Question bank — Linearization of nonlinear systems

False. It is only trustworthy in a tiny neighbourhood of the equilibrium, where the displacements $u=x-x^{\ast }$, $v=y-y^{\ast }$ are small enough that quadratic terms are negligible. Far away the discarded terms dominate and the picture can be completely different.

False. A symbolic $J(x,y)$ still contains $x,y$ and is not a constant-coefficient linear system. You must evaluate $J$ at a specific equilibrium to get numeric entries before eigenvalues mean anything.

True. Both eigenvalues have negative real part, so the point is hyperbolic; Hartman–Grobman then guarantees the nonlinear flow looks like the (stable node) linear flow nearby.

False. Pure imaginary eigenvalues mean $\text{Re}(\lambda )=0$, so the point is non-hyperbolic and the dropped quadratic terms can turn the "center" into a slow inward or outward spiral. You need Lyapunov or energy arguments to decide.

True. A saddle has real eigenvalues of opposite sign, so at least one mode grows like $e^{\lambda t}$ with $\lambda >0$ — that growing direction makes it unstable no matter what.

False. Setting $f=g=0$ is a system of equations that can have zero, one, or many solutions (the pendulum has infinitely many, at $x=n\pi$). Each equilibrium is analysed separately.

True. $\Delta >0$ with $\tau <0$ forces both eigenvalue real parts negative (their sum is $\tau <0$, product $\Delta >0$), so every mode decays.

True. Both eigenvalue real parts are now positive (sum $\tau >0$, product $\Delta >0$). It is an unstable node if ${\tau }^2-4\Delta \ge 0$ (real eigenvalues) and an unstable spiral if ${\tau }^2-4\Delta <0$ (complex eigenvalues).

True. The sign of the discriminant ${\tau }^2-4\Delta$ distinguishes them: negative gives complex eigenvalues (spiral), non-negative gives real eigenvalues (node). Both are stable if $\text{Re}(\lambda )<0$.

True — provided it is hyperbolic. A positive-real-part eigenvalue (hyperbolic) guarantees a growing direction survives in the nonlinear flow by Hartman–Grobman. If instead the "instability" came from a zero-real-part case, no such guarantee exists.

Error: incomplete root-finding. The factored second equation $y(-1+x)=0$ also allows $y=0$, which combined with the first equation opens up other solutions like $(0,0)$. You must exhaust every branch of $f=g=0$.

Error: labelling it 'stable'. A center has $\text{Re}(\lambda )=0$, so it neither decays nor grows — it is not asymptotically stable, and moreover linearization is inconclusive here for the true nonlinear behaviour.

Error: ignoring the zero eigenvalue. A zero eigenvalue means $\text{Re}(\lambda )=0$, making the point non-hyperbolic. Hartman–Grobman fails and stability cannot be read off the linearization at all.

Error: never substituting the equilibrium. Eigenvalues of a matrix full of variables are not the constant growth rates you need; substitute $(x^{\ast },y^{\ast })$ first, then solve the characteristic polynomial.

Error: it IS zero. By the very definition of an equilibrium, $f(x^{\ast },y^{\ast })=0$. If it weren't zero you weren't at an equilibrium and the whole linearization setup does not apply.

Error: rounding away hyperbolicity. Any nonzero real part, however small, makes the point hyperbolic — here an unstable spiral. "Basically zero" is not zero; the flow genuinely spirals outward.

Because near the equilibrium $u,v$ are small, so their squares are far smaller still ($0.01^2=0.0001$). The linear terms dominate, so keeping them captures the leading behaviour — this is the tangent-plane approximation.

Solutions are sums of $e^{\lambda t}\mathbf{v}$; the exponent $\lambda$ controls whether each mode grows or decays. The eigenvector $\mathbf{v}$ only sets the direction of that mode, not whether it survives in time.

The Jacobian collects the first partial derivatives $f_x,f_y,g_x,g_y$ — exactly the slopes of the tangent plane to each surface at the equilibrium. It is by construction the best linear approximation of the vector field there.

A zero real part is a knife-edge: the linear flow neither grows nor decays, so the tiny quadratic terms we threw away become decisive. Only when every mode strictly grows or strictly decays can we ignore them safely.

Linearization alone cannot certify it, but the pendulum conserves energy, and closed energy contours force closed orbits — an independent argument that upgrades the inconclusive linear center to a real one.

At the bottom the restoring force pulls a nudged pendulum back (oscillation → center); at the top a nudge is amplified by gravity, giving one growing and one shrinking direction (mixed-sign real eigenvalues → saddle).

An equilibrium needs both velocities to vanish at once; a point satisfying $f=0$ but $g\ne 0$ still moves. Only the common solutions of both equations are true fixed points.

At least one eigenvalue is zero (their product is $\Delta =0$), so the point is non-hyperbolic and linearization is inconclusive — a whole line or curve of near-equilibria may exist, needing deeper analysis.

The linearization is a center (closed elliptical loops), but $\text{Re}(\lambda )=0$ makes it non-hyperbolic, so it is inconclusive: the true nonlinear point could be a genuine center or a slow spiral. Confirm with an energy or Lyapunov argument, as with the pendulum's bottom.

It is an unstable improper node (both $\text{Re}(\lambda )>0$, so hyperbolic and unstable). If $J$ is diagonalizable it is an unstable star node; if $J$ is defective it is an unstable improper node whose trajectories flee outward along a single shear direction — no rotation involved.

Yes, but check whether $J$ is defective. A repeated negative eigenvalue keeps $\text{Re}(\lambda )<0$, so the point is hyperbolic and stable, and Hartman–Grobman applies. If $J$ is diagonalizable it is a star node; if $J$ is defective (non-diagonalizable, needing a Jordan block) it is an improper node whose trajectories bend along a single eigendirection by real-axis shear — there is no rotation or spiralling, the geometry differs but the stability is unchanged, so inspect the Jordan form.

The Jacobian requires partial derivatives to exist; if $f$ or $g$ is non-smooth there, no Jacobian exists and standard linearization simply does not apply — you need other methods.

Worse. A double-zero eigenvalue is highly degenerate: not even a rotation is guaranteed, and it is not genuinely imaginary (which would need a nonzero $\omega$). This strongly non-hyperbolic case requires Lyapunov or centre-manifold techniques.

It is a saddle (opposite signs → unstable), but "magnitude" only sets how fast trajectories move along each direction, not the qualitative type. One growing direction is enough for instability.

Not from linearization alone — it is non-hyperbolic. For the classic Lotka–Volterra model a conserved quantity independently proves the orbits really are closed, but a generic system with a linear center could secretly spiral.