Solving "Square" PDE

Question

I'm trying to find all solutions to:

$$f_{xx}+2f_{xy}+f_{yy} = 0$$

I noticed that if we let $D = \frac{\partial}{\partial x} + \frac{\partial}{\partial y}$ then we are looking for the kernel of $D^2$. I'm not sure where to go from here though.

Answer 1

Make a change of variables:

$$u=x-y\\v=x+y$$

Then

$$\frac{\partial}{\partial x}=\frac{\partial}{\partial u}+\frac{\partial}{\partial v}\\ \frac{\partial}{\partial y}=-\frac{\partial}{\partial u}+\frac{\partial}{\partial v} $$

and the equation becomes much simpler to solve

$$4\frac{\partial^2 f}{\partial v^2}=0\iff f(u,v)=vG(u)+H(u)$$

where $G,H$ are arbitrary functions. Back-substituting yields the most general solution for the PDE:

$$f(x,y)=(x+y)G(x-y)+H(x-y)$$