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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?

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This is an excellent question. The answer is "no." The way to see this is via a calibration argument.


Background on Calibrations

Def: Let $(M^n, g)$ be a Riemannian manifold. A calibration on $M$ is a $p$-form $\varphi \in \Omega^p(M)$ satisfying:

  1. $d\varphi = 0$, and
  2. $|\varphi(v_1, \ldots, v_p)| \leq 1$ for every orthonormal set $\{v_1, \ldots, v_p\}$ in $T_xM$.

An oriented $p$-dim subspace $V \subset T_xM$ is calibrated by $\varphi$ iff $\varphi(v_1, \ldots, v_p) = 1$ for some oriented orthonormal basis $\{v_1, \ldots, v_p\}$ of $V$.

An oriented $p$-dim submanifold $N^p \subset M^n$ is calibrated by $\varphi$ iff each tangent space $T_xN \subset T_xM$ is calibrated by $\varphi$.

The following theorem is due to F. Reese Harvey and H. Blaine Lawson (1982):

Fundamental Theorem on Calibrations: Let $(M^n, g)$ be a Riemannian manifold and $\varphi$ a calibration. Let $N, N' \subset M$ be two compact, oriented, $p$-dim submanifolds with $\partial N = \partial N'$. If $N$ is calibrated by $\varphi$, then $\text{Area}(N) \leq \text{Area}(N')$.

Proof: Using that $N$ is calibrated first, then Stokes' Theorem second, then the definition of calibration third, we have $$\text{Area}(N) = \int_{N} \varphi = \int_{N'} \varphi \leq \text{Area}(N'). \ \ \ \lozenge $$


Application: Graphical Minimal Surfaces in $\mathbb{R}^3$

Let $u \colon \overline{\Omega} \to \mathbb{R}$ be such that $\Sigma := \text{Graph}(u)$ is a minimal surface in $\mathbb{R}^3$. Regarding $\Sigma$ as the level set $\{v(x,y) = 0\}$, where $v(x,y) := z - u(x,y)$, we see that a unit normal vector field to $\Sigma$ is $$N_u = \frac{\nabla v}{\Vert \nabla v \Vert} = \frac{(-u_x, -u_y, 1)}{\sqrt{1 + (u_x)^2 + (u_y)^2}}.$$ Define the $2$-form $\varphi_u \in \Omega^2(\mathbb{R}^3)$ via $$\varphi_u(X,Y) = \det(X,Y, N_u) = (X \times Y) \cdot N_u.$$ The following exercise, together with the Fundamental Theorem above, gives the result:

Exercise: (a) The $2$-form $\varphi_u$ is a calibration on $\mathbb{R}^3$. That is:

  1. $d\varphi_u = 0$, and
  2. $|\varphi_u(X,Y)| \leq 1$ for every orthonormal set $\{X,Y\}$ in $T_x\mathbb{R}^3$.

(b) The surface $\Sigma = \text{Graph}(u)$ is calibrated by $\varphi_u$. That is: If $\{X,Y\}$ is an oriented orthonormal set having $X,Y \in T_x\Sigma$, then $\varphi_u(X,Y) = 1$.

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    $\begingroup$ Nice to see you back, Jesse. :) $\endgroup$ – Ted Shifrin Jun 1 '17 at 1:53
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    $\begingroup$ @TedShifrin: Thanks, Ted! :-) $\endgroup$ – Jesse Madnick Jun 1 '17 at 2:40
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    $\begingroup$ Wonderful answer, @JesseMadnick. Thanks for sharing this. $\endgroup$ – Edu Jun 2 '17 at 9:01
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    $\begingroup$ The "usual" form of Stokes' theorem won't work because the region that $N$ and $N'$ bound could be something wild (and it isn't hard to come up with examples where there's no clear "in between" region). What is the correct form that one should apply? $\endgroup$ – Ryan Unger Oct 7 '17 at 3:27
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    $\begingroup$ I suspect the answer might contain "see Federer" or "see Giaquinta-Modica-Soucek"... $\endgroup$ – Ryan Unger Oct 7 '17 at 3:40
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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:

enter image description here

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 cumming to the center from both sides. (Projection of an area minimizer to $(x,y)$ plane will enter in the gaps between half-discs.)

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