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\begin{itemize}
    \item[] For example $\phi = \frac{q}{2\pi} \ln
        r$ implies $\bf{u} = \frac{q}{2\pi r}
        \bf{e}_r$, radial flow $u_r \propto
        \frac{1}{r}$. Flux across any circle $r =
        a$ is $2\pi a u_r$ = q (indepdendent of
        $a$). Called a \emph{2D point source} of strength
        $q$ (line source in 3D). For example $\phi
        = \frac{\kappa}{2\pi}\theta$ implies
        $\bf{u} = \nabla \phi = \frac{\kappa}{2\pi
        r}\bf{e}_\theta$ circular flow $\propto
        \frac{1}{r}$. Circulation $\oint \bf{u}
        \dd \bf{dl}$ around a circle $r = a$ is
        $2\pi a \frac{\kappa}{2\pi a} = \kappa$
        independent of $a$ -- see section 2.5. A
        \emph{point vortex} of circulation
        (strength) $\kappa$. (line vortex in 3D).
        \\
        eg $\phi = U r\cos\theta$ uniform flow
        again \\
        eg $\phi = \frac{U\cos\theta}{r}$ 2D
        dipole \\
        eg Uniform flow past a cylinder with
        circulation $\kappa$:
        \[ \nabla^2 \phi = 0 \text{ in $r > a$}
        \qquad \phi \to Ur\cos\theta \text{ as $r
        \to \infty$} \]
        \[ \dfrac{\phi}{r} = 0 \text{ on $r = 0$}
        \qquad \oint \bf{u} \cdot \bf{dl} =
        [\phi]_{r = a} = \kappa \]
        -- condition needed to get a unique
        solution.
        \[ \implies \phi = U\cos\theta \left( r +
        \frac{a^2}{r} \right) +
        \frac{\kappa}{2\pi} \theta \]
        \[ \implies \bf{u} = \nabla \phi = \left(
        \pfrac{\phi}{r}, \frac{1}{r}
        \pfrac{\phi}{\theta} \right) \]
        \[ = \left( U\cos\theta \left( 1 -
        \frac{a^2}{r^2} \right), -U\sin\theta
        \left( 1 + \frac{a^2}{r^2} \right) +
        \frac{\kappa}{2\pi r} \right) \]
        \begin{center}
            \includegraphics[width=0.8\linewidth]
            {images/4c3887e2b36411ed.png}
        \end{center}
        For the right diagram: this is like a
        tennis ball with top spin (consider change
        of reference frame).
\end{itemize}

\subsection{Pressure in Potential Flow with Potential Forces}

Non-linear equation of motion has been reduced to
linear Laplace for the kinematics. But still have
non-linearity in the dynamic boundary condition
(pressure). Momentum equation:
\[ \rho \left( \pfrac{\bf{u}}{t} + (\bf{u} \cdot
\nabla)\bf{u} \right) = -\nabla p + \bf{f} \]
Potential forces $\bf{f} = -\nabla \chi$. Previous
identity:
\[ (\bf{u} \cdot \nabla)\bf{u} = \nabla \left(
\half u^2 \right) - \bf{u} \times \bf{\omega} \]
($\bf{u} \times \bf{\omega}$ is zero because
irrotational flow). Potential flow
\[ \pfrac{\bf{u}}{t} = \nabla \pfrac{\phi}{t} \]
Thus the momentum equation reduces to
\[ \nabla \left( \rho \pfrac{\phi}{t} + \half \rho
u^2 + p + \chi \right) = 0 \]
\[ \implies \boxed{\rho \pfrac{\phi}{t} + \half
\rho u^2 + p + \chi = f(t), \qquad
\text{independent of $\bf{x}$}} \]
``unsteady Bernoulli''. Notes:
\begin{enumerate}[(1)]
    \item $f(t)$ is actually irrelevant, since we
        can add arbitrary $g(t)$ to $\phi$. -- its
        the independence of $\bf{x}$ that matters.
    \item Not the same as steady Bernoulli! -- see
        handout
    \item For steady, irrotational flow, $H$ is
        constant everywhere.
\end{enumerate}

\subsection{Force on Translating Sphere and Cylinder}

\subsubsection{Steadym Translating Sphere}

For steady motion, it is most convenient to use
the frame of reference moving with the sphere
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/531e45d0b36711ed.png}
\end{center}
From section 4.2, uniform flow past stationary
sphere is given by
\[ \phi = U \cos\theta \left( r + \frac{a^3}{2r^2} \right) \]
and on $r = a$, $u_r = 0$, $u_\theta =
-\frac{3}{2} U\sin\theta$. Apply either form of
Bernoulli: ($\pfrac{\phi}{t} = 0$, $\chi = 0$) to
obtain
\[ \half \rho \left( -\frac{3}{2} U\sin\theta
\right)^2 + p(a, \theta) = \half \rho U^2 +
p_{\infty} \]
i.e. $p(a, \theta) = p_\infty + \half \rho U^2
\left( 1 - \frac{a}{4} \sin^2 \theta \right)$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/6d3c8404b36711ed.png}
\end{center}
Pressure distribution is symmetric force-aft and
round equator. $\implies$ The net force on the
sphere is zero!

\myskip
This surprising result is \emph{d'Alembert's
paradox}. In fact, there is no drag ($\equiv$ the
net force parallel to the motion) on any steadily
moving body in unbounded potential flow: KE is
constant, we've neglected viscosity / friction. So
conservation of energy gives no work done.

\begin{note*}[Non-examinable]
    effect of viscosity on translating sphere. See
    experimentally:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/49c57232b36811ed.png}
    \end{center}
    and find $F = 0.4 \half \rho U^2 (\pi a^2)$ --
    see handout.
\end{note*}

\noindent
However, potential flow is good slippery bubbles,
for rapid acceleration of a rigid particle, or
small amplitude oscillations.
