TITLE: Is there a minimal graph in $\mathbb{R}^3$ which is not area-minimizing?
QUESTION [11 upvotes]: Let $\Omega\subset\mathbb{R}^2$ be an open subset such that $\partial\Omega$ is a closed, simple curve. 
I'm trying to find an example of an $u:\overline{\Omega}\to\mathbb{R}$ such that $\Sigma:=\text{graph}(u)$ is a minimal surface and, yet, there exists another minimal surface $\Sigma'$ with $\partial\Sigma'=\partial\Sigma$ and $\text{Area}(\Sigma')<\text{Area}(\Sigma)$.
Does such an example exist?

REPLY [7 votes]: The calibration argument works perfectly if $\bar\Omega$ is convex;
if not, then there are counterexamples.
If $\bar\Omega$ is not convex, then the minimizer can escape from $\bar\Omega\times \mathbb R$ (where the form $\varphi_u$ is defined). In this case we have no control on its area.
For example, a minimal graph $z=u(x,y)$ might not minimize area if $u$ is defined on domain $\bar\Omega\subset \mathbb R^2$ as on the picture:

Indeed, imagine that boundary values are zero except the horizontal lines where they depend only on the $x$-coordinate. In this case (for suitable $u$) it is cheaper to take a disc in the $(x,y)$-plane with two thin belts approaching the center from both sides. (Projection of an area minimizer to $(x,y)$ plane will enter in the gaps between half-discs.)