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Logistic Growth Model and Stability Analysis

\[ 22. \quad \frac{dy}{dt} = \alpha y (1-y), \quad y(0) = y_0 \]
  • \( y \): proportion of infected
  • \( x + y = 1 \)
  • \( x \): susceptible

a) Equilibrium and Stability

Equilibrium points occur where \( \frac{dy}{dt} = 0 \):

\[ y = 0, \quad y = 1 \]

Phase line analysis for \( y' \):

Phase line diagram for y' showing equilibrium points at 0 and 1 with arrows indicating stability direction.
  • \( y = 0 \) is unstable
  • \( y = 1 \) is asymptotically stable

b) Solving the Differential Equation

Separating variables:

\[ \frac{1}{y(1-y)} dy = \alpha dt \]

Integrating both sides using partial fractions:

\[ \ln y - \ln(1-y) = \alpha t + K \] \[ \ln \left( \frac{y}{1-y} \right) = \alpha t + K \]
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Exponentiating both sides to solve for \( y \):

\[ \frac{y}{1-y} = e^{\alpha t + K} = C e^{\alpha t} \]

Using the initial condition \( y(0) = y_0 \):

\[ \frac{y_0}{1-y_0} = C \]

Substituting \( C \) back into the equation:

\[ \frac{y}{1-y} = \left( \frac{y_0}{1-y_0} \right) e^{\alpha t} \]

Solving for \( y \) algebraically:

\[ y = (1-y) \left( \frac{y_0}{1-y_0} \right) e^{\alpha t} \] \[ y = \left( \frac{y_0}{1-y_0} \right) e^{\alpha t} - \left( \frac{y_0}{1-y_0} e^{\alpha t} \right) y \] \[ y + \left( \frac{y_0}{1-y_0} \right) e^{\alpha t} y = \frac{y_0}{1-y_0} e^{\alpha t} \] \[ y \left[ 1 + \left( \frac{y_0}{1-y_0} \right) e^{\alpha t} \right] = \frac{y_0}{1-y_0} e^{\alpha t} \]

The final solution for the proportion of infected is:

\[ y = \frac{\frac{y_0}{1-y_0} e^{\alpha t}}{1 + \frac{y_0}{1-y_0} e^{\alpha t}} \]

As time approaches infinity:

\[ \lim_{t \to \infty} y = 1 \]
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17. Differential Equations: Gompertz Model

a) Consider the differential equation:

\[ \frac{dy}{dt} = r y \ln\left( \frac{K}{y} \right), \quad y(0) = y_0 \]

Substitution:

Let \( u = \ln\left( \frac{y}{K} \right) \)

Eliminate \( \frac{dy}{dt} \):

\[ \frac{du}{dt} = \frac{1}{\frac{y}{K}} \cdot \frac{y'}{K} = \frac{K}{y} \cdot \frac{y'}{K} = \frac{y'}{y} \]

Rearranging for \( y' \):

\[ y' = y u' \]

Substitute back into the original equation:

\[ y u' = r y \ln\left( \frac{K}{y} \right) \]

Simplify by dividing by \( y \) and using logarithmic properties:

\[ u' = -ru \]

Note on Logarithms:

\[ \ln\left( \frac{K}{y} \right) = \ln\left( \frac{y}{K} \right)^{-1} = -\ln\left( \frac{y}{K} \right) = -u \]

Solving the first-order linear differential equation for \( u \):

\[ u = C e^{-rt} \]

Substitute back for \( u \):

\[ \ln\left( \frac{y}{K} \right) = C e^{-rt} \]

Apply initial condition \( y(0) = y_0 \) to find \( C \):

\[ \ln\left( \frac{y_0}{K} \right) = C \]
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Final Solution Derivation

Substitute the constant \( C \) back into the equation for \( y \):

\[ \ln\left( \frac{y}{K} \right) = \ln\left( \frac{y_0}{K} \right) e^{-rt} \]

Exponentiate both sides to solve for \( \frac{y}{K} \):

\[ \frac{y}{K} = e^{\ln\left( \frac{y_0}{K} \right) e^{-rt}} \]

Multiply by \( K \) to isolate \( y \):

\[ y = K e^{\ln\left( \frac{y_0}{K} \right) e^{-rt}} \]
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2.6 Exact Equations

An exact equation has the form

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

such that \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] or \[ M_y = N_x \]

has solution \[ \psi(x, y) = C \]

where \[ \psi_x = M \quad \text{and} \quad \psi_y = N \]

It's called exact because the exact (or total) derivative of \[ \psi(x, y) = C \] is

\[ \frac{\partial \psi}{\partial x} \frac{dx}{dx} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = 0 \]

\[ \psi_x + \psi_y y' = 0 \longrightarrow \psi_x dx + \psi_y dy = 0 \]

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From continuity of \( \psi(x, y) \), we know

\[ (\psi_x)_y = (\psi_y)_x \]

\[ \implies M_y = N_x \]

\( \psi(x, y) \): potential function

Example

Solve \[ (2xy^2 + 2y) + (2x^2y + 2x)y' = 0 \]

\[ (2xy^2 + 2y)dx + (2x^2y + 2x)dy = 0 \]

M

N

Exact? Need \( M_y = N_x \)

\[ M_y = 4xy + 2 \]

\[ N_x = 4xy + 2 \]

So, yes, it is exact.

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Solving Exact Differential Equations

The solution is given by the implicit form:

\[ \psi(x,y) = C \]

where the partial derivatives satisfy:

\[ \psi_x = M \quad , \quad \psi_y = N \]

Step 1: Integrate \( M \) with respect to \( x \)

Pick \( M \) to integrate with respect to \( x \), treating \( y \) as a constant:

\[ \psi(x,y) = \int M \, dx = \int (2xy^2 + 2y) \, dx \]\[ = x^2y^2 + 2xy + h(y) \]

Note: \( h(y) \) is a function of \( y \) or a constant (it vanishes when taking the partial derivative with respect to \( x \)).

Step 2: Determine \( h(y) \)

We know that:

  • \( \psi_x = M \)
  • \( \psi_y = N \)

Taking the partial derivative of our expression for \( \psi \) with respect to \( y \):

\[ \psi_y = 2x^2y + 2x + \frac{dh}{dy} = N \]\[ = 2x^2y + 2x \]
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Comparing the terms from the previous step:

\[ \text{so } \frac{dh}{dy} = 0 \implies h = k \quad (\text{constant}) \]

Substituting \( h \) back into the expression for \( \psi(x,y) \):

\[ \psi(x,y) = x^2y^2 + 2xy + k \]

General Solution

The solution is defined by \( \psi(x,y) = C \):

\[ x^2y^2 + 2xy + k = C \]\[ x^2y^2 + 2xy = C - k \]
\[ x^2y^2 + 2xy = C \]

Note: The constant \( C \) depends on the Initial Condition (IC).

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Example: Solving an Exact Differential Equation

\[ (e^x \sin y - 2y \sin x) + (e^x \cos y + 2 \cos x + 3)y' = 0 \]

Step 1: Identify M and N

Let:

\[ M = e^x \sin y - 2y \sin x \]\[ N = e^x \cos y + 2 \cos x + 3 \]

Step 2: Test for Exactness

Check if \( M_y = N_x \):

\[ M_y = e^x \cos y - 2 \sin x \]\[ N_x = e^x \cos y - 2 \sin x \]

Since \( M_y = N_x \), the equation is exact.

Step 3: Find the Potential Function \( \psi \)

The solution is of the form \( \psi = C \), where \( \psi_x = M \) and \( \psi_y = N \).

\[ \psi = \int M \, dx = \int (e^x \sin y - 2y \sin x) \, dx \]\[ \psi = e^x \sin y + 2y \cos x + h(y) \]

Step 4: Solve for \( h(y) \)

Differentiate \( \psi \) with respect to \( y \) and set it equal to \( N \):

\[ \psi_y = N \]\[ e^x \cos y + 2 \cos x + h'(y) = e^x \cos y + 2 \cos x + 3 \]\[ h'(y) = 3 \]\[ h(y) = 3y \]
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Final Potential Function

\[ \psi(x, y) = e^x \sin y + 2y \cos x + 3y \]

General Solution

Solution:

\[ e^x \sin y + 2y \cos x + 3y = C \]

If the equation is not exact, use other methods:

  • Separable
  • Linear
  • Homogeneous
  • Bernoulli