Ocean waves of maximal amplitude

Transcription

1 Ocean waves of maimal amplitude C.H. van der Meer March 6 Applied Analysis and Mathematical Physics Group Department of Applied Mathematics University of Twente Graduation committee: prof. dr. E. van Groesen dr. G.A.M. Jeurnink dr. ir. O. Bokhove

2 Contents Introduction From linear dispersive wave equation to Maimal Spatial Amplitude 4. Setting Simple eample: Gaussian spectrum The influence of a phase difference Various descriptions for the maimum of the envelope Spectra based on field data Largest possible wave given certain momentum and energy. Formulation of the optimization problem Single peak solutions Verification by Fourier transformation Periodic solutions Physical constraints related to differential equations 3 3. Linear evolution of the largest possible wave Linear evolution of a soliton Comparison of cornered and smooth initial profiles Bibliography 44 Appendices 45

3 Introduction A freak wave is a wave that appears seemingly random and that is significantly larger than the surrounding waves. It is a phenomenon that occurs on many oceans around the world. These waves, that are known to have damaged and sunk some very large ships, will appear very sudden and disappear just a short time after. Tales of freak waves were once dismissed as seafaring myths. But nowadays the height of the sea is continually monitored at many places such as oil rigs. One of these offshore platforms, the Draupner platform in the Norwegian section of the North Sea, has recorded a freak wave at 5:4 on st January 995 [7]. (see figure ) We assume these freak waves originate from several smaller waves with different wavelengths that travel with different velocities. These smaller waves may at some point combine to one very big wave and almost immediately dissolve into smaller waves again. In this report I ll try to find a description for the maimum wave height for all times, hoping to understand some of the phenomena of the ocean a bit better. There are several partial differential equations that model wind generated water waves. In this report I will look at a few of them. For eample in chapter I ll use linear dispersive partial differential equations. The solution of such an equation, η (, t) with one spatial variable and one time variable t, represents the wave height. A function called Maimal Spatial Amplitude, MSA (t) = ma η (, t), is used to describe the maimum of the wave height at a fied time t. Though this formulation is simple, it is often not possible to get an eplicit epression for the MSA. In chapter I ll try to find some descriptions for the MSA using several mathematical analytical methods. Figure : This plot shows a freak wave at approimately 7 seconds after 5: hours

4 The linear dispersive differential equations I use in the first chapter have several constants of motion including momentum and energy. In chapter I will not use the differential equations to model water waves eactly, but instead I will only prescribe the amount of energy and momentum in the waves. Then I will look for the largest wave that satisfies these energy and momentum constraints. The mathematical solution is obtained by solving the ordinary differential equation that follows from Lagrange multiplier rule. We also look for periodic solutions of this differential equation. In chapter 3 the largest wave from the previous chapter will be used as an initial condition in the linear dispersive differential equations I used in the first chapter. The solutions will be represented using the computer program Matlab. In this chapter I will also look at the linear evolution of the soliton solution of the Korteweg de Vries equation. 3

5 Chapter From linear dispersive wave equation to Maimal Spatial Amplitude. Setting First we investigate initial value problems for PDE s which are of first order in time, linear and dispersive: t η (, t) = p ( )η (, t) (.) η (, ) = η () (.) with p () a polynomial. The dispersion relation for modes e i(k ωt) of the equation is then The PDE can then formally be given by Ω (k) = ip (ik) (.3) t η + iω ( i ) η = (.4) With the initial condition η (, ) given in a Fourier integral, the solution of the initial value problem is represented also by a Fourier integral η (, t) = g (k) e i(k Ω(k)t) dk (.5) η (, ) = g (k) e ik dk (.6) With an initial spectrum concentrated around k the epression for η (, t) can be simplified using a Taylor series for Ω (k) around k. We need at least second order to get a variation in the maimum water height. Take Ω (k) = Ω (k ) + Ω (k ) (k k ) + Ω (k ) (k k ) then η (, t) becomes η (, t) = = e ik g (k) e i(k Ω(k )t Ω (k )(k k )t Ω (k )(k k ) t) dk k t g (κ + k ) e i(κ Ω (k )κt Ω (k )κ t) dκ (.7) Ω(k ) 4

6 which is a combination of a carrier wave e ik A (, t) = Ω(k ) t k and an amplitude g (κ + k ) e i(κ Ω (k )κt Ω (k )κ t) dκ (.8) With the transformation to a moving frame: { ξ = Ω (k ) t τ = t (.9) the amplitude becomes A (ξ, τ) = g (κ + k ) e i(κξ Ω (k )κ τ) dκ (.) Remember η (, t) was given in the form of a Fourier integral (.5) and as the solution of a PDE (.4). Since the amplitude is given in a similar integral, it can also be seen as the solution of a PDE: τ A + i Ω (k ) ( i ξ ) A = τ A i Ω (k ) ξξ A = (.) The envelope of η (, t) is defined by: envelope (ξ, τ) = A (ξ, τ) (.) Since the solution is smaller than the envelope for every ξ and τ, the Maimal Spatial Amplitude (MSA) is smaller than the maimum of the envelope for every τ: MSA (τ) = ma where MSE stands for Maimal Spatial Envelope. {η (, t)} ma {envelope (ξ, τ)} = MSE (τ) (.3) ξ. Simple eample: Gaussian spectrum In this section I will illustrate the material of the previous section for certain PDE and initial condition. Use a Gaussian spectrum g (k) = σ π e (k k ) σ (.4) 5

7 for the initial condition (.6) η (, ) = = The solution η (, t) then becomes η (, t) = e ik g (k) e ik dk σ π e (k k ) with amplitude (using the coordinate transformation (.9)) A (ξ, τ) = σ π eiκξ κ ( σ e ik dk (.5) Ω(k ) t k A (, t) (.6) = e + iσ Ω (k )τ ξ σ /(+iσ Ω (k )τ) σ +iω (k )τ) dκ (.7) This is a comple solution. Therefore I have to take the real part of η (, t) before plotting and before calculating the Maimal Spatial Amplitude. The envelope of the solution η (, t) is: envelope (ξ, τ) = A (ξ, τ) (.8) +(σ Ω (k )τ) = e ξ σ / ( + (σ Ω (k ) τ) ) 4 (.9) This envelope is maimal at ξ = for all τ (first derivative = and second derivative ). If at time t the carrier wave has a top in ξ = (which corresponds to = Ω (k ) t), the maimum of η (, t) is the same as the maimum of the envelope. If for a given t the carrier wave has no top in ξ =, the solution η (, t) has it s maimum a short distance away from = Ω (k )t, and therefore the maimum of η (, t) is a bit smaller than the maimum of the envelope (see figure.). We can use the maimum of the envelope, which I denote MSE (τ), as the general form of the MSA (see figure.): MSA (τ) ( + (σ Ω (k )τ) ) = MSE (τ) (.) 4 In order to make some plots we need to specify the PDE. For the plots below (figure. -.3) I used the linear version of the Korteweg-de Vries equation (see [3]) This PDE has the dispersion relation t η + η + 6 η = (.) Ω (k) = k 6 k3 (.) For the initial conditions I used the Gaussian spectrum (.4) with k = and σ =. 6

8 Figure.: R (η (, t)) (solid) and the envelope (dotted) at times between t = 6 (bottom) and t = (top) t Figure.: MSE (t) (solid) and a sketch of MSA (t) (dotted). 7

9 Figure.3: MSE (t) over a longer period. t As figure. and.3 show, η (, t) attains the highest maimum if t = and =. This will be the case in general, because the term e i(k Ω(k)t) in.5 is always smaller or equal to, and it is eactly for all values of k only if and t are both zero, resulting in the highest value of the wave height: η (, ) = g (k)dk. In the remainder of this report I will not look at the actual MSA (t) anymore, because the function MSE (t) gives a good description of the general form of MSA (t) and is easier to calculate..3 The influence of a phase difference In this section I ll start with a different formula for η (, t), but I ll follow the same steps as in section., so the results from both sections can be compared. As seen in the eample from the previous section all the different waves had a maimum in = at t =. All these maimums in eactly the same place at the same time give the MSA it s maimum at t =. In general the waves will have a phase difference though. This is modeled by adding the term e iθ(k) to formula (.5): η (, t) = g (k)e iθ(k) e i(k Ω(k)t) dk (.3) You can see that choosing θ (k) = gives the same formula as formula (.5). Therefore choosing θ (k) = in the results of this section, should give the same results as in the previous section. 8

10 In section. the dispersion relation Ω (k) was approimated by the second-order Taylor epansion. We can do the same for the phase difference θ (k) θ (k ) + θ (k )(k k ) + θ (k )(k k ), which will result in an approimation to the real solutions only if the spectrum g (k) is concentrated around k. The solution η (, t) now becomes η (, t) = e ik Ω(k ) k t e iθ(k ) g (κ + k ) e i(κ Ω (k )κt Ω (k )κ t+θ (k )κ+ θ (k )κ ) dκ (.4) This is again a combination of a carrier wave and an amplitude. By using the transformation (.9) the amplitude simplifies to A (ξ, τ) = g (κ + k ) e i κ(ξ+θ (k )) Ω (k )κ τ θ (k ) Ω (k ) dκ (.5) Comparing this with equation (.) shows that the amplitude is now shifted in space and time, as is the envelope, which is the absolute value of this amplitude. Taking the maimum over all ξ for each τ results in a function MSE (τ) which is just shifted in time compared to the same function in the previous section: ( ) MSE (τ) = MSE θ= τ θ (k ) (.6) Ω (k ) But remember we assumed the spectrum to be concentrated around k = k in order to use the Taylor epansions, so this result might not be true for other spectra. For some specific phases, however, this result of a MSE shifted in time, is valid for any spectrum. For eample if the phase is a constant θ (k) = α: The term e iα can be put in front of the integral. And since the absolute value e iα =, the envelope of the wave height with a constant phase is the same as the envelope of the wave height with no phase. Another possibility is a linear phase θ (k) = βk. In this case the solution.3 will be η (, t) = g (k) e iβk e i(k Ω(k)t) dk (.7) = g (k) e i(k(+β) Ω(k)t) dk (.8) (.9) Therefore a linear phase is the same as a translation of the spatial variable. So the MSE (t) stays the same. Finally if the phase is a multiple of the dispersion relation θ (k) = γω (k). Then the solution can be written as η (, t) = g (k)e iγω(k) e i(k Ω(k)t) dk (.3) = g (k)e i(k Ω(k)(t γ)) dk (.3) (.3) 9

11 This formula shows that a phase which is a multiple of the dispersion relation, gives the same results as a linear translation of the time variable. Therefore the MSE will be shifted in time with respect to the case with zero phase. Of course any linear combinations of these three types of phases also result in a shifted Maimal Spatial Envelope: MSE θ=α+βk+γω(k) (t) = MSE θ= (t γ) (.33).4 Various descriptions for the maimum of the envelope If the envelope has it s maimum at ξ = for all τ, various descriptions of this maimum can be found by epressing A (, τ) in different ways. With a small change in notation the amplitude (.) becomes A (ξ, τ) =  (κ)e i(κξ ν(κ)τ) dκ (.34) Now rewrite A (, τ) using the fact that  (κ) is the Fourier transform of A (ξ, ) A (, τ) =  (κ) e iν(κ)τ dκ (.35) = ( ) A (ξ)e iκξ dξ e iν(κ)τ dκ (.36) ( ) = A (ξ) e i(κξ+ν(κ)τ) dκ dξ (.37) = A (ξ)s (ξ, τ)dξ (.38) Similar to η (, t) and A (ξ, τ) in section., S (ξ, τ) = e i(κξ+ν(κ)τ) dκ is also the solution of a PDE τ S + iν (i ξ ) S = (.39) with initial condition S (ξ, ) = e iκξ dκ = δ (ξ) (.4) If from ν (κ), κ can be written as a function of ν, the epression (.35) can also be

12 rewritten to A (, τ) = Â (κ) e iν(κ)τ dκ (.4) = iντ dκ (ν) Â (κ (ν)) e dν (.4) dν = ( ) A (ξ)e iκ(ν)ξ iντ dκ (ν) dξ e dν (.43) dν ( ) dκ (ν) = A (ξ) dν e i(κ(ν)ξ+ντ) dν dξ (.44) = A (ξ)g(ξ, τ)dξ (.45) Just like S (ξ, τ) also G (ξ, τ) = dκ(ν) dν e i(κ(ν)ξ+ντ) dν is the solution of a PDE ξ G + iκ (i τ )G = (.46) dκ (ν) G (, τ) = dν e iντ dν (.47) So if the envelope has it s maimum at ξ = for all τ, it can be epressed as the absolute value of an integral A (ξ)f (ξ, τ)dξ. Where f (ξ, τ) is the solution of the initial value problem (.39) and (.4) or boundary value problem (.46) and (.47)..5 Spectra based on field data In this section I will look at the behavior of the maimum amplitude when using spectra other than the Gaussian spectrum I used in section.. In particular the Pierson- Moskowitz and the JONSWAP spectrum which can be found for instance in the book of S.R.Massel [5]. These spectra are based on theoretical discoveries combined with field data of wind generated surface waves on different seas., hence more realistic than a Gaussian spectrum. Because these are frequency spectra the Fourier integral should be taken over ω instead of k. Another important difference is that these spectra are power spectra, but the Fourier integral needs an amplitude spectrum. Therefore I should use the square root of these power spectra. η (, t) = S (ω)e i(k(ω) ωt) dω (.48) where K (ω) is the inverse of the dispersion relation Ω (k) = k tanh (k). This is the eact dispersion relation; the one I used in section. is an approimation of this one. I don t consider any phase in this equation. Therefore the highest maimum of η (, t) will be obtained when and t are both zero. The value of this maimum is η (, ) = S (ω)dω. The absolute value η (, t) is the envelope of η (, t). Taking the maimum over t of this envelope results in the Maimal Temporal Envelope MTE ().

13 The spectrum proposed by Pierson and Moskowitz in 964 is [6] S (ω) = αg ω 5 e B( g ωu )4 (.49) where α = 8. 3, B =.74 and U is the wind speed at a height of 9.5 m above the sea surface. This spectrum was proposed for fully-developed sea. I calculated the normalized spectra with different values of the wind speed U and I plotted these normalized spectra and the corresponding MT E () (see figure.4 and.5) U= U=5 U=.5 S(ω) ω Figure.4: Pierson-Moskowitz spectrum

14 .8 U= U=5 U=.6.4 MTE() Figure.5: MT E () corresponding to Pierson-Moskowitz spectrum =.879. Substi- The eperimental spectra given by Pierson and Moskowitz yield Uωp g tution into (.49) leads to S (ω) = αg ω 5 e 5 4 ( ω ωp ) 4 (.5) The JONSWAP spectrum etends this form of the Pierson-Moskowitz spectrum to include fetch-limited seas. JONSWAP stands for Joint North Sea Wave Project. It is a wave measurement program carried out in 968 and 969 in the North Sea [4]. The JONSWAP spectrum model that is based on this program takes the form (ω ωp) S (ω) = αg ω 5 e 5 4 ( ω ωp ) 4 γ e σ ω p (.5) The mean JONSWAP spectrum yields γ = 3.3, σ =.7, σ =.9 and ( ). gx α =.76 (.5) U ( ).33 gx (.53) ω p = 7π g U U For comparison with the Pierson-Moskowitz spectrum I plotted the normalized JON- SWAP spectra and the corresponding MTE () for a fully developed sea (X = km) with different wind velocities U (figure.6 and.7). 3

15 5 4.5 U= U= U= S(ω) ω Figure.6: JONSWAP spectrum for fully developed sea.8 U= U= U=3.6.4 MTE() Figure.7: Corresponding MT E () As you can see in figure.7, a lower wind speed results in a higher and sharper peak in the MTE (). Just like it did with the Pierson-Moskowitz spectrum (figure.5) I also plotted the JONSWAP spectrum and the MTE for fetch X ranging from X = 5 km (fetch-limited seas) to X = km (fully developed sea) with U = m/s (figure.8 and.9). 4

16 3.5 X=5e3 X=e3 X=e3 3.5 S(ω) ω Figure.8: JONSWAP spectrum for different fetch with wind velocity U = m/s.4. X=5e3 X=e3 X=e MTE() Figure.9: Corresponding MT E () This figure shows that on fetch limited seas the MTE has a higher and sharper peak than on fully developed seas. Now I will observe how a change in some of the other parameters effects the spectrum and the MTE (). First I will look at γ which describes the degree of peakedness of the JONSWAP spectrum. A higher value of γ results in a spectrum that is more peaked, and 5

17 the corresponding MTE () will be wider (see figure. and.). 5 γ= γ=3.3 γ=8 4 S(ω) ω Figure.: JONSWAP spectrum for different values of γ. γ= γ=3.3 γ=8.8.6 MTE() Figure.: Corresponding MT E () 6

18 You can see in figures. and. that a spectrum which is more peaked, due to a higher value of γ, results in a MTE which is less peaked (the dotted line). The width of the peak region will change when changing σ and σ : Smaller values of σ and σ result in a smaller peak of the spectrum (and a wider base after normalizing the spectrum). The corresponding MTE () will be more peaked (figure. and.3) σ =. and σ =.3 σ =.7 and σ =.9 σ =. and σ = S(ω) ω Figure.: JONSWAP spectrum for different values of σ and σ 7

19 ..8 σ =. and σ =.3 σ =.7 and σ =.9 σ =. and σ =..6 MTE() Figure.3: Corresponding MT E () For a comparison with the Gaussian spectrum I used in a previous section, I normalized a Gaussian spectrum, the Pierson-Moskowitz spectrum and the JONSWAP spectrum on a fully developed sea with a wind speed U = m/s. Net I calculated the MTE () for each of those three spectra, and plotted them in one figure (see figure.4 and.5) Gaussian Pierson Moskowitz JONSWAP 3.5 S(ω) ω Figure.4: various spectra 8

20 .8 Gaussian Pierson Moskowitz JONSWAP.6.4 MTE() Figure.5: MTE () corresponding to various spectra As you can see in this last picture, the tail of the JONSWAP and the Pierson-Moskowits spectrum results in a sharper peak for the MTE (). 9

21 Chapter Largest possible wave given certain momentum and energy The differential equations I used in the previous chapter have several constants of motion including momentum and energy: I (η) = η d (.) H (η) = ( η) d (.) So the momentum and energy of the initial condition are the same as the momentum and energy of the wave height at any other time. In this chapter I want to investigate which wave has the largest wave hight given only the momentum and energy. I have to specify both the momentum and energy, because if I only specify one of them, the solution with the largest wave height has a maimum which goes to infinity.. Formulation of the optimization problem Consider the following problem with two constraints: momentum and energy. { } ma ma η () η η () d = γ, ( η ()) d = γ (.3) In this problem there are three functionals F (η) = ma {η ()} (.4) G (η) = η () d (.5) G (η) = ( η ()) d (.6)

22 We can translate this constrained variational problem into a differential equation by calculating the variational derivative for each functional and applying Lagrange multiplier rule. The first variation of G (η) is δg (η; v) = d dǫ G (η + ǫv) ǫ= = d (η (t) + ǫv ()) d dǫ ǫ= = η () v () d (.7) Which is the L inner product of v (t) and the variational derivative δg (η) = η (.8) The first variation of G (η) is δg (η; v) = d ( η () + ǫ v ()) d dǫ ǫ= = η () v () d = η () v () d + [ η () v ()] (.9) if η () goes to for ±, so does v () and so the boundary term vanishes. The related variational derivative is δg (η) = η (.) In order to calculate the first variational of the functional F (η), I have to rewrite it as follows: F (η + ǫv) = ma {(η + ǫv)()} = (η + ǫv)( (ǫ)) (.) where (ǫ) is the value of for which η () + ǫv () has its maimum. The first variation of F (η) is the derivative of F (η + ǫv) with respect to ǫ at ǫ =. For continuous functions η and v this is δf (η; v) = d dǫ (η + ǫv)( (ǫ)) = v ( ()) ǫ= (.)

23 In order to calculate the variational derivative, the first variation must be in the form of an integral δf (η; v) = v ( ()) = v () δ ( ()) d (.3) This first variation is again the inner product of v () and the variational derivative δf (η) = δ ( ) (.4) where = () is the value of for which η () has its maimum. Applying Lagrange multiplier rule to the three variational derivatives calculated above gives a differential equation. Single peak solutions σδf (η) = λ δg (η) + λ δg (η) (.5) σδ ( ) = λ η () λ η () (.6) In this section I want to find single solutions of the differential equation (.6). In section.4 I ll look for periodic solutions. The differential equation derived in the previous section is σδ ( ) = λ η () λ η () (.7) And the solution to this differential equation should of course satisfy the constraints in the constrained variational problem (.3) Because the first constraint η () d = γ < we must assume the boundary con- η () = and lim η () =. ditions lim η () d = γ (.8) ( η ()) d = γ (.9) If σ = the differential equation reduces to λ η () λ η () =. The only solution to this differential equation that satisfies the boundary conditions above is the trivial solution η () =. Therefore I take σ = (for any other value I can divide the entire differential equation by σ.). For < we have the differential equation λ η () λ η () =. Because of the first boundary condition, the solution to this differential equation is η () = c e Õ λ λ ( ) for < (.)

24 where λ and λ are both positive or both negative. In a similar way for > : η () = c e Õ λ λ ( ) for > (.) η () must be continuous in =, so c = c. Now for the δ-function it holds that δ ( a )d = for any a < and b >. Therefore b = b a λ η () λ η () d b = λ η () d + λ ( η (a) η (b)) a b ( Õ λ λ Õ (a λ = λ η () d + λ c e ) + e λ λ a λ (b ) ) (.) for any a < and b >. In the limit where a and b go to, the integral term tends to λ zero, and this equation reduces to = 4λ λ c, so c must be c = ± 4 λ λ, where the negative sign is used when λ and λ are negative. It turns out that a negative c produces a negative solution η () with a minimum in =. Therefore the solution we are looking for is the one with the positive value of c. Now substitute this value for c and the same value for c into the solution (.) and (.): Õ λ η (t) = e λ 4 (.3) λ λ Substituting this solution in the constraints gives two equations from which λ and λ can be epressed in γ and γ. The first constraint (.8) gives: γ = η () d Õ = e λ Õ ( ) λ λ 4 d + e ( ) λ λ λ 4 d λ λ = e Õ λ λ ( ) 6λ λ d + Õ e λ ( λ ) d 6λ λ Õ λ Õ = e ( λ ) λ λ e ( ) λ 3λ λ λ λ 3λ λ λ = ± (.4) 6λ λ λ 3

25 Similar for the second constraint (.9): γ = ± (.5) 6λ λ λ where the negative sign is used when λ and λ are negative. Combining equations (.4) and (.5) gives the following epressions for λ and λ : λ = ± γ/4 4γ 3/4 (.6) λ = ± γ/4 (.7) 4γ 3/4 with λ and λ both positive or both negative. The solution (.3) can now be written in terms of γ and γ. η () = (γ γ ) /4 e γ γ / (.8).6.4. η Figure.: η () for γ = (solid), γ = 3 (dashed) and γ = (dotted). In this plot I used = 5, but any other value of results only in a linear translation along the -ais. As you can see in this plot, a higher value of γ gives a solution η () with a smaller, but higher peak. This can be derived from the formulas as well. First the height of the peak is the maimum of.8: ma (η ()) = η ( ) = (γ γ ) /4 (.9) and second the verte angle, which is a measure for the width of the peak, is ( ) verte angle = arctan γ /4 γ 3/4 (.3) 4

26 .3 Verification by Fourier transformation We can check whether the calculations in the previous sections are correct using Fourier transform on the constrained variational problem we started with (.3), on the differential equation we found halfway through (.6) and on the solution we found in the end (.8). To simplify the calculations I ll take = in this section. In that case the constrained variational problem (.3) is ma η { ma η () The differential equation (.6) reduces to And the solution (.8) reduces to η () d = γ, ( η ()) d = γ } (.3) δ () = λ η () λ η () (.3) η () = (γ γ ) /4 e γ γ / (.33) If we epress the constrained variational problem in the Fourier transform ˆη of η, and solve this problem, we should get a solution which is the Fourier transform of the solution in the previous section. Let s take a look at the three functionals of the constrained variational problem. The first one is ma η () = η (). Epressed in ˆη this is η () = ˆη (k) e ik dk = ˆη (k) dk (.34) The second functional can be epressed in ˆη using Parseval s theorem. η () d = ˆη (k) dk (.35) For the last functional the calculations can be found in appendi A. ( η ()) d = k ˆη (k) dk (.36) With these three functionals the constrained variational problem (.3) can be epressed in ˆη { } ma ˆη (k)dk ˆη ˆη (k) dk = γ, k ˆη (k) dk = γ (.37) Now we need to find the variational derivatives of the three functionals and apply Lagrange multiplier rule. The variational derivative of ˆη (k) dk is. Assuming ˆη is 5

27 real, the variational derivative of ˆη (k) dk is ˆη (k) and the variational derivative of k ˆη (k) dk is kˆη (k). Applying Lagrange multiplier rule gives the following equation = λ ˆη (k) + λ k ˆη (k) (.38) This is indeed the Fourier transform of the differential equation (.3). The solution of this equation is ˆη (k) = (.39) (λ + λ k ) By substituting this solution in the constraints we get: γ = ± 6λ λ λ (.4) γ = ± 6λ λ λ (.4) where the negative sign is used when λ and λ are negative. From these two equations λ and λ can be epressed in γ and γ : λ = ± γ/4 4γ 3/4 λ = ± γ/4 4γ 3/4 (.4) (.43) with λ and λ both positive or both negative. Since we want to find the maimal solution, not the minimal one, we have to use the positive λ and λ. Then the solution ˆη epressed in γ and γ is ˆη (k) = (γ γ ) 3/4 γ + γ k (.44) The inverse Fourier transform of this solution is η () = (γ γ ) /4 e γ γ / (.45) which is indeed the solution (.33) we found in solving the constrained variational problem in η. 6

28 .4 Periodic solutions In this section I ll try to find periodic solutions of the differential equation (.6) σδ ( ) = λ η () λ η () (.46) For periodic solutions there is of course more than one where η attains it s maimum. I can choose = to simplify the calculations. I ll try to find a solution in the interval [, X] where X is the period of the solution η (). Because I choose =, the differential equation reduces to λ η () λ η () = (.47) in the interior of the interval [, X]. If λ and λ have the same sign, the solution of this differential equation is η () = c e s + c e s (.48) where s = λ /λ. This is a solution on one period, which I can etend to the entire real line, but for the etended solution to be continuous I must hold that η () = η (X) (.49) If σ =, and λ and λ have the same sign, the only possible periodic solution of the differential equation.46 is η () =. Therefore I can take σ =. We assumed the maimum to be at =. At this maimum the delta-function in the differential equation (.46) is not equal to zero, so for the solution (.48) to satisfy the differential equation at = it must satisfy the following equation = = + + δ () d λ η () d + = λ ( η (+) η ( )) λ η () d = λ ( η () η (X)) (.5) With these two equations I can find epressions for c and c, and the solution (.48) becomes ( ( e s X η () = λ s (e sx ) cosh s X )) (.5) λ with s = λ. 7

29 If I substitute this solution in the two constraints X X η () d = γ (.5) ( η ()) d = γ (.53) I get two equation which implicitly epress λ and λ in γ and γ. γ = esx + e sx sx 6λ λ s (e sx ) (.54) γ = esx + e sx sx + 6λ s (esx ) (.55) λ with s = λ. Now for some plots I choose X = π and γ =. η Figure.: η () for γ = (solid), γ = 3 (dashed) and γ = (dotted). As you can see in this plot, for a fied period X = π and a fied value of γ =, a higher value of γ results in a higher maimum of the solution. The net plot shows the relation between this maimum and γ. 8

30 replacemen 3.5 ma η () γ Figure.3: The maimum of η () as a function of γ. If I make a plot of the maimum of the solution on the entire real line, as I calculated in section., it will be the same as figure.3 for large values of γ only. For small values of γ the maimum of the solution on the entire real line is a little bit smaller than the maimum for the periodic case. In the beginning of this section I assumed that λ and λ have the same sign. If I assume they have a different sign, I can find a solution η () which has a lover maimum than the solution in this section. Therefore it is not the solution of the constrained variational problem (.3) { ma ma η () η η () d = γ, ( η ()) d = γ } but it is a solution of { crit η ma η () η () d = γ, ( η ()) d = γ } Calculations and plots for this case can be found in appendi B. 9

31 Chapter 3 Physical constraints related to differential equations In chapter we used a linear dispersive differential equation to model ocean waves (.4) t η + iω ( i ) η = (3.) This differential equations has several constants of motion including momentum and energy: I (η) = η d (3.) H (η) = ( η) d (3.3) So the momentum and energy of the initial condition are the same as the momentum and energy of the wave height at any other time. In chapter I found the wave with the highest possible maimum which satisfied certain momentum and energy by solving the following constrained problem { ma η ma η () I (η) = γ, H (η) = γ } (3.4) In this chapter I will use the solution to this problem as an initial condition for the differential equation. 3. Linear evolution of the largest possible wave The two constraints in problem (3.4) represent momentum and energy. I (η) = η d (3.5) H (η) = ( η) d (3.6) 3

32 I will show that these constraints are constants of motion for the differential equation (3.) with dispersion relation Ω (k) = k 6 k3. Taking the partial derivative of I (η): ( ) t (I (η)) = t η d (3.7) = ηη t d (3.8) The term η t can be substituted using the differential equation (3.) ( t (I (η)) = η η + ) 6 η d (3.9) = Using integration by parts on the second term we get ηη 3 ηη d (3.) t (I (η)) = ηη d + 3 η η d + [ = η + 6 (η ) ] 3 ηη [ ] 3 ηη (3.) (3.) This is equal to zero for the solutions of the optimization problem in section., because η and all it s derivatives go to zero for ±. Therefore I (η) is a constant of motion. Similarly for the second constraint we have ( ) t (H (η)) = t ηd (3.3) = = η η t d (3.4) η (η + ) 6 η d (3.5) [ 3 η η d + ] 3 η η (3.6) = [ ] η + [ = η + 6 (η ) ] 3 η η (3.7) = (3.8) So the second constraint is also a constant of motion. Since the two constraints in the optimization problem (3.4) are constants of motion for the differential equation (3.), we can use the optimal solution (.8) of the constrained problem as an initial condition. η (, ) = (γ γ ) 4 e γ γ Net is a plot of this initial condition for I (η) = γ = and H (η) = γ. (3.9) 3

33 .6 γ = γ =3 γ =.4. η(,) Figure 3.: Initial condition In this plot, and all the following plots in this section, the solid line corresponds to the initial condition with γ =, the dashed line corresponds to γ = 3 and the dotted line to γ =. In chapter the solution of the differential equation is η (, t) = ˆη (k) e i(k Ω(k)t) dk (3.) where ˆη (k) is the Fourier transform of the initial condition η (, ). For the optimal solution of the constrained problem the Fourier transform was calculated in section.3: In figure 3. is a plot of this Fourier transform with γ =. ˆη (k) = (γ γ ) 3/4 γ + γ k (3.) 3

34 .8 γ = γ =3 γ =.6.4. ˆη (k) k Figure 3.: Spectrum The term Ω (k) in equation (3.) is the dispersion relation. In the following plots I ll tanh k use the eact dispersion relation Ω (k) = k. k We can t work out the integral in equation (3.) to get an eplicit formula for the wave height. Still we are able to make some plots (with the use of Matlab) of the solution η (, t) at different times. I added a constant for the solution at t and at t = so that the plots at different times do not overlap each other. 33

35 .5 γ = γ =3 γ =.5.5 η(,t) Figure 3.3: η (, t) for t = (bottom), t = (middle) and t = (top) The phase velocity for these waves is V ph (k) = Ω(k) tanh k =. This means that short k k waves, which have a large wavenumber k, travel slow and large waves travel fast. You can also see this in the figure 3.3. For t < the short waves are close to zero and the long waves are further away to the left. But since these long waves travel faster, they will catch up with the short waves, so that they all add up at =, generating a freak wave. This is called phase focussing. We are really interested in the Maimal Spatial Envelope, MSE (t), which is for every t the maimum over all of the envelope of the solution: MSE (t) = ma η (, t) (3.) For the η (, t) I plotted in figure 3.3 the MSE is plotted in figure

36 .6 γ = γ =3 γ =.4. MSE(t) t Figure 3.4: Maimal Spatial Envelope 3. Linear evolution of a soliton The differential equation (3.) I used in the previous section and in chapter is the linearized KdV-equation. The KdV-equation in normalized form (see [3]): t η = ( η + η ) (3.3) has several constants of motion, including: I (η) = η () d (3.4) This KdV-equation has many soliton solutions: η H (η) = η3 d (3.5) 3 η (, ) = 3V ) (3.6) ( cosh V ( V t) where V denotes the velocity of the soliton. Now look at what happens if I use the constants of motion as constraints in the optimization problem (3.4) { } ma ma η () I (η) = γ, H (η) = γ (3.7) η 35

37 This constraint problem can be solved in a similar way as the one in chapter, resulting in a differential equation: σδf (η) = λ δg (η) + λ δg (η) (3.8) For very special values of γ and γ the σ will be zero. The solution of this problem is then a single soliton: η () = 3V ) (3.9) ( cosh V In general σ, but the left hand side of the differential equation is equal to zero for. So the solution of this problem is part of a soliton on > and also on <, combined at = to a continuous but non-smooth function. I will look at the smooth soliton first.. V=/ V= V=3/ η(,) Figure 3.5: Initial condition: soliton profile This smooth soliton will travel undisturbed with velocity V. Therefore the maimum will be the same for all time, so the Maimal Spatial Amplitude will be a constant. It is also interested to see what will happen if we use this soliton as an initial condition for the linear differential equation we used in the previous section. I made some plots again with Matlab. In all of the plots in this section the solid line corresponds to the soliton with V =, the dashed line to V = and the dotted one to V = 3. 36

38 First I plot the spectrum. 7 V=/ V= V=3/ ˆη (k) k Figure 3.6: Spectrum of the soliton profile Net is the solution η (, t) at different times for the linear evolution according to the dispersion relation Ω (k) = k tanh k k : 4 V=/ V= V=3/ 3 η(,t) Figure 3.7: η (, t) for t = (bottom), t = (middle) and t = (top) 37

39 And finally a plot of the Maimal Spatial Envelope:. V=/ V= V=3/.8.6 MSE(t) t Figure 3.8: Maimal Spatial Envelope In figure 3.7 you can see the phase focussing again, which I spoke about in the previous section (figure 3.3) In figure 3.8 you can see, the MSE is now not a constant. But it also does not descent as fast as the MSE in the previous section. The net section gives a better comparison of the two. 3.3 Comparison of cornered and smooth initial profiles For a better comparison of the solutions in the last two sections I will choose certain values for the different constants (γ, γ and V ) in such way that the asymptotic behavior of the initial condition and the momentum are the same in both cases. The solution in section 3. has momentum η d = γ. I choose the same value for the momentum of the soliton: γ = η d (3.3) 9V = ( )d (3.3) 4 cosh 4 V = 6V 3 (3.3) 38

40 The initial condition (3.9) in section 3. has asymptotic behavior e (γ /γ ) /. The soliton, which is used as initial condition in section 3., is for large values of approimately η () ( 3V e V ) = 6V e V for >> (3.33) So the asymptotic behavior is e V. This is the same as for the initial condition of section 3. if γ = γ V. Net are some plots. In these plots the dashed line corresponds to the soliton and the solid line to the initial condition of section 3., which is the optimal solution of chapter. In these pictures I used V =.75 and, to have the same momentum and asymptotic behavior for both initial conditions, γ = 6V 3/ =.9 and γ = γ V = opt. sol. Ch soliton.4. η(,) Figure 3.9: Initial condition 39

41 5 4.5 opt. sol. Ch soliton ˆη (k) k Figure 3.: Spectrum corresponding to the initial condition in figure opt. sol. Ch soliton η(,t) Figure 3.: η (, t) for t = (bottom), t = (middle) and t = (top) 4

42 .8 opt. sol. Ch soliton.6.4 MSE(t) t Figure 3.: Maimal Spatial Envelope In figure 3. you can see that the shape of the soliton changes much slower than the shape of the solution from chapter. As a result the MSE has a much slower descent. The soliton is a smooth solution of the constrained problem 3.7. But this problem also has cornered solutions consisting of part of a soliton for > and also for <, combined at = to a continuous but non-smooth function. The dotted line in the following plot shows such a function. The solid line is the solution of chapter again. The constants (γ, γ and V ) again are chosen in a way that the momentum and asymptotic behavior of these two initial conditions are the same. 4

43 .6 linear DE nonlinear DE.5.4 η(,) Figure 3.3: Initial condition.4 linear DE nonlinear DE. ˆη (k) k Figure 3.4: Spectrum corresponding to the initial condition in figure 3.3 4

44 linear DE nonlinear DE.5.5 η(,t) Figure 3.5: η (, t) for t = (bottom), t = (middle) and t = (top).65 linear DE nonlinear DE MSE(t) t Figure 3.6: Maimal Spatial Envelope 43

45 Bibliography [] Marcel Crok. Monstergolven bestaan. Natuurwetenschap en techniek, pages 7, juli/augustus 4. [] E. van Groesen. Lecture notes: Applied Analytical Methods. University of Twente, March. [3] E. van Groesen and B. van de Fliert. Lecture notes: Advanced Modelling in Science. University of Twente, October. [4] K. Hasselmann, T. P. Barnett, E. Bouws, H. Carlson, D. E. Cartwright, K. Enke, J. A. Ewing, D. E. Hasselmann H. Gienapp, P. Kruseman, A. Meerburg, P. Müller, D. J. Olbers, K. Richter, W. Sell, and H. Walden. Measurements of wind-wave growth and swell decay during the Joint North Sea Wave Project (JONSWAP). Deutsche Hydrographische Zeitschrift, A:p 95, 973. [5] S.R. Massel. Ocean surface waves: their physics and prediction, volume of Advanced series on ocean engineering. World scientific, 996. [6] W.J. Pierson and L. Moskowitz. A proposed spectral form for fully developed wind seas based on the similarity theory of S.A. Kitaigorodskii. Journal of Geophysical Research, 69:p58 59, 964. [7] D.A.G. Walker, P.H. Taylor, and R. Eatock Taylor. The shape of large surface waves on the open sea and the Draupner New Year wave. Applied Ocean Research, 6:p73 83, 4. 44

46 Appendi A In this appendi I ll show how to get equation (.36) in chapter. Which means I ll show how to epress ( t η (t)) dt in the Fourier transform ˆη. First we need to epress the derivative of η in the Fourier transform ˆη. ( ) t η (t) = t ˆη (ω)e iωt dω = iωˆη (ω)e iωt dω (A.) This equation states that iωˆη (ω) is the Fourier transform of t η (t). iωˆη (ω) = t η (t) e iωt dt Multiplication of equation (A.) with t η (t) yields ( t η (t)) = iωˆη (ω) t η (t) e iωt dω Now take the integral over t on both sides of this equation ( t η (t)) dt = iωˆη (ω) t η (t) e iωt dωdt = ( ) iωˆη (ω) t η (t)e iωt dt dω (A.) (A.3) (A.4) The part between the brackets is the comple conjugate of the right hand side of equation (A.). Therefore ( t η (t)) dt = iωˆη (ω)iωˆη (ω)dω = ω ˆη (ω) dω (A.5) 45

47 Appendi B In section.4 I found periodic solutions of the differential equation (.46) σδ ( ) = λ η () λ η () (B.) where I assumed that λ and λ have the same sign. Now I will try to find solutions when λ and λ have a different sign. Similar to section.4 I can choose =, which means that the solution η () attains it s maimum at =. I prescribe the period to be of length X. In the interior of the interval [, X] the differential equation reduces to λ η () λ η () = (B.) The solution of this equation is η () = c cos (s) + c sin (s) (B.3) where s = λ /λ. I can use three equations to find the values for the constants c, c and s: one from the periodicity and two from the constraints. η () = η (X) X X η () d = γ (B.4) (B.5) ( η ()) d = γ (B.6) From the first equation (B.4) we get c = c sin(sx). Using this epression for c cos(sx) and the second equation (B.5) it follows that c = ± γ s cos(sx). Now the third equation (B.6) sx+sin(sx) reduces to γ = γ s sx sin (sx) (B.7) sx + sin (sx) I can t find an eplicit epression for s from equation (B.7), but If I choose a value for X, I can make a plot of s versus γ γ. In the following plot I used X = π. Other values of X result in a similar plot. 46

48 PSfrag 5 5 γ γ s Figure B.: Relation between s and γ γ γ γ s Figure B.: Relation between s and γ γ, zoomed in on the first bump in figure B. As you can see in this plot, for some values of γ γ, s can have two or three different values. Now I want to plot the solution η (). If I take X = π, γ = and s = then also γ =. Then the solution becomes η () = ± sin () with [, π]. The solution with π the minus-sign gives a maimum at =, as we stated at the beginning of this chapter. 47

49 We can etend the solution η () = sin () on the interval [, π] to the entire real line, π which gives the following plot: η Figure B.3: Solution η () if X = π, γ = and γ =. The solution with the plus-sign has a minimum at =, but the maimum at = π is higher than the maimum in the previous plot. η Figure B.4: Solution with the plus-sign if X = π, γ = and γ =. In the net plot I use s = 3, so γ = 9. In this case the solution with the minus-sign is η () = sin(3). π η Figure B.5: Solution η () if X = π, γ = and γ = 9. As you can see, this solution has a local maimum at =, but it s global maimum is attained at some time. If I want larger values of γ, I have to choose larger values for s and more local etrema will appear in one period. The net plot shows the maimum η () as a function of γ. The solid line corresponds to the solution η () with a minimum in = and the dotted line (which is the same as the solid line for higher values of γ ) corresponds to the solution with a maimum in =. 48

50 PSfrag.8 ma.6 η () γ.4 Figure B.6: The maimum of the solution η () versus γ. The solid line corresponds with the solution with a plus-sign, the dotted one with a minus-sign ma.8 η () γ Figure B.7: The maimum of the solution η () versus γ zoomed in on the first bump in the previous picture If the solution η () I discussed in this section is the solution of the constrained variational problem (.3), the line in the previous plot is the value-function V (γ ) of that constrained variational problem (with γ = ). This value function must be increasing, but the line in the plot is not. Therefore the solution I found in this section is not the 49

51 solution of the constrained variational problem (.3) { } ma ma η () η η () d = γ, ( η ()) d = γ (B.8) but it is a solution of { crit η ma η () η () d = γ, ( η ()) d = γ } (B.9) Besides, whatever values I choose for γ and γ, the solution in section.4 has always a higher maimum than the solution in this appendi. 5