Differential operator that must be a constant multiple of the Laplacian Can you help me show that if $L$ is a partial differential operator given by 
$$L=a{{\partial^2}\over{\partial{x^2}}}+b{{\partial^2}\over{\partial{y^2}}}+c{{\partial^2}\over{\partial{x}\partial{y}}}$$ $a$,$b$,$c$ real numbers,
which commutes with all rotations $p_{\theta}(z)=ze^{i\theta}$, meaning, whenever $f$ is at least $C^2$ on $\Bbb C$, $(Lf)\circ p_{\theta} = L(f \circ p_{\theta})$ for any $\theta$, then $L$ must be a constant multiple of the Laplacian?
I tried directly calculating each side of the commutativity condition, which got me somewhere, but I can't seem to close it out. Essentially, I need to show that $c$=0 and $b=a$. What is the best way to do this? Picking certain $f$ and certain $\theta$ and deriving a contradiction perhaps?
There is a note in the problem telling me that I can also write $p_{\theta}(x,y)=(x\cos(\theta)-y\sin(\theta),x\sin(\theta)+y\cos(\theta))$, which is the form I have been trying to work with.
 A: We calculate derivatives of the composition $f\circ p_\theta$ using the chain-rule to find (below $C=\cos(\theta)$ and $S=\sin\theta$)
$$\frac{\partial}{\partial x} [f\circ p_\theta] = [Cf_x + Sf_y]\circ p_\theta$$
$$\frac{\partial}{\partial y} [f\circ p_\theta] = [Sf_x - Cf_y]\circ p_\theta$$
and
$$\frac{\partial^2}{\partial x\partial y} [f\circ p_\theta] = [-SCf_{xx} + (C^2-S^2)f_{xy} + SCf_{yy}]\circ p_\theta$$
$$\frac{\partial^2}{\partial y^2}[f\circ p_\theta] = [S^2f_{xx} - 2CSf_{xy} + C^2f_{yy}]\circ p_\theta$$
$$\frac{\partial^2}{\partial x^2} [f\circ p_\theta] = [C^2f_{xx} + 2CSf_{xy} + S^2f_{yy}]\circ p_\theta$$
Putting this togeather gives us
$$L[f\circ p_\theta] = \left[(aC^2+bS^2- cSC)f_{xx} + (aS^2+bC^2 + cSC)f_{yy} \\+ (2CS(a-b) + c(C^2-S^2))f_{xy}\right]\circ p_\theta$$
We also have
$$L[f]\circ p_\theta = \left[(a)f_{xx} + (b)f_{yy} + (c)f_{xy}\right]\circ p_\theta$$
In order to have $L[f\circ p_\theta] = L[f]\circ p_\theta$ for all $\theta,x,y$ we either need $f_{xx} = f_{yy} = f_{xy} = 0$ (for which $f = Ax + By$) or
$$a = aC^2+bS^2 - cSC$$
$$b = aS^2+bC^2 + cSC$$
$$c = 2CS(a-b) + c(C^2-S^2)$$
Take $\theta = \pi/2$ in the last equation to get $c=0$ and $a=b$ which implies that $L$ is a constant multiplicum of the Laplacian.
