# Method of characteristics for system of linear transport equations

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 $$v_0$$?

• A great solution to method of characteristics for a system of PDEs is given here. Nov 15, 2018 at 3:47

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).