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Is there such a thing as "method" independence?

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Old   February 14, 2020, 15:06
Default Is there such a thing as "method" independence?
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I read somewhere that someone was claiming that you can have a mesh independent solution for some flow problem using the first order upwind scheme, but that it might still be giving a wrong result, because it was not "method" independent i.e. discretization scheme independent. Accordingly, this person claims that if you repeated a mesh independence using second order upwind scheme, you would get a different value at mesh convergence.

Is this true, and if so, how can first order upwind scheme provide mesh independent value that is still incorrect?
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Old   February 14, 2020, 16:10
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If the numerical scheme is consistent, its solution converges to the solution of the PDE for h->0. A first order scheme will likely nerd a finer mesh than thesecond order one, but both will converge to thesame solution (for smooth problems!) or one has a bug.
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Old   February 14, 2020, 16:13
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Quote:
Originally Posted by vesp View Post
If the numerical scheme is consistent, its solution converges to the solution of the PDE for h->0. A first order scheme will likely nerd a finer mesh than thesecond order one, but both will converge to thesame solution (for smooth problems!) or one has a bug.
That is my understanding too....
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Old   February 14, 2020, 16:17
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Quote:
Originally Posted by vesp View Post
If the numerical scheme is consistent, its solution converges to the solution of the PDE for h->0. A first order scheme will likely nerd a finer mesh than thesecond order one, but both will converge to thesame solution (for smooth problems!) or one has a bug.
What do you mean by "smooth problems?"

Are you talking about discontinuities that would cause false diffusion to become significant, because mesh independence study should take care of that.

If you are talking about discontinuities causing second order scheme to cause overshoots and undershoots and be worse than first order, then yes, that makes sense too.
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Old   February 14, 2020, 16:39
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As addressed above, the Lax equivalence theorem ensures that a liner scheme that is consistent and stable converges towards the solution of the PDE, no matter about first, second or higher order of accuracy.
Maybe the issue is discussed in the framework of non linear problems?
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Old   February 14, 2020, 16:41
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Quote:
Originally Posted by arkie87 View Post
What do you mean by "smooth problems?"

Are you talking about discontinuities that would cause false diffusion to become significant, because mesh independence study should take care of that.

If you are talking about discontinuities causing second order scheme to cause overshoots and undershoots and be worse than first order, then yes, that makes sense too.



"smooth problems" means that all the infinite terms (the derivatives) in the local truncation error are bounded and of unitary order of magnitude.
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Old   February 14, 2020, 16:51
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As addressed above, the Lax equivalence theorem ensures that a liner scheme that is consistent and stable converges towards the solution of the PDE, no matter about first, second or higher order of accuracy.
Maybe the issue is discussed in the framework of non linear problems?
What is intended by "linear scheme"?

Why would a non-linear problem be different e.g. NS equations?
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Old   February 14, 2020, 16:59
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What is intended by "linear scheme"?

Why would a non-linear problem be different e.g. NS equations?

Linear scheme means that the process that produces the numerical solution is based on a linear transformation, that is f= A.x. For example the stationary time integration method, like FTUS, FTCS or other, can be written as


f^n+1 = (A)^n+1 . f^0


You can make non linear a scheme also for a linear PDE.

Of course, a non linear PDE generates a non linear algebric system
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