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Discontinuous Galerkin: Physical to Reference Frame transformation of 1D advection eq |
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#1 |
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Tom-Robin Teschner
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I have a question regarding the Discontinuous Galerkin (DG) discretisation of the 1D advection equation of the form
![]() ![]() ![]() If I discretise the advection equation in strong form using the standard DG method I arrive at (in the physical frame with test function ![]() ![]() where ![]() ![]() ![]() ![]() Here and in the previous equation, the subscripts E and F denote the element and its face, respectively. If I transform the above mass and stiffness matrix into reference frame, I end up with ![]() ![]() where J is the jacobian due to the mapping which cancels itself out in the stiffness matrix. Now, for the right-hand side we proceed in a similar fashion, for which we can write in reference frame: ![]() So far I am able to follow the discussion, however, when reading the book of Hesthaven and Warburton "Nodal Discontinuous Galerkin Methods", they drop the Jacobian from the right hand side, i.e. they have ![]() I am not quite sure how we can simplify the above expression (I am probably just missing something obvious here) so that the Jacobian vanishes (at least in 1D, as is here the case). Note: The Jacobian in the case turns out to be ![]() Last edited by t.teschner; January 8, 2019 at 09:15. |
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#2 |
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Dear Teschner,
the RHS term on the right side comes from the flux part. You miss the division of the jacobian. (see stiffness matrix part). Your first equation is no good idea to start your transformation. Regards |
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#3 |
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Tom-Robin Teschner
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I corrected the typo in the first equation (this is the same equation as Hesthaven and Warburton use, good idea or not, I am just trying to understand why the jacobian is missing in their case).
But the division of the jacobian should come from the ![]() ![]() |
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#5 |
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Edit:
I have to correct/extend my last post. Consider the 1D case: The transformation of ![]() ![]() The 2D/3D case will be different. A good explanation is given in the book of D. Kopriva " Implementing Spectral methods". In general is holds: Surface: ![]() Volume: ![]() where ![]() ![]() Regards Last edited by Eifoehn4; January 9, 2019 at 15:32. |
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#6 |
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Tom-Robin Teschner
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I agree with your comments about the equation (I would also write it more generally as a surface integral), but, again, I am just following the notation in the book of Hesthaven and Warburton to be consistent.
Can you be more specific why in 1D the transformation of ds and dx are different (and unity for the 1D case?)? This seems to be exactly the point where I am stuck on. Or if you could point me to the section/pages in the book of Kopriva that would also help. Thanks for your help so far. |
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#7 |
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I have to admit the 1D case is a little bit strange.
Consider the fundamental theorem of calculus which is nothing else as the divergence theorem in 1D ![]() You can do this also for the derivative ![]() You can do this also for an indefinite integral ![]() or with range ![]() If you compare it with ![]() you can see, that a surface integral in 1D is nothing else than the evaluation of its integral kernel at its boundaries. Your equation ![]() will become ![]() with ![]() ![]() or transformed ![]() with ![]() ![]() I think the answer to your question is, that the integral over ![]() Kopriva page 234. Cheers |
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#9 |
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Thanks for making the effort. This all makes sense and the Kopriva book also gave a good alternative view.
It doesn't explain why the Jacobian is not there (still), but I did some more reading in the Hesthaven and Warburton book, in different places they use a sort of hybrid scheme, where the left-hand side (i.e. ![]() ![]() I will think about this a bit more but thanks for your help. |
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#10 |
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so it seems the answer was easier than I thought, and we are probably talking about the same thing, just in different ways. the detail I overlooked is that in 1D the volume integral is a line (so it has a jacobian) but the surface integral is a point (so it doesn't have a jacobian) ...
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#12 |
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And it is even easier. Not only the jacobian vanishes. Formally you simply have no integral for RHS in 1D.
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