Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

I'm reading a paper about the fundamental solution of the wave operator in $\mathbb{R}^3$. The author said that the fundamental solution equals $$cH(t)\delta(t^2-|x|^2)=c/2tH(t)\delta(t-|x|)$$ where c is a constant depends on $n$. I don't quite understand how the equality is drived.

share|cite|improve this question
Maybe a citation would help someone answer? – anon Feb 26 '12 at 4:03
up vote 3 down vote accepted


The Dirac delta distribution of a continuously differentiable function $g(t)$ is given by (see Wikipedia) : $\displaystyle \delta(g(t))=\sum_i \frac{\delta(t-t_i)}{|g'(t_i)|}$ when the $g$ function has simple roots $t_i$.
This may be deduced from $g(t) \thicksim g'(t_i)(t-t_i)$ as $t\to t_i$ and $\delta(a\cdot(t-t_i))=\frac{\delta(t-t_i)}{|a|}$.

Applying this to $g(t)=t^2-x^2$ with $x$ any real :
$$ \delta(t^2-x^2)=\frac{\delta(t-x)}{|2t|}+\frac{\delta(t+x)}{|2t|}=\frac{\delta(t-|x|)}{2|x|}+\frac{\delta(t+|x|)}{2|x|}$$

We may conclude using $H(t)\cdot \delta(t+|x|)=0\ $ (because $t+|x|=0$ for $t\lt 0$ only where $H(t)=0$) and get : $$ H(t)\delta(t^2-x^2)=\frac{H(t)\delta(t-|x|)}{2|x|}$$

(you may write $t$ or $|x|$ at the denominator since the $\delta$ distributions cancel all other contributions!).

Note that I get an additional $2$ at the denominator (I think it's needed!).

share|cite|improve this answer
Yeah, there should be a 2 at the denominator. Great solution, thx – CC_Azusa Feb 26 '12 at 16:29

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.