minimal graphs converge to a harmonic function

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Suppose $\{u_i\}$ is a sequence of positive $C^2$-functions on the unit ball $B,$ satisfying the minimal surface equation, i.e., $${\rm div}\left(\frac{Du_i}{\sqrt{1+|Du_i|^2}}\right)=0$$ for all $i.$ Define another sequence $$w_i:=\frac{u_i}{u_i(0)}.$$ The problem is as follows. If $u_i$ converges to $0$ uniformly to $0$ on any compact subset of $B,$ with uniform $C^2$-bound (that is, $|u_i|+|Du_i|+|{\rm Hess}_{u_i}|<\varepsilon$ for all $i$), will a subsequence of $w_i$'s converge to a "harmonic" function?

Since the minimal surface equation for $u$ could be written as $$\sum_{i,j}\left(\delta_{ij}-\frac{D_iuD_ju}{1+|Du|^2}\right)D_{ij}u=0,$$ I think it suffices to prove that $|Du_i|$ converges to $0$ also. However, I even don't know if this is correct.

Any suggestion or comments are appreciated!

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Finally I somehow get a satisfactory explanation. The equation could be written as \begin{equation}\label{MSE} \sum_{i,j}\left(\delta_{ij}-\frac{D_iuD_ju}{1+|Du|^2}\right)D_{ij}u=0. \end{equation} Thus we attempt to show a subsequence of $Du_i$ converges to $0.$ For any $\varphi\in C^\infty_0(B),$ we have $$0=\int \varphi u_i \text{div}\frac{Du_i}{\sqrt{1+|Du_i|^2}}=\int\left\langle \frac{Du_i}{\sqrt{1+|Du_i|^2}},u_iD\varphi+\varphi Du \right\rangle.$$ Thus $$\int \frac{\varphi|Du_i|^2}{\sqrt{1+|Du_i|^2}}\le \int\frac{|u_i||Du_i||D\varphi|}{\sqrt{1+|Du_i|^2}}\le \varepsilon $$ for any $i=i(\varphi,\varepsilon)$ sufficiently large. Take $\varphi=1$ on $\overline{B_r(x)},$ and we derive (based on the a priori bound of $Du_i$) $$C\sup_{\overline{B_r(x)}}|Du_i|^2\le \varepsilon,$$ so $Du_i\to 0$ as $i\to\infty.$

Finally we show that $w_i$'s converge subsequentially (to a harmonic function by above). The the equation is uniformly elliptic since $|Du|$ is counded. Since $w_i$ is positive and $w_i(0)=1,$ the Harnack inequality implies $$\|w_i\|_{C^0(K)}\le C(K)$$ for any $K\subset\subset B,$ and the Schauder estimate implies $$\|w_i\|_{C^{2,\alpha}(K)}\le C(K)$$ also. The Arzela-Ascoli theorem then implies the subsequential convergence.