Eigenspaces of 2x2 Jacobian matrix with change of variable

Question

Let
$$ A= \begin{pmatrix} v & \rho\\ v\left(v+2\rho\right) & \rho\left(2v+\rho\right) \end{pmatrix} $$ where $\rho>0$, $v=\alpha-\rho$ with $\alpha\in\mathbb{R}$.

Show that $A$ has two distinct eigenvalues $\lambda_1(\rho,v)<\lambda_2(\rho,v)$ and compute a basis of eigenvectors assoiated to these eigenvalues.

My work We have $$\det(\lambda I_2-A)=\lambda^2-(v+\rho(2v+\rho))\lambda+v\rho(v-\rho)$$ The discriminant is $$\Delta=v^2+4\rho^2v^2+\rho^4+4\rho^3v+6v\rho^2$$

I cannot seem to find a way to show that it is strictly positive. Even if I assume it is, I do not know how to find easily the eigenvectors (apart from resolving a system which will take a very long time). There has to be something that I am missing.

PS: this comes from an hyperbolic PDE where I need to show the system hyperbolicity: $$ \left\{ \begin{array}{l} \partial_t \rho+\partial_x(\rho(\alpha-\rho))=0\\ \partial_t(\rho\alpha)+\partial_x(\rho\alpha(\alpha-\rho))=0 \end{array} \right. $$ and we chose the change of variables ${\bf u} =(\rho,\alpha\rho)\to {\bf v} = (\rho,v)$ with $v=\alpha-\rho$.

Answer 1

The initial system reads \begin{aligned} \rho_t + (\rho (\alpha-\rho))_x &= 0 \\ (\rho\alpha)_t + (\rho\alpha (\alpha - \rho))_x &= 0 \end{aligned} for the conserved variables ${\bf u} = (\rho, \rho\alpha)$. Now we introduce the new vector of conserved variables $ {\bf v} = (\rho, \alpha-\rho) $. With $v=\alpha-\rho$, the first equation of our system rewrites as $$ \rho_t + (\rho v)_x = 0 $$ and the second one becomes $$ (\rho (\rho+v))_t + (\rho v (\rho+v))_x = 0 \, . $$ Using the first equation and the product rule, we may rewrite the latter as $$ v_t + (v-\rho) v_x = 0 \, , $$ where we have also used $\rho\neq 0$. Therefore, the Jacobian matrix of the flux such that ${\bf v}_t + A {\bf v}_x = \bf 0 $ reads $$ A = \begin{bmatrix} v & \rho\\ 0 & v-\rho \end{bmatrix} $$ which eigenvalues are the diagonal entries. They are distinct provided $\rho>0$, and the system is then strictly hyperbolic.

Answer 2

Note that we can write $$ \Delta=v^2+4\rho^2v^2+\rho^4+4\rho^3v+6v\rho^2\\ = (1+4\rho^2)v^2+(4\rho^3+6\rho^2)v + \rho^4 $$ Consider the polynomial $p(v) = \Delta$. We see that this is a quadratic function with positive coefficients, so it will hold that $p(v) \geq 0$ for all $v$ if and only if its discriminant is either zero or negative. We find the discriminant of $p$ to be $$ \Delta_v = (4\rho^3 + 6\rho^2)^2 - 4\rho^4(1 + 4 \rho^2) = 48\rho^5 + 32\rho^4. $$ For positive $\rho$, $\Delta_v$ is positive. So, there must exist a value of $v$ for which $\Delta$ is negative, which in turn means that there are complex eigenvalues, contrary to your supposition.