The Central Idea
The standard logistic differential equation is
\[ \boxed{ \frac{dP}{dt} = rP\left(1-\frac{P}{K}\right) }. \]Here:
- \(P(t)\) is the changing quantity;
- \(r\) is the intrinsic growth-rate parameter;
- \(K\) is the carrying capacity or limiting scale.
The factor
\[ 1-\frac{P}{K} \]reduces growth as \(P\) approaches \(K\).
Why Exponential Growth Is Not Always Enough
Exponential growth uses
\[ P'=rP. \]This says that the relative rate
\[ \frac{P'}{P}=r \]remains constant.
The model contains no mechanism for crowding, saturation, limited resources, competition, or other constraints.
Logistic growth modifies the relative rate:
\[ \boxed{ \frac{P'}{P} = r\left(1-\frac{P}{K}\right) }. \]The Limiting Factor
The logistic correction factor is
\[ 1-\frac{P}{K}. \]If \(P\) is very small relative to \(K\), then
\[ \frac{P}{K}\approx0, \]so
\[ 1-\frac{P}{K}\approx1. \]Therefore,
\[ P'\approx rP. \]Early logistic growth can therefore resemble exponential growth.
What Happens Near Carrying Capacity?
When
\[ P\approx K, \]the factor
\[ 1-\frac{P}{K} \]approaches zero.
Therefore,
\[ P'\approx0. \]Growth slows as the system approaches the carrying capacity.
What Happens Above Carrying Capacity?
If
\[ P>K, \]then
\[ 1-\frac{P}{K}<0. \]Assuming \(r>0\) and \(P>0\),
\[ P'<0. \]Thus the model pushes the state downward toward \(K\).
Equilibrium Solutions
Equilibria occur where
\[ P'=0. \]For
\[ P'=rP\left(1-\frac{P}{K}\right), \]this gives
\[ rP\left(1-\frac{P}{K}\right)=0. \]Therefore,
\[ \boxed{P=0} \]and
\[ \boxed{P=K} \]are equilibrium solutions.
Stability of the Equilibria
Assume \(r>0\).
If
\[ 0so solutions move upward.
If
\[ P>K, \]then
\[ P'<0, \]so solutions move downward.
Therefore, \(P=K\) attracts nearby positive solutions and is stable.
The Zero Equilibrium
Near
\[ P=0, \]positive solutions satisfy
\[ P'>0. \]They move away from zero.
Thus, for \(r>0\), the equilibrium
\[ P=0 \]is unstable with respect to positive perturbations.
Phase-Line Interpretation
The logistic equation is autonomous:
\[ P'=f(P). \]We can summarize its qualitative behavior with sign information:
\[ P<0:\quad P'<0, \] \[ 0In many applications, only the physically meaningful region \(P\ge0\) is considered.
Expanding the Logistic Equation
The logistic equation may also be written
\[ P' = rP-\frac{r}{K}P^2. \]This shows the model as a competition between:
- a positive linear growth term \(rP\);
- a negative quadratic limiting term \(\frac{r}{K}P^2\).
The quadratic term becomes increasingly important as \(P\) grows.
Solving the Logistic Equation
Begin with
\[ \frac{dP}{dt} = rP\left(1-\frac{P}{K}\right). \]Rewrite
\[ 1-\frac{P}{K} = \frac{K-P}{K}. \]Therefore,
\[ \frac{dP}{dt} = \frac{r}{K}P(K-P). \]Separate variables:
\[ \frac{dP}{P(K-P)} = \frac{r}{K}\,dt. \]Partial Fractions
Decompose
\[ \frac{1}{P(K-P)}. \]We have
\[ \frac{1}{P(K-P)} = \frac{1}{K}\left( \frac{1}{P} + \frac{1}{K-P} \right). \]Therefore,
\[ \frac{1}{K} \int \left( \frac{1}{P} + \frac{1}{K-P} \right) dP = \frac{r}{K}\int dt. \]Integrating the Logistic Equation
Integrate:
\[ \frac{1}{K} \left( \ln|P| - \ln|K-P| \right) = \frac{r}{K}t+C. \]Multiply by \(K\):
\[ \ln\left| \frac{P}{K-P} \right| = rt+C_1. \]Exponentiating gives
\[ \frac{P}{K-P} = Ae^{rt}. \]The Explicit Logistic Solution
Solve
\[ \frac{P}{K-P}=Ae^{rt} \]for \(P\):
\[ P=Ae^{rt}(K-P). \]Therefore,
\[ P+Ae^{rt}P = KAe^{rt}. \]Factor:
\[ P(1+Ae^{rt}) = KAe^{rt}. \]Thus,
\[ P = \frac{KAe^{rt}}{1+Ae^{rt}}. \]Equivalently,
\[ \boxed{ P(t)=\frac{K}{1+Be^{-rt}} }. \]Using the Initial Value
Suppose
\[ P(0)=P_0. \]Then
\[ P_0 = \frac{K}{1+B}. \]Solve for \(B\):
\[ 1+B=\frac{K}{P_0}. \]Therefore,
\[ B=\frac{K-P_0}{P_0}. \]The initial-value solution is
\[ \boxed{ P(t) = \frac{K} {1+\left(\frac{K-P_0}{P_0}\right)e^{-rt}} }. \]Checking the Initial Value
At \(t=0\),
\[ e^{-r(0)}=1. \]Therefore,
\[ P(0) = \frac{K} {1+\frac{K-P_0}{P_0}}. \]Combine the denominator:
\[ 1+\frac{K-P_0}{P_0} = \frac{K}{P_0}. \]Thus,
\[ P(0)=P_0. \]Long-Term Behavior of the Explicit Solution
For \(r>0\),
\[ e^{-rt}\to0 \]as \(t\to\infty\).
Therefore,
\[ P(t) = \frac{K}{1+Be^{-rt}} \to \frac{K}{1} = K. \]Hence
\[ \boxed{ \lim_{t\to\infty}P(t)=K }. \]Example: Construct a Logistic Model
Suppose a population has carrying capacity
\[ K=10{,}000, \]intrinsic growth parameter
\[ r=0.3, \]and initial population
\[ P_0=1000. \]Then
\[ B = \frac{10000-1000}{1000} = 9. \]Therefore,
\[ \boxed{ P(t) = \frac{10000}{1+9e^{-0.3t}} }. \]Early Growth in the Example
When \(P\ll10000\),
\[ 1-\frac{P}{10000} \approx1. \]Therefore,
\[ P'\approx0.3P. \]The early portion of the logistic trajectory behaves approximately like exponential growth with rate parameter \(0.3\).
The S-Shaped Logistic Curve
When
\[ 0The process has three broad stages:
- slow growth when the quantity is initially small;
- rapid growth during the middle stage;
- slowing growth as the quantity approaches \(K\).
Where Growth Is Fastest
The logistic rate is
\[ P' = rP\left(1-\frac{P}{K}\right). \]Regard the right side as a function of \(P\):
\[ f(P) = rP-\frac{r}{K}P^2. \]This is a downward-opening quadratic.
Its maximum occurs at
\[ \boxed{P=\frac{K}{2}}. \]Maximum Growth Rate
Substitute
\[ P=\frac{K}{2} \]into the rate equation:
\[ P' = r\left(\frac{K}{2}\right) \left( 1-\frac12 \right). \]Therefore,
\[ \boxed{ P'_{\max}=\frac{rK}{4} }. \]The system grows fastest when it reaches half its carrying capacity.
The Inflection Point
For the standard logistic curve with \(0
This point is the inflection point of the logistic solution curve.
Before it, the growth rate is increasing.
After it, the growth rate is decreasing.
Deriving the Concavity Change
Differentiate
\[ P' = rP\left(1-\frac{P}{K}\right) \]with respect to \(t\):
\[ P'' = rP' \left(1-\frac{P}{K}\right) - \frac{rP}{K}P'. \]Factor:
\[ P'' = rP' \left( 1-\frac{2P}{K} \right). \]For a growing solution with \(P'>0\), the sign of \(P''\) changes when
\[ 1-\frac{2P}{K}=0. \]Therefore,
\[ P=\frac{K}{2}. \]Time of the Inflection Point
Starting from
\[ P(t)=\frac{K}{1+Be^{-rt}}, \]set
\[ P=\frac{K}{2}. \]Then
\[ \frac{K}{2} = \frac{K}{1+Be^{-rt}}. \]Therefore,
\[ 1+Be^{-rt}=2, \]so
\[ Be^{-rt}=1. \]Hence
\[ \boxed{ t_{\text{inflection}} = \frac{\ln B}{r} }. \]Inflection Time in Terms of the Initial Value
Since
\[ B=\frac{K-P_0}{P_0}, \]the inflection time is
\[ \boxed{ t_{\text{inflection}} = \frac{1}{r} \ln\left( \frac{K-P_0}{P_0} \right) }. \]This formula applies directly when the inflection occurs after the chosen initial time.
Relative Growth Rate
Divide the logistic equation by \(P\):
\[ \frac{P'}{P} = r\left(1-\frac{P}{K}\right). \]The relative growth rate is not constant.
Instead, it decreases linearly with \(P\).
At
\[ P=0, \]the relative rate approaches \(r\).
At
\[ P=K, \]it becomes zero.
Interpreting the Parameter \(r\)
The parameter \(r\) controls the intrinsic speed of growth.
Larger positive values of \(r\) generally produce faster transitions from low values toward carrying capacity.
The units of \(r\) are reciprocal time, such as
\[ \frac{1}{\text{year}} \]or
\[ \frac{1}{\text{day}}. \]Interpreting the Parameter \(K\)
The parameter \(K\) has the same units as \(P\).
In a population model, \(K\) may represent a carrying capacity.
In a market-adoption model, it may represent a saturation level.
In a production or storage model, it may represent an effective capacity.
Its interpretation depends on the system being modeled.
Logistic Models Are Not Only for Populations
Logistic behavior can appear whenever growth is self-reinforcing at low levels but increasingly constrained at high levels.
Applications may include:
- population growth;
- technology adoption;
- market penetration;
- spread of information;
- limited-capacity resource use;
- biological growth;
- certain learning curves;
- bounded diffusion processes.
Logistic Adoption
Suppose \(A(t)\) is the number of adopters of a technology and \(K\) is the maximum potential market.
A simple logistic model is
\[ A' = rA\left(1-\frac{A}{K}\right). \]Early adoption can accelerate as more users participate, while later growth slows as the remaining market shrinks.
Logistic Growth versus Exponential Growth
Exponential growth:
\[ P'=rP. \]Logistic growth:
\[ P'=rP\left(1-\frac{P}{K}\right). \]The key distinction is the limiting factor
\[ 1-\frac{P}{K}. \]Exponential growth has no finite upper equilibrium, while logistic growth has the stable equilibrium \(P=K\).
Estimating \(K\) from Structural Information
Sometimes carrying capacity is supplied directly by the context.
In other situations, it must be estimated from data or long-term behavior.
If observations appear to level off near a stable plateau, that plateau may provide an estimate of \(K\).
The quality of this estimate depends on the quality and time span of the data.
Estimating \(r\) When \(K\) Is Known
Suppose \(K\), \(P_0\), and \(P(t_1)=P_1\) are known.
Start from
\[ P_1 = \frac{K}{1+Be^{-rt_1}}, \]where
\[ B=\frac{K-P_0}{P_0}. \]Rearranging:
\[ \frac{K}{P_1}-1 = Be^{-rt_1}. \]Thus
\[ e^{-rt_1} = \frac{\frac{K}{P_1}-1}{B}. \]Formula for \(r\)
Take natural logarithms:
\[ -rt_1 = \ln\left( \frac{\frac{K}{P_1}-1}{B} \right). \]Therefore,
\[ \boxed{ r = -\frac{1}{t_1} \ln\left( \frac{\frac{K}{P_1}-1} {\frac{K-P_0}{P_0}} \right) }. \]This allows a logistic rate parameter to be fitted from an additional measurement when \(K\) is known.
Example: Estimate \(r\)
Suppose
\[ K=1000, \qquad P_0=100, \]and
\[ P(5)=300. \]First,
\[ B=\frac{1000-100}{100}=9. \]Then
\[ 300 = \frac{1000}{1+9e^{-5r}}. \]Rearranging gives
\[ \frac{1000}{300}-1 = 9e^{-5r}. \]From here, logarithms determine \(r\).
Linearizing the Logistic Model
From
\[ \frac{P}{K-P}=Ae^{rt}, \]take natural logarithms:
\[ \ln\left(\frac{P}{K-P}\right) = rt+\ln A. \]Therefore, if \(K\) is known and the logistic model is appropriate, a plot of
\[ \ln\left(\frac{P}{K-P}\right) \]against \(t\) should be approximately linear.
The Logit Transformation
If
\[ p=\frac{P}{K}, \]then \(p\) represents the fraction of capacity reached.
The expression
\[ \ln\left(\frac{p}{1-p}\right) \]is called the logit.
For a logistic trajectory,
\[ \boxed{ \ln\left(\frac{p}{1-p}\right) = rt+C }. \]Nondimensionalizing the Logistic Equation
Let
\[ p=\frac{P}{K}. \]Then
\[ P=Kp \]and
\[ P'=Kp'. \]Substitute into
\[ P'=rP\left(1-\frac{P}{K}\right): \] \[ Kp' = rKp(1-p). \]Divide by \(K\):
\[ \boxed{ p'=rp(1-p) }. \]A Universal Scaled Form
Introduce scaled time
\[ \tau=rt. \]Then
\[ \frac{dp}{d\tau}=p(1-p). \]This dimensionless form shows that many logistic systems have the same mathematical structure after appropriate scaling.
When the Initial Value Is Above \(K\)
If
\[ P_0>K, \]then
\[ P'<0. \]The solution decreases toward \(K\).
The carrying capacity is therefore a stable equilibrium from both sides in the positive region.
When the Initial Value Equals \(K\)
If
\[ P_0=K, \]then
\[ P'=rK(1-1)=0. \]Therefore,
\[ \boxed{P(t)=K} \]for all \(t\).
When the Initial Value Is Zero
If
\[ P_0=0, \]then
\[ P'=0. \]Therefore,
\[ \boxed{P(t)=0} \]is an equilibrium solution.
The standard explicit formula containing division by \(P_0\) must be handled separately in this case.
Common Error: Treating \(K\) as the Initial Value
In the logistic model,
\[ P_0 \]is the initial value, while
\[ K \]is the limiting equilibrium.
They play different roles and need not be equal.
Common Error: Assuming Growth Stops Abruptly at \(K\)
Logistic growth does not increase normally and then suddenly stop.
The rate decreases continuously as \(P\) approaches \(K\).
In the ideal model,
\[ P(t)\to K \]asymptotically.
Common Error: Calling \(r\) the Actual Growth Rate
In
\[ P'=rP\left(1-\frac{P}{K}\right), \]\(r\) is not the actual instantaneous absolute growth rate.
The actual rate is
\[ P'. \]The actual relative growth rate is
\[ \frac{P'}{P} = r\left(1-\frac{P}{K}\right). \]Common Error: Assuming Maximum Growth Occurs at \(K\)
At
\[ P=K, \]the growth rate is zero.
The maximum growth rate occurs at
\[ \boxed{P=\frac{K}{2}}. \]Common Error: Confusing \(P'\) with \(P'/P\)
The absolute growth rate is
\[ P' = rP\left(1-\frac{P}{K}\right). \]The relative growth rate is
\[ \frac{P'}{P} = r\left(1-\frac{P}{K}\right). \]These are different quantities with different units and interpretations.
Common Error: Using a Logistic Model Without Evidence of Saturation
A logistic curve should not be selected only because it has a familiar S-shape.
The model assumes:
- a meaningful limiting scale exists;
- growth is approximately proportional to the current state at low levels;
- the relative growth rate declines roughly linearly with the state;
- the parameters remain reasonably stable over the modeled period.
These assumptions should be checked against the system and the data.
Common Error: Assuming Carrying Capacity Is Permanently Fixed
In real applications, carrying capacity may change over time.
Technology, infrastructure, resource availability, policy, environmental conditions, or market size can change the effective value of \(K\).
The basic logistic model treats \(K\) as constant, so a changing capacity may require a more advanced model.
A Reliable Logistic-Model Workflow
- Identify the changing quantity \(P(t)\).
- Identify the time variable and units.
- Determine whether a limiting scale \(K\) is reasonable.
- Write \(P'=rP(1-P/K)\).
- Find the equilibrium solutions \(P=0\) and \(P=K\).
- Use sign analysis to determine qualitative behavior.
- Use \(P(t)=K/(1+Be^{-rt})\) for non-equilibrium solutions.
- Use the initial value to determine \(B\).
- Use additional data to estimate \(r\) or \(K\) when necessary.
- Identify the inflection level \(P=K/2\).
- Check the long-term limit.
- Interpret parameters and units in context.
- Assess whether logistic assumptions are reasonable.
Example: Full Logistic Workflow
A population has carrying capacity
\[ K=5000, \]growth parameter
\[ r=0.2, \]and initial population
\[ P_0=500. \]First compute
\[ B = \frac{5000-500}{500} = 9. \]Therefore,
\[ \boxed{ P(t)=\frac{5000}{1+9e^{-0.2t}} }. \]Analyze the Full Logistic Example
The equilibria are
\[ P=0 \]and
\[ P=5000. \]Since
\[ 0The maximum growth rate occurs at
\[ P=2500. \]Its value is
\[ P'_{\max} = \frac{rK}{4} = \frac{0.2(5000)}{4} = 250. \]Therefore,
\[ \boxed{ P'_{\max}=250 } \]population units per unit time.
Practice 1: Identify the Equilibria
Find the equilibrium solutions of
\[ P'=0.4P\left(1-\frac{P}{800}\right). \]Practice 2: Determine Direction of Change
For
\[ P'=0.4P\left(1-\frac{P}{800}\right), \]determine whether \(P\) increases or decreases when:
- \(0
- \(P>800\).
Practice 3: Construct the Explicit Model
Let
\[ K=1000, \qquad r=0.5, \qquad P_0=100. \]Construct the logistic solution.
Practice 4: Maximum Growth Level
If the carrying capacity is
\[ K=2400, \]at what value of \(P\) is the logistic growth rate greatest?
Practice 5: Maximum Growth Rate
Suppose
\[ r=0.3 \]and
\[ K=2000. \]Find the maximum possible logistic growth rate.
Practice 6: Long-Term Limit
For
\[ P(t)=\frac{500}{1+4e^{-0.2t}}, \]find
\[ \lim_{t\to\infty}P(t). \]Practice 7: Initial Value
For
\[ P(t)=\frac{1000}{1+9e^{-0.4t}}, \]find \(P(0)\).
Practice 8: Inflection Level
A logistic model has carrying capacity
\[ K=6000. \]At what state value does the solution change concavity?
Practice 9: Relative Growth Rate
For
\[ P'=0.2P\left(1-\frac{P}{1000}\right), \]find the relative growth rate when
\[ P=250. \]Practice 10: Compare Exponential and Logistic Growth
Explain the structural difference between
\[ P'=rP \]and
\[ P'=rP\left(1-\frac{P}{K}\right). \]Practice Answers
Practice 1
Set
\[ 0.4P\left(1-\frac{P}{800}\right)=0. \]Therefore,
\[ \boxed{P=0} \]and
\[ \boxed{P=800}. \]Practice 2
If
\[ 0are positive, so
\[ \boxed{P'>0}. \]The quantity increases.
If
\[ P>800, \]then
\[ 1-\frac{P}{800}<0, \]so
\[ \boxed{P'<0}. \]The quantity decreases.
Practice 3
Compute
\[ B = \frac{1000-100}{100} = 9. \]Therefore,
\[ \boxed{ P(t)=\frac{1000}{1+9e^{-0.5t}} }. \]Practice 4
Maximum logistic growth occurs at
\[ P=\frac{K}{2}. \]Therefore,
\[ \boxed{P=1200}. \]Practice 5
Use
\[ P'_{\max}=\frac{rK}{4}. \]Therefore,
\[ P'_{\max} = \frac{0.3(2000)}{4} = 150. \]Thus,
\[ \boxed{P'_{\max}=150}. \]Practice 6
Since
\[ e^{-0.2t}\to0, \]we have
\[ \boxed{ \lim_{t\to\infty}P(t)=500 }. \]Practice 7
Set \(t=0\):
\[ P(0) = \frac{1000}{1+9} = 100. \]Therefore,
\[ \boxed{P(0)=100}. \]Practice 8
The inflection level is
\[ \frac{K}{2}. \]Therefore,
\[ \boxed{P=3000}. \]Practice 9
The relative growth rate is
\[ \frac{P'}{P} = 0.2\left(1-\frac{P}{1000}\right). \]At
\[ P=250, \]this becomes
\[ 0.2\left(1-\frac{250}{1000}\right) = 0.2(0.75) = 0.15. \]Therefore,
\[ \boxed{ \frac{P'}{P}=0.15 }. \]Practice 10
Exponential growth uses
\[ P'=rP, \]so the relative growth rate remains constant.
Logistic growth uses
\[ P'=rP\left(1-\frac{P}{K}\right), \]so the relative growth rate decreases as \(P\) approaches \(K\).
The logistic model therefore includes a limiting equilibrium:
\[ \boxed{P=K}. \]What to Remember
The standard logistic equation is
\[ \boxed{ P'=rP\left(1-\frac{P}{K}\right) }. \]Its major features are:
- \(P=0\) and \(P=K\) are equilibria;
- for \(r>0\), \(P=K\) is stable in the positive region;
- early growth resembles exponential growth when \(P\ll K\);
- growth slows as \(P\) approaches \(K\);
- the maximum growth rate occurs at \(P=K/2\);
- the maximum rate equals \(rK/4\);
- the standard solution is \(P=K/(1+Be^{-rt})\);
- the parameter \(r\) controls growth speed;
- the parameter \(K\) sets the limiting scale.
The central structural idea is
\[ \boxed{ \text{growth} = \text{current size} \times \text{remaining capacity effect} }. \]Logistic models preserve self-reinforcing growth while introducing a mechanism that prevents unlimited expansion.
Where We Go Next
First-order models describe systems whose current rate depends on the current state. Many physical systems require a second derivative because acceleration, curvature, or another second-order effect is essential.
The next lesson, Second-Order Equations, introduces differential equations involving
\[ y'', \]including homogeneous equations, characteristic equations, complementary solutions, and the foundations needed for oscillation and forcing models.