J.J. Stoker - Water waves. The mathematical theory with applications (796980), страница 34
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as a symbol for a limit process in which theof the d-function until now, butpressure is first distributed over a segment the length of which isconsidered to grow small while the total pressure is maintained at theconstant value one. By inserting this expression for p in (6.7.3) andeliminating the quantity rj by making use of (6.7.2) the free surfacecondition is obtained in the formgVy(6.7.6)+ Vii - -i(*>wt*.6(x)e9t>0.Our problem nowwhichconsists in finding a solution <fj(x, y\ t) of (6.7.1)behaves properly at oo, and which satisfies the free surfacecondition (6.7.6) and the initial conditions (6.7.4).We proceed to solve the initial value problem by making use of theFourier transform applied to the variable x. The result of transforming(6.7.1)is-s*$(6.7.7)inwhich<p(s,y\ t) is theof the conditions at oo.*>areall+ ^,, =0,transform of y(x, y\The boundedof the formt)and use has been madesolutions of (6.7.7) for y< 0,WATER WAVES178(6.7.8)y(s, y;The transformnowis= A(s; t)e' vt).applied to the boundary condition (6.7.6),with the result:(6.7.9)g<p v+ fc, = - -4= ^7and on substitution ofA +g,A =(6.7.10)Theinitialttconditions (6.7.4)^(*;0)(6.7.11)The<p(s, 0; t)from(6.7.8)1=j,we0,findico-^ 7now="*, for<"".furnish for^(*;0)=A (s;t)the conditions0.solution of (6.7.10) subject to the initial conditions (6.7.11)iA(s;(6.7.12)fco r*=:t)=ta> (** T )-_^~VZnQ Jois,sinVgsrdr.Vgswe insert the last expression for A(s; t) in (6.7.8) and applythe inverse transform to obtain the following integral representationFinally,forour solution(6.7.13)<p(x 9qp(x, y; t)y9=t):ia)f000rcJoe* y/^uoU-t)cos sx^_JosinVgsrdrds.VgsThe fact that the solution is an even function of x has been used here.Our object now is to study the behavior of this solution as / -> oo.Since y is negative (we do not discuss here the limit as y -> 0,the behavior on the free surface) the integral with respect to si.e.converges well and there is no singularity on the positive real axis ofthe complex $-plane.
However, the passage to the limit t -> oo is morereadily carried out by writing the solution in a different form inwhich a singularity a pole, in fact then appears on the real axisof the 0-plane. (It seerns, indeed, likely that such an occurrence wouldbe the rule in any considerations of the present kind since the limitwould not usually be a function having a Fouriertransform, and one could expect that the limit function would somehow appear as a contribution in the form of a residue at a pole. ) It isfunction ast-> ooUNSTEADY MOTIONS179convenient to deform the path of integration in the $-plane into thepath L indicated in the accompanying figure.
The path L lies on thes-Fig. 0.7.1.planePath of integrationin s-planepositive real axis except for a semicircle in the upper half-plane ceno) 2 /g. By Cauchy's integral theorem this leavestered at the point s=the functionWe nowtpgiven in (6.7.13) unchanged.replace sinVgsr in (6.7.13) by exponentials and carryout the integration on i to obtain2p(a?,0;0(6.7.14)-~\eJLVgs Vgscos sxcb.*Q2 VgsVgs""IWenow to consider the three items in the bracket separately,we see, two of them do indeed have a singularity at s = o> 2 /gwishand, asis by-passed through our choice of the path L. The first twoitems arc rather obviously the result of the initial conditions andhence could be expected to pro\ide transients which die out aswhicht-> oo.
Thisisin fact the case, ascan be seen easilyin the followingway That branch of \/s is taken which is positive on the positive realaxis, and we operate always in the right half-plane. If, in addition,s is in the upper half-plane it follows that i(Vl&s db <*>) h as its real part:negative (o> being real). Consider now the contribution furnished bythe uppermost item in the square brackets. Since the exponential hasa negative real part on the semi-circular portion of the path L it isoo this part of the path makes a contributionclear that as t -++that tends to zero.
The remaining portions of L, whichlieon therealWATER WAVES180axis,are then readily seen to make contributions which die outcan be seen easily by integration by parts, for example,like 1/t: thisby application of knownorresultsabout Fourier transforms. Thesquare brackets has no singularity on the realthesothataxis,path L can be taken entirely on the real axis; thus,within accordancethe remarks just made concerning the similarmiddle itemin thesituation for the first item,outlike 1/t.Thuspresentation forfor largeit istclear that this contribution also dieswe obtain thefollowing asymptotic re-99:(6.7.15)L Ss~Actually, the right hand side is the solution of the steady stateproblem as obtained, for example, in the paper of Lamb [L.2] andby a different method by us in section 4.3 (although in a different form)when the condition at oo is the radiation condition stating that cpbehaves like an out-going progressing wave.
The steady state solutionas obtained in section 4.3 actually was a little more awkward to obtaindirectly through use of the radiation condition than it was to obtainthe solution (6.7.13) of the initial value problem. In particular, theasymptotic behavior of an integral representation had to be investi-gated in the former case also before the radiation condition could beused. Thus we have seen in this special case that the radiation condition can be replaced by boundedness conditions (in the space variables, that is) if one treats an appropriate initial value problem insteadof the steady state problem.Even though notstrictly necessarysince (6.7.15)knownistofurnish the desired steady state solution it is perhaps of interest toshow directly that the right hand side of (6.7.15) has the behaviorone expects for an out-going progressing wave when x -> + o.
Theprocedure is the same as that used in discussing (6.7.14): The factorcos sxisreplacedby exponentialsfir<p(x,y;t)~-e*\--\ J_L_fn(6.7.16)to obtainp*v e isx\*JL&-<*_1cfcCe sve~ lKX"I?-<fr+^2mJ L g*-a>* ]I-.as above one sees that the first integral makesa contribution that tends to zero as x ->oo. The second integral2is treated by deforming the path L over the pole s = o> /g into a pathwhich consists of the positive real axis except for a semi-circlein the lower half-plane. The contribution of the second integral thenBy the same argument+MUNSTEADY MOTIONS181Mconsists of the residue at the pole plus the integral over the pathBut the contribution of the latter integral is, once more, seen to tendto zero as x ->oo because of the factor e~ i8X Thus y(x, y; t) behaves+for large..x as follows:o>(6.7.17)<p(x,y)~/o> 22V(I)egt[e\ g\X(Ot]',SQand this does in fact represent a progressing wave in the positive^-direction which, in addition, has the wavelength 2jrg/o> 2 appropriateto a progressing sine wave with the frequency co in water of infinitedepth.6.8.
Justification of theMethod of Stationary PhaseIn section 6.5 the method of stationary phase was used (and it willbe used again later on) to obtain approximations of an asymptoticcharacter for the solutions of a variety of problems when thesesolutions are givenby means of=(6.8.1)integrals of theformCJaand the objectconstant kin soisisto obtainlarge. Sincemany importantan approximation valid when the realof such approximate formulasit seems worth while to give a mathema-we make usecases,the method of stationary phase, following a procedure due to Poincard.
The presentation given here is based upon thetical justification ofpresentation given by Copson [C.5].PoineareS's proof requires the assumption that<p(z)andy>(z)areregular analytic functions of the complex variable z in a domain<b of the real axis in its interior.xcontaining the segment S: a^(In what follows, we assume a and b to be finite, but an extension tothe case of infinite limits would not be difficult.) In addition <p(z) isassumed to be real when z is real. These conditions are more restricis necessary for the validity of the final result. For example,the function y might also depend on &, provided that y(x, k) is notstrongly oscillatory, or singular, for large values of k.
The assumptiontive thanis also not indispensable. However, these generalizawould complicate both the formulation and proof of our theoremwithout changing their essentials; consequently we do not considerof analyticitytionsthemhere.WATER WAVES182be shown thatthe main contributions to I(k) arise from thenearthat is, nearthosevaluesof x for which (p'(x)points ofthe points of stationary phase. The term of lowest order in the asymptotic development of I(k) with respect to k will then be obtained onIt will=Sthe basis of this observation. Kelvin himself offered a heuristic argument (cf. sec. 5 above) indicating why such a procedure should yieldthe desired result.Since<p'(z) isregular in thezeros are isolated.domain containingHence S can be dividedS,itfollows thatitsnumberofinto a finitesegments on which (p(z) has either one stationary point or no stationarypoint.
We shall show first that the contribution to I(k) from a segmentcontaining no stationary point is of order l/k. Next it will be shownthat a segment containing any given point of stationary phase can befound such that the contribution to the integral furnished by thesegment is of lower order than 1 /k and a formula for this contributionwill be derived. It turns out that this contribution of lowest order9independent of the length of the segment containing the pointof stationary phase, provided only that the segment has been chosenshort enough.
Once these facts have been proved, it is clear that theislowest order contributions to the integral arc to be found by addingthe contributions arising at each of the points of stationary phase.cSuppose, then, that (p(x) has no stationary point on a segmentxd of S. We may write^^-dxfJc-^Qinc^x^dbysince <p'(x)then leads to the result-^X]-hypothesis. Integration_.ik<p'(d)with tp^x)=ik(p'(c)d(y/y)-Sinceiby parts(ikj c<CdxJcbecause of the fact that kq>(x) is real, it follows that the integral in theabove expression is bounded.
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