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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.

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Maybe a citation would help someone answer? –  anon Feb 26 '12 at 4:03
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1 Answer 1

up vote 2 down vote accepted

EDIT

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!).

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Yeah, there should be a 2 at the denominator. Great solution, thx –  CC_Azusa Feb 26 '12 at 16:29
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