Why Equilibria Can Behave Differently
A logistic differential equation has equilibrium populations at which the rate of change is zero. But being an equilibrium does not, by itself, tell us whether nearby populations will move toward it or away from it. Stability answers that question.
The key is to compare the sign of the rate just below an equilibrium with its sign just above. A positive rate means the population increases; a negative rate means it decreases. If both directions point toward the equilibrium, nearby solutions move toward it. If they point away, nearby solutions move away.
Equilibria and Stability
Consider the logistic equation
where \(P(t)\) is the population at time \(t\), \(r\) is the growth-rate parameter, and \(K\) is the carrying capacity. In the usual population setting, \(r>0\) and \(K>0\). The right-hand side is the rate function, which we can write as \(f(P)=rP(1-P/K)\).
Setting each factor of the logistic rate equal to zero gives the two equilibria:
An equilibrium is stable if solutions that start sufficiently close to it remain close. It is asymptotically stable if it is stable and nearby solutions also approach it as time increases. An equilibrium is unstable if arbitrarily nearby initial values can lead to solutions that move away from it.
For populations, we usually consider \(P\ge 0\). Stability at \(P=0\) can also be discussed on the full real line; in this lesson, the sign analysis will make clear what happens from both sides.
Sign Analysis for Positive Growth
Assume \(r>0\) and \(K>0\). Since \(r\) is positive, it does not change the sign of the rate. The factors \(P\) and \(1-P/K\) determine that sign.
| Population range | Sign of \(P\) | Sign of \(1-P/K\) | Sign of \(dP/dt\) | Direction |
|---|---|---|---|---|
| \(P<0\) | Negative | Positive | Negative | Decreases |
| \(0<P<K\) | Positive | Positive | Positive | Increases |
| \(P>K\) | Positive | Negative | Negative | Decreases |
Near \(P=0\), a population just above zero increases, so it moves away from zero. On the full real line, a value just below zero decreases, also moving away. Thus \(P=0\) is unstable when \(r>0\). In the population domain \(P\ge 0\), the relevant direction is the one-sided motion from above: small positive populations grow away from zero.
Near \(P=K\), a value just below \(K\) increases toward \(K\), while a value just above \(K\) decreases toward \(K\). The rate points inward on both sides, so \(P=K\) is asymptotically stable.
Worked Example: Classifying Both Equilibria
Worked Example: Classifying Both Equilibria
Consider \( \frac{dP}{dt}=0.4P(1-P/100) \), where \(P\) is measured in thousands of organisms and \(t\) is measured in days. Find the equilibria, determine the direction of motion on either side, and classify each equilibrium.
First set the rate equal to zero:
The factors give \(P=0\) or \(1-P/100=0\), so \(P=100\). The equilibria are \(0\) and \(100\) thousand organisms.
For a value between the equilibria, take \(P=20\). Substitution gives
The rate is positive, so a solution at \(P=20\) increases. For a value above the carrying capacity, take \(P=130\):
The rate is negative, so a solution at \(P=130\) decreases. These calculations confirm that solutions between \(0\) and \(100\) move upward, and solutions above \(100\) move downward.
Therefore, small positive populations move away from \(0\), making \(P=0\) unstable. Values on either side of \(100\) move toward it, making \(P=100\) asymptotically stable. The units of the rates are thousands of organisms per day.
Why the Sign Argument Establishes Attraction
A sign chart indicates the direction of motion. To justify that the inward motion really leads to the equilibrium, consider a positive initial population \(P(0)=P_0\) with \(0<P_0<K\). As long as \(P\) is between \(0\) and \(K\), the rate is positive, so \(P(t)\) increases. It cannot cross the equilibrium \(K\): if it reached \(K\), the solution from that time onward would have the same initial value as the constant solution \(P(t)=K\). Uniqueness for this differential equation would force the two solutions to agree, so the original solution cannot pass through \(K\).
Consequently, \(P(t)\) increases while remaining below \(K\), and therefore has a limit \(L\le K\). If \(L<K\), then \(f(L)=rL(1-L/K)>0\), because \(L\ge P_0>0\) and \(L<K\). By continuity of \(f\), the rate would remain bounded below by some positive number once \(P(t)\) is sufficiently close to \(L\). The population would then continue increasing at a rate bounded away from zero, which contradicts its convergence to the finite limit \(L\). Thus \(L=K\).
If \(P_0>K\), the rate is negative, so \(P(t)\) decreases. Uniqueness prevents it from crossing \(K\), and it remains bounded below by \(K\). It therefore has a limit \(L\ge K\). If \(L>K\), then \(f(L)<0\); by continuity, the rate would eventually stay below a negative constant, contradicting convergence to \(L\). Therefore \(L=K\) from above as well.
A Direction-Arrow Proof of Stability
The sign analysis can be organized as a phase line: place the equilibria on a line of population values and draw an arrow in each interval to show the direction of motion. For \(r>0\), the directions on the population domain are
The arrow between \(0\) and \(K\) points toward \(K\); the arrow to the right of \(K\) also points toward \(K\). This makes the inward behavior at \(K\) visible at a glance. At zero, the arrow to the right points away from the equilibrium, which is why zero is unstable for nonnegative populations.
There is also a precise way to see the stability distinction. Near \(K\), solutions move toward \(K\) and do not cross it, so their distance from \(K\) does not increase. Starting sufficiently close to \(K\) therefore keeps a solution close, and the argument above shows it approaches \(K\). By contrast, any positive initial value below \(K\) eventually moves toward \(K\), not toward zero. Thus even a very small positive perturbation from zero does not remain close to zero for all future time; zero is unstable.
Worked Example: Checking Motion Close to Carrying Capacity
Worked Example: Checking Motion Close to Carrying Capacity
Suppose \( \frac{dP}{dt}=0.2P(1-P/50) \), with \(P\) measured in hundreds of cells. Determine the motion for populations of \(48\) and \(52\), and use the signs to classify the carrying-capacity equilibrium.
The equilibria satisfy
so they are \(P=0\) and \(P=50\). At \(P=48\),
This positive rate means the population increases toward \(50\). At \(P=52\),
This negative rate means the population decreases toward \(50\). Substituting the two populations into the rate function verifies the signs on both sides, so \(P=50\) is asymptotically stable.
When the Growth-Rate Parameter Is Negative
The classification depends on the parameters. If \(K>0\) but \(r<0\), the factor \(r\) reverses every nonzero sign in the sign chart. For \(0<P<K\), the rate is negative, so populations decrease toward zero. For \(P>K\), the rate is positive, so populations increase away from \(K\). In this case, \(P=0\) is asymptotically stable for positive populations, and \(P=K\) is unstable.
Worked Example: Reversed Stability When Growth Is Negative
Consider \( \frac{dP}{dt}=-0.3P(1-P/40) \), where \(P\ge 0\). Find the equilibria and check the direction of motion below and above \(40\).
The rate is zero when \(P=0\) or \(1-P/40=0\), so the equilibria are \(0\) and \(40\). At \(P=20\),
The rate is negative, so the population decreases toward zero. At \(P=50\),
The rate is positive, so the population increases away from \(40\). These substitutions confirm that zero is attracting for positive populations and \(40\) is repelling. The negative parameter reverses the usual stability pattern.
Common Mistake: Assuming Above Capacity Means Growth
A frequent error is to assume that a population above its carrying capacity must still increase because it is large. The logistic rate depends on both factors, and above \(K\) the factor \(1-P/K\) is negative. For example, with \(r=0.2\), \(K=50\), and \(P=60\), the correct calculation is
The negative rate means the population decreases. The sign of the factor \(1-P/K\) must be checked; it is not positive when \(P>K\).
Why Stability Matters
Equilibria describe population levels at which the model predicts no instantaneous change. Stability adds information about what happens after a small disturbance. In the usual logistic model, the carrying capacity attracts nearby positive populations, while the zero equilibrium does not. This distinction helps interpret the model: an equilibrium can be a possible constant solution without being a population level that nearby solutions approach.
For sign analysis, use a consistent sequence:
Set the rate function equal to zero.
Choose a population value on each side of the equilibria, or inspect the factors directly.
A positive rate means increasing population; a negative rate means decreasing population.
Motion toward an equilibrium indicates stability; motion away indicates instability.
Check Your Understanding
Use sign analysis to answer each question. Assume \(K>0\) unless a question specifies otherwise.
- For \( \frac{dP}{dt}=0.5P(1-P/80) \), identify both equilibria and classify them.
- For \(r>0\), what is the sign of the logistic rate when \(P>K\), and what direction does a solution move?
- For \(r>0\), explain why a small positive population moves away from the zero equilibrium.
- How does the stability of the equilibria change when \(r<0\)?
- What does the logistic equation predict about equilibria when \(r=0\)?
