Long-Run Limits of Logistic Populations

Why the Long-Run Population Matters

A logistic model describes population growth when growth is limited by available resources. Its carrying capacity is the population level the model associates with that limit. The explicit solution lets us ask a precise question: as time continues to increase, what value does the population approach?

For positive growth rate, the answer depends on the carrying capacity, not on whether the population starts below or above it. A population below capacity approaches it from below; one above capacity approaches it from above. A population that begins exactly at capacity stays there.

The Logistic Model and Its Solution

Let \(P(t)\) be a population at time \(t\), let \(r\) be its growth rate, and let \(K\) be its carrying capacity. In this tutorial, assume \(r>0\), \(K>0\), and \(t\geq 0\). The logistic differential equation is

For a positive initial population \(P(0)=P_0>0\), the logistic solution is

The constant \(A\) is determined by the initial condition. Substituting \(t=0\) gives \(e^0=1\), so

Thus the formula has the required initial population. The growth rate \(r\) controls the exponential factor, while the constant \(A\) records how the initial population compares with \(K\).

Why the Exponential Factor Disappears

Because \(r>0\), increasing \(t\) makes \(-rt\) decrease without bound. Consequently, \(e^{-rt}\) approaches zero. The denominator in the logistic formula therefore approaches \(1+A\cdot 0=1\), leaving \(K\) in the numerator.

Here is a justification that also checks the formula is defined for all \(t\geq0\). Since \(A=K/P_0-1\), we have \(A>-1\). If \(A\geq0\), then \(1+A e^{-rt}\geq1>0\). If \(-1<A<0\), then \(0<e^{-rt}\leq1\), and multiplication by the negative number \(A\) gives \(A e^{-rt}\geq A\). Thus \(1+A e^{-rt}\geq1+A=K/P_0>0\). The denominator never vanishes on the time interval considered.

The side from which the population approaches \(K\) can be seen by subtracting \(K\) from the solution:

The denominator is positive, and \(K e^{-rt}>0\). Therefore the sign of \(P(t)-K\) is the opposite of the sign of \(A\). When \(P_0<K\), \(A>0\), so \(P(t)<K\). When \(P_0>K\), \(-1<A<0\), so \(P(t)>K\). When \(P_0=K\), \(A=0\), so \(P(t)=K\) at every time. This establishes the approach from below, the approach from above, and the constant case directly from the solution.

Worked Example: Starting Below Carrying Capacity

Worked Example: Starting Above Carrying Capacity

Worked Example: Starting at Carrying Capacity

A Zero Initial Population

The formula with \(A=(K-P_0)/P_0\) requires \(P_0>0\), because it divides by \(P_0\). If the model allows \(P_0=0\), handle that case separately. The constant function \(P(t)=0\) satisfies the differential equation: its derivative is zero, and \(r(0)(1-0/K)=0\). Its long-run limit is therefore \(0\), not \(K\). This is why the positive-initial-population condition matters when stating the theorem.

A Common Sign Mistake

For a population above capacity, it is easy to reverse the subtraction in \(A\). For example, with \(K=800\) and \(P_0=1200\), an incorrect calculation is

That sign would produce \(P(0)=800/(1+1/3)=600\), which is not the specified \(1200\). The correct constant is \(A=(K-P_0)/P_0=-1/3\), giving \(P(0)=800/(1-1/3)=1200\). Checking the initial value is a quick way to catch this error before drawing a conclusion about the population.

What the Limit Does—and Does Not—Say

The limit identifies the population value the solution approaches over long times. It does not say that a positive initial population instantly becomes \(K\), nor does it say that the population equals \(K\) at some finite time. In the formula, \(e^{-rt}\) is positive at every finite \(t\), so when \(A\neq0\), the difference \(P(t)-K\) remains nonzero at every finite time. It tends to zero as time increases.

The conclusion also uses the condition \(r>0\). If the growth-rate parameter were not positive, the exponential factor would not have the same limiting behavior, and this argument would not establish the stated result. For the model assumptions here, the three positive-initial-population cases can be summarized as follows:

The main calculation is the same in each positive case: use \(r>0\) to make the exponential term tend to zero, then evaluate the resulting quotient. The initial population determines the value and sign of \(A\), which tells whether the solution starts below, at, or above capacity.