6. Transform solution
By applying the Laplace transform technique
\(\overset{\overline{}}{f}\left(x,s\right)\mathcal{=L}\left[f\left(x,t\right)\right]=\int_{0}^{\infty}{f\left(x,t\right)e^{-\text{st}}\ \text{dt}},\ \ \text{Re}(s)>0\)
into Eqs. (12)-(14) and using (16), we get
\(\frac{d\overset{\overline{}}{e}}{\text{dx}}-\frac{d\overset{\overline{}}{\theta}}{\text{dx}}=s^{2}\overset{\overline{}}{u}+\frac{e^{-\text{ωx}}}{s}\ \)(20)
\(\mathcal{l}_{q}\overset{\overline{}}{\theta}+\varepsilon\mathcal{l}_{q}\overset{\overline{}}{e}=\mathcal{l}_{\theta}\nabla^{2}\overset{\overline{}}{\theta}\)(21)
\({\overset{\overline{}}{\sigma}}_{\text{xx}}=\overset{\overline{}}{e}-\overset{\overline{}}{\theta}\)(22)
where
\(\mathcal{l}_{q}=s\left(1+\tau_{q}^{\alpha}s^{\alpha}\right),\ \ \mathcal{l}_{\theta}=\left(1+\tau_{\theta}^{\beta}s^{\beta}\right),\ \ \nabla^{2}=\frac{d^{2}}{dx^{2}}\)(23)
One can show that Eq. (20) becomes
\(\nabla^{2}\overset{\overline{}}{e}-s^{2}\overset{\overline{}}{e}=\nabla^{2}\overset{\overline{}}{\theta\ }-\frac{\omega}{s}e^{-\text{ωx}}\)(24)
From which together with Eq. (21), we obtain
\(\left[\nabla^{4}-A\nabla^{2}+B\right]\overset{\overline{}}{\theta}=-\alpha_{1}e^{-\text{ωx}}\)(25)
where
\(A=\frac{\mathcal{l}_{q}}{\mathcal{l}_{\theta}}\left(1+\varepsilon\right)+s^{2},\ \ B=s^{2}\frac{\mathcal{l}_{q}}{\mathcal{l}_{\theta}},\ \alpha_{1}=\frac{\text{ωε}\mathcal{l}_{q}}{s\mathcal{l}_{\theta}}\)
The homogenous solution \({\overset{\overline{}}{\theta}}_{h}\) of Eq.
(25) has the form:
\({\overset{\overline{}}{\theta}}_{h}(x,s)=A_{1}(s)e^{-m_{1}x}+A_{2}(s)e^{-m_{2}x}\)(26)
and \(m_{1}\) and \(m_{2}\) are given by
\(m_{1,2}=\frac{1}{2}\sqrt{\left[\frac{\mathcal{l}_{q}}{\mathcal{l}_{\theta}}\left(1+\varepsilon\right)+s^{2}\right]+\sqrt{\left[\frac{\mathcal{l}_{q}}{\mathcal{l}_{\theta}}(1+\varepsilon)+s^{2}\right]^{2}-4s^{2}\frac{\mathcal{l}_{q}}{\mathcal{l}_{\theta}}}}\)(28)
On the other hand, the particular solution\({\overset{\overline{}}{\theta}}_{p}\) of Eq. (24) given by
\({\overset{\overline{}}{\theta}}_{p}(x,s)=A_{3}e^{-\omega x}\) (29)
Where\(A_{3}=\frac{-\alpha_{1}}{\omega^{4}-A\omega^{2}+B}\text{\ .}\)
Then, the general solution of the function\(\overset{\overline{}}{\theta}\) has the form
\(\overset{\overline{}}{\theta}\left(x,s\right)=A_{1}e^{-m_{1}x}+A_{2}e^{-m_{2}x}+A_{3}e^{-\omega x}\)(30)
From Eqs. (21) and (30), we get
\(\overset{\overline{}}{e}\left(x,s\right)=\sum_{i}^{2}{\left[\frac{1}{\varepsilon}\left(\frac{\mathcal{l}_{\theta}}{\mathcal{l}_{q}}m_{i}^{2}-1\right)\right]A_{i}e^{-m_{i}x}}+\frac{1}{\varepsilon}\left(\frac{\mathcal{l}_{\theta}}{\mathcal{l}_{q}}\omega^{2}-1\right)A_{3}e^{-\omega x}\)(31)
Hence, the displacement \(\overset{\overline{}}{u}\) can be expressed as
\(\overset{\overline{}}{u}\left(x,s\right)=\sum_{i}^{2}{\left[\frac{1}{m_{i}\varepsilon}\left(1-\frac{\mathcal{l}_{\theta}}{\mathcal{l}_{q}}m_{i}^{2}\right)\right]A_{i}e^{-m_{i}x}}+\frac{1}{\text{ωε}}\left(1-\frac{\mathcal{l}_{\theta}}{\mathcal{l}_{q}}\omega^{2}\right)A_{3}e^{-\omega x}\)(32)
In view of Eq. (22) and using Eqs. (30) and (31), we have
\({\overset{\overline{}}{\sigma}}_{\text{xx}}=\sum_{i}^{2}{\left[\frac{1}{\varepsilon}\left(\frac{\mathcal{l}_{\theta}}{\mathcal{l}_{q}}m_{i}^{2}-\varepsilon-1\right)\right]A_{i}e^{-m_{i}x}}+\frac{1}{\varepsilon}\left(\frac{\mathcal{l}_{\theta}}{\mathcal{l}_{q}}\omega^{2}-\varepsilon-1\right)A_{3}e^{-\omega x}\ \)(33)
After applying Laplace transform, the boundary conditions (18) become
\(\text{\ \ \ \ \ }\par
\begin{matrix}\left.\ \overset{\overline{}}{\theta\ }\left(x,s\right)\right|_{x=0}=\overset{\overline{}}{G}\left(s\right)=\frac{{t_{0}\theta}_{0}}{t_{0}s+1}\\
\left.\ \overset{\overline{}}{u}\left(x,s\right)\right|_{x=0}=0\\
\end{matrix}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\) (34)
From Eqs. (30) and (31) taking into account the above conditions, we
obtain
\(\par
\begin{matrix}\end{matrix}\) (30)
Finally, from the above system we can determine the parameters \(A_{1}\)and \(A_{2}\) in the Laplace transform domain and hence the physical
fields of the medium.
To have the solutions of the studied fields in the physical domain, we
use a proper and effective numerical method depending on a Fourier
series expansion [26]. In this method, any function\(\overset{\overline{}}{\mathcal{M}}\left(r,s\right)\) in the Laplace
domain can be reversed to the time domain as
\(\mathcal{M}\left(r,t\right)=\frac{e^{\text{ct}}}{t}\left(\frac{1}{2}\overset{\overline{}}{\mathcal{M}}\left(r,c\right)+\text{Re}\sum_{n=1}^{m}{\overset{\overline{}}{\mathcal{M}}\left(r,c+\frac{\text{inπ}}{t}\right)\left(-1\right)^{n}}\right)\)(31)
where \(m\) is a finite number of terms, Re is the real part
and \(i\) is imaginary number unit. For faster convergence, numerous
numerical experiments have shown that the value of \(c\) fulfills the
relation \(\text{ct}\cong 4.7\) [26].