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If I have a system of pde $$\begin{cases} u_t+v_x=0\\ v_t+u_x=0\\ u(x,0)=u_0(x), v(x,0)=v_0(x)\end{cases}$$ how to extend the idea of method of characteristics to this situation? How do I parametrize the initial data and write the explicit solution to the problem for all $x$ and $t$ in terms of the functions $u_0$ and $v_0$?

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    $\begingroup$ A great solution to method of characteristics for a system of PDEs is given here. $\endgroup$
    – mattos
    Nov 15, 2018 at 3:47

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The system rewrites as ${\bf u}_t + {\bf M}\, {\bf u}_x = {\bf 0}$ with ${\bf u} = (u,v)^\top$. We diagonalize the matrix as ${\bf M} = {\bf S}\, {\bf J}\, {\bf S}^{-1}$ where $\bf J$ is diagonal. Setting ${\bf v} = {\bf S}^{-1}{\bf u}$, one obtains a diagonal system ${\bf v}_t + {\bf J}\, {\bf v}_x = {\bf 0}$ which rows can be solved independently by using the method of characteristics. Then, $\bf u$ is deduced from ${\bf u} = {\bf S}\,{\bf v}$. Here, we find \begin{aligned} u(x,t)&= \frac{1}{2}\big(u_0(x-t)+u_0(x+t)\big)+\frac{1}{2}\big(v_0(x-t)-v_0(x+t)\big)\\ v(x,t)&= \frac{1}{2}\big(v_0(x-t)+v_0(x+t)\big)+\frac{1}{2}\big(u_0(x-t)-u_0(x+t)\big) \end{aligned} This method works only for linear first-order systems ${\bf u}_t + {\bf M}\, {\bf u}_x = {\bf 0}$, which matrix $\bf M$ can be diagonalized in $\Bbb R$ (linear hyperbolic systems).

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