Fractional Elliptic Equations

Fractional Elliptic Equations Alassane Niang ([email protected]) African Institute for Mathematical Sciences (AIMS) Senegal Supervised by: Mouhamed M. Fall [email protected] African Institute for Mathematical Sciences, Senegal

15 June 2014 Submitted in Partial Fulfillment of a Masters II at AIMS

Abstract The fractional Laplacian is a nonlocal operator generated by a Lèvy process. The Lèvy processes occur widely in physics, chemistry, biology, mathematical finance, etc. Recently the stable Lèvy processes, which give rise to equations with the fractional Laplacian, have attracted lots of interest. In this project, we will revise the basic notions related to fractional elliptic problems: functional setting, existence results, maximum principles and regularity estimates.

R´ esum´ e (Fran¸cais) Le Laplacien fractionnaire est un opérateur non local généré par un processus de Lévy. Les processus de Lévy sont très répandus en physique, chimie, biologie, mathématiques financières, etc. Récemment, les processus Lévy stables, qui donnent lieu à des équations avec le Laplacien fractionnaire, ont attiré beaucoup d’intérët. Dans ce projet, nous allons réviser les notions de base relatives aux equations elliptiques fractionnaires: espaces fonctionnels, résultats d’existence, principes du maximum et estimations de régularité.

Declaration I, the undersigned, hereby declare that the work contained in this essay is my original work, and that any work done by others or by myself previously has been acknowledged and referenced accordingly.

Alassane Niang, 15 May 2014. i

Contents Abstract

i

Introduction

1

1 Preliminaries

4

1.1

Fractional Sobolev Spaces W s,p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

1.2

Fractional Laplacian, via the Gagliardo norm and Fourier transform, and its properties

8

1.3

Sobolev inequality and Compact Embedding involving W s,p . . . . . . . . . . . . . . . . 15

1.4

Other Properties of the Fractional Laplacian . . . . . . . . . . . . . . . . . . . . . . . . 20

1.5

Definition of (−∆)s in the space of tempered distribution S 0 . . . . . . . . . . . . . . . 23

2 Variational Formulation of the Elliptic Equation (−∆u = f )

.

27

2.1

Existence and Uniqueness of a Weak Solution . . . . . . . . . . . . . . . . . . . . . . . 27

2.2

Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.3

Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3 Variational Formulation of the Fractional Elliptic Equations ((−∆)s u = f )

37

3.1

Existence and Uniqueness of a Weak Solution . . . . . . . . . . . . . . . . . . . . . . . 39

3.2

Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.3

Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Conclusion

57

References

61

ii

Introduction Fractional differential equations (FDEs) are for many authors the generalization of partial differential equations (PDEs). More precisely, they are the limit cases of the PDEs because of their non-local property. In fact, the differential operators at a given point x depends on the values of the applied function in a small ball about x. But the Fractional Laplacian depends on all values of the function in Rn . Note that, the term fractional doesn’t lead to fraction or fractional number, it is a kind of generalisation of real-order differential operator. Non-local equations have many important applications in different relevant subjects; such as mathematical finance with the price of assets which can be modelled by Levy processes with jump (see [41]), nonlocal minimal surfaces [8, 10], ultra-relativistic limits of quantum mechanic [17], non-local electrostatics [21, 23, 35], the thin obstacle problem [38, 31] etc. See also [12, 38] for further motivations. Non-local equations, in particular fractional differential equations can be elliptic, parabolic or hyperbolic. This is a vague classification since each type of these equations has its specific characterizations. The elliptic equations are the one that doesn’t depends on time, they are called the time-independent equations. We gave many general statements, now to fix some ideas, we consider the following elliptic equation called Poisson’s equation, − ∆u = f,

(0.0.1)

where f ∈ L2 (Rn ) is the nonhomogenous term or the source function and u ∈ H 2 (Rn ). Physically, u(x) can be the density or the concentration of some quantity at a position x. −∆ is the diffusion term in an isotropic medium. Equation (0.0.1) is a second-orderP differential equation since the operator that has the highest order is the Laplace’s operator −∆ = − j ∂j2 . We consider the similar equation of order 2s that involves the so-called fractional Laplacian (−∆)s with s > 0, (−∆)s u = f.

(0.0.2)

The first concern face to a PDE is to give it a sense in a suitable functional space. That is classically to check eventual existence and unicity of a solution in that space. So, in that spirit, does the equation (0.0.2) make sense in some spaces? Maybe some kind of Sobolev spaces. And can even the so-called fractional Laplacian be explicitly defined? By definition, (−∆)s u(x) = C(n, s) lim

Z

ε→0+

Rn \Bε (x)

u(x) − u(y) dy, x ∈ Rn , |x − y|n+2s

where C(n, s) := π −(2s+n/2) Γ(s+n/2) (see [12, Section 3]). Now, if we look at the estimate of the Γ(−s) classic laplacian −∆u(x) = C(n) lim

ε→0+

1 εn+2

Z (u(x) − u(y))dy Bε (x)

where C(n) := π −n/2 Γ(n/2)n(n + 1)(see Lemma 2.2.4 for the proof), we see notable similarities between the two operators. But there is a difference that make those two operators to be sometimes 1

Page 2 very far from being similar, that’s the non-locality. In fact, −∆ depends on the values in Bε (x) but (−∆)s provides all values outside the ball Bε (x). This is also why numerical solutions of equations involving (−∆)s are still an open problem. For the readers that are familiar to the Fourier transform (see Definition 1.0.5), we state the following formula for u ∈ H 2 (Rn ) F (−∆u)(ξ) = |ξ|2 F (u)(ξ).

(0.0.3)

This fact shows interesting regularity properties of differential elliptic equations. In particular, provided u to be smooth enough and compactly supported in Rn , if we consider the equation (0.0.1) in Rn , we have the following Z

2

f dx = Rn

n Z X n i,j=1 R

Z n Z X ∂2u ∂2u ∂2u ∂2u dx = dx = |D2 u|2 dx, 2 2 ∂x ∂x ∂x ∂x n ∂xi ∂xj i j i j Rn i,j=1 R

thus, we see that the L2 -norm of f can estimate the L2 -norm of D2 u. And since −∆u = f =⇒ −∆

∂u ∂xi

=

∂f , ∂xi

if we repeat the previous argument, we have that the first derivative of f can estimate the third order derivative of u and so on up to infinity. These show that we can expect u ∈ H01 to be in H m+2 when f is in H m , m ∈ N. That means u gain two more derivatives compared to f . Further, if m = ∞, we have that u ∈ H m , m = 1, 2, ..., thus u ∈ C ∞ . The above computation is not a proof, since dealing with classic solution we assumed that u was smooth enough and compactly supported. Whereas with the weak solution u ∈ H01 (Ω), the previous computations cannot be carried out. What leads to the regularity of weak solution (see [14, Section 6.3 Regularity], [19, Section F] for documentation). If we come back to our so-called fractional Laplacian applying the Fourier transform, our first idea is to replace 2 by 2s in (0.0.3) as follow: F ((−∆)s u)(ξ) = |ξ|2s F (u)(ξ). This shows that the fractional Laplacian is a pseudo-differential operator. We can see also that it is certainly defined for Cc∞ (Rn ) functions. Now, we come to the main goal of our project. Consider, ( (−∆)s u = f u=0

in Ω, in Rn \Ω.

(0.0.4)

We want to study this fractional elliptic problem, seeing analysis of some notions such as weak solution, well-posedness, maximum principle and regularity which is the most focused notion. This equation is similar to (0.0.1) which has generally Dirichlet condition. In (0.0.4), the diffusion term is the fractional

Page 3 Laplacian operator. The most notable difference is the constraints on the surface ∂Ω for the classic elliptic problem against the volume-constraint for the fractional one in Rn \Ω. In fact, the restriction from the space H s (Ω) onto ∂Ω is not defined in the fractional Sobolev space H s (Ω) when s ∈ [0, 1/2]. Since the restriction to ∂Ω is not defined for functions that are not smooth enough. The importance such case s ∈ [0, 1/2] is that if we want to have solution with jump discontinuity, we should deal with spaces such as H s for s ∈ [0, 1/2] (see [44] and [27, Volume Constraint 1.1]). In the stochastic sense, a path of a sample for a symmetric process with jump is not continuous, it can jump at an exterior point of the bounded domain. The set of those points is the volume-constraint, it is the exterior domain(see [44] and [27, Volume Constraint 3.4]). We wish the reader is more oriented than disoriented, given our lack of precision. Chapter 1 is devoted to the preliminaries mainly referenced in [12]. We will define the fractional Sobolev spaces W s,p and H s , p = 2 with Gagliardo approach. Fractional Laplacian and its important properties will be discussed via Gagliardo norm and Fourier transform, i.e. with the Riesz Potential. We will finally deal with Sobolev inequality and Sobolev embedding where a proof of the compact embedding theorem involving W s,p , derived from [12], will be discussed. In chapter 2, we will study the classic Poisson’s equation. We check the well-posedness, prove the weak and strong maximum principle and give some regularity results derived from [14, Section 6.3 Regularity] and [19, Gilbarg Trudinger]. This is to make some comparison with (−∆)s u = f . Chapter 3 will be devoted to the study of fractional elliptic equations of the form (−∆)s u = f . We will discuss the well-posedness, prove the weak and strong maximum principles which are a result of Fall and Sven(see [15]). We’ll also prove some regularity estimate up to the boundary for the distributional solution of the equation (−∆)s u = f , this also is a result of Fall and Tobias(see [16]).

1. Preliminaries In this chapter, we shall give some definitions and preliminaries notions of fractional Sobolev spaces W s,p and its corresponding Hilbert spaces H s when p = 2. We also define the fractional Laplacian via Gagliardo norm and using Fourier transform with the Riesz potential. After, we give some properties of (−∆)s among other, continuity. The last section will be devoted to the Sobolev inequalities and Sobolev embedding involving W s,p in particular the compact embedding theorem. Most of the tools in this chapter is mainly referenced in [12]. First, let us give some basic tools referenced in [14, 6]. Notice that for our elementary inequalities, we refer to [14, B.2 Elementary inequalities]. 1.0.1 Definition (Schwartz Spaces). Schwartz spaces noted S are spaces of C ∞ functions that decrease rapidly, its topology is defined by the following semi-norms X |Dα ϕ(x)|, N = 0, 1, 2, . . . , pN (ϕ) = sup (1 + |x|)N x∈Rn

|α|≥N

where ϕ ∈ S (Rn ). Thus S (Rn ) = {ϕ ∈ C ∞ (Rn ) such that pN (ϕ) < +∞}. This space has the nice property that the Fourier transform is an isomorphism from S (Rn ) to itself. So by duality, one can define the Fourier transform for elements of S 0 , the dual space of S . It is also the space of tempered distribution. An other important property is that for 1 ≤ p ≤ +∞, S (Rn ) ⊂ Lp (Rn ), Lp is the space of p-integrable functions. Remark also that the space of smooth functions compactly supported in Rn : Cc∞ (Rn ) is a subspace of S (Rn ). Since for Ω an open set in Rn , Cc∞ (Ω) is dense in Lp (Ω) (see [6, Corollary 4.23]) then S (Rn ) is dense in Lp (Rn ). See [22, 34, 39] for more details. 1.0.2 Definition (Convolution). Let f, g : Rn −→ R be measurable functions. We call the convolution of f and g noted f ∗ g by Z f ∗ g(x) := f (x − y)g(y)dy, ∀x ∈ Rn , Rn

such that y 7−→ f (x − y)g(y) is integrable for any x ∈ Rn . the convolution product has important properties combined with mollifier and the Fourier transform. 1.0.3 Definition (Mollifier). We call (ρn )n≥1 a sequence mollifiers if it is a sequence of functions on Rn such that Z ρn ∈ Cc∞ (Rn ), supp ρn ⊂ B(0, 1/n), ρn (y)dy = 1 and ρn ≥ 0, in Rn . Rn

1.0.4 Theorem. Let f ∈ Lp (Rn ) with 1 ≤ p < +∞. Then ρn ∗ f −→ f as n → +∞ in Lp (Rn ). 4

Page 5 We refer to [6] for more tools. 1.0.5 Definition (Fourier Transform). Let ϕ ∈ S , the Fourier transform of ϕ is defined by Z 1 e−iξx˙ ϕ(x)dx, ξ ∈ Rn , F ϕ(ξ) = n (2π) 2 Rn note that one can extend F from S (Rn ) to S 0 (Rn ). The following definition also holds by a simple scaling argument, Z e−2iπξ·x ϕ(x)dx, ξ ∈ Rn . F ϕ(ξ) = Rn

1.0.6 Theorem (Plancherel). Let ϕ ∈ L1 (Rn ) ∩ L2 (Rn ), kϕkL2 (Rn ) = kϕk b L2 (Rn ) = kϕk ˇ L2 (Rn ) .

Proof. We have that kϕk2L2 (Rn ) =

Z

ϕ(x)ϕ(x)dx ¯ = ϕ ∗ ϕ(0) Z −1 = F (F (ϕ ∗ ϕ))(0) = F (ϕ ∗ ϕ)(ξ)dξ, Rn Z = ϕ(ξ) b ϕ(ξ)dξ b = kϕk b 2L2 (Rn ) . Rn

Rn

the proof is similar for ϕ. ˇ 1.0.7 Properties. Let ϕ ∈ Cc∞ (Rn ), then we have that 1) α ϕ(ξ) = (iξ)α ϕ, [ D b

(1.0.1)

f[ ∗ g(ξ) = fb(ξ)b g (ξ).

(1.0.2)

2)

Proof. Since ϕ ∈ Cc∞ (Rn ). For 1), α ϕ(ξ) = [ D

1 n (2π) 2

Z

e−ix.ξ Dα ϕ(x)dx

Rn

Z (−1)|α| = Dxα (e−ix.ξ )ϕ(x)dx n (2π) 2 Rn Z (iξ)α = e−ix.ξ ϕ(x)dx n (2π) 2 Rn = (iξ)α ϕ(ξ). b

Page 6

For 2), thanks to Fubini’s theorem (see [6, Theorem 4.5]), we have

Z f (x − y)g(y)dydx e−2iπξ.x n n R R Z Z e−2iπξ.(x−y) e−2iπξ.y f (x − y)g(y)dydx = n n ZR RZ −2iπξ.(x−y) e f (x − y)dx e−2iπξ.y g(y)dy = Rn Rn Z b e−2iπξ.y g(y)dy = fb(ξ)b g (ξ). = f (ξ) Z

f[ ∗ g(ξ) =

Rn

The following result is very useful as we will see in the forthcoming section 1.2. 2

1.0.8 Theorem. Let f (x) = e−πδ|x| where a > 0. Then we have n 2 fb(ξ) = δ − 2 e−π|ξ| /δ .

Proof. Remark that f ∈ Cc∞ (Rn ), by the scaling x 7→ δ −1/2 x and by the fact that n Y

f (x) =

2

e−π|xi | ,

i=1

we may do the proof for n = 1 and δ = 1. Set 2

f1 (x) := e−π|x| , x ∈ R, we get that, Z fb1 (ξ) =

e

−2iπxξ −π|x|2

e

π(iξ)2

Z

dx = e

R

2

e−π(x+iξ) dx.

R

If we shift the domain of integration from 0 to −ξ in the imaginary axis, we obtain Z 2 2 −π(ξ)2 b f1 (ξ) = e e−πx dx = e−π(ξ) . R

For the definitions of some functional spaces, Lebesgue integral etc., we refer to [1, pages 9 − 18]. 1.0.9 Definition (Regularity of Set). Let k ∈ N and α ∈ (0, 1]. Ω ⊂ Rn is of class C k,α if there exists M > 0 such that for any x ∈ ∂Ω, there exist a ball B = Br (x), r > 0 and an isomorphism ϕ : Q −→ B such that : ¯ ϕ ∈ C k,α (Q),

¯ ϕ−1 ∈ C k,α (B),

ϕ(Q+ ) = B ∩ Ω,

ϕ(Q0 ) = B ∩ ∂Ω

Section 1.1. Fractional Sobolev Spaces W s,p

Page 7

and −1 k ϕ kC k,α (Q) kC k,α (B) ¯ +kϕ ¯ ≤ M.

where Q is a cylinder Q := {x = (x0 , xn ) ∈ Rn−1 R : |x0 | < 1 and |xn | < 1}, and Q+ := {x = (x0 , xn ) ∈ Rn−1 R : |x0 | < 1 and 0 < xn < 1} Q0 := {x ∈ Q : xn = 0}. see [12, Section 1].

1.1

Fractional Sobolev Spaces W s,p

In this section, we give the definition of the fractional Sobolev space W s,p . Let Ω be an open set in Rn . For any real s > 0 and any p ∈ [1, ∞), we define the fractional Sobolev Spaces W s,p (Ω). They are also called Aronszajn, Gagliardo or Slobodeckij spaces,. Let s ∈ (0, 1). For any p ∈ [1, +∞), we define W s,p (Ω) as follows ( ) |u(x) − u(y)| ∈ Lp (Ω × Ω) ; W s,p (Ω) := u ∈ Lp (Ω) : n +s |x − y| p

(1.1.1)

thus W s,p (Ω) is an intermediary Banach space between Lp (Ω) and W 1,p (Ω), endowed with the natural norm Z 1 Z Z p |u(x) − u(y)|p p k u kW s,p (Ω) := |u| dx + dxdy . (1.1.2) n+sp Ω Ω Ω |x − y| where the term Z Z [u]W s,p (Ω) := Ω

Ω

|u(x) − u(y)|p dxdy |x − y|n+sp

1

p

.

is the so-called Gagliardo (semi)norm of u. 0

It is worth noticing that, as in the classical case with s being an integer, the space W s ,p (Ω) is continuously embedded in W s,p (Ω) when s ≤ s0 , see [12, Proposition 2.1]. The same happens also when s0 = 1, provided ∂Ω to be C 0,1 with bounded boundary, see [12, Proposition 2.2]. 1.1.1 Remark. If s ≥ 1, the definition (1.1.1) doesn’t hold. Suppose Ω is connected, any measurable function u : Ω −→ R such that Z Z |u(x) − u(y)|p dxdy < +∞ n+sp Ω Ω |x − y| is actually constant (see, [5, Proposition 2]). And this is strictly related to the fact that Z Z Z 1 |u(x) − u(y)|p p dxdy = C lim (s − 1) |∇u|p dx n+sp s→1− Ω Ω Ω |x − y| for a suitable positive constant C (see, [5, Corollary 4]).

(1.1.3)

Section 1.2. Fractional Laplacian, via the Gagliardo norm and Fourier transform, and its properties Page 8 0

1.1.2 Remark. Suppose Ω of class C 0,1 , we have that W s ,p (Ω) ⊆ W s,p (Ω) for s ≤ s0 and p ∈ [1, +∞) with s, s0 > 1 (see [12, Corollary 2.3]). 1.1.3 Theorem. For any s > 0, the space Cc∞ (Rn ) of smooth functions with compact support is dense in W s,p (Rn ). Proof. see [1, Adams Sobolev Spaces Theorem 7.38]. That is any function in W s,p (Rn ) can be approximated by a sequence of smooth functions compactly supported in Rn . For more general Ω ⊂ Rn , one can have this result with the aid of an extension operator that can extend function from Ω to Rn where Ω must be smooth enough (see [1]). Let Ds,p (Ω) = Cc∞ (Ω)kkW s,p . From theorem 1.1.3, one can have Ds,p (Rn ) = W s,p (Rn ), which is not true for a general set Ω ⊂ Rn . All previous results become more interesting when p = 2, since the spaces W s,2 (Ω) and Ds,2 (Ω) turn out to be Hilbert spaces. We will keep the same notation for Ds,2 (Ω) and for W s,2 (Ω), we will write H s (Ω). These spaces are strictly related to the fractional Laplacian operator (−∆)s (see Proposition 1.2.10) that we are going to introduce in the next section.

1.2

Fractional Laplacian, via the Gagliardo norm and Fourier transform, and its properties

In this section, we will define the fractional Laplacian, via the Gagliardo norm and the Fourier transform in Fourier space. We will also discuss its relation with the Riesz potential that we will give the definition. We will prove that the following definitions are equivalence: Z u(x) − u(y) s dy, (−∆) u(x) = C(n, s)P.V. n+2s Rn |x − y| 1 (−∆) u(x) = − C(n, s) 2 s

Z Rn

x ∈ Rn ,

u(x + y) − 2u(x) + u(x − y) dy, |y|n+2s

(−∆)s u(ξ) = F −1 |ξ|2s F (u)(ξ) ,

x ∈ Rn ,

ξ ∈ Rn .

Fractional Laplacian via Gagliardo norm Let u ∈ S and s ∈ (0, 1), we define (−∆)s as Z Z u(x) − u(y) u(x) − u(y) (−∆)s u(x) = C(n, s)P.V. dy = C(n, s) lim dy. n+2s n+2s ε→0 |x − y| n + R C Bε (x) |x − y|

(1.2.1)

Section 1.2. Fractional Laplacian, via the Gagliardo norm and Fourier transform, and its properties Page 9 where P.V. is an abbreviation commonly used to say ”in the (Cauchy) principal value sense”. It is necessary if s ≥ 12 . C(n, s) is a positive constant that depends only on n and s, precisely given by C(n, s) = π 2s+n/2

Γ(s + n/2) Γ(−s)

(1.2.2)

(see Remark 1.2.3). Note that the factor C(n,s) degenerates when s → 1 since the Gamma function is a meromorphic function with simple pole at each negative integer. 1.2.1 Remark. Due to the singularity of the Kernel, the right hand-side of (1.2.1) is not well defined in general. In the case s ∈ (0, 1/2) the integral in (1.2.1) is not really singular near x. Let u ∈ S then every Dα u is bounded so u is Lipschitz, we have Z Z Z |u(x) − u(y)| |u(x) − u(y)| |u(x) − u(y)| dy = dy + dy n+2s n+2s n+2s Rn |x − y| BR |x − y| C BR |x − y| Z Z |x − y| 1 ∞ n dy + kuk dy ≤ C L (R ) n+2s n+2s |x − y| |x − y| BR C BR Z Z 1 1 ≤ C dy + dy , n+2s−1 n+2s BR |x − y| C BR |x − y| using polar coordinates ρz = y − x with |y − x| = ρ, Z R Z Z 1 |u(x) − u(y)| 1 dy ≤ C dρ + dy n+2s 2s n+2s Rn |x − y| 0 ρ C BR |x − y| up to relabeling the positive constant C that depends only on n and the norm L∞ (Rn ) of u at some steps.

Riesz Potential The Riesz kernel Iα , with α ∈ (0, n), is defined as Iα (x) :=

1 |x|α−n γ(n, α)

where n

γ(n, α) = Π 2 −α

Γ( α2 ) . Γ( n−α 2 )

(1.2.3)

(1.2.4)

T he Riesz potential of a given function f , Iα (f ) is its convolution with the Riesz kernel, i.e. Z 1 f (y) Iα (f ) := Iα ∗ f (x) = dy. (1.2.5) γ(n, α) Rn |x − y|n−α γ(n, α)−1 in (1.2.3) is chosen to simplify computation via Fourier transform. 1.2.2 Proposition. Let α ∈ (0, n). Then F (Iα (x)) = |y|−α , where the equality is in the distributional sense.

(1.2.6)

Section 1.2. Fractional Laplacian, via the Gagliardo norm and Fourier transform, and its properties Page 10 Proof. We follow the proof in [12, Proposition 3.2] Z Z 1 −α |y| ϕ(y)dy = |x|α−n F ϕ(x)dx γ(n, α) Rn Rn

(1.2.7)

for any ϕ ∈ Cc∞ (Rn ). Note that (1.2.7) is equivalent to (1.2.6). 2

2

We have F (e−πδ|x| ) = δ −n/2 e−π|y| /δ , for any ϕ ∈ Cc∞ (Rn ) (see theorem 1.0.8) then Z Z π|y|2 2 − −n e−πδ|x| F ϕ(x)dx, δ 2 e δ ϕ(y)dy = Rn

Rn

so if we multiply by δ Z +∞ Z

n−α−2 2

Rn

0

and integrate over (0, +∞) with respect to δ, we get that Z +∞ Z π|y|2 n−α−2 2 − α+2 − δ 2 e−πδ|x| F ϕ(x)dxdδ. δ 2 e δ ϕ(y)dydδ =

(1.2.8)

Rn

0

Consider the left-hand side of the equality above, by the change variable t = |y|2 /δ, we have Z +∞ Z Z +∞ Z π|y|2 α+2 α−2 δ − 2 e− δ ϕ(y)dydδ = |y|−α t 2 e−πt ϕ(y)dydt 0 Rn 0Z Rn = A |y|−α ϕ(y)dy

(1.2.9)

Rn

where A :=

R +∞

t

0

α−2 2

e−πt dt.

For the right-hand side of (1.2.8), it is similar. If we put τ = |x|2 δ, it follows Z +∞ Z Z +∞ Z (n−α−2 n−α−2 2 δ 2 e−πδ|x| F ϕ(x)dxdδ = |x|α−n τ 2 ) e−πτ F ϕ(x)dxdτ 0 Rn 0Z Rn = B |x|α−n F ϕ(x)dx (1.2.10) Rn

where B :=

R +∞ 0

τ

n−α−2 2

e−πτ dτ .

We get the relation (1.2.7) from (1.2.9) and (1.2.10) by showing that A/B = γ(n, α). Indeed, by a simple change of variable η = πt Z +∞ Z α−2 −α −πt A= t 2 e dt = π 2 0

and Z B=

+∞

τ 0

n−α−2 2

0

e

−πτ

dτ = π

−n+α 2

Z

+∞

η

+∞

α α α η 2 −1 e−η dη = π − 2 Γ( ) 2

n−α −1 2

0

Therefore, from (1.2.11) and (1.2.12), we get that A/B = γ(n, α).

e−η dη = π

n−α 2

Γ(

n−α ). 2

(1.2.11)

(1.2.12)

Section 1.2. Fractional Laplacian, via the Gagliardo norm and Fourier transform, and its properties Page 11 Now, consider the following operator for any function f ∈ S I2s (f ) = (−∆)−s f,

s ∈ (0, n/2).

The fractional Laplacian (−∆)s can be defined as follow, for any s ∈ (0, 1) (−∆)s (u) = −∆(I2−2s (u)) (see [20]) and, by the properties of the Riesz operator (see, e.g., [26, Chapter I]), it follows that (−∆)s (u) = −I−2s (u)

(1.2.13)

This is the reason why one call the fractional Laplacian: the Riesz fractional derivative. 1.2.3 Remark. Now, we can justify the choice of the constant factor C(n, s) in (1.2.1) and in (1.2.2). We note that C(n, s) does not depend on the function u. Thus, if we take a function u such that for some x ∈ Rn , u(x) = 0, we have Z u(y) s (−∆) u(x) = −C(n, s)P.V. dy = −C(n, s)γ(n, −2s)I−2s (u). |x − y|n+2s n R Using (1.2.13) and the definition of C(n, s) in (1.2.2), we have that −I−2s u = −C(n, s)γ(n, −2s)I−2s u, so C(n, s) =

1 γ(n,−2s) .

The following Lemma show that we may write the singular integral in (1.2.1) as a weighted second order differential quotient without the P.V. 1.2.4 Lemma. Let s ∈ (0, 1) and let (−∆)s be the Fractional Laplacian Operator defined in (1.2.1). Then, for any u ∈ Cc∞ (Rn ), Z u(x + y) − 2u(x) + u(x − y) 1 s dy, ∀x ∈ Rn . (1.2.14) (−∆) u(x) = − C(n, s) 2 |y|n+2s Rn Proof. The equivalence of the definitions (1.2.1) and (1.2.14) immediately follow by the standard changing variable formula. Indeed, by set z = y − x and z 0 = x − y, we have Z Z 1 u(x + z) − u(x) 1 u(x − z 0 ) − u(x) 0 (−∆)s u(x) = − C(n, s)P.V. dz − C(n, s)P.V. dz 2 |z|n+2s 2 |z 0 |n+2s Rn Rn Z 1 u(x + z) − 2u(x) + u(x − z) dz. = − C(n, s)P.V. 2 |z|n+2s Rn This definition is useful to get rid of the singularity of the integral at the origin. For u ∈ Cc∞ (Rn ), second order Taylor expansion gives u(x + y) − 2u(x) + u(x − y) ≤ kD2 ukL∞ (B1 (0)) , |y|2

Section 1.2. Fractional Laplacian, via the Gagliardo norm and Fourier transform, and its properties Page 12 which implies that kD2 ukL∞ (B1 (0) u(x + y) − 2u(x) + u(x − y) ≤ , |y|n+2s |y|n+2s−2

(1.2.15)

we note that |y|−n−2s+2 is integrable near 0 for s ∈ (0, 1). Therefore, with the fact that u has compact support, the Lebesgue integral (1.2.14) yields.

An approach via Fourier transform We define the equivalent space of H s (Rn ) via the Fourier transform. Z s n 2 n 2s 2 ˆ H (R ) = u ∈ L (R ) : (1 + |ξ| )|F u(ξ)| dξ < +∞ .

(1.2.16)

Rn

The above definition holds for any real s ≥ 1 but it is not the case via the Gagliardo norm in (1.1.2). We will prove that one can see the fractional Laplacian (−∆)s as a pseudo-differential operator of symbol |ξ|2s . The proof can be found in many papers (see, e.g., [42, Chapter 16], [43, Section 3]). We will follow [43], in which is shown how singular integrals naturally arise as a continuous limit of discrete long jump random walks. First, we prove some useful lemma. 1.2.5 Lemma. Let u ∈ Cc∞ (Rn ), we have u(x + y) − 2u(x) + u(x − y) (x, y) 7−→ ∈ L1 (R2n ), |y|n+2s

Proof. Considering the fact that u ∈ Cc∞ (Rn ) (u and all its derivatives are bounded), we have Z Z |u(x + y) − 2u(x) + u(x − y)| |u(x + y) − 2u(x) + u(x − y)| dy = dy n+2s |y| |y|n+2s n R B1 Z |u(x + y) − 2u(x) + u(x − y)| dy + |y|n+2s Rn \B1 Z Z 2 −n−2s+2 2 = kD ukL∞ |y| dy + kD ukL∞ |y|−n−2s+2 dy B1 Rn \B1 ! Z Z |y|−n−2s+2 dy +

≤ C B1

|y|−n−2s+2 dy

Rn \B1

< +∞ where the positive constant C depends on kk∞ of D2 u. 1.2.6 Lemma. Let s ∈ (0, 1). There exists a positive dimensional C = C −1 (n, s) > 0 constant that depends on s, that is the normalization constant we’ve seen above, such that Z 1 − cos(ξ.y) dy = C|ξ|2s , ∀ξ ∈ Rn . n+2s |y| Rn

Section 1.2. Fractional Laplacian, via the Gagliardo norm and Fourier transform, and its properties Page 13 Proof. We consider here the rotation R(|ξ|e1 ) = ξ, where e1 is the first direction vector in Rn . Set z = (z1 , z2 , · · · , zn ) = |ξ|RT y, where RT is the transpose of R. We have (ξ · y) = (R(|ξ|e1 ) · y) = (|ξ|e1 · RT y) = (e1 · |ξ|RT y) = (e1 · z) = z1 . This result allows us to write that Z Z 1 − cos(ξ · y) 1 − cos(z1 ) −n dy = |ξ| dz n+2s n+2s |y| n |z/|ξ|| Rn ZR 1 − cos(z1 ) = |ξ|2s dz, |z|n+2s Rn R 1) and Rn 1−cos(z dz, computed by Jarohs Sven (see [24], is equal to the constant C(n, s)−1 = C seen |z|n+2s in (1.2.2). Since 1 − cos(z1 ) 1 ≤ n+2s−2 ∈ L1 (Rn ), n+2s |z| |z| we have that

Z Rn

1 − cos(ξ.y) dy = C|ξ|2s . |y|n+2s

1.2.7 Proposition. Let s ∈ (0, 1) and let (−∆)s : S → L2 (Rn ) be the fractional Laplacian operator defined by (1.2.1). Then, for any u ∈ Cc∞ (Rn ), F ((−∆)s u) = |ξ|2s (F u), ∀ξ ∈ Rn .

(1.2.17)

Proof. Using the definition in Lemma 1.2.4 via the weighted second order differential quotient in (1.2.14), set 1 1 (−∆)s u(x) = L u(x) = − C(n, s) 2

Z Rn

u(x + y) − 2u(x) + u(x − y) dy, ∀x ∈ Rn . |y|n+2s

Considering the linear operator L , we want to prove F (L u) = C(n, s)|ξ|2s (F u),

(1.2.18)

for suitable positive constant C depending only on s and n, where ξ is denoted the frequency variable. In view of Lemma 1.2.5, we can get the Fourier tranform in x into the integral in y by the Fubini and Tonelli’s Theorem (see [6, page 91]) in (1.2.18) and we get Z 1 F (u(x + y) − 2u(x) + u(x − y)) F (L u) = − dy 2 Rn |y|n+2s Z 1 eiξy + e−iξy − 2 = − dy(F u)(ξ) 2 Rn |y|n+2s Z 1 − cos(ξ.y) = dy(F u)(ξ) = C −1 (n, s)|ξ|2s F u(ξ). (1.2.19) |y|n+2s Rn Then by Lemma 1.2.5, we obtain the relation (1.2.18).

Section 1.2. Fractional Laplacian, via the Gagliardo norm and Fourier transform, and its properties Page 14 ˆ n ) defined in In the following proposition, we will state and prove the equivalence between space H(R (1.2.16) and the one defined the Gagliardo norm (see (1.1.1)). 1.2.8 Proposition. Let s ∈ (0, 1). Then the fractional Sobolev space H s (Rn ) defined in (1.1.2), for ˆ s (Rn ) defined in (1.2.16). In particular, for any u ∈ H s (Rn ) p = 2, coincides with the space H ZZ Z |u(x) − u(y)|2 [u]H s (Rn ) = C dxdy = C |ξ|2s |F u(ξ)|2 dξ, n+2s |x − y| 2n n R R for a suitable positive constant C depending only on n and s.

Proof. Let y ∈ Rn . Thanks to Plancherel’s Theorem 1.0.6 and the changing variable formula z = x − y, we get

Z Rn

Z Rn

Z Z |u(z + y) − u(y)|2 |u(x) − u(y)|2 dx dy = dzdy |x − y|n+2s |z|n+2s Rn Rn ! Z Z u(z + y) − u(y) 2 dy dz = |z|n/2+s Rn Rn

Z

u(z + .) − u(.) 2

=

2 n dz

n/2+s n |z| R L (R ) 2 Z

u(z + .) − u(.)

=

F

2 n dz |z|n/2+s Rn L (R ) Z Z |eiξz − 1|2 |F u(ξ)|2 dξdz = n+2s |z| n n R Z RZ (1 − cos ξ.z) = 2 |F u(ξ)|2 dzdξ |z|n+2s Rn Rn Z = 2C1 |ξ|2s |F u(ξ)|2 dξ, Rn

where we use Lemma 1.2.6 in the last equality, the positive constant C1 :=

R

Rn

1−cos(z1 ) dz. |z|n+2s

ˆ s holds only for p = q = 2 thanks to Plancherel 1.2.9 Remark. The equivalence of the spaces H s and H Formula. In fact, we cannot go forward and backward between an Lp and an Lq spaces via Fourier transform (see, e.g., the sharp inequality in [2] for 1 < p < 2 and q = p/(p − 1)). That’s why the general fractional space defined via Fourier transform for 1 < p < ∞ and s > 0 does not coincide with W s,p (Rn ) via the Gagliardo norm (see, e.g., [46] for more discussion). Now, we can state the relation between the fractional Laplacian operator (−∆)s and the fractional Sobolev space H s .

Section 1.3. Sobolev inequality and Compact Embedding involving W s,p .

Page 15

1.2.10 Proposition. Let s ∈ (0, 1) and let u ∈ H s (Rn ). Then, s

[u]H s (Rn ) = Ck(−∆) 2 )ukL2 (Rn ) ,

(1.2.20)

for a suitable positive constant depending only on s and n.

Proof. The equality in (1.2.20) follows by combining Proposition 1.2.7 and Proposition 1.2.8. Indeed, s

s

k(−∆) 2 uk2L2 (Rn ) = kF ((−∆) 2 u)k2L2 (Rn ) = Ck|ξ|s F uk2L2 (Rn ) = C[u]2H s (Rn ) .

1.3

Sobolev inequality and Compact Embedding involving W s,p .

We will state the theorem of Sobolev-type inequality, for the proof one can find the original one in the Appendix of [37] and it is for p = 2 (see Theorem 7 there). When p = 2 and s ∈ [1/2, 1), some of the statements may be strenghtened (see, [5]). For a more comprehensive treatement of fractional Sobolev-type inequalities, see [29, 30, 1, 42, 7] and the references therein.

Sobolev inequality 1.3.1 Theorem (Sobolev inequality). Let s ∈ (0, 1) and p ∈ [1, +∞) such that sp < n. Then there exists a positive constant C = C(n, s, p) such that, for any measurable and compactly supported function f : Rn → R, we have Z Z |f (x) − f (y)|p kf kLp∗ (Rn ) ≤ C dxdy. (1.3.1) n+sp Rn Rn |x − y| where p∗ = p(n, s) is the so-called ”fractional critical exponent” and it is equal to np/(n − sp). Consequently, the space W s,p (Rn ) is continuously embedded in Lq (Rn ) for any q ∈ [p, p∗].

Proof. A complete self-contained proof is in [12, Theorem 6.5]. It is important to know that if Ω is a general open subset of Rn , the statement in Theorem 1.3.1 still holds for W0s,p (Ω), namely W0s,p (Ω) ,→ Lp∗ (Ω). Indeed, the functions in W0s,p (Ω) can be extended on all Rn , by setting f equal to zero outside Ω. Whereas, the above embedding is not generally true for W s,p because we cannot always extend a function of W s,p (Ω) to a function of W s,p (Rn ) unless one make some regularity assumptions in Ω.


Page 16

1.3.2 Theorem. Let s ∈ (0, 1) and p ∈ [1, +∞) such that sp < n. Let Ω ⊆ Rn be an extension domain for W s,p . Then there exists a positive constant C = C(n, p, s, Ω) such that, for any f ∈ W s,p (Ω), we have kf kLq (Ω) ≤ Ckf kW s,p (Ω) , (1.3.2) for any q ∈ [p, p∗]; i.e., the space W s,p (Ω) is continuously embedded in Lq (Ω) for any q ∈ [p, p∗]. If, in addition, Ω is bounded, then the space W s,p (Ω) is continuously embedded in Lq (Ω) for any q ∈ [1, p∗]. Proof. See [12, Theorem 6.7].

1.3.3 Remark. A simple change of variable shows that in the critical case q = p∗, the constant C in Theorem 1.3.2 does not depend on Ω. Thus it coincides with the 1/p-power of C in (1.3.1). When sp = n. Notice that when sp → n the critical exponent p∗ goes to ∞. In this case, f belongs to Lq for any q if f is in W s,p (see theorem 6.9 and theorem 6.10 in [12]).

Compact Embedding Theorem We want to state and prove the following compact embedding theorem 1.3.4 for the fractional space W s,p in bounded domains. The proof is self-contained, the main one is inspired in the classical RieszFrechet-Kolmogorov Theorem (see [36, 25]). It doesn’t require, as it is said in [12], the use of Besov or other interpolation spaces, nor Fourier Transform and semigroup flows (see, for instance [11, Theorem 1.5]). For p = q = 2, see [33, Lemma 6.11]. 1.3.4 Theorem. Let s ∈ (0, 1), p ∈ [1, +∞), q ∈ [1, p], Ω ⊂ Rn be a bounded extension domain for W s,p and T be a bounded subset of Lp (Ω). Suppose that Z Z |f (x) − f (y)|p sup dxdy < +∞. (1.3.3) n+sp f ∈T Ω Ω |x − y| Then T is pre-compact in Lq (Ω). Proof. We follow the proof in [12, Theorem 7.1]. We want to prove that T is totally bounded in Lq (Ω). That is, for any ε ∈ (0, 1) there exist β1 , . . . , βM ∈ Lq (Ω) such that for any f ∈ T there exists j ∈ {1, . . . , M } that satisfies kf − βj kLq (Ω) ≤ ε.

(1.3.4)

By the assumption, Ω is an extension domain, so for f ∈ W s,p (Ω), we can find an extension function of f : f˜ ∈ W s,p (Rn ) such that kf˜kW s,p (Rn ) ≤ Ckf kW s,p (Ω) . Since Ω is bounded, for any cube Q that contains Ω, we can have kf˜kW s,p (Q) ≤ kf˜kW s,p (Rn ) ≤ Ckf kW s,p (Ω) . Thus with the fact that Q is a bounded open set, f˜ ∈ Lq (Q) for any q ∈ [1, p].

(1.3.5)


C0 := 1 + sup kf˜kLq (Q) + sup f ∈T

|f˜(x) − f˜(y)| dxdy < +∞, |x − y|n+sp

Z Z

f ∈T

Q

Page 17

Q

from (1.3.3) and (1.3.5). Now, for any ε ∈ (0, 1), set 1



s

ε

ρ = ρε := 

1 q

2C0 n

n+sp 2p

and

 < 1,

n

ερ q η = ηε := . 2 Now, we cover Q with a collection of disjoints cubes Q1 , . . . , QN of size ρ that is N [

Ω⊆Q=

Qj .

j=1

For any x ∈ Ω, we denote j(x) the unique index in 1, . . . , N such that x ∈ Qj(x) . Also, for any f ∈ T , let

Z

1

P (f )(x) :=

|Qj(x) |

(1.3.6)

f˜(y)dy.

Qj(x)

Notice that P (f ) is constant we denote qj (f ), for any j ∈ {1, . . . , N }, and P (f + g) = P (f ) + P (g) for any f, g ∈ T . We can define R(f ) := ρn/q (q1 (f ), . . . , qN (f )) ∈ RN , thus, we see that R(f + g) = R(f ) + R(g). Consider the spatial q-norm in RN as  kvkq := 

N X

1 q

q

|vj |

, ∀v ∈ RN .

j=1

We have the following,

kP (f )kqLq (Ω)

=

N Z X j=1

≤

Qj ∩Ω

N Z X j=1

|P (f )(x)|q dx

|P (f )(x)|q dx = ρn

Qj

= kR(f )kqq ≤

N X

|qj (f )|q

j=1

kR(f )kqq . ρn

(1.3.7)

Section 1.3. Sobolev inequality and Compact Embedding involving W s,p . Now using Hölder inequality with p =

kR(f )kqq =

=

q q−1 ,

Page 18

we have

q Z N X 1 ρn |qj (f )|q = ρn f˜(y)dy |Qj | Qj ρn j=1 Z q N X 1 ˜(y)dy f Qj ρn(q−1)

N X

j=1

≤ ρ

n(1−q)

N X j=1

=

N Z X j=1

1 |Qj |1−q

Z

|f˜(y)|q dy =

|f˜(y)|q dy

Qj

Z

Qj

Q

|f˜(y)|q dy = kf˜kqLq (Q) ≤ C0 , ∀f ∈ T .

So, sup kR(f )kqq ≤ C0 .

f ∈T

Accordingly, the set R(T ) is bounded in RN (with respect to the q-norm of RN ). Since Rn is a finite dimensional space, R(T ) is totally bounded. Therefore, there exist b1 , . . . , bM ∈ RN such that M [

R(T ) ⊆

Bη (bi ),

(1.3.8)

i=1

where Bη are balls of radius η taken in the q-norm of RN . For any i ∈ {1, . . . , M }, bi is represented as bi = (bi,1 , . . . , bi,N ) ∈ RN . for any x ∈ Ω, we set βi (x) := ρ−n/q bi,j(x) where j(x) is as in (1.3.6). βi is constant on Qj . Indeed, let x ∈ Qj , P (βi )(x) =

1 |Qj |

Z

−n q

ρ

−n q

bi,j dx = ρ

bi,j

Qj

= βi (x),

(1.3.9)

so qj (βi ) = ρ−n/q bi,j . Thus, n

R(βi ) = ρ q (q1 (βi ), · · · , qN (βi )) = (bi,1 , · · · , bi,N ) = bi.

(1.3.10)


Page 19

Furthermore, for any f ∈ T kf −

P (f )kqLq (Ω)

=

N Z X j=1

|f (x) − P (f )(x)|q dx

Qj ∩Ω

q Z 1 = f˜(y)dy dx f (x) − |Qj | Qj j=1 Qj ∩Ω Z q N Z X 1 ˜(y)dy dx = f (x) − f |Qj |q Qj j=1 Qj ∩Ω "Z #q N Z 1 X ≤ |f (x) − f˜(y)|dy dx. ρnq Qj ∩Ω Qj N Z X

(1.3.11)

j=1

We have that x, y ∈ Qj , then |x − y|2 ≤ nρ2 =⇒ ρ−n ≤ n

n+sp 2

ρsp |x − y|n+sp .

Using Hölder inequality and Jensen inequality (see [28, 2.2 Theorem]) with the previous equality with p and p/(p − 1), (1.3.11) becomes "Z #q N Z 1 X q kf − P (f )kLq (Ω) = |f (x) − f˜(y)|dy dx ρnq Q ∩Ω Q j j j=1 #q "Z N Z p q(p−1) X 1 p ˜ p |f (x) − f (y)| dy dx ≤ |Q | j ρnq Q Q ∩Ω j j j=1 #q Z "Z p 1 ˜(y)|p dy dx ≤ |f (x) − f ρnq/p Q Qj #q Z "Z p ˜(y)|p q q |f (x) − f 1 ( n+sp ) (n+sp) 2 p p ≤ n ρ dy dx n+sp ρnq/p Qj |x − y| Q #q Z "Z ˜ p ˜(y)|p q | f (x) − f ( n+sp ) sq 2 p ≤ n ρ dy dx n+sp Q Q |x − y| "Z Z #q p ˜(x) − f˜(y)|p q | f ( n+sp ) sq 2 p ≤ n ρ dydx (1.3.12) n+sp Q Q |x − y| εq . 2q Jensen inequality was used in (1.3.12) since t 7→ |t|q/p is concave for any fixed p and q such that q/p ≤ 1. ≤ C0 n

( n+sp ) pq sq 2

ρ

=

Accordingly, if we recall (1.3.7) and (1.3.9), we have for any j ∈ {1, . . . , M } that

kf − βj kLq (Ω) ≤ kf − P (f )kLq (Ω) + kP (βj ) − βj kLq (Ω) + kP (f − βj )kLq (Ω) ≤

ε kR(f ) − R(βj )kq + . 2 ρn/q

(1.3.13)

Section 1.4. Other Properties of the Fractional Laplacian

Page 20

Now, let f ∈ T if we recall (1.3.8) and take j ∈ {1, . . . , M } such that f ∈ Bη (bj ). Then, we have kR(f ) − bj kq ≤ η. Thus, from (1.3.10) and (1.3.13) we get ε kR(f ) − bj kq ε η ≤ + n/q = ε. + n/q 2 2 ρ ρ

kf − βj kLq (Ω) ≤

(1.3.14)

1.3.5 Corollary. Let s ∈ (0, 1) and p ∈ [1, +∞) such that sp < n. Let q ∈ [1, p∗), Ω ⊆ Rn be a bounded extension domain for W s,p and T be a bounded subset of Lp (Ω). Suppose that Z Z |f (x) − f (y)|p dxdy < +∞. sup n+sp f ∈T Ω Ω |x − y| Then T is pre-compact in Lq (Ω). Proof. The case q ∈ [1, p] is proven in Theorem 1.3.4. Now for any q ∈ (p, p∗), set t = t(p, p∗, q) ∈ (0, 1) such that 1/q = t/p + (1 − t)/p∗. Consider βj with j ∈ {1, . . . , N } as in theorem 1.3.4 and let f ∈ T . Thanks to Hölder inequality with p/(tq) and p ∗ /((1 − t)q), we obtain Z kf − βj kLq (Ω) =

qt

|f − βj | |f − βj |

q(1−t)

1 q

dx

Ω

Z ≤

p

t/p Z

|f − βj | Ω

p∗

(1−t)/p∗

|f − βj | dx Ω

t = kf − βj k1−t Lp∗ (Ω) kf − βj kLp (Ω) t ˜ t ≤ Ckf − βj k1−t W s,p (Ω) kf − βj kLp (Ω) ≤ Cε ,

where the relation (1.3.14) and the continuous embedding in Theorem 1.3.2) was used in the last inequality. 1.3.6 Remark. As in the classical case s = 1 (or, more generally, when s is an integer), also in the fractional case, the lack of compactness for the critical embedding (q = p∗) happens, because of translation and dilation invariance (see [32] for more details in this direction, for any 0 < s < n/2). Remark that in Theorem 1.3.4 and Corollary 1.3.5, the regularity assumption on Ω is necessary (see the counterexample in [12, Section 9 Example 9.2]).

1.4

Other Properties of the Fractional Laplacian

In this section, we shall discuss some properties of (−∆)s : convolution product of a function with (−∆)s and prove continuity of (−∆)s ϕ for any ϕ ∈ Cc∞ (Rn ). We will also state its relation with the Kelvin tranform. 1.4.1 Proposition. Let ψ, ϕ ∈ Cc∞ (Rn ), we have


Page 21

1. Translation. If we set ϕa (x) = ϕ(x + a), a ∈ R, ((−∆)s ϕa )(x) = ((−∆)s ϕ)(x + a), x ∈ Rn . 2. Scaling. If we set ϕλ (x) = ϕ(λx), λ ∈ R, ((−∆)s ϕλ )(x) = λ2s ((−∆)s ϕ)(λx).

(1.4.1)

3. Product. s

s

s

Z

(−∆) (ϕψ)(x) = ϕ(x)(−∆) ψ(x)+ψ(x)(−∆) ϕ(x)−C(n, s) Rn

(ϕ(x) − ϕ(y))(ψ(x) − ψ(y)) dy. |x − y|n+2s

4. Convolution. If we take ρn a regularization sequence, ρn ∗ ((−∆)s ϕ)(x) = ((−∆)s (ρn ∗ ϕ))(x), ∀x ∈ Rn .

(1.4.2)

Note that the same holds for the inverse of (−∆)s . Proof. Let ψ, ϕ ∈ Cc∞ (Rn ). For 1), we set the change of variable z = y + a, ϕa (x) − ϕa (y) dy |x − y|n+2s Z ϕ(x + a) − ϕ(y + a) = C(n, s)P.V. dy n+2s Rn |(x + a) − (y + a)| Z ϕ(x + a) − ϕ(z) = C(n, s)P.V. dz n+2s Rn |(x + a) − z| = ((−∆)s ϕ)(x + a).

((−∆)s ϕa )(x) = C(n, s)P.V.

Z

Rn

For 2), with ϕλ (x) = ϕ(λx), λ ∈ R by a simple change of variable z = λy, Z ϕλ (x) − ϕλ (y) s dy ((−∆) ϕλ )(x) = C(n, s)P.V. |x − y|n+2s n ZR ϕ(λx) − ϕ(λy) = C(n, s)P.V. dy |x − y|n+2s Rn Z ϕ(λx) − ϕ(z) −n = λ C(n, s)P.V. dz |x − z/λ|n+2s n ZR ϕ(λx) − ϕ(z) = λ2s C(n, s)P.V. dz n+2s Rn |λx − z| = λ2s ((−∆)s ϕ)(λx).

For 3) with x ∈ Rn , the matter is to do nothing, i.e., to add and remove the same value, Z ϕ(x)ψ(x) − ϕ(x)ψ(y) + ϕ(x)ψ(y) − ϕ(y)ψ(y) s ((−∆) ϕψ)(x) = C(n, s)P.V. |x − y|n+2s Rn Z ϕ(x)ψ(x) − ϕ(x)ψ(x) + ϕ(y)ψ(x) − ϕ(y)ψ(x) + ϕ(x)ψ(y) − ϕ(y)ψ(y) = ϕ(x(−∆)s ψ(x) + d |x − y|n+2s Rn Z (ϕ(x) − ϕ(y))(ψ(x) − ψ(y)) s s = ϕ(x)(−∆) ψ(x) + ψ(x)(−∆) ϕ(x) − C(n, s) dy. |x − y|n+2s Rn


Page 22

For 4), using (1.0.2), we have F (ρn ∗ ((−∆)s ϕ))(ξ) = ρ cn (ξ)F ((−∆)s ϕ)(ξ) = ρ cn (ξ)(|ξ|2s ϕ) b = |ξ|2s ρ cn (ξ)ϕ(ξ) b s = |ξ|2s ρ\ n ∗ ϕ(ξ) = F ((−∆) (ρn ∗ ϕ))(ξ).

for the inverse of the fractional Laplacian, the proof is similarly the same.

Fractional Kelvin transform We define the Kelvin transform and prove its important property with the fractional Laplacian. 1.4.2 Definition. Let s ∈ (0, 1) and consider the mapping x 7→ x∗ = x/|x|2 , x ∈ Rn \{0}. For a function u : Rn 7−→ R ∪ {−∞, +∞}, the Kelvin transform is defined as u∗ (x) := |x|2s−n u(x∗ ) = |x|2s−n u(x/|x|2 ), x ∈ Rn \{0}. x∗ is the inversion of the point x ∈ Rn \ 0 with respect to the unit sphere S1 (0). From this, we can see that the interior of the sphere is transform to the exterior one and vice versa. Thus, we come to the important property of the Kelvin transform which is that: it can transform a s-harmonic function in unbounded domain to a s-harmonic function in bounded domain. We have also the following feature |x − y| =

|x∗ − y ∗ | , x, y ∈ Rn \ {0}. |x∗ ||y ∗ |

(1.4.3)

1.4.3 Proposition. Let s ∈ (0, 1) and u be a smooth function in Rn \{0}. Then (−∆)s u∗ (x) = |x|−n−2s (−∆)s u(x∗ ), ∀x 6= 0Rn . Proof. Let x0 any fixed point in Rn \{0}. We observe that if we set v ∗ = u∗ − u∗ (x0 ) and use the fact that (−∆)s |x|2s−n = 0, we can suppose u(x∗0 ) = u∗ (x0 ) = 0. Indeed, (−∆)s v ∗ (x) = (−∆)s u∗ (x) − (−∆)s (u∗ (x0 ))(x) = (−∆)s u∗ (x) − u(x∗0 )(−∆)s |x|2s−n = (−∆)s u∗ (x). Therefore, s ∗

Z

(−∆) u (x0 ) = C(n, s)P.V. Rn

−u∗ (y) dy. |x0 − y|n+2s

Using the relation (1.4.3) and the changing variable formula z = y ∗ = y/|y|2 ⇒ dz = |y|−2n dy, we get

Section 1.5. Definition of (−∆)s in the space of tempered distribution S 0 .

s ∗

Page 23

−|y|2s−n u(y ∗ ) ∗ ∗ n+2s (|x0 ||y |) dy ∗ ∗ n+2s Rn |x0 − y | Z −u(y ∗ ) −2n = |x0 |−2s−n C(n, s)P.V. dy ∗ − y ∗ |n+2s |y| |x n R 0 Z −u(z) dz = |x0 |−2s−n (−∆)s u(x∗0 ). = C(n, s)P.V. ∗ n+2s Rn |x0 − z| Z

(−∆) u (x0 ) = C(n, s)P.V.

1.5

Definition of (−∆)s in the space of tempered distribution S 0 .

In this section, we will define the fractional Laplacian of a function u belonging to some weighted Lp space. We recall the fractional Laplacian (−∆)s in its nonlocal representation in the real space, Z ϕ(x) − ϕ(y) (−∆)s ϕ(x) = C(n, s)P.V. dy. n+2s n R |x − y| 1.5.1 Lemma. Let Ω ⊂ Rn be a bounded open set. Then there exists a constant C = C(n, s, Ω) > 0 such that ∀ϕ ∈ Cc∞ (Ω) Z kϕkC 2 (Ω) ϕ(x) − ϕ(y) dy ≤ C , ∀x ∈ Rn , ε ≤ 1. n+2s |x−y|>ε |x − y| 1 + |x|n+2s Proof. Ω is bounded, then there exists R0 > 0 such that Ω ⊂ B(0, R0 ). Let ε ≤ 1, noting that ϕ(x) = 0 if x ∈ Rn \B(0, R0 ), we have Z Z ϕ(x) − ϕ(y) |ϕ(y)| dy ≤ dy n+2s n+2s |x−y|>ε |x − y| |x−y|>ε |x − y| Z |ϕ(y)| dy ≤ |x − y|n+2s 0 |y|ε |x − y| 2 BR 0 1 = kϕkC 2 (Ω) |BR0 |(R0 + |x|)−n−2s 2 kϕkC 2 (Ω) |BR0 | ≤ (R0 + 12 |x|)n+2s kϕkC 2 (Ω) |BR0 | ≤ 2n+2s (2R0 )n+2s + |x|n+2s 2n+2s kϕkC 2 (Ω) |BR0 | ≤ C(n, s, R0 ) 1 + |x|n+2s where C(n, s, R0 ) = min(1, (2R0 )n+2s ). Thus for every x ∈ B(0, R), Z kϕkC 2 (Ω) ϕ(x) − ϕ(y) dy . ≤ C(n, s, R0 ) n+2s |x−y|>ε |x − y| 1 + |x|n+2s

(1.5.1)

Now consider x ∈ B(0, R). Note that with an integration by part, Z 1 Z 1 Z 1 d d ϕ(y) − ϕ(x) = ϕ(ty + (1 − t)x)dt = ϕ(x + t(y − x))dt = ∇ϕ(x + t(y − x)).(y − x)dt. 0 dt 0 dt 0 Therefore, Z |x−y|>ε

ϕ(x) − ϕ(y) =− |x − y|n+2s

Z

1Z

∇ϕ(x + t(y − x)). 0

|x−y|>ε

y−x dtdy. |x − y|n+2s

(1.5.2)

Since |x + t(y − x)| < R convexity of the ball, we remark that, |x − y| ≤ Note that

2R R + |x| ≤ , t ∈ (0, 1). t t

" # −n−2s+2  n 2 X −n−2s+2 2   ∇y (|x − y| ) = ∇y (xi − yi ) i=1

n X −n − 2s + 2 −n−2s+2 ∂j (|x−y| )= ×(−2)(xj −yj ) (xi − yi )2 2

"

# −n−2s+2 −1 2

= (n+2s−2)(xj −yj )|x−y|−n−2s ,

i=1

so, ∇y (|x − y|−n−2s+2 ) = (n + 2s − 2)

x−y |x − y|n+2s

then, we get from (1.5.2) with an integration by part, Z Z 1Z ϕ(x) − ϕ(y) = c ∇ϕ(x + t(y − x)).∇y (|x − y|−n−2s+2 )dtdy n+2s |x−y|>ε |x − y| 0 |x−y|>ε Z 1 Z 1 = −c ∆y ϕ(x + t(y − x))|x − y|−n−2s+2 dtdy t 0 |x−y|>ε Z 1Z + c ∇y ϕ(x + t(y − x)).(y − x)|x − y|−n−2s+2 dtdν 0

|x−y|=ε


Page 25

where c = (n + 2s − 2)−1 . Now let us compute the second integral term we note A. Set εσ = y − x, then εn−1 dσ = dν, Z 1Z A := ∇y ϕ(x + t(y − x)).(y − x)|x − y|−n−2s+2 dtdν |x−y|=ε

0

A = ε1−2s

1

Z

Z

0

note,

∇ϕ(x + tεσ).σdσdtk

t S1

1

Z

D2 ϕ(x + tτ εσ)[σ] · σdτ = ∇ϕ(x + tεσ) − ∇ϕ(x)

tε 0

therefore, A=ε

1−2s

1Z

Z

∇ϕ(x).σdσdt + ε 0

2−2s

0

S1

since, 1Z

Z

Z ∇ϕ(x).σdσdt =

0

by oddness of σ 7−→

1Z 1Z

Z

S1

0

n 1X i=1

0

D2 ϕ(x + tτ εσ)[σ].σdσdtdτ,

S1

∂ϕ ∂xi

Z

σ i dσdt = 0,

S1

σi,

then Z 1 Z 2−2s A=ε t 0

1Z

0

D2 ϕ(x + tτ εσ)[σ].σdσdtdτ ≤ ε2−2s kϕkC 2 (Ω) |S1 |.

(1.5.3)

S1

Compute Z B :=

1

Z

∆ϕ(x + t(y − x))|x − y|−n−2s+2 dtdy.

t |x−y|>ε

0

R+|x| t

≤ 2R t , by doing an integration, we have that Z 1 Z 1 B ≤ ε−n−2s+2 ∆y ϕ(x + t(y − x))dtdy 0 t |x−y|>ε Z 1 Z 1 −n−2s+2 = ε ∆y ϕ(x + t(y − x))dtdy 0 t |x−y|=ε Z 1 Z 2−2s = ε t ∇ϕ(x + tεσ)dσdt

With the fact that ε < |x − y| ≤

0

Z ≤

Z

S1

t 0

≤

1

S1 2−2s (2R)

2R t

2−2s ∇ϕ(x + tεσ)dσdt

|S1 |kϕkC 2 (Ω) (1.5.4) 2s with the same change of variable as the previous estimate. From the estimates in (1.5.3) and (1.5.4) (2R)2−2s and C1 := c|S1 | + 1 , we obtain 2s Z ϕ(x) − ϕ(y) dy ≤ C1 kϕkC 2 (Ω) |x−y|>ε |x − y|n+2s 1 + Rn+2s kϕkC 2 (Ω) 1 + Rn+2s kϕkC 2 (Ω) ≤ C(R, n, s) , ∀x ∈ B(0, R) 1 + |x|n+2s = C1

(1.5.5)


Page 26

combining (1.5.1) and (1.5.5), we obtain the desired result. By this result, we have that for any ϕ ∈ Cc∞ (Rn ), (−∆)s ϕ is continuous and for a suitable constant C = C(n, s, Ω) > 0, kϕkCc2 |(−∆)s ϕ(x)| ≤ C , ∀x ∈ Rn . 1 + |x|n+2s Now, put L1s

:=

Z

n

u:R →R:

Rn

|u(x)| dx < +∞ . 1 + |x|n+2s

1.5.2 Definition. Let Ω be an open set, provided u ∈ L1s , the distribution (−∆)s u is defined by Z u(−∆)s ϕdx, ∀ϕ ∈ Cc∞ (Ω). < (−∆)s u, ϕ >= Rn

Saying that (−∆)s u = f in D0 (Ω), is equivalent to the very weak formulation Z Z s f ϕdx, ∀ϕCc∞ (Ω). u(−∆) ϕdx = Rn

Ω

1.5.3 Definition. We define Ds,2 (Ω) = Cc∞ (Ω) with respect to the norm

k kH s

Z Z kukDs,2 (Ω) := If u ∈ Ds,2 (Ω) ⊂ L1s satisfies

R2n

the completion of Cc∞ (Ω) which is a Hilbert space

(ϕ(x) − ϕ(y))2 dxdy |x − y|n+2s

12 .

(−∆)s u = f in D0 (Ω),

we have the weak formulation Z f ϕdx, ϕ ∈ D(Ω),

< u, ϕ >Ds,2 (Ω) = Ω

where ZZ < u, ϕ >Ds,2 (Ω) := C(n, s) R2n Z := |ξ|2s u bϕdξ. b Rn

(u(x) − u(y))(ϕ(x) − ϕ(y)) dxdy |x − y|n+2s

2. Variational Formulation of the Elliptic Equation (−∆u = f ) In this chapter, we study elliptic problems of second order: the Poisson’s equation, on a bounded domain Ω ⊂ Rn . In particular, we shall do the weak formulation, prove existence and uniqueness of the solution. Finally, we study the weak and strong maximum principle and discuss about regularity estimates.

2.1

Existence and Uniqueness of a Weak Solution

In this section, we do the weak formulation and prove the existence of a unique weak solution for the equation (2.1.1) by Lax-Milgram theorem. Let Ω ⊂ Rn an open set bounded and f ∈ L2 (Ω). Consider the following ( −∆u = f in Ω 1 u ∈ H0 (Ω).

(2.1.1)

Let v our test function, by multiplying (2.1.1) and integrating by part in Ω, we get Z Z (−∆)u · vdx = f vdx, v ∈ H01 (Ω) Ω Z ZΩ Z ∂u ∇u · ∇vdx − v· dσ = f vdx, v ∈ H01 (Ω) ∂η ∂Ω Ω Ω or v ∈ H01 (Ω) then

Z v· ∂Ω

∂u dσ = 0, ∂η

therefore, we get the weak formulation of the equation (2.1.1): Z Z ∇u · ∇vdx = f vdx, v ∈ H01 (Ω). Ω

Ω

First, we need to define some notions and state Lax-Milgram’s theorem. 2.1.1 Definition. Let Ω be an open set of Rn , H a Hilbert space and a : H(Ω) × H(Ω) −→ R be a bilinear form. a is said to be: 1) continuous if there is a positive constant C such that |a(u, v)| ≤ CkukH(Ω) kvkH(Ω) ; 2) coercive if there is a constant α > 0 such that a(u, u) ≥ αkuk2H(Ω) . 27

Section 2.1. Existence and Uniqueness of a Weak Solution

Page 28

2.1.2 Theorem. Assume that a is a continuous coercive bilinear form on H. Then, given any l ∈ H ∗ the dual space of H, there exists a unique element u ∈ H such that a(u, v) = l(v), ∀v ∈ H. Proof. See [6, Corollary 5.8]. 2.1.3 Proposition. Let Ω ⊂ Rn an open set bounded and f ∈ L2 (Ω). Consider the equation ( −∆u = f in Ω 1 u ∈ H0 (Ω). There exists a unique weak solution for this equation. Proof. We want to apply Lax-Milgram’s theorem. Consider the bilinear a as

Z ∇u · ∇vdx

a(u, v) = Ω

and the linear form l as

Z

f vdx, v ∈ H01 (Ω).

l(v) = Ω

The proof will be finished if we prove that a is coercive and continuous and l is continuous. We have from Holder inequality that Z |l(v)| = f vdx ≤ kf kL2 (Ω) kvkL2 (Ω) , Ω

thanks to Poincaré’s inequality (see [6, Corollary 9.19]), we have the continuity of l in H01 (Ω. That is |l(v)| ≤ CkvkH01 (Ω) , where C = C(Ω) > 0. For the bilinear form a, using Holder inequality again, we have Z |a(u, v)| ≤ |∇u||∇v|dx ≤ k∇ukL2 (Ω) k∇vkL2 (Ω) = kukH01 (Ω) kvkH01 (Ω) . Ω

Therefore a is continuous in H01 (Ω). For the coercitivity of a, we first show that a(u, u) < +∞. By using Young’s inequality (see [14, B.2 Elementary inequalities]), we have kuk2H 1 (Ω) 0

Z = a(u, u) =

1 1 f udx ≤ kf k2L2 (Ω) + kuk2L2 (Ω) . 2 2 Ω

If we use Poincaré’s inequality and do some rearrangement, we get

Section 2.2. Maximum Principle

Page 29

a(u, u) = kuk2H 1 (Ω) ≤ C 0 kf k2L2 (Ω) < +∞. 0

where

C0

is a positive constant.

Then the coercitivity follows by the fact that 1 a(u, u) = kuk2H 1 (Ω) ≥ kuk2H 1 (Ω) . 0 0 2

The equation (2.1.1) admit a unique solution.

2.2

Maximum Principle

The maximum principle is an important characteristic of solution of second order elliptic equations. It lets us say that a function solution of an equation in a domain reaches its maximum on the boundary of that domain. Let Ω a bounded open subset of Rn .

Weak Maximum Principle The weak maximum principle is applied in the problems of uniqueness and continuous dependence of the solutions on boundaries. It says that the maximum of the function is reached on the boundary of the domain, but can possibly be achieved within the domain. 2.2.1 Proposition. Consider the following equation ( −∆u ≥ 0 in Ω, u≥0

u ∈ H 1 (Ω) on ∂Ω.

(2.2.1)

then u ≥ 0 in Ω. Proof. Define

( u+ := max(u(x), 0) u− := max(−u(x), 0)

thus, u(x) = u+ (x) − u− (x), x ∈ Ω. We define ( ∇u, ∇u+ = 0 and

( ∇u, ∇u− = 0

u≥0 otherwise. u≤0 otherwise.

(2.2.2)


Page 30

We have u+ , u− ∈ H 1 (Ω). Indeed, if u ≥ 0 or u ≤ 0 in Ω, nothing to do. kuk2H 1 (Ω) = kuk2L2 (Ω) + k∇uk2L2 (Ω) Z Z = |u|2 dx + |∇u|2 dx Ω Z ZΩ + − 2 (u − u ) dx + (∇u+ − ∇u− )2 dx = Ω Z Z Z Z Z ZΩ − 2 + 2 + − − 2 + 2 |∇u | dx + −2 ∇u+ ∇u− dx |∇u | dx + |u | dx + −2 u u dx + |u | dx + = Ω

Ω

Ω

Ω

Ω

Ω

since their supports are disjoint, u+ (x)u− (x) = 0 and ∇u+ (x)∇u− (x) = 0, for all x ∈ Ω. Therefore, +∞ > kuk2H 1 (Ω) = ku+ k2H 1 (Ω) + ku− k2H 1 (Ω) , then ku+ kH 1 (Ω) < +∞ and ku− kH 1 (Ω) < +∞. Note that if u ≥ 0, then u = u+ , this implies u− = 0. Test u− Z

∇u.∇u− dx ≥ 0, u− ∈ H01

Ω

thus,

Z

+

−

Z

∇u .∇u dx − Ω

(∇u− )2 dx ≥ 0.

Ω

Therefore, Z − (∇u− )2 dx = −k∇u− k2H 1 ≥ 0 =⇒ k∇u− k2H 1 = 0 =⇒ u− is constant on Ω 0

Ω

0

hence, u− = 0 in Ω since u− = 0 on ∂Ω. ¯ and u ∈ C 2 (Ω) such that 2.2.2 Proposition. Consider that −∆u < 0 in Ω. Let x0 ∈ Ω u(x0 ) := max u(x) ¯ x∈Ω

then x0 ∈ ∂Ω. Proof. By contradiction, suppose x0 ∈ Ω, i.e., x0 is an interior point of Ω. By the condition of the second order, the relation (2.2.3) gives (Hessu)(x0 ) ≤ 0 ⇒ T r (Hessu(x0 )) ≤ 0 ∆u(x0 ) ≤ 0. Contradiction, therefore x0 ∈ ∂Ω. ¯ and u ∈ C 2 (Ω) such that 2.2.3 Proposition. Consider that −∆u ≤ 0 in Ω. Let x0 ∈ Ω u(x0 ) := max u(x) ¯ x∈Ω

then x0 ∈ ∂Ω.

(2.2.3)


Page 31

Proof. Set vε (x) = u(x) + εecx1 , ∀ε > 0 ¯ c ∈ R and x1 a component of x. Then, we have where x ∈ Ω, ∆vε (x) = ∆u(x) + εc2 ecx1 , x ∈ Ω ⇒ −∆vε (x) < 0, in Ω

(2.2.4) (2.2.5)

now, we have the following problem ( −∆vε < 0 1 vε (x0 ) = maxx∈Ω¯ u(x) + εecx0 .

in Ω

where x10 is the first component of x0 and lim vε (x0 ) = u(x0 ) = max u(x) ¯ x∈Ω

ε−→0

hence, according to Proposition 2.2.2, we get the desired result.

Strong Maximum Principle As named, the strong maximum principle gives us strong statement as the solution function doesn’t achieve its maximum in the interior domain unless it is constant. We will state the strong maximum principle in the forthcoming Proposition 2.2.7. First, we need to prove some useful Lemma. 2.2.4 Lemma. Let x0 ∈ Ω and u ∈ C 2 (Ω), then −∆u(x0 ) = Cn lim

r−→0

where Cn =

Z

1 rn+2

(u(x0 ) − u(y))dy. Br (x0 )

2(n+2) |B1 | .

Proof. Since u ∈ C 2 (Ω), using Taylor expansion, we have 1 u(x) = u(x0 ) + ∇u(x0 )(x − x0 ) + Hessu(x0 )(x − x0 )(x − x0 ) + O(|x − x0 |3 ), 2 where Hessu is the Hessian of u, thus integrating over Br (x0 ), we obtain Z Z Z u(x)dx = u(x0 )dx + ∇u(x0 ) · (x − x0 )dx Br (x0 ) Br (x0 ) Br (x0 ) Z Z 1 + Hessu(x0 ) · (x − x0 ) · (x − x0 )dx + O(|x − x0 |3 )dx, 2 Br (x0 ) Br (x0 ) since Z ∇u(x0 ) · (x − x0 )dx = Br (x0 )

Z n X ∂u(x0 ) i=1

∂xi

Br (x0 )

(x − x0 )i dx.


Page 32

Set z = x − x0 , since z 7→ z i is odd, we get Z

z i dz = 0,

Br (x0 )

hence,

Z ∇u(x0 ) · (x − x0 )dx = 0,

(2.2.6)

Br (x0 )

and also Z n X ∂ 2 u(x0 ) Hessu(x0 ) · (x − x0 ) · (x − x0 )dx = (x − x0 )i (x − x0 )j dx, ∂x ∂x i j Br (x0 ) Br (x0 )

Z

i,j=1

with what we set just above and the fact that z 7→ z i is odd, we obtain Z

z i z j dz = 0, ∀i 6= j.

Br (x0 )

Therefore X ∂ 2 u(x0 ) Z (z i )2 dz, 2 ∂x Br (x0 ) i i,j=1

Z Hessu(x0 ) · (x − x0 ) · (x − x0 )dx = Br (x0 )

(2.2.7)

set z = tθ, polar coordinates representation of z, then |θ| = 1 and t > 0. Lebesgue measure is given by dz = tn−1 dtdσ(θ) and

Z

r

Z

i 2

(z ) dz = Br (x0 )

t

n−1 2

Z

t dt

0

(θi )2 dσ(θ).

S n−1

For some notions in polar coordinate and Lebesgue, we refer to [18, Chapter 0, Definitions and notations]. since θ ∈ S n−1 , then

n X (θi )2 = 1 i=1

we get the following Z

i 2

(z ) dz = Br (x0 )

= = = =

n Z rn+2 1 X (θi )2 dσ(θ) n+2n n−1 i=1 S Z n X rn+2 1 (θi )2 dσ(θ) n + 2 n S n−1 i=1 Z n+2 r 1 dσ(θ) n + 2 n S n−1 rn+2 |S n−1 | rn+2 = |B1 (x0 )| n+2 n n+2 r2 |Br (x0 )| n+2

(2.2.8)


Page 33

combining (2.2.6), (2.2.7) and (2.2.8), we obtain n

1 X ∂ 2 u(x0 ) r2 u(x0 )dx + u(x)dx = |Br (x0 )| + |Br (x0 )|O(|r|3 ) 2 ∂x2i n + 2 Br (x0 ) Br (x0 ) Z

Z

i=1

which implies that ∆u(x0 ) +

1 2(n + 2) r2 3 O(r ) = r2 2(n + 2) |Br (x0 )|

Z (u(x) − u(x0 ))dx, Br (x0 )

then, we have that

Z 2(n + 2) 1 ∆u(x0 ) = lim (u(x) − u(x0 ))dx r−→0 r2 |Br (x0 )| Br (x0 ) Z 2(n + 2) 1 (u(x) − u(x0 ))dx = lim r−→0 r n+2 |B1 | Br (x ) 0 Z 1 = Cn lim n+2 (u(x) − u(x0 ))dx, r−→0 r Br (x0 ) where Cn =

2(n+2) |B1 | ,

we multiply the last line by −1, we get that −∆u(x0 ) = Cn lim

r−→0

1 rn+2

Z (u(x0 ) − u(y))dy. Br (x0 )

This average of the classic Laplacian shows its most notable difference with the fractional one (see remark 3.2.3). 2.2.5 Lemma (Mean value property). Let x ∈ Ω, u ∈ C 2 (Ω) superharmonic, i.e., −∆u ≥ 0 in Ω and BR (x) ⊂⊂ Ω a ball of radius R > 0 centered at x. Then Z 1 u(x) ≥ u(y)dy. |BR (x)| BR (x) 2.2.6 Remark. Note that if −∆u = 0 in Ω, then we have the following classical mean value property Z 1 u(x) = u(y)dy. |BR (x)| BR (x) The proof is similar to the following one(proof of the above Lemma) or see [45]. Proof. We call Br (x) and Sr (x) respectively the ball of radius r centered at x and its corresponding sphere. Br (x) := {y ∈ Rn : |x − y| < r} Sr (x) := {y ∈ Rn : |x − y| = r}


Page 34

Let 0 < r ≤ R, from Green’s identities with v a test function and integrating over the ball Br (x), we obtain Z Z Z ∂u ∆u.vdy = − ∇u.∇vdy + .vdσ Br (x) Br (x) Sr (x) ∂ν where ν is the outward normal vector, ν points out of Br (x) and ∂Br (x) = Sr (x) known. Now, set v ≡ 1 then

Z

Z

Br (x)

Z

∂u dσ = ∂ν

∆udy = Sr (x)

∇u.νdσ, Sr (x)

since ν = σr , it follows Z ∆udy = r

−1

Z ∇u(σ).σdσ Sr (x)

Br (x)

= r−1

Z ∇u(x + σ).σdσ, Sr (0)

σ r

with the normalization z = Z

⇒ rn−1 dz = dσ, we get Z ∆udy = rn−1 ∇u(x + rz).zdz

Br (x)

S1 (0)

So according to (2.2.9) the function r 7→ estimates

R

!

Z

d = rn−1 dr

u(x + rz)dz

≤ 0.

(2.2.9)

S1 (0)

S1 (0) u(x + rz)dz

Z

is non-increasing, then we get the following

Z u(x + rz)dz ≤

lim

u(x + rz)dz

r−→0+

S1 (0)

Z

S1 (0)

u(x)dz = |S n−1 |u(x)

= S1 (0)

= n|B1 (0)|u(x).

(2.2.10)

Thus, using the polar coordinates, we obtain that Z

Z

R

u(y)dy = BR (x)

=

0 Rn

r Z

n−1

Z dr

u(x + rz)dz S1 (0)

u(x + rz)dz n S1 (0) Rn ≤ n|B1 (0)|u(x) n = Rn |B1 (0)|u(x) = |BR (0)|u(x)

= |BR (x)|u(x).


Page 35

We get the third line by using (2.2.10). Therefore 1 u(x) ≥ |BR (x)|

Z u(y)dy. BR (x)

Now let us state the strong maximum principle proposition. 2.2.7 Proposition. Let Ω ⊆ Rn be an open and connected set, B ⊂⊂ Ω strongly included in Ω, i.e., the closure of B is compact in Ω and u ∈ C 2 (Ω). Now, Consider   −∆u ≥ 0, in Ω; (2.2.11) u > 0, in B;   u≥0 in Ω, then u > 0 in Ω. Proof. We have that u > 0 in B ⊂⊂ Ω. Let x ∈ Ω such that ∃r > 0 : Br (x) ∩ B 6= ∅ then ∃y ∈ Br (x0 ) : u(y) > 0, so, 1 u(x0 ) ≥ |Br (x0 )|

Z u(x)dx > 0, since u ≥ 0 in Ω. Br (x0 )

therefore, for any x ∈ Ω such that Br (x) ∩ B 6= ∅, u(x) > 0. Let’s define AB := {x ∈ Ω : ∃r > 0 : Br (x) ∩ B 6= ∅} since Ω is connected, for any x in Ω, the ball Br (x) such that Br (x) ∩ B 6= ∅ always exists. Then AB = Ω.The proof is complete.

Section 2.3. Regularity

2.3

Page 36

Regularity

After we study the well-posedness of the classic Poisson’s equation with boundary values, that is to prove existence and uniqueness of the weak solution, it is worth to know if the solution is smooth or not. Thus, this leads us to the regularity matter. Note that we study the regularity of a function to be able to approximate it to an other function we can represent numerically in the computer. In this section, we will give some regularity estimate of the weak solution of Poisson’s equation. We shall not give the proof of the result since it is not actually the aim of this present work but can help in understanding regularity estimates concerning the fractional Poisson’s equation. We refer to [19, 6.3 The Dirichlet Problem] for more discussion. Let Ω ⊂ Rn be a bounded open set. We have the following interior estimate theorem. 2.3.1 Theorem. Let u be a C 2 (Ω) solution of −∆u = f in an open set Ω, f ∈ C α (Ω). Then u ∈ C 2,α (Ω). Proof. See [19, Lemma 6.16] by observing that the required assumptions on the coefficients of the operator are verified. ¯ Then the Dirichlet 2.3.2 Theorem. Assume that Ω is C 2,α bounded domain in Rn and f ∈ C α (Ω). Problem ( −∆u = f u=0

in on ∂Ω,

¯ has a unique solution u ∈ C 2,α (Ω). Proof. See [19, Theorem 6.14] noting that the required assumptions on the coefficients are verified. This theorem state a global regularity under the assumption that Ω is sufficiently smooth.

3. Variational Formulation of the Fractional Elliptic Equations ((−∆)su = f ) In this chapter, we study Elliptic problems of order 2s with s ∈ (0, 1) on a bounded domain Ω ⊂ Rn . We shall do the weak formulation and prove existence and uniqueness of the weak solution of (−∆)u = f . Then, we will study the maximum principle and discuss about the regularity estimates. First of all, we will define some basic notions and fix some notations we’ll use through this chapter. Let Ω ⊂ Rn be a bounded open set with Lipschitz boundary, s ∈ (0, 1). Consider the bilinear form Ds,2 (Ω) : Cc∞ (Ω) × Cc∞ (Ω) =⇒ R Z Z (u , v) 7−→ R2n

(u(x) − u(y))(v(x) − v(y)) dxdy. |x − y|n+2s

which is a scalar product on Cc∞ (Ω). We recall the hilbert space Ds,2 (Ω) the completion of Cc∞ (Ω). 3.0.3 Lemma. If Ω is a bounded Lipschitz open set, Ds,2 (Ω) = {u ∈ H s (Rn ), such that u = 0 in Rn \Ω} .

Proof. Since Ω is Lipschitz, Cc∞ (Ω) is dense in H0s (Ω) with respect to the Ds,2 (Ω)-norm. Then H0s (Ω) ⊆ Ds,2 (Ω) (see [12, Theorem 2.4] by extending). kk[]

Let u ∈ Ds,2 (Ω), since Ds,2 (Ω) = Cc∞ (Ω) s,2 , there exists a sequence (un ) ⊂ Cc∞ (Ω) such that un −→ u a.e. in Rn . Thus for a.e. x ∈ Rn \Ω and ε > 0, there exists nε > 0 such that for any n > nε , |u(x)| = |un (x) − u(x)| < ε, ∀ε > 0 then u = 0 in x ∈ Rn \Ω. We have also that Ω is bounded, so u ∈ H s (Rn ). Therefore, u ∈ H0s (Ω) which implies that Ds,2 (Ω) ⊆ H0s (Ω). For s = 1, Ds,2 (Ω) = H01 (Ω). The norm in Ds,2 (Ω) is Z Z kukDs,2 (Ω) =

R2n

|u(x) − u(y)|2 dxdy |x − y|n+2s

12 ,

and this norm is equivalent to the H s -norm according to the following proposition. 3.0.4 Proposition. Let s ∈ (0, 1) and Ω a bounded open subset of Rn . Let u : Ω −→ R be a measurable function compactly supported. Then, there exists a positive constant C = C(n, s, Ω) > 0 depending o n, s and Ω such that kukL2 (Ω) ≤ CkukDs,2 (Ω) . 37

(3.0.1)

Page 38 Proof. Consider the following estimate, ZZ R2n

|u(x) − u(y)|2 dxdy ≥ |x − y|n+2s

Since

Z

Z

Z

2

u (x)

|x − y|−n−2s dxdy

(3.0.2)

Rn \Ω

Ω

2s

|x − y|−n−2s ≥ C(n, s)|Ω|− n .

(3.0.3)

Rn \Ω

Indeed, here we follow th proof in [12, Lemma 6.1]. Set ρ :=

|Ω| ωn

1

n

.

Notice that Ω and the ball Bρ of radius ρ have the same measure. Thus it follows that |(C Ω ∩ Bρ (x)| = |Bρ (x)| − |Ω ∩ Bρ (x)| = |Ω| − |Ω ∩ Bρ (x)| = |Ω ∩ C Bρ (x)|. Therefore, Z CΩ

dy |x − y|n+2s

Z

Z dy dy + n+2s n+2s (C Ω)∩Bρ (x) |x − y| (C Ω)∩C Bρ (x) |x − y| Z Z dy dy + n+2s n+2s (C Ω)∩Bρ (x) ρ (C Ω)∩C Bρ (x) |x − y| Z |C Ω ∩ Bρ (x)| dy + n+2s n+2s ρ (C Ω)∩C Bρ (x) |x − y| Z |Ω ∩ C Bρ (x)| dy + n+2s n+2s ρ (C Ω)∩C Bρ (x) |x − y| Z Z dy dy + n+2s |x − y| |x − y|n+2s Ω∩C Bρ (x) (C Ω)∩C Bρ (x) Z dy n+2s C Bρ (x) |x − y|

= ≥ = = ≥ =

If we use polar coordinates centered at x, i.e., rz = y − x with r = |x − y|, Z ∞ Z dy = |S | r−2s−1 dr 1 n+2s |x − y| ρ CΩ = |S1 |

2s ρ−2s = C(n, s)|Ω|− n . 2s

Therefore, the relations (3.0.2) and (3.0.3) give the desired result. Now let Ω be a bounded open set with Lipschitz boundary, u ∈ Ds,2 (Ω) and f ∈ L∞ (Ω), we consider the following problem ( (−∆)s u = f in Ω (3.0.4) u=0 in Rn \Ω. ∗

We observe that u ∈ Ds,2 (Ω) ⊂ L2 ⊂ L1s implies that (−∆)s u can be define in the sense of distribution as in Section 1.5.


3.1

Page 39

Existence and Uniqueness of a Weak Solution

In this section, we are interested in the existence and the uniqueness of a weak solution of the above Dirichlet problem, that is the well-posedness of the equation. The problem (3.0.4) is weakly formulated as follows: Z R2n

(v(x) − v(y))(u(x) − u(y)) dydx − |x − y|n+2s

Z f (x)v(x)dx = 0,

v ∈ Ds,2 (Ω).

(3.1.1)

Ω

After we establish the weak formulation of the problem, we want to check if the latter is well-posed, i.e., there exists a unique solution depending continuously on f . 3.1.1 Proposition. Consider the following variational problem P Find u ∈ Ds,2 (Ω) such that , J(u, u) = minv∈Ds,2 (Ω) J(v, v) , where

Z J(u, v) =< u, v >Ds,2 (Ω) −

f vdx. Ω

Then, P admits a unique solution. We need to prove some results before the proof of the Proposition, 3.1.2 Lemma. Let J be such that Z J(u, v) =< u, v >Ds,2 (Ω) −

f vdx. Ω

Then J is continuous.

Proof. Let J(u, v) = a(u, v) − l(v) such that Z (u(x) − u(y))(v(x) − v(y)) a(u, v) = dydx |x − y|n+2s R2n Z l(v) = f vdx. Ω

We have

Z |l(v)| = f vdx ≤ kf kL2 (Ω) kvkL2 (Ω) , Ω

so the linear form l is continuous.

(3.1.2)


Page 40

Using Cauchy-Schwartz inequality (see [14, B.2 Elementary inequalities]) and Proposition 3.0.4, we have Z (u(x) − u(y))(v(x) − v(y)) |a(u, v)| = dydx n+2s |x − y| 2n ZR |u(x) − u(y)||v(x) − v(y)| ≤ dydx |x − y|n+2s 2n R ≤ C(n, s, Ω)kukL2 (Ω) kvkL2 (Ω) ≤ C(n, s, Ω)kukDs,2 (Ω) kvkDs,2 (Ω) , up to relabeling the constant C(n, s, Ω) depending only on n, s and Ω. Hence the bilinear form a is continuous. Therefore J is continuous. 3.1.3 Lemma. Let (un )n≥1 ⊂ Ds,2 (Ω) be a minimizing sequence, i.e., lim J(uk , uk ) =

k→+∞

inf

v∈Ds,2 (Ω)

J(v, v) = I(Ω)

(3.1.3)

then, (uk )k≥1 is bounded in Ds,2 (Ω). Proof. We know that (3.1.3) implies the following 1 I(Ω) ≤ J(ϕ, ϕ) < I(Ω) + , ∀ k ≥ 1, ϕ ∈ Ds,2 (Ω) k

(3.1.4)

Let ϕ ∈ Cc∞ (Ω), with Cauchy-Schwartz inequality and (3.1.4), we get that Z I(Ω) ≤ J(ϕ, ϕ) = kϕk2Ds,2 (Ω) − f ϕdx Ω

≤ kϕk2Ds,2 (Ω) + kf kL2 (Ω) kϕkL2 (Ω) I(Ω) < +∞

(3.1.5)

Since I(Ω) ≤ J(uk , uk ) < I(Ω) + k1 , ∀ k ≥ 1. and J(uk , uk ) =

kuk k2Ds,2 (Ω)

Z −

f uk dx.

(3.1.6)

Ω

Using Cauchy-Schwartz inequality and Young’s inequality (see [14, B.2 Elementary inequalities]), we have Z f uk dx ≤ kf kL2 (Ω) kuk kL2 (Ω) Ω

≤

1 1 kf k2L2 (Ω) + kuk k2L2 (Ω) 2 2

Combining (3.1.6) and (3.1.7), we get J(uk , uk ) ≥ kuk k2Ds,2 (Ω) −

1 kf k2L2 (Ω) + kuk k2L2 (Ω) 2

kuk k2Ds,2 (Ω) ≤ J(uk , uk ) +

1 kf k2L2 (Ω) + kuk k2L2 (Ω) 2

hence


Page 41

then (3.1.4) and (3.1.5) with Proposition 3.0.4 gives us that 1 1 + kf k2L2 (Ω) + kuk k2L2 (Ω) k 2 1 1 < I(Ω) + + kf k2L2 (Ω) + C(n, s, Ω)kuk kDs,2 (Ω) k 2

kuk k2Ds,2 (Ω) < I(Ω) +

where C(n, s, Ω) is a positive constant depending only on n, s and Ω. So, C(n, s, Ω)kuk k2Ds,2 (Ω) < I(Ω) +

1 1 + kf k2L2 (Ω) −→ R ∈ R, as k → +∞ k 2

where R = I(Ω) + 12 (kf k2L2 (Ω) , up to relabeling the constant positive C(n, s, Ω). Then (uk )k≥1 is bounded in Ds,2 (Ω). Proof. Proposition 3.1.1 With the regularity assumptions of Ω, the compact embedding Theorem 1.3.4 with the fractional H s norm tells us that Ds,2 (Ω) is compactly embedded (pre-compact) in L2 (Ω). Since (uk )k≥1 is bounded in Ds,2 (Ω) (Lemma 3.1.3), then there exists a subsequence (ukl )l≥1 of (uk )k≥1 such that ukl * u ∈ Ds,2 (Ω) (weak convergence) and also ukl −→ u ∈ L2 (Ω) that implies a pointwise convergence in Ω and a weak one, i.e., ukl * u ∈ L2 (Ω), as l −→ +∞ Notice that we have here three different convergences with different subsequence, but we keep the same notation. With the pointwise convergence, we get the same limit by unicity of the limit. Therefore, we get that Z J(ukl , ukl ) = so, Z R2n

R2n

|ukl (x) − ukl (y)|2 dydx − |x − y|n+2s

|ukl (x) − ukl (y)|2 dydx < |x − y|n+2s

Z

1 f ukl dx ≤ I(Ω) + , l Ω

∀l > 0.

Z

1 f ukl dx + I(Ω) + , l Ω

∀l > 0.

If we apply Fatou’s lemma(see [6, Lemma 4.1]), we have

Z

Z |ukl (x) − ukl (y)|2 dydx ≤ lim inf f ukl dx + I(Ω) l−→+∞ Ω |x − y|n+2s Z |u(x) − u(y)|2 dydx ≤ lim inf f ukl dx + I(Ω). n+2s l−→+∞ Ω R2n |x − y|

lim inf R2n l−→+∞ Z

Since (ukl )≥1 converge weakly to u in L2 (Ω), we have Z Z Z f ukl dx = lim f ukl dx = f udx lim inf l−→+∞ Ω

l−→+∞ Ω

Ω

Section 3.1. Existence and Uniqueness of a Weak Solution then Z R2n

|u(x) − u(y)|2 dydx ≤ |x − y|n+2s

Page 42

Z f udx + I(Ω). Ω

Therefore, Z J(u, u) = R2n

|u(x) − u(y)|2 dydx − |x − y|n+2s

Z f udx ≤ I(Ω). Ω

So, J(u, u) ≤ I(Ω)

=⇒

J(u, u) = I(Ω).

J has a minimizer u ∈ Ds,2 (Ω). To prove the uniqueness of that solution, let u1 , u2 ∈ Ds,2 (Ω) be solutions of (3.0.4), i.e., ( (−∆)s u1 = f in Ω u1 = 0 in Rn \Ω. and

( (−∆)s u2 = f u2 = 0

set w = u1 − u2 then we have

in Ω in Rn \Ω.

( (−∆)s w = 0 in Ω w=0 in Rn \Ω.

(3.1.7)

The weak formulation of the equation (3.1.7), with w the test function, is Z |w(x) − w(y)|2 dydx = 0, w ∈ Ds,2 (Ω) n+2s |x − y| 2n R which implies with inequality (3.0.1) that kwk2L2 (Ω) ≤ C(n, s, Ω)kwk2Ds,2 (Ω) = 0 =⇒ kwk2L2 (Ω) = 0 hence w = 0 a.e. in Ω, u1 = u2 a.e. in Ω. This finishes the proof. 3.1.4 Remark. Remark that we use the variational method to prove existence and uniqueness of the weak solution of equation (3.0.4). While we could use Lax-Milgram to do the same as we did with the classical Poisson’s equation. 3.1.5 Proposition. Equation (3.0.4) has a unique solution that depends continuously on the data f ∈ L2 (Ω). Proof. Thanks to Proposition 3.1.1, the equation (3.0.4) has a unique solution. To show that the solution depends continuously on the data f , let u1 , u2 ∈ Ds,2 (Ω) be solutions of the same type of equation as (3.0.4) with data f1 , f2 ∈ L2 (Ω) respectively. Note that Z < u, v >Ds,2 (Ω) = f vdx = l(v) Ω


Page 43

given by the weak formulation. Using inequality (3.0.1), we get ku1 − u2 kDs,2 (Ω) =

< u1 − u2 , u1 − u2 >Ds,2 (Ω) ku1 − u2 kDs,2 (Ω)

=

l(u1 − u2 ) ku1 − u2 kDs,2 (Ω)

≤

ku1 − u2 kL2 (Ω) kf1 − f2 kL2 (Ω) ku1 − u2 kDs,2 (Ω)

≤ C(n, s, Ω)

ku1 − u2 kDs,2 (Ω) kf1 − f2 kL2 (Ω) ku1 − u2 kDs,2 (Ω)

≤ C(n, s, Ω)kf1 − f2 kL2 (Ω) , we are done.

3.2

Maximum Principle

The maximum principle for s-harmonic (even s-superharmonic) function is a powerful tool in the study of fractional elliptic problems.

Weak Maximum Principle The weak maximum principle of a s-superharmonic function u tells us that the non-negativity of the solution function u in the whole space Rn except in the domain implies that u is non-negative in Ω. That is what we shall prove in the following proposition. 3.2.1 Proposition. Let w ∈ Ds,2 (Ω), consider the following problem ( (−∆)s w ≥ 0 in Ω w≥0 in Rn \Ω.

(3.2.1)

Then w ≥ 0 in Ω. Proof. First, show that w+ , w− ∈ Ds,2 (Ω) if w ∈ Ds,2 (Ω). If w ≥ 0, w− ≡ 0 ∈ Ds,2 (Ω) and w+ = w ∈ Ds,2 (Ω). If w ≤ 0, w+ ≡ 0 ∈ Ds,2 (Ω) and w− = w ∈ Ds,2 (Ω). Otherwise, we have that Z Z Z Z (w(x) − w(y))2 (w+ (x) − w− (x) − w+ (y) + w− (y))2 2 kwkDs,2 (Ω) = dydx = dydx |x − y|n+2s |x − y|n+2s R2n R2n Z Z (w+ (x) − w+ (y))2 + (w− (x) − w− (y))2 + 2(w+ (x) + w+ (y))(w− (y) − w− (x)) = dydx |x − y|n+2s R2n = kw+ k2Ds,2 (Ω) + kw− k2Ds,2 (Ω) Z Z −w+ (x)w− (x) − w+ (y)w− (y) + w+ (x)w− (y) + w− (x)w+ (y) + 2 dydx |x − y|n+2s R2n


Page 44

with the fact that, ∀x ∈ Rn , w+ (x)w− (x) = 0, since supp(w+ ) ∩ supp(w− ) = ∅ and w+ (x)w− (y) + w+ (y)w− (x) ≥ 0. Then kw+ k2Ds,2 (Ω) + kw− k2Ds,2 (Ω) ≤ kwk2Ds,2 (Ω) < +∞, therefore, kw+ k2Ds,2 (Ω) < +∞ and kw− k2Ds,2 (Ω) < +∞. Now, let w− be our test function, the weak formulation of (3.2.1) gives us Z (w(x) − w(y))(w− (x) − w− (y)) dydx ≥ 0 |x − y|n+2s R2n we have that

(w(x) − w(y))(w− (x) − w− (y)) dydx |x − y|n+2s R2n Z (w+ (x) − w− (x) − w+ (y) + w− (y))(w− (x) − w− (y)) dydx |x − y|n+2s R2n Z Z (w− (x) − w− (y))2 (w+ (x) − w+ (y))(w− (x) − w− (y)) − dydx + dydx n+2s |x − y| |x − y|n+2s R2n R2n Z Z (w− (x) − w− (y))2 w+ (x)w− (x) + w+ (y)w− (y) − dydx + dydx |x − y|n+2s |x − y|n+2s R2n R2n Z w+ (x)w− (y) + w+ (y)w− (x) dydx, |x − y|n+2s R2n Z

= = = −

since w+ (x)w− (x) = 0, ∀x ∈ Rn , we get Z − R2n

or,

w+ (x)w− (y)

+

(w− (x) − w− (y))2 dydx − |x − y|n+2s w+ (y)w− (x) Z − R2n

Z R2n

w+ (x)w− (y) + w+ (y)w− (x) dydx ≥ 0 |x − y|n+2s

≥ 0, ∀x, y ∈ Rn . Then

(w− (x) − w− (y))2 dydx = −kw− k2Ds,2 (Ω) ≥ 0 |x − y|n+2s

so kw− kDs,2 (Ω) = 0 =⇒ w− = 0 a.e. in Ω, w ≥ 0 in Ω.

Strong Maximum Principle If a s-superharmonic function u in Ω is positive in any bounded subset of Rn with positive measure, the strong maximum principle state that it is positive in Ω. In the next result, we will use some barrier to set positivity. This barrier will be constructed from a function denoted ϕB satisfying


Page 45

( (−∆)s ϕB (x) = 1 ϕB (x) = 0

in B in Rn \B.

(3.2.2)

This function is computed in Dyda, Bogdan and Fall-Jarohs (see [4, 13, 15]) and one have 2 ϕB (x) = C(n, s)( RB − |x|2

s

,

where RB is the radius of the ball B centered at 0. 3.2.2 Proposition. Let Ω be an open set. Suppose there exists G ⊂ Rn bounded, |G| > 0 such that  s in Ω   (−∆) w ≥ 0  ess inf G w > 0 in G (3.2.3) w≥0 in Rn    w ∈ Ds,2 (Ω) . Then w > 0 in Ω. 3.2.3 Remark. When s = 1, Ω must be connected so that the strong maximum principle holds, but here it is not necessary. Look at the two definitions below to get the matter; with the classical Laplacian, the integration is over the ball while in the fractional Laplacian the integration is in the whole space except in the ball. Thus, it is not worth to suppose connectivity.

1 − ∆u(x0 ) = lim r−→0 |Br (x0 )|

Z

1 (−∆) u(x0 ) = lim r−→0 |Br (x0 )|

Z

s

u(x0 ) − u(y)dy Br (x0 )

Rn \Br (x0 )

u(x0 ) − u(y) dy. |x0 − y|n+2s

(3.2.4)

Proof. Proposition 3.2.2 Let B a ball in Ω such that B ⊂⊂ Rn \ G. Define the characteristic function ( 1 if x ∈ G χG (x) = 0 otherwise. Now we want to construct the barrier, set vα (x) = ϕB (x) − αχG (x), x ∈ B.

(3.2.5)

Let us compute (−∆)s vα to choose α such that (−∆)s vα ≤ 0. Let x ∈ B, we have Z ϕB (x) − ϕB (y) χG (x) − χG (y) (−∆) vα (x) = C(n, s)P.V. dy − αC(n, s)P.V. dy n+2s |x − y| |x − y|n+2s Rn Rn Z −χG (y) = (−∆)s ϕB (x) − αC(n, s) dy, since x ∈ B |x − y|n+2s n R Z 1 = 1 + αC(n, s) dy, n+2s G |x − y| s

Z

Section 3.3. Regularity since x ∈ B, we have that

Page 46 1 |x−y|n+2s

∈ L1 (G). Thus, if (−∆)s vα (x) ≤ 0 then

α≤

−1 R C(n, s) G

Let’s choose α 0 on G, there exists δ > 0 such that w ≥ δvα , on Rn \B since vα ≥ 0 on G where δ=

ess inf G w . ess supG vα

Therefore, we have the following problem ( (−∆)s (w − δvα ) ≥ 0 on w − δvα ≥ 0 on

B Rn \B.

So, by the weak maximum principle (Proposition 3.2.1), we obtain that w ≥ δvα on B, then w ≥ ϕB on B. Therefore, w > 0 on B since ϕB > 0. So, for any x ∈ Ω, w(x) > 0, since we took any ball B in Ω and also w > 0 in G. We recall a very weak maximum principle due to Silvestre which was proved fr n > 2s (see [38]). 3.2.4 Proposition (Silvestre 2007). Let Ω ⊂⊂ Rn be an open set and u be a lower semi-continuous ¯ such that (−∆)s u ≥ 0 in Ω and u ≥ 0 in Rn \Ω, then u ≥ 0 in Rn . Moreover, if u(x) = 0 function in Ω for one point inside Ω then u ≡ 0 in Rn .

Proof. See [38, Proposition 2.2.8]. The maximum principles are very useful for the study of differential equations, they enables us to characterize the equations without solving them. They are very used to estimate regularity.

3.3

Regularity

In this section, we study the regularity of the equation (3.0.4) by using the Riesz potential and the Green function to estimate the regularity up to the boundary for a distributional solution of the equation (−∆)s u = f . Before we prove some regularity results, we need to recall the definition of some spaces. Recall Z |u(x)| L1s := u : Rn −→ R such that dx < +∞ , n+2s Rn 1 + |x|


Page 47

and Ω ⊂ Rn be an open set, we say u : Ω −→ R is in C 0,α , i.e., Hölder continuous with exponent α ∈ (0, 1) if kukC 0,α (Ω) := sup x6=y x,y∈Ω

|u(x) − u(y)| < +∞. |x − y|α

Interior H¨ older estimates for the distributional solution to the equation (−∆)s u = f . s 3.3.1 Lemma. Let f ∈ L∞ (B1 ), and let u ∈ L1s ∩ L∞ loc (B1 ) such that (−∆) u = f in B1 . Then for r ∈ (0, 1) and every α ∈ (0, min(1, 2s)), there exists a constant C(n, s, r, α) > 0 such that kukC 0,α (Br ) ≤ C(n, s, r, α) kukL∞ (Br ) + kf kL∞ (B1 ) . (3.3.1)

Proof. Let η ∈ Cc∞ (Rn ) non-negative such that, η ≡ 1 on Br , η ≡ 0 on Rn \B1 , and η ≤ 1 on Rn . Consider the Riesz Kernel (see relation (1.2.3) in Section 1.2).  2s−n , n > 2s;  C(n, s)|x − y| 1 1 I2s (x, y) = π log |x| , n = 2s = 1;   2s−n −C(1, s)|x − y| , n = 1 < 2s, for a positive normalization constant. Recall the inverse of the fractional Laplacian which is the Riesz Potential when n > 2s : Z u(y) −s dy = I2s ∗ u(x), (−∆) u(x) = C(n, s) n−2s Rn |x − y|

(3.3.2)

where C(n, s) = π 2s+n/2 Γ(s+n/2) Γ(−s) (see Section 1.2). Then, for

Z v(x) := Rn

I2s (x, y)(ηf )(y)dy, ∀x ∈ Rn ,

we claim that (−∆)s v(x) = η(x)f (x), ∀x ∈ Rn . Indeed, for any x, y ∈ Rn , we can see I2s (x, y) as 0 (x − y) and then we get that I2s Z v(x) = I2s (x, y)(ηf )(y)dy n R Z 0 = I2s (x − y)(ηf )(y)dy Rn

0 = I2s ∗ (ηf ) = I2s ∗ (ηf )

= (−∆)−s (ηf )(x) hence (−∆)s v(x) = η(x)f (x), ∀x ∈ Rn . Let a, b > 0, α ∈ (0, 1) and m ∈ R such that m + α > 0. We claim that


Page 48

|a−m − b−m | ≤ Cm,α |m||a − b|α max a−(m+α) , b−(m+α) ,

(3.3.3)

where Cm,α > 0 is a positive constant that depends only on m and α and also |lnb − lna| ≤

1 |a − b|α max a−α , b−α . α

(3.3.4)

Indeed, if we use H¨ older inequality. For the inequality (3.3.3), let c > 1, we have the following estimate Z c 1 −m t−m−1 dt (1 − c ) = m 1 Z c α Z ∞ 1−α −m−1 1−α ≤ dt t dt 1

1

i∞ 1 − α h −m−α = (c − 1) t 1−α −m − α 1 1−α 1−α = (c − 1)α m+α 1−α 1 (c − 1)α ≤ m+α α

1−α

1 set Cm,α := ( m+α )1−α . Thus, we have that

|1 − c−m | ≤ Cm,α |m||c − 1|α then if we take c := b/a, a > 0, we obtain −m −m a b − a α − b ≤ Cm,α |m| a this implies that a−m |a−m − b−m | ≤ Cm,α |b − a|α a−m−α ≤ Cm,α |b − a|α max(b−m−α , a−m−α ). up to relabeling the constant Cm,α . For (3.3.4) with the same c > 1, we that Z c lnc = t−1 dt 1 Z c α Z ≤ dt 1

∞

t

−1 1−α

1−α dt

1

i −α ∞ 1 − α h 1−α = (c − 1) t −α 1 1−α 1−α = (c − 1)α α 1 ≤ (c − 1)α , α α

1−α


Page 49

with c = b/a, a > 0, we have |lnb − lna| ≤ ≤

1 |b − a|α a−α α 1 |b − a|α max(b−α , a−α ), α

Now, using the triangle inequality, we see that ||x − z| − |y − z|| ≤ |x − y|, also we know that max(|x − z|, |y − z|) ≤ |x − z| + |y − z|. Therefore, we get that |x − z|−m − |y − z|−m ≤ Cm,α |x − y|α |x − z|−m−α + |y − z|−m−α .

(3.3.5)

And also, |ln|x − z| − ln|y − z|| ≤

1 |x − y|α |x − z|−α + |y − z|−α . α

(3.3.6)

If n > 2s, with α ∈ (0, min(1, 2s)) and x, y ∈ Br , Z v(x) − v(y) = C(n, s) |x − z|2s−n − |y − z|2s−n (ηf )(z)dz, Rn

using the inequality (3.3.5) with m = n − 2s, we can write Z |v(x) − v(y)| ≤ C(n, s, α) |x − y|α (|x − z|2s−n−α + |y − z|2s−n−α )|(ηf )(z)|dz, n R Z ≤ C(n, s, α)|x − y|α 2|x − z|2s−n−α |(ηf )(z)|dz, Rn |y−z| > 1, using the fact that the support of η supp(η) ⊆ B1 ⊂ B(x, 2) ball of radius 2 centered since |x−z| at x ∈ Br , we obtain Z α |v(x) − v(y)| ≤ C(n, s, α)|x − y| |x − y|2s−n−α |(ηf )(y)|dy n ZR α ≤ C(n, s, α)|x − y| |x − y|2s−n−α dykf kL∞ (B1 ) B(x,2) Z = C(n, s, α)|x − y|α |y|2s−n−α dykf kL∞ (B1 ) B(0,2) Z n α = 2 C(n, s, α)|x − y| |y|2s−n−α dykf kL∞ (B1 ) B1

up to relabeling the positive constant C(n, s, α) that depends on the dimension, s and α. Using polar coordinates y = rx, r > 0, we get that Z Z Z 1 |S1 | 2s−n−α |y| dy = rn−1+2s−n−α drdσ(x) = 2s −α B1 S1 0


Page 50

then, |v(x) − v(y)| ≤ C(n, s, α)|x − y|α kf kL∞ (B1 ) . Consequently, kvkC 0,α (B1 ) ≤ C(n, s, α)kf kL∞ (B1 )

(3.3.7)

where C(n, s, α) is a positive constant depending on the dimension, s and α. If n = 1 = 2s, x, y ∈ Br , the inequality (3.3.6) let us write that Z 1 |v(x) − v(y)| ≤ |log |x| − log |y|| |ηf (z)|dz π R Z 1 α ≤ |x − y| |x|−α + |y|−α |ηf (z)|dz απ ZR 1 α |x|−α |(ηf )(z)|dz, ≤ 2 |x − y| απ R since

|y| |x|

> 1. |v(x) − v(y)| ≤ 2

1 |x − y|α απ

Z

2

−2

|x|−α dzkf kL∞ (B1 ) ,

= Cα |x − y|−α kf kL∞ (B1 ) , then we get that kvkC 0,α (B1 ) ≤ Cα kf kL∞ (B1 )

(3.3.8)

where Cα a positive constant depending on α. If n = 1 > 2s, x, y ∈ Br , using again (3.3.5) with m = 1 − 2s and α ∈ (0, min(1, 2s)), Z |v(x) − v(y)| ≤ C1,s C(s, α) |x − y|α (|x − z|2s−1−α + |y − z|2s−n−α )|(ηf )(z)|dz, ZR ≤ Cs,α |x − y|α 2|x − z|2s−1−α |(ηf )(z)|dz, R Z α ≤ Cs,α |x − y| |x − z|2s−1−α |(ηf )(z)|dz R Z α ≤ Cs,α |x − y| |x − z|2s−1−α dzkf kL∞ (B1 ) ]x−2,x+2[ 1 α 2s−1−α

Z

= 2Cs,α |x − y|

|z|

−1 α

dzkf kL∞ (B1 )

= C(s, α)|x − y| kf kL∞ (B1 ) , consequently, kvkC 0,α (B1 ) ≤ C(n, s, α)kf kL∞ (B1 )

(3.3.9)

where C(s, α) a positive constant depending on s and α. Therefore, combining (3.3.7), (3.3.8) and (3.3.9), we get for every s ∈ (0, 1) and α ∈ (0, min(1, 2s)) the following estimate kvkC 0,α (B1 ) ≤ C(n, s, α)kf kL∞ (B1 ) (3.3.10)


Page 51

up to relabeling the positive constant C(n, s, α). Now, let’s take w := u − v, then w is s-harmonic in B1 , i.e., (−∆)s w = 0 in B1 . According to [3, Lemma 3.2], for every r0 ∈ (0, r), |∇w(x)| ≤ C(n, r0 , s) kwkL∞ (B1 ) − w(x) , ∀x ∈ Br0 , so we get that

|∇w(x)| ≤ C(n, s, r0 ) ku − vkL∞ (B1 ) − (u − v)(x) ≤ C(n, s, r0 ) kukL∞ (Br ) + kvkL∞ (Br ) + kukL∞ (Br ) + kvkL∞ (Br ) = C(n, s, r0 ) kukL∞ (Br ) + kvkL∞ (Br ) < +∞.

(3.3.11)

Hence w is Lipschitz in Br and then H¨ older also in Br . Indeed, |w(x) − w(y)| |x − y| |w(x) − w(y)| |w(x) − w(y)| = ≤ (2r)1−α < +∞, α α |x − y| |x − y| |x − y| |x − y| we took α ∈ (0, min(1, 2s)). Thus from (3.3.11), we get kwkC 0,α (Br ) ≤ C(n, s, r0 ) kukL∞ (Br ) + kvkL∞ (Br )

(3.3.12)

And finally, using the inequalities (3.3.10) and (3.3.12), we have that kukC 0,α (Br ) = kw + vkC 0,α (Br ) ≤ kwkC 0,α (Br ) + kvkC 0,α (Br ) ≤ C(n, s, r0 ) kukL∞ (Br ) + kvkL∞ (Br ) + C(n, s, α)kf kL∞ (B1 ) ≤ C(n, s, r0 ) kukL∞ (Br ) + C(n, s, α)kf kL∞ (B1 ) ≤ C(n, s, r, α) kukL∞ (Br ) + kf kL∞ (B1 ) , where C(n, s, r, α) is a positive constant depending only on n, s, r, and α. The distributional solution u of the equation (−∆)s u = f is Hölder continuous locally, that is the interior regularity. And this result will help to prove the regularity of the solution up to the boundary.

Boundary Regularity. The green function gives explicitly the solution of the Poisson’s equation involving the fractional Laplacian. Before we prove the regularity of the solution up to the boundary in the forthcoming Proposition 3.3.3. We will give some properties of the green function and use some of them to prove the Proposition 3.3.3. Let Ω be a C 1,1 bounded domain. Consider the Green function G in Ω. For any function f ∈ Lp (Ω) providing 2sp > n, we put Z Gf (x) := G(x, y)f (y)dy, G satisfies G(x, y) = 0 when x ∈

Rn \Ω

or y ∈

Ω Rn \Ω.


Page 52

3.3.2 Properties. Let ϕ ∈ Cc∞ (Rn ), then the following equalities are true (G(−∆)s ϕ) (x) = ϕ(x), ∀x ∈ Rn , (−∆)s (Gϕ)(x) = ϕ(x), ∀x ∈ Ω. For the Green function G, we have the following estimate (see [4, 40])  s s 2s−n min d (x)d (y) , 1  n > 2s, C|x − y|  2s  |x−y|   1 1 2 2 (y) n = 1 = 2s, C log 1 + d (x)d G(x, y) ≤ |x−y|    (2s−1)/2 s s  (x)d (y)  C|x − y|2s−1 min (d(x)d(y))2s−1 , d |x−y| n = 1 < 2s 2s |x−y|

(3.3.13)

where d is the distance function to Rn \Ω and C is a positive constant depending only on the dimension, s and Ω. Thus following results hold. 3.3.3 Proposition. Let f ∈ L∞ (Ω), then • If n ≥ 2s |Gf (x)| ≤ Cds (x)kf kL∞ (Ω) ,

(3.3.14)

|Gf (x)| ≤ Cd(2s−1)/2 (x)kf kL∞ (Ω) ,

(3.3.15)

• If n = 1 < 2s, we have that

where C > 0 is the constant that depends only on n, s and Ω. Furthermore, we get that kGf kC 0,β (Ω) ¯ ≤ Ckf kL∞ (Ω) ,

(3.3.16)

for β ∈ (0, s] if n ≥ 2s and β ∈ (0, 2s−1 2 ] for n = 1 < 2s.

Proof. Assume that f ≥ 0, without loss of generality (−f ≥ 0, one can take the absolute value). We have the following. For every x ∈ Ω, Z Z Z G(x, y)f (y)dy = G(x, y)f (y)dy + G(x, y)f (y)dy. Ω

Ω∩B(x,d(x)/2)

Ω∩(B(x,diam(Ω))\B(x,d(x)/2))

We prove (3.3.14). For n > 2s, using the first estimate of (3.3.13) with polar coordinates Z Z G(x, y)f (y)dy ≤ C |x − y|2s−n f (y)dy, Ω∩B(x,d(x)/2)

B(x,d(x)/2)

Z

Z

≤ Ckf kL∞ (Ω) S1

=

d(x) 2

r2s−1 drdσ,

0

Ckf kL∞ (Ω) 2s d (x) = C(Ω, s)ds (x)kf kL∞ (Ω) . s2s+1

(3.3.17)


Page 53

By the first estimate of (3.3.13), with the fact that d(y) ≤ |x − y| + d(x) ≤ 3|x − y| since y ∈ / B(x, d(x)/2) s Z Z d (x)ds (y) G(x, y)f (y)dy ≤ |x − y|2s−n min , 1 f (y)dy, |x − y|2s Ω∩(B(x,diam(Ω))\B(x,d(x)/2) B(x,diam(Ω))\B(x,d(x)/2) Z ds (x)ds (y) dy, ≤ Ckf kL∞ (Ω) |x − y|2s−n |x − y|2s B(x,diam(Ω))\B(x,d(x)/2) Z |x − y|s s s dy, ≤ 3 Cd (x)kf kL∞ (Ω) |x − y|2s−n |x − y|2s B(x,diam(Ω))\B(x,d(x)/2) Z s ≤ Cd (x)kf kL∞ (Ω) |x − y|s−n dy, = Cds (x)kf kL∞ (Ω)

B(x,diam(Ω)) Z diam(Ω)

Z

S1

rs−1 drdσ,

0

= Cds (x)kf kL∞ (Ω)

(3.3.18)

up to relabeling the positive constant C. Thus, (3.3.17) and (3.3.18) give the desired result for n > 2s. For n = 1 = 2s, note that for any a, b > 0, Z ba Z ba ba dt log(1 + )dt = ba log 2 + ba = a(2b log 2), t t + ba 0 0 1

with the second estimate of (3.3.13), by setting a = d 2 and b = 2diam(Ω) > d(y), ∀y ∈ Ω, we have that ! 1 1 Z d(x)/2 Z d(x)/2 d 2 (x)d 2 (y) G(x, y)f (y)dy ≤ C log 1 + dy, |x − y| 0 0 ! 1 1 Z d 12 (x)diam(Ω) d 2 (x)diam 2 (Ω) ≤ Ckf kL∞ (Ω) log 1 + dy, |x − y| 0 1

= Cd 2 (x)kf kL∞ (Ω) (2diam(Ω) log 2), 1

= Cd 2 (x)kf kL∞ (Ω) ,

(3.3.19)

where the positive constant C has changed. Using the fact that d(y) ≤ |x − y| + d(x) ≤ 3|x − y|, we have that ! 1 1 d 2 (x)d 2 (y) G(x, y)f (y)dy ≤ C log 1 + f (y)dy, |x − y| Ω∩(B(x,diam(Ω))\B(x,d(x)/2) B(x,diam(Ω)) ! 1 1 1 Z 3 2 d 2 (x)|x − y| 2 ≤ Ckf kL∞ (Ω) log 1 + dy, |x − y| B(x,diam(Ω)) Z diam(Ω) 1 1 1 ≤ Ckf kL∞ (Ω) log 1 + 3 2 d 2 (x)|y|− 2 ,

Z

Z

0

Z

diam(Ω)

≤ Ckf kL∞ (Ω)

1

1

1

3 2 d 2 (x)y − 2 dy,

0 1 2

= Cd (x)kf kL∞ (Ω)

(3.3.20)


Page 54

up to relabeling the positive constant C. Therefore, by (3.3.19) and (3.3.20), we obtain (3.3.14) for n ≥ 2s. Now, we prove (3.3.15), for n = 1 < 2s using the last estimate of (3.3.13), Z

Z G(x, y)f (y)dy ≤ C

Ω

2s−1

|x − y|2s−1

Ω

(d(x)d(y)) 2 |x − y|2s−1

≤ C|Ω|(diamΩ)

2s−1 2

kf kL∞ (Ω) .

Finally, let us prove (3.3.16). For n ≥ 2s, using (3.3.14), we obtain |Gf (x)| ≤ Cds (x)kf kL∞ (Ω) and |Gf (y)| ≤ Cds (y)kf kL∞ (Ω) , ∀x, y ∈ Rn if we assume that 2|x − y| ≥ max(d(x), d(y)), we have that |Gf (x) − Gf (y)| ≤ |Gf (x)| + |Gf (y)| ≤ C(ds (x) + ds (y))kf kL∞ (Ω) , ∀x, y ∈ Rn , ≤ 2s+1 C|x − y|s kf kL∞ (Ω) , = C|x − y|s kf kL∞ (Ω) , ∀x, y ∈ Rn

(3.3.21)

where the value of the constant C > 0 has been changed a little bit. For n = 1 < 2s, using (3.3.15) and the same argument as above(case n ≥ 2s, we obtain that |Gf (x) − Gf (y)| ≤ 2(2s+1)/2 C|x − y|

2s−1 2

kf kL∞ (Ω) = C|x − y|

2s−1 2

kf kL∞ (Ω) , ∀x, y ∈ Rn . (3.3.22)

Now, assume that |x − y| < d(y) and set r := d(y) such that y + rz ∈ Ω, ∀z ∈ B(0, 2). Put 2 2 F (z) := Gf (y + rz), z ∈ B(0, 2). Let us consider the case n ≥ 2s, by (3.3.14) |F (z)| = |Gf (y + rz)| ≤ Cds (y + rz)kf kL∞ (Ω) ≤ Cds (y)kf kL∞ (Ω) ≤ Crs kf kL∞ (Ω) .

(3.3.23)

Remark that (−∆)s F (z) = r2s (−∆)s Gf (z) = r2s f (z) in B(0, 2), then by the interior regularity Lemma 3.3.1 and (3.3.23), we get F ( x−y ) − F (0) r ≤ kF kC 0,α (B(0,1) , α ∈ (0, 1) |x−y|α α r

≤ C(n, s) kF kL∞ (B(0,1)) + r2s kf kL∞ (Ω) , ≤ C(n, s) Crs kf kL∞ (Ω) + r2s kf kL∞ (Ω) , = C(n, s, Ω)rs kf kL∞ (Ω) , where C(n, s, Ω) is a positive constant depending only on n, s and Ω. Therefore, we can conclude that |Gf (x) − Gf (y)| ≤ C(n, s, Ω)rs−α |x − y|α kf kL∞ (Ω) , providing α ∈ (0, s]. This, with the estimate in (3.3.21), we obtain (3.3.16) for n ≥ 2s. For n = 1 < 2s, the proof is similar when n ≥ 2s.


Page 55

3.3.4 Proposition. Let Ω be a C 1,1 bounded domain. Let f ∈ L∞ (Ω) and u solves the equation ( (−∆)s u = f, in Ω u ∈ Ds,2 (Ω). Then

Z u(x) = Gf (x) =

G(x, y)f (y)dy. Ω

In particular, u ∈ C(Rn ) and there exists a positive constant C depending only on Ω such that |u(x)| ≤ Cds (x)kf kL∞ (Ω) . This proposition shows that u is continuous up to the boundary. Proof. We can assume that f ≥ 0 (by decomposing f = f + − f − ), then u ≥ 0. Set Z v(x) = G(x, y)f (y)dy, Ω

and, Ωn :=

1 x ∈ Ω : d(x) > n

.

Consider the symmetric mollifier ρn (see Chapter 1 (1.0.3)), we denote vn = ρn ∗ v and fn = ρn ∗ f thus Z ρn ∗ v(x) = ρn (x − y)v(y)dy, Ωn Z = ρn (x − y)(−∆)−s f (y)dy, Ωn

= ρn ∗ (−∆)−s f (x) = (−∆)−s ρn ∗ f (x), so, (−∆)s vn (x) = fn (x), ∀x ∈ Ωn , further, vn ∈ Cc∞ (Rn ) and vn is non-negative. Let un solution to

( (−∆)s un = fn , un ∈ Ds,2 (Ωn ).

in Ωn

By the weak maximum principle in (see Proposition 3.2.1), we have that vn ≥ un in Rn since un = 0 in Rn \Ωn . Therefore, if we take the limit, we have that v ≥ u in Rn . Indeed, ρn ∗ f −→ f in L2 (Ω) (see Theorem 1.0.4) and un −→ u = 0 in Rn \Ω as n −→ +∞. Therefore, un −→ u ∈ Ds,2 (Ω). Take, Ω− n

:=

1 x ∈ R : d(x) > − n n

Consider u− n solution to the equation ( (−∆)s u− in Ω− n = f, n s,2 (Ω− ). u− ∈ D n n

,


Page 56

s,2 − e’s inequality. Then, there For n0 ≥ 1 fixed, (u− n )n≥n0 is bounded in D (Ωn0 ) (see (3.1.3) by Poincar´ − that converge strongly in L2 (Ω− ) is a subsequence of (un )n≥n0 , we keep the same notation, (u− n≥n n0 ). n 0 − n − − n − s,2 And un = 0 in R \Ωn , thus u = 0 in R \Ω. Therefore, u = u ∈ D (Ω). − − − − Note that u− n is continuous in Ωn , since Gf − un is s-harmonic in Ωn (it is smooth in Ωn ). 0 − Now, see that for every n ≥ n0 , we have (−∆)s (u− n − v) ≥ 0 in D (Ω), un − v is continuous up to the − n boundary of Ω by Proposition 3.3.3 and un − v ≥ 0 in R \Ω. Thus, thanks to the result of Silvestre n − n (see Proposition 3.2.4), u− n − v ≥ 0 in R . So, if we take the limit, we get that v ≤ u = u in R .

We finally can say that u = v in Rn .

Conclusion This project had the aim to study the basic notions related to the fractional elliptic equations. We defined some suitable functional spaces, prove some existence result and maximum principles. Then we finally discuss some regularity estimates. We did a little comparison between fractional equations and the classic one (differential) whose difference comes from the operators involved, namely respectively the fractional Laplacian and the classic one. We see that apart from the non local property that distinguishes them, they are very similar. This comparison is to facilitate the understanding of the reader familiar with the classic Laplace’s operator. If we borrow a sentence in the introduction of Hitchikers’guide(see [12]) modified a little bit, we would say that ”the advantage of this comparison is to be able to get more quickly from one place to another when the place you arrived at had probably become, as a result of this, very similar to place you had left.” We showed that the strong maximum principle implies directly uniqueness of the solution of fractional Poisson’s equation. The weak maximum principle showed the dependence of the solution on the data by the regularity proposition. We proved the interior Hölder estimates of the distributional solution and use the result with some additional assumptions on the domain to prove global regularity estimates of the distributional solution for (−∆)s u = f . Although many interesting works have been done and are being done on this area. And it remains many other exciting things to do. Some questions that people are asking still don’t have answer. Among the following, some could be the future work in which we will investigate. We can mention the regularity of the obstacle problem of a general integro-differential operator (see [9]), numerical solution to some fractional elliptic equations arising from finance and relativistic physic, Agmon-Douglas-Nirenberg regularity type result (see [6]).

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Acknowledgements First of all, I give thanks to Allah, the Almighty. I thank my supervisor Mouhamed Moustapha Fall who, despite his very busy schedule, has spared no effort to properly supervise my work and guide me in a very rigorous way. Thank you very much for your encouragements Professor. I thank all the students, tutors and staff of AIMS-Senegal, I don’t forget Amadou Tall. I thank my tutor Bernadette Faye who always took the time to read my work and correct my errors. I thank all members of Chair Mathematics. I thank all members of my family and especially my father, my mother, my brother and sisters who always supported me and did everything for my studies. My big brother Elhadj Abdoulaye Thiam, I don’t forget you for your useful help and encouragements. Ignace, you are in the set. I thank my friends: les deux Laye, Doudou, Bousso, Mbengue, Sofia etc. I cannot finish without thanking all members of AIMS family. A big thank to everybody who has supported and encouraged me directly or indirectly.

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References [1] Robert A. Adams and John J. F. Fournier. Sobolev Spaces, Volume 140, Second Edition. Academic Press, Amsterdam; Boston, 2 edition edition, July 2003. [2] William Beckner. Inequalities in fourier analysis. The Annals of Mathematics, 102(1):159, July 1975. [3] Krzysztof Bogdan, T. Kulczycki, and Adam Nowak. Gradient estimates for harmonic and $q$harmonic functions of symmetric stable processes. Illinois Journal of Mathematics, 46(2):541– 556, 2002. Mathematical Reviews number (MathSciNet)MR1936936, Zentralblatt MATH identifier1037.31007. [4] Krzysztof Bogdan and Byczkowski Tomasz. Potential theory of schrodinger operator based on fractional laplacian. 20:293–335, 2000. [5] Haim Brezis. How to recognize constant functions. connections with sobolev spaces. Russian Mathematical Surveys, 57(4):693–708, August 2002. [6] Haim Brezis. Functional Analysis, Sobolev Spaces and Partial Differential Equations. Springer New York, New York, NY, 2010. [7] Ham Brezis and Mironescu Petru. Gagliardo-nirenberg, composition and products in fractional sobolev spaces. 1(4):387–404, 2001. [8] Luis A. Caffarelli, J.-M. Roquejoffre, and O. Savin. Nonlocal minimal surfaces. arXiv:0905.1183 [math], May 2009. [9] Luis A. Caffarelli, Sandro Salsa, and Luis Silvestre. Regularity estimates for the solution and the free boundary of the obstacle problem for the fractional laplacian. Inventiones mathematicae, 171(2):425–461, February 2008. [10] Luis A. Caffarelli and Enrico Valdinoci. Uniform estimates and limiting arguments for nonlocal minimal surfaces. Calculus of Variations, 41(1):203–240, 2009. [11] Athanase Cotsiolis and Nikolaos K. Tavoularis. Best constants for sobolev inequalities for higher order fractional derivatives. Journal of Mathematical Analysis and Applications, 295:225–236, July 2004. [12] Eleonora Di Nezza, Giampiero Palatucci, and Enrico Valdinoci. Hitchhiker’s guide to the fractional sobolev spaces. arXiv:1104.4345 [math], April 2011. [13] Bartlomiej Dyda. Fractional calculus for power functions and eigenvalues of the fractional laplacian. Fractional Calculus and Applied Analysis, 15(4):536–555, December 2012. [14] Lawrence C. Evans. Partial Differential Equations: Second Edition. American Mathematical Society, Providence, R.I, 2 edition edition, March 2010. [15] Mouhamed Moustapha Fall and Sven Jarohs. Overdetermined problems with fractional laplacian. arXiv preprint arXiv:1311.7549, 2013.

59

REFERENCES

Page 60

[16] Mouhamed Moustapha Fall and Tobias Weth. Monotonicity and nonexistence results for some fractional elliptic problems in the half space. arXiv preprint arXiv:1309.7230v1, page 24, September 2013. [17] Charles L. Fefferman and Rafael de la Llave. Relativistic stability of matter (i). Revista Matemtica Iberoamericana, 2(1-2):119–213, 1986. [18] G. B. Folland. Introduction to Partial Differential Equations. Princeton University Press, 1995. [19] David Gilbarg and Neil S. Trudinger. Elliptic Partial Differential Equations of Second Order. Springer-Verlag Berlin and Heidelberg GmbH & Co. K, dition : 2nd revised edition edition, October 2013. [20] Rudolf Gorenflo and Francesco Mainardi. Random Walk Models for Space-Fractional Diffusion Processes. 1998. [21] Andreas Hildebrandt, Ralf Blossey, Sergej Rjasanow, Oliver Kohlbacher, and Hans-Peter Lenhof. Electrostatic potentials of proteins in water: a structured continuum approach. Bioinformatics (Oxford, England), 23(2):e99–103, January 2007. PMID: 17237112. [22] Lars Hrmander. The Analysis of Linear Partial Differential Operators I - Distribution Theory and Fourier Analysis. 1990. [23] R Ishizuka, S-H Chong, and F Hirata. An integral equation theory for inhomogeneous molecular fluids: the reference interaction site model approach. The Journal of chemical physics, 128(3):034504, January 2008. PMID: 18205507. [24] Sven Jarohs. A remark on different definitions of the fractional laplacian and its constants. page 8, 2013. [25] Andrej N. Kolmogorov. ber Kompaktheit der Funktionenmengen bei der Konvergenz im Mittel. Weidmann, 1931. [26] Naum S. Landkof. Foundations Of Modern Potential Theory. Springer-Verlag New York, Inc, Berlin; New York, January 1973. [27] R. B. Lehoucq, K. Zhou, Q. Du, and M. Gunzburger. A non-local vector calculus, non-local volumeconstrained problems, and non-local balance laws. Mathematical Models and Methods in Applied Sciences, 2011. [28] Elliott H. Lieb and Michael Loss. Analysis. American Mathematical Society, Providence, RI, 2 edition edition, March 2001. [29] Jacques Louis Lions. Théorèmes de trace et d’interpolation. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, 13(4):389–403, 1959. [30] Jacques Louis Lions and Enrico Magenes. Problémes aux limites non homogenes et applications / by j. l. lions, magenes. SERBIULA (sistema Librum 2.0). [31] Emmanouil Milakis and Luis Silvestre. Regularity for the nonlinear signorini problem. Advances in Mathematics, 217(3):1301–1312, February 2008.

REFERENCES

Page 61

[32] Giampiero Palatucci and Adriano Pisante. Improved sobolev embeddings, profile decomposition, and concentration-compactness for fractional sobolev spaces. arXiv:1302.5923 [math], February 2013. [33] Giampiero Palatucci, Ovidiu Savin, and Enrico Valdinoci. Local and global minimizers for a variational energy involving a fractional norm. Annali di matematica pura ed applicata, 192(4):673718, 2013. [34] Michael Reed. Methods of Modern Mathematical Physics: Functional analysis. Gulf Professional Publishing, 1980. [35] Scott Ridgway and Boland Mercedes. Continuum equations for dielectric response to macromolecular assemblies at the nano scale. Journal of Physics A: Mathematical and General, 37(41):9791, 2004. [36] Marcel Riesz. Sur les ensembles compacts de fonctions sommables. Acta Szeged Sect., 6:136–142, 1933. [37] Ovidiu Savin and Enrico Valdinoci. Density estimates for a nonlocal variational model via the sobolev inequality. arXiv:1103.6205 [math], March 2011. [38] Luis Silvestre. Regularity of the obstacle problem for a fractional power of the laplace operator. Communications on Pure and Applied Mathematics, 60(1):67–112, January 2007. [39] Elias M. Stein and Rami Shakarchi. Fourier Analysis: An Introduction. Princeton University Press, Princeton, April 2003. [40] Kulczcki Tadeusz. Properties of green function of symmetric stable processes. 17:334–364, 1997. [41] Peter Tankov and Rama Cont. Financial Modelling with Jump Processes. Chapman and Hall/CRC, Boca Raton, Fla, 1 edition edition, December 2003. [42] Luc Tartar. An Introduction to Sobolev Spaces and Interpolation Spaces. Springer, Berlin ; New York, 2007 edition edition, June 2007. [43] Enrico Valdinoci. From the long jump random walk to the fractional laplacian. arXiv preprint arXiv:0901.3261, 2009. [44] Marcus Webb. Analysis and Approximation of a Fractional Differential Equation. PhD thesis, Masters Thesis, Oxford University, Oxford, 2012. [45] Michel Willem. Analyse fonctionnelle elementaire. Vuibert, Paris, 2003. [46] William P. Ziemer. Weakly Differentiable Functions: Sobolev Spaces and Functions of Bounded Variation. Springer, New York, 1989 edition edition, October 1989.