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\textbf{Question}

Incompressible viscous fluid of constant density $\rho$ and
constant kinematic viscosity $\nu$ occupies the region $$\{
-\infty < x < \infty, -\infty < y < \infty \}$$ above an
impermeable plane situated at $z=0$. This plane creates flow in
the region $z>0$ by moving with a speed $U\cos (\omega t)$
parallel to itself, where $U$ and $\omega$ are both constant.
There are no body forces and no pressure gradient.

\begin{description}
%Question 3i
\item{(i)}
What flow would be produced if this experimetn was carried out
with an $inviscid$ fluid?

%Question 3ii
\item{(ii)}
By seeking time dependent solutions to the Navier-Stokes equations
of the form $$\un{q} = \left ( \begin{array}{c} RE | f(z) \exp
(i\alpha t)|\\ 0 \\ 0
\end{array} \right )$$
where $f$ is a suitably chosen function, $\alpha$ is a suitably
chosen constant and $RE$ signifies that the real part of the
expression is to be taken, show that $$u = U\exp \left ( -z
\sqrt{\frac{\omega}{2\nu}} \right ) \cos \left ( \omega t
-z\sqrt{\frac{\omega}{2\nu}} \right ).$$

%Question 3iii
\item{(iii)}
Sketch the velocity profile for this flow at time $t=0$.

\end{description}



\textbf{Answer}

$$\\$$
\begin{center}
\epsfig{file=311-bottomplate.eps, width=80mm}
\end{center}

(Assume wlog that the plane moves parallel to the x-axis)

\begin{description}
%Question 3i
\item{(i)}
For inviscid fluid there would be NO MOTION AT ALL since an
inviscid fluid can sustain slip at the walls.

%Question 3ii
\item{(ii)}
Seek a solution $\underline{q}=(u(z,t),0,0)$. Then
$div(\underline{q})=0$ so that mass conservation is satisfied.

The momentum equations give
\begin{eqnarray*} u_t=uu_x+vu_y+wu_z & = & 0 +
\nu(u_{xx}+u_{yy}+u_{zz})\\ 0 + uv_x +vv_y +wv_z & = & 0 + \nu(0 +
0+0)\\ 0+uw_x+vw_y+ww_z & = & 0 +\nu(0+0+0) \end{eqnarray*}

i.e. The second two give just $0=0$ and the first reduces to

$u_t=\nu u_{zz}$ with $u(0,t)=U\cos\omega t$.

Now trying $u=f(z)e^{i\alpha t}$ (and taking real parts
throughout) we see first from the B/C that $\alpha = \omega$,
$f(0)=U$. Also

$\displaystyle f(z)i\alpha e^{i\alpha t} = \nu f'' e^{i\alpha t}
\Rightarrow f''-\frac{i\alpha}{\nu}f=0$

$\Rightarrow f=Ae^{\lambda_1z}+Be^{\lambda_2z} \ \
(\lambda^2=i\alpha/\nu = i\omega/\nu)$

Thus $\displaystyle\lambda = \sqrt{\frac{\omega}{\nu}}\sqrt{i}$.
Now $i=e^{i\frac{\pi}{2}+2k\pi} \Rightarrow
\sqrt{i}=e^{i\frac{\pi}{4}+k\pi}$

i.e. the two roots of i are

$\begin{array}{lrcccl} & e^{i\frac{\pi}{4}} & = &
\cos\frac{\pi}{4}+i\sin\frac{\pi}{4} & = &
\frac{1}{\sqrt2}+\frac{i}{\sqrt2}\\ and & e^{5i\frac{\pi}{4}} & =
& \cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4} & = &
\frac{-1}{\sqrt2}-\frac{i}{\sqrt2}
\end{array}$

So for finiteness as $z\to +\infty$ need $\displaystyle
\lambda_2=(\frac{-1}{\sqrt2}-\frac{i}{\sqrt2})\sqrt{\frac{\omega}{\nu}}$

$\Rightarrow f=A\exp((-1-i)z\sqrt{\frac{\omega}{2\nu}})$ now
$f(0)=0 \Rightarrow A=U$

$\Rightarrow$ \begin{eqnarray*} u & = & U\exp \left
[(-1-i)z\sqrt{\frac{\omega}{2\nu}}+i\omega t \right ]
\\ & = & U\exp \left ( -z\sqrt{\frac{\omega}{2\nu}} \right ) \exp \left( i \left ( \omega
t-z\sqrt{\frac{\omega}{2\nu}} \right ) \right )
\end{eqnarray*}

%Question 3iii
\item{(iii)}

\begin{center}
\epsfig{file=311-bottomplate2.eps, width=70mm}
\end{center}

(For full marks must show alternation, decay and periodicity)
\end{description}



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