Lectures 11-12: Gravity waves
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Linear equations
Plane waves on deep water
Waves at an interface
Waves on shallower water
Water waves
z
air
x
0
water
The free surface of a liquid in equilibrium in a
gravitational field is a plane. If the surface is disturbed,
motion will occur in the liquid. This motion will be
propagated over the whole surface in the forms of
waves, called gravity waves.
Let us consider waves on the surface of deep water. We
neglect viscosity, as there are no solid boundaries, at
which it could cause marked effects; and we also neglect
compressibility and surface tension.
The governing equations are
d iv v  0
v
t
 v    v  
p

 g
Small-amplitude waves
The physical parameters characterising the wave-motion are the
amplitude of oscillations of fluid particles, a, the wavelength, λ, and the
period of oscillations, T.
a
The velocity of a fluid particle,v  . The significant change of the
T
velocity occurs at a distance λ.

v
a
 2.
The unsteady term can be estimated as
t
T
2
a


The non-linear term can be estimated as v   v 
.
2

T


This means, v   v
a


v t

a
 1
For the small-amplitude waves, when 
, the non-linear term
is negligibly small.
Irrotational motion
The linearised Navier-Stokes equation is


v
p 



g

0
   constant
t

t
For the oscillatory motion, the average position of a fluid particle is
z=0, the average velocity is 0. The average vorticity must be also 0.
As the vorticity is time-independent, it must be 0 at every moment (to
make the average value being 0).

  0 -- small-amplitude wave motion is an irrotational flow

Hence, the velocity field can be represented as v    .
The velocity potential φ is determined by the Laplace equation:
  0
Boundary
conditions: p
v
z 
 p0
z  

on the surface (ζ is the surface elevation above
the flat position), pressure is atmospheric.
waves on deep water, the fluid velocity is
bounded everywhere.
1. We use Bernoulli’s equation to rewrite the first boundary condition in
terms of φ.
On the surface,
Here, we neglect the term containing
v2, as it originates from the non-linear
p0
  

 g   constant


term, which is small for the small
  t  z 
amplitude waves.
The velocity potential φ is a technical variable that does not
 have the
physical meaning and is needed only to find the velocity ( v    ).
The velocity does not change if the potential is redefined as follows
 p0

    c t     
 constant t
 

This allows us to rewrite the Bernoulli’s equation as an equation for
the elevation ζ :
1   
  

g   t  z 
2. On the surface,
vz

  



t
  z  z 
3. This results in the following boundary condition on the surface
  1   

0


2
  z g  t  z 
2
4. Using the Taylor’s series over small ζ, the leading term of the
above equation is
  1   

0


2
 z g t  z 0
2
Finally, we have got the following mathematical problem
Equation:
Boundary
conditions:
  0
  1   

0


2
 z g t  z 0
2
v
z  

Plane wave solution
Let us seek the solution in the form of a single plane wave,
  f  z  cos  t  kx  Here,   2 
T
k 
Illustration:
http://www.youtube.com/
watch?v=aKGgsLHN1dc
c 
2


k
-- the circular frequency
(1/T would be the
regular frequency)
-- the wavenumber
-- the wave speed
Substitution into the Laplace equation gives
2
f   k f  0
 f  c 1e
kz
 c 2e
 kz
As the velocity is
bounded,
c2  0
Hence,
  c 1e
kz
cos  t  kx

Or, in terms of velocity,
vx 
vz 

x

z
  kc 1e
 kc 1e
kz
kz
sin  t  kx ,
cos  t  kx

Wave dispersion
Applying the boundary condition on the liquid’s surface, we obtain the
following dispersion relation
2
The waves of


g
k 
0
   kg  c 

different lengths
g
k
k
travel at different
speeds (see the
video).
Video: http://www.youtube.com/watch?v=lWi_KpBy8kU (note that long
waves travel faster)
Phase and group velocities
c 
U 

k
g

d
dk
is called the phase velocity, the velocity of travelling of
any given phase of a wave.
k

1
g
2
k
is called the group velocity, this is the velocity of the
motion of the wave packet.
The red dot moves with the phase velocity, while the green dot moves
with the group velocity.
Waves at an interface
z
Consider gravity waves at an interface
between two very deep liquid layers. The
density of the lower liquid is ρ1 and the density
of the upper liquid is ρ2.
ρ2 >ρ1
ρ2
x
0
ρ1
Motion is irrotational. The governing equations are

v
(lower liquid) 1    1 ,   1  0

(upper liquid) v 2    2 ,   2  0
Boundary conditions:
at infinity, z   , fluids’ velocities are bounded
At interface, z   ,
v z ,1  v z , 2
p1  p 2
We need to re-write the boundary conditions on the interface in
terms velocity potential.
Applying Bernoulli’s equation at an interface, we have
p1
 1 

 g   constant


1
  t  z 
p2
  2 

 g   constant


2
  t  z 
or, redefining φ1 and φ2
p1
 1 

 g  0


1
  t  z 
p2
  2 

 g  0


2
  t  z 
The pressure is continuous at interface, i.e.
  1 

  2 

 1 
 g     2 
 g 


  t  z 

  t  z 

The z-component of the velocity is continuous as well, i.e.

 1 
  2 





t
  z  z 
  z  z 
As a result, we have two following boundary conditions at interface
 1
 2
1 
 2 
1  2  g
 2 
g
2



t

z

t

z

 z 

 z 
2
2
 1 
  2 





  z  z 
  z  z 
Expanding over powers of small ζ and leaving only the linear terms,
we finally obtain
 1
 2
1 
 2 
1  2  g
 2 
g
2



t

z

t

z

 z 0

 z 0
2
2
 1 
  2 




 z  z 0  z  z 0
Now, seek solution in the form of a plane wave
 1  f 1  z  cos kx   t 
 2  f 2  z  cos kx   t 
Solving the Laplace equations, we get
 f 1  A 1e
2
 k f 2  f 2  0
 f 2  A 2e
The boundedness of the velocities leads to
f 1  A 1e
That is,
kz
2
 k f 1  f 1  0
kz
 1  A 1e
 2  B 2e
and
kz
f 2  B 2e
cos kx   t 
 kz
cos kx   t 
 kz
kz
 B 1e
 kz
 B 2e
 kz
Let us now use the boundary conditions at the interface
 1    gk A 1   2    gk B 2 
2
A1   B 2
2
kA 1   kB

2
2







gk



 gk
1
2

2
The resultant dispersion relation
 
2
1   2
1   2

gk
The phase and group speeds are
c 

k

1   2 g
1   2 k
;
U 
d
dk

1
1   2 g
2
1   2 k
The fluid velocity (A1 cannot be determined from the linear equations):
v x ,1 
v z ,1 
1
x
1
z
  kA 1e
 kA 1e
kz
kz
sin kx   t ;
cos kx   t ;
v x ,2 
v z ,2 
 2
x
 2
z
 kA 1e
 kA 1e
 kz
 kz
sin kx   t 
cos kx   t 
Water/air interface
ρ2<< ρ1
dispersion relation and the expressions for the phase and
group velocities become as those for the gravity waves on
a free surface of deep water
 
gk ,
c 

k


k
,
U 
d
dk

1

2
k
Waves on shallow water
z
The liquid is now bounded by a rigid wall.
0
x
-h
Boundary
conditions:
z  h :
z  :
The fluid velocity is determined by the
Laplace equation:
  0

0
z
p  p0
-- no fluid penetration through the
wall
-- pressure is atmospheric
The boundary condition at a free surface can be rewritten
as
  1   

0


2
 z g t  z 0
2
Seek the velocity potential in the form of a plane wave,
  f  z  cos kx   t 
Solving the Laplace equation, we obtain
   A cosh k  z  h    B sinh k  z  h   cos kx   t 
Using the boundary condition at the wall gives B=0.
That is,   A cosh k  z  h   cos kx   t 
Using the boundary condition at an interface produces the dispersion
2
relation

2
k sinh kh  
cosh kh   0    kg tanh kh 
g
Consequently, the

phase and group
c 

velocities are
k
U 
d
dk
g
k

tanh kh

kh


tanh

kh




2 k tanh kh  
cosh kh  
g
Deep water and long waves
Let us find the dispersion relation and the phase and group velocities
for the cases (a) of very deep water (h/λ>>1) and (b) of long waves
(h/λ<<1).
(a) Deep water, h/λ>>1, or kh>>1


tanh kh   1  

c 


k
 
g
k
,
kg
U 
d
dk

1
g
2
k
(b) Long waves, h/λ<<1, or kh<<1


tanh kh   kh  

c 


k
 k
gh ,
gh
U 
d
dk

gh
Appendix: solution of the
amplitude equation
(1) f   k 2 f  0
This is the linear ordinary differential equation with
constant coefficients.
Seeking solution in the exponential form, f  e mz , gives the auxiliary
equation,
m k
2
2
0
The roots of the auxiliary equation: m   k
There are two linearly independent solutions of equation (1),
(2) e kz and e  kz
Since equation (1) is linear, any linear combination of solutions (2) is a
solution of (1). The general solution can be written as
(3) f  Ae
kz
 Be
 kz
Here, A and B are unknown arbitrary constants
(to be determined from the boundary conditions).
Let us show that the general solution of (1) can be also written in the
following form
(4) f  A 1 cosh kz   B 1 sinh kz
 A1 and B1 are the arbitrary constants,
different from A and B.
Using the definitions of hyperbolic functions,
f  A1
e
kz
e
2
 kz
 B1
e
kz
e
and re-defining the constants
 kz
2
A 

A1  B 1
2
A1  B 1
2
e
,
kz

B 
A1  B 1
2
A1  B 1
2
e
 kz
,
we can prove that (4) is equivalent to (1).
Similarly, it can be easily proved that the general solution of (1) can be
also written in the forms
(5) f  A 2 cosh k ( z  h )   B 2 sinh k ( z  h ) 
or
(6) f  A 3 cosh k ( z  h )   B 3 sinh k ( z  h ) 
Conclusion:
- (3), (4), (5), and (6) are the equivalent forms of the general solution of
equation (1); all these forms are the linear combinations of basis
solutions (2). All these expressions include two unknown arbitrary
constants to be determined from the boundary equations, but the final
solution (with determined constants) is unique.
- It is recommended to chose such a form of the general solution that
can simplify derivations. For instance, (5) should be convenient when
one of the boundary conditions is imposed at z=-h, where f (  h )  A 3 ,
which immediately gives one constant, A3. (3), (4), and (6) can be also
used and should produce the same final expression, but intermediate
derivations can be lengthier.