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\begin{document}
{\noindent\bf MAE224A: Environmental Fluid Mechanics}
\hfill Fall Quarter 2010
\bigskip

\hfill \textrm{http://maecourses.ucsd.edu/~sllewell/MAE224A\_2010/}
\bigskip

\begin{center}
\bf\LARGE Solution III
\end{center}

\paragraph{1.} 
Consider an element of arc  $\mathrm{d} s$ on the free surface: the surface tension and inner pressure $p$ must be balanced by the pressure outside of the interface, $p_a$. The balance of forces perpendicular to the arc requires
\[
-p_a \,\mathrm{d} s + p\,\mathrm{d} s+\sigma \,\dd\theta = 0.
\]
The pressure difference is therefore given by
\[
p_a - p = \sigma \ts = \sigma \exx
\]
in two dimensions, where the curvature $\kappa$ is defined as
\[
\kappa = \ts = \frac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}}
\]
and where the free surface is parameterized by $x = x(t)$, $y =
y(t)$. For a two-dimensional curve with $y = \eta(x)$, the curvature
becomes
\[ 
\kappa = \frac{\partial^2\eta/\partial x^2}{[ 1+ (\partial
  \eta/\partial x) ^2] ^{3/2}} \approx \exx
\]
for small slopes. If the atmospheric pressure is taken to be 0, the
linearized dynamic condition at the surface is
\[
p = -\sigma \exx \qquad \mbox{at $z=0$}.
\]
This may be combined with the linearized Bernoulli equation
\[
\pt + \frac{p}{\rho}+ gz = 0
\]
to give
\[
\pt = \frac{\sigma}{\rho} \exx -g\eta.
\]
The solution to the Laplace equation is as in class, namely
\[
\phi = \frac{a\omega}{k}\frac{\cosh k(z+H)}{\sinh kH}\sin(kx-\omega t),
\]
with free surface,
\[
\eta = a \cos(kx -\omega t).
\]
The new dynamic condition at $z=0$ gives the dispersion relation
\[
\omega^2 = {k}\left( g+ \frac{\sigma  k^2}{\rho} \right ) \tanh kH.
\]
For the three-dimensional case, the $k^2$ term is replaced by $k^2 +
l^2$.  The group velocity $c_g$ is given by $\partial \omega
/ \partial k$, so
\[
c_g = \frac{(g/k)^{1/2}}{2} \frac{ [ 1 + 3T(kH)^2] \tanh kH + kH[1 +
  T(kH)^3] \mathrm{sech^2}kH}{\{[1 + T(kH)^2] \tanh kH \}^{1/2}},
\]
where $T = \sigma/\rho H^2$ is a non-dimensional version of surface
tension. In the limit of small $kH$, $c_g \approx \sqrt{gH}$, the
standard shallow water case (assuming $T$ is not ridiculously
large). For large $kH$ and $T$ not very small, $c_g \approx (3/2)
\sqrt{\sigma k/\rho}$, corresponding to capillary waves on deep water.


\paragraph{2} Using the Lagrangian description for fluid motion,
consider a fluid particle at $(x_0+\lambda,z_0+\zeta)$ with mean
position $(x_0,z_0)$. Then
\[ 
u = \xit, \qquad w = \zt,
\]
while the velocities are found by taking the gradient of the velocity
potential above:
\[
u = a\omega \frac{\cosh k(z+H)}{\sinh kH}\cos(kx-\omega t), \qquad w =
a\omega \frac{\sinh k(z+H)}{\sinh kH}\sin(kx-\omega t).
\]
For small-amplitude waves, the motion of the particle is small and
consequently the velocity of a particle along its path is
approximately equal to the velocity at the mean position $(x_0,z_0)$,
so that
\[
\xit = a\omega \frac{\cosh k(z_0+H)}{\sinh kH}\cos(kx_0-\omega t),
\qquad \zt = a\omega \frac{\sinh k(z_0+H)}{\sinh kH}\sin(kx_0-\omega
t).
\]
Integrating in time gives
\[
\lambda = -a\frac{\cosh k(z_0+H)}{\sinh kH}\sin(kx_0-\omega t), \qquad
\zeta = a\frac{\sinh k(z_0+H)}{\sinh kH}\cos(kx_0-\omega t)
\]
(the constant of integration is irrelevant). Using the identity
$\sin^2 x + \cos^2 x = 1$, the particle path may be written as
\[
\frac{\lambda^2}{\left [\cosh k(z_0+H)/\sinh kH \right]^2} +
\frac{\zeta^2}{[\sinh k(z_0+H)/\sinh kH]^2} = a^2;
\]
these are ellipses. In the deep water case, 
\[ 
\frac{\cosh k(z_0+H)}{\sinh kH} \simeq \frac{\sinh k(z_0+H)}{\sinh kH}
\simeq e^{kz_0},
\]
that the particle orbits are
\[
\lambda = -ae^{kz_0}\sin(kx_0-\omega t), \qquad \zeta =
ae^{kz_0}\cos(kx_0-\omega t).
\]
The equation for the particle paths becomes
\[
\zeta^2+\lambda^2=a^2e^{2kz_0}.
\]
i.e.~circles whose radius is equal to the amplitude $a$ at the surface
and decreases exponentially with depth. For shallow water, the
approximations
\[\cosh k(z_0+H) \simeq 1, \qquad \sinh k (z_0+H) \simeq k(z_0+H),
\qquad \sinh kH \simeq kH
\]
lead to
\[
\lambda = -\frac{a}{kH}\sin(kx_0-\omega t), \qquad \zeta = a\left( 1+
  \frac{z_0}{H} \right) \cos(kx_0-\omega t).
\]
These are thin ellipses described by 
\[
(\lambda kH)^2 + \frac{\zeta^2}{(1+ z_0/H)^2} = a^2.
\]
In this limit, the semi-major axis, $a/kH$, is depth-independent, and
the semi-minor axis decreases linearly to 0 at the bottom
boundary. Therefore, the particle orbits are ellipses that get
progressively thinner for particle near the bottom. 


\paragraph{3} We need to solve Laplace's equation with the
usual boundary conditions in $x > 0$. However we are now imposing a
velocity $u = \partial\phi/\partial x$ on $x = 0$ (for inviscid flow
we can only impose the normal velocity). We seek a separated solution
in the form $\phi = X(x) Z(z) \ee^{-\ci\omega t}$. Laplace's equation
gives
\[
X'' - \kappa^2 X, \qquad Z'' + \kappa^2 Z = 0.
\]
The linearized boundary conditions for $Z$ are $Z' = 0$ and $z = -H$ and
$-\omega^2 Z + g Z' = 0 $ at $z = 0$. We take $Z = \cos{\kappa(z+H)}$ to
satisfy the first of these. The second then gives
\[
\tan{\kappa H} = - \frac{\omega^2}{g\kappa}.
\]
Plotting the two sides of this equation shows that there are
infinitely positive roots $\kappa_n$. These correspond to modes $Z_n$ that decay
for large $x$, i.e.~trapped near the wall. However, there is also a
mode with imaginary $\kappa = \ci k$ with $Z = \cosh{k(z+H)}$ and
\[
\tanh{kH} = \frac{\omega^2}{gk}.
\]
Here we must take the mode with positive $k = k_0$ to obtain waves that
propagate away from the wall. Hence
\[
\phi = A_0 \ee^{\ci(k_0x -\omega t)} Z_0(z) + \sum_{n=1}^\infty A_n
\ee^{-\kappa_n x - \ci\omega t} Z_n(z).
\]
The coefficients are found by matching the velocity at the wall to the
forcing:
\[
u =  \ci k_0 A_0 Z_0(z) - \sum_{n=1}^\infty \kappa_n A_n Z_n(z).
\]
The vertical modes come from a Sturm--Liouville problem, so we may
take them to be orthonormal. Then the boundary condition may be
multiplied by $Z_m^*$ and integrated to give
\[
\ci k_0 A_0 = \int_{-H}^0 u(z) Z_0(z) \,\dd z, \qquad -\kappa_n A_n =
\int_{-H}^0 u(z) Z_n(z) \,\dd z.
\]
Only the portion of $u$ that projects onto the mode $Z_0$ (essentially
the lowest mode) gives a propagating response.

\end{document}
