We consider the minimal surface equation $$
(1+|\nabla u|^2) \, \Delta u=\sum\_{i,j=1}^n\partial\_iu \, \partial\_ju \, \partial\_{ij}u\quad\hbox{in $B\_1\subset\mathbb R^n.$}
$$
If $u\in C^2(B\_1)$ is a positive solution of above equation, does the Harnack inequality hold in $B\_{1/2}$? That is, is there a constant $C>0$, which is only dependent on $n$, such that $$\sup\_{B\_{1/2}}u\leq C\inf\_{B\_{1/2}}u?$$


If above equation is uniformly elliptic, then Harnack inequality for the uniformly elliptic equation concludes that the constant $C$ may be dependent on the elliptic constants. However, can we derive the Harnack inequality for the minimal surface equation without the uniformly elliptic condition?



Harnack inequality for the minimal surface equation
The minimal surface equation is uniformly elliptic, at least in the sense that its linearization at any solution is uniformly elliptic. It will be convenient to rewrite the equation as
$$\nabla \cdot \left(\frac{\nabla u}{\sqrt{1 + |\nabla u|^2}}\right) = 0$$
(this is the standard way that the equation is written on euclidean space, anyways). We now have the coefficient $A\_{ij} = (1 + |\nabla u|^2)^{-1/2} \delta\_{ij}$, and the PDE is $Pu := \nabla \cdot (A\nabla u) = 0$.


By elliptic regularity, the solution $u$ is smooth, so there exists $\varepsilon > 0$ such that on $B\_{3/4}$ (say) we have $|\nabla u| < \varepsilon^{-1}$. On $\{|\nabla u| < 1\} \cap B\_{3/4}$ we have $A\_{ii} > 1/10$, and on $\{|\nabla u| \geq 1\} \cap B\_{3/4}$ we have $A\_{ii} > \varepsilon^{1/2}/10$. Therefore $P$ is a uniformly elliptic operator and we can apply the usual proof of Harnack's inequality.


EDIT 2: Here's an explicit example. Recall that the catenoid is the surface of revolution corresponding to the catenary
$$z = f(x) := \varepsilon \cosh\left(\frac{x}{\varepsilon}\right)$$
and it is minimal. Taking only half of the surface of revolution, we get a minimal graph $z = u(x, y)$ with $u > 0$. Consider the restriction of $u$ to the ball $B$ of radius $1/4$ centered on $(x, y) = (1, 0)$. Then the graph of $u$ contains the the catenary, so $\inf\_B u \leq f(0.75)$ and $\sup\_B u \geq f(1.25)$. Now we estimate the ratio
$$\frac{\sup\_B u}{\inf\_B u} \geq \frac{f(1.25)}{f(0.75)} = \frac{\cosh(1.25/\varepsilon)}{\cosh(0.75/\varepsilon)}$$
which blows up as $\varepsilon \to 0$.


