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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 defining feature is memory: a derivative can depend on earlier values of the unknown function. Constant delays, time-dependent delays, and delays depending on the unknown function belong to this wider class. The retarded delay differential equation below isolates a single positive constant delay.^[ref]

The Malthusian delay differential equation is a simple linear member of this class. The method of steps for a delay differential equation constructs its future from a history function of a delay differential equation, and the characteristic equation of a delay differential equation studies its exponential function modes. The exact polynomial steps for a delay equation below make that construction explicit without approximating the solution of a differential equation. Applications include population dynamics and feedback control; the delayed negative feedback model and feedback delay explain why the timing of a return matters as well as its sign.

This delay differential equation replaces instantaneous logistic inhibition by a population value one delay earlier. A positive history function of a delay differential equation gives positive solutions while they exist. The equilibria are zero and one. Linearization at one gives $x^{\prime }(t)=-x(t-\tau )$, with characteristic equation $\lambda =-e^{-\lambda \tau }$. Delayed sign reversal permits overshoots around one and, at the delayed logistic Hopf threshold, a periodic branch.

A feedback delay is a time interval between an event and its returning influence. Multiple biochemical steps can introduce an effective lag; a fixed delay is an idealization, not a universal description of signal transduction. To isolate timing, let $x$ be a signed displacement from an equilibrium point and use normalized dimensionless variables:The local removal is $-x$ and the returning influence is negative at every delay. At $d=0$ the ordinary differential equation reduces to the decay already illustrated under signaling feedback. For $d>0$ a delay differential equation requires a history on $[-d,0]$, not just an initial condition at one instant. Negative values of $x$ are displacements, not negative chemical concentrations.

An exponential function $x=e^{\lambda t}$ gives the characteristic equation of a delay differential equationPutting $\lambda =i\omega$ and equating real and imaginary parts gives $\cos (\omega d)=-1/2$ and $\omega =2\sin (\omega d)$. The first crossing with $\omega >0$ therefore occurs atFor this particular mathematical model, the equilibrium point is asymptotically stable for $0\le d<d_c$. No nonnegative real delayed feedback characteristic root can satisfy the equation; the first loss of asymptotic stability is the indicated pair of imaginary delayed feedback characteristic roots. Differentiating the characteristic equation of a delay differential equation along a root yieldsThus the pair moves into growth as the delay increases through $d_c$. The neutral mode at $d_c$ is a periodic function, but its amplitude depends on history: this linear mathematical model does not select an attracting limit cycle.

The figure plots $x(t)=e^{at}\cos (bt)$ using a delayed feedback characteristic root $\lambda =a+ib$ and that same formula on the history interval. At $d=0.7$ the displayed mode decays; at $d=1.5$ it grows. These are exact modes up to numerical root accuracy, not step responses with identical histories. Molecular bounds and a changed return law would be needed to interpret what happens after an unstable biological response becomes large.

For the delay differential equationthe equilibrium $x=1$ is locally asymptotically stable for every fixed finite delay if $\alpha >0$. Set $x=1+u$; the exact perturbation equation isThe delayed perturbation enters only through the quadratic product, so the linearization is the ordinary differential equation $u^{\prime }=-\alpha u$ with characteristic exponent $-\alpha$.

Positive $\alpha$ therefore gives exponential linear decay and local asymptotic stability of the smooth retarded equation. Negative $\alpha$ gives a growing linear mode and instability. At $\alpha =0$, every history becomes constant after the initial time; a nearby initial value need not approach $x=1$. Thus that case is not asymptotically stable.

This example shows why introducing a delay need not produce a Hopf bifurcation: the delay must actually enter the linearized feedback before it can change the linear stability threshold.

For the constant-coefficient model $x^{\prime }(t)=-kx(t-\tau )$ with $k>0$ and $\tau \ge 0$, the zero equilibrium point is asymptotically stable if and only if $k\tau <\pi /2$. For positive $\tau$, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For $k\tau >\pi /2$, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function $x(t)=e^{st}$, allowing a complex number $s$. The characteristic equation isFor positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set $s=i\omega$ with $\omega >0$ and use Euler's formula:The smallest positive crossing is $\omega =k$ and $\omega \tau =\pi /2$. Thus at ${\tau }_c=\pi /(2k)$ the complex conjugate roots are $s=\pm ik$. In fact, with the matching initial history of delayed feedback,solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts $2\pi /k=4{\tau }_c$. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed $k$ givesThis calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if $s=\alpha +i\beta$ with $\alpha \ge 0$ and $\beta >0$, the characteristic equation implies $\cos (\beta \tau )\le 0$ and $0<\beta \le k$, hence $k\tau \ge \beta \tau \ge \pi /2$. A real nonnegative root is impossible because $s+k{e}^{-s\tau }>0$ there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

The tradeoff is concrete. Without feedback delay, increasing $k$ shortens the exponential decay time constant $1/k$. With known $\tau >0$, the same proportional control must satisfy $k<\pi /(2\tau )$. For $\tau =0.2\mathrm{s}$, the threshold is $k\approx 7.854{\mathrm{s}}^{-1}$: $k=5{\mathrm{s}}^{-1}$ is stable and $k=8{\mathrm{s}}^{-1}$ is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing $k$ or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

A feedback delay means a controller reacts to an earlier plant output. An accurate measurement can still be stale: when the plant output has already crossed the reference signal, a controller using the earlier value may continue pushing it away from the target. In a pure constant-delay model, $y_m(t)=y(t-\tau )$ with $\tau \ge 0$. This isolates timing from measurement noise; it does not describe every sampling or communication mechanism. ^[ref]

For the ideal scalar controlled plant $\dot{x}=u$, a zero reference signal, no control disturbance, and proportional control with constant feedback gain $k>0$, the equations areWith $\tau =0$, this is an ordinary differential equation with exponential decay $x(t)=x(0)e^{-kt}$. With $\tau >0$, it is a delay differential equation, requiring an initial history of delayed feedback. The delay stability of proportional integrator control depends on the product $k\tau$: even this simple controlled plant loses asymptotic stability at $k\tau =\pi /2$.

For positive feedback delay $\tau$, an initial condition at a single instant is insufficient. Prescribe a continuous function $\varphi$ throughout $[-\tau ,0]$: for $0\le t\le \tau$, the derivative depends on $x(t-\tau )=\varphi (t-\tau )$. A right derivative at $t=0$ is understood; the earlier history need not itself obey the subsequent control law. Once this first interval is computed, it supplies the delayed values for the next interval. This successive integration is the method of steps for a delay differential equation. ^[ref]

For the constant initial history of delayed feedback $\varphi (t)=1$, direct integration givesFor $k=1$, $\tau =1.2$, the plant output crosses zero at $t=1$, while its measured output remains positive until $t=2.2$. At $t=1.4$, the delayed command therefore increases the magnitude of the actual deviation. This short exact calculation explains the first diagram without relying on a numerical approximation.

The scalar equation $n^{\prime }(t)=rn(t-\tau )$ makes the present growth rate depend on the population one delay time earlier. A history function on $[-\tau ,0]$ is required in place of a single initial value.

This is one linear delay differential equation, rather than the entire delayed-evolution class.

Negative feedback closes a causal loop whose net effect opposes an initiating change. In signal transduction, activation followed by a downstream inhibition of the earlier activating stage is one example of negative feedback. Loop sign alone does not establish asymptotic stability or complete signaling adaptation; see the worked signaling feedback example.

A useful biological example is chemotaxis feedback regulation: an Escherichia coli biological cell adjusts its response to a sustained attractant instead of remaining at its immediate response. Increased CheA activity promotes CheB phosphorylation; CheB reverses receptor methylation, opposing their activation. After attractant addition, reduced activity instead favors receptor methylation by CheR, restoring activity. This is a return from receptor activity to the machinery regulating that activity. The response need not return exactly at every dose. ^[ref]

A motivating measurement is the chemotaxis FRET assay. CheY and CheZ carrying yellow and cyan fluorescent tags were observed in immobilized Escherichia coli biological cells on a coverslip in a flow chamber. Adding and removing attractant changed the yellow-to-cyan emission ratio, reporting the association of phosphorylated CheY with CheZ. This measures a molecular readout upstream of motor behavior. Comparisons with biological cells lacking CheR, CheB, or both tested the dependence on receptor modification; these perturbations also change starting activity and sensitivity, so they are not pure changes of one feedback gain. ^[ref]

The later chemotaxis feedback regulation experiment used only the Tar receptor and nonmetabolizable methylaspartate, supplied and removed by flow. Long recordings separated slow recovery from incomplete signaling adaptation; correction for bleaching addressed drift in the optical readout. Activity recovered incompletely after sufficiently large additions, with detectable imprecision already at $10\mu \mathrm{M}$ in that preparation. Restricting the receptor type and preventing consumption of the stimulus help interpret the perturbation, but limit transfer to other receptors and conditions. The readout infers CheA activity through CheY-CheZ association, rather than directly measuring every molecular step. ^[ref]

Negative feedback can also overshoot or sustain oscillations if its return arrives too late. The worked signaling feedback example gives the contrasting instantaneous case; the delayed negative feedback model below isolates the effect of timing. Loop sign describes causation, while asymptotic stability describes what happens to displacements over time.

For a constant solution $u(t)\equiv U$, its delayed value is also $U$. The equation becomes $0=U(1-U)$, so the only constant equilibria are $\boxed{U=0\text{ and }U=1}$. The delay differential equation requires the same constant history to realize each equilibrium.

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:

The delayed influx $rI(t-\tau )S(t-\tau )$ represents animals exposed one incubation period earlier who enter the infective individual compartment now. They die at the instantaneous rate $aI(t)$ once in that compartment. The displayed bookkeeping removes an animal from $S$ at the same delayed time that it enters $I$, preserving $S+I+R$.

For this bookkeeping, $S$ must include the exposed animals which have not yet completed incubation period. If $S$ were instead defined as strictly unexposed susceptible individuals, its loss term would be the current exposure rate and a separate latent compartment would be needed. The calculations here follow the printed delay differential equation. A full initial-value problem for a delay differential equation also needs a history on $[-\tau ,0]$; the following small-delay expansion of an epidemic transfer rate concerns a smooth compatible history and the regular regime assumed by the question.

Skip Thompson is a mathematician whose work includes the numerical solution of delay differential equations. His authored account explains equations in which the present rate of change depends on earlier values of the solution, with applications including population dynamics and epidemiology.^[ref] For example, $y^{\prime }(t)=-ay(t-\tau )$ with $\tau >0$ requires an initial history on $[-\tau ,0]$, rather than only one initial value. A numerical method must therefore approximate and retrieve past values as integration proceeds. Delayed feedback can also create oscillations that are absent from a corresponding equation without delay.

A mathematician.^[ref]

For the delay differential equationthe equilibrium $x=1$ is locally asymptotically stable forThe positivity of $\alpha$ is part of this statement.

Writing $x=1+u$ gives the linearization $u^{\prime }=\alpha [u(t)-2u(t-T)]$ and characteristic equationAt a purely imaginary root $\lambda =i\omega$, real and imaginary parts require $\cos (\omega T)=1/2$ and $\omega =2\alpha \sin (\omega T)$. The first crossing has $\omega =\sqrt{3}\alpha$ and $\omega T=\pi /3$, yielding $T_c$.

Implicit differentiation gives, at that crossing,Thus the roots cross toward instability as the delay increases. The imaginary-root condition is the linear onset needed for a possible Hopf bifurcation; the nonlinear bifurcation type requires further coefficients. At $\alpha =0$ the dynamics are stationary, while negative $\alpha$ is already unstable at zero delay.

Zdzislaw Jackiewicz is a mathematician working in numerical analysis whose work includes methods for ordinary differential equations, delay differential equations and Volterra integral equations. His authored account of general linear methods describes a framework combining internal stage evaluations with a vector of information passed between successive time steps.^[ref]

For $y^{\prime }=f(y)$, a Runge-Kutta method passes an approximation to $y$ between steps while computing several intermediate stages. A linear multistep method instead retains values from several earlier times. General linear methods put both patterns in one matrix formulation, allowing accuracy and numerical stability to be designed together. His Arizona State University profile identifies this subject and the construction of implicit-explicit methods as research interests.^[ref]