Fractional Heat Conduction with Heat Absorption in a Solid with a Spherical Cavity under Time-Harmonic Heat Flux

Abstract

The central-symmetric time-fractional heat conduction equation with heat absorption is investigated in a solid with a spherical hole under time-harmonic heat flux at the boundary. The problem is solved using the auxiliary function, for which the Robin-type boundary condition with a prescribed value of a linear combination of a function and its normal derivative is fulfilled. The Laplace and Fourier sine–cosine integral transformations are employed. Graphical representations of numerical simulation results are given for typical values of the parameters.

1. Introduction

Fractional calculus has many advantages in various areas of investigation [1,2,3,4,5,6]. Abstract integro-differential equations have been discussed in fundamental monographs [7,8].

The classical theory of heat conduction is based on the assumption that the heat flux vector $\mathbf{q}(\mathbf{x},t)$ at a point $\mathbf{x}$ at time t is proportional to the temperature gradient $\mathrm{grad}T(\mathbf{x},t)$ at the same point $\mathbf{x}$ and at the same time t. This is the content of the standard (local) Fourier law

$$\mathbf{q}(\mathbf{x},t)=-k\mathrm{grad}T(\mathbf{x},t),$$

(1)

where k is the thermal conductivity coefficient.

The time-nonlocal generalization of the Fourier law describing “long-tail” memory is formulated as [9,10]

$$\mathbf{q}(\mathbf{x},t)=-\frac{k}{\mathrm{\Gamma }\left(\alpha \right)}\frac{\partial }{\partial t}{\int }_0^t{\left(t-\tau \right)}^{\alpha -1}\mathrm{grad}T(\mathbf{x},\tau )\mathrm{d}\tau ,0<\alpha \le 1,$$

(2)

$$\mathbf{q}(\mathbf{x},t)=-\frac{k}{\mathrm{\Gamma }(\alpha -1)}{\int }_0^t{\left(t-\tau \right)}^{\alpha -2}\mathrm{grad}T(\mathbf{x},\tau )\mathrm{d}\tau ,1<\alpha \le 2,$$

(3)

where $\mathrm{\Gamma }\left(\alpha \right)$ is the gamma function and can be written in terms of the Riemann–Liouville time-fractional integrals and derivatives:

$$\mathbf{q}(\mathbf{x},t)=-k{D}_{RL}^{1-\alpha }\mathrm{grad}T(\mathbf{x},t),0<\alpha \le 1,$$

(4)

$$\mathbf{q}(\mathbf{x},t)=-k{I}^{\alpha -1}\mathrm{grad}T(\mathbf{x},t),1<\alpha \le 2.$$

(5)

Here,

$$I^{\alpha }f\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^t{\left(t-\tau \right)}^{\alpha -1}f\left(\tau \right)\mathrm{d}\tau ,\alpha >0,$$

(6)

is the Riemann–Liouville fractional integral, and

$$D_{RL}^{\alpha }f\left(t\right)=\frac{1}{\mathrm{\Gamma }(m-\alpha )}\frac{{\mathrm{d}}^m}{\mathrm{d}{t}^m}\left[{\int }_0^t{\left(t-\tau \right)}^{m-\alpha -1}f\left(\tau \right)\mathrm{d}\tau \right],m-1<\alpha <m,$$

(7)

is the Riemann–Liouville fractional derivative of order $\alpha$ [11,12]. The Caputo fractional derivative is defined as

$$D_C^{\alpha }f\left(t\right)=\frac{1}{\mathrm{\Gamma }(m-\alpha )}{\int }_0^t{\left(t-\tau \right)}^{m-\alpha -1}\frac{{\mathrm{d}}^{m}f\left(\tau \right)}{\mathrm{d}{\tau }^m}\mathrm{d}\tau ,m-1<\alpha <m.$$

(8)

Recall the equations of the Laplace transform for fractional integrals and derivatives [11,12]:

$$\mathcal{L}\left\{I^{\alpha }f\left(t\right)\right\}=\frac{1}{s^{\alpha }}{f}^{*}\left(s\right),$$

(9)

$$\mathcal{L}\left\{D_{RL}^{\alpha }f\left(t\right)\right\}=s^{\alpha }{f}^{*}\left(s\right)-\sum _{k=0}^{m-1}{D}^k{I}^{m-\alpha }f\left(0^{+}\right)s^{m-1-k},m-1<\alpha <m,$$

(10)

$$\mathcal{L}\left\{D_C^{\alpha }f\left(t\right)\right\}=s^{\alpha }{f}^{*}\left(s\right)-\sum _{k=0}^{m-1}{f}^{\left(k\right)}\left(0^{+}\right)s^{\alpha -1-k},m-1<\alpha <m,$$

(11)

where the asterisk marks the transform and s denotes the Laplace transform variable.

The constitutive equations for the heat flux (4) and (5) in combination with the law of conservation of energy lead to the time-fractional heat conduction equation with the Caputo derivative

$$\frac{{\partial }^{\alpha }T}{\partial t^{\alpha }}=a\mathrm{\Delta }T,0<\alpha \le 2,$$

(12)

where a can be taken as a counterpart of the thermal diffusivity coefficient, and the Caputo time-derivative $D_C^{\alpha }f\left(t\right)$ is denoted as

$$D_C^{\alpha }f(\mathbf{x},t)\equiv \frac{{\partial }^{\alpha }f(\mathbf{x},t)}{\partial t^{\alpha }}.$$

The generalization of Equation (12)

$$\frac{{\partial }^{\alpha }T}{\partial t^{\alpha }}=a\mathrm{\Delta }T-bT,0<\alpha \le 2,$$

(13)

has been considered in several publications [13,14,15,16,17,18]. Two special versions of Equation (13) with integer values of the time derivative are well known. The parabolic equation

$$\frac{\partial T}{\partial t}=a\mathrm{\Delta }T-bT$$

(14)

describes, for example, bioheat transfer and lateral heat or mass exchange in a thin plate [19,20,21,22,23], whereas the hyperbolic Klein–Gordon equation

$$\frac{{\partial }^{2}T}{\partial t^2}=a\mathrm{\Delta }T-bT$$

(15)

appears in quantum field theory, nonlinear optics, and solid state physics [24,25].

The solutions of the parabolic Equation (14) and time-fractional Equation (13) have different medical applications: hypothermia, ablation, thermotherapy (see, for example, [26,27,28,29,30] and the references therein).

Heat conduction in a medium with a spherical cavity was investigated in the literature under different generalized theories [31,32,33,34,35,36]. In this paper, we study the central-symmetric time-fractional heat conduction equation with heat absorption (13) in a solid with a spherical hole of radius R under time-harmonic heat flux boundary conditions. The problem is solved using the auxiliary function $v=rT$, for which the boundary condition of Robin type with the given boundary value of a linear combination of a function and its normal derivative is fulfilled. The Laplace and Fourier sine–cosine integral transformations are employed. Graphical representations of numerical simulation results are presented for typical values of the parameters. The present investigation extends and evolves the results of previous studies [37,38,39,40].

2. Statement and Solution of the Problem

Consider the central symmetric Equation (13) in an infinite medium with a spherical hole with radius R

$$\frac{{\partial }^{\alpha }T}{\partial t^{\alpha }}=a\left(\frac{{\partial }^{2}T}{\partial r^2}+\frac{2}{r}\frac{\partial T}{\partial r}\right)-bT,R<r<\infty ,0<t<\infty ,0<\alpha \le 2,$$

(16)

with zero initial conditions

$$\begin{array}{ccc}\hfill t=0:& & T=0,0<\alpha \le 2,\hfill \end{array}$$

(17)

$$\begin{array}{ccc}\hfill t=0:& & \frac{\partial T}{\partial t}=0,1<\alpha \le 2,\hfill \end{array}$$

(18)

and time-harmonic heat flux at the boundary

$$\begin{array}{ccc}\hfill r=R:& & D_{RL}^{1-\alpha }\frac{\partial T}{\partial r}=-g_0{\mathrm{e}}^{i\omega t},0<\alpha \le 1,\hfill \end{array}$$

(19)

$$\begin{array}{ccc}\hfill r=R:& & I^{\alpha -1}\frac{\partial T}{\partial r}=-g_0{\mathrm{e}}^{i\omega t},1<\alpha \le 2,\hfill \end{array}$$

(20)

where $g_0$ is a constant and $\omega$ denotes the angular frequency.

The zero condition at infinity is also assumed:

$$\underset{r\to \infty }{lim}T(r,t)=0.$$

(21)

The auxiliary function $v=rT$ and the auxiliary spatial variable $x=r-R$ are used. Then, instead of Equations (16)–(21), we obtain

$$\frac{{\partial }^{\alpha }v}{\partial t^{\alpha }}=a\frac{{\partial }^{2}v}{\partial x^2}-bv,0<x<\infty ,0<t<\infty ,0<\alpha \le 2,$$

(22)

$$\begin{array}{ccc}\hfill t=0:& & vs.=0,0<\alpha \le 2,\hfill \end{array}$$

(23)

$$\begin{array}{ccc}\hfill t=0:& & \frac{\partial v}{\partial t}=0,1<\alpha \le 2,\hfill \end{array}$$

(24)

$$\begin{array}{ccc}\hfill x=0:& & D_{RL}^{1-\alpha }\left(\frac{\partial v}{\partial x}-\frac{1}{R}vs.\right)=-g_{0}R{\mathrm{e}}^{i\omega t},0<\alpha \le 1,\hfill \end{array}$$

(25)

$$\begin{array}{ccc}\hfill x=0:& & I^{\alpha -1}\left(\frac{\partial v}{\partial x}-\frac{1}{R}vs.\right)=-g_{0}R{\mathrm{e}}^{i\omega t},1<\alpha \le 2,\hfill \end{array}$$

(26)

$$\underset{x\to \infty }{lim}v(x,t)\mathrm{is}\mathrm{bounded}.$$

(27)

The 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:

$$s^{\alpha }{v}^{*}=a\frac{{\partial }^2{v}^{*}}{\partial x^2}-b{v}^{*},$$

(28)

$$x=0:\frac{\partial v^{*}}{\partial x}-\frac{1}{R}{v}^{*}=-g_{0}R\frac{s^{\alpha -1}}{s-i\omega },0<\alpha \le 2,$$

(29)

$$\underset{x\to \infty }{lim}{v}^{*}(x,s)\mathrm{is}\mathrm{bounded}.$$

(30)

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]:

$$\mathcal{F}\left\{f\left(x\right)\right\}=\tilde{f}\left(\xi \right)={\int }_0^{\infty }K\left(x,\xi \right)f\left(x\right)\mathrm{d}x,$$

(31)

$${\mathcal{F}}^{-1}\left\{\tilde{f}\left(\xi \right)\right\}=f\left(x\right)=\frac{2}{\pi }{\int }_0^{\infty }K\left(x,\xi \right)\tilde{f}\left(\xi \right)\mathrm{d}\xi ,$$

(32)

with the kernel

$$K\left(x,\xi \right)=\frac{\xi cos\left(x\xi \right)+(1/R)sin\left(x\xi \right)}{\sqrt{{\xi }^2+1/R^2}}.$$

(33)

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:

$$\mathcal{F}\left\{\frac{{\mathrm{d}}^{2}f\left(x\right)}{\mathrm{d}{x}^2}\right\}=-{\xi }^2\tilde{f}\left(\xi \right)-\frac{\xi }{\sqrt{{\xi }^2+1/R^2}}\left(\frac{\mathrm{d}f\left(x\right)}{\mathrm{d}x}-\frac{1}{R}f\left(x\right)\right){|}_{x=0}.$$

(34)

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:

$${\tilde{v}}^{*}(\xi ,s)=a{g}_{0}R\frac{\xi }{\sqrt{{\xi }^2+1/R^2}}·\frac{1}{s-i\omega }·\frac{s^{\alpha -1}}{s^{\alpha }+a{\xi }^2+b}.$$

(35)

Using the convolution theorem and the following formula [11,12]

$${\mathcal{L}}^{-1}\left\{\frac{s^{\alpha -1}}{s^{\alpha }+c}\right\}=E_{\alpha }\left(-c{t}^{\alpha }\right),$$

(36)

where $E_{\alpha }\left(z\right)$ is the Mittag–Leffler function

$$E_{\alpha }\left(z\right)=\sum _{n=0}^{\infty }\frac{z^n}{\mathrm{\Gamma }(\alpha n+1)},\alpha >0,z\in \mathbb{C},$$

(37)

after inverting both integral transforms, we obtain

$$\begin{array}{ccc}\hfill v(x,t)& =& \frac{2a{g}_0{R}^2}{\pi }{\int }_0^{\infty }\frac{\xi }{R^2{\xi }^2+1}\left[R\xi cos\left(x\xi \right)+sin\left(x\xi \right)\right]\mathrm{d}\xi \hfill \\ & \times & {\int }_0^t{\mathrm{e}}^{i\omega (t-\tau )}{E}_{\alpha }\left[-\left(a{\xi }^2+b\right){\tau }^{\alpha }\right]\mathrm{d}\tau \hfill \end{array}$$

(38)

and finally

$$\begin{array}{ccc}\hfill T(r,t)& =& \frac{2a{g}_0{R}^2}{\pi r}{\int }_0^{\infty }\frac{\xi }{R^2{\xi }^2+1}\left\{R\xi cos\left[(r-R)\xi \right]+sin\left[(r-R)\xi \right]\right\}\mathrm{d}\xi \hfill \\ & \times & {\int }_0^t{\mathrm{e}}^{i\omega (t-\tau )}{E}_{\alpha }\left[-\left(a{\xi }^2+b\right){\tau }^{\alpha }\right]\mathrm{d}\tau .\hfill \end{array}$$

(39)

Figure 1, Figure 2, Figure 3 and Figure 4 present the real part of the solution (39) for different values of the parameters appearing in this equation. In numerical simulations, the following non-dimensional quantities marked by the bar are used:

$$\overline{T}=\frac{R^{1-2/\alpha }}{g_0{a}^{1-1/\alpha }}T,\overline{r}=\frac{r}{R},\overline{t}=\frac{a^{1/\alpha }}{R^{2/\alpha }}t,\overline{\omega }=\frac{R^{2/\alpha }}{a^{1/\alpha }}\omega ,\overline{b}=\frac{R^2}{a}b.$$

(40)

3. Quasi-Steady-State Approach

In this section, we present the quasi-steady-state approach for two special cases represented by integer values of the order of time derivative. In this approach, only the boundary conditions are imposed; the initial conditions are not taken into account, but the solution is written as the product of the time-harmonic expression and unknown function of spatial variables:

$$T\left(\mathbf{x},t\right)=U\left(\mathbf{x}\right){\mathrm{e}}^{i\omega t}.$$

(41)

3.1. Classical Heat Conduction ($\alpha =1$)

The problem is described by the equations

$$\frac{\partial T}{\partial t}=a\left(\frac{{\partial }^{2}T}{\partial r^2}+\frac{2}{r}\frac{\partial T}{\partial r}\right)-bT,R<r<\infty ,0<t<\infty ,$$

(42)

$$r=R:\frac{\partial T}{\partial r}=-g_0{\mathrm{e}}^{i\omega t},$$

(43)

$$\underset{r\to \infty }{lim}T(r,t)=0.$$

(44)

under the assumption

$$T\left(r,t\right)=U\left(r\right){\mathrm{e}}^{i\omega t}.$$

(45)

Hence,

$$i\omega U=a\left(\frac{{\mathrm{d}}^{2}U}{\mathrm{d}{r}^2}+\frac{2}{r}\frac{\mathrm{d}U}{\mathrm{d}r}\right)-bU,R<r<\infty ,$$

(46)

$$r=R:\frac{\mathrm{d}U}{\mathrm{d}r}=-g_0,$$

(47)

$$\underset{r\to \infty }{lim}U\left(r\right)=0.$$

(48)

Employing the auxiliary function $V=rU$ and the auxiliary spatial variable $x=r-R$, we get

$$\frac{{\mathrm{d}}^{2}V}{\mathrm{d}{x}^2}-\frac{b+i\omega }{a}V=0,0<x<\infty ,$$

(49)

$$x=0:\frac{\mathrm{d}V}{\mathrm{d}x}-\frac{1}{R}V=-g_{0}R,$$

(50)

$$\underset{x\to \infty }{lim}V\left(x\right)\mathrm{is}\mathrm{bounded}.$$

(51)

The general solution of Equation (49) has the form

$$V\left(x\right)=A{\mathrm{e}}^{-x\sqrt{(b+i\omega )/a}}+B{\mathrm{e}}^{x\sqrt{(b+i\omega )/a}},$$

(52)

where the integration constants are determined from the boundary conditions (50) and (51):

$$A=\frac{g_{0}R}{\sqrt{(b+i\omega )/a}+1/R},B=0.$$

(53)

The final result reads

$$T(r,t)=\frac{g_{0}R}{\sqrt{(b+i\omega )/a}+1/R}\frac{1}{r}{\mathrm{e}}^{-(r-R)\sqrt{(b+i\omega )/a}}{\mathrm{e}}^{i\omega t}.$$

(54)

Now, we analyze the particular case of the solution (39) for $\alpha =1$. Equation (35) for the value $\alpha =1$ becomes:

$${\tilde{v}}^{*}(\xi ,s)=a{g}_{0}R\frac{\xi }{\sqrt{{\xi }^2+1/R^2}}·\frac{1}{s-i\omega }·\frac{1}{s+a{\xi }^2+b},$$

(55)

and after inversion of the Laplace transform, we get

$$\tilde{v}(\xi ,t)=a{g}_{0}R\frac{\xi }{\sqrt{{\xi }^2+1/R^2}}·\frac{1}{a{\xi }^2+b+i\omega }\left[{\mathrm{e}}^{i\omega t}-{\mathrm{e}}^{-(a{\xi }^2+b)t}\right].$$

(56)

Next, we invert the Fourier sine–cosine transform and use the partial fraction decomposition

$$\frac{1}{\left({\xi }^2+1/R^2\right)\left[{\xi }^2+(b+i\omega )/a\right]}=\frac{1}{(b+i\omega )/a-1/R^2}\left[\frac{1}{{\xi }^2+1/R^2}-\frac{1}{{\xi }^2+(b+i\omega )/a}\right]$$

(57)

and integrals (A1)–(A4) from Appendix A. After some transformations, we obtain

$$\begin{array}{ccc}\hfill T(r,t)& =& \frac{g_{0}R}{\sqrt{(b+i\omega )/a}+1/R}\frac{1}{r}{\mathrm{e}}^{-(r-R)\sqrt{(b+i\omega )/a}}{\mathrm{e}}^{i\omega t}\hfill \\ \hfill & +& \frac{g_0}{(b+i\omega )/a-1/R^2}\frac{1}{r}\{exp\left(\frac{r-R}{R}+\frac{at}{R^2}-bt\right)\mathrm{erfc}\left(\frac{\sqrt{at}}{R}+\frac{r-R}{2\sqrt{at}}\right)\hfill \\ \hfill & -& \frac{R}{2}\left[\sqrt{(b+i\omega )/a}-1/R\right]{\mathrm{e}}^{-(r-R)\sqrt{(b+i\omega )/a}}\mathrm{erfc}\left(\sqrt{(b+i\omega )t}-\frac{r-R}{2\sqrt{at}}\right){\mathrm{e}}^{i\omega t}\hfill \\ & -& \frac{R}{2}\left[\sqrt{(b+i\omega )/a}+1/R\right]{\mathrm{e}}^{(r-R)\sqrt{(b+i\omega )/a}}\mathrm{erfc}\left(\sqrt{(b+i\omega )t}+\frac{r-R}{2\sqrt{at}}\right){\mathrm{e}}^{i\omega t}\},\hfill \end{array}$$

(58)

where $\mathrm{erfc}\left(z\right)$ is the complementary error function.

The first summand in Equation (58) coincides with the quasi-steady-state expression (54) and the remaining terms describe the transition regime.

Figure 5 compares the real parts of the solutions (54) and (58) for the values of heat absorption parameter $\overline{b}=0$ and $\overline{b}=2$.

3.2. Ballistic Heat Conduction ($\alpha =2$)

The quasi-steady state approach is described by the equations

$$\frac{{\partial }^{2}T}{\partial t^2}=a\left(\frac{{\partial }^{2}T}{\partial r^2}+\frac{2}{r}\frac{\partial T}{\partial r}\right)-bT,R<r<\infty ,0<t<\infty ,$$

(59)

$$r=R:\frac{\partial T}{\partial r}=-i\omega g_0{\mathrm{e}}^{i\omega t},$$

(60)

$$\underset{r\to \infty }{lim}T(r,t)=0.$$

(61)

under the assumption

$$T\left(r,t\right)=U\left(r\right){\mathrm{e}}^{i\omega t}.$$

(62)

Hence,

$$-{\omega }^{2}U=a\left(\frac{{\mathrm{d}}^{2}U}{\mathrm{d}{r}^2}+\frac{2}{r}\frac{\mathrm{d}U}{\mathrm{d}r}\right)-bU,R<r<\infty ,$$

(63)

$$r=R:\frac{\mathrm{d}U}{\mathrm{d}r}=-i\omega g_0,$$

(64)

$$\underset{r\to \infty }{lim}U\left(r\right)=0.$$

(65)

Making use of the auxilliary function $V=rU$ and the auxilliary spatial variable $x=r-R$, we get

$$\frac{{\mathrm{d}}^{2}V}{\mathrm{d}{x}^2}-\frac{b-{\omega }^2}{a}V=0,0<x<\infty ,$$

(66)

$$x=0:\frac{\mathrm{d}V}{\mathrm{d}x}-\frac{1}{R}V=-i\omega g_{0}R,$$

(67)

$$\underset{x\to \infty }{lim}V\left(x\right)\mathrm{is}\mathrm{bounded}.$$

(68)

We confine our consideration to the case $b>{\omega }^2$. Then,

$$V\left(x\right)=A{\mathrm{e}}^{-x\sqrt{(b-{\omega }^2)/a}}+B{\mathrm{e}}^{x\sqrt{(b-{\omega }^2)/a}},$$

(69)

and from the boundary conditions (67) and (68), it follows that

$$A=\frac{i\omega g_{0}R}{\sqrt{(b-{\omega }^2)/a}+1/R},B=0,$$

(70)

and consequently,

$$T(r,t)=\frac{i\omega g_{0}R}{\sqrt{(b-{\omega }^2)/a}+1/R}\frac{1}{r}{\mathrm{e}}^{-(r-R)\sqrt{(b-{\omega }^2)/a}}{\mathrm{e}}^{i\omega t}.$$

(71)

The general solution for ballistic heat conduction can be obtained from Equation (35) with $\alpha =2$:

$${\tilde{v}}^{*}(\xi ,s)=a{g}_{0}R\frac{\xi }{\sqrt{{\xi }^2+1/R^2}}·\frac{1}{s-i\omega }·\frac{s}{s^2+a{\xi }^2+b}.$$

(72)

After inverting the Laplace transform

$$\tilde{v}(\xi ,t)=a{g}_{0}R\frac{\xi }{\sqrt{{\xi }^2+1/R^2}}{\int }_0^t{\mathrm{e}}^{i\omega (t-\tau )}cos\left(\sqrt{a{\xi }^2+b}\tau \right)\mathrm{d}\tau$$

(73)

and after evaluation of the integral in Equation (73), we obtain

$$\begin{array}{ccc}\hfill \tilde{v}(\xi ,t)& =& i\omega a{g}_{0}R\frac{\xi }{\sqrt{{\xi }^2+1/R^2}}\frac{1}{a{\xi }^2+b-{\omega }^2}{\mathrm{e}}^{i\omega t}\hfill \\ & +& a{g}_{0}R\frac{\xi }{\sqrt{{\xi }^2+1/R^2}}\frac{1}{a{\xi }^2+b-{\omega }^2}\hfill \\ & \times & \left[-i\omega cos\left(\sqrt{a{\xi }^2+b}t\right)+\sqrt{a{\xi }^2+b}sin\left(\sqrt{a{\xi }^2+b}t\right)\right].\hfill \end{array}$$

(74)

Inversion of the Fourier sine–cosine transform gives

$$\begin{array}{ccc}\hfill v(x,t)& =& \frac{2i\omega a{g}_{0}R}{\pi }{\int }_0^{\infty }\frac{{\xi }^{2}cos\left(x\xi \right)+(\xi /R)sin\left(x\xi \right)}{\left({\xi }^2+1/R^2\right)\left(a{\xi }^2+b-{\omega }^2\right)}\mathrm{d}\xi {\mathrm{e}}^{i\omega t}\hfill \\ & +& \frac{2a{g}_{0}R}{\pi }{\int }_0^{\infty }\frac{{\xi }^{2}cos\left(x\xi \right)+(\xi /R)sin\left(x\xi \right)}{\left({\xi }^2+1/R^2\right)\left(a{\xi }^2+b-{\omega }^2\right)}\hfill \\ & \times & \left[-i\omega cos\left(\sqrt{a{\xi }^2+b}t\right)+\sqrt{a{\xi }^2+b}sin\left(\sqrt{a{\xi }^2+b}t\right)\right]\mathrm{d}\xi .\hfill \end{array}$$

(75)

Partial fraction decomposition

$$\frac{1}{\left({\xi }^2+1/R^2\right)\left[{\xi }^2+\left(b-{\omega }^2\right)/a\right]}=\frac{1}{(b-{\omega }^2)/a-1/R^2}\left[\frac{1}{{\xi }^2+1/R^2}-\frac{1}{{\xi }^2+(b-{\omega }^2)/a}\right]$$

(76)

after using integrals (A1) and (A2) from Appendix A leads to

$$\begin{array}{ccc}\hfill v(x,t)& =& \frac{i\omega g_{0}R}{\sqrt{(b-{\omega }^2)/a}+1/R}{\mathrm{e}}^{-x\sqrt{(b-{\omega }^2)/a}}{\mathrm{e}}^{i\omega t}\hfill \\ & -& \frac{2g_{0}R}{\pi \left[R^2(b-{\omega }^2)/a-1\right]}{\int }_0^{\infty }\{\frac{cos\left(x\xi \right)-R\xi sin\left(x\xi \right)}{{\xi }^2+1/R^2}\hfill \\ & -& \frac{\left[R^2(b-{\omega }^2)/a\right]cos\left(x\xi \right)-R\xi sin\left(x\xi \right)}{{\xi }^2+(b-{\omega }^2)/a}\}\hfill \\ & \times & \left[-i\omega cos\left(\sqrt{a{\xi }^2+b}t\right)+\sqrt{a{\xi }^2+b}sin\left(\sqrt{a{\xi }^2+b}t\right)\right]\mathrm{d}\xi .\hfill \end{array}$$

(77)

Reverting in Equation (77) to the variable r and temperature $T(r,t)$

$$\begin{array}{ccc}\hfill T(r,t)& =& \frac{i\omega g_{0}R}{\sqrt{(b-{\omega }^2)/a}+1/R}\frac{1}{r}{\mathrm{e}}^{-(r-R)\sqrt{(b-{\omega }^2)/a}}{\mathrm{e}}^{i\omega t}\hfill \\ & -& \frac{2g_{0}R}{\pi \left[R^2(b-{\omega }^2)/a-1\right]}\frac{1}{r}{\int }_0^{\infty }\{\frac{cos\left[\right(r-R\left)\xi \right]-R\xi sin\left[\right(r-R\left)\xi \right]}{{\xi }^2+1/R^2}\hfill \\ & -& \frac{\left[R^2(b-{\omega }^2)/a\right]cos\left[(r-R)\xi \right]-R\xi sin\left[(r-R)\xi \right]}{{\xi }^2+(b-{\omega }^2)/a}\}\hfill \\ & \times & \left[-i\omega cos\left(\sqrt{a{\xi }^2+b}t\right)+\sqrt{a{\xi }^2+b}sin\left(\sqrt{a{\xi }^2+b}t\right)\right]\mathrm{d}\xi ,\hfill \end{array}$$

(78)

we see that the first summand in the expression (78) coincides with the quasi-steady-state solution. For $x<\sqrt{a}t$, the integrals in Equation (78) cannot be evaluated analytically and should be calculated numerically. For $x>\sqrt{a}t$, the integrals in Equation (78) can be evaluated analytically using Equations (A5)–(A8) from Appendix A. In this case, Equation (78) leads to

$$T(r,t)=0\mathrm{for}r>R+\sqrt{a}t.$$

(79)

Hence, there is the wave front at $r=R+\sqrt{a}t$. Recall that in the Klein–Gordon Equation (59), the coefficient a can be taken as $a=c^2$, where c is the wave velocity.

Figure 6 shows the real parts of solutions (71) and (78) taking the non-dimensional quantities (40) for $\alpha =2$.

4. Concluding Remarks

We have studied time-fractional heat conduction with heat absorption in an infinite medium with a spherical hole under time-harmonic heat flux at its boundary. In numerical simulations, we have utilized different values of non-dimensional parameters to show the distinguishing properties of solutions.

It should be pointed out that the quasi-steady-state approach can be considered only for integer orders of the time derivative $\alpha =1$ and $\alpha =2$ and cannot be studied for fractional orders of the Caputo derivative because for a non-integer $\alpha$ [42],

$$\frac{{\mathrm{d}}^{\alpha }}{\mathrm{d}{t}^{\alpha }}{\mathrm{e}}^{i\omega t}\ne {\left(i\omega \right)}^{\alpha }{\mathrm{e}}^{i\omega t}.$$

(80)

It is observable from the figures that at different times, the absorption parameter $\overline{b}$ can cause a change in the sign of the temperature on the cavity surface. Figure 1 shows that for $0\le \alpha <1$, the temperature at the cavity surface is larger and for $1<\alpha \le 2$, it is smaller than that in the case of the standard heat conduction with $\alpha =1$. The situation changes and inverts at larger distances from the cavity, and there is a point at some distance $\overline{r}$ where the temperature $\overline{T}$ is approximately the same for all values of $\alpha$. In medical applications, it is possible to choose the necessary combination of the values of $\omega$, b, and $\alpha$ according to the goals of the procedure. In the literature, there is a discussion about the optimal choice of the order of the time-derivative $\alpha$ in the fractional heat conduction model for bilayered spherical tissue in the hyperthermia experiment [43].

It is evident from Figure 5 that for the parabolic Equation ($\alpha =1$), the process is dissipative itself and the absorption parameter $\overline{b}$ significantly decreases the difference between the quasi-steady-state solution and the general solution involving the transition process. In the case of the hyperbolic Klein–Gordon Equation ($\alpha =2$), the difference between the quasi-steady state approach and the general solution describing the transition regime is more visible (see Figure 6). In particular, the quasi-steady-state solution does not have a wave front. For $\alpha =2$, the wave front at $r=R+\sqrt{a}t$ (or in terms of non-dimensional quantities, at $\overline{r}=1+\overline{t}$) is also seen in Figure 1 for different values of non-dimensional time $\overline{t}$.

Our future work will focus on studying the space-time-fractional equations with the Caputo time derivative and Riesz fractional operator (fractional Laplacian).