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?
Best Answer
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).