Solution in the Semi-infinite Domains
We now solve Laplace's equation in the semi-infinite domains [math]\displaystyle{ \Omega
^{-}=\left\{ x\lt -l,\;-1\leq z\leq 0\right\} }[/math] and [math]\displaystyle{ \Omega ^{+}=\left\{
x\gt l,\;-H\leq z\leq 0\right\} }[/math] which are shown in Figure (\ref{fig_region}).
Since the water depth is constant in these regions we can solve Laplace's
equation by separation of variables. The potential in the region [math]\displaystyle{ \Omega
^{-} }[/math] satisfies the following equation
[math]\displaystyle{
\left.
\begin{matrix}
\nabla ^{2}\phi =0,\;\;\mathbf{x}\in \Omega ^{-}, \\
\phi _{n}-\nu \phi =0,\;\;z=0, \\
\phi _{n}=0,\;\;z=-1, \\
\phi =\tilde{\phi}\left( z\right) ,\;\;x=-\,l, \\
\lim\limits_{x\rightarrow -\infty }\phi \left( x,z\right) =\cosh \left(
k_{t}^{\left( 1\right) }\left( z+1\right) \right) e^{ik_{t}^{\left( 1\right)
}x} \\
+R\,\cosh \left( k_{t}^{\left( 1\right) }\left( z+1\right) \right)
e^{-ik_{t}^{\left( 1\right) }x},
\end{matrix}
\right\} (13)
}[/math]
where [math]\displaystyle{ \mathbf{x}=\left( x,z\right) \ }[/math]and [math]\displaystyle{ \tilde{\phi}\left( z\right) }[/math] is
an arbitrary continuous function[math]\displaystyle{ . }[/math]\ Our aim is to find the outward normal
derivative of the potential on [math]\displaystyle{ x=-l }[/math] as a function of [math]\displaystyle{ \tilde{\phi}\left(
z\right) }[/math].
We solve equation (13) by separation of variables \cite{Liu82,
Hazard} and obtain the following expression for the potential in the region [math]\displaystyle{
\Omega ^{-}, }[/math]
[math]\displaystyle{ \begin{matrix}
\phi \left( x,z\right) &=&\cosh \left( k_{t}^{\left( 1\right) }\left(
z+1\right) \right) \,e^{ik_{t}^{\left( 1\right) }x}+R\cosh \left(
k_{t}^{\left( 1\right) }\left( z+1\right) \right) \,e^{-ik_{t}^{\left(
1\right) }x} \\
&&+\sum_{m=1}^{\infty }\left\langle \tilde{\phi}\left( z\right) ,\tau
_{m}^{\left( 1\right) }\left( z\right) \right\rangle \tau _{m}^{\left(
1\right) }\left( z\right) e^{k_{m}^{\left( 1\right) }\left( x+l\right) }.
(14)
\end{matrix} }[/math]
The functions [math]\displaystyle{ \tau _{m}^{\left( 1\right) }\left( z\right) }[/math] ([math]\displaystyle{ m\geq 1) }[/math] are
the orthonormal modes given by
[math]\displaystyle{
\tau _{m}^{\left( 1\right) }\left( z\right) =\left( \frac{1}{2}+\frac{\sin
\left( 2k_{m}^{\left( 1\right) }\,\right) }{4k_{m}^{\left( 1\right) }}
\right) ^{-\frac{1}{2}}\cos \left( k_{m}^{\left( 1\right) }\left(
z+1\right) \right) ,\;\;m\geq 1.
}[/math]
The evanescent eigenvalues [math]\displaystyle{ k_{m}^{\left( 1\right) } }[/math] are the positive real
solutions of the Dispersion Relation for a Free Surface
[math]\displaystyle{
-k_{m}^{\left( 1\right) }\,\tan \left( k_{m}^{\left( 1\right) }H_{j}\right)
=\nu ,\;\;m\geq 1, (15)
}[/math]
ordered by increasing size. The inner product in equation (13) is
the natural inner product for the region [math]\displaystyle{ -1\leq z\leq 0 }[/math] given by
[math]\displaystyle{
\left\langle \tilde{\phi}\left( z\right) ,\tau _{m}^{\left( j\right) }\left(
z\right) \right\rangle =\int_{-1}^{0}\tilde{\phi}\left( z\right) ,\tau
_{m}^{\left( j\right) }\left( z\right) dx. (16)
}[/math]
The reflection coefficient is determined by taking an inner product of
equation (14) with respect to [math]\displaystyle{ \cosh \left( k_{t}^{\left( 1\right)
}\left( z+1\right) \right) . }[/math] This gives us the following expression for [math]\displaystyle{ R }[/math],
[math]\displaystyle{
R=\frac{\left\langle \tilde{\phi}\left( z\right) ,\cosh \left(
k_{t}^{\left( 1\right) }\left( z+1\right) \right) \right\rangle }{\frac{1}{2
}+\sinh \left( 2k_{t}^{\left( 1\right) }\,\right) /4k_{0}^{\left( 1\right) }}
e^{-ik_{t}^{\left( 1\right) }l}-e^{-2ik_{t}^{\left( 1\right) }l}.
(17)
}[/math]
The normal derivative of the potential on the boundary of [math]\displaystyle{ \Omega ^{-} }[/math] and [math]\displaystyle{
\Omega }[/math] [math]\displaystyle{ \left( x=-l\right) }[/math] is calculated using equation (14) and
we obtain,
[math]\displaystyle{
\left. \phi _{n}\right| _{x=-l}=\mathbf{Q}_{1}\tilde{\phi}\left( z\right)
-2ik_{t}^{\left( 1\right) }\cosh \left( k_{t}^{\left( 1\right) }\left(
z+1\right) \right) \,e^{-ik_{t}^{\left( 1\right) }l},
}[/math]
where the outward normal is with respect to the [math]\displaystyle{ \Omega }[/math] domain. The
integral operator [math]\displaystyle{ \mathbf{Q}_{1} }[/math] is given by
[math]\displaystyle{ \begin{matrix}
\mathbf{Q}_{1}\tilde{\phi}\left( z\right) &=&\sum_{m=1}^{\infty
}k_{m}^{\left( 1\right) }\left\langle \tilde{\phi}\left( z\right) ,\tau
_{m}^{\left( 1\right) }\left( z\right) \right\rangle \,\tau _{m}^{\left(
1\right) }\left( z\right) (18) \\
&&+ik_{t}\frac{\left\langle \tilde{\phi}\left( z\right) ,\cosh \left(
k_{t}^{\left( 1\right) }\left( z+1\right) \right) \right\rangle \cosh \left(
k_{t}^{\left( 1\right) }\left( z+1\right) \right) }{\frac{1}{2}+\sinh
\left( 2k_{t}^{\left( 1\right) }\,\right) /4k_{t}^{\left( 1\right) }}.
\end{matrix} }[/math]
We can combine the two terms of equation (18) and express [math]\displaystyle{ \mathbf{Q}_{1} }[/math] as
[math]\displaystyle{
\mathbf{Q}_{1}\tilde{\phi}\left( z\right) =\sum_{m=0}^{\infty }k_{m}^{\left(
1\right) }\left\langle \tilde{\phi}\left( z\right) ,\tau _{m}^{\left(
1\right) }\left( z\right) \right\rangle \,\tau _{m}^{\left( 1\right) }\left(
z\right) (19)
}[/math]
where [math]\displaystyle{ k_{0}^{\left( 1\right) }=ik_{t}^{\left( 1\right) } }[/math] and
[math]\displaystyle{
\tau _{0}^{\left( 1\right) }\left( z\right) =\left( \frac{1}{2}+\frac{\sin
\left( 2k_{0}^{\left( 1\right) }\right) }{4k_{0}^{\left( 1\right) }}\right)
^{-\frac{1}{2}}\cos \left( k_{0}^{\left( 1\right) }\left( z+1\right)
\right) .
}[/math]
As well as providing a more compact notation, equation (19)
will be useful in the numerical calculation of [math]\displaystyle{ \mathbf{Q}_{1}. }[/math]
Similarly, we now consider the potential in the region [math]\displaystyle{ \Omega ^{+} }[/math] which
satisfies
[math]\displaystyle{
\left.
\begin{matrix}
\nabla ^{2}\phi =0,\;\;\mathbf{x}\in \Omega ^{+}, \\
\phi _{n}-\nu \phi =0,\;\;z=0, \\
\phi _{n}=0, \;\;z=-H, \\
\phi =\tilde{\phi}\left( z\right) ,\;\;x=\,l, \\
\lim\limits_{x\rightarrow -\infty }\phi \left( x,z\right) =T\cosh \left(
k_{t}^{\left( 2\right) }\left( z+H_{2}\right) \right) e^{ik_{t}^{\left(
2\right) }x}.
\end{matrix}
\right\} (20)
}[/math]
Solving equation (20) by separation of variables as before we obtain
[math]\displaystyle{
\left. \phi _{n}\right| _{x=l}=\mathbf{Q}_{2}\tilde{\phi}\left( z\right) ,
}[/math]
where the outward normal is with respect to the [math]\displaystyle{ \Omega }[/math] domain. The
integral operator [math]\displaystyle{ \mathbf{Q}_{2} }[/math] is given by
[math]\displaystyle{
\mathbf{Q}_{2}\tilde{\phi}\left( z\right) =\sum_{m=0}^{\infty }k_{m}^{\left(
2\right) }\left\langle \tilde{\phi}\left( z\right) ,\tau _{m}^{\left(
2\right) }\left( z\right) \right\rangle \tau _{m}^{\left( 2\right) }\left(
z\right) . (21)
}[/math]
The orthonormal modes [math]\displaystyle{ \tau _{m}^{\left( 2\right) } }[/math] are given by
[math]\displaystyle{
\tau _{m}^{\left( 2\right) }\left( z\right) =\left( \frac{H}{2}+\frac{\sin
\left( 2k_{m}^{\left( 2\right) }\,H\right) }{4k_{m}^{\left( 2\right) }}
\right) ^{-\frac{1}{2}}\cos \left( k_{m}^{\left( 2\right) }\left(
z+H\right) \right) ,
}[/math]
The eigenvalues [math]\displaystyle{ k_{m}^{\left( 2\right) } }[/math] are the positive real solutions [math]\displaystyle{
\left( m\geq 1\right) }[/math] and positive imaginary solutions [math]\displaystyle{ \left( m=0\right) }[/math]
of the dispersion equation
[math]\displaystyle{
-k_{m}^{\left( 2\right) }\,\tan \left( k_{m}^{\left( 2\right) }H\right) =\nu.
}[/math]
The inner product is the same as that given by equation (16)
except that the integration is from [math]\displaystyle{ z=-H }[/math] to [math]\displaystyle{ z=0. }[/math] The transmission
coefficient, [math]\displaystyle{ T, }[/math] is given by
[math]\displaystyle{
T=\frac{\left\langle \tilde{\phi}\left( z\right) ,\cosh \left(
k_{t}^{\left( 2\right) }\left( z+1\right) \right) \right\rangle }{\frac{1}{2
}+\sinh \left( 2k_{t}^{\left( 2\right) }H\right) /4k_{t}^{\left( 2\right) }}
e^{-ik_{t}^{\left( 2\right) }l}. (22)
}[/math]
