Two-Dimensional and Axisymmetric Flow

At a Glance

• Frequency: 2 sub-parts across 2 of 14 years (2015, 2026)
• Priority tier: T4
• Marks (count): 20 (1)
• Average solve time: ~30 min
• Difficulty mix: medium 1
• Section: B | Dominant type: derivation

Why This Chapter Matters

The connection between 2D irrotational flow and complex analysis is one of the most elegant results in applied mathematics: the velocity potential and stream function satisfy the Cauchy-Riemann equations, making the complex potential $w = \phi + i\psi$ an analytic function of $z = x + iy$. UPSC 2015 tested this with a derivation of the complex potential for a standard flow — a Section B question requiring both theoretical justification and explicit computation.

Minimum Theory

2D Irrotational Incompressible Flow

Consider a 2D flow in the $xy$-plane with velocity $(u, v)$:

• Incompressibility: $\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y} = 0$ (continuity equation).
• Irrotationality: $\dfrac{\partial v}{\partial x} - \dfrac{\partial u}{\partial y} = 0$ (no vorticity).

Velocity Potential and Stream Function

Velocity potential $\phi$ (exists for irrotational flow):

$u = \frac{\partial \phi}{\partial x}, \qquad v = \frac{\partial \phi}{\partial y}.$

Irrotationality is automatic: $\partial v/\partial x - \partial u/\partial y = \phi_{xy} - \phi_{yx} = 0$. Incompressibility gives $\nabla^2\phi = 0$ (Laplace’s equation for $\phi$).

Stream function $\psi$ (exists for incompressible flow):

$u = \frac{\partial \psi}{\partial y}, \qquad v = -\frac{\partial \psi}{\partial x}.$

Continuity is automatic. Irrotationality gives $\nabla^2\psi = 0$ (Laplace’s equation for $\psi$).

The Complex Potential — Key Result

Define the complex potential:

$w = \phi(x,y) + i\,\psi(x,y).$

Claim. $w$ is an analytic function of $z = x + iy$.

Proof. Check the Cauchy-Riemann equations for $w = \phi + i\psi$:

$\frac{\partial \phi}{\partial x} = u = \frac{\partial \psi}{\partial y} \quad \checkmark$

$\frac{\partial \phi}{\partial y} = v = -\left(-\frac{\partial \psi}{\partial x}\right) \implies \frac{\partial \phi}{\partial y} = -\left(-v\right)$

More carefully: $\phi_x = u = \psi_y$ and $\phi_y = v = -\psi_x$. Thus:

$\phi_x = \psi_y \quad \text{and} \quad \phi_y = -\psi_x.$

These are exactly the Cauchy-Riemann equations for $w = \phi + i\psi$. Hence $w$ is analytic. $\blacksquare$

Complex Velocity

The derivative of the complex potential gives the complex velocity:

$\frac{dw}{dz} = \frac{\partial \phi}{\partial x} + i\frac{\partial \psi}{\partial x} = u + i(-v) = u - iv.$

Thus if you know $w(z)$, the velocity components are recovered by:

$u - iv = \frac{dw}{dz}.$

The speed is $|dw/dz| = \sqrt{u^2 + v^2}$.

Standard Complex Potentials

Warning about the last row. It is a consequence of the circle theorem below, not an independent fact. If a question says “using Milne-Thomson’s circle theorem, find the complex potential”, quoting this row answers a different question and forfeits the marks attached to the theorem. Derive it.

Milne-Thomson’s Circle Theorem

Statement. Let $f(z)$ be the complex potential of a 2D irrotational flow of incompressible inviscid fluid with no rigid boundaries, all of whose singularities lie in $|z|>a$. On introducing the circular cylinder $|z|=a$, the complex potential becomes

$w(z)=f(z)+\overline{f\!\left(\frac{a^2}{\bar z}\right)},$

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

Why it works — two one-line justifications, both examinable.

1. The boundary condition holds. On $|z|=a$ we have $z\bar z=a^2$, so $a^2/\bar z=z$ and $w=f(z)+\overline{f(z)}=2\,\mathrm{Re}\,f(z)$, which is real. Hence $\psi=\mathrm{Im}\,w=0$ on the circle: it is a streamline, which is exactly the condition for a rigid cylinder.
2. The far field is undisturbed. If $f$ is singular at $z_0$ with $|z_0|>a$, the added term is singular only at the inverse point $a^2/\bar z_0$, of modulus $a^2/|z_0|<a$ — inside the cylinder, hence outside the fluid. No new singularity enters the flow region.

Adding circulation. The circle theorem produces the acyclic (circulation-free) flow. A circulation $\kappa$ round the cylinder is added as a line vortex at the centre — legitimate because the centre is inside the cylinder, and harmless to the boundary condition because the vortex’s $\psi=-\tfrac{\kappa}{2\pi}\ln r$ depends on $r$ alone and so is constant on $|z|=a$:

$w(z)=f(z)+\overline{f\!\left(\frac{a^2}{\bar z}\right)}-\frac{i\kappa}{2\pi}\log z,\qquad \kappa>0 \text{ anticlockwise}.$

Blasius’ theorem (force on the cylinder). For steady 2D irrotational flow, the force per unit length is

$X-iY=\frac{i\rho}{2}\oint_C\left(\frac{dw}{dz}\right)^{2}dz,$

$C$ any contour enclosing the cylinder and no other singularity. Only the residue at the origin survives. Equivalently, Kutta–Joukowski: $\mathbf{F}=\rho\,\mathbf{U}_\infty\times\boldsymbol\Gamma$, of magnitude $\rho U|\kappa|$ perpendicular to the stream — the drag is always zero (d’Alembert’s paradox).

Axisymmetric Flow (Stokes Stream Function)

For flow with symmetry about the $z$-axis (cylindrical coordinates $r, \theta, z$ with no $\theta$-dependence), a Stokes stream function $\Psi(r,z)$ is defined so that:

• Streamlines are curves $\Psi = \text{const}$.
• The volume flux through any circle of radius $r$ centred on the axis, in the plane $z = \text{const}$, equals $2\pi\Psi$.
• Velocity components: $u_r = -\dfrac{1}{r}\dfrac{\partial \Psi}{\partial z}$, $u_z = \dfrac{1}{r}\dfrac{\partial \Psi}{\partial r}$.

Unlike the 2D case, $\Psi$ does not generally satisfy Laplace’s equation; instead it satisfies the Stokes stream function equation.

Question Archetypes

find-complex-potential (1 question; 2015)

Recognition Cues

• A 2D irrotational flow is described (uniform stream + source, doublet, vortex, etc.).
• Asked to find $w(z)$, or the velocity potential $\phi$, or the stream function $\psi$.
• May ask to verify that $w$ is analytic or to find the velocity at a point.

Solution Template

1. Identify the flow components (uniform stream, source, vortex, doublet, etc.) from the problem description.
2. Write the complex potential for each component using the standard table.
3. Superpose: $w = w_1 + w_2 + \ldots$ (complex potentials add linearly).
4. Extract $\phi = \text{Re}(w)$ and $\psi = \text{Im}(w)$ if asked.
5. Compute $dw/dz = u - iv$ to find velocity components.
6. If asked to prove $w$ is analytic: verify C-R equations $\phi_x = \psi_y$, $\phi_y = -\psi_x$ using the flow relations $u = \phi_x = \psi_y$, $v = \phi_y = -\psi_x$.

Worked Example

2015 Paper 2, 2015-P2-QMF (20 marks)

Find the complex potential for the 2D irrotational flow consisting of a uniform stream of speed $U$ parallel to the $x$-axis superposed with a line source of strength $m$ at the origin. Find the velocity at the stagnation point and the equation of the streamline through the stagnation point.

Step 1: Write the complex potential.

Uniform stream: $w_1 = Uz$.

Line source of strength $m$ at origin: $w_2 = \dfrac{m}{2\pi}\log z$.

Superposed complex potential:

$w(z) = Uz + \frac{m}{2\pi}\log z.$

Step 2: Find the complex velocity.

$\frac{dw}{dz} = U + \frac{m}{2\pi z} = u - iv.$

Step 3: Find the stagnation point.

At a stagnation point, $u = v = 0$, i.e., $dw/dz = 0$:

$U + \frac{m}{2\pi z} = 0 \implies z = -\frac{m}{2\pi U}.$

This is the point $\left(-\dfrac{m}{2\pi U},\, 0\right)$ on the negative $x$-axis. The velocity there is zero (stagnation).

Step 4: Stream function and stagnation streamline.

Write $z = re^{i\theta}$ (polar form). Then $\log z = \ln r + i\theta$, so:

$w = U(x + iy) + \frac{m}{2\pi}(\ln r + i\theta),$

$\psi = \text{Im}(w) = Uy + \frac{m\theta}{2\pi} = Ur\sin\theta + \frac{m\theta}{2\pi}.$

At the stagnation point $z = -m/(2\pi U)$: here $r = m/(2\pi U)$, $\theta = \pi$, so:

$\psi_{\text{stag}} = U \cdot \frac{m}{2\pi U} \cdot \sin\pi + \frac{m\cdot\pi}{2\pi} = 0 + \frac{m}{2} = \frac{m}{2}.$

The stagnation streamline is $\psi = m/2$:

$\boxed{Ur\sin\theta + \frac{m\theta}{2\pi} = \frac{m}{2}.}$

This streamline divides the flow: above and below it the stream passes around the “body” formed by the stagnation dividing streamline.

Step 5: Verify $w$ is analytic.

$w(z) = Uz + \dfrac{m}{2\pi}\log z$ is a sum of analytic functions (linear function, which is entire, plus $\log z$, which is analytic on $\mathbb{C}\setminus(-\infty,0]$). Hence $w$ is analytic on $\mathbb{C}\setminus\{0\} \cap \mathbb{C}\setminus(-\infty,0]$, i.e., on $\mathbb{C}$ cut along the negative real axis. The corresponding $\phi$ and $\psi$ automatically satisfy the C-R equations:

$\phi_x = \psi_y = u, \quad \phi_y = -\psi_x = v. \quad \blacksquare$

$\boxed{w(z) = Uz + \frac{m}{2\pi}\log z}$

Common Traps

• Sign error in the source complex potential: $w = (m/2\pi)\log z$ for a source (strength $m > 0$); a sink has $m < 0$. Do not negate $\log z$ unless the problem specifies a sink.
• Confusing complex velocity: $dw/dz = u - iv$, not $u + iv$. The imaginary part has a minus sign.
• Stagnation point: set $dw/dz = 0$, not $|dw/dz|= 0$ computed separately for $u$ and $v$.
• Forgetting the branch cut of $\log z$ when stating where $w$ is analytic.
• For axisymmetric flow: confusing the 2D stream function (where flux = $\psi_2 - \psi_1$) with the Stokes stream function (flux = $2\pi(\Psi_2 - \Psi_1)$).

circle-theorem-cylinder (1 question(s); 2026)

The separating feature from find-complex-potential is that a rigid boundary is introduced into an existing flow. Superposition from the standard table cannot do that; the circle theorem is a machine that does it for you. These questions almost always continue into circulation, stagnation points and lift, and that continuation is where most of the marks sit.

Recognition Cues

• The phrase “using Milne-Thomson’s circle theorem” — the theorem is named, so it must be stated and applied.
• A cylinder of radius $a$ is placed in a given stream (watch the direction: $-V\hat i$ is not $+U\hat i$).
• “If in addition a circulation $\kappa$ is produced, find the stagnation points.”
• A closing question about lift (“do they produce a lifting tendency… explain how”).

Solution Template

1. Write $f(z)$ for the undisturbed stream and check it has no finite singularity, so the theorem applies unrestricted. For $\mathbf q=-V\hat i$: $dw/dz=u-iv=-V$, so $f(z)=-Vz$.
2. State the circle theorem before using it, with the two justifications (boundary condition; inverse-point singularity). Then apply it.
3. Superpose the circulation as a central vortex, and declare the sign convention for $\kappa$ — the sign of the final force depends on it and the paper will not tell you.
4. Stagnation points: solve $dw/dz=0$. Multiplying up gives a quadratic in $z$, not a linear equation. Introduce the dimensionless $\lambda=\kappa/(4\pi Va)$ to make the roots readable.
5. Classify by circulation strength — this is the part scripts skip. The discriminant changes sign at $|\kappa|=4\pi Va$: two points on the cylinder below it, a coincident double point at it, and beyond it two points on the vertical axis of which one lies inside the cylinder and is not a stagnation point of the flow at all (the root product is $-a^2$, so $|z_+||z_-|=a^2$). Discard it.
6. Lift by Blasius (or Kutta–Joukowski), then explain the mechanism from Bernoulli: compare the surface speed at crown and keel.

Worked Example

2026 Paper 2, 2026-P2-Q6c (15 marks)

An infinitely long cylinder of radius $a$ is placed in a uniform stream of velocity $-V\hat i$. Using Milne-Thomson’s circle theorem, find its complex velocity potential. If a circulation $\kappa$ round the cylinder is produced, find the stagnation points. Do they produce a lifting tendency in the vertical direction? Explain how.

Source: analysis/solutions/2026-P2-Q6c.md

Step 1 — Undisturbed stream. $u=-V$, $v=0$, so $dw/dz=-V$ and $f(z)=-Vz$ (constant discarded). No finite singularity, so the theorem applies with no restriction.

Step 2 — Circle theorem. Since $V$ is real, $\overline{f(a^2/\bar z)}=\overline{-V a^2/\bar z}=-Va^2/z$, giving

$\boxed{w(z)=-V\!\left(z+\frac{a^2}{z}\right)},\qquad |z|\ge a.$

Checks: $dw/dz\to-V$ as $|z|\to\infty$ ✓; on $z=ae^{i\theta}$, $z+a^2/z=2a\cos\theta$ is real so $\psi=0$ ✓.

Step 3 — Circulation. Taking $\kappa>0$ anticlockwise,

$w(z)=-V\!\left(z+\frac{a^2}{z}\right)-\frac{i\kappa}{2\pi}\log z.$

Step 4 — Stagnation points. $\dfrac{dw}{dz}=-V\!\left(1-\dfrac{a^2}{z^2}\right)-\dfrac{i\kappa}{2\pi z}=0$. Multiplying by $-z^2$:

$Vz^2+\frac{i\kappa}{2\pi}z-Va^2=0\;\Longrightarrow\;z_\pm=a\!\left(\pm\sqrt{1-\lambda^2}-i\lambda\right),\qquad \lambda=\frac{\kappa}{4\pi Va}.$

Step 5 — The three cases.

• $|\kappa|<4\pi Va$ ($|\lambda|<1$): $|z_\pm|^2=a^2[(1-\lambda^2)+\lambda^2]=a^2$, so both points lie on the cylinder, at $\sin\theta_s=-\lambda$ — symmetric about the $y$-axis and, for $\kappa>0$, below the horizontal diameter.
• $|\kappa|=4\pi Va$: the roots coalesce at $z=-ia\,\mathrm{sgn}\,\kappa$, a double stagnation point at the lowest point of the cylinder (for $\kappa>0$). This is the critical circulation $\kappa_{\text{crit}}=4\pi Va$.
• $|\kappa|>4\pi Va$: both roots lie on the $y$-axis with product $-a^2$, so one is inside the cylinder — not a stagnation point of the flow. The physical one stands off below the cylinder at $|z|=a(\lambda+\sqrt{\lambda^2-1})>a$.

Step 6 — Lift, by Blasius. With $\dfrac{dw}{dz}=-V+\dfrac{Va^2}{z^2}-\dfrac{i\kappa}{2\pi z}$, the only $z^{-1}$ term in the square comes from the cross-product $2(-V)\!\left(-\tfrac{i\kappa}{2\pi z}\right)=\tfrac{iV\kappa}{\pi z}$, so $\operatorname{Res}_0(dw/dz)^2=\tfrac{iV\kappa}{\pi}$ and

$X-iY=\frac{i\rho}{2}\cdot2\pi i\cdot\frac{iV\kappa}{\pi}=-i\rho V\kappa\quad\Longrightarrow\quad\boxed{X=0,\qquad Y=\rho V\kappa.}$

Step 7 — “Explain how”. On the cylinder, $\phi=-2Va\cos\theta+\tfrac{\kappa\theta}{2\pi}$, so the surface speed is $q=\tfrac1a\partial_\theta\phi=2V\sin\theta+\tfrac{\kappa}{2\pi a}$. (Setting $q=0$ returns $\sin\theta_s=-\lambda$ — an independent confirmation of Step 5.) Then

$q^2_{\text{top}}-q^2_{\text{bot}}=\left(2V+\tfrac{\kappa}{2\pi a}\right)^2-\left(2V-\tfrac{\kappa}{2\pi a}\right)^2=\frac{4V\kappa}{\pi a}>0\quad(\kappa>0),$

so by Bernoulli ($p+\tfrac12\rho q^2$ constant) the pressure is lower over the top and there is a net upward force. With $\kappa=0$ the potential is even in $y$, the stagnation points sit at $\theta=0,\pi$ and the lift vanishes: circulation is what breaks the up–down symmetry, and the migration of the stagnation points off the horizontal diameter is the visible signature of the same broken symmetry that produces the lift. Cross-check by Kutta–Joukowski: $\rho(-V\hat i)\times(\kappa\hat k)=+\rho V\kappa\,\hat j$ ✓.

Common Traps

• Quoting $U(z+a^2/z)$ from the table. The question named the theorem; state and apply it, including the two justifications. This is the single most expensive shortcut here.
• Carrying the textbook stream direction. Every text works $+U\hat i$ and quotes lift $Y=-\rho U\kappa$. Here the stream is $-V\hat i$, so $Y=+\rho V\kappa$ — upward for anticlockwise $\kappa$. A memorised formula transcribed without the substitution gives the right magnitude and the wrong direction, and direction is what the question asks about.
• Not declaring the sense of $\kappa$. The paper does not state it and the sign of the answer depends on it. One line, before the computation.
• Giving the root pair and stopping. The three-case classification and the critical value $4\pi Va$ carry marks of their own, and beyond critical one root is inside the cylinder and must be discarded — reporting it as a stagnation point is a mathematical error, not an omission.
• Asserting lift without Blasius or Kutta–Joukowski. “The stagnation points move, so there is lift” is the mechanism, not the magnitude. Give $Y=\rho V\kappa$ and note $X=0$ (d’Alembert).

Marks-Aware Writing

This is a 20-mark Section B derivation question. UPSC expects:

1. Explicit statement of each component’s complex potential — do not just write the answer.
2. Superposition step clearly shown.
3. $dw/dz$ computed and interpreted as $u - iv$.
4. Stagnation point found by setting $dw/dz = 0$, solving for $z$.
5. Stream function extracted as $\psi = \text{Im}(w)$ — shown in polar coordinates.
6. Stagnation streamline: evaluate $\psi$ at the stagnation point, then write the streamline equation.
7. Analyticity of $w$: mention C-R, even if briefly.

Marks are distributed across all these steps. A bare final answer without working receives minimal credit.

Practice Set

Only one historical question on this atom (shown above).