Linearization and Phase Plane Analysis of Nonlinear Systems

MATH 341 — Notes 12

University of Scranton

2026-06-16

Goals

What We Will Cover

Critical points of nonlinear systems — definition and location via nullclines
Jacobian matrix — derived from the Taylor expansion; the linearized system
Behavior near a critical point — the nonlinear system inherits stability from its linearization (with one exception)
Asymptotic stability criterion: \(\text{tr}\,J < 0\) and \(\det J > 0\)
Nullclines and direction fields — tools for the global phase diagram
Seven-step procedure for sketching phase diagrams
Three worked examples — two critical points, three critical points, and a degenerate case where linearization fails

Note

Section 5.1 of (Logan 2015).

Tip

New vocabulary: critical point, Jacobian matrix, linearization, perturbation, nullcline, degenerate critical point.

From Scalar to Planar

Linearization for a Scalar Equation (Review)

For \(x' = f(x)\): equilibrium \(x_e\) satisfies \(f(x_e) = 0\).

Perturb: \(x(t) = x_e + u(t)\), expand by Taylor’s theorem:

\[u' = f(x_e) + f'(x_e)\,u + \tfrac{1}{2}f''(x_e)\,u^2 + \cdots\]

Since \(f(x_e)=0\) and higher terms are small:

\[u' \approx f'(x_e)\,u \implies u(t) = u(0)\,e^{f'(x_e)t}.\]

Sign of \(f'(x_e)\)	Behavior	Stability
\(< 0\)	Perturbations decay	Asymptotically stable
\(> 0\)	Perturbations grow	Unstable
\(= 0\)	No information	Must go to higher order

We now carry out the same program for two-dimensional nonlinear systems.

Critical Points of Nonlinear Systems

Consider the nonlinear autonomous system

\[x' = f(x,y), \qquad y' = g(x,y). \tag{5.1–5.2}\]

Definition: Critical Point

A critical point (equilibrium) \((x_e, y_e)\) satisfies \[f(x_e, y_e) = 0 \quad\text{and}\quad g(x_e, y_e) = 0\] simultaneously. Critical points represent constant solutions.

Tip

How to find them: the set \(f=0\) defines the \(x\)-nullclines; the set \(g=0\) defines the \(y\)-nullclines. Critical points are intersections of \(x\)-nullclines with \(y\)-nullclines.

The Jacobian Matrix

Perturb: \(x = x_e + u\), \(y = y_e + v\), where \(u,v\) are small. Taylor-expand the right sides about \((x_e, y_e)\):

\[u' = \underbrace{f(x_e,y_e)}_{=\,0} + f_x u + f_y v + \cdots \approx f_x u + f_y v,\] \[v' = \underbrace{g(x_e,y_e)}_{=\,0} + g_x u + g_y v + \cdots \approx g_x u + g_y v.\]

In matrix form: \(\mathbf{z}' = J\,\mathbf{z}\), where

Definition: Jacobian Matrix

\[J(x_e,y_e) = \begin{pmatrix} f_x(x_e,y_e) & f_y(x_e,y_e) \\ g_x(x_e,y_e) & g_y(x_e,y_e) \end{pmatrix}. \tag{5.4}\]

The linearized system \(\mathbf{z}' = J\mathbf{z}\) is a linear constant-coefficient system — everything from Chapter 4 applies.

Behavior Near a Critical Point

Key Results (assuming \(\det J \neq 0\))

Asymptotically stable: If \(J\) has eigenvalues with negative real parts, then \((x_e,y_e)\) is asymptotically stable for the nonlinear system. Equivalent conditions: \[\text{tr}\,J(x_e,y_e) < 0 \quad\text{and}\quad \det J(x_e,y_e) > 0. \tag{5.5}\]
Unstable: If \(J\) has an eigenvalue with positive real part, then \((x_e,y_e)\) is unstable.
Exceptional case: If \(J\) has purely imaginary eigenvalues (center for the linearization), the nonlinear system may be a center or a spiral — linearization is inconclusive.
Borderline case: Equal eigenvalues preserve stability/instability but the orbital shape may differ (e.g., node → spiral).

Tip

In practice: compute \(\text{tr}\,J\) and \(\det J\) at each equilibrium. This determines stability without solving the characteristic equation.

Nullclines and the Phase Diagram

Nullclines

\(x\)-nullclines (\(f(x,y)=0\)): vector field is vertical (straight up or down)
\(y\)-nullclines (\(g(x,y)=0\)): vector field is horizontal (straight left or right)
Intersections of \(x\)- and \(y\)-nullclines are critical points

Seven-Step Procedure

Find critical points; classify by \(J\) (eigenvalues, or tr/det test)
Draw \(x\)- and \(y\)-nullclines
Determine sign of \(f\) and \(g\) in each region: N/S/E/W → NE/NW/SE/SW
Sketch representative orbits
Find separatrix directions from eigenvectors of \(J\) at saddles/nodes
If possible, divide and integrate to find orbits \(y=y(x)\)
Use software for accurate phase portrait

Worked Examples

Example 5.1 — Setup

System:

\[x' = -x + xy, \qquad y' = -4y + 8xy.\]

Critical points. Factor each equation:

\[x(-1+y)=0 \implies x=0 \text{ or } y=1.\] \[4y(-1+2x)=0 \implies y=0 \text{ or } x=\tfrac{1}{2}.\]

Case \(x=0\): second equation gives \(y=0\). \(\implies (0,0)\).
Case \(y=1\): second equation gives \(x=\tfrac{1}{2}\). \(\implies (\tfrac{1}{2},1)\).

Jacobian:

\[J(x,y) = \begin{pmatrix}-1+y & x \\ 8y & -4+8x\end{pmatrix}.\]

Example 5.1 — Classification

At \((0,0)\):

\[J(0,0) = \begin{pmatrix}-1 & 0 \\ 0 & -4\end{pmatrix}, \quad \lambda = -1,\,-4.\]

\(\text{tr}\,J = -5 < 0\), \(\det J = 4 > 0\) \(\implies\) asymptotically stable node. \(\checkmark\)

At \((\tfrac{1}{2},1)\):

\[J\!\left(\tfrac{1}{2},1\right) = \begin{pmatrix}0 & \tfrac{1}{2} \\ 8 & 0\end{pmatrix}, \quad \lambda^2-4=0, \quad \lambda=\pm 2.\]

\(\det J = -4 < 0\) \(\implies\) unstable saddle point. \(\checkmark\)

Nullclines. \(x\)-nullclines: \(x=0\) and \(y=1\). \(y\)-nullclines: \(y=0\) and \(x=\tfrac{1}{2}\).

Example 5.1 — Phase Portrait

Example 5.2 — Three Critical Points

System: \(x' = x - x^3\), \(y' = 2y\). Equilibria: \((0,0)\), \((\pm 1, 0)\).

\[J(x,y) = \begin{pmatrix}1-3x^2 & 0 \\ 0 & 2\end{pmatrix}.\]

Point	\(J\)	Eigenvalues	Type
\((0,0)\)	\(\begin{pmatrix}1&0\\0&2\end{pmatrix}\)	\(1,\,2\)	Unstable node
\((\pm 1,0)\)	\(\begin{pmatrix}-2&0\\0&2\end{pmatrix}\)	\(-2,\,2\)	Saddle

When Linearization Fails

Example 5.3 — Degenerate Critical Point

System: \(x' = y^2\), \(y' = -\tfrac{2}{3}x\). Only critical point: \((0,0)\).

\[J(0,0) = \begin{pmatrix}0 & 0 \\ -\tfrac{2}{3} & 0\end{pmatrix}, \quad \det J = 0, \quad \lambda = 0,\,0.\]

\(\det J = 0\): Linearization Does Not Apply

When both eigenvalues are zero, the standard linear classification fails. We must use a different strategy.

Alternative — Divide and Integrate:

\[\frac{dy}{dx} = \frac{-\tfrac{2}{3}x}{y^2} \implies 3y^2\,dy = -2x\,dx \implies y^3 = -x^2 + C.\]

Orbits are the curves \(y = (C - x^2)^{1/3}\). Since \(x' = y^2 \ge 0\), all orbits move to the right.

Example 5.3 — Phase Portrait

Note

All orbits move to the right since \(x' = y^2 \ge 0\). The critical point is not of standard type — unusual orbital behavior is common for nonlinear systems.

Summary

Key Takeaways

A critical point \((x_e, y_e)\) satisfies \(f(x_e,y_e)=0\) and \(g(x_e,y_e)=0\). Critical points are intersections of \(x\)-nullclines with \(y\)-nullclines.
The Jacobian matrix \(J(x_e,y_e) = \begin{pmatrix}f_x & f_y \\ g_x & g_y\end{pmatrix}\) evaluated at the equilibrium gives the coefficient matrix of the linearized perturbation equations \(\mathbf{z}' = J\mathbf{z}\).
The nonlinear system inherits stability from its linearization in all non-exceptional cases. The key test is condition (5.5): \(\text{tr}\,J < 0\) and \(\det J > 0\) \(\implies\) asymptotically stable.
Exception (center): If \(J\) has purely imaginary eigenvalues, the nonlinear system may be a center or a spiral — linearization cannot distinguish these. Higher-order analysis or a Lyapunov function is required.
Degenerate case (\(\det J = 0\)): Linearization fails. Divide and separate variables to find orbits \(y=y(x)\) when possible, or use numerical methods.
For the phase diagram: find nullclines, classify all critical points via \(J\), determine direction field in each region, and sketch representative orbits. Eigenvectors of \(J\) give the directions of separatrices at saddle points.

Tip

Looking Ahead: §5.2 applies linearization to nonlinear mechanics: Newton’s law \(mx''=F(x,x')\) written as \(x'=y\), \(y'=F(x,y)/m\). The critical points correspond to equilibrium positions, and their type describes whether the motion is oscillatory or monotone.

Note

Next: Nonlinear Mechanics — Logan §5.2.

References

Logan, J David. 2015. A First Course in Differential Equations, Third Edition.