The problem:
Consider the following boundary-value problem for the function $\rho : \mathbb{R}^{+} \to \mathbb{R}$ with boundary conditions $\lim_{x\to \infty}\rho(x) \to 1$ and $\lim_{x\to 0}\rho(x) \to \infty$:
$$\rho^{\prime \prime}-\frac{\rho^{\prime 2}}{\rho}+\frac{\rho^{\prime}}{x}+\frac{\kappa^2}{x^2 \rho}-\rho^2 f(\rho)=0 $$
where $\kappa \in \mathbb{R}$ is a constant and
$$ f(\rho) = \frac{1}{2}e^{-(\rho-1)}(\rho-1)(3-\rho) $$
Define the "energy" of the solution
$$ E = \int _0^\infty x \left(\frac{\rho'}{\rho}\right)^2 dx $$
By playing around numerically, I am led to the following conjecture: $E =O(\kappa)$ for small $\kappa$.
A very quick analysis:
I will not attempt to prove that a solution exists, nor do I ask for such an argument. In the following I assume that a solution exists.
The asymptotic form at small $x$ is $\rho \sim -\kappa \ln(\alpha x)$ for some constant $\alpha$. Assuming this approximate form for $x<x_I$, the energy is at least (roughly)
$$ \int_0 ^{x_I} \frac{1}{x(\ln(\alpha x))^2} = - \frac{1}{\ln(\alpha x_I)} \sim \frac{\kappa}{\rho} $$
- Numerically, it appears that $\rho$ is monotonically decreasing. Assuming this is true, then it follows $\rho>1$, such that the $x<x_I$ contribution to the energy is $O(\kappa)$.
- The large $x$ asymptotic form is $\rho \sim 1+ \frac{\kappa^2}{x^2} +O(\frac{1}{x^3})$. Assuming this is valid for $x>x_O$, the large $x$ contribution to the energy is to leading order
$$ \int_{x_O} ^{\infty} x(2\kappa^2/x^3)^2 dx = \left(\frac{\kappa^2}{x_O^2}\right)^2 $$
- Depending on when the large $x$ asymptotic form becomes valid, this contribution may be $O(1)$ (or larger) or $O(\kappa)$, though my numerics suggest this contribution is at most $O(\kappa^2)$, such that in total $E = O(\kappa)$.
In summary: is it true that $E=O(\kappa)$ for small $\kappa$? Can it at least be proven that $E=O(\kappa^n)$ for some $n>0$?
