# How to interpret $\delta(x\pm ct)$ as the solution to the wave equation?

Consider the equation:

$$\left(\Delta-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)u(x,t)=0,\qquad(x,t\in\mathbb{R})\qquad(\star)$$

where $c>0$ is a constant.

Then I believe we have general solutions of the form

$$u(x,t)=\delta(x\pm ct),$$

which are plane waves.

From a more rigorous perspective, how should I interpret this? Does it mean that $(\star)$ has no regular solutions which exist as functions and therefore require us to extend to the theory of distributions? In this case, I suppose we would have to weight $\delta$ against a test function $\varphi\in\mathscr{S}(\mathbb{R})$ so that

$$u(x,t)=\langle T_{\pm ct\ast} \delta_x,\varphi\rangle=\langle\delta_x,T_{\pm ct}^\ast\varphi\rangle=\langle\delta,\varphi\circ T_{\pm ct}\rangle,$$

for all $\varphi\in\mathscr{S}(\mathbb{R})$, where $\delta_x:\mathscr{S}(\mathbb{R})\to\mathbb{C}$ which maps $\langle\delta_x,\varphi\rangle=\varphi(x)$ is the Dirac measure at $x\in\mathbb{R}$, $T_{\pm ct\ast}\in\operatorname{End}(\mathscr{S}'(\mathbb{R}))$ is the extension of the continuous linear operator $T_{\pm ct}^\ast\in\operatorname{End}(\mathscr{S}(\mathbb{R}))$, where $T_{\pm ct}:\mathbb{R}\to\mathbb{R}$ is the translation operator which maps $x\mapsto x\pm ct$?

Of course, another way of interpreting this is to take the Fourier transform of $(\star)$ with respect to the $t$ variable, which yields

$$\left(\Delta+\frac{\omega^2}{c^2}\right)u(x,t)=0,$$

so that

$$u(x,\omega)=e^{\pm i\omega x/c},$$

which is well defined as a function.

I suppose that I am basically asking how one interprets plane waves in a mathematically rigorous manner?

We need not invoke generalized functions here. Note that the general solution to the wave equation

$$\left(\Delta -\frac1{c^2}\frac{\partial^2}{\partial t^2}\right)u(x,t)=0 \tag1$$

is given by

$$u(x,t)=f(x-ct)+g(x+ct)$$

for any arbitrary pair of twice differentiable functions $f$ and $g$.

Upon taking the Fourier Transform of $(1)$, we obtain

$$\left(\Delta +\frac{\omega^2}{c^2}\right)U(x,\omega)=0 \tag2$$

which has general plane wave solutions

$$U(x,\omega)=\frac1cF(-\omega)e^{-i\omega x/c}+\frac1cG(\omega)e^{i\omega x/c}$$

• Thanks for the answer! One of the signs in your expression for $u(x,t)=f(x+ct)+g(x-ct)$ is incorrect though and in equation (2) you should have a $+$ sign instead of a $-$ sign as the Fourier transform of the double derivative yields an $i^2=-1$. Commented Jun 16, 2017 at 16:44
• You're welcome. And yes, there were two typographical (sign) errors which have been corrected. Commented Jun 16, 2017 at 16:45
• Jason, feel free to up vote and accept an answer as you see fit of course. Commented Jun 16, 2017 at 17:36

Rather than being the general solution, this is something more like a fundamental solution - $G(x,t) = \frac12\left(\delta(x+ct) + \delta(x-ct)\right)$ is the (generalized) solution corresponding to the initial data $G(x,0) = \delta(x), G_t(x,0) = 0$. You should be able to check that this is true if you interpret the PDE in terms of distributional derivatives.

This means we can write the general solution for initial data $u(x,0) = u_0(x), u_t(x,0) = 0$ as $$u(x,t) = \int G(y,t) u_0(y) dy = \frac12\left(u_0(ct) + u_0(-ct)\right).$$

If you want to allow non-zero initial velocity then you can generalise this to the full d'Alembert formula. Note that this gives a smooth solution $u$ for any smooth initial data $u_0$, so we certainly don't need to talk about distributional solutions in the sense you suggested.