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TurbJet September 2, 2018 22:46

What's the mathematical meaning of characteristic decomposition of Euler equations
 
Greetings,

When people apply nonreflecting BCs for compressible Euler equations, typically what we do is studying the characteristic decomposition of the compressible Euler equations, break the coefficient matrix into a diagonal matrix.

But what baffles me is that, what's the mathematical meaning of this decomposition? What do those variables (namely the eigenvalues and eigenvectors from the decomposition) mean mathematically? Why in this way can we impose BCs?

Thanks.

FMDenaro September 3, 2018 03:16

Quote:

Originally Posted by TurbJet (Post 705067)
Greetings,

When people apply nonreflecting BCs for compressible Euler equations, typically what we do is studying the characteristic decomposition of the compressible Euler equations, break the coefficient matrix into a diagonal matrix.

But what baffles me is that, what's the mathematical meaning of this decomposition? What do those variables (namely the eigenvalues and eigenvectors from the decomposition) mean mathematically? Why in this way can we impose BCs?

Thanks.






From a mathematical point of view, by using the characteristic method you can reduce the PDE system to an ODE system with initial value problem.
In terms of linear algebra, the original matrix is pre and post multiplicated my matrices of eigenvectors to get a diagonal matrix where only eigenvalue appear. Of course, the resolved variables are not the original one but the transformed one. That is also at the origin of the Riemann invariant for omo-entropic flows. Such variables are called invariant because they remain constant along the characteristic curves, allowing for an exact mathematical solution. However, this can be done exactly only for particular 1D assumption.

TurbJet September 3, 2018 10:11

Quote:

Originally Posted by FMDenaro (Post 705090)
From a mathematical point of view, by using the characteristic method you can reduce the PDE system to an ODE system with initial value problem.
In terms of linear algebra, the original matrix is pre and post multiplicated my matrices of eigenvectors to get a diagonal matrix where only eigenvalue appear. Of course, the resolved variables are not the original one but the transformed one. That is also at the origin of the Riemann invariant for omo-entropic flows. Such variables are called invariant because they remain constant along the characteristic curves, allowing for an exact mathematical solution. However, this can be done exactly only for particular 1D assumption.

Thanks for the reply! And do you have any idea about the mathematical or physical meaning of left & right eigenvectors? This is what most people have used when they do characteristic decomposition.

FMDenaro September 3, 2018 11:40

Quote:

Originally Posted by TurbJet (Post 705162)
Thanks for the reply! And do you have any idea about the mathematical or physical meaning of left & right eigenvectors? This is what most people have used when they do characteristic decomposition.


It is a linear algebra topic. Consider the matrix A and the following eigenvalues problems where the right and left eigenvectors appear




(A- lambda(k) I).rk = 0
lk^T .(A- lambda(k) I) = 0

so that you get a basis in R^N since lj^T.rk = 0.

After determining the eigenvectors, you can now build the matrix R =[r1,r2,...rN] and L =R^-1 such that



D = L.A.R

is the diagonal matrix having the eigenvalues as entries.
Therefore, the eigenvectors are used to build rotation matrices in such a way to express the diagonal form. Practically, you are rewriting the original matrix in terms of the principal axes





TurbJet September 3, 2018 21:07

Quote:

Originally Posted by FMDenaro (Post 705169)
It is a linear algebra topic. Consider the matrix A and the following eigenvalues problems where the right and left eigenvectors appear




(A- lambda(k) I).rk = 0
lk^T .(A- lambda(k) I) = 0

so that you get a basis in R^N since lj^T.rk = 0.

After determining the eigenvectors, you can now build the matrix R =[r1,r2,...rN] and L =R^-1 such that



D = L.A.R

is the diagonal matrix having the eigenvalues as entries.
Therefore, the eigenvectors are used to build rotation matrices in such a way to express the diagonal form. Practically, you are rewriting the original matrix in terms of the principal axes





So the sole purpose of using left/right eigenvectors is just to find out the principle axis of the matrix? Do the vectors themselves have some physical meaning?

FMDenaro September 4, 2018 03:34

the eigenvectors are tangent to the integral curves in the the variable space, have a look to the book of Leveque

Martin Hegedus September 4, 2018 14:13

The eigenvectors can be thought of as filters. The state variables are dotted with the eigenvector to get the variable of interest.

The physical variables of interest are entropy, vorticity, and the +/- acoustic waves.

At this time, I'm going to probably do some fudging. We always seem to use the term "eigenvector", however, I view the first three eigenvectors (the ones for entropy and vorticity which are multiplied by velocity) as an eigentensor. Not sure if eigentensor is an actual mathmatical term, but that's what it seems, at least to me, to be.


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