Delta distributions with nonlinear arguments

Question

I am confused by the use of nonlinear arguments with the Dirac $\delta$ distribution that I am encountering in the literature. This looks like a widespread use, but for concreteness let us focus on a single example. In this paper by Tataru and Geba, the very first formula is $$\tag{1} K(t, x)=c_n 1_{t>0} \begin{cases} (t^2-x^2)_+^{-\frac{n-1}{2}} & n \text{ even} \\ \delta^{\left(\frac{n-3}{2}\right)} (t^2-x^2) & n\text{ odd} \end{cases}$$

Big question. How to interpret this formula?

According to the linked paper, formula (1) is the solution to this problem: \begin{equation} \begin{cases} \frac{\partial^2 u}{\partial t^2}-\Delta u =0 \\ u(0)=0,\ \frac{\partial u}{\partial t}(0)=\delta(x). \end{cases} \end{equation} To gain some insight, I tried to restrict myself to the $n=3$ case, where Kirchhoff's formula tells us that $$K(t, x)=c_3 1_{t>0} \frac{1}{t} d\sigma_t, $$ where $d\sigma_t$ is the surface measure on the sphere of radius $t$.

Small question. Equating Kirchhoff's formula and formula (1) I infer that the following identity should hold true: $$\tag{?} \delta(t^2-\lvert x\rvert^2)=\frac{1}{t}d\sigma_t, \qquad x \in \mathbb{R}^3,\ t>0.$$ Can we prove (?) directly?

Understanding the identity (?) could be a good start towards the understanding of the more general formula (1), in which the superscript ${n-3\over 2}$ does not vanish.

P.S. The answer to the questions in this post might lie somewhere in Chapter 5 of the book "The Fourier transform and its applications" by Bracewell.

Indeed, the general formula for solution has derivative of order $(n-3)/2$. I have not yet succeeded at matching the exponents of $t$, however. — Post No Bulls, Dec 17 '13 at 08:44
@paulgarrett: I agree that it must be a derivative of some kind. — Giuseppe Negro, Dec 17 '13 at 08:56
Sergiu Klainerman's notes contain a very clear description of the fundamental solution to the wave equation. — Giuseppe Negro, Jul 03 '16 at 22:42
Related (this answer by robjohn contains a very clear physicist's explanation of the pullback formula for the Dirac delta). — Giuseppe Negro, Oct 13 '16 at 16:17

score 2 · Accepted Answer · answered Feb 23 '15 at 11:26

I suppose (key word here is suppose) that if we have the function $f(t,x)$ on $\Bbb R\times \Bbb R^d$ such that the equation $f(t,x)$ gives a manifold $M$ of dimension $d$ without border (problems arise when we have degenerations like $t^2-x^2=0$ in the point $(0,0)$), then we can define $\delta(f(t,x))$ as a measure on this manifold $M$.

If we need $\delta^{(k)}(f(t,x))$, then we can define it as $(\delta^{(k)}(f(t,x)),\phi) =(-1)^k (\delta (f(t,x)),\frac{\partial^k\phi}{\partial n^k}) $ with $\frac{\partial^k\phi}{\partial n^k}$ being the derivative of $\phi$ in the direction of normal vector to $M$.

Delta distributions with nonlinear arguments

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