# 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.