Exercises — Stability of equilibria — stable, unstable, saddle, spiral, centre

WHAT: The matrix is diagonal, so the two straight coordinate axes are the eigen-directions and the eigenvalues are simply the diagonal entries: ${\lambda }_1=-3$, ${\lambda }_2=-1$. WHY that shortcut: for a diagonal matrix $\det (J-\lambda I)=(-3-\lambda )(-1-\lambda )=0$ factors instantly. Classify: $\tau =-3+(-1)=-4<0$, $\Delta =(-3)(-1)=3>0$, $D=(-4{)}^2-4(3)=4>0$. Real, distinct, both negative $\Rightarrow$ stable node (sink). Everything nearby slides in.

$\tau =0+0=0$, $\Delta =(0)(0)-(2)(-2)=4>0$. Since $\tau =0$ and $\Delta >0$, eigenvalues are pure imaginary: $\lambda =\pm \sqrt{-4}=\pm 2i$. Centre — closed loops, neutrally stable (not asymptotically stable!).

Opposite signs $\Rightarrow$ product $\Delta ={\lambda }_1{\lambda }_2=(4)(-1)=-4<0$. Whenever $\Delta <0$ you have a saddle — attracts along the ${\lambda }_2$ eigen-direction, repels along the ${\lambda }_1$ one. Always unstable.

WHAT: partial derivatives give $J=\left(\begin{array}{cc}{\partial }_x(2x+y)& {\partial }_y(2x+y)\\ {\partial }_x(x+2y)& {\partial }_y(x+2y)\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right).$ $\tau =4$, $\Delta =(2)(2)-(1)(1)=3$, $D=16-12=4>0$. Real distinct, both positive (since $\tau >0,\Delta >0,D>0$) $\Rightarrow$ unstable node (source). Eigenvalues $\lambda =\frac{4\pm 2}{2}=3,1$.

$J=\left(\begin{array}{cc}-1& 4\\ -2& -5\end{array}\right)$. $\tau =-1+(-5)=-6<0$, $\Delta =(-1)(-5)-(4)(-2)=5+8=13>0$, $D=36-52=-16<0$. Complex eigenvalues with $\mathrm{Re}=\tau /2=-3<0$ $\Rightarrow$ stable spiral. It rotates inward.

$J=\left(\begin{array}{cc}0& 1\\ -4& -4\end{array}\right)$. $\tau =-4$, $\Delta =(0)(-4)-(1)(-4)=4$, $D=16-16=0$. $D=0$ is the boundary case: repeated eigenvalue $\lambda =\tau /2=-2$ (both negative) $\Rightarrow$ degenerate / critically-damped stable node. Physically this is critical damping — the fastest return with no overshoot.

Find fixed points: $\dot{y}=0\Rightarrow y=0$; $\dot{x}=x(1-x)=0\Rightarrow x=0$ or $x=1$. So two points: $(0,0)$ and $(1,0)$. Jacobian in general: $J=\left(\begin{array}{cc}1-2x& 0\\ 0& -2\end{array}\right)$. At $(0,0)$: $J=\left(\begin{array}{cc}1& 0\\ 0& -2\end{array}\right)$, $\Delta =-2<0\Rightarrow$ saddle. At $(1,0)$: $J=\left(\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right)$, $\tau =-3<0,\Delta =2>0,D=9-8=1>0\Rightarrow$ stable node. Why evaluate separately: the same nonlinear system can host different types — the Jacobian depends on where you sit.

Fixed points: $y=0$ and $x-x^3=x(1-x^2)=0\Rightarrow x=0,\pm 1$. Three points: $(0,0),(1,0),(-1,0)$. Jacobian: $J=\left(\begin{array}{cc}0& 1\\ 1-3x^2& 0\end{array}\right)$. At $(0,0)$: $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, $\tau =0,\Delta =-1<0\Rightarrow$ saddle (the hilltop between the two wells). At $(\pm 1,0)$: $1-3(1)=-2$, so $\left(\begin{array}{cc}0& 1\\ -2& 0\end{array}\right)$, $\tau =0,\Delta =2>0\Rightarrow$ centre (each well bottom). Linear says centre; since $\tau =0$ this is borderline — but here the system is conservative (energy $E=\frac{1}{2}{y}^2-\frac{1}{2}{x}^2+\frac{1}{4}{x}^4$ is constant), so the centres are genuine.

Fixed points: $y=0$, $\sin x=0\Rightarrow x=0$ or $x=\pi$. $J=\left(\begin{array}{cc}0& 1\\ -\cos x& -\frac{1}{2}\end{array}\right)$. At $(0,0)$: $\left(\begin{array}{cc}0& 1\\ -1& -\frac{1}{2}\end{array}\right)$, $\tau =-\frac{1}{2}<0$, $\Delta =1>0$, $D=\frac{1}{4}-4=-\frac{15}{4}<0\Rightarrow$ stable spiral (hanging-down rest, now decays because of friction). At $(\pi ,0)$: $-\cos \pi =1$, $\left(\begin{array}{cc}0& 1\\ 1& -\frac{1}{2}\end{array}\right)$, $\Delta =(0)(-\frac{1}{2})-(1)(1)=-1<0\Rightarrow$ saddle (upright, still unstable — damping can't save a balancing broomstick).

$J=\left(\begin{array}{cc}\mu & -1\\ 1& \mu \end{array}\right)$. $\tau =2\mu$, $\Delta ={\mu }^2-(-1)(1)={\mu }^2+1>0$ always. $D=4{\mu }^2-4({\mu }^2+1)=-4<0$ always $\Rightarrow$ always complex eigenvalues (always spiral or centre).

• (a) Stable spiral: need $\tau <0\Rightarrow \mu <0$.
• (b) Centre: need $\tau =0\Rightarrow \mu =0$.
• (c) Unstable spiral: $\tau >0\Rightarrow \mu >0$. As $\mu$ sweeps up through $0$, the eigenvalues' real part $\mathrm{Re}\lambda =\mu$ crosses zero — the fixed point switches stability. That crossing is the seed of a Hopf bifurcation.

Reverse Vieta: we need $\tau ={\lambda }_1+{\lambda }_2=8$ and $\Delta ={\lambda }_1{\lambda }_2=15$. Any matrix with that trace/determinant works. Simplest diagonal choice: $J=\left(\begin{array}{cc}3& 0\\ 0& 5\end{array}\right).$ Check: $\tau =8>0$, $\Delta =15>0$, $D=64-60=4>0$ — real, both positive $\Rightarrow$ unstable node. ✔ (A non-diagonal answer such as $\left(\begin{array}{cc}4& 1\\ 1& 4\end{array}\right)$ also has $\tau =8,\Delta =15$.)

With $\tau =-2$ fixed, $\Delta =k$ moving:

• $k<0$: $\Delta <0\Rightarrow$ saddle.
• $k=0$: $\Delta =0$ — a non-isolated / line of fixed points (degenerate). Transition value $k=0$.
• $0<k<1$: $\Delta >0,\tau <0,D=4-4k>0\Rightarrow$ stable node.
• $k=1$: $D=4-4(1)=0$ — degenerate stable node (critical). Transition value $k=1$.
• $k>1$: $D<0$, $\tau <0\Rightarrow$ stable spiral. So the ordered path is: saddle $\to$ (line) $\to$ stable node $\to$ (degenerate) $\to$ stable spiral. The parabola $D=0$, i.e. $\Delta ={\tau }^2/4=1$, is the node–spiral divide.

$J(0,0)=\left(\begin{array}{cc}0& 1\\ -1& -b\end{array}\right)$, so $\tau =-b$, $\Delta =1$, $D=b^2-4$.

• $b=0$: $\tau =0,\Delta =1\Rightarrow$ centre (undamped pendulum, closed orbits).
• $0<b<2$: $\tau <0$, $D<0\Rightarrow$ stable spiral (underdamped — swings while decaying).
• $b=2$: $D=0\Rightarrow$ degenerate stable node (critical damping).
• $b>2$: $D>0$, both eigenvalues real negative $\Rightarrow$ stable node (overdamped — creeps home without swinging). Exact threshold: the spiral→node switch is at $b=2$ (where $\Delta =b^2/4$, i.e. $b^2=4\Delta =4$). This is the same story as the Damped harmonic oscillator: underdamped, critical, overdamped.

Fixed points. $\dot{x}=x(-1+y)=0\Rightarrow x=0$ or $y=1$.

• If $x=0$: $\dot{y}=y=0\Rightarrow y=0$. Point $(0,0)$.
• If $y=1$: $\dot{y}=1-x^2=0\Rightarrow x=\pm 1$. Points $(1,1)$ and $(-1,1)$. General Jacobian: $J=\left(\begin{array}{cc}-1+y& x\\ -2x& 1\end{array}\right)$.
• At $(0,0)$: $\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$, $\Delta =-1<0\Rightarrow$ saddle.
• At $(1,1)$: $\left(\begin{array}{cc}0& 1\\ -2& 1\end{array}\right)$, $\tau =1$, $\Delta =(0)(1)-(1)(-2)=2>0$, $D=1-8=-7<0\Rightarrow$ unstable spiral.
• At $(-1,1)$: $\left(\begin{array}{cc}0& -1\\ 2& 1\end{array}\right)$, $\tau =1$, $\Delta =(0)(1)-(-1)(2)=2>0$, $D=1-8=-7<0\Rightarrow$ unstable spiral. Two mirror-image unstable spirals flanking a central saddle.

Why linearisation is untrustworthy here: Hartman–Grobman guarantees the linear picture matches the nonlinear flow only at hyperbolic fixed points — those with $\mathrm{Re}(\lambda )\ne 0$ for every eigenvalue. Here $\lambda =\pm i$ has $\mathrm{Re}=0$: the point is non-hyperbolic, exactly the excluded case. Tiny nonlinear terms can bend the neutral circles slowly inward (weak stable spiral) or outward (weak unstable spiral). What settles it: build a Lyapunov function $V$ (an energy-like $V>0$ with $V(0)=0$) and inspect $\dot{V}$ along trajectories. $\dot{V}<0$ ⇒ truly stable; $\dot{V}=0$ ⇒ genuine centre / conserved orbits; $\dot{V}>0$ ⇒ unstable. The phase portrait then confirms visually.