A Study on Fractional Diffusion—Wave Equation with a Reaction

Abstract

An analytical method for solving the fractional diffusion–wave equation with a reaction is investigated. This approach is based on the Laplace transform and fractional series method. An analytical derivation for the proposed method is presented. Examples are given to illustrate the efficiency of the method. The obtained solutions are very close to the exact solutions. Based on this study, we think that the obtained method is promising, and we hope that it can be implemented to other physical problems.

1. Introduction

Recently, fractional calculus has been gaining considerable attention. Fractional differential equations (FDEs) have been adapted as mathematical models [1,2]. Several applications for FDEs can be found in models of viscoelastic behavior [3,4,5], anomalous diffusion [6], compartment models [4], economics [7], epidemiology [8], dynamics of particles [9], biology [10,11], and signal and image processing [12]. From a modeling perspective, algebraic expressions involving functions of system parameters are considered to be the most sought-after way of representing the solution. By generalizing the approaches of solving differential equations using the fractional derivative, substantial results have been proven on FDEs.

There are a number of methods to choose for solving these problems, such as the Laplace transform method [13], Homotopy Analysis method [14], and collocation-shooting method [15]. Researchers have proposed several new integral transform methods to find an analytical solution of FDEs, such as Sumudu [16], Elzaki [17], Laplace [18], Mahgoub [19], and Natural [20]. In order to study fractional differential (FD) systems, the Adomian decomposition method was used with different transform methods such as Sumudu [10,21,22], Elzaki [11,23], Laplace [12,24,25], Mahgoub [13,26], and Natural [14,27]. In [28], Alaroud solved a system of fractional ordinary differential equations. He took the Laplace transform for the residual first, and then he solved the system. Then, he took the inverse Laplace transform to find the solution of the original problem. In [29], they used the Laplace residual method to solve the fractional delay ordinary differential equation. In this paper, we study a class of fractional partial differential equations. We take the Laplace transform with respect to the variable x; then, we use the series method to find the solution of the new problem. After that, we take the inverse Laplace transform. For more references, see [30,31,32,33,34,35,36,37]. Let us consider the following problem:

$${D^{2\alpha }}_t\mu ={\mu }_{xx}+{\mu }_{yy}+g(x,y,t),0<x,y<\infty ,t>0.$$

(1)

If $\alpha =\frac{1}{2}$, Equation (1) is called a diffusion equation, and when $\alpha =1$, Equation (1) is called a wave equation. In this paper, we are interested in studying the class of problem (1) when $\frac{1}{2}<\alpha <1$. This type of problem is called the fractional diffusion–wave equation (FDWE). Thus, FDWE is defined in the sense of generalized diffusion, which has fractional derivative. The diffusion process is asymmetric or stochastic, but it is determined on a molecule scale, so a fractional modification is needed. The derivation of the FDWE starts by considering the Cauchy problem in order to force the solution to have the main properties of the probability density in the variables $x,y$, evolving in time t. Then, we take the Laplace transform in terms of the variable t and the Fourier transform for the variables $x,y$ to generate the FDWE. Complete details of the derivation of FDWE can be found in [38,39]. Let us start with some definitions and some notations. Since our interest is in studying problems with variables in $\mathbb{R}$, we have limited our definitions to those variables for $\eta \in (0,1]$. We recall first a few definitions of fractional calculus for $\eta \in (0,1]$.

Definition 1 

([1]). The Caputo fractional derivative of h is defined as

$$D^{\eta }h\left(y\right)=\frac{1}{\mathsf{\Gamma }(1-\eta )}{\int }_0^y{(y-s)}^{-\eta }{h}^{\prime }\left(s\right)ds,$$

where $y>0$.

Some of the properties of a Caputo derivative are

• $D^{\eta }c=0$, where c is any constant.
• $D^{\eta }{y}^{\gamma }=\frac{\mathsf{\Gamma }(\gamma +1)}{\mathsf{\Gamma }(\gamma -\eta +1)}{y}^{\gamma -\eta }$, where $\gamma \ge \eta$.
• The Laplace transform of Caputo fractional derivative for $k-1<\alpha <k$ is

  $$\mathcal{L}\left\{D^{\alpha }\mu \left(t\right)\right\}=\frac{s^k\mathcal{L}\left\{\mu \left(s\right)\right\}-s^{k-1}\mu \left(0\right)-s^{k-2}{\mu }^{(k-1)}\left(0\right)}{s^{k-\alpha }}.$$

Now, we describe the series solution method for solving fractional differential equations. First, we need to introduce some definitions and theorems.

Definition 2 

([40]). The fractional power series about $z=z_0$ is given by

$$\sum _{i=0}^{\infty }{a}_i{(z-z_0)}^{i\alpha },\alpha \in \mathbb{R}-\mathbb{Z}$$

where $0\le j-1<\alpha <j,z\le z_0$.

For simplicity, we call this a fractional series, and it is denoted by FS. In case the function $\mu$ has FS at $z=z_0$, then we can rewrite $\mu$ as follows.

Theorem 1.

Let $\mu \left(z\right)$ have an FS about $z=z_0$ with

$$\mu \left(z\right)=\sum _{i=0}^{\infty }{a}_i{(z-z_0)}^{i\alpha },z_0\le z<z_0+\delta ,$$

(2)

where δ is the radius of convergence. If $D^{i\alpha }\mu \left(z\right)\in C(z_0,z_0+\delta )$ for $i=0,1,2,\dots ,$ then

$$a_i=\frac{D^{i\alpha }\mu \left(z_0\right)}{\mathsf{\Gamma }(1+i\alpha )}.$$

(3)

Proof. 

Assume that $\mu$ has an FS about $z=z_0$. At $z=z_0$, Equation (2) implies that $a_0=\mu \left(z_0\right)$, which satisfies Equation (3). For any $i\in \mathbb{N}$, Equation (2) implies that

$$D^{i\alpha }\mu \left(z\right)=\sum _{k=i}^{\infty }\frac{\mathsf{\Gamma }(k\alpha +1)}{\mathsf{\Gamma }(1+(k-i\left)\alpha \right)}{a}_k{(z-z_0)}^{(k-i)\alpha }.$$

(4)

When $z=z_0$, Equation (4) gives

$$D^{i\alpha }\mu \left(z_0\right)=\frac{\mathsf{\Gamma }(i\alpha +1)}{\mathsf{\Gamma }\left(1\right)}{a}_i,$$

(5)

or

$$a_i=\frac{D^{i\alpha }\mu \left(z_0\right)}{\mathsf{\Gamma }(i\alpha +1)}.$$

(6)

□

Let us consider the following fractional differential equation

$$D^{\alpha }\mu \left(z\right)=s(\mu ,z).$$

Let us assume that the solution is given in the FS form by

$$\mu \left(z\right)=\sum _{i=0}^{\infty }{a}_i{(z-z_0)}^{i\alpha }.$$

If the function s is linear with respect to $\mu$, then we can substitute the FS of $\mu$ in the fractional differential equation and compare the coefficients to find $a_i$. Otherwise, one can use the residual to find $a_i$ as follows:

$$D^{(i-j)\alpha }Re{s}_i\left(z_0\right)=0,$$

for $i=j+1,j+2,\dots ,0\le j-1<\alpha <j$, and

$$Re{s}_i\left(z\right)=D^{\alpha }\left(\sum _{i=0}^{\infty }{a}_i{(z-z_0)}^{i\alpha }\right)-s(\sum _{i=0}^{\infty }{a}_i{(z-z_0)}^{i\alpha },z).$$

For more details, see [40,41]. In some cases, we can find the closed form of the solution, while in others, we can approximate the solution by the first few tears. In the fractional partial differential equations, as in Equation (7), there are two ways to deal with this type of problems. The first approach is to use the FS approach, and in this case, the coefficients will be functions of $x,y$. Then, we solve the system of partial differential equations. Another approach is to use the Laplace or Fourier transforms first; then, we implement the FS approach. Homotopy perturbation methods are used commonly with the Laplace transform as in [42]. In this paper, we combine the FS with the Laplace transform method. Since the variables $x,y\in (0,\infty )$, we implement the Laplace transform for the variables of x and y, which yields the fractional ordinary differential equation. Then, we solve it using FS. As a special case, in [22], the authors used the fractional power series to solve

$$z^{\alpha }{D}_z^{\alpha }\mu +q\mu =0.$$

The Cauchy product of two FSs is given in the following definition. This definition helps us to avoid using the residue approach.

Definition 3.

The product of two FSs has the form

$$\left(\sum _{i=o}^{\infty }{a}_i{z}^{i\alpha }\right)\left(\sum _{j=o}^{\infty }{b}_j{z}^{j\alpha }\right)=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{a}_i{b}_j{z}^{(i+j)\alpha },\alpha \in \mathbb{R}-\mathbb{Z}.$$

In recent work in [25], the authors introduced the generalized Cauchy product fractional power series to solve a class of FDEs with constant coefficients. In case we do not have the product of fractional series, we use the generalized fractional powers series method to overcome this difficulty. This type of difficulty appears in several fractional order differential equations.

2. Method of Solution

Here, the Laplace transform is applied to solve a class of fractional-time diffusion—wave equations with a reaction analytically. We used Mathematica to do all calculations. In case a closed solution is not possible to calculate, we take the first few terms to approximate the solution.

Consider the following problem with respect to the variables x and y:

$$\begin{array}{c}{D}_t^{2\alpha }\mu (x,y,t)=a_1{\mu }_{xx}(x,y,t)+a_2{\mu }_{yy}(x,y,t)-a_3{\mu }_x(x,y,t)\hfill \\ -a_4{\mu }_y(x,y,t)-a_5\mu (x,y,t)+\theta (x,y,t),\hfill \end{array}$$

(7)

such that

$$\begin{array}{ccc}& & \mu (x,y,0)=g_1(x,y),{\mu }_t^{\alpha }(x,y,0)=g_2(x,y),0<x<\infty ,0<y<\infty ,\hfill \end{array}$$

(8)

$$\begin{array}{ccc}& & \mu (0,y,t)=h_1(y,t),{\mu }_x(0,y,t)=h_2(y,t),0<y<\infty ,0<t<T,\hfill \end{array}$$

(9)

$$\begin{array}{ccc}& & \mu (x,0,t)=r_1(x,t),{\mu }_y(x,0,t)=r_2(x,t),0<x<\infty ,0<t<T,\hfill \end{array}$$

(10)

where $\frac{1}{2}<\alpha \le 1,a_1,a_2>0,a_3,a_4,a_5\ge 0$ are constants, $\theta (x,y,t),g_1(x,y),g_2(x,y)$, $h_1(y,t),h_2(y,t),r_1(x,t),r_2(x,t)$ are continuous functions, and $D^{2\alpha }$ is the Caputo derivative.

Since the problem is symmetric with respect to the variables x and y, we can take the Laplace transform with respect to either x or y. Without loss of generality, take the Laplace transformation with respect to x to Equation (7) to get

$$\begin{array}{ccc}& & D_t^{2\alpha }{\mathsf{{\rm Y}}}_1(s,y,t)=a_1(s^2{\mathsf{{\rm Y}}}_1(s,y,t)-s{h}_2(y,t)-h_1(y,t))+\hfill \\ & & a_2\frac{{\partial }^2{\mathsf{{\rm Y}}}_1(s,y,t)}{\partial y^2}-a_3(s{\mathsf{{\rm Y}}}_1(s,y,t)-h_1(y,t))-a_4\frac{\partial {\mathsf{{\rm Y}}}_1(s,y,t)}{\partial y}\hfill \\ & & -a_5{\mathsf{{\rm Y}}}_1(s,y,t)+{\mathsf{\Phi }}_1(s,y,t),\hfill \end{array}$$

(11)

where $Y_1(s,y,t)={\mathcal{L}}_x\left({\mu }_1(x,y,t)\right)$ and ${\mathsf{\Phi }}_1(s,y,t)={\mathcal{L}}_x\left(\theta (x,y,t)\right)$. Take the Laplace transformations of Equation (11) with respect to y to get

$$\begin{array}{ccc}& & D_t^{2\alpha }\mathsf{{\rm Y}}(s,z,t)=a_1(s^2\mathsf{{\rm Y}}(s,z,t)-s{H}_2(z,t)-H_1(z,t))+a_2[z^2\mathsf{{\rm Y}}(s,z,t)-z{R}_2(s,t)\hfill \\ & & -R_1(s,t)]-a_3[s\mathsf{{\rm Y}}(s,z,t)-H_1(z,t)]-a_4[z\mathsf{{\rm Y}}(s,z,t)-R_1(z,t)]\hfill \\ & & -a_5\mathsf{{\rm Y}}(s,z,t)+\mathsf{\Phi }(s,z,t)\hfill \end{array}$$

(12)

where $Y(s,z,t)={\mathcal{L}}_x\left(Y_1(s,y,t)\right)$, $H_1(z,t)={\mathcal{L}}_y\left(h_1(y,t)\right)$, $H_2(z,t)={\mathcal{L}}_y\left(h_2(y,t)\right)$, $R_1(s,t)={\mathcal{L}}_x\left(r_1(x,t)\right)$, and $R_2(s,t)={\mathcal{L}}_x\left(r_2(x,t)\right)$. We can rewrite the previous equations as

$$\begin{array}{c}\hfill D_t^{2\alpha }\mathsf{{\rm Y}}(s,z,t)=Q(s,z)\mathsf{{\rm Y}}(s,z,t)+\mathsf{\Omega }(s,z,t)\end{array}$$

where

$$\begin{array}{ccc}\hfill & & Q(s,z)=a_1{s}^2+a_2{z}^2-a_{3}s-a_{4}z-a_5,\hfill \end{array}$$

(13)

$$\begin{array}{c}\mathsf{\Omega }(s,z,t)=-a_{1}s{H}_2(z,t)-a_1{H}_1(z,t)-a_{2}z{R}_2(s,t)-a_2{R}_1(s,t)\hfill \\ +a_3{H}_1(z,t)+a_4{R}_1(z,t)+\mathsf{\Phi }(s,z,t).\hfill \end{array}$$

(14)

Let the approximate solution of problem (12) be given by

$$\begin{array}{c}\hfill \mathsf{{\rm Y}}(s,z,t)=\sum _{n=0}^{\infty }{\beta }_n(s,z)t^{\alpha n}.\end{array}$$

(15)

Then, the fractional derivative of the solution in (15) is given as

$$\begin{array}{ccc}& & D_t^{2\alpha }\mathsf{{\rm Y}}(s,z,t)=\sum _{n=2}^{\infty }\frac{\mathsf{\Gamma }(1+n\alpha )}{\mathsf{\Gamma }(1+n\alpha -2\alpha )}{t}^{n\alpha -2\alpha }{\beta }_n(s,z)\hfill \\ & & =\sum _{n=2}^{\infty }\frac{\mathsf{\Gamma }(1+n\alpha )}{\mathsf{\Gamma }(1+\alpha (n-2\left)\right)}{t}^{\alpha (n-2)}{\beta }_n(s,z)\hfill \\ & & =\sum _{k=0}^{\infty }\frac{\mathsf{\Gamma }(1+\alpha (k+2\left)\right)}{\mathsf{\Gamma }(1+k\alpha )}{t}^{k\alpha }{\beta }_{k+2}(s,z).\hfill \end{array}$$

(16)

Let the series expansion of function in Equation (14) be given by

$$\mathsf{\Omega }(s,z,t)=\sum _{k=0}^{\infty }{\gamma }_k(s,z)t^{\alpha k}.$$

Then, the right hand side of Equation (12) can be written as

$$\begin{array}{ccc}& & Q(s,z)\mathsf{{\rm Y}}(s,z,t)+\mathsf{\Omega }(s,z,t)=\sum _{k=0}^{\infty }Q(s,z){\beta }_k(s,z)t^{\alpha k}+\sum _{k=0}^{\infty }{\gamma }_k(s,z)t^{\alpha k}.\hfill \end{array}$$

Therefore, Equation (12) can be written in the series format as

$$\begin{array}{c}\hfill \sum _{k=0}^{\infty }\frac{\mathsf{\Gamma }(1+\alpha (k+2\left)\right)}{\mathsf{\Gamma }(1+k\alpha )}{t}^{k\alpha }{\beta }_{k+2}(s,z)=\sum _{k=0}^{\infty }Q(s,z){\beta }_k(s,z)t^{\alpha k}+\sum _{k=0}^{\infty }{\gamma }_k(s,z)t^{\alpha k}\end{array}$$

(17)

which implies that

$$\begin{array}{c}\hfill \frac{\mathsf{\Gamma }(1+\alpha (k+2\left)\right)}{\mathsf{\Gamma }(1+k\alpha )}{\beta }_{k+2}(s,z)=Q(s,z){\beta }_k(s,z)+{\gamma }_k(s,z)\end{array}$$

(18)

or

$$\begin{array}{ccc}& & {\beta }_{k+2}(s,z)=\frac{\mathsf{\Gamma }(1+k\alpha )}{\mathsf{\Gamma }(1+\alpha (k+2\left)\right)}Q(s,z){\beta }_k(s,z)+\hfill \end{array}$$

(19)

$$\begin{array}{ccc}& & \frac{\mathsf{\Gamma }(1+k\alpha )}{\mathsf{\Gamma }(1+\alpha (k+2\left)\right)}{\gamma }_k(s,z).\hfill \end{array}$$

Take the Laplace transform with respect to x and then with respect to y to get

$$\mathsf{{\rm Y}}(s,z,0)=G_1(s,z),D^{\alpha }\mathsf{{\rm Y}}(s,z,0)=G_2(s,z)$$

(20)

where $G_1(s,z)={\mathcal{L}}_x({\mathcal{L}}_y\left(g_1(x,y)\right)$, and $G_2(s,z)={\mathcal{L}}_x({\mathcal{L}}_y\left(g_2(x,y)\right)$. Equations (12) and (15) imply that

$${\beta }_0(s,z)=G_1(s,z),{\beta }_1(s,z)=\frac{G_2(s,z)}{\mathsf{\Gamma }(1+\alpha )}.$$

Using Equation (14), one can generate $\{{\beta }_k(s,z):k=2,3,\dots \}$. Then, take the inverse Laplace transform with respect to s and then with respect to z of Equation (12) to find the analytical solution $u(x,y,t)$ for Problems (7)–(10).

3. Numerical Results

We demonstrate the feasibility of the method by presenting two symmetric examples.

Example 1.

Consider the following problem:

$$\begin{array}{ccc}& & D_t^{2\alpha }u(x,y,t)=u_{xx}(x,y,t)+u_{yy}(x,y,t)+\left(\frac{6t^{3-2\alpha }}{\mathsf{\Gamma }(4-2\alpha )}-t^3\right)e^x,\hfill \\ & & \frac{1}{2}<\alpha \le 1,x,y>0,0<t<T,\hfill \end{array}$$

(21)

such that

$$\begin{array}{ccc}& & u(x,y,0)=y+1,D_t^{\alpha }u(x,y,0)=0,x,y\ge 0,\hfill \end{array}$$

(22)

$$\begin{array}{ccc}& & u(0,y,t)=t^3+y+1,u_x(0,y,t)=t^3+y+1,y\ge 0,0\le t\le T,\hfill \\ & & u(x,0,t)=t^3{e}^x+1,u_y(x,0,t)=1,x\ge 0,0\le t\le T.\hfill \end{array}$$

(23)

Let $\alpha =\frac{3}{4}$. Then, the exact solution is

$$u(x,y,t)=t^3{e}^x+y+1.$$

Let $\mathsf{{\rm Y}}(s,z,t)={\mathcal{L}}_y\left\{{\mathcal{L}}_x\left\{u(x,y,t)\right\}\right\}$. Then, following the method described in the previous section and by taking the Laplace transform with respect to x first and then with respect to y, we get

$$D_t^{\frac{3}{2}}\mathsf{{\rm Y}}(s,z,t)=Q(s,z)\mathsf{{\rm Y}}(s,z,t)+\mathsf{\Omega }(s,z,t)$$

where

$$\begin{array}{ccc}& & Q(s,z)=s^2+z^2,\hfill \\ & & \mathsf{\Omega }(s,z,t)=\frac{8}{\sqrt{\pi }z(s-1)}-\frac{(z+1)(z^2+s^2+s)}{s{z}^2}{t}^{\frac{3}{2}}+\frac{5+s^2+z^2}{z(1-s)}{t}^3.\hfill \end{array}$$

(24)

Then, by comparing the coefficients of $t^{\alpha t}$ for $k=0,1,2,\dots$, the values of ${\gamma }_i$ are

$$\begin{array}{ccc}& & {\gamma }_0(s,z)=\frac{8}{\sqrt{\pi }z(s-1)},{\gamma }_2(s,z)=-\frac{(z+1)(z^2+s^2+s)}{s{z}^2},\hfill \\ & & {\gamma }_4(s,z)=\frac{5+s^2+z^2}{z(1-s)},{\gamma }_k(s,z)=0,k\in \{1,3,5,6,\dots \}.\hfill \end{array}$$

(25)

Using Mathematica, we take the inverse Laplace transform with respect to s first and then with respect to z to get

$${\beta }_0(s,z)={\mathcal{L}}_y^{-1}\left\{{\mathcal{L}}_x^{-1}\left\{u(x,y,0)\right\}\right\}=\frac{z+1}{s{z}^2},{\beta }_1(s,z)={\mathcal{L}}_y^{-1}\{{\mathcal{L}}_x^{-1}\left\{D_t^{\alpha }u(x,y,0)\right\}\}=0.$$

Using Equations (12) and (14), we get

$$u(x,y,t)={\mathcal{L}}_y^{-1}\left\{{\mathcal{L}}_x^{-1}\left\{\sum _{k=0}^{\infty }{\beta }_k(s,z)t^{\frac{3k}{4}}\right\}\right\}$$

where

$$\begin{array}{ccc}& & {\mathcal{L}}_y^{-1}\left\{{\mathcal{L}}_x^{-1}\left\{{\beta }_0(s,z)\right\}\right\}=1+y,\hfill \end{array}$$

(26)

$$\begin{array}{ccc}& & {\mathcal{L}}_y^{-1}\left\{{\mathcal{L}}_x^{-1}\left\{{\beta }_4(s,z)\right\}\right\}=e^x,\hfill \\ & & {\mathcal{L}}_y^{-1}\left\{{\mathcal{L}}_x^{-1}\left\{{\beta }_k(s,z)\right\}\right\}=0,k\in \{1,2,3,5,6,\dots \}.\hfill \end{array}$$

(27)

Thus, the analytical solution is

$$u_n(x,y,t)=u(x,y,t)=1+y+e^x{t}^3,n\ge 4$$

which is the same as the exact solution.

Example 2.

Consider the following problem:

$$D_t^{2\alpha }u(x,y,t)=u_{xx}(x,y,t)-2u_x(x,y,t)+u_{yy}(x,y,t)-u_y(x,y,t)+u(x,y,t)+f(x,y,t),$$

such that

$$\begin{array}{ccc}& & u(x,y,0)=0,D_t^{\alpha }u(x,y,0)=0,x,y\ge 0,\hfill \end{array}$$

(28)

$$\begin{array}{ccc}& & u(0,y,t)=t^{7/2}(y+1),u_x(0,y,t)=t^{7/2}(y+1),y\ge 0,0\le t\le T,\hfill \\ & & u(x,0,t)=t^{7/2}(x^2+1),u_y(x,0,t)=t^{7/2},x\ge 0,0\le t\le T,\hfill \end{array}$$

(29)

where $\frac{1}{2}<\alpha \le 1$ and

$$f(x,y,t)=\left(\frac{\mathsf{\Gamma }\left(\frac{9}{2}\right)(x^2+y+1)}{\mathsf{\Gamma }(\frac{9}{2}-2\alpha )t^{2\alpha }}-4+4x-x^2-y\right)t^{\frac{7}{2}}.$$

Let $\alpha =\frac{7}{8}$. Then, the exact solution is

$$u(x,y,t)=t^{7/2}(x^2+y+1).$$

Let $\mathsf{{\rm Y}}(s,z,t)={\mathcal{L}}_y\left\{{\mathcal{L}}_x\left\{u(x,y,t)\right\}\right\}$. Then, following the method described in the previous section and by taking the Laplace transform with respect to x first and then with respect to y, we get

$$D_t^{\frac{3}{2}}\mathsf{{\rm Y}}(s,z,t)=Q(s,z)\mathsf{{\rm Y}}(s,z,t)+\mathsf{\Omega }(s,z,t)$$

where

$$\begin{array}{ccc}& & Q(s,z)=s^2-2s+z^2+1,\hfill \end{array}$$

(30)

$$\begin{array}{ccc}& & \mathsf{\Omega }(s,z,t)=\frac{105\sqrt{\pi }(2z+s^2+s{z}^2)}{16\mathsf{\Gamma }\left(\frac{11}{4}{s}^3{z}^2\right)}{t}^{\frac{7}{4}}\hfill \\ & & +\frac{4sz+(s^3-s^4)(1+z)-2(z+z^3)-s^2(1+4z+z^2+z^3)}{s^3{z}^2}{t}^{\frac{7}{2}}.\hfill \end{array}$$

(31)

Then, by comparing the coefficients of $t^{\alpha t}$ for $k=0,1,2,\dots$, the values of ${\gamma }_i$ are

$$\begin{array}{ccc}& & {\gamma }_2(s,z)=\frac{105\sqrt{\pi }(2z+s^2+s{z}^2)}{16\mathsf{\Gamma }\left(\frac{11}{4}{s}^3{z}^2\right)},\hfill \end{array}$$

(32)

$$\begin{array}{ccc}& & {\gamma }_4(s,z)=\frac{4sz+(s^3-s^4)(1+z)-2(z+z^3)-s^2(1+4z+z^2+z^3)}{s^3{z}^2},\hfill \\ & & {\gamma }_k(s,z)=0,k\in \{0,1,3,5,\dots \}.\hfill \end{array}$$

(33)

Using Mathematica, we take the inverse Laplace transform with respect to s first and then with respect to z to get

$${\beta }_0(s,z)={\mathcal{L}}_y^{-1}\{{\mathcal{L}}_x^{-1}\left\{u\right(x,y,0\left)\right\}\}=0,{\beta }_1(s,z)={\mathcal{L}}_y^{-1}\left\{{\mathcal{L}}_x^{-1}\left\{D_t^{\alpha }u(x,y,t)\right\}\right\}=0.$$

Using Equations (12) and (14), we get

$$u(x,y,t)={\mathcal{L}}_y^{-1}\left\{{\mathcal{L}}_x^{-1}\left\{\sum _{k=0}^{\infty }{\beta }_k(s,z)t^{\frac{7k}{8}}\right\}\right\}$$

where

$$\begin{array}{ccc}& & {\mathcal{L}}_y^{-1}\left\{{\mathcal{L}}_x^{-1}\left\{{\beta }_4(s,z)\right\}\right\}=1+y+x^2,\hfill \\ & & {\mathcal{L}}_y^{-1}\left\{{\mathcal{L}}_x^{-1}\left\{{\beta }_k(s,z)\right\}\right\}=0,k\in \{0,1,2,\dots \}.\hfill \end{array}$$

(34)

Thus, the analytical solution is

$$u_n(x,y,t)=u(x,y,t)=(1+y+x^2)t^{\frac{7}{2}},n\ge 4$$

which is the same as the exact solution.

4. Conclusions

In this paper, we present an analytical approach for solving a fractional diffusion—wave equation with a reaction. This approach is based on the Laplace transform and the fractional series approach. We test the proposed approach by considering two numerical examples. We can summarize the novelty of this paper and future work in the following points:

• The main purpose of our study is to construct the Laplace transform series method for solving fractional diffusion equation with respect to the Caputo derivatives.
• We solve two examples to show the efficiency of the proposed method.
• We notice that the proposed method gave the exact solutions in the first and second example after a few steps, which gives us numerical evidence that the proposed method is efficient.
• Based on this study, we think that the suggested approach is promising and suitable for other physical problems such as fractional kinetics and anomalous transport.