I'm working on an unassessed course problem,
A cylinder with radius $a_1$ moves parallel to its axis with constant positive velocity $U$ inside a stationary coaxial cylinder with radius $a_2(> a_1)$. The region between the cylinders is filled with incompressible fluid. Assume that body forces can be ignored and that the pressure gradient in the direction parallel to the axis is zero. If the velocity in the fluid is in the direction of the axis, determine how it varies in the radial direction.
I'm used to solving problems where (in)viscidity and/or steadiness are given, but here neither is. It would be convenient if I could reason as follows. Does this make sense? \begin{align} & \begin{aligned} \text{Let}&\\ & \vec{F}\text{ be body force per unit mass,} \\ & \vec{F}_D\text{ be drag per unit area,} \\ & \vec{u}\text{ be fluid velocity,} \\ & p\text{ be pressure,} \\ & \mu\text{ be viscosity;} \end{aligned} \\[1em] & \text{we have }\vec{F}=\vec{0},\;\frac{\partial p}{\partial z}=0\text{, but }0<u_z \\ \therefore\;&\text{we must have }\vec{F}_D\neq\vec{0}\text{ on the cylinder wall(s)} \tag{$\ast$} \\ \therefore\;&\text{we must have }0<\mu \\ \therefore\;&\text{by the no-slip condition, we must have } (u_z)_{r=a_1}=U\text{, a constant} \\ \therefore\;&\text{we must have }\frac{\partial u_z}{\partial t}=0 \\ \therefore\;&\text{by Navier-Stokes, (etc.)} \end{align} It's mainly $(\ast)$ I'm wondering about.
For steady, unidirectional axial flow driven by the motion of the inner cylinder and without an imposed axial pressure gradient, the Navier-Stokes equations -- expressed in terms of cylindrical coordinates -- reduce to
$$\frac{\mu}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u_z}{\partial r} \right) = 0$$
Integrating twice, we get the general solution
$$u_z = A \log r +B,$$
where $A$ and $B$ are integration constants. The usual no-slip boundary conditions imposed at the inner and outer cylinder surfaces are sufficient to obtain a unique solution. These are
$$\left. u_z\right|_{r=a_1} = U, \quad\left. u_z\right|_{r=a_2} = 0,$$
yielding the solution
$$A = \frac{U}{\log \cfrac{a_1}{a_2}}, \quad B= -\frac{U\log a_2}{\log \cfrac{a_1}{a_2}}$$
Hence,
$$u_z = U \frac{\log \frac{r}{a_2}}{\log \frac{a_1}{a_2}}$$
What assumptions lead to this solution? First, we are given that there are no body forces, the pressure gradient parallel to the axis is zero, and the fluid velocity has only a non-zero axial component $u_z$.
With $u_r = u_\theta = 0$ as assumed, the continuity equation $\nabla \cdot \mathbf{u} = 0$ reduces to $\frac{\partial u_z}{\partial z } = 0$. At this stage we can not rule out dependence in the angular direction, and we have $u_z = u_z(r,\theta)$.
The axial component of the Navier-Stokes equations then reduces to
$$\rho \frac{\partial u_z}{\partial t} = \frac{\mu}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u_z}{\partial r} \right) + \frac{\mu}{r^2}\frac{\partial^2 u_z}{\partial \theta^2},$$
and the angular component reduces to
$$-\frac{\partial p}{\partial \theta} + \frac{2\mu}{r^2} \frac{\partial u_z}{\partial \theta}= 0$$
With no angular pressure gradient, it follows that $\frac{\partial u_z}{\partial \theta} = 0$ and $u_z = u_z(r)$ -- leading to the above solution in steady state.
If you do not make the steady-state assumption, then a solution to the parabolic eqution for $u_z(r,t)$ requires that you impose an initial condition such as an impulsive start at $t = 0+$.