# Ngô Quốc Anh

## September 30, 2013

### The Lichnerowicz equation under some variable changes

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 4:54

Let us consider the so-called Lichnerowicz equation $-\Delta_g u + hu = fu^{2^\star-1}+au^{-2^\star-1} \quad u>0$

on $(M,g)$, a Riemannian manifold of dimension $n \geq 3$. Here $h$, $f$, and $a$ are smooth function with $a \geq 0$.

• We first use the the following variable change $\displaystyle v=\log u \quad u=e^ v.$

Clearly, $\displaystyle\Delta v = \frac{\Delta u}{u} - \frac{|\nabla u|^2}{u^2}$

and $\displaystyle |\nabla v|^2 = \frac{|\nabla u|^2}{u^2}.$

Therefore, we can write $\displaystyle -\Delta v =-\frac{\Delta u}{u} +|\nabla v|^2.$

Using this rule, we can rewrite the equation as follows $\displaystyle \boxed{-\Delta v = -h+fu^{2^\star-2}+au^{-2^\star-2}+|\nabla v|^2=-h+fe^{(2^\star-2)v}+ae^{-(2^\star+2)v}+|\nabla v|^2. }$

## September 15, 2013

### Some integral identities on manifolds with boundary

In this note, I summary several useful integral identities on Riemannian manifolds with boundary.

1. Suppose that $f$ is a function and $X$ is a $1$-form, then $\displaystyle\boxed{\int_M {f\text{div}Xd{v_g}} = - \int_M {\left\langle {\nabla f,X} \right\rangle_g d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}}.}$

To prove this, we write everything in local coordinates as follows $\begin{array}{lcl} \displaystyle\int_M {f \text{div} Xd{v_g}} &=& \displaystyle\int_M {f{\nabla _i}{X^i}d{v_g}} \hfill \\ &=& \displaystyle - \int_M {{\nabla _i}f{X^i}d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}} \hfill \\ &=& \displaystyle - \int_M {\left\langle {\nabla f,X} \right\rangle_g d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}}\end{array}$

as claimed.

2. Using the previous identity, we can prove the following $\displaystyle \boxed{\int_M {{{\left\langle {X,\nabla (\text{div} X)} \right\rangle }_g}d{v_g}} = - \int_M {|\text{div} X|_g^2d{v_g}} + \int_{\partial M} {\text{div} X{{\left\langle {X,\nu } \right\rangle }_g}d{\sigma _g}}}$

where $X$ is again a vector field on $M$. To prove this, we simply apply the previous identity with $f$ replaced by $\text{div}(X)$ to get the desired result.
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