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.


\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.

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