+ denotes the set of positive integer numbers, L2 (Ω) is the Banach space of all measurable functions on Ω ⊂ Rd (d ≥ 1) with the inner product (·, ·) and norm k · k. Denote by B(X) the space of all bounded linear operators from normed linear space X into X. L∞ (0, b; L2 (Ω)) is the Banach space of all essentially bounded and measurable functions v equipped with the norm kvk∞ = esssupt∈[0,b] kv(t)k. We recall the following spectral problem ( − ∆ϕn (x) = λn ϕn (x),

x ∈ Ω, x ∈ ∂Ω,

ϕn (x) = 0,

(3.81)

where n ∈ N+ and the eigenvalues satisfy 0 < λ1 ≤ λ2 ≤ ... ≤ λn ≤ ..., with λn → ∞ as n → ∞. And the corresponding set of eigenfunctions {ϕn }∞ n=1 ⊂ 2 H01 (Ω). Obviously, {ϕn }∞ is an orthonormal basis of L (Ω). n=1 Let C([0, b]; L2 (Ω)) stand for the Banach space of all continuous functions defined on [0, b] with the norm kvkC([0,b];L2 (Ω)) = sup kv(t)k. t∈[0,b]

In the following, for 0 < µ < 1, we introduce the weighted continuous function space of all continuous functions defined on (0, b], which has the following form C µ ([0, b]; L2 (Ω)) := {v ∈ C((0, b]; L2 (Ω))|tµ v(t) ∈ C([0, b]; L2 (Ω))}, and it is easy to see that C µ ([0, b]; L2 (Ω)) is a Banach space equipped with the norm kvkµ = sup tµ kv(t)k. t∈[0,b]

190

Fractional Partial Differential Equations

For σ ≥ 0, we denote the space Hσ (Ω) as follows: ∞ n o X 2 Hσ (Ω) = v ∈ L2 (Ω) : λ2σ j |(v, ϕj )| < ∞ , j=1

and the norm of Hσ (Ω) is defined by   21 ∞ X 2 kvkHσ (Ω) =  λ2σ . j |(v, ϕj )| j=1

Lemma 3.29. [279, Lemma 4.2] Assume β, θ and C are positive constants, then, for ξ > 0, we have ( l1 (θ, C)β θ , 0 < θ < 1, βξ 1−θ F (ξ) = ≤ βξ + C l2 (θ, C, λ1 )β, θ ≥ 1, ξ ≥ λ1 , where l1 (θ, C) = (1 − θ)1−θ θθ C −θ and l2 (θ, C, λ1 ) = C −1 λ1 1−θ . 3.6.3

Existence of Solutions

Let u be a solution of problem (3.80), and let u(t, x) be expanded as follows u(t, x) =

∞ X

(u(t, ·), ϕj )ϕj (x).

j=1

From [39] and [278, Equation (10)], the solution of problem (3.80) can be described as ! Z b ∞ X Pα,j (t) u(t, x) = (g, ϕj ) − Pα,j (b − s)(f (s, u(s)), ϕj )ds ϕj (x) P (b) 0 j=1 α,j (3.82) ∞ Z t X + Pα,j (t − s)(f (s, u(s)), ϕj )dsϕj (x), j=1

0

where Z

∞

Pα,j (t) =

e−ξt Kj (ξ)dξ,

0

and Kj (ξ) =

λj ξ α sin (απ) γ . α π (−ξ + λj γξ cos (απ) + λj )2 + (λj γξ α sin (απ))2

For any v ∈ L2 (Ω), we denote two operators Pα (t)v =

∞ X j=1

Pα,j (t)(v, ϕj )ϕj (x),

Sα (t)v =

∞ X Pα,j (t) (v, ϕj )ϕj (x). P (b) j=1 α,j

(3.83)

Fractional Rayleigh-Stokes Equations

191

We use u(t) to denote the spatial function u(t, ·), then the equation (3.82) can be rewritten as Z b Z t u(t) = Sα (t)g − Sα (t) ◦ Pα (b − s)f (s, u(s))ds + Pα (t − s)f (s, u(s))ds, (3.84) 0

0

where ◦ stands for the composite operation. Noting that if u is an exact solution of problem (3.80), then u is at least once derivative with respect to time, this is a little bit constrained and higher smoothness is required for the given data. For this reason, a suitable definition of mild solutions for problem (3.80) is given below. Definition 3.10. For µ ∈ (0, 1), a function u is called a mild solution of problem (3.80), if u ∈ C µ ([0, b]; L2 (Ω)) and it satisfies the integral equation (3.84). Lemma 3.30. [278, Lemma 2.2] For α ∈ (0, 1) and t ∈ [0, b], we have the following estimates C2 C1 Pα,j (t) ≤ , Pα,j (b) ≥ , 1 + λj t1−α λj where C1 =

Γ(1 − α) + 1, γπ sin(απ)

C2 =

γ sin(απ)e−b 3π(α + 1)(γ 2 + 1 +

1 . ) λ21

In the following, we give some properties of operators Pα (t) and Sα (t). Lemma 3.31. [409, Lemma 3.1, Lemma 3.2] The operator Pα (t) is compact and continuous in the uniform operator topology for t > 0. For t ∈ [0, b] and v ∈ C µ ([0, b]; L2 (Ω)), we define an operator as follows: Z b (Qα v)(t) = Sα (t) ◦ Pα (b − s)v(s)ds. 0

Lemma 3.32. Let 1 − α < µ < 1. Then Qα : C µ ([0, b]; L2 (Ω)) → C µ ([0, b]; L2 (Ω)) is a completely continuous operator. Proof. For any n ∈ N+ and t ∈ (0, b], let Ωn = span{ϕ1 , ..., ϕn } and define Qnα : (0, b] × L2 (Ω) → Ωn as follows: Z b n (Qα v)(t) = Sαn (t) ◦ Pαn (b − s)v(s)ds, v ∈ C µ ([0, b]; L2 (Ω)), 0

where Pαn (t)v =

n X j=1

Pα,j (t)(v, ϕj )ϕj (x),

Sαn (t)v =

n X Pα,j (t) (v, ϕj )ϕj (x). P (b) j=1 α,j

Clearly, Ωn is a finite dimensional subspace of L2 (Ω). In the sequel, we choose a bounded subset Λr = {v ∈ C µ ([0, b]; L2 (Ω)) : kvkµ ≤ r} of C µ ([0, b]; L2 (Ω)), here r

192

Fractional Partial Differential Equations

is a positive constant. Then we need to prove that the set {(Qnα v)(t) : v ∈ Λr } is relatively compact in C µ ([0, b]; L2 (Ω)) for every n ∈ N+ . For any v ∈ Λr , in view of Lemma 3.30, we have kSαn (t) ◦ Pαn (b − s)vk   21 2 n  X P (t) α,j = Pα,j (b − s) |(v(s), ϕj )|2  P (b) α,j j=1  n  X  ≤ j=1

C1 λj C1 · · 1−α 1 + λj t C2 1 + λj (b − s)1−α

2

 21

(3.85)

|(v(s), ϕj )|2 

C12

(b − s)α−1 kv(s)k. C2 It yields that ≤

k(Qnα v)(t)k

Z

b

n n = Sα (t) ◦ Pα (b − s)v(s)ds

0

Z b ≤ kSαn (t) ◦ Pαn (b − s)v(s)kds 0

(3.86)

Z C12 b ≤ (b − s)α−1 kv(s)kds C2 0 C 2 Γ(α)Γ(1 − µ) α−µ b r, ≤ 1 C2 Γ(1 − µ + α) which implies that k(Qnα v)kµ ≤

C12 Γ(α)Γ(1 − µ) α b r. C2 Γ(1 − µ + α)

This derives the set {(Qnα v)(t) : v ∈ Λr } is uniformly bounded. Hence, by virtue of the Arzel` a-Ascoli theorem, it remains to prove that µ n {t (Qα v)(t) : v ∈ Λr } is equicontinuous for t ∈ [0, b]. In order to achieve this aim, we need to check that limt→0+ tµ (Qnα v)(t) R b exists and is finite. Indeed, for any v ∈ Λr , when t = 0, we have (Qnα v)(0) = 0 Sαn (0) ◦ Pαn (b − s)v(s)ds, from (3.86), we know that C 2 Γ(α)Γ(1 − µ) α−µ k(Qnα v)(0)k ≤ 1 b r < ∞, C2 Γ(1 − µ + α) which implies lim tµ (Qnα v)(t) = 0.

t→0+

Thus, for t1 = 0, t2 ∈ (0, b] and any v ∈ Λr , it is easy to see that ktµ2 (Qnα v)(t2 ) − tµ1 (Qnα v)(t1 )k ≤ tµ2 → 0,

C12 Γ(α)Γ(1 − µ) α−µ b r C2 Γ(1 − µ + α) as

t2 → 0+ .

Fractional Rayleigh-Stokes Equations

193

Moreover, for 0 < t1 < t2 ≤ b and any v ∈ Λr , we have ktµ2 (Qnα v)(t2 ) − tµ1 (Qnα v)(t1 )k

Z

Z b

b

µ

µ Sαn (t2 ) ◦ Pαn (b − s)v(s)ds − t1 = t2 Sαn (t1 ) ◦ Pαn (b − s)v(s)ds

0 0

Z

b

≤tµ2 (Sαn (t2 ) − Sαn (t1 )) ◦ Pαn (b − s)v(s)ds

0

Z

b

Sαn (t1 ) ◦ Pαn (b − s)v(s)ds + (tµ2 − tµ1 )

0 =:I1 + I2 . By using the inequalities 1 − e−t ≤ t, e−t ≤

1 t

for t > 0, from (3.83), we have

2

k(Sαn (t2 ) − Sαn (t1 )) ◦ Pαn (b − s)v(s)k 2 n  X |Pα,j (t2 ) − Pα,j (t1 )| Pα,j (b − s) |(v(s), ϕj )|2 = P (b) α,j j=1 2 Z n  X C1 ∞ −ξt2 −ξt1 α−1 ≤ |e −e |Kj (ξ)dξ(b − s) |(v(s), ϕj )|2 C 2 0 j=1 2 X 2  n Z ∞ C1 (b − s)α−1 e−ξt1 ξ(t2 − t1 )Kj (ξ)dξ |(v(s), ϕj )|2 ≤ C2 0 j=1   n  2 Z α−1 2 X C1 (b − s) t2 − t1 ∞ ≤ Kj (ξ)dξ |(v(s), ϕj )|2 C2 t 1 0 j=1   2  α−1 2 t2 − t1 C1 (b − s) kv(s)k2 , ≤ C2 t1 R∞ where we use the fact 0 Kj (ξ)dξ = 1. Therefore, we get I1 ≤

tµ2 (t2 − t1 ) C1 Γ(α)Γ(1 − µ) α−µ b r → 0, t1 C2 Γ(α + 1 − µ)

as

t2 → t1 .

From (3.86), we have I2 =(tµ2 − tµ1 )k(Qnα v)(t1 )k ≤

Γ(α)Γ(1 − µ) α−µ C12 µ (t − tµ1 ) b r → 0, C2 2 Γ(1 − µ + α)

as

t2 → t1 .

Thus, ktµ2 (Qnα v)(t2 ) − tµ1 (Qnα v)(t1 )k → 0 as t2 → t1 , which implies that the set {tµ (Qnα v)(t) : v ∈ Λr } is equicontinuous for t ∈ [0, b]. Consequently, for every n ∈ N+ , Qnα is a compact operator.

194

Fractional Partial Differential Equations

In the sequel, we prove the compactness of operator Qα . From the fact that Qnα is compact, it requires to verify Qnα → Qα as n → ∞ in C µ ([0, b]; L2 (Ω)). Indeed, for any v ∈ C µ ([0, b]; L2 (Ω)), we have

Z

b

k(Qα v)(t) − (Qnα v)(t)k = (Sα (t) ◦ Pα (b − s) − Sαn (t) ◦ Pαn (b − s))v(s)ds

0   12 2 Z b ∞  X P (t) α,j  ≤ Pα,j (b − s) |(v(s), ϕj )|2  ds P (b) α,j 0 j=n+1 Z b 1 C12 (b − s)α−1 kv(s)kds ≤ C2 1 + λn+1 t1−α 0 C2 Γ(α)Γ(1 − µ) α−µ 1 ≤ 1 b kvkµ . C2 1 + λn+1 t1−α Γ(α + 1 − µ) Therefore, from the fact that λn+1 → ∞ as n → ∞, we have kQα v − Qnα vkµ ≤

C12 Γ(α)Γ(1 − µ) 2α−1 1 b kvkµ C2 Γ(α + 1 − µ) λn+1

→0,

as

n → ∞.

This implies the operator Qα is compact in C µ ([0, b]; L2 (Ω)). µ 2 Finally, we prove the continuity of Qα . In fact, let {vm }∞ m=1 ⊂ C ([0, b]; L (Ω)) µ 2 and v ∈ C ([0, b]; L (Ω)), which satisfy limm→∞ vm = v. Similar to the process of (3.85), we have kSα (t) ◦ Pα (b − s)(vm (s) − v(s))k ≤

C12 (b − s)α−1 kvm (s) − v(s)k, C2

(3.87)

which implies that

Z

b

k(Qα vm )(t) − (Qα v)(t)k = Sα (t) ◦ Pα (b − s)(vm (s) − v(s))ds

0

2 Z b C ≤ 1 (b − s)α−1 kvm (s) − v(s)kds C2 0 C 2 Γ(α)Γ(1 − µ) α−µ b kvm − vkµ . ≤ 1 C2 Γ(α + 1 − µ) Hence kQα vm − Qα vkµ ≤

C12 Γ(α)Γ(1 − µ) α b kvm − vkµ → 0, C2 Γ(α + 1 − µ)

as m → ∞.

This means that Qα is continuous on C µ ([0, b]; L2 (Ω)). Combined with the above arguments, we conclude that Qα is completely continuous. The proof is completed.

Before giving our main result, we need the following hypothesis:

Fractional Rayleigh-Stokes Equations

195

(H1) f (t, v) is measurable for t ∈ (0, b], f (t, v) is continuous for v ∈ L2 (Ω) and there exists a constant Cf > 0 such that kf (t, v)k ≤ Cf kvk. Theorem 3.14. Assume g ∈ L2 (Ω) and (H1) holds. If µ ∈ (1 − α, 1), then the problem (3.80) has at least one mild solution in C µ ([0, b]; L2 (Ω)) provided that C12 Cf Γ(α)Γ(1 − µ) α C1 Cf b + b < 1. C2 Γ(α + 1 − µ) 1−µ

(3.88)

Proof. For any u ∈ C µ ([0, b]; L2 (Ω)), we define an operator as follows (T u)(t) = Sα (t)g − (Qα f )(t) + (Gα f )(t), where t

Z

Pα (t − s)f (s, u(s))ds.

(Gα f )(t) = 0

Clearly, proving the existence of solutions to the problem (3.80) is equivalent to proving that the operator equation T u = u has at least one fixed point in C µ ([0, b]; L2 (Ω)). In view of inequality (3.88), let us choose a positive constant r such that −1  C1 α−1+µ C 2 Cf Γ(α)Γ(1 − µ) α C1 Cf b − b b kgk ≤ r, 1− 1 C2 Γ(α + 1 − µ) 1−µ C2 and we denote a bounded closed and convex subset Br = {v ∈ C µ ([0, b]; L2 (Ω)) : kvkµ ≤ r} of C µ ([0, b]; L2 (Ω)). Next, we will check that T is a completely continuous operator and T maps Br into Br . From Lemma 3.30, for any v ∈ C µ ([0, b]; L2 (Ω)), we first estimate the operator Gα as follows: Z t C1 1−µ (3.89) t kvkµ . k(Gα v)(t)k ≤ C1 kv(s)kds ≤ 1 −µ 0 Claim I. The set {(T u)(t) : u ∈ Br } is relatively compact in C µ ([0, b]; L2 (Ω)). From Lemma 3.32 and the assumption of f , Qα is completely continuous from C µ ([0, b]; L2 (Ω)) into C µ ([0, b]; L2 (Ω)), and for any u ∈ C µ ([0, b]; L2 (Ω)), it is easy to see that f (t, u) ∈ C µ ([0, b]; L2 (Ω)), which implies that {(Qα f )(t) : u ∈ Br } is relatively compact in C µ ([0, b]; L2 (Ω)). We only need to prove that {(Gα f )(t) : u ∈ Br } is relatively compact in C µ ([0, b]; L2 (Ω)). From (3.89) and (H1), we have Z t tµ k(Gα f )(t)k ≤C1 tµ kf (s, u(s))kds 0 Z t ≤C1 Cf tµ ku(s)kds 0

C1 Cf tr, ≤ 1−µ

196

Fractional Partial Differential Equations

it yields that C1 Cf br, 1−µ which implies that Gα is a uniformly bounded operator. Now we prove the set {tµ (Gα f )(t) : u ∈ Br } is equicontinuous for t ∈ [0, b]. For t1 = 0, 0 < t2 ≤ b, from (H1), we have

Z t2

µ µ µ

Pα (t2 − s)f (s, u(s))ds kt2 (Gα f )(t2 ) − t1 (Gα f )(t1 )k = t2

kGα f kµ ≤

0

C1 Cf t2 r → 0, ≤ 1−µ

as t2 → 0+ .

For 0 < t1 < t2 ≤ b, we have ktµ2 (Gα f )(t2 ) − tµ1 (Gα f )(t1 )k

Z t2

Z t1

µ

µ

= t2 Pα (t2 − s)f (s, u(s))ds − t1 Pα (t1 − s)f (s, u(s))ds

0

Z0

t1 −δ

≤ tµ2 (Pα (t2 − s) − Pα (t1 − s))f (s, u(s))ds

0

Z t1

µ

+ t2 (Pα (t2 − s) − Pα (t1 − s))f (s, u(s))ds

t1 −δ t2

Z

µ

(t − tµ ) + Pα (t2 − s)f (s, u(s))ds 1

2

Z

µ +

t2

0

t1

t1

Pα (t1 − s)f (s, u(s))ds

= : I1 + I2 + I3 + I4 . Obviously, from Lemma 3.30, (H1) and Lemma 3.31, we have Z t1 −δ I1 ≤tµ2 kPα (t2 − s) − Pα (t1 − s)kB(L2 (Ω)) kf (s, u(s))kds 0

Cf tµ2 (t1 − δ)1−µ ≤ 1−µ →0,

sup

kPα (t2 − s) − Pα (t1 − s)kB(L2 (Ω)) r

s∈[0,t1 −δ]

as t2 → t1 .

From Lemma 3.30 and (H1), in the case δ → 0, we have Z t1 µ I2 ≤t2 kPα (t2 − s) − Pα (t1 − s)kB(L2 (Ω)) kf (s, u(s))kds t1 −δ

≤2C1 Cf tµ2

Z

t1

ku(s)kds t1 −δ

 2C1 Cf tµ2  1−µ t1 − (t1 − δ)1−µ r → 0. 1−µ As for I3 , one has Z t2 I3 ≤tµ2 kPα (t2 − s)kB(L2 (Ω)) kf (s, u(s))kds ≤

t1

Fractional Rayleigh-Stokes Equations

≤C1 Cf tµ2

Z

197

t2

ku(s)kds t1

≤

C1 Cf tµ2 1−µ )r → 0, − t1−µ (t 1 1−µ 2

as t2 → t1 .

And it is easy to see that I4 ≤(tµ2 − tµ1 )

t1

Z

kPα (t1 − s)kB(L2 (Ω)) kf (s, u(s))kds Z t1 ku(s)kds ≤C1 Cf (tµ2 − tµ1 ) 0

0

≤

C1 Cf t1−µ 1 1−µ

(tµ2 − tµ1 )r → 0,

as t2 → t1 .

Therefore, ktµ2 (Gα f )(t2 ) − tµ1 (Gα f )(t1 )k → 0,

as t2 → t1 ,

which implies that the set {tµ (Gα f )(t) : u ∈ RBr } is equicontinuous on [0, b]. t− From Lemma 3.31, we know that {Pα () 0 Pα (t − s)f (s, u(s))ds : u ∈ Br } is relatively compact, then we have

Z t−

Pα ()

P (t − s)f (s, u(s))ds − (G f )(t, u) α α

0

Z t

Z t−





Pα (t − s)f (s, u(s))ds Pα (t − s)f (s, u(s))ds + ≤kPα () − IkB(L2 (Ω))

t−

0


C1 Cf 1−µ C1 Cf (t − )1−µ r + t − (t − )1−µ r ≤kPα () − IkB(L2 (Ω)) 1−µ 1−µ → 0,

as  → 0.

Thus, the set {(Gα f )(t) : u ∈ Br } is relatively compact in C µ ([0, b]; L2 (Ω)). Therefore, the set {(T u)(t) : u ∈ Br } is relatively compact in C µ ([0, b]; L2 (Ω)). Claim II. T maps Br into Br . For any u ∈ Br , from the above discussion, we get that tµ (Qα f )(t) and tµ (Gα f )(t) is continuous for t ∈ [0, b]. It remains to show the continuity of tµ Sα (t)g. At first, from the expression of Sα (t) and Lemma 3.30, we can estimate kSα (t)gk as follows C1 α−1 kSα (t)gk ≤ t kgk, (3.90) C2 it yields that tµ kSα (t)gk ≤

C1 α−1+µ t kgk → 0, C2

as t → 0+ ,

which implies that limt→0+ tµ Sα (t)g exists and is finite. Thus, for t1 = 0, 0 < t2 ≤ b, we have ktµ2 Sα (t2 )g − tµ1 Sα (t1 )gk → 0,

as t2 → t1 .

198

Fractional Partial Differential Equations

From the expression of Kj (ξ), we can easy to check that Kj (ξ) ≤ For 0 < t1 < t2 ≤ b, we estimate Pα,j (t) in the following Z t2 Z ∞ e−ξt ξKj (ξ)dξdt |Pα,j (t2 ) − Pα,j (t1 )| = t1

1 πλj γξ α sin (απ) .

0

Z t2 Z ∞ 1 e−ξt ξ 1−α dξdt ≤ πλj γ sin (απ) t1 0 Γ(2 − α) = (tα−1 − t2α−1 ). (1 − α)πλj γ sin (απ) 1 By virtue of Lemma 3.30, (3.90) and combines with the above results, we have ktµ2 Sα (t2 )g − tµ1 Sα (t1 )gk ≤tµ2 kSα (t2 )g − Sα (t1 )gk + (tµ2 − tµ1 )kSα (t1 )gk ≤

C1 α−1 µ Γ(2 − α) (tα−1 − t2α−1 )kgk + t (t2 − tµ1 )kgk 1 C2 (1 − α)πγ sin (απ) C2 1

→ 0,

as t2 → t1 .

Thus we obtain that tµ Sα (t)g is continuous for t ∈ [0, b]. This implies that T u ∈ C µ ([0, b]; L2 (Ω)). For any u ∈ Br , from (3.87), (3.89) and (3.90), we have k(T u)(t)k ≤kSα (t)gk + k(Qα f )(t)k + k(Gα f )(t)k C1 Cf 1−µ C 2 Cf Γ(α)Γ(1 − µ) α−µ C1 α−1 b kukµ + t kukµ t kgk + 1 C2 C2 Γ(α + 1 − µ) 1−µ C1 C1 Cf 1−µ C 2 Cf Γ(α)Γ(1 − µ) α−µ ≤ tα−1 kgk + 1 b r+ t r, C2 C2 Γ(α + 1 − µ) 1−µ

≤

which implies that tµ k(T u)(t)k ≤

C 2 Cf Γ(α)Γ(1 − µ) α−µ µ C1 Cf C1 α−1+µ b rt + tr. t kgk + 1 C2 C2 Γ(α + 1 − µ) 1−µ

It yields that kT ukµ ≤

C1 α−1+µ C 2 Cf Γ(α)Γ(1 − µ) α C1 Cf b kgk + 1 b r+ br ≤ r. C2 C2 Γ(α + 1 − µ) 1−µ

Thus, we conclude that T maps Br into Br . Claim III. T is a continuous operator of u ∈ Br . From Lemma 3.32, it is easy to see that Qα is continuous in Br . In the sequel, we only need to prove the continuity of Gα . Indeed, let {un }∞ n=1 ⊂ Br and u ∈ Br , assume that limn→∞ un = u, from hypothesis (H1), for t ∈ (0, b], we have kf (t, un (t)) − f (t, u(t)))k → 0,

as n → ∞

and kf (s, un (s)) − f (s, u(s)))k ≤ 2kf (s, u(s)))k ≤ 2Cf s−µ kukµ ,

Fractional Rayleigh-Stokes Equations

199

it is easy to see that s−µ is integrable on (0, t], thus by the Lebesgue’s dominated convergence theorem, we have Z t kf (s, un (s)) − f (s, u(s)))kds → 0, as n → ∞. 0

On the other hand, we get Z t Z t kPα (t − s)(f (s, un (s)) − f (s, u(s)))kds ≤ C1 k(f (s, un (s)) − f (s, u(s)))kds, 0

0

which implies that Gα is continuous in Br . Thus T is continuous of u in Br . From the above arguments, we conclude that T is completely continuous, by applying the Schauder fixed point theorem, we conclude that the operator T has a fixed point on C µ ([0, b]; L2 (Ω)), which implies that the problem (3.80) has at least one mild solution. 3.6.4

Quasi-boundary Value Method

In this subsection, in view of inequality (3.90), one can check that Sα (·)g does not belong to L∞ (0, b; L2 (Ω)), and then the solution u of the problem (3.80) does not belong to L∞ (0, b; L2 (Ω)), it implies that the problem (3.80) is ill-posed in the sense of Hadamard. In order to solve this puzzle, the quasi-boundary value method is used to regularize the problem (3.80), and the corresponding regularized problem is given in the following:  α   ∂t v − (1 + γ∂t )∆v = f (t, x, v), x ∈ Ω, t ∈ (0, b), x ∈ ∂Ω, t ∈ (0, b),

v(t, x) = 0,

 

δ

(3.91)

x ∈ Ω,

v(b, x) + β(δ)v(0, x) = g (x),

where β(δ) > 0 is a regularization parameter and the noise data g δ satisfies kg δ − gk ≤ δ.

(3.92)

In what follows, we give the definition of mild solutions for the problem (3.91), and then, the well-posed results are obtained. Assume that uδβ is a solution of the problem (3.91), from [409, Definition 3.1], under the initial condition uδβ (0), uδβ can be described as ∞ ∞ Z t X X uδβ (t, x) = Pα,j (t)(uδβ (0), ϕj )ϕj (x) + Pα,j (t − s)(f (s, uδβ (s)), ϕj )dsϕj (x), j=1

j=1

0

(3.93) substituting t = b into (3.93), we have ∞ ∞ Z X X δ δ uβ (b, x) = Pα,j (b)(uβ (0), ϕj )ϕj (x) + j=1

j=1

b

Pα,j (b − s)(f (s, uδβ (s)), ϕj )dsϕj (x).

0

By virtue of uδβ (b, x) + β(δ)uδβ (0, x) = g δ (x) in (3.91), we obtain ! Z b 1 (uδβ (0), ϕj ) = (g δ , ϕj ) − Pα,j (b − s)(f (s, uδβ (s)), ϕj )ds , β(δ) + Pα,j (b) 0 (3.94)

200

Fractional Partial Differential Equations

which implies ∞ X uδβ (t, x) =

! Z b Pα,j (t) δ δ (g , ϕj ) − Pα,j (b − s)(f (s, uβ (s)), ϕj )ds ϕj (x) β(δ) + Pα,j (b) 0 j=1 ∞ Z t X + Pα,j (t − s)(f (s, uδβ (s)), ϕj )dsϕj (x). j=1

0

(3.95) Let Sαβ (t)v =

∞ X j=1

Pα,j (t) (v, ϕj )ϕj , β(δ) + Pα,j (b)

for v ∈ L2 (Ω),

stand for uδβ (t, ·), then (3.95) can be rewritten as Z t Z b Pα (t − s)f (s, uδβ (s))ds. Sαβ (t) ◦ Pα (b − s)f (s, uδβ (s))ds + uδβ (t) = Sαβ (t)g δ −

and let

uδβ (t)

0

0

(3.96) According to the above discussion, we introduce the following definition of mild solutions of the problem (3.91). Definition 3.11. A function uδβ is called a mild solution of the problem (3.91), if uδβ belongs to L∞ (0, b; L2 (Ω)) and it satisfies the integral equation (3.96). Before giving our main results, we provide the following constraint of f . 0

(H2) For any u, v ∈ L2 (Ω), there exists a constant Cf > 0 such that 0

kf (t, u) − f (t, v)k ≤ Cf ku − vk,

for t ∈ (0, b].

Theorem 3.15. Assume that (H2) holds. For every δ > 0 and g δ ∈ L2 (Ω), the problem (3.91) exists a unique solution in L∞ (0, b; L2 (Ω)) provided 0

C12 Cf bα 0 + C1 Cf b < 1. C2 α In addition, the solution depends continuously on g δ .

(3.97)

Proof. For any v ∈ L∞ (0, b; L2 (Ω)), let Z b Z t (T v)(t) = Sαβ (t)g δ − Sαβ (t) ◦ Pα (b − s)f (s, v(s))ds + Pα (t − s)f (s, v(s))ds. 0

0

Clearly, the operator T has a unique fixed point which means that the problem (3.91) has a unique solution. At the beginning, we prove that T maps L∞ (0, b; L2 (Ω)) into L∞ (0, b; L2 (Ω)). For any v ∈ L∞ (0, b; L2 (Ω)), from Lemma 3.30 and (H2), we have 0 Z Z t C12 Cf b 0 C1 k(T v)(t)k ≤ kg δ k + (b − s)α−1 kv(s)kds + C1 Cf kv(s)kds β(δ) C2 0 0 0

C12 Cf bα 0 C1 ≤ kg δ k + kvk∞ + C1 Cf tkvk∞ , β(δ) C2 α

Fractional Rayleigh-Stokes Equations

201

which implies that kT vk∞

C1 ≤ kg δ k + β(δ)

! 0 C12 Cf bα 0 + C1 Cf b kvk∞ . C2 α

Thus we get that T maps L∞ (0, b; L2 (Ω)) into itself. For any v, w ∈ L∞ (0, b; L2 (Ω)), in virtue of (H2) and (3.97), we prove that T is a contraction operator. k(T v)(t) − (T w)(t)k

Z

b

Sαβ (t) ◦ Pα (b − s) (f (s, v(s)) − f (s, w(s))) ds ≤

0

Z t

Pα (t − s) (f (s, v(s)) − f (s, w(s))) ds +

0

0

C12 Cf ≤ C2 0

b

Z

0

(b − s)α−1 kv(s) − w(s)kds + C1 Cf

0 α

Z

t

kv(s) − w(s)kds 0

C12 Cf b 0 kv − wk∞ + C1 Cf tkv − wk∞ , ≤ C2 α thus we have kT v − T wk∞ ≤ 0

C 2 C bα

! 0 C12 Cf bα 0 + C1 Cf b kv − wk∞ . C2 α

0

From the fact that 1C2fα +C1 Cf b < 1, by using the contraction mapping principle, we conclude that the problem (3.91) exists a unique solution in L∞ (0, b; L2 (Ω)). In the sequel, we prove the continuous dependence of the solution uδβ on the data g δ . Denote uδβ and u ˜δβ are two regularization solutions corresponding to the δ δ conditions g and g˜ , respectively. From Lemma 3.30 and (H2), for every δ > 0, we have kuδβ (t) − u ˜δβ (t)k

Z

b

β δ δ − g˜ )k + Sα (t) ◦ Pα (b − s)(f (s, uβ (s)) − f (s, u ˜β (s)))ds

0

Z t

+ Pα (t − s)(f (s, uδβ (s)) − f (s, u ˜δβ (s)))ds

0 0 Z C12 Cf b C1 kg δ − g˜δ k + (b − s)α−1 kuδβ (s) − u ˜δβ (s)kds ≤ β(δ) C2 0 Z t 0 + C1 Cf kuδβ (s) − u ˜δβ (s)kds 0 ! 0 C12 Cf bα 0 C1 δ δ ≤ kg − g˜ k + + C1 Cf b kuδβ − u ˜δβ k∞ , β(δ) C2 α ≤kSαβ (t)(g δ

δ

202

Fractional Partial Differential Equations

which implies that !−1 0 C12 Cf bα 0 C1 1− − C1 Cf b kg δ − g˜δ k. C2 α β(δ)

kuδβ − u ˜δβ k∞ ≤ Particularly, kuδβ (0) − u ˜δβ (0)k

Z

b

Sαβ (0) ◦ Pα (b − s)(f (s, uδβ (s)) − f (s, u ˜δβ (s)))ds ≤kSαβ (0)(g δ − g˜δ )k +

0 0

C 1 C f bα δ 1 kg δ − g˜δ k + kuβ − u ˜δβ k∞ β(δ) C2 α   !−1 0 0 C12 Cf bα C12 Cf bα 0 1   kg δ − g˜δ k. 1+ 1− ≤ − C1 Cf b β(δ) C2 α C2 α

≤

Thus we get the desired conclusion. The proof is completed. Let ( Cβ =

l1 (θ, C2 )β θ (δ),

0 < θ < 1,

l2 (θ, C2 , λ1 )β(δ),

θ ≥ 1,

where l1 , l2 are defined in Lemma 3.29. In the sequel, the convergence analysis result is given under a priori condition. Theorem 3.16. Assume the conditions in Theorem 3.15 hold, and the regularization parameter β(δ) satisfies the following condition: δ = lim β(δ) = 0. δ→0 δ→0 β(δ)

(3.98)

lim

For θ > 0, assume that u(0) ∈ Hθ (Ω). Then 0

kuδβ

C12 Cf bα 0 1− − C1 Cf b C2 α

− uk∞ ≤

!−1 

 C1 δ + Cβ ku(0)kHθ (Ω) . β(δ)

Proof. From Theorem 3.15, we know that there exists a unique regularization solution uδβ ∈ L∞ (0, b; L2 (Ω)), assume that u is a solution of the problem (3.80), for every δ > 0, from (H2) and (3.92), it is not difficult to check that uδβ − u ∈ L∞ (0, b; L2 (Ω)). In fact, denote Z b Z t uβ (t) = Sαβ (t)g − Sαβ (t) ◦ Pα (b − s)f (s, u(s))ds + Pα (t − s)f (s, u(s))ds. 0

0

It is easy to see that kuδβ (t) − u(t)k ≤ kuδβ (t) − uβ (t)k + kuβ (t) − u(t)k.

Fractional Rayleigh-Stokes Equations

203

In virtue of Lemma 3.30 and (H2), we estimate kuδβ (t) − uβ (t)k in the following: kuδβ (t) − uβ (t)k

Z

b

β δ Sα (t) ◦ Pα (b − s)(f (s, uβ (s)) − f (s, u(s)))ds − g)k +

0

Z t

Pα (t − s)(f (s, uδβ (s)) − f (s, u(s)))ds +

0 0 Z C12 Cf b C1 ≤ (b − s)α−1 kuδβ (s) − u(s)kds kg δ − gk + β(δ) C2 0 Z t 0 kuδβ (s) − u(s)kds + C1 Cf 0 ! 0 C12 Cf bα 0 C1 δ + + C1 Cf b kuδβ − uk∞ . ≤ β(δ) C2 α ≤kSαβ (t)(g δ

In the sequel, we estimate kuβ (t) − u(t)k. From Lemma 3.29 and Lemma 3.30, we have

X  

∞ Pα,j (t) Pα,j (t) − kuβ (t) − u(t)k =

j=1 β(δ) + Pα,j (b) Pα,j (b)

! Z b

× (g, ϕj ) − Pα,j (b − s)(f (s, u(s)), ϕj )ds ϕj (x)

0   12 2 ∞  X β(δ)λj ≤ |(u(0), ϕj )|2  β(δ)λ + C j 2 j=1 ≤Cβ ku(0)kHθ (Ω) , which implies that kuβ (t) − u(t)k ≤ Cβ ku(0)kHθ (Ω) . Based on the previous discussion, we obtain ! 0 C12 Cf bα 0 C1 δ δ kuβ − uk∞ ≤ + + C1 Cf b kuδβ − uk∞ + Cβ ku(0)kHθ (Ω) , β(δ) C2 α from (3.97), we conclude that 0

kuδβ

− uk∞ ≤

C12 Cf bα 0 − C1 Cf b 1− C2 α

!−1 

 C1 δ + Cβ ku(0)kHθ (Ω) . β(δ)

Thus we get the desired conclusion and the convergence result is also obtained. Moreover, from (3.98), we can easy to see that uδβ → u in L∞ (0, b; L2 (Ω)) as the regularization parameter δ → 0. The proof is completed.

204

3.7

Fractional Partial Differential Equations

Notes and Remarks

The results in Section 3.1 due to Zhou and Wang [409]. The main results in Section 3.2 are adopted from He, Zhou and Peng [163]. Section 3.3 is taken from Wang, Alsaedi, Ahmad and Zhou [346]. The contents in Section 3.4 are taken from Wang, Zhou, Alsaedi and Ahmad [348]. The results in Section 3.5 due to Peng and Zhou [296]. Section 3.6 is from Wang, Zhou and He [347].

Chapter 4

Fractional Fokker-Planck Equations

In this chapter, we study the time-fractional Fokker-Planck equations which can be used to describe the subdiffusion in an external time and space-dependent force field F (t, x). In Section 4.1, we present some results on existence and uniqueness of mild solutions allowing the “working space” that may have low regularity. Secondly, we analyze the relationship between “working space” and the value range of α when investigating the problem of classical solutions. Finally, by constructing a suitable weighted H¨ older continuous function space, the existence of classical solutions is derived without the restriction on α ∈ ( 21 , 1). In Section 4.2, a time-fractional Fokker-Planck initial-boundary value problem is considered, the spatial domain Ω ⊂ Rd , where d ≥ 1, has a smooth boundary. Existence, uniqueness and regularity of a mild solution u are proved under the hypothesis that the initial data u0 lies in L2 (Ω). For 1/2 < α < 1 and u0 ∈ H 2 (Ω) ∩ H01 (Ω), it is shown that u becomes a classical solution of the problem. Estimates of time derivatives of the classical solution are derived. 4.1 4.1.1

Operator Method Introduction

In the past few decades, the researches on anomalous diffusion problems with external force fields have been extensively investigated, and then the corresponding mathematical and physical model leads to the time-fractional Fokker-Planck equations. Metzler et al. in [264,265] showed how fractional Fokker-Planck equations for the description of anomalous diffusion in external fields F (x), can be derived from the generalized master equation, which shows that the probability density u(t, x) for a particle to be at time t and position x obeys a time-fractional Fokker-Planck equation of the form ∂t u − ∂t1−α (κα uxx − (F u)x ) = 0, where κα > 0 is a constant, ∂t = ∂∂t and ∂t1−α is the Riemann-Liouville fractional partial derivative of order 1 − α ∈ (0, 1) defined by ∂ ∂t1−α u(t, x) = 0 Dt−α u(t, x), t > 0, ∂t 205

206

Fractional Partial Differential Equations

with 0 Dt−α the Riemann-Liouville fractional integral given by Z t 1 −α D u(t, x) = (t − s)α−1 u(s, x)ds, t > 0, 0 t Γ(α) 0 where Γ(·) is the Gamma function. It is worth mentioning that Barkai et al. [34] generalized the continuous time random walk theory to include the effect of space dependent jump probabilities. When the mean waiting time diverges they also derived a fractional Fokker-Planck equation. Subsequently, Henry et al. [166] considered the more general case when F = F (t, x), which was formally equivalent to the subordinated stochastic Langevin equations for time- and space-dependent forces. In this section, we study the following initial-boundary value problems of timefractional Fokker-Planck equations in an open bounded domain Ω ⊂ Rd (d ≥ 1) with C 2 boundary:  1−α 1−α   ∂t u(t, x) − ∇ · (∂t κα ∇u − F ∂t u)(t, x) = f (t, x), (t, x) ∈ (0, T ) × Ω, u(t, x) = 0, (t, x) ∈ (0, T ) × ∂Ω,   u(0, x) = u0 (x), x ∈ Ω. (4.1) It is generally known that the equation (4.1) would resolve itself into time-fractional diffusion equations when F (t, x) = 0, which has been the focus of many studies due to its significant application in subdiffusive model of anomalous diffusion processes. Now, the researches on time-fractional Fokker-Planck equations were mainly focused on numerical analysis, see [173, 218]. However, its basic theoretical works are far from sufficient. McLean et al. [260] investigated the initial boundary value problem of a class of linear reaction convection-diffusion equations including the equation (4.1), and the well-posedness of weak solutions were obtained by novel energy methods in combination with a fractional Gronwall’s inequality and properties of fractional integrals. Further, McLean et al. [261] obtained the regularity theory of weak solutions of a linear time-fractional, advection-diffusion-reaction equation. Recently, Le et al. [219] used the Galerkin approximation method to analyze the existence and uniqueness of mild solutions of the equations (4.1), and then the existence and regular estimates of the classical solution were obtained when α ∈ ( 21 , 1). Compared with the diversity of the methods to time-fractional diffusion equations, the current methods of time-fractional Fokker-Planck equations are relatively single, which are basically based on the Galerkin approximation and energy estimates, combined with fractional calculus theory. In fact, the research ideas of time-fractional diffusion equations can be basically divided into two aspects: the one is to deduce the form of mild solutions, and then analyze the properties of solutions via estimating the solution operator; the other is to establish energy estimates of approximate solutions, and then address the procedure of taking limits to obtain the properties of solutions. However, the researchers only use the second idea to make qualitative analysis of time-fractional

Fractional Fokker-Planck Equations

207

Fokker-Planck equations, which is insightful work. It’s such a pity that the existence of classical solutions has remained elusive when α ∈ (0, 21 ], the fundamental reason is a stubborn “working space”. Considering the advantage of the first idea that the “working space” can be selected diversely, we first deduce the form of mild solutions to the objective problems expressed by Mittag-Leffler functions, and then use the properties of Mittag-Leffler functions to discuss that of solutions. The main novelties of our work are as follows. • The representation of mild solutions without fractional derivative operators is given, which allows the “working space” of mild solutions that may have low regularity. • The influence of the “working space” on the value range of α is analyzed when considering the existence of classical solutions. • The existence of classical solutions is obtained without any restriction on α by constructing a suitable weighted H¨ older continuous function space. The remainder of the section is organized as follows. In the next subsection, we introduce some notations and lemmas that will be used throughout the section. In Subsection 4.1.3, firstly we give some operators and secondly we deduce the representation of mild solutions without fractional derivative operators. In Subsection 4.1.4, we derive the well-posedness and space-regularity of mild solutions. The well-posedness of classical solutions under the restriction on α ∈ ( 12 , 1) is derived in Subsection 4.1.5. Finally, the existence and uniqueness of classical solutions are also obtained in a weighted H¨ older continuous function space when α ∈ (0, 1). 4.1.2

Preliminaries

Throughout the section, we assume that the external force field F = (F1 , ..., Fd )T , F ∈ W 1,1 (0, T ; L2 (Ω)) denotes the space of all integrable functions Fi : (0, T ) → L2 (Ω) such that Fi0 ∈ L1 (0, T ; L2 (Ω)). F ∈ W 1,∞ (0, T ; Ω) denotes the space of all ¯ with the norm functions whose divergence ∇ · F is continuous on [0, T ] × Ω kF kW 1,∞ := kF k∞ +

|∇ · F (t, x)|,

max (t,x)∈[0,T ]×Ω

where kF k∞ := max

max

1≤i≤d (t,x)∈[0,T ]×Ω

|Fi (t, x)|.

The assumption on F will be given later. Let A = −∆ and {λk } be the set of eigenvalues of the operator A with a homogeneous boundary condition, let {ek } denote the complete orthonormal system of eigenfunctions which forms an orthogonal basis of L2 (Ω) such that Aek = λk ek , x ∈ Ω; ek |∂Ω = 0, k = 1, 2, ..., where 0 < λ1 ≤ λ2 ≤ ... ≤ λk ≤ ... and limk→∞ λk = +∞.

208

Fractional Partial Differential Equations

For all θ ≥ 0, the fractional power operator Aθ possesses the following representation: ∞ ∞ n o X X 2 Aθ v = λθk (v, ek )ek , D(Aθ ) = v ∈ L2 (Ω) : λ2θ |(v, e )| < ∞ , k k k=1

k=1

and that D(Aθ ) is a Hilbert space with the norm: kvkD(Aθ ) =

∞ X

2 λ2θ k |(v, ek )|

 21

.

k=1

By duality, we can also set H −1 (Ω) = (H01 (Ω))∗ , D(A−θ ) = (D(Aθ ))∗ . Then D(A−θ ) is a Hilbert space endowed with the norm kvkD(A−θ ) = P  21 ∞ −2θ |(v, ek )|2 . We have D(Aθ ) ⊂ H 2θ (Ω) for θ > 0. For the details k=1 λk of the Sobolev spaces with fractional powers H 2θ (Ω), see Fujiwara [129] for exam1 ple. In particular, D(A 2 ) = H01 (Ω) and D(A) = H01 (Ω) ∩ H 2 (Ω), see Sakamoto and Yamamoto [313] and references therein. Let us recall the Mittag-Leffler function Eα,β (z) =

∞ X k=0

zk , β ∈ R, z ∈ C. Γ(αk + β)

Then there exists a positive constant Mα such that Eα,β (−t) ≤

Mα , for t ≥ 0. 1+t

(4.2)

In particular, when β = 1 it is so called one parameter the Mittag-Leffler function and that for α = β = 1 it is the exponential function. Lemma 4.1. [217] If v : [0, T ] → H 1 (Ω) and F ∈ W 1,∞ (0, T ; Ω), then k∇ · (F (t)v(t))k ≤ kF kW 1,∞ kv(t)kH 1 (Ω) . Let X be Banach space. The space of H¨ older continuous functions on [0, T ] with the exponent θ ∈ (0, 1) is denoted by C θ ([0, T ]; X) which has the representation form C θ ([0, T ]; X) = {u ∈ C([0, T ]; X) :

sup 0≤t1 α2 , we get that Z t+h 1 1 I4 (t, h) ≤ kA 2 Pα (t + h − s)A− 2 ∇ · (F (s)w(s))kds t Z t 1 1 1 + k[A 2 Pα (t + h − s) − A 2 Pα (t − s)]A− 2 ∇ · (F (s)w(s))kds 0

Z

−1

t+h

≤CMα κα 2

t − 12 2CMα κα

+

2−α

α

(t + h − s) 2 −1 kF (s)w(s)kds Z

t

α

0

αp−2 −1 ≤CMα κα 2 Bα,p h 2p (1

where Bα,p =



2(p−1) αp−2

α

[(t − s) 2 −1 − (t + h − s) 2 −1 ]kF (s)w(s)kds +

2 )kF kLp (0,T ;L2 (Ω)) kwkC([0,T ];L2 (Ω)) , 2−α

 p−1 p

and we have used the inequality Z t  p−1 p p α α [(t − s) 2 −1 − (t + h − s) 2 −1 ] p−1 ds 0

Z ≤

t

 p−1 p p p α α [(t − s)( 2 −1) p−1 − (t + h − s)( 2 −1) p−1 ]ds

0 αp−2

αp−2

αp−2

=Bα,p [h 2(p−1) + t 2(p−1) − (t + h) 2(p−1) ]

p−1 p

.

Therefore, combined with the estimates of Ii (t, h), i = 1, ..., 4, we conclude that kHw(t + h) − Hw(t)k → 0 as h → 0+ . Similar arguments can show that kHw(t + h) − Hw(t)k → 0 as h → 0− . Consequently, we obtain that Hw ∈ C([0, T ]; L2 (Ω)) for any w ∈ C([0, T ]; L2 (Ω)). Step 2. H has a unique fixed point. Indeed, for any w1 , w2 ∈ C([0, T ]; L2 (Ω)), we have kHw1 (t) − Hw2 (t)k

Z t

0

Q (t − s)∇ · (F (s)(w (s) − w (s))ds ≤ α 1 2

0

Z t



+ Pα (t − s)∇ · (F (s)(w1 (s) − w2 (s))ds

0

Z

−1 ≤CMα κα 2 [

t

α

(t − s) 2 kF 0 (s)(w1 (s) − w2 (s))kds

0

Z +

t

α

(t − s) 2 −1 kF (s)(w1 (s) − w2 (s))kds]

0 −1

αp−2

α

≤CMα κα 2 kw1 − w2 kC([0,T ];L2 (Ω)) [t 2 kF 0 kL1 (0,T ;L2 (Ω)) + t 2(p−1) kF kLp (0,T ;L2 (Ω)) ]. Then one can choose a T1 ∈ (0, T ) small enough which ensures that −1

α

αp−2

CMα κα 2 [T12 kF 0 kL1 (0,T ;L2 (Ω)) + T12(p−1) kF kLp (0,T ;L2 (Ω)) ] < 1.

Fractional Fokker-Planck Equations

215

It follows that H has a fixed point, thus the equation (4.10) has a unique solution in C([0, T1 ]; L2 (Ω)). Now, we will deal with the continuation of the solution to the interval [0, T ]. Let us make the assumption that we have obtained the solution u ¯ of the equation (4.10) on the interval [0, Tl ] for some Tl > 0. For Tl+1 > Tl , it is tempting to prove the solution on [Tl , Tl+1 ]. To do this, we introduce the complete space  ¯T = u ∈ C([0, Tl+1 ]; L2 (Ω)) : u(t) = u E ¯(t) for t ∈ [0, Tl ] , l+1

¯T , then u ∈ Let u ∈ E l+1 ¯T . C([0, Tl+1 ]; L2 (Ω)). According to the previous proof, we have that Hu ∈ E l+1 ¯T Next, we will show that the operator H is also a strict contraction on E l+1 when Tl+1 − Tl is sufficient small. We shall rewrite H as the following form: with the distance kukE¯T

l+1

= kukC([Tl ,Tl+1 ];L2 (Ω)) .

Hw(t) Tl

Z =Qα (t)u0 +

Qα (t − s)[f (s)+∇ · (F 0 (s)w(s))]ds

0

Z

Tl

−

Z

t

Pα (t − s)∇ · (F (s)w(s))ds + 0

Z

Qα (t − s)[f (s) + ∇ · (F 0 (s)w(s))]ds

Tl t

−

Pα (t − s)∇ · (F (s)w(s))ds. Tl

¯T , we have For u1 , u2 ∈ E l+1 Z

t

Qα (t − s)∇ · (F 0 (s)(u1 (s) − u2 (s)))ds

Hu1 (t) − Hu2 (t) = Tl

Z

t

−

Pα (t − s)∇ · (F (s)(u1 (s) − u2 (s)))ds. Tl

This follows from Lemma 4.2 that kHu1 − Hu2 kC([Tl ,Tl+1 ];L2 (Ω)) −1

αp−2

α

≤BF CMα κα 2 ku1 − u2 kC([Tl ,Tl+1 ];L2 (Ω)) [(Tl+1 − Tl ) 2 + (Tl+1 − Tl ) 2(p−1) ], where BF = sup{kF 0 kL1 (0,T ;L2 (Ω)) , kF kLp (0,T ;L2 (Ω)) }. Therefore kHu1 − Hu2 kE¯T

2−α

l+1

−1

≤ BF CMα (1 + T 2(p−1) )κα 2 ku1 − u2 kE¯T

αp−2

l+1

(Tl+1 − Tl ) 2(p−1) .

Moreover, we can choose one    − 2(p−1) 2−α αp−2 − 21 2(p−1) Tl+1 ∈ Tl , Tl + BF CMα κα (1 + T ) , which ensures that −1

2−α

αp−2

0 < BF CMα κα 2 (1 + T 2(p−1) )(Tl+1 − Tl ) 2(p−1) < 1. ¯T , this also shows that the Hence, the operator H is a strict contraction on E l+1 equation (4.10) has a unique solution on the interval [Tl , Tl+1 ]. We proceed to repeat the process on the intervals [Tl+1 , Tl+2 ], ..., until the equation (4.10) has a unique solution on the interval [0, T ]. The claim then follows.

216

Fractional Partial Differential Equations

We now show how to transform the solution w of the integral equation (4.10) to the mild solution of the equation (4.1) as long as the former exists. Theorem 4.1. Under the conditions of Lemma 4.3, we proceed to obtain that w ∈ L2 (0, T ; H01 (Ω)) and the equation (4.1) has a unique mild solution u ∈ L2 (0, T ; H −1 (Ω)), which satisfies Z t Z t u(t) =Sα (t)u0 + Sα (t − s)f (s)ds+ Sα (t − s)∇ · (F 0 (s)0 Ds−α u(s))ds 0 0 (4.15) Z t −α 0 − ∇ · (F (t)0 Dt u(t)) − Sα (t − s)∇ · (F (s)0 Ds−α u(s))ds. 0

Proof. First, we denote t

Z

Pα (t − s)∇ · (F (s)w(s))ds.

H(t) = 0

d Using the Young’s inequality of the convolution and dt Eα,1 (−λtα ) α−1 α −λt Eα,α (−λt ) for λ > 0, it follows that

Z t

2 1

α−1 α −2

λk (t − s) Eα,α (−λk κα (t − s) )λk (∇ · (F (s)w(s)), ek )ds

0

1 ≤ 2 κα =

=

L2 (0,T )

Z

T

0

 Z −1 2 λk |(∇ · (F (s)w(s)), ek )| ds

T α−1

|λk κα s

2 Eα,α (−λk κα s )|ds α

0

T

Z

1 [1 − Eα,1 (−λk κα T α )]2 κ2α

0

−α 2 λ−1 k |(∇ · (F (s)0 Ds u(s)), ek )| ds,

(4.16) which yields kHk2 2 1 L (0,T ;D(A 2 ))

Z t

2

− 12

= APα (t − s)A ∇ · (F (s)w(s))ds

2 0

≤

1 κ2α

L (0,T ;L2 (Ω))

Z 0

∞ T X

2 λ−1 k |(∇ · (F (s)w(s)), ek )| ds

k=1

C2 ≤ 2 kF k2L2 (0,T ;L2 (Ω)) kwk2C([0,T ];L2 (Ω)) < ∞. κα Moreover, in view of Lemma 4.2, we have kw(t)kH01 (Ω)   Z t Z t α α − 21 − 21 0 2 2 (t − s) kf (s)kds + κα kF (s)w(s)kds ≤Mα κα t ku0 k + 0

0

1 + kH(t)k D(A 2 )   α α − 12 − 12 0 2 2 ≤Mα κα t ku0 k + t kf kL1 (0,T ;L2 (Ω)) + κα kF kL1 (0,T ;L2 (Ω)) kwkC([0,T ];L2 (Ω))

+ kH(t)k

1

D(A 2 )

.

Fractional Fokker-Planck Equations

217

Therefore, w ∈ L2 (0, T ; H01 (Ω)). We can take the Riemann-Liouville fractional derivative of order α of the equation (4.10) and invoke (4.3), it holds that Z t Z t α ∂t w(t) =Sα (t)u0 + Sα (t − s)f (s)ds+ Sα (t − s)∇ · (F 0 (s)w(s))ds 0 0 Z t Sα0 (t − s)∇ · (F (s)w(s))ds, − ∇ · (F (t)w(t)) − 0

where we have used the following formula Z t  Pα (t − s)∇ · (F (s)w(s))ds ∂tα 0 Z s Z t 1 ∂ = Pα (s − τ )∇ · (F (τ )w(τ ))dτ ds (t − s)−α Γ(1 − α) ∂t 0 0 Z tZ t ∂ 1 (t − s)−α Pα (s − τ )∇ · (F (τ )w(τ ))dsdτ = Γ(1 − α) ∂t 0 τ Z t ∂ = Sα (t − s)∇ · (F (s)w(s))ds ∂t 0 Z t =∇ · (F (t)w(t)) + Sα0 (t − s)∇ · (F (s)w(s))ds, 0

where Sα0 (t)v =

∞ X

−λk κα tα−1 Eα,α (−λk κα tα )(v, ek )ek (x).

(4.17)

k=1

It follows from w = 0 Dt−α u and Lemma 4.2 that (4.15) holds and we have Z t Z t ku(t)kH −1 (Ω) ≤Mα ku0 k + Mα kf (s)kds + Mα kF 0 (s)0 Ds−α u(s)kds 0

+ +

0 −α kF (t)kL2 (Ω) k0 Dt u(t))kL2 (Ω)

Z t

−α − 12 0

A Sα (t − s)∇ · (F (s)0 Ds u(s))ds

.

0

In view of (4.16), it is easy to show that u ∈ L2 (0, T ; H −1 (Ω)). Next we verify under what conditions mild solutions may have improved space regularity. Theorem 4.2. (i) Suppose that F ∈ W 1,1 (0, T ; L2 (Ω)) ∩ L∞ (0, T ; H01 (Ω)). Then the mild solution u ∈ L2 (0, T ; L2 (Ω)) and 0 Dt−α u ∈ C([0, T ]; H01 (Ω)). (ii) Suppose that F ∈ W 1,1 (0, T ; L2 (Ω)) ∩ W 1,∞ (0, T ; Ω). Then 0 Dt−α u ∈ L2 (0, T ; H 2 (Ω)).

218

Fractional Partial Differential Equations

Proof. The proof of (i). It is obvious that the equation (4.10) has a unique mild solution w ∈ L2 (0, T ; H01 (Ω)) ∩ C([0, T ]; L2 (Ω)) under the assumptions of F ∈ W 1,1 (0, T ; L2 (Ω)) ∩ L∞ (0, T ; H01 (Ω)). Moreover,  α α − 12 1 k0 Dt−α u(t)k ≤M κ t 2 ku0 k + t 2 kf kL1 (0,T ;L2 (Ω)) α α D(A 2 )  −1 + κα 2 kF 0 kL1 (0,T ;L2 (Ω)) k0 Dt−α ukC([0,T ];L2 (Ω)) −1

Z

+ Mα κα 2 0

t

α

(t − s) 2 −1 kF (s)kH 1 (Ω) k0 Ds−α u(s))kH01 (Ω) ds.

It deduces from the generalized Gronwall’s inequality that  α α −1 k0 Dt−α u(t)kH01 (Ω) ≤Mα κα 2 t 2 ku0 k + t 2 kf kL1 (0,T ;L2 (Ω))  α −1 + κα 2 kF 0 kL1 (0,T ;L2 (Ω)) k0 Dt−α ukC([0,T ];L2 (Ω)) E α2 ,1 (µt 2 ), −1

where µ = Mα κα 2 kF kL∞ (0,T ;H 1 (Ω)) , and then 0 Dt−α u ∈ L∞ (0, T ; H01 (Ω)). Furthermore, similar to the proof of Theorem 4.1, we also easily know that 0 Dt−α u ∈ C([0, T ]; H01 (Ω)). From Lemma 4.2, it follows that Z t Z t α − 21 (t − s)− 2 kF 0 (s)0 Ds−α u(s)kds ku(t)k ≤Mα ku0 k + Mα kf (s)kds + Mα κα 0 0 Z t + kF (t)kH 1 (Ω) k0 Dt−α u(t))kH01 (Ω) + k Sα0 (t − s)∇ · (F (s)0 Ds−α u(s))dsk. 0

The similar arguments to (4.16) lead to the property that

Z t

2

0 −α

S (t − s)∇ · (F (s) D u(s))ds 0 s α

0

≤

1 κ2α

L2 (0,T ;L2 (Ω))

Z 0

∞ T X

|(∇ · (F (s)0 Ds−α u(s)), ek )|2 ds

k=1

1 ≤ 2 k∇ · (F (·)0 Dt−α u(·))k2L2 (0,T ;L2 (Ω)) κα 1 ≤ 2 kF kL2 (0,T ;H 1 (Ω)) k0 Dt−α ukC([0,T ];H01 (Ω)) . κα Then we can easily derive u ∈ L2 (0, T ; L2 (Ω)). The proof of (ii). We immediate calculate that Z t kw(t)kD(A) ≤kQα (t)u0 kD(A) + kQα (t − s)f (s)kD(A) ds 0 Z t + kQα (t − s)∇ · (F 0 (s)w(s))kD(A) ds 0

Fractional Fokker-Planck Equations

219

Z t

Pα (t − s)∇ · (F (s)w(s))ds +

0

D(A)

=:H1 (t) + H2 (t). From Lemma 4.1 and Lemma 4.2, we can easily know that Z t H1 (t) =kQα (t)u0 kD(A) + kQα (t − s)f (s)kD(A) ds 0 Z t + kQα (t − s)∇ · (F 0 (s)w(s))kD(A) ds 0   Z t Z t −1 ≤Mα κα ku0 k + kf (s)kds + kF (s)kW 1,∞ kw(s)kH 1 (Ω) ds . 0

0

2

It follows that H1 (·) ∈ L (0, T ; D(A)). On the other hand, in view of (4.16) and (4.17), we obtain that H2 (·) ∈ L2 (0, T ; D(A)). Therefore the desired conclusion follows. Remark 4.1. In fact, the result (i) of Theorem 4.2 is also obtained in [260, Theorem 4.1 and Theorem 4.2], which used the energy arguments combined with the generalized Gronwall’s inequality. 4.1.5

Classical Solutions in Case of α ∈ ( 21 , 1)

In this subsection, we prove the existence of classical solutions for the equation (4.1). First, we shall show that the equation (4.10) has a unique solution which belongs to AC([0, T ]; L2 (Ω)). To do this, we introduce the space  XT = w ∈ C([0, T ]; H01 (Ω)) : w0 ∈ L1 (0, T ; L2 (Ω)) , equipped with the norm kwkXT = kwkC([0,T ];H01 (Ω)) + kw0 kL1 (0,T ;L2 (Ω)) . The following result shows that the solution of the integral equation (4.10) belongs to XT . Lemma 4.4. Let u0 ∈ L2 (Ω) and f ∈ L1 (0, T ; L2 (Ω)). Suppose that F ∈ W 1,1 (0, T ; L2 (Ω)) ∩ W 1,∞ (0, T ; Ω). Then the equation (4.10) has a unique solution w ∈ XT . Proof. Consider the operator H : XT → XT given as (4.11). Then it is welldefined. In fact, let w ∈ XT , then Hw ∈ C([0, T ]; H01 (Ω)). Further, we immediately take the first derivative of Hw with respect to t and use (4.3) to obtain Z t  (Hw)0 (t) =Pα (t)u0 + ∂t Qα (t − s)f (s)ds 0 Z t  + ∂t Qα (t − s)∇ · (F 0 (s)w(s))ds 0 Z t  − ∂t Pα (t − s)∇ · (F (s)w(s))ds . 0

220

Fractional Partial Differential Equations

Then we can infer from Qα (0) = 0 that  Z t Z t Pα (t − s)f (s)ds. Qα (t − s)f (s)ds = ∂t 0

0

Besides, in view of w(0) = 0, it holds that  Z t Pα (t − s)∇ · (F (s)w(s))ds ∂t 0

Z

t

Pα (s)∂t [∇ · (F (t − s)w(t − s))]ds =Pα (t)∇ · (F (0)w(0)) + 0 Z t = Pα (t − s)∇ · (F 0 (s)w(s) + F (s)w0 (s))ds. 0

Thus (Hw)0 (t) =Pα (t)u0 +

Z

t

Z Pα (t − s)f (s)ds −

0

t

Pα (t − s)∇ · (F (s)w0 (s))ds.

0

Using Lemma 4.2 again, one can have  Z t k(Hw)0 (t)k ≤Mα tα−1 ku0 k + (t − s)α−1 kf (s)kds 0  Z t α −1 0 2 + kF kW 1,∞ (t − s) kw (s)kds .

(4.18)

0

It follows from the Young’s inequality of the convolution that (Hw)0 ∈ L1 (0, T ; L2 (Ω)). Using the Banach fixed point method and the continuation of the solution as we proved in Lemma 4.3, one also obtains that the equation (4.10) has a unique solution in XT . Our next result shows the existence of the mild solution of the equation (4.1). Theorem 4.3. Let u0 ∈ L2 (Ω) and f ∈ L1 (0, T ; L2 (Ω)). Suppose that F ∈ W 1,1 (0, T ; L2 (Ω)) ∩ W 1,∞ (0, T ; Ω). Then the equation (4.1) has the mild solution u ∈ L2 (0, T ; L2 (Ω)) ∩ L1 (0, T ; H01 (Ω)) such that ∂t1−α u ∈ L1 (0, T ; L2 (Ω)), which satisfies Z t Z t u(t) =Sα (t)u0 + Sα (t − s)f (s)ds − Sα (t − s)∇ · (F (s)∂s1−α u(s))ds. (4.19) 0

0

Proof. Similar to the proof of Theorems 4.1 and 4.2, the equation R t (4.1) has the mild solution u ∈ L2 (0, T ; L2 (Ω)) denoted as (4.15). In view of τ Sα0 (t − s)vds = [Sα (t − τ ) − Sα (0)]v and Sα (0)v = v for v ∈ L2 (Ω), we obtain that Z t Sα (t − s)∇ · (F 0 (s)0 Ds−α u(s))ds 0 Z t Z s Z t 0 0 −α = Sα (t − s)∇ · (F (τ )0 Dτ u(τ ))dτ ds + ∇ · (F 0 (s)0 Ds−α u(s))ds. 0

0

0

Fractional Fokker-Planck Equations

221

Then using the integration by parts and (4.8), one can infer from 0 Ds−α u(s) s=0 = 0 and Z s (F 0 (τ )0 Dτ−α u(τ ))dτ s=0 = 0 0

that t

Z s h i Sα0 (t − s) ∇ · (F (s)0 Ds−α u(s)) − ∇ · (F 0 (τ )0 Dτ−α u(τ ))dτ ds 0 0 Z Z t i ∂ h s ∇ · (F (τ )∂τ1−α u(τ ))dτ ds = Sα (t − s) ∂s 0 0 Z s h i t − Sα (t − s) ∇ · (F (s)0 Ds−α u(s)) − ∇ · (F 0 (τ )0 Dτ−α u(τ ))dτ 0 0 Z t Sα (t − s)∇ · (F (s)∂s1−α u(s))ds − ∇ · (F (t)0 Dt−α u(t)) = 0 Z t +∇· (F 0 (τ )0 Dτ−α u(τ ))dτ. Z

0

Combined with above equalities, (4.15) would be rewritten as Z t Z t u(t) =Sα (t)u0 + Sα (t − s)f (s)ds − Sα (t − s)∇ · (F (s)∂s1−α u(s))ds. 0

0

Using Lemma 4.2 again, we can deduce that α −1 ku(t)kH01 (Ω) ≤Mα κα 2 t− 2 ku0 k

+

+

Mα κ−1 α kF kW 1,∞

−1 Mα κα 2

t

Z

α

(t − s)− 2 kf (s)kds

0

Z

(4.20)

t −α

(t − s)

k∂s1−α u(s)kds.

0

It follows from the Young’s inequality for the convolution that u ∈ L1 (0, T ; H01 (Ω)). Our next result shows the precise conditions under which we have a regular solution. Theorem 4.4. Suppose that f ∈ L2 (0, T ; L2 (Ω)) and F ∈ W 1,1 (0, T ; L2 (Ω))∩ W 1,∞ (0, T ; Ω). If we restrict α ∈ ( 12 , 1), then the mild solution u of the equation (4.1) has improved regularity. (i) If u0 ∈ H01 (Ω), then u ∈ L2 (0, T ; D(A)) and ∂t1−α u ∈ L2 (0, T ; H01 (Ω)). 1 (ii) If u0 ∈ D(Aγ0 ) for γ0 = max{ 21 , γ1 } and γ1 ∈ [0, 2α ), then u ∈ C([0, T ]; D(Aγ1 )). (iii) If u0 ∈ D(A), then ∂t1−α u ∈ L2 (0, T ; D(A)). Proof. From Lemma 4.4, we know that Z t Z t 1−α ∂t u(t) =Pα (t)u0 + Pα (t − s)f (s)ds − Pα (t − s)∇ · (F (s)∂s1−α u(s))ds. 0

0

222

Fractional Partial Differential Equations

If α ∈ ( 12 , 1), then it follows from (4.18) and the generalized Gronwall’s inequality (see e.g. [165, Lemma 7.1.1]) that k∂t1−α u(t)k ≤ C ∗ (α, kF kW 1,∞ T )tα−1 , which also shows that ∂t1−α u ∈ L2 (0, T ; L2 (Ω)). The proof of (i). Since

Z t

1

2 P (t − s)∇ · (F (s)∂ 1−α u(s))ds A α s

0

Z t

1−α − 21

APα (t − s)A ∇ · (F (s)∂s u(s))ds =

, 0

by the same way as we derived (4.16), one can get

Z t

2

− 12 1−α

AP (t − s)A ∇ · (F (s)∂ u(s))ds α s

2 0

≤

1 κ2α

L (0,T ;L2 (Ω))

Z

∞ T X

0

Z

1−α λ−1 u(s)), ek )|2 ds k |(∇ · (F (s)∂s

k=1 T

1 kF (t)∂t1−α u(t)k2 dt κ2α 0 1 ≤ 2 kF k2L∞ (0,T ;L2 (Ω)) k∂t1−α uk2L2 (0,T ;L2 (Ω)) . κα

≤

1

Consequently, D(A 2 ) = H01 (Ω) and the Young’s inequality for the convolution yield that r T 2α−1 2 −1 α 1−α ku0 kH01 (Ω) + Mα κα 2 T 2 kf kL2 (0,T ;L2 (Ω)) k∂t ukL2 (0,T ;H01 (Ω)) ≤Mα 2α − 1 α 1−α + kF kL∞ (0,T ;L2 (Ω)) k∂t ukL2 (0,T ;L2 (Ω)) . On the other hand, according to (4.19) and Lemma 4.2, we immediately estimate Z t − 21 − α −1 2 (t − s)−α kf (s)kds ku(t)kD(A) ≤Mα κα t ku0 kH01 (Ω) + Mα κα 0 Z t −1 + Mα κα kF kW 1,∞ (t − s)−α k∂s1−α u(s)kH01 (Ω) ds. 0

Using the Young’s inequality for the convolution again, one can obtain that u ∈ L2 (0, T ; D(A)). The proof of (ii). For t ∈ [0, T ], let h > 0 be such that t + h ∈ [0, T ]. We use 1 ) of (4.12) to derive the case β = 1, γ = γ1 ∈ (0, 2α k[Sα (t + h) − Sα (t)]vkD(Aγ1 ) ≤

1 Mα κ−γ α (t−αγ1 − (t + h)−αγ1 )kvk, for v ∈ L2 (Ω). αγ1 (4.21)

In addition, it is easy to know that k[Sα (t + h) − Sα (t)]u0 kD(Aγ1 ) → 0 as h → 0+ for u0 ∈ D(Aγ1 ), and then according to the fundamental arguments, it follows that u ∈ C([0, T ]; D(Aγ1 )). Particularly, if γ1 = 0, it also holds that u ∈ C([0, T ]; L2 (Ω)).

Fractional Fokker-Planck Equations

223

The proof of (iii). By the same way as we derived (4.16), one can get

2

Z t

2

Z t





APα (t − s)f (s)ds = Pα (t − s)f (s)ds

2

2 2 0

L (0,T ;D(A))

0

L (0,T ;L (Ω))

1 ≤ 2 kf k2L2 (0,T ;L2 (Ω)) κα and

2

Z t

1−α

Pα (t − s)∇ · (F (s)∂s u(s))ds

2

0

L (0,T ;D(A))

1 ≤ 2 k∇ · (F (·)∂t1−α u(·))k2L2 (0,T ;L2 (Ω)) κα 1 ≤ 2 kF k2W 1,∞ k∂t1−α uk2L2 (0,T ;H 1 (Ω)) . 0 κα Further, Pα (·)u0 ∈ L2 (0, T ; D(A)) for u0 ∈ D(A), then ∂t1−α u ∈ L2 (0, T ; D(A)). In fact, the decay estimates of u and ∂t1−α u can be obtained when we use the following condition instead of f ∈ L2 (0, T ; L2 (Ω)). (Q) Assume that f ∈ L1 (0, T ; L2 (Ω)) satisfies kf (t)k ≤ Mf tϑ for some ϑ ∈ ( α2 −1, 0) and a.e. t ∈ [0, T ]. Remark 4.2. Let α ∈ ( 12 , 1), u0 ∈ H01 (Ω). Assume that F ∈ W 1,1 (0, T ; L2 (Ω)) ∩ W 1,∞ (0, T ; Ω) and (Q) hold. Then we also have u ∈ Lq (0, T ; H 2 (Ω)) for q ∈ [1, α2 ) and ∂t1−α u ∈ L2 (0, T ; H01 (Ω)). Moreover, there exist C0 = C(α, kF kW 1,∞ , T ) and C1 = C ∗ (α, kF kW 1,∞ , T ) such that k∂t1−α u(t)k

α

1

D(A 2 )

≤C0 tα−1 , ku(t)kD(A) ≤ C1 t− 2 .

Proof. Using Lemma 4.2 again, we can deduce from u0 ∈ H01 (Ω) that Z t α − 12 1−α α−1 1 (t − s) 2 −1 kf (s)kds ≤Mα t ku0 kH01 (Ω) + Mα κα k∂t u(t)k D(A 2 ) 0 Z t α − 21 (t − s) 2 −1 k∇ · (F (s)∂s1−α u(s))kds + Mα κα 0 α α −1 (ku0 kH01 (Ω) + Mf κα 2 B( , 1 + ϑ)T 1+ϑ− 2 ) 2 Z t α − 12 −1 + Mα κα kF kW 1,∞ (t − s) 2 k∂s1−α u(s)kH01 (Ω) ds.

α−1

≤Mα t

0

It follows from the generalized Gronwall’s inequality (see e.g. [165, Lemma 7.1.1]) that there exists C(α, kF kW 1,∞ , T ) such that k∂t1−α u(t)k −1

1

D(A 2 )

≤aC(α, kF kW 1,∞ T )tα−1 ,

−1

α

where a = κα 2 Mαα (ku0 kH01 (Ω) + Mf κα 2 B( α2 , 1 + ϑ)T 1+ϑ− 2 ).

224

Fractional Partial Differential Equations

On the other hand, in view of (4.19), we also obtain Z t − 12 − α −1 2 ku(t)kD(A) ≤Mα κα t ku0 kH01 (Ω) + Mα κα (t − s)−α kf (s)kds 0 Z t −1 + Mα κα kF kW 1,∞ (t − s)−α k∂s1−α u(s)kH01 (Ω) ds −α 2

≤Mα t

0 − 12 [κα ku0 kH01 (Ω)

α

1+ϑ− 2 + Mf κ−1 α B(1 − α, 1 + ϑ)T

α

+ C0 T 2 kF kW 1,∞ B(1 − α, α)]. Therefore, it follows that u ∈ Lq (0, T ; H 2 (Ω)) for q ∈ [1, α2 ). We now verify under what conditions a mild solution may become classical for as long as the former exists. Theorem 4.5. Let α ∈ ( 12 , 1), u0 ∈ D(A). Assume that f ∈ L2 (0, T ; L2 (Ω)) and F ∈ W 1,1 (0, T ; L2 (Ω)) ∩ W 1,∞ (0, T ; Ω). Then the mild solution u of the equation (4.1) obtained in Theorem 4.3 is also a classical solution. Proof. We differentiate (4.19) and use (4.17) to obtain that Z t 0 ∂t u(t) =Sα (t)u0 + f (t) + Sα0 (t − s)f (s)ds − ∇ · (F (t)∂t1−α u(t)) 0 Z t − Sα0 (t − s)∇ · (F (s)∂s1−α u(s))ds.

(4.22)

0

Similar to the proof of (4.16), it follows from Theorem 4.4(i) and u0 ∈ D(A) that ∂t u ∈ L2 (0, T ; L2 (Ω)). Further, we can immediately from −∆ek = λk ek and Sα0 (t)v = −κα APα (t)v for v ∈ L2 (Ω) that ∂t u(t, x) − κα ∂t1−α ∆u + ∇ · (F ∂t1−α u) = f (t, x), (t, x) ∈ (0, T ) × Ω. This combined with Theorem 4.4 shows that the claim follows. Remark 4.3. In Theorems 4.4 and 4.5, it needs to turn out that ∂t1−α u ∈ L2 (0, T ; L2 (Ω)), so we can’t get rid of the restriction on α ∈ ( 12 , 1), which verifies the results in [217] from another perspective. 4.1.6

Classical Solutions in Case of α ∈ (0, 1)

In this subsection, we study the existence and uniqueness of classical solutions in weighted H¨ older continuous function spaces, in which the restriction on α ∈ ( 21 , 1) can be removed. We introduce a working space for 0 < θ < η < min{ α2 , 1 − α},  ¯ T = w ∈ C([0, T ]; L2 (Ω)) : w0 ∈ F η,θ ((0, T ]; L2 (Ω)) , X

Fractional Fokker-Planck Equations

225

whose norm is kwkX¯ T = kwkC([0,T ];L2 (Ω)) + kw0 kF η,θ . Clearly, it is a Banach space. In view of the singular terms in integral equations and the space F η,θ ((0, T ]; L2 (Ω)), our analysis makes heave use of the following inequality. Lemma 4.5. Let µ ∈ (0, 1), ν,σ ∈ [0, 1) and 0 ≤ t1 < t2 ≤ T . The following inequality holds:  σ  Z t1 2 π 2 ν µ−1 (t1 − s)µ−1 (t2 − s)ν−1 s−σ ds ≤ + t−σ . 1 t2 (t2 − t1 ) sin(µπ) 1 − σ 0 Proof. We split the integral into two parts: Z t1 (t1 − s)µ−1 (t2 − s)ν−1 s−σ ds 0

Z =

t1 2

sµ−1 (t2 − t1 + s)ν−1 (t1 − s)−σ ds +

0 ν ≤2σ t−σ 1 t2

Z

Z

t1 t1 2

sµ−1 (t2 − t1 + s)ν−1 (t1 − s)−σ ds

t1 2

sµ−1 (t2 − t1 + s)−1 ds Z t1 ν µ−1 + 2t−1 t (t − t ) (t2 − t1 )1−µ sµ (t2 − t1 + s)−1 (t1 − s)−σ ds. 2 1 1 2 0

t1 2

According to Z ∞ sµ−1 (t2 − t1 + s)−1 ds =

π (t2 − t1 )µ−1 and (t2 − t1 )1−µ sµ ≤ t2 − t1 + s, sin(µπ) 0   2 2σ π ν µ−1 + 1−σ t−σ . then the above integral is bounded by sin(µπ) 1 t2 (t2 − t1 ) The following result shows that the solution of the integral equation (4.10) be¯T . longs to X Lemma 4.6. Let u0 ∈ L2 (Ω) and f ∈ F η,θ ((0, T ]; L2 (Ω)). Suppose that F ∈ W 1,1 (0, T ; L2 (Ω)) ∩ W 1,∞ (0, T ; Ω) ∩ C θ ([0, T ]; L2 (Ω)). Then the equation (4.10) ¯T . has a unique solution w ∈ X Proof. Consider the operator H : XT → XT given as (4.11). Then it is welldefined. In fact, let w ∈ XT , then Hw ∈ C([0, T ]; L2 (Ω)). It suffices to prove (Hw)0 ∈ F η,θ ((0, T ]; L2 (Ω)). To do this, let us consider the operator Z t Z t (Hw)0 (t) =Pα (t)u0 + Pα (t − s)f (s)ds − Pα (t − s)∇ · (F (s)w0 (s))ds 0

0

= : H1 (t) + H2 (t) + H3 (t). We only need to show Hi ∈ F η,θ ((0, T ]; L2 (Ω)) for i = 1, 2, 3. Noting that, from the definition of Pα (t), it is clear to know limt→0 t1−η Pα (t)u0 exists for η < α2 . Further, for 0 ≤ t1 < t2 ≤ T , in view of (4.12), one can see Mα Mα ku0 k (t2 − t1 )1−α ku0 k(t1α−1 − tα−1 )≤ kPα (t2 )u0 − Pα (t1 )u0 k ≤ . 2 1−α 1 − α (t1 t2 )1−α (4.23)

226

Fractional Partial Differential Equations

It implies that kH1 (t2 ) − H1 (t1 )k Mα ku0 k α−η  t1 θ  t2 − t1 1−α−θ t1−η+θ 1 ≤ t (t2 − t1 )θ 1−α 1 t2 t2 Mα ku0 k α−η ≤ T . 1−α Therefore, we have lim

sup

t2 →0 0≤t1 0 was defined in Lemma 4.12. 1,2

Lemma 4.20. Let m be a positive integer. Let vm (t) be the absolutely continuous solution of (4.60) that is guaranteed by Lemma 4.15. Then for almost all t ∈ [0, T ], one has Z t  t1−α  k∇vm (t)k2 + κα ρα tα−1 k∆vm k2 ds ≤ (C10 + C11 )ku0 k2H 2 (Ω) + C10 kgk2L2 , κα ρα 0 (4.68) Z

t 0 2 2 kvm k ds ≤ (C10 + C11 )CR ku0 k2H 2 (Ω) + C10 kgk2L2

(4.69)

0

and Z

t

k∂s1−α ∆vm (s)k2 ds ≤

0

 1  (C10 + C11 )ku0 k2H 2 (Ω) + C10 kgk2L2 , 2 κα

(4.70)

where   C10 := 3 1 + kFk21,∞ (C8 T + C9 Γ(α)T 1−α ) and C11 :=

2 (κ2α + kFk21,∞ ) 2α−1 6CR T . (2α − 1)Γ(α)2

Proof. Take the inner product of both sides of (4.60) with −∆vm ∈ Wm and integrate by parts with respect to x to get 1 d k∇vm k2 + κα h∂t1−α ∆vm , ∆vm i =h∇ · (F(t)∂t1−α vm ), ∆vm i − hg(t), ∆vm i 2 dt

 − κα ∆u0m − ∇ · (F(t)u0m ), ∆vm ωα (t). Integrating in time and noting that, by Lemma 4.12, Z t Z t α−1 2 ρα t k∆vm k ds ≤ h∂s1−α ∆vm , ∆vm i ds, 0

0

we obtain Z t 1 k∇vm (t)k2 + κα ρα tα−1 k∆vm k2 ds 2 0 Z Z t
1 t h 2

∇ · F(s)∂s1−α vm 2 + kg(s)k2 ≤3 k∆vm k ds + 4 0 0 i


2 + κα ∆u0m − ∇ · F(s)u0m ωα (s)2 ds, with a free parameter  > 0. Choosing  = κα ρα tα−1 /6 and recalling Lemma 4.13 yields Z t 2 α−1 k∇vm (t)k + κα ρα t k∆vm k2 ds 0

252

Fractional Partial Differential Equations

Z t  3 2 2 2 2 ≤ ds kg(s)k + 2κ k∆u k ω (s) 0m α α κα ρα tα−1 0 Z t  3kFk21,∞ 1−α 2 2 2 + ds k∂ v k 1 (Ω) + 2ku0m kH 1 (Ω) ωα (s) m s H κα ρα tα−1 0   Z t t1−α 1−α 2 2 2 2 k∂s vm kH 1 (Ω) ds , C11 ku0 kH 2 (Ω) + 3kgkL2 + 3kFk1,∞ ≤ κ α ρα 0 by Lemma 4.18. Invoking Corollary 4.1, we have  Z t Z t ωα (t − s) 1−α 2 1−α 2 1−α 2 k∂s ∇vm k ds k∂s vm k + k∂s vm kH 1 (Ω) ≤ ωα (t) 0 0   ≤ C8 t + C9 Γ(α)t1−α ku0 k2H 2 (Ω) + kgk2L2 ,

(4.71)

and the bound (4.68) follows. In a similar fashion, we next take the inner product of both sides of (4.60) 0 ∈ Wm and integrate by parts with respect to x to obtain with vm


0  0 0 0 kvm (t)k2 + κα hJ α ∇vm , ∇vm i = − ∇ · F(t)∂t1−α vm , vm


 0 0 + hg(t), vm i − ∇ · F(t)u0m − κα ∆u0m , vm ωα (t) 


1 2 0 ≤3kvm (t)k2 + kg(t)k2 + ∇ · F(t)∂t1−α vm 4 

2
+ ∇ · F(t)u0m − κα ∆u0m ωα2 (t) . Choosing  = 1/6 and invoking Lemma 4.13 gives 0 2 0 0 kvm k + 2κα hJ α ∇vm , ∇vm i ≤3kg(t)k2 + 3kFk21,∞ k∂t1−α vm k2H 1 (Ω)
+ 6 κ2α + kFk21,∞ ku0m k2H 2 (Ω) ωα2 (t).

Integrating both sides of the inequality in time and invoking Lemma 4.8, we deduce that Z t 0 2 kvm k ds ≤ 3kgk2L2 (0,T ;L2 ) + 3kFk21,∞ k∂t1−α vm k2L2 (0,t;H 1 (Ω)) + C11 ku0m k2H 2 (Ω) . 0

(4.72) The second result (4.69) now follows from (4.71), (4.72) and Lemma 4.18. Using similar arguments, take the inner product of both sides of (4.60) with −∂t1−α ∆vm ∈ 0 Wm , integrate by parts with respect to x and note that ∂t1−α ∆vm = J α ∆vm to obtain 0 0 h∇vm , J α ∇vm i + κα k∂t1−α ∆vm k2

=h∇ · (F(t)∂t1−α vm ), ∂t1−α ∆vm i − hg(t), ∂t1−α ∆vm i

 − κα ∆u0m − ∇ · (F(t)u0m ), ∂t1−α ∆vm ωα (t)
3  κα

∇ · F(t)∂t1−α vm 2 + kg(t)k2 ≤ k∂t1−α ∆vm k2 + 2 2κα 


2 + κα ∆u0m − ∇ · F(t)u0m ωα2 (t) .

Fractional Fokker-Planck Equations

253

Now integrate in time, invoking Lemma 4.8 and using (4.71), to deduce that Z t k∂s1−α ∆vm (s)k2 ds 0 Z t


3

∇ · F(s)∂s1−α vm 2 + kg(s)k2 ≤ 2 κα 0 


2 + κα ∆u0m − ∇ · F(s)u0m ωα2 (s) ds Z t  3 kg(s)k2 + 2κ2α k∆u0m k2 ωα2 (s) ds ≤ 2 κα 0 Z  3kFk21,∞ t  1−α 2 2 2 + ω (s) ds, + 2ku k k∂ v k 1 1 0m H (Ω) α m H (Ω) s κ2α 0 and (4.70) now follows by (4.71) and Lemma 4.18, which completes the proof of the lemma. Inequality (4.71) may also be derived (with a different constant factor) by applying (4.45) to (4.70). We remark that the function α 7→ ρα is monotone increasing for α ∈ (0, p 1), with ρα → 1 as α → 1. Thus, ρ1/2 < ρα < 1 for 1/2 < α < 1, with ρ1/2 = 2π/27 = 0.48240 . . . . 4.2.6.2

The classical solution

In the sequel, by using the method of compactness, we show that there is a subsequence of {vm }∞ m=1 such that the sum of its limit and the initial data satisfies the equation (4.27) almost everywhere. Theorem 4.9. Assume that α ∈ (1/2, 1), u0 ∈ H 2 (Ω) ∩ H01 (Ω), F ∈ W 1,∞ ((0, T ) × Ω) and g ∈ L2 (0, T ; L2 ). Then there exists a unique classical solution of (4.27), in the sense of Definition 4.6, such that   sup ku(t)k2H 1 (Ω) + ku0 k2L2 + k∂t1−α uk2L2 (0,T ;H 2 ) ≤ C12 ku0 k2H 2 (Ω) + kgk2L2 , (4.73) 0≤t≤T

where C12

  T 1−α 1 2 := (C10 + C11 ) 1 + CR + + 2 . κα ρα κα

Proof. From Lemma 4.20 we obtain   0 2 sup kvm (t)k2H 1 (Ω) + kvm kL2 + k∂t1−α vm k2L2 (0,T ;H 2 ) ≤ C12 ku0 k2H 2 (Ω) + kgk2L2 ,

0≤t≤T

(4.74) ∞ 1 2 which shows that the sequence {vm }∞ is bounded in L (0, T ; H )∩L (0, T ; H2 ∩ m=1 1 0 ∞ 2 2 H0 ) and that the sequence {vm }m=1 is bounded in L (0, T ; L ). Since the embeddings H 2 ,→ H 1 ,→ L2 are compact, it follows from Lemma 4.14 that there exists a ∞ subsequence of {vm }∞ m=1 (still denoted by {vm }m=1 ) such that vm → v strongly in C([0, T ]; L2 ) ∩ L2 (0, T ; H 1 ).

(4.75)

254

Fractional Partial Differential Equations

Furthermore, from the upper bounds of {vm }∞ m=1 we have vm → v weakly in L∞ (0, T ; H 1 ) ∩ L2 (0, T ; H 2 ∩ H01 ) 0 and vm → v 0 weakly in L2 (0, T ; L2 ).

(4.76)

By virtue of Lemma 4.11, the strong convergence in (4.75) implies that J α vm → J α v strongly in C([0, T ]; L2 ) ∩ L2 (0, T ; H 1 ). This, together with Corollary 4.1 and (4.70), yields ∂t J α vm → ∂t J α v weakly in L∞ (0, T ; L2 ) ∩ L2 (0, T ; H 2 ).

(4.77)

Multiplying both sides of (4.60) by a test function ξ ∈ L2 (0, T ; L2 ), integrating over (0, T ) × Ω and noting that Πm is a self-adjoint operator on L2 (Ω), we deduce that 0 hvm , ξiL2 (0,T ;L2 ) − κα h∂t1−α ∆vm , ξiL2 (0,T ;L2 ) + h∇ · (F∂t1−α vm ), Πm ξiL2 (0,T ;L2 )   =hg, Πm ξiL2 (0,T ;L2 ) − hωα ∇ · (Fu0m ) − κα ∆u0m , Πm ξiL2 (0,T ;L2 ) .

Now let m → ∞ in this equation and recall (4.76) and (4.77). We get hu0 , ξiL2 (0,T ;L2 ) − κα h∂t1−α ∆u, ξiL2 (0,T ;L2 ) + h∇ · (F∂t1−α u), ξiL2 (0,T ;L2 ) =hg, ξiL2 (0,T ;L2 ) ,

(4.78)

for all ξ ∈ L2 (0, T ; L2 ), where u := v + u0 . From (4.75)–(4.77), we have u0 ∈ L2 (0, T ; L2 )

and ∂t1−α u ∈ L2 (0, T ; H 2 ).

Hence, it follows from (4.78) that u satisfies (4.27) a.e. in (0, T ) × Ω. Taking the limit as m → ∞ in (4.74), we obtain (4.73). The uniqueness of the solution u follows from (4.73), which completes the proof of the theorem. Remark 4.6. It follows from the uniqueness in Theorems 4.8 and 4.9 that the mild solution will become the classical solution when α ∈ (1/2, 1) and u0 ∈ H 2 (Ω) ∩ H01 (Ω). Furthermore, the continuous dependence of both the mild and classical solutions on the initial data u0 follows from (4.46) and (4.73). 4.2.7

Regularity of Classical Solution

Recall that 1/2 < α < 1 and that in general C = C(Ω, κα , F, T ). From Lemma 4.22 onwards, we allow C = C(Ω, κα , F, T, q), where q appears in the statements of our results below. From Theorem 4.9, for almost every (t, x) ∈ (0, T ) × Ω the solution u(t, x) satisfies (4.27). Using the identity ∂t1−α u = (J α u)0 (t) = (J α u0 )(t) + u(0)ωα (t), we rewrite (4.27) as   u0 − ∇ · (κα ∇J α u0 − FJ α u0 ) = g(t) + ∇ · κα ∇u0 − F(t)u0 ωα (t). (4.79)

Fractional Fokker-Planck Equations

255

From this equation and the fact that J α φ(0) = 0 for any function φ ∈ L2 (0, T ; L2 ), we deduce that u0 (t) = O(tα−1 ) when t is close to 0. By letting z(t, x) := tu0 (t, x), we have z(0) = 0. The regularity of z is examined in the following lemma. RT Lemma 4.21. Assume that 0 ktg 0 (t)k2 dt is finite. Then, the function z defined above satisfies sup kz(t)k2H 1 (Ω) + kz 0 k2L2 + k∂t1−α zk2L2 (0,T ;H 2 )

0≤t≤T

(4.80)


≤C13 ku0 k2H 2 (Ω) + kgk2L2 + kg1 k2L2 , for some constant C13 . Proof. For any t > 0, multiplying both sides of (4.79) by t and using the elementary identity t(J α u0 )(t) = (J α z)(t) + α(J α+1 u0 )(t) = (J α z)(t) + α (J α u)(t) − u0 ωα+1 (t) α




(4.81)

α

= (J z)(t) + α(J u)(t) − u0 tωα (t), we obtain a differential equation for z:
z − ∇ · (J α κα ∇z − FJ α z) = tg(t) + α∇ · κα ∇J α u − FJ α u . Differentiating both sides of this equation with respect to t and noting that J α z 0 = ∂t1−α z, we have ¯ x), z 0 − ∇ · (∂t1−α κα ∇z − F∂t1−α z) = G(t,

(4.82)

where
¯ := g + tg 0 + α∇ · κα ∇∂t1−α u − F0 (t)J α u − F(t)∂t1−α u . G Applying Lemma 4.13 and letting g1 (t) = tg 0 (t), we find that  ¯ 2 2 ≤4 kgk2 2 + kg1 k2 2 + α2 (κα + kFk1,∞ )2 k∂t1−α uk2 2 kGk L L L L (0,T ;H 2 )  + α2 kF0 k21,∞ kJ α uk2L2 (0,T ;H 1 ) , with Lemma 4.16 and Theorem 4.9 implying that
2 2 ¯ 2 2 ≤ C ku0 k2 2 kGk L H (Ω) + kgkL2 + kg1 kL2 , for some constant C. Thus, applying Theorem 4.9 to equation (4.82) with initial data z(0) = 0, we deduce the bound (4.80). From Theorem 4.9, for almost every (t, x) ∈ (0, T ) × Ω, we have the identity ut − ∇ · (∂t1−α κα ∇u)) = f, (F∂t1−α u)

(4.83)

where f := g − ∇ · ∈ L2 (0, T ; H 1 ). The regularity of solutions to problem (4.83) subject to the initial condition u0 ∈ H 2 (Ω) ∩ H01 (Ω) was studied in [258]. In order to apply [258, Theorem 5.7], we need at least an upper bound

256

Fractional Partial Differential Equations

Rt for 0 skf 0 (s)kds which is proved in the following lemma. Here and subsequently, notation such as f 0 and f (j) indicates time derivatives, and we denote higherorder fractional derivatives by ∂tj−α u := ∂tj−1 ∂t1−α u = (J α u)(j) for j ∈ {1, 2, . . .} and 0 < α < 1. Lemma 4.22. Let u be the solution of (4.27) and f := g − ∇ · (F∂t1−α u). Then, for q ∈ {0, 1, 2, . . . }, F ∈ W q,∞ (0, T ; L2 (Ω)), and for any t ∈ (0, T ], there is a constant C = C(Ω, κα , F, q) such that   Z t q Z t X 2q (q) 2 2 2j (j) 2 s kf (s)k ds ≤ C ku0 kH 2 (Ω) + s kg (s)k ds . (4.84) 0

j=0

0

Proof. Inequality (4.84) holds for q = 0 by virtue of (4.73) (with t playing the role of T ) because
2 kf (t)k2 ≤ C kg(t)k2 + k∂t1−α ukH 1 (Ω) . For the case q = 1, we note first that


2 t2 kf 0 (t)k2 = t2 g 0 (t) − ∇ · F0 (t)∂t1−α u + F(t)∂t2−α u
≤ Ct2 kg 0 (t)k2 + k∂t1−α uk2H 1 (Ω) + k∂t2−α uk2H 1 (Ω) . By (4.81), we have (J α z)(t) = t(J α u0 )(t) − α(J α+1 u0 )(t), and differentiating with respect to t gives ∂t1−α z = t(J α u0 )0 (t) − (α − 1)(J α u0 )(t) = t∂t2−α u − (α − 1)∂t1−α u, where we used the identities (J α u0 )(t) = ∂t1−α u − u0 ωα (t) and (α − 1)ωα (t) = α−1 α−1 . Thus, Γ(α) t t∂t2−α u = ∂t1−α z + (α − 1)∂t1−α u. (4.85) Hence, by Theorem 4.9 and Lemma 4.21 (with t again playing the role of T ),   Z t Z t   2 2−α 2 2 2 2 0 2 s k∂s ukH 2 (Ω) ds ≤ C ku0 kH 2 (Ω) + kg(s)k + s kg (s)k ds , (4.86) 0

0

implying that the desired inequality (4.84) holds for q = 1. Multiply both sides of (4.85) by t and then differentiate with respect to t, obtaining t2 ∂t3−α u = ∂t1−α z + t∂t2−α z + (α − 1)∂t1−α u + (α − 3)t ∂t2−α u. (4.87) ¯ Since z satisfies (4.82) — an equation similar to (4.27a) but with a different source G and with z(0) = 0 — we get an estimate for z corresponding to (4.86): Z t Z t   2 ¯ ¯ 0 (s)k2 ds. kG(s)k + s2 kG s2 k∂s2−α zk2H 2 (Ω) ds ≤ C 0

0

This inequality, together with (4.73), (4.80), (4.86) and (4.87), yields  Z t Z t  3−α 4 2 2 s k∂t ukH 2 (Ω) ds ≤C ku0 kH 2 (Ω) + kg(s)k2 0 0   2 0 2 4 00 2 + s kg (s)k + s kg (s)k ds , which implies the desired inequality (4.84) for q = 2. The general case follows by iterating the arguments above, cf. [260].

(4.88)

Fractional Fokker-Planck Equations

257

We can now prove regularity estimates for the classical solution u. Theorem 4.10. Let gj (t) := tj g (j) (t) for j = 1, 2, . . . . For q ∈ {1, 2, . . . }, F ∈ W q,∞ (0, T ; L2 (Ω)) and for any t ∈ (0, T ],   q+1 X tq k∆u(q) (t)k ≤ Ct−(α−1/2) ku0 kH 2 (Ω) + kgj kL2 j=0

and q

(q)

t ku

1/2

(t)k ≤ Ct

 ku0 kH 2 (Ω) +

q X

 kgj kL2 .

j=0

Proof. By (4.83), it follows from [258, Theorem 4.4] with r = 2 and ν = α, and from [258, Theorem 5.6] with r = 0, µ = 2 and ν = α, that   q+1 Z t X j (j) q (q) −α s kf (s)k ds . t k∆u (t)k ≤ C ku0 kH 2 (Ω) + t j=0

0

Similarly, from [258, Theorem 4.4] with r = 2 and ν = α, and from [258, Theorem 5.4] with r = µ = 0, and   q Z t X tq ku(q) (t)k ≤ C tα ku0 kH 2 (Ω) + sj kf (j) (s)k ds . j=0

0

The theorem follows by Lemma 4.22 since Z t Z t 1/2 j (j) 1/2 s kf (s)k ds ≤ t s2j kf (j) (s)k2 ds . 0

0

Corollary 4.3. Let η > 1/2. If kg (j) (t)k ≤ M tη−1−j f or 0 ≤ j ≤ q + 1, F ∈ W q,∞ (0, T ; L2 (Ω)) and t ∈ (0, T ], then
tq k∆u(q) (t)k ≤ C t−(α−1/2) ku0 kH 2 (Ω) + M tη−α and
tq ku(q) (t)k ≤ C t1/2 ku0 kH 2 (Ω) + M tη . Proof. The assumption on g ensures that kgj k ≤ M tη−1/2 . The alternative and longer analysis in [260, Theorems 6.2 and 6.3] shows that these bounds can be improved to
tq k∆u(q) (t)k ≤ C ku0 kH 2 (Ω) + M tη−α and
tq ku(q) (t)k ≤ C tα ku0 kH 2 (Ω) + M tη , for any α ∈ (0, 1) and η > 0. 4.3

Notes and Remarks

The contents in Section 4.1 are taken from Peng and Zhou [294]. The results of Section 4.2 are due to Le, McLean and Stynes [219].

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Chapter 5

Fractional Schr¨ odinger Equations

In this chapter, we firstly study the linear fractional Schr¨odinger equation on a Hilbert space, with a fractional time derivative of order 0 < α < 1, and a selfadjoint generator A. Using the spectral theorem we prove existence and uniqueness of strong solutions, and we show that the solutions are governed by an operator solution family {Uα (t)}t≥0 . Moreover, we prove that the solution family Uα (t) converges strongly to the family of unitary operators e−itA , as α approaches to 1. Next, we apply the tools of harmonic analysis to study the Cauchy problem for time-fractional Schr¨ odinger equation. Some fundamental properties of two solution operators are estimated. The existence and a sharp decay estimate for solutions of the given problem in two different spaces are addressed. Finally, we study fractional Schr¨ odinger equations with potential and optimal controls. Existence, uniqueness, local stability and attractivity, and data continuous dependence of mild solutions are also presented respectively. Further, we study the optimal control problems for the controlled fractional Schr¨ odinger equations with potential. Existence and uniqueness of optimal pairs for the standard Lagrange problem are presented. 5.1 5.1.1

Linear Equations on Hilbert Space Introduction

The Schr¨ odinger equation is the basic equation of quantum mechanics, it describes the evolution in time of a quantum system. More recently, Laskin has introduced the fractional Schr¨ odinger equation, as a result of extending the Feynman path integral, the resulting equation is a fundamental equation in fractional quantum mechanics [213–215]. Furthermore, Laskin [213] states that “the fractional Schr¨odinger equation provides us with a general point of view on the relationship between the statistical properties of the quantum mechanical path and the structure of the fundamental equations of quantum mechanics”. Naber [274] introduced and examined some properties of the time-fractional Schr¨ odinger equation, ( C α α 0 Dt u(t) = (−i) Au(t), (5.1) u(0) = u0 , 259

260

Fractional Partial Differential Equations π

in which (−i)α = e−iα 2 . It was shown in [274] that the above equation (5.1) is equivalent to the usual Schr¨ odinger equation with a time dependent Hamiltonian. On the other hand, it was point out that the so-called quantum comb model [35,38, 180,181], leads to a time-fractional Schr¨ odinger equation with α = 12 . The equation (5.1) describes non-Markovian evolution in Quantum Mechanics. As a result this system has memory. Different aspects of the time-fractional Schr¨odinger equation have already been studied. Particular solutions were sought in [38, 106, 274] and numerical analysis performed in [128]. Nevertheless, to the best of our knowledge, there are no results in the literature which show in full generality the existence and uniqueness of solutions to the abstract Schr¨ odinger equation on a Hilbert space. The purpose of this section is to consider the abstract fractional evolution equation (5.1) on a Hilbert space H, in which A is a positive self adjoint operator on H, α and ∂∂tαu is the Caputo fractional derivative of order α ∈ (0, 1). We show that A generates a family of bounded operators {Uα (t)}t≥0 which are defined by the functional calculus of A via the Mittag-Leffler function when evaluated at A. Moreover if u0 belongs to the domain of A then we show that u(t) = Uα (t)u0 is the unique strong solution of problem (5.1). We also study the problem of the continuous dependence on α for Uα (t), and we show that lim Uα (t) = e−itA ,

α→1−

where e−itA is the unitary group whose infinitesimal generator corresponds to the self adjoint operator A. Thus, we recover in the limit as α → 1 the classical Theorem of Stone. In Subsection 5.1.2, we introduce the notations and give the definition of strong solution to the fractional Schr¨ odinger equation. Moreover, we formulate and prove some technical but very crucial lemma. The main result about existence and uniqueness of solution is shown in Subsection 5.1.3. The properties of the solution operator are formulated and proven in Subsection 5.1.4. We give an example in Subsection 5.1.5. 5.1.2

Preliminaries

We recall the definition of the Riemann-Liouville fractional integral by the convolution product, Z t 1 (t − s)α−1 f (s)ds, J α f (t) = Γ(α) 0 for a given locally integrable function f defined on the half line R+ = [0, ∞) and taking values on a Banach space X. Henceforth we use the notation, J α f (t) = (gα ∗ f )(t), tα−1 Γ(α) ,

(5.2) α+β

where gα (t) = for α > 0, t > 0. Then the following property holds: J f= α β J J f , for α, β > 0, f is suitable enough. Assume that u ∈ C([0, ∞); X) and that

Fractional Schr¨ odinger Equations

261

the convolution g1−α ∗ u belongs to C 1 ((0, ∞); X). Then the Caputo fractional derivative of order α ∈ (0, 1), can be interpreted as d C α (g1−α ∗ u)(t) − u(0)g1−α (t). (5.3) 0 Dt u(t) = dt ∂αu α Henceforth we shall denote the Caputo derivative either by C 0 Dt u(t) or ∂tα (t), indistinctly. Let X be a Banach space, and suppose that u0 ∈ X and ω ∈ C. If 0 < α < 1, then the equation C α 0 Dt u(t)

= ωu(t), u(0) = u0

(5.4)

has a unique solution given by u(t) = u0 Eα (ωtα ), see [41, 146, 197]. Moreover, the uniqueness of the solution of (5.4) follows by the uniqueness theorem for the Laplace transform. Let A be a densely defined self-adjoint operator on a Hilbert space H, and let 0 < α < 1. For a given u0 ∈ H, we study the following equation of fractional order α, ( C α α 0 Dt u(t) = (−i) Au(t), t > 0, (5.5) u(0) = u0 . We first introduce the notion of strong solution for the abstract fractional Cauchy problem (5.5). Definition 5.1. Let 0 < α < 1. Assume that u0 ∈ D(A). A function u is called a strong solution of (5.5) if u ∈ C(R+ , D(A)) and g1−α ∗ u belongs to C 1 ((0, ∞); H), and (5.5) holds for all t > 0. We will show that the strong solution of (5.5) is determined by the functional calculus for a self-adjoint operator when it is applied to the Mittag-Leffler function. Moreover, the following lemma will give us the necessary bounds we need in the proof of the qualitative properties of the solution operator. In order to prove the next lemma we recall from [147] that the Mittag-Leffler function has the following representation for α ∈ (0, 1], Z ∞ 1 1/α Eα (z) = Kα (r, z)dr + ez , | arg(z)| < π/α and z 6= 0, (5.6) α 0 in which 1/α

Kα (r, z) = −

e−r z sin(πα) . 2 πα(r − 2rz cos(πα) + z 2 )

(5.7)

Lemma 5.1. (a) Let α0 ≤ α < 1/2, in which α0 > 0. Then there is a positive constant M (α) such that for all t ≥ 0, sup |Eα ((−it)α ω)| ≤ M (α). ω≥0

(5.8)

262

Fractional Partial Differential Equations

(b) There is M > 0 such that for all t ≥ 0 and all α ∈ [1/2, 1), sup |Eα ((−it)α ω)| ≤ M.

(5.9)

ω≥0

Proof. First we show (5.8) that is α0 ≤ α < 1/2. We notice that it suffices to prove assertion (5.8) for t = 1. Indeed, let us assume that (5.8) holds for t = 1, then for any t > 0, we have |Eα ((−it)α ω)| = |Eα ((−i)α tα ω)| ≤ M (α). To begin we assume ω ≥ 1/α0 . Next we recall that (−i)α = e−iαπ/2 . Then we proceed to estimate |Kα (r, (−i)α ω)| for arbitrary ω ≥ 1/α0 . Thus, 1/α

B e−r ω |Kα (r, (−i) ω) | ≤ , πα |(r2 − 2r(a − ib)ωA + ω 2 (A − iB)| α

(5.10)

where A = cos(πα), B = sin(πα), a = cos(απ/2), b = sin(απ/2) and these quantities are all positive since 0 < α < 1/2. Next we set u(r) and v(r) the real and imaginary parts respectively of the denominator on the right hand side of (5.10), that is u(r) = r2 − 2raωA + ω 2 A and v(r) = 2rbωA − ω 2 B. Hence, 1/α

|Kα (r, (−i)α ω) | ≤

1/α

B e−r ω Be−r ω ≤ . 2 2 πα |(r − 2r(a − ib)ωA + ω (A − iB)| π α | u(r) |

(5.11)

On the other hand the quadratic u(r) = r2 − 2raωA + ω 2 A is positive for all real r and its minimum equals to ω 2 A(1 − a2 A) > 0 since a2 A < 1, and A > 0. But then, u(r) = r2 − 2raωA + ω 2 A ≥ ω 2 A(1 − a2 A) > 0. Hence the right side of (5.11) turns out to be less than or equals to 1/α

Be−r ω . π α ω 2 A(1 − a2 A)

(5.12)

Therefore, from (5.11) and (5.12) follows that 1/α

|Kα (r, (−i)α ω) | ≤

e−r , πA(1 − a2 A)

since 0 < α0 ≤ α < 1/2 and ω ≥ 1/α0 . Furthermore    

e−r for r > 1, 2 |Kα (r, (−i)α ω)| ≤ πA(1 −1 a A)   for r ≤ 1.  πA(1 − a2 A) Therefore, from (5.13) we obtain that the integral Z ∞ Kα (r, (−i)α ω)dr 0

(5.13)

Fractional Schr¨ odinger Equations

263

is bounded independently of ω ≥ α10 . But then, it follows from the integral representation (5.6) that there is a bound M1 (α) such that sup |Eα ((−i)α ω)| ≤ M1 (α).

(5.14)

ω≥1/α0

Now if ω ≤ 1/α0 , then we have that 1 . α0 Thus, from the very definition of the Mittag-Leffler function, we obtain that  k 1 ∞ X α0 |Eα ((−i)α ω)| ≤ = Eα (1/α0 ). Γ(αk + 1) |(−i)α ω| ≤

k=0

Moreover, by the Stirling’s formula Γ(x) =

√

1

θ

2πxx− 2 e−x+ 12x ,

where θ ∈ [0, 1] and for x large enough, we have  k 1  k α0 e2 e ≤√ . Γ(αk + 1) 2π α0α0 +1 k α0 Hence, by the Lebesgue’s theorem, we obtain that the map [α0 , 1] 3 α 7→ Eα (1/α0 ) is continuous. Therefore, there exists M (α0 ) such that sup |Eα ((−i)α ω)| ≤

sup

Eα (1/α0 ) = M (α0 ).

(5.15)

α∈[α0 ,1/2]

ω≤1/α0

Now, the proof of assertion (5.8) follows from (5.14) together with (5.15). Next we show (5.9). First we assume that ω ≥ 2 under the condition 1/2 ≤ α < 1 from the hypothesis. Again it suffices to prove assertion (5.9) for t = 1. We notice that A ≤ 0, and B, a, and b are all positive. Thus |v(r)| = |2rbωA − Bω 2 | = −2rbωA + Bω 2 ≥ Bω 2 > 0. Hence, 1/α

|Kα (r, (−i)α ω) | ≤

1/α

1/α

e−r B ω e−r e−r ≤ ≤ πα | v(r) | παω π

.

Furthermore,  −r  e for r > 1, |Kα (r, (−i)α ω)| ≤ 1π   for r ≤ 1. π Hence, reasoning as in the proof of (5.8) we obtain that there is a positive constant M which in this case does not depend on the value of α ∈ [1/2, 1), so that sup |Eα ((−i)α ω)| ≤ M. ω≥2

Now by an application of the same argument as in (5.15), we can show that there is a M2 > 0 independent of α ∈ [1/2, 1) such that sup |Eα ((−i)α ω)| ≤ M2 . ω≤2

Thus the proof of (5.9) now follows from these last two inequalities.

264

5.1.3

Fractional Partial Differential Equations

Existence of Dynamics

In this part of our section, we state and prove our principal assertion. Theorem 5.1. Let H be a Hilbert space and let A be a positive self-adjoint operator on H. Then there exists a unique strong solution to the problem ( C α α 0 Dt u(t) = (−i) Au(t), t > 0, (5.16) u(0) = u0 , u0 ∈ D(A). Moreover, there is a measure space (Ω, µ), a measurable function a on Ω and a unitary map W : L2 (Ω) → H such that the unique solution of the problem (5.16) has the following representation u(t) = W (Eα ((−it)α a(· ))W −1 u0 . Proof. Let us recall that because of the spectral theorem for a self-adjoint operator A : D(A) ⊂ H → H, there exists a measure space (Ω, µ), and a Borel measurable function a and a unitary map W : L2 (Ω, µ) → H such that the following diagram commutes L2 (Ω, µ)

Ma

W −1

W

 H

/ L2 (Ω, µ) O

A

/H

for each f ∈ L2 (Ω, µ) such that W f ∈ D(A). Moreover, if f ∈ L2 (Ω, µ) is given, then W f ∈ D(A) if and only if Ma f ∈ L2 (Ω, µ), see e.g. [309,327], where Ma f (x) = a(x)f (x). Thus, the spectral theorem ensures us that there exists a unitary map W from L2 (Ω) onto H such that W −1 AW ϕ(ξ) = a(ξ)ϕ(ξ),

ξ ∈ Ω.

(5.17)

Now the proof of the theorem falls naturally into two parts. Uniqueness. Let us assume that u is a strong solution to the problem (5.16). We define v(t, ξ) = (W −1 u(t))(ξ). Then it follows from (5.17) that (W −1 Au(t))(ξ) = W −1 AW v(t)(ξ) = a(ξ)v(t, ξ).

(5.18)

Let us observe that g1−α ∗ v ∈ C 1 ((0, ∞); L2 (Ω)). Indeed, we shall show that   d d g1−α ∗ v = W −1 g1−α ∗ u . dt dt Next we set Θ(t) = g1−α ∗ v. Then using the fact that W is an isometry, we get that

Θ(t + h) − Θ(t)

−1 d

− W ( g1−α ∗ u)

2 h dt L (Ω)

Fractional Schr¨ odinger Equations

265

 

−1 g1−α ∗ u(t + h) − g1−α ∗ u(t)

d

W = − g ∗ u(t) 1−α

2 h dt L (Ω)

g1−α ∗ u(t + h) − g1−α ∗ u(t)

d = − g1−α ∗ u(t)

→ 0, as h → 0, h dt H where the convergence follows from the assumptions on the function u. Moreover, we can easily check that derivative Θ0 is a continuous function, thus we obtain that g1−α ∗ v ∈ C 1 ((0, ∞); L2 (Ω)). Furthermore, by the continuity of W , we obtain   d d g1−α ∗ u = g1−α ∗ W −1 u. W −1 dt dt Thus, from the definition of the Caputo derivative, denote v0 = W −1 u0 , we obtain that   α d −1 ∂ u −1 W (t) = W g1−α ∗ u − u0 g1−α ∂tα dt d = g1−α ∗ W −1 u − v0 g1−α dt ∂ α v(t, · ) = . ∂tα Now if we apply W −1 to both sides of equation (5.16), then we obtain the following equation on L2 (Ω),  α  ∂ v(t, · ) = (−i)α a(· )v(t, · ), t > 0 ∂tα (5.19)  v(0, · ) = v0 , where u0 ∈ D(A). Now, it follows that the unique solution of the above fractional differential equation (5.19) is given by v(t, ξ) = Eα ((−it)α a(ξ))v0 . Since (W −1 u(t))(ξ) = Eα ((−it)α a(ξ))v0 , we get that u is given by u(t) = W (Eα ((−it)α a(· ))W −1 u0 ),

u0 ∈ D(A).

(5.20)

This finishes with the proof of the uniqueness property. Existence. Next, we shall show that u(t) given by formula (5.20) is indeed a strong solution to the initial value problem (5.16). First of all, we prove that u ∈ C(R+ , D(A)). We need to show that u(t) ∈ D(A), for all t ≥ 0. For this purpose let us recall that Ah = W (a(· )(W −1 h)(· )),

for h ∈ D(A).

Thus, by the spectral theorem, we know that h ∈ D(A) if and only if a(· ) (W −1 h)(· ) ∈ L2 (Ω), see [309, 327]. Hence, u0 ∈ D(A) if and only if a(ξ)(W −1 u0 )(ξ) belongs to L2 (Ω). But then, from the fact that ξ 7→ Eα ((−it)α a(ξ)) is bounded by Lemma 5.1, it follows that the function a(ξ)(W −1 u(t))(ξ) = Eα ((−it)α a(ξ))a(ξ)(W −1 u0 )(ξ),

ξ∈Ω

266

Fractional Partial Differential Equations

is in L2 (Ω) for all t ≥ 0 and effectively we get that u(t) ∈ D(A). Moreover, since the mapping t 7→ Eα ((−it)α a(ξ)) is continuous, the map u is continuous. Indeed, let us take t0 , t ∈ R+ , then we have

 

W (Eα ((−i(t + t0 ))α a(· )) − Eα ((−it0 )α a(· )))W −1 u0 H

 

α α −1 . = Eα ((−i(t + t0 )) a(· )) − Eα ((−it0 ) a(· )) W u0 L2 (Ω)

Since (Eα ((−i(t + t0 ))α a(ξ)) − Eα ((−it0 )α a(ξ)) is bounded by Lemma 5.1. Hence, there exists Mα such that (Eα ((−i(t + t0 ))α a(ξ)) − Eα ((−it0 )α a(ξ))(W −1 u0 )(ξ) ≤ Mα |(W −1 u0 )(ξ)|. Thus, by an application of the Lebesgue’s dominated convergence theorem the proof of the continuity of the function u defined in (5.20) is finished. Next, we prove that the map Z t 1 Φ(t) = (t − s)−α u(s)ds Γ(1 − α) 0 belongs to C 1 ((0, ∞); H). For this purpose we consider the following mapping Z t 1 (t − s)−α Eα ((−is)α a(ξ))ds. φ(t) = Γ(1 − α) 0 Once more by the definition of the Caputo derivative we get φ0 (t) = (−i)α a(ξ)Eα ((−it)α a(ξ)) +

1 1 . Γ(1 − α) tα

Now, we shall show that Φ0 (t) = W φ0 (t)W −1 u0 .

(5.21)

Let us notice that   φ(t + h) − φ(t) lim − φ0 (t) W −1 u0 = 0. h→0 h Moreover, by the mean value theorem and Lemma 5.1, we have !   Z 1 t+h 0 φ(t + h) − φ(t) 0 −1 0 −1 = − φ (t) W u φ (s)ds − φ (t) W u 0 0 h h t −1 ≤ C(cα (t) + a(ξ)|)|W u0 , for some constants C and cα (t) independent on h. Since u0 ∈ D(A), we have a(ξ)W −1 u0 , W −1 u0 belongs to L2 (Ω). Moreover,

Φ(t + h) − Φ(t)

0 −1

− W φ (t)W u 0

h H

  (5.22)

φ(t + h) − φ(t) 0 −1

− φ (t) W u = W . 0

h H

Fractional Schr¨ odinger Equations

267

Since W is unitary, it follows that





W φ(t + h) − φ(t) − φ0 (t) W −1 u0

h

H

 

φ(t + h) − φ(t) − φ0 (t) W −1 u0 =

2 .

h L (Ω)

(5.23)

Therefore from (5.22) and (5.23), we obtain

Φ(t + h) − Φ(t)

0 −1

− W φ (t)W u0

h



H 

φ(t + h) − φ(t)

= − φ0 (t) W −1 u0

2 . h L (Ω) Hence by the Lebesgue’s dominated convergence theorem, we have

 

φ(t + h) − φ(t) 0 −1

→ 0, as h → 0. − φ (t) W u0

2

h L (Ω)

Thus, the proof of (5.21) is complete and hence we have the differentiability of the function Φ. Furthermore, arguing as above, we get that Φ0 ∈ C((0, ∞); H). It remains to prove that the function u defined in (5.20) satisfies equation (5.16). In order to show this last claim we compute the Caputo derivative of the u. Thus, C α 0 Dt u(t)

= Φ0 (t) −

1 u0 Γ(1 − α) tα

= W (−i)α a(ξ)Eα ((−it)α a(ξ))W −1 u0 = W (−i)α a(ξ)W −1 W Eα ((−it)α a(ξ))W −1 u0 = (−i)α Au(t), and the whole proof of Theorem 5.1 is now finished. Remark 5.1. Let A be a self-adjoint operator. Then we shall denote by Uα (t) the corresponding solution operator family given by Theorem 5.1. To be more explicit Uα (t)φ = W (Eα ((−it)α a(· ))W −1 )φ, 5.1.4

φ ∈ H, t ≥ 0.

Properties of the Solution Operator

In this subsection, we study the properties of the solution operator Uα . Proposition 5.1. The family {Uα (t)}t≥0 satisfies (i) Uα (t) is strongly continuous for t ≥ 0 and Uα (0) = I; (ii) Uα (t)(D(A)) ⊆ D(A) and AUα (t)x = Uα (t)Ax, for all x ∈ D(A), t ≥ 0. Proof. (i) This follows from the proof of Theorem 5.1. (ii) Using similar consideration as in the proof of Theorem 5.1 we get that Uα (t)(D(A)) ⊆ D(A).

268

Fractional Partial Differential Equations

Next, the commutation property [Uα (t), A] = 0 on D(A) follows from the fact that A = W Ma(· ) W −1 ,

Uα (t) = W MEα ((−it)α a(· )) W −1 , t ≥ 0.

Thus AUα (t)φ = Uα (t)Aφ,

for all φ ∈ D(A), t ≥ 0.

Next, we state some further properties of the solution operator Uα . Proposition 5.2. Let α ∈ (0, 1). Then the solution operator enjoys the following properties. (i) Uα (t)∗ = W Eα ((it)α a(· ))W −1 ,

t > 0.

(ii) Uα (t)Uα (t)∗ = Uα∗ (t)Uα (t) = W |Eα ((it)α a(· ))|2 W −1 , t > 0. (iii) Let e−itA be the unitary group generated by the self-adjoint operator A. Then lim Uα (t)φ = e−itA φ,

α→1−

for every

φ ∈ H,

and

t ≥ 0.

Remark 5.2. In the paper of Dong and Xu [106], it has been pointed out that the quantity kUα (t)u0 k is not conserved during the evolution. Proof. (i) Let us take φ, ψ ∈ H. Using the fact that W is a unitary operator, we get (Uα (t)ψ, φ)H = (W Eα ((−it)α a(· ))W −1 ψ, φ)H = (Eα ((−it)α a(· ))W −1 ψ, W −1 φ)L2 (Ω) Z = Eα ((−it)α a(x))W −1 ψ(x)W −1 φ(x)dx Ω Z = W −1 ψ(x)Eα ((it)α a(x))W −1 φ(x)dx Ω

= (W −1 ψ, Eα ((it)α a(·))W −1 φ)L2 (Ω) = (ψ, W Eα ((it)α a(·))W −1 φ)H . Hence, we obtain Uα (t)∗ = W Eα ((it)α a(· ))W −1 . The proof of (ii) follows from the very definition of Uα (t). Next, we show (iii). We will prove that lim kUα (t)φ − e−itA φkH = 0,

α→1−

t ≥ 0,

Since, W is an isometry and e−itA = W e−ita(ξ) W −1 ,

φ ∈ H.

(5.24)

Fractional Schr¨ odinger Equations

269

we have that kUα (t)φ − e

−itA

φk2H

Z 2 2 = Eα ((−it)α a(ξ)) − e−ita(ξ) (W −1 φ)(ξ) dξ.

(5.25)

Ω

According to Lemma 5.1 part (b), for each t > 0 the function |Eα ((−it)α a(ξ))| is bounded independently of ξ ∈ Ω and α ∈ [1/2, 1). But then, there is M such that for all α ∈ [1/2, 1), and ξ ∈ Ω |Eα ((−it)α a(ξ)) − e−ita(ξ) | ≤ M. Hence |Eα ((−it)α a(ξ)) − e−ita(ξ) ||(W −1 φ)(ξ)| ≤ M |(W −1 φ)(ξ)|. Moreover   lim− Eα ((−it)α a(ξ)) − e−ita(ξ) = 0

α→1 −1

2

and W φ ∈ L (Ω). Then the dominated convergence theorem applies to (5.25) when α → 1− , and thus the proof of (iii) is finished. 5.1.5

An Example

We consider A = −∆, the Laplacian operator on L2 (Rn ). Then by the spectral theorem, we have Au := F −1 (|ξ|2 F)u,

for u ∈ D(A) := S(Rn ).

Next we find the strong solution of the following fractional Schr¨odinger equation. Suppose that 0 < α < 1 and consider the initial value problem  α  ∂ u (t, x) = (−i)α (−∆)u(t, x), t > 0, x ∈ Rn , ∂tα (5.26)  u(0, · ) = g( · ) ∈ C0∞ (Rn ). We will show that the strong solution of (5.26) is defined by a convolution kernel which is given by the Fourier transform in the distributional sense of the MittagLeffler function. To prove this claim we first recall some basic facts. We denote by S(Rn ) and by S 0 (Rn ) the Schwartz space and the space of tempered distributions respectively. Let ϕ be a function of S(Rn ). Then we recall that the action of the dilation operator on ϕ is defined as ϕλ (x) = ϕ(λx), λ ∈ R, x ∈ Rn . Furthermore the action on the Fourier transform F is 1 1 (Fϕ)λ = n Fϕ1/λ and Fϕλ = n (Fϕ)1/λ , λ > 0. (5.27) λ λ 1 If u is a distribution then we recall that huλ , ϕi = n hu, ϕ1/λ i, and the same λ identities as (5.27) are also verified. Next we set e(ξ) = Eα ((−i)α |ξ|2 ),

ξ ∈ Rn .

270

Fractional Partial Differential Equations

Thus, etα/2 (ξ) = Eα ((−it)α |ξ|2 ). We see that the hypotheses of Theorem 5.1 are satisfied. Hence the strong solution of (5.26) is given by u(t, x) = F −1 (etα/2 (Fg))(x).

(5.28)

We notice that the function ξ 7→ etα/2 (ξ) is bounded for each t ≥ 0 by Lemma 5.1. Thus etα/2 defines a tempered distribution by integration. But then, F(etα/2 ) is also a tempered distribution. Now if u ∈ S 0 (Rn ) and ϕ ∈ S(Rn ) then we have that u ∗ ϕ ∈ C ∞ (Rn ) ∩ S 0 (Rn ) and (Fu)(Fϕ) = F(u ∗ ϕ), see [230]. And then, we have that F −1 (etα/2 (Fg)) = (F −1 (etα/2 ) ∗ g) as tempered distributions. Moreover F −1 (etα/2 ) =

1 tnα/2

(F −1 e)1/tα/2 .

Hence taking into account the above considerations we can represent the solution of (5.26) as u(t, x) =

1 tnα/2

((F −1 e)1/tα/2 ∗ g)(x).

Since etα/2 ∈ C ∞ ∩ L∞ , we have that etα/2 (Fg) ∈ S. Thus, from formula (5.28), we get that the function x 7→ u(t, x) belongs to the Schwartz space for each t ≥ 0. Using Proposition 5.2 and the above considerations, we close the section with the following observation. Proposition 5.3. Let φ ∈ S(Rn ), then 1

((F tnα/2 5.2 5.2.1

−1

e)1/tα/2

1 ∗ φ) −→ n α→1 (4πit) 2 L2

Z

ei

|·−y|2 4t

φ(y)dy.

Rn

Nonlinear Schr¨ odinger Equation Introduction

The nonlinear Schr¨ odinger equation received a great deal of attention due to its extensive applications in quantum mechanics, optics, seismology and plasma physics [179, 286]. For more details, see the monographs by Bourgain [53], Cazenave [65] and the references therein. In this section, we consider the following nonlinear time-fractional Schr¨odinger equation ( i∂tα u(t, x) + ∆u(t, x) + λ|u(t, x)|p u(t, x) = 0, t ∈ R+ , x ∈ RN , (5.29) u(0, x) = u0 (x), x ∈ RN ,

Fractional Schr¨ odinger Equations

271

where ∂tα denotes the Caputo fractional derivative of order α ∈ (0, 1), λ ∈ C, p ≥ 0 and some assumptions on p will be specified later. Let v : [0, ∞) × RN → RN . The Caputo fractional derivative of order α for the function v and is defined by Z t 1 ∂ ∂tα v(t, x) = (t − s)−α v(s, x)ds, t > 0. Γ(1 − α) 0 ∂s In this section, we make use of the tools of harmonic analysis to study the existence and decay estimate of solutions for (5.29). The section is organized as follows. In Subsection 5.2.2, we give the concept of solutions for problem (5.29) and discuss some properties of the operators appearing in the solution via the tools of harmonic analysis. Subsection 5.2.3 is concerned with the global existence of solutions for problem (5.29). It is also shown that the solution continuously depends on the initial value and has a sharp decay estimate. Finally, we switch our discussion to Sobolev spaces. 5.2.2

Solution Operators

In this subsection, we give an integral formula which is formally equivalent to (5.29). We denote by Lq (RN ) (q > 0) the space of q-integral functions with the norm k · kq . L denotes the Laplace transform. F and F −1 represent the Fourier s transform and its inverse respectively. Let Hqs := F −1 [(1 + |ξ|2 ) 2 F(Lq (RN ))] and H˙ qs := F −1 [|ξ|s F(Lq (RN ))] denote the Sobolev space and the homogeneous s Sobolev space equipped with the norms kf kHqs := kF −1 [(1 + |ξ|2 ) 2 F(f )]kq and kf kH˙ s := kF −1 [|ξ|s F(f )]kq . In the sequel, ∗ denotes the convolution. q Now we study the formulation of mild solutions to equation (5.29). Let us recall the Wright-type function ∞ X (−θ)k Mα (θ) = k!Γ(1 − α(1 + k)) k=0

and Mittag-Leffler functions ∞ X Eα,1 (t) = k=0

∞

X tk tk , Eα,α (t) = . Γ(αk + 1) Γ(αk + α) k=0

To achieve our purpose, we transform the Cauchy problem (5.29) into an integral equation in the following lemma. Lemma 5.2. If u satisfies problem (5.29), then we have Z t u(t, x) = Sα (t)u0 (x) + iλ (t − τ )α−1 Pα (t − τ )|u(τ, x)|p u(τ, x)dτ, 0

where the solution operators Sα (t) and Pα (t) are given by   Z ∞ i| · |2 α −N −N 2 2 Sα (t)ϕ(x) = (4πit ) θ Mα (θ) exp( )dθ ∗ ϕ (x), 4θtα 0   Z ∞ i| · |2 α −N 1− N 2 2 Pα (t)ϕ(x) = (4πit ) αθ Mα (θ) exp( )dθ ∗ ϕ (x). 4θtα 0

(5.30)

272

Fractional Partial Differential Equations

Proof. Applying the space Fourier transform to (5.29), we get ( i∂tα F(u)(t, ξ) − |ξ|2 F(u)(t, ξ) + λF(|u|p u)(t, ξ) = 0, t > 0, F(u)(0, ξ) = F(u0 ),

(5.31)

which, on taking temporal Laplace transform, leads to the equation sα−1 iλ L[F(u0 )] + α L[F(|u|p u)](s, ξ). α 2 s + i|ξ| s + i|ξ|2

L[F(u)](s, ξ) =

(5.32)

Taking inverse Laplace transform of (5.32) together with the convolution theorem α−1 and the facts that L[Eα,1 (−ztα )](s) = ssα +z and L[tα−1 Eα,α (−ztα )](s) = sα1+z for z ∈ C, we obtain F(u)(t, ξ) =Eα,1 (−i|ξ|2 tα )F(u0 ) Z t (t − τ )α−1 Eα,α (−i|ξ|2 (t − τ )α )F(|u|p u)(τ, ξ)dτ. + iλ 0

Applying the inverse Fourier transform to the above equation yields u(t, x) =F −1 [Eα,1 (−i|ξ|2 tα )F(u0 )] Z t   + iλ (t − τ )α−1 F −1 Eα,α (−i|ξ|2 (t − τ )α )F(|u|p u)(τ, ξ) dτ. 0

According to the convolution theorem for Fourier transform, we obtain F −1 [Eα,1 (−i|ξ|2 tα )F(u0 )] = {F −1 [Eα,1 (−i|ξ|2 tα )]} ∗ u0 (x),   F −1 Eα,α (−i|ξ|2 tα )F(|u|p u)(τ, ξ)

(5.33)

(5.34) ={F −1 [Eα,α (−i|ξ|2 tα )]} ∗ [|u(t, x)|p u(t, x)]. R∞ Using Eα,1 (−z) = 0 Mα (θ) exp(−zθ)dθ for z ∈ C and the Fubini’s theorem, (5.33) takes the form Z ∞  1 N Z −1 2 α exp(ix · ξ) Mα (θ) exp(−θitα |ξ|2 )dθdξ F [Eα,1 (−i|ξ| t )] = 2π RN 0 Z  1 N Z ∞ Mα (θ)dθ exp(ix · ξ − θitα |ξ|2 )dξ = 2π N 0 R Z ∞  i|x|2  N α −N − =(4πit ) 2 θ 2 Mα (θ) exp dθ. 4θtα 0 R∞ Notice that Eα,α (−z) = 0 αθMα (θ) exp(−zθ)dθ for z ∈ C. Using a similar argument for (5.34) yields Z ∞  i|x|2  N N dθ. F −1 [Eα,α (−i|ξ|2 tα )] = (4πitα )− 2 αθ1− 2 Mα (θ) exp 4θtα 0 Consequently, we have Z u(t, x) =Sα (t)u0 (x) + iλ 0

t

(t − τ )α−1 Pα (t − τ )|u(τ, x)|p u(τ, x)dτ.

Fractional Schr¨ odinger Equations

273

Now we introduce the concept of mild solutions with the aid of the above lemma. Definition 5.2. For T > 0, by a mild solution of equation (5.29) on [0, T ] corresponding to an initial value u0 , we mean that u satisfies integral equation (5.30). Lemma 5.3. [65] For t 6= 0, let the operator T (t) be defined by  i| · |2  N ∗ ϕ, T (t)ϕ(x) = (4πit)− 2 exp 4t 0 for all ϕ ∈ S(RN ). If q ∈ [2, ∞] and t 6= 0, then T (t) maps Lq (RN ) continuously to Lq (RN ) and that 1

0

1

kT (t)ϕkq ≤ (4πt)−N ( 2 − q ) kϕkq0 , for ϕ ∈ Lq (RN ). Before proceeding further, we state the fundamental estimates for the solution operators Sα (t) and Pα (t), which will play an important role in proving the existence results. Lemma 5.4. If q ∈ [2, ∞] satisfies N ( 12 − 1q ) < 1 and t 6= 0, then Sα (t) and Pα (t) 0

map Lq (RN ) continuously to Lq (RN ) and there exists a constant C1 such that 1

1

1

1

kSα (t)ϕkq ≤ C1 t−αN ( 2 − q ) kϕkq0 and kPα (t)ϕkq ≤ C1 t−αN ( 2 − q ) kϕkq0 . Proof. By the definitions of Sα (t) and Pα (t), it follows that Z ∞ Sα (t)ϕ(x) = Mα (θ)T (θtα )ϕ(x)dθ, 0 Z ∞ Pα (t)ϕ(x) = αθMα (θ)T (θtα )ϕ(x)dθ. 0

From Lemma 5.3, we have Z kSα (t)ϕkq ≤

∞

Mα (θ)kT (θtα )ϕkq dθ Z ∞ 1 1 1 1 θ−N ( 2 − q ) Mα (θ)dθ · kϕkq0 ≤(4πtα )−N ( 2 − q ) 0

0 1

1

≤C1 t−αN ( 2 − q ) kϕkq0 and Z

∞

αθMα (θ)kT (θtα )ϕkq dθ Z ∞ 1 1 α −N ( 12 − q1 ) θ1−N ( 2 − q ) Mα (θ)dθ · kϕkq0 ≤α(4πt )

kPα (t)ϕkq ≤

0

0

≤C1 t

−αN ( 21 − q1 )

kϕkq0 ,

where   1 1 1 1 Γ(1−N ( 1 − 1 )) Γ(2−N ( 1 − 1 )) C1 = max (4π)−N ( 2 − q ) Γ(1−αN (21 −q1 )) , α(4π)−N ( 2 − q ) Γ(1+α−αN2 ( 1 q− 1 )) . 2

q

2

q

274

Fractional Partial Differential Equations

Lemma 5.5. If q ∈ [2, ∞] satisfies N ( 21 − 1q ) < 1 and t 6= 0, then there exists a constant C1 such that 1

1

1

1

kSα (t)ϕkHqs ≤ C1 t−αN ( 2 − q ) kϕkHqs0 and kPα (t)ϕkHqs ≤ C1 t−αN ( 2 − q ) kϕkHqs0 . Proof. Fix t 6= 0 and let w ∈ S(RN ). Then F −1 [wF(Sα (t)ϕ)] =F −1 [wEα,1 (−i|ξ|2 tα )F(ϕ)] =F −1 {Eα,1 (−i|ξ|2 tα )FF −1 [wF(ϕ)]} =Sα (t)(F −1 [wF(ϕ)]). In view of Lemma 5.4, we get 1

1

kF −1 [wF(Sα (t)ϕ)]kq ≤ C1 t−αN ( 2 − q ) kF −1 [wF(ϕ)]kq0 , for 2 ≤ q ≤ ∞. s

Choosing w = (1 + |ξ|2 ) 2 or w = |ξ|s , the results follows immediately from the definitions of the various Sobolev norms. On the other hand, we can proceed analogously with the operator Pα (t) to obtain the desired results. Remark 5.3. Note that the estimates given in the above lemma can be derived for the norms of H˙ qs (RN ) and H˙ qs0 (RN ). 5.2.3

Existence

This subsection is concerned with the existence of solutions for equation (5.29). In the sequel, we need the assumption that p satisfies p0 < p
2 (p0 < p < ∞ if N ≤ 2), N −2

(5.35)

where p0 is the positive root of the equation N x2 + (N − 2)x − 4 = 0. Let γ=

4 − p(N − 2) . 2p(p + 2)

Denote by Xp the space consisting of all Bochner measurable functions u : (0, +∞) → Lp+2 (RN ) such that sup tαγ ku(t)kp+2 < ∞. t>0

Obviously Xp is a complete metric space equipped with the norm kukXp = sup tαγ ku(t)kp+2 . t>0

Fractional Schr¨ odinger Equations

275

h  Let η ∈ 0, α1 − γ(p + 1) . For convenience, we set     Np Bα = B α 1 − , 1 − αγ(p + 1) , 2(p + 2)   Bη = B α 1 −

   Np , 1 − α γ(p + 1) + η . 2(p + 2)

Notice that Bα ≤ Bη . Theorem 5.2. Let p satisfy (5.35). If there exists ε > 0 such that kSα (t)u0 kXp ≤ ε,

(5.36)

then the equation (5.29) admits a unique solution u(t, x) ∈ Xp satisfying kukXp ≤ 2ε. Proof. Set Ωε = {u ∈ Xp : kukXp ≤ 2ε}. It is easy to see that Ωε is a complete metric space when equipped with the metric d(u, v) = sup tαγ ku(t) − v(t)kp+2 = ku − vkXp . t>0

For u, v ∈ Ωε , it follows from the estimate
||u|p u − |v|p v| ≤ (1 + p) |u|p + |v|p |u − v| and the H¨ older’s inequality that
k|u|p u − |v|p vk p+2 ≤ Cku − vkp+2 kukpp+2 + kvkpp+2 .

(5.37)

p+1

Let an operator Θ be defined by Z t Θu(t, x) = Sα (t)u0 (x) + iλ (t − τ )α−1 Pα (t − τ )|u(τ, x)|p u(τ, x)dτ. 0

From Lemma 5.4, we obtain Z

t

(t − τ )α−1 kPα (t − τ )|u(τ )|p u(τ )kp+2 dτ Z t Np ≤kSα (t)u0 kp+2 + |λ|C1 (t − τ )α[1− 2(p+2) ]−1 ku(τ )|p u(τ )k p+2 dτ p+1 0 Z t Np ≤kSα (t)u0 kp+2 + |λ|CC1 (t − τ )α[1− 2(p+2) ]−1 ku(τ )kp+1 p+2 dτ.

kΘu(t)kp+2 ≤kSα (t)u0 kp+2 + |λ|

0

0

Setting M (t) = sup τ αγ ku(τ )kp+2 , 0≤τ ≤t

276

Fractional Partial Differential Equations

we conclude that tαγ kΘu(t)kp+2 ≤tαγ kSα (t)u0 kp+2 + |λ|CC1 tαγ (M (t))p+1

Z

t

Np

(t − τ )α[1− 2(p+2) ]−1 τ −αγ(p+1) dτ

(5.38)

0 Np

≤ε + |λ|CC1 tα[1−γp− 2(p+2) ] (M (t))p+1

Z

1

Np

(1 − τ )α[1− 2(p+2) ]−1 τ −αγ(p+1) dτ.

0 4−p(N −2) 2p(p+2)

and p satisfies (5.35), we get Np Np < 1, γp + = 1. αγ(p + 1) < 1, 2(p + 2) 2(p + 2) Thus it follows that sup tαγ kΘu(t)kp+2 ≤ ε + |λ|Bα CC1 sup(M (t))p+1 , On the other hand, as γ =

t>0

(5.39)

t>0

that is, kΘu(t)kXp ≤ ε + |λ|Bα CC1 kukp+1 Xp . 1

1 Then we choose ε < ( 2p+2 |λ|B ) p such that ε + |λ|Bα CC1 (2ε)p+1 ≤ 2ε. This α CC1 implies that kΘu(t)kXp ≤ 2ε for u ∈ Ωε , that is, the operator Θ maps Ωε into itself. Following the method of derivation of (5.38), we further obtain from Lemma 5.4 and (5.37) that tαγ kΘu(t) − Θv(t)kp+2 Z t Np ≤λC1 tαγ (t − τ )α[1− 2(p+2) ]−1 k|u(τ )|p u(τ ) − |v(τ )|p v(τ )k p+2 dτ p+1 0 Z t
Np ≤|λ|CC1 tαγ (t − τ )α[1− 2(p+2) ]−1 ku(τ ) − v(τ )kp+2 ku(τ )kpp+2 + kv(τ )kpp+2 dτ 0 Z t Np ≤2|λ|CC1 (2ε)p d(u, v)tαγ (t − τ )α[1− 2(p+2) ]−1 τ −αγ(p+1) dτ 0

=2|λ|Bα CC1 (2ε)p d(u, v). With the choice of ε, we have 1 d(u, v), 2 which shows that Θ is a strict contraction on Ωε . Thus Θ has a fixed point u, which is the unique solution of (5.29). kΘu − ΘvkXp ≤ 2|λ|Bα CC1 (2ε)p d(u, v)
0 tαγ (1 + t)αη kSα (t)(u0 − u00 )kp+2 < ∞ for α[γ(p + 1) + η] < 1, η > 0, then ku(t) − u0 (t)kp+2 ≤ Ct−αγ (1 + t)−αη .

Fractional Schr¨ odinger Equations

277

Proof. Since u and u0 are mild solutions of the equation (5.29) with the initial values u0 and u00 , we can argue as in the proof of Theorem 5.2 to deduce that tαγ ku(t) − u0 (t)kp+2 ≤tαγ kSα (t)(u0 − u00 )kp+2 Z t Np (t − τ )α[1− 2(p+2) ]−1 k|u(τ )|p u(τ ) − |u0 (τ )|p u0 (τ )k p+2 dτ + |λ|C1 tαγ p+1

0 αγ

≤t

kSα (t)(u0 −

u00 )kp+2

p

αγ

+ 2|λ|Bη CC1 (2ε) sup t

0

ku(t) − u (t)kp+2 .

t>0 p

Further, the choice of ε ensures that 2|λ|Bη CC1 (2ε) < 12 . Thus sup tαγ ku(t) − u0 (t)kp+2 ≤ 2 sup tαγ kSα (t)(u0 − u00 )kp+2 , t>0

t>0

that is, ku − u0 kXp ≤ 2kSα (t)(u0 − u00 )kXp . On the other hand, we find that Np

(1 + t)αη (t − τ )α[1− 2(p+2) ]−1 ku(τ ) − u0 (τ )kp+2 ku(τ )kpp+2 + ku0 (τ )kpp+2 Np 1 + t αη ≤2(2ε)p ( ) (t − τ )α[1− 2(p+2) ]−1 τ −αγ(p+1) 1+τ × sup {τ αγ (1 + τ )αη ku(τ ) − u0 (τ )kp+2 }




0≤τ ≤t Np

≤2(2ε)p tαη (t − τ )α[1− 2(p+2) ]−1 τ −αγ(p+1)−αη × sup {τ αγ (1 + τ )αη ku(τ ) − u0 (τ )kp+2 }, 0≤τ ≤t 1+t αη ) ≤ ( τt )αη . Therefore, we have where we have used the inequality ( 1+τ

tαγ (1 + t)αη ku(t) − u0 (t)kp+2 ≤tαγ (1 + t)αη kSα (t)(u0 − u00 )kp+2 + 2|λ|CC1 (2ε)p tα(η+γ) sup {τ αγ (1 + τ )αη ku(τ ) − u0 (τ )kp+2 } 0≤τ ≤t

Z ×

t

(t − τ )

Np α[1− 2(p+2) ]−1

τ −α[γ(p+1)+η] dτ.

0

In view of α[γ(p + 1) + η] < 1 and (5.39), one can obtain sup{tαγ (1 + t)αη ku(t) − u0 (t)kp+2 } t>0

≤ sup{tαγ (1 + t)αη kSα (t)(u0 − u00 )kp+2 } t>0

+ 2|λ|Bη CC1 (2ε)p sup{τ αγ (1 + τ )αη ku(τ ) − u0 (τ )kp+2 }. t>0

Using 2|λ|Bη CC1 (2ε)p < 1 together with supt>0 tαγ (1 + t)αη kSα (t)(u0 − u00 )kp+2 < ∞ leads to the estimate sup{tαγ (1 + t)αη ku(t) − u0 (t)kp+2 } ≤ C2 , t>0

which implies that ku(t) − u0 (t)kp+2 ≤ C2 t−αγ (1 + t)−αη .

278

Fractional Partial Differential Equations

Now we discuss the analogue results in Sobolev spaces. To achieve the purpose, we firstly need the estimate for the nonlinear operator |u|p u on H˙ qs (RN ). Let q = N (p+2) N +ps . Then we have the following estimate:
k|u|p u − |v|p vkH˙ s0 ≤ Cku − vkH˙ s kukpH˙ s + kvkpH˙ s , (5.40) q

q

q

q

where s satisfies n No N 0 ≤ s < min p, , if p ∈ / 2N+ ; 0 ≤ s < , if p ∈ 2N+ . q q Set   1 2 N γ= − +s , 2 p q and denote by Xq the space consisting of all Bochner measurable functions u : (0, +∞) → H˙ qs (RN ) such that sup tαγ ku(t)kH˙ s < ∞. q

t>0

It is obvious that Xq is a complete metric space endowed with the norm kukXq = sup tαγ ku(t)kH˙ s . t>0

q

Secondly, we need to impose a second restriction on s, that is, N p+2 N p+2  − , − . (5.41) s ∈ ∆p := 2 p 2 p(p + 1) Therefore the values of q and γ together with (5.41) ensure that N 2 Np < 1, γp + 1− = 1. αγ(p + 1) < 1, 2(p + 2) 2 q Let us set  if 0 < p < 1,   {0} ∩ ∆p , Λp = [0, p) ∩ ∆p , if p ≥ 1 and p ∈ / 2N+ ,   [0, ∞) ∩ ∆p , if p ≥ 1 and p ∈ 2N+ . Note that p is called to be admissible if the set Λp is nonempty. Since the proofs of the following results are similar to that of Theorems 5.2 and 5.3, we just state the results. Theorem 5.4. Let p be admissible and s ∈ Λp . If there exists ε > 0 such that kSα (t)u0 kXq ≤ ε, (5.42) then the problem (5.29) admits a unique solution u(t, x) ∈ Xq satisfying kukXq ≤ 2ε. Theorem 5.5. Let u0 and u00 satisfy (5.42). Assume that u and u0 are mild solutions of the equation (5.29) with the initial values u0 and u00 , respectively. Then ku − u0 kXq ≤ 2kSα (t)(u0 − u00 )kXq . αγ Moreover, if supt>0 t (1 + t)αη kSα (t)(u0 − u00 )kH˙ s < ∞ for α[γ(p + 1) + η] < 1, q η > 0, then ku(t) − u0 (t)kH˙ s ≤ Ct−αγ (1 + t)−αη . q

5.3

Notes and Remarks

The results in Section 5.1 due to G´ orka, Prado and Trujillo [152]. The contents in Section 5.2 are taken from Peng, Zhou and Ahmad [299].

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Index

approximation(s), 9, 16, 30, 42, 48, 49, 60, 62, 115, 206, 234, 235, 239

fractional Rayleigh-Stokes equation(s), 115, 117, 173, 174, 176, 182 fractional Schr¨ odinger equation(s), 259, 260, 270

backward problem, 115 Besov-Morrey space(s), 115, 173–175 blow-up, 115, 125, 138, 140, 143, 154, 173

global solution(s), 16, 115, 132, 134, 142, 174, 182, 183

Cauchy problem(s), 17, 234, 259, 261, 271 classical solution, 16–18, 32, 35, 39, 65, 125, 161, 205–207, 211, 219, 224, 233, 234, 236, 253, 254, 257 compactness, 63, 64, 115, 119, 167, 189, 194, 238, 245, 253 continuation, 76, 115, 125, 138, 143, 154, 155, 158, 159, 215, 220, 228 convergence Analysis, 202

H¨ older continuity, 16, 39–41, 47, 227 initial value problem(s), 11, 15, 265, 269 Laplace transform, 11, 12, 117, 118, 124, 128, 131, 145, 146, 163, 261, 271, 272 local solution(s), 125, 137 mild solution(s), 15–18, 20–22, 28, 29, 31, 32, 35, 37, 40–42, 46, 49, 50, 60, 76, 87, 91, 94, 102, 115, 118, 119, 121, 124, 125, 131, 132, 134, 135, 138–141, 143, 145, 146, 149, 150, 153–155, 158, 159, 173, 179, 186–189, 191, 195, 199, 200, 205–207, 210–212, 216–218, 220, 221, 224, 228, 231, 233, 236, 244, 254, 259, 271, 273, 276–278 Mittag-Leffler function(s), 1, 9, 66, 92, 207, 208, 235, 260, 261, 263, 269, 271

energy method(s), 16, 76, 206 existence, 15–18, 21, 40–43, 46, 49, 52, 62, 71, 76, 78, 80, 87, 91, 96, 99, 102, 103, 115, 117–119, 125, 132, 149, 159–161, 163, 164, 173, 174, 186, 188, 189, 195, 205–207, 212, 219, 220, 224, 233, 234, 239, 240, 244, 259, 260, 265, 271, 273, 274 Fourier transform(s), 11, 12, 63, 69, 70, 77, 83, 91, 93, 269, 271, 272 fractional calculus, 1, 2, 11, 116, 125, 206 fractional diffusion equation(s), 125, 206 fractional Fokker-Planck equation(s), 205–207 fractional Navier-Stokes equations, 16, 17, 40, 48, 62, 78, 80, 88, 91, 125

optimal control, 16, 62, 73, 259 priori estimate, 165 regularity, 15–18, 32, 115, 117, 124, 160, 161, 163, 170, 205–207, 210, 217, 221, 231, 233–235, 239, 244, 255, 257 regularization, 188, 189, 199, 201–203 301

302

Fractional Partial Differential Equations

semigroup, 7, 12–14, 18, 41, 50, 124, 126, 127, 173, 174, 176, 250 Sobolev space(s), 17, 18, 41, 49, 76, 77, 91, 161, 175, 208, 271, 278 solution operator(s), 91, 94, 96, 104, 115, 118, 124–126, 131, 140, 143, 173, 174, 176, 179, 189, 206, 259–261, 267, 268, 271, 273 uniqueness, 15–18, 21, 40–42, 49, 62, 72, 76, 103, 115, 118, 146, 160, 161, 163,

169, 174, 205–207, 212, 224, 233, 234, 239, 247, 254, 259–261, 265 weak solution(s), 15, 16, 62, 65, 72, 73, 115, 160, 161, 163, 164, 167, 169, 170, 206 well-posedness, 15, 115, 124, 125, 132, 135, 143, 173, 174, 182, 183, 189, 206, 207, 211 Wright-type function, 1, 11, 92