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\textbf{Question}

A viscous fluid has constant density $\rho$ and constant kinematic
viscosity $\nu$. Non-dimensionalise the steady Navier-Stokes equations
with suitable scalings to show that, when the Reynolds number $Re$
satisfies
$$Re \ll 1,$$
the non-dimensional equations of motion become, to leading order, the
``slow-flow'' equations
\begin{eqnarray*}
\nabla p & = & \nabla^2 \un{q}\\
\nabla.\un{q} & = & 0
\end{eqnarray*}
where $p$ and $\un{q}$ denote the fluid pressure and velocity
respectively. Give an example of a fluid mechanics scenario wher such
a ``slow-flow'' approximation would apply.

The fluid is contained between two plates. Cartesian coordinates
$(x,z)$ are used. The plates are situated at $z=0$ and $z=\delta h(x)$
in dimensionless variables where $h(x)$ may be regarded as a given
function and $\delta \ll 1 $. Show how, by further scaling the slow
flow equations according to
\begin{eqnarray*}
z & = & \delta Z\\
w & = & \delta W\\
p & = & \frac{P}{\delta^2},
\end{eqnarray*}
the two-dimensional ``lubrication theory'' equations
\begin{eqnarray*}
P_x & = & u_{ZZ}\\
P_Z & = &0\\
u_x + W_Z & = & 0
\end{eqnarray*}
may be derived. Given the conditojs under which this limit is valid.

The top plate is held fixed, and the bottom plate is moved with a
non-dimensional speed of $1$ in the positive $x$-direction. Show that
the pressure satisfies Reynolds' equation
$$\frac{d}{dx} \left ( \frac{h^3}{6} \frac{dP}{dx} \right ) -
\frac{dh}{dx} = 0$$
and give suitable boundary conditions for this differential
equation. Suppose now that $h(x)=a+bx$ where $a$ and $b$ are
constants. WITHOUT SOLVING the equation comment briefly on whether $b$
should be positive or negative if the top plate is to support a load.

\textbf{Answer}

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

Since slow flow, we set $\underline{x}=L\underline{\overline{x}}$,
$p=(\mu U/L)\overline{p}$,
$\underline{q}=U\overline{\underline{q}}$, $\Rightarrow$

$\left. \begin{array}{c} \displaystyle
\frac{U^2}{L}(\overline{\underline{q}}.\nabla)\overline{\underline{q}}=-\frac{1}{\rho
L} \overline{\nabla}\overline{p}\left ( \frac{\mu U}{L} \right ) +
\frac{\nu U}{L^2} \overline{\nabla}^2 \overline{\underline{q}}\\
\overline{\nabla}.\overline{\underline{q}}=0 \end{array} \right
\}$

Multiplying the momentum equation by $\nu U/L^2$

$\displaystyle \Rightarrow
\frac{LU}{\nu}(\overline{\underline{q}}.\overline{\nabla})\overline{\underline{q}}=
-\overline{\nabla}\overline{p}+
\overline{\nabla}^2\overline{\underline{q}}$, but $\displaystyle
\frac{LU}{\nu}=Re$

Thus $Re(\overline{\underline{q}}.\nabla)\overline{\underline{q}}=
-\overline{\nabla}\overline{p}+
\overline{\nabla}^2\overline{\underline{q}}$, $\Rightarrow$ for
$Re \ll 1$ we get, to lowest order,
$\overline{\nabla}\overline{p}=\overline{\nabla}^2\overline{\underline{q}}$,
$\overline{\nabla}.\overline{\underline{q}}=0$

Any decent flow example will do, eg sperm swimming, treacle
flowing, tar running down a telegraph pole.

\begin{center}
$\begin{array}{c} \epsfig{file=311-flowgraph.eps, width=40mm}
\end{array}
\ \ \
\begin{array}{c}
\rm{We\ have\ (drop\ bars)}\\
\left. \begin{array}{c}
p_x=u_{xx}+u_{zz}\\
p_z=w_{xx}+w_{zz}\\
u_x+w_z=0
\end{array} \right \}
\end{array}$
\end{center}


So further scale $\displaystyle \frac{1}{\delta^2}P$, $w=\delta
W$, $z=\delta Z$.

$\Rightarrow$ $\left. \begin{array}{c} \displaystyle
\frac{1}{\delta^2}P_x=u_{xx}+\frac{1}{\delta^2}U_{ZZ}\\
\displaystyle \frac{1}{\delta}P_Z=\delta
W_{xx}+\frac{1}{\delta}W_{ZZ}\\ u_x+W_Z=0 \end{array} \right \}$

Now to leading order as $\delta\to 0$, we must get $$P_x=u_{ZZ}, \
\ P_Z=0, \ \ u_x+W_Z=0$$

Need $\delta \ll 1$, $\delta^2 Re \ll 1$.

Now the top plate is fixed and the bottom is moved with speed 1.

we have $\displaystyle u_{ZZ}=P_x \Rightarrow
u=\frac{Z^2P_x}{2}+AZ+B$

Now $u=0$ on $Z=h$, $U=1$ on $Z=0$

$\displaystyle \Rightarrow B=1, \ \ o=\frac{h^2P_x}{2}+Ah+1$

and so

$\begin{array}{ccl} u & = & \displaystyle 1+\frac{Z^2
P_x}{2}+Z\left ( -\frac{1}{h}-\frac{hP_x}{2} \right )\\ & = &
\displaystyle \frac{Z(Z-h)P_x}{2}+\left ( 1-\frac{Z}{h} \right )
\end{array}$

$\displaystyle \int_h^0 (u_x+W_Z) \,dZ =0 \Rightarrow \int_0^h u_x
\,dZ =0$ (since $W=0$ on $z=0,h$)

$\displaystyle \Rightarrow \frac{\partial}{\partial x}\int_0^h u
\,dZ = 0$ (since $u=0$ on $Z=h$)

$\displaystyle \frac{d}{dx} \left [ \frac{Z^3}{6}P_x
-\frac{Z^2h}{4}P_x+Z-\frac{Z^2}{2h} \right ]_0^h \Rightarrow
\frac{d}{dx}\left [ -\frac{h^3}{12}P_x+\frac{h}{2} \right ] = 0$


$\displaystyle \Rightarrow \frac{d}{dx} \left ( \frac{h^3}{6}P_x
\right ) - \frac{dh}{dx}=0$

We impose $P=0$ at $x=0,1$

When $h=a+bx$ need $b<0$ for $P>0$ so that fluid is being forced
into a NARROWING gap.

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