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.
Best Answer
Some simple observations at least rule out certain types of "nice" solutions. First of all, if $f=1$ anywhere, then $\Delta f$ must be singular. Moreover, if either $f>1$ or $f<1$, then $\Delta f$ has a sign opposite to $f-1$. So, if $f>1$, there cannot be a minimum of $f$, and if $f<1$, there cannot be a maximum. What qualitative properties do you require or expect?