# Ngô Quốc Anh

## March 5, 2014

### Uniformly upper Bound for Positive Smooth Solutions To The Lichnerowicz Equation In R^N

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 16:16

In this note, we are interested in the following Lichnerowicz type equation in $\mathbb R^n$

$\displaystyle -\Delta u =-u^q+\frac{1}{u^{q+2}}, \quad q>0.$

As we have already seen from the previous note that solutions for the above equation are always bounded from below for certain $q$.

Theorem (Brezis). Any solution of the Lichnerowicz equation with $q>0$ satisfies $u\geqslant 1$ in $\mathbb R^n$.

Remarkably, we are able to prove that solutions for the Lichnerowicz type equation are also bounded from above if we require $q>1$ instead of $q>0$. This result is basically due to L. Ma and X. Xu, see this paper.

Theorem (Ma-Xu). Any solution of the Lichnerowicz type equation with $q>1$ is uniformly bounded from above in $\mathbb R^n$.

Combining the two theorem above, we conclude that any solution of the Lichnerowicz equation, i.e. $q=(n+2)/(n-2)$, is uniformly bounded in $\mathbb R^n$.

The idea of the proof for Ma-Xu’s theorem is as follows: Denote

$\displaystyle f(u)=u^{-q-2} - u^q.$

Fix $x_0 \in \mathbb R^n$ but arbitrary, we then look for a positive radial super-solution $v(x)=v(|x|)>0$ of the Lichnerowicz type equation on the ball $B_R(x_0)$ with positive infinity boundary condition for some $R$ to be specified. This is equivalent to finding $v$ in such a way that

$\displaystyle\begin{array}{rcl}\displaystyle-\Delta v &\geqslant& f(v), \qquad\text{ in } B_R(x_0),\\ v&\equiv& +\infty, \qquad\text{ on }\partial B_R(x_0).\end{array}$

## 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

$\displaystyle \boxed{-\Delta v = -h+fu^{2^\star-2}+au^{-2^\star-2}+|\nabla v|^2=-h+fe^{(2^\star-2)v}+ae^{-(2^\star+2)v}+|\nabla v|^2. }$

## April 1, 2013

### A proof of the uniqueness of solutions of the Lichnerowicz equations in the compact case

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 4:38

In this small note, we aim to derive some uniqueness property of solutions of the following PDE

$\displaystyle -a\Delta_g u +\text{Scal}_g u =|\sigma|_g^2 u ^{-2\kappa-3}-b\tau^2 u^{2\kappa+1}$

on a compact manifold $(M,g)$ where $\kappa=\frac{2}{n-2}$, $a=2\kappa+4$, and $b=\frac{n-1}{n}$.

We assume that $\phi_1$ and $\phi_2$ are solutions of the above PDE. Setting $\phi=\frac{\phi_2}{\phi_1}$. We wish to prove that $\phi=1$.

Let us consider the following trick basically due to David Maxwell, see this paper. Let $\widetilde g= \phi_1^{2\kappa}g$. Then the well-known formula for the Laplace-Beltrami operator $\Delta_g$, which is

$\displaystyle {\Delta _g}u = \frac{1}{{\sqrt {\det g} }}{\partial _i}(\sqrt {\det g} {g^{ij}}{\partial _j}u)$

helps us to write

$\displaystyle {\Delta _{\widetilde g}}u = \phi _1^{ - 2\kappa } \Big({\Delta _g}u + \frac{2}{{{\phi _1}}}{\nabla _g}{\phi _1}{\nabla _g}u \Big).$

Consequently,

$\displaystyle {\Delta _g}\phi = \phi _1^{2\kappa }{\Delta _{\widetilde g}}\phi - \frac{2}{{{\phi _1}}}{\nabla _g}{\phi _1}{\nabla _g}\phi .$

We now calculate $- a{\Delta _g}{\phi _2} + {R_g}{\phi _2}$ as follows.

## April 11, 2012

### Existence results for the Einstein-scalar field Lichnerowicz equations on compact Riemannian manifolds

Filed under: Luận Văn, PDEs, Riemannian geometry — Tags: , — Ngô Quốc Anh @ 2:18

A couple of days ago, I got an acceptance for publication in Advances in Mathematics journal that makes me feel so exciting because of the prestige of the journal. This is part of my PhD thesis in NUS under the supervision of professor Xu. Besides, this is joint work with him.

The work looks like simple, I mean, we just try to solve the following PDE

$\displaystyle {\Delta _g}u + hu = f{u^{{2^\star} - 1}} + \frac{a}{{{u^{{2^\star} +1}}}}, \quad u>0,$

where $\Delta_g=-{\rm div}_g(\nabla_g)$ is the Laplace-Beltrami operator, $2^\star=\frac{2n}{n-2}$ is the critical Sobolev exponent, $M$ is a compact manifold without boundary of dimension $n \geqslant 3$, and $h$, $f$, $a \geqslant 0$ are smooth functions. In our work, the above PDE is numbered as (1.2). I don’t want to mention the physical background of the equation, in a few words, this equation is motivated by the Hamiltonian constraint equations of General Relativity through the so-called conformal method. Apparently, the important and frequently studied prescribing scalar curvature equation is just a particular case.

In this work, we focus on the negative Yamabe-scalar field invariant, that is, $h<0$. Our result basically consists of two theorems.

In the first result, we consider the case that $f$ may change its sign, we prove

Theorem 1.1. Let $(M,g)$ be a smooth compact Riemannian manifold without the boundary of dimension $n \geqslant 3$. Assume that $f$ and $a \geqslant 0$ are smooth functions on $M$ such that $\int_M f dv_g<0$$\sup f > 0$, $\int_M a dv_g >0$, and $|h| < \lambda _f$ where $\lambda_f$ is given in (2.1) below. Let us also suppose that the integral of $a$ satisfies

$\displaystyle\int_M {ad{v_g}} < \frac{1}{n-2}{\left( {\frac{{n - 1}}{n-2}} \right)^{n - 1}}{\left( {\frac{{|h|}}{{\int_M {|{f^ - }|d{v_g}} }}} \right)^n}\int_M {|{f^ - }|d{v_g}}$

where $f^-$ is the negative part of $f$. Then there exists a number $C > 0$ to be specified such that if

$\displaystyle\frac{{\sup {f }}}{{\int_M {{|f^ -| }d{v_g}} }}

Equation (1.2) possesses at least two smooth positive solutions.

In the next result, we consider the case that $f \leqslant 0$. In this case, we are able to get a complete characterization of the existence of solutions. More precisely, we prove

Theorem 1.2. Let $(M,g)$ be a smooth compact Riemannian manifold without boundary of dimension $n \geqslant 3$. Let $h<0$ be a constant, $f$ and $a$ be smooth functions on $M$ with $a \geqslant 0$ in $M$, $f \leqslant 0$ but not strictly negative. Then Equation (1.2) possesses one positive solution if and only if $|h|<\lambda_f$.

As one can see, the above theorem does not allow $f$ to be strictly negative. Fortunately, our approach can cover this case too. This is the last remark in the paper as we prove the following: if $f<0$ then Equation (1.2) always possesses one positive solution, I mean, without any condition on $f$ except the condition $f<0$.

It is important to note that in the case $f \leqslant 0$, the solution is always unique by using the monotone trick.

## June 8, 2011

### Uniformly boundedness of positive smooth solutions to the Lichnerowicz equation in R^n

Filed under: PDEs — Tags: , — Ngô Quốc Anh @ 7:26

Let us consider the following so-called Lichnerowicz equation in $\mathbb R^n$

$\displaystyle -\Delta u =-u^q+\frac{1}{u^{q+2}}, \quad q>0.$

Recently, Brezis [here] proved the following

Theorem. Any solution of the Lichnerowicz equation with $q>0$ satisfies $u\geqslant 1$ in $\mathbb R^n$.

Let us study the trick used in his paper.

Proof. Let

$f(t)=t^q-t^{q+2}, \quad t>0.$

Then $\Delta u =f(u)$. Fix any point $x_0 \in \mathbb R^n$ and consider the function

$u_\varepsilon(x)=u(x)+\varepsilon |x-x_0|^2, \quad \varepsilon>0, x \in \mathbb R^n.$

Since $u_\varepsilon(x) \to \infty$ as $|x|\to \infty$, $\min_{\mathbb R^n}u_\varepsilon(x)$ is achieved at some $x_1$. We have

$0\leqslant \Delta u_\varepsilon(x_1)=\Delta u(x_1)+2\varepsilon n=f(u(x_1))+2\varepsilon n.$

By definition,

$u(x_1)+\varepsilon |x_1-x_0|^2=u_\varepsilon(x_1) \leqslant u_\varepsilon(x_0)=u(x_0).$

Thus,

$u(x_1)\leqslant u(x_0).$

Since $f$ is increasing we deduce that

$f(u(x_1))\leqslant f(u(x_0)).$

Therefore,

$0\leqslant \Delta u_\varepsilon(x_1) \leqslant f(u(x_1))+2\varepsilon n \leqslant f(u(x_0))+2\varepsilon n.$

By sending $\varepsilon \to 0$ we deduce that $f(u(x_0)) \geqslant 0$. In other words, $u(x_0) \geqslant 1$.

As can be seen, he only uses the fact that $f$ is monotone increasing in his argument, therefore, this approach can be used for a wider class of nonlinearity.

## June 19, 2010

### Existence of global super-solutions of the Lichnerowicz equations

Filed under: PDEs — Tags: , — Ngô Quốc Anh @ 19:16

This entry devotes the existence of global super-solutions to the Lichnerowicz equations in the study of the Einstein equations in general relativity. This terminology plays an important role in the study of non-CMC case of the Einstein equations in vacuum case.

This terminology first introduced in 2009 by a paper due to  M. Holst, G. Nagy and G. Tsogtgerel published in Comm. Math. Phys. [here]. In that paper, the solvability comes from the existence of both global sub- and global super-solutions together with some new fixed-point arguments which we have already discussed before [here]. Maxwell recently got a significant result by relaxing the existence of global sub-solutions [here] so that the existence of global super-solutions is enough to guarantee the solvability of the Einstein equations in non-CMC case.

In the vacuum case, the classification depends on the sign of the Yamabe invariants.

In the conformal method, in three dimensions the study of the Einstein equations becomes the study of existence of solution $(\varphi, W)$ to a coupled system:

$\displaystyle -8\Delta \varphi + R\varphi=-\frac{2}{3}\tau^2\varphi^5+|\sigma+\mathbb LW|^2\varphi^{-7}$

and

$\displaystyle {\rm div} \mathbb LW=\frac{2}{3}\varphi^6d\tau$.

The first equation is usually called the Lichnerowicz equation

$\displaystyle -8\Delta \varphi + R\varphi=-\frac{2}{3}\tau^2\varphi^5+|\beta|^2\varphi^{-7}$

where $\beta$ is a symmetric $(0,2)$-tensor.

Definition. We say $\varphi_+$ is a super-solution of the Lichnerowicz equation if

$\displaystyle -8\Delta \varphi + R\varphi=-\frac{2}{3}\tau^2\varphi^5+|\beta|^2\varphi^{-7}$.

Now we have

Definition. We say $\varphi_+$ is a global super-solution of the Lichnerowicz equation if whenever

$0<\varphi \leqslant \varphi_+$

then

$\displaystyle -8\Delta \varphi + R\varphi=-\frac{2}{3}\tau^2\varphi^5+|\sigma+\mathbb LW_\varphi|^2\varphi^{-7}$

where $W_\varphi$ is a solution of second equation obtained from $\varphi$.

We are now in a position to derive main results.

## November 11, 2009

### A non-existence result for positive solutions to the Lichnerowicz equation in R^N

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: , — Ngô Quốc Anh @ 22:36

In this topic, adapted from a paper due to Li Ma and Xingwang Xu published in Comptes Rendus Mathematique we shall give a non-existence result concerning the following Lichnerowicz equation in $\mathbb R^N$

$\Delta u + R(x) u + A(x) u^{-p-1} + B(x) u^{p-1}=0$, $u>0$ on $\mathbb R^N$

where $R(x) \geq 0$, $A(x) \geq 0$, and $B(x)$ are given smooth functions of $x \in \mathbb R^N$. To be precise, we obtain the following

Theorem. Suppose $A:=A(x) \geq 0$, $B := B(x) \geq 0$, and $R(x) \geq 0$. Let $\beta = \frac{p+1}{2p}$. Assume that

$\displaystyle \int_0^{ + \infty } {\left( {\int_{B\left( {0,r} \right)} {{A^{1 - \beta }}{B^\beta }dx} } \right){r^{1 - N}}dr} = +\infty .$

Then there exists no positive solution to the above Lichnerowicz equation.

Let us denote the integral

$\displaystyle\frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {f\left( x \right)dS_x}$

by $\overline f$. We call $\overline f$ the average of $f$ on the sphere $S(0,r)$ of radius $r$, or sphere mean of a function around the origin.

Proof. Note that a simple calculation shows us that

$\displaystyle\frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {f\left( x \right)d{S_x}} = \frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {f\left( {rx} \right)d{S_x}}$.

Therefore

$\displaystyle {\overline u ^\prime }= \frac{d}{{dr}}\overline u = \frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\frac{d}{{dr}}u\left( {xr} \right)d{S_x}} =\frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\sum\limits_{k = 1}^N {{x_i}{u_{{x_i}}}} d{S_x}}$.

Since on the sphere $S(0,1)$, $x=(x_1,...,x_N)$ is also the outer normal vector, therefore

$\displaystyle\frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\sum\limits_{k = 1}^N {{x_i}{u_{{x_i}}}\left( {xr} \right)} d{S_x}} = \frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\nabla u\left( {xr} \right) \cdot {n_x}d{S_x}}$

Thus by the divergence theorem, one gets

$\displaystyle \frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\nabla u\left( {xr} \right)\cdot {n_x}d{S_x}} = \frac{r}{{{\omega _n}}}\int_{B\left( {0,1} \right)} {\Delta u dx} = \frac{1}{{{\omega _n}{r^{N-1}}}}\int_{B\left( {0,r} \right)} {\Delta udx}$.

Hence

$\displaystyle{\overline u ^\prime } = \frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{B\left( {0,r} \right)} {\Delta udx}$.

Differentiating once more yields

$\displaystyle{\overline u ^\prime }^\prime = \frac{d}{{dr}}{\overline u ^\prime } = - \underbrace {\frac{{N - 1}}{{{\omega _n}{r^N}}}\int_{B\left( {0,r} \right)} {\Delta udx} }_{\frac{{N - 1}}{r}\overline u'} + \frac{1}{{{\omega _n}{r^{N - 1}}}}\frac{d}{{dr}}\left( {\int_{B\left( {0,r} \right)} {\Delta udx} } \right)$.

Since

$\displaystyle\frac{d}{{dr}}\left( {\int_{B\left( {0,r} \right)} {\Delta udx} } \right) =\int_{\partial B\left( {0,r} \right)} {\Delta udx}$

then

$\displaystyle{\overline u ^\prime }^\prime = - \frac{{N - 1}}{r}\overline u' + \frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {\Delta udx} = - \frac{{N - 1}}{r}\overline u' + \overline {\Delta u}$.

Thus

$\displaystyle\overline {\Delta u} = {\overline u ^\prime }^\prime + \frac{{N - 1}}{r}\overline u'$

Therefore, taking this average operation we have

$\displaystyle - {\overline u ^\prime }^\prime - \frac{{N - 1}}{r}\overline u' = \overline {R(x)u} + \overline {A(x){u^{ - p - 1}} + B(x){u^{p - 1}}}$.

Since for each fixed $x\in \mathbb R^N$,

$\displaystyle\begin{gathered} A{u^{ - p - 1}} + B{u^{p - 1}} = \frac{{2p}}{{p - 1}}\frac{{p - 1}}{{2p}}A{u^{ - p - 1}} + \frac{{2p}}{{p + 1}}\frac{{p + 1}}{{2p}}B{u^{p - 1}} \\\qquad\quad\geq \frac{{p - 1}}{{2p}}A{u^{ - p - 1}} + \frac{{p + 1}}{{2p}}B{u^{p - 1}}. \\ \end{gathered}$

Then by using the general Cauchy inequality, one gets

$\displaystyle\frac{{p - 1}}{{2p}}A{u^{ - p - 1}} + \frac{{p + 1}}{{2p}}B{u^{p - 1}} \geqslant {\left( {A{u^{ - p - 1}}} \right)^{\frac{{p - 1}}{{2p}}}}{\left( {B{u^{p - 1}}} \right)^{\frac{{p + 1}}{{2p}}}} = {A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}$.

Thus,

$\displaystyle\overline {A{u^{ - p - 1}} + B{u^{p - 1}}} \geq \overline {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}}$.

It turns out that

$\displaystyle - {\left( r^{N - 1}\overline u ' \right)^\prime } \geq {r^{N - 1}}\left( {\overline {R(x)u}+ \overline {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}} } \right)$,

which implies that

$\displaystyle - {r^{N - 1}}\overline u ' \geq\frac{1}{\omega _n}\int_{B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}dx} + \frac{1}{\omega _n}\int_{B\left( {0,r} \right)} {R(x)udx} \geq \frac{1}{\omega _n}\int_{B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}dx}$

after an integration. This is because, by definition of the sphere mean,

$\displaystyle\begin{gathered}{r^{N - 1}}\left( {\overline {R(x)u} + \overline {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}} } \right) = {r^{N - 1}}\left( {\frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {R(x)ud{S_x}} + \frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}d{S_x}} } \right) \\\qquad\quad\;\;\;= \frac{1}{{{\omega _n}}}\left( {\int_{\partial B\left( {0,r} \right)} {R(x)ud{S_x}} + \int_{\partial B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}d{S_x}} } \right) \\\qquad\qquad\qquad\qquad\;\;\;= \frac{1}{{{\omega _n}}}\frac{d}{{dr}}\left[ {\int_0^r {\left( {\int_{\partial B\left( {0,s} \right)} {R(x)ud{S_x}} + \int_{\partial B\left( {0,s} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}d{S_x}} } \right)ds} } \right] .\\\end{gathered}$

Dividing both sides by $r^{N-1}$ and integrating this inequality over $[0, r_0]$, we have

$\displaystyle \overline u (0) \geq \overline u (0) - \overline u ({r_0}) \geqslant \int_0^{{r_0}} {\left( {{r^{1 - N}}\int_{B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}dx} } \right)dr}$.

Sending $r_0 \to \infty$ we have

$\displaystyle\overline u (0) \geq \int_0^{ + \infty } {\left( {{r^{1 - N}}\int_{B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}dx} } \right)dr}$,

which is impossible by our assumption. The proof is complete.