Let $\Gamma$ be a prescribed $n2$ dimensional set and assume $S \subset R^n$ is a minimal hypersurface with respect to some smooth metric $g$ on $R^n$, and $\partial S= \Gamma$. Is $S$ is stable with respect to variations in the metric $g$? I believe the stability problem for minimal surfaces has to be well understood but I can not find any references.

2$\begingroup$ I case $n=2$, (it seems that) you are asking if any geodesic is stable. This is obviously not true for general $g$. $\endgroup$– Anton PetruninNov 21, 2014 at 4:29

2$\begingroup$ You shoudl state more precisely what you mean by "stable" here. This term is ambiguous, especially in the present context. $\endgroup$– Benoît KloecknerNov 21, 2014 at 9:57

$\begingroup$ I roughly mean if $g'$ is "close" to $g$ then the minimal surface with respect to to $g'$ is "close" to S. $\endgroup$– A random mathematicianNov 21, 2014 at 10:07

$\begingroup$ Is the question: do solutions to the plateau problem in $(M,g)$ depend continuously on the metric $g$ ? If so, the answer will depend a lot on wether or not the Plateau problem has a unique solution and in which topology you measure the change in $g$ and the solution. Do you have a more precise question regarding the topologies involved ? $\endgroup$– Thomas RichardNov 21, 2014 at 10:58

$\begingroup$ Of course, the answer to your question is false in general. One possibility to recover something is that if you assume that the second variation operator has no zero eigenvalues for the Dirichlet problem, then the inverse function theorem should allow you to smoothly vary $S$ with $g$ for small variations of $g$. $\endgroup$– Otis ChodoshNov 21, 2014 at 11:39
2 Answers
Now that your comment has clarified your question, we can answer it: The answer is 'no'. There is the following wellknown example:
Consider the following family of circles: $C_\lambda$ is defined as $x^2+y^2 = 1$ and $z = \lambda$. Let $\lambda>0$ be fixed and orient $C_{\lambda}$ counterclockwise and orient $C_\lambda$ clockwise. Then for $\lambda$ sufficiently small, there will be a minimal surface of rotation, a catenoid, of the form $x^2+y^2 = c^2\cosh(z/c)$ for some $0<c<1$ that passes through the two circles. In fact, there will be two such values of $c$ when $\lambda$ is sufficiently small, and hence two such minimal surfaces with the same boundary. Only the 'outer' catenoid is the actual minimizer; the 'inner' catenoid is not stable in the usual sense of minimal surfaces.
However, when $\lambda$ increases beyond a certain value $\lambda_0 \approx 1.0318$, there is no value of $c$ that works. As $\lambda$ passes $\lambda_0$, the minimal surface (i.e., 'soap film') with these circles as boundary 'pops' and goes away. In fact, when $\lambda$ is sufficiently large, the minimal surface with these circles as boundary is the union of the two obvious discs in the planes $z = \pm\lambda$. Thus, the minimal surface at the value $\lambda_0$ is not stable in your sense.
Now, I have been moving the circles, but, obviously, I could have just moved the metric instead, considering the family $g_t = \mathrm{d}x^2+\mathrm{d}y^2+(1{+}t)\ \mathrm{d}z^2$, and letting the circles $C_{\pm\lambda_0}$ stay fixed. Thus, there are perturbations of the metric for which this particular minimal surface is not stable in your sense.
The answer is (trivially) no.
The reason is that $S$ is never going to even be stationary for arbitrary changes in the metric*.
*Proof: If $\phi$ is a nonnegative cutoff of compact support satisfying $0\leq \phi \leq 1$ and $\phi(p)=1$ for some $p\in S$ then $\frac{d}{dt}_{t=0} Area_{(1+t\phi)g}(S)>0$

$\begingroup$ The problem asks for the stability of the minimal surface. Of course the minimal surface will not be stationary for arbitrary changes in the metric. $\endgroup$ Nov 21, 2014 at 7:12