Graphics Reference
In-Depth Information
With this choice,
φ
now satisfies the Laplace equation in the interior of the
fluid and the bottom boundary condition. Let's write out what we have
for this Fourier mode so far:
φ
(
x, y, z, t
)=
φ
ij
(
t
)
e
√
−
12
π
(
ix
+
jz
)
/L
e
ky
+
e
−
2
kH
e
−ky
.
(13.1)
All that's left to determine is the time dependence of the potential
φ
ij
(
t
)
and the height field
h
ij
(
t
), and the only equations we have left are the
boundary conditions at
y
=0:
∂φ/∂t
=
gh
and
∂h/∂t
=
∂φ/∂y
.These
two boundary equations at the free surface become (after cancelling out
the common
e
√
−
12
π
(
ix
+
jz
)
/L
factor):
∂ φ
ij
∂t
−
gh
ij
,
(1 +
e
−
2
kH
)=
−
(13.2)
∂h
ij
∂t
φ
ij
(
k
ke
−
2
kH
)
.
=
−
Now, differentiate the second equation with respect to time, and replace
the
∂ φ
ij
/∂t
term with the first equation, to get
∂
2
h
ij
∂t
2
e
−
2
kH
kg
1
−
1+
e
−
2
kH
h
ij
.
=
−
(13.3)
This is another simple ordinary differential equation, with general solution
consisting of yet more Fourier sinusoids,
e
√
−
1
ω
k
t
and
e
−
√
−
1
ω
k
t
where the
wave frequency
ω
k
(how fast the wave is going up and down, no relation
at all to vorticity) is given by
kg
1
e
−
2
kH
1+
e
−
2
kH
.
−
ω
k
=
(13.4)
Before going on to the full solution, and accompanying numerical method,
it's instructive to pause a moment and reinterpret this height field solution.
Writing out the one of the components of the height field gives
h
(
x, z, t
)=
e
−
√
−
1
ω
k
t
e
√
−
12
π
(
ix
+
jz
)
/L
=
e
√
−
1(
k·
(
x,z
)
−ω
k
t
)
=
e
√
−
1
k·
[(
x,z
)
−c
k
tk
]
,
(13.5)
where
k
=
k/k
is the unit-length direction of the wave (normal to the crests
and troughs) and
c
k
is the wave speed, defined as
ω
k
k
c
k
=
g
(1
(13.6)
e
−
2
kH
)
k
(1 +
e
−
2
kH
)
.
−
=
