# Solve the wave equation explicitly on a helf-line with b.c.

A question from my pde homework:

Let $\alpha$ be constant, $\alpha \neq -1.$ Consider the wave equation on $x>0$, $t>0$ with the following data: $$u_{tt}-u_{xx}=0 \;\;\;\;\text{for x>0, t>0}\\u_t = \alpha u_x\;\;\;\; \text{at x=0} \\ u = f(x) \;\;\;\;\text{at t=0} \\ u_t = g(x) \;\;\;\;\text{at t=0}$$

Assume that $f$ and $g$ vanish near $x=0$. Give a formula for $u$. (Hint: start with $u=F(x+t)+G(x-t)$; find $F$ and $G$.) Why must $\alpha = -1$ be excluded? Can you do something even if $f$ and $g$ don't vanish near $x=0$?

First of all, how should I think of the condition $u_t = \alpha u_x$? I'm having a hard time intuitively picturing that.

Also, to employ the hint, I think should look along the line $x=t$, where $G$ is constant, and see how the function changes. But how can I get some information about this?

Ideas: I could try odd extension, but I don't see how that makes $G$ easier to find. I'm thinking of the way we can determine the solution to a wave equation in a bounded domain iteratively by looking on rectangles with sides parallel to $x=t$ and $x=-t$ (convenient because we know that the mixed derivative in these two directions is always zero). None of this is bearing fruit so far.

Suppose we have the wave equation in the semi plane: $$u_{tt}- c^2 u_{xx}=0 \\ u(x,0)=f(x) \\ u_t(x,0)=g(x)$$
$$u(x,t)=\frac{f(x+ct)+f(x-ct)}{2} + \frac{1}{2}\int_{x-ct}^{x+ct}g(\tau)d\tau$$ Then $$u_t(x,t)=\frac{f'(x+ct)-f'(x-ct)}{2} c + \frac{c}{2}\left(g(x-ct)+g(x+ct)\right)$$ $$u_x(x,t)=\frac{f'(x+ct)+f'(x-ct)}{2}+ \frac{1}{2}\left(g(x+ct)-g(x-ct)\right)$$ If we impose the condition $u_t(0,t)=\alpha u_x(0,t)$ we obtain $$\left(f'(ct)-f'(-ct)+ g(ct)+g(-ct)\right)\,c=\alpha \left(f'(ct)+f'(-ct) + g(ct)-g(-ct)\right)$$ Rearranging the terms in the above equation we get $$(f'(ct)-f'(-ct))c-(g(ct)-g(-ct))\alpha=\alpha(f'(ct)+f'(-ct))-c (g(ct)+g(-ct))$$ In the left hand side there is an odd function and in the right an even function. Therefore these terms should be both zero $$(f'(ct)-f'(-ct))c-(g(ct)-g(-ct))\alpha=0$$ and $$\alpha(f'(ct)+f'(-ct))-c (g(ct)+g(-ct))=0$$ These equations shows how to make the extension of $f$ and $g$ from $\mathbb{R}^+$ to $\mathbb{R}$.
• Great advice! I come up with $$\left[ \begin{matrix} 1 & \alpha \\ \alpha & 1 \end{matrix} \right] \left[ \begin{matrix} f'(-t) \\ g(-t) \end{matrix} \right] = \left[ \begin{matrix} f'(t) + \alpha g(t) \\ -\alpha f'(t) - g(t) \end{matrix} \right].$$ One thing I'm wondering about: I get that both $\alpha =1$ and $\alpha = -1$ are inconsistent, not just $\alpha = -1$. What am I missing? – Eric Auld Nov 16 '13 at 22:15