I have been solving a coupled PDE system analytically and I need to find the inverse Laplace transform of $(1)$ and get $T(x,y)$. $s$ is the Laplace domain variable and $\alpha, \beta, \gamma, T_{fi}, A , d$ are constants.
$$ \mathcal{L_x}(T)=\frac{As(s+\alpha) - \beta T_{fi}}{s(s^2 - \beta + \alpha s)}+ \frac{\gamma (e^{\sigma y}+e^{\sigma(2d-y)})}{s\sigma (1-e^{2\sigma d})} \tag 1 $$ where
$$\sigma = \sqrt{\frac{\beta s - \alpha s^2 - s^3}{s+\alpha}}$$
I have tried the Wolfram alpha widget but it fails to do the job. The problem comes especially with the second term. Can anyone suggest a way to handle the inverse transformation of $(1)$ ?
ORIGINS
I have been trying to solve $(1.1)$, and the Laplace transform mentioned in equation $(1)$ comes from $(1.1)$ subjected to the given boundary conditions.
The term $A$ is $T(0,y)$, which is temporarily being treated as a constant $$ \nabla^2 T - \beta T + \beta\Bigg[\alpha e^{-\alpha x} \Bigg(\int_0^x e^{\alpha s}T(s,y)\mathrm{d}s+\frac{T_{fi}}{\alpha}\Bigg)\Bigg]=0 \tag {1.1} $$ (1.1) is dictated by the following boundary conditions: $$ \frac{\partial T}{\partial x} \vert_{x=0} = \frac{\partial T}{\partial x} \vert_{x=L} = \frac{\partial T}{\partial y} \vert_{y=d} = 0 , \frac{\partial T}{\partial y} \vert_{y=0}=\gamma $$
Intermediate steps between $(1.1)$ to $(1)$
Taking Laplace transform of $(1.1)$ w.r.t. $x$ $$ s^2 \mathcal{L_x}T(x,y) - \color{red}{sT(0,y)} - \color{green}{\frac{\partial T(0,y)}{\partial x}} + \mathcal{L_x}\Bigg(\frac{\partial^2 T}{\partial y^2}\Bigg)-\\ \beta \mathcal{L_x}T(x,y) + \frac{\alpha \beta}{\alpha +s}\mathcal{L_x}T(x,y) + \frac{\beta}{\beta +s} T_{fi} = 0 \tag 2 $$ $T(0,y)$ is an unknown and we denote it with the letter $A$ for the rest of this analysis. $$ \mathcal{L_x}\Bigg(\frac{\partial^2 T}{\partial y^2}\Bigg)=\frac{\partial^2}{\partial y^2}\mathcal{L_x}(T(x,y)) $$ Equation $(2)$ becomes $$ \frac{\partial^2}{\partial y^2}\mathcal{L_x}(T)+\Bigg(s^2 - \beta + \frac{\alpha \beta}{\alpha +s}\Bigg)\mathcal{L_x}(T)-sA+\frac{\beta T_{fi}}{\alpha +s}=0 \tag 3 $$ Solve $(3)$ (an O.D.E) to find $\mathcal{L_x}(T)$ $\color{Blue}{\Rightarrow}$ Use $y$ B.C.(s) to determine the constants $\color{Blue}{\Rightarrow}$ Find $\color{black}{T=\mathcal{L_x^{-1}}(T)}$ $\color{Blue}{\Rightarrow}$ Use the $x=L$ B.C. to determine $A$
Solving $(3)$ gives $$ \mathcal{L_x}(T)=\frac{As(s+\alpha) - \beta T_{fi}}{s(s^2 - \beta + \alpha s)}+C_1 e^{\sigma y} + C_2 e^{-\sigma y} \tag 4 $$ where, $$\sigma = \sqrt{\frac{\beta s - \alpha s^2 - s^3}{s+\alpha}}$$ The $y$ boundary conditions become:
$\frac{\partial T}{\partial y} \vert_{y=d} = 0 , \frac{\partial T}{\partial y} \vert_{y=0}=\gamma \color{Blue}{\Rightarrow} \frac{\partial \mathcal{L_x}(T)}{\partial y} \vert_{y=d} = 0,\frac{\partial \mathcal{L_x}(T)}{\partial y} \vert_{y=0}=\gamma$
Utilizing these conditions $C_2=C_1 e^{2\sigma d}$ and $C_1=\frac{\gamma}{s\sigma (1-e^{2\sigma d})}$
Substituting in $(4)$ gives us $$ \mathcal{L_x}(T)=\frac{As(s+\alpha) - \beta T_{fi}}{s(s^2 - \beta + \alpha s)}+ \frac{\gamma (e^{\sigma y}+e^{\sigma(2d-y)})}{s\sigma (1-e^{2\sigma d})} \tag 1 $$
For separation of variables I assumed the following ansatz
$$ T(x,y)=\sum_{k=0}^{\infty}f_k(y)\cos(\frac{k\pi x}{L})=f_0(y)+\sum_{k=1}^{\infty}f_k(y)\cos(\frac{k\pi x}{L}) $$
alpha
andbeta
a stationary phase approximation might be appropriate. $\endgroup$