Consider the following non-linear generalization of the Laplace equation $$\Delta \phi - \frac{\sum_i (\partial_i \phi)^2}{2 \phi} = 0$$ I am looking for spherically symmetric solutions, so I assume that $\phi(x^i) = \phi(r)$ and express in spherical coordinates $$\phi'' + \frac{2}{r} \phi' - \frac{(\phi')^2}{2 \phi} = 0$$ This yields the general spherically symmetric solution $$\phi(r) = \frac{A (1 - B r)^2}{r^2}$$ where $A,B$ are some constants.
Now, apart from the constant solution ($B\to \infty$ while $AB$ kept finite), there are no non-singular solutions, they all have a singularity at $r=0$. This is quite in analogy to the solution of the usual Laplace equation, where we have the solution $\phi_{Laplace} = C/r + D $ and the singular term corresponds to the fact that there is a distributional "source" for the field $\phi_{Laplace}$, $$\phi''_{Laplace} + \frac{2}{r} \phi'_{Laplace} = - 4 \pi C \delta(r)$$ In the case of the Laplace equation you prove this by integrating the Laplace equation over a ball around $r=0$, using Gauss' theorem on the left-hand side, which makes it the flux of $\vec{\nabla} \phi_{Laplace}$ through a sphere around $r=0$. Since this flux is non-zero and only a delta function on the right hand side can do that, we are finished with our proof and we also obtain the correct coefficient in front of the delta.
However, in my case I cannot use any method like that because of the non-linear term. Similarly, it does not help to use the Fourier transform because the non-linear term is not tractable. Do you have any idea on how to understand the "source" of this general solution to the non-linear Laplace equation?
Thankfully in this case we can find a simple function of $\phi$ that satisfies a linear differential equation (we can spot this from the relationship between the general solutions, for example): provided that $\phi \neq 0$, we can rewrite your equation as $$ \phi^{1/2}\Delta(\phi^{1/2}) = 0. $$