Worked examples — Phase space — trajectories, phase portraits

This page is a drill through every case a phase-space problem can hand you. We build straight on the parent topic. If a symbol shows up, it was earned there or is earned here.

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The scenario matrix

Every cell below is covered by at least one worked example. The label like (C1) tags the example that hits it.

Case class	Specific scenario	Covered by
Sign of $V^{\prime \prime }$	Minimum of $V$ → center (stable loop)	C1, C6
	Maximum of $V$ → saddle (unstable)	C2, C6
	$V^{\prime \prime }=0$ → degenerate / non-hyperbolic equilibrium (cusp)	C7
Quadrant of flow arrow	All four quadrants of $(q,p)$ for the oscillator	C1
Full level-curve family	Hyperbolae ($E\ne 0$) filling out a saddle portrait	C2
Zero / degenerate input	Fixed point ($\dot{q}=\dot{p}=0$), zero-velocity turning point	C1, C3
	Degenerate fixed point where the usual $V^{\prime \prime }$ test fails	C7
Energy sign / level	Below-, at-, above-separatrix energy	C3
Limiting behaviour	$E\to 0$ (tiny loop), $E\to \infty$ (huge loop), $t\to \infty$ decay	C4, C5
Non-conservative	Damping → focus / node; sign of $\dot{E}$; all 3 damping regimes	C5
Real-world word problem	Pendulum: swing vs. whirl threshold in real numbers	C3
Exam-style twist	Given only $V(q)$, sketch the whole portrait	C6
Numeric readout	Period / max momentum from an ellipse	C4

Reference figure — the oscillator flow arrows, with labelled $q$–$p$ axes and quadrant markers:

The horizontal axis is position $q$, the vertical axis is momentum $p$. The four quadrants are marked QI (both positive), QII ($q<0,p>0$), QIII (both negative), QIV ($q>0,p<0$); the four orange dots are the points evaluated in C1.

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C1 — All four quadrants of the oscillator flow (cells: center, four quadrants, zero input)

Forecast: before reading on — at the rightmost point $(1,0)$, which way does the dot move next, up or down? Guess now.

1. Write the flow. Hamilton's equations with $m=\omega =1$ give $\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m}=p,\dot{p}=-\frac{\partial H}{\partial q}=-m{\omega }^{2}q=-q.$ Why this step? The arrow at a point IS $(\dot{q},\dot{p})$; these two formulas ARE the arrow field.

2. Evaluate at $(1,0)$ (in QI/QIV boundary, the "east" point): $(\dot{q},\dot{p})=(0,-1)$ — pointing straight down into QIV. Why this step? This is the "east" point of the loop; the dot must head south to circle clockwise.

3. Evaluate at $(0,1)$ (top): $(\dot{q},\dot{p})=(1,0)$ — pointing right into QI. At $(-1,0)$ (west): $(0,1)$ — pointing up into QII. At $(0,-1)$ (bottom): $(-1,0)$ — pointing left into QIII. Why this step? Right at top, up at left, left at bottom — that is exactly a clockwise circulation (look at the labelled arrows in the figure).

4. Origin. At $(0,0)$: $\dot{q}=0,\dot{p}=0$. The arrow has zero length → nothing moves → this is a fixed point, and since $V=\frac{1}{2}{q}^2$ has a minimum here, it is a center.

Verify: Speed along the loop $=\sqrt{{\dot{q}}^2+{\dot{p}}^2}=\sqrt{p^2+q^2}$. On the unit circle $q^2+p^2=1$ this equals $1$ everywhere — constant, matching a perfect circle traversed at uniform angular rate for the isotropic case $m=\omega =1$.

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C2 — Inverted oscillator: saddle and its full hyperbolic family (cells: maximum of $V$ → saddle, full level-curve family)

Forecast: at $(1,0)$, will the dot spiral, loop, or shoot away? And what shape are the orbits that are NOT the two special lines? Guess before step 5.

1. Write the flow. With $m=\lambda =1$, $\dot{q}=\frac{\partial H}{\partial p}=p,\dot{p}=-\frac{\partial H}{\partial q}=+m{\lambda }^{2}q=+q.$ Why this step? Note the crucial sign flip versus C1: because $V$ is a maximum, $-\partial V/\partial q=+q$ pushes the state away from $q=0$ instead of back toward it.

2. Origin is a fixed point: at $(0,0)$, $\dot{q}=\dot{p}=0$. And $V^{\prime \prime }(0)=-1<0$ (a maximum), so we expect instability. Why this step? Fixed point needs $p=0$ and $V^{\prime }(q)=0$; the sign of $V^{\prime \prime }$ decides its character.

3. Evaluate at $(1,0)$: $(\dot{q},\dot{p})=(0,+1)$ — pointing straight up, then the growing $p$ pushes $q$ further out. The state runs away → the origin is a saddle (answer to first forecast: shoots away). Why this step? Unlike a center (which curves the arrow back into a loop), here the arrow feeds the escape — hyperbolic, not elliptic, flow.

4. The two straight-line orbits (eigen-directions). Try $p=\pm q$. On $p=+q$: $\dot{q}=p=q$ and $\dot{p}=q=p$, so the state moves along the line $p=q$ outward — the unstable direction (grows like $e^{+t}$). On $p=-q$: $\dot{q}=p=-q$ and $\dot{p}=q=-p$, so it moves along $p=-q$ inward to the origin — the stable direction (decays like $e^{-t}$). These two crossing lines form the separatrix of the saddle. Why this step? A saddle always has exactly one incoming and one outgoing straight-line orbit.

5. The full family of orbits ($E\ne 0$). Every trajectory lies on a contour $H=\frac{1}{2}(p^2-q^2)=E=\text{const}$. Rearranged, $p^2-q^2=2E$ — the equation of a hyperbola.

   • $E>0$ (energy above the hilltop): $p^2-q^2=2E>0$, hyperbolae opening up/down (they cross the $p$-axis, never touching $q=0$ on the $q$-axis). The dot swoops in from far, bends near the origin, and escapes — it never reaches the top.
   • $E<0$: $p^2-q^2=2E<0$, i.e. $q^2-p^2=2|E|$, hyperbolae opening left/right (they cross the $q$-axis). The dot comes in with small momentum, is turned back by the hill, and retreats.
   • $E=0$: the degenerate hyperbola $p^2=q^2$, i.e. the two straight lines $p=\pm q$ — exactly the separatrix from step 4. Look at the figure: the two red separatrix lines divide the plane into four regions, and one nested hyperbola sits in each region, hugging the lines as $E\to 0$ and bulging out as $|E|$ grows. Why this step? This is the answer to the second forecast — the whole portrait is a family of hyperbolae asymptotic to the separatrix, filling every point off the two lines.

Verify: Along $p=q$, $H=\frac{1}{2}(q^2-q^2)=0$ — the separatrix sits exactly at $E=0$, the energy of the hilltop $V(0)=0$ ✓. Pick $E=\frac{1}{2}$: contour $p^2-q^2=1$ passes through $(0,1)$ — plug in: $1-0=1$ ✓, an up/down hyperbola as claimed. Contrast with C1: same origin, opposite sign of $V^{\prime \prime }$, loops replaced by hyperbolae.

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C3 — Pendulum: swing-vs-whirl threshold with real numbers (cells: energy levels, separatrix, word problem, zero velocity)

Forecast: guess whether $6\text{ m/s}$ is enough to go over the top. Above or below threshold?

1. Energy at the bottom ($\theta =0,\cos 0=1$): $E=\frac{1}{2}m{l}^2{\dot{\theta }}_0^2-mgl=\frac{1}{2}m{v}_0^2-mgl.$ Why this step? $l{\dot{\theta }}_0=v_0$ so $l^2{\dot{\theta }}_0^2=v_0^2$; substitute Energy Conservation with the launch state.

2. Energy needed at the top ($\theta =\pi ,\cos \pi =-1$, zero speed): $E_{\text{sep}}=0-mgl(-1)=+mgl.$ Why this step? The separatrix is the just barely reaches the top orbit — kinetic energy is zero there, so all the energy is the raised potential $V(\pi )=+mgl$ (the hilltop value from the figure).

3. Set equal, solve threshold speed: $\frac{1}{2}m{v}_0^2-mgl=mgl\Rightarrow v_0^2=4gl\Rightarrow v_0=2\sqrt{gl}.$ Numerically $v_0=2\sqrt{9.8\times 1}=6.261\text{ m/s}$. Why this step? Cancel $m$; the threshold does not depend on mass — a light or heavy bob whirls at the same launch speed. The kinetic energy at the bottom must supply the full $2mgl$ climb, giving $\frac{1}{2}m{v}_0^2=2mgl$.

4. Decide for $v_0=6$: since $6<6.261$, energy is below the separatrix → the pendulum swings back (libration), a closed loop in phase space. Why this step? Below-separatrix energy cannot reach $\theta =\pi$; the bob turns around at some ${\theta }_{\max }<\pi$ where $\dot{\theta }=0$ (a zero-velocity turning point).

Verify: Units: $\sqrt{gl}=\sqrt{({\text{m/s}}^2)(\text{m})}=\text{m/s}$ ✓. Sanity: $6.261>6$, so "just short" → swings, consistent with intuition that you need a hair more push to loop the loop.

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C4 — Ellipse readout: period and max momentum (cells: limiting values, numeric readout)

Forecast: does a bigger energy make the loop go around faster? Guess yes/no.

1. Semi-axes from the parent's ellipse form $\frac{q^2}{a^2}+\frac{p^2}{b^2}=1$: $a=\sqrt{\frac{2E}{m{\omega }^2}},b=\sqrt{2mE}.$ Why this step? Setting $H=E$ and dividing by $E$ gave exactly these axis lengths in the parent derivation.

2. Plug numbers: $a=\sqrt{\frac{2\cdot 9}{2\cdot 9}}=\sqrt{1}=1\text{ m},b=\sqrt{2\cdot 2\cdot 9}=\sqrt{36}=6\text{kg⋅m/s}.$ Why this step? $m{\omega }^2=2\cdot 9=18$; $2E=18$; ratio $=1$. Max momentum is $6$.

3. Period. For SHM $T=\frac{2\pi }{\omega }=\frac{2\pi }{3}\approx 2.094\text{ s}$, independent of $E$. Why this step? The angular frequency $\omega$ sets the going-around rate; the ellipse scales with $E$ but the time to circle it does not. So bigger loop ≠ faster circulation — answer to the forecast is no.

4. Limits. $E\to 0$: $a,b\to 0$, the loop shrinks to the center point. $E\to \infty$: $a,b\to \infty$, an ever-larger ellipse, but still period $T=2\pi /3$. Why this step? Confirms the degenerate ($E=0$) and unbounded ($E\to \infty$) ends of the family.

Verify: Consistency check — the ellipse area is $\pi ab=\pi \cdot 1\cdot 6=6\pi$, and the action $\oint pdq=2\pi E/\omega =2\pi \cdot 9/3=6\pi$ ✓ (they must match; this is the link to adiabatic invariants and Liouville's Theorem).

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C5 — Damped oscillator: sign of $\dot{E}$ and ALL three damping regimes (cells: non-conservative, $t\to \infty$ decay, focus vs. node)

Forecast: does energy leak at a constant rate, or faster when moving fast? And which regime spirals in phase space vs. which just slides straight to the origin? Guess.

1. Differentiate $E$: $\dot{E}=\dot{x}\ddot{x}+{\omega }_0^{2}x\dot{x}.$ Why this step? Chain rule term-by-term on the two energy pieces.

2. Substitute $\ddot{x}=-2\gamma \dot{x}-{\omega }_0^{2}x$: $\dot{E}=\dot{x}(-2\gamma \dot{x}-{\omega }_0^{2}x)+{\omega }_0^{2}x\dot{x}=-2\gamma {\dot{x}}^2.$ Why this step? The ${\omega }_0^{2}x\dot{x}$ terms cancel exactly, leaving a pure loss term.

3. Sign: ${\dot{x}}^2\ge 0$ and $\gamma >0$, so $\dot{E}=-2\gamma {\dot{x}}^2\le 0$ — energy never increases, and drains fastest where the bob moves fastest (answer to first forecast: not constant). Why this step? The dissipation rate scales with ${\dot{x}}^2$, biggest at loop-crossings of the $q$-axis.

4. Fraction after one period (underdamped, $\gamma =0.1$). From the envelope, amplitude $\propto e^{-\gamma t}$, so energy $\propto e^{-2\gamma t}$. Over $T=2\pi /{\omega }_0=2\pi$: $\frac{E(T)}{E(0)}=e^{-2\gamma T}=e^{-2(0.1)(2\pi )}=e^{-0.4\pi }\approx \mathrm{0.2846.}$ Why this step? Energy is amplitude-squared; the spiral crosses ever-smaller ellipses, so after one loop about $28.5\%$ remains.

5. Convert to decibels. A decibel is a logarithmic level unit: $\text{level}=10{\log }_{10}(\text{ratio})$ dB. So $10{\log }_{10}(0.2846)=10\times (-0.5459)\approx -5.46\text{ dB}.$ Why this step? The prompt asks for a decibel estimate — the log turns the multiplicative energy fraction into an additive level drop of about $-5.5$ dB per swing.

6. Classify the fixed point in all three regimes (roots $s=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$, both roots have negative real part → always attracting):

   • Underdamped $\gamma <{\omega }_0$: complex roots → the phase point circulates while shrinking → a stable focus (spiral sink). This is our $\gamma =0.1$ case: an inward spiral.
   • Critically damped $\gamma ={\omega }_0$: repeated real root → a stable degenerate (improper) node; trajectories come in tangent to one line, no spiralling.
   • Overdamped $\gamma >{\omega }_0$: two real negative roots → a stable node; trajectories slide straight in along the two eigen-directions, no loop at all. Why this step? This answers the second forecast: only the underdamped regime spirals; critical and overdamped go straight to the origin. See Stability and Fixed Points for the general focus-vs-node criterion.

Verify: $e^{-0.4\pi }\approx e^{-1.2566}\approx 0.2846$, between $0$ and $1$ ✓; $10{\log }_{10}(0.2846)\approx -5.46$ dB, negative as a loss must be ✓. Regime boundary check: discriminant ${\gamma }^2-{\omega }_0^2$ is $-0.99<0$ for $\gamma =0.1,{\omega }_0=1$ (complex → focus ✓), $=0$ at $\gamma =1$ (critical), $>0$ for $\gamma =2$ (real → node ✓).

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C6 — Exam twist: full portrait from $V(q)$ alone (cells: min→center, max→saddle, sketch task)

Forecast: how many fixed points, and how many are stable?

1. Fixed points $=$ extrema of $V$: $V^{\prime }(q)=q^3-q=q(q^2-1)=0\Rightarrow q=0,q=+1,q=-1$, all with $p=0$. Why this step? A fixed point needs $\dot{p}=-V^{\prime }(q)=0$ AND $\dot{q}=p/m=0$, so $p=0$ and $V^{\prime }=0$.

2. Classify by $V^{\prime \prime }(q)=3q^2-1$:

   • $q=\pm 1$: $V^{\prime \prime }=3-1=2>0$ → minima → centers (stable loops in each well).
   • $q=0$: $V^{\prime \prime }=-1<0$ → maximum → saddle (unstable). Why this step? Minima of $V$ give oscillatory closed orbits; a maximum gives the hyperbolic saddle flow (exactly the C2 mechanism).

3. Separatrix. The saddle at $q=0$ sits at energy $E=V(0)=0$. The curve $H(q,p)=0$ is the figure-eight separatrix: two loops, one around each well, kissing at the origin. Why this step? $E=0$ is exactly the energy that just reaches the central hilltop with zero speed.

4. Above and below. For $E<0$ (below the hilltop, above the well bottoms $V_{\min }=-\frac{1}{4}$): the dot is trapped in one well → a small loop. For $E>0$: a single large loop encircles both wells. Why this step? Enough energy to clear the central hill lets the bob visit both wells → one big orbit.

Verify: $V_{\min }=V(1)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$ and $V(0)=0$. Barrier height $=0-(-\frac{1}{4})=\frac{1}{4}$ J·(unit) $>0$ ✓, confirming the wells are genuinely separated by a hill.

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C7 — The degenerate case: when the $V^{\prime \prime }$ test fails (cells: $V^{\prime \prime }=0$, non-hyperbolic equilibrium)

Forecast: at $q=0$ the potential is flat to second order — is the equilibrium stable, unstable, or something in between? Guess before step 4.

1. Fixed point. $V^{\prime }(q)=q^2=0\Rightarrow q=0$; with $p=0$ this is the only equilibrium. Why this step? Same rule as always — $\dot{q}=p/m=0$ and $\dot{p}=-V^{\prime }(q)=0$.

2. The second-derivative test fails. $V^{\prime \prime }(q)=2q$, so $V^{\prime \prime }(0)=0$ — neither positive nor negative. The "min → center, max → saddle" dichotomy of C6 gives no answer here. Why this step? $V^{\prime \prime }=0$ means the quadratic approximation is flat; the fixed point is non-hyperbolic, and linearization cannot classify it. We must look at $V$ itself, not its curvature.

3. Shape of the potential. $V=\frac{1}{3}{q}^3$ is negative for $q<0$ and positive for $q>0$, passing through zero with a flat inflection at the origin. So the origin is a min from one side and a max from the other — a one-sided hill. Why this step? The direction of the restoring force $-V^{\prime }(q)=-q^2\le 0$ is the real story: it always pushes toward negative $q$, never back toward the origin from the left.

4. Character: a cusp / degenerate saddle-node. Because the force $-q^2$ is $\le 0$ everywhere, a state nudged to $q>0$ is pushed back toward $0$ (looks stable on the right), but a state at $q<0$ is pushed further negative (unstable on the left). The two behaviours meet at the origin in a cusp: trajectories approach along the $q<0$ side and are repelled along the... wait — approach from the right, flee to the left. The equilibrium is therefore half-stable (degenerate) — the phase curves form a cusp, not the clean crossing lines of a true saddle nor the closed loops of a center (answer to forecast: neither stable nor unstable — half-stable). Why this step? This is exactly why the matrix lists $V^{\prime \prime }=0$ separately: the hyperbolic classification (center/saddle) does not apply, and you must reason from the global shape of $V$.

Verify: Force $-V^{\prime }(q)=-q^2$: at $q=+0.5$ it is $-0.25<0$ (pushes left, toward $0$ — stable side); at $q=-0.5$ it is $-0.25<0$ (also pushes left, i.e. away from $0$ — unstable side). Same-sign force on both sides is the algebraic signature of a half-stable degenerate point ✓. Energy contours $H=\frac{p^2}{2}+\frac{1}{3}{q}^3=E$ have a cusp at the origin, not a crossing.

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Recall Quick self-test — each answer names the deciding algebra/step

Threshold launch speed for a pendulum to whirl ::: $v_0=2\sqrt{gl}$ from $\frac{1}{2}m{v}_0^2=2mgl$ (must supply the full top-to-bottom climb $2mgl$). Does bigger oscillator energy shorten the period? ::: No — $T=2\pi /\omega$ has no $E$ in it; the ellipse scales but the circulation time does not (C4 step 3). Rate of energy loss in a damped oscillator ::: $\dot{E}=-2\gamma {\dot{x}}^2\le 0$ (the ${\omega }_0^{2}x\dot{x}$ terms cancel; C5 step 2), largest where speed is largest. Energy fraction and dB after one damped period ($\gamma =0.1,{\omega }_0=1$) ::: $e^{-0.4\pi }\approx 0.285$, i.e. $10{\log }_{10}(0.285)\approx -5.46$ dB (C5 steps 4–5). The three damping regimes and their fixed-point types ::: Underdamped $\gamma <{\omega }_0$ → stable focus (spiral); critical $\gamma ={\omega }_0$ → degenerate node; overdamped $\gamma >{\omega }_0$ → stable node (C5 step 6). Full off-axis orbits of an inverted oscillator ::: Hyperbolae $p^2-q^2=2E$ — $E>0$ open up/down, $E<0$ open left/right, $E=0$ the two separatrix lines (C2 step 5). Fixed-point type at a maximum of $V$ ::: A saddle: sign flip $-\partial V/\partial q=+q$ gives run-away flow with stable line $p=-q$ and unstable line $p=+q$ (C2 step 4). Energy value of the double-well separatrix here ::: $E=V(0)=0$ (the hilltop value; C6 step 3). What happens when $V^{\prime \prime }=0$ at a fixed point? ::: The min/max test fails — the point is non-hyperbolic; reason from the global shape of $V$. For $V=\frac{1}{3}{q}^3$ it is a half-stable cusp (C7).