J.J. Stoker - Water waves. The mathematical theory with applications (796980), страница 21
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(Actually,when combined with the factor e ikz with z once more the spacevariable, the result is a harmonic function yielding a plane wave inwater of infinite depth and satisfying the free surface condition).(5.4.1)l9One can obtain agreatmany moresolutionsby multiplying theabove special solution by an analytic function g() and integratingalong a path P in the complex -plane:/(*, 5)(5.4.4)By--.'2m J{Pze ^+Tfg(C)d.appropriate choices of the analytic function g() and the path P,to satisfy the boundary conditions and the conditionwe might hopeat oo. This does, indeed, turn out to be the case.Still another way to motivate taking (5.4.4) as the starting pointof our investigation is the following.
It would seem reasonable tolook for solutions of (5.4.1) in the form of the exponential functions=+exp {mxly}. However, since we wish to work with analyticfunctions of complex variables it would also seem reasonable to expressxx and y in terms of zxiy, and this would lead toiy and z<p=m-4- z\___*! I(/%j\2/+U-/Z[\=_ Z\2\11/).InorderthatthisfunctionWATER WAVES98(which(5.4.1)!isclearly analytic in zwe mustrequire thatand z separately) be a solution of2k 2 = 0, and this leads at-f- Zm*{kz+24z\1,withanar-jbitrary parameter, as one can readily verify.
The method used byPeters [P.6] to arrive at a representation of the form (5.4.4) is bettermotivated though perhaps more complicated, since he operates with(5.4.1) in polar coordinates, applies the Laplace transform withrespect to the radius vector, transforms the resulting equation tothe Laplace equation, and eventually arrives at (5.4.4).One of the paths of integration used later on is indicated in Figure5.4.2.
The essential properties of this parth are: it is symmetricalwith respect to the real axis, goes to infinity in the negative direction-Fig. 5.4.2.The pathPplanein the f -planeof the real axis, enters the origin tangentially to the real axis andfrom the left, and contains in the region lying to the left of it anumber of poles of g(). (The path is assumedin the manner indicated so that the term z/factor will not make the integral diverge). Ourto enter the originin the exponentialdiscussion will takethe following course: We shall assume g() to be defined in the -planeslit along the negative real axis (and also on occasion on a Riemannsurface obtained by continuing analytically over the slit).
The choicePof the symmetrical pathleads to a functional equation for g()useofthethroughboundary conditions (5.4.2)! and (5.4.3)!, and viceversa a solution g() of the functional equation leads to a functionWAVES ON SLOPING BEACHES AND PAST OBSTACLESy)9>(#>=99^te /(**) satisfying the boundary conditions. (By theand Ste we mean, of course, that the imaginary, or real,symbols Jmpart of what followsto be taken.
)isWe seek a solution of the functionalequation which is defined and regular in the slitmost poles in the left half-plane (including certain-plane, with atfirstorder poleson the negative imaginary axis), and dying out at oo like 1/f. Oncesuch a function has been found, the prescribed conditions at oo willbe seen to follow by deforming the path P over the poles into a pathon the two edges of the slit along the negative real axis: the residuesat the poles on the negative imaginary axis clearly would yield contributions of the type{k2---- ~z\irz H4r} ,>0,which areeasily seen to be(-tr)Jof the desired type at oo, while the remaining poles and the integralover the deformed path will be found to yield contributions that tendto zeroWewhen+3te z ->oo.program by expressing the boundary conditionsbegin this(5.4.3 ) t in terms of the function f(z, z).
The first ofthese conditions will be satisfied if the following condition holds:(5.4.2 )jand^M(fz(5.4.2)iasone readilyn gradifSte {n<p==/}&t+ if) =n0,be satisfiedwill(5.4.3) 1the unit normal at the bottom surface,and the latter is given by+ /) sin{(/z real, positive,0,The conditionsees.with0,grad/5+{<*>-(fzfz)cos>}=ifi.e.0,or finally, in the form(5.4.3);Sm {f e-Upon making(5.4.5)i(0zSm-ift>/-<?}-z0,=re~ ta>r9use of (5.4.4) in (5.4.2)( the result~2mf c*t+*[CIJp4+ tlg(C)C>0.is-0,Jz real, positive,while (5.4.3 )[ yields(5.4.6)Jm.f2m Jpe* +?[V**I~ ^ e^4CJg(C)dCzTosatisfy theboundary conditionthat g(f ) satisfies the condition(5.4.5) itis==re-*",0,r>0.sufficient to requireWATER WAVES1002k//__+!r(5.4.7)The proof="ig()f real, positive.0,as follows: If (5.4.7) holds, then the integrand G(z, z, )in (5.4.5) is real for real z and real positive f Hence G takes on valueswhich are themselves conjugate, by theG, G at conjugate pointsis.,Schwarzshownreflection principle.
Since the pathin Figure 5.4.2, it follows that d takesPsymmetrical, ason values atthatis,are negative conjugates. Thus the integral (1 /2m)zisaandrealnew(5.4.7) holds. In consideringvariable s=e~ia)Gd\to obtain forzre~realwhenwe first introducenext (5.4.6)=isi<0the condition.replacing (5.4.6):Jm(5.4.8)ir^r\*L_Le**2ni J p/--k2=~\4 *Jg(seia>)e iu ds>r real.0,PHere P'is the path obtained by rotating(and the slit in theasofclockwiseaboutthewell,course)-planeorigin through the anglea). If g behaves properly at oo, and if the rotation of P' can beaccomplished without passing over any poles of the integrand, wemay deform P' back to P and obtainJm{(5.4.8)'r**+ ~Tp\sL-the same argument as before we(5.4.6) will be satisfied provided thatByJmg^e}?* =(5.4.9)Thusg()9boundaryconditions.stillknow from0,r real.see that the conditionsatisfies the conditionf real, positive.0,satisfiesg()nowthe conditionsthe function f(z z) constructed by(5.4.9),satisfythe functionif=1 g(se im ) e i(0 ds4*Jits(5.4.7)andaid will satisfy theAs we have already remarked, g() mustat oo, for example.
In addition,other conditionsearlier discussions in thisweand the preceding chapter thatnecessary to find two solutions (p(x, y) and <pi(x y)of our problemwhich are "out of phase at oo", in order that a linear combinationit is9them with appropriate timefactors will lead to a solution havingthe form of an arbitrary progressing wave at oo. In this connectionwe observe that if the path l of integration (as shown in FigureofPPPonly in reversal(it differs fromof direction of the portion in the upper half-plane), and if we defineits real part:<pi(x, y) as the imaginary part of / 1 (^, z) instead of5.4.3)istaken instead of the pathWAVES ON SLOPING BEACHES AND PAST OBSTACLES(5.4.10)x y)9= JmG(z)d2,9101=Gthe same integrand as before, then <pi(x, y), by the sameargument as above, will satisfy the boundary conditions providedthat the function g() also in this case satisfies the conditions (5.4.7)with-Fig. 5.4.3.The pathPlplanein the f -plane(5.4.9). It seems reasonable to expect that the integral over P lbehave the same as the integral over P when @te z is large andpositive (since the poles in the lower half-plane alone determine thisbehavior and the paths P and P l differ only in the upper halfplane) except that a factor i will appear, and hence that y> and <p lwill differ in phase at + oo (in the variable #, that is) by 90.
Thisdocs indeed turn out to be the case.Thus to satisfy the boundary conditions for both types of standingwave solutions we have only to find a function g() satisfying the conthe lastditions (5.4.7) and (5.4.9) which behaves properly at oocondition being needed in order that the path of integration can berotated in the manner specified in deriving (5.4.8)'. To this end wederive a functional equation for g() by making use of these con-andwillditions.(5.4.11)From(5.4.7)we have,U ~ ~ + i\ g(0 =while from (5.4.9)we haveclearly:~^-(?)*')real > Positive,WATER WAVES102ia)(5.4.12)g(C)*-= g(C*2iaa>>''^al, positive,,both by virtue of the reflection principle.
Eliminating g() from thetwo equations we obtain(5.4.13)real and positive, butThis functional equation was derived forsince g() is analytic it is clear that it holds throughout the domainof regularity of g(); it i g the basic functional equation for g(), asolution of which will yield the solution of our problem.
Of course,this equation is only a necessary condition that must be fulfilled ifthe boundary conditions are satisfied; later on we shall show that thesolution of it we choose also satisfies the condition (5.4.11 ), and hencethe condition (5.4.12) will also be satisfied since (5.4.13) holds,We proceed now to find a solution g() of (5.4.13) which has all ofthe desired properties needed to identify (5.4.4) and (5.4.10) asfunctions furnishing the solution of our problem, as has been done byPeters in the paper cited above.Wewhichtherefore proceed to treat the functional equation (5.4.13),is easily put in the form:2g( eZi<n<,tI,/__ *!4*<"(5.4.14)){(Cwith r 1>2The numbersItisr lt2 are real since->i)(->--- -we know that klies.betweenand1.convenient to set(5.4.15),(f,(Cin-which h(),real axis.likeg(),is^^ 2 )defined in the -plane slit along the negativewill have poles in the left half-plane, butThe function &()=*g> ++ Wi)(t=ir l and fir 2 of g(C) will be found toonly the poles at fcontribute a non- vanishing residue of f(z) for die z ->oo, and"this in turn would guarantee that f(z) behaves at GO on the free+surface like Ae~**.
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