Use local linearization without claiming a global portrait

Try

Consider x′=x−x³ and y′=−2y. An equilibrium must make both rates zero. Solve x(1−x²)=0 and y=0, obtaining three points: (−1,0),(0,0),(1,0). Do not stop after finding the origin; nonlinear equations can have several equilibria with different local behavior.

Draw a one-dimensional sign chart for the x equation and note that the y coordinate always decays toward zero. Predict which of the three points attracts nearby states and which repels in at least one direction.

Predict

Near an equilibrium, a differentiable vector field can be approximated by its derivative matrix, the Jacobian. Predict what happens if an eigenvalue of that matrix is zero: does the linear approximation necessarily determine whether higher-order terms push inward or outward? Compare x′=−x³ and x′=x³ at zero before drawing a conclusion.

Experiment

For the two-dimensional example, the Jacobian is diagonal with entries 1−3x² and −2. Evaluate it at each equilibrium and compare with the sign chart.

Equilibrium	Eigenvalues	Local conclusion
(−1,0)	−2,−2	Asymptotically stable
(0,0)	1,−2	Unstable saddle
(1,0)	−2,−2	Asymptotically stable

At the origin, the y axis contains decaying trajectories, but a small nonzero x component moves away toward one of the outer equilibria. Sketch points on both sides of the y axis to show why one stable direction does not make the whole equilibrium stable.

Understand

For a continuously differentiable autonomous system near an equilibrium, eigenvalues with strictly negative real parts imply local asymptotic stability. A positive real part implies instability. A hyperbolic equilibrium has no eigenvalues with zero real part; its linearization captures the local qualitative behavior in the standard linearization theory. These are local conclusions, not descriptions of all trajectories arbitrarily far away.

Zero real parts require further analysis. Both x′=−x³ and x′=x³ have derivative zero at the origin. In the first equation, positive states decrease and negative states increase, so nearby states approach zero. In the second, states move away on both sides. The identical zero linearization cannot distinguish them. Failure of the linear test is not evidence of neutral stability.

Nonlinear systems can also have attracting cycles. Let r²=x²+y² and consider x′=x(1−r²)−y, y′=y(1−r²)+x. For r positive, differentiation gives r′=r(1−r²), while the angular rate is one. Radius increases below one and decreases above one, producing an attracting periodic orbit r=1. The origin is an unstable equilibrium. A trajectory on the cycle keeps moving; the cycle is not a collection of equilibrium points.

This calculation is exact for the stated synthetic vector field. A plotted spiral alone would not prove a periodic orbit or its stability. Radial dynamics supply the missing argument, and the restriction r positive avoids dividing by zero at the origin.

Apply

Write a local-analysis report: list all equilibria, compute the Jacobian, classify eigenvalues, state the conclusion’s neighborhood scope, and identify inconclusive cases. For the cubic scalar examples, replace an inconclusive linear test with a sign argument.

For the cycle example, distinguish attraction to an orbit from convergence to a fixed state. Explain why a linear model around the origin cannot describe the global attracting circle. Complete the practice with assumptions and counterexamples rather than treating a numerical portrait as proof.

Continue

You now have a connected set of first-order, second-order, system, transform, and local nonlinear methods. Independent modeling, method selection, numerical implementation, and reviewed mixed assessments remain necessary evidence. Boundary-value problems, Fourier methods, partial differential equations, and broader nonlinear theory require additional study; this unit alone does not establish full undergraduate course equivalence.