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Old   April 25, 2000, 18:47
Default Flux Splitting: Derivation
  #1
Abhijit Modak
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Does anyone know how to prove that the split fluxes for Van Leer, Sterger-Warming, Liou-Stephen and Zha-Bilgen do satisfy the condition that df+/du > 0 and df-/du < 0? I tried the derivation; but the conserved variable form of the split flux vectors is too complex. Are there any tricks? Sincerely Abhijit
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Old   April 26, 2000, 10:59
Default Re: Flux Splitting: Derivation
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B O Bamkole
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Why, if I may ask do you need the terms df+/du and df-/du. If you need to use them in your implicit matrix you can get away with the approximation that A(+) = df+/du and A(-) = df-/du. If you require them in formulating some sort of higher-order correction to your basic first orderexplicit fluxes I am not sure that the condition you seek exists i.e all terms in both df+/du and df-/du are unconditionally positive and negative respectively.

As you know the basis of both flux-difference and flux-splitting schemes is that the fluid is homogeneous to degree one. In this case it is readily shown that F = A.Q where A is the Jacobian given by

A =(dF/dQ) and Q is the vector of conserved variables. The Jacobian can then be written as

A=L.E.R where L and R are matrices of the left and right eigenvectors of A and E is a diagonal matrix whose terms are the eigenvalues of A i.e E =diag(U, U+C,U-C) for a 1-D system. You can write E as the sum of positve and negative terms

E=E(+) + E(-) where E(+) = 0.5( E + abs(E)) and E(-) = 0.5( E - abs(E)).

E(+) and E(-) written in this form will always consist of only positive and negative terms.

Finally you have F as the sum of two 'split' fluxes F=F(+) + F(-) in which F(+) = L.E(+).R.Q and F(-) = L.E(-).R.Q

My point is that in both these expressions it is only for E(+) and E(-) that you know the signs.

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Old   April 27, 2000, 01:35
Default Re: Flux Splitting: Derivation
  #3
Abhijit Modak
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I think I need to clarify my question. For Euler equations we can select F+ = A+*u and F-=A-*u , A+ and A- having positive and negative eigenvalues respectively. So we do actaully achieve what we want i.e. F+ captures right running waves and F- the left running. But if we go back to the requirements that we define for flux splitting i.e. F = F+ + F- and dF+/du > 0 and dF-/du < 0 then we are not sure that the conditions on the derivative are satisfied. This is so because A+ = A+(u) and A-=A-(u). So there is the functional dependence on u in the wave split matrices. Therefore the derivative of F+ or F- is not merely A+ or A- respectively. If we want to show this for flux splitting schemes mentioned in my original question then we get very complex expressions for the Jacobian matrices and there eigenvalues. As correctly pointed out in the first reply this may be irrelevent as we have already captured the right waves. But I think at least from academic point of view, I would like to know the answer. Very Sincerely Abhijit
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Old   April 27, 2000, 06:35
Default Re: Flux Splitting: Derivation
  #4
B O Bamkole
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Hello again,

I am still not sure that I understand these two conditions:

df+/du > 0, and df-/du < 0.0.

I am familiar with the following conditions given in Section 20.2.2 of Hirsch for flux-splitting are given:

1 The fluxes are homogeneous functions of degree one in u ,

2.The flux jacobian A=df/du = A(+)+A(-)=df+/du+df-/du,

3 df+/du != A(+) , df-/du != A(-).

Thus one check on the correctness of your final derivatives notwithstanding their complexity is that they sum up to A.

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Old   April 27, 2000, 12:56
Default Re: Flux Splitting: Derivation
  #5
Abhijit Modak
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The equations are: F=Au i.e. fluxes are homogeneous functions of degree one. Now in wave speed split form: A=A+ + A-.(A+ has positive eigenvalues and A- has negative.) To write in the flux split form we may choose F+ = (A+)*u and F- = (A-)*u. We get F=F+ + F- as required. But we are not sure that dF+/du > 0 and dF-/du < 0. This is because the terms in A+ and A- are functions of u and hence dF+(-)/du is not merely A+(-) but one additional term. i.e. Since F+ = (A+(u))*u, dF+/du = A+(u) + (dA+/du)*u. In the 1981 paper by Van Leer he says that the fisrt eigenvalue for the flux vectors in his splitting is zero and there is a quadratic equation for the other 2. I am not able to derive that. The calculations for other spliiting schemes mentioned in the original question is too tedious. Very Sincerely Abhijit

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Old   May 2, 2000, 04:13
Default Re: Flux Splitting: Derivation
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B O Bamkole
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Hello again,

Sorry to keep going on about this. By the statement 'But we are not sure that dF+/du > 0 and dF-/du < 0' do you mean the eigenvalues of these two matrices are positive and negative respectively?
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