Fluid kinematics — Eulerian vs Lagrangian description: Worked examples

Everything on this page rests on one boxed tool. Let us restate it in plain words before using a single symbol.

Why do we even need the convective piece and not just $\partial / \partial t$ ? Because a particle can sit in a field that never changes in time, yet be swept into a region where the field has a different value. The tool below (Figure s01) shows why: the same number can grow along a trajectory even when it is frozen at every fixed dot.

Figure — Fluid kinematics — Eulerian vs Lagrangian description

The scenario matrix

Below is every case class this topic can throw. Each worked example names the cell(s) it covers.

Cell	What makes it special	Covered by
A. Steady, positive convective	$\partial / \partial t = 0$ , particle speeds up	Ex 1
B. Unsteady + convective, both add	local and convective same sign	Ex 2
C. Opposite signs (they fight)	local up, convective down (or vice versa)	Ex 3
D. Degenerate: uniform field	$\nabla ϕ = 0$ ⇒ convective $= 0$	Ex 4
E. Degenerate: zero velocity (rest point)	$u = v = w = 0$ ⇒ convective $= 0$	Ex 4
F. Full 2-D / 3-D, all components	every $u \partial_{x} + v \partial_{y} + w \partial_{z}$ nonzero	Ex 5
G. Lagrangian → Eulerian conversion	eliminate the label $a$	Ex 6
H. Limiting behaviour	let a parameter → 0 or → ∞, check trend	Ex 7
I. Real-world word problem	translate English → field, then apply tool	Ex 8
J. Exam twist: vector acceleration, sign trap	$ϕ = v$ , watch a negative	Ex 9

Prereqs live in Steady vs unsteady flow (what " $\partial / \partial t = 0$ " means) and Streamlines, pathlines and streaklines (what a particle's path is). The tool itself powers the Continuity equation, Euler's equation of motion and Navier-Stokes equations downstream.

Worked Examples

Worked example Ex 1 — Cell A: steady flow, particle still accelerates

A converging pipe has steady 1-D flow $u (x) = 3 (1 + 2 x)$ m/s ( $x$ in m), $v = w = 0$ , no time dependence. Find the acceleration $a_{x}$ of a particle at $x = 1$ m.

Forecast: the flow is "steady" — will $a_{x}$ be zero? Guess first.

Local term. $\partial u / \partial t = 0$ . Why this step? "Steady" means the field is frozen in time at every fixed point, so this term dies. (See Steady vs unsteady flow.)
Convective term. $u \frac{\partial u}{\partial x}$ . Compute $\partial u / \partial x = 3 \cdot 2 = 6$ . And $u (1) = 3 (1 + 2) = 9$ . Why this step? The particle rides into faster fluid; the rate it speeds up = (its speed) × (spatial slope of speed).
Add. $a_{x} = 0 + 9 \times 6 = 54$ m/s $^{2}$ . Why this step? $a_{x} = D u / D t$ ; here only convective survives.

Verify: general formula $a_{x} = u \partial_{x} u = 3 (1 + 2 x) \cdot 6 = 18 (1 + 2 x)$ ; at $x = 1$ gives $18 \times 3 = 54$ ✓. Units: $(m/s) (1/s) = m/s^{2}$ ✓. Steady ≠ zero acceleration.

Worked example Ex 5 — Cell F: full 2-D, every component alive

Velocity field $u = x$ , $v = - y$ (a "stagnation flow", steady). Find the acceleration components $a_{x}, a_{y}$ at the point $(2, 3)$ .

Forecast: which term survives when the flow is steady but 2-D?

Set $ϕ = u$ . $a_{x} = \partial_{t} u + u \partial_{x} u + v \partial_{y} u$ . Steady ⇒ $\partial_{t} u = 0$ . Here $\partial_{x} u = 1$ , $\partial_{y} u = 0$ . Why? Acceleration is the material derivative of velocity itself ( $ϕ = v$ ).
$a_{x} = 0 + x \cdot 1 + (- y) \cdot 0 = x$ . At $(2, 3)$ : $a_{x} = 2$ m/s $^{2}$ .
Set $ϕ = v$ . $a_{y} = u \partial_{x} v + v \partial_{y} v$ . Here $\partial_{x} v = 0$ , $\partial_{y} v = - 1$ .
$a_{y} = x \cdot 0 + (- y) (- 1) = y$ . At $(2, 3)$ : $a_{y} = 3$ m/s $^{2}$ . Why the $+ y$ ? Two negatives ( $v < 0$ moving toward $y = 0$ , and $v$ decreasing with $y$ ) multiply to a positive push.

Verify: $a_{x} = x = 2$ , $a_{y} = y = 3$ ✓. Note both are nonzero from the convective term alone — steady, yet accelerating in every direction. See Euler's equation of motion for what force supplies this.

Worked example Ex 6 — Cell G: Lagrangian → Eulerian conversion

Particle paths given by $x = a e^{2 t}$ (label $a$ = position at $t = 0$ ). Find the velocity as an Eulerian field $u (x, t)$ and its acceleration.

Forecast: the answer must contain no $a$ — Eulerian fields only know current position.

Lagrangian velocity. $u = \frac{\partial x}{\partial t}_{a} = 2 a e^{2 t}$ . Why? Differentiate the path holding the label fixed — you are riding the particle.
Eliminate the label. From $x = a e^{2 t}$ , $a = x e^{- 2 t}$ . Substitute: $u = 2 (x e^{- 2 t}) e^{2 t} = 2 x$ . Why? An Eulerian field cannot mention which particle; invert the path to trade $a$ for $(x, t)$ .
Acceleration, Eulerian way. $a_{x} = \partial_{t} u + u \partial_{x} u = 0 + (2 x) (2) = 4 x$ . Why cross-check? Lagrangian directly: $a = \partial_{t}^{2} x ∣_{a} = 4 a e^{2 t} = 4 x$ — must match.

Verify: $u (x, t) = 2 x$ ; both routes give $a_{x} = 4 x$ ✓. Leaving $a$ in the field is the classic error (see parent's mistake list).

Worked example Ex 8 — Cell I: real-world word problem

A cyclist rides at $6$ m/s along a straight road. Air temperature rises $0.2$ K for every metre eastward and the whole afternoon is warming at $3$ K/hour. The cyclist heads east. What warming rate does the cyclist's thermometer read (in K/s)?

Forecast: translate English into fields first — then it's just the tool.

Build the field. $\partial T / \partial x = 0.2$ K/m. Time rate: $3$ K/hour $= 3/3600$ K/s $= 8.33 \times 1 0^{- 4}$ K/s. Speed $u = 6$ m/s east. Why? Convert everything to SI (K, m, s) before combining — units must agree.
Local: $\partial T / \partial t = 8.33 \times 1 0^{- 4}$ K/s.
Convective: $u \partial_{x} T = 6 \times 0.2 = 1.2$ K/s.
Add: $D T / D t = 0.000833 + 1.2 = 1.2008$ K/s. Why? The cyclist is the moving parcel; both terms apply.

Verify: convective $= 1.2$ dominates completely; the slow afternoon warming ( $< 0.001$ K/s) is negligible over one ride. $D T / D t \approx 1.20$ K/s ✓. Lesson: motion into a spatial gradient usually beats slow global drift.

Worked example Ex 9 — Cell J: exam twist, vector acceleration with a sign trap

Unsteady flow $u (x, t) = x - 2 t$ , $v = 0$ , $w = 0$ . Find $a_{x}$ at $x = 1$ , $t = 1$ . (Watch the signs.)

Forecast: there is a local and a convective term, and $u$ itself is negative here — many students drop a sign.

Local: $\partial u / \partial t = - 2$ . Why the minus? $u$ literally decreases with time ( $- 2 t$ ) — genuine unsteadiness.
Convective: $u \partial_{x} u$ . Here $\partial_{x} u = 1$ , and at $(1, 1)$ : $u = 1 - 2 = - 1$ . Why compute $u$ at the point? The convective term uses the actual local velocity, which is $- 1$ (moving in $- x$ ), not $+ 1$ .
Combine: $a_{x} = - 2 + (- 1) (1) = - 3$ m/s $^{2}$ . Why? Both terms are negative here — the particle decelerates/reverses.

Verify: $a_{x} = \partial_{t} u + u \partial_{x} u = - 2 + (- 1) (1) = - 3$ ✓. Trap avoided by plugging the signed $u = - 1$ , not its magnitude.

Recall Self-test: name the cell, then solve

Ex 1 covers which cell, and why isn't $a_{x} = 0$ ? ::: Cell A (steady but convective); the particle moves into faster fluid, $a_{x} = 54$ m/s $^{2}$ . In Ex 3, what flips the sign of $D T / D t$ from $+ 4$ to $- 2$ ? ::: Raising wind speed to $u = 4$ makes convective $- 12$ overcome local $+ 10$ . Two independent ways the convective term becomes zero (Ex 4)? ::: Uniform field ( $\nabla ϕ = 0$ ) OR zero velocity ( $v = 0$ ). In Ex 6 why must the answer contain no $a$ ? ::: Eulerian fields depend only on current $(x, t)$ ; the particle label $a$ must be eliminated via the path equation. In Ex 9 what value of $u$ goes into the convective term at $(1, 1)$ ? ::: The signed local velocity $u = - 1$ , not $+ 1$ .