Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In our class notes we are asked to verify the following equality:

$$E(t,x)\ast (g(x)\delta(t)) = t\int_{\omega\in S^2}{\frac{g(x-t\omega)}{4\pi}dS(\omega)}$$ where $E(t,x)=\frac{1}{4\pi|x|}\delta(t-|x|)$ is the fundamental solution to the wave equation in $\mathbb{R}^3$.

My problem with this is that I am not sure how to deal with the convolution of these two distributions.

First, seeing as they both have compact support, then their convolution (as distributions) is well defined. Moreover, since $f(x)\delta_{a}(x)\ast g(x)\delta_{b}(x) = f(a)g(b)\delta_{a+b}(x)$, where $\delta_{a}(x)=\delta(x-a)$, then I would conclude that $$E(t,x)\ast(g(x)\delta(t)) = \frac{g(x)}{4\pi|x|}\delta(t-|x|).$$ However this is still only a distribution, and I cannot figure out how this would be equivalent to the given solution.

Second, if we appeal to the integral definition of the convolution, then we have $$E(t,x)\ast(g(x)\delta(t)) = \frac{1}{4\pi}\int_{\mathbb{R}^3}\int_{-\infty}^{\infty}{\frac{1}{|y|}\delta(s-|y|)g(x-y)\delta(t-s)}dsdy$$ which gives us an integral involving two delta distributions, again leaving me at a loss for how to proceed.

Any help in clarifying this equality would be greatly appreciated.

share|cite|improve this question
I realize my first approach was incorrect, as I was considering it as only a convolution in the $t$ variable, and not also in the $x$ variables. The correct method is below. – Patch Mar 5 '12 at 23:24
up vote 1 down vote accepted

So I finally figured this out, and figured I'd post the solution for future reference:

Going with the integral definition of convolution, we have \begin{align} E(t,x)\ast(g(x)\delta(t)) &= \frac{1}{4\pi}\int_{\mathbb{R}^3}\int_{-\infty}^{\infty}{\frac{1}{|y|}g(x-y)\delta(s-|y|)}\delta(t-s)dsdy\\ &= \frac{1}{4\pi}\int_{\mathbb{R}^3}|y|^{-1}g(x-y)\left(\int_{-\infty}^{\infty}{\delta(s-|y|)\delta(t-s)}ds\right)dy\\ &= \frac{1}{4\pi}\int_{\mathbb{R}^3}|y|^{-1}g(x-y)\left(\delta_{|y|}(t)\ast\delta_{0}(t)\right)dy\\ &= \frac{1}{4\pi}\int_{\mathbb{R}^3}|y|^{-1}g(x-y)\delta(t-|y|)dy\\ &= \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{\infty}r^{-1}g(x-\tilde{y})\delta(t-r)r^{2}\sin\phi dr d\theta d\phi\\ &= \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}\sin\phi\left(\int_{0}^{\infty}rg(x-\tilde{y})\delta(t-r)dr\right) d\theta d\phi\\ &= \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}\sin\phi\left\langle\delta(t-r),rg(x-\tilde{y})\right\rangle d\theta d\phi\\ &= \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}tg(x-t\omega)\sin\phi d\theta d\phi\\ \end{align} where $\omega\in S^{2}$, and since $\delta(t-r)=\delta_{t}(r)$. But $\sin\phi d\theta d\phi = dS(\omega)$, so we get the final equality $$\frac{t}{4\pi}\int_{\omega\in S^{2}}{g(x-t\omega)dS(\omega)}.$$

share|cite|improve this answer
You should clarify your $\tilde{y}$. – Vobo Mar 7 '12 at 20:18
By $\tilde{y}$ I just mean $y$ expressed in spherical coordinates; also $x$ is fixed inside the integral, hence no tilde. – Patch Mar 9 '12 at 11:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.