The aforementioned model equations are solved applying first the Hankel transform, followed by Laplace transform, the decoupling technique and a conventional ODE solution technique. The Hankel transform of
\(c(\rho ,x,\tau )\) is defined as (c.f. Carslaw and Jaeger
1953; Chen et al.
2011; Crank
1975; Sneddon
1972)
$$\begin{aligned} c_{H,j}(\lambda _{n},x,\tau )&=H[c_j(\rho ,x,\tau )] \\ & =\int \limits _{0}^{1} c_j(\rho ,x,\tau )J_{0}\left( \lambda _{n}\rho \right) \rho d\rho ,\quad j=1,2. \end{aligned}$$
(28)
After applying the above transformation on Eqs. (
17) and (
19) with respect to coordinate
\(\rho\), we obtain
$$\begin{aligned} \frac{\partial c_{H,1}}{\partial \tau }+ \frac{\partial c_{H,1}}{\partial x}= & {} \frac{1}{Pe_{z}}\frac{\partial ^2 c_{H,1}}{\partial x^2}-\frac{\lambda _{n}^2}{Pe_{\rho }}c_{H,1} \\&-\xi \left( c_{H,1}-c_{pH,1}|_{\rho _p=1}\right) \,, \end{aligned}$$
(29)
$$\begin{aligned} \frac{\partial c_{H,2}}{\partial \tau }+ \frac{\partial c_{H,2}}{\partial x}= & {} \frac{1}{Pe_{z,T}}\frac{\partial ^2 c_{H,2}}{\partial x^2}-\frac{\lambda _{n}^2}{Pe_{\rho ,T}}c_{H,2} \\&-\xi _T \left( c_{H,2}-c_{pH,2}|_{\rho _p=1}\right) \,. \end{aligned}$$
(30)
Here,
\(c_{H,j}(\lambda _{n},x,\tau )\) and
\(c_{pH,j}(\lambda _{n},x,\tau )\) denote the zeroth-order finite Hankel transforms of
\(c_j(\rho ,x,\tau )\) and
\(c_{p,j}(\rho _p,\rho ,x,\tau )\), respectively. Thus, the Hankel transformation eliminates the radial derivatives from the partial differential equations (PDEs). Next, we apply the Laplace transform defined as (c.f Sneddon
1972)
$$\begin{aligned} \bar{c}_{H,j}(\lambda _n,x,s)=\int \limits _{0}^{\infty }{e^{-st}}c_{H,j}(\lambda _n,x,\tau )dt, \quad t \ge 0, \end{aligned}$$
(31)
on Eqs. (
29) and (
30) with respect to
\(\tau\) for zero initial concentration of the solute to obtain
$$\begin{aligned}&\frac{1}{Pe_{z}}\frac{\partial ^2 \bar{c}_{H,1}}{\partial x^2}-\frac{\partial \bar{c}_{H,1}}{\partial x}-\left( s+\frac{\lambda _{n}^2}{Pe_{\rho }}\right) \bar{c}_{H,1} \\&\quad -\xi \left( \bar{c}_{H,1}-\bar{c}_{pH,1}|_{\rho _p=1}\right) =0\,, \end{aligned}$$
(32)
$$\begin{aligned}&\frac{1}{Pe_{z,T}}\frac{\partial ^2 \bar{c}_{H,2}}{\partial x^2}-\frac{\partial \bar{c}_{H,2}}{\partial x}-\left( s+\frac{\lambda _{n}^2}{Pe_{\rho ,T}}\right) \bar{c}_{H,2} \\&\quad -\xi _T \left( \bar{c}_{H,2}-\bar{c}_{pH,2}|_{\rho _p=1}\right) =0\,. \end{aligned}$$
(33)
Subsequently, applying the Hankel and Laplace transformations on the Danckwerts boundary conditions given in Eqs. (
27a) and (
27c), we obtain
$$\begin{aligned}&c_{H,j}(\lambda _{n},x=0,\tau )-\frac{1}{Pe_j}\frac{\partial c_{H,j}(\lambda _{n},x=0,\tau )}{\partial x} \\&\quad =\left\{ \begin{array}{ll} c^{\mathrm{inj}}_jF(\lambda _{n})\,,\quad {\text{if}}\,\,\, 0\le \tau \le \tau _{\mathrm{inj}}\,, \\ 0\,,\quad \qquad \quad \,\,\, {\text{if}}\,\, \tau >\tau _{\mathrm{inj}}\,, \end{array}\right. \end{aligned}$$
(34)
$$\begin{aligned}&\left. \frac{\partial c_{H,j}(\lambda _{n},x,\tau )}{\partial x}\right| _{x=1} = 0\,, \qquad j=1,2, \end{aligned}$$
(35)
and
$$\begin{aligned}&\bar{c}_{H,j}(\lambda _{n},x=0,s)-\frac{1}{Pe_j}\frac{\partial \bar{c}_{H,j}(\lambda _{n},x=0,s)}{\partial x} \\&\quad =\frac{c^{\mathrm{inj}}_jF(\lambda _{n})}{s}\left( 1-e^{-s\tau _{\mathrm{inj}}}\right) \,, \end{aligned}$$
(36)
$$\begin{aligned}&\left. \frac{\partial \bar{c}_{H,j}}{\partial x}\right| _{x=1} = 0\,, \qquad j=1,2. \end{aligned}$$
(37)
Here,
\(\bar{c}_{H,j}\) denotes the concentration after Hankel and Laplace transformation for
\(j=1\) and temperature for
\(j=2\), respectively.
Injecting through the inner region,
\(F(\lambda _{n})\) is given as
$$\begin{aligned}&F(\lambda _{n})=\left\{ \begin{array}{ll} \frac{\tilde{\rho }^2}{2}\,,\qquad \qquad \quad \,\, {\text{if}}\,\,\,\, \lambda _{n}=0\,,\\ \frac{\tilde{\rho }}{\lambda _{n}}J_{1}\left( \lambda _{n}\tilde{\rho }\right) \,,\qquad {\text{if}}\,\,\, \lambda _{n}\ne 0\,, \end{array}\right. \end{aligned}$$
(38)
and injecting through the outer region,
\(F(\lambda _{n})\) is given as
$$\begin{aligned}&F(\lambda _{n})=\left\{ \begin{array}{ll} \left( \frac{1}{2}-\frac{\tilde{\rho }^2}{2}\right) \,,\qquad \quad \,\, {\text{if}}\,\,\,\, \lambda _{n}=0\,,\\ -\frac{\tilde{\rho }}{\lambda _{n}}J_{1}\left( \lambda _{n}\tilde{\rho }\right) \,,\qquad {\text{if}}\,\,\, \lambda _{n}\ne 0\,. \end{array}\right. \end{aligned}$$
(39)
Next, using Eq. (
22) in Eqs. (
18) and (
20) and after rephrasing, we obtain
$$\begin{aligned} \eta \frac{\partial ^2}{\partial \rho _p^2} (\rho _pc_{p,1})= \, &a^*_1\frac{\partial }{\partial \tau } (\rho _pc_{p,1}) +a^*_2\frac{\partial }{\partial \tau } (\rho _pc_{p,2}), \end{aligned}$$
(40)
$$\begin{aligned} \eta _T\frac{\partial ^2}{\partial \rho _p^2} (\rho _pc_{p,2})= \, &a^{*}_3\frac{\partial }{\partial \tau } (\rho _pc_{p,1}) +a^{*}_4\frac{\partial }{\partial \tau } (\rho _pc_{p,2}), \end{aligned}$$
(41)
where
$$\begin{aligned} a^*_1= \, &\epsilon _p+a^{\mathrm{ref}}(1-\epsilon _p),\quad a^{*}_4=\epsilon _p+(1-\epsilon _p)\frac{\rho ^Sc^S_p}{\rho ^Lc^L_p} \\&+(1-\epsilon _p)\left( a^{\mathrm{ref}}c^{\mathrm{ref}}\frac{(\Delta H_{\mathrm{A}})^2}{R_g\rho ^Lc^L_p(T^{\mathrm{ref}})^2}\right) , \\ a^*_2= \, &(1-\epsilon _p)\left( a^{\mathrm{ref}}c^{\mathrm{ref}}\frac{\Delta H_{\mathrm{A}}}{R_g(T^{\mathrm{ref}})^2}\right) , \quad \\ a^{*}_3= \, &(1-\epsilon _p)\left( a^{\mathrm{ref}}\frac{\Delta H_{\mathrm{A}}}{\rho ^Lc^L_p}\right) . \end{aligned}$$
(42)
After applying the Hankel and Laplace transforms on Eqs. (
40) and (
41), we obtain
$$\begin{aligned} \frac{d^2}{d\rho _p^2}(\rho _p {\bar{c}}_{pH,1})&=\alpha _1(s)\rho _p \bar{c}_{pH,1}+\alpha _2(s)\rho _p {\bar{c}}_{pH,2}\,, \end{aligned}$$
(43)
$$\begin{aligned} \frac{d^2}{d\rho _p^2}(\rho _p {\bar{c}}_{pH,2})&=\alpha _3(s)\rho _p \bar{c}_{pH,1}+\alpha _4(s)\rho _p {\bar{c}}_{pH,2}\,, \end{aligned}$$
(44)
where,
\(\alpha _1(s)=\frac{a^*_1s}{\eta }\),
\(\alpha _2(s)=\frac{a^*_2s}{\eta }\),
\(\alpha _3(s)=\frac{a^{*}_3s}{\eta _T}\), and
\(\alpha _4(s)=\frac{a^{*}_4s}{\eta _T}\).
Representing Eqs. (
43) and (
44) in matrix notation, we obtain
$$\begin{aligned} {\frac{d^2}{d \rho _p^2}}\left( \begin{array}{c} \rho _p\bar{c}_{pH,1} \\ \rho _p\bar{c}_{pH,2} \\ \end{array} \right)&= \left( \begin{array}{cc} \alpha _1(s) &{} \alpha _2(s) \\ \alpha _3(s) &{} \alpha _4(s) \\ \end{array} \right) \left( \begin{array}{c} \rho _p\bar{c}_{pH,1} \\ \rho _p\bar{c}_{pH,2} \\ \end{array} \right) . \end{aligned}$$
(45)
We have a coupled system of equations above and the eigen-decomposition technique is used to decouple the system in order to find the solutions by using the boundary conditions in Eq. (
27d). Hence, Let
B represent the coefficient matrix of the system as
$$\begin{aligned} B=\left( \begin{array}{cc} \alpha _1(s) &{} \alpha _2(s) \\ \alpha _3(s) &{} \alpha _4(s) \\ \end{array} \right) . \end{aligned}$$
(46)
The above matrix has two distinct eigenvalues which are given as
$$\begin{aligned} \lambda _{1,2} =\,&\frac{1}{2}(\alpha _1(s)+\alpha _4(s)) \\&\pm \frac{1}{2}\sqrt{(\alpha _1(s)-\alpha _4(s))^2+4\alpha _2(s)\alpha _3(s)}\,, \end{aligned}$$
(47)
and the corresponding distinct eigenvectors
$$\begin{aligned} x_1=\left[ \begin{array}{c} \lambda _1-\alpha _4(s)\\ \alpha _3(s) \\ \end{array} \right] , \quad x_2=\left[ \begin{array}{c} \lambda _2-\alpha _4(s)\\ \alpha _3(s) \\ \end{array} \right] . \end{aligned}$$
(48)
Therefore, based on the above eigenvalues, the transformation matrix
A is obtained as
$$\begin{aligned} A = \left( \begin{array}{cc} \lambda _1-\alpha _4(s) &{} \lambda _2-\alpha _4(s) \\ \alpha _3(s) &{} \alpha _3(s)\\ \end{array} \right) . \end{aligned}$$
(49)
Using the matrix
A, we can formulate the following linear transformation as
$$\begin{aligned} \left[ \begin{array}{c} \bar{c}_{pH,1} \\ \bar{c}_{pH,2} \\ \end{array} \right] =\left( \begin{array}{cc} \lambda _1-\alpha _4(s) &{} \lambda _2-\alpha _4(s) \\ \alpha _3(s) &{} \alpha _3(s)\\ \end{array} \right) \left[ \begin{array}{c} \bar{b}_{pH,1} \\ \bar{b}_{pH,2} \\ \end{array} \right] . \end{aligned}$$
(50)
We use the linear transformation above on Eq. (
45) to obtain
$$\begin{aligned} {\frac{d^2}{d \rho _p^2}}\left( \begin{array}{c} \rho _p\bar{b}_{pH,1} \\ \rho _p\bar{b}_{pH,2} \\ \end{array} \right)&= \left( \begin{array}{cc} \lambda _1 &{} 0 \\ 0 &{} \lambda _2 \\ \end{array} \right) \left( \begin{array}{c} \rho _p\bar{b}_{pH,1} \\ \rho _p\bar{b}_{pH,2} \\ \end{array} \right) . \end{aligned}$$
(51)
Now, we have a system of ODEs in Eq. (
51) with the following explicit solutions
$$\begin{aligned} \bar{b}_{pH,1} = \, &\frac{1}{\rho _p}[C_1e^{\sqrt{\lambda _1}\rho _p}+C_2e^{-\sqrt{\lambda _1}\rho _p}], \\ \bar{b}_{pH,2} = \, &\frac{1}{\rho _p}[D_1e^{\sqrt{\lambda _2}\rho _p}+D_2e^{-\sqrt{\lambda _2}\rho _p}]. \end{aligned}$$
(52)
From Eqs. (
27d) and (
50), we obtain the boundary conditions
$$\begin{aligned}&\left. {\frac{\partial \bar{b}_{pH,1}}{\partial \rho _p}}\right| _{\rho _p=0}=0 , \quad \left. {\frac{\partial \bar{b}_{pH,2}}{\partial \rho _p}}\right| _{\rho _p=0}=0. \end{aligned}$$
(53)
By using the above boundary conditions, we get from Eq. (
52)
\(C_1=-C_2\) and
\(D_1=-D_2\). Therefore, Eq. (
52) reduces to
$$\begin{aligned} \bar{b}_{pH,1} =\frac{2C_1}{\rho _p}\sinh (\sqrt{\lambda _1}\rho _p), \qquad \bar{b}_{pH,2} =\frac{2D_1}{\rho _p}\sinh (\sqrt{\lambda _2}\rho _p). \end{aligned}$$
(54)
By using the values of
\(\bar{b}_{pH,1}\) and
\(\bar{b}_{pH,1}\) in the transformation given by Eq. (
50), we obtain the following results
$$\begin{aligned} \bar{c}_{pH,1}= \, &\left( \lambda _1-\alpha _4(s)\right) \left[ \frac{2C_1}{\rho _p}\sinh (\sqrt{\lambda _1}\rho _p)\right] \\&+ \left( \lambda _2-\alpha _4(s)\right) \left[ \frac{2D_1}{\rho _p}\sinh (\sqrt{\lambda _2}\rho _p)\right] , \\ \bar{c}_{pH,2}= \, &\alpha _3(s)\left[ \frac{2C_1}{\rho _p}\sinh (\sqrt{\lambda _1}\rho _p)\right] \\&+\alpha _3(s)\left[ \frac{2D_1}{\rho _p}\sinh (\sqrt{\lambda _2}\rho _p)\right] . \end{aligned}$$
(55)
By using the boundary conditions of Eq. (
27d) at
\(\rho _p=1\) in Eq. (
55), we obtain
$$\begin{aligned} C_1= \, &-\frac{1}{2\sinh (\sqrt{\lambda _1})}\left[ \frac{\phi _1(s)}{\upsilon (s)}\bar{c}_{H,1} -\frac{\phi _2(s)}{\upsilon (s)}\bar{c}_{H,2}\right] \,, \end{aligned}$$
(56)
$$\begin{aligned} D_1= \, &\frac{1}{2\sinh (\sqrt{\lambda _2})}\left[ \frac{\phi _3(s)}{\upsilon (s)}\bar{c}_{H,1} -\frac{\phi _4(s)}{\upsilon (s)}\bar{c}_{H,2}\right] , \end{aligned}$$
(57)
where
$$\begin{aligned} \phi _1(s)= \, &\zeta [\zeta _T-1 +\sqrt{\lambda _1}\coth (\sqrt{\lambda _1})], \\ \phi _2(s)= \, &\frac{(\lambda _1-\alpha _4(s))}{\alpha _3(s)}\zeta _T[\zeta -1 +\sqrt{\lambda _1}\coth (\sqrt{\lambda _1})], \\ \phi _3(s)= \, &\zeta [\zeta _T-1 +\sqrt{\lambda _2}\coth (\sqrt{\lambda _2})], \\ \phi _4(s)= \, &\frac{(\lambda _2-\alpha _4(s))}{\alpha _3(s)}\zeta _T[\zeta -1 +\sqrt{\lambda _2}\coth (\sqrt{\lambda _2})], \\ \upsilon (s)= \, &(\lambda _2-\alpha _4(s))[\zeta -1 +\sqrt{\lambda _2}\coth (\sqrt{\lambda _2})][\zeta _T-1 \\&+\sqrt{\lambda _1}\coth (\sqrt{\lambda _1})] \\&-(\lambda _1-\alpha _4(s))[\zeta -1 +\sqrt{\lambda _1}\coth (\sqrt{\lambda _1})][\zeta _T-1 \\&+\sqrt{\lambda _2}\coth (\sqrt{\lambda _2})]. \end{aligned}$$
(58)
By using these values in Eq. (
55), we get
$$\begin{aligned}\left. {\bar{c}_{pH,1}}\right| _{\rho _p=1}&=\left[ (\lambda _2-\alpha _4(s))\frac{\phi _3(s)}{\upsilon (s)}\right. \\&\quad \left. -(\lambda _1-\alpha _4(s))\frac{\phi _1(s)}{\upsilon (s)} \right] \bar{c}_{H,1} \\&\quad +\left[ {{(\lambda _1-\alpha _4(s))}\frac{\phi _2(s)}{\upsilon (s)}-{(\lambda _2-\alpha _4(s))}\frac{\phi _4(s)}{\upsilon (s)} }\right] \bar{c}_{H,2}, \end{aligned}$$
(59)
$$\begin{aligned}&\left. {\bar{c}_{pH,2}}\right| _{\rho _p=1}=\alpha _3(s)\left[ {\frac{\phi _3(s)}{\upsilon (s)}-\frac{\phi _1(s)}{\upsilon (s)} }\right] \bar{c}_{H,1} \\&\quad +\alpha _3(s)\left[ {\frac{\phi _2(s)}{\upsilon (s)}-\frac{\phi _4(s)}{\upsilon (s)} }\right] \bar{c}_{H,2}. \end{aligned}$$
(60)
Putting the values of Eqs. (
59) and (
60) in Eqs. (
32) and (
33), we get the system of equations
$$\begin{aligned}&{\frac{d^2}{d x^2}} { \begin{pmatrix} \bar{c}_{H,1} \\ \bar{c}_{H,2} \end{pmatrix}}-\left( \begin{array}{c} {Pe_{z}} \\ {Pe_{z,T}} \\ \end{array} \right) \frac{d}{dx}\left( \begin{array}{c} \bar{c}_{H,1} \\ \bar{c}_{H,2} \\ \end{array} \right) \\&\quad =\left( \begin{matrix} \gamma _1(s,\lambda _n) &{} \psi _2(s)\\ \psi _3(s) &{} \gamma _2(s,\lambda _n) \end{matrix} \right) { \begin{pmatrix} \bar{c}_{H,1} \\ \bar{c}_{H,2} \end{pmatrix} }. \end{aligned}$$
(61)
Here,
$$\begin{aligned} \gamma _1(s,\lambda _n)= \, &Pe_{z}\left( s+\frac{\lambda _n^2}{Pe_{\rho }}\right) +\psi _1(s), \quad \\ \gamma _2(s,\lambda _n)= \, &Pe_{z,T}\left( s+\frac{\lambda _n^2}{Pe_{\rho ,T}}\right) +\psi _4(s) \end{aligned}$$
(62)
and
$$\begin{aligned} \psi _1(s)&=Pe_{z}\xi \left[ 1-{(\lambda _2-\alpha _4(s))\frac{\phi _3(s)}{\upsilon (s)}+(\lambda _1-\alpha _4(s))\frac{\phi _1(s)}{\upsilon (s)} }\right] \,, \end{aligned}$$
(63)
$$\begin{aligned} \psi _2(s)&=-Pe_{z}\xi \left[ {(\lambda _1-\alpha _4(s))}\frac{\phi _4(s)}{\upsilon (s)}-{(\lambda _2-\alpha _4(s))}\frac{\phi _2(s)}{\upsilon (s)}\right] \,,\end{aligned}$$
(64)
$$\begin{aligned} \psi _3(s)&=-\alpha _3(s)Pe_{z,T}\xi _{T}\left[ \frac{\phi _3(s)}{\upsilon (s)}-\frac{\phi _1(s)}{\upsilon (s)}\right] \,,\end{aligned}$$
(65)
$$\begin{aligned} \psi _4(s)&=Pe_{z,T}\xi _{T}\left[ 1+\alpha _3(s)\left( {\frac{\phi _2(s)}{\upsilon (s)}-\frac{\phi _4(s)}{\upsilon (s)} }\right) \right] . \end{aligned}$$
(66)
In a similar manner, we decouple the system in Eq. (
61) by using the following transformation
$$\begin{aligned} \left[ \begin{array}{c} \bar{c}_{H,1} \\ \bar{c}_{H,2} \\ \end{array} \right] =\left( \begin{array}{cc} \lambda _3-\gamma _2(s,\lambda _n) &{} \lambda _4-\gamma _2(s,\lambda _n) \\ \psi _3(s) &{} \psi _3(s)\\ \end{array} \right) \left[ \begin{array}{c} \bar{b}_{H,1} \\ \bar{b}_{H,2} \\ \end{array} \right] , \end{aligned}$$
(67)
where,
$$\begin{aligned} \lambda _{3,4}= \, &\frac{1}{2}(\gamma _1(s,\lambda _n)+\gamma _2(s,\lambda _n)) \\&\pm \frac{1}{2}\sqrt{(\gamma _1(s,\lambda _n)-\gamma _2(s,\lambda _n))^2+4\psi _2(s)\psi _3(s)}\,. \end{aligned}$$
(68)
Thus, we get the form
$$\begin{aligned} {\frac{d^2}{d x^2}} { \begin{pmatrix} \bar{b}_{H,1} \\ \bar{b}_{H,2} \end{pmatrix}}-\left( \begin{array}{c} {Pe_{z}} \\ {Pe_{z,T}} \\ \end{array} \right) \frac{d}{dx}\left( \begin{array}{c} \bar{b}_{H,1} \\ \bar{b}_{H,2} \\ \end{array} \right) =\left( \begin{matrix} \lambda _3 &{} 0\\ 0 &{} \lambda _4 \end{matrix} \right) { \begin{pmatrix} \bar{b}_{H,1} \\ \bar{b}_{H,2} \end{pmatrix} }. \end{aligned}$$
(69)
The solutions of Eq. (
69) are given as
$$\begin{aligned} \bar{b}_{H,1} (x,s)= \, &C_3e^{m_1x}+C_4e^{m_2x},\quad \bar{b}_{H,2}(x,s) \\ = \, &D_3e^{n_1x}+D_4e^{n_2x}. \end{aligned}$$
(70)
Here,
$$\begin{aligned} m_{1,2} = \, &\frac{1}{2}{Pe_z} \pm {\frac{1}{2}\sqrt{Pe_z^2-4\lambda _{3}}}\,, \quad \\ n_{1,2} = \, &\frac{1}{2}{Pe_{z,}T} \pm {\frac{1}{2}\sqrt{Pe_{z,T}^2-4\lambda _{4}}}. \end{aligned}$$
(71)
In Eq. (
71),
\(m_1\) and
\(n_1\) are obtained by choosing the plus sign and
\(m_2\) and
\(n_2\) are obtained by considering the minus sign. Next, we use the two types of axial boundary conditions to find the integration constants
\(C_{3}\),
\(C_4\),
\(D_3\) and
\(D_{4}\).
By utilizing the transformation in Eq. (
67), we can transform the boundary conditions in Eq. (
70). After following a similar procedure of analytical solution derivation discussed in Qamar et al. (
2017), we obtain the following Hankel-Laplace domains solutions for Dirichlet boundary conditions from Eqs. (
36) and (
37) for
\(Pe_i\rightarrow \infty\) and
\(x=\infty\) as
$$\begin{aligned}&\overline{c}_{H,1}(\lambda _n,x,s)=\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) c_{1}^{\mathrm{inj}}F(\lambda _{n})}{(\lambda _3-\lambda _4)s}\bigg [(\lambda _3-\gamma _{2}(s,\lambda _n))e^{m_2x}- (\lambda _4-\gamma _{2}(s,\lambda _n))e^{n_2x}\bigg ] \\&\quad +\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) c_{2}^{\mathrm{inj}}F(\lambda _{n})(\lambda _3-\gamma _2(s,\lambda _n))(\lambda _4-\gamma _2(s,\lambda _n))}{\psi _3(s)(\lambda _3-\lambda _4)s}\bigg [e^{m_2x}- e^{n_2x}\bigg ], \end{aligned}$$
(72)
$$\begin{aligned} \overline{c}_{H,2}(\lambda _n,x,s)&=\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) \psi _3(s)c_{1}^{\mathrm{inj}}F(\lambda _{n})}{(\lambda _3-\lambda _4)s}\bigg [e^{m_2x}- e^{n_2x}\bigg ] \\&\quad +\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) c_{2}^{\mathrm{inj}}F(\lambda _{n})}{(\lambda _3-\lambda _4)s}\bigg [(\lambda _4-\gamma _{2}(s,\lambda _n))e^{m_2x}- (\lambda _3-\gamma _{2}(s,\lambda _n))e^{n_2x}\bigg ]. \end{aligned}$$
(73)
Analytical Laplace and Hankel inversions are not possible for the above equations. Thus, to get back the solutions in the actual coordinates, we use the numerical Laplace and Hankel inversions (Durbin
1974). In the Appendix
1 of this manuscript, we have presented a detailed derivation of the Durbin method (Durbin
1974). The Matlab software was used to implement this technique.
Next, applying the transformation in Eq. (
67) in a similar manner and following the solution procedure presented in Qamar et al. (
2017), we obtain the following Hankel-Laplace domains solutions for the Danckwerts boundary conditions
$$\begin{aligned}&\overline{c}_{H,1}(\lambda _n,x,s) \\&\quad =\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) c_{1}^{\mathrm{inj}}F(\lambda _n)}{(\lambda _3-\lambda _4)s}\bigg [\nu _{1}(\lambda _3-\gamma _{2}(s,\lambda _n))(m_1e^{m_2x}-m_2e^{m_1(x-1)+m_2})\bigg ] \\&\qquad -\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) c_{1}^{\mathrm{inj}}F(\lambda _n)}{(\lambda _3-\lambda _4)s}\bigg [ \nu _{2}(\lambda _4-\gamma _{2}(s,\lambda _n))(n_1e^{n_2x}-n_2e^{n_1(x-1)+n_2})\bigg ] \\&\qquad -\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) c_{2}^{\mathrm{inj}}F(\lambda _n)(\lambda _3-\gamma _2(s,\lambda _n))(\lambda _4-\gamma _2(s,\lambda _n))}{\psi _3(s)(\lambda _3-\lambda _4)s}\bigg [\nu _{1}(m_1e^{m_2x}-m_2e^{m_1(x-1)+m_2})\bigg ] \\&\qquad -\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) c_{2}^{\mathrm{inj}}F(\lambda _n)(\lambda _3-\gamma _2(s,\lambda _n))(\lambda _4-\gamma _2(s,\lambda _n))}{\psi _3(s)(\lambda _3-\lambda _4)s}\bigg [ \nu _{2}(n_1e^{n_2x}-n_2e^{n_1(x-1)+n_2})\bigg ], \end{aligned}$$
(74)
$$\begin{aligned}&\overline{c}_{H,2}(\lambda _n,x,s) \\&\quad =\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) \psi _3(s)c_{1}^{\mathrm{inj}}F(\lambda _n)}{(\lambda _3-\lambda _4)s}\bigg [\nu _{1}(m_1e^{m_2x}-m_2e^{m_1(x-1)+m_2})\bigg ]\\ & \qquad - \frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) \psi _3(s)c_{1}^{\mathrm{inj}}F(\lambda _n)}{(\lambda _3-\lambda _4)s}\bigg [ \nu _{2}(n_1e^{n_2x}-n_2e^{n_1(x-1)+n_2})\bigg ] \\&\qquad +\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) c_{2}^{\mathrm{inj}}F(\lambda _n)}{(\lambda _3-\lambda _4)s}\bigg [\nu _{1}(\lambda _4-\gamma _{2}(s,\lambda _n))(m_2e^{m_1(x-1)+m_2}-m_1e^{m_2x})\bigg ] \\&\qquad +\frac{\left( {1-e^{-s\tau _{\mathrm{inj}}}}\right) c_{2}^{\mathrm{inj}}F(\lambda _n)}{(\lambda _3-\lambda _4)s}\bigg [ \nu _{2}(\lambda _3-\gamma _{2}(s,\lambda _n))(n_1e^{n_2x}-n_2e^{n_1(x-1)+n_2})\bigg ]. \end{aligned}$$
(75)
Here,
$$\begin{aligned} \nu _{1}= \, &\frac{{Pe_{z}}}{m_1(Pe_{z}-m_2)+m_2(m_1-Pe_{z})e^{m_2-m_1}},\end{aligned}$$
(76)
$$\begin{aligned} \nu _{2}= \, &\frac{{Pe_{z,T}}}{n_1(Pe_{z,T}-n_2)+n_2(n_1-Pe_{z,T})e^{n_2-n_1}}. \end{aligned}$$
(77)
To get back solutions in the actual coordinates, we again apply the numerical Laplace and Hankel inversions to the above equations (Durbin
1974) as discussed in the Appendix
1 of this manuscript.