I need to show that the Riemann invariant $R = \frac{1}{2} (u^2+v^2) + \int \frac{c(p)^2}{p} dp $ is conserved along the characteristic $dy/dx = v/u$. My system of equations are: \begin{aligned} (pu)_x + (pv)_y &= 0 \\ p(uu_x + vu_y) + c(p)^2p_x &= 0 \\ p(uv_x +vv_y)+c(p)^2p_y &= 0 \end{aligned}
The system is hyperbolic if $u^2 + v^2 > c^2$ where $c=c(p)$ (cf. this post). The left eigenvector corresponding to the eigenvalue $\lambda = dy/dx = v/u$ is $(0,1,v/u)$. To find the Riemann invariant I do: $$ \begin{pmatrix} 0 & 1 & v/u \end{pmatrix} \begin{pmatrix} u & p & 0 \\ c^2 & pu & 0 \\ 0 & 0 & pu \end{pmatrix} \frac{d}{dx}\begin{pmatrix} p \\ u \\ v\end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} $$ which gives me $$ \begin{pmatrix} c^2 & pu & pv \end{pmatrix} \frac{d}{dx}\begin{pmatrix} p \\ u \\ v\end{pmatrix}= \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} $$ I am stuck on the final part as to how this implies that $\frac{d}{dx}[R] =0$. I am struggling to see how $\frac{d}{dx} \int \frac{c^2(p)}{p}dp = \frac{c^2(p) {dp}/{dx}}{p}$ which would make $\frac{d}{dx}[R] =0$.