Worked examples — Phase plane analysis — trajectories, critical points

Two numbers do all the classifying. Let $J$ be the Jacobian at a critical point. Write

• $\tau =\mathrm{tr}J={\lambda }_1+{\lambda }_2$ (the trace — sum of the diagonal),
• $\Delta =\det J={\lambda }_1{\lambda }_2$ (the determinant — cross-multiplied difference),

and the eigenvalues are the two roots of ${\lambda }^2-\tau \lambda +\Delta =0$, i.e. ${\lambda }_{1,2}=\frac{1}{2}\left(\tau \pm \sqrt{{\tau }^2-4\Delta }\right).$ The quantity under the root, $D={\tau }^2-4\Delta$, is the discriminant: if $D\ge 0$ the roots are real (straight-line flow), if $D<0$ they are complex (rotating flow). Keep $(\tau ,\Delta ,D)$ in your head — everything below is just reading off their signs. See Trace–determinant plane for the full map.

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The scenario matrix

Every 2-D linear (or linearized) critical point falls into exactly one cell below. Each column is a distinct sign/degeneracy case; the last two rows are the "story" and "exam" cells the syllabus loves.

Cell	Signs of $(\tau ,\Delta ,D)$	Type	Worked in
C1	$\Delta <0$	Saddle (unstable)	Ex 1
C2	$\Delta >0,\tau <0,D>0$	Stable node (sink)	Ex 2
C3	$\Delta >0,\tau >0,D>0$	Unstable node (source)	Ex 3
C4	$\Delta >0,\tau <0,D<0$	Stable spiral	Ex 4
C5	$\Delta >0,\tau >0,D<0$	Unstable spiral	Ex 4 (contrast)
C6	$\Delta >0,\tau =0$	Center (pure imaginary) — borderline	Ex 5
C7	$D=0$ (repeated root)	Degenerate / star node	Ex 6
C8	$\Delta =0$	Degenerate: a whole line of critical points	Ex 7
C9	Multiple critical points, nonlinear	Classify each separately	Ex 8 (word problem)
C10	Exam twist: $\Delta >0,\tau <0$ but nonlinear center suspected	Linearization inconclusive	Ex 9

The picture above is the whole matrix drawn as one map: horizontal axis is $\tau$, vertical axis is $\Delta$, the red parabola is $D=0$. Memorize the map, not the ten cells. Trace your eye: cross below the black horizontal line ($\Delta <0$) and you are always in the saddle zone; above it, moving left ($\tau <0$) is stable and right ($\tau >0$) is unstable, while crossing up over the red parabola switches you from node (below, real roots) to spiral (above, rotating roots).

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Ex 1 — Cell C1: the saddle ($\Delta <0$)

Forecast: two "3"s off-diagonal are large — guess the cross-coupling wins and this is unstable. Saddle or source?

1. Critical point. Set $x+3y=0$ and $3x+y=0$. Why this step? equilibrium needs both rates zero. Adding/subtracting gives only $(0,0)$.
2. Jacobian. Already linear, so $J=\left(\begin{array}{cc}1& 3\\ 3& 1\end{array}\right)$. Why? $f_x=1,f_y=3,g_x=3,g_y=1$.
3. Two numbers. $\tau =1+1=2$, $\Delta =(1)(1)-(3)(3)=-8$. Why? trace = diagonal sum, det = cross-product difference.
4. Read the sign. $\Delta <0$ $\Rightarrow$ saddle, always unstable — no need for $D$. Why? $\Delta ={\lambda }_1{\lambda }_2<0$ forces opposite signs, so one direction grows, one shrinks.

Verify: eigenvalues from ${\lambda }^2-2\lambda -8=0\Rightarrow (\lambda -4)(\lambda +2)=0\Rightarrow \lambda =4,-2$. Opposite signs confirm the saddle. ✓

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Ex 2 — Cell C2: stable node ($\Delta >0,\tau <0,D>0$)

Forecast: both diagonal terms are $-3$ (pulling inward). Guess: everything drains to origin — stable node.

1. Critical point. $-3x+y=0,x-3y=0\Rightarrow$ only $(0,0)$. Why? both rates must vanish.
2. Jacobian. $J=\left(\begin{array}{cc}-3& 1\\ 1& -3\end{array}\right)$. Why? partials of the linear right-hand side.
3. Three numbers. $\tau =-6$, $\Delta =9-1=8$, $D={\tau }^2-4\Delta =36-32=4>0$. Why $D$ now? $\Delta >0$ means same-sign eigenvalues — need $D$ to tell node (real, $D\ge 0$) from spiral ($D<0$).
4. Classify. $D>0$ real roots, $\tau <0$ both negative $\Rightarrow$ stable node (sink). Why? both eigenvalues negative means every direction decays.

Verify: ${\lambda }^2+6\lambda +8=0\Rightarrow (\lambda +2)(\lambda +4)=0\Rightarrow \lambda =-2,-4$. Both real and negative. ✓

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Ex 3 — Cell C3: unstable node ($\Delta >0,\tau >0,D>0$)

Forecast: flip all the signs of Ex 2 — same shape, opposite time direction. Guess: source.

1. Jacobian. $J=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$ (linear, read partials). Why first? linear system → constant Jacobian.
2. Three numbers. $\tau =6$, $\Delta =9-1=8$, $D=36-32=4>0$. Why? same recipe as Ex 2.
3. Classify. $D>0$ real, $\tau >0$ both positive $\Rightarrow$ unstable node (source). Why? both eigenvalues positive → every direction grows.

Verify: ${\lambda }^2-6\lambda +8=0\Rightarrow (\lambda -2)(\lambda -4)=0\Rightarrow \lambda =2,4$. Positive reals. ✓ (Exact mirror of Ex 2 — this is why time-reversal turns sinks into sources.)

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Ex 4 — Cells C4 & C5: stable vs unstable spiral ($D<0$)

Forecast: off-diagonal terms $\pm 2$ are equal-magnitude opposite-sign — that's rotation. Sign of the diagonal decides in vs out.

1. Jacobians. (a) $J_a=\left(\begin{array}{cc}-1& -2\\ 2& -1\end{array}\right)$, (b) $J_b=\left(\begin{array}{cc}1& -2\\ 2& 1\end{array}\right)$. Why? read partials.
2. Numbers (a). $\tau =-2$, $\Delta =1+4=5$, $D=4-20=-16<0$. Why $D$? to detect rotation ($D<0$ ⇒ complex roots).
3. Numbers (b). $\tau =+2$, $\Delta =1+4=5$, $D=4-20=-16<0$.
4. Classify. $D<0$ in both ⇒ spirals. (a) $\tau <0$ ⇒ stable spiral (inward). (b) $\tau >0$ ⇒ unstable spiral (outward). Why? real part of the eigenvalue $=\tau /2$ sets grow/decay; imaginary part sets the swirl.

The two panels below show exactly this. In panel (a) the red trajectory starts far out and coils inward to the black dot at the origin — that is $\tau <0$ draining energy every loop. In panel (b) it starts near the origin and flings outward, the same swirl run in reverse because $\tau >0$. Notice both spin the same way (the rotation comes from the $\pm 2$ off-diagonals, unchanged); only the in/out sense flips with the sign of $\tau$.

Verify: (a) ${\lambda }^2+2\lambda +5=0\Rightarrow \lambda =-1\pm 2i$, $\mathrm{Re}<0$. (b) ${\lambda }^2-2\lambda +5=0\Rightarrow \lambda =1\pm 2i$, $\mathrm{Re}>0$. ✓

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Ex 5 — Cell C6: a center ($\tau =0,\Delta >0$)

Forecast: no diagonal terms at all — nothing pulls in or pushes out. Guess: closed loops, a center.

1. Jacobian. $J=\left(\begin{array}{cc}0& -4\\ 1& 0\end{array}\right)$. Why? $f_x=0,f_y=-4,g_x=1,g_y=0$.
2. Numbers. $\tau =0$, $\Delta =0-(-4)(1)=4>0$, $D=0-16=-16<0$. Why note $\tau =0$? it sits exactly on the borderline between stable and unstable spirals.
3. Classify. $\tau =0,\Delta >0$ ⇒ center: eigenvalues pure imaginary $\Rightarrow$ neutral closed orbits, neither growing nor decaying. Why "neutral"? $\mathrm{Re}\lambda =0$ means no exponential factor at all.

Verify: ${\lambda }^2+4=0\Rightarrow \lambda =\pm 2i$, purely imaginary. The trajectories are ellipses $x^2+4y^2=\text{const}$ (from $\frac{dy}{dx}=\frac{x}{-4y}\Rightarrow 4ydy=-xdx$, integrate). ✓

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Ex 6 — Cell C7: repeated eigenvalue ($D=0$, star node)

Forecast: both variables decay at the same rate independently. Guess: everything shrinks straight toward the origin.

1. Jacobian. $J=\left(\begin{array}{cc}-2& 0\\ 0& -2\end{array}\right)$. Why? the two equations are uncoupled.
2. Numbers. $\tau =-4$, $\Delta =4$, $D=16-16=0$. Why flag $D=0$? it's the exact boundary — a repeated root.
3. Classify. $\lambda =-2$ (double), and $J$ is already diagonal ⇒ stable star node: trajectories are straight rays into the origin. Why star and not spiral? $D=0$ with a diagonalizable $J$ gives radial lines; if $J$ were not diagonalizable you'd get an improper node.

Verify: $(\lambda +2{)}^2=0\Rightarrow \lambda =-2,-2$. Trajectory shape: $\frac{dy}{dx}=\frac{-2y}{-2x}=\frac{y}{x}\Rightarrow y=cx$, straight lines. ✓

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Ex 7 — Cell C8: $\Delta =0$, a whole line of critical points

Forecast: the second equation is just twice the first — degenerate. Guess: not an isolated point but an entire line at rest.

1. Critical points. $x+y=0$ and $2x+2y=0$ are the same equation. Why? row 2 = 2 × row 1. So every point on the line $y=-x$ is a critical point.
2. Jacobian. $J=\left(\begin{array}{cc}1& 1\\ 2& 2\end{array}\right)$, so $\tau =3$, $\Delta =(1)(2)-(1)(2)=0$. Why compute these two? $\tau$ and $\Delta$ are the only inputs the classifier needs, and here $\Delta =0$ is the tell-tale of degeneracy — one eigenvalue must be zero.
3. Classify. $\Delta =0$ ⇒ one eigenvalue is $0$. Why? $\det J={\lambda }_1{\lambda }_2=0$ forces a zero eigenvalue. Solve ${\lambda }^2-3\lambda =0\Rightarrow \lambda =0,3$.
4. Meaning. The $\lambda =0$ direction is along the line of equilibria (no motion). The $\lambda =3>0$ direction pushes points away from the line ⇒ line is unstable. Why this reading? motion happens only in the non-zero-eigenvalue direction.

Verify: ${\lambda }^2-3\lambda =0\Rightarrow \lambda =0,3$. One zero (the line), one positive (repulsion). ✓

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Ex 8 — Cell C9: word problem with multiple critical points

Forecast: guess there are several rest states — extinction, rabbits-only, and a coexistence point where both survive; coexistence is often a spiral or center in such models.

1. Critical points. $\dot{y}=0\Rightarrow y=0$ or $x=1$. Why factor? products are zero when a factor is zero.
   • $y=0$: then $x(3-x)=0\Rightarrow x=0$ or $x=3$. Points $(0,0),(3,0)$.
   • $x=1$: then $1\cdot (3-1-2y)=0\Rightarrow 2-2y=0\Rightarrow y=1$. Point $(1,1)$.
2. General Jacobian. With $f=x(3-x-2y)=3x-x^2-2xy$ and $g=xy-y$: $J=\left(\begin{array}{cc}3-2x-2y& -2x\\ y& x-1\end{array}\right).$ Why symbolic first? one formula evaluated at each point.
3. At coexistence $(1,1)$. $J=\left(\begin{array}{cc}3-2-2& -2\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}-1& -2\\ 1& 0\end{array}\right)$. Why this point? it's the biologically interesting one.
4. Classify $(1,1)$. $\tau =-1$, $\Delta =(-1)(0)-(-2)(1)=2$, $D=1-8=-7<0$. Complex with $\mathrm{Re}=\tau /2=-\frac{1}{2}<0$ ⇒ stable spiral: populations oscillate but settle to coexistence. Why spiral? $D<0$ rotation, $\tau <0$ decay.

Verify: at $(1,1)$: ${\lambda }^2+\lambda +2=0\Rightarrow \lambda =\frac{-1\pm \sqrt{-7}}{2}$, $\mathrm{Re}=-1/2$. And $(0,0)$: $J=\left(\begin{array}{cc}3& 0\\ 0& -1\end{array}\right)\Rightarrow \lambda =3,-1$ (saddle); $(3,0)$: $J=\left(\begin{array}{cc}-3& -6\\ 0& 2\end{array}\right)\Rightarrow \lambda =-3,2$ (saddle). ✓

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Ex 9 — Cell C10: exam trap — linearization inconclusive

Forecast: the parent note warned: centers are borderline. Guess: the cubic term secretly decides, so "center" is wrong for $a\ne 0$.

1. Linearize. Drop cubic terms: $\dot{x}=-y,\dot{y}=x$, $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Why drop cubics? near origin $x^2+y^2$ is negligible.
2. Numbers. $\tau =0$, $\Delta =1$, $D=-4<0$ ⇒ pure imaginary $\lambda =\pm i$ ⇒ linear center. Why suspicious? $\mathrm{Re}\lambda =0$ exactly — the higher-order terms are no longer negligible.
3. Test with radius. Let $r^2=x^2+y^2$. Then $\frac{1}{2}\frac{d(r^2)}{dt}=x\dot{x}+y\dot{y}=x(-y+ax{r}^2)+y(x+ay{r}^2)=a{r}^2(x^2+y^2)=a{r}^4$. So $\dot{r}=a{r}^3$. Why compute $\dot{r}$? it directly reveals inward/outward drift the linear picture hid. See Stability and Lyapunov functions.
4. Verdict. $a>0$: $\dot{r}>0$ ⇒ unstable spiral. $a<0$: $\dot{r}<0$ ⇒ stable spiral. $a=0$: genuine center. So linearization is inconclusive — the cubic term is the decider. Why does this matter? it's the classic exam warning about relying on linearization for centers (and how limit cycles can appear).

Verify: $\dot{r}=a{r}^3$ so $\mathrm{sign}(\dot{r})=\mathrm{sign}(a)$. Linear eigenvalues $\pm i$ regardless of $a$. ✓

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Recall One-line decision procedure (collapse me and recite)

Compute $\Delta$: if $<0$ ⇒ saddle, done. Else compute $\tau$ (stable if $<0$) and $D={\tau }^2-4\Delta$ (node if $\ge 0$, spiral if $<0$). If $\tau =0,\Delta >0$ ⇒ linear center — but check nonlinear terms before believing it.