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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.

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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 $\mathbb{R}$.

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    $\begingroup$ 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? $\endgroup$ – Eric Auld Nov 16 '13 at 22:15
  • $\begingroup$ 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...." $\endgroup$ – Username Unknown Sep 22 '19 at 22:03

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