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UPSC 2026 Maths Optional Paper 2 Q8c — Step-by-Step Solution

20 marks · Section B

Navier-Stokes equation for a viscous fluid · Mechanics & Fluid Dynamics · asked 4× in 14 yrs · Read the full method →

Question

A viscous incompressible fluid is filled between two concentric cylinders of radius $a$, $b$ ($b > a$). The flow is steady and no body forces are taken into consideration. Discuss and formulate the velocity of the fluid if:

(i) the inner cylinder is given angular velocity $\Omega$ while the outer one is held at rest;

(ii) the outer cylinder is rotated with angular velocity $\Omega$ while the inner one is held at rest.

Technique

This is steady circular Couette flow. Posit a purely azimuthal, axisymmetric field $\mathbf{u} = \bigl(0, v(r), 0\bigr)$ in cylindrical polars; continuity is then satisfied identically and every inertial (convective) term in the $\theta$-momentum equation vanishes, so that equation collapses to the linear ODE $\dfrac{d}{dr}\!\left(\dfrac1r\dfrac{d(rv)}{dr}\right) = 0$ — the same Cauchy–Euler operator as Q8(a) with $n=1$, giving $v = C_1r + C_2/r$. The two cases differ only in the pair of no-slip conditions. The one fact worth flagging aloud is that $\mu$ cancels out of that ODE: the steady profile is independent of the viscosity, which controls the torque, not the shape.

Solution

Step 1 — Geometry, hypotheses, and the assumed velocity field

Take cylindrical polar coordinates $(r,\theta,z)$ with the $z$-axis along the common axis of the two cylinders. The fluid occupies the annulus

The question does not spell out the hypotheses its answer needs, so state them:

1. Incompressible, Newtonian fluid of constant density $\rho$ and constant dynamic viscosity $\mu > 0$.
2. Steady flow: $\partial/\partial t \equiv 0$ (given).
3. No body forces: $\mathbf{F} = \mathbf{0}$ (given). In particular gravity is ignored.
4. Cylinders infinitely long (or, equivalently, end effects neglected), so nothing depends on $z$: $\partial/\partial z \equiv 0$, and there is no axial flow, $u_z = 0$.
5. Axisymmetric flow: $\partial/\partial\theta \equiv 0$, the geometry and the driving being independent of $\theta$.
6. Purely azimuthal flow: $u_r = 0$. The motion is driven solely by the tangential drag of a rotating wall; no radial or axial pressure gradient is imposed, and no fluid is injected or withdrawn.
7. No-slip at both solid walls.
8. The flow is laminar and stable — see the stability caveat in Step 9.

Hence we look for a solution of the form

This form is posited, not derived; the logic below is that on substituting $(\ast)$ into the full Navier–Stokes system every equation is satisfied, so $(\ast)$ is an exact solution of the governing equations together with the boundary conditions — which is what “formulate the velocity” asks for.

Step 2 — Continuity is satisfied identically

The equation of continuity for an incompressible fluid in cylindrical polars is

With $u_r = u_z = 0$ this reduces to $\dfrac{1}{r}\dfrac{\partial u_\theta}{\partial\theta} = 0$, i.e.

So continuity itself forces $u_\theta$ to be independent of $\theta$ — assumption 5 is not needed for the azimuthal component, it is a consequence. With assumption 4 ($\partial/\partial z = 0$) we are left with $u_\theta = v(r)$ exactly as in $(\ast)$, and continuity holds identically.

Step 3 — Reduce the Navier–Stokes equations

The Navier–Stokes equations in cylindrical polars, with $\mathbf{F} = \mathbf{0}$, are:

$r$-component

This is nothing but the centripetal balance: the radial pressure gradient supplies the centripetal acceleration $v^2/r$ of the circling fluid. It contains no viscosity and does not involve $v$‘s derivatives, so it determines $p$ once $v$ is known and places no constraint on $v$.

$\theta$-component

Every inertial term vanishes — each one contains $u_r$, $u_z$, or a $\theta$-derivative. So the $\theta$-equation reduces to

$z$-component reduces to $\dfrac{\partial p}{\partial z} = 0$, consistent with $p = p(r)$. (With gravity along $-\hat z$ one would simply add $-\rho g z$ to $p$.)

Step 4 — The reduced equation is independent of the viscosity — and why

Since $\mu \neq 0$ it may be divided out of (10):

This is worth pausing on. Mathematically: all the inertial terms vanished identically, so the only surviving term in the whole $\theta$-momentum balance is the viscous one; $\mu$ is then an overall multiplicative factor on a homogeneous equation and cancels. Equally decisive is that the boundary conditions are kinematic (prescribed wall velocities), not dynamic (prescribed stresses) — had a torque been prescribed instead, $\mu$ would re-enter through the boundary data.

Physically: in the steady state the net viscous torque on every fluid annulus is zero — torque is simply transmitted unchanged from wall to wall. Viscosity determines how large that transmitted torque is (Step 9: torque $\propto \mu$) and how long the flow takes to spin up from rest, but it does not determine the shape of the steady profile, which is fixed by the geometry and the two wall speeds alone. Two fluids of wildly different viscosity in the same apparatus at the same $\Omega$ have identical steady velocity profiles and require very different torques to sustain them.

Note also that (11) is exactly the Cauchy–Euler radial equation $r^2R'' + rR' - n^2R = 0$ of part (a) with $n = 1$ — the same operator, and the same reason: both are the radial part of a two-dimensional harmonic problem.

Step 5 — General solution and its structure

Integrating (11) once,

(the same $r^{\pm1}$ pair the indicial equation $m^2 = 1$ of part (a) supplies).

The two pieces have clean identities. The axial vorticity of $(\ast)$ is

so the first integral of (11) says precisely that the vorticity is uniform across the gap. Thus:

• $C_1 r$ is rigid-body rotation with angular velocity $C_1$ and vorticity $2C_1$;
• $C_2/r$ is the irrotational (free/potential) vortex, vorticity zero, circulation $2\pi C_2$.

Every circular Couette flow is a superposition of these two.

Step 6 — Case (i): inner cylinder rotating at $\Omega$, outer at rest

No-slip boundary conditions. A point of the inner cylinder ($r=a$) moves with speed $a\Omega$ in the $\theta$-direction; the outer wall is stationary:

Substituting in (12) — better, in $rv = C_1r^2 + C_2$, which is linear in the constants:

Subtracting, $C_1(a^2 - b^2) = a^2\Omega$, so

Therefore

Check against the boundary conditions.

The angular velocity of the fluid is $\dfrac{v}{r} = \dfrac{a^2\Omega}{b^2-a^2}\left(\dfrac{b^2}{r^2} - 1\right)$, which decreases monotonically from $\Omega$ at $r=a$ to $0$ at $r=b$: the fluid is dragged round by the inner wall and progressively retarded by the outer one. The vorticity $2C_1 < 0$ is opposite in sign to the rotation — a genuinely counter-intuitive but correct feature of this flow.

Step 7 — Case (ii): outer cylinder rotating at $\Omega$, inner at rest

No-slip: $v(a) = 0$, $v(b) = b\Omega$. From $rv = C_1r^2 + C_2$,

Therefore

Check against the boundary conditions.

Note the structural mirror: case (i) is $\dfrac{a^2\Omega}{b^2-a^2}\left(\dfrac{b^2}{r}-r\right)$, case (ii) is $\dfrac{b^2\Omega}{b^2-a^2}\left(r-\dfrac{a^2}{r}\right)$ — the roles of $a$ and $b$, and of the rigid-rotation and free-vortex parts, are interchanged. Here $C_1 > 0$: the vorticity has the same sign as the rotation.

Step 8 — The pressure field

The velocity was obtained without the pressure: the $\theta$-momentum equation (10) decoupled from $p$ because $\partial p/\partial\theta = 0$ by axisymmetry. So $p$ is not needed to answer “formulate the velocity”. It is needed for $(\ast)$ to be an exact Navier–Stokes solution, and equation (9) delivers it. With $v = C_1r + C_2/r$,

so, integrating,

with $(C_1,C_2)$ read off from Step 6 or Step 7 as appropriate. The additive constant is arbitrary — as always for an incompressible flow, only pressure differences are determined. Since $v \ne 0$ in the gap, $dp/dr > 0$: pressure increases outwards, which is the correct sense for fluid moving in circles.

Step 9 — Discussion: limits, torque, and stability

(a) Case (i) with $b \to \infty$ — a single cylinder spinning in an unbounded fluid. Fixing $r$ and letting $b\to\infty$,

Equivalently $C_1 \to 0$, so the rigid-rotation part disappears and only the irrotational free vortex survives, of circulation $\Gamma = 2\pi a^2\Omega$. A remarkable conclusion: a cylinder rotating steadily in an unbounded viscous fluid generates a flow that is irrotational everywhere in the fluid, even though the fluid is viscous. Viscosity is not absent — it is what transmits the torque and it is what fixes the constant $a^2\Omega$ — but the steady field it produces has zero vorticity. The speed decays like $1/r$, so the disturbance is felt arbitrarily far out.

(b) Case (ii) with $a \to 0$ — no inner cylinder. For fixed $r > 0$,

rigid-body rotation of the entire fluid with the outer cylinder. This is exactly right: with nothing to hold the fluid back, the steady state is one in which the fluid turns as a solid body, so there is no shear anywhere, no viscous stress and no dissipation — the state a bucket of liquid reaches long after you start spinning it. The limit is singular in the expected way: for $a > 0$ the no-slip condition $v(a) = 0$ still holds, but the departure from $\Omega r$ is confined to $r = O(a)$, a region that shrinks to nothing.

(c) Torque — where the viscosity does enter. For this flow the only non-zero shearing stress is

The torque per unit axial length transmitted across the cylindrical surface of radius $r$ is $G = 2\pi r\cdot r\cdot\tau_{r\theta} = -4\pi\mu C_2$ — independent of $r$, as steady torque balance requires. Hence the torque per unit length that must be applied to keep the motion going has magnitude

in both cases, the sense being opposite. This is the Couette (Margules) viscometer formula: measuring $G$ at known $a,b,\Omega$ yields $\mu$ — and it is precisely because the profile is $\mu$-independent while the torque is $\mu$-proportional that the instrument works.

(d) Stability — why the two cases are genuinely different physically. The two velocity fields above are exact steady solutions, but only case (ii) is unconditionally realised. By Rayleigh’s circulation criterion, an inviscid rotating flow is stable to axisymmetric disturbances iff the square of the angular momentum $|rv|$ increases outwards, $\dfrac{d}{dr}(rv)^2 \ge 0$. Using $rv = C_1r^2 + C_2$, $\dfrac{d}{dr}(rv)^2 = 4C_1 r\,(rv)$, and $rv > 0$ on $(a,b)$ in both cases:

• Case (i): $C_1 < 0$, so $\frac{d}{dr}(rv)^2 < 0$ — angular momentum decreases outwards, and the flow is Rayleigh-unstable. Above a critical Taylor number the profile breaks down into Taylor vortices. The formula of Step 6 is therefore the observed flow only for sufficiently small $\Omega$ (or sufficiently large $\mu$, which is stabilising).
• Case (ii): $C_1 > 0$, so $\frac{d}{dr}(rv)^2 > 0$ — Rayleigh-stable at every $\Omega$ against axisymmetric disturbances.

This asymmetry is almost certainly why the question asks for the two cases side by side. $\qquad\blacksquare$

Answer