Questions: Linearization and the Jacobian Matrix

Question 1

For the system ẋ = x² + y, ẏ = x - y, what is the Jacobian matrix at the fixed point (0, 0)?

• A. [[0, 1], [1, -1]]
• B. [[1, 1], [1, -1]]
• C. [[0, 0], [1, -1]]
• D. [[2x, 1], [1, -1]] evaluated at (0,0), which is [[0, 1], [1, -1]]

Answer: (D)

The Jacobian is [[∂f₁/∂x, ∂f₁/∂y], [∂f₂/∂x, ∂f₂/∂y]] = [[2x, 1], [1, -1]]. At (0,0), this gives [[0, 1], [1, -1]]. Options A and D describe the same correct matrix. The eigenvalues of this matrix ((-1 ± √5)/2) determine local stability: one positive and one negative eigenvalue means (0,0) is a saddle point.

Question 2

The Hartman-Grobman theorem guarantees that linearization gives the correct qualitative picture near a fixed point. Under what condition does this guarantee fail?

• A. When the fixed point is a saddle — saddles are too sensitive to nonlinear perturbation
• B. When any eigenvalue of the Jacobian has zero real part — the fixed point is non-hyperbolic
• C. When the system has more than two dimensions
• D. When the nonlinear terms are not polynomial

Answer: (B)

Hartman-Grobman requires the fixed point to be hyperbolic: all eigenvalues must have nonzero real parts. When an eigenvalue has zero real part (Re(λ) = 0), the linear and nonlinear systems can have qualitatively different phase portraits. A linear center might become a nonlinear spiral; a zero eigenvalue might correspond to a bifurcation. Dimensionality and the form of nonlinear terms don't affect the theorem's applicability — only the eigenvalue condition matters.

Question 3

Linearization around a fixed point tells you the exact quantitative behavior of the nonlinear system in a neighborhood of that point.

• T. True
• F. False

Answer: False

Linearization gives qualitative (topological) equivalence, not quantitative agreement. The Hartman-Grobman theorem says there exists a homeomorphism (continuous but not necessarily smooth map) between the nonlinear and linear phase portraits near a hyperbolic fixed point. This means the topology of trajectories is the same — same number of attracting/repelling directions, same type of fixed point — but distances, speeds, and angles may differ. For quantitative predictions, you need higher-order terms or numerical methods.

Question 4

A three-dimensional system has a fixed point whose Jacobian has eigenvalues -2, -1 + 3i, and -1 - 3i. Describe the local behavior near this fixed point.

Model answer: All three eigenvalues have negative real parts, so the fixed point is asymptotically stable. The real eigenvalue -2 gives exponential decay along one direction. The complex conjugate pair -1 ± 3i produces oscillatory decay (spiraling) in the plane spanned by the corresponding eigenvectors. Locally, trajectories spiral inward while also decaying along the third direction, creating a stable spiral node — trajectories approach the fixed point along helical paths.

In three dimensions, you can have richer combinations than in 2D. Here, one eigendirection is a pure exponential decay (the real eigenvalue) and a two-dimensional eigenplane has spiraling decay (the complex pair). The spiral frequency is 3 (from the imaginary part) and the decay rate is 1 (from the real part). Because all real parts are negative, the fixed point is a global attractor in its neighborhood.