PHY 711 Classical Mechanics and
Mathematical Methods
10-10:50 AM MWF Olin 103
Plan for Lecture 34:
Chapter 10 in F & W:
Surface waves
-- Non-linear contributions and soliton
solutions
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PHY 711 Fall 2014 -- Lecture 34
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PHY 711 Fall 2014 -- Lecture 34
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Surface waves in an incompressible fluid
General problem
including
p0
non-linearities
z
z
h
y
x
Within fluid:
0  z  hz
 1 2

 2 v  g  z  h   constant
t
 2  0
At surface:
v  v ( x, y, z , t )   ( x, y, z , t )
with z  z  x, y, t 
z  h z
dz z
z
z

 vx
 vy
dt
t
x
y
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   ( x, y , z , t )
where vx , y  vx , y  x, y, h  z , t 
PHY 711 Fall 2014 -- Lecture 34
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p0
z
z
h
y
x
Further simplifications; assume trivial y - dependence
    x, z , t 
z  z  x, t 
Within fluid :
0  z  h z
 dz
At surface :
v z  x, z  h  z , t   

z
dt
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Non-linear effects in surface waves:
p0
z
z
h
y
z=0
x
Dominant non-linear effects  soliton solutions
 30 x  ct 
z ( x, t )  0 sech 

 h 2h 
2
0  constant
 0 
where c 
 gh 1 

1  0 / h
2
h


gh
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Detailed analysis of non-linear surface waves
[Note that these derivations follow Alexander L. Fetter and
John Dirk Walecka, Theoretical Mechanics of Particles and
Continua (McGraw Hill, 1980), Chapt. 10.]
We assume that we have an incompressible fluid:   constant
Velocity potential: ( x, z, t ); v( x, z, t )  ( x, z, t )
The surface of the fluid is described by z=h+z(x,t). It is
assumed that the fluid is contained in a structure
(lake, river, swimming pool, etc.) with a structureless
bottom defined by the z = 0 plane and filled to an
equilibrium height of z = h.
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Defining equations for (x,z,t) and z(x,t)
where 0  z  h  z ( x, t )
Continuity equation:
v  0
 2  ( x, z , t )  2  ( x, z , t )


0
2
2
x
z
Bernoulli equation (assuming irrotational flow) and gravitation
potential energy
2
2

 ( x, z , t ) 1   ( x, z , t )   ( x, z , t )  

 
 
   g ( z  h)  0.
t
2 
x
z
 
 
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Boundary conditions on functions –
Zero velocity at bottom of tank:
 ( x,0, t )
 0.
z
Consistent vertical velocity at water surface
v z ( x, z , t ) t z  h  z
dz
z

 v  z 
.
dt
t
 ( x, z , t )  ( x, z , t ) z ( x, t ) z ( x, t )



t z  h z  0
z
x
x
t
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Analysis assuming water height z is small relative to
variations in the direction of wave motion (x)
Taylor’s expansion about z = 0:

z 2  2
z 3  3
z 4  4
 ( x, z , t )   ( x,0, t )  z
( x,0, t ) 
( x,0, t ) 
( x,0, t ) 
( x,0, t )
2
3
4
z
2 z
3! z
4! z
Note that the zero vertical velocity at the bottom ensures
n


that all odd derivatives
vanish from the
(
x
,0,
t
)
z n
Taylor expansion . In addition, the Laplace equation allows
us to convert all even derivatives with respect to z
to derivatives with respect to x.
z 2  2
z 4  4
Modified Taylors expansion:  ( x, z, t )  ( x,0, t ) 
( x,0, t ) 
( x,0, t )
2 x 2
4! x 4
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Check linearized equations and their solutions:
Bernoulli equations -Bernoulli equation evaluated at z  h  z ( x, t )
 ( x, h, t )

 gz ( x, t )  0
t
Consistent vertical velocity at z  h  z ( x, t )
 ( x, z , t ) z ( x, t )


t z  h z  0
z
t
Using Taylor's expansion results to lowest order
 ( x, z , t )
 2 ( x,0, t )

h
z
x 2
Decoupled equations:
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( x, h, t )
( x,0, t )


t
t
 2 ( x,0, t )
 2 ( x,0, t )
 gh
.
2
2
t
x
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Analysis of non-linear equations -- keeping the lowest
order nonlinear terms and include up to 4th order
derivatives in the linear terms. Let  ( x, t )   ( x,0, t )
Approximate form of Bernoulli equation evaluated at surface: z  h  z
2
2
2

 (h  z )  
1    
 


     (h  z ) 2    gz  0
2
t
2
tx
2  x  
x  


2
3
 h 2  3
1   


    gz  0.
2
t
2 tx
2  x 
2
Approximate form of surface velocity expression :

  h3  4 z

 0.
 (h  z ( x, t ))  
4
x 
x  3! x
t
The expressions keep the lowest order nonlinear terms
and include up to 4th order derivatives in the linear terms.
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 h 2  3
1   
Coupled equations: 


   gz  0.
2
t
2 tx
2  x 

  h3  4 z

 0.
 (h  z ( x, t ))  
4
x 
x  3! x
t
2
Traveling wave solutions with new notation:
u  x  ct
 ( x, t )   (u ) and z ( x, t )   (u )
Note that the wave “speed” c will be consistently
determined
d  (u ) ch d  (u ) 1  d  (u ) 
c

 
  g (u )  0.
3
du
2 du
2  du 
2
3
2
d 
d  (u )  h3 d 4  (u )
d (u )
c
 0.
 (h   (u ))

4
du 
du  6 du
du
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Integrating and re-arranging coupled equations
d  (u ) ch 2 d 3  (u ) 1  d  (u ) 
c

 
  g (u )  0.
3
du
2 du
2  du 
2
g
h2
1
g
h2 g
g2 2
2
  '      ''' (  ')    
 '' 3 
c
2
2c
c
2c
2c
d 
d  (u )  h3 d 4  (u )
d (u )
c
 0.
 (h   (u ))

4
du 
du  6 du
du
h3
 (h   )  '  ''' c  0
6
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Integrating and re-arranging coupled equations – continued -Expressing modified surface velocity equation in terms of (u):
2
2
3
 g

h g
g 2
hg
(h   )    
 '' 3   
 '' c  0
2c
2c
 c
 6c
3
gh
gh
g
gh  2


 1  2   2  '' 2 1  2   0
c 
3c
c  2c 

h
3
2
 hg 
 1  2  (u )   ''(u )   (u )   0.
c 
3
2h

2
Note: c  gh 
2
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Solution of the famous Korteweg-de Vries equation
Modified surface amplitude equation in terms of 
h2
3
2
 hg 
 1  2  (u )   ''(u )   (u )   0.
c 
3
2h

Soliton solution
 30 x  ct 
z ( x, t )   ( x  ct )  0 sech 

h
2
h


2
 0 
c
 gh 1 

1  0 / h
2
h


gh
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where 0 is a constant
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Relationship to “standard” form of Korteweg-de Vries equation
New variables:
  20 ,
x
3 x
,
2h h
and
t 
3 ct
.
2h 20 h
Standard Korteweg-de Vries equation

  3
 6
 3  0.
t
x x
Soliton solution:
 

 ( x , t )  sech 
( x   t ) .
2
 2


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PHY 711 Fall 2014 -- Lecture 34
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More details
Modified surface amplitude equation in terms of  :
h 2 
3
2
 hg 
1  2  (u )   (u )   (u )   0.
c 
3
2h


gh 
d
Some identities: 0  1  2 ;
 c ;
h
c
t
du
Derivative of surface amplitude equation:
 d

.
x du
0
h2
3
 '  '''  '  0.
h
3
h
Expression in terms of x and t:
0 
h 2  3 3 


 
 0.
3
ch t
3 x
h x
Expression in terms of x and t :

  3
 6
 3  0.
t
x x
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Summary
Soliton solution
 30 x  ct 
z ( x, t )   ( x  ct )  0 sech 

 h 2h 
2
 0 
c
 gh 1 

1  0 / h
 2h 
gh
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where 0 is a constant
PHY 711 Fall 2014 -- Lecture 34
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Photo of canal soliton http://www.ma.hw.ac.uk/solitons/
Some links:
Website – http://www.ma.hw.ac.uk/solitons/
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