minimal graphs converge to a harmonic function 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!
 A: 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.
