Two classic nonlinear systems show what linearization can and cannot tell us: rabbits and foxes, and a swinging pendulum.
Key idea
Lotka–Volterra: $x' = x(a - by)$ (prey), $y' = y(-c + dx)$ (predators). Coexistence equilibrium $\left(\frac{c}{d}, \frac{a}{b}\right)$; its linearization is a center with angular frequency $\sqrt{ac}$, and the populations cycle on closed orbits.
Pendulum: $\theta'' + \frac{g}{L}\sin\theta = 0$, or $\theta' = \omega$, $\omega' = -\frac{g}{L}\sin\theta$. Hanging down $(0, 0)$ is a center (a spiral sink with damping); balanced upright $(\pi, 0)$ is a saddle. Small swings have period $2\pi\sqrt{L/g}$.
Pendulum: $\theta'' + \frac{g}{L}\sin\theta = 0$, or $\theta' = \omega$, $\omega' = -\frac{g}{L}\sin\theta$. Hanging down $(0, 0)$ is a center (a spiral sink with damping); balanced upright $(\pi, 0)$ is a saddle. Small swings have period $2\pi\sqrt{L/g}$.
Why it works
For Lotka–Volterra, $J = \begin{bmatrix} a - by & -bx \\ dy & -c + dx \end{bmatrix}$; at $\left(\frac{c}{d}, \frac{a}{b}\right)$ the diagonal vanishes and $J = \begin{bmatrix} 0 & -\frac{bc}{d} \\ \frac{ad}{b} & 0 \end{bmatrix}$, with eigenvalues $\pm i\sqrt{ac}$. A center alone is inconclusive, but this model has a conserved quantity, $dx - c\ln x + by - a\ln y$, constant along trajectories, so the orbits are closed. For the pendulum, $J = \begin{bmatrix} 0 & 1 \\ -\frac{g}{L}\cos\theta & 0 \end{bmatrix}$; at $\theta = \pi$, $\cos\theta = -1$ gives $\Delta < 0$, a saddle.
Worked example: Coexistence
Find the coexistence equilibrium of $x' = x(2 - 0.5y)$, $y' = y(-3 + 0.25x)$.
- $2 - 0.5y = 0$ gives $y = 4$; $-3 + 0.25x = 0$ gives $x = 12$.
Answer: $(12, 4)$: 12 prey and 4 predators (in thousands, say)
Worked example: The cycle's period
For that model, estimate the period of small oscillations around $(12, 4)$.
- $ac = 2\cdot 3 = 6$, so the angular frequency is $\sqrt{6}$.
- Period $\approx \frac{2\pi}{\sqrt{6}}$.
Answer: about 2.57 time units
Worked example: A damped pendulum
Classify $(0, 0)$ and $(\pi, 0)$ for $\theta' = \omega$, $\omega' = -4\sin\theta - \omega$.
- $J = \begin{bmatrix} 0 & 1 \\ -4\cos\theta & -1 \end{bmatrix}$.
- At $(0, 0)$: $\tau = -1$, $\Delta = 4$, $\tau^{2} - 4\Delta = -15 < 0$: spiral sink.
- At $(\pi, 0)$: $\Delta = -4 < 0$: saddle.
Answer: hanging down: spiral sink (it swings and settles); upright: saddle (unstable)
Common mistake
Concluding from a linear center that a nonlinear system cycles, or that it does not. For Lotka–Volterra the closed orbits come from the conserved quantity, not from the linearization; add a little crowding term to the prey and the center becomes a spiral sink.