\documentclass[a4paper,12pt]{article}
\newcommand{\un}{\underline}
\usepackage{epsfig}
\begin{document}
\parindent=0pt

\textbf{Question}

A viscous incompressilbe fluid with constant density $\rho$ and
constant dynamic viscosity $\mu$ flows unsteadily in the
$(x,y)$-plane. There are no body foces. Show that when the Reynolds
number $Re$ is much less than one the flow is governed by the SLOW
FLOW EQUATIONS
\begin{eqnarray*}
\nabla p & = & \mu \nabla^2 \un{q}\\
\textrm{div} \un{q} & = & 0
\end{eqnarray*}
where $p$ and $\un{q}$ denote respectively the pressure and velocity
of the flow.

Show further that, for two-dimensional flow, if a stream function
$\phi (x,y)$ is defined in the normal way, then $\phi$ satisfies the
biharmonic equation
$$\nabla^4 \phi = 0$$

In terms of plane polar coordinates $(r,\theta)$, a wedge of i
ncreasing angle is formed by hinging two infinte plates $\theta = \pm
\Omega t$ at $r=0$. The plates thus move with angular velocities $\pm
\Omega$. The value of $\Omega$ is chosen so that the plates move
slowly apart and slow viscous flow takes place between them. The
velocity of the fluid is denoted by $\un{q} = u \un{e}_r + v
\un{e}_{\theta}$ where $\un{e}_r$ and $\un{e}_{\theta}$ are unit
vectors in the $r$ and $\theta$ directions respectively.

Given that the stream function $\phi (r, \theta)$ may be defined by
\begin{eqnarray*}
u & = & \frac{1}{r}\phi_{\theta}\\
v & = & -\phi_r
\end{eqnarray*}
and that, in spherical polar coordinates
$$\nabla^2 \phi = \phi_{rr} + \frac{1}{r} \phi_r + \frac{1}{r^2}
\phi_{\theta \theta}$$
verify that
$$\phi = r^2 \left [ A_1(t) \sin 2\theta + A_2(t) \theta \right ]$$
is a suitable stream function for the flow, where $A_1(t)$ and
$A_2(t)$ are functions that should be determined. Show further that
the mass flow passing across any arc $r=a$ is independent of time.

\newpage
\textbf{Answer}

Start with unsteady Navier-Stokes.

$\left. \begin{array}{rcl} \displaystyle
\underline{q}_t+(\underline{q}.\nabla)\underline{q} & = &
\displaystyle  \frac{-1}{\rho}\nabla p +
\nu\nabla^2\underline{q}\\ \displaystyle div(\underline{q}) & = &
0 \end{array} \right \}$

Non-dimensionalize by setting
$\underline{x}=L\overline{\underline{x}}$,
$\underline{q}=U\overline{\underline{q}}$, $p=\left ( \frac{\mu
U}{L} \right ) \overline{p}$, where L and U are a representitive
length and speed in the flow. Also set $t=\left ( \frac{L}{U}
\right ) \overline{t}$.

$\Rightarrow \left. \begin{array}{rcl} \displaystyle
\frac{U^2}{L}(\overline{\underline{q}}_{\overline{t}}+(\overline{\underline{q}}.\overline{\nabla})\overline{\underline{q}})
& = & \displaystyle \frac{-\mu U}{\rho L^2}
\overline{\nabla}\overline{p} + \frac{\nu
U}{L^2}\overline{\nabla}^2\overline{\underline{q}} \\
\overline{\nabla}.\overline{\underline{q}} & = & 0 \end{array}
\right \}$

The momentum equation now becomes
$Re[\overline{\underline{q}}_{\overline{t}}+
(\overline{\underline{q}}.\overline{\nabla})\overline{\underline{q}}
] = -\overline{\nabla}\overline{p} + \overline{\nabla}^2
\overline{\underline{q}}$

So for $Re \ll 1$ we have to leading order (re-dimensionalize)

$\left. \begin{array}{l} \nabla p=\mu \nabla^2\underline{q}\\
div(\underline{q})=0 \end{array} \right \}$

Now if we define $u=\psi_y$, $v=-\psi_x$ then
$\div(\underline{q})$.

Also since $curl(grad(p))\equiv 0$, we have
$\nabla^2 curl(\underline{q})$.

Now, $curl(\underline{q}) = \left | \begin{array}{ccc}
\underline{i} & \underline{j} & \underline {k}\\ \displaystyle
\frac{\partial}{\partial x} & \displaystyle
\frac{\partial}{\partial y} & \displaystyle
\frac{\partial}{\partial z}\\ \psi_x & -\psi_y & 0
\end{array} \right | $ $=$ $\left ( \begin{array}{c} 0\\ 0\\
-\psi_{xx} -\psi_{yy} \end{array} \right )$ $=$ $\left (
\begin{array}{c} 0\\ 0\\ -\nabla^2\psi \end{array} \right
)$

$\Rightarrow \nabla^2(-\nabla^2\psi)=0$, so $\psi$ satisfies the
biharmonic equation $$\nabla^4\psi=0$$

\begin{center}
$\begin{array}{c} \epsfig{file=311-cornerflow2.eps, width =30mm}
\end{array}
\ \ \
\begin{array}{c}
\rm{Now\ } \psi=r^2[A_1\sin2\theta+A_2\theta]
\end{array} $
\end{center}

\begin{eqnarray*}
\nabla^2\psi & = &
2[A_1\sin2\theta+A_2\theta]+[A_1\sin2\theta+A_2\theta]+[-4A_1\sin2\theta]\\
& = & 4A_2\theta
\end{eqnarray*}
But now $\nabla^2(4A_2\theta0=0$.

So certainly the given $\psi$ satisfies the biharmonic equation.

Boundary conditions, (symmetric so need only look at
$\theta=\Omega t$)

At $\theta = \Omega t$ the plate velocity is
$0\tilde{\underline{e}}_r + \Omega\tilde{\underline{e}}_{\theta}r$

$\Rightarrow$ we need $$u=0,\ \ v=r\Omega \ \ \rm{at} \
\theta=\Omega t$$

Thus $\psi_{\theta}=0$, $\psi_{r}=-r\Omega$ at $\theta=\Omega t$

$\Rightarrow$ \begin{eqnarray*} r^2[A_1(t)2\cos 2\Omega t+A_2(t)]
& = & 0\\ 2r[A_1(t)\sin 2\Omega t +A_2\Omega t] & = & -r\Omega
\end{eqnarray*}

Solving these $\Rightarrow$ \begin{eqnarray*} \displaystyle A_1(t)
& = & \frac{-\Omega}{2[\sin 2\Omega t -2\Omega t\cos 2\Omega t]}\\
A_2(t) & = & \frac{\Omega\cos 2\Omega t}{[\sin 2\Omega t -2\Omega
t\cos 2\Omega t]} \end{eqnarray*}

$\displaystyle \psi=\frac{r^2}{(\sin 2\Omega t-2\Omega t\cos
2\Omega t)}\left [ \frac{-\sin 2\theta}{2}+\theta\cos 2\Omega t
\right ]$

Mass flow
\begin{eqnarray*}
\rho\int_{\theta=\Omega t}^{\Omega t}
ur\,d{\theta} & = & \rho\int_{-\Omega t}^{\Omega t} \psi_{\theta}
\,d{\theta}\\
& = & \rho [\psi (r_1,\Omega t) -\psi (r_1,-\Omega t)]
\end{eqnarray*}

\begin{eqnarray*}
& = & \frac{\rho r^2\Omega}{(\sin 2\Omega t-2\Omega
t\cos 2\Omega t)}\left [ \frac{-\sin 2\Omega t}{2} +\Omega t\cos
2\Omega t \right.\\
& & \left. -\frac{\sin 2\Omega t}{2}+\Omega t\cos 2\Omega 4 \right
]\\
& = & -\rho r^2 \Omega
\end{eqnarray*}
$\Rightarrow$ independent of t.


\end{document}