← 2026 Paper 2

UPSC 2026 Maths Optional Paper 2 Q6c — Step-by-Step Solution

15 marks · Section B

Two-Dimensional and Axisymmetric Flow · Mechanics & Fluid Dynamics · asked 2× in 14 yrs · Read the full method →

Question

Suppose that there is an infinitely long cylinder of radius '$a$' placed in a uniform stream having velocity $-V\hat{i}$. Then using Milne-Thomson’s circle theorem, what will be its complex velocity potential? If in addition, a circulation round the cylinder of $k$ is produced, find out the stagnation points. Do they produce a lifting tendency in the vertical direction? Explain how.

Technique

The circle theorem is a machine: give it the potential of the undisturbed stream, and it returns the potential with the cylinder inserted — so the only real work is writing $f(z)$ for the stream $-V\hat i$ correctly (note the minus: this is the mirror image of the textbook $+U\hat i$ case, and it flips the sign of the lift at the end). Circulation is then superposed as a line vortex at the centre, which leaves $|z|=a$ a streamline because its $\psi$ is a function of $r$ alone. Stagnation points come from $dw/dz=0$, a quadratic in $z$ whose discriminant changes sign at $|k|=4\pi Va$ — the classification by circulation strength is where the marks that most scripts leave on the table are. The lift follows from Blasius (or Kutta–Joukowski), and the physical “how” from Bernoulli applied to the surface speed.

Solution

Throughout: two-dimensional, steady, irrotational flow of an inviscid incompressible fluid of density $\rho$, in the complex plane $z=x+iy$; the cylinder’s cross-section is the circle $|z|=a$ and the fluid occupies $|z|\ge a$. The complex potential is $w(z)=\phi+i\psi$, with the fundamental relation

Take $V>0$, so that the stream runs in the negative $x$-direction.

Step 1 — The undisturbed stream.

The stream velocity is $\mathbf q=-V\hat i$, i.e. $u=-V$, $v=0$. Hence $\dfrac{dw}{dz}=-V$ and, ignoring an irrelevant additive constant,

This has no singularity in the finite plane, so the circle theorem applies with no restriction. (Written out: $f=-V(x+iy)$ gives $\phi=-Vx$, $\psi=-Vy$ — streamlines $y=\text{const}$ traversed in the sense of decreasing $x$. Correct.)

Step 2 — State Milne-Thomson’s circle theorem before using it.

Circle theorem. Let $f(z)$ be the complex potential of a two-dimensional irrotational flow of incompressible inviscid fluid in which there are no rigid boundaries and all of whose singularities lie in $|z|>a$. Then on introducing the circular cylinder $|z|=a$ into the flow, the complex potential becomes

and $|z|=a$ is a streamline of the new flow.

Two one-line justifications, both worth marks:

(i) The boundary condition holds. On $|z|=a$ we have $z\bar z=a^{2}$, so $a^{2}/\bar z=z$ and therefore

i.e. the circle is the streamline $\psi=0$ — exactly the condition for a rigid cylinder.

(ii) The flow far away is unchanged. If $f$ is singular at $z_0$ with $|z_0|>a$, the added term is singular only at the inverse point $a^2/\bar z_0$, whose modulus $a^2/|z_0|<a$ — inside the cylinder, hence outside the fluid. So no new singularity is introduced into $|z|>a$.

Step 3 — Apply the theorem.

With $f(z)=-Vz$ and $V$ real, $\overline{f\!\left(\dfrac{a^2}{\bar z}\right)}=\overline{\left(-V\dfrac{a^{2}}{\bar z}\right)}=-V\dfrac{a^{2}}{z}$. Therefore

Two checks. As $|z|\to\infty$, $\dfrac{dw}{dz}\to-V$, recovering the stream $-V\hat i$. On $z=ae^{i\theta}$, $z+\dfrac{a^2}{z}=ae^{i\theta}+ae^{-i\theta}=2a\cos\theta$, so $w=-2Va\cos\theta$ is real and $\psi=0$. $\checkmark$

Step 4 — Superpose the circulation.

The circle theorem produces the acyclic (circulation-free) flow. A circulation $k$ round the cylinder is obtained by placing a line vortex at the centre $z=0$ — which is inside the cylinder, hence legitimate.

Convention (state it, because the sign of the final force depends on it): $k$ is the circulation taken positive in the anticlockwise sense. The corresponding potential is

Indeed, with $z=re^{i\theta}$, $w_v=-\dfrac{ik}{2\pi}\ln r+\dfrac{k\theta}{2\pi}$, so

and $\displaystyle\oint q_\theta\,r\,d\theta=k$ round any circuit enclosing the origin. Because $\psi_v$ depends on $r$ alone, it is constant on $|z|=a$, so adding it does not disturb the boundary condition. Hence

Check on the circulation. For any circuit $C$ round the cylinder, $\displaystyle\oint_C\frac{dw}{dz}\,dz=\Gamma+iQ$ ($\Gamma$ = circulation, $Q$ = flux). Here $\dfrac{dw}{dz}=-V+\dfrac{Va^{2}}{z^{2}}-\dfrac{ik}{2\pi z}$ has residue $-\dfrac{ik}{2\pi}$ at $z=0$, so the integral is $2\pi i\left(-\dfrac{ik}{2\pi}\right)=k$: circulation $k$, zero flux (no source). $\checkmark$

Step 5 — Stagnation points.

Stagnation points are the zeros of $\dfrac{dw}{dz}$ lying in the fluid $|z|\ge a$:

Multiplying by $-z^{2}$ (legitimate, since $z\neq0$ in the fluid):

This quadratic has

Introduce the dimensionless circulation

so that (5) becomes the compact

Step 6 — Classify by the strength of the circulation. (This is the part answers skip.)

(i) Weak circulation, $|k|<4\pi Va$ (i.e. $|\lambda|<1$). The square root is real and

so both stagnation points lie on the cylinder, at $z=ae^{i\theta_s}$ with

They are placed symmetrically about the $y$-axis and, for $k>0$, below the horizontal diameter (third and fourth quadrants). For $k=0$, $\theta_s=0,\pi$: the points $z=\pm a$, the front and rear generators of the plain symmetric flow.

(ii) Critical circulation, $|k|=4\pi Va$ ($|\lambda|=1$). The two roots coalesce:

a single double stagnation point at the lowest point of the cylinder when $k>0$ (highest when $k<0$). The critical value is

(iii) Strong circulation, $|k|>4\pi Va$ ($|\lambda|>1$). Now $\sqrt{a^2-k^2/16\pi^2V^2}=ia\sqrt{\lambda^{2}-1}$ is imaginary and

so both roots lie on the $y$-axis. Their product is $-a^{2}$, hence $|z_+|\,|z_-|=a^{2}$: one is inside the cylinder (it is the inverse point, not part of the fluid and not a physical stagnation point) and one is outside. For $k>0$ the physical one is

a single stagnation point standing off the cylinder, directly below it; the streamline through it closes on itself, trapping a body of fluid that is carried round with the cylinder.

Figure to draw (small, half a text-column, and it earns its space because it makes the classification and the lift argument visible at a glance): draw the circle $|z|=a$ with centre $O$ and the two axes. Along the top and bottom draw three horizontal streamlines with arrowheads pointing in the $-x$ direction, labelled "$-V\hat i$"; crowd the lines above the cylinder and space them out below it. Draw a curved arrow anticlockwise round the circle labelled “circulation $k>0$”. Mark the two stagnation points $S_1,S_2$ on the lower half of the circle, symmetric about the $y$-axis, with the angle $\theta_s$ from the positive $x$-axis to $OS_2$ and the annotation $\sin\theta_s=-k/(4\pi Va)$. Add a thick upward arrow through $O$ labelled $Y=\rho Vk$. In a small inset beside it, draw the same circle with a single point below it on the $y$-axis, labelled ”$|k|>4\pi Va$: stagnation point leaves the cylinder”. Do not add streamlines inside the circle — there is no flow there.

Step 7 — The force on the cylinder: Blasius’ theorem.

Blasius’ theorem. For steady two-dimensional irrotational flow, the force per unit length on a cylinder is given by

where $C$ is any contour enclosing the cylinder and no other singularity.

From (3),

Only the coefficient of $z^{-1}$ in the square is needed (all other terms integrate to zero). Squaring, the $z^{-1}$ term arises solely from the cross-product of $-V$ with $-\dfrac{ik}{2\pi z}$:

(the terms $\frac{Va^2}{z^2}\cdot(-V)$, $\frac{Va^2}{z^2}\cdot\left(-\frac{ik}{2\pi z}\right)$ and the two squares contribute only $z^{-2},z^{-3},z^{-4}$). Hence $\operatorname{Res}_{z=0}\left(\dfrac{dw}{dz}\right)^{2}=\dfrac{iVk}{\pi}$ and

Comparing real and imaginary parts:

$X=0$ is d’Alembert’s paradox: no drag. The entire force is transverse to the stream — i.e. vertical.

Step 8 — “Explain how”: the physical mechanism, from Bernoulli.

On the cylinder $z=ae^{i\theta}$, equation (3) gives

(Setting $q=0$ in (9) returns $\sin\theta=-\dfrac{k}{4\pi Va}$ — an independent confirmation of (7), obtained without touching the quadratic (4).)

Bernoulli’s equation for the steady irrotational flow gives $p+\tfrac12\rho q^{2}=p_{\infty}+\tfrac12\rho V^{2}$, so where the fluid is faster, the pressure is lower. Compare the crown and the keel:

For $k>0$ this is positive: the flow is faster over the top and slower underneath, hence $p_{\text{top}}<p_{\text{bot}}$, hence a net upward force. That is the mechanism.

Integrating it out confirms (8) without Blasius. The pressure force on an element is $-p\,\hat n\,a\,d\theta$ with outward normal $\hat n=(\cos\theta,\sin\theta)$, so

(the constant part of $p$ contributes nothing, since $\int_0^{2\pi}\sin\theta\,d\theta=0$). With (9),

and $\int_0^{2\pi}\sin^{3}\theta\,d\theta=\int_0^{2\pi}\sin\theta\,d\theta=0$, $\int_0^{2\pi}\sin^{2}\theta\,d\theta=\pi$. Hence

The same computation with $\cos\theta$ in place of $\sin\theta$ gives $X=0$, since $\int\sin^{2}\theta\cos\theta=\int\sin\theta\cos\theta=\int\cos\theta=0$ over a period. $\checkmark$

Step 9 — Answer the question asked.

Yes — provided $k\neq0$, and the tendency is exactly vertical. The reasoning, in the order an examiner wants it:

1. With $k=0$ the potential (2) is even in $y$: the flow is symmetric about the $x$-axis, the stagnation points sit at $\theta=0,\pi$ on the horizontal diameter, the pressure distribution is up–down symmetric, and $Y=0$. No lift.
2. Circulation destroys that symmetry, and the migration of the stagnation points off the horizontal diameter to $\sin\theta_s=-k/(4\pi Va)$ is precisely the visible signature of the asymmetry. Their displacement and the lift are two faces of the same broken symmetry — which is the sense in which the stagnation points “produce” the lift.
3. Quantitatively — the Kutta–Joukowski theorem. In vector form, for circulation $\boldsymbol\Gamma=k\hat k$ (anticlockwise) in a stream $\mathbf U_\infty$,

in agreement with (8). The magnitude is $\rho V|k|$ per unit length, the direction is perpendicular to the stream — vertical — upward when $k>0$ (anticlockwise), downward when $k<0$. 4. Carry the sign of the stream. For the usual textbook stream $+V\hat i$ the same anticlockwise $k$ gives $Y=-\rho Vk$, i.e. downward. Reversing the stream to $-V\hat i$, as this paper does, reverses the lift. Quoting the book formula without this reversal is the single commonest way to lose the last marks here. 5. Validity across the classification. Equation (8) holds for every $k$: the lift does not care whether the stagnation points are on the cylinder or standing off below it. What changes at $|k|=4\pi Va$ is only their location. $\qquad\blacksquare$

Answer