Worked examples — Linearization of nonlinear systems

Before we start, one reminder of the vocabulary, so no symbol arrives unexplained:

Recall The four numbers that decide everything

From the Jacobian $J=\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right)$ evaluated at an equilibrium, we build:

• Trace $\tau =f_x+g_y$ — the sum of the diagonal. Think "total tendency to grow or shrink."
• Determinant $\Delta =f_x{g}_y-f_y{g}_x$ — the "twisting/scaling" strength.
• Eigenvalues: $\lambda =\frac{\tau \pm \sqrt{{\tau }^2-4\Delta }}{2}$.
• The discriminant $D={\tau }^2-4\Delta$ decides real vs complex: $D>0$ real, $D<0$ complex, $D=0$ repeated. Two shortcuts we lean on repeatedly: $\tau ={\lambda }_1+{\lambda }_2$ (sum) and $\Delta ={\lambda }_1{\lambda }_2$ (product).

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The scenario matrix

Every equilibrium you will ever classify falls into one of these cells. The right column names the example on this page that hits it.

#	Cell (what makes it special)	Eigenvalue signature	Example
C1	Real, both negative	${\lambda }_1,{\lambda }_2<0$	Ex 1 stable node
C2	Real, both positive	${\lambda }_1,{\lambda }_2>0$	Ex 2 unstable node
C3	Real, opposite signs	${\lambda }_1<0<{\lambda }_2$	Ex 3 saddle
C4	Complex, negative real part	$\alpha \pm i\beta ,\alpha <0$	Ex 4 stable spiral
C5	Complex, positive real part	$\alpha \pm i\beta ,\alpha >0$	Ex 5 unstable spiral
C6	Degenerate: pure imaginary ($\tau =0$)	$\pm i\beta$	Ex 6 center — the inconclusive case
C7	Degenerate: a zero eigenvalue ($\Delta =0$)	$\lambda =0$ present	Ex 7 non-hyperbolic line
C8	Word problem with two equilibria of different type	mixed	Ex 8 Lotka–Volterra
C9	Exam twist: classify from $\tau ,\Delta$ alone + repeated-root ($D=0$) boundary	boundary $D=0$	Ex 9 parameter sweep

The rule that guides every example is the boundary map below — memorise its shape and you can classify before touching a single eigenvalue.

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Ex 1 — Cell C1: stable node (both eigenvalues negative)

Forecast: Both variables have a big negative self-term ($-2x$, $-2y$) and weak coupling. Guess: everything decays → stable node. Predict two real negative eigenvalues.

1. Find the equilibrium. Set both to zero: $-2x+y=0$ and $x-2y=0$. Solving gives $x=y=0$. Why this step? Equilibria are the only points where linearization is even defined — they are where the constant term of the Taylor expansion vanishes.
2. This system is already linear, so $J$ is the coefficient matrix, constant everywhere. Here $f=-2x+y$ so $f_x=-2,f_y=1$; $g=x-2y$ so $g_x=1,g_y=-2$: $J=\left(\begin{array}{cc}-2& 1\\ 1& -2\end{array}\right).$ Why? When $f,g$ are already linear, the partials are the constant coefficients — no evaluation needed.
3. Trace and determinant. $\tau =-2+-2=-4$, $\Delta =(-2)(-2)-(1)(1)=4-1=3$. Why? These two numbers locate us on the $\tau$–$\Delta$ map in the figure.
4. Discriminant and eigenvalues. $D={\tau }^2-4\Delta =16-12=4>0$ → real. $\lambda =\frac{-4\pm 2}{2}=-1,-3$. Why $D$ first? $D$'s sign tells us "node vs spiral" before we even compute the roots.
5. Classify. Both real and negative → stable node. Why this reads as a stable node? $\Delta >0$ ⇒ eigenvalues share a sign (product positive); $\tau <0$ ⇒ that shared sign is negative (sum negative); $D>0$ ⇒ real, not spiralling. Negative + real = pure decay along two directions = stable node.

Verify: on the figure this is the left interior region (below the parabola, left of centre). Sum of eigenvalues $-1+-3=-4=\tau$ ✔, product $(-1)(-3)=3=\Delta$ ✔. Forecast confirmed.

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Ex 2 — Cell C2: unstable node (both eigenvalues positive)

Forecast: Positive self-terms $+3x,+3y$; the $3x^2$ is a higher-order nonlinearity that dies near the origin. Guess: unstable node, two real positive eigenvalues.

1. Equilibrium at origin? $f(0,0)=0$, $g(0,0)=0$ ✔ (recall $f=3x+y$, $g=x+3x^2+3y$). Why this step? We must confirm $(0,0)$ is actually a fixed point before expanding around it.
2. Jacobian, general. $f_x=3,f_y=1,g_x=1+6x,g_y=3$. Why compute $g_x=1+6x$? The $3x^2$ term contributes $6x$ to the derivative — but at the origin it will vanish, showing exactly why quadratic terms don't affect linear stability.
3. Evaluate at $(0,0)$. $g_x=1+0=1$, so $J=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right).$ Why substitute? A Jacobian still containing $x$ is not a linear system — this is the "steel-man" trap from the parent note.
4. Numbers. $\tau =6,\Delta =9-1=8,D=36-32=4>0$. $\lambda =\frac{6\pm 2}{2}=4,2$.
5. Classify. Both real and positive → unstable node. Why this reads as an unstable node? $\Delta >0$ ⇒ same sign; $\tau >0$ ⇒ that sign is positive; $D>0$ ⇒ real. Positive + real = growth along two directions = unstable node. It is Ex 1 mirrored across $\tau =0$.

Verify: sum $4+2=6=\tau$ ✔, product $4\cdot 2=8=\Delta$ ✔. On the figure: right interior region. Notice the $3x^2$ never entered — proof that dropping quadratics was legitimate at a hyperbolic point.

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Ex 3 — Cell C3: saddle (opposite signs)

Forecast: The strong cross-coupling ($2y$, $2x$) usually beats the diagonal → I expect one growing and one decaying direction → saddle.

1. Origin is an equilibrium: both RHS vanish at $(0,0)$ ✔ ($f=x+2y$, $g=2x+y-y^3$).
2. Jacobian. $f_x=1,f_y=2,g_x=2,g_y=1-3y^2$.
3. At $(0,0)$: $g_y=1-0=1$, so $J=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$. Why? The $-y^3$ contributes $-3y^2$, which is zero at the origin.
4. Numbers. $\tau =2,\Delta =1-4=-3$. Here's the key insight: $\Delta <0$. Why does $\Delta <0$ guarantee a saddle? $\Delta ={\lambda }_1{\lambda }_2$. A negative product forces the two eigenvalues to have opposite signs — one positive, one negative — no matter what $\tau$ is. That is a saddle by definition.
5. Eigenvalues explicitly: $D=4+12=16$, $\lambda =\frac{2\pm 4}{2}=3,-1$.
6. Classify: saddle (unstable).

Verify: product $3\cdot (-1)=-3=\Delta$ ✔ (negative ⇒ opposite signs, as predicted). Sum $3+(-1)=2=\tau$ ✔. On the figure this sits below the $\Delta =0$ line — the entire lower band is saddle territory.

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Ex 4 — Cell C4: stable spiral (complex, negative real part)

Forecast: Diagonal is negative (decay), but the antisymmetric off-diagonal ($+4y$ vs $-x$) creates rotation. Guess: stable spiral — spirals inward.

1. Linear already, so with $f=-x+4y$, $g=-x-y$: $J=\left(\begin{array}{cc}-1& 4\\ -1& -1\end{array}\right)$.
2. Numbers. $\tau =-2,\Delta =(-1)(-1)-(4)(-1)=1+4=5$.
3. Discriminant. $D={\tau }^2-4\Delta =4-20=-16<0$. Why does $D<0$ mean spiral? $\sqrt{D}$ becomes imaginary, so $\lambda =\alpha \pm i\beta$. The imaginary part $\beta$ is rotation; the real part $\alpha =\tau /2$ is growth/decay.
4. Eigenvalues. $\lambda =\frac{-2\pm \sqrt{-16}}{2}=-1\pm 2i$.
5. Classify. Real part $\alpha =-1<0$ → stable spiral. Trajectories rotate (from $\beta =2$) while shrinking (from $\alpha =-1$).

Verify: sum $(-1+2i)+(-1-2i)=-2=\tau$ ✔; product $(-1{)}^2-(2i{)}^2=1+4=5=\Delta$ ✔. On the map ($\tau <0,D<0$) this is above the parabola on the left — the stable-spiral region. In figure s02, watch the blue trajectory: it winds around the equilibrium (the $\beta$ rotation) while its radius shrinks toward the white dot (the $\alpha =-1$ decay). That geometric spiral is the sign of $\alpha$ made visible.

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Ex 5 — Cell C5: unstable spiral (complex, positive real part)

Forecast: Same rotation as Ex 4 but positive diagonal → outward spiral, unstable spiral.

1. With $f=x+4y$, $g=-x+y$: $J=\left(\begin{array}{cc}1& 4\\ -1& 1\end{array}\right)$.
2. $\tau =2,\Delta =1-(4)(-1)=1+4=5$.
3. $D=4-20=-16<0$ → complex.
4. $\lambda =\frac{2\pm 4i}{2}=1\pm 2i$.
5. Classify. $\alpha =+1>0$ → unstable spiral. Only the sign of the real part flipped from Ex 4; the whole character changed. Why this reads as unstable spiral? $D<0$ puts us above the parabola (spiral); $\tau >0$ puts us on its right half, so $\alpha =\tau /2>0$ — the radius grows each turn.

Verify: sum $=2=\tau$ ✔, product $1+4=5=\Delta$ ✔. On figure s01: same height as Ex 4 but mirrored to the right of $\tau =0$. Identical $\Delta$, opposite $\tau$ — the map's vertical axis is the entire difference between stable and unstable spirals.

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Ex 6 — Cell C6 (degenerate): center — where linearization can LIE

Forecast: Pure rotation, no decay term at all → center (closed loops). But: the parent warned that centers are inconclusive for nonlinear systems.

1. With $f=y$, $g=-x$: $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$.
2. $\tau =0,\Delta =0\cdot 0-(1)(-1)=1$. Why is $\tau =0$ the alarm bell? $\tau =0$ with $\Delta >0$ gives $\lambda =\pm i\sqrt{\Delta }$ — pure imaginary, zero real part. Zero real part = non-hyperbolic = Hartman–Grobman does not apply.
3. $D=0-4=-4<0$; $\lambda =\pm i$. Pure imaginary → center for the linearization.
4. The honest conclusion. For this exact linear system the origin truly is a center. But add any nonlinear term, e.g. $\dot{x}=y,\dot{y}=-x-x^{2}y$: the linearization is still $\pm i$, yet the extra $-x^{2}y$ drains energy and the real trajectory spirals inward. The linear picture lied.

Verify: sum $i+(-i)=0=\tau$ ✔, product $i\cdot (-i)=1=\Delta$ ✔. On the map this sits exactly on the vertical $\tau =0$ axis — the razor-thin center line. Because $\tau =0$, you must reach for a Lyapunov function or energy argument — never trust a center from linearization alone.

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Ex 7 — Cell C7 (degenerate): a zero eigenvalue

Forecast: $\dot{y}=-y$ pulls $y$ to zero (stable direction), but $\dot{x}=x^2$ is neither linearly stable nor unstable — its derivative vanishes at $0$. Expect a zero eigenvalue → non-hyperbolic.

1. Origin is a fixed point: $0^2=0$, $-0=0$ ✔ ($f=x^2$, $g=-y$).
2. Jacobian. $f_x=2x,f_y=0,g_x=0,g_y=-1$.
3. At $(0,0)$: $f_x=0$, so $J=\left(\begin{array}{cc}0& 0\\ 0& -1\end{array}\right)$.
4. Numbers. $\tau =-1,\Delta =0$. Why does $\Delta =0$ matter? $\Delta ={\lambda }_1{\lambda }_2=0$ forces at least one eigenvalue to be zero. Zero real part ⇒ non-hyperbolic ⇒ linearization inconclusive.
5. $\lambda =0,-1$.
6. The truth from the nonlinear term. $\dot{x}=x^2\ge 0$ always: $x$ grows for $x>0$ and creeps toward $0$ for $x<0$. So the $x$-axis is stable from the left, unstable from the right — a "saddle-node"-like semi-stable behaviour the linear zero could never reveal.

Verify: sum $0+(-1)=-1=\tau$ ✔, product $0\cdot (-1)=0=\Delta$ ✔. On the map, $\Delta =0$ is exactly the horizontal axis — the boundary line where classification breaks down.

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Ex 8 — Cell C8 (word problem): predator–prey, two different equilibria

Forecast: Coexistence means $x>0$ and $y>0$. If it's stable, both populations settle; if it's a spiral, they oscillate. Ecology suggests damped oscillation → guess stable spiral or stable node.

1. Equilibria. Here $f=x(3-x-2y)$ and $g=y(x-1)$. From $g=0$: $y=0$ or $x=1$. Coexistence needs $y\ne 0$, so $x=1$. Plug into $f=0$: $1(3-1-2y)=0\Rightarrow 2-2y=0\Rightarrow y=1$. Coexistence point $(1,1)$. Why solve BOTH factors? The parent's third mistake — forgetting an equilibrium — happens when you find one root and stop; $f=g=0$ can have several solutions (here also $(0,0)$ and $(3,0)$), each with its own stability. We deliberately chase the coexistence root by keeping $y\ne 0$.
2. Jacobian, general. $f=3x-x^2-2xy$, so $f_x=3-2x-2y,f_y=-2x$. $g=xy-y$, so $g_x=y,g_y=x-1$. Why expand the products first? Differentiating $x(3-x-2y)$ directly is error-prone; multiplying out makes each partial mechanical.
3. Evaluate at $(1,1)$. $J=\left(\begin{array}{cc}3-2-2& -2\\ 1& 1-1\end{array}\right)=\left(\begin{array}{cc}-1& -2\\ 1& 0\end{array}\right).$ Why substitute the specific point? A Jacobian still containing $x,y$ is not a linear system — we need pure numbers.
4. Numbers. $\tau =-1+0=-1,\Delta =(-1)(0)-(-2)(1)=0+2=2$. Why these first? They locate us on the map and give the stability sign directly.
5. Discriminant. $D={\tau }^2-4\Delta =1-8=-7<0$ → complex. $\lambda =\frac{-1\pm i\sqrt{7}}{2}$.
6. Classify. Real part $-\frac{1}{2}<0$ → stable spiral. The two populations spiral into coexistence: damped oscillation, ecologically sensible. Why do these signs force a stable spiral? $D<0$ ⇒ complex eigenvalues ⇒ spiral (not a node); $\tau <0$ ⇒ real part $\tau /2<0$ ⇒ the spiral shrinks ⇒ stable. On figure s01 this lands above the parabola, just left of the vertical axis — the same neighbourhood as Ex 4.

Verify: sum of eigenvalues $=-1=\tau$ ✔; product $={\left(\frac{-1}{2}\right)}^2+{\left(\frac{\sqrt{7}}{2}\right)}^2=\frac{1}{4}+\frac{7}{4}=2=\Delta$ ✔. Forecast (damped oscillation) confirmed. Populations both positive at $(1,1)$ ✔ — a physically valid equilibrium.

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Ex 9 — Cell C9 (exam twist): the $D=0$ boundary and a parameter sweep

Forecast: For small $a$ the damping is weak — rotation likely (spiral). For large $a$ the damping dominates — maybe a node. Expect a boundary at $D=0$ where a repeated real root appears.

1. With $f=-x+y$, $g=-x-ay$: $J=\left(\begin{array}{cc}-1& 1\\ -1& -a\end{array}\right)$.
2. Trace and determinant. $\tau =-1-a,\Delta =(-1)(-a)-(1)(-1)=a+1$. Why lead with $\tau ,\Delta$? The exam trick: you can decide stability from signs alone. Here $a>0$ ⇒ $\tau =-1-a<0$ and $\Delta =a+1>0$ ⇒ always stable, for every positive $a$.
3. Node vs spiral: the discriminant. $D={\tau }^2-4\Delta =(1+a{)}^2-4(a+1)=(a+1)(a+1-4)=(a+1)(a-3).$ Why factor? It exposes the sign change cleanly. Since $a+1>0$, the sign of $D$ is the sign of $(a-3)$.
4. Critical value. $D=0$ at $a=3$.
   • $0<a<3$: $D<0$ → stable spiral.
   • $a=3$: $D=0$ → repeated real eigenvalue $\lambda =\tau /2=-2$ (both roots equal). This is the boundary case — see step 5.
   • $a>3$: $D>0$ → stable node.
5. What the repeated root means (star vs improper node). When $D=0$ the two eigenvalues coincide at $\lambda =\tau /2$. Two sub-cases exist:
   • If $J$ at that point is a scalar multiple of the identity ($J=\lambda I$), every direction is an eigenvector → a star (proper) node: trajectories run dead-straight into the origin.
   • Otherwise (only one independent eigenvector) → an improper (degenerate) node: trajectories bend as they come in, sharing a single tangent line. Here at $a=3$, $J=\left(\begin{array}{cc}-1& 1\\ -1& -3\end{array}\right)\ne -2I$, so it has only one eigenvector → an improper stable node. Why check this? "Repeated eigenvalue" alone does not tell you the shape — you must test whether the eigenvector space is 2D (star) or 1D (improper). The eigenvector count is the deciding tool.
6. Limiting behaviour (honest asymptotics). Consider $a\to \infty$. The eigenvalues satisfy ${\lambda }_1+{\lambda }_2=-1-a$ and ${\lambda }_1{\lambda }_2=a+1$. The fast root grows like ${\lambda }_1\approx -a$ (it carries almost the whole trace). The slow root is then ${\lambda }_2=\frac{\Delta }{{\lambda }_1}\approx \frac{a+1}{-a}\to -1$ as $a\to \infty$ — so the slow eigenvalue tends to $-1$, not to $0$. Both stay negative: an increasingly anisotropic stable node. At the other extreme $a\to 0^{+}$: $\tau \to -1,\Delta \to 1,D\to -3<0$ — a stable spiral with real part $\tau /2\to -\frac{1}{2}$.

Verify: at the boundary $a=3$: $\tau =-4,\Delta =4$, so $D=16-16=0$ ✔, repeated eigenvalue $\lambda =\tau /2=-2$ ✔, and $J=\left(\begin{array}{cc}-1& 1\\ -1& -3\end{array}\right)$ has a single eigenvector (geometric multiplicity 1) ✔ → improper node. At $a=1$: $\tau =-2,\Delta =2,D=4-8=-4<0$ → stable spiral ✔. At $a=5$: $\tau =-6,\Delta =6,D=36-24=12>0$ → stable node ✔.

Wrap-up (the exam moral). The whole family is stable for every $a>0$ — stability never changes. What the parameter sweep changes is only the geometry of approach: for $0<a<3$ the state spirals in (oscillatory settling), at $a=3$ it sits exactly on the $D=0$ parabola as a degenerate improper node, and for $a>3$ it decays straight in without overshoot. The exam skills demonstrated: read stability from the signs of $\tau ,\Delta$ instantly, use $D$ to pin the shape, and — on the $D=0$ boundary — count eigenvectors to separate star from improper node.

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Recall One-line self-test for every cell

Cover the answers. Sign of $\Delta <0$ tells you? ::: Saddle (opposite-sign real eigenvalues) — regardless of $\tau$. $\tau <0,\Delta >0,D>0$? ::: Stable node. $\tau >0,\Delta >0,D<0$? ::: Unstable spiral. $\tau =0,\Delta >0$? ::: Center — linearization inconclusive. $\Delta =0$? ::: A zero eigenvalue — non-hyperbolic, inconclusive. $D=0$ with $\tau <0$ and one eigenvector? ::: Degenerate (repeated-root) improper stable node. $D=0$ with $J=\lambda I$? ::: Star (proper) node — every direction an eigenvector.