Why Logistic Models Have Equilibria
A population that grows rapidly at first may eventually slow as resources become limited. A logistic differential equation represents this pattern: growth depends both on the current population and on how far it is below a limiting level.
Let \(P(t)\) denote a population at time \(t\). The standard logistic model is
Here \(k\) is the growth-rate parameter and \(K\) is the carrying capacity. The population is usually assumed to be nonnegative. If \(P\) is measured in individuals and \(t\) in years, then \(k\) has units of inverse years, while \(K\) has units of individuals. The right-hand side has units of individuals per year, as a population rate should.
For the logistic equation, an equilibrium value makes at least one factor on the right-hand side zero. Thus \(P=0\) and \(P=K\) are the equilibrium values. These give the constant solutions \(P(t)=0\) and \(P(t)=K\).
How the Initial Population Determines Growth or Decline
For a positive population, \(kP\) is positive, so the sign of \(\frac{dP}{dt}\) is determined by \(1-\frac{P}{K}\). Comparing the initial population \(P(0)\) with \(K\) therefore gives the direction of change.
| Initial population | Sign of \(\frac{dP}{dt}\) | Model prediction |
|---|---|---|
| \(0<P(0)<K\) | Positive | Population increases |
| \(P(0)=K\) | Zero | Population remains constant |
| \(P(0)>K\) | Negative | Population decreases |
| \(P(0)=0\) | Zero | Population remains at zero |
This sign analysis distinguishes an equilibrium value from a carrying capacity’s role in a particular solution. If the initial population is below \(K\), it grows toward \(K\); if it is above \(K\), it declines toward \(K\). In either case, \(K\) is an equilibrium, but a solution starting elsewhere is not constant.
Deriving the Solution and Its Long-Term Limit
The sign chart gives the direction of change, but an explicit solution makes the long-term conclusion precise. For a positive initial population, define \(Q(t)=\frac{K}{P(t)}-1\). This substitution is valid when \(P(t)>0\). Differentiating and using the logistic equation gives
Thus \(Q\) satisfies an exponential decay equation. Its solution is \(Q(t)=Q(0)e^{-kt}\), where \(Q(0)=\frac{K}{P(0)}-1\). Substituting back and writing \(P_0=P(0)>0\) gives the logistic solution:
This formula is defined for all \(t\geq 0\) when \(P_0>0\): if \(P_0\leq K\), the coefficient of \(e^{-kt}\) is nonnegative; if \(P_0>K\), it is between \(-1\) and \(0\), so the denominator remains positive. As \(t\) increases, \(e^{-kt}\) approaches zero. Consequently,
This also shows why \(K\) is approached rather than reached at a finite time. For positive finite \(t\), \(e^{-kt}>0\); the exponential term becomes zero only in the limit as \(t\) tends to infinity. If \(P_0=K\), the coefficient is already zero and the solution is the constant equilibrium \(P(t)=K\).
Worked Examples
Worked Example: A Population Below Carrying Capacity
Suppose \(\frac{dP}{dt}=0.4P\left(1-\frac{P}{100}\right)\) and \(P(0)=20\). Find the equilibria, determine whether the population initially grows or declines, and state its long-term limit.
Step 1: Find the equilibria. Set the right-hand side equal to zero:
Step 2: Determine the initial direction. Evaluate the rate at \(P=20\):
The rate is positive, so the population initially increases. The carrying capacity is \(K=100\), and \(20<100\), in agreement with the sign analysis.
Step 3: Write the solution and limit. The coefficient in the solution is \(\frac{100}{20}-1=4\), so
At \(t=0\), this gives \(P(0)=100/(1+4)=20\). As \(t\to\infty\), \(e^{-0.4t}\to0\), so \(P(t)\to100\).
Check: Differentiating the solution gives \(\frac{dP}{dt}=\frac{160e^{-0.4t}}{(1+4e^{-0.4t})^2}\). Substitution into the model gives \(0.4P(1-P/100)=\frac{160e^{-0.4t}}{(1+4e^{-0.4t})^2}\), so the solution satisfies the differential equation.
Worked Example: A Population Above Carrying Capacity
Suppose \(\frac{dP}{dt}=0.2P\left(1-\frac{P}{120}\right)\) and \(P(0)=150\). Determine the initial direction of change and the long-term behavior.
Step 1: Identify the equilibrium values.
Step 2: Evaluate the initial rate.
The initial rate is negative, so the population declines. Since \(150>120\), this is the expected direction: the population is above the carrying capacity \(K=120\).
Step 3: Find the solution and limit. Here \(\frac{K}{P_0}-1=\frac{120}{150}-1=-0.2\). Therefore,
At \(t=0\), the denominator is \(1-0.2=0.8\), giving \(P(0)=120/0.8=150\). As \(t\to\infty\), the denominator approaches \(1\), so \(P(t)\to120\) from above.
Check: Let \(D=1-0.2e^{-0.2t}\). Differentiation gives \(D'=0.04e^{-0.2t}\) and \(P'=-4.8e^{-0.2t}/D^2\). Substitution into the model gives \(0.2(120/D)(1-1/D)=-4.8e^{-0.2t}/D^2\), matching the derivative.
Worked Example: Starting at an Equilibrium
Suppose \(\frac{dP}{dt}=0.5P\left(1-\frac{P}{80}\right)\) and \(P(0)=80\). Decide whether the population changes and verify the result.
Step 1: Identify the carrying capacity. The equation has \(K=80\), so \(P=80\) is an equilibrium value. Setting \(P=80\) in the rate gives
The population has zero rate of change at this value. The constant function \(P(t)=80\) is therefore the solution for this initial condition, and its long-term limit is \(80\).
Check: The explicit-solution coefficient is \(\frac{80}{80}-1=0\). Thus \(P(t)=80/(1+0e^{-0.5t})=80\), which has derivative zero and satisfies the model.
Equilibrium Stability and a Common Pitfall
The sign of the rate on either side of an equilibrium shows what nearby solutions do. For \(0<P<K\), the rate is positive, so populations move upward toward \(K\). For \(P>K\), the rate is negative, so populations move downward toward \(K\). Thus positive solutions on either side move toward \(K\): the carrying-capacity equilibrium is stable for positive populations. At \(P=0\), a small positive population has positive rate and moves away from zero, so the zero equilibrium is unstable in the population model.
In a contextual response, state what the parameters and limit mean in the situation. The carrying capacity is the population level toward which this model predicts positive populations will move, not a guarantee that real populations remain exactly there. The prediction depends on the logistic model’s assumptions, including constant parameters and the absence of additional effects not represented by the equation.
Check Your Understanding
For each question, explain your reasoning using the logistic equation and the initial value.
- For \(P'=0.3P(1-P/60)\), find both equilibrium values.
- For the equation in question 1 with \(P(0)=15\), is the population initially increasing or decreasing? What is its long-term limit?
- For \(P'=0.1P(1-P/200)\) with \(P(0)=250\), calculate the initial rate and interpret its sign.
- Why is \(P(t)=K\) a solution when the initial population equals the carrying capacity?
- For a positive initial population, what feature of the explicit solution explains why the population approaches \(K\) as \(t\to\infty\)?
