Property of Dirac delta function in $\mathbb{R}^n$

Question

How does one prove the following identity?

$$\int _Vf(\pmb{r})\delta (g(\pmb{r})) \, d\pmb{r}=\int_S \frac{f(\pmb{r})}{\left|\operatorname{grad} g(\pmb{r})\right|} \, d\sigma$$

where $S$ is the surface inside $V$ where $g(\pmb{r})=0$ and it is assumed that $\operatorname{grad} g(\pmb{r})\neq 0$. Thanks.

Edit: I have proved a one-dimensional version of this formula:

$$\delta (g(x))=\sum _a \frac{\delta (x-a)}{\left|g'(a)\right|}$$

where $a$ goes through the zeroes of $g(x)$ and it is assumed that at those points $g'(a)\neq 0$. the integral can be divided into into a sum of integrals over small intervals containing the zeros of $g(x)$. In these intervals $g(x)$ can be approximated by $g(a)+(x-a)g'(a)=(x-a)g'(a)$ since $g(a)=0$. Thus

$$\int_{-\infty}^\infty f(x)\delta (g(x)) \, dx=\sum _a \int _{a-\varepsilon}^{a+\varepsilon }f(x)\delta ((x-a)g'(a)) \, dx$$

Using the property $\delta (kx)=\frac{\delta (x)}{|k|}$, it follows that

$$\int_{-\infty}^\infty f(x)\delta (g(x)) \, dx=\sum _a \frac{f(a)}{\left|g'(a)\right|}$$

This is the same result we would have obtained if we had written $\sum _a \frac{\delta (x-a)}{\left|g'(a)\right|}$ instead of $\delta (g(x))$ as a factor of the integrand.

Answer 1

Try replacing $\delta(x)$ with $\varphi_\epsilon(x)=\varphi(x/\epsilon)/\epsilon$, where $\varphi$ is a positive function of compact support and whose integral is $1$. For such $\varphi$, $\lim_{\epsilon\to 0}\;\varphi_\epsilon\to\delta$ in the sense of distributions. Near points $\pmb{r}\in S$, $g(\pmb{x})=(\pmb{x}-\pmb{r})\cdot \nabla g(\pmb{r})+o(\pmb{x}-\pmb{r})$.

On $S$, $\nabla g=\pmb{n}|\nabla g|$, where $\pmb{n}$ is the surface normal to $S$. So near $\pmb{r}\in S$, $$ \begin{align} \varphi_\epsilon(g(\pmb{x}))&=\varphi((\pmb{x}-\pmb{r})\cdot \nabla g(\pmb{r})/\epsilon)/\epsilon+o(\pmb{x}-\pmb{r})\\ &=\varphi((\pmb{x}-\pmb{r})\cdot \pmb{n}/\epsilon')/\epsilon'/|\nabla g(\pmb{r})|+o(\pmb{x}-\pmb{r})\\ &=\varphi_{\epsilon'}((\pmb{x}-\pmb{r})\cdot \pmb{n})/|\nabla g(\pmb{r})|+o(\pmb{x}-\pmb{r}) \end{align} $$ where $\varphi_{\epsilon'}((\pmb{x}-\pmb{r})\cdot \pmb{n})$ is an approximation of surface measure on $S$ near $\pmb{r}$.

Thus, $\delta(g(\pmb{r}))\;d\pmb{r}=\;\displaystyle{\frac{d\sigma}{|\nabla g(\pmb{r})|}}$ where $d\sigma$ is surface measure on $S$.

Answer 2

By Taylor series $g(\mathbf{x}) = g(\mathbf{r}) + \vec{\mathrm{grad} g(\mathbf{r})}.(\mathbf{x}-\mathbf{r}) + o(\vert \mathbf{x}-\mathbf{r} \vert)$ as a new coordinate in the vicinity of the surface, where $g(\mathbf{r})=0$. Change basis using $\mathbf{n}_1 = \frac{\vec{\mathrm{grad} g(\mathbf{r})}}{\vert{\mathrm{grad} g(\mathbf{r})}\vert}$ as a first vector, and remaining $\mathbf{n}_i$ for $i=2, \ldots, n$ are chosen by Gram orthogonalization procedure. Let $t_i$ be coordinates in this system, $\mathbf{r} = \sum_i t_i \mathbf{n}_i$. Then $dV_x = dx_1 \wedge d x_2 \wedge \ldots \wedge d x_n = \vert J \vert dt_1 \wedge d t_2 \wedge \ldots \wedge d t_n = dV_t$.

$$ \int f(\mathbf{r}) \delta( g(\mathbf{r})) dV_x = \int f(\mathbf{r}) \delta( \vert \mathrm{grad} g(\mathbf{r}) \vert t_1 ) dV_t = \int f(\mathbf{r}) \frac{1}{\vert \mathrm{grad} g(\mathbf{r}) \vert }\delta( t_1 ) dV_t $$

Integration overt $t_1$ produces $d \sigma$.

This is a little hand-wavy, but gives you an idea.

Answer 3

What you are quoting is a general statement about pull-backs of distributions. Since I am not entirely sure of your background, I won't try to give a detailed explanation here. Rather, I will refer you to Chapter 7 of Friedlander and Joshi's Introduction to the Theory of Distributions.