Solution to PDE wave equation using d'Alembert's formula and given that $f(x)$ and $g(x)$ are both odd functions.

Question

Show that if $f(x)$, $g(x)$ are both odd functions about $x_0$, then the solution of the wave equation, $u_{tt}=c^2u_{xx}$, is also odd about $x_0$. J. d'Alembert's solution is \begin{equation}u(x,t)=\frac{1}{2}(f(x-ct)+f(x+ct))+\frac{1}{2c}\int_{x-ct}^{x+ct}g(\xi)d\xi\end{equation} In other words, $u((x_0-x),t)=-u((x_0+x),t)$. Particularly $u(x_0,t)=0$ for all $t>0$. I have already shown that, given $f$ and $g$ are both odd, then the solution of the wave equation satisfies $u(-x,t)=-u(x,t)$. Particularly, $u(0,t)$ for all $t>0$. How do I show this is true around $x_0-x$, $x_0+x$?

Answer 1

The proof is very similar to the case $x_0 = 0$. Indeed, we have $$ 2\, u(x_0\pm x,t) = f(x_0\pm x-ct) + f(x_0\pm x+ct) + \int_{x_0\pm x-ct}^{x_0\pm x+ct} \frac{g(\xi)}{c} d\xi, $$ Let us introduce the change of variables (CoV) $\xi = x_0+\zeta$, such that \begin{aligned} \int_{x_0-x-ct}^{x_0-x+ct} \frac{g(\xi)}{c} d\xi &= \int_{-x-ct}^{-x+ct} \frac{g(x_0+\zeta)}{c} d\zeta & & \small\text{ [CoV $\text d \xi = \text d \zeta$]}\\ &= \int_{-x-ct}^{-x+ct} -\frac{g(x_0-\zeta)}{c} d\zeta & & \small\text{[oddness of $g$]}\\ &= \int_{x_0+x+ct}^{x_0+x-ct} \frac{g(\eta)}{c} d\eta & & \small\text{[CoV $\text d \eta = -\text d \zeta$]}\\ & = -\int_{x_0+x-ct}^{x_0+x+ct} \frac{g(\eta)}{c} d\eta , & &\small\text{[switching $\textstyle\int$ bounds]} \end{aligned} where we have successively used the oddness of $g$ and introduced $\eta = x_0-\zeta$. Using the oddness of $f$ at $x_0$, it follows immediately that \begin{aligned} 2\, u(x_0- x,t) &= -f(x_0+ x+ct) - f(x_0+ x-ct) - \int_{x_0+ x-ct}^{x_0+ x+ct} \frac{g(\eta)}{c} d\eta \\ &= -2\, u(x_0+x, t), \end{aligned} which means that $u$ is odd at $x_0$ for all $t$. Otherwise, one could have simply set $F(x) = f(x_0+x)$ and $G(x) = g(x_0+x)$ which are odd functions of $x$ at the origin, and reuse the result obtained for the d'Alembert formula about $x_0=0$.