How to solve a PDE with time dependent domain?

Question

I am concerning if it is possible to solve a PDE of time dependent domain. My PDE is as follow:

Assuming \begin{align} v&=\frac{1}{2}\left\{\frac{1}{\epsilon}-\frac{1}{u}\right\}\\ \theta(u)&=(\alpha-\beta)\epsilon^{2}\left\{\frac{1}{\epsilon^{2}}-\frac{1}{u^{2}}\right\}+(\alpha-\beta)\epsilon\left\{\frac{1}{\epsilon}-\frac{1}{u}\right\}\\ \end{align} 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)&=e^{-z},\quad W(v,\theta(u))=0,\quad \theta(u)\leq z<\infty,\quad 0\leq v<\frac{1}{\epsilon}\\ \end{split} \end{align}

Is it possible to solve it?

Answer 1

Some remarks. Not a full answer.

First, as I said in the comments the condition at $W(0,0)$ is not clear, as it could be $W(0,0)=1$ or $W(0,0)=0$ because for $v=0$ we have $\theta=0$ as well.

Second, it makes sense to reduce the number of parameters and simplify the conditions by setting:

$$2( \alpha-\beta)=a$$

$$v=\frac{1}{\epsilon}t,\qquad t \in (0,1)$$

$$\theta=ah(t)$$

$$h(t)=3t-2t^2$$

$$z=ay,\qquad z \in [h(t),\infty)$$

$$h(0)=0, \qquad h(1)=1$$

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Then the equation becomes:

$$\frac{\partial W(t,y)}{\partial t}= \frac{1}{\epsilon a^2} \frac{\partial^{2} W(t,y)}{\partial y^{2}}$$

$$W(0,y)=e^{-ay}$$

$$W(t,3t-2t^2)=0$$

Here the contradition between the initial and "boundary" conditions becomes even more apparent.

As for the latter condition, it just states that for:

$$y(t)=3t-2t^2 \Rightarrow W(t,y)=0$$

The domain of the equaion looks like this:

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While the above doesn't tell anything about the method of solution, I think it simplifies the problem statement.

As for the question of the OP "Is it possible to solve it?" I asnwer "Yes" and reference a 1997 paper (in Russian, unfortunately), which is available here which introduces a general method of solving exactly this kind of problems.