DG method, very small time step

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 April 17, 2013, 05:34 DG method, very small time step #1 New Member   Join Date: Mar 2013 Posts: 14 Rep Power: 13 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 !

April 17, 2013, 05:38
#2
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cfdnewbie
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Quote:
 Originally Posted by Wenjun 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 !

yeah, DG has a timestep of 1/(2n+1), where n is the ansatz degree. Which order is your ansatz?

April 17, 2013, 05:43
#3
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Quote:
 Originally Posted by cfdnewbie yeah, DG has a timestep of 1/(2n+1), where n is the ansatz degree. Which order is your ansatz?
I am a new guy of DG. I use the second-order element and the third-order TVD Runge-Kutta method. I set dx = 0.1, how should I choose the time step?

 Tags dg method, time step size