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\title{Small particles in a viscous fluid}
\author{John Hinch}

\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \titlepage
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Small particles in a viscous fluid}

\blue{Course in three parts}

\vspace{20pt}

\blue{1. A quick course in micro-hydrodynamics}

\vspace{15pt}

\blue{2. Sedimentation of particles}

\vspace{15pt}

\blue{3. Rheology of suspensions}

\vspace{20pt}
Good textbook for parts 1 \& 2: 

{\em A Physical Introduction to Suspension Dynamics} \break
by Elisabeth Guazzelli, Jeffrey F. Morris and Sylvie Pic \break
(Cambridge Texts in Applied Mathematics 2012).

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{A quick course in micro-hydrodynamics}
 \tableofcontents[hideallsubsections]
\end{frame}

\section{Stokes equations}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{A quick course in micro-hydrodynamics}
 \tableofcontents[currentsection,hideothersubsections]
\end{frame}

\subsection{Continuum mechanics}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Continuum mechanics}

Continuum description: mass density $\rho(\bfx,t)$, velocity $\bfu(\bfx,t)$, etc.

\rightline{\smallgray{Eulerian, not Lagrangian}}

\pause
\bigskip
Mass conservation
\[
 \pd{\rho}t + \nabla\cd(\rho\bfu) = 0
\]
\pause
But for constant density, flows are incompressible
\[ 
 \nabla\cd\bfu = 0.
\]

\pause
\bigskip
Forces in continuum description:
\centerline{ volume forces $\bfF$ and surface tractions $\sigma_{ij}n_j$. }

\pause
\medskip
So Cauchy momentum equation
\[
 \rho\left(\pd{\bfu}t + \bfu\cd\nabla \bfu\right) = \nabla\cd\bsi + \bfF
\]

\end{frame}

\subsection{Navier-Stokes equation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
  \frametitle{Navier-Stokes equation}

Newtonian viscous fluids
\[
 \sigma_{ij} = -p\delta_{ij} + 2\mu e_{ij}, 
\]
with  pressure $p(\bfx,t)$ determined globally by incompressibility 
\break\smallgray{rather than locally by an equation of state $p(\rho)$}

\pause
and strain-rate
\[
 e_{ij} = \tfrac12\left(\pd{u_i}{x_j} + \pd{u_j}{x_i} \right)
\]
\gray{Note this is the most general relationship between $\sigma$ and $\nabla\bfu$ which is linear, instantaneous and isotropic.}

\pause
\bigskip
Hence the Navier-Stokes equation (momentum for a Newtonian viscous fluid)
\smallgray{assuming $\mu$ constant}.
\[
  \rho\left(\pd{\bfu}t + \bfu\cd\nabla \bfu\right) 
    = -\nabla p + \mu\nabla^2\bfu + \bfF
\]

\pause
\medskip
Boundary conditions
\[
 \bfu(\bfx) \qword{given, or} \bsi\cd\bfn(\bfx) \qword{given}
\]

\end{frame}

\subsection{Small Reynolds number}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Small Reynolds number}

For a flow $U$ over distances $L$, the Reynolds number is
\[
 Re = \frac{\vert \rho \bfu\cd\nabla\bfu \vert}{\vert\mu\nabla^2\bfu\vert}
  \pause  = \frac{\rho U^2/L}{\mu U /L^2}
   \pause = \frac{UL}{\nu}
\]
where $\nu = \mu/\rho$ is the kinematic viscosity.

%\bigskip
%Also for rapidly oscillating flows $Re_2=\omega L^2/\nu$, and in thin films
%$Re_3=Uh^2/\nu L$.

\pause
\medskip
Small Reynolds number, $Re\ll1$ if \pause
\begin{itemize}
 \item small $U$, e.g. $1\,{\rm cm/day}$ in oil reservoirs, \pause
 \item small $L$, e.g. $10\,\mu{\rm m}$ bacteria, \pause
 \item large $\nu$, e.g. $10^8\,{\rm m^2/s}$ molten glass \pause
 \item e.g. $1\,\mu{\rm m}$ water droplet falls under gravity in air at 
$0.1\,{\rm mm/s}$, so $Re=10^{-5}$. 
\end{itemize}

\end{frame}

\subsection{Stokes equations}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Stokes equations}

If $Re\ll1$, then Stokes flow (also called ``creeping flow'')
\begin{gather*}
 0  = -\nabla p + \mu\nabla^2\bfu + \bfF \\
 \text{with}\quad
    \nabla\cd\bfu = 0.
\end{gather*}

\pause
\bigskip
Note: Stokes theory for $Re\ll1$ usually works for $Re<2$.

\end{frame}

\section{Simple properties}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{A quick course in micro-hydrodynamics}
 \tableofcontents[currentsection,hideothersubsections]
\end{frame}

\subsection{Linear and instantaneous}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Simple properties}

1. \blue{Linear} and \blue{Instantaneous}

Dropping nonlinear $\bfu\cd\nabla\bfu$
and time-dependent $\partial\bfu/\partial t$ 
leaves Stokes equations linear and instantaneous.

\pause
\bigskip
E.g. Rigid particle translation at $\bfU(t)$ in unbounded fluid

Flow $\bfu(\bfx,t)$ \blue{linear} \& \blue{instantaneous} in $\bfU(t)$, 
also $\bsi(\bfx,t)$, hence drag force
\[
 \bfF(t) = {\bf A}\cd\bfU(t)
\]
with ${\bf A}$ depending on size, shape, orientation and viscosity.

\pause
\subsection{Reversible in time}
\bigskip
2. \blue{Reversible in time}

Apply force $\bfF(t)$ in $0\le t\le t_1$. 

Now reverse force and its history, i.e. 
$\bfF(t)= -\bfF(2t_1-t)$ in $t_1\le t\le 2t_1$

\pause
Then flow $\bfu(\bfx,t)$ and its history reverses.

\pause
Hence all fluid particles return to starting position. 

\pause
\medskip
Hence cannot swim at $Re\ll1$ by reversible flapping.

\end{frame}
\subsection{Reversible in space}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Simple properties}

3. \blue{Reversible in space}:

Linearity $+$ symmetry of geometry $\Rightarrow$ certain parts of $\bfu$ vanish.

\pause
\medskip
E.g. 1. A sphere sedimenting next to a vertical wall does not migrate towards or away from the wall at $Re\ll1$.

\pause
\medskip
E.g. 2. Two equal spheres fall without separating. \pause (Spin?)

\pause
\medskip
E.g. 3. An ellipsoid \gray{(particle with three perpendicular planes of symmetry)} falls under gravity without rotation in an unbounded flow.

\pause
\medskip
E.g. 4. Two rigid spheres in a shear flow \gray{(possibly unequal, possibly next to a rigid wall)} resume their original undisturbed streamlines after a collision.

\end{frame}
\section{Flow past a sphere}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{A quick course in micro-hydrodynamics}
 \tableofcontents[currentsection,hideothersubsections]
\end{frame}

\subsection{The solution}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Flow past a sphere}

Uniform flow $\bfU$ past a rigid sphere of radius $a$.

\pause
\bigskip
1. \blue{The solution} -- more important than derivation \pause
\[
 \bfu  =  \bfU \left(1 - \frac{3a}{4r} - \frac{a^3}{4r^3} \right)
    + \bfx (\bfU\cd\bfx)\left(-\frac{3a}{4r^3} + \frac{3a^3}{4r^5}\right), 
\]
\[
 p  = - \frac{3a\mu\bfU\cd\bfx}{2r^3} \qword{and}
  \bsi \cd\bfn\vert_{r=a}  = \frac{3\mu}{2a}\bfU .
\]

\pause
\bigskip
Hence the Stokes drag on the sphere is
\[
\int_{r=a} \bsi \cd\bfn \,dS =
 4\pi a^2 \, \frac{3\mu}{2a}\bfU = 6\pi\mu a\bfU .
\]

\end{frame}
\subsection{Method 1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Flow past a sphere}

2. \blue{Solution method 1}

The linearity of the Stokes equations means that $\bfu(\bfx)$ must be
linear in $\bfU$.

\pause
Further, the problem has spherical symmetry about the centre of the
sphere, which  take as the origin. 

\pause
The velocity and pressure fields must therefore take the forms
\begin{eqnarray*}
 \bfu(\bfx) & = & \bfU f(r) + \bfx (\bfU\cd\bfx) g(r), \\
 p(\bfx) & = & \mu (\bfU\cd\bfx) h(r),
\end{eqnarray*}
where $r = |\bfx|$, and $f$, $g$ and $h$ are functions of scalar $r$ to
be determined.

\pause
\bigskip
Now
\[
 \pd{u_i}{x_j} = U_ix_j f'/r +  \delta_{ij}U_nx_n g +
   x_iU_j g + x_ix_jU_nx_n g'/r.
\]

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Flow past a sphere}
 \framesubtitle{solution method 1}

Contracting $i$ with $j$, we have the incompressibility condition
\[
 0 = \nabla\cd\bfu = U_nx_n(f'/r + 4g + rg') .
\]
\pause
Differentiating again for momentum equation
\begin{eqnarray*}
 \mu\nabla^2 u_i & = & \mu U_i\left(f''+ 2f'/r + 2g\right) +
         \mu x_iU_nx_n\left(g'' + 6g'/r\right) \\
 \nabla_ip & = & \mu U_i h + \mu x_iU_nx_n h'/r
\end{eqnarray*}

\pause
\medskip
Hence the governing equations give
\[
 f'/r + 4g + rg' = 0,\quad
 f''+ 2f'/r + 2g = h \qword{and}
 g'' + 6g'/r = h'/r .
\]
\pause
Eliminating $h$ and then $f$ yields
\[
 r^2g''' + 11rg'' + 24 g' = 0.
\]
\pause
Solutions of the form $g = r^\alpha$. 

\pause
\medskip
Substituting, one finds $\alpha =
0$, $-3$ and $-5$, with associated 
$f = -(\alpha+4)r^{\alpha+2}/(\alpha+2)$ and
$h = -(\alpha+5)(\alpha +2) r^\alpha$.

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Flow past a sphere}
 \framesubtitle{solution method 1}

Hence the general solution of the assumed form linear in $\bfu$ is
\begin{align*}
 \bfu(\bfx) & =  
   \bfU \left(-2Ar^2 + B + Cr^{-1} -\tfrac13Dr^{-3} \right) + 
 \bfx (\bfU\cd\bfx)\left( A + Cr^{-3} + Dr^{-5} \right), \\
 p(\bfx) & =  \mu (\bfU\cd\bfx) \left(-10A + 2Cr^{-3}  \right).
\end{align*}
\pause
We shall need the stress exerted across a spherical surface with
unit normal $\bfn = \bfx/r$
\[
 \bsi \cd\bfn = 
  \bfU \left(-3Ar + 2Dr^{-4} \right) + 
 \bfx (\bfU\cd\bfx)\left( 9Ar^{-1} - 6Cr^{-4} - 6Dr^{-6} \right)
\]

\pause
\medskip
Applying the boundary conditions on the rigid sphere and for the far
field, we find the coefficients
\[
 A = 0, \qquad B=1, \qquad C = -\tfrac34a \qword{and} D = \tfrac34a^3
\]

\medskip
\blue{For solution given earlier}

\end{frame}
\subsection{Method 2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Flow past a sphere}

2. \blue{Solution method 2}

\bigskip
Use a Stokes streamfunction for the axisymmetric flow
\[
 u_r = \frac{1}{r^2\sin\theta}\pd\Psi\theta \qword{and}
 u_\theta = - \frac{1}{r\sin\theta}\pd\Psi{r} .
\]

\pause
\medskip
The vorticity equation (curl of the momentum equation, to eliminate
the pressure) is then at low Reynolds numbers
\[
 {\cal D}^2 {\cal D}^2 \Psi = 0 \qword{where}
 {\cal D}^2 = \pd{^2}{r^2} + \frac{\sin\theta}{ r^2}\pd{}{\theta}
   \left(\frac{1}{\sin\theta}\pd{}{\theta} \right) .
\]

\pause
\medskip
The uniform flow at infinity has $\Psi = \frac12Ur^2\sin^2\theta$, 
so one tries $\Psi = F(r)\sin^2\theta$, 
and finds $F = Ar^4 + Br^2 + Cr + D/r$.

\end{frame}
\subsection{Method 3}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Flow past a sphere}

3. \blue{Solution method 3}

\bigskip
One can show (Papkovich-Neuber) that the general solution of the 
Stokes equation can be
expressed in terms of a vector harmonic function $\bph(\bfx)$
(i.e.~$\nabla^2 \bph = 0$) 
\[
\bfu = 2\bph - \nabla(\bfx\cd\bph) \quad p = - 2\mu\nabla\cd\bph . 
\]
\[
 \sigma_{ij} = 2\mu 
  \left(\delta_{ij}\pd{\phi_n}{x_n} - x_k\pd{^2\phi_k}{x_i \partial x_j}\right).
\]

\pause
\medskip
Linearity and spherical symmetry then give
\[
 \bph = A\bfU\frac{1}{r} + B \bfU\cd\nabla \nabla \frac{1}{r} ,
\]
with coefficients $A$ and $B$ to be determined by applying the
boundary conditions.

\end{frame}

\subsection{Method 4}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Flow past a sphere}

4. \blue{Solution method 4}

\bigskip
The pressure and vorticity are harmonic functions.

\medskip
Using linearity and spherical symmetry, they must take the form
\[
 p = \mu A \bfU\cd\bfx/r^3 \qword{and}
 \nabla\wedge\bfu = B \bfU\wedge\bfx/r^3 .
\]

\medskip
The final step to $\bfu$ is tedious.

\end{frame}

\subsection{Sedimenting sphere}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Sedimentation of a rigid sphere}

Force balance, with densities $\rho_s$ of sphere and $\rho_f$ of fluid
\begin{align*}
   0 &\quad =& \rho_s\tfrac{4\pi}3 a^3g &\quad- &\rho_f\tfrac{4\pi}3 a^3g &\quad -& 6\pi\mu a\bfU \\
 \text{no inertia} & & \text{weight} && \text{buoyancy} & & \text{Stokes drag}
\end{align*}

\pause
\bigskip
So Stokes settling velocity
\[
 \bfU = \frac{2\Delta\rho a^2g}{9\mu}
\]

\pause
\bigskip
E.g. $1\,\mu{\rm m}$ sphere, $\Delta\rho=10^3\,{\rm kg\,m^{-3}}$, 
water $\mu = 10^{-3}\,{\rm Pa\,s}$ 

gives $U=2\mu{\rm m/s}$, i.e. falls through diameter in a second.

\pause
(Check $Re=10^{-6}$)

\pause
\bigskip
Drag on a fluid sphere (Student exercise!)
\[
 \bfF = - 2\pi\tfrac{2\mu_f +3\mu_s}{\mu_f + \mu_s}\mu_f a U
\]

\end{frame}

\subsection{Rotating sphere}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Rotation of a rigid sphere}

Sphere rotating at angular velocity $\boldsymbol{\Omega}$.

Flow
\[
 \bfu(\bfx) = \boldsymbol{\Omega}\wedge\bfx \frac{a^3}{r^3}
\]
(A potential flow, so satisfies Stokes equations.)

\pause
\medskip
Hence couple on sphere -- student exercise!
\[
 {\bf G} = 8\pi\mu a^3 \boldsymbol{\Omega}
\]

\end{frame}

\subsection{Flow past an ellipsoid}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Stokes flow past an ellipsoid}

For principle semi-diameters $a_1, a_2, a_3$ Oberbeck (1876) found \pause
\[
 \text{Force}\quad F_1 = - \frac{16\pi\mu U_1}{L + a_1^2K_2}
 \qword{and Couple}
   G_1 = - \frac{16\pi\mu(a_2^2 + a_3^2)}{3(a_2^2K_2 + a_3^2K_3)}
\]
\pause where
\[{
 L = \int_0^\infty \frac{d\lambda}{\Delta(\lambda)} 
  \qword{and}
 K_i =  \int_0^\infty \frac{d\lambda}{(a_i^2+\lambda)\Delta(\lambda)}
}\]
with $\Delta^2 =(a_1^2+\lambda)(a_3^2+\lambda)(a_3^2+\lambda)$.

\pause
\medskip
For \blue{a disk} $a_1 \ll a_2=a_3$
\[
 F_1 \sim 16\pi\mu a_2U_1, \quad F_2 \sim \tfrac{32}{3}\mu a_2 U_2,
 \quad G_i \sim \tfrac83\mu a_2\Omega_i
\]

\pause
\medskip
For \blue{a rod} $a_1 \gg a_2=a_3$, where $\ln = \ln\tfrac{2a_1}{a_2}$
\[
 F_1 \sim \frac{4\pi\mu a_1 U_1}{\ln - \tfrac12}, \quad
 F_2 \sim \frac{8\pi\mu a_1 U_2}{\ln + \tfrac12}, \quad
 G_1 \sim \tfrac{16}{3}\pi\mu a_1a_2\Omega_1, \quad
 G_2 \sim \frac{\tfrac83\pi\mu a_1^3\Omega_2}{\ln - \tfrac12}
\]

\pause
\bigskip
\blue{Important conclusion} Drag $\approx 6\pi\mu$ with largest diameter.

\end{frame}

\section{More simple properties}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
  \frametitle{A quick course in micro-hydrodynamics}
  \tableofcontents[currentsection,hideothersubsections]
\end{frame}

\subsection{A useful result}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{1. A useful result}

Let $\bfu^S(\bfx)$ be a Stokes flow with $\bfF=0$ in $V$, and let $\bfu(\bfx)$ be any other incompressible flow, then
\[
 \int_V 2\mu e^S_{ij}e_{ij}\,dV = \int_S \sigma^S_{ij}n_ju_i\, dA.
\]

\pause
\bigskip Because
\begin{gather}
 2\mu e^S_{ij} = \sigma^S_{ij} + p^S\delta_{ij} \qword{and} 
 p^S\delta_{ij}e_{ij} = p^S\nabla\cd\bfu = 0 \\
 \qword{so} 2\mu e^S_{ij}e_{ij} = \sigma^S_{ij}e_{ij}.
\end{gather}
\pause
And
\begin{gather}
 \sigma^S_{ij} = \sigma^S_{ji} \qword{so} \\
  \sigma^S_{ij}e_{ij} = \sigma^S_{ij}\pd{u_i}{x_j}
   = \pd{}{x_j}\left(\sigma^S{ij} u_i\right) - 
       \underset{=\bfF=0}{\pd{\sigma^S_{ij}}{x_j}}u_i
\end{gather}
\pause
Hence result by divergence theorem.

\end{frame}
\subsection{Minimum dissipation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{2. Minimum dissipation}

Let $\bfu(\bfx)$ and $\bfu^S(\bfx)$ be two incompressible flows in $V$, both satisfying the same boundary condition $u=u^S=\bfU(\bfx)$ give on $S$.

\pause
Let $\bfu^S$ also satisfy the Stokes equation with $\bfF=0$ in $V$.

\pause
Then
\begin{gather*}
 \int_V2\mu e_{ij}e_{ij}\,dV = \int_V2\mu e^S_{ij}e^S_{ij}\,dV \\
  + \underset{\longleftarrow{\rm positive}\longrightarrow}{\int_V2\mu(e_{ij}-e^S_{ij})(e_{ij}-e^S_{ij})\,dV}
  + \int_V4\mu e^S_{ij}(e_{ij}-e^S_{ij})\,dV.
\end{gather*}
\pause
The last integral is of the form of the useful result
\[
 \int_V4\mu e^S_{ij}(e_{ij}-e^S_{ij})\,dV = 
 \int_S2\sigma^S_{ij}n_j(u_i-u^S_i)\,dA = 0 \qword{by bc}
\]

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{2. Minimum dissipation}

Hence
\[
 \int_V2\mu e_{ij}e_{ij}\,dV \ge \int_V2\mu e^S_{ij}e^S_{ij}\,dV,
\]
\pause
i.e. the Stokes flow $\bfu^S(\bfx)$ has the minimum dissipation out of all incompressible flows satisfying the boundary condition

\pause
\bigskip
Hence e.g. drag larger at non-zero Reynolds number.

\pause
\bigskip \blue{Warning: Same geometry.} Cannot select geometry by minimum dissipation.

\end{frame}
\subsection{Uniqueness}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{3. Uniqueness}

If $\bfu^1(\bfx)$ and $\bfu^2(\bfx)$ are two Stokes flows in $V$ satisfying the same boundary conditions, then minimum dissipation gives
\[
 \int_V2\mu(e^1_{ij}-e^2_{ij})(e^1_{ij}-e^2_{ij})\,dV = 0
\]
\pause
Hence
\[
 e^1_{ij}-e^2_{ij} = 0 \qword{in $V$,}
\]
\pause
i.e. $\bfu^1-\bfu^2$ is strainless, 
\pause
i.e. a solid body translation $+$ rotation, 
\pause
i.e. zero by the boundary conditions. 
\pause
Hence
\[
 \bfu^1(\bfx) = \bfu^2(\bfx) \qword{in $V$.}
\]
\pause
Hence Stokes flows are unique.

\end{frame}
\subsection{Geometric bounding}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{4. Geometric bounding}

Rigid cube, sides of length $2L$, moving at $\bfU$, drag $\bfF$.

\pause
Let $\bfu^S(\bfx)$ be Stokes flow outside cube, $V$. 

\pause
Then dissipation
\[
 \int_V2\mu e^S_{ij}e^S_{ij}\,dV
 =\text{rate of working by surface forces} \pause
 = -\bfU\cd\bfF.
\]

\pause
Cube just contained by sphere radius $a=\sqrt3L$, also moving at $\bfU$.

\pause
\bigskip
Define second flow
\[
 \bfu(\bfx) = \begin{cases}
                \text{the Stokes flow for sphere} & \text{outside sphere,} \\
                \bfU & \text{in gap.}
              \end{cases}
\]

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{4. Geometric bounding}

For this second flow
\begin{gather*}
 \int_V2\mu e_{ij}e_{ij}\,dV =  \int_{r>a}2\mu e_{ij}e_{ij}\,dV
 \qword{because $e=0$ in gap,} \\
 = \text{rate of working by sphere}
 = 6\pi\mu\sqrt3L \bfU\cd\bfU
\end{gather*}

\pause
Hence minimum dissipation bounds drag $\bfF$ on cube
\[
 -\bfF\cd\bfU \le 6\pi\mu\sqrt3L\bfU\cd\bfU
\]

\pause
Similarly for sphere just contained inside cube
\[
 6\pi\mu L\bfU\cd\bfU \le -\bfF\cd\bfU
\]

\pause
\bigskip
\blue{Student exercises}: bound for tetrahedron (not so tight).

\end{frame}
\subsection{Reciprocal theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{5. Reciprocal theorem}

For the same volume $V$, let $\bfu_1$ be the Stokes flow 
with volume forces ${\bf f}_1$ satisfying boundary conditions 
$\bfu_1 = \bfU_1$.

\pause
Let $\bfu_2$ be Stokes flow in $V$ with ${\bf f_2}$ and $\bfU_2$.

\pause
\bigskip
Then by the useful result
\begin{gather*}
 \int_V \bfu_1\cd{\bf f}_2\,dV + \int_S \bfU_1\cd\bsi_2\cd\bfn\,dA
 = \int_V2\mu {\bf e}_1:{\bf e}_2\,dV \\
 = \int_V \bfu_2\cd{\bf f}_1\,dV + \int_S \bfU_2\cd\bsi_1\cd\bfn\,dA
\end{gather*}
\pause
i.e. work done by one velocity on the forces of the other is vice versa.

\pause
\smallgray{Greens theorem in any other subject}

\end{frame}
\subsection{Symmetry of resistance matrix}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{6. Reciprocal theorem -- application to resistance matrix}

General rigid body motion in fluid at rest a infinity, 

translating at $\bfU(t)$ and rotating \smallgray{(about a selected point)} $\boldsymbol{\Omega}(t)$.

\pause
By linearity and instantaneity, the force $\bfF(t)$ and couple ${\bf G}(t)$ 
\smallgray{(about the same selected point)}
\[
 \begin{pmatrix} \bfF \\ {\bf G} \end{pmatrix} =
\begin{pmatrix} {\bf A} \; {\bf B} \\ {\bf C}\; {\bf D} \end{pmatrix} 
\begin{pmatrix} \bfU \\ \boldsymbol{\Omega} \end{pmatrix}
 \]

\pause
The Reciprocal theorem gives for the two rigid body motions
\[
 \bfU_1\cd\bfF_2 + \boldsymbol{\Omega}_1\cdot{\bf G}_2 =
 \bfU_2\cd\bfF_1 + \boldsymbol{\Omega}_2\cdot{\bf G}_1
\]
\pause
True all $\bfU_1$ etc, so
\[
 {\bf A} = {\bf A}^T,\quad {\bf B} = {\bf C}^T \qword{and} {\bf D} = {\bf D}^T.
\]

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{6. Reciprocal theorem -- application to resistance matrix}

$ {\bf B} = {\bf C}^T $ means
\[
 \text{force due to rotation} = \text{couple due to translating.}
\]

\pause
${\bf A} = {\bf A}^T$ \& ${\bf D} = {\bf D}^T$ for symmetric cube means
\[
 {\bf A} \;\&\; {\bf D} \qword{diagonal, and} {\bf B} = {\bf C}^T = 0 
\]

\pause
Thus drag on a cube is parallel to velocity, \break also for symmetric tetrahedron.

\pause
\bigskip
Need ``corkscrew'' feature for ${\bf B} \neq 0$.

\end{frame}
\subsection{Faxen's formula}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{7. Reciprocal theorem -- application to Faxen's formula}

\pause
A force-free sphere placed in arbitrary flow $\bfu^\infty(\bfx)$ moves with what velocity ${\bf V}$? 
Forces which generate $\bfu^\infty$ kept constant.

\pause
\medskip
Let $\bfu^+$ be Stokes flow with sphere inserted in $\bfu^\infty$.

\pause
Then disturbance flow is $\bfu_1 = \bfu^+ - \bfu^\infty$.

\pause
Now same forces for $\bfu^+$ \& $\bfu^\infty$, so ${\bf f}_1=0$. 
Also $\bfu_1\to0$ far from sphere.

\pause
Let $\bfu_2(\bfx)$ be flow outside a sphere translating at $\bfU_2$ 
with ${\bf f}_2=0$.

\pause
\medskip
Applying Reciprocal theorem
\[
 \int \bfu_1\cd\underset{=0}{{\bf f}_2}\,dV 
  + \int \bfu_1\cd\underset{=-3\mu\bfu_2/a}{\bsi_2\cd\bfn}\,dA =
 \int \bfu_2\cd\underset{=0}{{\bf f}_1}\,dV 
   + \int \underset{=\bfU_2}{\bfu_2}\cd\bsi_1\cd\bfn\,dA.
\]
\pause
Now look at RHS and then LHS, using  $\bfu_1 = \bfu^+ - \bfu^\infty$.

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{7. Reciprocal theorem -- application to Faxen's formula}

\[
 {\rm RHS} = \bfU_2\cdot\left(
  \int\bsi^+\cd\bfn\,dA - \int\bsi^\infty\cd\bfn\,dA
        = 0 - 0 \right) = 0,
\]
as both integrals are force on sphere.

\pause
Hence
\[
 {\rm LHS} = -\frac{3\mu}{a}\bfU_2\cdot
 \left(\int\underset{=\bf V}{\bfu^+}\,dA - \int\bfu^\infty\,dA \right)
 = {\rm RHS} = 0
\]

\pause
\bigskip
For all $\bfU_2$, so velocity of sphere inserted into $\bfu^\infty(\bfx)$ is
\[
 {\bf V} = \frac1{4\pi a^2}\int_{r=a}\bfu^{\infty}(\bfx)\,dA
\]

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{More simple properties}
 \framesubtitle{7. Reciprocal theorem -- application to Faxen's formula}

\[
 {\bf V} = \frac1{4\pi a^2}\int_{r=a}\bfu^{\infty}(\bfx)\,dA
\]

\pause
\bigskip
Finally use a Taylor series
\[
 \bfu^\infty(\bfx) = \bfu^\infty(0) + \bfx\cd\nabla\bfu^\infty\vert_0 +
   \tfrac12\bfx\bfx:\nabla\nabla\bfu^\infty\vert_0 + \cdots 
\]

\pause
\bigskip
Integrating over the sphere, the odd terms vanish by symmetry, \pause so
\[
 {\bf V} = \bfu^\infty(0) + \tfrac{a^2}6\nabla^2\bfu^\infty\vert_0
\]

\pause
with higher even terms vanishing by 
$\nabla^{2n}(\text{Stokes equations}) = 0$.

\end{frame}

\section{Greens function}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{A quick course in micro-hydrodynamics}
 \tableofcontents[currentsection,hideothersubsections]
\end{frame}

\subsection{Stokeslet}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Greens function for Stokes equations}
 \framesubtitle{or `Stokeslet'}

For a point momentum source
\begin{gather*}
 \nabla\cd\bfu = 0 \\
 0 = - \nabla p + \mu \nabla^2\bfu + \bfF\delta(\bfx)
\end{gather*}

\pause
\bigskip
\blue{Solution} -- more important than derivation
\begin{gather*}
 \bfu(\bfx) = \bfF\cd{\bf G}(\bfx) = 
   \frac1{8\pi\mu}\left(\bfF\frac1r + (\bfF\cd\bfx)\bfx\frac1{r^3} \right)\\
 \bsi(\bfx) = \bfF\cd{\bf K}(\bfx) 
      = - \frac3{4\pi}\bfF\cd\bfx\bfx\bfx\frac1{r^5}
\end{gather*}
\pause 
{\bf G} is called the `Oseen tensor'.

\pause
\bigskip
Already seen this in flow past a sphere:

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Greens function for Stokes equations}
 \framesubtitle{Far field for a sphere}


Far from the sphere $r\gg a$, the flow is\pause
\[
 \bfu  =  \bfU \left(\gray{1} - \frac{3a}{4r} - \gray{\frac{a^3}{4r^3}} \right)
   + \bfx (\bfU\cd\bfx)\left(-\frac{3a}{4r^3} + \gray{\frac{3a^3}{4r^5}}\right), 
\]

\pause
But the drag is $\bfF=-6\pi\mu a\bfU$, i.e.
\[
 \bfu - \bfU \sim 
 \frac1{8\pi\mu}\left(\bfF\frac1r + (\bfF\cd\bfx)\bfx\frac1{r^3} \right)
\]

\pause
\bigskip
Hence far-field due to force is universal, independent of particle shape.

\end{frame}

\subsection{Integral representation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Greens function for Stokes equations}
 \framesubtitle{Integral representation}

To solve
\begin{gather*}
 \nabla\cd\bfu = 0 \\
 0 = - \nabla p + \mu \nabla^2\bfu + {\bf f}(\bfx)
\
\end{gather*}
with boundary conditions on $\bfu(\bfx)$ or $\bsi(\bfx)\cd\bfn$.

\pause
\bigskip
Use the Reciprocal theorem (Greens theorem) with $\bfu_1$ for the unknown flow and $\bfu_2$ for the Greens function for point source at $\bfx'$ \pause
\begin{gather*}
 \bfu(\bfx') 
  = \underset{\text{forces in $V$}}{\int_V{\bf G}(\bfx-\bfx')\cd{\bf f}\,dV} \\
 + \int_S\biggl(\underset{\raisebox{-10pt}{\text{\small forces on $S$}}}{{\bf G}(\bfx-\bfx')\cd\bsi(\bfx)\cd\bfn}
    - \underset{\raisebox{-10pt}{\text{\small dipoles on $S$}}}{({\bf K}(\bfx-\bfx')\cd\bfn)\cd{\bfu}} \biggr)\,dA
\end{gather*} 

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Greens function for Stokes equations}
 \framesubtitle{Integral representation -- Boundary integral Method}

Letting $\bfx'$ in $V$ tend onto the surface $S$ yields an integral equation for the unknown $\bfu$ \gray{(or $\bsi\cd\bfn$)} on $S$ in terms of the known $\bsi\cd\bfn$ \gray{(or $\bfu$)} on $S$.

\pause
\medskip
Delicate limit $\bfx'\to S$: 
$\int {\bf K}\cd\bfn \to +\frac12\bfu$ for $\bfx'$ in $V$, \pause
$-\frac12\bfu$ for $\bfx'$ outside $V$, more complex at corners on $S$.

\pause
\bigskip
Basis of numerical {\em Boundary Integral Method}.

\pause
\medskip
Advantage: fewer points on surface than in volume, \pause and no infinity.

\pause
\medskip
Disadvantages:
Special attention needed in numerical evaluation of singular integrals, 
\pause and 
there are often eigensolutions, e.g. constant pressure induces no flow.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Greens function for Stokes equations}
 \framesubtitle{Integral representation - for a suspended drop}

Extension to a drop of viscosity $\lambda\mu$ surrounded by a fluid of viscosity $\mu$.

\pause
\medskip
One knows the jump across the interface of the normal viscous stress
\[
 \bsi\cd\bfn = \gamma \kappa \bfn
\]
with surface tension $\gamma$ and surface curvature $\bfn$.

\pause
\bigskip
Adding $\lambda$ times the integral equation for the interior to that for the exterior, one reduce the stress contribution to its difference \pause
\begin{gather*}
 \tfrac12(1+\lambda)\bfu(\bfx') = \bfu^\infty(\bfx') \\
 - \int_S {\bf G}(\bfx-\bfx')\cd\gamma\kappa\bfn\,dA
 - (1-\lambda)\int_S {\bf K}(\bfx-\bfx')\cd\bfn\cd\bfu\,dA
\end{gather*}

\end{frame}

\subsection{Slender-body theory}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Greens function for Stokes equations}
 \framesubtitle{Slender-body theory}

For slender bodies, approximate the distribution of forces $\bsi\cd\bfn$
on the surface $S$ \pause by a distribution of forces ${\bf f}(s)$ 
along the centreline ${\bf X}(s)$ in $-L\le s\le L$ \pause
\[
 \bfu(\bfx') = \bfu^\infty(\bfx') 
  + \int_{-L}^L {\bf G}({\bf X}(s)-\bfx')\cd{\bf f}(s)\,ds
\]

\pause
Satisfy the boundary condition by evaluating at distance $\epsilon R(s_0)$ from centreline at $s=s_0$, either numerically 

\pause
\bigskip
or asymptotically for
\[
 {\bf f}(s_0) \sim \frac{2\pi\mu}{\ln\frac{L}{R}}
  \left(2{\bf I} - {\bf X}'{\bf X}' \right) \cd
  \left({\bf U}(s_0) - \bfu^\infty({\bf X}(s_0)) \right)
\]

\end{frame}

\section{Effect of small inertia}
\subsection{Whitehead paradox}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Effects of small inertia on flow past a sphere}
 \framesubtitle{Whitehead paradox}

Flow $\bfU$ past a sphere of radius $a$.

\pause
At $Re\ll1$, first approximation is given by Stokes flow.

\pause
Attempt to find correction as regular perturbation fails.

\pause
In far field
\[
 u = U + O\left(\frac{Ua}{r} \right), \qword{so}
 \rho\bfu\cd\nabla\bfu = O\left(\frac{\rho U^2 a}{r^2} \right)
\] 
\pause
A correction $\bfu_2$ forced by this
\[
 -\nabla p_2 + \mu\nabla^2\bfu_2 = \rho\bfu\cd\nabla\bfu\vert_\text{Stokes}
\]
\pause
would give
\[
 u_2 = O(\rho U^2 a/\mu)
\]
which does not decay.

\end{frame}

\subsection{Oseen equation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Effects of small inertia on flow past a sphere}
 \framesubtitle{Oseen equation}

Look more carefully at Stokes solution. In far field \pause
\[
 \text{Viscous terms} = O\left(\frac{\mu U a}{r^3} \right),
 \quad\text{while inertial terms} = O(\left(\frac{\rho U^2a}{r^2} \right)
\]
\pause
so at $r = \mu/U$ inertial terms no longer small.

\pause
\bigskip
Fortunately in far field $\bfu = \bfU + \bfu'$ with $\bfu'$ small, 
\pause so can linearise Navier-Stokes
\[
 \rho\bfU\cd\nabla\bfu' = -\nabla p' + \mu\nabla^2\bfu' + \bfF\delta(\bfx)
\]
where in far field sphere appears a point force.

\pause
\bigskip
Solve by Fourier transforms \pause or representation
\[
 \bfu' = \nabla\phi + \nu\nabla\chi - \bfU\chi 
 \qword{and} p' = -\rho\bfU\cd\nabla\phi.
\]

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Effects of small inertia on flow past a sphere}
 \framesubtitle{Oseen equation solved}

Find
\[
 \phi = - \frac{3a\nu}{2r} \text{\smallgray{(point volume source)}}
 \qword{and} 
 \chi = \frac{3a}{2r}e^{\left(\frac{\bfU\cd\bfx}{2\nu}-\frac{Ur}{2\nu}\right)}
\]

\pause
\bigskip
Look nearer to sphere $a \ll r \ll \nu/U$, 
\pause \quad $r^{-2}$ terms cancel and \pause
\[
 \bfu' \sim \text{$\frac1r$ Stokeslet} 
 +  \text{$\bfU\frac{3Ua}{4\nu}$ uniform flow}
\]
\pause
Hence drag increases by $1 + \frac38Re$.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{Effects of small inertia on flow past a sphere}
 \framesubtitle{Oseen wake}

Look far from sphere $r \gg \nu/U$ near downstream axis
\[
 \bfu' \sim -\bfU\frac{3a}{2z}e^{-\frac{U(x^2+y^2)}{4\nu z}}
\]
\pause
i.e. a \blue{wake} diffusing to $r=\sqrt{\nu(t=z/U)}$ with mass flux deficit
\[
 \int\rho u'\,dxdy = -6\pi\mu Ua
 \quad\text{=momentum deficit $/U$}
\]
\pause
\gray{Helpful idea at higher $Re$}

\pause
\medskip
Impacts on time-dependent flows -- Basset history wrong

\pause
\bigskip
Missing mass flux in wake goes to point source $\phi$-flow.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 \frametitle{A quick course in micro-hydrodynamics}
 \tableofcontents[hideallsubsections]
\end{frame}

\end{document} 

