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2.1.GENERALSOLUTION TO WAVE EQUATION 1.138J/2.062J/18.376J,WAVEPROPAGATION

Fall,2006MIT NotesbyC.C.Mei CHAPTERTWO

ONEDIMENSIONALWAVES INSIMPLESYSTEMS

GeneralsolutiontowaveequationIt is easy to verify by direct substitution that the most general solution of the onedimensionalwaveequation:

∂ 2φ∂t2 =c2∂ 2φ

∂x2 (1.1)canbesolved by

φ(x,t) = f (x−ct) + g(x+ct) (1.2)wheref (ξ ) and g(ξ )arearbitraryfunctionsof ξ . In the x, t(space,time)planef (x−ct)is constant along the straight line x−ct= constant. Thus to the observer (x, t) who moves at the steady speed c along the positivwe x-axis., the function f is stationary.Thus to an observer moving from left to right at the speed c, the signal describedinitially byf (x) at t=0 remainsunchanged in formas t increases, i.e., f propagatestotherightatthespeedc. Similarlyg propagatestotheleftatthespeedc. Thelinesx−ct=constantandx+ct=contantarecalled thecharacteristic curves(lines)alongwhichsignalspropagate. Notethatanotherwayof writing(1.2)is

φ(x, t) = F (t−x/c) + G(t+x/c) (1.3)Letusillustratesanapplication.

Branchingof arteriesReferences: YCFung:Biomechanics,Circulation. Springer1997M.J.Lighthill:Waves in Fluids ,Cambridge1978.

2



22.2 BRANCHINGOF ARTERIES Recallthegoverningequationsforpressureand velocity

∂ 2 p 2∂ 2 p=c (2.1)∂t2 ∂x2

∂ 2u 2∂ 2u=c (2.2)

∂t2 ∂x2Thetwoarerelated bythemomentumequation

∂u ∂pρ =− (2.3)

∂t ∂x Thegeneralsolutionsare:

p= p+(x−ct) + p−(x+ct) (2.4)u=u+(x−ct) + u−(x+ct) (2.5)

Since∂p ∂x = p+ + p−,

andρ

∂u−=−ρcu +ρcu

∂t + −

whereprimes

indicated

ordinary

differentiation

with

repect

to

the

argument.

Equation

(2.3)canbesatisfied if

p+ =ρcu+, p− =−ρcu− (2.6)DenotethedischargebyQ=uAthen

ZQ± =u±A=± p± (2.7)where

p± ρcZ =

± = (2.8)Q± A

istheratioof pressuretofluxrateand iscalltheimpedance . It isthepropertyof thetube.

Now we examine the effects of branching; Referening to figure x, the parent tubesbranches into two charaterized by wave speeds C 1 and C 2 and impedaces Z 1 and Z 2.Anincidentwaveapproachingthe junctionwillcausereflectioninthesametube

p= pi(t−x/c) + pr(t+x/c) (2.9)



2.2.WAVES DUE TO INITIALDISTURBANCES 4and transmitted wavesinthebranchesare p1(t−x/c1) and p2(t−x/c2). Atthe junctionx=0weexpectthecontinuityof pressureand fluxes,hence

pi(t) + pr(t) = p1(t) = p2(t) (2.10) pi− pr p1 p2

= + (2.11)Z Z 1 Z 2

DefinethereflectioncoefficientRtobetheamplituderatioof reflectedwavetoincidentwave,then

1 − 1 + 1 pr(t) Z Z 1 Z 2

R= =

(2.12) pi(t) 1 + 1 + 1

Z Z 1 Z 2Similarlythetranmissioncoefficientsare

p1(t) p2(t) Z 2T = = = (2.13)

pi(t) pi(t) 1 + 1 + 1Z Z 1 Z 2

Notethatbothcoefficientsareconstantsdependingonlyontheimpedances. Hencethetransmitted wavesaresimilarinformtotheincidentwavesexceptsmallerbythefactorT . Thetotalwaveontheincidencesideishoweververydifferent.

3 Waves inan infinitedomaindueto initialdisturbances

Recallthegoverningequationforone-dimensionalwavesinatautstring∂ 2u 2∂ 2u−c = 0, −∞< x < ∞. (3.1)∂t2 ∂x2

Lettheinitialtransversedisplacementandvelocitybegivenalongtheentirestringu(x,0)=f (x) (3.2)

∂u (x, t) = g(x), (3.3)

∂t wheref (x) and g(x)arenon-zeroonlyinthefinitedomainof x. Atinfinitiesx→ ±∞, u and∂u/∂tarezeroforanyfinitet. Theseconditionsarebestdisplayedinthespace-timediagramasshowninFigure1.



52.2.WAVES DUE TO INITIALDISTURBANCES t

2u =c u

tt xx

x

u=f(x) u =g(x)t

Figure1: Summaryof theinitial-boundary-valueproblemIn (3.1.1) the highest time derivative is of the second order and initial data are

prescribed for u and ∂u/∂t. Initial conditions that specify all derivatives of all orderslessthanthehighestinthedifferentialequationarecalledtheCauchyinitialconditions.

Recallthatthegeneralsolutionisu=φ(ξ ) + ψ(η) = φ(x+ct) + ψ(x−ct), (3.4)

whereφ

and

ψ

are

so

far

arbitrary

functions

of

the

characteristic

variables

ξ =

x

+

ct

andη=x−ctrespectively.

Fromtheinitialconditionswegetu(x,0) = φ(x) + ψ(x) = f (x)

∂u (x,0) = cφ(x)−cψ(x) = g(x). (3.5)

∂t Thelastequationmaybeintegrated withrespecttox

1 xφ−ψ= g(x)dx +K, (3.6)c x0

whereK isanarbitraryconstant. Nowφandψcanbesolved from(3.1.6)and (3.1.7)asfunctionsof x,

1 1 xφ(x) = [f (x) + K ] + g(x)dx

2 2c x01 1 x

ψ(x) = [f (x)−K ]− g(x)dx,2 2c x0



62.2.WAVES DUE TO INITIALDISTURBANCES

x,t

c c

cc

x

(

0

influenceRange of

1 1

1 1

)

x-ct x+ct x

t

Domain ofdependence

Figure2:

Domain

of

dependence

and

range

of

influence

whereK andx0 aresomeconstants. Replacingtheargumentsof φbyx+ctand of ψbyx−ctand substitutingtheresultsinu, we get

1 1 x−ctu(x, t) = f (x−ct)− g dx

2 2c x01 1 x+ct

+ f (x+ct) + g dx2 2c x0

1 1 x+ct= [f (x

−ct) +

f (x

+

ct)]

+

g(x

)dx

,

(3.7)2 2c x−ctwhich is d’Alembert’s solution to the homogeneous wave equation subject to generalCauchyinitialconditions.

Toseethephysicalmeaning,letusdrawinthespace-timediagramatriangleformedby two characteristic lines passing through the observer at x, t, as shown in Figure 2.Thebaseof thetrianglealongtheinitialaxist=0beginsatx−ctand endsatx+ct.Thesolution(3.1.9)dependsonthe initialdisplacementat justthetwocornersx−ctand x+ct, and on the initial velocity only along the segment from x−ct to x+ct.Nothing outside the triangle matters. Therefore, to the observer at x, t, the domain of dependence is the base of the characteristic triangle formed by two characteristicspassingthroughx, t. Ontheotherhand,thedataatanypointxontheinitiallinet= 0 mustinfluenceallobserversinthewedgeformedbytwocharacteristicsdrawnfromx,0intotheregionof t > 0;thischaracteristicwedgeiscalledtherange of influence .

Letusillustratethephysicaleffectsof initialdisplacementandvelocityseparately.



72.2.WAVES DUE TO INITIALDISTURBANCES u, t

xO

Figure3: WavesduetoinitialdisplacementCase(i): Initialdisplacementonly: f (x)= 0 and g(x)=0. Thesolutionis

1 1u(x, t) = f (x−ct) + f (x+ct)

2 2and isshownforasimplef (x) inFigure3atsuccessivetimesteps. Clearly,theinitialdisturbanceissplitintotwoequalwavespropagatinginoppositedirectionsatthespeedc. Theoutgoingwavespreservetheinitialprofile,althoughtheiramplitudesarereducedbyhalf.

Case(ii): Initialvelocityonly: f (x) = 0,andg(x)=0. Considerthesimpleexamplewhere

g(x) = g0 when |x|< b, and= 0 when |x|>0.

ReferringtoFigure4,wedividethex∼tdiagramintosixregionsbythecharacteristicswithBandC lyingonthexaxisatx=

−band +b,respectively. Thesolutioninvarious

regionsis:u= 0

inthewedgeABE ;u= 1 x+ct

g0dx = go(x+ct+b)

2c −b 2cinthestripEBIF ;

1 x+ctu= godx =got

2c x−ct



8

4

3.2.REFLECTION FROM THE FIXED END u, t

I

E F G H

A B O C D x

Figure4: WavesduetoinitialvelocityinthetriangleBC I ;

u= 12c b

−bg0dx = gob

cinthewedgeF IG;

u= 12c b

x−ctg0dx = go

2c(b−x+ct)inthestripGICH ; and

u= 0 inthewedgeHCD. Thespatialvariationof u isplotted forseveral instants inFigure4. Notethatthewave fronts inbothdirectionsadvanceatthespeedc. IncontrasttoCase(i),disturbancepersistsforalltimeintheregionbetweenthetwofronts.

Reflection fromthefixedendof astringLetususethed’Alembertsolutiontoaprobleminahalf infinitedomainx > 0. Consideralongand tautstringstretched fromx=0toinfinity. Howdodisturbancesgeneratedneartheleftend propagateastheresultof initialdisplacementand velocity?

Attheleftboundaryx=0mustnowaddtheconditionu= 0, x = 0, t > 0. (4.1)

Inthespace-timediagram letusdrawtwocharacteristicspassingthroughx, t. For anobserverintheregionx > ct,thecharacteristictriangledoesnot intersectthetime



3.4.FORCED WAVES IN AN INFINTE DOMAIN 9axisbecausetisstilltoosmall. Theobserverdoesnotfeelthepresenceof thefixedendatx=0,hencethesolution(3.1.9)foraninfinitelylongstringapplies,

1 1 x+ctu= [f (x+ct) + f (x−ct)]+ g(τ )dτ, x > ct. (4.2)

2 2c x−ctBut for x < ct, this result is no longer valid. To ensure that the boundary conditionis satisfied we employ the idea of mirror reflection. Consider a fictitious extension of thestringto−∞< x ≤0. If onthesidex < 0the initialdataare imposedsuchthatf (x) = −f (−x), g(x) = −g(−x), then u(0, t)=0 is assured bysymmetry. We nowhaveinitialconditionsstatedovertheentirexaxis

u(x,0)=F (x) and ut(x, 0)=G(x) − ∞ < x < ∞,where f (x) if x > 0

F (x) = −f (−x) if x < 0

g(x) if x > 0G(x) =

−g(−x) if x < 0.TheseconditionsaresummarizedinFigure5. Hencethesolutionfor0< x < ct is

1 1 0 x+ctu = [F (x+ct) + F (x−ct)] + + G(x)dx

2 2c x−ct 01 1 0 x+ct

= [f (x+ct)−f (ct−x)] + + g(x)dx2 2c ct−x 0

= 1[f (x+ct)−f (ct−x)] + 1 ct+x

g(x)dx. (4.3)2 2c ct−x

Thedomainof dependenceisshownbythehatchedsegmentonthexaxisinFigure6.

5 Forcedwaves inan infinitedomainConsidertheinhomogeneouswaveequation

∂ 2u ∂ 2u=c2 +h(x, t) t > 0, |x|<∞,

∂t2 ∂x2



103.4.FORCED WAVES IN AN INFINTE DOMAIN

t

u=0 2u =c u

tt xx

x

u=-f(-x) u=f(x)

u =-g(-x) u =g(x)t t

Figure5: Initial-boundary-valueproblemandthemirrorreflection x+ct=x +ct

0 0 t

x-ct=x -ct 0 0

( x ,t0)

0

x( x -ct ,0) ct -x ,0) ( x +ct ,0)

0 0 O (0 0 0 0

Figure6: Reflectionfromafixed end



¯

¯

113.4.FORCED WAVES IN AN INFINTE DOMAIN whereh(x, t)represents forcing. Becauseof linearity,wecantreattheeffectsof initialdataseparately. Letusthereforefocusattentiononlytotheeffectsof persistentforcingand lettheinitialdatabezero,

∂u u(x,0)=0, = 0, (5.1)

∂t t=0Theboundaryconditionsare

u→ 0, |x| → ∞. (5.2)¯LettheFouriertransformof anyfunctionf (x) and its inverse f (α)bedefinedby

∞

f (α) = f (x)e−iαxdx (5.3)−∞

1 ∞f (x) = f (α)eiαxdα (5.4)

2π −∞Thetransformedwaveequationisnowanordinarydifferentialequationforthetransformof u(x, t),i.e.,u(α, t),

d2u+c2α2u =h t > 0

dt2whereh(α, t)denotesthetransformof theforcingfunction. Theinitialconditionsforuare:

¯ du(α,0)u(α,0)=f (α) = 0, = g(α) = 0

dtLet us hidethe parametric dependenceonα for thetime being. The general solutiontothetheinhomogeneoussecond-orderordinarydifferentialequationis

t h(τ )u =C 1u1(t) + C 2u2(t) + [u1(τ )u2(t)− u2(τ )u2(t)]dτ, (5.5)

W 0whereu1 and u2 arethehomogeneoussolutions

u1 =e−iαct

u2 =eiαct

andW istheWronskianW = u1u2 − u2u1 = 2iαc=constant.

ThetwoinitialconditionsrequirethatC 1 =C 2 =0,hencetheFouriertransformist

u = h(α, τ )eiαc(t−τ )− e−iαc(t−τ )

dτ. (5.6)

2iαc0



2.6. STRINGEMBEDDED IN AN ELASTIC SUROUNDING 12Theinversetransformis

1 t 1 ∞ h(α, τ ) iα(x+c(t−τ ))−eiα(x−c(t−τ ))

u(x, t) = dτ e dα (5.7)2π 0 2c −∞ iαTheintegrand canbewrittenasanintegral:

1 t 1 ∞ x+c(t−τ )u= dτ dα h(α, τ ) dξ eiαξ

2π 0 2c −∞ x−c(t−τ )Byinterchangingtheorderof integration

u= 1

tdτ

x+c(t−τ )

dξ 1

∞dα eiαξh(α, τ )

2c 0 x−c(t−τ ) 2π −∞and usingthedefinitionof Fourierinversion,weget

1 t x+c(t−τ )u(x, t) = dτ dξ h(ξ, τ ) (5.8)

2c 0 x−c(t−τ )Theright-hand sideistheintegrationof hoverthecharacteristictriangledefinedby

thetwocharacteristicspassingthrough(x, t). inthex−tplane. Thustheobserver isaffected onlybytheforcinginsidethecharacteristictriangle.

Fornon-zeroinitialdatau(x,0)=f (x) and ut(x,0)=g(x), we get by linear super

positionthefullsolutionof D’Alambert1 1 x+ct

u(x, t) = [f (x+ct) + f (x−ct)]+ dξg(ξ )2 2c x−ct

1 t x+c(t−τ )+ dτ dξ h(ξ, τ ), (5.9)

2c 0 x−c(t−τ )Thedomainof dependenceisentirelywithinthecharacteristictriangle.

6 Stringembeded inanelasticsuroundingReference: Graff :Waves in Elastic Solids

If the lateral motion of the string is restrained by elastic springs along the entirelength,thegoverningequationcanbefound from§1.1byreplacingtheexternalpressurebytheelasticrestoringforce−KV perunitlength,

∂ 2V ∂ 2V ρ =T −KV (6.1)

∂t2 ∂x2



132.6. STRINGEMBEDDED IN AN ELASTIC SUROUNDING whichcanbewrittenas

1∂ 2V ∂ 2V K c2o ∂t

2 =∂x

2−

T

V (6.2)where

T co = (6.3)

ρ6.1 MonochromaticwavesFor any linearized wave problem, if the range of the spatial corrdinate x is (−∞,∞),andallcoefficientsareindependentof x, t,thenthefirsttaskistoexaminethephysicsof

sinusoidal

wave

train

of

the

form:

V (x, t) = |A| cos(kx− iωt− φA) = Aeikx−iωt (6.4)

whereA=|A|eiφA isacomplexnumberwithmagnitude|A|and phaseangleφA. After examining the physical meaning of this special type of waves, it is possible to use theprincipleof superpositiontoconstructmoregeneralsolutions. It iscustomarytoomitthesymbol=”therealpartof”forthesakeof brevity,i.e.,

V (x, t) = Aeikx−iωt

(6.5)Firstof allafewdefinitionsaboutsinusoidalwavesingeneral. Weshallcall

θ(x, t) = kx− ωt (6.6)thewave phase . Clearlythetrigonometricfunctionisperiodicinphasewiththeperiod2π. In the x, tplane,V hasaconstantvaluealongalineof contantphase. Inparticular,θ= 2nπ,(n= 0,1,2,· · · )correspond tothewavecrestswhereV =|A| isthegreatest.On the other hand, θ = (2n+ 1)π,(n = 0,1,2,· · · ) correspond to the wave troughswhere V =−|A| is the smallest. |A| is half of the separation between adjacent crestsand troughs and is called thewave amplitude ; we also call A the complex amplitude.

∂θClearly∂x represents the number of phase lines per unit distance, i.e., the density of

phaselines,atagiveninstant;itiscalled thewavenumber ,∂θ

wavenumber=k= (6.7)∂x



142.6. STRINGEMBEDDED IN AN ELASTIC SUROUNDINGOntheotherhand−∂θ represenststhenumberof phase linespassingacrossafixedx∂t perunittime;itiscalled thewave frequency .

∂θ wavefrequency =ω=− (6.8)

∂t Tostaywithaparticularlineof contantphase,sayacrest,onemusthave

dθ=kdx−ωdt= 0 namelyonemustmoveatthephase velocity ,

dx

ω

c= = (6.9)dt θ=constant kNowbacktothestring. Substituting(6.10)in(6.1)weget

K ω2−k2− + V =0 (6.10)

T c2o

or K

ω=co k2 + (6.11)T

Thephasespeed isω

K 1/2c= =co 1 + > co (6.12)|k| T k2

See figure 6.1. Note that, due the stiffening by the lateral support, the phase speedisalwaysgreaterthanco,and decreasesmonotonicallywiththewavenumber. Longerwaves(small|k|)arefaster,whileshorterwaves(larger|k|)areslower. As|k|increases,capproachesthefinitelimitco fortheshortestwaves. Ingeneralasinusoidalwavewhosephasevelocitydependsonthewavelength, i.e.,ω isanonlinear functionof k, iscalledadispersive wave ,and (6.11)oritsequivalent(6.20)iscalled thedispersion relation .

An interestingphysical feature fordispersivewaves in generalcanbe found bysu

perposingtwotrainsof sighltlydifferentfrequenciesandwavenumbers:V =A ei(k+x−ω+t) +Aei(k−x−ω−t) (6.13)

wherek± =k±k, ω± =ω(k±), k k. (6.14)



152.6. STRINGEMBEDDED IN AN ELASTIC SUROUNDING

Figure 7: Phase and group velocities of a string in elastic surrounding. Point of Sta

tionaryphasek0.Letusapproximateω± by

dωω± =ω(k)

±k +O(k2)

dkthen

V =Aeikx−iωt e−ik(x−(dω/dk)t) +e−ik(x−(dω/dk)t) =A(x, t)eikx−iωt (6.15)where

A= 2 cos k(x−(dω/dk)t) (6.16)Thefactorexp(ikx−iωt)iscalled thecarrier wave andA(x, t) the envelope . Thus the result isa sinusoidal wavetrainwith a slowly varying envelope which has avery longwavelength2π/k 2π/kand movesatthegroupvelocity

dωgroupvelocity=cg = (6.17)

dkwhichisingeneraldifferentfromthephasevelocityfordispersivewaves.

Inourstringproblem,thegroupvelocityiseasilyfound fromthedispersionrelation(6.11)

cok co2 T k cg co2cg =

= = , hence = <1 (6.18)k2 +K/T c ρω c c2



162.6. STRINGEMBEDDED IN AN ELASTIC SUROUNDINGThus the group velocity is always smaller than the phase velocity, and increases withthe wavenumber from 0 for the longest wave to the finite limit co (equal to the phasevelocity)fortheshortestwaves.

Whenω > ωc where K

ωc =co . (6.19)T

iscalled thecutoff frequencyωc,k isreal 1/2ω2 K

k=±|k|, with |k|= − (6.20)c2 T o

being the real and positive square root. V (x, t) is a propagating wave, with the plus(minus)signcorrespondingtoright- (left-)goingwave. If ω < ωc,k=±iκ isimaginary,withκbeingthepostiveroot: 1/2

K ω2κ = − (6.21)

T c2o

Then eikx = e∓κx. For boundedness one must choose the minus (plus) sign for x > 0(x < 0). Oscillationsarelocalized orevanecent ;thereisnowaveradiation.

As a simple application, (6.4) is the response in a semi-infinite string forced tooscillateattheleftendx = 0,

V (0, t) = {Ae−iωt}. (6.22)If ω > ωc, then

V (x, t) = Aei|k|x−iωt (6.23)where |k| isdefinedby(6.20). Therequirementthatwavesduetoa localdisturbancecanonlyradiateoutwards iscalledtheradiation condition . Morewillbesaid on it inAppendixB.

If ω < ωc,wemustrequireboundednessatinfinitysothatV (x, t) = Ae−κx−iωt, x > 0 (6.24)

At the cutoff frequency, k = 0, V is constant in x; the infinitely long string wouldoscillate in unison. This absurd result signfies the breakdown of the the linearized theory.



2.6. STRINGEMBEDDED IN AN ELASTIC SUROUNDING 176.2 EnergytransportAlong

aunit

length

of

the

string

the

kinetic

energy

density

is,

2ρ ∂V

KE = (6.25)2 ∂t

The potential energy in any segment dx of the string is the work needed to deform itfromstaticequilibrium. Thepartduetolengtheningof thestringagainsttensionis

2 2∂V T ∂V

T (ds− dx) = T 1 + dx− T dx ≈ dx∂x 2 ∂x

Addingthepartagainstthesprings,thetotalpotentialenergyperunitlengthis 2T ∂V K

P E = + V 2 (6.26)2 ∂x 2

Wenowcalculatetheirtimeaverages. If twotime-harmonicfunctionsarewritteninthecomplexform:

a= Ae−iωt , b = Be−iωt (6.27)thetimeaverageof theirproductisgivenby

2πωω t+ 1 1ab≡ abdt= (AB∗) = (A∗B) (6.28)

2π t 2 2Usingthisrecipe,theperiod-averagedenergydensitiesare,

KE = ρ (−iωAeiθ)2 = ρω2|A|2

2 4P E = T (ikAeıθ)2 +K

(Aeıθ)2 = T k2 +K |A|2

2 2 4 4whereθ

≡ |k

|x

−ωtisthewavephase. Since

T K ω2 = k2 +|rho T

wehaveρ

P E = ω2|A|24

Thusthetotalenergyisequallypartitioned betweenkineticanpotentialenergies,and E =KE +P E = ρ

ω2|A|2 (6.29)2



2.7.DISPERSION FROM ALOCALIZED DISTURBANCE 18Nowtheaveraged rateof energy influxacrossanystationxcanbecalculated from

theproductof transversecomponentof thetensionandthetransversevelocity,∂V ∂V T ρ T k −T =−T (ikV )(−iωV ) = kω|A|2 = ω2|A|2 =Ecg∂x ∂t 2 2 ρω

This power input is transported from left to right by the wave. Hence the speed of energy transport is the group velocty. This result is quite general for many physicalproblemsand isnotlimited tothesprings.

7 Dispersion froma localized initialdisturbanceThesolutionformonochromaticwavesalreadyshowsthatwavesof differentwavelengthsmove at different velocities. What then is the consequence of an initial disturbance?Since a general initial disturbance bounded in space can be represented by a Fourierintegralwhichamountstothesumof infinitelymanysinusoidswithawidespectrum,weshallemploythetoolsof Fouriertransform.

Inadditiontothegoverningequation:1∂

2

V ∂

2

V

KV

= − , − ∞ < x < ∞, t > 0 (7.1)c2 ∂t2 ∂x2 T weaddtheinitial(Cauchy)conditions

∂V (x,0)V (x,0)=f (x), = 0 (7.2)

∂t LetusdefinetheFouriertransformanditsinverseby

∞

∞

f (k) = e−ikxf (x)dx, f (x) = 1eikxf (k)dk (7.3)

2π−∞ −∞Thetransformsof (7.1)is

1d2V KV =−k2V − , t > 0 (7.4)

c2 dt2 T and of theintitialconditionsare,

¯ ¯ dV (k,0)V (k, 0)=f (k), = 0 (7.5)

dt



192.7.DISPERSION FROM ALOCALIZED DISTURBANCE Thesolutionforthetranformis

¯ ¯V

(k, t) =

f (k)cos

ωt

(7.6)

where

K ω=co k2 + (7.7)

T TheFourierinversionis

1 ∞V (x, t) = dkf (k)eikxcosωt (7.8)

2π −∞Anyrealfunctionf (x)canbeexpressed asthesumof anevenand anodd function

¯of

x.

For

simplicity

let

us

assume

that

f (x) is even in

x

so

that

f (k)

is

real

and even

in

k, then

1 ∞V (x, t) = dkf (k)cos kxcosωt

π 0whichcanbemanipulated to

V (x, t) = 1 ∞

dkf (k) eikx−iωt +eikx+iωt dk (7.9)2π 0

The first term in the integand represents the right-going wave while the second, left-

going. Eachpartcorrepondstoasuperpositionof sinusoidalwavetrainsovertheentire¯rangeof wavenumbers,withinthesmallrange(k, k+dk)theamplitudeisf (k)/2. The

¯function f (k)/2 is called the Fourier amplitude spectrum. In general explicit evalua

tion of the Fourier integrals is not feasible. We shall therefore only seek approximateinformation. Themethod of stationaryphaseisparticularlyusefulhere. Itaimsattheasymptoticapproximationof theintegral b

I (t) = F (k)eitφ(k)dk (7.10)a

forlarget. Letusfirstgiveaquickderivationof themathematicalresult. AssumethatF (k), φ(k)areordinary functionsof k. If t is large,thenask increasesalongthepathof integrationboththerealand imaginarypartsof theexponentialfunction

cos(tφ(k))+isin(tφ(k))oscillatesrapidlybetween-1to+1,unlessthereisapointof stationaryphaseko within(a,b)sothat

dφ(ko)=φ(ko) = 0, a < ko < b. (7.11)

dk



2.7.DISPERSION FROM ALOCALIZED DISTURBANCE 20

Figure8: Neighborhood of StationaryPhaseThenimportantcontributiontotheFourierintegralcomesonlyfromtheneighborhoodof ko. Nearthepointof stationaryphase,weapproximatethephaseby

φ(k) = φ(ko) + 1(k− ko)2φ(ko) + · · · 2

and theintegralby bI (t)≈ F (ko)eitφ(ko) exp it

(k− ko)2φ(ko) dk2a

With an error of O(1/t), we also replace the limits of the last integral by ±∞; the justificationisomitted here. Nowitisknownthat

∞

e±itk2dk= π

e±iπ/4t−∞

ItfollowsthatI (t) =

bF (k)eitφ(k)dk≈ F (ko)eitφ(ko)±iπ/4

2π 1/2

+O

1, if ko ∈ (a, b),

t|φ(ko)| ta(7.12)

where the sign is + (or−) if φ(ko) is positive (or negative). It can be shown that if thereisnostationarypointintherange(a,b),thentheintegralI (t) is small

1I (t) = O , if ko ∈/ (a, b). (7.13)

t



212.7.DISPERSION FROM ALOCALIZED DISTURBANCE Letusapplythisresulttotheright-goingwave

it(kx/t−iω)

dkV +(x, t) = 1

∞

dkf (k)e (7.14)4π 0For a fixed x/t=constant,wehave

K φ(k) = kx/t−ω(k), with x/t=ξ, ω=co k2 + (7.15)

T Foranobservertravellingatthespeedx/t lessthanthemaximumω(k),i.e.,co, there isastationarypointko attherootof

x/t=ω(ko) = T ko

= coko(7.16)

ρω(ko) k2 +K o T

Therootko increasesfromzeroforlongwavestothemaximumco forveryshortwaves.If theobserverisfasterthanco,thereisnostationarypoint.

Sinceφ(ko) = −ω(ko)<0, (7.17)

thefinalresultisthatitφ(ko)−iπ/4V +(x, t) ∼ 2

1π f (ko) tω

2(

π

ko)e=

1 f (ko) 2π

eikox−ω(ko)t−iπ/4 (7.18)2π tω(ko)

¯Thetransformf (k)dependsonthespecificprofileof theinitialdisturbance. Forexam

ple,if Sb

f (x) = (7.19)π(x2 +b2)

whichhasanareaS and characteristicwidthb,theFouriertransformisf (k) = Se−|k|b (7.20)

To a specific observer identified by the speed of travel x/t < co, the approximateresultcanbeviewed asasimpleharmonicwavetrainwithwavenumberko, frequencyω(ko)and amplitude

f (ko) 2πA= (7.21)2π tω(ko)



222.7.DISPERSION FROM ALOCALIZED DISTURBANCE

Figure9: Snapshotsof wavefield onthesidex > 0.Since ω(k) increases from 0 with increasing k to the finite maximum T/ρco = co, an observer faster than co sees no waves. However any observer slower than co is accom

panied byatrainof progressivesinusoidalwaves. The localwavelengthko issuchthatthe group velocty matches the observer’s speed. The faster the observer, the shorterthe waves. If asnapshot is taken, thentheshortestwave, whosephasevelocity is thelowestareseenatthefrontwhichmovesasfastasthecrestsof theshortestwaves. The

¯crestsof longerwavesadvancemuchfasterthanthelocalenvelope. Becausef (k) is the greatestatk=0,theenvelopeof thelongestwaveswhichstaysnearthesource, isthebiggest. Theenvelopeof theshortestwavesislowerinamplitudeandspreadsoutwiththewave front. Theentiredisturbanceattenuates intimeast−1/2. Seefigure7 foranoverview.

Note also that very near the wave front, cg → co; the second derivative φ(ko) = −ω(ko)vanishes. Hencetheasymptoticformulabreaksdown. Abetterapproximationisneeded,butisomittedhere. (SeeC.C.Mei,1989,Applied Dynamics of Ocean Surface Waves ). Finallyweexaminethepropagationof waveenergyinthistransientproblem.



232.8. SCATTERINGOF SINUSOIDALWAVES Using(7.21)thelocalenergydensityis:

1 ρω2 f (ko)2E

= 2ρω

2

|A|2

= 4πtω(ko)At any given t, the waves between two observers moving at slightly different speeds,cg(k1) and cg(k2),i.e.,betweentwopointsx1/t=cg(k1) and x2/t=cg(k2)areessentiallysimpleharmonicsothatthetotalenergyis

x2 x2 ρω2(f (ko))2dxE = dx

4πt ω(ko)x1 x1Sincex=ω(ko)tforfixedt, we have

dx=

ω

(ko)dko

tNowforx2 > x1,k2 > k1,itfollowsthat x2 k2 ρω2

dxE = dko (f (ko))2 =constant4πx1 k1

Therefore the total energy between two observers moving at the local group velocityremainsthesameforalltime. Inotherwords,wavesaretransportedbythelocalgroupvelocityevenintransientdispersion.

8 Scatteringof sinusoidalwavesIf alongalongrod therearesomelocalized inhomogeneities,anincomingtrainof sinu

soidalwaveswillbepartlyreflectedandpartlytransmitted. Thescatteredsignalstellussomethingaboutthescatterer. Todeterminethescatteringproperties foraknownscattereriscalled thescatteringproblem. Todeterminethescattererfromthescatteringdataiscalledtheinversescatteringproblem. Weshallonlystudytheformer.

Variousmathematicaltechniquesareneededfordifferentcases: (i)Weakscattererscharacterized bysmallamplituderelativetothewavelength,orslowvariationwithinawavelength,(ii)Strongscatterersif theirdimensionsarecomparabletothewavelength.

8.1 WeakscatteringLetthelongrod haveaslightlynonuniformcrosssection,

S (x) = S o(1+a(x)), 1. (8.1)



242.8. SCATTERINGOF SINUSOIDALWAVES wherea(x)diminishestozeroatx∼ ±∞. Thewaveequationreads

∂ 2u ∂ ∂u

ρS (x) =E S (x) (8.2)∂t2 ∂x ∂x Forsinusoidalwaves

u(x, t) = U (x)e−iωt

thespatialfactorisgoverned byd dU

S +ρω2SU = 0 (8.3) dx dx

ThenthesolutioncanbesoughtbytheperturbationmethodU (x) = U 0(x) + U 1(x) + 2U 2(x) + · · · (8.4)

Substitutingthisintothegoverningequationd d

ES o (1+a(x)) U 0(x) + U 1(x) + 2U 2(x) + · · · dx dx

+ρω2S o(1+a(x)) U 0(x) + U 1(x) + 2U 2(x) + · · · = 0 (8.5) and equatingthecoefficientsof likepowersof tozero,wegetfromthetermsof orderO(0):

ES od2U 0

+ρω2S 0U 0 = 0 (8.6) dx2

ThesolutionissimplytheincidentwaveU 0 =Aeikx, with k=ω ρ/E (8.7)

AttheorderO() we get

ES od2U 1

+ρω

2

S 0U 1 +a(x)

ES o

d2U 0+

ρω

2

S 0U 0+

ES o

dadU 0= 0

∂x2 ∂x2 dx dx

or,d2U 1 dadU 0

+k2U 1 =− (8.8)dx2 dx dx

Thusthescattered waveisatmostof theorderand isgoverned byaninhomogeneousequation. Letusdefinethefundamentalsolution(Green’sfunction)by

d2G(x, x)+k2G(x, x) = δ (x− x) (8.9)

dx2



252.8. SCATTERINGOF SINUSOIDALWAVES G∼ outgoingwavesat± ∞ (8.10)

ItwillbeshowninAppendixAthatG(x, x) = eik|x−x|

(8.11)2ik

Bystraightforward differentiation,itcanbeverified thatU 1(x) = − ∞ da(x)dU 0(x)

G(x, x)dx =− 1 ∞ da(x)dU 0(x)eik|x−x|dx (8.12)

−∞ dx dx 2ik −∞ dx dx

SinceU 0(x) = eikx, we have = −

A ∞ da(x)eikx

eik|x−x

|dxU 1 2 −∞ dx

= −A eikx x da

dx +e−ikx ∞ da

e2ikxdx (8.13)2 −∞ dx

x dx

= −A eikxa(x) + e−ikx ∞ dae2ikxdx

2 dxx

= −A a(x)eikx +e−ikx ∞ d a(x)e2ikx

dx− a2ike2ikx dx

2 dxx ∞

= ikAe−ikx a(x)e2ikxdx (8.14)x

Far to the right we getU 1 = 0, x → ∞. (8.15)

Thusonthetransmissionsidemodificationof theincidentwavescanatmostbeof theorderO(2). Fartotheleftx∼ −∞, ∞

U ∼ =Ae−ikx ik a(x)e2ikxdx , x → −∞. (8.16)=U 1−∞

Letusdefinethereflectioncoefficientby ∞R=ik a(x)e2ikxdx, (8.17)

−∞Forinstanceif aisahumpof unittotaldimensionlessareaandlengthb,

ba(x) = (8.18)

π(x2 +b2)then

R=ikbe−2kb (8.19)



262.8. SCATTERINGOF SINUSOIDALWAVES

Figure10: Reflectioncoefficientgivenby(8.19).Hence reflection is small if kb is large, i.e., long and gentle obstacles are not effectivein reflecting short waves . For very long waves or a short obstacle, reflection is alsovanishingly small. The maximum reflection coefficient is R = iao/2 when kb= 1/2.Notethatfartotheright,U 1(x)→ 0 as x→ ∞. Thetransmissionvwaveis

U ∼=U 0 +U 1 =Aeik0x(1+O(2)). (8.20)ThetranmissioncoefficientisT = 1 + O(2). Thisisconsistentwiththegenerallawof energyconservation|R|2 +|T |2 =1tobeprovenlater.Homework: Checkif thegoverningequationatO(2)forU 2 is:

ES od2U 2

+ρω2S 0U 2 =ES oa

dadU 0 −dadU 1

(8.21)

∂x2 dx dx dx dxCarry out the solution to find the transmission coefficient T from the amplitude of U 0 +U 1 +2U 2 atx∼ ∞,and showthatenergyisconserved totheorderO(2).



2.8. SCATTERINGOF SINUSOIDALWAVES 278.2 Strongscatteringof longwater-wavesbyashelf Consider

an

ocean

bottom

with

astep-wise

variation

of

depth.

h1, x < −a;h= h2, −a < x < a; (8.22)

h3 =h1, x > a How does the step interupt the propagation of an incident wave of unit amplitudearrivingfromx∼ −∞?

Ineachzoneof contantdepth(i= 1, 2, 3),theshallowwaterequationsread:∂ζ i ∂ui

+hi =0 (8.23)∂t ∂x ∂ui ∂ζ i

+g =0 (8.24)∂t ∂x

Forsinusoidalwavesζ i =ηie−iωt, ui =U ie−iωt (8.25)

weget∂U i

−iωηi +hi =0 (8.26)

∂x ∂ηi−iωU i +g =0 (8.27)∂x

ord2ηi ω

+ki2ηi = 0, where ki =√ (8.28)dx2 ghi

Ata junction,thepressureand thefluxmustbeequal,hencedη1 dη2

η1 =η2, and h1 =h2 , x =−a; (8.29)dx dxdη2 dη3

η2 =η3, and h2 =h1 , x =a. (8.30)dx dx

Theformsof thesolutionsineachzoneof constantdepthare:η1 =eik1(x+a) +Re−ik1(x+a), x < −a; (8.31)η2 =Aeik2x +Be−ik2x, − a < x < a (8.32)

η3 =T eik1(x−a), x > a (8.33)



282.8. SCATTERINGOF SINUSOIDALWAVES Thereflectionand transmissioncoefficientsRandT aswellasAandBareyetunknown.

Applyingthematchingconditionsattheleft junction,wegettworelations1 + R=Ae−ik2a +Beik2a (8.34)

k1h1(1−R) = k2h2(Ae−ik2a−Beik2a). (8.35)Similarlythematchingconditionsatx=agives

Aeik2a +Be−ik2a =T (8.36)k2h2(Aeik2a

−Be−ik2a) = k1h1T. (8.37)

Thesefourequationscanbesolved togive4s

T =(1 +s)2e−2ik2a−(1−s)2e2ik2a (8.38)

−(1−s2)(e−2ik2a−e2ik2a)R=

(1 +s)2e−2ik2a−(1−s)2e2ik2a (8.39)A= T

e−ik2a(1+s) (8.40)2T

B= eik2a(1−s) (8.41)2

where k1h1 h1 c1

s= = = (8.42)k2h2 h2 c2

Theenergyassociated withthetranmsitted and reflected wavesare:4s2

|T |2 =4s2 + (1 −s2)2sin22k2a (8.43)

(1−s2)sin22k2a|R

|2 =

4s2 + (1 −s2)2sin22k2a (8.44)Itisevidentthat|R|2 +|T |2 = 1.

Overtheshelf thefreesurfaceisgivenby2s (1+s)eik2(x−a) + (1 −s)e−ik2(x−a)

η2 = (8.45)(1+s)2e−ik2a−(1−s)2eik2a

Recallingthetimefactore−iωt, we see that the free surface over the shelf consists of two wavetrainsadvancing inoppopsitedirections. Thereforealongtheshelf thetwowaves



2.8. SCATTERINGOF SINUSOIDALWAVES 29caninterfereeachotherconstructively,withthecrestsof onecoincidingwiththecrestsof theotheratthesamemoment. Atotherplacesthe interference isdestructive,withthecrestsof onewavetraincoincidingwiththetroughsof theother. Theenvelopeof energyontheshelf isgivenby

4s2 cos2k2(x− a) + s2sin2k2(x− a)|η|2 =4s2 + (1 − s)2sin22k2a (8.46)

Atthedownwaveedgeof theshelf,x=a,theenvelopeis4s2

|η|2 =4s2 + (1 − s)2sin22k2a (8.47)

Notethatthereflectionand transmissioncoefficientsareoscillatory ink2a. In par

ticulatr for 2k2a = nπ,n = 1,2,3..., that is 4a/λ = n, |R| = 0 and |T | = 1. The shelf istransparenttotheincidentwaves,correspoinidngtothemostconstructiveinter

ference and the strongest transmission Mininum transmision and maximum reflectionoccur when 2k2a= (n− 1/2)π, or 4a/λ=n− 1/2, when the interference isthe mostdestructive. Thecorrespondingtransimssionandreflectioncoefficientsare

min

|T

|2 =

4s2, max

|R

|2 = (1− s2)2

. (8.48)

(1+

s2

)2

(1+

s2

)2

SeeFigure11The features of interference can be explained physically. The incident wave train

consists of periodic crests and troughs. When one of the crests first strikes the leftedgeatx=a,partof theitistransmitted ontotheshelf and partisreflected towardsx ∼ −∞. After reaching the right edge at x = a, the tranmitted crest has a partreflected tothe leftand re-reflected bytheedgex=−atotherightagain. Whenthere-reflected crest arrives at the right edge the second time, its total travel distance is4a. If 4a is an integral multple of the wave length λ2, the crest is in phase with allthe other crests entering the shelf either before (the third, fourth, fifth, ... time) orafter. Thusallthecrestsreinforceoneanotherattherightedge. This isconstructiveinterference, leading to the strongest tranmission to the right x ∼ ∞. On the otherhand if 2k2a= (n− 1/2)π or4a/λ=n− 1/2,somecrestswillbe inoppositephasetosome other crests, leading to the most destructive interference at the right edge, andsmallesttransmission.



302.9.GENERALIDENIITIES ON SCATTRERING

0

0.2

0.4

0.6

0.8

1

Max|R|2

Min|T|2

0 2 4 6 8 10

s 2

0 2 4 6 8 10

2k a2

Figure 11: Transmission and reflection coefficients for a rectangular shelf. From Mei,C.C.,Applied Dynamics of Ocean Surface Waves.

From the results in the following exercise, the reflected wave suffers a reversal of phase at the discontinuity (a crest turns into a trough and vice versa). Hence at thereflectionedgeawavecresttranmittedontotheshelf returns fromx =a asatrough.Itisthentransmitted totheleftof x =−a astrouigh. Ontheotherhand,acrestthatisreflected atx =−a appearsasacrest. Thereforetheinterferenceisdestructive. TheresultingR,whichisthesumof multiplereflections,issmall.Homework: Scattering by a depth discontinuity: Consider the scattering of long

water waves by a depth discontinuity at x = 0, i.e., h = h1, −∞ < x < 0 and h =h2, 0< x < ∞. Showthatthetranmssionand reflectioncoefficientsare:

2 1−

(h2/h1)1/2T

=

, R =

(8.49)1 + (h2/h1)1/2 1 + (h2/h1)1/2

Thusthereflected wavesuffersareversalof phasewhenh2/h1 > 1,i.e,approachingthediscontinuityfromtheshallowside.

0

0.2

0.4

0.6

0.8

1

|R|2

= 2 / 2√

= 1 / 2

= 1 / 4s

s

s



2.9.GENERALIDENIITIES ON SCATTRERING 319 Generalidentitiesinscatteringof monochromatic

wavesbyarbitrary inhomogeneitiesScatteringdueto inhomogeneities, causedbynonuniformitieseither ingeometryor inmaterialproperties,requiresthesolutionof ordinarydifferentialequationswithvariablecoefficients. In general one must resort to numerical means. For the sake of checkingnumericalaccuracyand forgaining physical insight, identitieswhich mustbetrue areuseful. They are often deduced by arguments typical in the derivation or the use of Green’stheorem.

We illustratethese identitiesbytheexampleof an infinitely longrodwithvariablecrosssectionS (x),

∂ ∂u

ρS ∂ 2uS = (9.1)

∂x ∂x E ∂t2E and ρ are taken to be constants. S (x) is nonuniform only in a finite neighborhoodaroundx= 0. Elsewhere S =S o =constant. Considermonochromaticwaves

u(x, t) = U (x)e−iωt (9.2)

sothat

U (x)

is

governed

by

the

ordinary

differential

equation

ω2ρ

(SU )

+k2SU = 0, k2 = (9.3)E

Theboundaryconditionsdependonthesourceof theincidentwaves.Letusconsidertwoproblems: onewiththeincidentwavefromx∼ −∞;theother

withtheincidentwavefromx∼ ∞.If theincidentwavesarrivefromx∼ −∞,theasymptoticformof thesolutionmust

beU 1(x)∼ A−eikx +B−e−ikx =A− eikx +R1e−ikx , kx ∼ −∞ (9.4)

andU 1(x)∼ A+eikx =A−T 1eikx kx∼ ∞ (9.5)

whereT 1 andR1 arethetransmissionand reflectioncoefficientswhichareapartof theunknownsolutionof theleft-incidenceproblem. LetusdefinetheJostfunctionby

U 1f 1(x) = (9.6)

A−T 1



322.9.GENERALIDENIITIES ON SCATTRERINGthen

1 ikx R1 −ikxf 1(x)∼

e + e kx −

1 (9.7)T 1 T 1

andf 1(x)∼ eikx kx 1 (9.8)

Ontheotherhand,if theincidentwavesarrivefromx∼ ∞,theasymptoticformof thesolutionmustbe

U 2(x)∼ b−e−ikx =b+T 2e−ikx, kx ∼ −∞ (9.9)and

U 2(x)∼ a+eikx +b+e−ikx =b+ R2eikx +e−ikx , kx ∼ ∞ (9.10)whereT 2 andR2 aretheunknowntransmissionand reflectioncoefficientsof theright-

incidenceproblem. SimilarlywedefineanotherJostfunctionU 2

f 2 = (9.11)b+T 2

sothatitbehavesasf 2(x)

∼e−ikx, kx

∼ −∞ (9.12)

andf 2(x)∼ 1

e−ikx +R2eikx, kx ∼ ∞ (9.13)

T 2 T 2Sincebothf 1 andf 2 satisfy(9.3),wehave

(Sf 1)

+k2Sf 1 = 0, and (Sf 2)

+k2Sf 2 = 0, (9.14)Multiplyingthefirstbyf 2 and thesecond byf 1,thentakingthedifference,wegetafterpartialintegration, d

(S (f 1f − f 2f 1))=0,

dx 2implyingthefollowingWronskianidentity

W (f 1, f 2) = (f 1f 2 − f 2f 1) = C

(9.15)S (x)



2.10.REFRACTION IN ASLOWLY VARYINGMEDIUM 34Similarlybyusingtheasymptoticexpression(9.5)onthetransmissionside,weget

S (U U ∗− U ∗

U

) x∼∞=S o A+eikx (−ik)A∗

+e−ikx − S o A+∗e−ikx (ik)A+eikx

=−2ikS o|A+|2

Equatingthetwolimitsweconcludethat|A−|2 =|B−|2 +|A+|2 (9.19)

or|R1|2 +|T 1|2 = 1. (9.20)

Thisismerelyastatementof conservationflux: thefluxrateof totalscattered energy(reflected and transmitted) is equal to the flux rate of the incident wave energy; thespeed of energy tranport being the same on both sides of the scatterer. Clearly thesameenergyconservationmusthold fortheright-incidenceproblem.

|R2|2 +|T 2|2 =1 (9.21)If thescatteringproblemissolved numericallyby,say,finiteelements,itisnecesssary

thatthecomputed scatteringcoefficientsRandT satisfytheidentities(9.16)(9.20)and(9.21).Homework: Extend theidentities(9.16)and (9.20)whenS (x)→ S 1 askx→ −∞

andS (x)→ S 2 askx→ ∞, where S 1 andS 2 aretwodifferentconstants.

10 Refraction inaslowlyvaryingmediumLetusconsidertime-harmoniclongwavesinshallowwaterwaves. Thegoverningequa

tionisd

dηω2

h + η=0 (10.1)dx dx g

Consideraseadepthwhichvariesslowlywithinawavelength,i.e.,1 dh

=O(µ) 1 (10.2)khdx



352.10.REFRACTION IN ASLOWLY VARYINGMEDIUM Earlieranalysissuggeststhatreflectionisnegligiblysmall. Thusthesolutionisexpectedtobealocallyprogressivewavewithboththewavenunmberand amplitudevaryingmuchmoreslowlythanthewavephaseinx. Hencewetrythesolution

η=A(x)eiθ(x) (10.3)whereθ(x)− ωtisthephasefunctionand

dθk(x) = (10.4)

dxis the local wave number and A is the complex amplitude. Note the spatial rate of variationof thephasefunction,i.e.,thewavenumberisingneralnotsmall,thereforeθitself isnotaslowlyvaryingfunctionof x. Letuscalculatethefirstderivative:

dηdx =

ikA+dA

dx

eiθ

and assumedAdx =O(kL)−1 1

kAIn fact we shall assume each derivative of h, A or k is µ times smaller than kh,kAork2. Futhurmore,

d dη

ω2 dA

d dA

d(khA) ω2Ah + η= ik ikh+h + h +i + eiθ = 0

dx dx g dx dx dx dx gNotethatthecomplexamplitudeAcanbewrittenas

A=|A(x)|eiθA(x) (10.5)Thephaseof A,θA,isaslowlyvaryingfunctionof x. Itcanbeconsideredasapartof thewavephase,althoughitsspatialgradientismuchsmallerthank.

Nowlet’sexpandA=A0 +A1 +A2 +· · · (10.6)

withA1/A0 =O(µ), A2/A0 =O(µ2),· · · . From O(µ0)thedispersionrelationfollows:ω

ω2 =ghk2, or k=√ (10.7)gh



362.10.REFRACTION IN ASLOWLY VARYINGMEDIUM Thus the local wave number and the local depth are related to frequency accordingto thewellknowndispersionrelation for constantdepth. As the depthdecreases, thewavenumberincreases. Hencethelocalphasevelocity

ωc= = gh (10.8)

kalsodecreases.

From O(µ) we get, dA0 d(khA0)

ikh +i = 0 dx dx

or,aftermultiplicationbyA0,d (khA2

0) = 0dx

whichimpliesd

(kh|A0|2)=0 (10.9)dx

hencekh|A0|2 =C 2 =constant

or,gh|A0|

2=

constant

=

gh∞|A∞|

2(10.10)

Since in shallow water the group velocity equals the phase velocity, the above resultmeans that the rate of energy flux is the same for all x and is consistent with theoriginal assumption of unidirectional propagation. Furthermore, the local amplitudeincreaseswithdepthas

A0(x) h∞

1/4= (10.11)

A∞ hThisresultiscalled Green’slaw.

Insummary,theleadingordersolutionis 1/4 1/4 x ζ =A∞ h∞

eiθ−iωt =A∞ h∞exp i k(x)dx−iω (10.12)

h hHomework: Study the refraction of a train of obliquely incident waves towards aslopingbeach. Theseadepthvariesslowlyonlyinx,h=h(x).



372.10.REFRACTION IN ASLOWLY VARYINGMEDIUM A Solution fortheGreen functionLetusdividethedomainintotwoparts,correspondingtotwosidesof theconcentratedforcing. On the right side x > x0, we define G(x, x0) ≡ G+(x, x0) which must be governed by:

d2G+k2G+ = 0, x > x0 (A.1)

dx2 0To satisfy the radiation condition we take the sol,ution to beTo satisfy the radiationconditionwetake

G+ =Ceik0x (A.2)OntheleftsidewedefineG(x, x0) = G−(x, x0),whichmustbegoverned by

d2G−+k2G− = 0, x < x0 (A.3)

dx2 0

TosatisfytheradiationconditionwetakeG+ =De−ik0x (A.4)

Atx=x0 werequirecontinuity:G+(x0, x0) = G−(x), x)) (A.5)

HenceCeikox0 =De−ik0x0 (A.6)

Inadditionwegetanothermatchingconditionatx=x0 by integratingeq(8.8)acrosstheconcentrated forcing

x0+δ

x0+δd2G

+k02G dx= δ (x

−x0)dx= 1.

dx2

x0−δ x0−δTheleft-hand sidecanbeintegrated togivethesecond matchingcodition,

dG+ dG−− = 1 (A.7) dx dxx0 x0

implyingthatik0Ceik0x0 +ik0De−ik0xo = 1 (A.8)



2.10.REFRACTION IN ASLOWLY VARYINGMEDIUM 38Thus

−ik0x0 ik0x0e eC 0 = , D0 = , (A.9)

2ik0 2ik0Thefinalsolutionis

ik0(x−x0) −ik0(x−x0)e eG+ = , x − x0 >0; G− = , x − x0 <0.

2ik0 2ik0ormorecompactly,

eik0|x−x0|

G(x, x0) = (A.10)2ik0

Moreonone-dimensionalwaveswillbediscussedlater.

B OntheradiationconditionB.1 Formsof theradiationconditionIn the theory of diffraction and/or radiation of time-harmonic waves, an importantboundaryconditionistheso-calledradiationconditionduetoA.Sommerefeld:The disturbance due toeither diffraction by a finite scatterer,or radiation from a finite region,can only behave as outgoing waves at infinity.

LetusfirstwritetheunknownasΦ(x, t) = φ(x)e−iωt (B.1)

Inascattering(diffraction)problemweletφ=φI +φ (B.2)

where φI represents the incident wave (e.g., φI = Aoeikx), and φ denotes the distur

bance(scattered wave). Inaradiationproblem,φ representstheradiatedwaveduetomotion in the finite region. The radiation condition can be stated mathematically intwoequivalentforms,asfollows.

1. One-dimensionalpropagation:φ(x)→ A0B±e±ikx, as kx→ ±∞; (B.3)

or, ∂φ

lim ∓ ikφ → 0 (B.4)kx→±∞ ∂x



392.10.REFRACTION IN ASLOWLY VARYINGMEDIUM 2. Two-dimensionalpropagation:

ikrφ(r,θ,z )→ A0B(θ, z )√ e

, as kr→ ∞; (B.5)kror, √ ∂φ

lim kr − ikφ → 0 (B.6)kr→∞ ∂r

3. Three-dimensionalpropagation:±ikr

φ(r,θ,ψ)→ A0(θ, ψ)e , as kr→ ±∞; (B.7)kr

or, ∂φlim kr − ikφ → 0 (B.8)

kr→∞ ∂r Notethattheradiationcondition ismeantonly forwavesthat isstrictlysinusoidal intimefromancientpast. Fortransientdisturbances,theconditionshouldbechangedto

Φ(x, t)→ 0, as |x| → ∞, t < ∞. (B.9)Physically. if thesourceof thewavedisturbanceisinafiniteregion,itisnotpossible

for wave energy to flow inward from infinity. Mathematically, a sinuoidal wave is thelimit of a transient wave where the local and sinusouidal forcing begins at some timein the past, say t = 0. For all finite t, the boundary condition (B.9) holds. In orderto derive the behavior at very large time (the quasi steady state), one can solve aninitial-boundayvalueproblemundertheinitialconditions,

∂φφ, = 0, t = 0, ∀x (B.10)

∂t and

the

additional

boundary

condition

in

the

source

region,

e.g.,

∂φ

−iωt=n·Ve , t > 0. (B.11)∂n

and thentakethelimitof larget. TheexplicitsolutionfortheIBVPisnottrivial. Weshalldemonstrate foraone-dimensionalexample wherethe local energy supplygrowsslowlyincomparisonwiththewaveperiod.



2.10.REFRACTION IN ASLOWLY VARYINGMEDIUM 40B.2 Transientmotionof anelasticallysupportedstringConsider

asemi-infinte

string,

ρV tt− T V xx +KV = 0, , x > 0, t > 0 (B.12)

subjecttotheboundarycondition,V (0, t) = A0(t)e−iωt, t > 0 (B.13)

where1 and A0(t) increasesslowly fromA0(−∞) = 0 to A0(∞) = A =constantwiththetimerangeωt1 =O(1). Thusattheboundearythereissinusoidalforcingwithaslowlyincreasingamplitude. Atinfinitytheboundaryconditionis

V (∞, t)→ 0, t < ∞. (B.14)Let us use the perturbation method of mulitple scales. In vieew of (B.13), we in

troduce two time coordinates t and t1 = t. Suggested by the elementary example of wave groups, we expect that the envelope to be slowly varying in space as well. Letus therefore also introduce two space coordinates x, x1 = x1. We now pretend thatV =V (x, x1;t, t1)sothattheoriginaltimederivativesbecome,accordingtothechainrule,

∂ ∂V ∂V → + , (B.15)∂t ∂t ∂t1

∂ 2V ∂ ∂

∂V ∂V ∂ 2V ∂ 2V ∂ 2V → + + = + 2 +2 . (B.16)

∂t2 ∂t ∂t1 ∂t ∂t1 ∂t2 ∂t∂t1 ∂t21Likewisethespacederivativesbecome

∂ ∂V ∂V → + , (B.17)∂x ∂x ∂x1

∂ 2V

∂ ∂

∂V ∂V

∂ 2V ∂ 2V ∂ 2V → + + = + 2 +2 . (B.18)∂x2 ∂x ∂x1 ∂x ∂x1 ∂x2 ∂t∂x1 ∂x2

1

Wenextsubstitutetheseinto(B.12)V =V 0(x, x1;t, t1) + V 1(x, x1;t, t1) + 2V (x, x1;t, t1) + · · · (B.19)

andseparatedifferentorders. AtorderO(0)wehavethefollowingperturbationequa tion: O(0):

∂ 2V 0 ∂ 2V 0ρ − T +KV 0 =0 (B.20)

∂t2 ∂x2



412.10.REFRACTION IN ASLOWLY VARYINGMEDIUM AtorderO(),weget:

∂ 2V 1 ∂ 2V 1 ∂ 2V 0 ∂ 2V 0ρ − T +KV 1 =−2ρ + 2T (B.21)∂t2 ∂x2 ∂t∂t1 ∂x∂x1

Letustaketheleading-ordersolutiontobeasinusoidalprogressivewaveV 0 =A(x1, t1)ei(kx−ωt) (B.22)

where ρω2 K

k=± − . (B.23)T T

Letthesquarerootberealsothatthereispropagation. However,wedonotknowthesign

of

k,

hence

the

direction

of

propagation.

AttheorderO()Eqn. (B.21)canbewritten

∂ 2V 1 ∂ 2V 1

∂A ∂A ikx−iωtρ − T +KV 1 =−2i kT +ρω e . (B.24)

∂t2 ∂x2 ∂x1 ∂x1The forcingtermontheright isahomogeneoussolutionof the left,henceunboundedresonanceof V 1 mustoccurunless

∂A ∂A kT +ρω = 0, (B.25)

∂x1 ∂x1or ∂A Tk ∂A

+ =0 (B.26)∂t1 ρk∂x1

Thisistheconditionof secularity(solvability)andisafirstorderhyperbolicequation.Thegeneralsolutionis

x1A(x1, t1) = A t1− (B.27)

Tk/ρω Nowweapplytheboundarycondition(B.13)atx1 = 0,

A(t1) = A0(t1) (B.28)Hencetheenvelopeis

x1A(x1, t1) = A0 t1− (B.29)

Tk/ρω For this transient problem we expect that A(∞, t1) approaches zero when x→ ∞

for finite t1, according to (B.14). Since for large x1, A(x1, t1)→ A0(−x1/Tk/ρω), wemusttakethepositivesignfork inviewof thegivenbehaviourof A0. Thus

k > 0 (B.30)



422.10.REFRACTION IN ASLOWLY VARYINGMEDIUM implyingthatthesolutionisarightwardpropagatingwave. Thustheradiationconditionisderived.

NotefurtherthatT k

=cg (B.31)ρω

isthegroupvelocityinthestring,thusA(x1, t1) = A(t1−x1/cg) (B.32)

or,Aisaslowlyvaryingfunctionof x−cgt. Thusresultistrueforalldispersivewaves.

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