Streamlines, pathlines, streaklines: Worked examples

This is a practice child page for the parent topic. Before we start, one reminder of the three tools so no symbol is used unearned:

Recall The three curves in one breath

Streamline — freeze time, follow the arrows. ODE: $\frac{d x}{u} = \frac{d y}{v}$ with $t$ a constant.
Pathline — one particle's whole trip. ODE: $\frac{d x}{d t} = u$ , $\frac{d y}{d t} = v$ , integrate in time.
Streakline — the smoke plume: all particles that passed one point, seen now.

Here $u$ = the horizontal (x-direction) speed of the fluid, $v$ = the vertical (y-direction) speed. Together $v = (u, v)$ is the velocity field.

Every symbol below is defined the first time it appears. A "field" just means: give me a point $(x, y)$ and a time $t$ , I hand you back the two speeds $u$ and $v$ .

The scenario matrix

Think of this topic as a machine that can be fed different kinds of flow. Each row is a case class; the last column names the worked example that lands on it.

#	Case class	What makes it special	Example
C1	Steady, curved field	$\partial v / \partial t = 0$ → all three curves coincide	Ex 1
C2	Steady, straight-but-shearing field	streamlines curve even though flow never changes in time	Ex 2
C3	Unsteady field, all three differ	time appears explicitly in $v$	Ex 3
C4	Sign / quadrant bookkeeping	$u, v$ change sign; which way do arrows point in each quadrant	Ex 4
C5	Zero / degenerate input (stagnation point)	$u = 0$ and/or $v = 0$ ; division-by-zero in the ODE	Ex 5
C6	Limiting behaviour	$t \to \infty$ or a source switched on long ago	Ex 6
C7	Real-world word problem (smoke plume in gusty wind)	translate physical story into $u, v$	Ex 7
C8	Exam twist — reverse question	given streamlines, reconstruct a field & test steadiness	Ex 8

We will cover all eight.

Worked example Ex 1 — Steady curved field (C1): all three coincide

Field: $v = (u, v) = (- y, x)$ . Find the streamline, pathline (released at $(1, 0)$ at $t = 0$ ), and streakline (source at $(1, 0)$ ). Show they are the same curve.

Forecast: guess before reading — will they match? (This field has no $t$ in it, so...)

Step 1 — Streamline. Solve $\frac{d x}{- y} = \frac{d y}{x} \Rightarrow x d x = - y d y \Rightarrow x^{2} + y^{2} = C$ . Why this step? "Tangent parallel to $v$ " gives the ratio ODE with $t$ frozen; there is no $t$ to freeze here so nothing changes. Through $(1, 0)$ : $C = 1$ , a unit circle.

Step 2 — Pathline. Solve $\frac{d x}{d t} = - y, \frac{d y}{d t} = x$ . Differentiate the first: $\overset{x}{¨} = - \overset{y}{˙} = - x$ , so $x = cos t$ , $y = sin t$ (using $x (0) = 1, y (0) = 0$ ). Why this step? A particle follows the actual field over time; the coupled system is the harmonic-oscillator pair, whose solution is a circle traced at unit angular speed.

Step 3 — Streakline. A particle released at time $τ$ from $(1, 0)$ is $x = cos (t - τ), y = sin (t - τ)$ . Fix $t$ , sweep $τ$ : the point $(cos (t - τ), sin (t - τ))$ runs over the whole unit circle. Why this step? Steady flow means every released particle rides the identical circular path; the plume of all of them is that circle.

Verify: all three give $x^{2} + y^{2} = 1$ . ✔ This is the golden rule (C1: steady ⇒ identical).

Worked example Ex 2 — Steady shear (C2): curved streamlines from a "straight" flow

Field: $v = (u, v) = (y, 0)$ — fluid moves purely horizontally, but faster higher up (a shear layer). Find the streamlines.

Forecast: the flow is entirely sideways ( $v = 0$ ). Do you expect straight or curved streamlines?

Step 1 ODE: $\frac{d x}{y} = \frac{d y}{0}$ . Why this step? Plug $u = y, v = 0$ into $\frac{d x}{u} = \frac{d y}{v}$ .

Step 2 A denominator of $0$ means the numerator must be $0$ too: $d y = 0 \Rightarrow y = const$ . Why this step? $\frac{d y}{0}$ is only sensible as "the change in $y$ that keeps the ratio finite is zero." Physically $v = 0$ so nothing moves vertically.

Verify: streamlines are horizontal lines $y = c$ . Straight! (Contrast: a pathline here is also a straight line because the field is steady — Ex 1's rule again. C2 shows "straight arrows ⇒ straight streamlines," a sanity anchor before we complicate things.) ✔

Worked example Ex 3 — Unsteady field (C3): the three curves split apart

Field: $v = (u, v) = (1, t)$ . For a particle/source at the origin, find (a) streamline at $t = 2$ , (b) pathline released at $t = 0$ , (c) streakline observed at $t = 2$ .

Forecast: the field literally contains $t$ . Guess: will (a), (b), (c) be the same or three different curves?

Step 1 — Streamline at $t = 2$ . Freeze $t = 2$ : $\frac{d x}{1} = \frac{d y}{2} \Rightarrow d y = 2 d x \Rightarrow y = 2 x$ . Why this step? Streamlines are a snapshot; at $t = 2$ every arrow points in direction $(1, 2)$ , so the snapshot lines have slope $2$ .

Step 2 — Pathline from $(0, 0)$ at $t = 0$ . $\frac{d x}{d t} = 1 \Rightarrow x = t$ ; $\frac{d y}{d t} = t \Rightarrow y = \frac{1}{2} t^{2}$ . Eliminate $t = x$ : $y = \frac{1}{2} x^{2}$ . Why this step? One particle integrated in time; the vertical speed grows with $t$ , bending the trail into a parabola.

Step 3 — Streakline at $t = 2$ . Particle released at $τ$ from origin: $x = t - τ, y = \frac{1}{2} (t^{2} - τ^{2})$ . Put $t = 2$ : $x = 2 - τ, y = \frac{1}{2} (4 - τ^{2})$ . Eliminate $τ = 2 - x$ : $y = \frac{1}{2} (4 - (2 - x)^{2}) = 2 x - \frac{1}{2} x^{2}$ . Why this step? Different particles left at different $τ$ , each carrying a different "history" of the growing vertical speed; the plume seen now is $y = 2 x - \frac{1}{2} x^{2}$ .

Verify: three different curves — line $y = 2 x$ , parabola $y = \frac{1}{2} x^{2}$ , parabola $y = 2 x - \frac{1}{2} x^{2}$ . They agree only at the origin. ✔ (C3: unsteady ⇒ all differ.)

Worked example Ex 4 — Signs & quadrants (C4): reading arrow directions everywhere

Field: $v = (u, v) = (- y, x)$ again (rotational). Without solving anything, state the direction of flow in each of the four quadrants, and confirm the sense of rotation.

Forecast: counter-clockwise or clockwise? Guess from the point $(1, 0)$ first.

Step 1 — Quadrant I $(x > 0, y > 0)$ : $u = - y < 0$ (leftward), $v = x > 0$ (upward) ⇒ arrow points up-left. Step 2 — Quadrant II $(x < 0, y > 0)$ : $u = - y < 0$ (left), $v = x < 0$ (down) ⇒ down-left. Step 3 — Quadrant III $(x < 0, y < 0)$ : $u = - y > 0$ (right), $v = x < 0$ (down) ⇒ down-right. Step 4 — Quadrant IV $(x > 0, y < 0)$ : $u = - y > 0$ (right), $v = x > 0$ (up) ⇒ up-right. Why these steps? You never need calculus to find direction; just evaluate the two signed components. Signs of $u, v$ fully fix the arrow.

Verify: starting on the positive $x$ -axis $(1, 0)$ : $u = 0, v = 1$ → straight up. Up-left in QI, then continuing... this circulates counter-clockwise. Consistent with Ex 1's circles traced by $x = cos t, y = sin t$ . ✔

Worked example Ex 5 — Zero / degenerate input (C5): a stagnation point

Field: $v = (u, v) = (x, - y)$ (a "saddle"/stagnation flow, like fluid hitting a wall and splitting). Find the streamlines and interpret the origin.

Forecast: at the origin both $u$ and $v$ are... what? Can a particle sitting there ever move?

Step 1 At the origin $x = 0, y = 0 \Rightarrow u = 0, v = 0$ . Both zero ⇒ stagnation point (fluid at rest there). The ODE $\frac{d x}{u} = \frac{d y}{v}$ becomes $\frac{0}{0}$ — undefined, so the origin is a special degenerate point, not part of any single streamline. Why this step? Always test for zeros first: division-by-zero in the ratio ODE flags a stagnation point.

Step 2 Away from the origin: $\frac{d x}{x} = \frac{d y}{- y} \Rightarrow ln ∣ x ∣ = - ln ∣ y ∣ + C \Rightarrow x y = k$ . Why this step? Separate and integrate; $\int d x / x = ln ∣ x ∣$ is the natural tool because the field is linear in each coordinate.

Step 3 — Cover the degenerate lines. The value $k = 0$ gives the two axes $x = 0$ and $y = 0$ themselves (the incoming/outgoing streamlines that hit the stagnation point). Why this step? $k = 0$ is a legitimate limiting case of the family $x y = k$ ; without it the picture near the origin is incomplete.

Verify: streamlines are hyperbolas $x y = k$ , plus the axes at $k = 0$ ; at the origin $v = 0$ (stagnation). Check a point: at $(2, 1)$ , $v = (2, - 1)$ ; the hyperbola $x y = 2$ has slope $\frac{d y}{d x} = - y / x = - 1/2$ , and $\frac{v}{u} = - 1/2$ — matches. ✔

Worked example Ex 6 — Limiting behaviour (C6): where does an old streakline sit?

Field: $v = (1, t)$ (same as Ex 3), source at origin. Where is the streakline point corresponding to a very old release $τ \to - \infty$ , observed at $t = 2$ ? What does the far tail of the plume do?

Forecast: old smoke has travelled longest. Guess whether the tail runs off to $+ x$ or $- x$ , and up or down.

Step 1 Streakline point at $t = 2$ : $x = 2 - τ, y = \frac{1}{2} (4 - τ^{2})$ . Why this step? Reuse Ex 3's release formula, keep $τ$ symbolic to take a limit.

Step 2 Let $τ \to - \infty$ : $x = 2 - τ \to + \infty$ ; $y = \frac{1}{2} (4 - τ^{2}) \to - \infty$ (the $- τ^{2}$ dominates). Why this step? Limiting analysis: check which term wins as $τ$ grows in magnitude. Here $- τ^{2}$ beats everything.

Step 3 On the curve $y = 2 x - \frac{1}{2} x^{2}$ , as $x \to + \infty$ , $y \to - \infty$ — matches. Why this step? Cross-check the limit against the closed-form curve from Ex 3.

Verify: the oldest smoke sits far to the right and far below; the plume's tail curves downward. At a concrete old value $τ = - 4$ : $x = 6, y = \frac{1}{2} (4 - 16) = - 6$ , and indeed $2 (6) - \frac{1}{2} (36) = 12 - 18 = - 6$ . ✔ (C6.)

Worked example Ex 7 — Real-world word problem (C7): smoke in a gusty wind

Story: A chimney at the origin releases smoke continuously. Wind blows steadily east at $2 m/s$ , but a rising thermal adds an upward gust that grows linearly: vertical speed $= 0.5 t m/s$ (with $t$ in seconds, $t = 0$ when observation started). Find the shape of the visible smoke plume at $t = 4 s$ .

Forecast: steady wind + growing updraft — the plume is a straight line or a curve? Guess.

Step 1 — Translate to a field. $u = 2$ , $v = 0.5 t$ . Time appears ⇒ unsteady ⇒ this is a streakline problem (dye/smoke ⇒ streakline, per Flow visualization techniques (dye, smoke, PIV)). Why this step? Continuous injection from one fixed point is the definition of a streakline; identify the curve type before computing.

Step 2 — Pathline of the particle released at $τ$ from origin: $\frac{d x}{d t} = 2 \Rightarrow x = 2 (t - τ)$ ; $\frac{d y}{d t} = 0.5 t \Rightarrow y = 0.25 (t^{2} - τ^{2})$ . Why this step? Build the streakline out of individual pathlines, each starting at the origin at its release time $τ$ .

Step 3 — Observe at $t = 4$ , sweep $τ \in [?, 4]$ . $x = 2 (4 - τ) = 8 - 2 τ$ , $y = 0.25 (16 - τ^{2})$ . Eliminate $τ = (8 - x) /2 = 4 - \frac{x}{2}$ : $y = 0.25 (16 - (4 - \frac{x}{2})^{2}) = 0.25 (16 - 16 + 4 x - \frac{x ^{2}}{4}) = x - \frac{x ^{2}}{16} .$ Why this step? Fix the "now" ( $t = 4$ ), let release time vary — that is exactly what "all smoke seen now" means.

Verify: plume is the curve $y = x - \frac{x ^{2}}{16}$ (a downward parabola). Units: $x$ in metres, $y$ in metres. Nearest fresh puff ( $τ = 4$ ) sits at the origin $(0, 0)$ ✔; a puff released at $τ = 0$ sits at $x = 8, y = 0.25 (16) = 4$ , and $8 - \frac{64}{16} = 8 - 4 = 4$ ✔.

Worked example Ex 8 — Exam twist (C8): reverse-engineer the field

Twist: You are given that the streamlines are the circles $x^{2} + y^{2} = C$ and that the flow is steady. Propose a velocity field, and then test: is $v = (- y, x)$ the only answer? Is $v = (- 2 y, 2 x)$ also valid?

Forecast: must the field be unique, or can many fields share the same streamline pattern?

Step 1 — Direction constraint. Streamlines $x^{2} + y^{2} = C$ have slope $\frac{d y}{d x} = - \frac{x}{y}$ (differentiate implicitly). We need $\frac{v}{u} = - \frac{x}{y}$ . Why this step? The streamline slope must equal $v / u$ ; that is the only thing streamlines pin down — a ratio, not the magnitudes.

Step 2 — Test $v = (- y, x)$ : $\frac{v}{u} = \frac{x}{- y} = - \frac{x}{y}$ ✔. Test $v = (- 2 y, 2 x)$ : $\frac{v}{u} = \frac{2 x}{- 2 y} = - \frac{x}{y}$ ✔. Why this step? Both satisfy the ratio, so both trace the same circles — the field is not unique; only the direction is fixed, speed (and even a scalar function of position) is free.

Step 3 — Steadiness check. Neither field contains $t$ , so $\partial v / \partial t = 0$ ⇒ steady ⇒ by the golden rule, for each field the pathlines and streaklines are also these circles. Why this step? The exam wants you to notice that "same streamlines" ≠ "same flow": the two fields circulate at different speeds (one twice the other) yet look identical in a snapshot of directions.

Verify: at $(1, 0)$ , field 1 gives $v = (0, 1)$ (speed $1$ ); field 2 gives $v = (0, 2)$ (speed $2$ ). Both tangent to the unit circle, different magnitudes. ✔ Streamlines encode direction only.

Recall Cover-check: did we hit every cell?

Steady curved (C1) ::: Ex 1 Steady shear, straight streamlines (C2) ::: Ex 2 Unsteady, all three differ (C3) ::: Ex 3 Signs/quadrants (C4) ::: Ex 4 Zero/degenerate stagnation (C5) ::: Ex 5 Limiting behaviour (C6) ::: Ex 6 Real-world word problem (C7) ::: Ex 7 Exam reverse twist (C8) ::: Ex 8

Connections

Velocity field and material derivative
Steady vs unsteady flow
Continuity equation
Stream function ψ
Lagrangian vs Eulerian description
Flow visualization techniques (dye, smoke, PIV)

Worked examples — Streamlines, pathlines, streaklines

The scenario matrix

Connections

Concept Map