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


\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}} }


\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.


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