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.


\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

\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


\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



\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).


\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 ).


\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<u \in H^1(\mathbb R^n).

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.


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


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.

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