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\textbf{Question}

Viscous liquid of constant density $\rho$ and constant kinematic
viscosity $\nu$ is at rest in the region $o \le y \le h$ between two
rigid parallel plates. There are no body forces. At time $t=0$ the top
plate is set into motion parallel to its own plane with speed $U$ in
the direction of the $x$-axis and is maintained at this speed
thereafter. The plate at $y=0$ is held fixed and the is no applied
pressure gradient. Show that a flow solution of the form $\un{q}(x,t)
= (u(y,t), 0, 0)$ is possible provided $u$ satisfies
$$u_t = \nu u_{yy}$$
and give suitable boundary and initial conditions for this equation.


Using seperation of variables, or otherwise, show that a solution to
the governing partial differential equation is
$$u = C_1y +C_2 + \sum_{n=1}^{\infty} e^{-k_n^2t} \left ( A_n \sin
\frac{k_n}{\sqrt{\nu}} y +B_n \cos \frac{k_n}{\sqrt{\nu}} y \right )$$
where $A_n$, $B_n$, $C_1$, $C_2$ and $k_n$ are constants. By futher
imposing the boundary conditions, show that the solution for the flow
is given by
$$ u = \frac{Uy}{h} + \frac{2U}{\pi} \sum_{n=1}^{\infty}
\frac{(-1)^n}{n} \sin \left ( \frac{n\pi y}{h} \right ) \exp \left ( -
\frac{n^2 \pi^2 \nu t}{h^2} \right )$$

Explain briefly what you would expect the flow to look like for a very
viscous fluid.

[You may use, without proof, the fact that the Fourier sine series
representation of the function $\xi$ for $\xi \in [0,1]$ is given by
$$\xi = \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n\pi}\sin (n\pi\xi).]$$

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\textbf{Answer}

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Choose $\underline{q}=((u(y,t)),0,0)$ then $div(\underline{q})=0$

Navier-Stokes equations become:
\begin{eqnarray*} u_t+0 & = & -p_x/\rho + \nu (u_{xx}+ u_{yy}+
u_{zz})\\ 0 & = & p_y/\rho +0\\ 0 & = & -p_z/\rho + 0
\end{eqnarray*}

So since we are told that there are no pressure gradients
$$u_t=\nu u_{yy}$$

\begin{tabular}{r} Initial conditions:-\\ Boundary condtions:-\\
Also by no slip
\end{tabular}
$\begin{array}{l} u(y,0)=0\\ u(h)=0, \ (t<0), \ \ u(h)=U \ (t \ge
0)\\ u(0)=0
\end{array}$

Now to use seperation of variables, set $u=Y(y)T(t)$.

Then
\begin{eqnarray*}
YT' & = & \nu TY''\\ \Rightarrow T'/T & = & \nu Y''/Y
\end{eqnarray*}

By the standard separation of variables argument both sides must
be either a constant or zero. Thus either

$\nu Y''/Y=0 \ \ \Rightarrow Y=C_1y+C_2, \ \ T=constant$

or

$T'/T=-k^2$ (choose constant -ve so solutions don't grow at
$t=\infty$)

$$\Rightarrow T'+k^2t=0,\ \ \ T=Ae^{-k^2t}$$

Also $\displaystyle Y''+\frac{k^2}{\nu}Y=0 \ \ \Rightarrow \ \
Y=B\cos\frac{k}{\sqrt{\nu}}y+C\sin\frac{k}{\nu}y$.

Since the equation is linear, solutions may be added.
$$\displaystyle \Rightarrow u=C_1y+C_2+\sum_{n=1}^{\infty}
e^{-k_n^2t} \left ( A_n\sin\frac{k_n}{\sqrt{\nu}}y
+B_n\cos\frac{k_n}{\sqrt{\nu}}y \right) $$

(the term $n=0$ just gives 0 and constant-see later).

Now we have to impose the boundary conditions:-

$\begin{array}{rcl} u(0)=0 & \Rightarrow &\\ & & C_2=0, B_n=0 \ \
\forall n \end{array}$

Thus $$\displaystyle u=C_1y+\sum_{n=1}^{\infty} A_n
e^{-k_n^2t}\sin \left ( \frac{k_n}{\sqrt{\nu}}y \right ).£$$

Now the only way to have $u(y)=h \ \ \forall t \ge 0$ is to have
\begin{eqnarray*} \displaystyle c_1 & = &\frac{U}{h} \ \ \rm{and} \\ \displaystyle
\sin \left ( \frac{k_n}{\sqrt{\nu}} h \right ) & = & 0, \ \ \
\frac{k_nh}{\sqrt{\nu}} = n\pi \ \ \ (n \in \textbf{Z})
\end{eqnarray*}

\begin{eqnarray*} \Rightarrow k_n & = & \sqrt{\nu}n\pi /h \ \ (n \in
\textbf{Z})\\ u & = & \displaystyle \frac{Uy}{h}+
\sum_{n=1}^{\infty} A_n \sin \left ( \frac{n\pi y}{h} \right )
\exp \left ( -\frac{\nu n^2\pi^2}{h^2}r \right ) \end{eqnarray*}

Finally we need $u=0 \ \ \forall y$ at $t=0$.

$\displaystyle \Rightarrow 0 = \frac{Uy}{h} +\sum_{n=1}^{\infty}
a_n\sin \left (\frac{n\pi y}{h} \right )$.

From the result given in the question, for $\in [0,h]$ (set
$x=frac{y}{h}$) $$\displaystyle \frac{y}{h} = \sum_{n=1}^{\infty}
\frac{2(-1)^{n+1}}{n\pi}\sin \left (\frac{n\pi y}{h} \right) $$

\begin{eqnarray*} \displaystyle \Rightarrow A+n & = &
-\frac{u}{n\pi}2(-1)^{n+1}\\ u & = & \displaystyle \frac{Uy}{h} +
\sum_{n=1}^{\infty} \frac{U}{n\pi}2(-1)^n \sin\left (\frac{n\pi
y}{h} \right ) \exp \left (-\frac{n^2\pi^2 \nu t}{h^2} \right )
\end{eqnarray*}

When $\nu$ is very large, the exponential terms would be very
small for all but the smallest t. Thus for a very viscous fluid we
would expect $$u\sim\frac{Uy}{h} \ \ \rm{after\ a\ very\ short\
time}.$$

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