Proving iff - Mean curvature and Weingarten Map

Question

The parametrized surface: $$\textbf{R}\supset{U}\ni(x,y)\longmapsto (x,y,f(x,y))\in\textbf{R}^3$$ defined by the graph of a smooth function $z=f(x,y)$ I know that the Weingarten map:$$L:=II\circ I^{-1}$$ equals the Guassian curvature. Now I'm interested in calculating that the trace of L (which I believe is the mean curvature of this surface) and using that to prove that the surface has mean curvature zero if and only if $f$ satisfies the minimal surface equation, namely $$H=\frac{(1+f_y^2)f_{xx}-2f_xf_yf_{xy}+(1+f_x^2)f_{yy}}{2(1+f_x^2+f_y^2)^{3/2}}$$ I have attached what I have accomplished so far- I am not sure how the proof part of this is supposed to work, although I believe this is because I may have screwed up the calculations.

Answer 1

I think there is something wrong with your computation. When computing the first fundamental forrm you should get \begin{equation} \begin{pmatrix} E & F \\ F & G \end{pmatrix} = \begin{pmatrix} 1+f_x ^2 & f_x f_y \\ f_x f_y & 1+f_y ^2 \end{pmatrix} \end{equation} The second fundamental form is: \begin{equation} \begin{pmatrix} L & M \\ M & N \end{pmatrix} = \frac{1}{\sqrt{fx ^2+f_y ^2-1}} \begin{pmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{pmatrix} \end{equation} And the shape opratator $dn_p$ (I didn't use L since L is the element in position 11 of the second fundamental form) is: \begin{equation} dn_p = \frac{1}{(1+fx ^2+f_y ^2)^(3/2)} \begin{pmatrix} f_{xx}(1+f_y ^2)-f_x f_y f_{xy} & f_{xy}(1+f_y ^2)-f_x f_y f_{yy} \\ f_{xy}(1+f_x ^2)-f_x f_y f_{xx} & f_{yy}(1+f_x ^2)-f_x f_y f_{xy} \end{pmatrix} \end{equation} Now the mean curvature is defined as: \begin{equation} H=-\frac{1}{2} Tr(dn_p) \end{equation} Then \begin{equation} H=-\frac{1}{2 (1+f_x ^2+f_y ^2)^(3/2)} ( f_{xx}(1+f_y ^2)-2 f_x f_y f_{xy}+f_{yy}(1+f_x ^2) ) \end{equation}