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The idea is that at 𝑡∗ = 5 the whole wave bunches up together to form the delta
function. However this cannot be graphed because it is a singularity. The speed
of this wave 𝑐0 is about 70 cm s-1 . Notice how where the wave largest is about
𝑐0 𝑡∗, where the delta function would form
Here, the wave is dissipating. Notice from the beginning how the wave spike
travels to the right while the wave train seems to be travelling to the left.
Nonlinear-dispersive mechanism of the freak wave formation in shallow water
Joshua Moser, Christopher Wai, Diane Henderson, Vishal Vasan
Department of Mathematics, Eberly College of Science, The Pennsylvania State University
Penn State Applied Mathematics Summer REU 2014
Dispersion relation- The relation that explains the physics of the waves
𝜔2 = 𝑔𝑘(1 +
𝑇𝑘2
𝑔
) tanh 𝑘ℎ
The relation is dependent upon the wavelength of the waves, as shown in
𝑘. 𝑘 =
2𝜋
𝐿 𝑥
KdV Regime
The Korteweg-de Vries equation is a nonlinear partial differential equation
that has applications to water waves in shallow water.
𝑢 𝜏 + 6𝑢𝑢 𝜒 + 𝑢 𝜒𝜒𝜒 = 0
Small-amplitude waves in shallow water is a statement of weak
nonlinearity. 𝑢 𝜏 + 𝑢 𝜒𝜒𝜒 = 0
Linearized KdV and Fourier Transform
Solutions can be found using Fourier Transforms such that,
𝑢 𝜒, 𝜏 = න
−∞
∞
𝐴(𝑘, 𝜏)𝑒−𝑖𝑘𝜒 𝑑𝑘 .
The appropriate partial derivatives are,
𝑢 𝜏 = ‫׬‬−∞
∞ 𝜕𝐴
𝜕𝜏
𝑒−𝑖𝑘𝜒 𝑑𝑘 and 𝑢 𝜒𝜒𝜒 = ‫׬‬−∞
∞
𝐴 𝑘, 𝜏 (−𝑖𝑘)3 𝑒−𝑖𝑘𝜒 𝑑𝑘.
Plugging in, the ordinary differential equation and solution to this KdV
equation is,
𝜕𝐴
𝜕𝜏
− 𝐴 𝑘, 𝜏 𝑖𝑘 3= 0 and 𝐴 𝑘, 𝜏 = 𝐴0(𝑘)𝑒 𝑖𝑘3 𝜏
The general solution can then be expressed as,
𝑢 𝜒, 𝜏 = න
−∞
∞
𝐴0(𝑘)𝑒(𝑖𝑘𝜒+𝑖𝑘3 𝜏) 𝑑𝑘
Delta Function with Initial Condition
Next, using delta function initial condition of 𝑢 𝜒, 𝑜 = 𝛿 𝜒 and the
Fourier Transform, 𝐴0 𝑘 =
1
2𝜋
, making 𝑢 𝜒, 𝜏 =
1
2𝜋
‫׬‬−∞
∞
𝑒 𝑖𝑘𝜒+𝑖𝑘3 𝜏 𝑑𝑘,
which looks an awful lot like the Airy Integral.
Airy Integral
Making some simple substitutions, 𝜉 =
𝜒
3𝜏
1
3
and 𝐾 = 𝑘 3𝜏
1
3, 𝑢 𝜒, 𝜏 = 𝑓 𝜉, 𝜏 =
1
2𝜋
‫׬‬−∞
∞
𝑒
ቁ𝑖(𝜉𝐾+
1
3
𝐾3
𝑑𝐾 is formed to utilize what is known about Airy Integrals.
The last step is to change the scales to be in real space and real time, using the
substitutions of 𝜒 =
𝜀
ℎ
(𝑥 − 𝑐0 𝑡) and 𝜏 = 𝜀(
𝑐0
6ℎ
)𝑡.
In Real Time and Real Space
The concluding equation for the surface displacement is,
𝑢 𝑥, 𝑡 =
1
3𝜀
𝑐0
6ℎ
(𝑡)
1
3
𝐴𝑖
𝜀
ℎ
𝑥 − 𝑐0 𝑡
3𝜀
𝑐0
6ℎ
(𝑡)
1
3
=
3
2ℎ
𝜂
Considering Different Initial Conditions for the Rogue Wave
Instead of having the rogue wave occur at the origin all the time, we can use different
initial conditions which will yield an equation where we can choose the time when we
want the rogue wave to occur. Because we know the speed of the wave, we can choose
the correct time where the wave will occur at a position of our choice. Using the new
initial condition,
𝑢 χ, τ∗ = δ(χ)
We can do similar calculations as before to obtain the equation (the following in 𝜂),
𝜂 𝑥, 𝑡 =
2ℎ
3
1
3𝜀
𝑐0
6ℎ
(𝑡 − 𝑡∗)
1
3
𝐴𝑖
𝜀
ℎ
𝑥 − 𝑐0 𝑡
3𝜀
𝑐0
6ℎ
(𝑡 − 𝑡∗)
1
3
Observe that 𝑡 has been replaced with 𝑡 − 𝑡∗ with the exception of the 𝑡 in 𝜒
Plots of Wave Formation for 𝑡∗ = 5
𝑡 = 3 𝑡 = 4 𝑡 = 4.5
𝑡 = 4.9 𝑡 = 4.99𝑡 = 4.75
𝑡 = 5.01 𝑡 = 5.1 𝑡 = 5.25
𝑡 = 5.5 𝑡 = 6 𝑡 = 7
Generating the Wave and the Wave Paddle PDE
The wave is generated by a vertical wave paddle that moves back and forth to
displace water in a fashion that generates the desired wave. The differential
equation used to find the motion of the paddle is,
𝜂 𝐿(𝑡), 𝑡 + ℎ
𝑑𝐿
𝑑𝑡
= 𝑐0 𝜂 𝐿(𝑡), 𝑡 +
𝑐0
6
ℎ
𝜀
2
𝜂 𝑥𝑥(𝐿 𝑡 )
Where 𝐿(𝑡) is the 𝑥 position of the paddle with respect to time. While assuming
the initial condition,
𝐿 0 = 0
This differential equation can be solved numerically using the classical Runge-
Kutta method and the computed values can be input into the wave paddle.

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Rogue Waves Poster

  • 1. The idea is that at 𝑡∗ = 5 the whole wave bunches up together to form the delta function. However this cannot be graphed because it is a singularity. The speed of this wave 𝑐0 is about 70 cm s-1 . Notice how where the wave largest is about 𝑐0 𝑡∗, where the delta function would form Here, the wave is dissipating. Notice from the beginning how the wave spike travels to the right while the wave train seems to be travelling to the left. Nonlinear-dispersive mechanism of the freak wave formation in shallow water Joshua Moser, Christopher Wai, Diane Henderson, Vishal Vasan Department of Mathematics, Eberly College of Science, The Pennsylvania State University Penn State Applied Mathematics Summer REU 2014 Dispersion relation- The relation that explains the physics of the waves 𝜔2 = 𝑔𝑘(1 + 𝑇𝑘2 𝑔 ) tanh 𝑘ℎ The relation is dependent upon the wavelength of the waves, as shown in 𝑘. 𝑘 = 2𝜋 𝐿 𝑥 KdV Regime The Korteweg-de Vries equation is a nonlinear partial differential equation that has applications to water waves in shallow water. 𝑢 𝜏 + 6𝑢𝑢 𝜒 + 𝑢 𝜒𝜒𝜒 = 0 Small-amplitude waves in shallow water is a statement of weak nonlinearity. 𝑢 𝜏 + 𝑢 𝜒𝜒𝜒 = 0 Linearized KdV and Fourier Transform Solutions can be found using Fourier Transforms such that, 𝑢 𝜒, 𝜏 = න −∞ ∞ 𝐴(𝑘, 𝜏)𝑒−𝑖𝑘𝜒 𝑑𝑘 . The appropriate partial derivatives are, 𝑢 𝜏 = ‫׬‬−∞ ∞ 𝜕𝐴 𝜕𝜏 𝑒−𝑖𝑘𝜒 𝑑𝑘 and 𝑢 𝜒𝜒𝜒 = ‫׬‬−∞ ∞ 𝐴 𝑘, 𝜏 (−𝑖𝑘)3 𝑒−𝑖𝑘𝜒 𝑑𝑘. Plugging in, the ordinary differential equation and solution to this KdV equation is, 𝜕𝐴 𝜕𝜏 − 𝐴 𝑘, 𝜏 𝑖𝑘 3= 0 and 𝐴 𝑘, 𝜏 = 𝐴0(𝑘)𝑒 𝑖𝑘3 𝜏 The general solution can then be expressed as, 𝑢 𝜒, 𝜏 = න −∞ ∞ 𝐴0(𝑘)𝑒(𝑖𝑘𝜒+𝑖𝑘3 𝜏) 𝑑𝑘 Delta Function with Initial Condition Next, using delta function initial condition of 𝑢 𝜒, 𝑜 = 𝛿 𝜒 and the Fourier Transform, 𝐴0 𝑘 = 1 2𝜋 , making 𝑢 𝜒, 𝜏 = 1 2𝜋 ‫׬‬−∞ ∞ 𝑒 𝑖𝑘𝜒+𝑖𝑘3 𝜏 𝑑𝑘, which looks an awful lot like the Airy Integral. Airy Integral Making some simple substitutions, 𝜉 = 𝜒 3𝜏 1 3 and 𝐾 = 𝑘 3𝜏 1 3, 𝑢 𝜒, 𝜏 = 𝑓 𝜉, 𝜏 = 1 2𝜋 ‫׬‬−∞ ∞ 𝑒 ቁ𝑖(𝜉𝐾+ 1 3 𝐾3 𝑑𝐾 is formed to utilize what is known about Airy Integrals. The last step is to change the scales to be in real space and real time, using the substitutions of 𝜒 = 𝜀 ℎ (𝑥 − 𝑐0 𝑡) and 𝜏 = 𝜀( 𝑐0 6ℎ )𝑡. In Real Time and Real Space The concluding equation for the surface displacement is, 𝑢 𝑥, 𝑡 = 1 3𝜀 𝑐0 6ℎ (𝑡) 1 3 𝐴𝑖 𝜀 ℎ 𝑥 − 𝑐0 𝑡 3𝜀 𝑐0 6ℎ (𝑡) 1 3 = 3 2ℎ 𝜂 Considering Different Initial Conditions for the Rogue Wave Instead of having the rogue wave occur at the origin all the time, we can use different initial conditions which will yield an equation where we can choose the time when we want the rogue wave to occur. Because we know the speed of the wave, we can choose the correct time where the wave will occur at a position of our choice. Using the new initial condition, 𝑢 χ, τ∗ = δ(χ) We can do similar calculations as before to obtain the equation (the following in 𝜂), 𝜂 𝑥, 𝑡 = 2ℎ 3 1 3𝜀 𝑐0 6ℎ (𝑡 − 𝑡∗) 1 3 𝐴𝑖 𝜀 ℎ 𝑥 − 𝑐0 𝑡 3𝜀 𝑐0 6ℎ (𝑡 − 𝑡∗) 1 3 Observe that 𝑡 has been replaced with 𝑡 − 𝑡∗ with the exception of the 𝑡 in 𝜒 Plots of Wave Formation for 𝑡∗ = 5 𝑡 = 3 𝑡 = 4 𝑡 = 4.5 𝑡 = 4.9 𝑡 = 4.99𝑡 = 4.75 𝑡 = 5.01 𝑡 = 5.1 𝑡 = 5.25 𝑡 = 5.5 𝑡 = 6 𝑡 = 7 Generating the Wave and the Wave Paddle PDE The wave is generated by a vertical wave paddle that moves back and forth to displace water in a fashion that generates the desired wave. The differential equation used to find the motion of the paddle is, 𝜂 𝐿(𝑡), 𝑡 + ℎ 𝑑𝐿 𝑑𝑡 = 𝑐0 𝜂 𝐿(𝑡), 𝑡 + 𝑐0 6 ℎ 𝜀 2 𝜂 𝑥𝑥(𝐿 𝑡 ) Where 𝐿(𝑡) is the 𝑥 position of the paddle with respect to time. While assuming the initial condition, 𝐿 0 = 0 This differential equation can be solved numerically using the classical Runge- Kutta method and the computed values can be input into the wave paddle.