I asked the question on [math.stackexchange.com](https://math.stackexchange.com/questions/1090494/method-of-characteristics-of-a-system-of-first-order-pdes), 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. 


Could anyone please give some advice? 


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.



Method of characteristics of a system of first order pdes
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. 


Similar remarks apply to the second and fourth equations, so you really have four independent first order linear PDE for single unknowns, so the method of characteristics can be applied to each separately.


