The Question:
Suppose a circular cylinder of radius $a$ moves with constant velocity $U$ in the $x$-direction in a two-dimensional irrotational, incompressible flow whose velocity decays to zero at infinity.
The circulation around the cylinder is zero.
Let $\phi$ be the velocity potential of the flow. Show that
$$\frac{\partial \phi}{\partial r} = U\cos \theta \qquad \text{ on } \;\;r=a$$
where $(r,\theta)$ are the plane polar coordinates.
So basically, I don't really know how to put all the given information together.
How do I deduce anything about the velocity potential at $r=a$ if the cylinder moves with constant velocity?
Any hints would be much appreciated.
For potential flow, the fluid velocity is the gradient of the potential, $\mathbf{u} = \nabla \phi$.
The boundary condition at a solid (impermeable) surface is the no flux condition. If $\mathbf{n}$ is the unit normal vector and $\mathbf{U}$ is the velocity at a point on the surface the no-flux condition is expressed as
$$\nabla \phi \cdot \mathbf{n} = \mathbf{U} \cdot \mathbf{n}$$
Using polar coordinates, the velocity of the cylinder is $U \mathbf{e}_x = U \cos\theta \,\mathbf{e}_r - U \sin \theta \, \mathbf{e}_\theta$, and the normal vector is $\mathbf{n} = \mathbf{e}_r$.
Hence, the boundary condition at the surface, where $r = a$, is
$$\frac{\partial \phi}{\partial r} = \nabla \phi \cdot \mathbf{e}_r = [U \cos\theta \,\mathbf{e}_r - U \sin \theta] \cdot \mathbf{e}_r = U \cos \theta $$