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How does one prove the following identity?

$$\int _Vf(\pmb{r})\delta (g(\pmb{r}))d\pmb{r}=\int _S\frac{f(\pmb{r})}{|\text{grad} g(\pmb{r})|}d\sigma$$

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

$$\int _{-\infty }^{\infty }f(x)\delta (g(x))dx=\sum _a \int _{a-\epsilon }^{a+\epsilon }f(x)\delta \left((x-a)g'(a)\right)dx$$

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.

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1  
The $\delta(g(\mathbf r))$ is, hmm, funny. –  Mariano Suárez-Alvarez Aug 11 '11 at 16:40
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The fact that $\delta$ is not a function in the mathematical sense was recently pointed to you math.stackexchange.com/questions/56681/ux-deltax/56684#56684 as well as a rigorous way to avoid this problem in a specific case. You could try to adapt this argument to the question at hand. // And the usual: what have you tried, what are your definitions, where are you stuck, and so on. –  Did Aug 11 '11 at 16:47
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@Gortaur: as Didier recalls, $\delta$ is not a function, so compositions such as $\delta(g(\mathbf r))$ can only be treated with a certain amount of precision in the definitions. –  Mariano Suárez-Alvarez Aug 11 '11 at 17:02
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In this case I suggest you read seriously (and ponder for a while) sections 3.4 and 3.5 here: en.wikipedia.org/wiki/Dirac_delta_function –  Did Aug 11 '11 at 17:14
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Well, you seem to be reproducing the simple layer formula stated (with reference) at the end of section 3.5 of the WP page. –  Did Aug 11 '11 at 18:37

3 Answers 3

up vote 3 down vote accepted

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.

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You mean the vector basis $(\mathbf{n}_i)$ depends on the point $\mathbf{r}$? –  Did Aug 11 '11 at 18:48
    
So you use the coordinates of $\textbf{r}$ in a vector basis depending on $\textbf{r}$ itself? Hmmm... –  Did Aug 11 '11 at 18:59
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@Didier Kind of yes ( and I said it is hand-wavy), but the variable I integrate over ($t_1$) only depends on the point on the surface. Delta function there is pulling back a measure on a surface defined implicitly, and there was a reference to several books already. For a mathematical physics background the book of Nakahara, "Geometry, Topology and Physics" is excellent and discusses this in much details. –  Sasha Aug 11 '11 at 19:09
    
I have no doubt that Nakahara is excellent but the trouble is that I fail to see how to save your solution. –  Did Aug 11 '11 at 19:17
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@Didier Integration over $\mathbb{R}^n$ can be replaced with integration over $2\epsilon$ thin layer containing the surface, because delta function has no support outside of it. Then one would integrate across the layer which will fix $\epsilon =0 $. The key is to use Taylor formula, so that moving across the layer $g(\mathbf{r}) = \varepsilon \vert \mathrm{grad} g(\mathbf{r}) \vert$. The remaining measure must be $d \sigma$. –  Sasha Aug 11 '11 at 19:34

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 $S$.

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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.

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Or Laurent Schwartz - Théorie des distributions. Another option is Duistermaat and Kolk - Distributions: Theory and Applications. –  Jonas Teuwen Aug 11 '11 at 18:53

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