# Reference request: Minimal surface equation proof

The equation I'm refering to was posed by Lagrange and states that if $$S\subset \mathbb{R^3}$$ is a surface with zero mean curvature, ie $$H=0, \: \: \forall p \in S$$ and S is given as the graph of a function $$z(x,y)$$ then $$(1+z_x^2)z_{yy}-2z_xz_yz_{xy}+(1+z_y^2)z_{xx}=0$$ I'm looking for a proof of this. I thought I would find it in Do Carmo's Differential Geometry of Curves and Surfaces but it seems to not be on there.

I'm going to change $$x$$ and $$y$$ to $$u$$ and $$v$$, respectively, so that I can call my patch $$\textbf{x}$$ without confusion.
Parameterize the surface via the patch $$\textbf{x}$$ given by $$(u,v)\mapsto (u,v,z(u,v)).$$ We compute $$\textbf{x}_u=(1,0,z_u),$$ $$\textbf{x}_v=(0,1,z_v),$$ and obtain the components of the first fundamental form as \begin{align*}E&=\textbf{x}_u\cdot\textbf{x}_u=1+z_u^2\\F&=\textbf{x}_u\cdot \textbf{x}_v=z_uz_v\\ G&=\textbf{x}_v\cdot \textbf{x}_v=1+z_v^2.\end{align*} The unit normal is given by $$U=\frac{\textbf{x}_u\times \textbf{x}_v}{\sqrt{EG-F^2}}=\frac{(-z_{u},-z_{v},1)}{\sqrt{1+z_u^2+z_v^2}}.$$ Next, we find that $$\textbf{x}_{uu}=(0,0,z_{uu}),$$ $$\textbf{x}_{uv}=(0,0,z_{uv}),$$ and $$\textbf{x}_{vv}=(0,0,z_{vv}).$$ We can now compute the quantities \begin{align*} L&=\textbf{x}_{uu}\cdot U=\frac{z_{uu}}{\sqrt{1+z_u^2+z_v^2}}\\ M&=\textbf{x}_{uv}\cdot U=\frac{z_{uv}}{\sqrt{1+z_u^2+z_v^2}}\\ N&=\textbf{x}_{vv}\cdot U=\frac{z_{vv}}{\sqrt{1+z_u^2+z_v^2}}. \end{align*} Finally, we can just use the formula $$H=\frac{GL+EN-2FM}{2(EG-F^2)}=\frac{(1+z_v^2)z_{uu}+(1+z_u^2)z_{vv}-2(z_uz_v)z_{uv}}{2(1+z_u^2+z_v^2)^{3/2}},$$ which equal zero if and only if $$(1+z_v^2)z_{uu}+(1+z_u^2)z_{vv}-2(z_uz_v)z_{uv}=0,$$ as desired.
• I'm sorry, I was checking it more carefully and there seems to be a minor mistake as $U$ should be given by: $U = (-z_u, -z_v, 1)/\sqrt{1+z_u^2+z_v^2}$. (the numerator was wrong) – D. Brito Jul 26 at 16:02