Ngô Quốc Anh

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.

Proof. Two cases are possible

If 2 \leqslant p<\infty. Define

\displaystyle f=-{\rm div}(|\nabla u|^{p-2}\nabla u)

then for all v \in W_0^{1,p}(\Omega) we have

\displaystyle \int_\Omega |\nabla u|^{p-2}\nabla u \cdot \nabla vdx = \int_\Omega f v dx.

Here is the trick. Let t>0. Set

\displaystyle v=(u-t)^+ \in W_o^{1,p}.

Using this test function and observing that v is supported on the set \{u>t\} and that on this set v=u-t, we get

\displaystyle \int_{\{u>t\}} |\nabla u|^pdx = \int_{\{u>t\}}f(u-t)dx.

Thus, differentiating with respect to t, we get

\displaystyle -\frac{d}{dt}\int_{\{u>t\}} |\nabla u|^pdx = \int_{\{u>t\}}f dx.

Integrating this relation over the range of u, i.e. on [0,M] we get

\displaystyle\int_\Omega|\nabla u|^p dx =\int_0^M \left(\int_{\{u>t\}}fdx\right)dt.

For t such that |\nabla u| \ne 0 on the level set \{u=t\}, the Green theorem implies

\displaystyle \int_{\{u>t\}}fdx=-\int_{\{u=t\}}|\nabla u|^{p-2}\nabla u \cdot \nu d\sigma=\int_{\{u>t\}}|\nabla u|^{p-1}d\sigma

since on the level set \{u=t\} the tangential derivatives of u vanish and since u>t inside this surface, we have

\displaystyle -\nabla u \cdot \nu = |\nabla u|.

If 1\leqslant p<2. Due to the presence of singularity, we need to use an approximation technique. Let \varepsilon>0. Define

\displaystyle f_\varepsilon=-{\rm div}\left((|\nabla u|^2+\varepsilon)^\frac{p-2}{2}\nabla u\right)

then for all v \in W_0^{1,p}(\Omega) we have

\displaystyle \int_\Omega (|\nabla u|^2+\varepsilon)^\frac{p-2}{2}\nabla u \cdot \nabla vdx = \int_\Omega f_\varepsilon v dx.

Choose the same test function as in the previous case we arrive at

\displaystyle \int_{\{u>t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u|^2 dx =\int_0^M \left(\int_{\{u=t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| d\sigma\right)dt.

We now pass the limit as \varepsilon \to 0 on both sides. On the LHS, the integrand coverges pointwise to |\nabla u|^p; further, since 1 \leqslant p<2, we have

\displaystyle (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u|^2 \leqslant |\nabla u|^p

which is integrable. Thus, by the Dominated Convergence Theorem, we get

\displaystyle \lim_{\varepsilon \to 0} \int_\Omega  (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u|^2 dx = \int_\Omega |\nabla u|^pdx.

On the RHS, one again observes that

\displaystyle (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| \leqslant |\nabla u|^{p-1}

which is integrable on the set \{u=t\}. So again by the Dominated Convergence Theorem,

\displaystyle \int_{\{u=t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| d\sigma \to \int_{\{u=t\}}|\nabla u|^{p-1} d\sigma

as \varepsilon \to 0.

Further,

\displaystyle \int_{\{u=t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| d\sigma \leqslant \int_{\{u=t\}}|\nabla u|^{p-1} d\sigma

and

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

since u \in \mathcal D(\Omega). Thus another application of the Dominated Convergence Theorem yields

\displaystyle \lim_{\varepsilon \to 0}\int_0^M \left(\int_{\{u=t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| d\sigma\right)dt = \int_0^M \left(\int_{\{u=t\}} |\nabla u|^{p-1} d\sigma\right)dt.

The proof follows.

Remark. Concerning the Schwarz symmetrization, the following

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

holds.

See also: Symmetrization And Applications (Series in Analysis) by S. Kesavan.

2 Comments »

  1. In case 1, the second from the last equation, the right most integral should be \int_{u=t}, right?

    Comment by Chao — August 15, 2010 @ 19:13

    • You are right, Chao. The correct one reads as follows

      \displaystyle \int_{\{u>t\}}fdx=-\int_{\{u=t\}}|\nabla u|^{p-2}\nabla u \cdot \nu d\sigma=\int_{\{u=t\}}|\nabla u|^{p-1}d\sigma.

      Thanks for pointing out this misprint.

      Comment by Ngô Quốc Anh — August 15, 2010 @ 19:21


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