Ngô Quốc Anh

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.

Proof. Let \varepsilon>0. Define

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

Multiplying by (u-t)^+ and integrating by parts we get

\displaystyle \int_{\{u>t\}}\frac{|\nabla u|^2}{|\nabla u|^2+\varepsilon}dx = \int_{\{u>t\}}f(u-t)dx.

Differentiating with respect to t gives

\displaystyle -\frac{d}{dt}\int_{\{u>t\}}\frac{|\nabla u|^2}{|\nabla u|^2+\varepsilon}dx = \int_{\{u>t\}}fdx.

For a sufficiently small h, we can integrate the previous relation from t-h to t to get

\displaystyle \int_{\{t-h\leqslant u<t\}}\frac{|\nabla u|^2}{|\nabla u|^2+\varepsilon}dx = \int_{t-h}^t\left(\int_{\{u>\tau\}}fdx\right)d\tau=\int_{t-h}^t\left( \int_{\{u=\tau\}}\frac{|\nabla u|}{|\nabla u|^2+\varepsilon}d\sigma\right)d\tau.

We are now in a position to apply the Dominated Convergence Theorem to pass to the limit as \varepsilon\to 0 to get

\displaystyle \int_{\{t-h\leqslant u<t\}}dx =\int_{t-h}^t\left( \int_{\{u=\tau\}}\frac{1}{|\nabla u|}d\sigma\right)d\tau.

Note that

\displaystyle \int_{\{t-h\leqslant u<t\}}dx =\int_{\{u>t-h\}}dx-\int_{\{u>t\}}dx.

Thus, dividing by h and taking the limit as h\to 0 yields the desired result.

Remark. The following is also true

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

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