
I assume that you are asking for a proof of the WeierstrassEnneper representation theorem that, roughly speaking, tells you how to express solutions of the minimal surface equation in terms of holomorphic functions (of one variable). In outline, the classical proof is the following one: Assume that a solution $\phi:D\to\mathbb{R}$ to the above equation is specified on a simplyconnected domain $D\subset\mathbb{R}^2$.
Set $U = U(x,t) = \bigl(x,t,\phi(x,t)\bigr)$, note that
$$
dU = (1,0,\phi_x)\ dx + (0,1,\phi_t)\ dt
$$
and that the unit vector $N= (\phi_x, \phi_t, 1)/(1+{\phi_x}^2+{\phi_t}^2)^{1/2}$ satisfies $N\cdot dU = 0$.
Set $\nu = N\times dU$, i.e.,
$$
\nu = \frac{(\phi_x\phi_t,\ {\phi_x}^2{}1,\ \phi_t)\ dx
+ ({\phi_t}^2{+}1,\ \phi_x\phi_t,\ \phi_x)\ dt }
{(1+{\phi_x}^2+{\phi_t}^2)^{1/2}},
$$
and note that the above equation on $\phi$ is equivalent to the condition that $d\nu = 0$. Since $D$ is simply connected, it follows that there exists a function $V:D\to\mathbb{R}^3$ such that $dV = \nu$. ($V$ is unique up to an additive constant.) Obviously, $N\cdot dV = N\cdot \nu = 0$.
Set $W = U + i\ V:D\to \mathbb{C}^3$. A little vector algebra now shows that the image of the differential of $W$ at every point is a complex line in $\mathbb{C}^3$, i.e., that $W(D)\subset\mathbb{C}^3$ is, in fact, a(n embedded) holomorphic curve. Moreover, the tangent lines to this holomorphic curve are null with respect to the (complex linear) inner product on $\mathbb{C}^3$, i.e., if $\bigl(w_1(\zeta),w_2(\zeta),w_3(\zeta)\bigr)$ is a (local) holomorphic parametrization of $W(D)$, then $w_1'(\zeta)^2+w_2'(\zeta)^2+w_3'(\zeta)^2=0$.
Conversely, if one starts with a holomorphic null curve $W(\zeta) = \bigl(w_1(\zeta),w_2(\zeta),w_3(\zeta)\bigr)$ in $\mathbb{C}^3$, such that its projection to $\mathbb{R}^3$ can be written as a graph $\bigl(x,t,\phi(x,t)\bigr)$, then $\phi$ must satisfy the minimal surface equation.
There are various refinements if one doesn't ask that the surface be representable as a graph, and that leads to the WeierstrassEnneper representation theorem in its full glory.

answered Mar 30 '12 at 15:35

