Local Stability of Population Equilibria

Why Equilibrium Stability Matters

A steady state specifies a cell population that remains constant when it is exactly at equilibrium. In a model, the more practical question is often whether a small departure from that state fades or grows. Stability analysis answers this locally: it describes the response to sufficiently small perturbations, not necessarily the behavior of populations far from equilibrium.

For a system of autonomous differential equations, the Jacobian summarizes how the rates of change respond to small changes in the state variables. Its eigenvalues indicate whether small perturbations tend to decay, grow, or require further analysis. The method applies both to linear turnover models and, through local approximation, to nonlinear models such as logistic growth.

Definition: For an autonomous model \( \dot{x}=f(x) \), an equilibrium \(x^*\) satisfies \(f(x^*)=0\). It is locally stable if solutions starting sufficiently close to \(x^*\) remain close to it. It is locally asymptotically stable if it is locally stable and nearby solutions also converge to \(x^*\) as time increases. It is unstable if it is not locally stable.

Here \(x\) is a vector of population states, and \(f\) is assumed to have continuous first derivatives in a neighbourhood of the equilibrium. The Jacobian \(J(x^*)\) is the matrix of partial derivatives of the state-change rates with respect to the states, evaluated at \(x^*\). Its eigenvalues may be real or complex; their real parts govern whether the corresponding linearized modes decay or grow.

J(x^*)=\left[\frac{\partial f_i}{\partial x_j}(x^*)\right], \qquad \dot{\delta x}\approx J(x^*)\delta x, \qquad \delta x=x-x^*.

How the Linearization Gives a Stability Test

The linearized equation follows from a first-order Taylor expansion around the equilibrium. Since \(f(x^*)=0\), the constant term vanishes, leaving the Jacobian term plus a remainder that becomes small relative to the perturbation as \(\delta x\) approaches zero. The eigenvalues of \(J(x^*)\) therefore describe the leading-order dynamics near the equilibrium.

f(x^*+\delta x) = f(x^*)+J(x^*)\delta x+o(\|\delta x\|) = J(x^*)\delta x+o(\|\delta x\|).

For a linear mode with eigenvalue \(\lambda\), the perturbation has a factor proportional to \(e^{\lambda t}\). If \(\lambda\) is complex, this factor combines an oscillation with an envelope proportional to \(e^{\operatorname{Re}(\lambda)t}\). Thus a negative real part means decay for that mode, while a positive real part means growth.

Theorem: For a continuously differentiable autonomous model, if every eigenvalue of \(J(x^*)\) has negative real part, then \(x^*\) is locally asymptotically stable. If at least one eigenvalue has positive real part, then \(x^*\) is unstable. If no eigenvalue has positive real part but at least one has zero real part, the equilibrium is nonhyperbolic, and the Jacobian test alone does not determine stability. An equilibrium is hyperbolic when none of its Jacobian eigenvalues has zero real part.

The nonhyperbolic case is a limit of what this test can conclude, not a stability category. Nonlinear terms omitted by the first-order approximation may determine the outcome. A zero eigenvalue does not, by itself, mean either stable or unstable.

Worked Example: Stability in a Two-Compartment Turnover Model

Consider a hypothetical progenitor-to-mature-cell model with constant input \(u=120\) cells/day and transfer or loss rates \(a=0.4\) day\(^{-1}\) and \(b=0.2\) day\(^{-1}\):

\dot{P}=u-aP, \qquad \dot{M}=aP-bM.

At equilibrium, both rates of change are zero. The first equation gives \(P^*=u/a=120/0.4=300\) cells. The second gives \(M^*=aP^*/b=(0.4\times300)/0.2=600\) cells. The Jacobian is constant:

J= \begin{pmatrix} -a & 0\\ a & -b \end{pmatrix} = \begin{pmatrix} -0.4 & 0\\ 0.4 & -0.2 \end{pmatrix} \ \text{day}^{-1}.

Because this matrix is triangular, its eigenvalues are its diagonal entries: \(\lambda_1=-0.4\) day\(^{-1}\) and \(\lambda_2=-0.2\) day\(^{-1}\). Both are negative, so the equilibrium is locally asymptotically stable. In this affine linear model, perturbations from equilibrium satisfy the linearized equations exactly, and both modes decay; the conclusion holds globally for the mathematical system as well. The slower mode has a decay timescale of \(1/0.2=5\) days, while the faster mode has a timescale of \(1/0.4=2.5\) days.

Density Regulation: Stability Can Differ Across Equilibria

Density-regulated growth makes the net growth rate depend on population size. For a scalar model \(\dot{N}=f(N)\), the Jacobian is the single derivative \(f'(N^*)\). A negative derivative means small deviations are pulled back toward the equilibrium; a positive derivative means deviations grow. Different equilibria of the same model can therefore have different stability.

Worked Example: The Two Logistic Equilibria

Take the logistic model \(\dot{N}=rN(1-N/K)\), with hypothetical \(r=0.6\) day\(^{-1}\) and \(K=1000\) cells. Its equilibria solve \(rN(1-N/K)=0\), giving \(N^*_1=0\) and \(N^*_2=1000\) cells. The derivative is

f'(N)=r\left(1-\frac{2N}{K}\right).

At zero, \(f'(0)=0.6\) day\(^{-1}\), so the Jacobian has a positive eigenvalue and \(N=0\) is unstable. At carrying capacity, \(f'(K)=0.6(1-2)=-0.6\) day\(^{-1}\), so \(N=K\) is locally asymptotically stable. A small displacement from \(K\) is opposed by the local density feedback. This calculation identifies local behavior at each equilibrium; it does not assert that every initial population follows the same trajectory or that the model is appropriate for every biological setting.

For a scalar model, the local approximation has a particularly direct form. Writing \(N=N^*+\delta N\), the first-order equation is \(\dot{\delta N}\approx f'(N^*)\delta N\), whose solution is \(\delta N(t)\approx\delta N(0)e^{f'(N^*)t}\). This is why the sign of the derivative classifies a one-dimensional hyperbolic equilibrium.

When the Jacobian Is Not Enough

Suppose the derivative at an equilibrium is zero. The first-order approximation then predicts no change in the perturbation, even though higher-order terms may push it toward or away from equilibrium. This is not a defect in the calculation; it is a limitation of a linear approximation when the leading linear term vanishes.

Worked Example: A Nonhyperbolic Density-Feedback Equilibrium

Consider a hypothetical density-regulated model near \(K=500\) cells:

\dot{N}=-c(N-K)^3, \qquad c=2.0\times10^{-6}\ \text{cells}^{-2}\text{day}^{-1}.

At \(N^*=K\), the rate is zero. The Jacobian eigenvalue is the scalar derivative \(f'(N)=-3c(N-K)^2\), so \(f'(K)=0\). The equilibrium is nonhyperbolic: the Jacobian test alone cannot classify it. Set \(y=N-K\). Then \(\dot{y}=-cy^3\). For a nonzero initial displacement \(y(0)=y_0\), separation of variables gives

\frac{d}{dt}(y^{-2})=2c, \qquad y(t)=\frac{y_0}{\sqrt{1+2cy_0^2t}}.

The denominator grows with time, so \(y(t)\) approaches zero from either side. The equilibrium is locally asymptotically stable, but convergence is not exponential: the displacement decays at a rate proportional to \(t^{-1/2}\) for large \(t\). Reversing the sign, \(\dot{N}=+c(N-K)^3\), also gives a zero Jacobian eigenvalue, but nonzero displacements move away from \(K\); that equilibrium is unstable. These two models demonstrate why a zero eigenvalue cannot settle the question.

Practical Checks, Alternatives, and Limitations

A reliable calculation begins by confirming that the proposed state really is an equilibrium, then evaluating the Jacobian there. The eigenvalues must be interpreted through their real parts, with units consistent with the model’s time unit. In cell-count models, Jacobian eigenvalues typically have units of inverse time.

Check the equilibrium first. If \(f(x^*)\ne0\), the Jacobian there does not classify an equilibrium because there is no equilibrium at that state.
Check every eigenvalue. One negative eigenvalue does not make a multidimensional equilibrium stable if another eigenvalue is positive.
Do not ignore small real parts. A real part close to zero can indicate slow dynamics or sensitivity to parameter uncertainty. Rounding it to zero can change whether the result is decisive or inconclusive.
Distinguish local from global behavior. A local result applies near the equilibrium. It does not rule out other equilibria or describe all large perturbations.
Check model assumptions. The test here concerns a smooth, autonomous model with fixed parameters. It does not by itself classify a system with explicitly time-varying rates.

For an exactly linear system, eigenvalues give a complete account of stability. In a nonlinear model, numerical trajectories can help reveal behavior over a chosen range of initial conditions, but a simulation is not a proof of local stability: it samples only particular starting points and times. In one dimension, a phase-line sign check can clarify whether the vector field points toward or away from an equilibrium. When the Jacobian has zero-real-part eigenvalues, nonlinear analysis is required; the cubic example above shows how direct examination of higher-order terms can resolve a case. Each approach has limits: simulations depend on numerical choices, phase lines are simplest in one dimension, and linearization is inconclusive at nonhyperbolic equilibria.

Common pitfall: “All eigenvalues are nonpositive, so the equilibrium is stable.” This conclusion is not justified if any eigenvalue has zero real part. The correct procedure is to classify the equilibrium as nonhyperbolic and examine the nonlinear terms or use another appropriate stability argument. Conversely, a zero eigenvalue alone does not establish instability.

Key takeaway: If every Jacobian eigenvalue has negative real part, the equilibrium is locally asymptotically stable; if any has positive real part, it is unstable. If an eigenvalue has zero real part, the linearization test is inconclusive and further analysis is needed.

Check Your Understanding

Try each question, then compare your reasoning with the solutions.

A linear one-compartment turnover model has \(\dot{N}=q-kN\), with \(q=90\) cells/day and \(k=0.3\) day\(^{-1}\). Find its equilibrium and classify it.
A two-state model has Jacobian eigenvalues \(-0.5\) day\(^{-1}\) and \(+0.1\) day\(^{-1}\). What does the eigenvalue test conclude?
What is the stability conclusion from a Jacobian eigenvalue of zero alone?
For \(\dot{N}=rN(1-N/K)\) with \(r>0\), what is the eigenvalue at \(N=K\), and what does its sign imply?

Solutions. (1) Set \(q-kN^*=0\), giving \(N^*=90/0.3=300\) cells. The Jacobian is \(f'(N^*)=-k=-0.3\) day\(^{-1}\), so the equilibrium is locally asymptotically stable. (2) The positive eigenvalue means at least one perturbation mode grows; the equilibrium is unstable. (3) The equilibrium is nonhyperbolic, and the Jacobian alone is inconclusive; nonlinear terms must be examined. (4) \(f'(K)=r(1-2)=-r<0\), so the equilibrium at carrying capacity is locally asymptotically stable.