Ngô Quốc Anh

May 24, 2013

A proof of the uniqueness of solutions of the Lichnerowicz equations in the compact case with boundary

Filed under: PDEs — Ngô Quốc Anh @ 1:39

In this note, a method introduced by David Maxwell was considered. More precisely, we proved that

\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) without boundary where \kappa=\frac{2}{n-2}, a=2\kappa+4, and b=\frac{n-1}{n} admits at most one solution.

In this note, we consider the case when (M,g) has boundary \partial M and together with our PDE above, we have the following Neumann boundary condition

\displaystyle \partial_\nu u+\frac{1}{\kappa}H_g u=-\frac{1}{2\kappa}\Theta_-u^{\kappa+1}

where \nu is the outward normal vector field on \partial M.

As before, 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 at least \phi is constant. Recall that the following holds

\displaystyle - a{\Delta _{\widetilde g}}(\phi - 1) = |{\sigma _1}|_{\widetilde g}^2({\phi ^{ - 2\kappa - 3}} - \phi ) + b{\tau ^2})(\phi - {\phi ^{2\kappa + 1}})

where \widehat g=\phi_1^\frac{4}{n-2}g is our conformal change. For the detailed calculation, we refer the reader to the previous notes.

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May 20, 2013

Super polyharmonic property of solutions: The case of single equations

Filed under: PDEs — Ngô Quốc Anh @ 9:39

Recently, I have read a paper by Chen and Li published in the journal CPAA [here or here] about the super polyharmonic property of solutions for some partial differential systems.

In this notes, we consider their result in a very particular case – the single equations. We shall prove the following.

Theorem 2.1. Let p be a positive integer and q>1. For each positive solution u of

\displaystyle (-\Delta)^p u \geqslant u^q

in \mathbb R^n, there holds

\displaystyle (-\Delta)^i u >0 \quad i=\overline{1,p-1}.

Proof. For simplicity, we write

\displaystyle v_i = (-\Delta)^i u \quad i=\overline{1,p-1}.

We must show that v_i >0 for all i. For simplicity, we divide the proof into two steps.

Step 1. Proving v_{p-1}>0.

Assume the contradiction, we then have two possible cases

Case 1. There is some x^0 \in \mathbb R^n such that v_{p-1}(x^0)<0.

Case 2. v_{p-1} \geqslant 0 and there is a point x_0 such that v_{p-1}(x_0)=0.

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