The shallow water equations written as:
$$h_t+vh_x+hv_x=0$$
$$v_t+gh_x+vv_x=0$$
Can be written as
$$h_t+(vh)_x=0$$
$$v_t+(gh + \frac{v^2}{2})_x=0$$
Put in a system of conservation laws $P_t+Q_x=0$ with $P=(h,hv)^T$ requires that they are in the form
$$h_t+(vh)_x=0$$
$$(vh)_t+(\frac{v^2}{2}+\frac{gh^2}{2})_x=0$$
How is the second PDE derived? It says that it is from conservation of momentum, so is it derived from a physical concept or by $\frac{d}{dt}\int^b_a P_k dx = \left[ Q_k \right]^b_a$?
Then once in this form $P_t+Q_x=0$ with $Q=(vh, \frac{v^2}{2}+\frac{gh^2}{2})^T$ to show that this is hyperbolic provided that $h>0$ I need to show that $Q=AP$ and $A$ has eigenvalues $\lambda_{\pm}=v\pm\sqrt{gh}$ In the old representation it was clear to see that:
$A=\left(
\begin{array}{cc}
v & g \\
h & v
\end{array}
\right)$
giving the correct eigenvalues, but in the new one I can not see how to do that.
Best Answer
Firstly, I believe your conservational form is incorrect: You want to put it in a on a form $u_t + f(u)_x = 0$. I did this in class a while back for the same equations, and found the following form, with $m=vh$:
$\left(\begin{array}{cc}h\\m\end{array}\right)_t + \left(\begin{array}{cc}m\\\frac{m^2}{h}+\frac{gh^2}{2}\end{array}\right)_x = 0$
As you see, where you have $\frac{v^2}{2}$, I have $\frac{m^2}{h} = \frac{v^2h^2}{h} = v^2h$. For a full derivation, see the report we wrote at the time, available here. It's right at the beginning of the report, in section 1.2. A little later in the report we also linearize the problem and find the eigenvalues, so hopefully you can also have a few of your questionmarks regarding that sorted out.
As to why this form is called "conservational", I must admit I don't remember the details, and they're not in that report either. However, the book we used ("Finite Volume Methods for Hyperbolic Methods" by LeVeque, it's cited with full info in the report) had a good explanation, so if you can get your hands on a copy (maybe it's part of the preview on Google Books?) I can recommend it.