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 $D$?



comparison principle for the minimal surface equation
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
\begin{equation}
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).
\end{equation}
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.


