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- Laurent Schwartz Theorie Des Distributions Djvu Files
- Theorie Des Distributions
- Laurent Schwartz Theorie Des Distributions Djvu File Online
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How does one prove the following identity?
$$int _Vf(pmb{r})delta (g(pmb{r}))dpmb{r}=int _Sfrac{f(pmb{r})}{|text{grad} g(pmb{r})|}dsigma$$
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.
becko
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$begingroup$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 $varphi$, $lim_{epsilonto 0};varphi_epsilontodelta$ 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, $delta(g(pmb{r}));dpmb{r}=;displaystyle{frac{dsigma}{|nabla g(pmb{r})|}}$ where $dsigma$ is surface measure on $S$.
robjohn♦robjohn278k2828 gold badges329329 silver badges659659 bronze badges
$endgroup$$begingroup$What you are quoting is a general statement about pull-backs of distributions. Since I am not entirely sure of your background, I won't try to give a detailed explanation here. Rather, I will refer you to Chapter 7 of Friedlander and Joshi's Introduction to the Theory of Distributions.
Laurent Schwartz Theorie Des Distributions Djvu Files
By Taylor series $g(mathbf{x}) = g(mathbf{r}) + vec{mathrm{grad} g(mathbf{r})}.(mathbf{x}-mathbf{r}) + o(vert mathbf{x}-mathbf{r} vert)$ as a new coordinate in the vicinity of the surface, where $g(mathbf{r})=0$. Change basis using $mathbf{n}_1 = frac{vec{mathrm{grad} g(mathbf{r})}}{vert{mathrm{grad} g(mathbf{r})}vert}$ as a first vector, and remaining $mathbf{n}_i$ for $i=2, ldots, n$ are chosen by Gram orthogonalization procedure. Let $t_i$ be coordinates in this system, $mathbf{r} = sum_i t_i mathbf{n}_i$. Then $dV_x = dx_1 wedge d x_2 wedge ldots wedge d x_n = vert J vert dt_1 wedge d t_2 wedge ldots wedge d t_n = dV_t$.
$$ 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 $$
Integration overt $t_1$ produces $d sigma$.
This is a little hand-wavy, but gives you an idea.
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$begingroup$The question is that:if $u$ is a distribution on real line, and first distributive derivative of it is a positive distribution, is $u$ indeed a locally integrable function? And if the second derivative is positive, is $u$ a convex function on real line?
stephenkkstephenkk
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Theorie Des Distributions
$begingroup$Yes to both. The proof is in the book of Laurent Schwartz. Theorie des Distributions. Hermann, Paris, 1950–51. The theorems you want are on page 54. The proof is a bit sketchy and it is in French.. sorry!For the first, you first prove that if a derivative of a distribution is constant then the distribution is actually a constant function. Then you prove that if a distribution is nonnegative, that is $T(phi)ge 0$ for all $phige 0$, then $T$ is given by $T(phi)=int_Rphi ,dmu$ for some measure $mu$. For this you use Hahn-Banach and Riesz representation theorem.Next if a distribution $S$ has nonnegative derivative $S'$ then $S'(phi)=int_Rphi ,dmu$. You consider the function $f(x)=int_0^x,dmu$ and prove that its derivative in the sense of distributions is exactly $S'$. Hence $S'-f'=0$ in the sense of distributions and so $S-f$ is a constant. The convex case is similar.
Edit Given $phi$ by Fubini's theorem you have$$int_{mathbb{R}} phi'(x)f(x),dx=int_{mathbb{R}} phi'(x)int_0^x1,dmu(t),dx=int_{mathbb{R}} int_t^inftyphi'(x),dx,dmu(t)=-int_{mathbb{R}} phi(t),dmu(t)$$
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