Linearization of nonlinear systems

What we are studying

HOW we linearize — derive from scratch

We want to know how a small disturbance behaves. Let $u = x - x^*, \qquad v = y - y^*$ be tiny displacements from the equilibrium. Then $\dot u = \dot x = f(x,y)$.

Taylor expand $f$ about $(x^*,y^*)$ (this is the whole trick): $f(x,y) = \underbrace{f(x^*,y^*)}_{=0} + f_x\,u + f_y\,v + \underbrace{\tfrac12 f_{xx}u^2+\dots}_{\text{higher order}}$

• Why is the first term 0? Because $(x^*,y^*)$ is an equilibrium — by definition $f(x^*,y^*)=0$.
• Why drop the quadratic terms? Near equilibrium $u,v$ are small, so $u^2, uv, v^2$ are tiny-squared — negligible compared to the linear terms. This is the 80/20 move: keep the part that dominates.

Doing the same for $g$:

Classifying the equilibrium

Solve $\det(J-\lambda I)=0$ for eigenvalues $\lambda_{1,2}$. With trace $\tau = f_x+g_y$ and determinant $\Delta = f_xg_y - f_yg_x$: $\lambda = \frac{\tau \pm \sqrt{\tau^2 - 4\Delta}}{2}$

Worked Example 1 — a simple nonlinear pair

System: $\dot x = x - x^2 - xy,\quad \dot y = -y + xy$ (a Lotka–Volterra-style model).

Step 1 — find equilibria. Set both to 0.

• Why? Equilibria are where the analysis lives. $x(1-x-y)=0$ and $y(-1+x)=0$. From the 2nd: $y=0$ or $x=1$.
• $y=0$: first gives $x(1-x)=0 \Rightarrow (0,0),(1,0)$.
• $x=1$: first gives $1-1-y=0\Rightarrow y=0$ (already have it). So equilibria: $(0,0)$ and $(1,0)$.

Step 2 — Jacobian (general). $J=\begin{pmatrix}1-2x-y & -x\\ y & -1+x\end{pmatrix}$ Why these entries? $f_x=\partial_x(x-x^2-xy)=1-2x-y$, etc.

Step 3 — evaluate at $(0,0)$. $J=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\Rightarrow \lambda=1,\,-1.$ Opposite signs → saddle (unstable). Why? One mode grows ($e^{t}$), one decays.

Step 4 — evaluate at $(1,0)$. $J=\begin{pmatrix}-1&-1\\0&0\end{pmatrix}\Rightarrow \lambda=-1,\,0.$ A zero eigenvalue → non-hyperbolic, linearization inconclusive (steel-man flag below).

Worked Example 2 — the pendulum

$\ddot\theta + \sin\theta = 0$. Write as a system with $x=\theta,\ y=\dot\theta$: $\dot x = y,\qquad \dot y = -\sin x.$

Step 1 — equilibria. $y=0$ and $\sin x=0\Rightarrow x=n\pi$. Two physical cases: $(0,0)$ = hanging down, $(\pi,0)$ = balanced upright.

Step 2 — Jacobian. $J=\begin{pmatrix}0&1\\-\cos x&0\end{pmatrix}$ Why $-\cos x$? $\partial_x(-\sin x)=-\cos x$.

Step 3 — bottom $(0,0)$: $J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, $\lambda=\pm i$. Pure imaginary → center (predicted oscillation). Physically right! But formally linearization is inconclusive — energy conservation confirms it really is a center.

Step 4 — top $(\pi,0)$: $\cos\pi=-1$, so $J=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\lambda=\pm1$. Saddle → unstable. Of course: balancing a pendulum upright is unstable. ✔

Common mistakes (Steel-man + fix)

Active-recall flashcards

Connections

• Phase Plane Analysis — where trajectories live.
• Eigenvalues and Eigenvectors — engine of classification.
• Stability of Equilibria — node/saddle/spiral/center catalogue.
• Taylor Series — the approximation we truncate.
• Lyapunov Functions — rescue tool for non-hyperbolic cases.
• Hartman-Grobman Theorem — the legal justification.
• Lotka-Volterra Predator Prey — classic application.

Concept Map

Hinglish (regional understanding)