# Ngô Quốc Anh

## January 31, 2011

### Stereographic projection, 2

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 19:15

This is a sequel to this topic where we have recalled several properties of the stereographic projection $\pi : \mathbb S^n \to \mathbb R^n$. Recall that by the following transformation

$\displaystyle v(x)=u(\pi^{-1}(x))\left( \frac{2}{1+|x|^2}\right)^\frac{n-2}{2}, \quad x \in \mathbb R^n$

we know that

$\displaystyle -\Delta_g u(\xi) + \frac{n(n-2)}{4}u(\xi) = K(\xi)u(\xi)^\frac{n+2}{n-2} \quad \text{ on } \mathbb S^n$

and

$\displaystyle -\Delta v(x) = K(\pi^{-1}(x))v(x)^\frac{n+2}{n-2} \quad \text{ on } \mathbb R^n$

are equivalent in the weak sense.

The way to see it comes from the following identities

$\displaystyle \int_{{\mathbb{S}^n}} {|\nabla u(\xi ){|^2} + \frac{{n(n - 2)}}{4}u{{(\xi )}^2} = \int_{{\mathbb{R}^n}} {|\nabla v(x){|^2}} }$

and

$\displaystyle \int_{{\mathbb{S}^n}} {|u(\xi ){|^{\frac{{2n}}{{n - 2}}}} = \int_{{\mathbb{R}^n}} {|v(x){|^{\frac{{2n}}{{n - 2}}}}} }$

where $u \in H^1(\mathbb S^n)$.

Now, we would like to mention the fact that this projection can be used to classify solutions to the following fourth-order elliptic equation coming from the $Q$-curvature problem

$\Delta^2u-c_n\Delta u+d_nu=Ku^\frac{n+4}{n-4}$

on the $n$-sphere $\mathbb S^n$ where

$c_n =\frac{1}{2}(n^2-2n-4), \quad d_n=\frac{n-4}{16}n(n^2-4)$

and $K$ is a given function defined on $\mathbb S^n$. To be exact, up to a constant, the following function

$\displaystyle\frac{{{\beta _n}}}{{{2^{\frac{{n - 4}}{2}}}}}{\left( {\frac{\lambda }{{1 + \frac{{{\lambda ^2} - 1}}{2}(1 - \cos d(x,a))}}} \right)^{\frac{{n - 4}}{2}}}$

will solve the PDE where

$\displaystyle {\beta _n} = {\left( {(n - 4)(n - 2)n(n + 2)} \right)^{\frac{{n - 4}}{8}}}.$

Similarly, the following function

$\displaystyle {c_n}{\left( {\frac{\lambda }{{{\lambda ^2} + 1 + (1 - {\lambda ^2})\cos d(x,a)}}} \right)^{\frac{{n - 2}}{2}}}$

will solve the following PDE

$\displaystyle - \Delta u + \frac{{n(n - 2)}}{4}u + K{u^{\frac{{n + 2}}{{n - 2}}}} = 0$

on the $n$-sphere.