I have been reading Dierkes-Hildebrandt's, "Minimal Surfaces" recently and I came across this proof for showing that a minimal surface has zero mean curvature at all points. I had a doubt regarding this proof.
Given a regular $C^2$ surface $X:\Omega\to \mathbb{R}^3$, we define a variation for $X$ as:
$$Z:\Omega\times(-\varepsilon, \varepsilon)\to \mathbb{R^3}\ \text{with}\ Z(\omega,0)=X(\omega)$$
Now we consider the taylor expansion of $Z$ around $(\omega,0)$ to get:
$$Z(\omega, \varepsilon)=X(\omega)+\varepsilon Y(\omega)+\varepsilon^2R(\omega,\varepsilon)$$
where $Y(\omega)=\frac{\partial}{\partial \varepsilon}Z(\omega, \varepsilon)|_{\varepsilon=0}\ $ is the first variation of family of surfaces $Z(\cdot, \varepsilon)$
Now we write $Y(\omega)$ as:
$$Y(\omega)=\eta^\beta(\omega)X_{u^{\beta}}(\omega)+\lambda(\omega)N(\omega)$$
where $X_{u^\beta}=\frac{\partial}{\partial u^\beta}X$.
Now we calculate $\sqrt{|Z_u|^2|Z_v|^2-\langle Z_u,Z_v\rangle^2}$, and by utilizing Gauss, Codazzi, and Weingarten equations, we get:
$$\sqrt{|Z_u|^2|Z_v|^2-\langle Z_u,Z_v\rangle^2}=\mathcal{W}+\varepsilon[(\eta^1\mathcal{W})_u+(\eta^2\mathcal{W})_v-2H\mathcal{W}\lambda]+\cdots$$
where $\mathcal{W}=\sqrt{\mathcal{EG}-\mathcal{F^2}}$ and $H$ is the mean curvature.
Next we define the first variation of the area functional $A_\Omega$ on $\Omega$ at $X$ in the direction of $Y$ as:
$$\delta A_\Omega(X,Y)=\frac{d}{d\varepsilon}A_\Omega(Z(\cdot,\varepsilon))|_{\varepsilon=0}$$
Thus:
$$\delta A_\Omega(X,Y)=\int_\Omega [(\eta^1\mathcal{W})_u+(\eta^2\mathcal{W})_v-2H\mathcal{W}\lambda]\ du \ dv \\ = \int_{\partial{\Omega}}\mathcal{W}(\eta^1du-\eta^2dv)-\int_{\Omega}2H\mathcal{W}\lambda\ du \ dv \\ =-\int_{\Omega}2H\mathcal{W}\lambda\ du \ dv \\ =-\int_{X}2\langle Y,N\rangle H\mathcal{W}\ du \ dv$$.
Since $\lambda$ can be chosen as an arbitrary $C^1$ function, the first variation vanishes for all vector fields $Y$ iff the mean curvature $H$ is identically zero.
Now I had a few doubts. Firstly, how do we assume that $\lambda$ can be an arbitrary $C^1$ function?
Secondly, why does $\int_{\partial{\Omega}}\mathcal{W}(\eta^1du-\eta^2dv)$ vanish?
I apologize if I am missing something trivial. Please help
THANKS
So basically you are choosing the variational vector field $$ Y(\omega)=\frac{\partial}{\partial \epsilon}|_{\epsilon=0} Z(\omega, \epsilon) $$ suitably by requiring that when we write $$ Y(\omega)=\eta^\beta(\omega)X_{u^\beta}(\omega) + \lambda(\omega)N(\omega), $$ $\lambda$ can be an arbitrary $C^1$ function, and $\eta^1$ and $\eta^2$ vanish on the boundary $\partial \Omega$. That is why the boundary term $$\int_{\partial \Omega} {\mathcal W}(\eta^1du-\eta^2dv)=0,$$ and $\lambda$ can be arbitrary $C^1$.
So it all boils down to given a variational vector field $Y(\omega)$, do we have a family of surface $Z(\omega,\epsilon)$ such that $Y=\frac{\partial}{\partial \epsilon}|_{\epsilon=0}Z$. I think this follows from the ODE theory. Given $Y(\omega)$, you can extend it to $Y(\omega,\epsilon)$ for $\epsilon$ in a small neighborhood of $0$ such that $Y(\omega,0)=Y(\omega)$, and then solve the ODE system $$ \begin{cases} \frac{\partial}{\partial \epsilon}Z(\omega, \epsilon)=Y(\omega, \epsilon),\\ Z(\omega, \epsilon)=X(\omega). \end{cases} $$ This guarantees that an arbitrary variational field $Y(\omega)$ comes from a family $Z(\omega, \epsilon)$ of surfaces.