Introduction

This topic explores logistic models that arise from differential equations, focusing on situations where growth is limited by a carrying capacity. The essential knowledge tells us that the model for logistic growth that arises from the statement "The rate of change of a quantity is jointly proportional to the size of the quantity and the difference between the quantity and the carrying capacity" is $\frac{dy}{dt}=ky(a-y)$. Additionally, the logistic differential equation and initial conditions can be interpreted without solving the differential equation, the limiting value (carrying capacity) of a logistic differential equation as the independent variable approaches infinity can be determined using the logistic growth model and initial conditions, and the value of the dependent variable in a logistic differential equation at the point when it is changing fastest can be determined using the logistic growth model and initial conditions.

Understanding Logistic Growth Models

The Logistic Differential Equation

The logistic model is represented by the differential equation:

$\frac{dy}{dt}=ky(a-y)$

where:

This equation states that the rate of change is proportional to both the current quantity and the remaining capacity for growth.

Comparing Exponential and Logistic Growth

Exponential Growth: $\frac{dy}{dt}=ky$ (unlimited growth)

Logistic Growth: $\frac{dy}{dt}=ky(a-y)$ (growth limited by carrying capacity)

Example 1:

A population grows logistically with carrying capacity 1000 and growth constant $k=0.02$. The differential equation is:

$\frac{dP}{dt}=0.02P(1000-P)$

Analyzing Logistic Models Without Solving

Determining the Carrying Capacity

The carrying capacity is the value $a$ in the differential equation $\frac{dy}{dt}=ky(a-y)$.

At the carrying capacity, $y=a$, so $\frac{dy}{dt}=ka(a-a)=0$.

This means growth stops when the population reaches the carrying capacity.

Example 2:

For $\frac{dN}{dt}=0.5N(200-N)$, the carrying capacity is $a=200$.

Behavior Analysis

When $0<y<a$: Both $y>0$ and $(a-y)>0$, so $\frac{dy}{dt}>0$ (increasing)

When $y=a$: $\frac{dy}{dt}=0$ (equilibrium)

When $y>a$: $y>0$ but $(a-y)<0$, so $\frac{dy}{dt}<0$ (decreasing)

Example 3:

For $\frac{dy}{dt}=2y(50-y)$ with carrying capacity 50:

Finding the Point of Maximum Growth Rate

The Inflection Point

The growth rate $\frac{dy}{dt}=ky(a-y)$ is maximized when its derivative equals zero.

Taking the derivative: $\frac{d^{2}y}{d{t}^2}=k\frac{d}{dt}[y(a-y)]=k[(a-y)-y]=k(a-2y)$

Setting $\frac{d^{2}y}{d{t}^2}=0$: $k(a-2y)=0$

Since $k\ne 0$: $a-2y=0$, so $y=\frac{a}{2}$

The maximum growth rate occurs at half the carrying capacity.

Example 4:

For $\frac{dP}{dt}=0.03P(800-P)$:

The maximum growth rate occurs when $P=\frac{800}{2}=400$.

The maximum growth rate is: $\frac{dP}{dt}=0.03(400)(800-400)=0.03(400)(400)=4800$

Interpreting the Inflection Point

At $y=\frac{a}{2}$:

Example 5:

A bacterial culture has carrying capacity 10,000 and follows logistic growth. The fastest growth occurs when the population reaches 5,000 bacteria.

Slope Fields and Solution Curves

Understanding Slope Field Behavior

For $\frac{dy}{dt}=ky(a-y)$:

Near $y=0$: Slopes are small (slow initial growth)

Near $y=\frac{a}{2}$: Slopes are steepest (maximum growth rate)

Near $y=a$: Slopes approach zero (growth slows near capacity)

Example 6:

For $\frac{dy}{dt}=y(4-y)$:

Solution Curve Characteristics

All logistic solution curves have:

Interpreting Initial Conditions

Effect of Different Starting Values

Case 1: $y(0)<\frac{a}{2}$ (below inflection point)

Case 2: $y(0)=\frac{a}{2}$ (at inflection point)

Case 3: $y(0)>\frac{a}{2}$ but $y(0)<a$ (above inflection point, below capacity)

Case 4: $y(0)>a$ (above carrying capacity)

Example 7:

For $\frac{dP}{dt}=0.1P(100-P)$ with carrying capacity 100:

Applications and Context

Population Dynamics

Logistic models are commonly used for:

Example 8:

A deer population in a forest follows $\frac{dP}{dt}=0.05P(300-P)$.

The forest can support a maximum of 300 deer. The population grows fastest when there are 150 deer.

Technology Adoption

The spread of new technologies often follows logistic patterns.

Example 9:

The adoption of smartphones in a market follows $\frac{dN}{dt}=0.2N(50-N)$, where $N$ is in millions of users.

The market saturation point is 50 million users, and adoption accelerates most rapidly at 25 million users.

Problem-Solving Strategies

Systematic Analysis Approach

Reading Information from the Differential Equation

From $\frac{dy}{dt}=ky(a-y)$:

Practice Section