# comparison principle for the minimal surface equation

Consider the (inhomogeneous) minimal surface equation for functions $$u,f:D\to \mathbb{R}$$ for some smooth domain $$D\subset \mathbb{R}^n$$

$$Lu:=\operatorname{div} \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}=f.$$

Then is it true that $$u_1\leq u_2$$ on $$\mathbb{R}^n\setminus D$$ and $$Lu_1\geq Lu_2$$ implies $$u_1\leq u_2$$ on $$D$$?

• The trick for these sorts of problems is to notice that $v = u_1 - u_2$ solves a linear PDE. The coefficients of this PDE will depend on $u_1,u_2$, but that's no problem. If I am not mistaken the '$c$-term' should be zero, so you can just apply your normal comparison principles. Commented Mar 24, 2023 at 22:09
• This answer is perhaps relevant when $D$ is bounded (and Lipschitz). Commented Mar 25, 2023 at 8:47

For the case that $$D$$ is bounded, as Leo Moos has noted, more details can be found in a textbook of minimal surface, cf. Lemma 1.26 in T. Colding, W. Minicozzi, A Course in Minimal Surfaces.
That is, let $$F(X)=\frac{X}{\sqrt{1+|X|^2}},$$ then $$$$F(\nabla u_1)-F(\nabla u_2)=\left(\int_0^1dF\big(\nabla u_2+t(\nabla u_1-\nabla u_2)\big) dt\right)(\nabla u_1-\nabla u_2).$$$$ From this, one can conclude that $$v=u_1-u_2$$ satisfies an equation of the form $$\operatorname{div}(a_{i,j}\nabla v)\leq 0,$$ where the matrix is defined as $$(a_{i,j})=\int_0^1dF(\nabla u_2+t(\nabla u_1-\nabla u_2)) dt.$$
In particular, for a unit vector $$V$$ and a vector $$X$$, we have $$dF(x)V=\frac{V}{\sqrt{1+|X|^2}}-\frac{\langle V,X\rangle}{(1+|X|^2)^{\frac{3}{2}}}X.$$ Thus, $$(1+|X|^2)^{\frac{3}{2}}\langle V, dF(X)V\rangle=(1+|X|^2)-\langle V,X\rangle^2\geq 1,$$ which means $$(a_{i,j})$$ is positive, therefore, the usual comparison principle gives the claim.