Exercises — Streamlines, pathlines, streaklines

(a) Streakline — many puffs, one source (the chimney), all seen now. (b) Pathline — one object (the balloon), its whole trajectory over time. (c) Streamline — a single instant's direction field; tangent-to-$\vec{v}$ everywhere. Mnemonic check: SNAP, TRAIL, SMOKE. (c)=SNAP, (b)=TRAIL, (a)=SMOKE.

False. Time is frozen — it is a constant parameter, not a variable of integration. We only integrate the spatial relationship at one instant. (Integrating over time is what a pathline does.)

They are all identical. A steady field ($\partial \vec{v}/\partial t=0$) never changes, so there is no "history" difference between a snapshot, one particle's trail, and a smoke plume. This is the golden rule from the parent note. See Steady vs unsteady flow.

Step 1 — write the ODE. $\frac{dx}{2}=\frac{dy}{3}$. Why: plug $u=2,v=3$ into $\frac{dx}{u}=\frac{dy}{v}$. Here both components are nonzero constants ($u=2\ne 0$, $v=3\ne 0$), so no degenerate points exist and dividing by $u$ and $v$ is legal everywhere — this is why the compact form is safe for this problem. Step 2 — cross-multiply. $3dx=2dy$. Why: multiply both sides by $2\cdot 3=6$. Because $u,v$ are constants we can move them past the differentials freely; this step is exactly the safe cross-multiplied form $vdx=udy$ with $v=3,u=2$. Step 3 — integrate. $3x=2y+C\Rightarrow y=\frac{3}{2}x+C^{\prime }$. Result: straight parallel lines of slope $\frac{3}{2}$ — exactly the direction of $\vec{v}=(2,3)$ (rise $3$ over run $2$).

What the figure shows. In the figure below, the blue arrows are the constant velocity $\vec{v}=(2,3)$ drawn at a grid of points — notice every arrow is identical (uniform flow). The orange lines are the streamlines $y=\frac{3}{2}x+C$; follow any one and you will see the blue arrows lie exactly along it (tangent = velocity). The red annotation marks the slope $v/u=3/2$: rise $3$ for every run $2$. This is the visual meaning of "tangent parallel to $\vec{v}$."

Step 1 — ODE. $\frac{dx}{x}=\frac{dy}{y}$, valid where $x\ne 0$ and $y\ne 0$. Step 2 — integrate both sides. $\ln |x|=\ln |y|+C$. Why: $\int \frac{dx}{x}=\ln |x|$; this is the natural log because the integrand is $1/x$. Step 3 — exponentiate. $|x|=e^C|y|\Rightarrow y=kx$ for constant $k$. Edge cases. On the $y$-axis $x=0$ so $u=0$: fall back to $vdx=udy\Rightarrow ydx=0\Rightarrow dx=0$, a vertical streamline — the $y$-axis itself. On the $x$-axis $y=0$ so $v=0$: $ydx=xdy\Rightarrow xdy=0\Rightarrow dy=0$, a horizontal streamline — the $x$-axis. At the origin both $u=v=0$: a stagnation point where all the radial lines meet. Result: straight lines through the origin (the $y=kx$ family plus the two axes). The flow pushes radially outward, so streamlines are the radial spokes. ✔ (matches Continuity equation intuition: fluid streaming out along rays.)

Step 1. $\frac{dx}{dt}=1\Rightarrow x=t$. Step 2. $\frac{dy}{dt}=2\Rightarrow y=2t$. Step 3 — eliminate $t$. $t=x$, so $y=2x$. Result: the line $y=2x$. Since the flow is steady, this pathline is also a streamline (slope $v/u=2$). ✔

Step 1 — freeze $t=2$, so $v=2$ is a constant. ODE: $\frac{dx}{1}=\frac{dy}{2}$. (Both $u=1$ and $v=2$ are nonzero, so no degenerate points at this instant.) Why: the golden rule of streamlines — treat $t$ as a fixed number, here $2$. Step 2 — integrate. $2x=y+C\Rightarrow y=2x+C$. Result: at $t=2$ the streamlines are straight lines of slope $2$. Compare this to the pathline (below) — they differ!

Step 1. $\frac{dx}{dt}=1\Rightarrow x=t$ (since $x_0=0,t_0=0$). Step 2. $\frac{dy}{dt}=t\Rightarrow y=\frac{1}{2}{t}^2$. Step 3 — eliminate $t$. $t=x$, so $y=\frac{1}{2}{x}^2$. Result: a parabola, not a straight line. So the pathline (curve) ≠ the $t=2$ streamline (line). This is the whole point of unsteady flow.

What the figure shows. The figure below overlays all three curves for this field on one axis so the split is visible at a glance: the blue straight line is the $t=2$ streamline $y=2x$ (a snapshot), the orange parabola is this pathline $y=\frac{1}{2}{x}^2$ (one particle's whole trail), and the green parabola is the streakline of L3.3. The red dot at $(2,2)$ is the single particle released at $\tau =0$ — notice the pathline and streakline both pass through it, because it is a real particle common to both constructions. Follow the curves from the gray origin outward and you can see how a snapshot (line) genuinely differs from a history (parabolas) in unsteady flow.

Step 1 — pathline of the particle released at time $\tau$ (start at origin at $t=\tau$): $x={\int }_{\tau }^{t}1d{t}^{\prime }=t-\tau ,y={\int }_{\tau }^t{t}^{\prime }d{t}^{\prime }=\frac{1}{2}(t^2-{\tau }^2).$ Step 2 — fix $t=2$, sweep $\tau \le 2$. $x=2-\tau ,y=\frac{1}{2}(4-{\tau }^2).$ Step 3 — eliminate $\tau$. From $x=2-\tau$: $\tau =2-x$. Then $y=\frac{1}{2}(4-(2-x{)}^2)=\frac{1}{2}(4-4+4x-x^2)=2x-\frac{1}{2}{x}^2.$ Result: the streakline is $y=2x-\frac{1}{2}{x}^2$, a different parabola from the pathline $y=\frac{1}{2}{x}^2$. All three curves through the origin now differ — proving the flow is unsteady. ✔ (This is the green curve in the L3.2 figure.)

Quick sanity checks: at $x=0$ (i.e. $\tau =2$, the just-released particle), $y=0$ ✔. The pathline released at $\tau =0$ is at $(x,y)=(2,2)$ when $t=2$; the streakline at $x=2$ gives $y=2(2)-\frac{1}{2}(4)=2$ ✔ — the two curves must meet at that shared particle.

Steadiness: $u=-y$ and $v=x$ contain no explicit $t$, so $\partial \vec{v}/\partial t=0$. The flow is steady. Golden rule: steady ⇒ pathlines = streamlines. Therefore the pathlines are the same concentric circles $x^2+y^2=R^2$. A particle at radius $R$ orbits on that circle forever. Edge note: at the origin $u=-0=0$ and $v=0$, so the origin is a stagnation point — the centre of the swirl, where the fluid does not move and no unique streamline direction exists. (This connects to Lagrangian vs Eulerian description: the Eulerian field and the Lagrangian trajectory agree here.)

Step 1 — what "constant $\psi$" means. Along a curve where $\psi$ doesn't change, its total differential is zero: $d\psi =\frac{\partial \psi }{\partial x}dx+\frac{\partial \psi }{\partial y}dy=0.$ Step 2 — substitute the definitions. $\frac{\partial \psi }{\partial x}=-v$ and $\frac{\partial \psi }{\partial y}=u$, so $-vdx+udy=0\Rightarrow vdx-udy=0\Rightarrow \frac{dx}{u}=\frac{dy}{v}.$ Step 3 — recognise it. This is exactly the streamline ODE — and note it arrives naturally in the safe cross-multiplied form $vdx-udy=0$, which stays valid even where $u=0$ or $v=0$. So lines of constant $\psi$ are streamlines. This is why the Stream function ψ is such a powerful shortcut — you draw streamlines just by contouring one scalar.

Step 1. Need $u=\frac{\partial \psi }{\partial y}=-y\Rightarrow \psi =-\frac{1}{2}{y}^2+f(x)$. Step 2. Need $v=-\frac{\partial \psi }{\partial x}=x\Rightarrow \frac{\partial \psi }{\partial x}=-x\Rightarrow f(x)=-\frac{1}{2}{x}^2+\text{const}$. Step 3. So $\psi =-\frac{1}{2}(x^2+y^2)$. Contours $\psi =\text{const}$ give $x^2+y^2=\text{const}$ — concentric circles, matching the parent note. ✔

Step 1 — get the slope of the curves. Differentiate $xy=C$: $y+x\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$. Step 2 — match to $\frac{dy}{dx}=\frac{v}{u}$. We need $\frac{v}{u}=-\frac{y}{x}$. The simplest choice: $u=x,v=-y$. Step 3 — verify. Streamline ODE $\frac{dx}{x}=\frac{dy}{-y}\Rightarrow \ln |x|=-\ln |y|+C\Rightarrow xy=\text{const}$. ✔ Edge cases: on the axes one component vanishes — the origin is again a stagnation point ($u=v=0$), and the axes themselves are the degenerate streamlines $C=0$. Result: $\vec{v}=(x,-y)$ (a "saddle" / straining flow) gives hyperbolic streamlines. Any scalar multiple $\vec{v}=(kx,-ky)$ works too — direction, not magnitude, sets streamline shape.

(a) Streamline at $t=2$ (freeze $t$, so $u=2$): $\frac{dx}{2}=\frac{dy}{1}\Rightarrow x=2y+C$. Through origin: $x=2y$, i.e. $y=\frac{1}{2}x$ — a straight line.

(b) Pathline (release at $\tau =0$): $\frac{dx}{dt}=t\Rightarrow x=\frac{1}{2}{t}^2$; $\frac{dy}{dt}=1\Rightarrow y=t$. Eliminate $t=y$: $x=\frac{1}{2}{y}^2$ — a sideways parabola. At $t=2$ the particle is at $(2,2)$.

(c) Streakline at $t=2$ from origin. Particle released at $\tau$: $x={\int }_{\tau }^2{t}^{\prime }d{t}^{\prime }=\frac{1}{2}(4-{\tau }^2)$, $y={\int }_{\tau }^{2}d{t}^{\prime }=2-\tau$. From $y=2-\tau$: $\tau =2-y$. Then $x=\frac{1}{2}(4-(2-y{)}^2)=\frac{1}{2}(4y-y^2)=2y-\frac{1}{2}{y}^2$ — a different parabola.

Count: three distinct curves (line, parabola $x=\frac{1}{2}{y}^2$, parabola $x=2y-\frac{1}{2}{y}^2$) — the flow is unsteady ($u=t$ depends on $t$), so all three separate. Shared-particle check: the pathline and streakline must meet at the particle released at $\tau =0$: pathline at $t=2$ gives $(2,2)$; streakline at $y=2$ gives $x=2(2)-\frac{1}{2}(4)=2$ ✔.

Not reliable in unsteady flow. A streakline is made of particles released at many past times, each carrying the imprint of the field as it was then. The curve's shape is a blend of history, not a snapshot of the present velocity. Only the tangent at the source point equals the current velocity direction (the just-released particle, $\tau \to t$). Everywhere else along the streak reflects older flow states. In steady flow the claim becomes true, because past and present fields are identical. This is the deep reason experiments (Flow visualization techniques (dye, smoke, PIV)) must be interpreted with care.

Distinct-in-unsteady, identical-in-steady ::: the three curves (streamline, pathline, streakline) What is frozen for a streamline ::: time $t$ What is swept for a streakline ::: release time $\tau$, at fixed observation time Constant-$\psi$ contours are ::: streamlines Safe form of the streamline ODE where a component is zero ::: $vdx=udy$ Point where $u=v=0$ ::: a stagnation point