Plateau's Problem from an annulus 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?
 A: Robert Bryant gives an excellent answer. Let me add some more comments.
Your question is solved by Douglas who was awarded the Fields medal due to his work on Plateau's problem. His paper, however, is a bit hard to read since it is written nearly 90 year ago. Here's a modern solution due to Jost. Douglas's result on your question can be roughly described as the following:    $\DeclareMathOperator{\area}{Area}$
Let $a_i=\inf\{\area(\Sigma_i)\mid \Sigma_i   \text{ is a disk spanned by } \gamma_i\}$. Suppose that there's an annulus $\Sigma$ spanned by $\gamma_1 \cup \gamma_2$ such that $$\area(\Sigma) < a_1+a_2,$$ then there's a minimal annulus spanned by $\gamma_1 \cup \gamma_2$. The criterion above is called the Douglas criterion, and the strict inequality is essential.
A final remark. To solve this question, you can't fix an annulus $A=\{1 < |z| < 2\}$ since conformal annuli are not conformally equivalent to each other. The key ingredients are the following.
Let $\Gamma= \gamma_1 \cup \gamma_2$. Let
\begin{equation}
\begin{split}
\mathcal{F}_\Gamma = \{ & u\mid u \in C(\bar{\Sigma}) \cap W^{1,2}(\Sigma, \mathbb{R}^3), \Sigma=\mathbb{S}^1 \times (0,s), \\
& u|_{\partial \Sigma}: \partial \Sigma \to \Gamma \text{ is a homeomorphism. }\}.
\end{split}
\end{equation}

*

*Take a sequence of functions $u_k \in \mathcal{F}_\Gamma, u_k: \Sigma_k=\mathbb{S}^1 \times (0, s_k) \to \mathbb{R}^3$ such that $E(u_k) \to \inf$, where $E(u_k)$ is the energy of $u_k$.

*Show that $\{s_k\}$ has positive lower bound. This is proved simply by Cauchy's inequality.

*Show that $\{s_k\}$ has finite upper bound. This is guranted by Douglas's criterion, which is the key idea of Douglas. Then we get a convergent subsequence $\Sigma_k \to \Sigma = \mathbb{S}^1 \times (0, s)$.

*Show that $u_k|_{\partial \Sigma}$ is equicontinuous. This is obtained by Douglas's criterion and Courant-Lebesgue's lemma. Therefore, we can suppose that $u_k|_{\partial \Sigma}$ converges uniformly.

*Replace each $u_k$ by the unique harmonic function with the same boundary value, which is denoted by $v_k$. Then $v_k$ is also an energy minimizing sequence and $v_k$ converges uniformly. The limit function is denoted by $v$.

*Show that $v$ attains the infimum of energy, and then attains the infimum of area.

A: 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 $2\pi$.
