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How do i show that $(1+u^2_x)u_{yy} -2u_xu_yu_{xy}+ (1+u_y)u_{xx}=0$ I have been thinking about it for a while i can't seem to have a starting point of how to go about this PDE problem.

\textbf{Proof:}\ Let $X(u,v)=(u,v,f(u,v))$ be the parametrization of the Surface M. Let's find the partial derivatives in terms of $u$ and $v$ $ X_u= (1,0,f_u)$ and $ X_v= (0,1,f_v)$ Let's find the second partial derivatives in terms of $u$ and $v$ $ X_{uu}= (0,0,f_{uu})$, $ X_{vv}= (0,0,f_{vv})$, $ X_{vu}= (0,0,f_{vu})$, and $ X_{uv}= (0,0,f_{uv})$ We will now find the Normal Vector, we will denote it as $\bf{N}$ $\bf{N}=$ $(-\frac{f_u}{ \sqrt{f^2_u+f^2_v+1}},\frac{f_v}{ \sqrt{f^2_u+f^2_v+1}},\frac{1}{ \sqrt{f^2_u+f^2_v+1}})$ $E =X_u \cdot X_u = 1+ f^2_u, F = X_u \cdot X_v = f_uf_v,$ and $ G =X_v \cdot X_v = 1+f^2_v$ $L =X_{uu} \cdot$ $\bf{N}=$ $ \frac{f_{uu}}{ \sqrt{f^2_u+f^2_v+1}}$, $M =X_{uv} \cdot$ $\bf{N}=$ $ \frac{f_{uv}}{ \sqrt{f^2_u+f^2_v+1}},$ and $N =X_{vv} \cdot$ $\bf{N}=$ $ \frac{f_{vv}}{ \sqrt{f^2_u+f^2_v+1}}$ Since $LG-2MF+NE=0$ We now obtain $\frac{(1+f^2_v)f_{uu}}{ \sqrt{f^2_u+f^2_v+1}} -2\frac{(f_uf_vF{uv})f_{uu}}{ \sqrt{f^2_u+f^2_v+1}}+\frac{(1+f^2_u)f_{vv}}{ \sqrt{f^2_u+f^2_v+1}}=0 \implies f_{uu}(1+f^2_v)-2f_uf_vf_{uv}+f_{vv}(1+f^2_{vv})= 0$ Which we desired.

Which is the same as $(1+u^2_x)u_{yy} -2u_xu_yu_{xy}+ (1+u_y)u_{xx}=0$

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    One needs a formula for $u$ (in terms of $x,y$) to do this at all. Once one has that, find all the partials and plug in. – 2017-01-05
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    Yeah you are correct i just solve it i forget that i was working with a parametrization – 2017-01-05
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    Just for completeness maybe put the formula in the question... – 2017-01-05

1 Answers 1

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If you want to derive the minimal surface equation, it is the Euler-Lagrange equation for the surface area functional

$$I(u) = \int_\Omega \sqrt{1+u_x^2+u_y^2} \, dxdy.$$

That is, minimal surfaces minimize (at least locally) surface area. The Euler-Lagrange equation is

$$\text{div}\left( \frac{\nabla u}{\sqrt{1+u_x^2 + u_y^2}}\right) = 0.$$

If you expand the divergence you will get the minimal surface equation you quoted.