# Solving linear second order hyperbolic PDE $\nabla \cdot \left( M \nabla u\right)=0$

Let $$\Omega \subset \mathbb{R}^{2}$$ be a bounded and connected domain with a smooth boundary $$\Gamma$$. Furthermore, let $$M$$ be a matrix-valued function, where the entries $$m_{ij}$$ are smooth and bounded for each $$i,j \in \{ 1,2\}$$. We also assume that $$M$$ is symmetric, i.e., $$m_{12}=m_{21}$$. Furthermore, we assume that $$M$$ is indefinite, i.e. it has one positive and one negative eigenvalue for every $$(x,y) \in \Omega$$ and at least one zero eigenvalue for $$(x,y) \in \Gamma$$.

Then I want to show that the PDE
\begin{align} \nabla \cdot \left( M \nabla u\right)=0, \qquad & \text{ in } \Omega, \\ u=g \qquad & \text{ on } \Gamma, \end{align} has a unique solution, where $$g$$ is some arbitrary smooth function.

This is a hyperbolic PDE in $$\Omega$$ and parabolic on the boundary $$\Gamma$$. This is clear since, by using the the product rule and $$u_{xy}=u_{yx}$$ we get, $$\nabla \cdot \left( M \nabla u\right) = m_{11}u_{xx} + m_{12}u_{xy} + m_{22}u_{yy} + \text{lower order terms}$$ Then $$M$$ is hyperbolic if $$m_{12}^{2}-m_{11}m_{22}>0$$. Which holds since,

$$M$$ is indefinite $$\implies$$ $$\operatorname{det}(M) < 0$$ $$\implies$$ $$m_{11}m_{22}-m_{12}^{2}<0$$ $$\Leftrightarrow$$ $$m_{12}^{2}-m_{11}m_{22}>0.$$

Solving the special case for $$g=0$$ (is $$u=0$$ the only solution), would already be helpful for me. There is extensive literature for similar problems when $$M$$ is positive (non-negative) PDE is (sub)-elliptic. I assume there should be some literature on my hyperbolic problem as well, however, I could not find any.