TITLE: Plateau's Problem from an annulus
QUESTION [6 upvotes]: Let $(M^n,g)$ be a complete Riemannian manifold with bounded geometry, that is, it has bounded curvature and positive injectivity radius. Given two disjoint smoothly embedded homotopically trivial closed curves $\gamma_1$ and $\gamma_2$, we consider the problem of minimizing annulus $\Sigma$ with $\partial \Sigma=\gamma_2\cup \gamma_2$.
More precisely, we consider all maps $u\in W^{1,2}(A,M)\cap C^0(\bar A,M)$, where $A:=\{z\in \mathbb C \,\mid 1 < |z| <2\}$, such that $u$ restricted to $\partial A$ are reparametrizations of $\gamma_1$ and $\gamma_2$. Here, we assume at least one such map $u$ exists.
Can we prove the existence of such $u$ whose image has the least area? Is there any reference for such existence?

REPLY [10 votes]: Such an annulus need not exist.  For example, consider two circles in $\mathbb{R}^3$ defined by $x^2+y^2 = 1$ and $z = \pm R$.  If $R$ is sufficiently large, then there cannot be a minimizing annulus (or, indeed, any minimizing connected surface) with these two circles as boundary.
The reason is the following:  First, one can certainly find an annulus with these two surfaces as boundary whose area is as close to $2\pi$ as desired:  Just take the two flat disks with these circles as boundary and join them by a very thin tube $x^2+y^2 = \epsilon^2>0$ and $|z|\le R$ for $\epsilon$ very small and smooth the result to get an annulus.
Meanwhile, if there were a connected minimal surface $A$ with these circles as boundary, then $A$ would have to pass through the plane $z=0$ at some point $p = (x_0,y_0,0)$, and hence the ball $B$ of radius $r<R$ centered on $p$ would not meet the two circles.  Then the monotonicity formula for minimal surfaces implies that the part of the surface $A$ inside the ball $B$ would have to have area at least $\pi r^2$. Letting $r$ approach $R$, one sees that the area of $A$ would be at least $\pi R^2$.
Thus, if $R^2>2$, then no connected minimal surface with these two circles as boundary can achieve the lower bound of $2\pi$.