= A solvable infrared renormalization flow
{title2=$dg/d\ell=\epsilon g-g^2,\quad\ell=\log b$}
Take the following bounded analytic example for a dimensionless positive coupling $g$, with fixed $\epsilon>0$ and $g(0)=g_0>0$:
$$
\frac{dg}{d\ell}=\epsilon g-g^2,\qquad
\ell=\log b=\log(L/a).
$$
Increasing $\ell$ means increasing observation length $L$, hence motion toward the <infrared fixed point>. This convention matters: if the inverse-length scale is $\mu=\mu_0e^{-\ell}$, the same trajectory has <renormalization-group beta function> $\beta_\mu(g)=\mu\,dg/d\mu=-\epsilon g+g^2$. The two derivatives have opposite signs.
To solve the equation, set $v=1/g$. It obeys $dv/d\ell=-\epsilon v+1$, a linear <ordinary differential equation>. Therefore
$$
g(\ell)=\frac{\epsilon}{1+(\epsilon/g_0-1)e^{-\epsilon\ell}}.
$$
For $\ell\geq0$ its denominator is positive, and $g$ stays between $g_0$ and $\epsilon$. Thus the trajectory is bounded on this domain. The <renormalization-group fixed points> are $g=0$ and $g_*=\epsilon$. The derivative of the length-flow right-hand side is $\epsilon-2g$: zero is repulsive from positive $g$, whereas $g_*$ has <eigenvalue> $-\epsilon$ and attracts positive trajectories. Linearization gives $g-g_*\propto e^{-\epsilon\ell}=b^{-\epsilon}$ near $g_*$. The inverse-length <beta function> has positive derivative at the same infrared-attractive point.
For the chosen $\epsilon=1/5$, take two starting models with $g_0=1/20$ and $g_0=3/5$. At $b=1024$, $e^{-\epsilon\ell}=1/4$, so their couplings are respectively $4/35$ and $6/25$. Both move toward $1/5$, from opposite sides. They are still visibly different: slow decay of an <irrelevant operator> can leave substantial corrections at accessible lengths. The case $g_0=0$ stays exactly zero and is outside this positive-coupling basin.
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