Transient processes in a gas / plate structure in the case of light gas loading
aa r X i v : . [ phy s i c s . c l a ss - ph ] J u l Exchange pulse related to phase synchronism in gas —plate structure
M.A. Mironov, A.V. Shanin, A.I. Korolkov, K.S. KniazevaJuly 28, 2020
The paper is motivated by a simple experiment that can be conducted without any equipment.Hit abruptly a thin ∼ exchange pulse .The nature of the exchange pulse can be roughly explained as follows. The kick radiatesa flexural wave in the ice (the water loading under the ice can be taken into account, but itdoesn’t make much difference). The waves in the ice are described by the dispersion diagram ,i. e. by a surface in the coordinates ( ω, k ), where ω is the temporal circular velocity, and k isthe spatial wavenumber vector in the horizontal plane. The dispersion diagram for the elasticplate is non-trivial: ω ∼ | k | . At the same time, the sound pulse is carried to the observerby a wave in the air. The dispersion diagram for the sound in air is well-known: ω = c | k | .So, there are two dispersion diagrams for two subsystems participating in sound generationand propagation. The intersection points of these dispersion diagrams correspond to phasesynchronism of waves in ice and air, i. e. the waves near this point have common frequency (itis usually called the frequency of “coincidence”) and wavenumber, and thus they can interactefficiently. A transient phenomenon born by these crossing points is a quasi-monochromaticpulse with the frequency close to that of the intersection point.The duration of the exchange pulse depends on the distance x between the source and theobserver, and on the group velocities v , v of waves in ice and air. Namely, the exchangepulse can be heard for times x/v < t < x/v . This can be explained physically as follows:the wave runs some distance 0 < x ′ < x in the fast medium, then it is radiated into the slowmedium, and then it runs the distance x − x ′ there. The range of travel times for all possible x ′ is [ x/v , x/v ]. Indeed, v = c , and a detailed study shows that v = 2 c at the point of phasesynchronism.The wave excitation problem can be solved easily by the Fourier transform with respect totime and the in-plane spatial coordinates. As the result, the wave pulse becomes described as1 2D or 3D Fourier integral (the 3D integral can be converted to the 2D one by taking a trivialintegral with respect to the angular variable). The aim of this paper is obtaining an asymptoticestimation of the 2D Fourier integral.Two physical situations are considered here. The first first one is a thin 2D waveguide witha rigid wall and a flexible wall. The second physical situation is a flexible plate loaded by gason one side. The problem is also 2D. In both cases the gas loading is assumed to be light, i. e.the density of gas tends to zero. Such an assumption simplifies the consideration a lot. Thephysical system becomes split into two subsystem: the heavy one (the plate), and the lightone (the gas). The denominator of the integrand of the Fourier integral is becomes expressedas a product of two functions, each of which can be treated as a complex dispersion functionfor an isolated subsystem. The zero sets of the dispersion functions are the branches of thedispersion diagrams. The idea to study the dispersion diagram as a 2D complex function wasfirst introduced in [1, 2], and then developed in [3, 4, 5].The interaction of waves in a plate and a surrounding liquid or gas is a well-studied topic.Numerous results have been obtained for the plate driven by a transversal harmonic concen-trated force. The formal solution in terms of the Fourier integral can be found in [6, 7]. A basicanalysis of the integrals with the help of the saddle point method is performed in [8]. Particu-larly, an asymptotic estimation of the Fourier integral in the far field zone for the frequencies“below coincidence” has been found. The “free waves” corresponding to the poles of the Fourierintegral are discussed in [9]. It is shown for the case of line harmonic excitation that one of thereal poles can contribute to the far field, and it represents a subsonic surface wave. The complexpoles can lead to leaky waves for a certain range of parameters. An asymptotic expression fora leaky field of a lightly loaded plate at frequencies “above coincidence” is presented. A lightlyloaded membrane excited by a concentrated force is studied in [10]. Asymptotic results coveringall range of parameters are derived. In [11] some asymptotic results for a highly loaded plateare obtained. In [12] an intermediate regime of loading is studied (significant loading), whenthe fluid loading is heavy enough to affect the surface waves, but the inertia of the plate stillcan be considered as negligible. In [13] the admittance of fluid loaded membrane is evaluatedasymptotically. In [14, 15] the dispersion equation is approximated by a rational function. Theacoustical radiated field is calculated under this approximation. A thick plate (elastic layer)immersed in the fluid is studied in [16, 17, 18]. In [19, 20, 21, 22] plates governing by moresophisticated equations of motion are studied.There are several works devoted to transient processes in fluid-loaded elastic plates [23, 24].In [25] the wave radiated by an plate excited by a pulse are studied. An expression for a firstarrival pulse is presented. The exchange pulse is not discussed. In [26] energy of the exchangepulse is estimated. In [27, 28, 29] transient processes are studied numerically.The current paper is organized as follows. In Section 2 the problems of wave excitation ina gas / plate waveguide and in a plate loaded by a gas halfspace are formulated and solvedformally by Fourier transfromation. In Section 3 the problem for the integral for the waveguideis computed rigorously by residue integration. In Section 4 the integral for plate loaded bygas is estimated. For this, the crossings of branches of the dispersion diagram and the saddlepoints on the dispersion diagrams are considered. In Appendix the standard integrals used for2he asymptotic study are described. The geometry of the problem is as follows. A gas layer occupies the domain 0 < z < H inthe ( x, z )-plane (see Fig. 1, left). An elastic plate made of an isotropic material is the layer − h < z <
0. Assume that h is small and that the bending waves in the plate are described bythe linear thin plate theory [30].The plate is stress-free at the surface z = − h , the contact conditions are fulfilled at thesurface z = 0, and the surface z = H is acoustically hard (impenetrable). The source is appliedto the plate at the point (0 ,
0) from above. The profile is the Dirac’s delta-function. z H source x - h gasplate z source x gasplate Fig. 1: Geometry of Problem 1 (left) and Problem 2 (right)Describe waves in the gas by the acoustic potential φ ( t, x, z ). The acoustic potential islinked with the pressure p and the particle velocity components ( v x , v z ) by the relations [6] p = ρ ∂φ∂t , v x = − ∂φ∂x , v z = − ∂φ∂z , (1)where ρ is the (constant) gas density. The governing equation in the gas is the wave equation (cid:18) ∂ ∂x + ∂ ∂z − c ∂ ∂t (cid:19) φ ( t, x, z ) = 0 , (2)where c is the (constant) wave velocity in the gas.Let ζ ( t, x ) be the vertical displacement of the plate. The equation of motion of the platecan be written as [30] (cid:18) T ∂ ∂x + ρ p h ∂ ∂t (cid:19) ζ ( t, x ) + p ( t, x,
0) = f δ ( t ) δ ( x ) . (3)3ere ρ p is the density of the plate, T = Eh − ν ) (4)is the flexural stiffness of the plate ( E is the Young’s modulus of the elastic material, and ν isthe Poisson’s ratio), f is the amplitude of excitation.The second term in the left-hand side of (3) is responsible for the gas loading, and theright-hand side corresponds to the point source excitation.The boundary conditions for the air layer are ∂φ∂z (cid:12)(cid:12)(cid:12)(cid:12) z = H = 0 , (5) ∂φ∂z (cid:12)(cid:12)(cid:12)(cid:12) z =0 = − ∂ζ∂t . (6)The aim is to find the pressure in the gas near the plate, i. e. the function p ( t, x, φ and ζ should be equal to zero for t < x and the Laplace transform with respect to t :˜ w ( ω, k ) = 14 π ∞ Z −∞ ∞ Z w ( t, x ) exp {− ikx + iωt } dt dx, Im[ k ] = 0 , Im[ ω ] > . (7)The inverse of this transform is w ( t, x ) = ∞ Z −∞ ∞ + iǫ Z −∞ + iǫ ˜ w ( ω, k ) exp { ikx − iωt } dω dk, (8)where ǫ is an arbitrary positive parameter.Using these transformations, one can write down the acoustic pressure near the plate in theform of a double integral p ( t, x,
0) = ρf π ∞ Z −∞ ∞ + iǫ Z −∞ + iǫ ω cos( γH ) exp { ikx − iωt } γ sin( γH ) ( T k − ρ p hω ) + ω ρ cos( γH ) dω dk, (9)where γ = γ ( ω, k ) = p ω /c − k . (10)The value of the square root (10) is selected in such a way that it has a positive imaginary parton the whole integration plane of (8). One can see that this choice leads to continuous γ onthe integration plane. 4et us make two simplifications of (9). First, let ρ be small, i. e. let the gas loading ofthe plate be light. Some discussion of the conditions of a light, heavy and significant loadingfor a plate can be found in [9, 11, 12]. The second term in the denominator can be neglectedcomparatively to the first one on the whole integration plane: p ( t, x, ≈ ρf π T ∞ Z −∞ ∞ + iǫ Z −∞ + iǫ ω cos( γH ) exp { ikx − iωt } ( k − σω ) γ sin( γH ) dω dk, (11)where σ = ρ p h/T. (12)The second simplification for (9) is to assume that the air waveguide is narrow, i. e. to take | γH | ≪
1. The assumption yields cos( γH ) ≈
1, sin( γH ) ≈ γH , and p ( t, x, ≈ − ρf π T H ∞ Z −∞ ∞ + iǫ Z −∞ + iǫ ω exp { ikx − iωt } ( k − σω )( k − ω /c ) dω dk. (13)Physically, only the piston mode is propagating in the acoustic waveguide. Representation (13)is one of the simplest integrals possessing an exchange mode term in its asymptotics.The denominator of (13) has two factors, related to the plate and to the air piston mode.Introduce the complex dispersion diagrams d c1 , d c2 as the sets d c1 = { ( ω, k ) ∈ C | k − σω = 0 } , d c2 = { ( ω, k ) ∈ C | k − ω /c = 0 } . (14)The points of d c1 correspond to the plate mode, namely to a function ζ ( t, x ) = ζ e ikx − ı ωt , which is a solution of the equation of motion for an unloaded plate without sources. Thesimilar statement is valid for the set d c2 and the gas layer with Neumann walls. Indeed, d c2 isthe dispersion diagram for piston modes in such a layer.Introduce also the real dispersion diagrams d = { ( ω, k ) ∈ R | k − σω = 0 } , d = { ( ω, k ) ∈ R | k − ω /c = 0 } . (15)These sets play an important role in asymptotic estimation of integrals.Let ( ω ∗ , k ∗ ) be a crossing point of the diagrams d c1 and d c2 : ω ∗ = c σ / , k ∗ = cσ / . (16)All intersection of the dispersion diagrams d and d are as follows: d c1 ∩ d c2 = { (0 , , ( ω ∗ , k ∗ ) , ( − ω ∗ , − k ∗ ) , ( ω ∗ , − k ∗ ) , ( ω ∗ , − k ∗ ) } . (17)The real dispersion diagrams d and d and the crossing point ( ω ∗ , k ∗ ) are sketched in Fig. 2.5 k w k * * d d Fig. 2: Real dispersion diagrams d , d and the crossing point ( ω ∗ , k ∗ ) Let the gas occupy the half-plane z > −∞ < x < ∞ and the plate (the same as in theprevious subsection) is attached to the half-plane along the line z = 0 (see Fig. 1, right). Thewave process in the gas is described by the wave equation (2), and the plate is described byequation (3). Condition (5) is omitted, and (6) remains valid. It is not necessary to introducea radiation condition, since for a causal solution it should be valid automatically.Let us look for the acoustic pressure in the gas near the plate, i. e. at z = 0. The solutionof the problem can be obtained using the same method as above: p ( t, x,
0) = − ρf π ∞ Z −∞ ∞ + iǫ Z −∞ + iǫ ω exp { ikx − iωt } iT ( k − σω ) γ ( ω, k ) + ρω dω dk. (18)If the air loading is light, representation (18) can be simplified as follows: p ( t, x,
0) = iρf π T ∞ Z −∞ ∞ + iǫ Z −∞ + iǫ ω exp { ikx − iωt } ( k − σω ) γ ( ω, k ) dω dk. (19) The integral (13) can be easily computed analytically. The integrand is a regular function onthe real axis of k and near this real axis. For simplicity, assume that contour integration in the k -plane coincides with the real axis almost everywhere, and bypasses above the points 0 , ± k ∗ .Denote this contour by R + . This choice of the contour is arbitrary, and result remains the sameif the contour passes below the points 0 , ± k ∗ .First, take the internal integral in the ω -domain using the residue method. The contourof integration should be closed in the lower half-plane, since the factor e − iωt decays there for t >
0. For a fixed k , there are four poles ω = ± kc , ω = ± σ − / k , and all of them fall within6he closed contour. The result is a p ( t, x,
0) = I ( t, x ) + I ( t, x ) + I ( t, x ) + I ( t, x ) , (20)where I , ( t, x ) = ∓ iρf c πT H Z R + exp { ik ( x ∓ ct ) } k ( k − k ∗ ) dk, (21) I , ( t, x ) = ± iρf c πT Hσ / Z R + exp { ikx ∓ iσ − / k t } k − k ∗ dk. (22)The integrals I , can be taken using the residue method: I , ( t, x ) = 0 , x > ± ct, (23) I , ( t, x ) = ± ρf c T Hσ (2 − exp { ik ∗ x ∓ iω ∗ t } − exp {− ik ∗ x ± iω ∗ t } ) , x < ± ct. (24)The integrals I , can be expressed through the Fresnel’s integrals: I ( t, x ) = ρf c T Hσ (cid:18) e ik ∗ x − iω ∗ t C (cid:18) σ / ( x − ct )2 t / (cid:19) − e − ik ∗ x − iω ∗ t C (cid:18) σ / ( x + 2 ct )2 t / (cid:19)(cid:19) , (25) I ( t, x ) = ρf c T Hσ (cid:18) e − ik ∗ x + iω ∗ t ¯ C (cid:18) σ / ( x − ct )2 t / (cid:19) − e ik ∗ x + iω ∗ t ¯ C (cid:18) σ / ( x + 2 ct )2 t / (cid:19)(cid:19) , (26)where C ( a ) for real a is the Fourier integral defined by (108) in Appendix A3. Representations(25) and (26) follow from the non-trivial formula (107).The bar sign denotes the complex conjugation here and below.Let us analyze the solution. Introduce the “formal velocity” V ≡ xt . (27)Consider the asymptotics x → ∞ , V = const. Introduce also the value η = c / x / σ / (28)and the function F ± ( x, V ) = V / √ πx / σ / ( V ∓ c ) exp (cid:26) i ( V ∓ c ) σ / x V + i π (cid:27) (29)Use the asymptotics (109) and (110) for C ( ξ ) for | ξ | ≫
1. As the result, if | V − c | ≫ η theasymptotics is as follows: p ( t, x, ≈ ρf c HT σ ( u ( x, t ) + u ( x, t ) + u ( x, t )) + C.C. (30)7here C.C. denotes the complex conjugated terms, u ( x, t ) is the pulse mainly attributed tothe plate wave: u ( x, t ) = e ik ∗ x − iω ∗ t F + ( x, x/t ) − e − ik ∗ x − iω ∗ t F − ( x, x/t ) , (31) u ( x, t ) is the pulse mainly attributed to the piston wave in the gas waveguide: u ( x, t ) = (cid:26) , < x/t < c, , x/t > c, (32)and u is what we call here the exchange pulse : u ( x, t ) = (cid:26) exp { ik ∗ x − iω ∗ t } , c < x/t < c, , < x/t < c or x/t > c, (33)Note that p ( t, − x,
0) = p ( t, x,
0) due to the problem symmetry.In the intermediate zone | V − c | ∼ η the argument of function C is of order of 1. Noasymptotics of C can be used in this zone, so one should use the function C itself. For a fixedlarge x , the width of the intermediate zone in the t variable is as follows:∆ t ∼ x / c / σ / . (34)Thus, this width grows as x / . It is important that this width grows, but slower than linearly.The exchange pulse is a purely monochromatic pulse with a smooth front at V ≈ c and anabrupt front at V = c . The values V = c and V = 2 c are, thus, the boundaries of the domainoccupied by the exchange pulse in the ( x, V ) plane. These values emerged in our research quitenaturally in the process of getting the explicit solution. We should note that 2 c and c are thevalues of the group velocity of the branches of the dispersion diagram at the crossing point( ω ∗ , k ∗ ). Below we discuss this feature in details and demonstrate that the exchange pulsealways have fronts linked to the group velocities of the interacting modes.Note also that the field is continuous at x = ct . Unfortunately, the integral (19) cannot be taken explicitly. So our next aim is to develop atechnique of evaluation of such an integral in a general situation. As above, we fix the value V = x/t and take x → ∞ to build the asymptotic procedure. The idea of the technique isquite standard: the surface of integration should be deformed in such a way that the integrandis exponentially small everywhere except neighborhoods of several “special points”. Thesepoints are the saddle points on the real dispersion diagrams and the crossing points of thedispersion diagrams. The integrals over the neighborhoods of the “special points” can be takenapproximately (asymptotically). Some typical integrals of this sort are listed in the Appendix.8 .1 Overview of the surface deformation procedure At each point of the branches d c j of the dispersion diagrams one can define the group velocity v gr = dωdk . (35)The group velocities at the points of the real branches d j are real.The saddle points on the real branches of the dispersion diagrams are the points at which v gr = V . Let us find the position of the saddle point on d , for which ω = σ − / k . The groupvelocity is v gr = 2 σ − / k. (36)The saddle point has coordinates ( ω s , k s ) with ω s = ω s ( V ) = σ / V , k s = k s ( V ) = σ / V . (37)The main statement of the estimation procedure is as follows: For almost all values of V , the terms of the field not decaying exponentially as x → ∞ are produced by the fragments of the integration surface that are located in neighborhoods ofeither the saddle points on the real branches, or the crossing points of the real branches of thedispersion diagrams. The idea of the proof is as follows. Consider the integral (19) as an integral of an analytic differential 2-form over some surface (smooth manifold) [31]: u ( x, V ) = Z Γ Ψ , (38)where Ψ = exp { ix ( k − ω/V ) } ( k − σω ) γ ( ω, k ) dω ∧ dk, (39)and Γ is an oriented manifold Im[ k ] = 0, Im[ ω ] = ǫ .According to the 2D Cauchy’s theorem, one can deform continuously the manifold Γ , andthe value of the integral should remain the same if the manifold does not cross the singular sets d c j of the integrand form during the course of deformation.To estimate the integral, one has to find a deformation of Γ into a new manifold Γ , suchthat the integrand is exponentially small almost everywhere, i. e. such that x (Im[ k ] − Im[ ω ] /V ) ≫ . (40)The neighborhoods on Γ where one cannot fulfill the inequality (40) are used to build theestimation of the wave field components.We consider only small deformations of Γ . Since x is large, a small deformation is enoughto fulfill (40). 9et the deformed integration manifold Γ be parametrized by real parameters ( ω ′ , k ′ ) asfollows: Γ : ω = ω ′ + if ( ω ′ , k ′ ) , k = k ′ + if ( ω ′ , k ′ ) , (41)where f , f are some smooth real functions. The deformation process can be described usinga parameter χ ∈ [0 , χ ) : ω = ω ′ + iχf ( ω ′ , k ′ ) + i (1 − χ ) ǫ, k = k ′ + iχf ( ω ′ , k ′ ) . (42)One can see that Γ(0) = Γ , Γ(1) = Γ . Let ( ω , k ) be a real point belonging to the dispersion diagram d j . Consider a small complexneighborhood of ( ω , k ). Let us describe the eligible deformations of Γ in this neighborhood,i. e. the deformations in which the manifold of integration does not cross d c j .Let the group velocity of the corresponding branch of the dispersion diagram at ( ω , k ) beequal to v . Note that a point ( ω, k ) in this neighborhood belongs to the complex branch of thedispersion diagram only if Im[ k ] ≈ Im[ ω ] v (43)(indeed, this is not a sufficient condition). This follows from the fact that k − k ≈ ω − ω v , (44)and (43) is the imaginary part of (44).Let the deformation of Γ in the neighborhood of ( ω , k ) be chosen in such a way that thefunctions f , f are locally constant. Consider the conditions (43) and (40) graphically. Let be V > v . Take the coordinate plane (Im[ ω ] , Im[ k ]). The fragment of the initial manifold Γ isshown in this plane as a single point ( ǫ, is shown also as a single point ( f ( ω , k ) , f ( ω , k )). The deformation process is the motionalong the segment connecting these points (see Fig. 3, left).The inequality (40) is fulfilled is the resulting point ( f , f ) is located above the line Im[ k ] =Im[ ω ] /V (with a margin of the height equal to 1 /x ). The line Im[ k ] = Im[ ω ] /V is shown bythe bold solid line in the figure.The manifold Γ( τ ) can hit the singularity only if the segment [Γ , Γ ] crosses the line onwhich (43) is valid. This line is shown dotted in the figure. This means that if the segment[Γ , Γ ] does not cross the dotted line, then the deformation is eligible.An inverse statement is much more subtle, but it also can be proven: If the segment [Γ , Γ ]does cross the dotted line in the diagram, then Γ( τ ) does cross the singularity set at somepoints, and the deformation is not eligible. 10 m[ ] k Im[ ] w Im[ ] =Im[ ] k V w /
Im[ ] =Im[ ] k w / v V > v ( ,0) e ( , ) f f Im[ ] k Im[ ] w V < v " " type deformation + " " type deformation - Im[ ] k Im[ ] w "0" type deformation GG G G G G G Fig. 3: Types of deformation of Γ One can see that the eligible deformation in the case
V > v looks like it is shown in Fig. 3,left. The point Γ belongs to the first quadrant of the coordinate plane. This deformation willbe referred to as a deformation of the “+” type.If V < v then the eligible deformation is as shown in Fig. 3, center. The point Γ nowbelongs to the third quadrant of the coordinate plane. This deformation will be referred to asthe deformation of the “ − ” type.For the “special points” (the saddle points or the crossing points of the dispersion diagrams)we need a deformation that does not obey (40), but still does not cross the singularities. Sucha deformation is called “0” type, and is shown in Fig. 3, right.Let v grow continuously from v < V to v > V . The point Γ then goes far into the firstquadrant. One can see that the “+” type deformation cannot be transformed continuouslyinto the “ − ” type deformation. Therefore, between the domains of “+” type deformationand “ − ” type deformation there should be a zone with “0” type deformation. This is a veryimportant conclusion, since the domains with “+” or “ − ” type deformation do not produce non-vanishing field components, while the domains with the “0” type can produce such components.Finally, if v <
0, only a “+” type deformation is eligible, and it looks as shown in Fig. 4.The deformation should be made carefully only near the real branches of the dispersiondiagram. We assume that an eligible smooth deformation can be easily found in the rest of the( ω ′ , k ′ )-plane, since there are no obstacles for deformation at such places. This is why, belowwe indicate the type of the deformation only for the neighborhoods of the real branches of thedispersion diagram. In this subsection we analyze the deformation of the integration surface for (19) in terms of thetypes of the deformation introduced above. Note that the singular sets of (13) and (19) are thesame, so the conclusions obtained here can be applied to (13) as well, and the exact solution is11 m[ ] k Im[ ] w Im[ ] =Im[ ] k V w /
Im[ ] =Im[ ] k w / vv < +G G Fig. 4: Deformation of Γ when v < d and d .They are crossing at five points (17). The group velocity takes all values from −∞ to ∞ on d .For each V = 0, there exist two saddle points on d . On d the group velocity takes values ± c only.There are five crossing points of the dispersion diagrams. The group velocity on d is equalto ± c at the non-zero crossing points.Let be V > c . Fig. 5, left, shows how the deformation type should be chosen in this case.The saddle points on d are marked by symbol “s”. The “+”, “ − ”, and “0” types of surfacedeformation are marked by corresponding signs. w k ss 0 -- ++++ + ++++ + +++ +++ V c > 2 d d + w k ss 0 -- +++ + ++++ + +++ +++ c < V c < 2 0 - - w * Fig. 5: Deformation diagram for
V > c (left), and for c < V < c (right)Consider the saddle point with positive ω and k . The group velocity on d is bigger than V to the right of the saddle point and is smaller than V to the left of the saddle point. Thus, near d , one should choose the “ − ” type of the surface deformation to the right, and “+” type ofthe deformation to the left. Since the deformation should be continuous, one should choose the120” type of the deformation in the neighborhood of the saddle point. Thus, the saddle pointson d produce non-vanishing components of the wave field.Indeed, there is some continuous transition between “0” type and the “+”/“ − ” types.All other branches are labeled with “+” type deformation. In particular, for each of fivecrossing points, all branches crossing at them are of the “+” type. Therefore, the vicinities ofthe crossing points can be shifted according to the “+” type, and the crossing points do notproduce non-vanishing field components.Let be c < V < c . Corresponding deformation diagram is shown in Fig. 5, right. Theneighborhood of the crossing point ( ω ∗ , k ∗ ) changes its status comparatively to Fig. 5, left.The branch d should be deformed according to the “ − ” type, while d should be deformedaccording to the “+” type. Since the deformation is continuous, the neighborhood of thecrossing point should be of the “0” type. Therefore, the the crossing point produces a non-vanishing field component. Besides, the saddle point on d (also deformed according to the“0” type) produces another field component.Consider the case 0 < V < c . The diagram is shown in Fig. 6. The neighborhood ofthe crossing point ( ω ∗ , k ∗ ) can now be deformed according to the “ − ” type, and thus it doesnot produce non-vanishing field components. The crossing point (0 ,
0) should be deformedaccording to the “0” type, since one branch crossing at this point is deformed according to the“ − ” type, while all other branches are deformed according to the “+” type. Besides, there aresaddle points on d that also produce non-vanishing components. w k ss 0 -- + + + +++++ +++ < V c -- ---- Fig. 6: Deformation diagram for 0 < V < c
When V ≈ c , the saddle point is close to the crossing point of the branches of the dispersiondiagram. This case produces an intermediate asymptotics and it should be considered in aspecial way. The diagram is shown in Fig. 7, left. One can see that the zone of “0” typedeformation covers both the crossing point and the saddle pointThe most sophisticated case is V ≈ c . In this case, the whole neighborhood of the branch d is tangential to the line k = ω/V (see Fig. 7, right), and thus it cannot be analyzed locally.Let us summarize this diagram consideration. The saddle point on d produces a non-vanishing component for all V , the crossing point (0 ,
0) produces a non-vanishing component13 k ss 0 -- +++ + ++++ + +++ +++ d d cV » w k ss 0 -- + + ++++++ +++ d d cV » -- Fig. 7: Deformation diagram for V ≈ c (left), and for V ≈ c (right)only for 0 < V < c , and the crossing point ( ω ∗ , k ∗ ) produces a non-vanishing component onlyfor c < V < c .One can see that corresponding wave components are analogous to u , u , and u from (30)for the integral (13). The last component is the exchange pulse. Thus, the qualitative analysisbased on deformation of the integration surface is in agreement with the exact solution. Remark.
The case
V /c ≈ According to the analysis of the surface integral deformation, if V is not close to 2 c , c , or 0,one can expect to obtain an asymptotic estimation of the field in the form p ( t, x, ≈ u ( t, x ) + ¯ u ( t, x ) + u ( t, x ) + u ( t, x ) + ¯ u ( t, x ) , (45)where the terms in the right are the following wave components: u is produced by the saddle point on d with positive ω and k ,¯ u is produced by the saddle point on d with negative ω and k , u is produced by the crossing point of dispersion diagrams at (0 , u is produced by the crossing point of dispersion diagrams at ( ω ∗ , k ∗ ),¯ u is produced by the crossing point of dispersion diagrams at ( − ω ∗ , − k ∗ ).The terms u and ¯ u are non-zero for all V , the term u is non-zero for 0 < V < c , theterms of the exchange pulse u and ¯ u are non-zero for c < V < c .14n the zone V ≈ c the estimation of the field can be found in the form p ( t, x,
0) = u c ( t, x ) + ¯ u c ( t, x ) , (46)where u c is the term produced by the crossing point ( ω ∗ , k ∗ ) and the neighboring saddle point.In the zone V ≈ c the estimation of the field is as follows: p ( t, x,
0) = u ( t, x ) + ¯ u ( t, x ) + u c ( t, x ) , (47)where u ( t, x ) is the saddle-point term introduced above, and u c ( t, x ) is the term produced bythe elongated zone located along the branch k = ω/c in Fig. 7, right.The width of the zone V ≈ c can be estimated using a standard reasoning based on theconcept of the “domain of influence” [32]. Namely, V belongs to the intermediate zone if thephase difference between the crossing point ( ω ∗ , k ∗ ) and the saddle point ( ω s , k s ) is of order of 1.This happens if x | ( k ∗ − ω ∗ /V ) − ( k s ( V ) − ω s ( V ) /V ) | ∼ . (48)Using (37), one can get the estimation | V − c | ∼ η , where η is defined by (28), so (34) is valid.To have a consistent set of asymptitotic expansions, the term u c should have asymptotics u c ≈ u for V − c ≫ η, (49) u c ≈ u + u for 2 c − V ≫ η. (50)Similarly, one can estimate the width of the intermediate zone V ≈ c . The condition is asfollows: the phase difference between the points (0 ,
0) and ( ω ∗ , k ∗ ) should be of order of 1: x | k ∗ − ω ∗ /V | ∼ , (51)resulting in | V − c | ∼ δv, δv = σ − / x − . (52)For consistency, function u c should have the following asymptotics: u c ≈ u + ¯ u for V − c ≫ δv, (53) u c ≈ u for c − V ≫ δv. (54) Estimation of u Approximate the algebraic factors of the integrand of (19) near ( ω s , k s ) as follows: ω ( k − σω ) p ω /c − k ≈ c σ √ V − c k − σ / ω / . Evaluation of corresponding integral is shown in Appendix A2. The parameter µ is equal to 1.The result of application of formula (93) is u ( t, x ) = ρf cV / π / T σ / x / √ V − c exp { ix ( k s − ω s /V ) + 3 πi/ } . (55)15 stimation of u Approximate the algebraic function in the integrand near the origin as follows: ω ( k − σω ) p ω /c − k ≈ iσ k − ω/c ) / ( k + ω/c ) / . (56)This approximation can be used if V is not very small comparatively to c . The integral ZZ exp { ikx − iωt } ( k − ω/c ) / ( k + ω/c ) / dω dk can be estimated as the integral (73) from Appendix A1. The parameters for the integral are µ = µ = 1 / ω † = 0, k † = 0, v = c , v = − c . Formula (88) can be applied. The result is u ( t, x ) = − ρf cVπT σx √ c − V . (57) Estimation of u Estimate the algebraic function in the integrand near the point ( ω ∗ , k ∗ ) as follows: ω ( k − σω ) p ω /c − k ≈ c / √ σ / p ω/c − k ( k − σ / ω / ) ≈ (58) c / √ iσ / p k − ω/c ( k − ω/ (2 c ) − c √ σ/ . To estimate the integral, one can use the method described in Appendix A1. This is theintegral of the type (73) with µ = 1, µ = 1 / ω † = ω ∗ , k † = k ∗ , v = 2 c , v = c . Formula(87) can be used. The result is u ( t, x ) = ρf c / V / exp { ik ∗ x − iω ∗ t + 3 πi/ } √ π / T σ / x / √ c − V . (59)
Estimation of u c Assume that V ≈ c . Use the first approximation (58). Apply the procedure of estimationdescribed in Appendix A3. The standard integral is (95) with ω † = ω ∗ , k † = k ∗ , v = 2 c , v = c , α = (8 σ / c ) − .The estimation of the integral is given by (115). The result is u c ( t, x ) = iρf c / / πT σ / x / exp { ik ∗ x − iω ∗ t } B (cid:18) x / σ / (2 c − V ) √ c / (cid:19) (60)with B given by (111) and (106).The asymptotics (112) and (113) provide (49) and (50).16 .5 Non-local estimation of (19) for V ≈ c Consider the case V ≈ c . Our aim is to build an estimation of the integral over a long spotmarked by “0” sign and located along the branch k = ω/c in Fig. 7, right. In other words, theaim is to compute u c from (47).Let be t = x/c + δt. (61)Change the order of integration in (19) and take the integral with respect to k for a fixed ω .For this, shift the integration contour into the upper half-plane since the factor e ikx decaysthere. Note that there are two poles and a branch point in the upper half-plane of k . Thepoles are k = k ( ω ) = σ / ω / and k = ik ( ω ), while the branch point is k = ω/c . After thedeformation, the contour will look as shown in Fig. 8. Re[ ] k Im[ ] k w / ck ( ) w k ( ) w ik ( ) w Fig. 8: Deformation of integration contour in the k -domainThe integral can be written as p ( t, x,
0) = u r1 ( t, x ) + u r2 ( t, x ) + u b ( t, x ) , (62)where the first two terms are the residue integrals: u r1 ( t, x ) = − ρf c πT σ / ∞ + iǫ Z −∞ + iǫ exp { ik ( ω ) x − iωx/c − iω δt ) }√ ω − ω ∗ dω, (63) u r2 ( t, x ) = − iρf c πT σ / ∞ + iǫ Z −∞ + iǫ exp {− k ( ω ) x − iωx/c − iω δt ) }√ ω + ω ∗ dω, (64)The branch point integral can be estimated for large x : u b ( x, t ) ≈ ρf e iπ/ c / √ π / T x / ∞ + iǫ Z −∞ + iǫ exp {− iω δt } ( ω − ω ∗ ) √ ω dω. (65)The contours for the first two integrals can be deformed as it is shown in Fig. 9 and estimatedfor large x . The result of the estimation for each integral is a sum of a saddle point term and17 e[ ] w Im[ ] w w * Re[ ] w Im[ ] w-w * Fig. 9: Contour integration for estimation of u r1 (left) and u r2 (right) in the ω -domainof the branch cut term. We are interested only in the branch cut term, since the saddle pointterms are equal to u and ¯ u for V = c . Denote the branch cut terms of u r1 and u r2 by u r1b and u r2b , respectively. Their estimations are u r1b ( t, x ) ≈ − ρf c / exp {− iω ∗ δt − iπ/ } √ T π / σ / x / , u r2b ( t, x ) = ¯ u r1b ( t, x ) (66)Estimate the integral (65). If δt < u b = 0. If δt > ω = ω ∗ and ω = − ω ∗ be u br1 and u br2 , respectively. Let thebranch cut integral be denoted by u bb : u b = u br1 + u br2 + u bb . (67) Re[ ] w Im[ ] w w * -w * e g - Fig. 10: Contour integration for estimation of (65) in the ω -domainA detailed computation yields u br1 ( t, x ) = − u r1b ( t, x ) , u br2 ( t, x ) = − u r2b ( t, x ) , (68) u bb ( t, x ) = ρf e iπ/ c / √ π / T x / Z γ − exp {− iω δt } ( ω − ω ∗ ) √ ω dω. (69)The last expression can be rewritten as u bb ( t, x ) = − ρf c / √ π / T x / σ / E ( σ / c δt ) , (70)18here E ( a ) = ∞ Z exp {− ξa } ( ξ + 1) √ ξ dξ. (71)Some properties of function E ( a ) are displayed in Appendix A4.Finally, u c = (cid:26) u r1b + ¯ u r1b , t < x/c,u bb , t > x/c (72)From (66) if follows that u r1b ( t, x ) = u ( t, x ) for V = c. This means that there is no intermediate zone before the front t = x/c , i. e. the representation p = u + u + C.C. remains valid for all t < x/c even for small x/c − t . Indeed, this guarantees(53).The property (117) can be used to establish validity of (54).The width of the intermediate zone is δt ∼ σ − / c − . This agrees with (52).The property (118) yields that the field u c is continuous at t = x/c . This feature clarifiesthe meaning of the asymptotics (70). One can see that the asymptotics (57) for u is singularat V = c . However, the intermediate zone matches u with the exchange pulse u + ¯ u withoutmaking the field singular.The scheme of all asymptotics is shown in Fig. 11. This scheme shows the zones of validityof asymptotics in the t -domain for a large fixed x . As we noted, the scheme does not coversmall values of V (i. e. large values of t ). The matching rules (49), (50), (53), (54) guaranteethe consistency of the scheme. x c / (2 ) x c / t uu + cc uu + uuuu +++ uuu ++ bb11 uuu ++ ? ~ -- c s ~ xc -- s exchange pulse Fig. 11: Zones of validity of different asymptotics for a fixed x Concluding remarks
The paper can be summarized as follows. The integrals (13) and (19) are analyzed. Bothintegrals describe non-stationary wave processes in two-component systems with a high densitycontrast between the subsystems. The expressions (13) and (19) are 2D Fourier integrals. Thedenominators of the integrands have zero sets that are crossing.The paper contains an exact computation of (13) and an asymptotic estimation of (19).In both cases we find a monochromatic pulse existing for x/ (2 c ) < t < x/c . The frequencyand the wavenumber of this pulse are equal to those of the point of phase synchronism of thesubsystems. That is why, we call it the exchange pulse.A general scheme of estimation of a 2D Fourier integral is presented. The procedure ofestimation is based on the fact that the field components are produced by saddle points on thebranches of the dispersion diagrams and by crossing points of the branches of the dispersiondiagrams. A set of “standard integrals” related to such points are given in Appendix.The work can be continued in three directions. First, one can consider avoiding–crossingsinstead of crossings of the dispersion diagrams. As it is known [12, 26], this is a more realisticsituation emerging when the contrast between subsystems is not very high. Second, one canstudy the field at the observation point not close to the horizontal surface. This leads toappearance of the factor exp { iγz } , which makes the estimation procedure different. Third,one can introduce and study more “standard integrals”. For example, finding the asymptoticsof (19) for V /c ≈ eferences [1] Philip Wayne Randles. Modal Representations for the High-Frequency Response of ElasticPlates . PhD thesis, 1969.[2] P.W. Randles and J. Mlklowitz. Modal representations for the high-frequency response ofelastic plates.
International Journal of Solids and Structures , 7(8):1031–1055, aug 1971.[3] A. V. Shanin. Precursor wave in a layered waveguide.
The Journal of the Acoustical Societyof America , 141(1):346–356, jan 2017.[4] A. V. Shanin, A. I. Korolkov, and K. S. Knyazeva. Multi-contour saddle point method ondispersion diagrams for computing transient wave field components in waveguides. In . IEEE, aug 2018.[5] A.V. Shanin, K.S. Knyazeva, and A.I. Korolkov. Riemann surface of dispersion diagramof a multilayer acoustical waveguide.
Wave Motion , 83:148–172, dec 2018.[6] P.M. Morse and K.U. Ingard.
Theoretical acoustics . New York: McGraw-Hill, 1968.[7] M.C. Junger and D. Feit.
Sound, structures, and their interaction , volume 225. MIT pressCambridge, MA, 1986.[8] L.Ya. Gutin. Sound radiation from an infinite plate excited by a normal point force.
Sov.Phys. Acoust , 10(4):369–371, 1965.[9] D.G. Crighton. The free and forced waves on a fluid-loaded elastic plate.
Journal of Soundand Vibration , 63(2):225–235, mar 1979.[10] D.G. Crighton. The green function of an infinite, fluid loaded membrane.
Journal of Soundand Vibration , 86(3):411–433, feb 1983.[11] D.G. Crighton. The modes, resonances and forced response of elastic structures underheavy fluid loading.
Philosophical Transactions of the Royal Society of London. Series A,Mathematical and Physical Sciences , 312(1521):295–341, oct 1984.[12] C.J. Chapman and S.V. Sorokin. The forced vibration of an elastic plate under significantfluid loading.
Journal of Sound and Vibration , 281(3-5):719–741, mar 2005.[13] D.G. Crighton. Approximations to the admittances and free wavenumbers of fluid-loadedpanels.
Journal of Sound and Vibration , 68(1):15–33, jan 1980.[14] D.T. DiPerna and D. Feit. An approximate analytic solution for the radiation from a line-driven fluid-loaded plate.
The Journal of the Acoustical Society of America , 110(6):3018–3024, dec 2001. 2115] D.T. DiPerna and D. Feit. An approximate green’s function for a locally excited fluid-loaded thin elastic plate.
The Journal of the Acoustical Society of America , 114(1):194–199,jul 2003.[16] S.I. Rokhlin, D.E. Chimenti, and A.H. Nayfeh. On the topology of the complex wavespectrum in a fluid-coupled elastic layer.
The Journal of the Acoustical Society of America ,85(3):1074–1080, mar 1989.[17] A. Freedman. Anomalies of the a0leaky lamb mode of a fluid-loaded, elastic plate.
Journalof Sound and Vibration , 183(4):719–737, jun 1995.[18] S.V. Sorokin. Analysis of time harmonic wave propagation in an elastic layer under heavyfluid loading.
Journal of Sound and Vibration , 305(4-5):689–702, sep 2007.[19] D. Feit. Pressure radiated by a point-excited elastic plate.
The Journal of the AcousticalSociety of America , 40(6):1489–1494, dec 1966.[20] A.D. Stuart. Acoustic radiation from submerged plates. I. influence of leaky wave poles.
The Journal of the Acoustical Society of America , 59(5):1160, 1976.[21] A.D. Stuart. Acoustic radiation from submerged plates. II. radiated power and damping.
The Journal of the Acoustical Society of America , 59(5):1170, 1976.[22] J.D. Smith. Symmetric wave corrections to the line driven, fluid loaded, thin elastic plate.
Journal of Sound and Vibration , 305(4-5):827–842, sep 2007.[23] E.B. Magrab and W.T. Reader. Farfield radiation from an infinite elastic plate excited by atransient point loading.
The Journal of the Acoustical Society of America , 44(6):1623–1627,dec 1968.[24] A.D. Stuart.
Acoustic radiation from a point excited inifinite elastic plate . PhD thesis,The Pensilvania State University, 1972.[25] Mackertich S.S. and S.I. Hayek. Acoustic radiation from an impulsively excited elasticplate.
The Journal of the Acoustical Society of America , 69(4):1021–1028, apr 1981.[26] A.V. Akol’zin and M.A. Mironov. Energy flows in media caused by a normally excitedplate near the frequency of coincidence (in russian).
Trudy nauchnoi scholy prof S. A.Rubaka , 2:90, 2001.[27] J.H. James. Sound radiation from infinite thin plate. Technical Report ARE TM(UHA)87508, Admiralty research establishment, 1987.[28] R. Scherrer, L. Maxit, J.-L. Guyader, C. Audoly, and M. Bertinier. Analysis of the soundradiated by a heavy fluid loaded structure excited by an impulsive force.
Internoise 2013 ,09 2013. 2229] A. Langlet, M. William-Louis, G. Girault, and O. Pennetier. Transient response of aplate–liquid system under an aerial detonation : Simulations and experiments.
Computers& Structures , 133:18–29, mar 2014.[30] L.D. Landau.
Theory of Elasticity , volume 7. Elsevier LTD, Oxford, 2004.[31] B.V. Shabat.
Introduction to complex analysis . American Mathematical Society, 1992.[32] V.A. Borovikov.
Uniform stationary phase method . Institution of Electrical Engineers,London, 1994.
Appendix. Standard local integrals
A1. Crossing of two singular sets
Consider the integral I ( x, V ) = ∞ Z −∞ ∞ + iǫ Z −∞ + iǫ exp { ix ( k − ω/V ) } ( k − ψ ( ω )) µ ( k − ψ ( ω )) µ dω dk (73)with ψ , ( ω ) = k † + ω − ω † v , , (74)where ω † , k † , v , v are some real values. One can see that the integrand of (73) has two singularsets (lines): k = ψ , ( ω ) . The lines are crossing at the point ( ω † , k † ). The values v and v are group velocities of the“dispersion diagrams” k = ψ , ( ω ) at the crossing point. Let be v > v . We do not assumethat v and v are positive.Let us find the asymptotics of I as V = const > x → ∞ .Real parameters µ , determine the type of singularities. We are particularly interested inthe cases µ = 1 (a polar set), or µ = 1 / · ) µ is positive real if the argument is positive real, and that thisfunction is continuous on the integration surface.Introduce the variables η , = k − ψ , ( ω ) = ( k − k † ) − ω − ω † v , . (75)The integral I can be rewritten as I ( x, V ) = v v exp { ix ( k † − ω † /V ) } v − v × (76)23 − iǫ/v Z −∞− iǫ/v exp (cid:26) ix ( V − v ) v η V ( v − v ) (cid:27) dη η µ ∞− iǫ/v Z −∞− iǫ/v exp (cid:26) ix ( v − V ) v η V ( v − v ) (cid:27) dη η µ . Note that the integration contour in the plane η j , j = 1 ,
2, passes below the real axis if v j > v j <
0. Using this fact, close the contours of integration inappropriate half-planes and obtain that I ( x, V ) = 0 if V < v or V > v . (77)If v < V < v , then I ( x, V ) = v v exp { ix ( k † − ω † /V ) } v − v I c (cid:18) µ , x ( V − v ) v ( v − v ) V (cid:19) I c (cid:18) µ , x ( v − V ) v ( v − v ) V (cid:19) , (78) I c ( µ, a ) ≡ Z γ exp { iaξ } ξ µ dξ, (79)where γ = γ + if a > γ = γ − if a <
0. Contours γ + and γ − are shown in Fig. 12. Re[ ] h Im[ ] h g + Re[ ] h Im[ ] h g - Fig. 12: Deformed contours of integration in the η -planeCompute I c ( µ, a ). If µ is not a positive integer then I c ( µ, a ) = e πi (1 − µ ) / (1 − e πiµ ) a µ − Γ(1 − µ ) if a > , (80) I c ( µ, a ) = e πi ( µ − / (1 − e − πiµ )( − a ) µ − Γ(1 − µ ) if a < , (81)Γ( · ) is the Gamma-function.If µ is a positive integer than I c ( µ, a ) = 2 πe πiµ/ a µ − ( µ − a > , (82) I c ( µ, a ) = − πe πiµ/ a µ − ( µ − a < . (83)In particular, for positive a I c (1 , a ) = 2 πi, (84)24 c (1 / , a ) = 2 e πi/ √ π √ a . (85)Let be v > V > v > µ = µ = 1. Then I ( x, V ) = 4 π v v exp { ix ( k † − ω † /V ) } v − v . (86)Let be v > V > v > µ = 1, µ = 1 /
2. Then I ( x, V ) = − π / v v / V / exp { ix ( k † − ω † /V ) − πi/ } x / ( v − v ) / ( v − V ) / . (87)Finally, let be v < < V < v , µ = µ = 1 /
2. Then I ( x, V ) = 4 π √− v v V exp { ix ( k † − ω † /V ) } x p ( v − V )( V − v ) . (88) A2. A saddle point on a singular set
Consider the integral I ( x, V ) = ∞ Z −∞ ∞ + iǫ Z −∞ + iǫ exp { ix ( k − ω/V ) } ( k − ψ ( ω )) µ dω dk, (89)where ψ ( ω ) = k s + ω − ω s V − α ( ω − ω s ) . (90)Parameters µ , ω s , k s , α are real. Parameter ǫ is small. One can see that the point ( ω s , k s ) is asaddle point of the integral on the dispersion diagram k = ψ ( ω ). Indeed, α = − d ψ ( ω s ) d ω s . Introduce the coordinates η = k − ψ ( ω ) = ( k − k s ) − ω − ω s V + α ( ω − ω s ) , ξ = ω − ω s . (91)After a deformation of the integration surface, obtain I ( x, V ) = exp { ix ( k s − ω s /V ) } ∞ Z −∞ Z γ + exp { ix ( η − αξ ) } η µ dη dξ. (92)The integral can be taken: I ( x, V ) = exp { ix ( k s − ω s /V ) } I c ( µ, x ) I a ( αx ) , (93)where I a ( ξ ) = (cid:26) exp {− iπ/ } p π/ξ, ξ > , exp { iπ/ } p − π/ξ, ξ < . (94)25
3. A saddle point near a crossing point of singular sets
Consider the integral I ( x, V ) = ∞ Z −∞ ∞ + iǫ Z −∞ + iǫ exp { ix ( k − ω/V ) } ( k − ψ ( ω )) µ ( k − ψ ( ω )) µ dω dk, (95) ψ ( ω ) = k † + ω − ω † v − α ( ω − ω † ) , (96) ψ ( ω ) = k † + ω − ω † v , (97)Let be v > v >
0. Assume that V ≈ v , i. e. the saddle point belongs to the branch k = ψ ( ω ).Introduce the variables η = k − ψ ( ω ) = ( k − k † ) − ω − ω † v + α ( ω − ω † ) , (98) η = k − ψ ( ω ) = ( k − k † ) − ω − ω † v . (99)Solve the equations η = δk − δωv + α ( δω ) , η = δk − δωv with respect to the variables δω = ω − ω † and δk = k − k † . To get convenient formulae, assumethat the term α ( δω ) is small, and the equation can be solved iteratively. After the seconditeration obtain δk ≈ v η − v η v − v − α v v ( v − v ) ( η − η ) , (100) δω ≈ v v v − v ( η − η ) − α v v ( v − v ) ( η − η ) . (101)Using these approximations, write the exponential factor of I in the formexp n ix (cid:16) k − ωV (cid:17)o ≈ exp n ix (cid:16) k † − ω † V (cid:17)o × (102)exp (cid:26) ix ( V − v ) v ( v − v ) V η + ix ( v − V ) v ( v − v ) V η − iαxv v ( V − v )( v − v ) V ( η − η ) (cid:27) . Since V ≈ v , the factor v − V in the second term is small, and the size of the integrationdomain in η is much bigger than the size in η . Thus, one can replace ( η − η ) by η :exp n ix (cid:16) k − ωV (cid:17)o ≈ exp n ix (cid:16) k † − ω † V (cid:17)o × (103)exp (cid:26) ix ( V − v ) v ( v − v ) V η + ix ( v − V ) v ( v − v ) V η − iαxv v ( V − v )( v − v ) V η (cid:27) . I can be written approximately as I ( x, V ) ≈ v v exp { ix ( k † − ω † /V ) } v − v I c (cid:18) µ , x ( V − v ) v ( v − v ) V (cid:19) I s (cid:18) µ , x ( v − V ) v ( v − v ) V , αxv v ( V − v )( v − v ) V (cid:19) , (104)where I s ( µ, a, b ) = Z γ s exp { iaξ − ibξ } ξ µ dξ = b ( µ − / ˆ I s ( µ, a/ √ b ) , (105)ˆ I s ( µ, ξ ) = Z γ s exp { i ( τ ξ − τ ) } τ µ dτ, (106)where the contour of integration of integration is shown in Fig. 13. Im[ ] tg Re[ ] t s Fig. 13: Contour γ s Function ˆ I s is rather complicated. We consider two cases. For µ = 1ˆ I s (1 , ξ ) = 2 πi (1 − C ( ξ/ , (107)where C is the Fresnel integral C ( ξ ) = 1 √ πi ∞ Z ξ e iζ dζ . (108)The well-known formula (107) can be proven by differentiation of (106) with respect to ξ .The Fresnel integral (108) has asymptotics for real aC ( ξ ) = exp { iξ + iπ/ } √ πξ , for ξ ≫ , (109) C ( − ξ ) = − exp { iξ + iπ/ } √ πξ + 1 , for ξ ≫ . (110)If µ = 1 / B ( ξ ) ≡ ˆ I s (1 / , ξ ) (111)can be expressed through the functions of the parabolic cylinder [32]. However, it is simpleto tabulate B ( ξ ) directly by using the definition (106). The real and imaginary part of thefunction computed numerically are plotted in Fig. 14.27
15 −10 −5 0 5 10 15−1.5−1−0.500.511.522.533.5 ξ B ( ξ ) Re[B]Im[B]
Fig. 14: Numerically computed function B ( ξ )Function B ( ξ ) has the following asymptotics for real ξ : B ( ξ ) ≈ r πξ exp { i ( ξ − π ) / } + 2 r πξ e πi/ for ξ ≫ , (112) B ( − ξ ) ≈ r πξ exp { i ( ξ + π ) / } for ξ ≫ , (113)The asymptotics (112) and (113) describe function B ( ξ ) well enough from | ξ | ∼
3. The maxi-mum value of B ( ξ ) is equal to 3.727.Finally, an approximation of (95) for µ = µ = 1 can be obtained by combining (104),(107), and (84) and taking V ≈ v : I ( x, V ) ≈ π v v exp { ix ( k † − ω † /V ) } v − v C (cid:18) x / ( V − v )2 α / v (cid:19) . (114)Similarly, for µ = 1, µ = 1 / I ( x, V ) ≈ πi ( αx ) / r v v v − v exp { ix ( k † − ω † /V ) } B (cid:18) x / ( v − V ) α / v (cid:19) . (115)28
4. A special function for the non-local estimation
Consider the integral defined for real ξ > E ( ξ ) = ∞ Z e − ξτ (1 + τ ) √ τ dτ. (116)The graph of this function is shown in Fig. 15. ξ E ( ξ ) Fig. 15: Numerically computed function E ( ξ )Function E has the following asymptotics for ξ ≫ E ( ξ ) ≈ √ π √ ξ . (117)We also note that E (0) = π √ ..