Stability Analysis

Stability Analysis

Finding an equilibrium is only half the story. We also need to know: will the system stay near the equilibrium, or drift away?

Linearization

For $\frac{dy}{dt} = f(y)$ near an equilibrium $y^*$:

$f(y) \approx f(y^*) + f'(y^*) \cdot (y - y^*) = f'(y^*) \cdot (y - y^*)$

since $f(y^*) = 0$. The small perturbation $\delta = y - y^*$ satisfies:

$\frac{d\delta}{dt} = f'(y^*) \cdot \delta$

This is just exponential growth/decay! The solution is $\delta(t) = \delta(0) \cdot e^{f'(y^*) \cdot t}$.

Stability Criterion

Example

For logistic growth $f(y) = y(1 - y)$:

• $f'(y) = 1 - 2y$
• At $y^* = 0$: $f'(0) = 1 > 0$ → unstable (population grows away from 0)
• At $y^* = 1$: $f'(1) = -1 < 0$ → stable (population returns to carrying capacity)

Numerical Derivative

We estimate $f'(y^*)$ using the central difference:

$f'(y^*) \approx \frac{f(y^* + \varepsilon) - f(y^* - \varepsilon)}{2\varepsilon}$

Your Task

Implement stability(f, x_eq, eps=1e-5) that classifies an equilibrium as "stable", "unstable", or "neutral".