Consider the Navier Stokes equations for the steady, two-dimensional flow of an inviscid, incompressible fluid:
\begin{equation} \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0, \end{equation} \begin{equation} u\frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+\frac{1}{\rho} \frac{\partial p}{\partial x}=0, \end{equation} \begin{equation} u\frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+\frac{1}{\rho} \frac{\partial p}{\partial y}=0, \end{equation}
I am interested to find the eigenvalues and the characteristic equation for this system. Naturally, the first step is to put the system into the matrix form. Let $w=(u,v,p)^T,$ then we have $$\begin{pmatrix}1&0&0\\u&0&\frac{1}{\rho}\\0&u&0\end{pmatrix}\frac{\partial}{\partial x}w+\begin{pmatrix}0&1&0\\v&0&0\\0&v&\frac{1}{\rho}\end{pmatrix}\frac{\partial}{\partial y}w=0$$ First matrix is invertible, and so we can write the above equation as $$\frac{\partial}{\partial x}w+\begin{pmatrix}0&1&0\\0&\frac{v}{u}&\frac{1}{u\rho}\\v\rho&-\rho u&0\end{pmatrix}\frac{\partial}{\partial y}w=0.$$ We can find the eigenvalues from the last matrix (denote it $A$), by solving $$det(A-\lambda I)=det \begin{pmatrix}-\lambda&1&0\\0&\frac{v}{u}-\lambda&\frac{1}{u\rho}\\v\rho&-\rho u&-\lambda\end{pmatrix}=0$$ $$\implies \lambda^2(\frac{v}{u}-\lambda)-\lambda+\frac{v}{u}=0$$ $$\implies \lambda_1=\frac{v}{u}, \lambda_{2,3}=\pm i.$$
Hence, we have one real eigenvalue of the system $\lambda=\frac{v}{u}$. What I am not sure of is how to find the corresponding characteristic equation, and if this is related to the characteristic polynomial of the matrix. My feeling is that the characteristic equation is $\frac{dy}{dx}=\frac{v}{u},$ but i don't know how to show this using the methods I know for just one PDE (not the system). Thank you!
