I'm studying Sherk surface which is the unique minimal surface with parametrization given by $\phi(x,y)=(x,y,f(x)+g(y))$. Using the mean curvature formula, is easy to show that this surface is minimal if and only if f,g satisfies: \begin{equation} \boxed{(1+f_x^2)g_{yy}+(1+g_y^2)f_{xx}=0} \end{equation} and I know (looking in bibliography) that a solution is \begin{equation} f(x)=\frac{1}{a}\log\cos(ax) \quad \quad g(y)=-\frac{1}{a}\log\cos(ay) \quad \quad-\pi/2<x,y<\pi/2 \end{equation} And this solution must be unique according to the bibliography. But I don't know how I can warrant that it happens. I don't remember too much about differential equations. There is some result which proof that, in this case, exists only this solution?
Or... there is some easy way to achieve this explicit solution?
Even more, it's is possible to show that the solutions must be periodic ?
Thanks in advance,
Carles.
Rewriting the equation as $$\frac{g_{yy}}{1+g_y^2} = \frac{f_{xx}}{1+f_x^2}$$ we see that the left side is independent of $x$, while the right side is independent of $y$. So, both sides are constant. Call this constant $a$. We are led to solve the ODE $$u'' = a(1+(u')^2)$$ Let $v=u'$ here. Then $v' = a(1+v^2)$. Separate the variables (let $t$ be the independent variable here) and integrate: $$\int \frac{dv}{1+v^2} = \int a\,dt$$ Hence $\arctan v = at+b$, $v = \tan(at+b)$, and $u = \frac{1}{a}\log \cos(at+b)$.
I don't think there is a way to show the solution is periodic without actually finding the solution first.