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Euler and Logistic Methods · Tutorial 588 of 999

From Limited Growth to the Logistic Equation

Translate a population’s limited-growth assumptions into the logistic equation and interpret each part of the model.

Advanced 10 min read

What You'll Learn

  • Express limited growth as a per-capita growth rate.
  • Explain why remaining capacity is represented by a fraction.
  • Derive the logistic equation from the population assumptions.
  • Interpret the growth-rate parameter and carrying capacity.
  • Check the model’s units and compare equivalent parameterizations.
  • Identify a common modeling error involving the capacity factor.

From Limited Growth to a Model

A population may grow rapidly when resources are plentiful, yet slow down as it approaches the amount its environment can support. To represent this idea with a differential equation, we must specify how the growth rate depends on both the current population and the remaining capacity.

The logistic equation is built from two assumptions: population growth is proportional to the population already present, and the per-capita growth rate is proportional to the fraction of capacity that remains available. Together, these assumptions produce a particular differential equation.

Modeling goal: Translate the stated assumptions into a rate rule for a population \(y(t)\), where \(t\) denotes time.

Define the Quantities and Assumptions

Let \(y(t)\) be the population at time \(t\). Let \(K\) be the environment’s carrying capacity: the population level used to represent the limit imposed by available resources. We assume \(K>0\). Population and capacity are measured in the same units, such as individuals.

When the population is below capacity, the amount of capacity not occupied is \(K-y\). But the phrase fraction of capacity still available refers to a ratio, not an amount. The available fraction is the unoccupied capacity divided by total capacity:

$$ \frac{K-y}{K}=1-\frac{y}{K}. $$

This fraction is dimensionless: the population units in the numerator cancel the population units in the denominator. For \(0\leq y\leq K\), it ranges from \(1\), when no capacity is occupied, to \(0\), when the population equals capacity.

We also assume that the per-capita growth rate decreases in direct proportion to this available fraction. Per-capita growth means the population’s rate of change divided by the current population, when \(y>0\). Let \(r>0\) be the maximum per-capita growth rate, in units of reciprocal time. Under the model, this is the per-capita rate approached when the population is small relative to capacity.

Definition: The logistic model assumes that for \(y>0\), the per-capita growth rate is \(r\) times the fraction of capacity remaining: $$ \frac{1}{y}\frac{dy}{dt}=r\left(1-\frac{y}{K}\right), $$ where \(r>0\) is the low-density per-capita growth-rate parameter and \(K>0\) is the carrying capacity.

Deriving the Logistic Equation

The assumptions specify the population’s per-capita rate. To obtain the rate of change of the whole population, multiply that per-capita rate by the population \(y\). This gives the logistic differential equation.

$$ \frac{dy}{dt} =y\left[r\left(1-\frac{y}{K}\right)\right] =ry\left(1-\frac{y}{K}\right). $$

The same derivation can be written as a sequence of modeling steps. Each step follows from one stated assumption or from the definition of per-capita growth.

1
Find the available fraction.
Capacity not occupied is \(K-y\), so the available fraction is \((K-y)/K=1-y/K\).
2
Write the per-capita rate.
Proportionality to the available fraction gives \(r(1-y/K)\).
3
Convert to total population growth.
Multiply the per-capita rate by \(y\), obtaining \(dy/dt=ry(1-y/K)\).

This derivation explains why the model has both factors \(y\) and \(1-y/K\). The factor \(y\) represents the assumption that more individuals produce more potential growth. The second factor reduces the per-capita rate as the available fraction shrinks.

Units and Parameter Meaning

Suppose \(y\) and \(K\) are measured in individuals and \(t\) in years. Then \(dy/dt\) has units of individuals per year. The ratio \(y/K\) has no units, so \(1-y/K\) is dimensionless. Therefore, \(r\) must have units of inverse years, and the right side has the required units:

$$ \left(\frac{1}{\text{year}}\right) (\text{individuals}) (1) = \frac{\text{individuals}}{\text{year}}. $$

The parameter \(r\) is not the population’s actual per-capita growth rate at every population size. The model makes that rate depend on \(y\): it is \(r(1-y/K)\). The capacity \(K\) sets the scale against which the current population is compared. A rate rule that omits the division by \(K\) generally has different units and does not use the fraction of capacity remaining.

Equivalent form: Expanding the logistic equation gives $$ \frac{dy}{dt}=ry-\frac{r}{K}y^2. $$ The coefficient of \(y^2\) is \(r/K\), not an independent carrying-capacity value. This form is algebraically equivalent to \(ry(1-y/K)\).

Worked Example: Modeling a Rabbit Population

Worked Example: Modeling a Rabbit Population

A habitat supports about \(1200\) rabbits. Suppose the low-density per-capita growth rate is \(0.4\) per year. Write the model, then find the predicted instantaneous growth rate when the population is \(300\) rabbits.

Step 1: Identify the parameters. The capacity is \(K=1200\) rabbits, and \(r=0.4\) per year. The model is

$$ \frac{dy}{dt}=0.4y\left(1-\frac{y}{1200}\right). $$

Step 2: Find the available fraction at \(y=300\).

$$ 1-\frac{300}{1200} =1-\frac{1}{4} =\frac{3}{4} =0.75. $$

Step 3: Find the per-capita rate. It is \(r\) times the available fraction:

$$ 0.4(0.75)=0.30\text{ per year}. $$

Step 4: Convert to total population growth. Multiply by the \(300\) rabbits present:

$$ \frac{dy}{dt} =300(0.30) =90\text{ rabbits per year}. $$

Check: Substitution directly into the model gives \(0.4(300)(1-300/1200)=120(0.75)=90\) rabbits per year, matching the result. The units are also consistent: \((1/\text{year})(\text{rabbits})(1)=\text{rabbits per year}\).

Worked Example: Turning a Verbal Rule into an Equation

Worked Example: Turning a Verbal Rule into an Equation

A microorganism population has capacity \(5000\) organisms. Its per-capita growth rate is \(0.6\) per day at low density, and the model assumes that this rate falls in proportion to the fraction of capacity remaining. Find the equation and the predicted population growth rate at \(y=1500\).

Step 1: Translate the low-density rate. The parameter is \(r=0.6\) per day. The capacity is \(K=5000\) organisms. Thus

$$ \frac{dy}{dt}=0.6y\left(1-\frac{y}{5000}\right). $$

Step 2: Compute the remaining fraction at \(y=1500\).

$$ 1-\frac{1500}{5000} =1-\frac{3}{10} =\frac{7}{10} =0.7. $$

Step 3: Compute per-capita growth.

$$ 0.6(0.7)=0.42\text{ per day}. $$

Step 4: Multiply by the current population.

$$ \frac{dy}{dt} =1500(0.42) =630\text{ organisms per day}. $$

Check: The equation gives \(0.6(1500)(0.7)=900(0.7)=630\) organisms per day. The per-capita rate has units \(1/\text{day}\), and multiplying by organisms produces organisms per day.

Worked Example: Converting an Unnormalized Coefficient

Worked Example: Converting an Unnormalized Coefficient

Suppose a model is written as \(dy/dt=a y(K-y)\), where \(K=500\) individuals and \(a=0.002\) per individual per week. Rewrite it in the standard logistic form and evaluate the predicted rate at \(y=100\).

Step 1: Identify the units. Since \(a y(K-y)\) must have units of individuals per week, \(a\) has units \(1/(\text{individual}\cdot\text{week})\). The product \(aK\) therefore has units \(1/\text{week}\).

Step 2: Factor out \(K\). Since \(K-y=K(1-y/K)\),

$$ a y(K-y) = a yK\left(1-\frac{y}{K}\right) =(aK)y\left(1-\frac{y}{K}\right). $$

Thus the standard parameter is \(r=aK\). Numerically,

$$ r=(0.002)(500)=1\text{ per week}. $$

Step 3: Write the standard equation and evaluate it.

$$ \frac{dy}{dt} =1y\left(1-\frac{y}{500}\right), $$

so at \(y=100\),

$$ \frac{dy}{dt} =100\left(1-\frac{100}{500}\right) =100\left(1-\frac{1}{5}\right) =100\left(\frac{4}{5}\right) =80\text{ individuals per week}. $$

Check: The original form gives \(0.002(100)(500-100)=0.002(100)(400)=0.002(40000)=80\) individuals per week. Both forms agree, and \(r=1\) per week has the units required by the standard equation.

Why the Fraction Matters

A common mistake is to use the unoccupied amount \(K-y\) where the model calls for the unoccupied fraction \((K-y)/K\). For example, with \(K=500\), \(y=100\), and \(r=1\) per week, an incorrect calculation might be

$$ \frac{dy}{dt} \overset{\text{wrong}}{=}ry(K-y) =(1)(100)(400) =40000. $$

This is wrong for two reasons. First, it uses the amount \(400\) rather than the fraction \(400/500=0.8\). Second, \(r\) has units of inverse weeks, so \(ry(K-y)\) has units of individuals squared per week, not individuals per week.

The correction is to divide the unoccupied amount by total capacity:

$$ \frac{dy}{dt} =ry\frac{K-y}{K} =(1)(100)\frac{400}{500} =100(0.8) =80\text{ individuals per week}. $$

Using the fraction makes the model dimensionally consistent and makes the per-capita rate equal to \(r\) times a proportion. The alternative form \(a y(K-y)\) can be valid, but its coefficient \(a\) must have different units and is not the same parameter as \(r\).

What the Assumptions Do—and Do Not—Say

The logistic equation is a model, not a consequence of the word “limited” alone. The equation follows when we assume that the per-capita rate is proportional to the fraction of capacity remaining. Other assumptions about resource use, interactions, or changing environments could produce different equations.

The standard interpretation takes \(r>0\), \(K>0\), and a population between zero and capacity when describing a growing population. Within that range, the available fraction is between zero and one. The equation gives an instantaneous rate at a given population; it does not, by itself, guarantee that real data follow the model. The assumptions should be appropriate to the situation and the time period being studied.

One practical benefit of the formulation is that its components can be checked separately. Ask whether growth should scale with the number already present, whether the per-capita rate should decline with crowding, and whether the decline should be linear in the remaining fraction. If those are reasonable approximations, the logistic equation is a coherent translation of them.

Modeling takeaway: “Growth is proportional to population” contributes the factor \(y\). “Per-capita growth is proportional to the fraction of capacity still available” contributes \(r(1-y/K)\). Multiplying them gives \(dy/dt=ry(1-y/K)\).

Check Your Understanding

Use the assumptions and definitions in this tutorial to answer the following questions.

  1. A population has capacity \(K=800\). What fraction of capacity remains available when \(y=200\)?
  2. For \(r=0.3\) per month, \(K=800\), and \(y=200\), what is the per-capita growth rate and the total population growth rate?
  3. Explain why the factor \(1-y/K\) is dimensionless.
  4. If an equation is written \(dy/dt=a y(K-y)\), what are the units of \(a\) when population is measured in individuals and time in days?
  5. State the two assumptions that combine to produce the logistic equation.

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