# Solutions for PDE $(1+u_x^2)v_{yy} -2u_xu_yv_{xy} + (1+u_y^2)v_{yy}$

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 $$0$$.

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