$u_{tt} = u_{xx}$ with unusual boundary conditions

Question

Solve using the reflection method $u_{tt} = u_{xx}$

With $ x< ct , t > 0$ for some $c \in \mathbb{R}$

And $u(ct,t) = 0, u(x,0) = f(x), u_t(x,0) = 0$

I thought of rotating the $t$ axis in order for the boundary condition to be $u(0,t') = 0$. Would that work? How to do that?

Answer 1

As already pointed out, a typo is suspected in the wording of the problem. Nevertheless, supposing that the EDP is : $$u_{tt}=u_{xx}$$ The general solution is on the form : $$u(x,t)=F(x+t)+G(x-t)$$ where F and G are any fonctions (of course, at least two times derivable).

The boundary conditions are : $$u(ct,t)=F\left(ct+t)\right)+G\left(ct-t)\right)=0$$ $$u(x,0)=F(x)+G(x)=f(x)$$ $$u_t(x,0)=F'(t)-G'(t)=0$$ So, $G(t)=F(t)+$constant

which is the same as : $G(x)=F(x)+c$

$F(x)+G(x)=2F(x)+c=f(x)$ $$F(x)=\frac{1}{2}(f(x)-c)$$ $$G(x)=\frac{1}{2}(f(x)+c)$$ $$u(x,t)=\frac{1}{2}(f(x+t)-c)+\frac{1}{2}(f(x-t)+c)$$ $$u(x,t)=\frac{1}{2}\left(f(x+t)+f(x-t)\right)$$

The remaining condition $u(ct,t)=0$ leads to : $$f(ct+t)+f(ct-t)=0$$ or, which is the same : $$f\left( (c+1)x\right)+f\left( (c-1)x\right)=0$$

This is a functional equation. In the general case, any function $f$ is not solution and the problem has no solution.

But in the particular cases of function $f$ which is solution of the functional equation, the solution of the problem is : $$u(x,t)=\frac{1}{2}\left(f(x+t)+f(x-t)\right)$$

Answer 2

This is a problem where the wave is reflected from a moving node. The solution should, according to d'Alembert's formula, have a "forward" moving wave $f(x-t)$ and a backward moving wave $f(x+t)$. If the node moves with the speed smaller than wave speed, which here is $1$, i.e. $|c|<1$, then due to boundary at right there should also be a reflected wave $g(x+t)$. Now from physical considerations we can guess how $g(x+t)$ looks. It looks like a Doppler-extended/compressed in spacetime backward-moving wave $f(x+t)$, namely

$$g(x+t)=-f\left(\frac{c-1}{c+1}(x+t)\right).$$

You can then check by substitution $x=ct$ that the solution $$u(x,t)=\frac12\left(f(x-t)+f(x+t)-f\left(\frac{c-1}{c+1}(x+t)\right)\right)$$ satisfies the condition at moving boundary. It also satisfies the original wave equation and both initial conditions.

In the equations above $f(q)$ should be taken to be $0$ for $q\ge0$.

If $c\ge1$, then the node won't add any reflections, and you can just use the two first terms in the solution. If $c<-1$, then this will lead to creation of shock wave, and the solution may be singular near the boundary.

As for your idea of rotating $t$ to fix the boundary, I think you actually can do it, in a certain sense of "rotation": perform a Lorentz boost in the direction of the moving node, solve the equation with these spacetime coordinates with the fixed boundary node, then boost back to get the final solution.

Answer 3

Hint:

Let $\begin{cases}x_1=x-ct\\t_1=t\end{cases}$ ,

Then $u_x=u_{x_1}(x_1)_x+u_{t_1}(t_1)_x=u_{x_1}$

$u_{xx}=(u_{x_1})_x=(u_{x_1})_{x_1}(x_1)_x+(u_{x_1})_{t_1}(t_1)_x=u_{x_1x_1}$

$u_t=u_{x_1}(x_1)_t+u_{t_1}(t_1)_t=u_{t_1}-cu_{x_1}$

$u_{tt}=(u_{t_1}-cu_{x_1})_t=(u_{t_1}-cu_{x_1})_{x_1}(x_1)_t+(u_{t_1}-cu_{x_1})_{t_1}(t_1)_t=-c(u_{x_1t_1}-cu_{x_1x_1})+u_{t_1t_1}-cu_{x_1t_1}=u_{t_1t_1}-2cu_{x_1t_1}+c^2u_{x_1x_1}$

$\therefore u_{t_1t_1}-2cu_{x_1t_1}+c^2u_{x_1x_1}=u_{x_1x_1}$

$u_{t_1t_1}-2cu_{x_1t_1}+(c^2-1)u_{x_1x_1}=0$

The conditions $u(ct,t)=0$ , $u(x,0)=f(x)$ and $u_t(x,0)=0$ become $u(0,t_1)=0$ , $u(x_1,0)=f(x_1)$ and $u_{t_1}(x_1,0)=0$ respectively and the problem converts to the problem of usual type.