A two-dimensional layer of viscous fluid lies between two rigid boundaries at $y=\pm L_{0}$. The boundary at $y=L_{0}$ oscillates in its own plane with velocity $\left(U_{0} \cos \omega t, 0\right)$, while the boundary at $y=-L_{0}$ oscillates in its own plane with velocity $\left(-U_{0} \cos \omega t, 0\right)$. Assume that there is no pressure gradient and that the fluid flows parallel to the boundary with velocity $(u(y, t), 0)$, where $u(y, t)$ can be written as $u(y, t)=\operatorname{Re}\left[U_{0} f(y) \exp (i \omega t)\right]$.

(a) By exploiting the symmetry of the system or otherwise, show that

$$f(y)=\frac{\sinh [(1+i) \Delta \hat{y}]}{\sinh [(1+i) \Delta]}, \text { where } \hat{y}=\frac{y}{L_{0}} \text { and } \Delta=\left(\frac{\omega L_{0}^{2}}{2 \nu}\right)^{1 / 2}$$

(b) Hence or otherwise, show that

where $\Delta_{\pm}=\Delta(1 \pm \hat{y})$.

(c) Show that, for $\Delta \ll 1$,

$$u(y, t) \simeq \frac{U_{0} y}{L_{0}} \cos \omega t$$

and briefly interpret this result physically.

$$\begin{aligned} & \frac{u(y, t)}{U_{0}}=\frac{\cos \omega t\left[\cosh \Delta_{+} \cos \Delta_{-}-\cosh \Delta_{-} \cos \Delta_{+}\right]}{(\cosh 2 \Delta-\cos 2 \Delta)} \\ & +\frac{\sin \omega t\left[\sinh \Delta_{+} \sin \Delta_{-}-\sinh \Delta_{-} \sin \Delta_{+}\right]}{(\cosh 2 \Delta-\cos 2 \Delta)}, \end{aligned}$$