Logistic Growth and Its Equilibrium

From Density-Limited Per-Capita Growth to a Population Equation

Exponential growth describes a population whose per-capita net growth rate remains constant. In many cell-population settings, however, the per-capita rate can decline as the population grows. Limited space, nutrients, or other resources are possible biological reasons. A simple way to represent this effect is to make the per-capita rate decrease linearly with population size. This assumption leads to the logistic growth equation.

Let $N(t)$ denote the number of cells at time $t$. The model assumes the per-capita net growth rate is highest at low density and falls to zero at a characteristic population size $K$. Here, “per-capita net growth rate” means the rate of change in population size divided by the current population size; it summarizes the combined effect of cell production and cell loss, not either process separately.

Definition: The logistic model is $$ \frac{dN}{dt}=rN\left(1-\frac{N}{K}\right), $$ where $r>0$ is the low-density per-capita net growth rate, with units of inverse time, and $K>0$ is the carrying-capacity parameter, with units of cells. Equivalently, for $N>0$, the per-capita net growth rate is $r(1-N/K)$.

The equation follows by multiplying the per-capita rate by the population size: $dN/dt=N\,g(N)$, where $g(N)=r(1-N/K)$. At low density, $N/K$ is small and the per-capita rate is close to $r$. At $N=K$, it is zero. Above $K$, the model assigns a negative per-capita net growth rate.

Assumptions and Meaning of the Parameters

The equation is a compact model, not a complete description of every cell-population mechanism. Its basic assumptions are:

The population can be represented by a single state variable, $N(t)$, rather than by distinct subpopulations with different growth behavior.
Population change is continuous and deterministic at the scale of interest. Treating cell counts as continuous is an approximation when counts are small.
The environment and model parameters remain constant over the time interval considered. In particular, $r$ and $K$ do not vary with time.
The population is closed with respect to cell movement: no immigration or emigration is represented separately.
The per-capita net growth rate declines linearly with $N$, reaching zero at $K$.

The parameter $r$ describes the low-density limit of the per-capita net rate, not necessarily a cell-division rate. If cell loss occurs, $r$ is the net of production and loss in that limit. The parameter $K$ is the population size at which this model predicts zero net change. It is not necessarily a physical maximum, a directly measured resource capacity, or a count that the population can never exceed. Its interpretation depends on whether the assumptions and time scale are appropriate.

For $N>0$, $r$ and $K$ imply a useful direction of change: if $N<K$, then $dN/dt>0$; if $N>K$, then $dN/dt<0$; and if $N=K$, then $dN/dt=0$. These signs describe the model’s predictions on either side of its equilibrium. They do not establish that a biological system follows the model.

Equilibria and the Positive Equilibrium

An equilibrium is a population size at which the derivative is zero. Setting the logistic equation equal to zero gives two mathematical equilibria: $N=0$ and $N=K$. The positive equilibrium, when $r>0$ and $K>0$, is $K$. The zero equilibrium is also a solution: if there are no cells, this model has no source term that could create them.

Result: For the logistic equation with $r>0$ and $K>0$, the positive equilibrium is $N^*=K$. At that population size, the modeled per-capita net growth rate is $r(1-K/K)=0$, so the total net rate is also zero.

This result is specific to the chosen rate function. It says that the population size at which the model predicts zero net change equals the parameter $K$. It does not say that every observed population will settle at $K$ regardless of its environment or history.

Worked Example: Finding the Positive Equilibrium

Suppose a hypothetical cell population is described by $dN/dt=0.35N(1-N/(2.4\times10^6))$, with time measured in weeks and $N$ measured in cells. Find both equilibria and interpret the positive one.

At equilibrium, the derivative is zero: $$ 0=0.35N\left(1-\frac{N}{2.4\times10^6}\right). $$

A product is zero if at least one factor is zero. Thus $N=0$, or $1-N/(2.4\times10^6)=0$, which gives $N=2.4\times10^6$ cells. The positive equilibrium is therefore $2.4\times10^6$ cells.

At that size, the per-capita net rate is $0.35(1-1)=0$ per week. For example, at $N=1.8\times10^6$ cells, the per-capita net rate is $0.35(1-1.8/2.4)=0.0875$ per week, and the total rate is $0.0875(1.8\times10^6)=157{,}500$ cells per week. The positive rate below $K$ is consistent with the model’s prediction of growth toward the equilibrium region.

Solving the Logistic Equation

For a positive initial population $N(0)=N_0>0$, the logistic equation has the solution below. The derivation uses separation of variables and partial fractions; it makes explicit how the initial condition determines the trajectory.

\begin{aligned} \frac{dN}{dt}&=rN\left(1-\frac{N}{K}\right),\\ \frac{dN}{N(1-N/K)}&=r\,dt,\\ \left(\frac{1}{N}+\frac{1}{K-N}\right)dN&=r\,dt,\\ \ln\left(\frac{N}{K-N}\right)&=rt+C. \end{aligned}

The partial-fraction identity holds because $1/[N(1-N/K)]=K/[N(K-N)]=1/N+1/(K-N)$. Integrating $1/(K-N)$ gives $-\ln|K-N|$, so the combined logarithm is $\ln|N/(K-N)|$. For a trajectory with $0<N_0<K$, the ratio is positive; applying the initial condition gives the equivalent solution below. The expression also yields the solution for $N_0>K$, where the denominator remains positive for $t\geq0$.

N(t)=\frac{K}{1+\left(\frac{K-N_0}{N_0}\right)e^{-rt}},\qquad N_0>0.

To verify the initial condition, set $t=0$: the denominator becomes $1+(K-N_0)/N_0=K/N_0$, so $N(0)=N_0$. If $N_0=0$, the separated derivation does not apply because it divides by $N$; the original differential equation instead gives the constant solution $N(t)=0$. The formula also shows that $r$ has units of inverse time, making $rt$ dimensionless, while $(K-N_0)/N_0$ is dimensionless.

Worked Example: Predicting a Cell Count at a Specified Time

For a hypothetical population, let $r=0.20$ per day, $K=500{,}000$ cells, and $N_0=100{,}000$ cells. Calculate the model-predicted population after 5 days.

First, the initial-condition coefficient is $(K-N_0)/N_0=(500{,}000-100{,}000)/100{,}000=4$. Also, $rt=(0.20\ \text{day}^{-1})(5\ \text{days})=1$. Substituting into the solution: $$ N(5)=\frac{500{,}000}{1+4e^{-1}} =\frac{500{,}000}{1+4(0.367879)} =\frac{500{,}000}{2.471516} \approx 202{,}305\ \text{cells}. $$

The model therefore predicts about $202{,}300$ cells at day 5. The prediction exceeds the initial $100{,}000$ cells and remains below $K=500{,}000$ cells, consistent with positive growth below the carrying-capacity parameter. This is a model calculation, not a claim about an observed experiment.

Estimating and Interpreting the Capacity Parameter

The per-capita form can help interpret a rate measurement when $r$ is known. Rearranging $g(N)=r(1-N/K)$ gives $K=N/[1-g(N)/r]$, provided $g(N)\ne r$. This algebra does not establish that the logistic model is appropriate; it only gives the value of $K$ implied by that observation under the model.

Worked Example: Inferring the Implied Carrying Capacity

In a hypothetical system, the low-density per-capita net rate is $r=0.50$ per day. At a population of $400{,}000$ cells, the measured per-capita net rate is $0.20$ per day. Under the logistic assumption, calculate the implied $K$.

Substitute the values into $g(N)=r(1-N/K)$: $$ 0.20=0.50\left(1-\frac{400{,}000}{K}\right). $$

Divide both sides by $0.50$: $0.40=1-400{,}000/K$. Thus $400{,}000/K=0.60$, and $K=400{,}000/0.60\approx666{,}667$ cells. This is the capacity parameter implied by the specified rate data and logistic form. A single rate measurement does not independently verify that per-capita growth declines linearly with cell number.

Diagnostics, Alternatives, and Limitations

A useful diagnostic is to compare observed per-capita net rates with population size. Under the logistic model, $g(N)=r-(r/K)N$ is a straight line in $N$, with intercept $r$, slope $-r/K$, and zero crossing $K$. Systematic curvature, a zero crossing that changes across conditions, or rates that depend strongly on time or treatment may indicate that a constant-parameter logistic model is not an adequate description. Measurements can also be noisy, especially when rates are estimated from short time intervals, so an apparent departure from a line should be assessed in light of measurement uncertainty.

The logistic model differs from exponential growth in one central assumption: exponential growth holds the per-capita net rate constant, whereas the logistic model makes it decline with density. Exponential growth can be a reasonable approximation when $N$ is small relative to $K$, because then $1-N/K$ is close to 1. Other density-dependent growth laws use different relationships between density and per-capita rate. Their forms and consequences are distinct choices, rather than automatic corrections to logistic growth.

The model has important limits. A single $K$ cannot represent changing resources or treatment effects unless these are explicitly added. The linear decline in per-capita rate is a simplifying assumption, not a universal biological law. The deterministic equation also does not represent random changes in small populations, and the one-state description does not distinguish cell types or mechanisms. If fitted trajectories repeatedly overshoot, change rate, or show systematic residual patterns, the issue may be the model structure rather than an arithmetic error.

Common Pitfalls

Calling $K$ a guaranteed maximum. The model has a positive equilibrium at $K$, but it can start above $K$, in which case the predicted rate is negative. Interpret $K$ as the model’s zero-net-growth population size.
Confusing $r$ with a division rate. In this equation, $r$ is the low-density per-capita net rate. It does not separately identify cell division and cell loss.
Dropping the initial-condition coefficient. The solution is not generally $K/(1+e^{-rt})$. The factor $(K-N_0)/N_0$ is needed to make $N(0)=N_0$.
Using inconsistent units. The product $rt$ must be dimensionless. If $r$ is per day, time must be expressed in days.
Treating a fitted $K$ as proof of mechanism. A model parameter can summarize a pattern without proving which biological resource or process causes density dependence.

Check Your Understanding

Work each problem from the model definition, then compare with the solutions below.

For $r=0.10$ per hour and $K=80{,}000$ cells, write the per-capita net growth rate at $N=20{,}000$ cells. State whether the total population is increasing or decreasing.
Find both equilibria of $dN/dt=0.4N(1-N/300{,}000)$. Which one is the positive equilibrium?
For $K=600{,}000$ cells and $N_0=150{,}000$ cells, identify the coefficient multiplying $e^{-rt}$ in the logistic solution.
Explain why a population observed above an estimated $K$ does not by itself contradict the logistic model.

Solutions. 1. $g(20{,}000)=0.10(1-20{,}000/80{,}000)=0.075$ per hour. Since this is positive and $N>0$, $dN/dt=Ng(N)>0$, so the population is increasing. 2. Setting the right-hand side to zero gives $N=0$ or $1-N/300{,}000=0$, so the equilibria are 0 and 300,000 cells; the positive equilibrium is 300,000 cells. 3. The coefficient is $(K-N_0)/N_0=(600{,}000-150{,}000)/150{,}000=3$. 4. $K$ is the model’s zero-net-growth size, not a hard upper bound. If $N>K$, the model predicts a negative rate, so a population can be above $K$ while declining.