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).

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