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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?

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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 $\varepsilon \to 0$.

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  • $\begingroup$ @user88544 The point I was trying to make is that on the linear level, Harnack's inequality depends on the ellipticity, so on the quasilinear level, the same thing had better be true. However, this was kind of hazy and worded confusingly by me, so I am sorry for that. I have added a more explicit counterexample. The Harnack inequalities for the catenoids blow up as $\varepsilon \to 0$. $\endgroup$
    – Aidan Backus
    Commented Aug 17, 2023 at 1:04
  • $\begingroup$ @user88544 A catenary curve is given by the graph of cosh, not inverse cosh -- I don't even know how inverse cosh would come up in this situation. $\endgroup$
    – Aidan Backus
    Commented Aug 17, 2023 at 15:45
  • $\begingroup$ I think by looking at pictures of half-catenoids it is pretty easy to geometrically see why this is a counterexample. As $\varepsilon \to 0$, the neck gets thinner and thinner, while the wider part gets fatter and fatter. The Harnack constant measures the ratio of the fatness to the thinness, and as $\varepsilon \to 0$, the neck shrinks to a point while the wide part becomes infinitely wide. $\endgroup$
    – Aidan Backus
    Commented Aug 17, 2023 at 15:47
  • $\begingroup$ Sorry, I misunderstood your example. You are right! $\endgroup$
    – user88544
    Commented Aug 18, 2023 at 0:34
  • $\begingroup$ Almost useless, since the constant of Harnack constant depends on the elliptic constant, and in your arguments, it depends on the function u itself. $\endgroup$
    – gaoqiang
    Commented Dec 2 at 12:42

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