13
$\begingroup$

If we employ the Method of Lines for discretization (separate time and space discretization) of hyperbolic PDEs we obtain after spatial discretization by our favorite numerical method (fx. Finite Volume Method) does it matter in practice which ODE solver we employ for the temporal discretization (TVD/SSP/etc)?

Some additional information added: The accuracy issue can be a problem for non-smooth problems. It is will-known that nonlinear hyperbolic PDEs can develop shocks in finite time despite the initial solution is smooth in which case accuracy can degrade to first order for high-order methods.

ODE Stability analysis is typically done based on linearization to obtain a linear semi-discrete system of ODEs of the form q_t = J q (with q a perturbation vector), where the eigenvalues of J should be scaled inside the absolute stability region of chosen time-stepping method. Alternative strategies is to use pseudospectra or possibly an energy method for stability analysis.

I understand that the motivation for TVD/SSP methods are to avoid spurious oscillations caused by the time-stepping methods which may result in unphysical behavior. Question is if experiences shows these types of time-stepping methods to be superior compared to, e.g., a classical work horse as explicit Runge-Kutta Method or others. Obviously, they should have better properties for classes of problem where solution may exhibit shocks. One could therefore argue that we should only employ these types of methods for time-integration.

| cite | improve this question | |
$\endgroup$
3
$\begingroup$

I don't know if you're still interested in an answer, but here I go anyway:

You already said you know about shock formation in nonlinear equations. That is exactly why you have to choose your time integrator carefully. It's no use applying a TVD spatial discretization when the time discretization is not - you will see the same oscillations you have probably seen with higher order numerical fluxes.

What it boils down to is that forward Euler works. You already mentioned SSP (strong stability preserving) in your question. This is a special class of Runge-Kutta methods that makes use of that. Basically, you have to choose the coefficients of the method in such a way that it can be written as a convex combination of Euler steps. That way, properties like TVD and such will be preserved.

There is a very good book on SSP methods by Gottlieb, Ketcheson and Shu called "Strong Stability Preserving Runge-Kutta and Multistep Time Discretizations" amazon link

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Correct me if I'm wrong, but forward Euler is almost definitely going to be unstable on a hyperbolic problem. No resolution of modes associated with pure imaginary eigenvalues. $\endgroup$ – Reid.Atcheson Mar 24 '13 at 4:51
  • $\begingroup$ @Reid.Atcheson: All monotone methods I know are based on forward Euler - upwind, Lax-Friedrichs, Godunov... It just depends on what you do in space. $\endgroup$ – Anke Mar 25 '13 at 12:14
  • $\begingroup$ Forward euler may be unstable in L2 norm if combined with a high order space scheme. Then you use 2-stage, 3-stage, etc. SSPRK schemes which are L2 stable. It is easier to prove TVD for forward Euler scheme. Using a SSPRK scheme then guarantees TVD for high order scheme also. The time step for TVD is smaller than that for L2 stability $\endgroup$ – cfdlab Jun 27 '16 at 4:01
2
$\begingroup$

Yes, it matters. The usual two things to be concerned about:

  1. Accuracy. Some ODE schemes are more accurate than others, higher order, and so on. The rule of thumb is to choose a method with order of accuracy similar to your spatial discretization.

  2. Stability. For hyperbolic problems you expect the operator to have pure imaginary eigenvalues, so you want an ODE solver which includes some part of the imaginary access in its stability domain. See for example Appendix G in Fornberg, A Practical Guide to Pseudospectral Methods.

With hyperbolic equations, some people want to insure that their solutions are always positive, so there are various kinds of filters and tricks to insure this. But I know almost nothing about this.

I am far from an expert, but I thought I'd try to answer since the question has been here for a while.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Hyperbolic system involves only real eigenvalues (distinct if it is strictly hyperbolic) and corresponding real eigen-vectors. $\endgroup$ – Subodh Mar 22 '13 at 15:22

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.