A two-dimensional system has a fixed point where the Jacobian has eigenvalues λ₁ = -3 and λ₂ = -1. What type of fixed point is this?
AUnstable node — both eigenvalues push trajectories away
BStable node — both eigenvalues are real and negative, so all trajectories approach the fixed point
CSaddle point — the eigenvalues have opposite signs
DStable spiral — the eigenvalues are complex with negative real part
Both eigenvalues are real and negative, so perturbations decay exponentially along both eigendirections. Trajectories approach the fixed point tangent to the slow eigendirection (λ₂ = -1, which decays more slowly). This is a stable node. If the eigenvalues were complex conjugates with negative real part, you'd get a stable spiral instead.
Question 2 Multiple Choice
A fixed point with eigenvalues λ = ±i (purely imaginary) is classified as a center in linear analysis. For a nonlinear system, can you conclude the fixed point is a center?
AYes — purely imaginary eigenvalues always guarantee a center, even in nonlinear systems
BNo — purely imaginary eigenvalues are a borderline case where nonlinear terms determine whether the fixed point is a true center, a stable spiral, or an unstable spiral
CNo — purely imaginary eigenvalues mean the fixed point is always unstable in the nonlinear case
DYes — but only if the system is Hamiltonian
Purely imaginary eigenvalues put us on the boundary between stable and unstable spirals. The linearization predicts closed orbits, but higher-order nonlinear terms can break the perfect periodicity, causing trajectories to slowly spiral in (stable) or out (unstable). This is precisely why Lyapunov's indirect method fails when eigenvalues have zero real part — the linearization is structurally unstable. Hamiltonian systems are special: their symplectic structure does guarantee centers persist, but this requires additional structural information beyond just eigenvalues.
Question 3 True / False
Lyapunov stability requires that trajectories starting near a fixed point stay near it forever, while asymptotic stability additionally requires that they converge to the fixed point.
TTrue
FFalse
Answer: True
Lyapunov stability (also called stability in the sense of Lyapunov) means: for any ε > 0, there exists δ > 0 such that trajectories starting within δ of x* remain within ε for all future time. Asymptotic stability adds convergence: trajectories not only stay close but actually approach x* as t → ∞. A center is Lyapunov stable but not asymptotically stable — orbits stay close (on closed loops) but never converge to the center. A stable node or spiral is asymptotically stable.
Question 4 Short Answer
Explain why a saddle point, despite being unstable, plays a crucial role in organizing the phase portrait of a dynamical system.
Think about your answer, then reveal below.
Model answer: The stable and unstable manifolds of a saddle point act as separatrices that divide phase space into regions with qualitatively different long-term behavior. Trajectories on the stable manifold approach the saddle; those on the unstable manifold depart from it. These manifolds form the boundaries of basins of attraction for other attractors, so the saddle determines which initial conditions flow to which attractor. The saddle itself is never the final destination (except for the measure-zero set on its stable manifold), but it shapes the global flow topology.
Consider a ball on a saddle-shaped surface: it rolls away in two directions but approaches along two others. In phase space, the stable manifold of a saddle often forms the boundary between the basin of attraction of two stable fixed points. A system that starts on one side of this separatrix ends up at one attractor; starting on the other side leads to the other. This makes saddle points the 'traffic directors' of dynamical systems.