### 6-freakwaves

```Freak Waves in Shallow
Water
Josh Moser
&
Chris Wai
Rogue waves are being reported more and more in today’s world.
Rogue waves are being reported more and more in today’s world.
Rogue waves are being
reported more and more
in today’s world.
Rogue waves are being reported more and more in today’s world.
,
The dispersion relation describes the physics of the waves
2

2 = (1 +
) tanh ℎ

The dispersion relation describes the physics of the waves
ℎℎ  =
2

wavelength,  .
and depends on
The Korteweg-de Vries equation is a nonlinear, partial
differential equation that has applications to water waves
+ 6 +  = 0
Small-amplitude waves in shallow water is a statement of weak
nonlinearity
+  = 0
∞
,  =
(, ) −
−∞
Appropriate partial derivatives of
∞
,  =
(, ) −
−∞
∞
=
−∞
∞
−

and
,  (−)3  −
=
−∞
Plugging into the linearized KdV equation to find the ordinary
differential equation
∞

−∞
−

−  ,

which is

3
= 0
−  ,   3 = 0
The solution of this ODE is  ,  = 0 ()
3
Plugging this solution into  ,
So that  ,  =
∞
(+ 3 )
()

−∞ 0
To find 0 , take the Fourier Transform of the initial condition (, 0)
So that
∞
−

()
−∞ 0
1
0 () =
2

∞
(, 0)
−∞
Which is particularly useful when considering an ideal
situation in which  ,  =
1 ∞

)
(,
0

2 −∞
1
2
∞
( − 0)
−∞

= 0 ()
1 (0)
1
=

=
2
2
1
0 () =
2
So now we have an equation that looks like an Airy Integral
1
,  =
2
1
=
2
∞

+ 3

−∞
∞
−∞
1
+  3
3

Now we can make simple substitutions to exploit what is known
=

3
1
3
=  3
1
,  =  ,  =
2
1
3
∞
−∞
1
+ 3
3

We know that this is in the form of the Airy Integral using slow time and
slow space scales
1
,  =  ,  =
2
,  =
1
3
∞
1 3
+
3

−∞
1 ()
3

Substitutions can be made to convert it back to real time and space
scales for the wavemaker

=
− 0
ℎ
0
=

6ℎ
Substitutions can be made to convert it back to real time and space
scales for the wavemaker
1
,  =
3
0
()
6ℎ
1
3

− 0
ℎ
3
0
()
6ℎ
1
3
3
=

2ℎ
Now consider different initial
conditions
• The initial condition  χ, 0 only considers what the
χ axis looks like when at τ = 0.
• Namely the initial condition  χ, 0 = δ(χ)
δ(χ)
χ, 0
τ=0
0
χ
• It is not useful to consider when the freak wave is
forms when it is at one end of the tank.
• So consider the initial condition  χ, τ∗ = δ(χ)
where τ∗ is an arbitrary value we can choose
• Because we know the speed the wave is travelling,
we can choose when we want the freak wave to
form and then calculate how far into the tank the
freak wave will occur.
δ(χ)
χ, τ∗
0
τ = τ∗
χ
∞
,  =
0 ()
−(+ 3 )

−∞
Applying the new initial condition  χ, τ∗ = δ(χ) we have,
∞
, ∗ = δ χ =
0 ()
−∞
−(+ 3 ∗ )

Then applying the Fourier Transform we have,
1
0 () =
2
∞
δ χ
3 )
(+
∗ χ

−∞
We know that the delta function has the property such that,
∞
δ  −     = ()
−∞
In the case above,  = 0
So evaluating,
∞
1
0  =
δ χ
2
−∞
1  3
∗
=

2
+ 3 ∗
χ
Now we know the identity of 0  under the initial conditions
χ, τ∗ = δ χ . So  ,  becomes,
1
,  =
2
∞

−∞
+ 3 (−∗ )

Using similar substitutions as earlier, we can rewrite this in Airy integral
form
=

1
3
3( − ∗ )
=  3( − ∗ )
Then just as before we have
1
,  =  ,  =
2
∞
−∞
1
3
1
+33

Finally in this form which is no different from earlier except that  has
been replace with  − ∗
,  =
1
3( − ∗ )
1 ()
3
Then using similar substitutions to convert back to real time and space

=
− 0
ℎ
0
=
( − ∗ )
6ℎ
,  =
1
0
3
( − ∗ )
6ℎ
1
3

− 0
ℎ
0
3
( − ∗ )
6ℎ
So,
2ℎ
,  =
3
1
0
3
( − ∗ )
6ℎ
1
3
1
3
3
=

2ℎ

− 0
ℎ
0
3
( − ∗ )
6ℎ
1
3
We know the speed of the wave and can pick a ∗ such that the freak
wave will occur somewhere reasonable in the tank.
Here are some plots to demonstrate what the wave surface should look like when ∗ = 5
=3
= 4.9
= 5.25
=4
= 4.99
= 5.5
= 4.5
= 5.01
=6
= 4.75
= 5.1
=7
How do we generate these waves?
• In the wave tank in the Pritchard laboratory
• The wave maker is a vertical paddle that moves backwards and
forwards. It is varying in  over time . Let us call the position of the
• Without loss of generality, let us assume a wave tank that extends
infinitely in one direction. Which are the conserved finite quantities in
this case?
Mass flux
Mass flux
(),  + ℎ
Mass flux
To consider flux, we must define the direction from the “inside” to the
“outside”. Namely, we must parameterize the function  and obtain a
vector function. We define (). And to parameterize,
= ,  ,  .
A vector function that is in a normal direction to  is then.
= − ,  , 1 .
Then the normal component of velocity is,
∙  ,  =  −   .
In general for water waves, the kinematic free surface boundary condition in one
space and one time dimension for an air-water interface is,

−
+
= 0   =

So by the kinematic free surface boundary condition,
=  −     =
Where  is the velocity potential and so  and  are the velocity of the water at
the position and time , ,  .
The mass flux “through” the wave paddle would be the height of the
fluid at () times the velocity of fluid in the  direction.

(),  + ℎ

Because water cannot go through the paddle, the water’s velocity at

=

= ()
So then the flux through the surface of the wave is given by the integral
of the change of the wave surface over time, for all of the wave surface.
Namely,
∞
()
∞
=
()
−
Now we can equate the mass flux at the paddle with that of the wave
surface to get the relation,

(),  + ℎ
=

∞
()

Referring back to,
+  = 0
3
,  =
(, )
2ℎ

=
− 0
ℎ
0
=

6ℎ
+
∞
0 ℎ
=  + 0  +
6
3
= 0
3
∞
0 ℎ
= −
0  +
6
()
()
∞
0 ℎ
= 0  (),  +
6
()

2
(  )

0 ℎ
(),  + ℎ
= 0  (),  +

6
2
(  )
In conclusion
• Lot’s of differential equations
• Very mathy
• Had a lot of fun
Things to do next and new questions to ask
• Find numerical solution and try to replicate results
• See how the data can be used to predict freak waves that may come
into the coast
• Find how the sea in real life translates to boundary and initial
conditions
• Find the set of the conditions that cause freak waves at the shore
• Predict and “control”
• Save lives
res14oncruiseshipinM_146AB/ship50footwave2.jpg
• http://graphics8.nytimes.com/images/2006/07/11/sci
ence/11wave.1.395.jpg