\documentclass[twoside]{article}
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\markboth{\hfil On the tidal motion around the earth  \hfil EJDE--2000/35}
{EJDE--2000/35\hfil  Ranis N. Ibragimov  \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.~{\bf 2000}(2000), No.~35, pp.~1--11. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
%
  On the tidal motion around the earth complicated by the circular
  geometry of the ocean's shape
\thanks{ {\em Mathematics Subject Classifications:} 35Q35, 76C99.
\hfil\break\indent
{\em Key words:} Cauchy-Poisson free boundary problem, shallow water theory,
\hfil\break\indent conformal mapping
\hfil\break\indent
\copyright 2000 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Submitted January 5, 2000. Published May 16, 2000.  \hfil\break\indent
Dedicated to Professor L. V. Ovsyannikov on his 80-th birthday.} }
\date{}
%
\author{ Ranis N. Ibragimov }
\maketitle

\begin{abstract}
 We study the Cauchy-Poisson free boundary problem on the stationary motion
 of a perfect incompressible fluid circulating around the Earth.
 The main goal is to find the inverse conformal mapping of the unknown
 free boundary in the hodograph plane onto some fixed boundary
 in the physical domain. The approximate solution to the problem is
 obtained as an application of this method.
 We also study the behaviour of tidal waves around the Earth. It is shown
 that on a positively curved bottom the problem admits two different high
 order systems of shallow water equations, while the classical problem for
 the flat bottom admits only one system.
\end{abstract}


\newtheorem{theorem}{Theorem}


\section{Introduction}

We study the Cauchy-Poisson problem on the stationary motion of a perfect
fluid which
has a free boundary and has a solid bottom represented by a circle with a
sufficiently large radius.  We have shown in [4]
that such problem can be associated with a two dimensional model to an
oceanic motion around the Earth since we consider strictly longitudinal
flow. Since the problem is a free boundary problem, the analysis is rather
difficult.

The permanent water waves have been considered in a great number of papers.
However, most researchers are concerned with fluid motion which is infinitely
deep and extends infinitely both rightward and leftward.  See Crapper [1],
Stoker [9] or Stokes [10] for the history.  Such problems are usually called
Stoke problem (if the surface tension is neglected) and Wilton's problem (if the
surface tension is taken into account).

We consider water waves for which ratio of depth of fluid above the circular
bottom to the radius of the circle is small (shallow water).  In these the
disturbance to the water does not penetrate unchanged to the bottom and the
effective inertia of the water is therefore reduced.

Our primary concern is to find the conformal mapping (for the Stoke problem) of
the unknown free boundary onto fixed one.  The resulting Dirichlet problem can
be solved numerically using Okamoto method [3].  The more detailed structure of
the bifurcation of solutions for the related problem was numerically computed by
Fujita at al.  [3].  The existence of nontrivial solutions for the analogous
reduced Dirichlet problem can be found in Okamoto [7], Ibragimov [4] as well as
in classical literature (see e.g., [8] or [9]).

The higher order shallow water equations in the non-stationary case are derived
in this paper.  It is shown that the present problem admits two different
systems of shallow water equations while the classical problem for the flat
bottom admits only one system (see [2]).

We note that papers [3], [6] and [7] are concerned with fluid whose surface
tension is taken into account. In fact, the surface tension plays the role
of a ''regulator'' of the problem which simplifies the analysis
substantially. Furthermore the nature of the problem requires the surface
tension to be neglected. Thus, the present paper represents more systematic
approach to the problem.

The present paper aims to investigate the problem by using a conformal
mapping which distinguishes our paper from [3], [4], [6] and [7].

\section{Basic Equations}

The analysis of this problem is performed in the following notation: $R$ is
the radius of the circle, $r$ is a distance from the origin, $\theta $ is a
polar angle, $h_0$ is the undisturbed level of the liquid above the circle
and $h=h(\theta )$ is the level of the disturbance of the free
boundary above the circle. For the sake of simplicity we assume that the
pressure is constant on the free boundary. The stream function $\psi =\psi
(r,\theta )$ defines the velocity vector, i.e.,
\[
v^r=-\frac{1}{r}\psi_\theta ,\quad v^\theta =\psi_r\,.
\]
Hence irrotational motion of an ideal incompressible fluid of the constant
pressure in the homogeneous gravity field $g=$const is described by the
stream function $\psi $ in the domain
\[
\Omega_h=\left\{ (r,\theta ): 0\leq \theta \leq 2\pi ,
 R\leq r\leq R+h_0+h(\theta )\right\}
\]
which is bounded by the bottom $\Gamma _{R}=\left\{ (r,\theta )
:r=R, \theta \in [0,2\pi ]\right\} $ and the free
boundary with equation $\Gamma_h=\left\{ (r,\theta )
:r=R+h_0+h(\theta ), \theta \in [0,2\pi ]
\right\} $. Note that $\psi $ is a harmonic function in $\Omega_h$, since
we assumed that the flow is irrotational. More specifically, we assume that
the fluid is incompressible and inviscid and that the flow is stationary.
Then the problem is to find the function $h(\theta )$ and the
stationary, irrotational flow beneath the free boundary
$r=R+h_0+h(\theta )$ given by the stream function $\psi $ which satisfy the
following differential equations
\begin{eqnarray}
&\Delta \psi =0(\mbox{in }\Omega_h),\qquad \psi =0(
\mbox{on }\Gamma _{R}),\qquad \psi =a\mbox{ (on $\Gamma_h$),}& \\
&\left| \nabla \psi \right| ^{2}+2gh=\mbox{constant}\quad (\mbox{ on }
\Gamma_h), &  \\
&\frac{1}{2}\int_0^{2\pi } (R+h_0+h(\theta))^{2}\, d\theta =\pi
(R+h_0)^{2},&
\end{eqnarray}
where the constant $a$  denotes the flow rate.
Equations (1)-(3) represent the free boundary Cauchy-Poisson problem in
which
the boundary $\Gamma_h$ is unknown as well as the stream function.

\section{The Inverse Transforms Principle}

\subsection*{Constant flow}

The exact solution
\begin{equation}
h\equiv 0 \quad \mbox{and}\quad \psi =\psi _0=a\log r
\end{equation}
of (1)-(3) corresponds to the constant flow with undisturbed free
boundary. The trivial solution (4) represents a flow whose streamlines are
concentric circles with the common center at the origin.

The following non-dimensional quantities are introduced:

\[
r=R+h_0r',\quad h=h_0h',\quad \psi
=a\psi ',\quad \epsilon =\frac{h_0}{R},\quad %
\mathcal{F}=\frac{h_0\sqrt{gh_0}}{a},
\]
where $\mathcal{F}$ is a Froude number and $R$ is used as a vertical scale.
We consider $\epsilon $ as the small parameter of the problem. After
dropping the prime, Equations (1)-(3) are written by $\psi ',h'$, and
$(r',\theta )$ as follows
\begin{eqnarray}
&\Delta _{(\epsilon )}\psi =0\quad \mbox{(in $\Omega_h$)}, &\\
&\psi =0 \mbox{(on $\Gamma _{R}$)}, &\\
&\psi =1\mbox{(on $\Gamma_h$)}, &\\
&\left| \nabla _{(\epsilon )}\psi \right| ^{2}+2\mathcal{F}
^{-2}h=\mbox{ constant } \quad \mbox{(on $\Gamma_h$)}, & \\
&\frac{1}{2} \int_0^{2\pi} (1+\epsilon+\epsilon h(\theta ))^{2}d\theta
=\pi (1+\epsilon )^{2}\quad \mbox{(on  $\Gamma_h$)}. &
\end{eqnarray}
Here the Laplace and gradient operators are given by
\[
\Delta _{(\epsilon )}=(\epsilon \partial
_{\theta })^{2}+[(1+\epsilon r)\partial_r]^{2},
\quad \nabla _{(\epsilon )}
=\Big(\frac{\epsilon \partial_\theta }{(1+\epsilon r)},\partial_r\Big),
\]
where the subscripts imply the differentiation.

We further consider the complex potential $\omega (\zeta )
=\phi +i\psi $ where $\zeta =(1+\epsilon r)e^{i\theta }$ is the
independent complex variable and $\phi (\zeta )$ is the
velocity potential which is characterized by the analyticity of $\phi
+i\psi $, i.e.,
\[
\phi_r=\frac{\epsilon \psi_\theta }{(1+\epsilon r)},
\quad \frac{\epsilon \phi_\theta }{(1+\epsilon r)}=-\psi_r.
\]
We note that the complex velocity $d\omega /d\zeta $ is a single-valued
analytic function of $\zeta $, although $\omega $ is not single-valued. In
fact, when we turn around the bottom $r=1$ once, $\phi $ increases by
\[
-\int_0^{2\pi} \psi_r(1,\theta )\,d\theta
\]
which has a positive sign by the maximum principle (Hopf's lemma). Hence,
if
we remove the width of annulus region $\theta =0$,
$r\in [1,1+\epsilon ]$, then at every point $(r,\theta )$,
$\omega (\zeta )$ is single-valued analytic function which maps
the rectangular (in the $\omega (\alpha )$-hodograph plane) domain with
$\phi$ in $[0,-2\pi/\log (1+\epsilon)]$ and $\psi$ in $[0,1]$ as
coordinates
onto the annulus
\[
\Gamma_h^{0}=\left\{ (r,\theta ):1<r<1+\epsilon ,
\theta \in [0,2\pi ]\right\} .
\]
We represent the constant flow (4) by
\begin{equation}
\omega (\zeta )=\phi +i\psi =\frac{i\log (1+\epsilon r)
-\theta }{\log (1+\epsilon )},
\end{equation}
where $r=\xi _0(\psi )$ and $\theta =\eta _0(\phi)$ transform the
rectangular domain in the hodograph plane onto $\Gamma_h^{0}$.
Consequently,
each conformal mapping by the function $\omega (\zeta )$ between hodograph
and physical planes represents an irrotational flow in the physical $\zeta
$
plane. Furthermore, (10) implies that
\begin{equation}
\eta _0(\phi )=-\phi \log (1+\epsilon )
\quad\mbox{and} \quad \xi _0(\psi )=\epsilon ^{-1}([1+\epsilon ]^{\psi }
-1).
\end{equation}

\subsection*{Reduction on the boundary}

Now a two-dimensional infinitesimal disturbance $\xi '$ and $\eta
'$ is superimposed on $\xi _0$ and $\eta _0$. Then the
resulting transform components are
\[
\xi =\xi _0+\xi ',\quad \eta =\eta _0+\eta '.
\]
The perturbed quantities $\xi '$ and $\eta '$ are assumed
to be small quantities so that the nontrivial solution is close to the
trivial one.

Then with (11) the inverse transform can be combined in form
\[
\log \zeta =\psi \log (1+\epsilon )-i\phi \log (
1+\epsilon )+\log (1+\frac{\epsilon \xi '}{
1+\epsilon \xi _0})+i\eta ',
\]
since $\log (1+\epsilon \xi _0)=\psi \log (
1+\epsilon )$. Consequently, the polar angle $\theta $ and radius
$r$ are given by equations
\[
\theta =-\phi \log (1+\epsilon )+\eta ',
\quad r=\epsilon ^{-1}[(1+\epsilon )^{\psi }(
1+\frac{\epsilon \xi '}{1+\epsilon \xi _0})-1].
\]

Since the motion is irrotational, we can decrease the dimension of the
problem by one. In other words, we introduce the boundary value for the
function $\xi $ and reduce the basic equations to the quantities which
arise
from the condition on the free boundary $\psi =1$. To this end we introduce
the regular function $f(\omega )=\alpha +i\beta $ such that
\[
(\alpha ,\beta )=\frac{1}{(\lambda ^2+(\eta ') 2)}_{\phi }
(\lambda _{\phi },-\eta _{\phi }' ).
\]
Then the nonlinear boundary condition (8) can be reduced to the
differential
equation of the conservation form for $f(\omega )$ by virtue of
the following

\begin{theorem}
Let the function $\lambda (\phi ,\psi )$ defined by
\begin{equation}
1+\frac{\epsilon \xi '}{1+\epsilon \xi _0}=e^{\lambda
(\phi ,\psi )}
\end{equation}
be differentiable.  We introduce the derivative operator $%
\partial _{\mathbf{n}}$ along the normal to $\psi =1$ by $\partial
_{\mathbf{%
n}}\mu =\lambda _{\psi }|_{\psi =1}$, where $\mu (\phi )
=\lambda (\phi ,1)$. Then on the free boundary, the
following three relations hold:
\begin{equation}
\beta _{\phi }=-\partial _{\mathbf{n}}\alpha =-\partial _{\phi
}\partial _{\mathbf{n}}\left\{ (\tau +\frac{1+\epsilon ^{2}}{2}
)e^{2\mu }\right\} ,
\end{equation}
where $\tau =(b-\mu \mathcal{F}^{-2})(1+\epsilon
)^{2}$, and $b=const$ is the Bernoulli constant.
\end{theorem}

\paragraph{Proof.} Since the velocity potential $\omega $ defined by  (10)
is analytic function, $\xi $ and $\eta $ are single-valued and they satisfy
the Cauchy-Riemann equations
\[
\frac{\epsilon }{1+\epsilon \xi }\xi _{\phi }'=\eta _{\psi
}',\quad \frac{\epsilon }{1+\epsilon \xi }\xi _{\psi
}'-\frac{\epsilon ^{2}(1+\epsilon \xi )^{-1}}{
(1+\epsilon \xi _0)}\xi _{0_{\psi }}=-\eta _{\phi}'
\]
which can be simplified as
\begin{equation}
\lambda _{\phi }=\eta _{\psi }',\mbox{ \qquad }\lambda _{\psi
}=-\eta _{\phi }'.
\end{equation}
 From (11), (14) and the representation $\zeta =(1+\epsilon \xi
')e^{i\eta '}$, it follows that
\begin{equation}
\left| \frac{d\zeta }{d\phi }\right| ^{2}=\epsilon ^{2}(
\epsilon ^{-1}+1)^{2}e^{2\lambda }\lambda _{\phi }^{2}+(
1+\epsilon \xi )^{2}\eta _{\phi }^{^{2}}
\end{equation}
since $\xi _0=1$ on the free boundary.

By virtue of the  (14) and (15) and presentation
$\eta _{\phi }=-\log (1+\epsilon )-\partial _{\mathbf{n}}\mu $,
the Bernoulli equation (8) takes the form
\[
\tau [e^{2\lambda }\lambda _{\phi }^{2}+\frac{1}{(
1+\epsilon )^{2}}((1+\epsilon \xi ')
^{2}(\log (1+\epsilon )+\partial _{\mathbf{n}}\mu
)^{2})]-\frac{1}{2}=0
\]
which can be transformed to the conservation law i.e.,
\begin{equation}
\alpha (\phi ,1)=\frac{\partial }{\partial \phi }(
\tau +\frac{(1+\epsilon ^{2})}{2})e^{2\mu }.
\end{equation}
Thus, the first equation in (13) holds due to analyticity of function $%
f(\omega )$ and the second equation in (13) follows from
the definition of the normal derivative operator $\partial _{\mathbf{n}}$.
Finally, the last equation is the consequence of the changing of the order
of differentiation $\partial _{\phi }\partial _{\mathbf{n}}\mu =\partial
_{\mathbf{n}}\mu _{\phi }$. \hfill$\diamondsuit$\smallskip


Note that function $\mu (\phi )$ is being found by
analyticity of $f(\omega )$ and thus transformation $\xi (
\phi ,\psi )$ is determined by definition (12) as
\[
\xi '=(\frac{1}{\epsilon }+\xi _0(\psi )
)(e^{\lambda (\phi ,\psi )}-1).
\]

\subsection*{Solution to the Dirichlet Problem in a fixed domain}

In view of Theorem 1, it follows from the definition for function $\alpha
(\phi ,1)$ and (16) that integrating (13) over $\phi $ along $\psi =1$
leads
to the following equation on the free boundary:
\[
4\tau e^{2\mu }[\log (1+\epsilon )+\partial _{\mathbf{n}}\mu ]
+\partial _{\mathbf{n}}([2\tau +1+\epsilon ^{2}]e^{2\mu })
-\epsilon \delta _0=0\,,
\]
where $\delta _0$ is the constant of integrating which represent the
horizontal impulse flow.

Finally, simplifying the last equation we arrive to the Dirichlet problem
in
the fixed domain
\begin{eqnarray}
\lambda _{\phi \phi }+\lambda _{\psi \psi }=0 \quad (0<\psi <1),&& \\
\lambda (\phi ,0)=0,\quad \lambda (\phi,1)=\mu ,&& \\
\mu \partial _{\mathbf{n}}\mu \mathcal{F}^{-2}-(b+\frac{1}{4}[1-
\mathcal{F}^{-2}]-\frac{\epsilon }{2})\partial _{\mathbf{n}}\mu
&&\nonumber\\
+\frac{\log (1+\epsilon )}{2}(\mu \mathcal{F}
^{-2}-b)+\frac{\delta e^{-2\mu }}{(1+\epsilon )^{2}}& =& 0\,,
\end{eqnarray}
where we denote $\delta =\frac{\delta _0}{8}.$

Now the problem (17)-(19) is reduced to finding of one function
$\mu (\phi )$ since if function $\mu $ is known then function
$\lambda (\phi ,\psi )$ is defined as the solution of the
mixed problem for the Laplace equation (17) and the boundary conditions
(18). In particular, $\lambda _{\psi }|_{\psi =1}$ can be considered as the
result of action of the operator $\partial _{\mathbf{n}}$ on function
$\mu$.
Namely, we represent $\lambda (\phi ,\psi )$ by the Fourier
series (see for example Ovsjannikov [8] or Stoker [9]). Then the dependence
between $n-th$ Fourier coefficients of functions $\lambda $, $\mu $ and $%
\partial _{\mathbf{n}}\mu $ is given by
\begin{equation}
[\lambda (\phi ,\psi )]_{n}=\frac{\sinh n\psi }{
\sinh n}\mu _{n},\quad [\mu (\phi )]_{n}=\mu _{n}\quad
[\partial _{\mathbf{n}}\mu (\phi)]_{n}=\mu _{n}\cot n\,,
\end{equation}
in which $\mu (\phi )=\mu _{n}e^{in\phi }$ (summation is
assumed). Thus problem (17)-(19) is written in terms of $[\mu (
\phi )]_{n}$ only. Since $[\partial _{\mathbf{n}}\mu ]_{n}$ are given by
(20),
we can represent the disturbance $\mu(\phi )$ by the expansion in series
 with respect to parameter $\epsilon $ (see also Ovsjannikov [8]).
Consequently we apply
the stretching transformation and expansion
\begin{equation}
(\mu ,\partial _{\mathbf{n}}\mu ,)= \sum_0^\infty
 \epsilon ^{i}\left\{ \epsilon (\overline {\mu }_{i},\epsilon
 (\overline {\partial} _{\mathbf{n}}
\overline{\mu })_{i})\right\} \quad (i=0,\infty)
\end{equation}
which is a characteristic of a shallow water. Substitution of
representation
(21) into (19) and elimination $\mathop{\rm mod}\epsilon ^{3}$ (neglecting
of the terms with $\epsilon ^{m}$, $m\geqslant 4)$ yields the approximate
solution of the form $[\mu _{i}]_{n}=[\mu _{i}(b,%
\mathcal{F})]_{n}$ as follows:
\begin{eqnarray}
&&\delta -\frac{b}{2}+
+\epsilon \big\{ -2\delta +\frac{b}{4}+\frac{\mathcal{F}^{-2}}{2}\mu
_{1}-\frac{1}{4}[1-\mathcal{F}^{-2}+4b](\partial _{
\mathbf{n}}\mu )_{1}-2\delta \mu _{1}\big\}  \nonumber\\
&&+\epsilon ^{2}\big\{ 3\delta -\frac{b}{6}-\frac{\mathcal{F}^{-2}}{4}\mu
_{1}+
\frac{\mathcal{F}^{-2}}{4}\mu _{2}+\frac{1}{2}(\partial _{\mathbf{n}
}\mu )_{1}-\frac{1}{4}[1-\mathcal{F}^{-2}+4b](
\partial _{\mathbf{n}}\mu )_{2} \nonumber\\
&&+4\delta \mu _{1}+2\delta \mu _{1}^{2}-2\delta \mu _{2}+\mu _{1}(
\partial _{\mathbf{n}}\mu )_{1}\big\}
+\epsilon ^{3}\big\{-4\delta +\frac{b}{8}+\frac{\mathcal{F}^{-2}}{6}\mu
_{1}-
\frac{\mathcal{F}^{-2}}{4}\mu _{2} \nonumber \\
&&+\frac{\mathcal{F}^{-2}}{2}\mu _{3}+\frac{1
}{2}(\partial _{\mathbf{n}}\mu )_{2}
-\frac{1}{4}[1-\mathcal{F}^{-2}+4b](\partial _{\mathbf{n}}\mu )_{3}
-6\delta \mu _{1}-4\delta \mu _{1}^{2}-\frac{4}{3}\delta \mu_{1}^{3}
\nonumber\\
&&+4\delta \mu _{2}-2\delta \mu _{3}+4\delta \mu _{1}\mu _{2}+\mu
_{1}(\partial _{\mathbf{n}}\mu )_{2}+(\partial _{\mathbf{n
}}\mu )_{1}\mu _{2}\big\}+o(\epsilon ^{4}).
\end{eqnarray}
Thus, in view of (20), Expression (22) represents the recurrent system of
algebraic equations for determination of all $[\mu _{i}]_{n}$,
where the horizontal impulse flow has asymptotic $\delta =\frac{b}{2}.$

The shape of the free boundary $h(\theta )$ can be determined
numerically using Okmaoto's method [3]. The existence of exact solution $%
(\psi ,h)$ can be established analytically by the Fixed Point
Theorem (see, for example, Okamoto \& Shoji [7] or Ibragimov [4]).



\section{Behavior of Tides waves}

\subsection*{\protect\bigskip Existence of stationary waves}

The main concern of this part is the evolution of tides around the Earth in
time $t$. In order to bring out the essential parameters of the problem,
the
dimensional fundamental equations, (1)-(2), are written in
non-stationary case as follows:
\begin{eqnarray*}
&\Delta (\psi )=0\quad (\mbox{in }\Omega_{h}),  &\\
&\psi_\theta =0\quad  (\mbox{on }\Gamma _{R}),  &\\
&rh_{t}+\psi_\theta
+\psi_rh_{\theta }=0\mbox{ \qquad }(\mbox{on }\Gamma_h), \\
&-h_{\theta }\psi _{t\theta }+r^{2}\psi _{tr}+\frac{r}{2}(\frac{\psi
_{\theta }^{2}}{r^{2}}+\psi_r^{2})_\theta +rgh_{\theta }=0\quad
 (\mbox{on }\Gamma_h), &
\end{eqnarray*}
where $\Delta =(\partial _{\theta \theta }+r^{2}\partial
_{rr}+r\partial_r).$

The perturbed quantities $h'$ and $\psi '$ are
interrelated as
\[
h=h',\quad \psi =-\frac{\gamma }{2\pi }\log r+\psi ',
\]
where $\gamma $ is the intensity of the vortex. For the small disturbances
we obtain the linear problem in the domain $D_0=\left\{ (r,\theta
):R\leqslant r\leqslant R+h_0, 0\leqslant \theta \leqslant
2\pi \right\} $ as follows
\begin{eqnarray}
&\Delta (\psi ')=0\quad (\mbox{in } D_0),& \\
&\psi_\theta '=0\quad  (\mbox{on }\Gamma _{R}), &\\
&h_{\theta }'+\frac{\psi_\theta '}{r}-\frac{\gamma
h_{\theta }'}{2\pi r^{2}}=0\quad  (\mbox{on }\Gamma_{h_0}),& \\
&r^{2}\psi _{tr}'-\frac{\gamma }{2\pi }\psi _{r\theta }^{\prime
}+rgh_{\theta }'=0\quad  (\mbox{on }\Gamma _{h_0})\,.&
\end{eqnarray}

Since Equations (23)-(26) are linear, the method of superposition is
applicable
(see also Friedrichs [2]). Hence it is sufficient to look for periodic
solutions of the form
\begin{equation}
(h',\psi ')=(H,\Psi (r))\exp \left\{ i(k\theta -wt)\right\}
\end{equation}
in which the wave number $k$ is a given real quantity and eigenvalues $w$
give the different modes of the tide's wave propagation. Substitution of
representation (27) into (23)-(26) leads to the expression
\[
\Psi (r)=c(r^{k}-R^{2k}r^{-k})
\]
and to the equations
$$\displaylines{ 
H(w+\frac{k\gamma }{2\pi r^{2}})-\frac{kc}{r}(
r^{k}-R^{2k}r^{-k})=0, \cr 
\frac{kg}{cr}H-(w+\frac{k\gamma }{2\pi r^{2}})(
kr^{k-1}+kR^{2k}r^{-k-1})=0 
}$$
in which $c$ is a constant of integrating.
Consequently, the determinantal equation for the longitudinal tide wave is
as follows 
\begin{equation}
w=\pm \sqrt{\frac{\frac{kg}{(R+h_0)}\big[(
R+h_0)^{k-1}-R^{2k}(R+h_0)^{-k-1}\big]}{(
R+h_0)^{k-1}+R^{2k}(R+h_0)^{-k-1}}}-\frac{k\gamma }{%
2\pi (R+h_0)^{2}}
\end{equation}

Thus surface tide waves (on the constant flow) are dispersive with two
different modes of propagation. Simplification of relation (28) shows that
the tide wave is propagated with a speed 
\[
a_0=\frac{w}{k}=\pm \sqrt{gh_0}\sqrt{\frac{\tanh [k\ln (
1+\epsilon )]}{kRh_0(1+\epsilon )}}-\frac{%
\gamma }{2\pi R^{2}(1+\epsilon )^{2}}. 
\]
Hence the condition of the existence of stationary tide waves ($a_0=0$) is 
\[
\left| \gamma 
\right| \leqslant 2\pi R\epsilon ^{-\frac{1}{2}}(
1+\epsilon )^{2}\sqrt{gh_0}.
\]

\subsection*{Splitting phenomena for shallow water equations}

We suppose that the parameter $\epsilon $ is infinitesimally small. So we
consider $R$ as the natural physical scale. Note that kinematic condition
can be written as the mass balance equation. Namely,
\begin{equation}
h_{t}+(R+h)^{-1}\partial_\theta  \int_R^{R+h} v^\theta dr=0
\end{equation}
since the radial velocity component is given by
\[
v^r=-r^{-1} \int_R^{R+h} v_{\theta }^\theta \,dr\,.
\]
Hence the mass balance equation (29) takes the form
\[
rh_{t}+\partial_\theta (uh)=0\quad (\mbox{on }%
\Gamma_h),
\]
where the average velocity $u(\theta ,t)$ is defined by
relation
\[
u(\theta ,t)=h^{-1} \int_R^{R+h} v^\theta (r,\theta ,t)\,dr\,.
\]
To go further, it is better to introduce an nondimensionalization here. We
put
\[
t=\frac{R}{U}t',\quad \psi =h_0U\psi ',\quad u=Uu',
\]
where $U$ is a unit of velocity. Hereafter index prime will be omitted.
Then
the impulse equation is written as
\begin{equation}
-\frac{\epsilon ^{2}h_{\theta }\psi _{t\theta }}{(1+\epsilon
h)^{2}}+\psi _{tr}+\frac{1}{2(1+\epsilon h)}\partial
_{\theta }(\frac{\epsilon ^{2}\psi_\theta ^{2}}{(
1+\epsilon h)^{2}}+\psi_r^{2})+\frac{h_{\theta }}{(
1+\epsilon h)}=0.
\end{equation}
We represent the stream function $\psi $ by the Lagrangian expansion (see
also Ovsjannikov [8] or Friedrichs [2])
\[
\psi =\sum_{i=0}^\infty \epsilon ^{i}\psi^{(i)}\,.
\]
Then the Laplace equation takes the form ($\mathop{\rm mod}\epsilon ^{2}$)
\begin{eqnarray}
\psi _{rr}^{(0)}+\epsilon (\psi _{rr}^{(1)
}+2r\psi _{rr}^{(0)}+\psi_r^{(0)}) &&\\
+ \epsilon ^{2}(\psi _{\theta \theta }^{(0)}+\psi
_{rr}^{(2)}+2r\psi _{rr}^{(1)}+r^{2}\psi
_{rr}^{(0)}+\psi_r^{(1)}+r\psi_r^{(0)})&=&0\,. \nonumber
\end{eqnarray}
Equation (31) represents the recurrent system of differential equations for
the
determination of $\psi ^{(i)}$ as the solution of the Cauchy
problem with boundary conditions $\psi (0,\theta ,t)=0$, $\psi
(1+h,\theta ,t)=uh$ for $\psi ^{(0)}$ and zero
boundary conditions for $\psi ^{(1)}$ and $\psi ^{(
2)}$. Hence function $\psi $ ($\mathop{\rm mod}\epsilon ^{2}$) is as
follows:
\begin{equation}
\psi =ur+\epsilon (u\frac{r^{2}}{2}-uh\frac{r}{2})
+\epsilon ^{2}(-u_{\theta \theta }\frac{r^{3}}{6}+uh\frac{r^{2}}{4}%
+u_{\theta \theta }h^{2}\frac{r}{6}-uh^{2}\frac{r}{4}).
\end{equation}
We use the Tailor expansion
\begin{equation}
(1+\epsilon h)^{-1}=1-\epsilon h+(\epsilon
h)^{2}+\cdot \cdot \cdot
\end{equation}
to write (30) as
\begin{equation}
\psi _{tr}+\frac{1}{2}(\epsilon ^{2}\psi_\theta ^{2}+\psi
_{r}^{2})_\theta -\epsilon ^{2}h_{\theta }\psi _{t\theta
}+(\epsilon h-1)(\frac{1}{2}\epsilon h(\psi
_{r}^{2})_\theta +\epsilon hh)_\theta +h_{\theta }=0\,.
\end{equation}

Multiply (30) by $(1+\epsilon h)^{2}$ and then use expansion
(33). Then (30) becomes
\begin{equation}
\psi _{tr}+\frac{1}{2}(\epsilon ^{2}\psi_\theta ^{2}+\psi
_{r}^{2})_\theta -\epsilon ^{2}h_{\theta }\psi _{t\theta
}+\epsilon h(\psi _{tr}(2+\epsilon h)+\frac{1}{2}(\psi
_{r}^{2})_\theta +h_{\theta })+h_{\theta }=0\,.
\end{equation}
If we substitute $\psi $ defined by (32) into (34), we obtain the
following equation of the shallow water theory
\begin{eqnarray}
u_{t}+uu_{\theta }+h_{\theta }+\epsilon (\frac{h}{2}u_{t}-\frac{u}{2
}h_{t}+u^{2}h_{\theta }-hh_{\theta })  &&\nonumber\\
+\epsilon ^{2}\bigg(hh_{\theta }u_{\theta t}-\frac{h^{3}}{3}u_{\theta
\theta t}+\frac{h^{2}}{4}u_{t}+\frac{h}{3}u_{\theta
\theta }h_{t}+hh_{\theta
}u_{\theta }^{2}  &&\\
+h^{2}u_{\theta }u_{\theta \theta }+\frac{3}{4}uu_{\theta }+hu^{2}h_{\theta
}-\frac{1}{3}u_{\theta }u_{\theta \theta }-\frac{1}{3}uu_{\theta \theta
\theta }+h^{2}h_{\theta }\bigg)&=&0\,. \nonumber
\end{eqnarray}
Consequently, the substitution of $\psi $ (32) into (35) yields
\begin{eqnarray}
u_{t}+uu_{\theta }+h_{\theta }+\epsilon \bigg(\frac{h}{2}u_{t}-\frac{u}{2}
h_{t}+\frac{h}{2}uu_{t}+\frac{u^{2}}{2}h_{\theta }+\frac{3}{2}huu_{\theta }
\bigg) && \nonumber \\
+\epsilon ^{2}\bigg(-hh_{\theta }u_{\theta t}-\frac{h^{2}}{3}u_{\theta
\theta t}+\frac{9}{4}h^{2}u_{t}+\frac{h}{3}h_{t}u_{\theta \theta
}-uhh_{t} && \\
 +hh_{\theta }u_{\theta }^{2} +h^{2}u_{\theta }u_{\theta \theta }
+\frac{3}{4}h^{2}uu_{\theta }+\frac{1}{4}u^{2}hh_{\theta }-\frac{1}{3}%
u_{\theta \theta \theta } && \nonumber\\
+\frac{3}{4}h^{2}u_{\theta }+\frac{1}{2}uh_{\theta
}+\frac{uh}{2}h_{\theta }\bigg)&=&0\,. \nonumber
\end{eqnarray}
Equatrions (36) and (37) supplied with the kinematic condition
\begin{equation}
\partial _{t}(\epsilon h^{2}+2h)+2\partial_\theta (
uh)=0
\end{equation}
represent two systems of the shallow water equations.

To verify that these two systems are different, we compute their first
integrals in the stationary case as follows
$$\displaylines{
h+\epsilon (2c^{2}h-\frac{1}{2}h^{2})+\epsilon ^{2}(
3c^{2}h^{2}-\frac{1}{3}h^{3})=J_{1},  \cr
h+\frac{1}{2}\epsilon c^{2}h+\frac{1}{2}\epsilon ^{2}(\frac{17}{4
}c^{2}h^{2}-\frac{1}{2}c^{2}h^{2}-ch) = J_{2},
}$$
where $c$, $J_{1}$, $J_{2}$ are constants of integration.
Obviously, $J_{1}\neq J_{2}$.

At first sight, it seems that the problem (23)-(26) does not have a unique
solution because of that fact. However, it can be shown that the solution
of
the problem is invariant with respect to the decomposition of the function
which represent the free boundary. Since it is not difficult, the proof is
omitted.

\paragraph{Acknowledgments}
I wish to express my gratitude to Prof. Makarenko N. of Novosibirsk State
University, Russia, for his helpful comments and advice through this study.
In addition, I would like to thank Prof. Ovsjnnikov L.V. for valuable
discussions on this subject on seminars at Lavryentiev's Institute of
Hydrodynamics, Russia.
I am grateful to Prof. Okamoto H. of Research Institute of Mathematical
Sciences (Kyoto University) for arranging my visit to Japan and encouraging
discussions.
This work was supported by NSERC research grant (Canada)

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\noindent{\sc Ranis N. Ibragimov} \\
Department of Mathematics and Statistics\\
University of Victoria \\
P.O. Box 3045 STN CSC \\
Victoria, B.C., V8P 2A1, Canada \\
e-mail: ranis@nbnet.nb.ca
\end{document}