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 $\displaystyle\frac{1}{ {\rm vol} (B)}\int_B {ud{v_g}} \leqslant \frac{1}{ {\rm vol} (B)}\int_B {|u|d{v_g}} \leqslant \frac{1}{{\sqrt { {\rm vol} (B)} }}{\left( {\int_B {|u{|^2}d{v_g}} } \right)^{1/2}}.$

Using the mononicity of the exponential function and the fact that $\|u_i\|_{H^1(B)}$ is bounded, we can conclude that $\displaystyle\int_B {e^{2u_i}dv_g}$

is bounded. Hence $\displaystyle\int_B {K_i e^{2u_i}dv_g}$

is bounded as well. We now let $v_i \in H^2 \cap H_0^1(B)$ be the unique solution of the auxiliary problem $\begin{array}{rcl} -\Delta v_i + k &=& K_i e^{2u_i} \quad \text{ on } B,\\v_i &=&0 \quad \text{ on } \partial B.\end{array}$

The standard Schauder estimate tells us that the sequence $\{v_i\}_i$ is bounded in $H^2(B)$. Hence in view of the Sobolev embedding, $|v_i| \leqslant C(B)$

for some constant $C>0$. Using the equation $-\Delta v_i + k = K_i e^{2u_i}$ and the one that $v_i$ solves, we conclude that the function $w_i = u_i - v_i$

is harmonic on $B$. It now suffices to show that $w_i$ is bounded from above. To do so, we use the mean value property of harmonics functions. For example, pick any $y \in B$ and choose a small ball $B_r(y) \subset B$. Then we can write $\begin{array}{lcl}{w_i}(y) &=&\displaystyle\frac{1}{{{\rm vol}({B_r}(y))}}\int_{{B_r}(y)} {{w_i}d{v_g}} \\ &\leqslant & \displaystyle\frac{1}{{{\rm vol}({B_r}(y))}}\left( {\int_{{B_r}(y)} {{u_i}d{v_g}} + \int_{{B_r}(y)} {|{v_i}|d{v_g}} } \right).\end{array}$

Each terms on the right most hand side can be estimated further as follows: $\displaystyle\int_{{B_r}(y)} {|{v_i}|d{v_g}} \le C + \log \left( {\int_B {{e^{|{v_i}|}}d{v_g}} } \right) \leqslant C$

and $\displaystyle\int_{{B_r}(y)} {{u_i}d{v_g}} \le \frac{1}{2}\log \left( {\int_{{B_r}(y)} {{e^{2{u_i}}}d{v_g}} } \right) \le C + \frac{1}{2}\log \left( {\int_B {{e^{2{u_i}}}d{v_g}} } \right) \leqslant C.$

Hence we have just shown that $w_i$ is pointwise bounded from above. It seems that this cannot be true for any sequence of functions rather than those solving the PDE.