% vim: tw=50
% 14/02/2023 10AM

\subsubsection*{Boundary Conditions at an Interface}

Consider two fluids in parallel viscous flow
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/6d3a397eac5011ed.png}
\end{center}
At $y = h$, $u_1 = u_2$ by no-slip. By continuity
of stress,
\begin{align*}
    \mu_1 \pfrac{u_1}{y}
    &= \mu_2 \pfrac{u_2}{y} \\
    p_1
    &= p_2
\end{align*}

\subsection{Unsteady Parallel Viscous Flow and
Viscous Diffusion}

\begin{example*}[Impulsively started flat plate]
    Consider a semi-infinite fluid domain $y > 0$,
    initially at rest, no applied pressure
    gradient and $\bf{f} = \bf{0}$. At $t = 0$,
    the boundary at $y = 0$ starts to move with
    velocity $(U, 0)$
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/c8804a1cac5011ed.png}
    \end{center}
    \begin{align*}
        \rho \pfrac{u}{t}
        &= -\ub{\pfrac{p}{x}}_{0} + \mu
        \pfrac[2]{u}{y} + \ub{f_x}_{0} \\
        u
        &= 0
        &&(t = 0, y \to \infty) \\
        u
        &= U
        &&\text{at } y = 0, t > 0 (\text{no-slip})
    \end{align*}
    Hence the velocity satisfies the
    \emph{diffusion equation}
    \[ \pfrac{u}{t} = \nu \pfrac[2]{u}{y} \]
    where the kenimatic velocity $\nu =
    \frac{\mu}{\rho}$ can be thought of as a
    diffusivity for momentum (or vorticity -- see
    later).

    \myskip
    For other solutions of the diffusion equation
    see Methods (for example separation of
    variables, Fourier Transforms, Green's
    functions etc) and Sheet 2 Questions 4 and 5.
    Here there is no externally imposed length
    scale for $y$ since $0 < y < \infty$. But we
    need a lengthscale! Can find a similarity
    solution using a scaling argument (see IA DEs,
    IA D\&R).
    \[ \pfrac{u}{t} = \nu \pfrac[2]{u}{y} \]
    suggests $\frac{u}{t} \sim \nu \frac{u}{y^2}$.
    $u(0) = U$ suggests $u \sim U$. Hence $y \sim
    (\nu t)^{\half}$.
    Try $u = U f(\eta)$ where $\eta =
    \frac{y}{(\eta t)^{\half}} \implies
    \pfrac{\eta}{t} = -\half \frac{\eta}{t}$
    \[ \implies -\half \eta f' = f'' \implies
    -\half \eta = \frac{f''}{f'} \]
    \[ \implies -\frac{1}{4} \eta^2 = \ln f' +
    \text{constant} \implies f = A + B
    \int_\eta^\infty e^{-\frac{1}{4} \eta'^2} \dd \eta' \]
    $f(\infty) = 0$, $f(0) = 1$ implies $A = )$,
    $B = \frac{1}{\sqrt{\pi}}$ hence $u = U \erfc
    \left( \frac{y}{2\sqrt{\nu t}} \right)$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/fe8711d4ac5211ed.png}
    \end{center}
    same shape -- self-similar -- but with width
    $\uparrow$ as $(\nu t)^{\half}$.
\end{example*}

\subsubsection*{Comments and Variations}

\begin{enumerate}[(1)]
    \item Pure dimensional analysis would also
        allow $\frac{y}{ut}$ or $\frac{yU}{\nu}$
        as dimensionless groups, but the equations
        show these are not relevant -- the
        equations and scaling arguments are
        better!
    \item Kinematic and dynamic viscosity
        \begin{center}
        \begin{tabular}{c|c|c|c}
            & $\rho$ & $\mu$ & $\nu$ \\
            \hline
            Water &
            $10^3 \mathsf{kgm^{-3}}$ &
            $10^{-3} p_{\text{as}}$ &
            $10^{-6} \mathsf{m^2 s^{-1}}$ \\
            Water &
            $10^3 \mathsf{kgm^{-3}}$ &
            $10^{-3} p_{\text{as}}$ &
            $10^{-6} \mathsf{m^2 s^{-1}}$ \\
            Water &
            $10^3 \mathsf{kgm^{-3}}$ &
            $10^{-3} p_{\text{as}}$ &
            $10^{-6} \mathsf{m^2 s^{-1}}$ \\
        \end{tabular}
        \end{center}
        \begin{enumerate}[(a)]
            \item $\nu_{\text{air}} \approx
                20\nu_{\text{water}}$ so motion
                spreads further / faster in air.
            \item Sheer stress exerted by fluid on
                the plate
                \[ \tau_s = \mu \pfrac{u}{y} =
                -\frac{\mu U}{(\nu t)^{\half}}
                f'(0) = -\frac{\mu Uj}{\sqrt{\pi
                \nu t}} \]
                \[ \left. \frac{\mu}{\sqrt{\nu}}
                \right|_{\text{water}} \approx 200
                \left. \frac{\mu}{\sqrt{\nu}}
                \right|_{\text{air}} \]
                so water exerts a much bigger
                stress.
        \end{enumerate}
    \item Now add a stationary boundary at $y = h$
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/5efbd2f0ac5511ed.png}
        \end{center}
        Have 2 relevant length scales $h$ and
        $(\nu t)^{\half}$. If $h \gg (\nu
        t)^{\half}$ expect boundary to have little
        (exponentially small) effect on the
        previous solution. As $t \to \infty$
        expect to approach steady Couette flow
        (linear profile). Deviations from this
        steady state decay like $e^{-kt}$ with $k
        \propto \frac{\nu}{h^2}$ and are small for
        $\nu t \gg h^2$ (Example Sheet 2, Question
        5) $\implies$ Characteristic timescale $T
        = \frac{h^2}{\nu}$ for diffusion accross
        the cell. For example $h = 10\mathsf{cm}$,
        $T = 0.07\mathsf{s}$ for golden syrup, but
        $3\mathsf{hrs}$ for water. $t \ll T$
        effects of viscosity ar econfined to a
        ``boundary layers'' $\delta \sim (\nu
        t)^{\half}$. $t \gg T$ almost steady,
        viscosity dominant everywhere.
    \item Simple oscillating boundary (Example
        Sheet 2 Question 4)
        \begin{center}
            \includegraphics[width=0.4\linewidth]
            {images/b3835c24ac5711ed.png}
        \end{center}
        Imposed timescale $\frac{1}{\omega}$
        during which velocity variation can
        diffuse $\left( \frac{\nu}{m}
        \right)^{\half}$
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/d08e7b5aac5711ed.png}
        \end{center}
\end{enumerate}
