Characteristic Equations for Linear DDE Stability

Why Stability Requires a Characteristic Equation

For an ordinary differential equation, a small perturbation from equilibrium evolves according to a finite-dimensional linear system. In a delay differential equation (DDE), the rate of change can also depend on an earlier state. A perturbation therefore feeds into its own future evolution after a delay, and stability is governed by an equation that is generally transcendental rather than polynomial.

The key task is to examine whether small deviations from equilibrium decay or grow. The characteristic equation describes possible exponential modes of the linearized system. Its roots, including complex roots, encode those modes. A root with positive real part implies a growing perturbation; if every root has negative real part, the linearized perturbations decay.

Linearizing Around an Equilibrium

Consider an autonomous scalar DDE with one fixed delay:

\dot{x}(t)=f\bigl(x(t),x(t-\tau)\bigr), \qquad \tau>0.

Here \(x(t)\) is a scalar state, \(\tau\) is a constant delay, and \(f\) is assumed continuously differentiable near the equilibrium of interest. A constant equilibrium \(x^*\) must satisfy \(f(x^*,x^*)=0\): both the current state and the delayed state equal \(x^*\) when the system is at equilibrium. The history function specifies the state over an interval of length \(\tau\), but does not change this equilibrium condition.

Definition: Let \(x^*\) satisfy \(f(x^*,x^*)=0\). Define \(a=\partial f/\partial x(t)\rvert_{(x^*,x^*)}\) and \(b=\partial f/\partial x(t-\tau)\rvert_{(x^*,x^*)}\). The first-order linearization in the perturbation \(u(t)=x(t)-x^*\) is \(\dot{u}(t)=a u(t)+b u(t-\tau)\).

The coefficients \(a\) and \(b\) are local sensitivities evaluated at the equilibrium, not necessarily the coefficients of the original nonlinear model. If \(x\) has units of concentration and time is measured in days, both derivatives have units of inverse days. For a smooth \(f\), the linearization follows from a first-order Taylor expansion: terms that are quadratic or higher in the perturbations are omitted.

Deriving the Characteristic Equation

To find exponential modes of the linearized DDE, substitute \(u(t)=C e^{\lambda t}\), where \(C\ne 0\) and \(\lambda\) may be complex. The delayed perturbation is \(u(t-\tau)=C e^{\lambda(t-\tau)}\). Substitution and cancellation of the nonzero factor \(C e^{\lambda t}\) give the characteristic equation.

\lambda=a+b e^{-\lambda\tau}, \qquad \Delta(\lambda)=\lambda-a-b e^{-\lambda\tau}=0.

The exponential term contains the unknown \(\lambda\), so this is a transcendental equation. For a positive fixed delay, it generally has infinitely many complex roots. Each root corresponds to a possible exponential mode; the largest real part among the roots determines the long-term behavior of the linearized perturbations.

Stability result: For a scalar retarded DDE with constant delay, the linearized system is asymptotically stable when every characteristic root has negative real part, and unstable when any characteristic root has positive real part. For a smooth nonlinear DDE, the linearization gives local exponential stability if all roots are strictly in the left half-plane; a root in the right half-plane implies local instability. If roots lie on the imaginary axis, the linearization alone does not settle nonlinear stability.

This linearized stability result is a local statement: it concerns sufficiently small deviations around the specified equilibrium, not every possible initial history or the model’s global behavior. The imaginary-axis case is non-hyperbolic, so higher-order nonlinear terms may matter. Determining parameter values at which such roots occur requires a separate analysis.

Worked Example: Delayed Negative Feedback With Delay-Independent Stability

Suppose a hypothetical concentration model is \(\dot{x}(t)=R-kx(t)-q x(t-\tau)\), where \(R=11\) nM/day, \(k=0.8\) per day, \(q=0.3\) per day, and \(\tau=2\) days. Find the equilibrium and assess its local stability.

At equilibrium, \(x(t)=x(t-\tau)=x^*\), so \(0=R-(k+q)x^*\). Thus \(x^*=R/(k+q)=11/(0.8+0.3)=10\) nM. The equilibrium condition is satisfied because \(11\) nM/day minus \(1.1\) per day times \(10\) nM equals zero.

The model is smooth, and its derivatives at the equilibrium are \(a=-k=-0.8\) per day and \(b=-q=-0.3\) per day. Its characteristic equation is therefore \(\lambda+0.8+0.3e^{-2\lambda}=0\), with \(\lambda\) in inverse days.

Here \(a=-k\), where \(k=0.8\) per day, and \(|b|=0.3\) per day. Since \(k>|b|\), the delay-independent condition derived below applies. The equilibrium is locally exponentially stable for this delay and, in fact, for any nonnegative fixed delay under these same coefficients.

A Useful Delay-Independent Stability Condition

Some models can be classified without finding their infinitely many roots. For \(\dot{u}(t)=-k u(t)+b u(t-\tau)\), where \(k>0\), a sufficient condition for stability at any fixed delay is \(k>|b|\). To justify this condition, suppose a characteristic root \(\lambda=\sigma+i\omega\) has \(\sigma\ge 0\). The characteristic equation implies \(\lambda+k=b e^{-\lambda\tau}\). Taking absolute values gives

|\lambda+k|=|b|e^{-\sigma\tau}\leq |b|.

But \(|\lambda+k|\geq \operatorname{Re}(\lambda+k)=\sigma+k\geq k\). These inequalities contradict \(k>|b|\). Therefore, there is no characteristic root with nonnegative real part; the linearized system is asymptotically stable. This is a sufficient condition, not a necessary one: failing the inequality does not by itself show instability.

Worked Example: A Positive Delayed Feedback With an Unstable Root

Consider a hypothetical smooth model whose linearization about an equilibrium is \(\dot{u}(t)=-1.0u(t)+1.5u(t-\tau)\), with both coefficients in inverse days and \(\tau=2\) days. The equilibrium is assumed to satisfy the nonlinear model’s equilibrium equation. Determine whether the linearization is stable.

The characteristic equation is \(\Delta(\lambda)=\lambda+1.0-1.5e^{-2\lambda}=0\), with \(\lambda\) in inverse days. For real \(\lambda\), \(\Delta\) is continuous. At zero, \(\Delta(0)=1.0-1.5=-0.5\) per day. As real \(\lambda\) tends to positive infinity, the exponential term tends to zero and \(\Delta(\lambda)\) tends to positive infinity.

Consequently, \(\Delta\) has a positive real root. Numerically, \(\lambda\) is approximately \(0.138\) per day: substituting gives \(0.138+1.0-1.5e^{-0.276}\), which is approximately zero. This growing exponential mode makes the equilibrium locally unstable. The argument establishes instability from one positive root; it does not require locating the other roots.

Worked Example: Positive Feedback That Still Satisfies the Stability Bound

Let \(x\) be a hypothetical concentration measured in nM, and consider \(\dot{x}(t)=3-0.6x(t)+0.4\bigl(x(t-\tau)-5\bigr)\), with rates in nM/day and coefficients \(0.6\) and \(0.4\) per day. Take \(\tau=5\) days. Check the equilibrium and classify its local stability.

Substituting \(x(t)=x(t-\tau)=5\) nM into the right-hand side gives \(3-0.6(5)+0.4(5-5)=3-3+0=0\) nM/day. Thus \(x^*=5\) nM is an equilibrium. The model is smooth, with derivatives \(a=-0.6\) per day and \(b=+0.4\) per day.

The characteristic equation is \(\lambda+0.6-0.4e^{-5\lambda}=0\). The delayed feedback is positive, but \(k=0.6\) per day exceeds \(|b|=0.4\) per day. The sufficient condition therefore shows that the equilibrium is locally exponentially stable. The sign of delayed feedback alone does not determine stability; its magnitude relative to the instantaneous decay also matters.

Checking the Result and Avoiding Common Errors

Characteristic-root analysis is useful because it distinguishes local growth from decay without relying on a particular initial history. Its conclusions are nevertheless sensitive to correct model specification. Before interpreting roots, verify that the equilibrium satisfies \(f(x^*,x^*)=0\), that both partial derivatives are evaluated at \((x^*,x^*)\), and that the delay and rate coefficients use compatible time units.

For numerical root calculations, check that each reported value has a small residual \(|\Delta(\lambda)|\), and inspect whether additional roots could have larger real parts. Finding several stable roots does not establish stability if an unexamined root lies farther to the right. The infinitely many roots also mean that a finite list of roots is not automatically a complete stability proof; use a justified criterion or a root-finding method that addresses the rightmost roots.

Using only the instantaneous derivative. A negative \(a\) does not settle stability because \(b e^{-\lambda\tau}\) also enters the characteristic equation.
Dropping the delay exponential. The delayed mode is multiplied by \(e^{-\lambda\tau}\); replacing it with \(1\) is valid only in the zero-delay limit.
Reversing the stability interpretation. A root with positive real part means an exponentially growing mode and local instability, not stability.
Treating a sufficient condition as necessary. If \(k\leq |b|\), the delay-independent bound gives no conclusion; it does not prove instability.
Trusting a simulation alone. A stable-looking finite-time trajectory cannot rule out slow growth, and a trajectory can depend on its chosen history. Simulation complements rather than replaces root analysis.

The main alternatives answer related but different questions. An ODE Jacobian check is appropriate when there is no delay, but it does not include the delayed feedback term. Direct DDE simulation, using a history function and a method such as the method of steps, illustrates trajectories for selected histories and time horizons but does not by itself certify stability for all small perturbations. Characteristic-root analysis targets local stability directly, though it depends on a correct linearization and can require careful numerical root computation. None of these local tests establishes global stability or makes an approximate model valid outside the neighborhood of its equilibrium.

When \(\tau=0\), the equation reduces to \(\dot{u}=(a+b)u\), with the single characteristic root \(\lambda=a+b\). For positive delay, the characteristic equation is generally transcendental and has infinitely many roots, so an ODE-style check of \(a+b\) is not a general substitute.

Check Your Understanding

Try each problem, then compare your reasoning with the solutions below.

A model is \(\dot{x}(t)=10-0.8x(t)-0.2x(t-\tau)\), with \(x\) in nM and rates in nM/day. Find its equilibrium and characteristic equation. Solution: At equilibrium, \(0=10-(0.8+0.2)x^*\), so \(x^*=10\) nM. The derivatives are \(a=-0.8\) and \(b=-0.2\) per day, giving \(\lambda+0.8+0.2e^{-\lambda\tau}=0\).
For \(\dot{u}=-0.7u-0.9u(t-\tau)\), is the delay-independent condition sufficient to conclude stability? Solution: No. Here \(k=0.7\) and \(|b|=0.9\) per day, so \(k>|b|\) is false. This sufficient condition does not classify the system; its failure is not proof of instability.
For \(\dot{u}=-0.4u+0.7u(t-\tau)\), where \(\tau>0\), show that the linearization is unstable. Solution: Its characteristic function on the real line is \(\Delta(\lambda)=\lambda+0.4-0.7e^{-\lambda\tau}\). Since \(\Delta(0)=-0.3\) per day and \(\Delta(\lambda)\) tends to positive infinity as real \(\lambda\) tends to positive infinity, continuity guarantees a positive real root. That root implies instability.
If \(\tau=0\), \(a=-0.9\) per day, and \(b=0.3\) per day, what is the characteristic root and its stability implication? Solution: The equation becomes \(\lambda=a+b=-0.9+0.3=-0.6\) per day. The root is negative, so perturbations decay exponentially and the linearized equilibrium is asymptotically stable.