DG method, very small time step

Wenjun · April 17, 2013, 06:34

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 !

cfdnewbie · April 17, 2013, 06:38

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?

Wenjun · April 17, 2013, 06:43

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?

April 17, 2013, 06:34	DG method, very small time step	#1
Wenjun 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 !

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