An inhomogeneous half-space containing a cavity is bonded to a homogeneous half-space. Waves are incident on the interface and the problem is to calculate the scattered waves. For a circular cavity in an exponentially graded half-space, it is shown how to solve the problem by constructing an appropriate set of multipole functions. These functions are singular on the axis of the cavity, they satisfy the governing differential equation in each half-space, and they satisfy the continuity conditions across the interface between the two half-spaces. Seven recent publications are criticized: They do not take proper account of the interface between the two half-spaces.

Introduction

Consider two half-spaces, x>0 and x<0, welded together along the interface at x=0. The left half-space (x<0) is homogeneous. The right half-space is inhomogeneous. If a wave is incident from the left, it will be partly reflected and partly transmitted into the right half-space. We assume that these fields can be calculated.

Suppose now that the right half-space contains a cavity or some other defect (see Fig.1). How are the basic fields described above modified by the presence of the cavity? In general, it is not easy to answer this question, as the associated mathematical problem is difficult, in general.

In some recent papers, Fang et al. claimed to solve a variety of such problems. All concern “exponential grading,” meaning that the material parameters are proportional to eβx for x>0, where β is a given constant. The papers concern antiplane shear waves (1,2,3,4), thermal waves (5,6), and shear waves in a piezoelectric material (7). All of these papers assume that the effect of the interface on the cavity can be found by introducing simple image terms, as if the interface were a mirror or a rigid wall. Unfortunately, this assumption is incorrect.

In this paper, we outline how the problems described above can be solved. We do this in the context of antiplane shear waves with exponential grading and a circular cavity. The main technical part concerns the derivation of suitable multipole potentials; these reveal the complicated image system.

The study of problems involving scatterers near boundaries or interfaces has a long history. For linear surface water waves interacting with a submerged circular cylinder, see the famous paper by Ursell (8). For plane-strain elastic waves in a homogeneous half-space with a buried circular cavity, see Ref. 9. There are also many papers on the scattering of electromagnetic waves by objects near plane boundaries; see, for example, Ref. 10.

Some problems involving objects near plane boundaries can be solved using images. However, determining the strength and location of the images may be difficult: Doing so will depend on the governing differential equations and on the conditions to be satisfied on the plane boundary. For two interesting examples where the location of the images is not obvious, we refer to Chap. 8 of Ting’s book (11) (construction of static Green’s functions in anisotropic elasticity) and a paper by Stevenson (12) (construction of Green’s function for the anisotropic Helmholtz equation in a half-space).

The basic scattering problem is formulated in Sec. 2. The reflection-transmission problem (for which the cavity is absent) is solved in Sec. 3. The solution of this problem gives the “incident” field that will be scattered by the cavity. To solve the scattering problem, we construct an appropriate set of multipole functions (Sec. 4A). Each of these satisfies the governing differential equations and the interface conditions, and is singular at the center of the circular cavity. Each multipole function is defined as a contour integral of Sommerfeld type; for a careful discussion of similar functions, see Refs. 13,9. In Sec. 4B, the multipole functions are combined so as to satisfy the boundary condition on the cavity, leading to an infinite linear system of algebraic equations. The far-field behavior of the multipole functions is deduced in Sec. 4C. Closing remarks are made in Sec. 5.

Formulation

We consider the antiplane deformations of two elastic half-spaces, bonded together. In terms of Cartesian coordinates (x,y), the half-space x<0 is homogeneous, the half-space x>0 is inhomogeneous (“graded”), and the interface is at x=0. The homogeneous region has shear modulus μ0 and density ρ0 (both constants). The inhomogeneous region has shear modulus μ(x) and density ρ(x) given by 
μ(x)=μ0e2βxandρ(x)=ρ0e2βx
1
where β is a constant. Thus, the material parameters are continuous across the interface.

It is not our purpose here to discuss whether any real materials can be well represented by the functional forms given in Eq. 1. Certainly, the choices in Eq. 1 do lead to some mathematical simplifications and they have been used in the past; see, for example, Refs. 14,15.

For time-harmonic motions, with suppressed time-dependence eiωt, the governing equation is 
τxzx+τyzy+uω2ρ(x)=0
where u(x,y) is the antiplane component of displacement and the stress components are given by 
τxz=μ(x)uxandτyz=μ(x)uy
Thus, in the homogeneous region, where we write u0 instead of u, we obtain the two-dimensional Helmholtz equation 
(2+k02)u0=0withk02=ρ0ω2/μ0
2
In the inhomogeneous region, we obtain 
2u+2βux+k02u=0
This equation is satisfied by writing 
u(x,y)=eβxw(x,y)
where w satisfies a different two-dimensional Helmholtz equation 
(2+k2)w=0withk2=k02β2
3
For simplicity, we assume that k02>β2 and write k=+k02β2.
The interface conditions require that the displacements and normal stresses be continuous, so that 
u0(0,y)=u(0,y)=w(0,y)
4
 
|u0x|x=0=|ux|x=0=|wx|x=0βw(0,y)
5

Incident Field

Suppose that a plane wave is incident on the interface from the homogeneous side. This wave is given by 
uin(x,y)=eik0(xcosα0+ysinα0)
where α0 is the angle of incidence, |α0|<π/2; α0=0 gives normal incidence. There will be a reflected wave ure and a transmitted wave utr. Evidently, 
ure(x,y)=Reik0(xcosα0+ysinα0),x<0
6
 
utr(x,y)=Teβxeik(xcosα+ysinα),x>0
7
where R, T, and α are to be found. Writing u0=uin+ure and u=utr, Eq. 4 gives 
1+R=Tandk0sinα0=ksinα
Then, Eq. 5 gives 
(1R)ik0cosα0=T(ikcosαβ)
Solving for R gives 
R=k0cosα0kcosαiβk0cosα0+kcosα+iβ=iβk0cosα0+kcosα
and then T=1+R.

For a simple check, put β=0; we obtain k=k0, α=α0, R=0, and T=1, as expected.

Scattering by a Buried Cavity

Next, we investigate how the wavefields of Sec. 3 are modified if there is a cavity in the inhomogeneous half-space, x>0. See Fig. 1.

We suppose that the cavity’s cross section is circular, with boundary 
(xb)2+y2=a2with0<a<b
We also introduce polar coordinates, (r,θ), so that 
x=b+rcosθandy=rsinθ
8
Thus, the cavity’s boundary is given by r=a, and the boundary condition is 
ur=0onr=a
9
where u is the total field in the inhomogeneous half-space.

To solve such a scattering problem, we write 

u0=uin+ure+v0,x<0
 
u=utr+v,x>0,r>a
where v0 solves Eq. 2, v=weβx, and w solves Eq. 3. Also, v0 must satisfy the Sommerfeld radiation condition and v must decay with x.

Multipole Functions

To represent the scattered field, we introduce functions ϕn of the form 
ϕn={eβx{Hn(1)(kr)einθ+Φn},x>0Ψn,x<0}
where Hn(1) is a Hankel function and n is an arbitrary integer. We require that Φn solves Eq. 3 and Ψn solves Eq. 2. In addition, Φn and Ψn are to be chosen so that ϕn satisfies the interface conditions, Eqs. 4,5.
The use of polar coordinates is convenient for handling the circular cavity but it is inconvenient when trying to impose the conditions at x=0. Therefore, we convert from polar coordinates to Cartesian coordinates using an integral representation; see the Appendix 0 for details. In particular, if we insert Eq. 8 in Eq. 4, we obtain the integral representation 
Hn(1)(kr)einθ=(1)nπi+πiek(bx)sinhτikycoshτenτdτforx<b,|y|<
10
Notice that this formula is valid on the interface x=0. The contour of integration in Eq. 10 is also described in the Appendix 0.
The form of Eq. 10 suggests using a similar integral representation for Φn, and so we write 
Φn(x,y)=(1)nπi+πiA(τ)ekxsinhτeikycoshτ+kbsinhτnτdτ,x>0
11
where A(τ) is to be found; Φn solves Eq. 3 automatically for any reasonable choice of A.
We shall also need a similar integral representation for Ψn(x,y) in x<0, where the wavenumber is k0. However, in order to match solutions across the interface at x=0, we shall require the same dependence on y as in Eq. 11. Thus, we consider 
Ψn(x,y)=(1)nπi+πiB(τ)exΔ(τ)eikycoshτ+kbsinhτnτdτ,x<0
12
where B(τ) is to be found, 
Δ(τ)=(k2cosh2τk02)1/2=(k2sinh2τβ2)1/2
and the square root is taken so that ReΔ0 on the contour. Notice that Eq. 2 is satisfied automatically for any reasonable choice of B.
We are now ready to enforce the interface conditions. Continuity of ϕn across x=0 gives 1+A=B whereas continuity of ϕn/x gives 
ksinhτβ+(ksinhτβ)A(τ)=Δ(τ)B(τ)
Hence 
A(τ)=ksinhτ+Δ+βksinhτΔβ
13
and 
B(τ)=2ksinhτksinhτΔβ
14
These formulas complete the construction of the multipole functions ϕn.

Note that when β=0, k=k0, Δ(τ)=ksinhτ, A=0, B=1, and Ψn=Hn(1)(kr)einθ, as expected.

Imposing the Boundary Condition

In the homogeneous half-space, we write 
u0=uin+ure+ncnϕn
where Σn denotes summation over all integers n. Similarly, in the graded half-space, we write 
u(r,θ)=utr+ncnϕn
Then, by construction, the governing partial differential equations and the interface conditions along x=0 are all satisfied. It remains to determine the coefficients cn using the boundary condition on r=a, Eq. 9; this gives 
ncn|ϕnr|r=a=|utrr|r=a
15
To proceed, we write both sides of this equation as Fourier series in θ. For the right-hand side, we have 
utr=eβxTbeikrcos(θα)=eβxTbmimJm(kr)eim(θα)
where Tb=Texp(ikbcosα) and Jn is a Bessel function. Also, we have the expansion 
eβx=eβbs(1)sIs(βr)eisθ
where In is a modified Bessel function. Hence, 
utr(r,θ)=eβbm(1)mUm(r)eimθ
where 
Um(r)=Tbs(i)sIms(βr)Js(kr)eisα
In a similar way, we obtain 
Φn(r,θ)=m(1)mfmnJm(kr)eimθ,0<r<b
with 
fmn=(1)nπi+πiA(τ)e2kbsinhτe(m+n)τdτ
Hence 
ϕn(r,θ)=eβbm(1)mVmn(r)eimθ
where 
Vmn(r)=(1)nImn(βr)Hn(1)(kr)+sfsnIms(βr)Js(kr)
16
Thus, Eq. 15 and orthogonality of {eimθ} give 
ncnVmn(a)=Um(a),allm
17
which is a linear system of algebraic equations for the coefficients cn.

Far-Field Behavior of Ψn

We should expect cylindrical waves in the homogeneous half-space. These arise from the far-field behavior of Ψn(x,y), for x<0. Thus, put 
x=RcosΘ,y=RsinΘ,|Θ|<π/2
Then, making the substitution kcoshτ=k0coshs in Eq. 12 gives Δ=k0sinhs and 
Ψn=1πi+πiBn(k0coshs;β)ek0Rsinh(siΘ)ds,|Θ|<π/2
18
where 
Bn(ζ;β)=2(ζ2k02)1/2exp{b(ζ2k2)1/2}(ζ2k2)1/2+β+(ζ2k02)1/2(ζ+(ζ2k2)1/2(k))n
the square roots being defined to have non-negative real parts.
The formula for Ψn, Eq. 18, is convenient for estimating Ψn when k0R1, as we can use the saddle-point method ((16), Chap. 8). There is one relevant saddle point at s=s0 where s0=i(12π+Θ). As coshs0=sinΘ and sinh(s0+iΘ)=i, the standard argument gives 
Ψn1πiBn(k0sinΘ;β)eik0Rexp{12ik0R(ss0)2}ds
19
 
2πk0Rei(k0Rπ/4)Bn(k0sinΘ;β)asR
20
where the contour of integration in Eq. 19 passes through the saddle point.

When β=0, we obtain Bn(k0sinΘ;0)=ineik0bcosΘeinΘ. Then, Eq. 20 agrees with the known far-field expansion of Hn(1)(kr)einθ, when one takes into account that θπΘ and rR+bcosΘ as R.

Near-Field Behavior of Φn

As the expression for Φn, Eq. 11, is similar to Eq. 10, it is reasonable to ask if Φn corresponds to a simple image term. To see that it does not, let us define polar coordinates centered at the mirror-image point, (x,y)=(b,0): x=b+rcosθ, y=rsinθ. Then, calculations similar to those described in the Appendix 0 show that 
Hn(1)(kr)ein(πθ)=(1)nπi+πiekxsinhτeikycoshτ+kbsinhτnτdτ
21
for |θ|<π/2. The integral on the right-hand side of Eq. 21 should be compared with the integral defining Φn, Eq. 11. For them to be equal, the function A(τ), defined by Eq. 13, would have to be constant: It is not, and it is not well approximated by a nonzero constant. Thus, it is not justified to replace Φn with a simple image term: We notice that Fang et al. (2) used image terms similar to those on the left-hand side of Eq. 21, with πθ replaced with θ.

Discussion

We have outlined how to solve the scattering problem for a cavity buried in a graded half-space; the result is the infinite linear algebraic system, Eq. 17. The system matrix is very complicated: One has to calculate (d/dr)Vmn(r) at r=a, where Vmn is defined by Eq. 16 as an infinite series of special functions with coefficients given as contour integrals. In principle, the system matrix could be computed but it is unclear whether this is a worthwhile exercise, given the limitations of the underlying model, with both shear modulus and density varying exponentially; see Eq. 1. However, it may be possible to extract asymptotic results from the exact system of equations for small cavities or for cavities that are far from the interface: This remains for future work.

Appendix: Integral Representations

As explained in Sec. 4A, we need to convert from polar coordinates to Cartesian coordinates in order to apply the interface conditions at x=0. This is done using certain integral formulas. Thus, from Ref. 17 (p. 178, Eq. 2), we have the integral representation 
Hn(1)(kr)=1πi+πiekrsinhwnwdw
A1
The integration is along any contour in the complex w-plane, starting at w= and ending at w=πi+. When w=ξ+iη, where ξ and η are real, |ekrsinhw|=ekrsinhξcosη. Thus, we can generalize Eq. 1 to 
Hn(1)(kr)=1πi+iη1+iη2ekrsinhwnwdw
A2
where the constants η1 and η2 must satisfy 
12π<η1<12πand12π<η2<32π
In other words, we have some flexibility in our choice of contour, flexibility that we shall exploit shortly.
Put w=τ+i(θπ). Then Eq. 2 becomes 
Hn(1)(kr)einθ=(1)nπi+iβ1+iβ2ekr(sinhτcosθ+icoshτsinθ)enτdτ
A3
where the constants β1 and β2 must satisfy 
12π<β1+θπ<12πand12π<β2+θπ<32π
In particular, the choices β1=0 and β2=π show that 
Hn(1)(kr)einθ=(1)nπi+πiekr(sinhτcosθ+icoshτsinθ)enτdτfor12π<θ<32π
A4
This is the integral representation that we use in Sec. 4A.
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