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.

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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.

(more…)

May 24, 2010

Co-area formula for gradient

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.

(more…)

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