We consider a homogeneous transversely isotropic magneto-thermoelastic medium initially at a uniform temperature T0, permeated by an initial magnetic field \( {\overrightarrow{H}}_0=\left(0,{H}_0,0\right) \) acting along the y-axis. The rectangular Cartesian co-ordinate system (x, y, z) having origin on the surface (z = 0) with z-axis pointing vertically into the medium is introduced. From the generalized Ohm’s law
$$ \overrightarrow{E}=-{\mu}_0{H}_0\left(-\ddot{w},0,\ddot{u}\right) $$
(14)
$$ \overrightarrow{j}=\left(-\frac{\partial Hy}{\partial z}-{\varepsilon}_0{\dot{E}}_x,0,-\frac{\partial Hy}{\partial x}-{\varepsilon}_0{\dot{E}}_z\right). $$
(15)
$$ \overrightarrow{F}=\left({\mu}_0{H}_0^2\left(\frac{\partial e}{\partial x}-{\varepsilon}_0{\mu}_0\ddot{u}\right),0,{\mu}_0{H}_0^2\left(\frac{\partial e}{\partial z}-{\varepsilon}_0{\mu}_0\ddot{w}\right)\right) $$
(16)
In addition, we consider the plane such that all particles on a line parallel to y-axis are equally displaced, so that the equations of displacement vector (u, v, w) and conductive temperature φ for transversely isotropic thermoelastic solid are given by
$$ u=u\left(x,z,t\right),v=0,w=w\left(x,z,t\right)\ and\ \varphi =\varphi \left(x,z,t\right). $$
(17)
Now, using the transformation on Eqs. (5) and (12) following Slaughter (2002) with the aid of (16) and (17) yields
$$ {C}_{11}\frac{\partial^2u}{\partial {x}^2}+{C}_{13}\frac{\partial^2w}{\partial x\partial z}+{C}_{44}\left(\frac{\partial^2u}{\partial {z}^2}+\frac{\partial^2w}{\partial x\partial z}\right)-{\beta}_1\frac{\partial }{\partial x}\left\{\varphi -\left({a}_1\frac{\partial^2\varphi }{\partial {x}^2}+{a}_3\frac{\partial^2\varphi }{\partial {z}^2}\right)\right\}+{\mu}_0{H}_0^2\left(\frac{\partial e}{\partial x}-{\varepsilon}_0{\mu}_0\ddot{u}\right)=\rho \frac{\partial^2u}{\partial {t}^2}, $$
(18)
$$ \left({C}_{13}+{C}_{44}\right)\frac{\partial^2u}{\partial x\partial z}+{C}_{44}\frac{\partial^2w}{\partial {x}^2}+{C}_{33}\frac{\partial^2w}{\partial {z}^2}-{\beta}_3\frac{\partial }{\partial z}\left\{\varphi -\left({a}_1\frac{\partial^2\varphi }{\partial {x}^2}+{a}_3\frac{\partial^2\varphi }{\partial {z}^2}\right)\right\}+{\mu}_0{H}_0^2\left(\frac{\partial e}{\partial z}-{\varepsilon}_0{\mu}_0\ddot{w}\right)=\rho \frac{\partial^2w}{\partial {t}^2}, $$
(19)
$$ {K}_1\frac{\partial^2\varphi }{\partial {x}^2}+{K}_3\frac{\partial^2\varphi }{{\partial z}^2}=\left(1+\chi {D}_{\chi}\right)\left[{T}_0\left({\beta}_1\frac{\partial \dot{u}}{\partial x}+{\beta}_3\frac{\partial \dot{w}}{\partial z}\right)+\rho {C}_E\left\{\dot{\varphi}-{a}_1\frac{\partial^2\dot{\varphi}}{{\partial x}^2}-{a}_3\frac{\partial^2\dot{\varphi}}{{\partial z}^2}\right\}\right] $$
(20)
where from (7)
$$ {\displaystyle \begin{array}{c}{\beta}_1=\left({C}_{11}+{C}_{12}\right){\alpha}_1+{C}_{13}{\alpha}_{3,}\\ {}{\beta}_3=2{C}_{13}{\alpha}_1+{C}_{33}{\alpha}_{3,}\\ {}e=\frac{\partial u}{\partial x}+\frac{\partial w}{\partial z}\end{array}} $$
And stress strain relations given by (5) after using (9) and (17) becomes
$$ {t}_{11}\left(\mathrm{x},\mathrm{z},\mathrm{t}\right)={C}_{11}\frac{\partial u}{\partial x}+{C}_{13}\frac{\partial w}{\partial z}-{\beta}_1\left\{\varphi -\left({a}_1\frac{\partial^2\varphi }{{\partial x}^2}+{a}_3\frac{\partial^2\varphi }{{\partial z}^2}\right)\right\}, $$
(21)
$$ {t}_{22}\left(\mathrm{x},\mathrm{z},\mathrm{t}\right)={c}_{12}\frac{\partial u}{\partial x}+{c}_{13}\frac{\partial w}{\partial z}-{\beta}_1\left\{\varphi -\left({a}_1\frac{\partial^2\varphi }{{\partial x}^2}+{a}_3\frac{\partial^2\varphi }{{\partial z}^2}\right)\right\}, $$
(22)
$$ {t}_{33}\left(\mathrm{x},\mathrm{z},\mathrm{t}\right)={C}_{13}\frac{\partial u}{\partial x}+{C}_{33}\frac{\partial w}{\partial z}-{\beta}_3\left\{\varphi -\left({a}_1\frac{\partial^2\varphi }{{\partial x}^2}+{a}_3\frac{\partial^2\varphi }{{\partial z}^2}\right)\right\}, $$
(23)
$$ {t}_{13}\left(\mathrm{x},\mathrm{z},\mathrm{t}\right)={C}_{44}\left(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}\right), $$
(24)
$$ {t}_{12}\left(\mathrm{x},\mathrm{z},\mathrm{t}\right)={t}_{23}\left(\mathrm{x},\mathrm{z},\mathrm{t}\right)=0. $$
(25)
We assume that medium is initially at rest. The undisturbed state is maintained at reference temperature T0. Then, we have the initial and regularity conditions given by
$$ u\left(x,z,0\right)=0=\dot{u}\left(x,z,0\right),w\left(x,z,0\right)=0=\dot{w}\left(x,z,0\right),\varphi \left(x,z,0\right)=0=\dot{\varphi}\left(x,z,0\right)\ for\ z\ge 0,-\infty <x<\infty, \kern0.5em u\left(x,z,t\right)=w\left(x,z,t\right)=\varphi \left(x,z,t\right)=0\ \mathrm{for}\ t>0\ \mathrm{when}\ z\to \infty . $$
To simplify the solution, the following dimensionless quantities are used:
$$ {\displaystyle \begin{array}{c}\left({x}^{\prime },{z}^{\prime}\right)=\frac{1}{L}\left(x,z\right),\left({u}^{\prime },{w}^{\prime}\right)=\frac{{\rho c}_1^2}{L{\beta}_1{T}_0}\left(u,w\right),\left({a}_1^{\prime },{a}_3^{\prime}\right)=\frac{1}{L^2}\left({a}_1,{a}_3\right),{\rho C}_1^2={C}_{11}\\ {}{\varphi}^{\prime }=\frac{\varphi }{T_0},\left({t}_{11}^{\prime },{t}_{13}^{\prime },{t}_{33}^{\prime}\right)=\frac{1}{\beta_1{T}_0}\left({t}_{11},{t}_{13},{t}_{33}\right),{t}^{\prime }=\frac{C_1}{L}t.\end{array}} $$
(26)
Making use of dimensionless quantities defined by (26) in Eqs. (18)–(20), after suppressing the primes, yields
$$ \left(1+{\delta}_5\right)\frac{\partial^2u}{\partial {x}^2}+\left({\delta}_4+{\delta}_5\right)\frac{\partial^2w}{\partial x\partial z}+{\delta}_2\left(\frac{\partial^2u}{\partial {z}^2}+\frac{\partial^2w}{\partial x\partial z}\right)-\frac{\partial }{\partial x}\left\{\varphi -\left({a}_1\frac{\partial^2\varphi }{\partial {x}^2}+{a}_3\frac{\partial^2\varphi }{\partial {z}^2}\right)\right\}=\left(\frac{\varepsilon_0{\mu}_0^2{H}_0^2}{\rho }+1\right)\frac{\partial^2u}{\partial {t}^2}, $$
(27)
$$ \left({\delta}_1+{\delta}_5\right)\frac{\partial^2u}{\partial x\partial z}+{\delta}_2\frac{\partial^2w}{\partial {x}^2}+\left({\delta}_3+{\delta}_5\right)\frac{\partial^2w}{\partial {z}^2}-\frac{\beta_3}{\beta_1}\frac{\partial }{\partial z}\left\{\varphi -\left({a}_1\frac{\partial^2\varphi }{\partial {x}^2}+{a}_3\frac{\partial^2\varphi }{\partial {z}^2}\right)\right\}=\left(\frac{\varepsilon_0{\mu}_0^2{H}_0^2}{\rho }+1\right)\frac{\partial^2w}{\partial {t}^2}, $$
(28)
$$ {K}_1\frac{\partial^2\varphi }{\partial {x}^2}+{K}_3\frac{\partial^2\varphi }{{\partial z}^2}=\left(1+\chi {D}_{\chi}\right)\left[\frac{\partial }{\partial t}\left({\delta}_6\frac{\partial u}{\partial x}+{\delta}_8\frac{\partial w}{\partial z}\right)+{\delta}_7\frac{\partial }{\partial t}\left\{\varphi -{a}_1\frac{\partial^2\varphi }{\partial {x}^2}-{a}_3\frac{\partial^2\varphi }{{\partial z}^2}\right\}\right], $$
(29)
where
$$ {\delta}_1=\frac{C_{13}+{C}_{44}}{C_{11}},{\delta}_2=\frac{C_{44}}{C_{11}},{\delta}_3=\frac{C_{33}}{C_{11}},{\delta}_4=\frac{C_{13}}{C_{11}},{\delta}_5=\frac{\beta_1{T}_0{\mu}_0{H}_0^2}{L{\rho}^2{C}_1^4},{\delta}_6=\frac{{L\beta}_1^2{T}_0}{\rho {C}_1},{\delta}_8=\frac{\mathrm{L}{\beta}_1{\beta}_3{T}_0}{\rho {C}_1},{\delta}_7=\rho {C}_E{C}_1L. $$
Laplace transforms is defined by
$$ \mathcal{L}\left[f\left(x,z,t\right)\right]=\underset{0}{\overset{\infty }{\int }}{e}^{- st}f\left(x,z,t\right) dt=\overset{\sim }{f}\left(x,z,s\right). $$
(30)
with basic properties
$$ L\left(\frac{\partial f}{\partial t}\right)=s\overline{f}\left(x,z,s\right)-f\left(x,z,0\right), $$
(31)
$$ L\left(\frac{\partial^2f}{\partial {t}^2}\right)={s}^2\overline{f}\left(x,z,s\right)- sf\left(x,z,0\right)-{\left(\frac{\partial f}{\partial t}\right)}_{t=0}. $$
(32)
Fourier transforms of a function f with respect to variable x with ξ as a Fourier transform variable is defined by
$$ \mathcal{F}\left\{\overset{\sim }{f}\left(x,z,s\right)\right\}=\hat{f}\left(\xi, z,s\right)=\underset{-\infty }{\overset{\infty }{\int }}\overset{\sim }{f}\left(x,z,s\right){e}^{i\xi x} dx. $$
(33)
With basic properties: if f(x) and first (n − 1) derivatives of f(x) vanish identically as x→ ± ∞ , then
$$ \mathcal{F}\left\{\frac{\partial^n\overset{\sim }{f}\left(x,z,s\right)}{\partial {x}^n}\right\}={\left( i\xi \right)}^n\hat{f}\left(\xi, z,s\right) $$
(34)
Applying Laplace and Fourier transforms defined by (31)–(34) on Eqs. (24)-(29) yields
$$ -\left(1+{\delta}_5\right){\xi}^2\hat{u}+\left({\delta}_4+{\delta}_5\right) i\xi D\hat{w}+{\delta}_2\left({D}^2\hat{u}+ i\xi D\hat{w}\ \right)- i\xi \left\{\hat{\varphi}-\left(-{a}_1{\xi}^2\hat{\varphi}+{a}_3{D}^2\hat{\varphi}\right)\right\}=\left(\frac{\varepsilon_0{\mu}_0^2{H}_0^2}{\rho }+1\right){s}^2\hat{u}, $$
(35)
$$ \left({\delta}_1+{\delta}_5\right) i\xi D\hat{u}-{\delta}_2{\xi}^2\hat{w}+\left({\delta}_3+{\delta}_5\right){D}^2\hat{w}-\frac{\beta_3}{\beta_1}D\left\{\hat{\varphi}-\left(-{a}_1{\xi}^2\hat{\varphi}+{a}_3{D}^2\hat{\varphi}\right)\right\}=\left(\frac{\varepsilon_0{\mu}_0^2{H}_0^2}{\rho }+1\right){s}^2\hat{w}, $$
(36)
$$ -{K}_1{\xi}^2\hat{\varphi}+{K}_3{D}^2\hat{\varphi}=\left(1+G\right)\left[s\left({\delta}_6 i\xi \hat{u}+{\delta}_8D\hat{w}\right)+{\delta}_7s\left\{\hat{\varphi}-\left(-{a}_1{\xi}^2\hat{\varphi}+{a}_3{D}^2\hat{\varphi}\right)\right\}\right], $$
(37)
where
$$ D=\frac{d}{dz},\mathrm{G}=\frac{\uptau_0}{\upchi}\left[\left(1-{\mathrm{e}}^{-\mathrm{s}\upchi}\right)\left(1-\frac{2\mathrm{b}}{\upchi \mathrm{s}}+\frac{2{\mathrm{a}}^2}{\upchi^2{\mathrm{s}}^2}\right)-\left({\mathrm{a}}^2-2\mathrm{b}+\frac{2{\mathrm{a}}^2}{\upchi \mathrm{s}}\right){\mathrm{e}}^{-\mathrm{s}\upchi}\right]. $$
These equations on further simplifications become
$$ \left[-\left(1+{\delta}_5\right){\xi}^2-\left(\frac{\varepsilon_0{\mu}_0^2{H}_0^2}{\rho }+1\right){s}^2+{\delta}_2{D}^2\right]\hat{u}\left(\xi, z,s\right)+\left[\left({\delta}_4+{\delta}_5+{\delta}_2\right) i\xi D\right]\hat{w}\left(\xi, z,s\right)+\left(-\mathrm{i}\xi \right)\left[1+{a}_1{\xi}^2-{a}_3{D}^2\right]\hat{\varphi}\left(\xi, z,s\right)=0, $$
(38)
$$ \left[\left({\delta}_1+{\delta}_5\right) i\xi D\right]\hat{u}\left(\xi, z,s\right)+\left[-{\delta}_2{\xi}^2+\left({\delta}_3+{\delta}_5\right){D}^2-\left(\frac{\varepsilon_0{\mu}_0^2{H}_0^2}{\rho }+1\right){s}^2\right]\hat{w}\left(\xi, z,s\right)-\frac{\beta_3}{\beta_1}D\left[1+{a}_1{\xi}^2-{a}_3{D}^2\right]\hat{\varphi}\left(\xi, z,s\right)=0, $$
(39)
$$ \left[\left(1+G\right)\mathrm{s}{\delta}_6 i\xi \right]\hat{u}\left(\xi, z,s\right)+\left[\left(1+G\right)\mathrm{s}{\delta}_8D\right]\hat{w}\left(\xi, z,s\right)+\left[-{K}_1{\xi}^2+{K}_3{D}^2+\left(1+G\right){\delta}_7s\left(1+{a}_1{\xi}^2-{a}_3{D}^2\right)\right]\hat{\varphi}\left(\xi, z,s\right)=0, $$
(40)
The above equations can be written as
$$ \left[{\zeta}_1+{\delta}_2{D}^2\right]\hat{u}\left(\xi, z,s\right)+\left[{\zeta}_2D\right]\hat{w}\left(\xi, z,s\right)+\left[{\upzeta}_3+{\upzeta}_4{D}^2\right]\hat{\varphi}\left(\xi, z,s\right)=0, $$
(41)
$$ \left[{\zeta}_5D\right]\hat{u}\left(\xi, z,s\right)+\left[{\zeta}_6+{\zeta}_7{D}^2\right]\hat{w}\left(\xi, z,s\right)+\left[{\zeta}_8D+{\zeta}_9{D}^3\right]\hat{\varphi}\left(\xi, z,s\right)=0, $$
(42)
$$ \left[{\zeta}_{10}\right]\hat{u}\left(\xi, z,s\right)+\left[{\zeta}_{11}\right]\hat{w}\left(\xi, z,s\right)+\left[{\zeta}_{12}+{\zeta}_{13}{D}^2\right]\hat{\varphi}\left(\xi, z,s\right)=0, $$
(43)
Equations (41)–(43) will have a non-trivial solution if determinant of coefficient matrix of \( \left(\hat{u},\hat{w},\hat{\varphi}\right) \) vanishes, and thus we obtain the characteristic equation as
$$ \left({AD}^6+{BD}^4+{CD}^2+E\right)\left(\hat{u},\hat{w},\hat{\varphi}\right)=0, $$
(44)
where
$$ {\displaystyle \begin{array}{c}D=\frac{d}{dz},\mathrm{A}=-{\updelta}_2{\upzeta}_{11}{\upzeta}_9+{\upzeta}_{13}{\updelta}_2{\upzeta}_7,\\ {}\mathrm{B}={\updelta}_2{\upzeta}_6{\upzeta}_{13}+{\upzeta}_7{\upzeta}_1{\upzeta}_{13}+{\updelta}_2{\upzeta}_7{\upzeta}_{12}-{\updelta}_2{\upzeta}_{11}{\upzeta}_8-{\upzeta}_1{\upzeta}_9{\upzeta}_{11}-{\upzeta}_5{\delta}_2{\upzeta}_{13}+{\delta}_2{\upzeta}_{10}{\upzeta}_9+{\upzeta}_{11}{\upzeta}_4{\upzeta}_5-{\upzeta}_{10}{\upzeta}_7{\upzeta}_4,\\ {}\begin{array}{c}\mathrm{C}={\upzeta}_{13}{\upzeta}_1{\upzeta}_6+{\updelta}_2{\upzeta}_6{\upzeta}_{12}+{\upzeta}_1{\upzeta}_7{\upzeta}_{12}-{\upzeta}_1{\upzeta}_{11}{\upzeta}_8+{\upzeta}_4{\upzeta}_8{\upzeta}_9-{\upzeta}_5{\delta}_2{\upzeta}_{12}-{\upzeta}_{10}{\delta}_2{\upzeta}_8+{\upzeta}_{11}{\upzeta}_3{\upzeta}_5-{\upzeta}_3{\upzeta}_7{\upzeta}_{10},\\ {}E={\upzeta}_{12}{\upzeta}_1{\upzeta}_6-{\upzeta}_{10}{\upzeta}_6{\zeta}_3,\\ {}\begin{array}{c}{\upzeta}_1=-\left(1+{\delta}_5\right){\xi}^2-\left(\frac{\varepsilon_0{\mu}_0^2{H}_0^2}{\rho }+1\right){s}^2,{\upzeta}_2=\left({\delta}_4+{\delta}_5+{\delta}_2\right) i\xi, {\upzeta}_3=- i\xi \left(1+{a}_1{\xi}^2\right),\\ {}{\upzeta}_4={a}_3 i\xi, {\upzeta}_5=\left({\delta}_1+{\delta}_5\right) i\xi, {\upzeta}_6=-{\delta}_2{\xi}^2-\left(\frac{\varepsilon_0{\mu}_0^2{H}_0^2}{\rho }+1\right){s}^2,{\upzeta}_7=\left({\delta}_3+{\delta}_5\right),\\ {}\begin{array}{c}{\upzeta}_8=-\left(1+{a}_1{\xi}^2\right)\frac{\upbeta_3}{\upbeta_1},{\upzeta}_9=\frac{\beta_3}{\beta_1}{a}_3,{\upzeta}_{10}=\left[\left(1+G\right)\mathrm{s}{\delta}_6 i\xi \right],{\upzeta}_{11}=\left(1+G\right)\mathrm{s}{\delta}_8,\\ {}\ {\upzeta}_{12}=\left[-{K}_1{\xi}^2+\left(1+G\right){\delta}_7s\left(1+{a}_1{\xi}^2\right)\right],{\upzeta}_{13}={K}_3-\left(1+G\right){\delta}_7s{a}_3\end{array}\end{array}\end{array}\end{array}} $$
The roots of Eq. (44) are ± λi, (i = 1, 2, 3). The solution of Eqs. (41)–(43) satisfying the radiation condition that \( \left(\hat{u},\hat{w},\hat{\varphi}\right)\to 0\ \mathrm{as}\ z\to \infty \) can be written as
$$ \hat{\mathrm{u}}\left(\xi, z,s\right)={\sum}_{\mathrm{j}=1}^3{A}_j{e}^{-{\lambda}_jz}, $$
(45)
$$ \hat{\mathrm{w}}\left(\xi, z,s\right)={\sum}_{j=1}^3{d}_j{A}_j{e}^{-{\lambda}_{\mathrm{j}}z}, $$
(46)
$$ \hat{\upvarphi}\left(\xi, z,s\right)={\sum}_{j=1}^3{l}_j{A}_j{e}^{-{\lambda}_jz}, $$
(47)
where Aj, j = 1, 2, 3 being undetermined constants and dj and lj are given by
$$ {d}_j=\frac{\delta_2{\zeta}_{13}{\lambda}_j^4+\left({\zeta}_{13}{\zeta}_1-{\upzeta}_{10}{\upzeta}_4+{\delta}_2{\zeta}_{12}\right){\lambda}_{\mathrm{j}}^2+{\zeta}_1{\zeta}_{12}-{\zeta}_{10}{\zeta}_3}{\left({\upzeta}_{13}{\upzeta}_7-{\upzeta}_1{\upzeta}_9\right){\lambda}_j^4+\left({\upzeta}_{13}{\zeta}_6+{\zeta}_{12}{\zeta}_7-{\zeta}_1{\zeta}_8\right){\lambda}_j^2+{\zeta}_{12}{\zeta}_6}, $$
(48)
$$ {l}_j=\frac{\delta_2{\upzeta}_7{\lambda}_j^4+\left({\delta}_2{\zeta}_6+{\zeta}_1{\upzeta}_7-{\delta}_2{\zeta}_5\right){\lambda}_{\mathrm{j}}^2+{\zeta}_6{\zeta}_1}{\left({\upzeta}_{13}{\upzeta}_7-{\upzeta}_1{\upzeta}_9\right){\lambda}_j^4+\left({\upzeta}_{13}{\zeta}_6+{\zeta}_{12}{\zeta}_7-{\zeta}_1{\zeta}_8\right){\lambda}_{\mathrm{j}}^2+{\zeta}_{12}{\zeta}_6}. $$
(49)
$$ {\hat{t}}_{11}\left(\xi, z,s\right)= i\xi \hat{u}\left(\xi, z,s\right)+{\delta}_4\mathrm{D}\hat{w}\left(\xi, z,s\right)-\frac{\beta_3}{\beta_1}\left[1+{a}_1{\xi}^2-{a}_3{D}^2\right]\hat{\varphi}\left(\xi, z,s\right), $$
(50)
$$ {\hat{t}}_{33}\left(\xi, z,s\right)={\delta}_4 i\xi \hat{u}\left(\xi, z,s\right)+{\updelta}_3\mathrm{D}\hat{w}\left(\xi, z,s\right)-\frac{\beta_3}{\beta_1}\left[1+{a}_1{\xi}^2-{a}_3{D}^2\right]\hat{\varphi}\left(\xi, z,s\right), $$
(51)
$$ {\hat{t}}_{13}\left(\xi, z,s\right)={\delta}_2\left(D\hat{u}\left(\xi, z,s\right)+\mathrm{i}\xi \hat{w}\left(\xi, z,s\right)\right). $$
(52)
Thus, using (45)–(47) in Eqs. (50)–(52) gives
$$ {\hat{t}}_{11}\left(\xi, z,s\right)= i\xi {\sum}_{\mathrm{j}=1}^3{A}_j{e}^{-{\lambda}_jz}+{\delta}_4\mathrm{D}{\sum}_{j=1}^3{d}_j{A}_j{e}^{-{\lambda}_{\mathrm{j}}z}-\left[1+{a}_1{\xi}^2-{a}_3{D}^2\right]{\sum}_{j=1}^3{l}_{\mathrm{j}}{A}_j{e}^{-{\lambda}_jz}, $$
(53)
$$ {\hat{t}}_{33}\left(\xi, z,s\right)={\delta}_4 i\xi {\sum}_{\mathrm{j}=1}^3{A}_j{e}^{-{\lambda}_jz}+{\updelta}_3\mathrm{D}{\sum}_{j=1}^3{d}_j{A}_j{e}^{-{\lambda}_{\mathrm{j}}z}-\frac{\beta_3}{\beta_1}\left[1+{a}_1{\xi}^2-{a}_3{D}^2\right]{\sum}_{j=1}^3{l}_{\mathrm{j}}{A}_j{e}^{-{\lambda}_jz}, $$
(54)
$$ {\hat{t}}_{13}\left(\xi, z,s\right)={\delta}_2\left(D{\sum}_{\mathrm{j}=1}^3{A}_j{e}^{-{\lambda}_jz}+\mathrm{i}\xi {\sum}_{j=1}^3{d}_j{A}_j{e}^{-{\lambda}_{\mathrm{j}}z}\right). $$
(55)
which on further simplification gives
$$ {\hat{t}}_{11}\left(\xi, z,s\right)=\left(\left( i\xi -{\delta}_4{\lambda}_1{d}_1-\left(\left[1+{a}_1{\xi}^2-{a}_3{\lambda_1}^2\right]{l}_1\right)\right){e}^{-{\lambda}_1z}{A}_1+\left( i\xi -{\delta}_4{\lambda}_2{d}_2-\left(\left[1+{a}_1{\xi}^2-{a}_3{\lambda_2}^2\right]{l}_2\right)\right){e}^{-{\lambda}_2z}{A}_2+\left( i\xi -{\delta}_4{\lambda}_3{d}_3-\left(\left[1+{a}_1{\xi}^2-{a}_3{\lambda_3}^2\right]{l}_3\right)\right){e}^{-{\lambda}_3z}{A}_3\right), $$
(56)
$$ {\hat{t}}_{33}\left(\xi, z,s\right)=\left(\left({\delta}_4 i\xi -{\delta}_3{\lambda}_1{d}_1-\frac{\beta_3}{\beta_1}\left(\left[1+{a}_1{\xi}^2-{a}_3{\lambda_1}^2\right]{l}_1\right)\right){e}^{-{\lambda}_1z}{A}_1+\left({\delta}_4 i\xi -{\delta}_3{\lambda}_2{d}_2-\frac{\beta_3}{\beta_1}\left(\left[1+{a}_1{\xi}^2-{a}_3{\lambda_2}^2\right]{l}_2\right)\right){e}^{-{\lambda}_2z}{A}_2+\left({\delta}_4 i\xi -{\delta}_3{\lambda}_3{d}_3-\frac{\beta_3}{\beta_1}\left(\left[1+{a}_1{\xi}^2-{a}_3{\lambda_3}^2\right]{l}_3\right)\right){e}^{-{\lambda}_3z}{A}_3\right), $$
(57)
$$ {\hat{t}}_{13}\left(\xi, z,s\right)=\left(-{\lambda}_1+ i\xi {d}_1\right){\delta}_2{A}_1{e}^{-{\lambda}_1z}+\left(-{\lambda}_2+ i\xi {d}_2\right){\delta}_2{A}_2{e}^{-{\lambda}_2z}+\left(-{\lambda}_3+ i\xi {d}_3\right){\delta}_2{A}_3{e}^{-{\lambda}_3z}. $$
(58)
which can be further simplified as
$$ {\hat{t}}_{11}\left(\xi, z,s\right)=\left({S}_1{A}_1+{S}_2{A}_2+{S}_3{A}_3\right) $$
(59)
$$ {\hat{t}}_{33}\left(\xi, z,s\right)=\left({M}_1{A}_1+{M}_2{A}_2+{M}_3{A}_3\right) $$
(60)
$$ {\hat{t}}_{13}\left(\xi, z,s\right)=\left({N}_1{A}_1+{N}_2{A}_2+{N}_3{A}_3\right) $$
(61)
where
$$ {S}_j=\left\{ i\xi -{\delta}_4{d}_j{\lambda}_j-{l}_j\left[\left(1+{a}_1{\xi}^2\right)-{a}_3{\lambda}_j^2\right]\right\}{e}^{-{\lambda}_jz} $$
(62)
$$ {N}_j={\delta}_2\left(-{\lambda}_j+ i\xi {d}_j\right){e}^{-{\lambda}_jz}, $$
(63)
$$ {M}_j=\left\{{\delta}_4 i\xi -{\delta}_3{d}_j{\lambda}_j-\frac{\beta_3}{\beta_1}{l}_j\left[\left(1+{a}_1{\xi}^2\right)-{a}_3{\lambda}_j^2\right]\right\}{e}^{-{\lambda}_jz}, $$
(64)