Problem with simple laplacian equation I would like to solve the following PDE:
$$ 
\partial_x^2 u + \partial_y^2 u = -\frac{2 x^2 (x^2-y^2)}{\left(x^2+y^2\right)^2}
$$
The right side comes from $ x^2 \partial_x^2 \log(x^2 +y^2) $. Switching the polar coordinates, the right side is deceptively simple:
$$
-2 \cos(\theta)^2 \cos(2 \theta) 
$$
In polar coordinates, the laplacian is:
$$
\partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2
$$
so it seems as though it should be fairly simple to find a solution. If I assume $ u(r, \theta) $ is like $ r^2 F(\theta) $, and try to solve for theta, I get something that is not periodic in $ \theta $. I am not sure if there is nonetheless a way to extract a meaningful solution. I'm at a bit of a loss for any other approaches. Any suggestions?
Edit: For $ F(\theta) $, I used mathematica to get:
$$
F(\theta) = \frac{1}{24} (-3 - 3 \cos(2 \theta) + \cos(4 \theta) - 6 \theta \sin(2 \theta))
$$
Plus of course any homogenous solution. It is the $ \theta \sin (2 \theta) $ term that makes me so sad.
 A: Let's not forget that for many purposes, logarithm is a monomial of degree $0$. That is, we should consider $r^2\log r$ alongside with $r^2$. Letting
$$u(r)=r^2F(\theta)+r^2\log r\,G(\theta) \tag1$$
we arrive at the system 
$$F''+4F+4G=-2\cos^2 \theta \, \cos 2\theta,\qquad G''+4G=0 \tag2$$
Solving these, we get a bunch of constants. One of them allows us to get rid of $\theta\sin 2\theta$; specifically, $$F(\theta)=\frac{1}{24} \cos4\theta, \qquad 
 G(\theta)=-\frac14\cos 2\theta \tag3$$ if I got the computations right. 

Remark: this example shows that solution of $\Delta u=f$ with bounded $f$ does not have locally bounded second-order derivatives in general. (E.g., $u_{rr}$ is unbounded here.) This is in contrast to the fact that for $1<p<\infty$ having $\Delta u\in L^p$ implies $u\in W_{\rm loc}^{2,p}$. Put another way, we witness the failure of the Riesz transforms to preserve $L^\infty$ the way they preserve $L^p$ for $1<p<\infty$. 
But if $\Delta u$ was Hölder- or Dini-continuous, we would have $u\in C^2$.
A: Let $\begin{cases}p=x+iy\\q=x-iy\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$
$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial y}=i\dfrac{\partial u}{\partial p}-i\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(i\dfrac{\partial u}{\partial p}-i\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(i\dfrac{\partial u}{\partial p}-i\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial y}+\dfrac{\partial}{\partial q}\left(i\dfrac{\partial u}{\partial p}-i\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial y}=-\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}-\dfrac{\partial^2u}{\partial q^2}=-\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}-\dfrac{\partial^2u}{\partial q^2}$
$\therefore\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}-\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}-\dfrac{\partial^2u}{\partial q^2}=-\dfrac{2\left(\dfrac{p+q}{2}\right)^2\left(\left(\dfrac{p+q}{2}\right)^2-\left(\dfrac{p-q}{2i}\right)^2\right)}{(pq)^2}$
$4\dfrac{\partial^2u}{\partial pq}=-\dfrac{2\times\dfrac{(p+q)^2}{4}\left(\dfrac{(p+q)^2}{4}+\dfrac{(p-q)^2}{4}\right)}{p^2q^2}$
$\dfrac{\partial^2u}{\partial pq}=-\dfrac{(p+q)^2((p+q)^2+(p-q)^2)}{32p^2q^2}$
$\dfrac{\partial^2u}{\partial pq}=-\dfrac{2(p+q)^2(p^2+q^2)}{32p^2q^2}$
$\dfrac{\partial^2u}{\partial pq}=-\dfrac{(p^2+2pq+q^2)(p^2+q^2)}{16p^2q^2}$
$\dfrac{\partial^2u}{\partial pq}=-\dfrac{p^4+2p^3q+2p^2q^2+2pq^3+q^4}{16p^2q^2}$
$\dfrac{\partial^2u}{\partial pq}=-\dfrac{p^2}{16q^2}-\dfrac{p}{8q}-\dfrac{1}{8}-\dfrac{q}{8p}-\dfrac{q^2}{16p^2}$
$u(p,q)=f(p)+g(q)+\dfrac{p^3}{48q}-\dfrac{p^2\ln q}{16}-\dfrac{pq}{8}-\dfrac{q^2\ln p}{16}+\dfrac{q^3}{48p}$
$u(x,y)=f(x+iy)+g(x-iy)+\dfrac{(x+iy)^3}{48(x-iy)}-\dfrac{(x+iy)^2\ln(x-iy)}{16}-\dfrac{x^2}{8}-\dfrac{y^2}{8}-\dfrac{(x-iy)^2\ln(x+iy)}{16}+\dfrac{(x-iy)^3}{48(x+iy)}$
