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2.1 Population Models

  • \( P(t) \): population as function of time
  • \( P(t_0) = P_0 \): initial population (\( t_0 \) is often \( t = 0 \))

Exponential growth / natural growth:

\[ \frac{dP}{dt} = kP \]

\( k \) = constant governing growth

  • when \( P \) is large \( \rightarrow \) fast growth
  • when \( P \) is small \( \rightarrow \) slow growth
\[ \frac{dP}{dt} = kP, \quad P(0) = P_0 \quad \text{is separable} \]

Solution: \( P(t) = P_0 e^{kt} \)

\( \rightarrow \) doesn't limit how big \( P \) can get (\( t \to \infty, P \to \infty \))

unrealistic since in reality population usually is limited by, for example, food. So there is a population cap.

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tweak the right side of \( \frac{dP}{dt} = kP \)

change to:

\[ \frac{dP}{dt} = [\beta - \delta]P \]

\( \beta \): birth parameter

\( \delta \): death parameter

both \( \beta \) and \( \delta \) can depend on \( P \)

For example:

\[ \beta = \beta_0 - \beta_1 P \]

\( \beta_0, \beta_1 \) positive constants

birth decreases linearly with \( P \)

(large \( P \to \) slow birth)

one simple way to model environmental limitation

  • keep \( \delta = \delta_0 \) (constant)
  • plug into \( \frac{dP}{dt} = [\beta - \delta]P \)
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The Logistic Equation

\[\begin{aligned} \frac{dP}{dt} &= (\beta_0 - \beta_1 P - \delta_0) P \\ &= (\beta_0 - \delta_0) P - \beta_1 P^2 = aP - bP^2 \end{aligned}\]

In the expression \(aP - bP^2\), \(aP\) represents the birth rate and \(bP^2\) represents the death rate.

This is called the logistic equation.

\[ = bP \left( \frac{a}{b} - P \right) \]

Most common form of logistic eq.

\[ \frac{dP}{dt} = kP(M - P) \]

Where \(M\) is the carrying capacity → the maximum population the environment can sustain.

Notice:

  • \(\frac{dP}{dt} = 0\) at \(P = 0\) and \(P = M\)
  • When \(0 < P < M\), \(\frac{dP}{dt} > 0\) (growth)
  • When \(P > M\), \(\frac{dP}{dt} < 0\) (decline)
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Example

A population is modeled by the logistic eq.

\[ \frac{dP}{dt} = aP - bP^2 = kP(M - P) \]
  • The initial population is 120 (\(t = 0\))
  • There are 8 births per month and 6 deaths per month initially (\(t = 0\))

Find:

  1. Carrying capacity
  2. Time to reach 95% of the carrying capacity
\[ \frac{dP}{dt} = aP - bP^2 \]

Where \(aP\) is the birth rate and \(bP^2\) is the death rate.

At \(t = 0\), \(P = 120\)

At \(t = 0\), 8 births/month \(= aP = a(120) \rightarrow a = \frac{1}{15}\)

At \(t = 0\), 6 deaths/month \(= bP^2 = b(120)^2 \rightarrow b = \frac{1}{2400}\)

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Logistic Growth Model: Solving the Differential Equation

\[ \frac{dP}{dt} = \frac{1}{15}P - \frac{1}{2400}P^2 = \frac{1}{2400}P \left( \frac{1/15}{1/2400} - P \right) \]\[ = \frac{1}{2400}P(160 - P) \]

Comparing this to the standard form \( KP(M - P) \):

So, the carrying capacity is 160.

Solving the Differential Equation

Now let's solve:

\[ \frac{dP}{dt} = \frac{1}{2400}P(160 - P) \]

Method: Separable / Bernoulli

\[ \frac{1}{P(160 - P)} dP = \frac{1}{2400} dt \]\[ \int \frac{1}{P(160 - P)} dP = \int \frac{1}{2400} dt \]

Applying partial fractions to the left-hand side:

\[ \frac{1}{P(160 - P)} = \frac{A}{P} + \frac{B}{160 - P} \]
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Partial Fraction Decomposition

\[ 1 = A(160 - P) + BP \]\[ 0P + 1 = (B - A)P + 160A \]

\( B - A = 0 \implies B = A \)

\( 160A = 1 \implies A = \frac{1}{160}, B = \frac{1}{160} \)

Integration

\[ \int \frac{1}{P(160 - P)} dP = \int \frac{1}{2400} dt \]\[ \int \left( \frac{1}{160} \cdot \frac{1}{P} + \frac{1}{160} \cdot \frac{1}{160 - P} \right) dP = \int \frac{1}{2400} dt \]\[ \frac{1}{160} \ln P - \frac{1}{160} \ln(160 - P) = \frac{1}{2400}t + C \]

Simplification and Initial Condition

Multiply by 160:

\[ \ln P - \ln(160 - P) = \frac{1}{15}t + C \]\[ \ln \left( \frac{P}{160 - P} \right) = \frac{1}{15}t + C \]\[ \frac{P}{160 - P} = Ce^{\frac{1}{15}t} \]

Using initial condition \( P(0) = 120 \):

\[ \frac{120}{40} = Ce^0 = C = 3 \]
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Solving for Population Over Time

Starting with the algebraic manipulation of the population equation:

\[ \frac{P}{160 - P} = 3e^{\frac{1}{15}t} \]
\[ P = (160 - P)(3e^{\frac{1}{15}t}) \]
\[ P = 480e^{\frac{1}{15}t} - 3Pe^{\frac{1}{15}t} \]
\[ P + 3Pe^{\frac{1}{15}t} = 480e^{\frac{1}{15}t} \]

General Solution for P:

\[ P = \frac{480e^{\frac{1}{15}t}}{1 + 3e^{\frac{1}{15}t}} \]

Calculating Time to Reach 95% Carrying Capacity

Given the carrying capacity is 160, we find 95% of that value:

  • 95% of 160 = \( (0.95)(160) \)

Setting the population equation equal to this value to solve for \( t \):

\[ (0.95)(160) = \frac{480e^{\frac{1}{15}t}}{1 + 3e^{\frac{1}{15}t}} \]

Solve for \( t \)...

\[ t \approx 28 \text{ months} \]

Slope field for the differential equation with population P on the y-axis and time t on the x-axis. Horizontal asymptotes are shown at P=160 and P=0. Solution curves approach 160 over time.