## 2 Statement and Solution of the Problem  
Consider the central symmetric Equation (13) in an infinite medium with a spherical hole with radius \(R\)  
\[\frac{\partial^{\alpha}T}{\partial t^{\alpha}}=a\bigg{(}\frac{\partial^{2}T} {\partial r^{2}}+\frac{2}{r}\,\frac{\partial T}{\partial r}\bigg{)}-bT,\quad R <r<\infty,\ \ 0<t<\infty,\ \ 0<\alpha\leq 2, \tag{16}\]  
with zero initial conditions  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]  
\[t=0:\qquad\frac{\partial T}{\partial t}=0,\quad 1<\alpha\leq 2, \tag{18}\]  
and time-harmonic heat flux at the boundary  
\[r=R:\qquad D^{1-\alpha}_{RL}\frac{\partial T}{\partial r}=-g_{0}\,\mathrm{e} ^{i\omega t},\quad 0<\alpha\leq 1, \tag{19}\]  
\[r=R:\qquad I^{\alpha-1}\frac{\partial T}{\partial r}=-g_{0}\,\mathrm{e}^{i \omega t},\quad 1<\alpha\leq 2, \tag{20}\]  
where \(g_{0}\) is a constant and \(\omega\) denotes the angular frequency. The zero condition at infinity is also assumed:  
\[\lim_{r\to\infty}T(r,t)=0. \tag{21}\]  
The auxiliary function \(v=rT\) and the auxiliary spatial variable \(x=r-R\) are used. Then, instead of Equations (16)-(21), we obtain  
\[\frac{\partial^{\alpha}v}{\partial t^{\alpha}}=a\frac{\partial^{2}v}{ \partial x^{2}}-bv,\quad 0<x<\infty,\ \ 0<t<\infty,\ \ 0<\alpha\leq 2, \tag{22}\]  
\[t=0:\qquad vs.=0,\qquad 0<\alpha\leq 2, \tag{23}\]  
\[t=0:\qquad\frac{\partial v}{\partial t}=0,\quad 1<\alpha\leq 2, \tag{24}\]  
\[x=0:\qquad D^{1-\alpha}_{RL}\bigg{(}\frac{\partial v}{\partial x}-\frac{1}{R} \,vs.\bigg{)}=-g_{0}R\,\mathrm{e}^{i\omega t},\quad 0<\alpha\leq 1, \tag{25}\]  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]0  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]1The initial boundary value problem will be solved employing the integral transform tools. Using the rules (9)-(11), applying the Laplace transform to Equations (22)-(27) gives:  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]2  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]3  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]4  
Provided the Robin boundary condition with the given boundary value of a linear combination of a function and its normal derivative, the Fourier sine-cosine transform is used [5,41]:  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]5  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]6  
with the kernel  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]7  
Here, the Fourier sine-cosine transform is denoted by the tilde, with \(\xi\) being the transformation variable. The Fourier sine-cosine transform of the second derivative of a function has the form:  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]8  
Applying the Fourier sine-cosine transform with respect to the auxiliary spatial variable \(x\) to Equation (28) while taking into consideration the boundary condition (29), we get the solution in the transform domain:  
\[t=0:\qquad T=0,\qquad 0<\alpha\leq 2, \tag{17}\]9  
Using the convolution theorem and the following formula [11,12]  
\[t=0:\qquad\frac{\partial T}{\partial t}=0,\quad 1<\alpha\leq 2, \tag{18}\]0  
where \(E_{\alpha}(z)\) is the Mittag-Leffler function  
\[t=0:\qquad\frac{\partial T}{\partial t}=0,\quad 1<\alpha\leq 2, \tag{18}\]1  
after inverting both integral transforms, we obtain  
\[t=0:\qquad\frac{\partial T}{\partial t}=0,\quad 1<\alpha\leq 2, \tag{18}\]2 \[t=0:\qquad\frac{\partial T}{\partial t}=0,\quad 1<\alpha\leq 2, \tag{18}\]3and finally  
\[t=0:\qquad\frac{\partial T}{\partial t}=0,\quad 1<\alpha\leq 2, \tag{18}\]4 \[t=0:\qquad\frac{\partial T}{\partial t}=0,\quad 1<\alpha\leq 2, \tag{18}\]5  
Figures 1-4 present the real part of the solution (39) for different values of the parameters appearing in this equation.