# Ngô Quốc Anh

## April 16, 2014

### Locally H^1-bounded implies pointwise upper bounded for the prescribed Gaussian curvature equations

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 15:58

I want to continue my previous post on the prescribed Gaussian curvature equations. Still borrowing the idea and technique introduced in the Struwe et al’ paper, today, I want to talk about how one can pass from locally $H^1$-bounded to pointwise bounded. As always, we are interested in solving the following PDE $\displaystyle -\Delta u + k = K_\lambda e^{2u}.$

For the sake of clarity, let say $K_\lambda \equiv K_i \searrow K$ as $i \to \infty$ and suppose for each $n$, $u_i$ solves the PDE, i.e. the following $\displaystyle -\Delta u_i + k = K_i e^{2u_i}$

holds. As we have already seen, the sequence of solution $\{u_i\}_i$ is $H^1$-bounded in the region $M_-=\{x \in M: K(x) <0\}$. We now show that such an $H^1$-boundedness can guarantee that $\{u_i\}_i$ is pointwise bounded from above in $M_-$. As we shall see later, perhaps, the argument used below only works for the sequence of solutions of the PDE.

To see this, it suffices to prove that $\displaystyle u_i \leqslant C(B)$

for any but fixed ball $B \subset \overline B \subset M_-$. To see this, we first observe from the Trudinger inequality that for each $p>2$ $\displaystyle\int_B {\exp (pu)d{v_g}} \leqslant c\exp \left[ {\eta \frac{{{p^2}}}{4} \int_B |\nabla u|^2 dv_g} + \frac{1}{{\rm vol}(B)}\int_B u dv_g\right]$

for some $\eta,c>0$. Note that

## April 15, 2014

### Locally H^1-bounded for the Palais-Smale sequences in the region when the Gaussian curvature candidate is negative

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 16:22

In 1993, Ding and Liu announced their result about the multiplicity of solutions to the prescribed Gaussian curvature on compact Riemannian $2$-manifolds of negative Euler characteristic. Their interesting result then published in the journal Trans. Amer. Math. Soc. in 1995, see this link.

Roughly speaking, by starting with the prescribed Gaussian curvature equation, i.e. $\displaystyle -\Delta u + k = Ke^{2u}$

when the Euler characteristic $\chi(M)$ is negative, i.e. $\displaystyle 2\pi \chi (M) = \int_M k e^{2u}dv_g <0,$

they perturbed $K$ using $\displaystyle K_\lambda = K+\lambda$

where $\lambda$ is a real number and the candidate function $K$ is assumed to be $\displaystyle \max_{x \in M} K(x)=0$.

Then they were interested in solving the following PDE $\displaystyle -\Delta u + k = K_\lambda e^{2u}.$

Their main result can be stated as follows:

Theorem (Ding-Liu). There exists a $\lambda^\star > 0$ such that

• the PDE has a unique solution for $\lambda \leqslant 0$;
• the PDE has at least two solutions if $0<\lambda<\lambda^\star$; and
• the PDE has at least one solution when $\lambda = \lambda^\star$.

## March 12, 2014

### H^1 boundedness of a sequence of solutions to the prescribing Gaussian curvature problem in the negative case

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 22:31

In a previous note, we showed how to prove an $H^1$ boundedness of a sequence of solutions $\{u_k\}_k$ to the following PDE $\displaystyle -\Delta u_k +\alpha_k = R(x)e^{u_k}$

in the negative case, i.e. $\alpha _k\searrow \alpha$ for some $\alpha <0$ as $k \to \infty$. In this note, we do not change $\alpha_0$, instead, we are going to change $R$. More precisely, we are interested in some $H^1$ boundedness of a sequence of solutions $\{u_k\}_k$ to the following PDE $\displaystyle -\Delta u_k +\alpha = R_k(x)e^{u_k}$

for some $R_k = R+\lambda_k$ with $\lambda_k \searrow 0$ as $k \to \infty$. This note is adapted from a recent preprint by Michael Struwe et al., see this.

As always, we assume that $\alpha <0$ is constant and $(M,g)$ is a closed, connected Riemannian surface with smooth background metric $g$. We further assume for the sake of simplicity that $\text{vol}(M,g)=1$.

Step 1. We claim for sufficiently large $k$ that $\displaystyle \int_M R_k^4 e^{u_k} \leqslant C(R)$

for some constant $C$ depending only on $\|R\|_{C^1}$ and on $(M,g)$.

Proof of Step 1. To prove this, we observe that $\displaystyle \int_M R_k =\int_M e^{-u_k} \big( -\Delta u_k +\alpha\big) = -\int_M |\nabla u_k|^2 e^{-u_k} + \alpha\int_M e^{-u_k},$

which implies, by $\alpha <0$, for large $k$ that $\displaystyle\int_M |\nabla u_k|^2 e^{-u_k}\leqslant \int_M R_k < 1+\int_M R.$