How to linearize system of equations with partial derivatives? I have a system of equations that I wish to linearize around an equilibrium point. The system is as follows:
We have two functions $h(x,t)$ and $u(x,t)$ and constants $d$ and $g$. We then have the following system of equations:
$$\frac{\partial h}{\partial t} + \frac{\partial (h+d)u}{\partial x} = 0$$
$$\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + g\frac{\partial h}{\partial x} =0$$
The equilibrium point I wish to linearize around is $h(x,t)=0; u(x,t)=0$. I have tried setting the system into the form $\mathbf{x'} = \mathbf{f}(\mathbf{x})$ where $\mathbf{x} = [h,u]^T$ and $\mathbf{f}$ denotes the rest of the system. Using regular linearization techniques only results in the system becoming zero, which is not what I expected. I suppose it has something to do with the derivatives to $x$ in $\mathbf{f}$, but I'm not sure what to do. What am I doing wrong? 
 A: These are the one-dimensional Saint-Venant equations. It is a non-linear system of first-order PDEs that can be rewritten as
\begin{aligned}
&\frac{\partial A}{\partial t} + \frac{\partial (Au)}{\partial x} = 0 \\
&\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + g\frac{\partial h}{\partial x} = 0
\end{aligned}
with $A = h+d$. The linearization attempt in OP is more appropriate for ODEs.
Let us linearize the previous PDE system around $(h,u) = 0$. Linearization around $(h,u) = 0$ is equivalent to linearization around $(A,u) = (d,0)$. Therefore, the quadratic terms $\partial_x(hu)$ and $u \partial_x u$ are discarded. The linearized Saint-Venant system reads:
\begin{aligned}
&\frac{\partial h}{\partial t} + d\frac{\partial u}{\partial x} = 0 \\
&\frac{\partial u}{\partial t} + g\frac{\partial h}{\partial x} = 0
\end{aligned}
With the state vector ${\bf x} = [h,u]^T$ introduced in OP, this can be written as a linear hyperbolic system $\partial_t {\bf x} + {\bf A}\, \partial_x {\bf x} = {\bf 0}$, where the matrix ${\bf A}$ depends on the constants $g$, $d$.
