# How to solve $f\frac{\partial^2f}{\partial x\partial y} = \frac{\partial f}{\partial x}\frac{\partial f}{\partial y}$

I want to solve the following non-linear PDE:

$$f\frac{\partial^2f}{\partial x\partial y} = \frac{\partial f}{\partial x}\frac{\partial f}{\partial y}.$$

I don't know much about solving PDE's, especially non-linear ones, so I'm not sure how to find all the solutions. It is easily verified that a function of the form $f(x, y) = g(x)h(y)$, with $g$, $h$ differentiable, is a solution, but are all solutions of this form?

• Perhaps if you change coordinates to $z=x+\mathrm{i}y$ and $\bar{z}=x-\mathrm{i}y$, you get something along the lines of Laplace's equation. That should be a good first step... – Alex Nelson Aug 2 '13 at 4:26

## 1 Answer

I think the answer is yes. Note that

$$\frac{\partial}{\partial y} \left ( \frac{1}{f} \frac{\partial f}{\partial x}\right ) = \frac\frac{\partial f}{\partial y} \frac{\partial f}{\partial x} - f \frac{\partial^2 f}{\partial y \partial x}}{f^2} =$$

so that

$$\frac{1}{f} \frac{\partial f}{\partial x} = \frac{\partial \log{f}}{\partial x}= g(x)$$

or

$$\log{f} = G(x) + r(y)$$

where $G(x) = \int^x dx' g(x')$. This implies that $f(x,y) = a(x) b(y)$ by taking logs on both sides. A similar analysis leads to the same conclusion with $x$ and $y$ reversed in the first equation above.