An Area Variation Question

Question

I have a question on a non-typical "area" variation question. Let (M,dA) be a 2 dimensional manifold and f be a smooth function. Let $\Gamma$ be a compact 2- dimensional submanifold in M whose boundary is $\gamma(s)$. Consider the integral

$\int_{\Gamma} fdA$

I want to compute a variation where $\gamma \rightarrow \gamma + \delta\gamma$ (that is I only perturn the boundary of the region). Let V be the vector field of the variation and $F_t$ be the map from $\Gamma\times I$ to $M$ giving the variation. If you directly try to apply the usual computations you get something like

$\int_{\Gamma} V(f)dA + \int_{\gamma} f (V*)ds - \int_{\Gamma} fH<N,V>dA$

where V* is the dual form associated to the vector field V, H is mean curvature and N is the normal direction. In this case I have several puzzlements. Since the ambient manifold is also 2-dim this means N is non-zero only in the boundary so the last term cancels. Now I would like to write everything in terms of $\delta\gamma$. However $V$ equals to $\delta\gamma$ only on the boundary. So I could not figure out what to do with the first term. Is this computation correct and if so what would one do with the first term to express in terms of $\delta\gamma$. In variational problems one in the final sets an equality which must hold for all variations. And V is a vector field localized on some nbd of the boundary curve we can find a sequence variations where the first terms goes to zero so it does not contribute.

Another approach would be to somehow calculate "area under the curve" of $\delta\gamma(\theta)$ with density f, but that is only makes sense where I can write the curve in polar coordinates as ($\theta$, $r(\theta)$)

Thanks

Answer 1

The first term is an awkward by-product of the way in which you implemented the variation. In addition to perturbing the boundary of $\gamma$, the field $V$ also moves other points around, and that has an effect on the integral of $\int_\Gamma f$ unless $f$ is a constant function. This effect is small when $V$ is localized near the boundary, because the area in which $V(f)$ is nonzero is small. In other words, what you said was correct:

And $V$ is a vector field localized on some nbd of the boundary curve we can find a sequence variations where the first terms goes to zero so it does not contribute.

I think the second approach you outlined is more efficient:

to somehow calculate "area under the curve"

You don't have to write the curve globally in polar coordinates, because the variational formula can be justified locally: by multiplying $\delta\gamma$ by a partition of unity subordinate to coordinate charts. It helps to assert from the beginning that $\delta\gamma$ is normal variation. (General variation can be decomposed in tangential ad normal components, and the tangential variation of boundary curve obviously has no effect on what it bounds.) Let's say that the variation of $\gamma$ is $\gamma+ \epsilon\phi n$ where $n$ is the outward normal to $\gamma$ and $\phi$ is a smooth real-valued function along $\gamma$. Then $$\frac{d}{d\epsilon}_{\,\big|\epsilon=0}\int_\Gamma f\,dA = \int_\gamma f \phi \tag1$$ which is the cleanest variational formula imaginable. As I mentioned earlier, (1) can be justified in local coordinates.