0
$\begingroup$

I want to solve the Euler equations in 1D numerically. I am currently using DG-Lax Friedrich flux splitting technique and RK-4 explicit time stepping scheme.

Now,using the idea of characteristics it is possible to decompose the Jacobian matrix into incoming and outgoing waves depending on the sign on the sign of the eigen values. Here I am using the idea of characteristics variables where I shift my problem into a characteristic variable. For eg . $ A = W \Lambda W^{-1}$,Then, $w = W^{-1}u$ ;, where u is the unkown of vectors and w is the characteristic variable.

I can solve the problem using both techniques, but I am struggling to find the equivalence. Where does the idea of characteristics come into picture in the the DG scheme ?

$\endgroup$
  • $\begingroup$ From the PDE point of view, these two approaches are equivalent, however from the numerical point of view they are not. This is essentially a consequence of the fact that the chain rule is not preserved by the discretisation, meaning that the variable transformation cannot be obtained discretely in the same way as in the continuous setting. Different formulations of the equations generally yield numerical schemes with different properties, and the particular choice may have a significant impact on the solution. See e.g. Pirozzoli, 'Numerical methods for high speed flows' for an overview. $\endgroup$ – ekkilop Nov 5 '17 at 10:32
  • $\begingroup$ If I understand correctly. you mean to say that till the point I use the concept of characteristics variable, all the approaches are equivalent ? In the concept of characteristics we decide the number of boundary conditions in one element based on the sign of the eigen values (incoming and outgoing). Where is this fact incorporated in a DG-flux splitting scheme. ? Thanks for the references, I will look into this. $\endgroup$ – Bullet Nov 7 '17 at 9:49
  • $\begingroup$ What I'm saying is that a (non-singular) variable transformation in the PDE yields an equivalent PDE. It doesn't matter if it's characteristic variables or some other choice, however characteristics are often suitable for the derivation of boundary conditions as you say. However, the discretisations of the original and the transformed PDEs will not be equivalent in the sense that there will in general not be a transform that takes you from one to the other. The two discretisations may possess different properties, useful in different circumstances. $\endgroup$ – ekkilop Nov 7 '17 at 10:20

Your Answer

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

Browse other questions tagged or ask your own question.