Stability Switches and Hopf Bifurcations in Scalar DDEs

From delayed feedback to a stability threshold

A fixed delay can change more than the timing of a response. In a negative-feedback system, the feedback signal may arrive late enough to reinforce deviations that it would otherwise correct. The equilibrium can then lose stability, and the resulting dynamics may show oscillations. The key question here is precise: for a scalar delayed negative-feedback model, at what delay can characteristic roots cross the imaginary axis, and in which direction do they cross?

Consider a perturbation \(y(t)\) from an equilibrium, with current-state decay and delayed negative feedback:

The constants \(a\) and \(b\) have units of inverse time; \(\tau\) has units of time. This is a linear model with constant coefficients and a single fixed delay. Its characteristic equation is obtained by substituting an exponential solution \(y(t)=e^{\lambda t}\):

The equilibrium is locally asymptotically stable for this linear system when every characteristic root \(\lambda\) has negative real part. A root with positive real part implies instability. A pair on the imaginary axis is a marginal case, so it is not asymptotically stable. The characteristic-equation approach is the starting point established in the earlier material on characteristic equations for linear DDE stability; here we use it to locate and classify stability switches.

Finding imaginary roots and critical delays

Set \(\lambda=i\omega\) in the characteristic equation. Using \(e^{-i\omega\tau}=\cos(\omega\tau)-i\sin(\omega\tau)\), equate real and imaginary parts:

Squaring and adding these equations gives \(a^2+\omega^2=b^2\). A positive frequency therefore requires \(b>a\), and then

The cosine must be negative and the sine positive, so the phase \(\theta=\omega\tau\), taken between \(0\) and \(2\pi\), lies in the second quadrant. Its value is

Each \(\tau_k\) is a critical delay. The corresponding imaginary roots are the conjugate pair \(+i\omega\) and \(-i\omega\). The period of the linear oscillation at the crossing is \(2\pi/\omega\). Since \(a/b\) is dimensionless, \(\theta\) is an angle in radians; dividing by \(\omega\), in inverse-time units, gives a delay in time units.

Why the crossing is from stable to unstable

An imaginary root identifies a possible stability boundary, but does not alone show that roots cross the boundary as \(\tau\) changes. To establish the direction, differentiate \(F(\lambda,\tau)=0\) implicitly along a characteristic root:

At \(\lambda=i\omega\), the characteristic equation gives \(b e^{-i\omega\tau}=-a-i\omega\). Substituting this relation into the derivative yields

For a complex quotient, multiply numerator and denominator by the conjugate of the denominator. The real part of the resulting numerator is \(\omega^2(1+a\tau)-a\omega^2\tau=\omega^2\), so

The denominator is positive and \(\omega>0\), so the crossing is transverse and directed into the right half-plane as \(\tau\) increases. The conjugate root crosses in the corresponding direction. Thus, for this model, each critical delay adds a pair of right-half-plane roots; it does not cause a later return to stability.

When no delay can destabilize this model

If \(b\leq a\), there is no nonzero imaginary-root solution: the frequency condition would require \(\omega^2=b^2-a^2\leq0\). A stronger argument shows that no characteristic root can have nonnegative real part for any \(\tau\geq0\). Suppose \(\lambda=x+iy\) with \(x\geq0\) were such a root. The characteristic equation implies

When \(b<a\), these inequalities contradict each other. When \(b=a\), equality would require \(x=0\), \(y=0\), and \(e^{-x\tau}=1\); however, \(\lambda=0\) is not a root because \(F(0,\tau)=a+b=2a>0\). Thus the system has no characteristic roots with nonnegative real part when \(b\leq a\), and its equilibrium remains stable for every nonnegative delay.

Using the thresholds and interpreting oscillations

For \(b>a\), the first threshold \(\tau_0\) separates a stable region from an unstable region. At exactly \(\tau_0\), an imaginary pair lies on the axis, so the linear system is not asymptotically stable. For \(\tau_0<\tau<\tau_1\), there is one conjugate pair in the right half-plane. At each subsequent threshold \(\tau_k\), another pair crosses into that half-plane. In particular, taking \(\tau=10\) days in the preceding \(a=0.5,\ b=0.7\) example gives \(\tau_0<10<\tau_1\), since \(\tau_1=(2.366399+2\pi)/0.489898\approx17.66\) days. The linearized equilibrium is unstable with one right-half-plane pair.

This is a conclusion about linear stability and an oscillatory mode, not a complete prediction of the nonlinear trajectory. The characteristic roots establish the frequency and growth or decay near the equilibrium. A nonlinear system can limit, amplify, distort, or otherwise alter the oscillations as the state moves away from that equilibrium. A characteristic pair crossing the imaginary axis is often described as a Hopf point or a candidate Hopf bifurcation. Establishing a nonlinear Hopf bifurcation and characterizing the nearby periodic solutions also requires a suitable smooth nonlinear model and appropriate nondegeneracy conditions. The linear calculation alone does not prove that a stable periodic orbit emerges.

A practical analysis should check the model assumptions and the root calculation, not only inspect a simulated time series. Confirm that the feedback sign is negative, that \(a\) and \(b\) use consistent inverse-time units, and that the equilibrium has been correctly linearized. Check that the derived \(\omega\) is real and positive before calculating \(\theta\) or \(\tau_k\). Near a threshold, small growth rates can make simulated trajectories look nearly periodic or nearly stationary for a long time; that appearance does not determine which side of the threshold the system occupies. A numerical characteristic-root calculation can check the roots on either side of a predicted threshold, while time-domain simulation can illustrate trajectories for specified history functions. Neither replaces the crossing calculation, and nonlinear simulation cannot by itself establish the linear root locations exactly.

The main alternatives answer different questions. The characteristic-equation calculation gives an analytic threshold and crossing direction for this scalar fixed-delay model. Numerical root finding is useful when a characteristic equation is too complicated for closed-form thresholds, but it needs checks for missed roots and sensitivity to initial guesses. Time simulation shows behavior for a chosen initial history, but a single trajectory may not reveal weak instability or distinguish a transient oscillation from a persistent one. These approaches complement each other; the analytic result applies only to the stated linearized model and its assumptions.