Normalization of eigenvectors of the Euler equations
 

Hello

I would like some help regarding the computation of eigenvalues for the implementation of Roe’s scheme for gas dynamics.

I don’t know what is the role to normalize eigenvectors (left and right) so they will be suitable to Roe’s scheme.

Thanks

Tali Neuman

p.s:

I succeeded to:


1. Compute eigenvalues from the jacobian matrix of the non conservative form of the Euler equations.
2. Compute the left/right eigenvectors for the jacobian matrix of the non conservative form of the Euler equations.
3. Transform those eigenvectors to be suitable to the jacobian matrix of the conservative form of the Euler equations.

you can find the answer in the book of LeVeque, remeber you have one degree of freedom in determining the components of the eigenvector

degree of freedom in determining the components of the eigenvector
 

Hello FMDenaro

Thanks for replying.
Can you please point on a specific place in the book where there is a reference to this issue?

Thanks a lot
Tali Neuman

have a look through the book, several parts can be useful for you, in particular tha anlysis of Euler equations

Choice of the number for the degree of freedom component
 

Can I choose any number for the degree of freedom component of each of the right / left eigenvectors for the Roe’s scheme to work?

a number if the components comes from the linear equations, otherwise they are functions

Choosing values for degree of freedom component
 

Thanks for replying
I would like to know if I understood well your answer:
For the jacobian matrix of the 1D Euler equations, there are three eigenvectors.
In each of them, there is one component which can be determined freely (=degree of freedom).
After choosing freely those components for the right / left eigenvectors, can I use them in the Roe’s scheme?

Thanks a lot

remember that the expressions of the eigenvalues depends on the type of variables you are using [rho,u,p], [rho,u,s].
In the book you will find several numerical methods

Regardless
 

for each eigenvalue lambda(k) you solve the system

[A-I lambda(k)]*r(k) = 0


therefore you see that one scalar component is free