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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?

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  • $\begingroup$ Where did this PDE come from? How do you expect the solution to behave? Usually problems like that originate in some practical context, hence you should have some idea what behavior to expect $\endgroup$ – Yuriy S Feb 21 '18 at 15:06
  • $\begingroup$ Your initial and boundary conditions are contradictory. For $v=0$ we have $\theta=0$. Substituting, from the initial condition we have $W(0,0)=1$ and from the boundary condition $W(0,0)=0$. Which is it? $\endgroup$ – Yuriy S Feb 21 '18 at 16:16
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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:

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

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

$$\theta=ah(t)$$

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

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

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


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:

enter image description here


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.

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  • $\begingroup$ Thanks. I cannot read Russian, I guess from the notes that it use Fourier transform. I would like to know how he do the Fourier transform on the boundary and initial condition. Are there any relevant reference? $\endgroup$ – will_cheuk Mar 2 '18 at 7:42
  • $\begingroup$ @will_cheuk, I only found this, so I don't know. What are you going to do about the contradiction in the initial conditions? $\endgroup$ – Yuriy S Mar 2 '18 at 11:17
  • $\begingroup$ Indeed, I am working on first hitting process and ultimately form corresponding PDE stated below. I haven't encountered solving heat equation with moving boundary and I would like to know how to tackle the problem. Indeed, I read some paper say "solution of heat equation in a domain with moving boundary obtained by hartree functions" which states that there exists an analytical method to solve the problem. Again, it is written in Russian, that's why I raise the question here. For the contradiction part, $W(0,z)=\begin{cases} e^{-z} & \text{$z>0$}\\ 0 & \text{$z=0$} \end{cases}$ $\endgroup$ – will_cheuk Mar 2 '18 at 11:31
  • $\begingroup$ @will_cheuk, so, if I understand you correctly, your function is not continuous at $z=0$? Then you can't use the initial conditions for $z=0$ at all... I'm not sure how this affects the solution $\endgroup$ – Yuriy S Mar 2 '18 at 14:19
  • $\begingroup$ i understand your concern. i have same concern at first when i derive the equation. just like pde kn a semiinfinite domain. $\endgroup$ – will_cheuk Mar 3 '18 at 0:55

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