Simulation of Standing and Propagating Sea Waves . . .
265
𝜙 into the boundary condition yields
𝜁 t = (if (x) − 1)
[
D 1 (x, z) ∗ F
−1
x {2𝜋uE(u)}
] ,
where f (x) = 𝜁 x ∕
√
1 + 𝜁 2
x − 𝜁 x . Applying Fourier transform to both sides of this
equation yields formula for coefficients E:
E(u) =
1
2𝜋u
F u
{
𝜁 t ∕
(
if (x) − 1∕
√
1 + 𝜁 2
x
)}
F u
{
D 1 (x, z)
}
Finally, substituting z for 𝜁 (x, t) and plugging resulting equation into (16) yields formula for 𝜙(x, z):
𝜙(x, z) = F
−1
x
⎧
⎪
⎨
⎪
⎩
e 2𝜋uz
2𝜋u
F u
{
𝜁 t ∕
(
if (x) − 1∕
√
1 + 𝜁 2
x
)}
F u
{
D 1 (x, 𝜁(x, t))
}
⎫
⎪
⎬
⎪
⎭
.
(17)
Multiplier e 2𝜋uz ∕(2𝜋u) makes a graph of a function to which Fourier transform
is applied asymmetric with respect to OY axis. This makes it difficult to apply FFT
which expects periodic function with nought on both ends of the interval. Using
numerical integration instead of FFT is not faster than solving the initial system of
equations with numerical schemes. This problem is alleviated by using formula (19)
for finite depth fluid with wittingly large depth h. This formula is derived in the
following section.
Formula for Finite Depth Fluid
On the sea bottom vertical fluid velocity component equals nought: 𝜙 z = 0 on z =
−h, where h—water depth. In this case equation v = −iu, which came from Laplace
equation, can not be neglected, hence the solution is sought in the following form:
𝜙(x, z) = F
−1
x
{(
C 1 e
2𝜋uz
+ C 2 e
−2𝜋uz
)
E(u)
} .
(18)
Plugging 𝜙 into the boundary condition on the sea bottom yields
C 1 e
−2𝜋uh
− C 2 e
2𝜋uh
= 0,
hence C 1 =
1
2
Ce 2𝜋uh and C 2 = −
1
2
Ce −2𝜋uh . Constant C may take arbitrary value here,
because after plugging it becomes part of unknown coefficients E(u). Plugging formulae for C 1 and C 2 into (18) yields
𝜙(x, z) = F
−1
x {cosh (2𝜋u(z + h)) E(u)} .