How to solve a system of PDE $u_t+u_x=v, v_t+v_x=-u$

Question

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.

Answer 1

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 $\begin{cases}u(t,x)=f(x-t)\sin\dfrac{x+t}{2}+g(x-t)\cos\dfrac{x+t}{2}\\v(t,x)=f_{x-t}(x-t)\sin\dfrac{x+t}{2}+g_{x-t}(x-t)\cos\dfrac{x+t}{2}+f(x-t)\cos\dfrac{x+t}{2}-g(x-t)\sin\dfrac{x+t}{2}\end{cases}$

Answer 2

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.