A circular cylinder can be represented exactly in two-dimensional incompressible potential flow by superposing

1. a uniform stream of speed $U$ in the positive $x$ direction;
2. a doublet centered at the origin.

Because both component potentials satisfy Laplace's equation away from the origin, their sum does as well.

Uniform flow plus a doublet

The uniform-flow potential and stream function are

$$\phi_U=Ur\cos\theta,$$

$$\psi_U=Ur\sin\theta.$$

For a doublet aligned with the $x$ axis,

$$\phi_D=\frac{\kappa}{2\pi r}\cos\theta,$$

$$\psi_D=-\frac{\kappa}{2\pi r}\sin\theta.$$

Therefore

$$\phi=\left(Ur+\frac{\kappa}{2\pi r}\right)\cos\theta,$$

$$\psi=\left(Ur-\frac{\kappa}{2\pi r}\right)\sin\theta.$$

Choose the doublet strength

$$\boxed{\kappa=2\pi Ua^2},$$

where $a$ will become the cylinder radius. Then

$$\boxed{\phi=U\left(r+\frac{a^2}{r}\right)\cos\theta},$$

$$\boxed{\psi=U\left(r-\frac{a^2}{r}\right)\sin\theta}.$$

Velocity field

Using

$$v_r=\frac{\partial\phi}{\partial r},$$

and

$$v_\theta=\frac1r\frac{\partial\phi}{\partial\theta},$$

we obtain

$$\boxed{v_r =U\left(1-\frac{a^2}{r^2}\right)\cos\theta},$$

$$\boxed{v_\theta =-U\left(1+\frac{a^2}{r^2}\right)\sin\theta}.$$

Far from the origin,

$$r\gg a,$$

so

$$v_r\to U\cos\theta,$$

$$v_\theta\to-U\sin\theta,$$

which is the original uniform flow.

Why $r=a$ behaves like a solid cylinder

At

$$r=a,$$

the radial velocity is

$$v_r(a,\theta)=0.$$

Thus no fluid crosses the circle $r=a$. The circle satisfies the stationary no-penetration boundary condition and can be interpreted as the surface of an impermeable cylinder.

The stream function also becomes

$$\psi(a,\theta)=0,$$

so the cylinder surface is itself a streamline.

Potential flow does not impose no slip. The tangential velocity on the cylinder surface is

$$\boxed{v_\theta(a,\theta)=-2U\sin\theta}.$$

Except at special points, fluid therefore slips tangentially along the idealized surface.

Stagnation points

A stagnation point has zero velocity. On the cylinder surface,

$$v_r=0$$

automatically, so stagnation also requires

$$v_\theta=-2U\sin\theta=0.$$

Hence

$$\boxed{\theta=0,\pi}.$$

The ideal cylinder has one stagnation point at the front and one at the rear.

Surface speed

The surface speed is

$$|\mathbf v|=|v_\theta| =2U|\sin\theta|.$$

It is zero at the front and rear stagnation points and reaches its maximum magnitude at the top and bottom,

$$\theta=\frac\pi2,\frac{3\pi}2,$$

where

$$\boxed{|\mathbf v|_{max}=2U}.$$

Pressure distribution from Bernoulli

For this steady, incompressible, irrotational, inviscid flow at constant elevation, Bernoulli's constant is the same throughout the connected flow region:

$$p+\frac12\rho v^2 =p_\infty+\frac12\rho U^2.$$

Define the pressure coefficient

$$\boxed{C_p=\frac{p-p_\infty}{\tfrac12\rho U^2}}.$$

On the surface,

$$\frac{v^2}{U^2}=4\sin^2\theta,$$

so

$$\boxed{C_p=1-4\sin^2\theta}.$$

At the stagnation points,

$$C_p=1,$$

while at the top and bottom,

$$C_p=-3.$$

The ideal pressure distribution is symmetric between the front and rear halves of the cylinder.

Worked numerical example

Air approaches a cylinder at

$$U=10,\mathrm{m/s}.$$

At

$$\theta=90^\circ,$$

the surface speed is

$$v=2U=20,\mathrm{m/s}.$$

Taking

$$\rho=1.2,\mathrm{kg/m^3},$$

Bernoulli gives

$$p-p_\infty =\frac12\rho(U^2-v^2).$$

Thus

$$p-p_\infty =\frac12(1.2)(100-400) =-180,\mathrm{Pa}.$$

So

$$\boxed{p=p_\infty-180,\mathrm{Pa}}.$$

The pressure is lower where the potential-flow speed is higher.

What this solution teaches

The cylinder solution is a canonical demonstration of potential-flow construction by superposition:

• Laplace's equation allows elementary solutions to be added;
• doublet strength is chosen to enforce no penetration on a desired body shape;
• the resulting velocity field determines pressure through Bernoulli;
• tangential slip remains because the model is inviscid.

This exact ideal solution is mathematically consistent, but its symmetric pressure field leads to a famous failure when predicting drag on a real cylinder.