J.J. Stoker - Water waves. The mathematical theory with applications (796980), страница 30
Текст из файла (страница 30)
Consider for this purpose the transform of d nfldx n and integrate by parts (which requires that d nf/dx nbe continuous):df-sieJIf the (nzero at(6.3.9)thatis,^l)-st derivative is to possess a transform itoo and hence we havedx ndx nByto~lthe transform of the n-th derivativeof the (nl)-st derivative.we obtain the resultmust tendis (is)times the transformrepeated application of this formulaWATER WAVES156=-(6.8.10)(isrlprovided that f(x) and its first n derivatives are continuous andthat all of these functions possess transforms.A rigorous justification of the transform technique used in thefollowing for solving problems involving partial differential equationsnot an entirely trivial affair (sec, for example, Courant-Hilbert[C.10], vol.
2, p. 202 ff.). Such a justification could be given, but weshall not carry it out here. Indeed, it would be reasonable to take theattitude that one may proceed quite formally provided that one verifies a posteriori that the solutions obtained in this way really satisfyisconditions of the problem. This is usually not too difficult to do,and, since the relevant uniqueness theorems are available, this courseallisperfectly satisfactory.Motions due to disturbances originating at the surfaceWe wish to determine first the motion in two dimensions due to thea^aapplication of an impulse over a segment of the surfacewhen the water is at rest in the equilibrium position.
Weat t =suppose the depth h of the water to be constant and that it extends to6.4.^xinfinity in the horizontal direction. The velocity potential 0(cc, y; t)must satisfy the following conditions. It must be a solution of theLaplace equation:(6.4.1)& xx+0yy =satisfying the-oo<o?<oo,0,-A^j/^0, *^0,boundary conditions(6.4.2)tt+ g$ y =0;y=y=-h,t0,>0and(6.4.3)Theyfirst ofsurface(6.1.6),(6.4.4)is=0,t^0.these conditions states that the pressure on the freet0.
As initial conditions we have, in view ofzero for(6.1.7),>and the assumed physical0(x, 0; 0)situation:= -- /(a?),Q(6.4.5)t(x, 0;0)=0,with I(x) the impulse per unit area applied to the free surface. InUNSTEADY MOTIONSwe must impose conditions attwo derivatives with respect to x,at oo in such athatThese are thatandand t should tend to zerooo.addition,its first157y,of these functions possess Fourier transforms with respect to x. This, in particular, requires that I(x) in(6.4.4) should vanish at oo.
Actually we consider only the specialwayallcase in whichT/.I(x)(6.4.6)== P(Ii.e.|x|const.,0,|'|xx||<a> a,the case in which a uniform impulse is applied to the segmenta, the remainder of the surface being left undisturbed.<solution 0(x, y; t) will now be determined by applying theFourier transform in x to the relations (6.4.1) (6.4.5) with the objectThe(as always in such problems) of obtaining a simpler problem for thetransform 0(s, y; t)(p(s, y; t).
Once the transform 99 has been foundthelatterby solvingproblem the inversion formula yields the solution=0.Webye~ i8Xbegin by applying the transform to (6.4.1), i.e. by multiplyingand integrating over the intervalooxoo; the result is- s*<p(8(6.4.7)9y; t)+<p yy (s,y; t)<-<in view of (6.3.10) and the assumed behavior ofat oo. (Clearly, italso necessary to suppose that the operation of differentiatingistwice with respect to y can be interchanged with the operation ofover the infinite interval.) This step already achievesintegratingone of the prime objects of the approach using a transform: the trans-form cpsatisfiesan ordinarydifferential equation(6.4.8)differential equation instead of the partialsatisfied by 0. The general solution of (6.4.7) isp(s, y,t)=A(s\t)e\*\v+B(s; t)e~^vterms of the arbitrary "constants" A(s; t) and B(s; t).
It is a simplematter to find the appropriate special solution that also satisfies thebottom condition (6.4.3), and from it to continue (just as is done inwhat follows) in such a way as to find the solution for water of uniformdepth. However, we prefer to take the case of infinite depth and toinoo. Theby the condition that V -+ when y ->forweobtainthatmusthavesoalsothisthenpropertyreplace (6.4.3)transform<p($,<py\ t) in this case the solutions(6.4.9)<p(s, !/;*)=A(s;t)e\*\*.WATER WAVES158The transformnext applied to the free surface condition (6.4.2) toisobtain(6.4.10)cp it+g(jp yand upon insertion of==y0,from<p($, 0; t)>0,t(6.4.9)wefind for A(s;thet)differential equationA +g\8\A=0,(6.4.11)t>0.tiFinally, the initial conditions must be taken into account.
The transform of (6.4.5) leads, evidently, to the condition0, andt (s; 0)=Athe solution of (6.4.11) satisfying this conditionA(s\(6.4.12)with a(s)have=q?(s, 0; 0)of I(x) as given3>(s, y; t)(6.4.4);0(8, y;*)0(x, y;P-t)= -7=\C(6.4.15)-(cf.r4~teand we have for(I/Q)!(S)I(s)e^v e isxwe have= 2Pcos (Vg\s\t) ds.forCac-\V2ncos sx dx=2 Pa:=-sin sa*!Jothe solution<Z>(#, t/; t)2Paas one can readily verify.(6.1.2)):=t).-a.(6.4.6))-a0(a?, y; t) =t)we/oo=eV2nJfinally for(6.4.16) ri(x;is,(6.4.4)of course, the transformthen leads immediately to the solution00fsin sa1-#=eSqil* v cos^ cosy(4X/,Vg* 0*.sa^rp Jo(from=From(6.4.4).7(*)*W*cos (Vg|*|(6.3.8)aV2nJand hencehence a(s)I=In our special caseT(*)by usingwhich I(s)y; t) the resultThe inversion formula(6.4.14)a(s) cos (Vg\s\t)(l/g)/(s) inby(f(s,(6.4.13)=to be determinedstill=t)isFor the-free surface elevation2 Pafpz limnQVg->oJo-^ e*we havesin sacossin(Vgst)Vsds.UNSTEADY MOTIONS159<One observes that the integrals converge well for all ybecause ofthe exponential factor e sy i.e.
everywhere except possibly on the freesurface. These formulas can now be used to obtain the solution for thecase of an impulse concentrated on the surface at x0; one need only-r->thatawhileinoosuchathatthe total imsupposeway2Patendsatofinite limit. For a unit total impulse we wouldpulse%=Pandthen obviously obtain for(6.4.17)(6.4.18)define(Wee s * cosKQ JoIf*=w(a?;---7-KQV8diverges for ylimsx cos (Vgst) ds,_e s v cos sx sin(Vgst)\/s ds.v^o Joas a limit for->since the integral obviously0.
This would, however, not be necessary in (6.4.16).)r\(x\ t)=_/10(x, y; t) =the formulas:77yoperating in the same way, one can easily obtain the solutionscorresponding to the case of an initial elevation of the free surface atBy=with no impulse applied. The only difference would be thatin (6.4.4) would be assumed to vanish whilewould bet in (6.4.5)different from zero. We simply give the result of such a calculation,timet0,but only for the limit case in which the initial elevationtrated at the origin. Forand r\ the formulas are:0(x, y;(6.4.19)(6.4.20)t)rj(x\ t)VS-^==nf*8e * cos(Vgst)JIf>ys_00- limconcen-ds/sx sinise*vcos sx cos(Vgst)ds.v-*0 JoThereisnodifficulty in treatingproblems having cylindrical sym-metry that are exactly analogous to the above two-dimensional cases.In these cases also one could begin with the solutions having symmetryof this type that are simple harmonic in the time (cf.
Chapter 3):(6.4.210(r,)with a 2distancegm(forVx*+z2j/; t)eMwater of infinite depth). Here the quantity r is thefrom the j/-axis, and J (mr) is the Bessel functionis regular at the origin. One could now build up moresolutionsby superposition of these solutions and satisfycomplicatedgiven initial conditions by using the Fourier-Bessel integral. This isof order zero thatthe method followed byLamb[L.3], p. 429.Instead of this procedure,WATER WAVES160one could make use of the Hankel transform in a fashion exactly analogous to the Fourier transform procedure used above (cf. Sneddon[S.ll], p.
290, and Hinze [H.15]). We content ourselves here withgiving the result for the velocity potential 0(r, y; t) and the surfaceelevation rj(r; t) due to the application of a concentrated unit impulseat the origin att(6.4.22)<P(r,y;t)(6.4.23)n(r; t)=0:=-1_/e syJ<>(sr) cos2nQJoNaturally=--_/oo1T(Vgst)slimZnQVg *-owe want to discussds,__eJ(sr) sin(Vgs12t)s*ds.Jothe character of the motions furnishedby the above relations, and in doing so we come upon a fact that holdsgood in all problems of this type: it is a comparatively straightforwardmatter to obtain an integral representation for the solution, but notalways an easy matter to carry out the details of the discussion of itsit is due to the factproperties.
The reason for this is not far to seekthat the solutions are given in terms of an integral over an integrandwhich is oscillatory in character and which changes rather rapidlyover even small intervals of the integration variable for importantranges in the values of the independent variables. Hence even a numerical integration would not be easy to carry out. The fact is that themotions are really of a complicated nature, as we shall see, and hencea mathematical description of them can be expected to present somedifficulties. Indeed, the phenomena under consideration here arcanalogous to the refraction and diffraction phenomena of physicaloptics and thus depend on intricate interference effects, which arefurther complicated in the present instances by the fact that the wavemotions are subject to dispersion, as we have seen in Chapter 3.Some insight into the nature of the solutions furnished by our formulas can be obtained by expanding the integrands in power seriesand integrating term by term (cf.
Lamb. [L.3], p. 385).* Thefor r/(x; t) as given by (6.4.20), for example, is found to be (for xItisclear that thereisa singularity for x=0,result>0):as one would expect.* Thesubsequent discussion in this section follows closely the presentationgiven by Lamb.UNSTEADY MOTIONSTheconverges for all values of the dimensionless quantitybut practically the series is useful only for small values ofseries2gt l2x,2161gt /2x,for small values ofi.e.t,or large values of x.One observesalsothat any particular "phase" of the disturbance such as a zero of rj>for examplemust propagate with a constant acceleration, since anysuch phase is clearly associated with a specific constant value of thequantity2gt /2x.It is in many respects more useful to find an asymptotic representation for the motion valid in the present case for large values ofthe quantity gt 2/2x, for which the power series are not very usefulbecause of their slow convergence.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
