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\subsection{The Vorticity Equation for Viscous Flow}

$\nabla \times (\nu \nabla^2 \bf{u}) = \nu
\nabla^2 \bf{w}$! Take curl of Navier-Stokes, and
use the previous derivation from Euler (Section
2.5.2) to obtain
\[ \boxed{\pfrac{\bf{w}}{t} + (\bf{u} \cdot
\nabla) \bf{w} = (\bf{w} \cdot \nabla)\bf{u} + \nu
\nabla^2 \bf{w}} \]
($(\bf{u} \cdot \nabla)\bf{w}$ is advection,
$(\bf{w} \cdot \nabla) \bf{u}$ is vortex
stretching, $\nu\nabla^2 \bf{u}$ is diffusion).

\myskip
For $R_e \gg 1$: Away from rigid boundaries,
vorticity is dominantly advected and amplified by
vortex stretching, and the final term is small. At
rigid boundaries vorticity is generated by the
no-slip boundary condition and diffuses a short
distance to form a boundary layer.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/681d39d8b1d111ed.png}
\end{center}

\myskip
For parallel flow:
\[ \bf{u} = (u(y, t), 0, 0) \implies \bf{\omega} = (0,
0, \omega(y, t)) \]
where $\omega = -\pfrac{\bf{u}}{t}$
\[ \pfrac{\omega}{t} = \nu \pfrac[2]{\omega}{t} \]
diffusion equation.

\newpage
\section{Inviscid Irrotational Flow}

AKA Potential Flow.

\subsection{The Velocity Potential}

In inviscid flow, if $\bf{\omega} = \nabla \times
\bf{u} = \bf{0}$ at $t = 0$ (irrotational flow)
then
\[ \Dfrac{\bf{\omega}}{t} = (\bf{\omega} \cdot
\nabla)\bf{u} \implies \nabla \times \bf{u} = 0
\quad \forall t \ge 0 \]
``irrotational flow remains irrotational'' in
inviscid flow.

\myskip
$\nabla \times \bf{u} = \bf{0}$ implies there
exists a \emph{velocity potential} $\phi(\bf{x},
t)$
\[ \boxed{\bf{u} = \nabla \phi} \]
(for example $\phi = \int^{\bf{x}} \bf{u} \cdot
\dd \bf{x}$). Note:
\begin{enumerate}[(1)]
    \item The $+$ sign (cf $\bf{f} = -\nabla
        \chi$)
    \item Can add any $f(t)$ without changing
        $\bf{u}$.
\end{enumerate}
Incompressibility $\nabla \cdot \bf{u} = 0$
implies
\[ \boxed{\nabla^2 \phi = 0} \]
Kinematic boundary condition $\bf{u} \cdot \bf{n}
= \bf{U} \cdot \bf{n}$ implies
\[ \boxed{\bf{n} \cdot \nabla \phi = \bf{U} \cdot
\bf{n}} \]
So solving the Euler equation for an irrotational
flow reduces to solving the more familiar, and
linear, Laplace's equation with Neumann boundary
conditions (or $\pfrac{\phi}{n}$) -- the solution
is non-zero because of the boundary conditions.
Important examples of irrotational flow are flows
starting from rest or with uniform flow upstream.

\subsection{Examples}

For simple boundaries (spheres, cylinders,
rectangular channel, half-space). We can build
solutions using separable solutions to $\nabla^2
\phi = 0$ in suitable coordinate systems cf
Methods.
\begin{itemize}
    \item Cartesians: for example uniform flow,
        $\bf{u} = \bf{U} \implies \phi = \bf{U}
        \cdot \bf{x} = U_1 x_1 + U_2 y + U_3 z$
        For example
        \[ \phi = Z(z)X(x) \]
        with
        \begin{align*}
            Z(z)
            &= \begin{cases}
                e^{\pm kz} \\
                \cosh kz \\
                \sinh kz
            \end{cases} \\
            X(x)
            &= \begin{cases}
                e^{\pm ikx} \\
                \cos kx \\
                \sin kx
            \end{cases}
        \end{align*}
        so that $\nabla^2 \phi = 0$. Corresponding
        flows are periodic in the $x$-direction,
        for example waves in section 5.
    \item Spherical coordinates: General
        axisymmetric solution to $\nabla^2 \phi =
        0$ in spherical coordinates is
        \[ \phi = \sum_{n \ge 0} (A_n r^n + B_n
        r^{n - 1}) P_n(\cos\theta) \]
        where $P_n$ is a Legendre polynomial. We
        will only need the first few simple modes.
        (for example $\phi = A_0 \implies \bf{u} =
        \bf{0}$) For example $\phi = \frac{B}{r}
        \implies \bf{u} = \nabla \phi =
        -\frac{B}{r^2} \bf{e}_r$ radial flow
        $\propto \frac{1}{r^2}$. Volume fluw
        across any sphere $r = a$ is
        \[ q = \int_{r = a} \bf{u} \cdot \bf{n}
        \dd A = 4\pi A^2 \left( -\frac{B}{a^2}
        \right) = -4\pi B \]
        independent of $a$ (by mass conservation).
        \[ \boxed{\phi = -\frac{q}{4\pi r}} \]
        gives a \emph{point source of strength
        (volume flux) $q$}. ($q < 0$ gives a
        \emph{point sink}). For example $\phi = U
        r\cos\theta = U_2$ implies $\bf{u} =
        \nabla \phi = U \bf{e}_z$ \emph{uniform
        flow again}. Another example $\phi =
        \frac{U\cos\theta}{r^2} \implies \bf{u} =
        \nabla \phi = \cdots \implies$
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/9d993d20b1d411ed.png}
        \end{center}
        Another example: Uniform flow past a
        stationary sphere:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/e6e529a2b1d511ed.png}
        \end{center}
        \[ \begin{cases}
            \nabla^2 \phi = 0 \text{ in $r > a$}
            & \text{irrotational / incompressible}
            \\
            \phi \to Ur\cos\theta \text{ as $r \to
            \infty$} & \text{far-field} \\
            \bf{u} \cdot \bf{n} = \pfrac{\phi}{r}
            = 0 \text{ on $r = a$} &
            \text{kinematic bc}
        \end{cases} \]
        Linear problem, forcing $\alpha \cos
        \theta = P_1 (\cos\theta)$. Try $\phi =
        U\cos\theta \left( r + \frac{B}{r^2}
        \right)$:
        \[ \left. \pfrac{\phi}{r} \right|_{r = a}
        = \left. U\cos\theta \left( 1 - \frac{2B}{r^2}
        \right) \right|_{r = a} \]
        \[ \implies B = \frac{a^3}{2} \]
        \[ \implies \phi = U\cos\theta \left( r +
        \frac{a^3}{2r^2} \right) \]
        \[ \implies \bf{u} = \nabla \phi = \left(
        \pfrac{\phi}{r}, \frac{1}{r}
        \pfrac{\phi}{\theta}, 0 \right) \]
        in sphericals
        \[ = \left( U\cos\theta \left( 1 -
        \frac{a^3}{r^3} \right) , -U\sin\theta
        \left( 1 + \frac{a^3}{2r^3} \right), 0 \right) \]
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/a2a296c0b1d611ed.png}
        \end{center}
    \item Cylindrical  Geometry / 2D Flow: General
        solution to $\nabla^2 \phi = 0$ in plane
        polars is
        \[ \phi = C_0 + A_0 \ln r + B_0\theta +
        \sum_{n \ge 1} (A_n r^n + B_n r^{-n})(C_n
        \cos n\theta + D_n \sin n\theta) \]
        -- again we will only need a few modes.
\end{itemize}
