1 Governing Equations For Waves On The Sea Surface

wherec sg T k3 kρ(2.61)Note that cg c when k km , and cgc, if k km(2.62)In the limit of pure capillary waves of k km , cg 3c/2. For pure gravity wavescg c/2 as in (2.58).3Wave resistance of a two-dimensional obstacleRef: Lecture notes on Surface Wave Hydrodynamics Theodore T.Y. WU, Calif. Inst.Tech.As an application of the information gathered so far, let us examine the wave resistance on a two dimensional body steadily advancing parallel to the free surface. Let thebody speed be U from right to left and the sea depth be constant.Due to two-dimensionality, waves generated must have crests parallel to the axis ofthe body (y axis). After the steady state is reached, waves that keep up with the shipmust have the phase velocity equal to the body speed. In the coordinate system fixedon the body, the waves are stationary. Consider first capillary -gravity waves in deepwater λ λ/λm O(1) and kh 1. Equating U c we get from the normalizeddispersion relationU 2 c2 11 λ 2λ (3.1)where U U/cm . Henceλ2 2c2 λ 1 0 (λ λ 1 )(λ λ 2 )which can be solved to give 1/2 λ 1 24 c c 1λ 2(3.2)andλ 1 121λ 2(3.3)

Thus, as long as c U 1 two wave trains are present: the longer gravity wavewith length λ 1 , and the shorter capillary wave with length λ 2 . Since cg 1 c U andcg 2 c U , and energy must be sent from the body, the longer gravity waves mustfollow, while the shorter capillary waves stay ahead of, the body.Balancing the power supply by the body and the power flux in both wave trains, wegetRc (c cg 1 )Ē1 (cg 2 c)Ē2(3.4)Recalling thatcg1 λ2 3 c2 λ2 1we find,1cg2 1 1 1 2c2λ 1! 1/λ 11 1/λ 2 λ 1/λ 22c2For the longer wave we replace cg /c by cg 1 /c and λ by λ 1 in the precedingequation, and use (3.2), yielding1 1/2cg 1 1 c 4 c (3.5)Similarly we can show that 1/2cgcg 2 1 1 c 4 1 1 c c SinceρgA211Ē1 1 22λ 1!ρgA21 11 λ 1 2 λ 1λ 1!(3.6) ρgA21 λ 2 c2 ,(3.7)we get finally 1 1 R ρg λ 2 A21 λ 1 A22 (c4 1)1/2 ρg λ 2 A21 λ 1 A22 (U 4 1)1/222(3.8)Note that when U 1, the two waves become the same; no power input from the bodyis needed to maintain the single infinite train of waves; the wave resistance vanishes.When U 1, no waves are generated; the disturbance is purely local and there isalso no wave resistance. To get the magnitude of R one must solve the boundary valueproblem for the wave amplitudes A1 , A2 which are affected by the size (relative to thewavelengths), shape and depth of submergence.13

Figure 3: Dependence of wave resistance on speed for pure gravity wavesWhen the speed is sufficiently high, pure gravity waves are generated behind thebody. Power balance then requires that cgρgA2R 1 Ē U21kh 2 sinh 2kh!(3.9)The wavelength generated by the moving body is given implicitly byU ghtanh khkh!1/2(3.10) gh the waves generated are very long, kh 1, cg c gh, and the wave resistance drops to zero. When U gh, the waves are very short, kh 1,When U R ρgA24(3.11)For intermediate speeds the dependence of wave resistance on speed is plotted in figure(3).14

4Narrow-banded dispersive waves in generalIn this section let us discuss the superposition of progressive sinusoidal waves with theamplitudes spread over a narrow spectrum of wave numbersζ(x, t) Z A(k) cos(kx ωt θA )dk 0Z A(k)eikx iωt dk(4.1)0whereA(k) is complex denotes the dimensionless amplitude spectrum of dimension(length)2 . The component waves are dispersive with a general nonlinear relation ω(k).Let A(k) be different from zero only within a narrow band of wave numbers centered atko . Thus the integrand is of significance only in a small neighborhood of ko . We thenapproximate the integral by expanding for small k k ko and denote ωo ω(ko ),ωo0 ω 0 (ko ), and ωo00 ω 00 (ko ), iko x iωo tZiko x iωo tZζ e e i kx i(ω ωo )tA(k) e0dk dk A(k) exp i kx i0oiko x iωo tn ωo0 k 1 ωo00 ( k)2 t · · ·2 A(x, t)ewhereA(x, t) Z 0(4.2) 1dk A(k) exp i kx i ωo0 k ωo00 ( k)2 t · · ·2(4.3)Although the integration is formally extends from 0 to , the effective range is onlyfrom ko ( k)m to ko ( k)m , i,.e., the total range is O(( k)m ), where ( k)m is thebandwidth. Thus the total wave is almost a sinusoidal wavetrain with frequency ωo andwave number ko , and amplitude A(x, t) whose local value is slowly varying in space andtime. A(x, t) is also called the envelope. How slow is its variation?If we ignore terms of ( k)2 in the integrand, (4.3) reduces toA(x, t) Z 0dk A(k) exp [i k(x ωo0 t)](4.4)Clearly A A(x ωo0 t). Thus the envelope itself is a wave traveling at the speed ωo0 .This speed is called the group velocity,cg (ko ) 15dωdk(4.5)ko

Figure 4: Envelope of waves with a rectangular band of wavenumbersNote that the characteristic length and time scales are ( km ) 1 and (ωo0 km ) 1 respectively, therefore much longer than those of the component waves : ko 1 and ωo 1 . In otherwords, (4.3) is adequate for the slow variation of Ae in the spatial range of km x O(1)and the time range of ωo0 km t O(1).As a specific example we let the amplitude spectrum be a real constant within thenarrow band of ko κ, ko κ,Zζ Ako κeikx iω(k)t dk,ko κκ ko(4.6)theniko x iωo tζ ko AeZκdξeiko ξ(x cg t) · · · κ2A sin κ(x cg t) iko x iωo t e Aeiko x iωo tx cg t(4.7)where ξ k ko /ko andA 2A sin κ(x cg t)(x cg t)(4.8)as plotted in figure (4).By differentiation, it can be verified that A A cg 0, t x16(4.9)

Multiplying (4.9) by A ,A A A cg A 0, t xand adding the result to its complex conjugate,A A A cg A 0, t xwe get A 2 A 2 cg 0 t x(4.10)We have seen that for a monochromatic wave train the energy density is proportionalto A 2 . Thus the time rate of change of the local energy density is balanced by the netflux of energy by the group velocity.Now let us examine the more accurate approximation (4.3). By straightforwarddifferentiation, we findZ"# iω 00 (ko ) A ( k)2 A(k)eiS dk iω 0 (ko ) k t20Z A (i k)A(k)eiS dk x0Z 2A2 ( k)A(k)eiS dk2 x0where1S k x ωo0 k t ωo00 ( k)2 t2(4.11)is the phase function. It can be easily verified that Aiω 00 2 A A ωo0 o t x2 x2(4.12)By keeping the quadratic term in the expansion, (4.12) is now valid for a larger spatialrange of ( k)2 x O(1). In the coordinate system moving at the group velocity, ξ x cg t, τ t, we easily find A A A(ξ, τ ) cg, t τ ξ A(ξ, τ ) A x xso that (4.12) simplifies to the Schrödinger equation: Aiω 00 2 A o 2 τ2 ξ17(4.13)

By manipulations similar to those leading to (4.10), we get A 2iω 00 A A o AA τ2 ξ ξ ξ!(4.14)Thus the local energy density is not conserved over a long distance of propagation.Higher order effects of dispersion redistribute energy to other parts of the envelope.For either a wave packet whose envelope has a finite length ( A( ) 0), or for aperiodically modulated envelope (A(x) A(x L)), we can integrate (4.14) to give Z A 2 dξ 0 τ(4.15)where the integration extends over the entire wave packet or the group period. Thusthe total energy in the entire wave packet or in a group period is conserved.5Radiation of surface waves forced by an oscillatingpressureWe demonstrate the reasoning which is typical in many similar radiation problems.The governing equations are 2 φ φxx φyy 0, z 0.(5.1)with the kinematic boundary conditionφz ζt , z 0(5.2)and the dynamic boundary conditionpa φt gζ 0ρ(5.3)where pa is the prescribed air pressure. Eliminating the free surface displacement weget(pa )t,ρLet us consider only sinusoidal time dependence:φtt gφz pa P (x)e iωt18z 0.(5.4)(5.5)

and assumeφ(x, z, t) Φ(x, z)e iωt , ζ(x, t) η(x)e iωt(5.6)then the governing equations become 2 Φ Φxx Φyy 0, z 0.Φz iωη, z 0(5.7)(5.8)andiωω2Φz Φ P (x), z 0.gρgDefine the Fourier transform and its inverse byf (α) Z iαxdx ef (x), 1f (x) 2πZ (5.9)dα eiαx f (α),(5.10) We then get the transforms of (5.1) and (5.4)Φ̄zz α2 Φ̄ 0,z 0(5.11)subject toiωω2Φ̄z Φ̄ P̄ (α),gρgThe solution finite at z for all α isz 0.(5.12)Φ̄ Ae α zTo satisfy the free surface condition α A ω2iω P̄ (α)P (α)A gρghenceA iω P̄ (α)ρg α ω 2 /gorZiω P̄ (α)1 ρgΦ dαeiαx e α z2π α ω 2 /gZZ iω 1 10 ,dαeiαx e α zdx0 e iαx P (x0 )ρg 2π α ω 2 /g Z iω Z 0100 1 dx P (x )dα eiα(x x ) e α zρg 2π α ω 2 /g19(5.13)

Letk ω2gZ (5.14)we can rewrite (5.13) asiωΦ ρgZ 1dx P (x )π 00dα eαz0cos(α(x x0 ))α k(5.15)The final formal solution isφ iω iωteρgZ dx0 P (x0 ) 1πZ dα eαz0cos(α(x x0 ))α k(5.16)If we choseP (x0 ) Po δ(x0 )theniωPo 1Φ G(x, z) ρg πZ 0dα eαz(5.17)cos(αx)α k(5.18)is clearly the response to a concentrated surface pressure and the response to a pressuredistribution (5.16) can be written as a superposition of concentrated loads over the freesurface,φ Z dx0 P (x0 )G(x x0 , z).(5.19) whereG(x, z, t) iωPo iωt 1 Z cos(αx)edα eαzρgπ 0α k(5.20)In these results, e.g., (5.20), the Fourier integral is so far undefined since the integrandhas a real pole at α k which is on the path of integration. To make it mathematicallydefined we can chose the principal value, deform the contour from below or from abovethe pole as shown in figure (5). This indefiniteness is due to the assumption of quasiFigure 5: Possible paths of integrationsteady state where the influence of the initial condition is no longer traceable. We mustnow impose the radiation condition that waves must be outgoing as x . This20

Figure 6: Closed contour in the upper half planecondition can only be satisfied if we deform the contour from below. Denoting thiscontour by Γ, we now manipulate the integral to exhibit the behavior at infinity, and toverify the choice of path. For simplicity we focus attention on G. Due to symmetry, itsuffices to consider x 0. Rewriting,iωPo iωt 1e(I1 I2 )ρg2π"#iωPo iωt 1 Ze iαxeiαxαz e dα eρg2π Γα k α kG(x, z, t) (5.21)Consider the first integral in (5.21). In order that the first integral converges forlarge α , we close the contour by a large circular arc in the upper half plane, as shownin figure (6), where α 0 along the arc. The termeiαx ei αx e αxis exponentially small for positive x. Similarly, for the second integral we must chosethe contour by a large circular arc in the lower half plane as shown in figure (7).Back to the first integral in (5.21)Zeiαx eαzI1 dαα kΓ(5.22)The contour integral isIeiαx eαzdαα kZeiαx eαz dαα kΓZeiαx eαz dαα kCZ 0eiαx eαzdα I1 0 α ki 21Zeiαx eαzdαα ki 0

Figure 7: Closed contour in the lower half planeThe contribution by the circular arc C vanishes by Jordan’s lemma. The left hand sideisLHS 2πieikxekz(5.23)by Cauchy’s residue theorem. By the change of variable α iβ, the right hand sidebecomesRHS I1 iZ0dβ HenceI1 2πieikx ekz iNow consider I2I2 ZdαΓZe βx eiβziβ k dβ0e βx eiβziβ k(5.24)e iαx eαzα k(5.25)and the contour integral along the contour closed in the lower half plane, Idαe iαx eαz I2 0 α kZ dα0e iαx eαzα kAgain no contribution comes from the circular arc C. Now the pole is outside thecontour hence LHS 0. Let α iβ in the last integral we getI2 iZ dβ0e βx e iβy iβ k(5.26)Adding the results (5.24) and (5.26).,Z!ie βx eiβz ie βx e iβzI1 I2 2πie e dβ iβ k iβ k0Z βxe 2πieikx ekz 2dβ 2(β cos βy k sin βy)β k20ikx kz 22(5.27)

Finally, the total potential is, on the side of x 0, ω iωt 1e(I1 I2 ) e iωtρg2πi()Zω iωt ikx kz 1 e βx ee e dβ 2(β cos βz k sin βz)ρgπ 0β k2G(x, z, t) (5.28)The first term gives an outgoing waves. For a concentrated load with amplitudePo , the displacement amplitude is Po /ρg. The integral above represent local effectsimportant only near the applied pressure. If the concentrated load is at x x0 , onesimply replaces x by x x0 everywhere.6The Kelvin pattern of ship waveThe action of the ship’s propellerHas a thrust patternTo which the ship reacts by moving forward,Which also results secondarily,In the ship’s bow elevated waves,And its depressed transverse stern wave,Which wave disturbances of the waterAre separate from the propeller’s thrust waves.–R.Buckminster Fuller, Intuition- Metaphysical Mosaic. 1972. Anyone flying over a moving ship must be intrigued by the beautiful pattern in theship’s wake. The theory behind it was first completed by Lord Kelvin, who inventedthe method of stationary phase for the task. Here we shall give a physical/geometricalderivation of the key results (lecture notes by T. Y. Wu, Caltech).Consider first two coordinate systems. The first r (x, y, z) moves with ship at theuniform horizontal velocity U. The second r0 (x0 , y 0 , z) is fixed on earth so that wateris stationary while the ship passes by at the velocity U. The two systems are related bythe Galilean transformation,r0 r Ut23(6.29)

A train of simple harmonic progressive waveζ {A exp[i(k · r0 ωt)]}(6.30)in the moving coordinates should be expressed asζ {A exp[ik · (r Ut) iωt]} {A exp[ik · r i(ω k.U)t]} {A exp[ik · r iσt]}(6.31)in the stationary coordinates. Therefore the apparent frequency in the moving coordinates isσ ω k·U(6.32)The last result is essentially the famous Doppler’s effect. To a stationary observer, thewhistle from an approaching train has an increasingly high pitch, while that from aleaving train has a decreasing pitch.If a ship moves in very deep water at the constant speed U in stationary water,then relative to the ship, water appears to be washed downstream at the velocity U.A stationary wave pattern is formed in the wake. Once disturbed by the passing ship,a fluid parcel on the ship’s path radiates waves in all directions and at all frequencies.Wave of frequency ω spreads out radially at the phase speed of c g/ω according tothe dispersion relation. Only those parts of the waves that are stationary relative to theship will form the ship wake, and they must satisfy the conditionσ 0,(6.33)i.e.,ω k · U, or c ωk ·Ukk(6.34)Referring to figure 8, let O, (x 0) represents the point ship in the ship-boundcoordinates. The current is in the positive x direction. Any point x1 is occupied bya fluid parcel Q1 which was disturbed directly by the passing ship at time t1 x1 /Uearlier. This disurbed parcel radiates waves of all frequencies radially. The phase ofwave at the frequency ω reaches the circle of radius ct1 where c g/ω by the deep waterdispersion relation. Along the entire circle however only the point that satisfies (6.34)24

Figure 8: Waves radiated from disturbed fluid parcelcan contribute to the stationary wave pattern, as marked by P . Since OQ1 x1 U t1 ,Q1 P ct1 and OP Ut1 · k/k, where k is in the direction of Q 1 P . It follows that4OP Q1 is a right triangle, and P lies on a semi circle with diameter OQ1 . Accountingfor the radiated waves of all frequencies, hence all c, every point on the semi circlecan be a part of the stationary wave phase formed by signals emitted from Q1 . Nowthis argument must be rectified because wave energy only travels at the group velocitywhich is just half of the phase velocity in deep water. Therefore stationary crests dueto signals from Q1 can only lie on the semi-circle with the diameter O1 Q1 OQ1 /2.Thus P1 instead of P is one of the points forming a stationary crest in the ship’s wake,as shown in figure 8.Any other fluid parcel Q2 at x2 must have been disturbed by the passing ship at timet2 x2 /U earlier. Its radiated signals contribute to the stationary wave pattern onlyalong the semi circle with diameter O2 Q2 OQ2 /2. Combining the effects of all fluidparcels along the x axis, stationary wave pattern must be confined inside the wedgewhich envelopes all these semi circles. The half apex angle βo of the wedge, which definesthe wake, is given bysin βo U t/4 1/3,3U t/4(6.35)hence βo sin 1 1/3 19.5 , see figure 9.Now any point P inside the wedge is on two semicircles tangent to the boundaryof the wedge, i.e., there are two segments of the wave crests intersecting at P : oneperpendicular to P Q1 and one to P Q2 , as shown in figure 9.Another way of picturing this is to examine an interior ray from the ship. In figure25

Figure 9: Wedge angle of the ship wakeP1ObM2M1P2O'QFigure 10: Geometrical relation to find Points of dependence(10), draw a semi circle with the diameter O0 Q OQ/2, then at the two intersections P1and P2 with the ray are the two segments of the stationary wave crests. In other words,signals originated from Q contribute to the stationary wave pattern only at the twopoints P1 and P2 , as shown in figure 10. Point Q can be called the point of dependencefor points P1 and P2 on the crests.For any interior point P there is a graphical way of finding the two points of dependence Q1 and Q2 . Referring to figure 10, 4O0 QP1 and 4O0 QP2 are both right triangles.Draw O1 M1 k QP1 and O2 M2 k QP2 where M1 and M2 lie on the ray inclined at the angle β. it is clear that OM1 OP1 /2 and OM2 OP2 /2, and 4M1 O0 P1 and 4M2 O0 P2are both right triangles. Hence O0 lies on two semi circles with diameters M1 P1 andM2 P2 .We now reverse the process, as shown in figure 11. For any point P on an interiorray, let us mark the mid point M of OP and draw a semi circle with diameter MP .The semi circle intersects the x axis at two points S1 and S2 . We then draw from P twolines parallel to MS1 and MS2 , the two points of intersection Q1 an Q2 on the x axisare just the two points of dependence.26

Figure 11: Points of dependenceLet 6 P Q1 O 6 MS1 O θ1 and 6 P Q2 O 6 MS2 O θ2 . thentan(θi β) P SiP Si 2 tan θiMSiP Qi /2i 1, 2.hencetan θi tan β1 tan θi tan βwhich is a quadratic equation for θi , with two solutions:2 tan θi tan θ1 tan θ2 1 q1 8 tan2 β4 tan β(6.36)They are real and distinct if1 8 tan2 β 0(6.37)These two angles define the local stationary wave crests crossing P , and they mustbe perpendicular to P Q1 and P Q2 . There are no solutions if 1 8 tan2 β 0, whichcorresponds to sin β 1/3 or β 19.5 , i.e., outside the wake. At the bou

c/ gh kh Figure 2: Phase speed of capillary-gravity waves in water of constant depth For gravity waves on deep water, kh 1, tanhkh 1. Hence ω q gk, c r g k (2.22) Thus longer waves travel faster. These are also called short gravity waves. If however the waves are very long or the