Ngô Quốc Anh

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 $\displaystyle\int_M |\nabla u_k|^2 e^{-u_k}\leqslant \int_M R_k < 1+\int_M R.$

We now multiply our PDE by $R_k^3$ and integrate by parts to find $\begin{array}{lcl}\displaystyle \int_M R_k^4 e^{u_k} &=&\displaystyle\int_M R_k^3 \big( -\Delta u_k +\alpha \big)\\&=&\displaystyle 3\int_M \langle \nabla u_k, \nabla R_k \rangle R_k^2+ \alpha\int_M R_k^3\\ &\displaystyle\leqslant & 3|\nabla R|\int_M |\nabla u_k| R_k^2 + C(R).\end{array}$

By the Young inequality, we can bound $\displaystyle C(R)\int_M |\nabla u_k| R_k^2 \leqslant \frac 12 \int_M R_k^4 e^{u_k} + C(R) \int_M |\nabla u_k|^2 e^{-u_k}.$

Thus, the claim follows.

Step 2. There holds $\displaystyle \int_M R_k^2 u_k \leqslant C(R).$

Proof of Step 2. A proof for this claim is quite obvious since $|u_k| \leqslant C e^{u_k/2}$ for some $C$. Else, we can use the Jensen and Holder inequalities as follows $\displaystyle \int_M R_k^2 u_k =2\|R_k\|_2^2 \int_M \log (e^{u_k/2})\frac{R_k^2}{\|R_k\|_2^2}\leqslant 2\|R_k\|_2^2 \log \left( \frac{\int_M R_k^2 e^{u_k/2}}{\|R_k\|_2^2}\right).$

Step 3. There exists a uniform constant $C>0$ such that $\displaystyle \left\|u_k-\int_M \frac{R_k^2}{\|R_k\|_2^2} u_k \right\|_2\leqslant C \|\nabla u_k \|_2.$

Proof of Step 3. This is somehow a Poincare type inequality with a modified weight $\int_M \frac{R_k^2}{\|R_k\|_2^2} u_k$ rather than $\int_M u_k$. However, since $R_k^2/\|R_k\|_2^2 dv$ is again a volume form, this conclusion follows.

Step 4. The sequence $\{u_k\}_k$ is bounded in the following sense $\displaystyle \|\nabla u_k\|_2^2 + |\alpha|\Big| \int_M u_k \Big| \leqslant 4 E_k(u_k) +C$

for some $C>0$ where $E_k$ is the energy given as follows $\displaystyle E_k(u_k)=\frac 12 \int_M |\nabla u_k|^2 + \int_M \alpha u_k - \int_M R_k e^{u_k}.$

Proof of Step 4. Clearly, $\displaystyle 2E_k(u_k)=\|\nabla u_k\|_2^2 + 2\alpha \int_M u_k - 2\int_M R_k e^{u_k}=\|\nabla u_k\|_2^2 + 2\alpha \int_M u_k - 2\int_M \alpha.$

Hence, $\displaystyle 2E_k(u_k)=\|\nabla u_k\|_2^2 +2\alpha \int_M \frac{R_k^2}{\|R_k\|_2^2} u_k+ 2\alpha \int_M \left( u_k - \int_M \frac{R_k^2}{\|R_k\|_2^2} u_k\right) - 2\int_M \alpha.$

This gives $\displaystyle 2E_k(u_k) \geqslant \|\nabla u_k\|_2^2+ 2\alpha \int_M \frac{R_k^2}{\|R_k\|_2^2} u_k - C\|\nabla u_k\|_2 - C.$

Furthermore, we know from $R_k\searrow R$ that $\displaystyle\int_M \frac{R_k^2}{\|R_k\|_2^2} u_k \leqslant C(R)$

for some constant $C(R)$. Hence we can select a sufficiently large constant $C(R)$ in such a way that $\displaystyle \alpha \int_M \frac{R_k^2}{\|R_k\|_2^2} u_k \geqslant |\alpha| \left| \int_M \frac{R_k^2}{\|R_k\|_2^2} u_k \right| - C(R) \geqslant |\alpha| \left| \int_M u_k \right| - |\alpha| \left| \int_M u_k - \int_M \frac{R_k^2}{\|R_k\|_2^2} u_k \right| - C.$

Using Step 3 and Cauchy-Schwartz inequality, we can verify that $\displaystyle E_k(u_k) \geqslant \frac 12 \|\nabla u_k\|_2^2+ |\alpha| \left|\int_M u_k\right| - C\|\nabla u_k\|_2 - C \geqslant \frac 14 \|\nabla u_k\|_2^2+\frac 14 |\alpha| \left|\int_M u_k\right| - C$

and our claim follows.

Using Step 1, we can conclude that the sequence of solution $\{u_k\}_k \subset H^1(M)$ is somehow bounded in $H^1(M)$ if the energies are also bounded.