Consider a unidirectional flow with dynamic viscosity $\mu$ along a straight rigid-walled channel of uniform cross-sectional shape $\mathcal{D}$ driven by a uniform applied pressure gradient $G$. Write down the differential equation and boundary conditions governing the velocity $w$ along the channel.

Consider the situation when the boundary includes a sharp corner of angle $2 \alpha$. Explain why one might expect that, sufficiently close to the corner, the solution should be of the form

$$w=(G / \mu) r^{2} f(\theta)$$

where $r$ and $\theta$ are polar co-ordinates with origin at the vertex and $\theta=\pm \alpha$ describing the two planes emanating from the corner. Determine $f(\theta)$.

If $\mathcal{D}$ is the sector bounded by the lines $\theta=\pm \alpha$ and the circular arc $r=a$, show that the flow is given by

$$w=(G / \mu) r^{2} f(\theta)+\sum_{n=0}^{\infty} A_{n} r^{\lambda_{n}} \cos \lambda_{n} \theta$$

where $\lambda_{n}$ and $A_{n}$ are to be determined.

[Note that $\int \cos (a x) \cos (b x) d x=\{a \sin (a x) \cos (b x)-b \sin (b x) \cos (a x)\} /\left(a^{2}-b^{2}\right)$.]

Considering the values of $\lambda_{0}$ and $\lambda_{1}$, comment briefly on the cases: (i) $2 \alpha<\frac{1}{2} \pi$;

(ii) $\frac{1}{2} \pi<2 \alpha<\frac{3}{2} \pi$; and (iii) $\frac{3}{2} \pi<2 \alpha<2 \pi$.