TITLE: Does a minimum area disk that is bounded by a cycle $C$ continuously deform in $R^3$ as $C$ moves in $R^3$?
QUESTION [6 upvotes]: Let $C_1=(v_1,v_2,\ldots,v_{i-1},v_i)$ and $C_2=(v_1,v_2,\ldots,v_{i-1},v'_i)$ be two cycles that are drawn in $R^3$ in the shape of an unknot (not knotted) with straight line segments as their edges (note that $C_1$ and $C_2$ differ from each other in one vertex: Vertex $v_i$ of $C_1$ is replaced with vertex $v'_i$ in $C_2$). 
Let $D_1$ and $D_2$ be two disks with minimum area that are bounded by $C_1$ and $C_2$ respectively and let $l$ be the line that passes through $v_i$ and $v'_i$. 
Also assume that:

$C_1$ has a unique minimum area disk,
we can choose $v'$ to be arbitrarily close to $v$.

Now assume that we start to move $v_i$ on $l$ towards $v'_i$ (we gradually change the position of $v_i$ towards $v'_i$). Under these conditions, can we guarantee that $D_1$ continuously deforms in $R^3$ until $D_1 = D_2$?
Any references that contains a theorem that proves the continuous deformation of such a surface?

REPLY [8 votes]: This is a standard question. 
Look at the following image from Morgan's "Geometric measure theory". It should convince you that the answer is no. The curve admits two area minimizing discs and it admits arbitrary small perturbations so that just one of them stay area minimizing.