= Solution
Write the <minimal surface equation for a graph> as $\operatorname{div}F(Du)=0$ in its <weak formulation>, where $F(p)=p/\sqrt{1+|p|^2}$. For negative increments use the same <difference quotient> convention, so
$$
w_h=\delta_{\ell,-h}u=\frac{u(x)-u(x-he_\ell)}h.
$$
For $\zeta\in C_c^1(\Omega'')$, the function $\delta_{\ell,h}\zeta$ is an admissible <test function> in $\Omega$. Commuting the first <partial derivative> with translation, then applying <discrete integration by parts>, gives
$$
0=\int_\Omega F_i(Du)\delta_{\ell,h}(D_i\zeta)
=-\int_{\Omega''}\delta_{\ell,-h}(F_i(Du))D_i\zeta.
$$
The <fundamental theorem of calculus along a line segment> gives the <averaged linearization of a nonlinear divergence-form equation>
$$
\delta_{\ell,-h}(F_i(Du))=A_h^{ij}(x)D_jw_h,
\qquad
A_h(x)=\int_0^1DF\bigl((1-t)Du(x-he_\ell)+tDu(x)\bigr)\,dt.
$$
Thus \b[the backward difference quotient solves a linear equation in divergence form]:
$$
\boxed{D_i(A_h^{ij}D_jw_h)=0\quad\text{weakly on }\Omega''.}
$$
To verify the <ellipticity of the minimal surface flux>, compute
$$
DF(p)=\frac{I}{\sqrt{1+|p|^2}}-\frac{p\otimes p}{(1+|p|^2)^{3/2}},
$$
and hence
$$
\frac{|\xi|^2}{(1+|p|^2)^{3/2}}
\leq \xi\cdot DF(p)\xi
\leq |\xi|^2.
$$
The <eigenvalue> parallel to $p$ is $(1+|p|^2)^{-3/2}$; every orthogonal <eigenvalue> is $(1+|p|^2)^{-1/2}$. The same lower and upper bounds pass to the averaged <symmetric matrix> $A_h$ whenever both endpoint <gradients> have norm at most $M$:
$$
\boxed{(1+M^2)^{-3/2}|\xi|^2\leq A_h^{ij}\xi_i\xi_j\leq|\xi|^2.}
$$
The printed $C^1(\Omega)$ hypothesis supplies such an $M$ on each <relatively compact subset>, uniformly for sufficiently small $h$. It therefore establishes a locally <uniformly elliptic operator>. A single lower bound on all of $\Omega''$ additionally requires bounded <gradients> there and at the shifted points; this is automatic for bounded $\Omega$ and fixed $h$, but is not supplied on an arbitrary open set.
Back to article page
