Chapter 9. Waves over Real Seabeds
We have to consider the followings in real sea:
- viscosity (fluid)
- Roughness, rigidity, permeability (seabed) Effect of viscosity
Inside the bottom boundary layer, - Viscosity is important.
- No-slip condition at bottom:
=0
∂
−∂
=
−
= h
x z
u φ
Linear laminar Navier-Stokes equation for viscous fluid:
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂
− ∂
∂ =
∂
2 2 2
1 2
z u x
u x
p t
u ν
ρ
z g w x
w z
p t
w ⎟⎟⎠−
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂
− ∂
∂ =
∂
2 2 2
1 ν 2
ρ
To evaluate relative sizes of various terms, introduce nondimensional variables:
x'=kx z'=kz t'=σt
ga p p
= ρ '
( )
1 ( , )' ,
' u w
w a
u = σ
By chain rule,
' '
' k x
x x x
x ∂
= ∂
∂
∂
∂
= ∂
∂
∂
2 2 2 2
2
' '
' '
' k x
k x k x x k x
k x x
x ∂
= ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
∂
∂
= ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
∂
∂
= ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
∂
∂
= ∂
∂
∂
' '
' t t
t t
t ∂
= ∂
∂
∂
∂
= ∂
∂
∂ σ
The x -momentum equation becomes
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂
− ∂
∂ =
∂
2 2 2 2 2
2
' ' '
' '
' '
'
z u x
a u x k
ga p k t
a u ρ ν σ
σ ρ
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂
− ∂
∂ =
∂
2 2 2 2 2
2 '
' '
' '
' '
'
z u x
u k
x p gk t
u
σ ν σ
where
2 1 2
1 2
2
2 1 (1);
−
− = = =
=
=
gk O C
C gk k C
gk
gk F
σ F
) 10 ( and
);
10
~ 10
1 ( 7 8 1 5
2
2 − −
−
− = =
=
=
=
= Ck O
C O k Ck
k
k ν
ν ν
ν σ
ν R
R
which implies that frictional stresses are negligible so that there is a slip boundary condition at the bottom (u≠0 at z=−h). However, physically, there is no flow at the bottom in the viscous flow. Hence, our argument above must be modified.
Near the bottom, u varies rapidly with z . Therefore, the vertical length scale must be different from the horizontal length scale. Let us take the vertical length scale as the boundary layer thickness, δ , so that
δ z'= z
and
' 1 '
' z z
z z
z ∂
= ∂
∂
∂
∂
= ∂
∂
∂
δ ;
2 2 2 2 2
' 1
z
z ∂
= ∂
∂
∂ δ
Using this scale but the same scales as before for other parameters,
2 2 2 2
2 2 2
' ' 1
' ' '
' '
'
z a u x
a u x k
ga p k t a u
∂ + ∂
∂ + ∂
∂
− ∂
∂ =
∂ σ
ν δ σ
ν ρ ρ
σ
2 2 2 2 2
2 '
' '
' 1 ' ' 1 '
'
z u x
u x
p t
u
∂ + ∂
∂ + ∂
∂
− ∂
∂ =
∂
σ δ
ν F R
The second term on the RHS is again very small compared with other terms, but the last term is of O(1) with δ ~ ν/σ . Now, the x -momentum equation can be approximated to
2
1 2
z u x
p t
u
∂ + ∂
∂
− ∂
∂ =
∂ ν
ρ
Assuming u=up +ur, where and , respectively, represent the potential and rotational part of the flow, we can separate the equation into
up ur
x p t
up
∂
− ∂
∂ =
∂
ρ 1
2 2
z u t
ur r
∂
= ∂
∂
∂ ν
each of which has a form of Euler equation and diffusion equation, respectively.
For the potential flow part, we have
) (
cosh ) (
cosh i kx t
p e
kh z h k
u gak σ
σ −
= +
For the rotational part, assume
)
) (
(z eikx t Af
ur = −σ
Substituting into the diffusion equation,
0 )
' '
(−iσAf −νAf ei(kx− tσ) =
or
0 '
'+i f =
f ν
σ
Solving this equation and keeping only the term that decays away from bed (or
as ),
0 ) (z →
f (h+ )z →∞
) ( 2 / ) 1 ( )
(
) /
(z Be i h z Be i h z f = −σ ν + = − − σ ν +
Time-out
De Moivre theorem: z1/n =n reiθ/n
) 1 ( 2 1 2 1 2 1 4
sin3 4
cos3
1 4
2 3 2 / 3
2 e e i i i
i = i = i = + =− + =− −
− π π π π
ν σ ν
ν σ σ ν
σ (1 ) (1 ) 2
2 / 1
/ i i i
i = − =− − =− −
−
Now ur becomes
)) ( 2 / (
) ( 2 /
) ( ) ( 2 / ) 1 (
z h t
kx i z h
t kx i z h i r
e Ae
e Ae
u
+ +
− +
−
− +
−
−
=
=
ν σ σ ν
σ
σ ν
σ
Using u=up +ur =0 at z=−h,
cosh 0
1 ei(kx− t) +Aei(kx− t) = kh
gak σ σ
σ
or
kh A gak
cosh 1
− σ
=
Therefore,
)) ( 2 / (
) ( 2 /
cosh
1 h z i kx t h z
r e e
kh
u =−gak − σ ν + −σ+ σ ν + σ
and
[
cosh ( )cos( ) cos( /2 ( ))]
cosh
) ( 2
/ kx t h z
e t kx z
h kh k
u= gak + −σ − − h+z −σ + σ ν +
σ σ ν
See Figure 9.1 for the horizontal velocity profile near the bed.
The vertical velocity in the bottom boundary layer can be obtained from the continuity equation:
x dz w u
z w x u
∂
−∂
=
∂
→
∂ = + ∂
∂
∂ 0
∫
∫
∫
∫
− − − − + =∂
− ∂
∂ =
− ∂
=
∂
⎟=
⎠⎞
⎜⎝
⎛ ∂ = =
= z h z
h z
h z
h z
h ds
x dz u
x w u
z w w w z
w( ) ( ) 0 Eq.(9.12)
Bed shear stress
) 4 / (
) 4 / (
) (
cosh
2 2 cosh
2 2
cosh )
(
π σ
π σ
σ
σ ν ρ
ν σ σ
μ
ν σ ν
σ σ
μ
μ μ
τ
−
−
−
−
−
=
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
=
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
≅ ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂ +∂
∂
= ∂
−
=
t kx i
t kx i
t kx i p r zx
kh e gak
kh e gak
e kh i
gak
z u z u x
w z h u
z
where the relationship, A(1−i)= 2Ae−iπ/4, was used by De Moivre theorem. In terms of flow velocity, we have
( )
max ) 4 / ( max
max
max 8 2 cosh
)
( ⎟⎟
⎠
⎞
⎜⎜⎝
=⎛
⎟⎠
⎜ ⎞
⎝
=⎛
⎟⎠
⎜ ⎞
⎝
=⎛
−
= −σ−π
σ ν ρ ρ
τzx ρ b b w b b eikx t
kh u gak
f u u
f u h
z
where = potential flow velocity outside boundary layer. Solving for the friction coefficient, ,
ub
f
ν σν ς
σ ν σ
σ σ
ν
/ 8 /
8 cosh
8 cosh
8
max max max
max
2 2 2
b b b
b
b kh u u
gak u
kh gak
f =u = = =
or
2 / 1
8
b
f =R ; / 2 / 103 ~104 for smooth bottom
max = <
= b ςb ν ςbσ ν
b u
R
See Figure 9.2 of textbook for linear relation between and for laminar flow.
f
log logRb
Energy dissipation
Mean rate of energy dissipation is given by (Fluid Dynamics, Batchelor)
kh E k
z h z d u
z h z d
w z
u x w x
u
h r
h D
2 sinh
2 /
) (
) ( 2
2
0
2 0
2 2
2
ν σ ν
ρν ρν ε
=
⎟ +
⎠
⎜ ⎞
⎝
⎛
∂
≅ ∂
⎪⎭ +
⎪⎬
⎫
⎪⎩
⎪⎨
⎧ ⎟
⎠
⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
= ∂
∫
∫
x AE ECg
D =
∂
−∂
= ( )
ε where A=νk σ /2ν /sinh2kh.
Consider waves on a flat bottom. Since Cg ≠ f(x), we have
C E A x
E
g
−
∂ =
∂
x C
A g
e E
E= 0 −( / )
x C
A g
e a
a2 = 02 −( / )
x kh C
k x
C
A g g
e a e
a
a ⎟
⎟
⎠
⎞
⎜⎜
⎝
−⎛
− =
= 2 sinh2
2 /
0 ) 2 / ( 0
ν σ ν
which states that the wave amplitude decays exponentially with x .
Turbulent boundary layer
Large waves + Rough bottom → Turbulent boundary layer (most cases in nature)
In turbulent boundary layer,
) / (ke b f
f = ς
as shown in Figure 9.2 and where
2d90
ke =
considering only skin friction without form drag due to ripples.
The bed shear stress is given by
t t f u
u f u h
z b b m
zx ρ ρ σ σ
τ cos cos
8 ) 8
( =− = = 2
where
t u
ub = mcosσ with
kh a kh um gak
sinh cosh
σ
σ =
=
Note that the time average of shear stress is zero. The mean rate of energy dissipation is not zero and is given by
3 3 3
4 / 0
3 3
4 / 0
3 3
2
sinh 6
6 3 2 2
4 8
) cos 4 / (
1 8
4 cos / 1 8
) 8 (
⎟⎠
⎜ ⎞
⎝
= ⎛
=
=
=
=
=
−
=
∫
∫
kh a f
f u f u
T tdt f u
T tdt f u u
f u u h
m m
T m
T m
b b b
zx D
σ π
ρ
π ρ π ρ
σ σ σ
ρ
ρ σ τ ρ
ε
Energy loss with distance
kh a f dx
EC d
D g
3 3 3
sinh 6 )
( σ
π ε =−ρ
−
=
On flat bottom,
kh a f dx
gCg da
3 3 3 2
6 sinh 2
1 σ
π ρ =− ρ
kh a f dx agCg da
3 3 3
6 sinh σ
− π
=
khdx gC
f a
da
g 3 3
2 6π sinh
− σ
=
Defining A= fσ3/(6πgCg sinh3kh) and integration give
1
1 Ax C
a=− +
−
Using the boundary condition a=a0 at x=0, we get C1=−1 a/ 0. Therefore
0
1 1
Ax a a=− −
−
Finally we have
x Aa a a
0 0
1+
=
which is not exponential decay with x . For small Aa0, we have
x
e Aa
a x Aa a
a≅ 0(1− 0 )≅ 0 − 0 ← exponential decay
Waves over viscous mud bottom
At the bottom, we need the kinematic and dynamic matching conditions:
φ χ
χ φ =− +
∂
−∂
∂ =
−∂
∂ =
∂ z h
z z
t 2 1 on
χ +
−
=
= p z h
p1 2 on
In Region 1, we can assume the solution as
(
Acoshk(h z) Bsinhk(h z))
ei(kx σt)φ1= + + + −
where the second term is included because the vertical velocity is no longer zero at the bottom. Using LDFSBC,
) sinh cosh
0 (A kh B kh
g
a =−iσ +
Using LCFSBC,
0 ) sinh cosh
( )
cosh sinh
(
2 + =
−
+ A kh B kh
kh g B
kh A
k σ
Multiplying g cosh/ kh on both sides,
0 ) tanh (
) tanh
(gk kh− 2 +B gk − 2 kh =
A σ σ
From LDFSBC,
kh kh B
iga kh
kh A i B
ga
cosh tanh cosh
sinh 0
0 − = −
= −
σ σ Plug in LCFSBC,
) tanh
( cosh tanh
) tanh
( 2 0 σ2
σ σ ⎟ −
⎠
⎜ ⎞
⎝
⎛ − +
=
− B kh gk kh
kh B iga
kh gk
) tanh cosh (
) tanh tanh
tanh
( 2 2 2 0 2 gk kh
kh kh iga
kh gk
kh gk
B − − + = σ −
σ σ σ
) tanh cosh (
cosh
1 0 2
2 gk kh
kh iga
Bgk kh = σ −
σ
) tanh cosh (
) tanh cosh (
cosh2 0 2 0 2
kh k gk
kh kh ia
kh gk iga gk
B= kh − = σ −
σ σ σ
Now,
) tanh cosh (
tanh cosh tanh
cosh
cosh )sinh tanh cosh (
cosh
0 2
2 2
2 0
0 2 0
kh k gk
kh ia
kh gk
kh kh gk k
kh ia
kh kh kh
k gk kh ia
kh A iga
σ σ σ σ
σ σ σ
−
=
⎟⎠
⎜ ⎞
⎝
⎛ − +
=
−
−
=
Note: On rigid bottom, B=0 or σ2 =gk tanhkh.
In Region 2, assuming infinite depth,
) ( ) (h z ikx t
k e
de σ
φ2 = + −
which indicates exponential decay as h+ z→−∞. The horizontal velocity is given by
r r
p u
u x u
u2 2 2 2 + 2
∂
−∂
= +
= φ
where is the boundary layer correction given by (cf. Eq. 9.9) u2r
) ) ( ( 2 / ) 1 ( 2
2 h z ikx t
i e
fe u r
ν σ
σ + −
= −
Assuming is small and using the linearized kinematic matching condition at the boundary:
wr
h z z
z
t =−
∂
−∂
∂ =
−∂
∂ =
∂χ φ2 φ1 on
Bk dk m
i =− =−
− σ 0
B d =
∴ ,
σ σ
σ
idk iBk
i
m0 = Bk =− =−
Applying dynamic matching condition:
χ +
−
=
= p z h
p1 2 on
χ ρ
φ ρ ρ φ ρ
ρ + + − − =− +
∂
= ∂
∂ −
∂ gh g z h z h
gz t
t1 1 2 2 1 2 ( ) on
1
where the last two terms represent hydrostatic pressures due to water and mud, respectively.
Linearizing,
h z t g
t g − =−
∂
= ∂
∂ −
∂ 1 1 2 2 2 on
1 φ ρ χ ρ φ ρ χ
ρ
0 2 2 0
1
1A gm i d gm
iσρ −ρ =−σρ −ρ
−
⎟⎠
⎜ ⎞
⎝⎛−
− +
= σρ ρ ρ σ
σρ A i B g iBk
i 1 2 ( 2 1)
⎥⎦
⎢ ⎤
⎣
⎡ ⎟+
⎠
⎜ ⎞
⎝⎛ −
− =
−
= 2 2
1 2 2
1 1 2 1
2 ( ) 1
σ σ
ρ ρ σ
ρ ρ ρ ρ
ρ gk gk
gkB B B
A
Now,
kh gk
kh gk
gk gk
B A
tanh
1 2 tanh
2 2
2 2
−
= −
⎟+
⎠
⎜ ⎞
⎝⎛ −
= σ
σ σ
σ ρ
ρ
1
where Eqs. (9.51) and (9.52) were used. Rearranging gives the dispersion relationship:
kh gk
kh gk gk
kh gk gk
tanh )
tanh (
) tanh (
1 2 2 2 2 2
1
2 σ σ
σ σ σ ρ
ρ ⎟ − + − = −
⎠
⎜ ⎞
⎝⎛ −
(
2 gk)
( 2 gktanhkh) gk( 2 gktanhkh) gk 2 4 tanhkh1
2 σ σ σ σ σ
ρ
ρ − − + − = −
(
1 tanh)
(1 tanh ) 1 ( )2tanh 01 2 2 1
2 4 1
2 ⎟⎟ =
⎠
⎜⎜ ⎞
⎝
⎛ −
+ +
−
+ kh gk kh gk kh
ρ σ ρ
ρ σ ρ ρ
ρ
0 tanh
1 tanh
) (
1 2 1
2 2
2 ⎥=
⎦
⎢ ⎤
⎣
⎡ ⎟⎟
⎠
⎜⎜ ⎞
⎝
⎛ −
⎟⎟−
⎠
⎜⎜ ⎞
⎝
⎛ +
−gk kh gk kh
ρ ρ ρ
σ ρ σ
We have two possible cases:
=gk
σ2 ← surface wave case
or
kh kh gk
tanh tanh 1
1 2 1 2 2
+
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛ −
= ρ ρ ρ ρ
σ ← interfacial wave case
Using Eqs. (9.52) and (9.59),
) tanh (
cosh
) tanh (
cosh
2 2 0 2
0
kh gk
kh
kh iBk gk
kh i
kB m
a
= −
⎟⎠
⎜ ⎞
⎝⎛−
−
=
σ σ σ σ
σ
For surface wave case,
sinh 1 cosh
1 )
tanh 1 (
0 cosh
0 = >
= −
= − ekh
kh kh
kh kh
gk
gk m
a
The free surface and interface are in phase, and the free surface amplitude is greater than the interface amplitude.
For interfacial wave case,
e kh
m
a −
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛ −
−
= 1
1 2 0
0
ρ ρ
0 0 / m
a depends of ρ2/ρ1 and kh . a0 /m0 <1 if ρ2 /ρ1<2 for kh=0 .
1 2/ρ
ρ can be larger for larger kh to satisfy a0 /m0 <1. Also the free surface and
interface are 180° out of phase because a0 /m0 <0 always.
Because the free surface and interface are out of phase, between them a quasi- bottom exists where there is no vertical flow. If we denote its vertical distance from the SWL as
180°
z0 so that
0
1 0 on z z
z = =−
∂
∂φ
the dispersion relationship becomes
0 2 =gktanhkz σ
Waves over rigid, porous bottom
At the bottom, we need the kinematic and dynamic matching conditions:
h z
z =w =−
∂
−∂φ1 2 on
h z p
p1 = 2 on =−
In the porous medium, we use the equation for unsteady Darcy’s flow:
2 2
2 1
1 u p
K t
u
n =− − ∇
∂
∂
ρ ν r r
where n = porosity and K = permeability constant. Using ν =μ/ρ, the above equation can be written as
2 2 K p2
t u u n
K =− − ∇
∂
∂
μ
ν r
r
Since ur ∝ei(kx−σt), the above equation becomes
2
2 2
2 K p
u K u
n
i =− − ∇
− ν μ
σ r r
12 9 ~10 10
~ − −
K m2 and ν ~10−6 m2/s. Therefore, . Thus, the preceding equation can be approximated to
6 3 ~10 10
~ ) /
(σK ν − −
2
2 K p
u =− ∇ r μ
Using the continuity equation, ∇ ur⋅ 2 =0, we get
2 =0
∇ p2
1
2
which is the governing equation in the porous medium.
As with the waves over a mud bottom, assume the solutions as
(
Acoshk(h z) Bsinhk(h z))
ei(kx σt)φ = + + + −
and
) ( ) (h z ikx t
k e
De
p = + −σ
which give non-zero vertical velocity at the bottom and exponential decay of pore pressure as h+ z→−∞.
As done for viscous mud bottom, LKFSBC and LDFSBC give
) tanh
cosh ( 2
0 gk kh
k kh
A ia σ
σ −
=
) tanh cosh ( 2
0 gk kh
k kh
B= ia σ −
σ
Dynamic matching condition at the bottom:
h z
t = p =−
∂
∂φ1 2 on ρ
or
D A
i =
− σρ ρσ i A D
=−
Kinematic matching condition at the bottom:
h z z
K p
z =−
∂
− ∂
∂ =
−∂ 1 2 on
μ φ
or
K kD Bk=− μ
−
K D B= μ
Now
kh gk
kh gk
K i i K
B A
tanh tanh 1
2 2
−
= −
=
⎟⎟⎠
⎜⎜ ⎞
⎝
− ⎛
= σ
σ ν
σ ρσ μ
which gives
) tanh (
) tanh
( 2 K gk 2 kh
kh gk
i σ
ν
σ − = σ −
The LHS and RHS are imaginary and real, respectively. Therefore we get
kh gk tanh
2 =
σ and σ2 =gk cothkh
which are in conflict. To satisfy the dispersion relationship, the wave number should be complex:
i
r ik
k k = +
The real part is related to the wave length, while the imaginary part determines the spatial damping:
) ( 0 ) ( 0
t x k i x k t
kx
i a e i e r
e
a σ σ
η= − = − −
In intermediate depth, where we can assume that Kσ /ν<<1 and (or no significant damping), Reid and Kajiura (1957) obtained
<<1 h ki
h k gk tanh
2 ≅
σ r r
h k h
k K k k
r r
r
i 2 sinh2
) / ( 2
≅ + σ ν
which is shown in Figure 9.6 of textbook.
In shallow water where kh <π /10, and are given by Eqs. (9.95) and (9.96), respectively.
ki kr