Ngô Quốc Anh

May 16, 2011

Some properties of the Yamabe equation in the null case

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 19:56

Let us consider the Yamabe equation in the null case, that is

\displaystyle -\Delta u = f u^\frac{n+2}{n-2}, \quad x \in M

where M is a compact manifold of dimension n without boundary. We assume that u>0 is a smooth positive solution.

Since the manifold is compact without the boundary, the most simple result is

\displaystyle\int_M f u^\frac{n+2}{n-2}=0

by integrating both sides of the equation. Now we prove that

\displaystyle\int_M f <0.

Indeed, multiplying both sides of the PDE with u^{-\frac{n+2}{n-2}} and integrating over M, one obtains

\displaystyle -\int_M(\Delta u) u^{-\frac{n+2}{n-2}} = \int_M f .

By the divergence theorem, the left hand side of the above equation is just

\displaystyle \int_M \nabla u \cdot \nabla(u^{-\frac{n+2}{n-2}})=-\frac{n+2}{n-2}\int_M u^\frac{4}{n-2}|\nabla u|^2.

Thus \int_M f \leqslant 0. If the equality occurs, one deduces that u is positive constant which does not satisfy the PDE. Therefore, \int_M f <0 as required. Consequently, \inf f<0.

Next we prove that

\displaystyle\int_M f u^\frac{2n}{n-2}>0.

Indeed, multiplying both sides of the PDE with u and integrating over M, one gets

\displaystyle - \int_M {(\Delta u)} u = \int_M f {u^{\frac{{2n}}{{n - 2}}}}.

Again by the divergence theorem,

\displaystyle -\int_M(\Delta u) u= \int_M |\nabla u|^2 \geqslant 0.

Thus

\displaystyle\int_M f u^\frac{2n}{n-2}>0

since u cannot be constant. Consequently, one sees that \sup f>0 is necessary. In other words, function f need to change sign.

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