## 3 Solution to the Problem  
In this section, the analytical solution of the fractional differential Equation (8) will be determined along with the initial and boundary conditions  
\[H(r,0)=0,\ r\in[R,\infty), \tag{11}\]  
\[\left.\frac{\partial H(r,t)}{\partial r}\right|_{r=R}=-\frac{Q}{2\pi RT_{f}} \tag{12}\]  
\[\sqrt{r}\frac{\partial H(r,t)}{\partial r}\to 0,\ \ \sqrt{r}\frac{\partial^{2}H(r,t)}{ \partial r^{2}}\to 0\ \ \ as\ \ r\rightarrow\infty. \tag{13}\]  
Applying the Laplace transform to Equations (8), (12) and (13), and by using the initial condition (11), we get the transformed problem  
\[\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial\overline{H}(r,s)}{ \partial r})=\frac{1}{D_{f}}[\frac{as^{\alpha}}{(1-\alpha)s^{\alpha}+\alpha} +\frac{bs^{\beta}}{(1-\beta)s^{\beta}+\beta}]\overline{H}(r,s),\ \alpha,\beta\in(0,1], \tag{14}\]  
\[\left.\frac{\partial\overline{H}(r,s)}{\partial r}\right|_{r=R}=-\frac{Q}{2 \pi RT_{f}}\frac{1}{s} \tag{15}\]  
\[\sqrt{r}\frac{\partial\overline{H}(r,s)}{\partial r}\to 0,\ \sqrt{r}\frac{ \partial^{2}\overline{H}(r,s)}{\partial r^{2}}\to 0,\ \as\ \ \ r\rightarrow\infty \tag{16}\]  
In order to find the solution to Equations (14)-(16) we use the Dirichlet-Weber transform defined as [42]  
\[\hat{\overline{H}}(\xi,s)=\int_{R}^{\infty}r\overline{H}(r,s)[J_{v}(r\xi)Y_{v }(R\xi)-J_{v}(R\xi)Y_{v}(r\xi)]dr, \tag{17}\]  
where \(J_{v}(\cdot)\), \(Y_{v}(\cdot)\) are Bessel functions of the first kind with order \(v\). The inverse formula of the integral transform (17) is [42]  
\[\overline{H}(r,s)=\int_{0}^{\infty}\frac{J_{v}(r\xi)Y_{v}(R\xi)-J_{v}(R\xi)Y_ {v}(r\xi)}{J_{v}^{2}(R\xi)+Y_{v}^{2}(R\xi)}\xi\cdot\hat{\overline{H}}(\xi,s)d\xi. \tag{18}\]  
Using the notation,  
\[\overline{\varphi_{0}}(s)=\frac{1}{D_{f}}[\frac{as^{\alpha}}{(1-\alpha)s^{ \alpha}+\alpha}+\frac{bs^{\beta}}{(1-\beta)s^{\beta}+\beta}] \tag{19}\]into Equation (14), we obtain the transformed equation,  
\[\frac{\partial^{2}\overline{H}(r,s)}{\partial r^{2}}+\frac{1}{r}\frac{\partial \overline{H}(r,s)}{\partial r}=\overline{\varphi}_{0}(s)\overline{H}(r,s) \tag{20}\]  
which can be written in the equivalent form  
\[\left.\frac{\partial H(r,t)}{\partial r}\right|_{r=R}=-\frac{Q}{2\pi RT_{f}} \tag{12}\]0  
Making the change of unknown function \(\overline{H}(r,s)\) into \(\overline{u}(r,s)=\frac{\partial\overline{H}(r,s)}{\partial r}\), Equation (21) becomes  
\[\left.\frac{\partial H(r,t)}{\partial r}\right|_{r=R}=-\frac{Q}{2\pi RT_{f}} \tag{12}\]1  
The function \(\overline{u}(r,s)\) has to satisfy the boundary condition  
\[\left.\frac{\partial H(r,t)}{\partial r}\right|_{r=R}=-\frac{Q}{2\pi RT_{f}} \tag{12}\]2  
\[\left.\frac{\partial H(r,t)}{\partial r}\right|_{r=R}=-\frac{Q}{2\pi RT_{f}} \tag{12}\]3  
Using the asymptotic expansions  
\[\left.\frac{\partial H(r,t)}{\partial r}\right|_{r=R}=-\frac{Q}{2\pi RT_{f}} \tag{12}\]4  
and the identity  
\[\left.\frac{\partial H(r,t)}{\partial r}\right|_{r=R}=-\frac{Q}{2\pi RT_{f}} \tag{12}\]5  
it easily turns out that  
\[\left.\frac{\partial H(r,t)}{\partial r}\right|_{r=R}=-\frac{Q}{2\pi RT_{f}} \tag{12}\]6  
Now, applying the Dirichlet-Weber transform to Equation (22), and using Equation (25) and the boundary condition (23), we obtain the transformed equation  
\[\left.\frac{\partial H(r,t)}{\partial r}\right|_{r=R}=-\frac{Q}{2\pi RT_{f}} \tag{12}\]7  
where \(k_{0}=\frac{Q}{\pi^{2}RT_{f}}\).