Worked examples — Hamilton-Jacobi equation

Before symbols fly, four plain-word reminders (every symbol used below is one of these):

• $H$ = the energy expressed in coordinates $q$ and momenta $p$ (the Hamiltonian).
• $S$ = Hamilton's principal function; its slope in a coordinate is the momentum there: $p_i=\partial S/\partial q_i$. Physically $S$ is the action $\int Ldt$.
• $W$ = the time-independent part of $S$ when energy is conserved: $S=W-Et$.
• $P$ = a new canonical momentum produced by the canonical transformation. The whole point of HJ is that these $P$'s are constants of the motion. In every 1-DOF example below we let the conserved energy itself play that role, writing $P=E$: energy is a constant, so it is the most natural "new momentum" to hand the machine.

The one recipe we repeat every single time:

1. Write $H(q,\partial S/\partial q,t)+\partial S/\partial t=0$.
2. If $H$ has no explicit $t$: put $S=W-Et$, giving $H(q,\partial W/\partial q)=E$.
3. Solve for each $\partial W/\partial q=W^{\prime }$, integrate to get $W$.
4. Choose the constant energy as the new momentum $P=E$, then set the new coordinate $Q=\partial S/\partial P=\text{const}$ — this constant equation is the trajectory.
5. Invert to get $q(t)$; check.

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The scenario matrix

Every problem the HJ method can throw belongs to one of these cells. The examples below are chosen so that together they hit all of them.

Cell	What makes it different	Example
A. Separable, unbounded 1-DOF	motion runs off to infinity; $E$ any sign of momentum	Ex 1 (free particle, both directions)
B. Separable, bounded 1-DOF	turning points, oscillation; sign of $p$ flips	Ex 2 (oscillator, full period + both branches)
C. Degenerate / zero input	$E=0$ or vanishing force — the limiting case	Ex 3 ($E=0$ free particle; constant-force limit)
D. Constant force (linear potential)	non-quadratic, unbounded one way, bounded other	Ex 4 (particle under gravity)
E. Multi-DOF additive separation	$S=\sum W_i(q_i)$; each coordinate its own constant	Ex 5 (2-D free particle)
F. Non-Cartesian / central force	separation in $r,\theta$; angular momentum as a constant	Ex 6 (central force, radial equation)
G. Real-world word problem	you must build $H$ from a story	Ex 7 (skier down a frictionless slope)
H. Exam twist — time-dependent $H$	$S=W-Et$ forbidden; must keep the $t$ term	Ex 8 (particle with a time-varying push)

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Ex 1 — Free particle, both directions (Cell A)

Forecast: guess before reading — a free particle should give a straight line $x=vt+x_0$ with $v=\pm \sqrt{2E/m}$. Which sign does the method deliver?

The figure plots both solutions: the red line is the right-going particle ($+v$) and the black line the left-going one ($-v$). Same energy, opposite slope — that is the story of this cell, and steps 1–4 build it.

1. Write HJ, separate time. $\frac{1}{2m}(W^{\prime }{)}^2=E$, so $W^{\prime }=\pm \sqrt{2mE}$. Why this step? Energy is conserved (no explicit $t$), so $S=W-Et$ and the PDE collapses to an algebraic equation for the slope $W^{\prime }=p$. The $\pm$ is the whole point of this cell: $p$ can be either sign at the same energy — momentum squared loses direction, we restore it here.
2. Integrate. $W=\pm \sqrt{2mE}x$, hence $S=\pm \sqrt{2mE}x-Et$. Why? Integrating the slope rebuilds the function; the sign carries which way the particle goes.
3. Constant new coordinate. Take $P=E$ (energy is the conserved new momentum). Then $Q=\frac{\partial S}{\partial E}=\pm \sqrt{\frac{m}{2E}}x-t=-t_0$. Why? We demanded $K=0$, so by Hamilton's equations $\dot{Q}=\partial K/\partial P=0$: $Q$ literally cannot change in time. A quantity frozen in time equals a constant, which we name $-t_0$ — and that equation is the trajectory in disguise.
4. Invert. Rearranging $\pm \sqrt{m/2E}x-t=-t_0$ for $x$ gives $x=\pm \sqrt{\frac{2E}{m}}(t-t_0)$. Why? We want position as a function of time, $x(t)$, so we isolate $x$. The bracket $(t-t_0)$ shows $t_0$ is just the instant the particle passes the origin.

Verify: velocity $\dot{x}=\pm \sqrt{2E/m}$, so $H=\frac{1}{2}m{\dot{x}}^2=E$ ✅. Take $E=8\text{ J},m=1\text{ kg}$: $v=\pm 4\text{ m/s}$. Units: $\sqrt{\text{J}/\text{kg}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$ ✅. The two signs are the two lines in the figure (right-going in red, left-going in black).

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Ex 2 — Harmonic oscillator, full period, both branches (Cell B)

Forecast: bounded means $x$ can only live where $E\ge \frac{1}{2}m{\omega }^2{x}^2$. Guess the turning points and the period before computing.

The figure shows the potential $V(x)$ (black parabola) cut by the energy line $E$; the red band on the axis is the allowed region $|x|\le x_{\max }$, and the two red dots where $V=E$ are the turning points. Watch that band as we compute — it is exactly the range the algebra permits.

1. Solve for the slope. $W^{\prime }=p=\pm \sqrt{2m(E-\frac{1}{2}m{\omega }^2{x}^2)}$. Why? Solving $H(x,p)=E$ for $p$. The square-root is real only when $E-\frac{1}{2}m{\omega }^2{x}^2\ge 0$, i.e. $|x|\le x_{\max }=\sqrt{2E/(m{\omega }^2)}$. Those two equalities are the turning points — the edges of the red allowed region in the figure.
2. Sign of $p$ names the branch. Moving right ($x$ increasing) uses $+$; after hitting $x_{\max }$ the particle turns and uses $-$. This is the bounded-case subtlety cell B exists to show.
3. Constant coordinate. With $P=E$, $Q=\frac{\partial S}{\partial E}=\int \frac{mdx}{\sqrt{2m(E-\frac{1}{2}m{\omega }^2{x}^2)}}-t=\text{const}.$ Why? Same $K=0$ logic; $\partial /\partial E$ acts under the integral.
4. Do the integral. Substituting $x=x_{\max }\sin \theta$ gives $\frac{1}{\omega }\arcsin (x/x_{\max })=t+\text{const}$. Why the $\arcsin$? Because the integrand is exactly the derivative of an arcsine — the geometry of a circle in phase space. In Action-Angle Variables this same integral becomes the angle variable.
5. Invert. $x(t)=x_{\max }\sin (\omega t+\varphi )$, amplitude $x_{\max }=\sqrt{2E/(m{\omega }^2)}$. Why? We solved the previous line for $x$ by taking the sine of both sides (sine undoes arcsine); the leftover constant becomes the phase $\varphi$. This gives position as an explicit function of time.
6. Period. One full loop (right branch and left branch) covers phase $2\pi$: $T=2\pi /\omega$. Why? The sine repeats when its argument $\omega t$ advances by $2\pi$; the elapsed time for that is $T=2\pi /\omega$. Both branches (out and back) together make exactly one such repeat.

Verify: at $x=x_{\max }$, $\dot{x}=x_{\max }\omega \cos (\pm \pi /2)=0$ — the particle stops at the turning point ✅. Energy: $\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2{x}_{\max }^2=E$ ✅. Numbers $m=1,\omega =2,E=8$: $x_{\max }=\sqrt{16/4}=2$, $T=\pi \approx 3.1416$ ✅.

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Ex 3 — Degenerate input: $E=0$ and the vanishing case (Cell C)

Forecast: zero energy usually means "sitting still," but does the method agree for both systems?

1. (a) Free particle, $E=0$. $W^{\prime }=\sqrt{2m\cdot 0}=0\Rightarrow p=0$, so $\dot{x}=p/m=0$: the particle is at rest anywhere. Why? $E=0$ with only kinetic energy forces $p=0$. This is the degenerate edge of Cell A — velocity collapses to zero.
2. (b) Oscillator, $E=0$. $x_{\max }=\sqrt{2\cdot 0/(m{\omega }^2)}=0$, so the only allowed point is $x=0$. Why? The allowed region $|x|\le x_{\max }$ shrinks to a single point — the oscillator sits at the bottom of its well, $x(t)\equiv 0$.

Verify: (a) $x(t)=x_0$ constant gives $\dot{x}=0$, $H=0$ ✅. (b) $x_{\max }=0\Rightarrow x(t)=0$, and $H=0$ ✅. Both degenerate limits are consistent with $E\to 0$ of Ex 1–2 (amplitude and speed both $\to 0$).

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Ex 4 — Constant force / linear potential (Cell D)

Forecast: this is projectile motion in 1-D — expect $z(t)$ quadratic, a parabola in time. Unbounded downward-in-energy but with a top turning point when thrown up.

The figure plots the resulting height $z(t)$ as a red downward parabola in time, with the black dot marking the peak where the particle momentarily stops. Steps 1–4 derive exactly that curve; glance back at the peak as we reach step 1's turning point.

1. Slope. $W^{\prime }=p=\pm \sqrt{2m(E-mgz)}$. Why? Solve $H=E$ for $p$ (isolate the momentum, since $W^{\prime }=p$ is always our first target). Real only for $z\le E/(mg)$ — the highest reachable height $z_{\max }=E/(mg)$, the single turning point (bounded above, unbounded below): the signature of a linear potential, and the peak in the figure.
2. Constant coordinate. With $P=E$, $Q=\frac{\partial S}{\partial E}=\int \frac{mdz}{\sqrt{2m(E-mgz)}}-t=\text{const}.$ Why? Because $K=0$ forces $\dot{Q}=0$, so $Q$ is frozen and equals a constant. Differentiating $S=\int W^{\prime }dz-Et$ by the new momentum $E$ pulls the $\partial /\partial E$ under the integral and drops a $-t$ from the $-Et$ term.
3. Integrate. $\int \frac{mdz}{\sqrt{2m(E-mgz)}}=-\frac{1}{g}\sqrt{\frac{2(E-mgz)}{m}}$. Why? Direct substitution $u=E-mgz$; the $\sqrt{u}$ antiderivative appears.
4. Set constant, solve. $-\frac{1}{g}\sqrt{\frac{2(E-mgz)}{m}}-t=-t_0$. Square and rearrange: $z(t)=\frac{E}{mg}-\frac{g}{2}(t-t_0{)}^2.$ Why? We want $z$ as a function of $t$, so we isolate the square-root, square to remove it, then solve the linear-in-$z$ equation. The subtracted $(t-t_0{)}^2$ term is what bends the trajectory into the parabola of the figure.

Verify: $\dot{z}=-g(t-t_0)$, $\ddot{z}=-g$ — constant downward acceleration ✅ (Newton's $F=-mg$). At $t=t_0$, $z=E/(mg)=z_{\max }$ and $\dot{z}=0$: the peak ✅. Numbers $E=20\text{ J},m=1\text{ kg},g=10{\text{ m/s}}^2$: $z_{\max }=2\text{ m}$ ✅.

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Ex 5 — Multi-DOF additive separation (Cell E)

Forecast: each direction is an independent free particle, so guess $S$ splits into an $x$-part plus a $y$-part, each carrying its own constant.

1. Assume the split. $S=W_x(x)+W_y(y)-Et$. Why? Because $H$ is a sum of independent pieces, the ansatz "one $W$ per coordinate" makes the PDE fall apart into separate equations — the core trick of multi-DOF HJ. Here $W_x^{\prime }\equiv d{W}_x/dx$ and $W_y^{\prime }\equiv d{W}_y/dy$.
2. Separate. $\frac{1}{2m}(W_x^{\prime }{)}^2+\frac{1}{2m}(W_y^{\prime }{)}^2=E$. Set $\frac{1}{2m}(W_x^{\prime }{)}^2={\alpha }_x$ (a constant). Then $\frac{1}{2m}(W_y^{\prime }{)}^2=E-{\alpha }_x\equiv {\alpha }_y$. Why introduce ${\alpha }_x$? An $x$-only term equals a ($y$+const) term for all $x,y$ only if each side is constant. Those constants ${\alpha }_x,{\alpha }_y$ are the new momenta — exactly the "$n$ constants" the parent note warned you not to forget.
3. Integrate. $W_x=\sqrt{2m{\alpha }_x}x$, $W_y=\sqrt{2m{\alpha }_y}y$.
4. Trajectory. $Q_x=\partial S/\partial {\alpha }_x=\sqrt{\frac{m}{2{\alpha }_x}}x-t=\text{const}$ gives $x=\sqrt{2{\alpha }_x/m}(t-t_0)$; likewise $y=\sqrt{2{\alpha }_y/m}(t-t_1)$. Why? Each $\alpha$ is a new momentum, so each $Q=\partial S/\partial \alpha$ is frozen ($\dot{Q}=0$) and equals a constant; isolating $x$ (and $y$) from those two constant equations gives the two uniform-motion lines.

Verify: speeds $v_x=\sqrt{2{\alpha }_x/m}$, $v_y=\sqrt{2{\alpha }_y/m}$ give $\frac{1}{2}m(v_x^2+v_y^2)={\alpha }_x+{\alpha }_y=E$ ✅. Numbers $m=1,{\alpha }_x=2,{\alpha }_y=3$: $v_x=2,v_y=\sqrt{6}\approx 2.449$, total $E=5$ ✅.

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Ex 6 — Central force, separation in $r,\theta$ (Cell F)

Forecast: $\theta$ never appears in $H$ (only $p_{\theta }$), so guess its momentum is conserved — that constant should be angular momentum.

1. Ansatz. $S=W_r(r)+W_{\theta }(\theta )-Et$, with $p_r=W_r^{\prime }\equiv d{W}_r/dr$, $p_{\theta }=W_{\theta }^{\prime }\equiv d{W}_{\theta }/d\theta$. Why? Same additive-separation idea as Ex 5, now in polar coordinates.
2. Isolate $\theta$. The $\theta$-dependence enters only through $p_{\theta }^2/r^2$. Since $H$ has no $\theta$ itself, $p_{\theta }=W_{\theta }^{\prime }$ must be constant; call it $\ell$ (angular momentum). So $W_{\theta }=\ell \theta$. Why this step? A coordinate absent from $H$ is cyclic; its momentum is automatically conserved — HJ makes this fall out as a separation constant.
3. Radial equation. Substitute $p_{\theta }=\ell$: $\frac{1}{2m}[(W_r^{\prime }{)}^2+\frac{{\ell }^2}{r^2}]+V(r)=E\Rightarrow W_r^{\prime }=\pm \sqrt{2m(E-V(r))-\frac{{\ell }^2}{r^2}}.$ Why? With $\theta$ handled, only $r$ remains — a 1-DOF problem with an effective potential $V_{\text{eff}}=V+{\ell }^2/(2m{r}^2)$. The $\pm$ matters here: the $+$ root is a particle moving outward ($r$ increasing), the $-$ root a particle moving inward ($r$ decreasing); a bound orbit uses both in turn, flipping sign at each radial turning point where the root vanishes.
4. Result. $S=\pm \int \sqrt{2m(E-V)-{\ell }^2/r^2}dr+\ell \theta -Et$; the two constants $(E,\ell )$ are the new momenta.

Verify (special case $V=0$, closest-approach turning point). For $V=0$ the turning point ($W_r^{\prime }=0$) sits at $r_0=\ell /\sqrt{2mE}$. Numbers $\ell =6,m=1,E=8$: $r_0=6/\sqrt{16}=1.5$ ✅. At that radius all energy is angular: ${\ell }^2/(2m{r}_0^2)=36/(2\cdot 2.25)=8=E$ ✅.

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Ex 7 — Real-world word problem: the frictionless slope (Cell G)

Forecast: energy conservation says $v=\sqrt{2gh}$, independent of the angle. Will HJ reproduce that?

The figure sets the scene: the red dot is the skier partway down the black slope of angle $\alpha$, with the vertical height $h$ marked. The single coordinate we use, $s$, runs along the slope from the start — step 1 turns this picture into a Hamiltonian.

1. Build $H$. Along the slope the drop per distance is $\sin \alpha$, so height $=h-s\sin \alpha$ and potential $V=mg(h-s\sin \alpha )$. Thus $H=\frac{p_s^2}{2m}+mg(h-s\sin \alpha )=E$. Why this step? The hardest part of a word problem is writing $H$: pick the natural coordinate ($s$ along the incline, as drawn) and express energy in it.
2. Start conditions fix $E$. At rest at the top: $s=0,p_s=0\Rightarrow E=mgh$. Why? $E$ is a number set by the initial state.
3. Slope of $W$. $W_s^{\prime }=p_s=\sqrt{2m(E-mg(h-s\sin \alpha ))}=\sqrt{2m^{2}gs\sin \alpha }$ (using $E=mgh$). Why? Solve $H=E$ for $p_s$; the $mgh$ cancels beautifully, leaving only the drop $s\sin \alpha$.
4. Speed. $v=p_s/m=\sqrt{2gs\sin \alpha }$. The vertical drop is $d=s\sin \alpha$, so $v=\sqrt{2gd}$.

Verify: at the bottom of a slope whose foot is at height $0$, total vertical drop $d=h$, giving $v=\sqrt{2gh}$ — angle cancels, matching energy conservation ✅. Numbers $g=10,h=5$: $v=\sqrt{100}=10\text{ m/s}$ ✅. Units $\sqrt{({\text{m/s}}^2)\cdot \text{m}}=\text{m/s}$ ✅.

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Ex 8 — Exam twist: explicitly time-dependent $H$ (Cell H)

Forecast: because energy is not conserved (the force does time-dependent work), the trick from Ex 1–7 is forbidden. Expect $p$ to grow like $\frac{1}{2}b{t}^2$.

1. Why the shortcut fails. $H$ contains $t$ explicitly, so $\partial S/\partial t\ne -\text{const}$: the split $S=W(x)-Et$ assumes constant energy and is not allowed here. This is exactly the trap an exam sets.
2. Full HJ. $\frac{1}{2m}(\frac{\partial S}{\partial x}{)}^2-btx+\frac{\partial S}{\partial t}=0$. Why? Keep the honest time term — with no conserved energy we must work with the full time-dependent PDE.
3. Ansatz linear in $x$. Try $S=f(t)x+g(t)$ so $p=\partial S/\partial x=f(t)$. Why this shape? The force is uniform in $x$ (same everywhere), so the momentum should depend on time alone; a function first-degree in $x$ is the simplest $S$ whose $x$-slope is $x$-independent.
4. Match. Substitute: $\frac{f^2}{2m}-btx+f^{\prime }x+g^{\prime }=0$. Collect the $x$-term: $f^{\prime }=bt\Rightarrow f=\frac{1}{2}b{t}^2+p_0$. The rest: $g^{\prime }=-f^2/(2m)$. Why? The equation must hold for all $x$, so the coefficient of $x$ and the $x$-free part must each vanish separately — one equation for $f$, one for $g$. Here $p_0$ is the integration constant (the momentum at $t=0$).
5. Momentum and velocity. From $p=f(t)=\frac{1}{2}b{t}^2+p_0$ and $\dot{x}=p/m$: $\dot{x}=\frac{1}{2m}b{t}^2+\frac{p_0}{m}.$ Why? The type-2 relation gives $p$ directly; dividing by $m$ turns momentum into velocity, since $p=m\dot{x}$ for this kinetic term.
6. Integrate to position. Integrating $\dot{x}$ once in time: $x(t)=x_0+\frac{p_0}{m}t+\frac{b}{6m}{t}^3.$ Why? We wanted $x(t)$, and velocity is $dx/dt$; integrating $t^2\to t^3/3$ produces the cubic, with $x_0$ the starting position. The growing $t^3$ term is the fingerprint of a force that itself grows with time.

Verify: Newton says $m\ddot{x}=F=bt$, so $\ddot{x}=bt/m$. From our $x(t)$: $\ddot{x}=\frac{b}{m}t$ ✅. Numbers $b=6,m=1,p_0=0,x_0=0$: $x(t)=t^3$, so $x(2)=8$ and $\ddot{x}(t)=6t\Rightarrow F=6t$, matching $bt$ ✅.

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Wrap-up: every cell filled

Recall Which example covered which scenario?

Unbounded 1-DOF, both signs of $p$ ::: Ex 1 (Cell A) Bounded oscillation, turning points, period ::: Ex 2 (Cell B) Degenerate zero-energy limit ::: Ex 3 (Cell C) Linear potential (constant force) ::: Ex 4 (Cell D) Multi-DOF additive separation ::: Ex 5 (Cell E) Central force, cyclic angle, angular momentum constant ::: Ex 6 (Cell F) Real-world story where you must build H ::: Ex 7 (Cell G) Explicitly time-dependent H — no S=W−Et ::: Ex 8 (Cell H)