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I am becoming frustrated in trying to understand the wave equation in the semiinfinte case: $ u_{tt} -c^2 u_{xx} =0 $ when $ x\geq 0 $

$u(x,0)=f(x) $

$ u_t(x,0)= g(x) $

and

$ u(0,t)=0 $ or $ u_x (0,t)=0$ . I know that in the first case we can expand $f,g$ as odd functions to $ (-\infty, \infty) $ and in the second case, we can expand $f,g$ as even functions.

The problem is that I can't understand why in the first case we need them to be odd but in the second case we need them to be even.... Can someone help me figure this out?

I couldn't find any good explanation online...

Thanks !

decarts
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  • Latex hint: to get $u_{xx}$, use braces around the subscripts; thus, to get $u_{xx}$, type u_{xx}, not u_xx. In these examples I obviously omitted the dollar signs. – Robert Lewis Aug 23 '13 at 07:51
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    I think the reason is as simple as observing that odd functions are zero at $x=0$ and even functions have zero slope at $x=0$. – Ron Gordon Aug 23 '13 at 08:24

1 Answers1

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Suppose we have the wave equation in the semi plane:

$$ u_{tt}- c^2 u_{xx}=0 \\ u(x,0)=f(x) \\ u_t(x,0)=g(x) $$

The solution is $$u(x,t)=\frac{f(x+ct)+f(x-ct)}{2} + \frac{1}{2}\int_{x-ct}^{x+ct}g(\tau)d\tau$$.

In particular $$u(0,t)=\frac{f(ct)+f(-ct)}{2} + \frac{1}{2}\int_{-ct}^{ct}g(\tau)d\tau = F(t) + G(t),$$

where (can be verified easily) $F:=\frac{f(ct)+f(-ct)}{2}$ is an even function and $G= \frac{1}{2}\int_{-ct}^{ct}g(\tau)d\tau$ is and odd function.

Note that if we have the aditional condition $u(0,t)=0$, then F and G must(should?) be zero. Then f and g must be odd functions. That is the reason why we should make an odd extension.

The case $u_x(0,t)=0$ is similar.

Pocho la pantera
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