# Method of characteristics of a system of first order pdes

I asked the question on math.stackexchange.com, but didn't get any reply. So, I asked it again here. Any suggestion or hint is welcome, and thank you for your attention.

Consider the system of first order PDEs

$\left\{ \begin{eqnarray} \frac{\partial}{\partial t} v_1 + \frac{\partial}{\partial x_1} p_1 + \eta(x_1) v_1 = 0 \\ \frac{\partial}{\partial t} v_2 + \frac{\partial}{\partial x_2} p_2 + \eta(x_2) v_2 = 0 \\ \frac{\partial}{\partial t} p_1 + \frac{\partial}{\partial x_1} v_1 + \eta(x_1) p_1 = f(t) \\ \frac{\partial}{\partial t} p_2 + \frac{\partial}{\partial x_2} v_2 + \eta(x_2) p_2 = 0 \end{eqnarray} \right.$

with initial condition $v_1 = v_2 = p_1 = p_2 = 0$ at $t=0$, where $v_1,v_2,p_1,p_2 : \mathbb{R}^3 \to \mathbb{R}$ are the unknown functions, $\eta, f: \mathbb{R} \to \mathbb{R}$ are given smooth functions.

I would like to consider this system of PDEs with the method of characteristics. I know the method for a single PDE, but I don't know how to find the characteristics of such a system.

Regards.

I could understand that this system is a hyperbolic one.

Take $U=(v_1,v_2,p_1,p_2)'$, then the system can be rewritten as

$\frac{\partial}{\partial t} U + \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 1 ~ & 0 \\ 0 & 0 & 0 & 0 \\ 1 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \frac{\partial}{\partial x_1} U + \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 1 \\ 0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) \frac{\partial}{\partial x_2} U + \left( \begin{array}{cccc} ~\eta(x_1) ~ & 0 ~ & 0 ~ & 0 \\ 0 & \eta(x_2) & 0 & 0 \\ 0 ~ & 0 ~ & \eta(x_1) ~ & 0 \\ 0 & 0 & 0 & \eta(x_2) \end{array} \right) U =\left( \begin{array}{cccc} 0 \\ 0 \\ f(t) \\ 0 \end{array} \right) .$

Take $B_1 = \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 1 ~ & 0 \\ 0 & 0 & 0 & 0 \\ 1 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)$, $B_2 = \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 1 \\ 0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 1 & 0 & 0 \end{array} \right)$, then define a matrix

$B (y_1,y_2)= y_1 B_1 + y_2 B_2 = \left( \begin{array}{cccc} ~0 ~ & 0 ~ & y_1 ~ & 0 \\ 0 & 0 & 0 & y_2 \\ y_1 ~ & 0 ~ & 0 ~ & 0 \\ 0 & y_2 & 0 & 0 \end{array} \right)$.

The eigenvalues of $B$ are $-y_1, y_1, -y_2, y_2$, and the corresponding eigenvectors are $(-1, 0, 1, 0)', (1, 0, 1, 0)', (0, -1, 0, 1)', (0, 1, 0, 1)'$.

Since the eigenvalues of $B$ are distinct and all real and the four eigenvectors are linearly independent, the system of first order PDEs is hyperbolic.

First of all, your system seems to uncouple quite strongly. The first and third equations only involve the unknowns $v_1$ and $p_1$ and the second and fourth equations only involve $v_2$ and $p_2$, so you are really talking about two independent problems, so you should separate them this way.
Second, if you add the first and third equations and set $z = p_1 + v_1$, this becomes the single PDE $$\frac{\partial z}{\partial t} + \frac{\partial z}{\partial x_1} + \eta(x_1)z = f(t),$$ so it's an equation for a single unknown, easily solved by the method of characteristics. Similarly, if you subtract the first equation from the third and set $w = p_1 - v_1$, then you get $$\frac{\partial w}{\partial t} - \frac{\partial w}{\partial x_1} + \eta(x_1)w = f(t),$$ again, an equation for a single unknown.