# Can I solve this time-independent PDE by adding a time derivative and marching in time?

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

• What is the domain in terms of Y and V? Do you know whether the PDE is hyperbolic, parabolic, or elliptic? – David Ketcheson Jul 8 '13 at 17:27
• 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. – phubaba Jul 9 '13 at 20:35
• Where are the boundary conditions at $Y=0$ and $Y=\infty$? – Hui Zhang Jul 12 '13 at 18:01

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.