I am wondering if there is an analog of the following theorem by Morgan and White:
Suppose $g\_E$ is the Euclidean metric on $\mathbf R^3$. If $\gamma$ is a closed $C^{k,\alpha}$ curve in $(\mathbf R^3,g\_E)$ that bounds a unique area minimizing surface, then every closed curve sufficiently closed to $\gamma$ in $C^{k,\alpha}$ also bounds a unique area minimizing surface.


Specifically, I am wondering what happens if we fix a curve, say a circle in $\mathbf R^3$, and let the metric change. Intuitively, when the metric is closed to $g\_E$, one expects that $\gamma$ still bounds a unique minimal surface. Is there any theorem that settled this question? More generally, if I consider the set of all circles on $\mathbf R^3$, do I expect that when the metric $g$ is closed to $g\_E$, all of them bound a unique area minimizing surface?


Any insight or reference would be very appreciated.



Minimal surface on $R^3$ with with non Euclidean metric
Below I sketch the proof of the following theorem:



> 
> Theorem: Suppose that $\Sigma^n\subset (M^{n+1},g\_0)$ is smooth and uniquely area-minimizing relative to it's boundary $\Gamma : = \partial\Sigma$ and is strictly stable (no nontrivial Jacobi fields with Dirichlet boundary conditions). Then if $g$ is a sufficiently small $C^{k,\alpha}$ perturbation of $g\_0$, there's a unique area minimizer with boundary $\Gamma$ for the metric $g$ (and it's still smooth and strictly stable).
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Remarks:


1. I think $k =1, \alpha \in (0,1)$ should suffice but I didn't think through each step very carefully to verify this.)
2. Note that this theorem actually proves the Euclidean one since if $\gamma,\gamma'$ are nearby boundaries, then you can find $\Phi : \mathbb{R}^n\to \mathbb{R}^n$ diffeomorphism (small) taking $\gamma'$ to $\gamma$. Then the $\gamma'$ solution (for $g\_E$) is a $\gamma$ solution for $\Phi^\*g\_E$.




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Proof: Suppose that $g\to g\_0$ and $\Sigma\_1,\Sigma\_2$ are distinct $g$-minimizers with boundary $\Gamma$ (singular is fine). Since $\Sigma$ is smooth then Allard's theorem (interior and boundary) imply that $\Sigma\_1,\Sigma\_2$ converge smoothly to $\Sigma$ all the way up to the boundary. In other words, there's $u\_1,u\_2$ on $\Sigma$ so that $\textrm{graph}(u\_i) = \Sigma\_i$ and $u\_i |\_{\partial\Sigma} = 0$ (where the graph is taken using the normal exponential map of $g\_0$) and $u\_i\to 0$ in some $C^{k',\alpha}$ space (which you have to work out depending on $k$). Write $H(u,g)$ for the mean curvature of the graph (as above) of $u$ with respect to $g$. Note that $H(u\_i,g) = 0$ by assumption. Thus,
$$
0 = H(u\_2,g)-H(u\_1,g) = D\_1H(u\_1,g) (u\_2-u\_1) + O(||u\_2-u\_1||\_{C^2}^2)
$$
(we used Taylor's theorem). The error is uniform with respect to the metric.


Now, divide by $||u\_2-u\_1||\_{C^2}$ to get that $w:=(u\_2-u\_1)/||u\_2-u\_1||\_{C^2}$ satisfies
$$
0 = D\_1H(u\_1,g) w + o(1)
$$
Using Schauder estimates we can get that $||w||\_{C^{2,\alpha}} = O(1)$ (you have to check that the $o(1)$ term is bounded in $C^\alpha$ and that $D\_1(u\_1,g)w$ is a uniformly elliptic operator. Note that as $g\to g\_0$ and $u\_1\to 0$, $D\_1H(u\_1,g)$ limits to the Jacobi operator on $\Sigma$ with respect to $g\_0$ (this is the definition of the Jacobi operator $J=D\_1H(0,g\_0)$. Moreover, by the $C^{2,\alpha}$ bounds for $w$, we get that the limit $w\_0$ is $C^2$ and nonzero and solves $Jw\_0 = 0$ (with Dirichlet boundary conditions).


This is a contradiction, proving uniqueness for $g$ minimizers, $g$ close to $g\_0$. Strict stability follows from the fact that the first Dirichlet eigenvalue of $J$ is continuous as the metric and minimizer varies.




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For your second question about the result being uniform for all disks, this cannot be true. Here is a strange proof (probably there is a simpler way):


Construct a non-flat $g$ on $\mathbb{R}^3$ rotationally symmetric with scalar curvature $> 0$ and so that $g=g\_E + O(r^{-1})$ at infinity (along with derivatives). This is called an asymptotically flat metric. It turns out that you can do this with $||g-g\_E||\_{C^k}$ as small as you like. (You can also do this while arranging that there are no closed minimal surfaces in $(\mathbb{R}^3,g)$). This can all be accomplished by looking at the scalar curvature of $dr^2 + \varphi(r)^2 g\_{S^2}$ and playing around with the resulting differential inequality.)


Thus, it suffices to prove that for this metric $g$ fixed, for $R$ sufficiently large, the circle of radius $R$ in the $xy$-plane has non-unique minimizers. Suppose the minimizer was unique. Then, by symmetry it must be the disk in the $xy$-plane. (The symmetry of the manifold already shows this is a minimal surface, but it might not be the minimizer; if not the minimizer then $z\mapsto -z$ takes the minimizer to a second one!).


If each disk of radius $R$ was a minimizer then the $xy$-plane will be a minimizer (on compact sets). However, the [proof](https://mathscinet.ams.org/mathscinet/article?mr=0526976) of the positive mass theorem shows that the scalar curvature would vanish along the plane. (Alternatively, see [here](https://arxiv.org/pdf/1510.07406.pdf) and the references contained within.)




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For your question in the comment, I think the following is true:



> 
> Theorem: Fix $R\_0>0$ and suppose that $||g-g\_E||\_{C^{k,\alpha}} \leq \epsilon=\epsilon(R\_0)$. Then there's a unique $g$-minimizer for any circle of radius $R\leq R\_0$.
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Proof: If not there's $g\to g\_E$ in $C^{k,\alpha}$ and circles $C$ with radius $R\leq R\_0$ where things fail. You can translate $C$ to the origin and rescale so that they have radius $1$. The rescaled metrics $g'$ still converge in $C^{k,\alpha}$ (this is where we used $R\leq R\_0$). Thus, the previous proof applies.


