# Ngô Quốc Anh

## June 3, 2010

### Lower bound for integral of exp(u)

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 21:51

As mentioned before, today I will derive a very short and beautiful proof concerning the lower bound of $\int\exp(u(x))dx$ where $u$, a positive solution to the following PDE

$\displaystyle -\Delta u =e^{u(x)}, \quad x \in \mathbb R^2$.

This proof I firstly learned from a paper published in Duke Math. J. in 1991 by W. Cheng and C. Li [here].

We assume

$\displaystyle \int_{\mathbb R^2} e^{u(x)}dx < \infty$.

Denote by $\Omega_t$ the following set

$\Omega_t = \{ x \in \mathbb R^2 : u(x)>t\}$.

It follows from this topic that

$\displaystyle \int_{\Omega_t} e^{u(x)}dx =-\int_{\Omega_t}\Delta u dx = \int_{\partial \Omega_t} |\nabla u|d\sigma$.

Also, it follows from this topic that

$\displaystyle -\frac{d}{dt}\int_{\Omega_t}dx =\int_{\partial \Omega_t} \frac{1}{|\nabla u|}d\sigma$.

Thus, by the Schwarz inequality and the isoperimetric inequality

$\displaystyle \left(\int_{\partial\Omega_t} \frac{1}{|\nabla u|}d\sigma\right)\left(\int_{\partial\Omega_t}|\nabla u| d\sigma\right) \geqslant |\partial\Omega_t|^2 \geqslant 4\pi |\Omega_t|$.

Hence

$\displaystyle -\left(\frac{d}{dt}\int_{\Omega_t}dx\right) \left(\int_{\Omega_t} e^{u(x)}dx\right)\geqslant 4\pi |\Omega_t|$.

So

$\displaystyle \frac{d}{dt}\left(\int_{\Omega_t} e^{u(x)}dx\right)^2=2\left(\int_{\Omega_t} e^{u(x)}dx\right)\frac{d}{dt}\left(\int_{\Omega_t} e^{u(x)}dx\right)$.

It is worth noticing that

$\displaystyle\frac{d}{dt}\left(\int_{\Omega_t} e^{u(x)}dx\right)=e^t\frac{d}{dt}\left(\int_{\Omega_t}dx\right)$

which yields

$\displaystyle \frac{d}{dt}\left(\int_{\Omega_t} e^{u(x)}dx\right)^2=2\left(\int_{\Omega_t} e^{u(x)}dx\right)e^t\frac{d}{dt}\left(\int_{\Omega_t}dx\right) \leqslant -8\pi e^t\int_{\Omega_t}dx$.

Integrating from $-\infty$ to $\infty$ gives

$\displaystyle -\left(\int_{\mathbb R^2} e^{u(x)}dx\right)^2\leqslant -8\pi\int_{\mathbb R^2}e^{u(x)}dx$

which implies

$\displaystyle \int_{\mathbb R^2} e^{u(x)}dx \geqslant 8\pi$.

## May 30, 2010

### Co-area formula for gradient, 2

Filed under: Giải Tích 6 (MA5205) — Tags: — Ngô Quốc Anh @ 20:29

Let us recall the following result

Theorem. Let $\Omega \subset \mathbb R^n$ be an open set and $u \in \mathcal D(\Omega)$. If $u \geqslant 0$ then for any $1 \leqslant p<\infty$, we have

$\displaystyle\int_\Omega|\nabla u|^p dx =\int_0^M \left(\int_{\{u=t\}}|\nabla u|^{p-1}d\sigma\right)dt$

where $M=\sup u$ over $\overline \Omega$.

The above result has been proven in this entry. If we chose $p=0$ then we would have

$\displaystyle\int_\Omega dx =\int_0^M \left(\int_{\{u=t\}}\frac{1}{|\nabla u|}d\sigma\right)dt$.

If we now replace $\Omega=\{u>t\}$, we get

$\displaystyle\int_{\{u>t\}}dx =\int_t^M \left(\int_{\{u=t\}}\frac{1}{|\nabla u|}d\sigma\right)dt$.

By differentiating with respect to $t$, we arrive at

$\displaystyle \frac{d}{dt}\int_{\{u>t\}}dx =-\int_{\{u=t\}}\frac{1}{|\nabla u|}d\sigma$.

In this entry, we shall prove the foregoing identity is actually true. This is the second interesting formula I have mentioned before. The proof, of course again, is based on a clever choice of test function.

## May 24, 2010

Filed under: Giải Tích 6 (MA5205) — Tags: — Ngô Quốc Anh @ 18:55

I found two interesting formulas related to co-area formula while reading some tricks done by Talenti regarding to the best constant of the Sobolev inequality. The first result is to derive a representation of

$\displaystyle \int_\Omega |\nabla u|^pdx$

and the second result is to deal with differentiation of level sets. Having all these stuffs, I will derive a very short and beautiful proof concerning the lower bound of $\int \exp(u) dx$ where $u$, a positive solution to the following PDE

$\displaystyle-\Delta u = e^u$

This proof I firstly learned from a paper published in Duke Math. J. in 1991 by W. Cheng and C. Li [here].

Co-area formula. Suppose that $\Omega$ is an open set in $\mathbb R^n$, and $u$ is a real-valued Lipschitz function on $\Omega$. Then, for an integrable function $g$

$\displaystyle\int_\Omega g(x) |\nabla u(x)| dx = \int_{-\infty}^\infty \left(\int_{\{u=t\}}g(x) dH_{n-1}(x)\right)dt$

where $H^{n-1}$ is the $(n-1)$-dimensional Hausdorff measure.

The Sard theorem. Let $f :\mathbb{R}^n \rightarrow \mathbb{R}^m$ be $C^k$, $k$ times continuously differentiable, where $k \geqslant \max\{n-m+1, 1\}$. Let $X$ be the critical set of $f$, the set of points $x$ in $\mathbb R^n$ at which the Jacobian matrix of $f$ has ${\rm rank} < m$. Then $f(X)$ has Lebesgue measure $0$ in $\mathbb R^m$.

The Sard theorem has some useful applications. For example, if $u \in \mathcal D(\Omega)$ the space of test functions where $\Omega \subset \mathbb R^n$, then for almost every $t$ in the range of $u$, we have that $|\nabla u|\ne 0$ on the level set $\{u=t\}$. Thus that level set will be an $(n-1)$-dimensional surface. Furthermore

$\{u=t\}=\partial \{u>t\}$

and

$|\{u=t\}|=0$.

Theorem. Let $\Omega \subset \mathbb R^n$ be an open set and $u \in \mathcal D(\Omega)$. If $u \geqslant 0$ then for any $1 \leqslant p<\infty$, we have

$\displaystyle\int_\Omega|\nabla u|^p dx =\int_0^M \left(\int_{\{u=t\}}|\nabla u|^{p-1}d\sigma\right)dt$

where $M=\sup u$ over $\overline \Omega$.