Lesson 9 (09/14/20) - Population Models (2.1)

Page 1

Lesson 9 (09/14/20) - Population Models (2.1)

Warmup: Find \( A \), \( B \) such that

\[ \frac{1}{P(2-P)} = \underbrace{\frac{A}{P} + \frac{B}{2-P}}_{\text{partial fractions}} \] \[ \begin{aligned} \frac{A}{P} + \frac{B}{2-P} &= \frac{A(2-P) + BP}{P(2-P)} \\ &= \frac{2A + P(-A+B)}{P(2-P)} \\ \text{want} \quad &= \frac{1}{P(2-P)} \end{aligned} \]

Compare coeff:

\[ \begin{aligned} 2A &= 1 \implies A = 1/2 \\ -A + B &= 0 \implies A = B \implies B = 1/2 \end{aligned} \] \[ \frac{1}{P(2-P)} = \frac{1}{2P} + \frac{1}{2(2-P)} \]

Page 2

1. Seperable:

\[ y' = f(y) \, g(x) \qquad \int \frac{1}{f(y)} \, dy = \int g(x) \, dx \]

2. Linear:

\[ y' + p(x)y = q(x) \] \[ \mu(x) = e^{\int p(x) \, dx} \rightsquigarrow \mu'(x) = \mu(x)p(x) \]

Multiply \( \mu(x) \)

\[ \underbrace{\mu(x)y' + \mu(x)p(x)y}_{(\mu(x)y(x))'} = \mu(x)q(x) \]

3. Homogeneous:

\[ y' = \phi\left(\frac{y}{x}\right) \] \[ v = \frac{y}{x} \quad \text{or} \quad y = xv \] \[ y' = xv' + v \] \[ xv' + v = \phi(v) \rightsquigarrow \text{Seperable.} \]

Page 3

4. Bernoulli

\[ y' + p(x)y = q(x)y^n \qquad n \neq 0, 1 \] \[ y^{-n}y' + p(x)y^{1-n} = q(x) \]

Substituted \( v = y^{1-n}, \quad v' = (1-n)y^{-n}y' \)

\[ \frac{1}{(1-n)} v' + p(x)v = q(x) \quad \text{is Linear} \]

5. Exact

\[ M(x,y) + N(x,y)\frac{dy}{dx} = 0 \]

check: \( M_y = N_x? \)

if it is then find \( \phi(x,y) \) such that

\( \phi_x = M \)
\( \phi_y = N \)

\( \phi(x,y) = c \) is general solution

Page 4

Population Models:

1. Constant growth/decay:

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

\( k \) is constant

\( k > 0 \implies \text{growth} \)

\( k < 0 \implies \text{decay} \)

Seperable \( \implies \frac{1}{P} \frac{dP}{dt} = k \implies \text{integrate} \)

\[ P(t) = P(0)e^{kt} \]

2. Birth Rate: \( \beta(t) \)

Death Rate: \( \delta(t) \)

Replace \( k \) with \( \beta(t) - \delta(t) \)

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

Page 5

Example

Birth rate = \( \beta(t) = 3t^2 \), death rate = \( \delta(t) = t \), \( P(0) = 750 \).

\[ \frac{dP}{dt} = [\beta(t) - \delta(t)] P, \quad P(0) = 750 \]

IVP:

\[ \frac{dP}{dt} = (3t^2 - t) P, \quad P(0) = 750 \] \[ \int \frac{1}{P} \, dP = \int (3t^2 - t) \, dt + K \] \[ \ln P = \left( t^3 - \frac{t^2}{2} \right) + K \] \[ P(t) = e^{t^3 - t^2/2} \cdot \underbrace{e^K}_{P(0)} \] \[ \leadsto P(t) = 750 \, e^{t^3 - t^2/2} \] \[ t^3 - \frac{t^2}{2} = t \left[ t^2 - \frac{t}{2} \right] > kt \quad \text{for large enough } t \ \& \ \text{any } k. \]

\( P(t) \) grows much faster than a constant growth.

Page 6

Logistic Model

Birth Rate \( = \beta(t) \leadsto \) decreases linearly as population increases \[ \beta(t) = a - b \, P(t), \quad b \neq 0. \]
Death Rate \( = \delta(t) = \text{Constant} = c \)

\[ \begin{aligned} \frac{dP}{dt} &= (\beta(t) - \delta(t)) P(t) = [a - b \, P(t) - c] P(t) \\ &= b \left[ \underbrace{\left(\frac{a - c}{b}\right)}_{M} - P(t) \right] P(t) \end{aligned} \]

Logistic equation:

\[ \frac{dP}{dt} = b(M - P)P, \quad P(0) = P_0 \]

Page 7

Example: Solve \(\frac{dP}{dt} = P(2-P)\), \(P(0) = 1\) with \(b = 1\), \(M = 2\)

\[ \int \frac{1}{P(2-P)} \, dP = \int dt + K \]

using partial fractions

\[ \int \frac{1}{2} \left[ \frac{1}{P} + \frac{1}{2-P} \right] \, dP = t + K \] \[ \frac{1}{2} [\ln P - \ln(2-P)] = t + K \] \[ \ln \left[ \frac{P(t)}{2 - P(t)} \right] = 2t + K \] \[ \frac{P(t)}{2 - P(t)} = e^{2t + K} = A e^{2t} \]

Page 8

\[ P(t) = (2 - P(t)) A e^{2t} = 2A e^{2t} - A P(t) e^{2t} \]

\[ P(t)[1 + A e^{2t}] = 2A e^{2t} \]

\[ P(t) = \frac{2A e^{2t}}{1 + A e^{2t}} \]

\[ P(0) = 1 \implies 1 = \frac{2A}{1 + A} \implies A = 1 \]

\[ P(t) = \frac{2e^{2t}}{1 + e^{2t}} \]

Page 9

\( P(t) = \frac{2e^{2t}}{1 + e^{2t}} \) What happes as \( t \rightarrow \infty \)?

\[ \lim_{t \rightarrow \infty} P(t) = \lim_{t \rightarrow \infty} \frac{2e^{2t}}{1 + e^{2t}} \]

L'Hopitals

\[ = \lim_{t \rightarrow \infty} \frac{4e^{2t}}{2e^{2t}} = 2. \]

limiting Population or CARRYING Capacity

Page 10

\[ \frac{dP}{dt} = b(M - P)P, \quad \underline{b > 0} \]

equilibrium solution: \( P = 0, \quad P = M \)

Direction field and solution curves for the differential equation dP/dt = b(M - P)P where b > 0. The vertical axis represents population P and the horizontal axis represents time t. Equilibrium lines are at P = M and P = 0. For P > M, slope field segments have negative slope with a red solution curve decreasing asymptotically toward P = M with a downward right arrow. For P < M (and P > 0), a red logistic solution curve starts near P = 0 and increases asymptotically toward P = M with an upward right arrow.

Visual Description: Direction field and solution curves for the differential equation dP/dt = b(M - P)P where b > 0. The vertical axis represents population P and the horizontal axis represents time t. Equilibrium lines are at P = M and P = 0. For P > M, slope field segments have negative slope with a red solution curve decreasing asymptotically toward P = M with a downward right arrow. For P < M (and P > 0), a red logistic solution curve starts near P = 0 and increases asymptotically toward P = M with an upward right arrow.

\( P > M \)
\( (M - P)P < 0 \)
\( \Rightarrow P'(t) < 0 \)

\( P = M \)

\( P < M \)
\( (M - P) > 0 \)
\( P' > 0 \)

\( P = 0 \)

\( M = \text{Limiting population or Carrying capacity} \)

Page 11

\[ \frac{dP}{dt} = b(M - P)P, \quad b < 0 \implies \text{signs of } \frac{dP}{dt} \text{ switch from previous case} \]

Direction field and trajectory plot for the differential equation dP/dt = b(M - P)P with b < 0. The vertical axis represents population P and the horizontal axis represents time t. An unstable equilibrium line is drawn at P = M and the horizontal axis line at P = 0. In the region P > M, M - P < 0, resulting in P' > 0 with blue slope segments tilted positively and a red trajectory curving upward away from P = M. In the region P < M, M - P > 0, resulting in P' < 0 with blue slope segments tilted negatively and a red trajectory curving downward toward P = 0.

Visual Description: Direction field and trajectory plot for the differential equation dP/dt = b(M - P)P with b < 0. The vertical axis represents population P and the horizontal axis represents time t. An unstable equilibrium line is drawn at P = M and the horizontal axis line at P = 0. In the region P > M, M - P < 0, resulting in P' > 0 with blue slope segments tilted positively and a red trajectory curving upward away from P = M. In the region P < M, M - P > 0, resulting in P' < 0 with blue slope segments tilted negatively and a red trajectory curving downward toward P = 0.

\(P > M\)
\(M - P < 0\)
\(P' > 0\)

\(P = M\)

\(P < M\)
\(M - P > 0\)
\(P' < 0\)

\(P = 0\)

\(M = \text{Threshold population}\)