CFD Lax Equivalence Theorem?

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 November 3, 1999, 13:49 CFD Lax Equivalence Theorem? #1 Amadou Sowe Guest   Posts: n/a In the thoery of numerical ODEs the Lax Equivalence Theorem says that given a well-posed IVP(initial-value problem) or IBVP (initial boundary-value problem) and a finite-difference scheme consistent with it stability and convergence are equivalent. (1) Is there anything like this in CFD where the governing equations are generally non-linear? (2) If not how can we justify using this theorem in CFD or generally speaking in PDEs? (3) Is there a relationship between this theorem and the fact that when our residuals go towards zero and the conclusion that our solution has converged? Thanks in advance to any 'takers'.

 November 3, 1999, 16:41 Re: CFD Lax Equivalence Theorem? #2 Praveen Chandrashekar Guest   Posts: n/a (1)There is no equivalent theorem for non-linear equations. In fact, the theory for non-linear pde's is extremely lacking. Existence cannot be proved for non-linear equations like the Euler equations. If we assume existence, then the Lax-Wendroff theorem tells us about the uniqueness of the numerical scheme for scalar conservation laws. Stability is also a difficult thing to prove in the non-linear case. People usually apply some linearisation and then prove the stability for the linearised equations. Such a procedure can only provide necessary conditions only. (2)Most of the numerical schemes and related concepts in cfd are first developed for the linear scalar case, and then extended in a straight-forward way to non-linear systems of equations. There is no reason why this should work, but it does almost all the time. In fact, it can be taken as an unwritten rule in cfd that a scheme which does not work in the linear scalar case will almost certainly fail in the non-linear case. Hence the results concerning linear equations are necessary conditions for non-linear equations, but may not be sufficient. (3)The decay of residuals in a numerical solution has nothing to do with uniqueness of the solution. Jameson has shown that the Euler equations can have different numerical solutions in the transonic case. Only for a scalar conservation law, the Lax-Wendroff theorem tells us that a numerical scheme that is consistent, conservative and stable will converge to a correct weak solution(for hyperblic equations).

 November 4, 1999, 12:05 Re: CFD Lax Equivalence Theorem? #3 Amadou Sowe Guest   Posts: n/a Thanks for your comments. I am familiar with the Lax-Wendroff scheme but not the LaxWendroff theorem. From your comments it looks like it might be related to the Lax-Equivalence theorem. Can you tell me what it is? In your third comment, you said that the decay of residuals in a numerical solution has nothing to do with the uniqueness of the solution. I do not agree with you entirely on this point. Let us assume we are solving the non-singular matrix equation Ax = b. Let us define e = x-x' where x is the exact solution and x' the approximate solution. Also, let the residual be defined as r = b - Ax' = A(x-x')= Ae. It looks like as the residual goes to zero e goes to zero. As e goes to zero x'=e+x approaches the exact solution. Thus in this case as the residual goes to zero our numerical solution approaches the unique solution. It is for this and a few more examples why I do not agree with you on point (3).

 November 4, 1999, 13:18 Re: CFD Lax Equivalence Theorem? #4 Praveen Chandrashekar Guest   Posts: n/a (1)That is true. But we are talking of systems of non-linear pde's. The corresponding numerical schemes are also non-linear. In such a case, the Lax Equivalence theorem is a neccesary condition, but not sufficient to ensure convergence. The example you gave, Ax=b, is of course linear. (2)The Lax-Wendroff theorem states that a numerical scheme for the scalar conservation law, du/dt + df/dx = 0, converges to a unique (weak) solution, if it is consistent, stable and conservative. There is no equivalent theorem for systems of conservation laws.

 November 4, 1999, 14:32 Re: CFD Lax Equivalence Theorem? #5 Amadou Sowe Guest   Posts: n/a Thanks again. You are right that my question was specific to non-linear pde's. However, your occasional references to linear methods coupled with your statement (3) were the reasons why I thought that your conclusion reflected in (3) included both linear and non-linear pde's. Hence the reason for my linear example. It does look like the Lax-Equivalence theorem could be used to weed out unstable or inconsistent schemes.

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