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#1 |
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New Member
Join Date: Mar 2013
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When I apply the Discontinous Galerkin method on the KdV equation, the time step can only take very small number, like 0.0001. Can anyone tell me it is normal or not. Thanks !
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#2 | |
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cfdnewbie
Join Date: Mar 2010
Posts: 557
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Quote:
yeah, DG has a timestep of 1/(2n+1), where n is the ansatz degree. Which order is your ansatz? |
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#3 |
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