# DiffEQ with various quadratic terms

I have the following DiffEQ and would like to solve it,

$$2F\left( \frac{{d}^{2}}{d{z}^{2}}F\right) -{\left( \frac{d}{dz}F\right) }^{2}+2F\left( \frac{{d}^{2}}{d{y}^{2}}F\right) -{\left( \frac{d}{dy}F\right) }^{2}+2F\left( \frac{{d}^{2}}{d{x}^{2}}F\right) -{\left( \frac{d}{dx}F\right) }^{2}=0$$

Any help would be appreciated, Thanks.

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You could have just said $F\,\Delta F=\frac{1}{2}\|\nabla F\;\|^2$. – anon Jan 26 '12 at 2:36

Note that constant $F$ works. Otherwise, assuming $F > 0,$ your condition is just $$\Delta \sqrt F = 0.$$ So you want some positive harmonic function $G$ and then $F = G^2.$ Similar for $F < 0, \; F = - H^2, \; \; \Delta \sqrt {-F} = 0.$
From Liouville's theorem, if you want a solution on all of $\mathbb R^3$ then $G$ and $F$ are constant. So, if you start with a nonconstant harmonic $G,$ such as linear, along a finite set of surfaces $G$ actually becomes $0$ and everything goes sideways.
EDIT, Thursday: Looking again, it is not so bad when the function is $0,$ as long as we are squaring, so the gradient is also $0.$ So, a solution, and possibly all solutions, are some real constant $c$ and harmonic $W,$ then $$F = c W^2.$$