I am working on a problem, and I ran into the following equation
$$\nabla^2f =f_{xx} + f_{yy} + f_{zz} = \frac{12 xy(x^2-y^2)}{(x^2 +y^2)^2}$$
I want to know if there exists a scalar field $f:\mathbb{R}^3 \to \mathbb{R}$ such that the relation above holds, and if so, if there's a way that I can find $f$ explicitly. Intuitively I know that if $f$ exists it isn't unique, since adding a constant to it would also result in a correct solution, but for this case, I'm just interested in finding one example that works.
I've basically only tried guessing and checking. I noticed that I could write the expression as $\frac{12 x^5 y}{(x^2 + y^2)^3} - \frac{12 x y^5}{(x^2 + y^2)^3}$, and I tried integrating each of the fractions separately with respect to $y$ and $x$ respectively to see if I got something similar, but that rapidly turned out to be impractical.
I'm not sure if there's a method to inverse a Laplace operator to find and explicit function. Could anyone tell me if there's a way to solve this? Thank you!
In cylindrical coordinates the equation simplifies down to
$$f_{rr}+\frac{1}{r}f_r + \frac{1}{r^2}f_{\theta\theta}+f_{zz} = 3\sin4\theta$$
We can pretend $f$ is independent of $z$ to simplify the equation further. So guess that $f$ is of the form
$$f(r,\theta, z) = kr^2\sin4\theta$$
This gives us the equation
$$\implies 2k+2k-16k = -12k = 3$$
Thus we have the solution
$$f(r,\theta,z) = -\frac{1}{4}r^2\sin4\theta = -r^2\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)$$
which makes it easier to convert to Cartesian
$$f(x,y,z) = \frac{xy(y^2-x^2)}{x^2+y^2}$$