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I'm in the middle of a proof and i'm trying to understand a step of the proof which does give me a hard time. The proof is about minimal surfaces and at the moment I'm trying to understand why the mean curvature has to vanish on a regular surface for it to be minimal. The question is not about differential geometry, but I'm trying to be as complete as possible. The book im using is from Dierkes, Hildebrandt,.. and its called Minimal Surfaces I (Page 54..).

First some notation and the conditions of the problem. I do adumbrate the situation of the proof very so little, but the question is strictly technical, so if you don't need it, skip ahead to the actual problem and find what you need in the notation section later on:

NOTATION:

Let $U$ be an open set in $\mathbb R^2$ and $w = (u,v) =(u^1,u^2) \in U$.

Let $X: U\rightarrow\mathbb R^3 $ be a regular surface of class $ C^2 $ with its spherical image defined by $ N = \frac {1}{\mathcal W} X_u \times X_v $.

$\mathcal W = (\mathcal E \mathcal G - \mathcal F^2)^{1/2}$ where $\mathcal E,\mathcal F, \mathcal G $ are the coefficients of the first fundamental form.

There is $Z:\;U\times(-\varepsilon,\varepsilon)\;\rightarrow \mathbb R^3 ,\;\varepsilon> 0$, which is a Variation of X of Class $C^2$, with the property that $Z(w,0)=X$ for all $w \in U$. By Taylor expansion we get: $Z(w,\varepsilon) =X(w)+Y(w)+\varepsilon^2 R(w,\varepsilon)$, where $R$ is a continous remainder term.

$Y(w)= \frac{\partial}{\partial\varepsilon}Z(w,\varepsilon)|_\varepsilon \in C^1(U,\mathbb R^3) $ is called the first variation of the family of surfaces $Z(.,\varepsilon)$.

For partial derivations we use $ {\partial X}/{\partial u^\beta}=X_{u^\beta}$.

Now we can write $Y(w) =\sum \eta^\beta(w)X_{u^\beta} + \lambda(w)N(w)$ with $\beta \in {{1,2}}$ and $w = (u,v) =(u^1,u^2)$ with functions $\eta^1,\eta^2,\lambda $ of class $C^1(U)$.

QUESTION:

Performing a partial integration, it follows that: $\int_U [\;(\eta^1\mathcal W)_u +(\eta^2 \mathcal W)_v\;]\;du dv= \int_{\partial U}\mathcal W(\eta^1 dv-\eta^2du)$

This is supposed to be partial integration with the help of the divergence theorem. But as i understood partial integration, there has to be a vectorfield there, which isn't, and a continous scalarfunction. The thing is, with what i got, i don't know where to start from. Any first hints to get me going would be very appreciated.

Thank you in advance,

marcel

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This is the first question i have posted. Is the formatting and the way i posted the question appropriate? I'm thankful for any form of criticism.Thanks! –  Marcel Aug 13 '12 at 16:21
    
+1 for using "Adumbrate" and for a well-formulated question. There's no need to sign your posts, as your name appears right at the bottom. –  Adam Saltz Aug 13 '12 at 17:55
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1 Answer

up vote 2 down vote accepted

There is a vector field here: it has components $\eta^1 W$ and $\eta^2 W$. Its divergence appears in the integral on the left. Its flux through the boundary is on the right. The flux is written using the counterclockwise parametrization of the boundary $u=u(t)$ and $v=v(t)$, which makes $(du,dv)$ a tangent vector and $(dv,-du)$ an outward normal vector. It's not the unit normal: it's the "normal of right size", meaning it already incorporates the length term $\sqrt{(du)^2+(dv)^2}$ in it.

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Thank you Mr. Kovalev.But i don't see the vectorfield quite yet.There is $ \eta^1 : U \rightarrow \mathbb R $ and $ \mathcal W : \mathbb R^2 \rightarrow \mathcal R$ And if derive them, $\mathcal W_u :\mathcal R^2 \rightarrow \mathcal R $ same for $\eta_u$ isn't it? –  Marcel Aug 14 '12 at 6:26
    
Sorry for the double post, just learning about the edit function and the site as a whole.What would the vectorfield look like?something like: $ \frac {\eta}{\mathcal W} $ ? –  Marcel Aug 14 '12 at 6:32
    
@Marcel The map $(x,y)\mapsto (\eta^1W, \eta^2 W)$ defines a vector field in the plane. We have our canonical basis in the plane, so vector fields can be identified with pairs of functions, in the same way as vectors can be represented by pairs of numbers. –  user31373 Aug 14 '12 at 12:31
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