Finding the Riemann Invariants of a system of 2 PDEs I've been asked to find the Riemann Invariants for the system:
$$
\begin{pmatrix} 
\cos(v) & 0 \\
0 & \cos(v)
\end{pmatrix}  
\begin{pmatrix} 
u_x \\
v_x 
\end{pmatrix} +
\begin{pmatrix} 
\sin(v) & -1 \\
-1 & \sin(v)
\end{pmatrix} 
\begin{pmatrix} 
u_y \\
v_y 
\end{pmatrix} = 0$$
It's easy to show that the two families of characteristic projections satisfy:
$$\frac{dy}{dx} = \tan(v) \pm \sec(v)$$
and that along these curves we have, for $\underline{u} = (u,v)$:
$$(1,\mp1) (\cos(v)\frac{\partial \underline{u}}{\partial x} +(\sin(v)\pm1)\frac{\partial \underline{u}}{\partial y}) = 0   $$
I'm only really having trouble with the last step, which is finding the Riemann invariants from this (i.e. the $R_\pm$ such that the above expression can be written as $\frac{d}{dx}R = 0$.)
Any help in seeing what these Riemann invariants are would be appreciated.
 A: Left-dividing by the matrix $\cos v \, \bf{I}$ for $\cos v \neq 0$, we rewrite the system as ${\bf U}_x + {\bf A}{\bf U}_y = {\bf 0}$ where
\begin{aligned}
{\bf A} &= \begin{pmatrix}
\tan v & -\sec v\\
-\sec v & \tan v
\end{pmatrix} \\[0.5em]
&= \frac12 \begin{pmatrix}
1 & -1\\
1 & 1
\end{pmatrix} \begin{pmatrix}
\tan v -\sec v & 0\\
0 & \tan v +\sec v
\end{pmatrix} \begin{pmatrix}
1 & 1\\
-1 & 1
\end{pmatrix}
\end{aligned}
and ${\bf U} = (u,v)^\top$. The eigenvectors of ${\bf A}$ are the vectors $(\pm 1, 1)^\top$, corresponding to the eigenvalues $\lambda_\mp({\bf U}) = \tan v \mp \sec v$. The corresponding Riemann invariants $R_\mp$ have their gradient orthogonal to the eigenvectors $(\pm 1, 1)^\top$, i.e.
$$
\begin{pmatrix}\pm 1\\ 1\end{pmatrix}
\cdot\nabla_{\bf U} R_\mp = 0 ,\qquad\text{e.g.}\qquad
\nabla_{\bf U} R_\mp = \begin{pmatrix}\mp 1\\ 1\end{pmatrix} .
$$
Therefore, we find $R_\mp = v \mp u$ up to some arbitrary constants. Note that
\begin{aligned}
(R_\mp)_x &= v_x \mp u_x \\
&= -(\tan v \pm \sec v) (v_y \mp u_y) \\
&= -\lambda_{\pm}({\bf U})\, (R_\mp)_y \, .
\end{aligned}
Characteristic curves are given by $dy/dx = \lambda_{\mp}({\bf U})$. Along a curve parametrized by $y(x)$, we have
$$
\frac{d}{dx}(R_\mp) = (R_\mp)_x + \frac{d y}{dx} (R_\mp)_y = \left( \frac{d y}{dx} -\lambda_{\pm}({\bf U}) \right) (R_\mp)_y
$$
so that the '$+$' Riemann invariant is constant along the other ('$-$') characteristic curve (and vice versa). Integral curves ${\bf U} = {\bf V}(\xi)$ are given by $d {\bf V}/d\xi = (\pm 1, 1)^\top$. Along such a curve, we have
$$
\frac{d}{d\xi}(R_\mp) = \nabla_{\bf U}R_\mp\cdot \frac{d{\bf V}}{d\xi} = 0
$$
so that the '$+$' Riemann invariant is constant along its ('$-$') integral curve (and vice versa).

E. Godlewski, P.-A. Raviart, Numerical Approximation of Hyperbolic Systems of Conservation Laws, Springer, 1996 doi:10.1007/978-1-4612-0713-9
