I wish to solve the following nonlinear PDE that I derived in statistical physics. (I was curious if I could include higher order terms into a model for heat transfer described in a homework problem.) Find a function $f:\mathbb R^3 \to \mathbb R$ parametrized in spherical coordinates s.t. $$(f - 1) \Delta f + f^2 = 0$$ where I have suppressed the argument $(r, \theta, \phi)$ to $f$ for brevity.
If I try a spherical harmonic, I have that $\Delta f = 0$, so that $f^2 = 0$ and $f$ vanishes identically. Likewise, if $f$ solves the Helmholtz equation, we have $\Delta f = -k^2 f$ and $-k^2 f(f-1) + f^2 = 0$ which also fails to yield a useful solution.
Is there an ansatz that I may choose to reduce the problem into a linear equation. I remember from quantum physics that a certain nonlinear PDE arising from the time independent Schrodinger equation in a Harmonic potential could be made linear by multiplying an unknown function by the asymptotic behavior. In that case, $f(u) = g(u) \exp\left(-\frac{u^2}{2}\right)$ was the required solution. Here, however, I am having difficulty getting an idea of the asymptotic behavior. Could I assume invariance with respect to $\theta$ and $\phi$ and try to find asymptotic behavior for $r$? Can this equation be transformed into a well known nonlinear PDE? (I had actually noticed that in the 1D rectangular case, the equation bares some resemblance to a Sturm-Liouville problem.)
An Instructive Mistake: I have been trying to investigate if the original equation satisfies the Painleve criterion via Mathematica, but have been unsuccessful in getting a definitive result. I am willing to broaden my search for a solution $f:\mathbb R^3 \to \mathbb C$. I recall a certain paper that I read while studying quantum physics, which made a rather bold claim (given that no stipulations were put on $G$) about solving certain classes of PDEs.
Let $G : \mathbb R_+ \to \mathbb R$ be any real valued function and consider the generalized Schrodinger equation $$i\hbar \frac{\partial \psi}{\partial t} + G(-\hbar^2\Delta)\psi = 0$$ will have a solution $$\psi(x,t) = e^{-itG(-\hbar^2\Delta)} \psi_0(x)$$ because $$\mathcal F(G(-\hbar^2 \Delta)\psi)(p) = G(|p|^2) \hat \psi(p)$$ acts as a Fourier multiplier.
With this in mind, I could manipulate my initial equation as
$$ \begin{align} (f-1)\Delta f + f^2 &= 0 \\ f\Delta f + f^2 &= \Delta f \\ \frac{\Delta f}{\Delta f + f} &= f \\ \underbrace{\left(\frac{\Delta}{\Delta + 1}\right)}_{G(\Delta)} f &= f \end{align} $$
and define $F(r,\theta,\phi,t) = \upsilon(t)f(r,\theta,\phi)$ after neglecting constants by
$$ \begin{align} F(r,\theta,\phi,t) &= \exp\left(-it\left(\frac{\Delta}{\Delta+1}\right)\right)f_0(r,\theta,\phi) \\ & = \exp\left(-itG(\Delta)\right)f_0(r,\theta,\phi) \end{align} $$
Could I treat the above like a flow to prove that constant phase shifts of the equation remain valid if we allow it to take complex values? I did not think that this result would be trivial because the powers of $f$ are not the same throughout the equation.
Correction: As @Zachary mentioned in the comments, this method fails because $G$ is a nonlinear function.
Update: I numerically integrated this equation under a variety of initial conditions and observed horrible ill-conditioning and poor stability. I have decided to accept the answer that no desirable solution to the equation exists.
