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A delay differential equation relates a derivative at the current argument to values at earlier arguments. For example, the Buchstab function obeys $\frac{d}{du}(uw(u))=w(u-1)$ for $u>2$.

The delayed positive term models recruitment depending on the population a time $T$ earlier; the quadratic negative term represents instantaneous density-dependent losses. Thus this delay differential equation delays recruitment rather than the strength of crowding acting on the present population.

With $x=1+y$, the linearized equation is $\dot{y}=\alpha y(t-T)-2\alpha y(t)$. Its characteristic equation of a delay differential equation is $\lambda +2\alpha =\alpha e^{-\lambda T}$. A root with nonnegative real part would satisfy $|\lambda +2\alpha |\ge 2\alpha$, whereas $|\alpha e^{-\lambda T}|\le \alpha$. This is impossible. Consequently the equilibrium is locally exponentially asymptotically stable for every $\alpha >0$ and $T>0$. There is no upper bound on $\alpha T$ in this model.

Put $x=1+y$ and retain linear terms. The delay differential equation is $\dot{y}(t)=-\alpha y(t-T)$, with characteristic equation of a delay differential equation $\lambda +\alpha e^{-\lambda T}=0$. Write $z=\lambda T$ and $q=\alpha T$. If a root has $\mathrm{Re}z\ge 0$, then $|z|=q{e}^{-\mathrm{Re}z}\le q$. Writing $z=a+ib$ gives $a=-q{e}^{-a}\cos b$. For $q<\pi /2$, $|b|<\pi /2$ and the right side is negative, contradicting $a\ge 0$. All characteristic roots therefore lie in the left half-plane.

An imaginary root $z=ib$ requires $\cos b=0$ and $b=q\sin b$. The first crossing is $q=\pi /2$, $z=\pm i\pi /2$. Implicit differentiation gives $dz/dq=-e^{-z}/(1+z)$, whose real part at the positive crossing is $(\pi /2)/(1+{\pi }^2/4)>0$. The later imaginary crossings occur at $q=\pi /2+2\pi k$ and also enter the right half-plane. Thus the equilibrium is locally exponentially asymptotically stable whenand unstable when $\alpha T>\pi /2$. At equality the linearized equation has neutral oscillatory modes; it is the Hopf bifurcation threshold. If stability is meant for the nonlinear equation itself, the borderline can also be resolved. In dimensionless time $s=t/T$, let $q_0=\pi /2$. A small center oscillation has the form $y=R\cos \theta +R^2(\cos 2\theta /5+\sin 2\theta /10)+O(R^3)$. Substitution at quadratic order gives these second-harmonic coefficients. At cubic order the fundamental forcing is $q_0{R}^3(\cos \theta /20+3\sin \theta /20)$. Projection onto the center mode divides the complex forcing by the characteristic derivative $1+i{q}_0$, yielding the Hopf normal form amplitude equationIts cubic coefficient is negative, so the equilibrium at equality is locally asymptotically stable, with algebraic rather than exponential decay. Thus the usual strictly decaying linear condition is $\alpha T<\pi /2$, while nonlinear local asymptotic stability includes $\alpha T=\pi /2$. The purely imaginary linear modes alone would not justify that borderline conclusion.

Put $c=e^{-\gamma }$, where $\gamma$ is the Euler--Mascheroni constant, and let $f(u)=w(u)-c$. The Buchstab function equation becomesThe given Gamma function bound implies $uf(u)\to 0$. Indeed,so $u/\Gamma (u+1)\to 0$.

Suppose first that $f\ge 0$ throughout an interval $[u,u+1]$, with $u\ge 1$. The delay differential equation and integration successively on $[u+1,u+2]$, $[u+2,u+3]$, and so on propagate nonnegativity to the entire tail $[u,\infty )$. In particular, $vf(v)$ is nonnegative and nondecreasing for $v\ge u+1$. Its limit is zero, so it is identically zero there. Thus $f=0$ on that tail.

The delay differential equation then propagates this equality backwards: if $f=0$ for $v\ge a$, then $f(v-1)=0$ for $v>\max (a,2)$. Repeating finitely many times would give $f=0$ on $(1,2)$, contradicting $f(v)=1/v-c$ there. Hence every interval $[u,u+1]$ contains a point where $f<0$.

If instead $f\le 0$ on $[u,u+1]$, the same propagation makes $vf(v)$ nonpositive and nonincreasing on the tail. Its limit zero again forces it to vanish, giving the same contradiction. Thus every such interval also contains a point where $f>0$. The oscillation of the Buchstab function is therefore strict on every unit interval: