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s + i
n =1 J n ( kr )( C 3 cos( n ϕ )+ C 4 sin( n ϕ ))
+
d k
d p exp
{
pt
}
0
s
i
( C 5 ( p , k ) cosh( kz )+ C 6 ( p , k ) sinh( kz )) .
×
(2.48)
Substitution of equation (2.48) into the boundary condition on the surface (2.30)
yields the relationship between the coefficients,
C 5 ( p , k ) p 2
C 6 ( p , k )=
gk .
(2.49)
We shall now write the integral representation for the function describing
the space-time law of motion of the basin floor,
r ( r )
η ϕ (
t ( t )
η
( r ,
ϕ
, t )=
η
ϕ
)
η
, t )=
s + i
H 0 ( p , k )+
s + i
J 0 ( kr ) k A 0
2
η
( r ,
ϕ
d k
dp exp
{
pt
}
d k
dp
0
s
i
0
s
i
n =1 J n ( kr ) k ( A n cos( n ϕ )+ B n sin( n ϕ )) H n ( p , k ) ,
×
exp
{
pt
}
(2.50)
where
π
1
π
η ϕ (
A n =
ϕ
) cos( n
ϕ
) d
ϕ
,
π
π
1
π
η ϕ (
ϕ
ϕ
ϕ
,
B n =
) sin( n
) d
π
1
2
H n ( p , k )=
r ( r )
t ( t ) exp
d t
d r
η
η
{−
pt
}
rJ n ( kr ) .
π
i
0
0
Substitution of formulae (2.48) and (2.50) into the boundary condition on the basin
floor (2.31) reveals that equality of the left-hand and right-hand parts is possible,
only when the following three conditions are fulfilled:
C 3 = A n ,
C 4 = B n ,
pH n ( p , k )
k sinh( kH )+ p 2
C 5 ( p , k )=
.
g k cosh( kH )
It is now possible to write out the resultant expression for the potential, which is
the solution of equation (2.43) with the boundary conditions (2.30) and (2.31)
 
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