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

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