A Clarification on Computing the Riemann Invariant for the Shallow Water Equation The shallow water wave equation is given by:
$$
\begin{bmatrix}
h \\
u 
\end{bmatrix} _t +
\begin{bmatrix}
u & h \\
g & u
\end{bmatrix}
\begin{bmatrix}
h \\
u 
\end{bmatrix} _x
$$
The eigenvalues and eigenvectors are: $\lambda_\pm = u \pm \sqrt{gh} \quad r_\pm =\begin{bmatrix}
h \\
\pm \sqrt{gh} 
\end{bmatrix}$
This document outlines how to compute the Riemann invariant. In particular, using the components of the eigenvectors (say for the positive one) they write: $$0 = h \, du + \sqrt{gh} \, dh = h\left(du + \sqrt{\frac g h}\,dh\right) = h\,\,d(u + 2\sqrt{gh})$$
I follow the computation, but in the eigenvector, it looks like the first coefficient is associated with the $h$-coordinate (in this case $h$) and the second is associated with the $u$-coordinate (in this case $\sqrt{gh}$). However when they write the differential, they associate $h$ with $du$ and $\sqrt{gh}$ with $dh$. This confuses me, why is the differential switched?
 A: The left eigenvectors are used, not the right ones. This is because you are looking for a function of $w=(h,u)$ that is constant on one of the curves where $\frac{dx}{dt}=\lambda$. If $f$ is such a function then on that curve you have by the chain rule
$$0=\frac{df}{dt}= [f_h \, f_u](w_t+\lambda w_x)=[f_h \, f_u](-A+\lambda)w_x.$$
This holds if the gradient of $f$ is a left eigenvector.

First note that $(\sqrt{gh} \, h)$ is a left eigenvector of
$\begin{bmatrix}
u & h \\
g & u 
\end{bmatrix}$
so that we have:
\begin{align}
0 &= \left[\sqrt{gh} \quad h\right] \left(
-\begin{bmatrix}
u & h \\
g & u 
\end{bmatrix}
+ \lambda\right) \begin{bmatrix} h \\ u
\end{bmatrix}_x
\\
& \text{continuing...} \\
&= h\left[\sqrt{\frac g h} \quad 1\right] \left(
\begin{bmatrix}
h \\
u 
\end{bmatrix}_t
+ \lambda \begin{bmatrix} h \\ u
\end{bmatrix}_x \right)
\end{align}
Letting $f_h = \sqrt{\frac g h}, f_u = 1, \lambda = \frac{dx}{dt}$ we have:
\begin{align}
&= h\left[f_h \quad f_u\right] \left(
\begin{bmatrix}
h \\
u 
\end{bmatrix}_t
+ \frac{dx}{dt} \begin{bmatrix} h \\ u
\end{bmatrix}_x \right) \\
&= h \frac{df}{dt}
\end{align}
Where $f = u + 2\sqrt{gh}$ is constant along characteristics given by $\lambda$ ($h$ is non-zero in general).
