8
$\begingroup$

I want to solve this PDE:

Currently I have some code that will automatically generate pde solutions for a very similar pde that includes a time derivative (partial d/ partial t) using an ADI method.

I'm wondering if there is a way to approximate the attached pde with a pde that includes a time derivative?

I know there are approaches for one dimensional pdes. For example in the attached pde if you remove all the Y derivatives, then the pde can be approximated by adding the time derivative and multiplying the diffusion term by a large number, using implicit stepping and using 1 step.

Any help would be appreciated, Thanks, Rob

$\endgroup$
  • $\begingroup$ What is the domain in terms of Y and V? Do you know whether the PDE is hyperbolic, parabolic, or elliptic? $\endgroup$ – David Ketcheson Jul 8 '13 at 17:27
  • $\begingroup$ it is a parabolic pde, domain of V is 0 to infinity, domain of Y is 0 to infinity. This is a variation on the heston pde popular in financial literature. $\endgroup$ – phubaba Jul 9 '13 at 20:35
  • $\begingroup$ Where are the boundary conditions at $Y=0$ and $Y=\infty$? $\endgroup$ – Hui Zhang Jul 12 '13 at 18:01
7
$\begingroup$

The equation is too complex for me to tell whether the answer is yes or no. But in general, here is what you need to consider: Assume you have any equation (algebraic, differential, or partial differential) $f(u)=0$ and you wonder whether you can find $u$ by considering the problem $\frac{dv(t)}{dt}+f(v(t))=0$ in hopes that $u = \lim_{t\rightarrow \infty} v(t)$. Then, the following is clear:

  • It is certainly true that $v_1(t)=u$ is a stationary solution of the time dependent equation.

  • But it is not clear whether it is a stable solution. This is important because if you start with a $v_2(t) \neq u$, then $v_2(t) \rightarrow v_1(t)=u$ only if $v_1(t)$ is a stable solution of the equation. In other words, without stability of $v_1(t)$, you can't expect the pseudo-time dependent problem to converge to the solution of the original, time independent problem.

  • That stability is not automatically guaranteed is easy to see. Consider, for example, the equation $f(u)=-\Delta u-h=0$. You can find the solution to this by solving the heat equation, $\frac{dv(t)}{dt}-\Delta v(t)=h$. But if you had started with $\tilde f(\tilde u)=\Delta \tilde u+h=0$ (which, of course, has exactly the same solution), you would not have been able to find the solution by solving $\frac{d\tilde v(t)}{dt}+\Delta \tilde v(t)=-h$ since this equation does not in general have a solution.

$\endgroup$
  • $\begingroup$ Gotcha, so depending on say the coefficients in my pde, as time increases, and depending on my terminal conditions, I will not reach the right solution. Could I then test the stability of the pde by prescribing some different terminal conditions and seeing whether I achieve stability or not? $\endgroup$ – phubaba Jul 9 '13 at 20:47
  • $\begingroup$ Do you mean initial conditions when you talk about terminal conditions? If you walk time forward, you can prescribe only initial conditions and observe terminal values. $\endgroup$ – Wolfgang Bangerth Jul 10 '13 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.