Fixed points and stability

Begin with two nearby starts

Supply a dimensionless state u that changes in time t, measured in seconds. Its exact teaching rule is du/dt = (u − u³)/T, with the positive time scale T = 1 s. The symbol du/dt means the instantaneous rate of change; it has units s⁻¹. A positive rate increases u and a negative rate decreases it. This invented model is not a measured apparatus.

At u = 0.5 the rate is (0.5 − 0.125)/(1 s) = 0.375 s⁻¹, so the state increases. At u = 0.9 it is 0.171 s⁻¹, also positive. Both starts move toward u = 1, where the rate is zero. Starting exactly at u = 0 also gives zero rate, yet a small positive start near zero moves away. “Zero rate here” and “nearby starts stay close” ask different questions.

Algebra, functions, graphs, and logarithms prepares the substitutions and graph reading. Follow the signs first if derivatives are new. Derivatives, Taylor series, and local approximation and Ordinary differential equations provide an optional return path for the last two sections; the needed rate and slope ideas are introduced here.

Find the constant solutions

Write f(u) = (u − u³)/T, so the rule is du/dt = f(u). It is autonomous: its rate depends on the current state, with no separate time argument. An equilibrium, also called a fixed point in this continuous-time setting, is a state u* for which the constant function u(t) = u* is a solution.

Set f(u*) = 0. Since T is positive, this means u*(1 − u*²) = 0. Therefore u* = −1, 0 or 1. These are all the equilibria in the allowed real state domain. The asterisk marks an equilibrium value; it is not multiplication.

The polynomial rate is continuously differentiable, which gives local existence and uniqueness for each finite initial state. In this model a solution cannot cross an equilibrium and continue through it: uniqueness would conflict with the constant solution through that same state. We will use this property together with the signs. Smoothness alone would not guarantee existence for all future time in every differential equation.

Read the rate graph and phase line

The upper graph plots f(u) against u. Its horizontal coordinate is state, not time. Where the curve is above its internal zero line, the state increases; below that line, it decreases. Its three zero crossings give the equilibria. The lower phase line puts the same equilibria on one state axis and records only the direction of change between them.

Factorization gives the signs throughout four open intervals: positive for u < −1, negative for −1 < u < 0, positive for 0 < u < 1, and negative for u > 1. The lower arrows consequently point right, left, right and left. They point toward −1 and 1 and away from 0. Their equal drawn lengths encode direction only; the upper graph carries the rate magnitude.

The table gives exact selected substitutions. The complete curve comes from the supplied polynomial, not a fit through seven observed values. The drawing shows −1.5 ≤ u ≤ 1.5; this is a display window, while the model permits every finite real initial state.

Define what stays close

An equilibrium is stable if every chosen closeness tolerance can be protected for all future time by choosing a sufficiently close initial state. In symbols: for every ε > 0, there is a δ > 0 such that |u(0) − u*| < δ implies |u(t) − u*| < ε for all t ≥ 0. The bars denote absolute distance. Both ε and δ are dimensionless here.

It is asymptotically stable if it is stable and all sufficiently close starts additionally approach it as t tends to infinity. Stable alone means staying close, not necessarily approaching. An unstable equilibrium fails the staying-close condition. None of these definitions requires the exact equilibrium solution itself to leave.

These are local statements about initial perturbations in the stated deterministic model. They do not promise resistance to continuing external forcing, measurement error or a changed evolution rule.

Establish the two attracting equilibria

Near u* = 1, a start below 1 but above 0 increases, and a start above 1 decreases. It cannot cross 1, so its distance from 1 never increases. For any ε > 0, choosing δ = min(ε, 1/2) keeps every start with |u(0) − 1| < δ within distance ε for all future time. This neighborhood excludes the unstable equilibrium at zero.

Such a solution is monotone and bounded between its starting value and 1, so it has a limit. That limit must have zero rate: a nonzero continuous rate near a proposed limit would keep moving the state by a definite amount per second. The only equilibrium in this neighborhood is 1. Thus the solution approaches 1, establishing asymptotic stability.

The same reasoning with signs reversed establishes asymptotic stability at −1. More broadly, every positive initial state tends to 1, every negative one tends to −1, and the exact zero state stays zero. Solutions remain bounded between their starting state and the relevant equilibrium. Local existence and the continuation property on bounded closed state intervals therefore give solutions for all t ≥ 0 in this particular model.

An equilibrium can be unstable

At u* = 0, the exact initial value u(0) = 0 gives the constant zero solution. However, any positive start in (0, 1) increases toward 1, while a negative start in (−1, 0) decreases toward −1. These motions depart from zero even when the starting distance is very small.

To test the definition, fix ε = 1/2. Whatever δ > 0 is proposed, choose u(0) = min(δ/2, 1/4). This is positive and closer to zero than δ. Because it approaches 1, it eventually exceeds 1/2. No positive δ protects the chosen ε for all future time, so zero is unstable.

This argument concerns neighboring initial states. It does not say that the exact zero solution spontaneously moves, and a finite display of several quiet samples would not establish the all-future-time stability condition.

Use a nonzero slope with conditions

The derivative f′(u*) is the local slope of the rate graph, not the rate f(u*) itself. It compares a small change in rate with a small change in the state. For a scalar autonomous equation with f continuously differentiable near an equilibrium, f′(u*) < 0 gives local asymptotic stability, and f′(u*) > 0 gives instability. This test is for continuous time, with the stated regularity and a nonzero slope.

The signs explain the test. A negative slope at a zero crossing makes f positive immediately to its left and negative immediately to its right: both directions point inward. A positive slope reverses those directions. Continuity of the derivative preserves this local pattern in a sufficiently small neighborhood; it is not a conclusion from distant samples.

For our polynomial, f′(u) = (1 − 3u²)/T. Its values at −1, 0 and 1 are −2 s⁻¹, +1 s⁻¹ and −2 s⁻¹. They agree with the phase-line arguments. A fuller linearization analysis is a later topic; the scalar slope rule here does not classify arbitrary systems with several state components.

When the slope is zero

Now separate two other supplied scalar models: du/dt = −u³/T and du/dt = +u³/T, still with T = 1 s. Both have the equilibrium u* = 0 and derivative zero there. In the minus model, negative states increase and positive states decrease toward zero. In the plus model, the two signs point away from zero. The zero derivative alone decides neither result.

For the minus model, distance from zero decreases and the limit is zero, so the equilibrium is asymptotically stable. For the plus model, arbitrarily small nonzero starts grow away and leave a fixed small neighborhood, so it is unstable. The direction reasoning examines the actual nonlinear terms rather than giving zero slope a stability label.

Keep the equation, allowed states, time direction and meaning of perturbation with each conclusion. Our graph explains one continuous scalar model; the small comparison explains a limit of one test. Numerical time steps, several-component systems and continuously driven disturbances require their own analysis.

Worked examples

Try it yourself

1. The main model has f(0) = 0 and the solution starting exactly at zero stays zero. Does that make the equilibrium stable?

   Show the answer and reasoning

   No. Zero is an unstable equilibrium.

   Stability tests all sufficiently close initial states for all future time. A positive start as small as desired in (0, 1) eventually exceeds the distance 1/2 from zero, while the exact zero solution remains constant.

2. In the main model, a start is u = −1.5. Find its local rate. Does an increasing state necessarily move away from the equilibrium at −1?

   Show the answer and reasoning

   Its rate is +1.875 s⁻¹. It increases toward −1.

   u³ = −3.375, so u − u³ = 1.875. Increasing a state below −1 reduces its distance from −1. Coordinate direction and distance from a selected equilibrium are different quantities.

3. For the main model near u* = −1, choose δ for a tolerance ε = 0.1. What happens to starts with |u(0) + 1| < δ?

   Show the answer and reasoning

   Choose δ = 0.1. All such starts remain within distance 0.1 from −1 for all t ≥ 0 and approach −1.

   The choice min(ε, 1/2) is 0.1 and keeps starts in (−1.1, −0.9), away from zero. Directions point toward −1 on both sides and uniqueness prevents crossing it, so distance does not increase. The only possible limiting equilibrium there is −1.

4. The lower arrow lengths are equal. Do they say that all states change equally fast, or that a state reaches the next equilibrium in the same elapsed time?

   Show the answer and reasoning

   Neither. They indicate the sign of change only.

   The lower axis is state u, not time. Rate magnitude is the upper ordinate f(u) in s⁻¹. For example, the upper table gives 0.375 s⁻¹ at u = 0.5 and 1.875 s⁻¹ at u = −1.5, although their phase-line directions are both right.

5. A scalar continuously differentiable rate has f(0) = 0 and f′(0) = 0. Must the equilibrium be stable? Use the two separate cubic models.

   Show the answer and reasoning

   No. The minus cubic is asymptotically stable and the plus cubic is unstable at zero.

   Both derivatives vanish at zero, but their nonlinear signs point toward or away from it. The zero slope leaves the derivative test inconclusive; the actual neighboring rate signs settle these particular cases.

6. Supply a separate rule du/dt = 0 for every real u, with all solutions constant. Is u* = 0 stable? Is it asymptotically stable?

   Show the answer and reasoning

   It is stable but not asymptotically stable.

   For any ε > 0 choose δ = ε. Every start within δ stays at the same distance forever. Any arbitrarily close nonzero start stays at its own value rather than approaching zero, so the required convergence fails. This separate zero-rate model illustrates the distinction between the two definitions.

What to carry forward

• An equilibrium is a constant solution; stability asks what happens to nearby initial states.
• Stable means sufficiently close starts stay close for all future time; asymptotically stable also requires convergence.
• For the supplied cubic rule, phase directions give asymptotically stable equilibria at −1 and 1 and an unstable equilibrium at 0.
• The nonzero-slope test requires a scalar continuously differentiable continuous-time rate; zero slope is inconclusive.
• Keep model domains and units visible: a rate-versus-state graph is not a time trajectory, and direction arrows do not encode rate magnitude.