Stability and Equilibrium

Section 2.2

Autonomous Equations

An Ordinary Differential Equation (ODE) is considered autonomous if there is no explicit dependence on the independent variable \(t\):

$$\frac{dy}{dt} = f(y)$$

Common population models like exponential growth and logistic growth are examples of autonomous equations:

Definition: Critical Point

A critical point of an autonomous equation is a value \(y = c\) (or \(P = c\)) such that \(f(c) = 0\). At this point, \(\frac{dy}{dt} = 0\), making the constant function \(y(t) \equiv c\) an equilibrium solution.

Classifying Stability

Stability describes how solution curves behave in the neighborhood of a critical point:

Worked Examples

Example 1: Basic Analysis

Given \(\frac{dy}{dt} = y - 1\):

  • The function is \(f(y) = y - 1\).
  • Critical point at \(y = 1\).
  • Behavior:
    • If \(y < 1 \implies y' < 0\) (decreasing away from 1).
    • If \(y > 1 \implies y' > 0\) (increasing away from 1).
  • Conclusion: The equilibrium \(y \equiv 1\) is unstable.

Example 2: Semi-stability

Given \(y' = y(y - 1)^2\):

  • Critical points at \(y = 0\) and \(y = 1\).
  • Testing Points:
    • \(y = -1 \implies y' = (-1)(-2)^2 = -4\) (Negative / Away from 0).
    • \(y = 0.5 \implies y' = (0.5)(-0.5)^2 = 0.125\) (Positive / Toward 1).
    • \(y = 2 \implies y' = (2)(1)^2 = 2\) (Positive / Away from 1).
  • Conclusion: \(y = 0\) is unstable, and \(y = 1\) is semi-stable.

Example 3: Logistic Form

Given \(y' = 3y(2 - y)\):

  • Critical points at \(y = 0\) and \(y = 2\).
  • Solution curves move "toward" \(y = 2\) from both sides (above and below).
  • Conclusion: \(y = 0\) is unstable; \(y = 2\) is stable.

Example 4: Complex Polynomial

Given \(\frac{dy}{dt} = (y - 1)(y + 2)(y^2 - 9)\):

Critical points are found at \(y = 1, -2, -3, 3\). Their classifications are as follows:

Critical Point (\(y\)) Classification
3 Unstable
1 Stable
-2 Unstable
-3 Stable