Review / Part II — Higher-Order & Nonlinear / Ch 13

Linearization & the Jacobian

Ch 13 — Part II — Higher-Order & Nonlinear

Near an equilibrium of a nonlinear system \(\mathbf{x}' = \mathbf{F}(\mathbf{x})\), the behavior is well approximated by the Jacobian \(J = D\mathbf{F}(\mathbf{x}^*)\), provided the equilibrium is hyperbolic (no eigenvalue on the imaginary axis).

Hartman–Grobman (informal)

If \(J\) has no purely-imaginary or zero eigenvalues, trajectories near \(\mathbf{x}^*\) look topologically like those of the linear system \(\mathbf{u}' = J \mathbf{u}\).

Algorithm

  1. Find equilibria by solving \(\mathbf{F}(\mathbf{x}^*) = \mathbf{0}\).
  2. Compute Jacobian \(J = \partial F_i / \partial x_j\).
  3. Evaluate \(J\) at \(\mathbf{x}^*\) and classify eigenvalues.
  4. Infer local portrait (node, spiral, saddle, etc.).

Centers are the edge case

When linearization gives purely imaginary eigenvalues, the nonlinear system may have a center, spiral, or more exotic behavior — need conserved quantities or Lyapunov functions.

Practice Problems

Problem 1hard
Find equilibria of \(x' = x - x^3 - xy,\; y' = -y + x^2 y\) and classify \((0,0)\) via the Jacobian.
Hint
Factor each equation and look at \((0,0)\).
Solution

From the second equation, \(y(x^2 - 1) = 0\). With \(y=0\), the first gives \(x(1-x^2)=0\Rightarrow x\in\{0,\pm 1\}\). So \((0,0), (\pm 1, 0)\) are all equilibria. The Jacobian is \(J = \begin{pmatrix}1 - 3x^2 - y & -x \\ 2xy & -1 + x^2\end{pmatrix}\). At \((0,0)\) we get \(\operatorname{diag}(1, -1)\) — eigenvalues \(+1, -1\), a saddle. (At \((\pm 1, 0)\) the Jacobian has a zero eigenvalue, so Hartman–Grobman fails and the test is inconclusive.)

Answer: \((0,0)\) is a saddle.

Problem 2medium
Linearize \(x' = y,\; y' = -\sin x\) (pendulum) at \((0,0)\) and \((\pi,0)\).
Hint
\(J = \begin{pmatrix}0 & 1 \\ -\cos x & 0\end{pmatrix}\).
Solution

At (0,0): \(\lambda^2 + 1 = 0\Rightarrow\pm i\) — linear center (nonlinear also center by conservation). At \((\pi,0)\): \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\), \(\lambda = \pm 1\), saddle.

Answer: Origin center/linear, \((\pi,0)\) saddle.