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


\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.


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