I am reading a proof on Bernstein's Theorem on Minimal Surfaces. On this proof it is claimed that if $u: \mathbb{R^2} \rightarrow \mathbb{R}$ sastisfies the minimal surface equation, ie: $$ (1+u_x^2)u_{yy} -2u_xu_yu_{xy} + (1+u_y^2)u_{xx} = 0 $$ Then both the functions $\psi_1 = \arctan(u_x)$ and $\psi_2 = \arctan(u_y)$ are solutions of the equation $$(1+u_x^2)v_{yy} -2u_xu_yv_{xy} + (1+u_y^2)v_{xx} = 0$$

Since the above equation is in someway "symmetric" (changing $x$'s into $y$'s and vice versa does not alter the equation) then I believe it is easily seen that if $\psi_1$ satisfies the equation then $\psi_2$ must also do so.

What I've done so far:

We have the following identities: $$ (\psi_1)_{xx} = \frac{u_{xxx}(1+u_x^2) - 2u_xu_{xx}^2}{(1+u_x^2)^2} $$ $$ (\psi_1)_{xy} = \frac{u_{xxy}(1+u_x^2) - 2u_xu_{xy}u_{xx}}{(1+u_x^2)^2} $$ $$ (\psi_1)_{yy} = \frac{u_{xyy}(1+u_x^2) - 2u_xu_{xy}^2}{(1+u_x^2)^2} $$ Which when we substitute correspondingly on the $v$'s in the equation and after using the hypothesis of the minimal surface equation we are left with $$ \frac{u_{xxx}(1+u_y^2) - 2u_xu_yu_{xxy}+u_{xyy}(1+u_x^2)+2u_{x}(u_{xx}u_{yy}-u_{xy}^2)}{(1+u_x^2)} $$ Which I cannot then, prove is equal to $0$.

  • $\begingroup$ You wrote :$$(1+u_x^2)u_{yy} -2u_xu_yu_{xy} + (1+u_y^2)u_{yy} = 0$$ Why didn't you write $$(2+u_x^2+u_y^2)u_{yy} -2u_xu_yu_{xy}=0 $$ ? $\endgroup$ – JJacquelin Oct 30 '19 at 8:01
  • $\begingroup$ This was a mistake. The second $u_{yy} $ should instead have been $u_{xx} $. I have correcred the equation in the question. $\endgroup$ – D. Brito Oct 30 '19 at 22:09

Take a partial derivative of the original equation with respect to $x$ $$ 2u_{x}u_{xx}u_{yy} + (1+u_x^2)u_{xyy} - 2u_{xx}u_{y}u_{xy} -2_{x}u_{xy}^2 - 2u_xu_yu_{xxy} + 2u_{y}u_{xy}u_{xx} + u_{xxx}(1+u_y^2)=0, $$ Upon some cancellation, the LHS is exactly the numerator of the last expression you obtained.


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