# 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.

## 1 Answer

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
• How do you know this? "In the left hand side there is an odd function and in the right an even function. Therefore these terms should be both zero...." – Username Unknown Sep 22 '19 at 22:03