1
$\begingroup$

This question is about approximate solutions to some classes of PDEs obtained using the "method of lines".

For example, for an initial-value problem given by a PDE on a circle, one can choose $n$ equally spaced points on the circle, and define an $n$-dimensional ODE system whose solution may approximate the solution of that PDE at the $n$ points (for matching initial conditions). The exact form of the ODEs may be derived from finite difference formulas, or from some other methods.

I think that, for example, if the PDE is given by the heat equation, then (for well-chosen ODEs) we may get a good approximation for all $t \ge 0$, but if the PDE is given by a simple first-order transport equation then the approximation seems to become worse for large $t$.

(Here is what I mean by "good approximation": for fixed $n$ we define some "total error" $E(n)$ and we want $lim_{n\to\infty} E(n) = 0$, or, if that is not possible, we want to have at least some upper bounds on $E(n)$. It is up to us to choose a useful/reasonable formula for the "total error".)

My question is: under what conditions (and for what classes of PDEs, on what types of domains, etc.) is it known that the solutions of the ODE system given by the "method of lines" does provide a good approximation of the solution of the PDE for all $t \ge 0$?

In other words, under what conditions (and for what classes of PDEs) is the difference between the PDE solution $u(x,t)$ (which we assume is defined for all $t \ge 0$) and the ODE system solution $U_n(x_j,t), 1 \le j \le n$ (also assumed to be defined for all $t \ge 0$) either convergent to zero (as $n\to\infty$), or at least bounded in some way (e.g., in $L^1$, $L^\infty$, etc.)?

Please note that I am asking this question for $t$ belonging to the whole positive interval, $t \in [0,\infty)$. Approximations on a bounded interval $t \in [0,T]$ are not as useful for me, unless you think they may be used to conclude something (either positive or negative) for the whole $[0,\infty)$.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.