The Diffusion Equation Involving the Dunkl–Laplacian Operator in a Punctured Domain

Abstract

The purpose of this paper is to establish solvability results for both direct and inverse problems associated with the diffusion equation involving the Dunkl–Laplacian operator in a punctured domain. We demonstrate the existence and uniqueness of solutions for both types of problems. Additionally, explicit formulas for the solutions to the considered problems are derived.

1. Introduction

In this paper, we are interested in studying direct and inverse problems for the diffusion equation

$${\mathcal{D}}_{0^{+},t}^{\gamma }u(t,x,y)-{\Lambda }_{\alpha ,x}^{2}u(t,x,y)+{\Delta }_{\theta ,y}u(t,x,y)=f(x,y)$$

(1)

in the domain $Q_T:=(0,T)\times \mathbb{R}\times {\mathsf{\Omega }}_0$, where ${\mathsf{\Omega }}_0=\mathsf{\Omega }\backslash \left\{y_0\right\}\subset {\mathbb{R}}^2$, $\mathsf{\Omega }$ is a domain in ${\mathbb{R}}^2$, $y_0$ is a fixed point in $\mathsf{\Omega }$, T is a fixed positive number, and ${\mathcal{D}}_{0^{+},t}^{\gamma }$ is the Caputo fractional derivative. In the x variable, our spatial operator is generated by the first-order singular differential–difference operator on $\mathbb{R}$ given by

$${\Lambda }_{\alpha }h(x)={\displaystyle \frac{\partial }{\partial x}}h(x)+{\displaystyle \frac{2\alpha +1}{x}}\left({\displaystyle \frac{h(x)-h(-x)}{2}}\right).$$

In the $y=(y_1,y_2)$ variable, we have the following operator:

$$\left\{\begin{array}{c}{\Delta }_{\theta }=-\Delta ,\hfill \\ D({\Delta }_{\theta })=\{u\in \mathcal{D}|{\theta }_1{\xi }^{-}(u)={\theta }_2{\xi }^{+}(u),{\theta }_1^2+{\theta }_2^2\ne 0\},\hfill \end{array}\right.$$

where $\Delta$ is

$$\Delta ={\displaystyle \frac{{\partial }^2}{\partial y_1}}+{\displaystyle \frac{{\partial }^2}{\partial y_2}}.$$

Classes of operators of the ${\Delta }_{\theta }$ type were studied in [1,2,3,4]. In [1], the authors showed the self-adjointness of ${\Delta }_{\theta }$. The operator ${\Lambda }_{\alpha }$ is called the Dunkl operator which was introduced in 1989 by C. Dunkl [5], where $\alpha \ge 1/2$. The Dunkl operator is associated with the reflection group ${\mathbb{Z}}_2$ on $\mathbb{R}$. The Dunkl operators are one of the important model examples of operators in pure mathematics and physics. A solution of the spectral problem generated by the Dunkl operator is called the Dunkl kernel $E_{\alpha }(ix\lambda )$, which is used to define the Dunkl transform ${\mathcal{F}}_{\alpha }$ [6]. The main properties of the Dunkl transform were given by M.F.E. de Jeu in 1993 [7]. For more information about harmonic analysis associated with the operator ${\Lambda }_{\alpha }$, we refer the readers to the papers [7,8,9,10].

Before starting the discussion of our results, we recall that in many physical problems, it is required to determine the coefficients or the right-hand side (the initial term, in the case of the diffusion equation) in the differential equation from some available information; these problems are known as inverse problems. Many authors consider the solvability problem for inverse problems for diffusion equations (see [11,12,13,14,15,16,17,18,19,20] and the references therein).

This paper is organized as follows. In Section 2, we collect some results about harmonic analysis associated with the operator ${\Delta }_{\theta }$, the Dunkl operator on $\mathbb{R}$, and the Caputo fractional operator. In Section 3, we prove our main Theorems 3 and 4 about the solvability of the direct and inverse problems associated with the Dunkl operator on $\mathbb{R}$.

We use the following classic notations: the spaces $C^1(\mathbb{R}),\mathcal{S}(\mathbb{R}),{\mathcal{S}}^{\prime }(\mathbb{R})$ are the space of continuously differentiable functions on $\mathbb{R}$, the space of Schwartz functions on $\mathbb{R}$, and the set of tempered distributions on $\mathbb{R}$, respectively.

2. Preliminaries

2.1. The Operator ${\Delta }_{\theta }$

In Equation (1), we have the operator

$$\left\{\begin{array}{c}{\Delta }_{\theta }=-\Delta ,\hfill \\ D({\Delta }_{\theta })=\{u\in \mathcal{D}|{\theta }_1{\xi }^{-}(u)={\theta }_2{\xi }^{+}(u)\}.\hfill \end{array}\right.$$

(2)

where ${\theta }_1,{\theta }_2$ are real numbers such that ${\theta }_1^2+{\theta }_2^2\ne 0$. Here, we denote by $\mathcal{D}$ the set of all functions

$$u(y)=u_1(y)+\alpha G(y,y_0),y\in {\mathsf{\Omega }}_0,$$

where $\alpha \in \mathbb{R}$, $u_1\in \mathbb{D}=\{u_1\in W_2^2(\mathsf{\Omega }),u_1{|}_{\partial \mathsf{\Omega }}=0\}$, $W_2^2(\mathsf{\Omega })$ is the Sobolev space, $\partial \mathsf{\Omega }$ is a boundary of the domain $\mathsf{\Omega }$, and $G(y,y_0)$ is a Green function of the Dirichlet problem for the Laplace operator.

In (2), the functionals ${\xi }^{-}(u)$ and ${\xi }^{+}(u)$ are given by the expressions

$$\begin{array}{c}{\xi }^{-}(u)=\underset{ϵ\to +0}{lim}{\int }_{y_0^{(1)}-ϵ}^{y_0^{(1)}+ϵ}\left[{\displaystyle \frac{\partial u(\xi ,y_0^{(2)}+ϵ)}{\partial \eta }}-{\displaystyle \frac{\partial u(\xi ,y_0^{(2)}-ϵ)}{\partial \eta }}\right]d\xi \hfill \\ \hfill +{\int }_{y_0^{(2)}-ϵ}^{y_0^{(2)}+ϵ}\left[{\displaystyle \frac{\partial u(y_0^{(1)}+ϵ,\eta )}{\partial \xi }}-{\displaystyle \frac{\partial u(y_0^{(1)}-ϵ,\eta )}{\partial \xi }}\right]d\eta \end{array}$$

and

$${\xi }^{+}(u)=u_0(y_0)=\underset{y\to y_0}{lim}\left(u(y)-{\displaystyle \frac{{\xi }^{-}(u)}{2}}G(y,y_0)\right),$$

where $y_0=(y_0^{(1)},y_0^{(2)})$.

We note that $\mathcal{D}$ is a separable Hilbert space and the operator ${\Delta }_{\theta }$ has a discrete positive spectrum ${\left\{{\mu }_{\xi }\right\}}_{\xi \in \mathcal{I}}$, and the corresponding system of eigenfunctions ${\left\{e_{\xi }\right\}}_{\xi \in \mathcal{I}}$ is a Riesz basis in $\mathcal{D}$, where $\mathcal{I}$ is a countable set. So, there is a unique expansion of the function $u\in \mathcal{D}$ in the form

$$u=\sum _{\xi \in \mathcal{I}}{u}_{\xi }{e}_{\xi },$$

where $u_{\xi }={(u,e_{\xi })}_{\mathcal{D}}$ is the Fourier coefficient for all $\xi \in \mathcal{I}$.

Theorem 1 

([1], Theorem 2, p. 9). The operator ${\Delta }_{\theta }$ is self-adjoint in the space $\mathcal{D}$.

2.2. The Dunkl Operator

The first-order singular differential–difference operator ${\Lambda }_{\alpha },\alpha \ge -1/2$, given by

$${\Lambda }_{\alpha }h(x)={\displaystyle \frac{d}{dx}}h(x)+{\displaystyle \frac{2\alpha +1}{x}}\left({\displaystyle \frac{h(x)-h(-x)}{2}}\right),h\in C^1(\mathbb{R})$$

called the Dunkl operator, associated with the reflexion group ${\mathbb{Z}}_2$ on $\mathbb{R}$. If $\alpha =-1/2$, the Dunkl operator turns into the ordinary differential operator ${\Lambda }_{-1/2}={\displaystyle \frac{d}{dx}}$.

For $\alpha \ge -1/2$ and $\lambda \in \mathbb{R}$, the spectral problem associated with the Dunkl operator

$$\left\{\begin{array}{c}{\Lambda }_{\alpha }h(x)-(i\lambda )h(x)=0,\hfill \\ h(0)=1.\hfill \end{array}\right.$$

has a unique solution $E_{\alpha }(ix\lambda )$ called the Dunkl kernel, given by

$$E_{\alpha }(ix\lambda )=j_{\alpha }(ix\lambda )+{\displaystyle \frac{ix\lambda }{2(\alpha +1)}}{j}_{\alpha +1}(ix\lambda ),x\in \mathbb{R},$$

where

$$j_{\alpha }(ix\lambda )=\Gamma (\alpha +1)\sum _{k=0}^{+\infty }{\displaystyle \frac{1}{k!}}{\displaystyle \frac{{(ix\lambda /2)}^{2k}}{\Gamma (k+\alpha +1)}}$$

is the normalized Bessel function of order $\alpha$.

Remark 1. 

For $\alpha =-{\displaystyle \frac{1}{2}}$, we have

$$\left\{\begin{array}{c}{\displaystyle \frac{d}{dx}}h(x)-(i\lambda )h(x)=0,\hfill \\ h(0)=1.\hfill \end{array}\right.$$

The solution of this problem is

$$E_{-1/2}(ix\lambda )=e^{ix\lambda }.$$

Theorem 2. 

Let $\alpha >-{\displaystyle \frac{1}{2}}$ and $\lambda \in \mathbb{R}$. Then, we have the following estimates for the Dunkl kernel:

$$\left|{\displaystyle \frac{d^k}{d{x}^k}}{E}_{\alpha }(x,\lambda )\right|\le {|\lambda |}^k\mathrm{and}\left|{\displaystyle \frac{d^k}{d{\lambda }^k}}{E}_{\alpha }(x,\lambda )\right|\le {|x|}^k$$

for all $k\in \mathbb{N}$ and $x\in \mathbb{R}$. In particular, we have

$$\left|E_{\alpha }(x,\lambda )\right|\le 1$$

(3)

for all $x,\lambda \in \mathbb{R}$, when $k=0$.

Definition 1. 

We denote by $L^p(\mathbb{R},{\mu }_{\alpha }),1\le p\le +\infty$, the space of measurable functions h on $\mathbb{R}$ such that

$${\parallel h\parallel }_{p,\alpha }={\left({\int }_{\mathbb{R}}{|h(x)|}^{p}d{\mu }_{\alpha }(x)\right)}^{{\displaystyle \frac{1}{p}}}<+\infty ,1\le p<+\infty ,$$

$${\parallel h\parallel }_{\infty }=\underset{x\in \mathbb{R}}{sup}|h(x)|<+\infty .$$

Here, ${\mu }_{\alpha }$ is the measure defined on $\mathbb{R}$ by

$$d{\mu }_{\alpha }(x)={\displaystyle \frac{{|x|}^{2\alpha +1}}{2^{\alpha +1}\Gamma (\alpha +1)}}dx,\alpha \ge -1/2.$$

For $h\in L^1(\mathbb{R},{\mu }_{\alpha })$, the Dunkl transform is defined by

$${\mathcal{F}}_{\alpha }(h)(\lambda )=\widehat{h}(\lambda ):={\int }_{\mathbb{R}}h(x)E_{\alpha }(-ix\lambda )d{\mu }_{\alpha }(x),\lambda \in \mathbb{R}.$$

(4)

This transform has the following properties ([7]):

(i)

For all $h\in \mathcal{S}(\mathbb{R})$, we have

$${\mathcal{F}}_{\alpha }({\Lambda }_{\alpha }h)(\lambda )=i\lambda {\mathcal{F}}_{\alpha }(h)(\lambda ),\lambda \in \mathbb{R}.$$

(5)

(ii)

For all $h\in L^1(\mathbb{R},{\mu }_{\alpha })$, the Dunkl transform ${\mathcal{F}}_{\alpha }$ is a continuous function on $\mathbb{R}$ satisfying

$$\parallel {\mathcal{F}}_{\alpha }{(h)\parallel }_{\infty }\le {\parallel h\parallel }_{1,\alpha }.$$

(iii)

($L^1$-inversion) For all $h\in L^1(\mathbb{R},{\mu }_{\alpha })$ with ${\mathcal{F}}_{\alpha }(h)\in L^1(\mathbb{R},{\mu }_{\alpha })$, we have

$$h(x)={\int }_{\mathbb{R}}{\mathcal{F}}_{\alpha }(h)(\lambda )E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ).$$

(6)

(iv)

${\mathcal{F}}_{\alpha }$ is a topological isomorphism on $\mathcal{S}(\mathbb{R})$ which extends to a topological isomorphism on ${\mathcal{S}}^{\prime }(\mathbb{R})$.

(v)

(Plancherel theorem) The Dunkl transform ${\mathcal{F}}_{\alpha }$ is an isometric isomorphism of $L^2(\mathbb{R},{\mu }_{\alpha })$. In particular,

$$\parallel {\mathcal{F}}_{\alpha }{(h)\parallel }_{2,\alpha }={\parallel h\parallel }_{2,\alpha }.$$

(7)

Notation ([10], p. 22). For $s\in \mathbb{R}$, we denote by

$$W_{\alpha }^{s,2}(\mathbb{R},{\mu }_{\alpha }):=\{h\in {\mathcal{S}}^{\prime }(\mathbb{R}){:\parallel h\parallel }_{W_{\alpha }^{s,2}(\mathbb{R},{\mu }_{\alpha })}^2={\int }_{\mathbb{R}}{(1+{\lambda }^2)}^s|{\mathcal{F}}_{\alpha }(h)(\lambda ){|}^{2}d{\mu }_{\alpha }(\lambda )<\infty \}$$

the usual Sobolev space on $\mathbb{R}$.

2.3. The Caputo Fractional Derivatives

Here, we give basic definitions from fractional calculus.

Definition 2 

([21], p. 69). Let $[a,b]$ be a finite interval on the real axis $\mathbb{R}$ and $-\infty <a<b<\infty$. The left and right Riemann–Liouville fractional integrals $I_{a^{+}}^{\gamma }$ and $I_{b^{-}}^{\gamma }$ of order $\gamma \in \mathbb{R}$$(\gamma >0)$ are defined by

$$I_{a^{+}}^{\gamma }\left[f\right](t):={\displaystyle \frac{1}{\Gamma (\gamma )}}{\int }_a^t{(t-s)}^{\gamma -1}f(s)ds,t\in (a,b],$$

(8)

and

$$I_{b^{-}}^{\gamma }\left[f\right](t):={\displaystyle \frac{1}{\Gamma (\gamma )}}{\int }_t^b{(s-t)}^{\gamma -1}f(s)ds,t\in [a,b),$$

(9)

respectively.

Definition 3 

([21], p. 70). The left and right Riemann–Liouville fractional derivatives ${\mathbb{D}}_{a^{+}}^{\gamma }$ and ${\mathbb{D}}_{b^{-}}^{\gamma }$ of order $\gamma \in \mathbb{R}$$(0<\gamma <1)$ are given by

$${\mathbb{D}}_{a^{+}}^{\gamma }\left[f\right](t):={\displaystyle \frac{d}{dt}}{I}_{a^{+}}^{1-\gamma }\left[f\right](t),\forall t\in (a,b],$$

(10)

and

$${\mathbb{D}}_{b^{-}}^{\gamma }\left[f\right](t):=-{\displaystyle \frac{d}{dt}}{I}_{b^{-}}^{1-\gamma }\left[f\right](t),\forall t\in [a,b),$$

(11)

respectively.

Definition 4 

([21], p. 91). The left and right Caputo fractional derivatives ${\mathcal{D}}_{a^{+}}^{\gamma }$ and ${\mathcal{D}}_{b^{-}}^{\gamma }$ of order $\gamma \in \mathbb{R}$$(0<\gamma <1)$ are defined by

$${\mathcal{D}}_{a^{+}}^{\gamma }\left[f\right](t):={\mathbb{D}}_{a^{+}}^{\gamma }[f(t)-f(a)],t\in (a,b],$$

(12)

and

$${\mathcal{D}}_{b^{-}}^{\gamma }\left[f\right](t):={\mathbb{D}}_{b^{-}}^{\gamma }[f(t)-f(b)],t\in [a,b),$$

(13)

respectively.

Definition 5 

([22], p. 18, Definition 3). Let X be a Banach space. We say that $u\in C^{\gamma }([0,T],X)$ if $u\in C([0,T],X)$ and ${\mathcal{D}}_t^{\gamma }u\in C([0,T],X)$.

Lemma 1 

([23]). For $0<\gamma <1$, the Mittag-Leffler-type function ${\mathbb{E}}_{\gamma ,\gamma }\left(-\lambda t^{\gamma }\right)$ satisfies

$$0\le {\mathbb{E}}_{\gamma ,\gamma }\left(-\lambda t^{\gamma }\right)\le {\displaystyle \frac{1}{\Gamma (\gamma )}},t\in [0,\infty ),0\le \lambda .$$

3. The Main Result

3.1. Direct Problem

The initial-value problem for the Dunkl-type heat operator ${\Delta }_k-{\partial }_t$ was considered by M. Rösler [24], p. 122. Also, readers can find more information about the direct problem for the heat equation associated with the Dunkl operators in Mejjaoli’s papers [25,26].

Let us introduce the following definitions:

Definition 6. 

1. 

The space $L^p(\mathbb{R},{\mu }_{\alpha };\mathcal{D})$ is the $L^p$ space on $\mathbb{R}$ given by the norm

$${\parallel f\parallel }_{L^p(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^p:={\int }_{\mathbb{R}}{\parallel \widehat{f}(\lambda ,·)\parallel }_{\mathcal{D}}^{p}d{\mu }_{\alpha }(\lambda )<+\infty ;$$

2. 

The space $W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})$ is the Sobolev space on $\mathbb{R}$ given by the norm

$${\parallel f\parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2:={\int }_{\mathbb{R}}(1+{\lambda }^2){\parallel \widehat{f}(\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )<+\infty ;$$

3. 

The space $C([0,T];L^p(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))$ is the space of all continuous functions $f(·,x,y)$ on $[0,T]$, such that

$${\parallel f\parallel }_{C([0,T];L^p(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))}:=\underset{0<t<T}{max}{\parallel f(t,·,·)\parallel }_{L^p(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}<+\infty ;$$

4. 

The space $C([0,T];W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))$ is the space of all continuous functions $f(·,x,y)$ on $[0,T]$, such that

$${\parallel f\parallel }_{C([0,T];W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))}:=\underset{0<t<T}{max}{\parallel f(t,·,·)\parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}<+\infty .$$

Definition 7. 

We define ${\mathcal{D}}^1$ as

$${\mathcal{D}}^1:=\{u\in \mathcal{D}:{\Delta }_{\theta }u\in \mathcal{D}\}.$$

In this subsection, we consider the direct problem for (1).

Problem 1. 

We aim to find a function u satisfying the equation

$${\mathcal{D}}_{0^{+},t}^{\gamma }u(t,x,y)-{\Lambda }_{\alpha ,x}^{2}u(t,x,y)+{\Delta }_{\theta ,y}u(t,x,y)=f(t,x,y),(t,x,y)\in Q_T,$$

under the condition

$$u(0,x,y)=g(x,y),(x,y)\in \mathbb{R}\times {\mathsf{\Omega }}_0.$$

Our first main result reads as follows.

Theorem 3. 

Let $f\in C([0,T];W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))\cap C([0,T];L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1))$ and $g\in W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };$ $\mathcal{D})\cap L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)$. Then, Problem 1 has a unique solution u in the space $C^{\gamma }([0,T];L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))$, given by

$$\begin{array}{cc}\hfill u(t,x,y)& =\sum _{\xi \in \mathcal{I}}({\int }_{\mathbb{R}}{\int }_0^t{(t-\tau )}^{\gamma -1}{\mathbb{E}}_{\gamma ,\gamma }\left(-({\lambda }^2+{\mu }_{\xi }){(t-\tau )}^{\gamma }\right)\hfill \\ \hfill & \times \widehat{f}(\tau ,\lambda ,·)E_{\alpha }(ix\lambda )d\tau d{\mu }_{\alpha }(\lambda ),e_{\xi }(·){)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}\widehat{g}(\lambda ,·){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y),\hfill \end{array}$$

where ${\mathbb{E}}_{\gamma ,1}(z)$ and ${\mathbb{E}}_{\gamma ,\gamma }(z)$ are Mittag-Leffler functions:

$${\mathbb{E}}_{\gamma ,1}(z):=\sum _{k=0}^{\infty }{\displaystyle \frac{z^k}{\Gamma (\gamma k+1)}},{\mathbb{E}}_{\gamma ,\gamma }(z):=\sum _{k=0}^{\infty }{\displaystyle \frac{z^k}{\Gamma (\gamma k+\gamma )}},$$

where $z,\gamma \in \mathbb{C}$ and $Re(\gamma )>0$.

Proof. 

First, we prove the existence of the solution of Problem 1. By using Dunkl transform ${\mathcal{F}}_{\alpha }$ (4) and (5) for Problem 1, we obtain

$${\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}(t,\lambda ,y)+{\lambda }^2\widehat{u}(t,\lambda ,y)+{\Delta }_{\theta ,y}\widehat{u}(t,\lambda ,y)=\widehat{f}(t,\lambda ,y),\lambda \in \mathbb{R},$$

(14)

$$\widehat{u}(0,\lambda ,y)=\widehat{g}(\lambda ,y),\lambda \in \mathbb{R}.$$

(15)

Since $\mathcal{D}$ is the separable Hilbert space and eigenfunctions ${\left\{e_{\xi }\right\}}_{\xi \in \mathcal{I}}$ of the operator ${\Delta }_{\theta }$ are the Riesz basis in $\mathcal{D}$, we seek the function $\widehat{u}$ in the form

$$\widehat{u}(t,\lambda ,y)=\sum _{\xi \in \mathcal{I}}{\widehat{u}}_{\xi }(t,\lambda )e_{\xi }(y),$$

(16)

where Fourier coefficients ${\widehat{u}}_{\xi }$ are unknown functions. Substituting (16) into Equations (14) and (15), we obtain

$${\mathcal{D}}_{0^{+},t}^{\gamma }{\widehat{u}}_{\xi }(t,\lambda )+({\lambda }^2+{\mu }_{\xi }){\widehat{u}}_{\xi }(t,\lambda )={\widehat{f}}_{\xi }(t,\lambda ),$$

(17)

$${\widehat{u}}_{\xi }(0,\lambda )={\widehat{g}}_{\xi }(\lambda ).$$

(18)

Here, the functions ${\widehat{f}}_{\xi }$ and ${\widehat{g}}_{\xi }$ are Fourier coefficients of the functions $\widehat{f}$ and $\widehat{g}$, respectively. The general solution [21] of Equation (17) is

$$\begin{array}{c}{\widehat{u}}_{\xi }(t,\lambda )={\widehat{g}}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\hfill \\ \hfill +{\int }_0^t{(t-\tau )}^{\gamma -1}{\mathbb{E}}_{\gamma ,\gamma }\left(-({\lambda }^2+{\mu }_{\xi }){(t-\tau )}^{\gamma }\right){\widehat{f}}_{\xi }(\tau ,\lambda )d\tau .\end{array}$$

(19)

Putting (19) into (16), we obtain the solution of the problems in (14) and (15), i.e.,

$$\begin{array}{c}\widehat{u}(t,\lambda ,y)=\sum _{\xi \in \mathcal{I}}\left({\int }_0^t{(t-\tau )}^{\gamma -1}{\mathbb{E}}_{\gamma ,\gamma }\left(-({\lambda }^2+{\mu }_{\xi }){(t-\tau )}^{\gamma }\right){\widehat{f}}_{\xi }(\tau ,\lambda )d\tau \right)e_{\xi }(y)\hfill \\ \hfill +\sum _{\xi \in \mathcal{I}}{\widehat{g}}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)e_{\xi }(y)\end{array}$$

(20)

or

$$\begin{array}{cc}\hfill \widehat{u}(t,\lambda ,y)& =\sum _{\xi \in \mathcal{I}}\left({\int }_0^t{(t-\tau )}^{\gamma -1}{\mathbb{E}}_{\gamma ,\gamma }\left(-({\lambda }^2+{\mu }_{\xi }){(t-\tau )}^{\gamma }\right){(\widehat{f}(\tau ,\lambda ,·),e_{\xi }(·))}_{\mathcal{D}}d\tau \right)e_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{(\widehat{g}(\lambda ,·),e_{\xi }(·))}_{\mathcal{D}}{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)e_{\xi }(y)\hfill \\ \hfill & =\sum _{\xi \in \mathcal{I}}{\left({\int }_0^t{(t-\tau )}^{\gamma -1}{\mathbb{E}}_{\gamma ,\gamma }\left(-({\lambda }^2+{\mu }_{\xi }){(t-\tau )}^{\gamma }\right)\widehat{f}(\tau ,\lambda ,·)d\tau ,e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\left(\widehat{g}(\lambda ,·){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y).\hfill \end{array}$$

Then, applying inverse Dunkl transform ${\mathcal{F}}_{\alpha }^{-1}$ (6), we obtain

$$\begin{array}{cc}\hfill u(t,x,y)& =\sum _{\xi \in \mathcal{I}}({\int }_{\mathbb{R}}{\int }_0^t{(t-\tau )}^{\gamma -1}{\mathbb{E}}_{\gamma ,\gamma }\left(-({\lambda }^2+{\mu }_{\xi }){(t-\tau )}^{\gamma }\right)\hfill \\ \hfill & \times \widehat{f}(\tau ,\lambda ,·)E_{\alpha }(ix\lambda )d\tau d{\mu }_{\alpha }(\lambda ),e_{\xi }(·){)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}\widehat{g}(\lambda ,·){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y).\hfill \end{array}$$

(21)

Consequently, the function u given by Formula (21) is a solution of Problem 1.

Now, we prove the convergence of the obtained infinite series corresponding to the functions $\widehat{u}(t,\lambda ,·)$, ${\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}(t,\lambda ,·)$, and ${\Delta }_{\theta }\widehat{u}(t,\lambda ,·)$. We can see the convergence of the infinite series for the function $\widehat{u}(t,\lambda ,·)$ (20) in the space $\mathcal{D}$, i.e.,

$$\begin{array}{cc}\hfill \parallel \widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2& \le \sum _{\xi \in \mathcal{I}}{\left|{\int }_0^t{(t-\tau )}^{\gamma -1}{\mathbb{E}}_{\gamma ,\gamma }\left(-({\lambda }^2+{\mu }_{\xi }){(t-\tau )}^{\gamma }\right){\widehat{f}}_{\xi }(\tau ,\lambda )d\tau \right|}^2\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\left|{\widehat{g}}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\right|}^2\hfill \\ \hfill & \le \sum _{\xi \in \mathcal{I}}{\left({\displaystyle \frac{1}{\Gamma (\gamma )}}{\int }_0^t|{\widehat{f}}_{\xi }(\tau ,\lambda )|d\tau \right)}^2+\sum _{\xi \in \mathcal{I}}{\left|{\widehat{g}}_{\xi }(\lambda )\right|}^2\hfill \\ \hfill & \le \sum _{\xi \in \mathcal{I}}{\left({\displaystyle \frac{1}{\Gamma (\gamma )}}{\int }_0^T|{\widehat{f}}_{\xi }(t,\lambda )|dt\right)}^2+{\parallel \widehat{g}(\lambda ,·)\parallel }_{\mathcal{D}}^2\hfill \\ \hfill & \lesssim \sum _{\xi \in \mathcal{I}}{\int }_0^T|{\widehat{f}}_{\xi }{(t,\lambda )|}^{2}dt+{\parallel \widehat{g}(\lambda ,·)\parallel }_{\mathcal{D}}^2\hfill \\ \hfill & ={\int }_0^T\parallel \widehat{f}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}dt+{\parallel \widehat{g}(\lambda ,·)\parallel }_{\mathcal{D}}^2\hfill \end{array}$$

from Lemma 1, Hölder’s inequality, and the inequality

$$0<{\mathbb{E}}_{\gamma ,1}(-z)<1,0<z,0<\gamma <1.$$

(22)

Consequently,

$$\parallel \widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2\lesssim {\int }_0^T\parallel \widehat{f}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}dt+{\parallel \widehat{g}(\lambda ,·)\parallel }_{\mathcal{D}}^2.$$

(23)

Here, for our convenience, we will write $U\lesssim W$, which means $U\le CW$ for some positive constant C independent of U and W. Applying the operator ${\Delta }_{\theta }$ to (20), we obtain

$$\begin{array}{cc}\hfill {\Delta }_{\theta ,y}\widehat{u}(t,\lambda ,y)& =\sum _{\xi \in \mathcal{I}}\left({\int }_0^t{(t-\tau )}^{\gamma -1}{\mathbb{E}}_{\gamma ,\gamma }\left(-({\lambda }^2+{\mu }_{\xi }){(t-\tau )}^{\gamma }\right){\widehat{f}}_{\xi }(\tau ,\lambda )d\tau \right){\Delta }_{\theta ,y}{e}_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\widehat{g}}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right){\Delta }_{\theta ,y}{e}_{\xi }(y)\hfill \\ \hfill & =\sum _{\xi \in \mathcal{I}}\left({\int }_0^t{(t-\tau )}^{\gamma -1}{\mathbb{E}}_{\gamma ,\gamma }\left(-({\lambda }^2+{\mu }_{\xi }){(t-\tau )}^{\gamma }\right){\mu }_{\xi }{\widehat{f}}_{\xi }(\tau ,\lambda )d\tau \right)e_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\mu }_{\xi }{\widehat{g}}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)e_{\xi }(y).\hfill \end{array}$$

Then, we have

$$\parallel {\Delta }_{\theta }\widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2\lesssim {\int }_0^T\parallel {\Delta }_{\theta }\widehat{f}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}dt+{\parallel {\Delta }_{\theta }\widehat{g}(\lambda ,·)\parallel }_{\mathcal{D}}^2.$$

(24)

To obtain the estimate for ${\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}(t,\lambda ,·)$, we rewrite Equation (14) as follows

$${\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}(t,\lambda ,y)=\widehat{f}(t,\lambda ,y)-{\lambda }^2\widehat{u}(t,\lambda ,y)-{\Delta }_{\theta ,y}\widehat{u}(t,\lambda ,y).$$

Then, we obtain

$$\begin{array}{cc}\hfill \parallel {\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2& \lesssim \parallel \widehat{f}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2+{\lambda }^2\parallel \widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2+{\parallel {\Delta }_{\theta }\widehat{u}(t,\lambda ,·)\parallel }_{\mathcal{D}}^2\hfill \\ \hfill & \lesssim \parallel \widehat{f}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2+{\lambda }^2\left({\int }_0^T\parallel \widehat{f}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}dt+{\parallel \widehat{g}(\lambda ,·)\parallel }_{\mathcal{D}}^2\right)\hfill \\ \hfill & +{\int }_0^T\parallel {\Delta }_{\theta }\widehat{f}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}dt+{\parallel {\Delta }_{\theta }\widehat{g}(\lambda ,·)\parallel }_{\mathcal{D}}^2\hfill \end{array}$$

by using (23) and (24). Thus, we can see the convergence for ${\widehat{u}}_t$ in the space $\mathcal{D}$.

Let us estimate the function u as follows

$${\parallel u(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2=\parallel \widehat{u}{(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2={\int }_{\mathbb{R}}{\parallel \widehat{u}(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )$$

$$\begin{array}{cc}\hfill & \lesssim {\int }_{\mathbb{R}}{\int }_0^T\parallel \widehat{f}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}dtd{\mu }_{\alpha }(\lambda )+{\int }_{\mathbb{R}}{\parallel \widehat{g}(\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )\hfill \\ \hfill & ={\int }_0^T{\parallel f(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^{2}dt+{\parallel g\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2,\hfill \end{array}$$

by using (7). So, it follows that

$${\parallel u\parallel }_{C([0,T];L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))}^2\lesssim {\parallel f\parallel }_{C([0,T];L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))}^2+{\parallel g\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2.$$

Finally, by estimating the function $u_t$, as given

$$\parallel {\mathcal{D}}_{0^{+},t}^{\gamma }{u(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2=\parallel {\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}{(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2={\int }_{\mathbb{R}}{\parallel {\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )$$

$$\begin{array}{cc}\hfill & \lesssim {\int }_{\mathbb{R}}\parallel \widehat{f}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )+{\int }_{\mathbb{R}}{\int }_0^T(1+{\lambda }^2){\parallel \widehat{f}(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}dtd{\mu }_{\alpha }(\lambda )\hfill \\ \hfill & +{\int }_{\mathbb{R}}(1+{\lambda }^2)\parallel \widehat{g}(\lambda ,·){\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )+{\int }_{\mathbb{R}}{\int }_0^T{\parallel {\Delta }_{\theta }\widehat{f}(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}dtd{\mu }_{\alpha }(\lambda )\hfill \\ \hfill & +{\int }_{\mathbb{R}}{\parallel {\Delta }_{\theta }\widehat{g}(\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )\hfill \\ \hfill & ={\parallel f(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\int }_0^T{\parallel f(t,·,·)\parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^{2}dt+{\parallel g\parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2\hfill \\ \hfill & +{\int }_0^T{\parallel f(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^{2}dt+{\parallel g\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2\hfill \end{array}$$

from Equation (7) in property (v), we have

$$\begin{array}{cc}\hfill \parallel {\mathcal{D}}_{0^{+},t}^{\gamma }{u\parallel }_{C([0,T];L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))}^2& \lesssim {\parallel f\parallel }_{C([0,T];W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))}^2+{\parallel g\parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2\hfill \\ \hfill & +{\parallel f\parallel }_{C([0,T];L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1))}^2+{\parallel g\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2.\hfill \end{array}$$

The existence is proved.

Now, we will prove the uniqueness of the solution. Let us suppose that $u_1$ and $u_2$ are two different solutions of Problem 1. Then, $u=u_1-u_2$ is the solution of the following problem:

$${\mathcal{D}}_{0^{+},t}^{\gamma }u(t,x,y)-{\Lambda }_{\alpha ,x}^{2}u(t,x,y)+{\Delta }_{\theta ,y}u(t,x,y)=0,(t,x,y)\in Q_T,$$

$$u(0,x,y)=0,(x,y)\in \mathbb{R}\times {\mathsf{\Omega }}_0.$$

This problem is a particular case of Problem 1 when $f(t,x,y)\equiv 0$ and $g(x,y)\equiv 0$. So, this problem has a unique solution (Theorem 3) given by expression (21). Consequently, if in (21) f and g equal zero, it will give us the trivial solution $u(t,x,y)\equiv 0$. This means $u_1\equiv u_2$. It contradicts the condition, showing the uniqueness of the solution of Problem 1. □

3.2. Inverse Problem

Problem 2. 

We aim to find a couple of functions $\{u,f\}$ satisfying Equation (1) under the conditions

$$u(0,x,y)=\varphi (x,y),(x,y)\in \mathbb{R}\times {\mathsf{\Omega }}_0$$

(25)

and

$$u(T,x,y)=\psi (x,y),(x,y)\in \mathbb{R}\times {\mathsf{\Omega }}_0.$$

(26)

Theorem 4. 

Assume that $\varphi ,\psi \in W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})\cap L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)$. Then, Problem 2 has a unique solution $u\in C([0,T];L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))\cap C^{\gamma }([0,T];L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))$, and $f\in L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})$, and these can be written in the forms

$$\begin{array}{cc}\hfill u(t,x,y)& =\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}{\displaystyle \frac{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\widehat{\psi }(\lambda ,·)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & -\sum _{\xi \in \mathcal{I}}({\int }_{\mathbb{R}}{\displaystyle \frac{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)\widehat{\varphi }(\lambda ,·)\hfill \\ \hfill & \times E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·){)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\widehat{\varphi }(\lambda ,·)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill f(x,y)& =\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}{\displaystyle \frac{{\lambda }^2+{\mu }_{\xi }}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\widehat{\psi }(\lambda ,·)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & -\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}{\displaystyle \frac{({\lambda }^2+{\mu }_{\xi }){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\widehat{\varphi }(\lambda ,·)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y).\hfill \end{array}$$

Proof. 

First of all, we start by proving the existence result. By using the Dunkl transform ${\mathcal{F}}_{\alpha }$ (4) on both of sides of (1) and initial and final time conditions (25) and (26), we obtain

$${\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}(t,\lambda ,y)+{\lambda }^2\widehat{u}(t,\lambda ,y)+{\Delta }_{\theta ,y}\widehat{u}(t,\lambda ,y)=\widehat{f}(\lambda ,y),(\lambda ,y,t)\in \mathbb{R}\times {\mathsf{\Omega }}_0\times (0,T),$$

(27)

$$\widehat{u}(0,\lambda ,y)=\widehat{\varphi }(\lambda ,y),(\lambda ,y)\in \mathbb{R}\times {\mathsf{\Omega }}_0,$$

(28)

$$\widehat{u}(T,\lambda ,y)=\widehat{\psi }(\lambda ,y),(\lambda ,y)\in \mathbb{R}\times {\mathsf{\Omega }}_0.$$

(29)

where Fourier coefficients ${\widehat{u}}_{\xi }$ are unknown functions. Substituting (16) into Equations (27)–(29), we obtain

$${\mathcal{D}}_{0^{+},t}^{\gamma }{\widehat{u}}_{\xi }(t,\lambda )+({\lambda }^2+{\mu }_{\xi }){\widehat{u}}_{\xi }(t,\lambda )={\widehat{f}}_{\xi }(\lambda ),\xi \in \mathcal{I},$$

(30)

with conditions

$${\widehat{u}}_{\xi }(0,\lambda )={\widehat{\varphi }}_{\xi }(\lambda ),\xi \in \mathcal{I}$$

(31)

and

$${\widehat{u}}_{\xi }(T,\lambda )={\widehat{\psi }}_{\xi }(\lambda ),\xi \in \mathcal{I}$$

(32)

Here, the functions ${\widehat{f}}_{\xi },{\widehat{\varphi }}_{\xi }$ and ${\widehat{\psi }}_{\xi }$ are Fourier coefficients of the functions $\widehat{f},\widehat{\varphi }$ and $\widehat{\psi }$, respectively. The general solution of the equation (30) is

$${\widehat{u}}_{\xi }(t,\lambda )={\displaystyle \frac{{\widehat{f}}_{\xi }(\lambda )}{{\lambda }^2+{\mu }_{\xi }}}\left(1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\right)+C_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right),$$

(33)

where the functions ${\widehat{f}}_{\xi }$ and $C_{\xi }$ are unknown functions. By using conditions (31) and (32) we can find them, as follows:

$${\widehat{u}}_{\xi }(0,\lambda )=C_{\xi }(\lambda )={\widehat{\varphi }}_{\xi }(\lambda ),$$

$${\widehat{u}}_{\xi }(T,\lambda )={\displaystyle \frac{{\widehat{f}}_{\xi }(\lambda )}{{\lambda }^2+{\mu }_{\xi }}}\left(1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)\right)+{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)={\widehat{\psi }}_{\xi }(\lambda ).$$

Then, ${\widehat{f}}_{\xi }$ is represented as

$${\widehat{f}}_{\xi }(\lambda )=({\lambda }^2+{\mu }_{\xi }){\displaystyle \frac{{\widehat{\psi }}_{\xi }(\lambda )-{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}.$$

Now, substituting the functions $C_{\xi }$ and ${\widehat{f}}_{\xi }$ into (33), we obtain a solution $\{{\widehat{u}}_{\xi },{\widehat{f}}_{\xi }\}$ of the problem (30)–(32), given by

$$\begin{array}{c}{\widehat{u}}_{\xi }(t,\lambda )={\displaystyle \frac{{\widehat{\psi }}_{\xi }(\lambda )-{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\left(1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\right)\hfill \\ \hfill +{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\end{array}$$

and

$${\widehat{f}}_{\xi }(\lambda )=({\lambda }^2+{\mu }_{\xi }){\displaystyle \frac{{\widehat{\psi }}_{\xi }(\lambda )-{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}.$$

Thus, it gives us

$$\begin{array}{cc}\hfill \widehat{u}(t,\lambda ,y)& =\sum _{\xi \in \mathcal{I}}{\displaystyle \frac{{\widehat{\psi }}_{\xi }(\lambda )-{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\left(1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\right)e_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)e_{\xi }(y)\hfill \end{array}$$

(34)

and

$$\widehat{f}(y,\lambda )=\sum _{\xi \in \mathcal{I}}({\lambda }^2+{\mu }_{\xi }){\displaystyle \frac{{\widehat{\psi }}_{\xi }(\lambda )-{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}{e}_{\xi }(y),$$

which are a solution $\{\widehat{u},\widehat{f}\}$ of problems (27)–(29).

Now, we check the convergence of the infinite series corresponding to the functions $\widehat{u}(t,\lambda ,·)$, ${\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}(t,\lambda ,·)$, ${\Delta }_{\theta }\widehat{u}(t,\lambda ,·)$, and $\widehat{f}(\lambda ,·)$ in the space $\mathcal{D}$. First, we check the convergence of the sum (34) corresponding to the function $\widehat{u}(t,\lambda ,·)$, as follows:

$$\begin{array}{cc}\hfill \parallel \widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2& \le \sum _{\xi \in \mathcal{I}}{\left|{\displaystyle \frac{{\widehat{\psi }}_{\xi }(\lambda )-{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\left(1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\right)\right|}^2\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\left|{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\right|}^2\hfill \\ \hfill & \lesssim \sum _{\xi \in \mathcal{I}}{\left|{\widehat{\psi }}_{\xi }(\lambda )\right|}^2+\sum _{\xi \in \mathcal{I}}{\left|{\widehat{\varphi }}_{\xi }(\lambda )\right|}^2.\hfill \end{array}$$

Thus, we obtain

$$\parallel \widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2\lesssim \parallel \widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+{\parallel \widehat{\varphi }(\lambda ,·)\parallel }_{\mathcal{D}}^2.$$

(35)

To establish the convergence of the sum corresponding to the function ${\Delta }_{\theta }\widehat{u}(t,\lambda ,·)$, we need to use

$$\begin{array}{cc}\hfill {\Delta }_{\theta ,y}\widehat{u}(t,\lambda ,y)& =\sum _{\xi \in \mathcal{I}}{\mu }_{\xi }{\displaystyle \frac{{\widehat{\psi }}_{\xi }(\lambda )-{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\hfill \\ \hfill & \times \left(1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\right)e_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\mu }_{\xi }{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)e_{\xi }(y).\hfill \end{array}$$

Consequently, we have

$$\begin{array}{c}\hfill \parallel {\Delta }_{\theta }\widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2\lesssim \sum _{\xi \in \mathcal{I}}{\left|{\mu }_{\xi }{\widehat{\psi }}_{\xi }(\lambda )\right|}^2+\sum _{\xi \in \mathcal{I}}{\left|{\mu }_{\xi }{\widehat{\varphi }}_{\xi }(\lambda )\right|}^2\end{array}$$

or

$$\parallel {\Delta }_{\theta }\widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2\lesssim \parallel {\Delta }_{\theta }\widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+{\parallel {\Delta }_{\theta }\widehat{\varphi }(\lambda ,·)\parallel }_{\mathcal{D}}^2.$$

(36)

Now, for $\widehat{f}(\lambda ,·)$, we obtain

$$\begin{array}{cc}\hfill \parallel \widehat{f}{(\lambda ,·)\parallel }_{\mathcal{D}}^2& ={\left|({\lambda }^2+{\mu }_{\xi }){\displaystyle \frac{{\widehat{\psi }}_{\xi }(\lambda )-{\widehat{\varphi }}_{\xi }(\lambda ){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\right|}^2\hfill \\ \hfill & \lesssim {\left|({\lambda }^2+{\mu }_{\xi }){\widehat{\psi }}_{\xi }(\lambda )\right|}^2+{\left|({\lambda }^2+{\mu }_{\xi }){\widehat{\varphi }}_{\xi }(\lambda )\right|}^2\hfill \\ \hfill & \lesssim {\lambda }^2\parallel \widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+\parallel {\Delta }_{\theta }\widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+{\lambda }^2\parallel \widehat{\varphi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+{\parallel {\Delta }_{\theta }\widehat{\varphi }(\lambda ,·)\parallel }_{\mathcal{D}}^2.\hfill \end{array}$$

Thus,

$$\parallel \widehat{f}{(\lambda ,·)\parallel }_{\mathcal{D}}^2\lesssim {\lambda }^2(\parallel \widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+\parallel \widehat{\varphi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2)+\parallel {\Delta }_{\theta }\widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+{\parallel {\Delta }_{\theta }\widehat{\varphi }(\lambda ,·)\parallel }_{\mathcal{D}}^2.$$

(37)

Finally, by rewriting Equation (27) as follows

$${\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}(t,\lambda ,y)=\widehat{f}(\lambda ,y)-{\lambda }^2\widehat{u}(t,\lambda ,y)+{\Delta }_{\theta ,y}\widehat{u}(t,\lambda ,y)$$

we obtain

$$\parallel {\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}{(t,\lambda ,·)\parallel }_{\mathcal{D}}^2\lesssim {\lambda }^2(\parallel \widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+\parallel \widehat{\varphi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2)+\parallel {\Delta }_{\theta }\widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+{\parallel {\Delta }_{\theta }\widehat{\varphi }(\lambda ,·)\parallel }_{\mathcal{D}}^2$$

(38)

from (35)–(37).

After calculations,

$$\begin{array}{cc}\hfill \widehat{u}(t,\lambda ,y)& =\sum _{\xi \in \mathcal{I}}{\displaystyle \frac{{\left(\widehat{\psi }(\lambda ,·),e_{\xi }(·)\right)}_{\mathcal{D}}-{\left(\widehat{\varphi }(\lambda ,·),e_{\xi }(·)\right)}_{\mathcal{D}}{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\hfill \\ \hfill & \times \left(1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\right)e_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\left(\widehat{\varphi }(\lambda ,·),e_{\xi }(·)\right)}_{\mathcal{D}}{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)e_{\xi }(y)\hfill \\ \hfill & =\sum _{\xi \in \mathcal{I}}{\left({\displaystyle \frac{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\widehat{\psi }(\lambda ,·),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & -\sum _{\xi \in \mathcal{I}}{\left({\displaystyle \frac{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)\widehat{\varphi }(\lambda ,·),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\left(\widehat{\varphi }(\lambda ,·){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \end{array}$$

and by using the inverse Dunkl transform ${\mathcal{F}}_{\alpha }^{-1}$ to $\widehat{u}$ and $\widehat{f}$, we have a couple of functions $\{u,f\}$ given by

$$\begin{array}{cc}\hfill u(t,x,y)& =\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}{\displaystyle \frac{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\widehat{\psi }(\lambda ,·)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & -\sum _{\xi \in \mathcal{I}}({\int }_{\mathbb{R}}{\displaystyle \frac{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)\widehat{\varphi }(\lambda ,·)\hfill \\ \hfill & \times E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·){)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & +\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })t^{\gamma }\right)\widehat{\varphi }(\lambda ,·)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \end{array}$$

(39)

and

$$\begin{array}{cc}\hfill f(x,y)& =\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}{\displaystyle \frac{{\lambda }^2+{\mu }_{\xi }}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\widehat{\psi }(\lambda ,·)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y)\hfill \\ \hfill & -\sum _{\xi \in \mathcal{I}}{\left({\int }_{\mathbb{R}}{\displaystyle \frac{({\lambda }^2+{\mu }_{\xi }){\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}{1-{\mathbb{E}}_{\gamma ,1}\left(-({\lambda }^2+{\mu }_{\xi })T^{\gamma }\right)}}\widehat{\varphi }(\lambda ,·)E_{\alpha }(ix\lambda )d{\mu }_{\alpha }(\lambda ),e_{\xi }(·)\right)}_{\mathcal{D}}{e}_{\xi }(y),\hfill \end{array}$$

(40)

which is a solution of Problem 2.

By using (35), we obtain

$$\begin{array}{cc}\hfill {\parallel u(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2& =\parallel \widehat{u}{(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2={\int }_{\mathbb{R}}{\parallel \widehat{u}(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )\hfill \\ \hfill & \lesssim {\int }_{\mathbb{R}}\parallel \widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )+{\int }_{\mathbb{R}}{\parallel \widehat{\varphi }(\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )\hfill \\ \hfill & ={\parallel \psi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel \varphi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2.\hfill \end{array}$$

Thus,

$${\parallel u\parallel }_{C([0,T];L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))}^2\lesssim {\parallel \psi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel \varphi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2.$$

Let us estimate the function f, as follows:

$${\parallel f\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2=\parallel \widehat{f}{\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2={\int }_{\mathbb{R}}{\parallel \widehat{f}(\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )$$

$$\begin{array}{cc}\hfill & \lesssim {\int }_{\mathbb{R}}\left({\lambda }^2(\parallel \widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+\parallel \widehat{\varphi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2)+\parallel {\Delta }_{\theta }\widehat{\psi }{(\lambda ,·)\parallel }_{\mathcal{D}}^2+{\parallel {\Delta }_{\theta }\widehat{\varphi }(\lambda ,·)\parallel }_{\mathcal{D}}^2\right)d{\mu }_{\alpha }(\lambda )\hfill \\ \hfill & \lesssim {\parallel \psi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel \varphi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+\parallel {\Delta }_{\theta }{\psi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel {\Delta }_{\theta }\varphi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2\hfill \\ \hfill & ={\parallel \psi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel \varphi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel \psi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2+{\parallel \varphi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2\hfill \end{array}$$

from (37) and (7). It follows that

$$\begin{array}{cc}\hfill {\parallel f\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2& \lesssim {\parallel \psi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel \varphi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2\hfill \\ \hfill & +{\parallel \psi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2+{\parallel \varphi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2.\hfill \end{array}$$

By using (7) and inequality (38), we have

$$\parallel {\mathcal{D}}_{0^{+},t}^{\gamma }{u(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2=\parallel {\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}{(t,·,·)\parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2={\int }_{\mathbb{R}}{\parallel {\mathcal{D}}_{0^{+},t}^{\gamma }\widehat{u}(t,\lambda ,·)\parallel }_{\mathcal{D}}^{2}d{\mu }_{\alpha }(\lambda )$$

$$\lesssim {\parallel \psi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel \varphi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel \psi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2+{\parallel \varphi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2.$$

Finally, we obtain

$$\begin{array}{cc}\hfill \parallel {\mathcal{D}}_{0^{+},t}^{\gamma }{u\parallel }_{C([0,T];L^2(\mathbb{R},{\mu }_{\alpha };\mathcal{D}))}^2& \lesssim {\parallel \psi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2+{\parallel \varphi \parallel }_{W_{\alpha }^{1,2}(\mathbb{R},{\mu }_{\alpha };\mathcal{D})}^2\hfill \\ \hfill & +{\parallel \psi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2+{\parallel \varphi \parallel }_{L^2(\mathbb{R},{\mu }_{\alpha };{\mathcal{D}}^1)}^2.\hfill \end{array}$$

The existence is proved.

Now, let us prove the uniqueness of the solution. Taking into account the property of Dunkl transform (iv), we see that a pair of functions $\{u,f\}$ is uniquely determined by Formulas (39) and (40). The uniqueness is proved. □