6-freakwaves
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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 ππ π2 = ππ(1 + ) tanh πβ π The dispersion relation describes the physics of the waves πΌπ π€βππβ π = 2π πΏπ₯ wavelength, πΏπ₯ . and depends on The Korteweg-de Vries equation is a nonlinear, partial differential equation that has applications to water waves π’π‘ + 6π’π’π + π’πππ = 0 Small-amplitude waves in shallow water is a statement of weak nonlinearity π’π + π’πππ = 0 ∞ π’ π, π = π΄(π, π)π −πππ ππ −∞ Appropriate partial derivatives of ∞ π’ π, π = π΄(π, π)π −πππ ππ −∞ ∞ π’π = −∞ ∞ ππ΄ −πππ π ππ ππ and π΄ π, π (−ππ)3 π −πππ ππ π’πππ = −∞ Plugging into the linearized KdV equation to find the ordinary differential equation ∞ π −∞ −πππ ππ΄ − π΄ π, π ππ ππ which is ππ΄ ππ 3 ππ = 0 − π΄ π, π ππ 3 = 0 The solution of this ODE is π΄ π, π = π΄0 (π)π ππ 3 π Plugging this solution into π’ π, π So that π’ π, π = ∞ (πππ+ππ 3 π) π΄ (π)π ππ −∞ 0 To find π΄0 , take the Fourier Transform of the initial condition π’(π, 0) So that ∞ −πππ π΄ (π)π −∞ 0 1 π΄0 (π) = 2π ππ ∞ π’(π, 0)π πππ ππ −∞ Which is particularly useful when considering an ideal situation in which π’ π, π = πΏ π 1 ∞ πππ ) π’(π, 0 π ππ 2π −∞ 1 2π ∞ πΏ(π − 0)π −∞ πππ = π΄0 (π) 1 ππ(0) 1 ππ = π = 2π 2π 1 π΄0 (π) = 2π So now we have an equation that looks like an Airy Integral 1 π’ π, π = 2π 1 π΄π π§ = 2π ∞ π πππ+ππ 3 π ππ −∞ ∞ −∞ 1 π π π§+ π 3 3 π ππ Now we can make simple substitutions to exploit what is known π= π 3π 1 3 πΎ = π 3π 1 π’ π, π = π π, π = 2π 1 3 ∞ −∞ 1 π ππΎ+ πΎ3 3 π ππΎ We know that this is in the form of the Airy Integral using slow time and slow space scales 1 π’ π, π = π π, π = 2π π π, π = 1 3π ∞ 1 3 π ππΎ+ πΎ 3 π −∞ 1 π΄π(π) 3 ππΎ Substitutions can be made to convert it back to real time and space scales for the wavemaker π π= π₯ − π0 π‘ β π0 π= π π‘ 6β Substitutions can be made to convert it back to real time and space scales for the wavemaker 1 π’ π₯, π‘ = 3π π0 (π‘) 6β 1 π΄π 3 π π₯ − π0 π‘ β 3π π0 (π‘) 6β 1 3 3 = π 2β Now consider different initial conditions • The initial condition π’ χ, 0 only considers what the χ axis looks like when at τ = 0. • Namely the initial condition π’ χ, 0 = δ(χ) δ(χ) π’ 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 π’ 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) π’ χ, τ∗ 0 τ = τ∗ χ ∞ π’ π, π = π΄0 (π)π −(πππ+ππ 3 π) ππ −∞ Applying the new initial condition π’ χ, τ∗ = δ(χ) we have, ∞ π’ π, π∗ = δ χ = π΄0 (π)π −∞ −(πππ+ππ 3 π∗ ) ππ Then applying the Fourier Transform we have, 1 π΄0 (π) = 2π ∞ δ χ 3π ) (πππ+ππ ∗ πχ π −∞ We know that the delta function has the property such that, ∞ δ π₯ − π π π₯ ππ₯ = π(π) −∞ In the case above, π = 0 So evaluating, ∞ 1 π΄0 π = δ χ π 2π −∞ 1 ππ 3 π ∗ = π 2π πππ+ππ 3 π∗ πχ Now we know the identity of π΄0 π under the initial conditions π’ χ, τ∗ = δ χ . So π’ π, π becomes, 1 π’ π, π = 2π ∞ π −∞ πππ+ππ 3 (π−π∗ ) ππ Using similar substitutions as earlier, we can rewrite this in Airy integral form π= π 1 3 3(π − π∗ ) πΎ = π 3(π − π∗ ) Then just as before we have 1 π’ π, π = π π, π = 2π ∞ −∞ 1 3 1 π ππΎ+3πΎ3 π ππΎ Finally in this form which is no different from earlier except that π has been replace with π − π∗ π π, π = 1 3(π − π∗ ) 1 π΄π(π) 3 Then using similar substitutions to convert back to real time and space π π= π₯ − π0 π‘ β π0 π= π (π‘ − π‘∗ ) 6β π’ π₯, π‘ = 1 π0 3π (π‘ − π‘∗ ) 6β 1 π΄π 3 π π₯ − π0 π‘ β π0 3π (π‘ − π‘∗ ) 6β So, 2β π π₯, π‘ = 3 1 π0 3π (π‘ − π‘∗ ) 6β 1 π΄π 3 1 3 3 = π 2β π π₯ − π0 π‘ β π0 3π (π‘ − π‘∗ ) 6β 1 3 We know the speed of the wave and can pick a π‘∗ 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 π‘∗ = 5 π‘=3 π‘ = 4.9 π‘ = 5.25 π‘=4 π‘ = 4.99 π‘ = 5.5 π‘ = 4.5 π‘ = 5.01 π‘=6 π‘ = 4.75 π‘ = 5.1 π‘=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 π₯ over time π‘. Let us call the position of the wave paddle in π₯, L(π‘) • 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 π πΏ(π‘), π‘ + β Mass flux To consider flux, we must define the direction from the “inside” to the “outside”. Namely, we must parameterize the function π and obtain a vector function. We define π(π₯). And to parameterize, π π₯ = π₯, π π₯, π‘ . A vector function that is in a normal direction to π is then. π π₯ = −ππ₯ π₯, π‘ , 1 . Then the normal component of velocity is, π β ππ₯ , ππ§ = ππ§ − ππ₯ ππ₯ . In general for water waves, the kinematic free surface boundary condition in one space and one time dimension for an air-water interface is, ππ ππ ππ ππ − + = 0 ππ π§ = π ππ‘ ππ§ ππ₯ ππ₯ So by the kinematic free surface boundary condition, ππ‘ = ππ§ − ππ₯ ππ₯ ππ π§ = π Where π is the velocity potential and so ππ§ and ππ₯ are the velocity of the water at the position and time π₯, π§, π‘ . The mass flux “through” the wave paddle would be the height of the fluid at πΏ(π‘) times the velocity of fluid in the π₯ direction. ππΏ π πΏ(π‘), π‘ + β ππ‘ Because water cannot go through the paddle, the water’s velocity at the paddle must match the velocity of the paddle at the paddle, ππΏ ππ₯ = ππ‘ ππ‘ π₯ = πΏ(π‘) 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, ∞ πΏ(π‘) ∞ ππ‘ ππ₯ = πΏ(π‘) ππ§ − ππ₯ ππ₯ ππ₯ Now we can equate the mass flux at the paddle with that of the wave surface to get the relation, ππΏ π πΏ(π‘), π‘ + β = ππ‘ ∞ πΏ(π‘) ππ‘ ππ₯ Referring back to, π’π + π’πππ = 0 3 π’ π, π = π(π, π) 2β π π= π₯ − π0 π‘ β π0 π= π π‘ 6β ππ + ππππ ∞ π0 β = ππ‘ + π0 ππ₯ + 6 π 3 ππ₯π₯π₯ = 0 3 ∞ π0 β ππ‘ ππ₯ = − π0 ππ₯ + 6 π πΏ(π‘) πΏ(π‘) ∞ π0 β ππ‘ ππ₯ = π0 π πΏ(π‘), π‘ + 6 π πΏ(π‘) ππ₯π₯π₯ ππ₯ 2 ππ₯π₯ (πΏ π‘ ) ππΏ π0 β π πΏ(π‘), π‘ + β = π0 π πΏ(π‘), π‘ + ππ‘ 6 π 2 ππ₯π₯ (πΏ π‘ ) 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 • http://myarchive.us/richc/2010/Deadlywavekilss2inju res14oncruiseshipinM_146AB/ship50footwave2.jpg • http://graphics8.nytimes.com/images/2006/07/11/sci ence/11wave.1.395.jpg • https://www.google.com/search?q=freak+waves&esp v=2&tbm=isch&source=lnms&sa=X&ei=8f_9U4S5HrS _sQS514LgCQ&ved=0CAgQ_AUoAw&biw=1280&bih= 923#facrc=_&imgdii=_&imgrc=GW7u_JXGbsENlM%25 3A%3B6HPoEPp_faOGvM%3Bhttp%253A%252F%252 Fwww.ipacific.com%252Fforum%252Findex.php%253 Faction%253Ddlattach%253Btopic%253D507.0%253B attach%253D408%3Bhttp%253A%252F%252Fwww.ip acific.com%252Fforum%252Findex.php%253Ftopic%2 53D507.0%3B350%3B250
