most recent 30 from http://mathoverflow.net 2013-05-22T00:22:51Z http://mathoverflow.net/feeds/question/96372 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96372/rate-of-change-of-mass-of-a-parameterized-region Jennifer Gao 2012-05-08T20:52:23Z 2012-05-09T10:33:17Z

<p>Let $R_t$ be a family of compact, simply connected regions in the plane defined by</p> <p><code>$R_t = \{x\in\mathbb{R}^2 : h(x) \leq t\}$</code></p> <p>for all $t$, where $h(x)$ is some nicely behaved smooth function. Suppose $f(x)$ is a probability density on $\mathbb{R}^2$ and define</p> <p>$M(t) = \iint_{R_t} f(x) dA$</p> <p>for all $t$. Is it true that</p> <p>$\frac{dM}{dt}|_{t=t_0} = \int_{\partial R_{t_0}} f(x) /\|\nabla h\| ds$</p> <p>where $ds$ denotes integration with respect to arc length? If not, what is the right expression for $\frac{dM}{dt}$? I assume this is some well-known first-year calculus-type problem but I can't find it stated in any context (it may very well be a common homework problem though I've not seen it).</p>

http://mathoverflow.net/questions/96372/rate-of-change-of-mass-of-a-parameterized-region/96425#96425 Bazin 2012-05-09T10:33:17Z 2012-05-09T10:33:17Z

<p>With $H$ the Heaviside function (characteristic function of $\mathbb R_+$), you have $$ M(t)=\int_{\mathbb R^n} f(x) H(t-h(x)) dx $$ and thus, at least formally, $$ \dot M(t)=\int_{\mathbb R^n} f(x) \delta_0(t-h(x)) dx= \int_{\mathbb R^n} f(x)\Vert\nabla h(x)\Vert ^{-1} \underbrace{\delta_0(t-h(x))\Vert\nabla h(x)\Vert dx}_{d\sigma} , $$ where $d\sigma$ is the Euclidean surface measure on {$x, h(x) =t$}. Here, it is important to assume that $h$ is say $C^1$ such that $dh\not=0$ at $h=t$ for $t$ in neighborhood of some distinguished value $t_0$. As a result, $$ \dot M(t)=\int_{\partial R(t)}f(x)\Vert\nabla h(x)\Vert ^{-1} d\sigma. $$ The previous computation can be justified by Green's formula: for $X$ a $C^1_c$ vector field on $\Omega$ ( an open subset of $\mathbb R^n$ with a $C^1$ boundary) $$ \int_\Omega \text{div} X\ dx=\int_{\partial \Omega} X\cdot \nu\ d\sigma, $$ where $\nu$ is the exterior unit normal to $\partial \Omega$, and $d\sigma$ is the Euclidean surface measure on $\partial \Omega$. That formula can be proven by showing $$ d\sigma=\lim_{\epsilon\rightarrow 0} \phi(\frac{\rho(x)}{\epsilon})\epsilon^{-1}\Vert\nabla \rho(x)\Vert\quad\text{(distribution sense),} $$ for $\phi\in C^\infty_c(\mathbb R),\int \phi(t) dt=1$, $\Omega=${$x, \rho(x)&lt;0$}, $d\rho\not=0$ at $\rho=0$.</p>