# Ngô Quốc Anh

## March 29, 2011

### The (original) Picone identity

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 17:02

For differentiable functions $v > 0$ and $u \geqslant 0$, the following Picone’s identity is well known

$\displaystyle {\left| {\nabla u - \frac{u}{v}\nabla v} \right|^2} = {\left| {\nabla u} \right|^2} - 2\frac{u}{v}\nabla u \cdot \nabla v + \frac{{{u^2}}}{{{v^2}}}{\left| {\nabla v} \right|^2} = {\left| {\nabla u} \right|^2} - \nabla \left( {\frac{{{u^2}}}{v}} \right) \cdot \nabla v \geqslant 0.$

The proof is very simple. For each partial derivative $\frac{\partial}{\partial x_i}$ we have

$\displaystyle\frac{\partial }{{\partial {x_i}}}\left( {\frac{{{u^2}}}{v}} \right) = \frac{1}{{{v^2}}}\left[ {\frac{{\partial ({u^2})}}{{\partial {x_i}}}v - {u^2}\frac{{\partial v}}{{\partial {x_i}}}} \right] = \frac{1}{{{v^2}}}\left[ {2u\frac{{\partial u}}{{\partial {x_i}}}v - {u^2}\frac{{\partial v}}{{\partial {x_i}}}} \right]$

which implies

$\displaystyle\nabla \left( {\frac{{{u^2}}}{v}} \right) = \frac{1}{{{v^2}}}\left[ {2uv\nabla u - {u^2}\nabla v} \right] = \frac{{2u}}{v}\nabla u - \frac{{{u^2}}}{{{v^2}}}\nabla v.$

Thus

$\displaystyle - \nabla \left( {\frac{{{u^2}}}{v}} \right) \cdot \nabla v = - \left[ {\frac{{2u}}{v}\nabla u - \frac{{{u^2}}}{{{v^2}}}\nabla v} \right] \cdot \nabla v = - 2\frac{u}{v}\nabla u \cdot \nabla v + \frac{{{u^2}}}{{{v^2}}}{\left| {\nabla v} \right|^2}.$

The Picone identity is very useful. We shall address this later on.

## March 26, 2011

### Asympotic behavior of integrals, 4

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 21:25

We consider the following PDE

$\Delta u = f(x), \quad x \in \mathbb R^2$.

By letting

$\displaystyle w(x) = \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\left[ {\log |x - y| - \log |y|} \right]f(y)dy}$

via the potential theory, we has already proved that

$u-w={\rm const.}.$

As such, the analysis of $w$ turns out to be the core of the studying of solutions to our PDE. As in this entry, we showed that the following limit

$\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \left[ {w(x) - \alpha \log |x|} \right] = -\frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\log |y|f(y)dy}$

exists for certain function $f$. Not just the behavior at the infinity, as a question proposed also in that entry, we can control the decay rate of

$\displaystyle {w(x) - \alpha \log |x| + \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\log |y|f(y)dy} }$

i.e. we need the fact

$\displaystyle\left| {w(x) - \alpha \log |x| + \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\log |y|f(y)dy} } \right| \leqslant \frac{{C\log |x|}}{{|x|}},\quad \forall |x| \geqslant 1$

for some positive constant $C$ where $w$ is a particular solution to

I do think this result is correct since it has been used once in a paper by X.X. Chen published in Calc. Var. Partial Differential Equations [here] but some idea is involved. I leave here as my own open question needed to be addressed in the future.

## March 23, 2011

### A proof of the uniqueness of the solution of the prescribing Gaussian curvature problem

Filed under: Uncategorized — Ngô Quốc Anh @ 22:58

Let us continue the problem of prescribing Gaussian curvature. Our PDE reads as the follows

$\displaystyle -\Delta u +K_0(x)=K(x)e^{2u}, \quad x \in M$

where $M$ is a compact manifold without the boundary. Today we show that if

$\displaystyle K(x) \leqslant 0$

then our PDE has unique solution.

Assume that $u_1$ and $u_2$ are solutions to the PDE, that is

$\displaystyle\begin{gathered} - \Delta {u_1} + {K_0}(x) = K(x){e^{2{u_1}}}, \hfill \\ - \Delta {u_2} + {K_0}(x) = K(x){e^{2{u_2}}}, \hfill \\ \end{gathered}$

By subtracting, we have

$\displaystyle - \Delta ({u_1} - {u_2}) = K(x)({e^{2{u_1}}} - {e^{2{u_2}}}).$

Multiplying both sides by $u_1-u_2$, integrating over $M$, and the using the integration by parts we arrive at

$\displaystyle\int_M {{{\left| {\nabla ({u_1} - {u_2})} \right|}^2}dv} = \int_M {K(x)\frac{{{e^{2{u_1}}} - {e^{2{u_2}}}}}{{{u_1} - {u_2}}}{{\left| {{u_1} - {u_2}} \right|}^2}dv} .$

Since $K(x) \leqslant 0$, it follows that

$\displaystyle\int_M {{{\left| {\nabla ({u_1} - {u_2})} \right|}^2}dv} \leqslant 0.$

In particular, $u_1 \equiv u_2$.

## March 20, 2011

### A note on the equation involving the prescribing Gaussian curvature problem

Filed under: Uncategorized — Ngô Quốc Anh @ 2:41

Let $M$ be a smooth and compact two dimensional Riemannian manifold. Let $g_o(x)$ be a metric on $M$ with the corresponding Laplace-Beltrami operator $\Delta$ and Gaussian curvature $K_o(x)$. Given a function $K(x)$ on $M$, can it be realized as the Gaussian curvature associated to the point-wise conformal
metric

$\displaystyle g(x) = e^{2u(x)}g_o(x)$

To answer this question, it is equivalent to solve the following semi-linear elliptic equation

$\displaystyle -\Delta u +K_0(x)=K(x)e^{2u}, \quad x \in M.$

In this entry, we summarize some basic steps in order to simplify the above PDE. We first let $\overline u=2u$, then our PDE becomes

$\displaystyle -\frac{1}{2}\Delta \overline u +K_0(x)=K(x)e^{\overline u}.$

Let $v$ be a solution of the following PDE

$\displaystyle -\Delta v=2K_0(x)-2\overline K_0$

where

$\displaystyle \overline K_0=\frac{1}{|M|}\int_M K_0(x)dv$

is nothing but the average of $K_0$ over $M$. The solvability of the foregoing PDE comes from the fact that

$\displaystyle\int_M (2K_0(x)-2\overline K_0)dv=0.$

We let $w=\overline u+v$. Then it is easy to verify that $w$ solves the following

$\displaystyle -\Delta w +2\overline K_0=2K(x)e^{-v}e^w.$

Finally, letting

$\alpha=2\overline K_0, \quad R(x)=2K(x)e^{-v(x)}$

we get

$\displaystyle -\Delta w +\alpha = R(x)e^w$

or by renaming $w$ by $u$

$\displaystyle -\Delta u +\alpha = R(x)e^u.$

The advantage of this equation is that here $\alpha$ is constant. To be precise, by the Gauss-Bonnet theorem, we have

$\displaystyle \alpha=\frac{4\pi}{|M|}\chi(M)$

where $\chi(M)$ is the characteristic of $M$.

## March 16, 2011

### Cofactor matrix has divergence-free rows

Filed under: PDEs — Ngô Quốc Anh @ 15:45

In this entry, we prove the following interesting result

Let $\mathbf{u} : \mathbb R^n \to \mathbb R^n$ be a smooth function. Then

$\displaystyle \sum_{i=1}^n (\mbox{cof}D\mathbf{u})_{i,x_i}^k=0$

for each $k=\overline{1,n}$ fixed.

For simplicity, let us write $\mathbf{u}=(u^1,...,u^n) \in \mathbb R^n$. Then

$\displaystyle D\mathbf{u} = \left( {\begin{array}{*{20}{c}} {u_{{x_1}}^1}&{u_{{x_2}}^1}& \cdots &{u_{{x_2}}^1} \\ {u_{{x_1}}^2}&{u_{{x_2}}^2}& \cdots &{u_{{x_2}}^2} \\ \vdots & \vdots & \ddots & \vdots \\ {u_{{x_1}}^n}&{u_{{x_2}}^n}& \cdots &{u_{{x_n}}^n} \end{array}} \right).$

## March 13, 2011

### Jacobi’s formula for the derivative of a determinant revisited

Filed under: Uncategorized — Ngô Quốc Anh @ 12:23

Last time, we discussed [here] Jacobi’s formula expresses the differential of the determinant of a matrix A in terms of the adjugate of A and the differential of A. The formula is

$\displaystyle d\mbox{det} (A) = \mbox{tr} (\mbox{adj}(A) \, dA)$.

A more useful formula is the following

$\displaystyle \frac{d}{dt}\mbox{det} (A+tB) = \mbox{det}(A+tB) \mbox{tr}\big((A+tB)^{-1}B\big)$.

Let us firstly reprove the Jacobi formula. Assuming $(A^{ij})$ is the cofactor matrix with respect to $A=(a_{ij})$. It then holds

$\mbox{det}(A)=\sum_{j}a_{ij}A^{ij}.$

Therefore,

$\displaystyle\frac{d}{{d{a_{ij}}}}(\det (A)) = \frac{d}{{d{a_{ij}}}}\left( {\sum\limits_k {{a_{ik}}{A^{ik}}} } \right) = {A^{ij}}.$

## March 10, 2011

### log K is harmonic implies that K is contant

Filed under: PDEs — Ngô Quốc Anh @ 18:29

In this short note we present a result in a paper due to Edward M. Fan [here]. To be precise, we prove

Given a sphere $(\mathbb S^2,g_0)$ with standard metric, if $K(x)>0$ and $\Delta \ln \big(K(x)\big)=0$, then $K$ must be a constant.

Proof. To prove the result, we shall use the following well-known formula

$\displaystyle \Delta \ln K=\frac{\Delta K}{K}-\frac{|\nabla K|^2}{K^2}.$

Therefore, the fact that $\ln K$ is harmonic implies that

$\displaystyle\frac{\Delta K}{K}=\frac{|\nabla K|^2}{K^2}.$

Since $K>0$, multiplying both sides by $K^2$, we get

$\displaystyle K\Delta K=|\nabla K|^2.$

Integrating both sides on $\mathbb S^2$ using standard volume form, we get

$\displaystyle \int_{\mathbb S^2}K\Delta K dv_{g_0}=\int_{\mathbb S^2}|\nabla K|^2dv_{g_0}.$

Now integration by parts shows that

$\displaystyle -\int_{\mathbb S^2}|\nabla K|^2dv_{g_0}=\int_{\mathbb S^2}|\nabla K|^2dv_{g_0}.$

This in turn shows that

$|\nabla K|=0.$

We conclude that $K(x)$ must be a constant function.

## March 7, 2011

### An identity involving inner product of gradients

Filed under: Uncategorized — Ngô Quốc Anh @ 22:41

In this short note, we prove the following identity

For any functions $\eta, u$, it holds

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = |\nabla (\eta u){|^2} - {u^2}|\nabla \eta {|^2}.$

The proof is elementary. By the product rule for the gradient, we know that

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = \nabla u \cdot \big((\eta u)\nabla \eta + \eta \nabla (\eta u)\big).$

Thus

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = \eta u\nabla \eta \cdot \nabla u + \eta \nabla u \cdot \nabla (\eta u).$

The term $\eta \nabla u \cdot \nabla (\eta u)$ can be rewritten as follows

$\displaystyle\eta \nabla u \cdot \nabla (\eta u) = \nabla (\eta u) \cdot \nabla (\eta u) - u\nabla \eta \cdot \nabla (\eta u) = |\nabla (\eta u){|^2} - u\nabla \eta \cdot \nabla (\eta u).$

We then have

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = \eta u\nabla \eta \cdot \nabla u + |\nabla (\eta u){|^2} - u\nabla \eta \cdot \nabla (\eta u).$

Keep in mind that

$\displaystyle\eta u\nabla \eta \cdot \nabla u - u\nabla \eta \cdot \nabla (\eta u) = - u\nabla \eta \cdot (\nabla (\eta u) - \eta \nabla u) = - u\nabla \eta \cdot (u\nabla \eta ).$

Therefore

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = |\nabla (\eta u){|^2} - {u^2}|\nabla \eta {|^2}.$

## March 3, 2011

### The Paneitz operator in any dimension

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 16:43

Let us recall from this topic the following fact: Let $(M,g)$ be a compact Riemannian $4$-manifold, and let ${\rm Ric}_g$ and $R_g$ denote the Ricci tensor and the scalar curvature of $g$, respectively. The so-called Paneitz operator $P_g$ acts on a smooth function $u$ on $M$ via

$\displaystyle {P_g^4}(u) = \Delta _g^2u + {\rm div}\left( {\frac{2}{3}{R_g} - 2{\rm Ric}_g} \right)du$

which plays a similar role as the Laplace operator in dimension two where $d$ is the de Rham differential. Associated to this operator is the notion of $Q$-curvature given by

$\displaystyle Q_g^4=-\frac{1}{6}(\Delta R_g - R_g^2 +3|{\rm Ric}_g|_g^2).$

Under the following conformal change

$\widetilde g = e^{2u}g$

passing from $Q_g^4$ to $Q_{\widetilde g}^4$ is easy through the following formula

$P_g^4 (u)+Q_g^4=Q_{\widetilde g}^4e^{4u}.$

## March 1, 2011

### The implicit function theorem: A PDE example

Filed under: Giải Tích 3, PDEs — Tags: — Ngô Quốc Anh @ 23:29

This entry devotes an existence result for the following semilinear elliptic equation

$-\Delta u + u = u^p+f(x)$

in the whole space $\mathbb R^n$ where $0.

Our aim is to apply the implicit function theorem. It is known in the literature that

Theorem (implicit function theorem). Let $X, Y, Z$ be Banach spaces. Let the mapping $f:X\times Y\to Z$ be continuously Fréchet differentiable.

If

$(x_0,y_0)\in X\times Y, \quad F(x_0,y_0) = 0$,

and

$y\mapsto DF(x_0,y_0)(0,y)$

is a Banach space isomorphism from $Y$ onto $Z$, then there exist neighborhoods $U$ of $x_0$ and $V$ of $y_0$ and a Frechet differentiable function $g:U\to V$ such that

$F(x,g(x)) = 0$

and $F(x,y) = 0$ if and only if $y = g(x)$, for all $(x,y)\in U\times V$.

Let us now consider

$X=L^2(\mathbb R^n), \quad Y=H_+^2(\mathbb R^n), \quad Z=L^2(\mathbb R^n)$.

Let us define

$F(f,u)=-\Delta u + u - u^p-f(x), \quad f \in X, \quad u \in Y, \quad x \in \mathbb R^n$.

It is not hard to see that Fréchet derivative of $F$ at $(f,u)$ with respect to $u$ in the direction $v$ is given by

${D_u}F(f,u)v = - \Delta v + v - p{u^{p - 1}}v$.

Since $-\Delta +I$ defines an isomorphism from $Y$ to $Z$, it is clear to see that our PDE is solvable for $f$ small enough in the $X$-norm.