Question bank — Stability of equilibria — stable, unstable, saddle, spiral, centre

False ::: $\Delta <0$ forces the two eigenvalues to have opposite signs, so one always grows — this is a saddle, which is unstable. Stability needs $\tau <0$ and $\Delta >0$ together.

False ::: $\tau <0$ only means the eigenvalues sum to a negative number. If $\Delta <0$ you still get a saddle (e.g. ${\lambda }_1=-3,{\lambda }_2=+1$ give $\tau =-2<0$ but one grows). You need $\Delta >0$ as well.

False ::: A centre is only neutrally (marginally) stable. A nudged trajectory settles onto a new closed orbit and never returns to the fixed point, so it fails the "converges as $t\to \infty$" test that asymptotic stability requires. See Lyapunov stability.

Mostly false ::: Complex eigenvalues mean rotation is present, but if the real part is exactly zero ($\tau =0$) you get a centre (closed loops), not a spiral. Only $\mathrm{Re}(\lambda )\ne 0$ gives an inward/outward spiral.

True, with a caveat ::: Both real and negative $\Rightarrow$ every mode decays $\Rightarrow$ stable node. But "both negative" already requires $\Delta ={\lambda }_1{\lambda }_2>0$, $\tau ={\lambda }_1+{\lambda }_2<0$, and $D\ge 0$. Why $D\ge 0$? Because the eigenvalues equal $\frac{\tau \pm \sqrt{D}}{2}$, and if $D<0$ that square root is imaginary — the roots would be a complex pair, not two real negatives. Real eigenvalues require a non-negative discriminant.

False ::: It is reliable only at hyperbolic points, where no eigenvalue has zero real part (Linearisation and Hartman–Grobman theorem). At a borderline centre ($\tau =0$) the discarded nonlinear terms can turn it into a slow spiral.

$t\to -t$) turns a stable node into an unstable node. True ::: Reversing time flips every eigenvalue's sign ($\lambda \to -\lambda$), so $\tau \to -\tau$ while $\Delta$ (a product) is unchanged. A stable node ($\tau <0$) becomes an unstable node ($\tau >0$); a saddle stays a saddle.

$D=0$. False ::: The spiral/centre region is $D<0$; the star and degenerate nodes live exactly on $D={\tau }^2-4\Delta =0$. The parabola is the borderline between real-eigenvalue nodes and complex-eigenvalue spirals in the Trace–determinant plane.

$J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ has $\tau =0,\Delta =1$, so it's a stable centre." Error ::: A centre is not stable in the asymptotic sense — call it a centre (neutrally stable). Saying "stable centre" wrongly implies trajectories return to the origin.

$(\pi ,0)$ has $\Delta =1>0$, so it's a stable node." Error ::: Recompute: $-\cos \pi =+1$, giving $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ with $\Delta =(0)(0)-(1)(1)=-1<0$ — a saddle. The sign slip on $\Delta$ flipped an unstable equilibrium into a fake stable one.

Error ::: The Jacobian matrix must be re-evaluated at each equilibrium separately — the partial derivatives depend on position (e.g. $-\cos x$ differs at $x=0$ vs $x=\pi$), so the same system can host a centre and a saddle.

$D<0$ means the eigenvalues are negative, so the point is stable." Error ::: $D<0$ makes eigenvalues complex, not "negative" — $D$ is under a square root, it is not $\lambda$ itself. Stability of the resulting spiral is decided by $\mathrm{Re}(\lambda )=\tau /2$, i.e. the sign of $\tau$.

$D<0$ with $\tau >0$ is still fine — spirals are bounded, so it can't blow up." Error ::: $D<0,\tau >0$ is an unstable spiral: $\mathrm{Re}(\lambda )=\tau /2>0$ means the amplitude grows while rotating, so trajectories spiral outward and run away. Boundedness is not guaranteed by rotation alone.

$\det (J-\lambda I)={\lambda }^2-\tau \lambda +\Delta$, the eigenvalues are $\tau$ and $\Delta$." Error ::: $\tau$ and $\Delta$ are the sum and product of the eigenvalues (Vieta), not the eigenvalues themselves. The roots are $\lambda =\frac{\tau \pm \sqrt{{\tau }^2-4\Delta }}{2}$.

Error ::: A saddle has a stable direction (the eigenvector with $\lambda <0$) along which special trajectories do approach. Only trajectories not perfectly aligned with it eventually escape along the unstable direction.

$\Delta$ tell us whether eigenvalues share a sign? Because $\Delta ={\lambda }_1{\lambda }_2$ ::: A product is positive only when both factors share a sign, and negative when they have opposite signs — that single number encodes "same sign or not" without solving for the roots.

Because $\tau ={\lambda }_1+{\lambda }_2=2\mathrm{Re}(\lambda )$ for complex pairs ::: The trace is (twice) the average real part, and the real part is the growth/decay exponent in $e^{\lambda t}$ — so $\tau <0$ means the average mode decays inward.

Because near a fixed point the flow is almost linear ::: A first-order Taylor expansion turns the curvy velocity functions $f,g$ into a constant matrix $J$ whose eigenvalues we can read off exactly, giving a proof of local behaviour rather than a picture.

Because light damping keeps $D=4{\beta }^2-4{\omega }_0^2<0$ ::: With $\beta <{\omega }_0$ the discriminant is negative, forcing complex eigenvalues, so rotation (oscillation) coexists with decay — that is exactly a stable spiral. See Damped harmonic oscillator.

$2\times 2$ real system have exactly one complex eigenvalue? Because the characteristic polynomial has real coefficients ::: Complex roots of a real polynomial come in conjugate pairs, so $\lambda$ and $\bar{\lambda }$ always appear together — you get two complex eigenvalues or none.

$\pm i\omega$ produce closed loops rather than a slow inward drift? Because there is zero real part ::: $e^{(0\pm i\omega )t}$ has constant magnitude and only rotates, so trajectories neither shrink nor grow — geometrically they trace ellipses forever, the signature of a centre.

parabola in the $(\tau ,\Delta )$ plane? Because that boundary is exactly $D=0$ ::: Setting ${\tau }^2-4\Delta =0$ gives $\Delta ={\tau }^2/4$, the equation of an upward parabola. Above it ($\Delta >{\tau }^2/4$) $D<0$ so you spiral; below it $D>0$ so you have real-eigenvalue nodes — see the sketch above and Trace–determinant plane.

$D={\tau }^2-4\Delta =0$ with $\tau <0$? A degenerate/star stable node ::: The eigenvalues coincide (repeated real $\lambda =\tau /2<0$). All modes still decay so it is stable, but trajectories approach along a single or shared direction rather than two distinct ones.

$\Delta >0,\tau >0,D<0$? An unstable spiral ::: Complex eigenvalues with positive real part $\tau /2>0$: trajectories rotate while their amplitude grows, spiralling outward away from the fixed point. It is the mirror image of the stable spiral (flip the sign of $\tau$).

$\Delta =0$ (with $\tau \ne 0$) signify? A whole line of equilibria ::: One eigenvalue is zero, so the fixed point is non-isolated: there is a neutral direction along which the system doesn't move, and the classification into node/spiral breaks down.

$\tau =0$ and $\Delta =0$? A doubly degenerate case ::: Both eigenvalues are zero; the linearisation is entirely uninformative and the fate is decided completely by the nonlinear terms — never trust the linear picture here.

Whether the nonlinear terms preserve it ::: Since $\tau =0$ is non-hyperbolic, Hartman–Grobman does not apply; the true system may be a weak spiral. A conserved quantity (energy) is the usual way to confirm a genuine centre.

Only measure-zero ones ::: Trajectories launched exactly on the stable eigenvector approach the fixed point, but any real-world nudge off that line eventually escapes — so a saddle counts as unstable overall.

$\Delta >0$, $\tau >0$, and $D=0$? A degenerate unstable node ::: Positive product, positive sum, zero discriminant give a repeated positive eigenvalue — every mode grows, so it repels; the "star" geometry just means the two directions have merged.