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non-linear terms in conservative or non-conservative form

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Old   May 24, 2012, 07:56
Unhappy non-linear terms in conservative or non-conservative form
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Hi,

When solving an equation that involves non-linear terms such as u\frac{\partial u}{\partial x} in Burger's equations. Will the results be different if the conservative form of the term \frac{1}{2}\frac{\partial u^{2}}{\partial x} was used instead. Does it result in differences in terms of accuracy and stability.

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Old   May 24, 2012, 12:28
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Hi,

When solving an equation that involves non-linear terms such as u\frac{\partial u}{\partial x} in Burger's equations. Will the results be different if the conservative form of the term \frac{1}{2}\frac{\partial u^{2}}{\partial x} was used instead. Does it result in differences in terms of accuracy and stability.

Thanks!
yes, the discrete equations can drive to very different results (e.g. different wave propagation), as advice you must use always the conservative form!
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Old   May 24, 2012, 12:39
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Thanks very much.

On a slightly different note, if the non-lineaity was in an unsteady term for instance \frac{\partial}{\partial t}\left(v^{2}\right), how would one go about discretizing with respect to time? Say, Explicit Euler was being used, would it be possible to just find v^2 at the next time step and take the square root of that or is there more to it due to the non-linearity? I think this would reduce the accuracy but I am not really sure.

Thanks again!
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Old   May 24, 2012, 12:46
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Thanks very much.

On a slightly different note, if the non-lineaity was in an unsteady term for instance \frac{\partial}{\partial t}\left(v^{2}\right), how would one go about discretizing with respect to time? Say, Explicit Euler was being used, would it be possible to just find v^2 at the next time step and take the square root of that or is there more to it due to the non-linearity? I think this would reduce the accuracy but I am not really sure.

Thanks again!
The equation for the kinetic energy equation has such a term and the equation is solved in time for the variable k=u^2 ....
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Old   May 24, 2012, 14:51
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Just as a quick addition to this: There's something called the "skew symmetric form" that avoids aliasing errors and consists of a combination of both formulations. In that sense, it might not always be advisable to use the conservative form alone, but on general, yes, use it rather than the non-con form!
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Old   May 31, 2012, 13:37
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Thanks!

Can I just ask more specifically why the conservative form is preferred. If you know any good reading material on this specific topic , please let me know.
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Old   May 31, 2012, 13:59
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Thanks!

Can I just ask more specifically why the conservative form is preferred. If you know any good reading material on this specific topic , please let me know.
any CFD book treating the FV approach is good for you, e.g. LeVeque, Ferziger & Peric, etc ...
then, many papers in journals are more specific in analyzing the properties of the conservative (divetgence) form in terms of accuracy and stability
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Old   June 1, 2012, 01:20
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If solutions are discontinuous, then conservative form should be used.

For smooth solutions, the skew symmetric form can be used; it conserves energy and thats one of the reasons it is used.
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