= Positive-equilibrium separatrix for a cubic profile equation
{title2=$f''=f^3-a^2f,\quad f(\infty)=a>0$}
Multiply the <ordinary differential equation> by $f'$ and integrate. A solution approaching the positive <equilibrium> $a$ has $f'\to0$: this follows from the equation and convergence of $f$, since a nonzero limiting derivative or energy would preclude convergence. Its <first integral> is therefore
$$
\frac12(f')^2=\frac14(f^2-a^2)^2.
$$
The branch tending to $a$ at positive infinity obeys $f'=(a^2-f^2)/\sqrt2$. Separation yields
$$
f(\eta)=a\frac{1-Ae^{-\sqrt2a\eta}}{1+Ae^{-\sqrt2a\eta}}.
$$
The constant solution is $A=0$. For $A>0$ this is a translated <hyperbolic tangent>; for $-1<A<0$ it is a regular <hyperbolic cotangent> branch on $\eta\geq0$. The condition $A>-1$ excludes both an interior pole and a divergent value at the finite endpoint. The opposite sign in the first integral moves away from the positive <equilibrium> as $\eta$ increases, except at the constant solution.
A useful integral consequence is
$$
\int_0^\infty[a-f(\eta)]\,d\eta=\sqrt2\log(1+A).
$$
Indeed the integrand is $2aAe^{-\sqrt2a\eta}/(1+Ae^{-\sqrt2a\eta})$, the negative of $\sqrt2$ times the derivative of $\log(1+Ae^{-\sqrt2a\eta})$. An integral deficit can therefore fix the translation parameter uniquely, even when the differential equation and far value leave it free.
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