# Why this equality must holds for minimal surfaces?

When minimizing a surface area with respect to a fixed volume $V$, I found in some notes that the parametrization $X: U \longrightarrow \mathbb{R}^3$ must satisfy the equality $\iint_U (2H - \lambda) \langle \varphi, N \rangle du dv = 0$ for all variations $\varphi$, where $N$ is the normal, $\lambda$ is some constant and $H$ is the mean curvature. I know that it's related to some constraint, so there's the Lagrange multiplier $\lambda$, however I didn't understand this well.

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I like to think of this in physical terms as a bubble enclosing a fixed amount of air. The forces on any patch of the bubble are $-2H\mathbf N$ due to the surface tension of the bubble, and $\lambda\mathbf N$ due to the pressure of the air inside. Then your equation merely states that the bubble is in equilibrium. Of course, this is not a rigorous mathematical explanation. –  Rahul Feb 3 '13 at 21:16
Yes, the physical picture is so nicely fits in with the mathematical counter parts. –  Narasimham Aug 14 '14 at 18:18

This is correct: the constraint is that the surface must enclose volume $V$. The test function $\varphi$ describes a variation of the surface: we compare the original parametrization $X:U\to\mathbb R^3$ again the perturbed $X+\epsilon\varphi$. We are interested in the first derivative with respect to $\epsilon$, evaluated at $\epsilon=0$. For the surface area this derivative turns out to be $\iint_U 2H\langle \varphi, N\rangle$, and for the volume it is simply $\iint_U \langle \varphi, N\rangle$. At a critical point the gradient of the objective (surface area) is a multiple of the gradient of the constraint (volume). Hence, $\iint_U (2H-\lambda)\langle \varphi, N\rangle =0$.
It is not surprising that only the normal component of $\varphi$ matters. It should be geometrically intuitive that variation in tangential directions has $o(\epsilon)$ effect on both surface area and the enclosed volume. For this reason, it is convenient to consider only normal variations: that is, $\varphi=\psi N$ where $\psi$ is a scalar function.