TITLE: How to solve a system of PDE $u_t+u_x=v, v_t+v_x=-u$
QUESTION [7 upvotes]: Solve the following initial value problem:
$$u_t+u_x=v,
\\v_t+v_x=-u,
\\u(0,x)=u_0(x), 
\\  
v(0,x)=v_0(x).$$
I did not learn any method to solve a system of PDE so I guess there is a "trick". So far we've learned how to solve PDE's of the form:
$$u_t+a(t,x)u_x=f(t,x,u)$$
I could really use help.
Thanks.

REPLY [2 votes]: The present system can be solved by the method of characteristics. Indeed, the two equations provide the same characteristic curves:

$t'(s) = 1$, letting $t(0)=0$ gives $t=s$;
$x'(s) = 1$, letting $x(0)=x_0$ gives $x=s+x_0$.

Hence, the characteristics are straight lines with slope 1 in the $x$-$t$ plane.
The first equation leads to $u'(s) = v(s)$ and the second equation leads to $v'(s) = -u(s)$, so that $u''(s) + u(s) = 0$ and $v''(s) + v(s) = 0$. These second-order differential equations can be solved by imposing the initial conditions $u(0) = u_0(x_0)$, $u'(0) = v_0(x_0)$ and $v(0) = v_0(x_0)$, $v'(0) = -u_0(x_0)$:
$$
\begin{aligned}
u(s) &= u_0(x_0) \cos s + v_0(x_0) \sin s\\
v(s) &= v_0(x_0) \cos s - u_0(x_0) \sin s
\end{aligned}
$$
Combining all equations deduced from the method of characteristics, we have
$$
\begin{aligned}
u(x,t) &= u_0(x-t) \cos t + v_0(x-t) \sin t\\
v(x,t) &= v_0(x-t) \cos t - u_0(x-t) \sin t
\end{aligned}
$$
For this type of linear system, it should be also possible to proceed by Fourier transform.