what does mean a zero eigenvalue in a PDE? I understand that in a PDE, the eigenvalues are some kind of speed of information propagation. For a hyperbolic system of PDE's, with three eigenvalues and one of them being zero, what does this mean? Is it the same than having a system of two equations?
This is the system of equations (SWE):
\begin{align}
\frac{\partial h}{\partial t} + \frac{\partial}{\partial x_i}(h \overline{u_i}) &= 0 \nonumber \\
%Depth averaged and filtered momentum equation
\frac{\partial (h{\overline{u_{i}}})}{\partial t}+\frac{\partial}{\partial x_{j}} \left(h {\overline{u_{i}}}\ {\overline{u_j}}\right)&=-\frac{1}{2} \frac{\partial g h^2}{\partial x_i}- gh\frac{\partial z}{\partial x_{i}}+ \frac{1}{\rho}(\overline{\tau_{\eta_i}}-\overline{\tau_{b_i}})
\label{swesimpl}
\end{align}
where $h,u_i$ and $u_j$ are the unknowns. 
I'm using and augmented version of Roe solver to solve numerically this system of PDE's. So the augmented version of the Roe solver provides estimates for the three wave speeds (the eigenvalues), and depending of the sign of such waves, there is an expression for the fluxes to compute the solution at the next time step, via finite volume method. I'm working in a two dimensional case, that means there are three unknowns (three eigenvalues). But for certain numerical simulation, sometimes I get a zero eigenvalue, the solver does not take into account that possibility. So I'm wondering if I could use the traditional solver (which was developed for a one dimensional case, i.e. two eigenvalues).
However, I'm not sure about the physical meaning of having one of the three eigenvalues being zero and if this can be considered to be equal to have only two eigenvalues...

If the eigenvalues are the wave propagation speeds, then a zero eigenvalue just says that one wave is stationary
