Classify linear motion beyond two distinct real modes

Try

For x′=−x−2y and y′=2x−y, compute the velocity at (1,0). It is (−1,2), pointing inward and upward. Predict whether nearby trajectories spiral clockwise or counterclockwise. Then test another point rather than inferring a complete portrait from one arrow.

The matrix has eigenvalues −1±2i. Their negative real part suggests decay, while the imaginary part suggests rotation. We can verify both features directly with radius and angle calculations.

Predict

Let R=x²+y² be squared distance from the origin. Predict whether R′ contains the cross terms xy after substituting the system. For comparison, predict what happens to R in the center system x′=−y,y′=x. The distinction between shrinking circles and constant-radius circles separates asymptotic stability from stability alone.

Experiment

For the spiral system, R′=2x(−x−2y)+2y(2x−y)=−2R. Thus radius r=√R decays as e^(−t) for nonzero trajectories. The angular rate is (xy′−yx′)/R=2, so rotation is counterclockwise at two radians per time unit.

System pattern	Eigenvalue information	Forward behavior
Two negative real eigenvalues	Both modes decay	Stable node, with possible repeated-root structure
One negative and one positive	One mode grows	Saddle
Negative real part, nonzero imaginary part	Decay and rotation	Stable spiral
Pure imaginary pair in a linear system	No exponential decay	Center

For the center example, R′=0. Draw three circles with different radii and arrows. They stay near the origin if they start near it, but do not approach it unless they begin there.

Understand

A repeated eigenvalue can produce a defective matrix, meaning there are too few independent eigenvectors for an eigenvector basis. Consider x′=x+y,y′=y. The matrix has only eigenvalue one and one eigenvector direction, the x axis. Solve the second equation first: y=be^t. Then x′−x=be^t, so multiplying by e^(−t) gives (e^(−t)x)′=b. Integration yields x=(a+bt)e^t.

The extra factor t is essential. For initial state (0,2), y=2e^t and x=2te^t, so x(1)/e=2. Writing only a multiple of e^t times the sole eigenvector would miss every solution with a nonzero initial y coordinate. This is the system analogue of a repeated characteristic root.

Stability concerns all sufficiently small perturbations, not one specially selected trajectory. A saddle with eigenvalues −2 and 3 has a decaying trajectory on its stable eigenvector line, yet the equilibrium is unstable because nearby states can contain a growing component. A portrait must represent those neighboring trajectories too.

These classifications concern constant linear systems. Purely imaginary eigenvalues in a nonlinear system do not automatically imply a center. Nonlinear terms may cause inward or outward motion that the linearization cannot detect.

Apply

For each row of the table, sketch several starting points and their qualitative paths. Include a special eigenvector trajectory and a nearby trajectory where relevant. Label the axes with state variables and show time direction with arrows.

Explain why “all eigenvalues have nonpositive real parts” is insufficient for asymptotic stability. A center does not decay, and a defective zero eigenvalue can introduce polynomial growth. Complete the practice by checking both a radial invariant and the defective-system formula directly.

Continue

Laplace transforms offer another way to solve initial-value problems, especially with switching inputs. The state-space picture remains useful for interpreting the transient modes that appear after inversion.

Use parameter families and boundary portraits to partition every real parameter value and resolve repeated, zero and imaginary eigenvalue boundaries.