I have recently learned about Discontinuous Galerkin method to solve differential equations and I was trying to implement it to solve Euler equation. For now, consider the standard Sod Shock Tube Case.

Without using any limiter I am getting an unstable scheme (even with CFL 0.01) and my solution blows up. Presently I am using 1st order polynomials in each cell using Lobatto nodes (in case someone needs to know). Also, I am calculating the interface flux using AUSM method.

Now I am not sure whether my code is wrong or does it behave like this because I haven't used any flux limiter?

  • $\begingroup$ A first order scheme should be fairly dissipative. On the other hand you do not state how it is blowing up. There can be a few reasons why it blows up. One is because of oscillations due to the shock becoming too large but a first order scheme should dissipate that. One is using say an Euler scheme for time stepping. Another is positivity being violated. That may occur because you are using AUSM. $\endgroup$ – Vikram Mar 15 '17 at 10:06
  • $\begingroup$ Yes, it is indeed because of positivity being violated. $\endgroup$ – Manish Mar 15 '17 at 17:03
  • $\begingroup$ Then try a positivity preserving Riemann solver like the simple Lax Friedrichs flux. If that solves it then you know what was causing the issue. $\endgroup$ – Vikram Mar 15 '17 at 17:32
  • $\begingroup$ Sure. I think that is a good idea. Will try it. $\endgroup$ – Manish Mar 15 '17 at 17:46

A finite element scheme (CG or DG) without stabilization is expected to result in a solution that violates the maximum principle, even for linear scalar problems. That is why sufficient dissipation/viscosity is needed in addition to the dissipation on the approximate Riemann solver on the interface. See the papers by Cockburn and Shu for a detailed discussion.

Various approaches exist to deal with this issue such as entropy viscosity, SUPG stabilization etc...

  • $\begingroup$ @user814 Could you provide a link for the papers? $\endgroup$ – ares Oct 22 '18 at 21:05

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.