How to solve a system of PDE $u_t+u_x=v, v_t+v_x=-u$ 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.
 A: Let $\partial_a$ denote the directional derivative in the $({1 \over \sqrt{2}},{1 \over \sqrt{2}})$ direction. Then your equations are
$$\partial_a u = {1 \over \sqrt{2}} v$$
$$\partial_a v = -{1 \over \sqrt{2}} u$$
Combining you get
$$\partial_a^2 u + {1 \over 2} u = 0$$
$$\partial_a^2 v + {1 \over 2} v = 0$$
These are easy to solve. Note that your initial conditions combined with the equations will also give expressions $\partial_a u(0,x)$ and $\partial_a v(0,x)$ which together with $u(0,x)$ and $v(0,x)$ will imply unique solutions.
A: 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.
A: Hint:
$u_{tt}+u_{xt}=v_t$
$u_{xt}+u_{xx}=v_x$
$\therefore v_t+v_x=u_{tt}+2u_{xt}+u_{xx}=-u$
$u_{tt}+2u_{xt}+u_{xx}+u=0$
Let $\begin{cases}p=x+t\\q=x-t\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial t}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial t}=\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial t}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial t}=\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial pq}-\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}-2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$
$\dfrac{\partial^2u}{\partial x\partial t}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}-\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial q^2}$
$\therefore\dfrac{\partial^2u}{\partial p^2}-2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}+2\left(\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial q^2}\right)+\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}+u=0$
$4\dfrac{\partial^2u}{\partial p^2}+u=0$
$u(p,q)=f(q)\sin\dfrac{p}{2}+g(q)\cos\dfrac{p}{2}$
$u(t,x)=f(x-t)\sin\dfrac{x+t}{2}+g(x-t)\cos\dfrac{x+t}{2}$
$u_t(t,x)=-f_t(x-t)\sin\dfrac{x+t}{2}+\dfrac{f(x-t)}{2}\cos\dfrac{x+t}{2}-g_t(x-t)\cos\dfrac{x+t}{2}-\dfrac{g(x-t)}{2}\sin\dfrac{x+t}{2}$
$u_x(t,x)=f_x(x-t)\sin\dfrac{x+t}{2}+\dfrac{f(x-t)}{2}\cos\dfrac{x+t}{2}+g_x(x-t)\cos\dfrac{x+t}{2}-\dfrac{g(x-t)}{2}\sin\dfrac{x+t}{2}$
$\therefore v(t,x)=f_{x-t}(x-t)\sin\dfrac{x+t}{2}+g_{x-t}(x-t)\cos\dfrac{x+t}{2}+f(x-t)\cos\dfrac{x+t}{2}-g(x-t)\sin\dfrac{x+t}{2}$
Hence $\begin{cases}u(t,x)=f(x-t)\sin\dfrac{x+t}{2}+g(x-t)\cos\dfrac{x+t}{2}\\v(t,x)=f_{x-t}(x-t)\sin\dfrac{x+t}{2}+g_{x-t}(x-t)\cos\dfrac{x+t}{2}+f(x-t)\cos\dfrac{x+t}{2}-g(x-t)\sin\dfrac{x+t}{2}\end{cases}$
