A Neumann problem for a diffusion equation with n-dimensional fractional Laplacian

We study an initial-boundary value problem for a n-dimensional stochastic diffusion equation with fractional Laplacian on R+n\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mathbb{R}_{+}^{n}$\end{document}. In order to prove existence and uniqueness, we generalize the Fokas method to construct the Green function for the associated linear problem and then we apply a fixed point argument. Also, we present an example where the explicit solutions are given.


Introduction
The classical diffusion phenomenon is governed by a second order linear partial differential equation, whose Green function is given by a Gaussian probability density function and which describes the movement of energy through a medium in response to a gradient of energy. On the other hand, the diffusion processes in various systems with complex structure, such as liquid crystals, glasses, polymers, biopolymers, and proteins, usually do not follow a Gaussian density, as a consequence the phenomenon is described by a fractional partial differential equation [7]. Dipierro et al., [4] have studied the asymptotic behavior of the solutions of the time-fractional diffusion equation.
There is some previous work for the initial-boundary value problem on the first quadrant R 2 + for fractional diffusion equations, where the Green function has been constructed and an integral representation of the solution was found [3,6]. In this note, we consider the equation where the operator α is defined via the Riesz fractional derivative, for each coordinate. Let us notice that the generalization of the Laplacian most commonly used [1,9] is different from the one we use in this work. However, Eq. (1) is an idealized version because many aspects are missing in the modeling; such as the inhomogeneity of the medium, external sources, and measurement errors.
Then a more realistic version is obtained by considering a stochastic version with additive noise. For example, Balanzario and Kaikina [2] studied the stochastic nonlinear Landau-Ginzburg equations on the half-line with Dirichlet white-noise boundary conditions, Shi and Wang [11] studied the solution for a stochastic fractional partial differential equation driven by an additive fractional space-time white noise. In Sanchez et al. [10], studied the stochastic version of (1) for the 2-dimensional case; however, the n-dimensional case on R n + := {x = (x 1 , . . . , x n ) : x j ≥ 0, j = 1, . . . n} has not been studied. In the present work we tackle this problem via the main ideas of the Fokas method (unified transform) [5], this method is a technique for solving initial-boundary value problems for partial differential equations. Moreover, it generates integral representation formulas for solutions, where the integrals converge uniformly on the boundary.

Preliminaries
Let us give some known definitions and results.

Definition 1
The n-dimensional Fourier-Laplace transform is defined as follows: . . n} and m(k j ) ≤ 0, k · x is the usual inner product, and its inverse is defined by u(x, t) = 1 (2π) n R n e ik·x u(k, t) dk.

Definition 2
The Riesz fractional operator is defined by Here, α ∈ (2, 3), x j ∈ R n + is the vector x, where the jth coordinate is y j , j = 1, . . . n.
Note that the operator, using integration by parts, D α x j can be represented in the following form [8]: then, for m(k l ) ≤ 0, Here, |k| α := n l=1 |k l | α and k [-l] ∈ C n is the k vector, where its lth coordinate is zero.
Proof The theorem follows from the linearity of the operator α and the well-known equation

Green function
We consider a linear problem for an evolution equation with initial condition u 0 and boundary conditions h j , j = 1, . . . , n, + is such that jth and lth coordinates, x l and x j , are equal to zero for j = l.
. Suppose that there exists some function u(x, t), which satisfies (2). Then u(x, t) has the following integral representation: where the Green operators are given by and the Green functions are Here, k α = n l=1 k α l .
Proof Applying Theorem 1 to Eq. (2), we obtain Now, we multiply the above equation by e |k| α t and integrate from 0 to t, for m(k l ) ≤ 0, where Now, we initially consider 2-dimensional case. Thus, Eq. (4) is expressed as Applying the inverse transform in (5) with respect to k 1 and moving the contour of integration for the terms with g 1 j in the integrand, we obtain where D + 1 = {k 1 ∈ C : 0 ≤ m(k 1 ) ≤ π 2α | e(k 1 )|}. Let us note the following: if we substitute k 1 by -k 1 , the functions g 1 j from Eq. (5) are invariant. Then, making this change of variables in (5), we get for m(-k 1 ), m(k 2 ) ≤ 0. Substituting g 1 2 from Eq. (7) in (6) and using the fact that by the Cauchy theorem, we obtain the following integral representation: Applying the inverse transform in (8) with respect to k 2 and moving the contour of integration for the terms with g 2 j in the integrand, we obtain where D + 2 = {k 2 ∈ C : 0 ≤ m(k 2 ) ≤ π 2α | e(k 2 )|}. Let us note the following: if we substitute k 2 by -k 2 , the functions g 2 j from Eq. (8) are invariant. Then, making this change of variables in (7), we get for m(k 1 ), m(k 2 ) ≥ 0. Substituting g 2 2 (|k| α , ±k [-2] , t) from Eq. (10) in (9) and using the fact that ∂D + 2 e ik 2 x 2 u(x 1 , -k 2 , t) dk 2 = 0, by the Cauchy theorem, we obtain the following integral representation: where r ∈ S 2 = {(±k 1 , ±k 2 )} and r [-l] is such that the lth coordinate is equal to zero. In Eq. (11) we have, after interchanging the integration order, integrals of the form R 2 R 2 + e ik·(x 1 ±y 1 ,x 2 ±y 2 )-|k| α t u 0 (y) dy dk, 2 u x 2 (±y 1 , 0, s) dy 1 ds dk.
We notice that all the integrals above are absolutely integrable, then using the Fubini theorem, after some simplifications, we arrive from Eq. (11) at the following equation: where the Green operators are given by and the Green functions are where k α = k α 1 + k α 2 . Now, following the previous arguments we can tackle the ndimensional case. This can be achieved, via mathematical induction over n, passing from Eq. (4) to Eq. (12), through the steps that we describe in the 2-dimensional case. Analogous to Eq. (11), we obtain an integral representation for u, where r ∈ S n = {(±k 1 , ±k 2 , . . . , ±k n )} and r [-l] is such that the lth coordinate is equal to zero. Interchanging the integrals in the above equation, by Fubini's theorem, we obtain the desired result.

Stochastic nonlinear problem
In order to state the problem, we define the Brownian sheetḂ on R n + × [0, T] on a complete probability space ( , F, F t , P), here F is a σ -algebra, {F t } t≥0 is a right-continuous filtration on ( , F) such that F 0 contains all P-negligible subsets and P is a probability measure. We consider a center Gaussian field B = {B(x, t)|x ≥ 0, t ≥ 0} with covariance function given by K (x, t), (y, s) = min{t, s}diag min{x 1 , y 1 }, . . . , min{x n , y n } .
We suppose that B generates a (F t , t ≥ 0)-martingale measure in the sense of Walsh [12]. Let the initial condition u 0 be F 0 × B(R n + ) measurable, where B(R n + ) is the Borelian σalgebra over R n + . Now, we consider the following initial-boundary value problem for a nonlinear equation: where x ∈ R n + , t > 0, α ∈ (2, 3), N is a Lipschitzian operator; i.e., |N u -N v| ≤ C|u -v|, C > 0, and the compatibility conditions h j (x [-j,-l] , t) = h l (x [-j,-l] , t) are satisfied. We understand the solutions for the problem (13) in the following sense: u is a solution if, for all x ∈ R n + and t > 0, the following equation is fulfilled: where the Green operators G I (t), G B l (t) are given in Eq. (3) and the Green function is (15) Theorem 2 Let the initial data u 0 (x) ∈ L 1 (R n + ) and the boundary data h j (x [-j] , t) ∈ C(R + ; L 1 (R n + )). Suppose that, for each T > 0, there exists a constant C > 0 such that, for each x ∈ R n + , t ∈ [0, T] and u, v ∈ R n , |N u -N v| ≤ C|u -v|, and for some p ≥ 1, Then, there exists a unique solution u(x, t) to Eq. (13). Moreover, for all T > 0 and p ≥ 1, Proof First, we define a Picard succession: where u 0 (x, t) = R n + G I (x, y, t)u 0 (y) dy.
Now, let us prove that {u n (x, t)} n≥0 converges in L p ( ). Using the fact that, for all t ≥ 0, G(x, t) from Eq. (15) is a probability density function with respect to x, we obtain, for n ≥ 2, and by (16) and Burkholder's inequality we have Then, by Gronwall's lemma we obtain Taking n → ∞ in L p ( ) at both sides of (17) shows that u(x, t) satisfies the problem (2). Finally, we have to prove the uniqueness of the solution. Let u and v be the two solutions of problem (2), then

Example
In this section, we consider an example for the case n = 2, with the initial condition In Fig. 1, we present the plot of the solution u(x, t) for t = 0.02, 0.1, 0.5, 1, and α = 2.5.