How to do laplace transform on time dependent domain?

Question

I want to solve \begin{align} \begin{split} \frac{\partial W(v,z)}{\partial v}&=\frac{\partial^{2} W(v,z)}{\partial z^{2}}\\ \end{split} \end{align} where \begin{align} \begin{split} W(0,z)&=\delta_{a}(z),\quad W(v,\theta(v))=0,\quad \theta(v)\leq z<\infty,\quad 0\leq v<\infty\\ \end{split} \end{align}

My aim is to solve the problem. I would like to have any standard reference talking about solving the above equation.

Currently, I use Laplace transform, which gives solving (partly)

\begin{align} \begin{split} s\hat{W}(s,z)-\delta_{a}(z)&=\frac{\partial^{2} \hat{W}(s,z)}{\partial z^{2}}\\ \end{split} \end{align}

Clearly, the above equation can be solved. Let's omit detailed derivation for the solution of $\hat{W}(s,x)$. I have tried to work out the solution for $\hat{W}(s,x)$, let say it is $Ae^{\sqrt{s}x}+Be^{-\sqrt{s}x}-\frac{1}{2s}e^{-|x-a|}$. I want to change it back by inverse Laplace transform and find corresponding $A$ and $B$ by matching the boundary condition. However, $A$ and $B$ should be constant, doing in this way may not be suitable.

I think we need to make use boundary values after integral transform. My question is: how to tackle the boundary value problem using Laplace transform?

Answer 1

Starting from your result (I didn't check it, the last term seems doubtful) : $$\hat{W}(s,z)=\hat A(s)e^{\sqrt{s}z}+\hat B(s)e^{-\sqrt{s}z}-\frac{1}{2s}e^{-|z-a|}.$$ Of course, no $x$ must appear in this equation.

The inverse Laplace transform (noted $L^{-1}$) involves convolution (noted $\star$) : $$W(x,z)=L^{-1}\left(\hat A(s)\right) \star L^{-1}\left(e^{\sqrt{s}z}\right) +L^{-1}\left(\hat B(s)\right) \star L^{-1}\left(e^{-\sqrt{s}z}\right)+ \frac12e^{-|z-a|} L^{-1}\left(\frac1s \right)$$

$$W(x,z)=A(x)\star L^{-1}\left(e^{\sqrt{s}z}\right) +B(x)\star L^{-1}\left(e^{-\sqrt{s}z}\right)+ \frac12e^{-|z-a|} L^{-1}\left(\frac1s \right)$$

$L^{-1}\left(\frac1s \right)=1$

$L^{-1}\left(e^{\sqrt{s}z}\right)=-\frac{z}{2\sqrt{\pi}x^{3/2}}\exp\left(-\frac{z^2}{4x} \right)$

$L^{-1}\left(e^{-\sqrt{s}z}\right)=\frac{z}{2\sqrt{\pi}x^{3/2}}\exp\left(-\frac{z^2}{4x} \right)$

$$W(x,z)=A(x)\star \left( -\frac{z}{2\sqrt{\pi}x^{3/2}}\exp\left(-\frac{z^2}{4x} \right)\right) +B(x)\star \left( \frac{z}{2\sqrt{\pi}x^{3/2}}\exp\left(-\frac{z^2}{4x} \right)\right)+\frac12e^{-|z-a|} $$

$$W(x,z)=C(x)\star \left( \frac{z}{2\sqrt{\pi}x^{3/2}}\exp\left(-\frac{z^2}{4x} \right)\right) +\frac12e^{-|z-a|} $$

One must not confuse the operator $\star$ with multiplication. Better, use the integral form : Eq.(1) in http://mathworld.wolfram.com/Convolution.html

The arbitrary function $C(x)$ has to be determined according to the boundary conditions. This is the main difficulty. Generally this leads to an integral equation which cannot always be analytically solved. All depends on the form of boundary condition. In the present case, I am afraid that is is not possible to go further without specific equation for the boundary condition $W(v,\theta(v))=0$.

Answer 2

I propose a different approach : \begin{align} \begin{split} \frac{\partial W(v,z)}{\partial v}&=\frac{\partial^{2} W(v,z)}{\partial z^{2}}\\ \end{split} \end{align} First, look for particular solutions on the form $W_p(v,z)=V(v)Z(z)$ $$\frac{V'}{V}=\frac{Z''}{Z}=\mu^2\quad ;\quad V(v)=e^{\mu^2v}\quad\;\quad Z(z)=e^{\mu z}$$ $\mu$ real or complex constant. And with $C$ any real or complex constant : $$W_p(v,z)=C e^{\mu^2v+\mu z}$$ The general solution of the PDE is any linear combination of the particular solutions : $$W(v,z)=\int f(\mu)e^{\mu^2v+\mu z}d\mu$$ $f(\mu)$ is an arbitrary function to be determined according to the boundary conditions.

The bounds of the integral are also arbitrary, to be determined according to the boundary conditions.

$W(0,z)=\delta_{a}(z)=\int f(\mu)e^{\mu z}d\mu$

$W(v,\theta(v))=0=\int f(\mu)e^{\mu^2v+\mu \theta(v)}d\mu$

The bounds of the integrals have to be chosen to make the whole consistent.

As guessed, it ends with integral equation. I hope that you will be more lucky with them than with whose from my preceding answer.