Stability of minimal surfaces Let $\Gamma$ be a prescribed $n-2$ dimensional set and assume $S \subset R^n$ is a minimal hyper-surface 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. 
 A: Now that your comment has clarified your question, we can answer it:  The answer is 'no'.  There is the following well-known 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.
A: 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 non-negative cutoff of compact support satisfying $0\leq \phi \leq 1$ and $\phi(p)=1$ for some $p\in S$ then $\frac{d}{dt}|_{t=0} Area_{(1+t\phi)g}(S)>0$
