Chapter 6. Wavemaker Theory
LCFSBC(Linearized Combined Free Surface Boundary Condition): Take time derivative of LDFSBC and substitute into LKFSBC.
0 1 on
2
2 =
∂
−∂
∂ =
∂ z
z t
g
φ φ
Since φ∝eiσt, ∂2φ/∂t2 =−σ2φ. Therefore,
0 on
0
2 = =
∂ −
∂ z
g z σ φ
φ (LCFSBC)
Lateral boundary condition:
(1) Kinematic boundary condition on wavemaker (2) Radiation condition: Waves outgoing at x=+∞
Paddle displacement is described by
z t x S sinσ
2 )
= (
z t x S t z x
F sinσ
2 ) ) (
, ,
( = −
F t n F
u ∇
∂
∂
= −
⋅ /
F k dz t
i dS
F n F
∇
−
∇ =
= ∇
sin ˆ 2
ˆ 1 σ
z t x S z t
t S dz dS
u w σ σ σ sinσ
2 ) on (
0 2 cos
) sin (
2 − = =
−
By Taylor series expansion about x=0
0 2 cos
) sin (
sin 2 2
) (
2 cos ) sin (
2
0 0
=
⎟ +
⎠
⎜ ⎞
⎝
⎛ − −
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛ − −
=
=
L
x x
z t t S dz dS u w
t x z
S
z t t S dz dS u w
σ σ σ
σ
σ σ σ
Assume S is small so that S=O(H). Then the linearized wavemaker KBC is
0 on
2 cos )
( =
∂ =
−∂
= S z t x
u φx σ σ
Referring to Table 3.1 of textbook, assume the solution as
(
Fe k hGez) (Hkx ktz IBx kCz)
Dz tE t
A t z x
x k x
ks s s s σ
σ σ
φ
cos sin
cos
cos ) )(
( ) sin(
) ( cosh )
, , (
+ +
+
+ +
+
− +
=
−
(
C k z D k z)
tBe t x k z h k A t z
x p p σ ksx s s σ
φ( , , )= cosh ( + )sin( − )+ − cos + sin cos
The first term satisfies BBC. For the second term to satisfy BBC, we need
(
sin( ) cos( ))
0 Be k C k h k D k h
z s s s s
x k h
z
s − − + −
−
=
∂ =
−∂ −
−
=
φ
0 ) cos(
)
sin(− + − =
−C ksh D ksh
h k C D=− tan s
(
cos cos sin sin)
cos ( )cos
sin tan
cos sin
cos
z h k E h k z k h k z h k
k C
z k h k C z k C z k D z k C
s s
s s
s s
s s
s s
s
+
=
−
=
−
= +
t z h k Be
t x k z h k A t z
x p p σ ksx s σ
φ( , , )= cosh ( + )sin( − )+ cos ( + )cos
∴ −
Applying LCFSBC,
0 on
0
2 = =
∂ −
∂ z
g z σ φ φ
{
cosh sin( ) cos cos}
0cos sin
) sin(
sinh
2
= +
−
−
−
−
−
−
t h k Be
t x k h k g A
t h k e
Bk t x k h k Ak
s x k p
p
s x k s p
p p
s s
σ σ σ
σ σ
h k gk h
g k h k k
h k gk
h g k
h k k
s s s
s s
p p
p p
p
tan 0
cos sin
tanh 0
cosh sinh
2 2
2 2
−
=
⇒
=
−
−
=
⇒
=
−
∴
σ σ σ σ
h h k
gk h
p p
tanh
2 =
σ
k h h
gk h
tan s
2 =−
σ
s
We have infinite number of roots, ks(n), n=1,2,L. Let us examine the first term:
h x e
h x e
h x e
e e ks x h x
3 for
009 . 0
2 for
043 . 0
for 208 . 0
2 / 3
2 / 2
/ )
1 (
=
=
=
=
=
=
=
=
=
≤
−
−
− −
−
π π π π
This term decays exponentially with the distance from the wavemaker, x. The other terms decay more rapidly with x. These terms are called evanescent modes. Now the velocity potential becomes
∑
∞=
− +
+
− +
=
1
)
( cos[ ( )( )]cos
) sin(
) ( cosh )
, , (
n
s x n k n p
p
p k h z k x t C e k n h z t
A t z
x σ s σ
φ
↑ ↑
progressive wave evanescent waves decaying exponentially with x
Ap and Cn(n=1 to∞) should be determined from WBC:
∑
∞= =
= +
+ +
−
∂ =
− ∂
0 1
2 cos ) cos (
)]
)(
( cos[
) ( cos
) ( cosh
n
s s
n p
p p x
z t t S z
h n k n k C t
z h k k
x A σ σ σ σ
φ
Examining variation over depth,
2 0
2
2 −k Z =
dz Z d
h dz z
dZ =0 on =−
0 on
0
2 = =
− Z z
g dz dZ σ
This is the Sturm-Liouville problem, which gives the orthogonality of eigenfunctions, or
n m dz
Z
hZm n = ≠
∫
−0 0 ifIn our problem, the eigenfunctions are and
. To find , multiply
) ( coshkp h+z L
, 2 , 1 )], )(
(
cos[ks n h+z n= Ap coshkp(h+z) on both side of WBC and integrate over depth:
∫
∫ ∑∫
−
∞
= −
−
+
=
+ +
+ +
−
0
1 0
0 2
) ( 2 cosh
) (
) ( cosh )]
)(
( cos[
) ( )
( cosh
h p
n h n s s p
h p p p
dz z h z k
S
dz z h k z
h n k n k C dz
z h k k
A σ
∫
∫
−
−
+ +
=−
∴ 0
2 0
) ( cosh
) ( 2 cosh
) (
h p
p
h p
p
dz z h k k
dz z h z k
S A
σ
Similarly,
∫
∫
−
−
+
= 0 +
2 0
)]
)(
( [ cos ) (
)]
)(
( 2 cos[
) (
h s
s
h s
n
dz z h n k n
k
dz z h n z k
S C
σ
For piston-type wavemaker, S
z S( )=
and for flap type wavemaker,
⎟⎠
⎜ ⎞
⎝⎛ +
= h
S z z
S( ) 1
where S = stroke at SWL.
Far from the wavemaker, where the evanescent modes are negligible,
) cos(
1 cosh )
2 cos( 0 A k h k x t
g t
t g x H k
p p
p z
p σ φ σ σ
η =− −
∂
= ∂
−
=
=
h k g A
H 2σ p cosh p
−
=
See versus in Figure 6.2 of textbook. See also Eqs. (6.25) and (6.26).
For small , S
H/ kph
h kp
⎪⎩
⎪⎨
⎧
= for flap wavemaker 2
emaker piston wav
for h k
h k S H
p p
Mean power (averaged over one wave period) required by the wavemaker is calculated by the energy flux away from the wavemaker:
∫ ∫
+ −=
= t T
t h d
g p udz
EC T
P 1 0
3-D Wavemaker
“Snake-type” wavemaker
The boundary value problem is
2 0
2 2 2 2 2
2 =
∂ +∂
∂ +∂
∂
=∂
∇ x y z
φ φ φ φ
h
z = z=−
∂
∂φ 0 on
0 on
0
2 = =
∂ −
∂ z
g z σ φ φ
0 on
) 2 cos(
)
( − =
∂ =
−∂ S z y t x
x σ λ σ
φ
Assume φ(x,y,z,t)= X(x)Z(z)Y(y,t). z -problem is the same as 2-D wavemaker problem. Thus,
∑
∞=
+ +
+
=
1
) )(
( cos )
( cosh )
(
n
s
n k n h z
C z
h k A z Z
with σ2 =gktanhkh and σ2 =−gks tanksh. Periodicity condition in y-direction gives
) cos(
) sin(
) ,
(y t C y t D y t
Y = λ −σ + λ −σ
Substituting these into the Laplace equation,
1 2 2 0
2
2 ±k −λ =
dx X d X
where + is for progressive wave, and − is for evanescent waves. The solutions are
x k
F x k
E
X = cos 2 −λ2 + sin 2 −λ2
x k x
k s
s
s Fe
Ee
X = 2+λ2 + − 2+λ2
Now the velocity potential is given by
∑
∞=
+
− −
+ +
− +
− +
=
1
2 2
) cos(
)]
)(
( cos[
) sin(
) ( cosh
2 2
n
x k s
n k n h z e y t
C
t y x k
z h k A
s λ σ
σ λ λ φ
λ
WBC gives
) 2 cos(
) (
) cos(
)]
( cos[
) cos(
) ( cosh
1
2 2 2 2 0
t z y
S
t y z
h k k
C
t y z
h k k
x A
n
s s
n x
σ λ σ
σ λ λ
σ λ φ λ
−
=
− +
+ +
− +
−
−
∂ =
−∂
∑
∞=
=
Use orthogonality to determine A and Cn.
Cylindrical wavemaker
Spiral wavemaker (Type III in text)
Boundary value problem:
1 0 1
2 2 2 2 2 2
2
2 =
∂ + ∂
∂ + ∂
∂ + ∂
∂
=∂
∇ r r r r z
φ θ
φ φ
φ φ
h
z = z=−
∂
−∂φ 0 on
0 on
0
2 = =
∂ −
∂ z
g z σ φ φ
KBC on cylinder wall:
) sin( t R
r= +ε θ −σ
0 ) sin(
) , ,
(r t =r−R− − t =
F θ ε θ σ
0 on
0 ) cos(
)
cos( − + − 2 − = =
= t F
r u u Dt t
DF
r ε θ σ
σ θ
σε θ
Neglecting the second order term,
0 on
)
cos( − =
−
∂ =
−∂
= t F
ur φr σε θ σ
Taylor series expansion about r=R gives
[ ]
[
cos( )]
sin( )[
cos( )]
0)
cos( sin( )
= +
−
∂ +
− ∂ +
− +
=
− +
=
=
− +
=
L
R r r
R r r
t R
r r
t r u
t t
u
t u
σ θ σε σ
θ ε σ
θ σε
σ θ
σε ε θ σ
R r
r =− − t =
∂
− ∂
∴ φ σεcos(θ σ ) on
Using separation of variable, the solution satisfying the radiation condition (outgoing waves as r→∞) is
∑
∞=
−
− + +
+
=
1
) ( 1
) ( )
1 (
1 ( )cosh ( ) ( )cos ( )
) , , , (
n
t i n
n n t
i p
p
pH k r k h z e C K k r k h z e
A t z
r θ θ σ θ σ
φ
Ap and Cn can be determined from orthogonality condition by integrating over z.
Plunger wavemaker (See Figure 6.8 of textbook.)
