Ladies, gentlemen, I need your help with this one
Consider the linear system of equations
$u_t+Au_x=0$
with $u=(u_1,…,u_n)^T$ and $A\in n\times n$ matrix with constant entries.
Derive the Rankine-Hugoniot jump conditions for this system of linear hyperbolic equations
$A(u_l-u_r) =s(u_l-u_r) $
and show that $u_l-u_r$ is the eigenvector of $A$ and $s$ the associated eigenvalue.
I've tried to diag $A$ and then to decouple the system, but this doesn't work.
Best Answer
Sure diagonalization works! Upon diagonalization, the proof is exactly the same as in the scalar case. The equality $A[\![u]\!]=s[\![u]\!]$ defines an eigenvalue problem, where $[\![u]\!]=u_l-u_r$ is the jump in $u$. Therefore, let us introduce the following eigendecomposition of the matrix $A$: $$ A = P \Lambda P^{-1} , $$ where the matrix $\Lambda = \text{diag}(\lambda_1 , \dots , \lambda_n)$ is diagonal and presumably real. Then, introducing $v = P^{-1} u$, we have: $$ v_t + \Lambda v_x = 0 , $$ which is a decoupled system of $n$ linear advection equations $v_{i,t} + \lambda_i v_{i,x} = 0$. The Rankine-Hugoniot condition implies $\lambda_i [\![v_i]\!] = s[\![v_i]\!]$ component-wise, i.e. $\Lambda [\![v]\!] = s[\![v]\!]$. Then, upon left-multiplication of the previous equation by $P$, we get the desired result $$A [\![u]\!] = s[\![u]\!] \, ,$$ where we have used $[\![v]\!] = P^{-1} [\![u]\!]$.