# how to solve a pde whose coefficient is the function itself

I am studying differential geometry, Walker metric in three dimension. I try to find the geodesic equations of a Walker manifold and I need to solve the following PDE. Unfortunately, I didn't take any PDE course and I don't know what kind of PDE it is and how to solve it.

$f\frac{\partial f}{\partial x}-\frac{\partial f}{\partial z}=0,\quad \frac{\partial f}{\partial y}=0$

where $f(x,y,z)$ is a continuous function.

Any suggestion would be greatly appreciated.

Also, can you advise me an elementary PDE book easy to follow and self-study.

• The geodesic equations are second order and quadratic in first derivatives. Are you sure you have the right system? – user26977 Jan 25 '16 at 19:10
• the first term cries out to be written as $\tfrac12 \frac{\partial}{\partial x} (f^2)$. – user66081 Jan 25 '16 at 19:12
• f is not the geodesic curve, it is the function of the Walker metric. I might be wrong but I got this PDE a few times. There is a system of PDEs and I need to solve this PDE as a part of the solution. – francesca Jan 25 '16 at 19:25

Take $f(x,y,z)= -\frac{x}{z}$. Then: $\frac{\partial f}{\partial x} = -\frac{1}{z}$, and $\frac{\partial f}{\partial z}= \frac{x}{z^2}$. Then: $f \frac{\partial f}{\partial x} = (-\frac{x}{z})(-\frac{1}{z})= \frac{x}{z^2}= \frac{\partial f}{\partial z}$ and $\frac{\partial f}{\partial y}=0$, as desired.