# Ngô Quốc Anh

## February 22, 2014

### An application of the Brezis-Li-Shafrir estimate to the limiting case of the prescribing Gaussian curvature problem in the negative case

Filed under: PDEs, Riemannian geometry — Ngô Quốc Anh @ 14:56

On a Riemannian surface $M$, let consider the following PDE

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

naturally arising from the prescribed Gaussian curvature problem. A simple variable change, one can assume that $\alpha$ is a negative constant, see this. It follows from a very well-known result due to Kazdan and Warner that it is necessary to have

$\displaystyle \int_M R(x) dx<0.$

In addition, Kazdan and Warner also showed that if $\int_M R(x) dx<0$ and $R$ changes sign, then there exists a number $\alpha_0 \in (-\infty, 0)$ such that the above PDE is solvable for all $\alpha > \alpha_0$ but not if $\alpha<\alpha_0$. In fact, the number $\alpha_0$ can be characterized as follows

$\displaystyle \alpha_0 = \inf\{\alpha : \text{the PDE is solvable}\}.$

This can be easily seen from the the following comprising property: If the PDE is solvable for some $\alpha_1$, it is also solvable for any $\alpha_2 \geqslant \alpha_1$.

However, Kazdan and Warner did not tell us what happens when $\alpha=\alpha_0$. In an attempt to see what really happens when $\alpha=\alpha_0$, Chen and Li made use of the Brezis-Li-Shafrir estimate to answer in the following way: The PDE is also solvable even when $\alpha=\alpha_0$. The purpose of this note is to talk about the beautiful Chen-Li argument, see this.

The idea is to approximate the equation for $\alpha_0$ by a sequence $\{\alpha_k\}_k$ of negative real numbers in the following sense $\alpha _k\searrow a_0$ as $k \to \infty$. Their proof consists of three steps as follows:

Step 1. Minimizing the associated energy functional, for each fixed $\alpha_k <0$,

$\displaystyle J_k (u) = \frac 12 \int_M |\nabla u|^2 dx +\alpha_k \int_M u dx - \int_M R(x) e^u dx$

to obtain a minimizer $u_k$ solving the equation.

Step 2. Show that the sequence of minimizers $\{u_k\}_k$ is bounded in the region where $R(x)$ is positively bounded away from $0$.

Step 3. Prove that $\{u_k\}_k$ is bounded in $H^1(M)$ and hence converges to a desired solution.

Since a proof for Step 1 is probably well-known, we omit it. However, the key point in this prove is to show that any solution for the PDE is bounded from below, say by a negative constant $A$. To show that it is also bounded from above, we consider the region where $R(x)$ is positively bounded away from $0$. We will use a $\sup + \inf$ inequality of Brezis, Li, and Shafrir.

Theorem (Brezis, Li, and Shafrir). Assume that $V$ is a Lipschitz function satisfying

$\displaystyle 0 < a\leqslant V (x)\leqslant b < 1$

and let $K$ be a compact subset of a domain ­ $\Omega$ in $\mathbb R^2$. Then any solution $u$ of the equation

$\displaystyle -\Delta u=V(x)e^u \quad x \in \Omega$

satisfies

$\displaystyle \sup_K u +\inf_K u\leqslant C \big( a,b,\|\nabla V\|_\infty, K, \Omega\big).$

To do so, we pick a point $x_0 \in M$ at which $R(x_0)>0$. Then we select $\varepsilon>0$ sufficiently small in such a way that

$\displaystyle R(x)\geqslant a >0 \quad \forall x \in \Omega = B_{2\varepsilon} (x_0).$

Then we let $v_k$ solve the following

$\begin{array}{rcl}-\Delta v_k -\alpha_k&=&0, \quad x\in \Omega,\\v_k &=&1, \quad x\in\partial \Omega.\end{array}$

Then we let

$w_k = u_k + v_k,$

which then implies that $w_k$ solves

$\displaystyle -\Delta w_k=R(x)e^{-v_k}e^{w_k}$

in $M$. By the Maximum Principle, the sequence $\{v_k\}_k$ is bounded from above in $M$ and from below in the small ball ­$\Omega \subset M$.

This can be seen as follows: Assume that $v_k$ achieves its maximum value at some $x_k \in \Omega$. Then there holds $-\Delta v_k (x_k) \geqslant 0$ which is impossible since $\alpha_k <0$ and $-\Delta v_k -\alpha_k =0$. Thus, $x_k \in \partial \Omega$ which gives an uniformly upper bound for $v_k$ in $M$. To obtain a lower bound for the sequence $\{v_k\}_k$, let consider

$\begin{array}{rcl}-\Delta (v_{k+1}-v_k)&=&\alpha_{k+1}-\alpha_k, \quad x\in \Omega,\\v_{k+1}-v_k &=&0, \quad x\in\partial \Omega.\end{array}$

Since $\alpha_{k+1}-\alpha_k >0$, the function $v_{k+1}-v_k$ achieves its minimum value at some point on the boundary $\partial \Omega$. Therefore, $v_{k+1} \geqslant v_k$ for all $k$ which gives a lower bound as desired.

Since the metric is locally point-wise conformal to the Euclidean metric, we can apply the Brezis-Li-Shafrir estimate to conclude that $\{w_k\}_k$ is also uniformly bounded from above in the smaller ball $B_\varepsilon (x_0)$. This tells us that $\{w_k\}_k$ is uniformly bounded in the region where $R(x)$ is positively bounded away from $0$, so is $\{u_k\}_k$.

In the final step, we prove that the sequence $\{u_k\}_k$ is bounded in $H^1(M)$. To do so, we first select $\delta >0$ small enough such that the set

$D=\{x \in M : R(x) > \delta\}$

is non-empty. From the previous step, we know that $\{u_k\}_k$ is uniformly bounded in $D$. We now introduce a function $K$ given as follows

$\begin{array}{rcl}K(x)&<&0, \quad x\in D,\\K(x) &=&0, \quad x\not\in D.\end{array}$

Then, it is well-known that for each $\alpha_k$, there exists a unique solution $v_k$ to the following equation

$\displaystyle -\Delta v_k +\alpha_k = K(x)e^{v_k}.$

xThen since $-\Delta v_k +\alpha_k \leqslant 0$ and $\alpha_k \to \alpha_0$, the sequence $\{v_k\}_k$ is bounded. Let $w_k = u_k - v_k$, we then know that

$\displaystyle -\Delta w_k = R(x) e^{u_k} - K(x)e^{v_k}.$

It then suffices to prove that the sequence $\{w_k\}_k$ is bounded in $H^1(M)$. Multiplying both sides of the preceding equation by $e^{w_k}$ and integrating the resulting equation over $M$, we arrive at

$\displaystyle \int_M |\nabla w_k|^2 e^{w_k} dx=\int_M R(x)e^{u_k + w_k} dx -\int_D K(x)e^{u_k} dx.$

On the other hand, since each $u_k$ is a minimizer of the functional $J_k$, the second order derivative $J_k''(u_k)$ is positive definite. That is to saying

$\displaystyle \left( \frac{d^2}{dt^2}J_k (u_k +t\phi)\right) \bigg|_{t=0} \geqslant 0$

for any $\phi \in H^1(M)$, or equivalently,

$\displaystyle \int_M \Big( |\nabla \phi|^2-R(x)e^{u_k}\phi^2\Big) dx \geqslant 0.$

Chosing $\phi = e^{w_k/2}$, we obtain

$\displaystyle \frac 14\int_M |\nabla w_k|^2 e^{w_k} dx\geqslant \int_M R(x)e^{u_k + w_k} dx.$

Hence, we have just shown that

$\displaystyle -\int_D K(x)e^{u_k} dx \geqslant \frac 34\int_M |\nabla w_k|^2 e^{w_k} dx.$

Noticing that $\{u_k\}_k$ is uniformly bounded in the region $D$ and $\{w_k\}_k$ is bounded from below in $M$, the preceding inequality tells us that $\int_M |\nabla w_k|^2dx$ is bounded, so does $\int_M |\nabla u_k|^2dx$. Therefore, it is necessary to control $\int_M u_k ^2dx$ from above. From now on, there is nothing to do with our PDE, the only thing we need is the fact that $\{u_k\}_k$ is bounded in $D$. Suppose that

$\displaystyle\int_M u_k ^2dx \to +\infty$

as $k\to \infty$. Making use of the Friedrichs inequality

$\displaystyle\int_M u_k ^2dx \leqslant C\left( \int_M |\nabla u_k|^2 dx + \left| \int_M u_k dx\right|^2\right),$

we conclude that, as $k \to \infty$,

$\displaystyle\int_M u_kdx \to +\infty.$

In view of the Poincare inequality and the fact that $\int_M |\nabla u_k|^2dx$ is bounded, the sequence $\{u_k - \int_M u_k dx\}_k$ is bounded in $H^1(M)$. Using the following elementary inequality

$\displaystyle \left| u_k - \int_M u_k dx \right| + |u_k|^2\geqslant \frac 12 \left| \int_M u_k dx \right|^2 \to +\infty,$

and the fact that $\{u_k\}_k$ is bounded in $D$, we know that

$\displaystyle \int_M \Big| u_k - \int_M u_k dx \Big|^2 dx \to +\infty$

as $k \to \infty$. This contradicts to the fact that the sequence $\{u_k - \int_M u_k dx\}_k$ is bounded in $H^1(M)$.