Ngô Quốc Anh

March 20, 2012

The Yamabe problem: The work by Hidehiko Yamabe

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 20:58

Following the previous post, we are interested in solving the following equation

\displaystyle - 4\frac{{n - 1}}{{n - 2}}{\Delta _g}\varphi + {\text{Sca}}{{\text{l}}_g}\varphi = {\text{Sca}}{{\text{l}}_{\widetilde g}}{\varphi ^{\frac{{n + 2}}{{n - 2}}}},

where \widetilde g=\varphi^\frac{4}{n-2}g (with \varphi \in C^\infty, \varphi>0) is a conformal metric conformally to g. In this entry, we introduce the Hidehiko Yamabe approach. His approach is variational. To keep his notation used, we rewrite the PDE as the following

\displaystyle -\Delta \varphi + R\varphi = C_0 \varphi^\frac{n+2}{n-2}.

Yamabe tried to minimize the following

\displaystyle {F_q}(u) = \frac{{\displaystyle\int_M {\left( {\frac{{4(n - 1)}}{{n - 2}}|\nabla u{|^2} + R{u^2}} \right)d{v_g}} }}{{{{\left( {\displaystyle\int_M {|u{|^q}d{v_g}} } \right)}^{\frac{2}{q}}}}}

over the Sobolev space H^1(M) where q \leqslant \frac{2n}{n-2}. Let us say

\displaystyle {\mu _q} = \mathop {\inf }\limits_{u \in {H^1}(M)} {F_q}(u).

In the first stage, he showed that

Theorem B. For any q<\frac{2n}{n-2}, there exists a positive function \varphi_q satisfying

\displaystyle -\Delta \varphi_q + R\varphi_q = \mu_q \varphi_q^\frac{n+2}{n-2}.

(more…)

March 10, 2012

An integral of 1/(1+|x|^2) over the whole Euclidean space

Filed under: Giải Tích 2, Giải Tích 3, Linh Tinh — Ngô Quốc Anh @ 2:46

Sometimes, we need a precise value for following

\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}}, \quad \alpha>\frac{n}{2}.

As such, I am going to calculate it and place the result here for future works.

In order to evaluate the above integral, we need to use the so-called co-area formula. We first write

\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \int_0^{ + \infty } {\left( {\int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} } \right)dr}.

Note that

\displaystyle \begin{gathered} \int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} = \frac{1}{{{{(1 + {r^2})}^\alpha }}}\int_{\partial {B_0}(r)} {dS} \hfill \\ \qquad\qquad\qquad= \frac{1}{{{{(1 + {r^2})}^\alpha }}}\text{Area}({B_0}(r)) = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}\end{gathered}.

Therefore,

\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)}}{{\Gamma (\alpha )}}

(more…)

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