Most real systems are nonlinear: $x' = f(x, y)$, $y' = g(x, y)$. We cannot usually solve them, but near an equilibrium they behave like a linear system, and phase portraits tell us what that linear system does. The tool is the Jacobian matrix of partial derivatives.
Key idea
At an equilibrium $(x^{*}, y^{*})$ (where $f = g = 0$), the linearization is $\mathbf{u}' = J\mathbf{u}$ with $$J = \begin{bmatrix} f_x & f_y \\ g_x & g_y \end{bmatrix}_{(x^{*}, y^{*})}.$$ If $J$ gives a saddle, node or spiral, the nonlinear system has the same type near the equilibrium. If $J$ gives a center, the test is inconclusive.
Why it works
Let $u = x - x^{*}$, $v = y - y^{*}$. The tangent plane approximation gives $f(x, y) \approx f(x^{*}, y^{*}) + f_xu + f_yv$ $= f_xu + f_yv$, since $f = 0$ at the equilibrium; similarly for $g$. So $(u, v)' \approx J(u, v)$, with errors of second order in the small displacements. When the eigenvalues have nonzero real parts, those small errors cannot change the qualitative picture (the Hartman–Grobman theorem). For a center, the real part is 0, and the tiny nonlinear terms can tip the orbits into spirals either way.
Worked example: Find the equilibria
Find the equilibria of $x' = x - y$, $y' = x^{2} - 4$.
- $x' = 0$ gives $y = x$; $y' = 0$ gives $x = \pm 2$.
Answer: $(2, 2)$ and $(-2, -2)$
Worked example: Linearize and classify
Classify both equilibria of that system.
- $J = \begin{bmatrix} 1 & -1 \\ 2x & 0 \end{bmatrix}$.
- At $(2, 2)$: $J = \begin{bmatrix} 1 & -1 \\ 4 & 0 \end{bmatrix}$, $\tau = 1$, $\Delta = 4$, $\tau^{2} - 4\Delta < 0$: spiral source.
- At $(-2, -2)$: $J = \begin{bmatrix} 1 & -1 \\ -4 & 0 \end{bmatrix}$, $\Delta = -4 < 0$: saddle.
Answer: $(2, 2)$ is a spiral source, $(-2, -2)$ is a saddle
Worked example: Competing species
For $x' = x(5 - x - 2y)$, $y' = y(4 - x - y)$, classify the equilibrium with $x, y > 0$.
- Both brackets vanish: $x + 2y = 5$ and $x + y = 4$, so $(x, y) = (3, 1)$.
- $J = \begin{bmatrix} 5 - 2x - 2y & -2x \\ -y & 4 - x - 2y \end{bmatrix}$; at $(3, 1)$: $\begin{bmatrix} -3 & -6 \\ -1 & -1 \end{bmatrix}$.
- $\Delta = 3 - 6 = -3 < 0$.
Answer: a saddle: coexistence is unstable, and one species eventually wins
Common mistake
Evaluating the Jacobian at the origin, or forgetting to plug in the equilibrium at all, and classifying a matrix that still contains $x$ and $y$. Each equilibrium has its own Jacobian. And a center of the linearization does not prove the nonlinear system has closed orbits.