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A stable domain $D$ in a minimal surface $S\subset \mathbb{R}^3$ is a domain for which the area-functional $A(t):=\int_{S_t}dS_t$ has non-negative second derivative, i.e. $A''(0)\geq 0$, for all compactly supported variatons on $D$. We consider normal variations of the form $F:S\times ]-\epsilon,\epsilon[\rightarrow \mathbb{R}^3: (p,t)\mapsto p+tf(p)N_p$, with $f:S\rightarrow \mathbb{R}$ a compactly supported function on $S$ and $N_p$ a local Gauss map on $S$. We denote $S_t$ the varied surface at time $t$, (so $S_0=S$).

It can be shown that a minimal surface is a critical point of the area functional, i.e. $A'(0)=0$, so therefore a stable domain in a minimal surface reaches a minimum of the area functional.

My question: Can we then say that a stable domain $D$ in a minimal surface $S$ is area minimizing among all surfaces with the same boundary $\partial D$ and contained in the cylinder $\partial D \times \mathbb{R}$? Or when and under which conditions is a stable domain in a minimal surface area minimizing among all surfaces with the same boundary $\partial D$?

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As far as I know, the concept of stable domain is no longer in use, at least in the field of minimal surfaces. Instead, in most of the research that I know, the stable minimal surfaces are defined as those which are area minimizing relative to nearby surfaces with the same boundary. Precisely those that one would expect to obtain in physical experiments.

But if one wants to define a stable domain for a minimal surface, it would be as follows:

Let $M$ be a two-dimensional, orientable and smooth manifold and $D\subset M$ a domain on $M$.

Let $X:M\to\mathbb{R}^3$ be a minimal immersion into the Euclidean space. It is known that $D$ is a critical point of the area of the induced metric, for all variations of $\bar{D}$ which keep $\partial D$ fixed. When this critical point is a minimum for all such variations, we say that $D$ is stable.

Answering your question, yes.

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The short answer to your question is no, a stable domain $D$ does not in general minimise area among all surfaces with the same boundary.

Let me elaborate: for simplicity assume that $D$ is orientable, so that the Gauss map is defined on the whole of $D$. Let $D_t$ be the surfaces resulting from a variation along $f \in C_c^1(D)$ and denote $A(D_t)$ their respective areas. At least for small $t$, the surfaces $D_t$ are going to be embedded in $\mathbb{R}^3$ also. The following definitions are standard: $D$ is called stable if $\frac{\mathrm{d^2}}{\mathrm{d}t^2} A(D_t) \geq 0$ for all variations $f \in C_c^1(D)$, and is called area-minimizing if it has the least area of all surfaces with the same boundary. (Sometimes this is also called absolutely area-minimizing to distinguish with the case where one restricts oneself to surfaces with the same genus as $D$, but this distinction is not important here.)

As $S$ is minimal, we have $\frac{\mathrm{d}}{\mathrm{d}t} A(D_t) = 0$. However, this does not show that the function $t \mapsto A(D_t)$ reaches a minimum at $t = 0$, even when $\frac{\mathrm{d^2}}{\mathrm{d}t^2} A(D_t) \geq 0$. The problem is roughly the same as for functions: the higher-order variations of $A(D_t)$ may dominate. Nobody is to say that $A(D_t)$ does not behave like $t^3$ near $0$, for instance.

Of course, the surfaces $D_t$ all have the same boundary as $D$, so this answers your second question in the negative: not only is a stable domain $D$ in general not area-minimizing among surfaces with the same boundary, but it might even be that $A(D_t) < A(D)$.

The question also asks whether stable domains are area-minimizing "in the cylinder $\partial D \times \mathbb{R}$". Though I understand the idea behind this, I do not really see a way of making sense of this. How would you embed this cylinder in $\mathbb{R}^3$? If for instance you chose to embed it using the normal to $S$ at every point in $\partial D$, you would generally get self-intersections. In any case, any reasonable definition of surfaces "in the cylinder $\partial D \times \mathbb{R}$" should include the surfaces $D_t$ constructed above, so the answer would still be no.

You also ask about conditions under which a stable minimal surface would automatically be area-minimizing. The questions seems to me a little bit like asking when a function $f: (-\epsilon,\epsilon) \to \mathbb{R}$ with $f'(0) = 0$ and $f''(0) \geq 0$ has a minimum at $0$. In other words, I do not know any non-trivial conditions on $D$ that would ensure this, but I would be happy to think about it some more if you could elaborate on what you were conjecturing.

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