Navier-Stokes equation for a viscous fluid

At a Glance

• Frequency: 4 sub-parts across 4 of 14 years (2014, 2023, 2025, 2026)
• Priority tier: T3
• Marks (count): 20 (3)
• Average solve time: ~17 min
• Difficulty mix: hard 2, medium 1
• Section: B | Dominant type: derivation

Why This Chapter Matters

Navier-Stokes questions carry 20 marks each and appear in Section B — the highest-value slot in the paper. Two distinct archetypes have appeared: deriving the parabolic velocity profile for channel (Poiseuille) flow by reducing and integrating the NS equations (2014), and verifying that a given velocity field satisfies the full NS equations and recovering the pressure distribution (2023, 2025). The verification type has appeared twice in the last three years and is likely to recur. Both types reduce to straightforward algebra once the student knows the Navier-Stokes equation, the continuity equation, and the structure of the inertia term — there is no need to know the full derivation of NS from first principles.

Minimum Theory

Navier-Stokes Equation. For an incompressible Newtonian viscous fluid with constant viscosity $\mu$:

$\rho\!\left[\frac{\partial\mathbf{v}}{\partial t} + (\mathbf{v}\cdot\nabla)\mathbf{v}\right] = -\nabla p + \mu\nabla^2\mathbf{v} + \rho\mathbf{F},$

together with the incompressibility condition $\nabla\cdot\mathbf{v} = 0$. The four terms on the two sides are: inertia (left), pressure gradient, viscous stress $\mu\nabla^2\mathbf{v}$, and body force. For steady flow $\partial_t = 0$.

Channel-Flow Reduction (Poiseuille/Couette–Poiseuille). For parallel flow $\mathbf{v} = (u(y), 0, 0)$: continuity gives $\partial u/\partial x = 0$ automatically; the convective term $(\mathbf{v}\cdot\nabla)\mathbf{v} = 0$; the $x$-momentum equation reduces to:

$\mu\frac{d^2u}{dy^2} = \frac{dp}{dx}.$

With $dp/dx = -P$ (constant, $P > 0$ drives flow) and no-slip at the walls, the solution is the parabolic (Poiseuille) profile. For Couette–Poiseuille flow (one wall moving at speed $U$), the profile has an additional linear term.

Circular Couette Reduction (rotating cylinders). A channel is not an annulus, and the Cartesian reduction above does not transfer. For purely azimuthal axisymmetric flow $\mathbf{v}=(0,\,v(r),\,0)$ in cylindrical polars $(r,\theta,z)$:

• Continuity $\dfrac1r\partial_r(ru_r)+\dfrac1r\partial_\theta u_\theta+\partial_z u_z=0$ reduces to $\partial u_\theta/\partial\theta=0$ — so $\theta$-independence of $u_\theta$ is deduced, not assumed.
• Every inertial term in the $\theta$-equation vanishes (each contains $u_r$, $u_z$, or a $\theta$-derivative), leaving only the viscous term:

$\frac{d}{dr}\!\left(\frac{1}{r}\frac{d(rv)}{dr}\right)=0\qquad\Longleftrightarrow\qquad \frac{d^2v}{dr^2}+\frac1r\frac{dv}{dr}-\frac{v}{r^2}=0.$

Note the $-v/r^2$ term: this is a Cauchy–Euler equation with indicial roots $m=\pm1$, so

$v(r)=C_1r+\frac{C_2}{r},$

the rigid-body rotation $C_1r$ (vorticity $2C_1$) plus the irrotational free vortex $C_2/r$ (vorticity $0$). Dropping the $-v/r^2$ term gives $r^2v''+rv'=0$ and loses $C_2/r$ entirely — the commonest fatal slip.

• The $r$-equation is not redundant: it reads $\dfrac{dp}{dr}=\dfrac{\rho v^2}{r}$, the centripetal balance, and determines $p$ once $v$ is known.
• $\mu$ cancels from the reduced equation, so the steady profile is independent of viscosity. Viscosity fixes the torque, not the shape: $\tau_{r\theta}=\mu r\dfrac{d}{dr}\!\left(\dfrac{v}{r}\right)=-\dfrac{2\mu C_2}{r^2}$, and the torque per unit length $G=2\pi r^2\tau_{r\theta}=-4\pi\mu C_2$ is independent of $r$, as steady balance requires.

Verification Approach. When a velocity field is given and you must “show it satisfies the equations of motion”: (1) check continuity; (2) compute the inertia and viscous terms; (3) substitute into each component of NS; (4) show the residual in each component is zero or is balanced by $\nabla p$; (5) integrate $\nabla p$ to find $p$, checking compatibility (mixed partial conditions).

Question Archetypes

navier-stokes-verification (2 question(s); 2023, 2025)

Recognition Cues — The question provides explicit velocity components $u(x,y)$, $v(x,y)$, $w = 0$ (or a profile $u(y)$) and asks you to “show that this satisfies the equation of motion” or “determine under what conditions” it is a solution. The task is substitution and verification, not derivation.

Solution Template

1. Continuity: compute $\nabla\cdot\mathbf{v}$ and show it equals zero.
2. Inertia term: for each component, compute $(\mathbf{v}\cdot\nabla)v_i = u\,\partial_x v_i + v\,\partial_y v_i + w\,\partial_z v_i$.
3. Viscous term: for each component, compute $\mu\nabla^2 v_i = \mu(\partial_{xx} + \partial_{yy} + \partial_{zz})v_i$.
4. NS equations: for each component, the equation is $\rho\cdot(\text{inertia}) = -\partial p/\partial x_i + \mu\nabla^2 v_i + \rho F_i$. If inertia and viscous terms are both zero, $\nabla p = \rho\mathbf{F}$. Otherwise, set $\partial p/\partial x_i =$ (viscous term) $-$ (inertia term).
5. Pressure: integrate $\partial p/\partial x_i$ for each $i$; check compatibility ($\partial_j(\partial_i p) = \partial_i(\partial_j p)$); assemble $p(x,y,z)$.

Worked Example(s)

2023 Paper 2, 2023-P2-Q8c (20 marks)

Determine under what conditions $u = c(x^2-y^2)$, $v = -2cxy$, $w = 0$ is a solution to Navier-Stokes with $B_x = 0 = B_y$, $B_z = -g$.

Step 1 — Continuity.

$u_x + v_y + w_z = 2cx + (-2cx) + 0 = 0. \checkmark$

Incompressibility is satisfied for any constant $c$.

Step 2 — Viscous terms. Each velocity component is harmonic:

$\nabla^2 u = \partial_{xx}[c(x^2-y^2)] + \partial_{yy}[c(x^2-y^2)] = 2c + (-2c) = 0.$

$\nabla^2 v = 0 \text{ similarly.}$

Viscous stresses vanish, so NS reduces to Euler form: $\rho(\mathbf{v}\cdot\nabla)\mathbf{v} = -\nabla p + \rho\mathbf{F}$.

Step 3 — Inertia terms.

$u\,u_x + v\,u_y = c(x^2-y^2)(2cx) + (-2cxy)(-2cy) = 2c^2x(x^2+y^2).$

$u\,v_x + v\,v_y = c(x^2-y^2)(-2cy) + (-2cxy)(-2cx) = 2c^2y(x^2+y^2).$

Step 4 — Conditions. From the NS equations:

$p_x = -2\rho c^2 x(x^2+y^2), \quad p_y = -2\rho c^2 y(x^2+y^2), \quad p_z = -\rho g.$

Compatibility check: $\partial_y p_x = -4\rho c^2 xy = \partial_x p_y$ ✓. The system is consistent for any constant $c$ and any viscosity $\mu$. The only condition needed is incompressibility — which already holds.

$\boxed{\text{Condition: incompressible (}\nabla\cdot\mathbf{v}=0\text{) — satisfied for any } c.}$

Step 5 — Pressure. Integrate $p_x$ in $x$:

$p = -2\rho c^2\left(\frac{x^4}{4} + \frac{x^2 y^2}{2}\right) + h(y,z).$

Differentiate in $y$ and match $p_y$: $h_y = -2\rho c^2 y^3 \Rightarrow h = -\tfrac{\rho c^2 y^4}{2} + k(z)$.

Match $p_z = -\rho g$: $k'(z) = -\rho g \Rightarrow k = -\rho gz + C$.

Noting $(x^2+y^2)^2 = x^4 + 2x^2y^2 + y^4$:

$\boxed{p = -\frac{\rho c^2}{2}(x^2+y^2)^2 - \rho g z + C.}$

This is the Bernoulli form $p = p_0 - \tfrac{1}{2}\rho|\mathbf{v}|^2 - \rho gz$ since $|\mathbf{v}|^2 = c^2(x^2+y^2)^2$ (by the same identity as MF-08).

2025 Paper 2, 2025-P2-Q7c (20 marks)

Show that $u(y) = (U/h)y - (hy/2\mu)(dp/dx)(1-y/h)$, $v = 0 = w$, $p = p(x)$, satisfies the equation of motion with no body force.

Step 1 — Continuity. $\partial u/\partial x = 0$ since $u$ depends only on $y$; $v = w = 0$. So $\nabla\cdot\mathbf{v} = 0$. ✓

Step 2 — Inertia term vanishes. $(\mathbf{v}\cdot\nabla)u = u\,\partial_x u + 0 + 0 = 0$ (since $u$ has no $x$-dependence). The convective acceleration is zero.

Step 3 — $x$-momentum. The NS $x$-equation reduces to $0 = -dp/dx + \mu\,d^2u/dy^2$.

Let $G = dp/dx$ (constant). Expand $u(y)$:

$u(y) = \frac{U}{h}y - \frac{Gh}{2\mu}y + \frac{G}{2\mu}y^2.$

Then:

$u''(y) = \frac{G}{\mu}, \quad \mu u''(y) = G = \frac{dp}{dx}.$

So $-dp/dx + \mu u'' = -G + G = 0$. Satisfied. ✓

Step 4 — $y$ and $z$ momentum. $v = w = 0$ and $p = p(x)$ gives $\partial p/\partial y = 0$ and $\partial p/\partial z = 0$, so both equations read $0 = 0$. ✓

$\boxed{u(y) = \frac{U}{h}y - \frac{hy}{2\mu}\frac{dp}{dx}\!\left(1-\frac{y}{h}\right) \text{ satisfies all NS equations and continuity.}}$

The balance is $\mu u'' = dp/dx$: viscous stress equals the pressure gradient. This is the plane Couette–Poiseuille profile (linear Couette part $+$ parabolic Poiseuille part).

Common Traps

• Not showing the inertia term vanishes. In the 2025 problem the convective acceleration is zero because $u = u(y)$ has no $x$-dependence. This must be stated explicitly; examiners award marks for it.
• Viscous terms vanishing (2023). The Laplacian $\nabla^2\mathbf{v} = 0$ because each component is harmonic (linear in second-order terms that cancel). This is the key condition that removes $\mu$ from the equations. State it after the computation.
• Checking only the $x$-momentum equation. The $y$ and $z$ equations (with $v = w = 0$) also need to be addressed — even if they reduce trivially to $0 = 0$, show the step.
• Pressure integration constants. When integrating $\nabla p$ for a 2D or 3D flow, the “constant” after integrating in $x$ is a function of $y$ and $z$, not a number. Carry it through and pin it down with the remaining equations.

navier-stokes-channel (1 question(s); 2014)

Recognition Cues — “Find the Navier-Stokes equation for steady laminar flow between two parallel plates” or “derive the velocity distribution for viscous flow in a channel”. You are expected to start from the full NS equation, apply the symmetry assumptions, reduce to an ODE, and integrate with no-slip boundary conditions.

Solution Template

1. Assume $\mathbf{v} = (u(y), 0, 0)$. State why: steady + parallel laminar flow.
2. Continuity: $\partial u/\partial x = 0$ ✓ (since $u$ depends only on $y$).
3. Convective term: $(\mathbf{v}\cdot\nabla)\mathbf{v} = 0$ (no $x$-gradient of $u$). State this makes the problem linear.
4. $y$- and $z$-momentum: $\partial p/\partial y = 0$, so $p = p(x)$ only.
5. $x$-momentum: $0 = -dp/dx + \mu\,d^2u/dy^2$. Set $dp/dx = -P$ (const).
6. Integrate twice: $u(y) = -Py^2/(2\mu) + C_1 y + C_2$.
7. Apply no-slip at $y = \pm h$: determine $C_1 = 0$, $C_2 = Ph^2/(2\mu)$.
8. State key results: $u_{\max} = Ph^2/(2\mu)$, volume flux $Q = 2Ph^3/(3\mu)$, mean velocity $\bar{u} = \tfrac{2}{3}u_{\max}$.

Worked Example(s)

2014 Paper 2, 2014-P2-Q8c (20 marks)

Find the Navier-Stokes equation for steady laminar flow of a viscous incompressible fluid between two infinite parallel plates.

Plates at $y = \pm h$; flow in the $x$-direction driven by pressure gradient $-\partial p/\partial x = P > 0$.

Step 1 — Assumptions and NS. With $\mathbf{V} = (u(y), 0, 0)$, steady flow:

$\rho\!\left[\frac{\partial\mathbf{V}}{\partial t} + (\mathbf{V}\cdot\nabla)\mathbf{V}\right] = -\nabla p + \mu\nabla^2\mathbf{V}.$

Steady: $\partial_t = 0$. Continuity: $\partial u/\partial x = 0$ ✓. Convective: $u\,\partial_x u = 0$.

Step 2 — Reduced equations.

• $x$: $0 = -\partial p/\partial x + \mu\,d^2u/dy^2$.
• $y$: $0 = -\partial p/\partial y$ $\Rightarrow$ $p = p(x)$.

Set $\partial p/\partial x = -P$:

$\frac{d^2u}{dy^2} = -\frac{P}{\mu}.$

Step 3 — Integrate.

$u(y) = -\frac{Py^2}{2\mu} + C_1 y + C_2.$

Step 4 — Boundary conditions. No-slip at $y = \pm h$:

$u(h) = 0:\quad -\frac{Ph^2}{2\mu} + C_1 h + C_2 = 0.$

$u(-h) = 0:\quad -\frac{Ph^2}{2\mu} - C_1 h + C_2 = 0.$

Subtract: $2C_1 h = 0 \Rightarrow C_1 = 0$. Add: $C_2 = Ph^2/(2\mu)$.

Step 5 — Profile and derived quantities.

$\boxed{u(y) = \frac{P}{2\mu}(h^2 - y^2).}$

Maximum at $y = 0$: $u_{\max} = Ph^2/(2\mu)$.

Volume flux per unit width:

$Q = \int_{-h}^h u\,dy = \frac{P}{2\mu}\!\left[h^2 y - \frac{y^3}{3}\right]_{-h}^h = \frac{P}{2\mu}\cdot\frac{4h^3}{3} = \frac{2Ph^3}{3\mu}.$

Mean velocity: $\bar{u} = Q/(2h) = Ph^2/(3\mu) = \tfrac{2}{3}u_{\max}$.

Common Traps

• Convective term is zero, but it must be shown. Write $(\mathbf{V}\cdot\nabla)\mathbf{V} = u\,\partial_x\mathbf{V} = 0$ (since $u$ has no $x$-dependence); this is what makes the problem linear.
• Sign of pressure gradient. Setting $\partial p/\partial x = -P$ with $P > 0$ gives flow in the $+x$ direction. Without the minus sign, $u$ would be negative everywhere.
• Symmetric BCs give $C_1 = 0$. Using only $u(h) = 0$ and forgetting $u(-h) = 0$ leaves $C_1$ undetermined.
• Volume flux formula. $Q = 2Ph^3/(3\mu)$, and the mean velocity is $(2/3)u_{\max}$, not $(1/2)u_{\max}$.

couette-annulus (1 question(s); 2026)

The geometry is an annulus, not a channel, and the driving is rotation, not a pressure gradient. Everything up to “posit the field, kill the terms with a reason, integrate, apply no-slip” transfers from navier-stokes-channel; the operator and the boundary conditions do not.

Recognition Cues

• “Viscous incompressible fluid between two concentric cylinders of radii $a$ and $b$.”
• One cylinder is given angular velocity $\Omega$ and the other held at rest — usually asked twice, once for each.
• “Steady, no body forces” is stated; “discuss and formulate” signals that a bare pair of formulae will not carry the marks.

Solution Template

1. State the hypotheses the question omits — infinitely long cylinders (or end effects neglected), axisymmetric, purely azimuthal, no-slip, laminar. Without them $\mathbf{v}=(0,v(r),0)$ is an assumption, not a deduction, and saying so is worth a mark. Note that $\partial u_\theta/\partial\theta=0$ does not need assuming — continuity gives it.
2. Reduce the $\theta$-equation term by term, with a reason attached to each vanishing term. This is the bulk of a 20-mark “formulate”.
3. Solve $\dfrac{d}{dr}\!\left(\dfrac1r\dfrac{d(rv)}{dr}\right)=0$ by integrating twice: $rv=C_1r^2+C_2$, so $v=C_1r+C_2/r$.
4. Convert angular velocity to a wall speed before applying no-slip: $v(a)=a\Omega$, not $v(a)=\Omega$.
5. Fit the constants separately in each case — do not assert case (ii) “by symmetry”, the constants genuinely differ.
6. Discuss: the $\mu$-cancellation, the pressure from the $r$-equation, the two limiting cases, and the torque.

Worked Example

2026 Paper 2, 2026-P2-Q8c (20 marks)

Viscous incompressible fluid fills the region between concentric cylinders of radii $a,b$ ($b>a$); steady flow, no body forces. Formulate the velocity if (i) the inner cylinder has angular velocity $\Omega$ and the outer is at rest; (ii) the outer is rotated with angular velocity $\Omega$ and the inner is at rest.

Source: analysis/solutions/2026-P2-Q8c.md

Reduction. With $\mathbf{v}=(0,v(r),0)$ continuity holds identically; every inertial term in the $\theta$-momentum equation contains $u_r$, $u_z$ or $\partial_\theta$ and so vanishes, leaving $\mu\dfrac{d}{dr}\!\left(\dfrac1r\dfrac{d(rv)}{dr}\right)=0$. Since $\mu\ne0$ it divides out, and integrating twice gives $v=C_1r+C_2/r$. The $r$-equation gives $dp/dr=\rho v^2/r$ and the $z$-equation $\partial p/\partial z=0$.

Case (i): inner spins. No-slip: $v(a)=a\Omega$, $v(b)=0$. Using $rv=C_1r^2+C_2$ (linear in the constants):

$C_1a^2+C_2=a^2\Omega,\qquad C_1b^2+C_2=0\;\Longrightarrow\;C_1=-\frac{a^2\Omega}{b^2-a^2},\quad C_2=\frac{a^2b^2\Omega}{b^2-a^2},$

$\boxed{v_{\text{(i)}}(r)=\frac{a^2\Omega}{b^2-a^2}\cdot\frac{b^2-r^2}{r}.}$

Check: $v(a)=a\Omega$ ✓, $v(b)=0$ ✓.

Case (ii): outer spins. No-slip: $v(a)=0$, $v(b)=b\Omega$:

$C_1=\frac{b^2\Omega}{b^2-a^2},\quad C_2=-\frac{a^2b^2\Omega}{b^2-a^2},\qquad\boxed{v_{\text{(ii)}}(r)=\frac{b^2\Omega}{b^2-a^2}\cdot\frac{r^2-a^2}{r}.}$

Note the mirror structure: the roles of $a$ and $b$, and of the rigid-rotation and free-vortex parts, are interchanged.

Discussion (this is what “discuss” is asking for).

• $\mu$ cancels. All inertial terms vanished, so the only surviving term is viscous and $\mu$ is an overall factor on a homogeneous equation. Physically: in the steady state the net viscous torque on every annulus is zero, so torque is transmitted unchanged from wall to wall; viscosity fixes how much torque, not the shape of the profile.
• Pressure. $p(r)=\rho\!\left(\tfrac12C_1^2r^2+2C_1C_2\log r-\tfrac{C_2^2}{2r^2}\right)+\text{const}$, increasing outwards — the correct sense for fluid moving in circles.
• Limits. Case (i) with $b\to\infty$ gives $v\to a^2\Omega/r$: a cylinder spinning in unbounded viscous fluid produces a flow that is irrotational everywhere in the fluid. Case (ii) with $a\to0$ gives $v\to\Omega r$: rigid-body rotation, no shear, no dissipation.
• Torque. $|G|=\dfrac{4\pi\mu a^2b^2\Omega}{b^2-a^2}$ per unit length in both cases, opposite in sense — the Couette (Margules) viscometer formula. The instrument works precisely because the profile is $\mu$-independent while the torque is $\mu$-proportional.
• Stability. By Rayleigh’s criterion, $\tfrac{d}{dr}(rv)^2=4C_1r(rv)$. Case (i) has $C_1<0$ — angular momentum decreases outwards, Rayleigh-unstable, and above a critical Taylor number the profile breaks into Taylor vortices. Case (ii) has $C_1>0$, stable. That asymmetry is almost certainly why the paper asks for the two cases side by side.

Common Traps

• Losing the $-v/r^2$ term. Solving $r^2v''+rv'=0$ instead of $r^2v''+rv'-v=0$ throws away $C_2/r$ and makes both boundary conditions unsatisfiable. The operator is $\tfrac{d}{dr}\!\left(\tfrac1r\tfrac{d(rv)}{dr}\right)$ — write it in that form and integrate it directly rather than expanding.
• Writing $v(a)=\Omega$. Angular velocity is not a speed. $v(a)=a\Omega$; the dimensional error destroys the answer.
• Doing case (i) and asserting case (ii) by symmetry. The constants differ: $C_1=-a^2\Omega/(b^2-a^2)$ versus $+b^2\Omega/(b^2-a^2)$. Both cases must be fitted.
• Dropping the radial equation. It is not redundant — it is what determines the pressure, via the centripetal term $\rho v^2/r$.
• Assuming $\theta$-independence. Continuity gives it; claiming it as a hypothesis wastes an available mark.

Marks-Aware Writing

20-mark derivation (channel flow). The examiner wants to see the full NS equation written out, the reduction steps stated with justification (steady, continuity, parallel flow), the ODE, the double integration, the application of both boundary conditions, and the final profile. State $u_{\max}$ and $Q$ since these are standard results that full-marks answers include. Expect to write 8–10 lines of algebra.

20-mark verification. The examiner wants to see each of the four steps: continuity, inertia calculation, viscous/Laplacian calculation, and pressure integration. For flows where the inertia or viscous term vanishes, state this explicitly and show the computation — do not skip the calculation because the answer is zero. Show the compatibility check when integrating $\nabla p$. A complete verification of all three momentum components (not just the $x$-component) is needed for full marks.

Practice Set