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% 16/02/2023 10AM

\begin{enumerate}[(1)]
    \setcounter{enumi}{4}
    \item Add a stationary boundary
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/b90a747aade211ed.png}
        \end{center}
        Compare either timescales
        $\frac{\nu}{h^2}$ and $\frac{1}{\omega}$
        or lengthscales $\left( \frac{\nu}{\omega}
        \right)^{\half}$ and $h$. Then
        dimensionless parameter $S = \frac{\omega
        h^2}{\nu}$ (Stokes number). $s \gg 1$
        looks like (4) with $\delta \sim \left(
        \frac{\nu}{\omega} \right)^{\half} \ll h$,
        $s \gg 1$ looks like Couette flow (linear
        profile) with amplitude $U(t)$.
\end{enumerate}
These examples illustrate that the inviscid
solution ($\nu \equiv 0$, $\implies u = 0$ and
slip at boundary) is \emph{not} uniformly the same
as the limit of ``small'' viscosity (smallness
means $\nu \ll \frac{h^2}{t}$ or $\ll \omega h^2$)
but the viscous effects only felt in thin boundary
layers $\delta \ll h$.

\subsection{The Navier-Stokes Equation}

In Part II it is shown that $\bf{\tau}(\bf{x}, t;
\bf{n}) = -p \bf{n} + \mu [(\bf{n} \cdot \nabla)
\bf{u} + (\nabla \bf{u}) \cdot \bf{n})]$ and hence
the flow satisfies the Navier-Stokes equation
\begin{equation*}
    \boxed{
    \begin{aligned}
        \rho \Dfrac{\bf{u}}{t}
        &= -\nabla p + \mu \nabla^2 \bf{u} +
        \bf{f} \\
        \nabla \cdot \bf{u}
        &= 0
    \end{aligned}
    }
\end{equation*}
This reduces to the Euler equation if $\mu = 0$
and to the parallel viscous flow equations if
$\bf{u} = (u(y, t), 0, 0)$. When is viscosity
important?

\subsubsection*{The Reynolds Number}

Suppose a flow has a characteristic lengthscale
$L$ and velocity scale $U$. For example
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/7d3ee2daade411ed.png}
\end{center}
Assume the characteristic timescale $T$ is
$\frac{L}{U}$ (or steady) and let the
characteristic scale of the pressure difference be
denoted by $P$ (to be found). Estimate the scale
of the terms in the Navier-Stokes Equation:
\begin{center}
\begin{tabular}{ccccccc}
    $\pfrac{\bf{u}}{t}$ & $+$ & $\bf{u} \cdot
    \nabla \bf{u}$ & $=$ & $-\frac{1}{\rho} \nabla
    p$ & $+$ & $\nu \nabla^2 \bf{u}$ \\
    $\sim \frac{U}{L / U}$ & & $\sim
    \frac{U^2}{L}$ & & $\sim \frac{p}{\rho L}$ & &
    $\sim \frac{\nu}{U}L^2$ \\
    $1$ & $:$ & $1$ & $:$ & $\frac{p}{\rho U^2}$ &
    $:$ & $\frac{\nu}{LU} \equiv \frac{1}{R_e}$
\end{tabular}
\end{center}
The Reynolds number $\boxed{R_e \equiv
\frac{UL}{\nu}}$ is a dimensionless parameter
describing the relative importance of inertia and
viscosity:
\[ \frac{\rho \Dfrac{\bf{u}}{t}}{\mu \nabla^2
\bf{u}} \sim R_e \]
If $R_e \ll 1$ then expect $\rho
\Dfrac{\bf{u}}{t} \ll \mu\nabla^2 \bf{u}$, inertia
negligible, and can approximate the N-S equation
by the Stokes equations
\begin{equation*}
    \boxed{
    \begin{aligned}
        0
        &= -\nabla p + \mu \nabla^2 \bf{u} +
        \bf{f} \\
        \nabla \cdot \bf{u}
        &= 0
    \end{aligned}
    }
\end{equation*}
Scaling for pressure $p \sim \frac{\mu U}{L}$ like
viscous stress.

\myskip
If $R_e \gg 1$ expect $\mu \nabla^2 \bf{u} \ll
\rho \Dfrac{\bf{u}}{t}$ and viscosity negligible
(except perhaps in thin boundary layers at rigit
boundaries). Approximate the N-S equation by the
Euler equation $\rho \Dfrac{\bf{u}}{t} = -\nabla p
+ \bf{f}$ ouside the boundary layer. Pressure
scaling $P \sim pU^2$ like Bernoulli.

\myskip
How thin are the boundary layers? For example
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/008b1eb8ade711ed.png}
\end{center}
On timescale $\frac{L}{U}$ viscous diffusion
affects the velocity over a distance $\delta \sim
\left( \nu \frac{L}{U} \right)^{\half}$ or
$\frac{\delta}{L} \sim \left( \frac{\nu}{LU}
\right)^{\half} = \frac{1}{R_e^{\half}}$.

\subsection{Stagnation Point Flow: An Illustrative Example}

Consider the 2D inviscid incompressible flow
$\bf{u} = (Ex, -Ey)$ in $y > 0$, $\psi = Exy$
and $p = p_0 - \half \rho E^2 (x^2 + y^2)$
(Bernoulli)
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/783575d0ade711ed.png}
\end{center}
Solves $\rho (\bf{u} \cdot \nabla) \bf{u} =
-\nabla p$, $v = 0$ on $y = 0$. Also satisfies the
N-S equation, but not the viscous no-slip boundary
condition $u = 0$ if $y = 0$ is a rigid boundary.
Fortunately, and unusually, there is an exact
solution for this case!

\myskip
Try / guess $\psi = Ex f(y)$ $\implies \bf{u} =
(Ex f', -Ef)$. Want $f(0) = f'(0) = 0$ and $f(y)
\approx y$ as $y \to \infty$. Substitute into
$(\bf{u} \cdot \nabla)\bf{u} = -\frac{1}{\rho}
\nabla p + \nu \nabla^2 \bf{u}$, $x$-component:
\begin{align*}
    \left(Ex f' \pfrac{}{x} - E f
    \pfrac{}{y}\right)Ex f'
    &= -\frac{1}{\rho} \pfrac{\phi}{x} + \nu
    \left( \pfrac[2]{}{x} + \pfrac[2]{}{y} \right)
    (Ex f') \\
    \implies E^2 x (f'^2 - ff'')
    &= -\frac{1}{\rho} \pfrac{p}{x} + \nu Exf'''
    \tag{1}
\end{align*}
Similarly, $y$-component:
\[ E^2 f'f = -\frac{1}{\rho} \pfrac{\phi}{y} - \nu
E f'' \tag{2} \]
\[ \pfrac{}{x} (2) \implies \frac{\partial^2
p}{\partial x \partial y} = 0 \implies p = X_{(x)}
+ Y_{(y)} \]
\[ (1) \implies \pfrac{p}{x} \propto x \]
$f' \to 1$ as $y \to \infty$ $\implies X(x) = p_0
- \half \rho E^2 x^2$
\[ \implies f'^2 - ff'' = 1 + \frac{\nu}{E} f''' \]
Rescale $f(y) = \delta F(\eta)$ where $\eta =
\frac{y}{\delta}$ and $\delta = \left(
\frac{\nu}{E} \right)^{\half}$. So $F''' = F'^2 -
FF'' - 1$, $F(0) = F'(0) = 0$, $F' \to 1$ as $\eta
\to \infty$. Solve numerically:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0093cea8ade911ed.png}
\end{center}
For $\eta \gtrsim 2$, $F' \approx 1$, $F \approx
\eta - 0.6$. $y \gtrsim w \sqrt{\frac{\nu}{E}}$,
$u \approx Ex$, $v \approx -E(y - 0.65)$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/2f92a396ade911ed.png}
\end{center}
