Scattering by a Cavity in an Exponentially Graded Half-Space

Martin, P. A.

doi:10.1115/1.3086585

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.

Figure 1

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The scattering problem. The half-plane on the left of $x = 0$ is homogeneous. The other half-plane is inhomogeneous. The circular cavity has radius $a$ ⁠. A plane wave is incident from the left.

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 $e^{- β x}$ 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

μ_{0}

and density

ρ_{0}

(both constants). The inhomogeneous region has shear modulus

μ (x)

and density

ρ (x)

given by

μ (x) = μ_{0} e^{2 β x} and ρ (x) = ρ_{0} e^{2 β x}

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

e^{- i ω t}

⁠, the governing equation is

\frac{\partial τ_{x z}}{\partial x} + \frac{\partial τ_{y z}}{\partial y} + u ω^{2} ρ (x) = 0

where

u (x, y)

is the antiplane component of displacement and the stress components are given by

τ_{x z} = μ (x) \frac{\partial u}{\partial x} and τ_{y z} = μ (x) \frac{\partial u}{\partial y}

Thus, in the homogeneous region, where we write

u_{0}

instead of

u

⁠, we obtain the two-dimensional Helmholtz equation

(\nabla^{2} + k_{0}^{2}) u_{0} = 0 with k_{0}^{2} = ρ_{0} ω_{2} / μ_{0}

2

In the inhomogeneous region, we obtain

\nabla^{2} u + 2 β \frac{\partial u}{\partial x} + k_{0}^{2} u = 0

This equation is satisfied by writing

u (x, y) = e^{- β x} w (x, y)

where

w

satisfies a different two-dimensional Helmholtz equation

(\nabla^{2} + k^{2}) w = 0 with k^{2} = k_{0}^{2} - β^{2}

3

For simplicity, we assume that

k_{0}^{2} > β^{2}

and write

k = + \sqrt{k_{0}^{2} - β^{2}}

⁠.

The interface conditions require that the displacements and normal stresses be continuous, so that

u_{0} (0, y) = u (0, y) = w (0, y)

4

{\frac{\partial u_{0}}{\partial x} |}_{x = 0} = {\frac{\partial u}{\partial x} |}_{x = 0} = {\frac{\partial w}{\partial x} |}_{x = 0} - β w (0, y)

5

Incident Field

Suppose that a plane wave is incident on the interface from the homogeneous side. This wave is given by

u_{in} (x, y) = e^{i k_{0} (x cos α_{0} + y sin α_{0})}

where

α_{0}

is the angle of incidence,

| α_{0} | < π / 2

⁠;

α_{0} = 0

gives normal incidence. There will be a reflected wave

u_{re}

and a transmitted wave

u_{tr}

⁠. Evidently,

u_{re} (x, y) = R e^{i k_{0} (- x cos α_{0} + y sin α_{0})}, x < 0

6

u_{tr} (x, y) = T e^{- β x} e^{i k (x cos α + y sin α)}, x > 0

7

where

R

⁠,

T

⁠, and

α

are to be found. Writing

u_{0} = u_{in} + u_{re}

and

u = u_{tr}

⁠, Eq. gives

1 + R = T and k_{0} sin α_{0} = k sin α

Then, Eq. gives

(1 - R) i k_{0} cos α_{0} = T (i k cos α - β)

Solving for

R

gives

R = \frac{k_{0} cos α_{0} - k cos α - i β}{k_{0} cos α_{0} + k cos α + i β} = \frac{- i β}{k_{0} cos α_{0} + k cos α}

and then

T = 1 + R

⁠.

For a simple check, put $β = 0$ ⁠; we obtain $k = k_{0}$ ⁠, $α = α_{0}$ ⁠, $R = 0$ ⁠, 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

{(x - b)}^{2} + y^{2} = a^{2} with 0 < a < b

We also introduce polar coordinates,

(r, θ)

⁠, so that

x = b + r cos θ and y = r sin θ

8

Thus, the cavity’s boundary is given by

r = a

⁠, and the boundary condition is

\frac{\partial u}{\partial r} = 0 on r = a

9

where

u

is the total field in the inhomogeneous half-space.

To solve such a scattering problem, we write

u_{0} = u_{in} + u_{re} + v_{0}, x < 0

u = u_{tr} + v, x > 0, r > a

where

v_{0}

solves Eq. ,

v = w e^{- β x}

⁠, and

w

solves Eq. . Also,

v_{0}

must satisfy the Sommerfeld radiation condition and

v

must decay with

x

⁠.

Multipole Functions

To represent the scattered field, we introduce functions

ϕ_{n}

of the form

ϕ_{n} = {\begin{array}{l} e^{- β x} {H_{n}^{(1)} (k r) e^{i n θ} + Φ_{n}}, & x > 0 \\ Ψ_{n}, & x < 0 \end{array}

where

H_{n}^{(1)}

is a Hankel function and

n

is an arbitrary integer. We require that

Φ_{n}

solves Eq. and

Ψ_{n}

solves Eq. . In addition,

Φ_{n}

and

Ψ_{n}

are to be chosen so that

ϕ_{n}

satisfies the interface conditions, Eqs. ,.

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. in Eq. , we obtain the integral representation

H_{n}^{(1)} (k r) e^{i n θ} = \frac{{(- 1)}^{n}}{π i} \int_{- \infty}^{\infty + π i} e^{k (b - x) sinh τ - i k y cosh τ} e^{- n τ} d τ for x < b, | y | < \infty

10

Notice that this formula is valid on the interface

x = 0

⁠. The contour of integration in Eq. is also described in the Appendix 0.

The form of Eq. suggests using a similar integral representation for

Φ_{n}

⁠, and so we write

Φ_{n} (x, y) = \frac{{(- 1)}^{n}}{π i} \int_{- \infty}^{\infty + π i} A (τ) e^{k x sinh τ} e^{- i k y cosh τ + k b sinh τ - n τ} d τ, x > 0

11

where

A (τ)

is to be found;

Φ_{n}

solves Eq. automatically for any reasonable choice of

A

⁠.

We shall also need a similar integral representation for

Ψ_{n} (x, y)

in

x < 0

⁠, where the wavenumber is

k_{0}

⁠. However, in order to match solutions across the interface at

x = 0

⁠, we shall require the same dependence on

y

as in Eq. . Thus, we consider

Ψ_{n} (x, y) = \frac{{(- 1)}^{n}}{π i} \int_{- \infty}^{\infty + π i} B (τ) e^{x Δ (τ)} e^{- i k y cosh τ + k b sinh τ - n τ} d τ, x < 0

12

where

B (τ)

is to be found,

Δ (τ) = {(k^{2} {cosh}^{2} τ - k_{0}^{2})}^{1 / 2} = {(k^{2} {sinh}^{2} τ - β^{2})}^{1 / 2}

and the square root is taken so that

Re Δ \geq 0

on the contour. Notice that Eq. is satisfied automatically for any reasonable choice of

B

⁠.

We are now ready to enforce the interface conditions. Continuity of

ϕ_{n}

across

x = 0

gives

1 + A = B

whereas continuity of

\partial ϕ_{n} / \partial x

gives

- k sinh τ - β + (k sinh τ - β) A (τ) = Δ (τ) B (τ)

Hence

A (τ) = \frac{k sinh τ + Δ + β}{k sinh τ - Δ - β}

13

and

B (τ) = \frac{2 k sinh τ}{k sinh τ - Δ - β}

14

These formulas complete the construction of the multipole functions

ϕ_{n}

⁠.

Note that when $β = 0$ ⁠, $k = k_{0}$ ⁠, $Δ (τ) = - k sinh τ$ ⁠, $A = 0$ ⁠, $B = 1$ ⁠, and $Ψ_{n} = H_{n}^{(1)} (k r) e^{i n θ}$ ⁠, as expected.

Imposing the Boundary Condition

In the homogeneous half-space, we write

u_{0} = u_{in} + u_{re} + \sum_{n} c_{n} ϕ_{n}

where

Σ_{n}

denotes summation over all integers

n

⁠. Similarly, in the graded half-space, we write

u (r, θ) = u_{tr} + \sum_{n} c_{n} ϕ_{n}

Then, by construction, the governing partial differential equations and the interface conditions along

x = 0

are all satisfied. It remains to determine the coefficients

c_{n}

using the boundary condition on

r = a

⁠, Eq. ; this gives

\sum_{n} c_{n} {\frac{\partial ϕ_{n}}{\partial r} |}_{r = a} = - {\frac{\partial u_{tr}}{\partial r} |}_{r = a}

15

To proceed, we write both sides of this equation as Fourier series in

θ

⁠. For the right-hand side, we have

u_{tr} = e^{- β x} T_{b} e^{i k r cos (θ - α)} = e^{- β x} T_{b} \sum_{m} i^{m} J_{m} (k r) e^{i m (θ - α)}

where

T_{b} = T exp (i k b cos α)

and

J_{n}

is a Bessel function. Also, we have the expansion

e^{- β x} = e^{- β b} \sum_{s} {(- 1)}^{s} I_{s} (β r) e^{i s θ}

where

I_{n}

is a modified Bessel function. Hence,

u_{tr} (r, θ) = e^{- β b} \sum_{m} {(- 1)}^{m} U_{m} (r) e^{i m θ}

where

U_{m} (r) = T_{b} \sum_{s} {(- i)}^{s} I_{m - s} (β r) J_{s} (k r) e^{- i s α}

In a similar way, we obtain

Φ_{n} (r, θ) = \sum_{m} {(- 1)}^{m} f_{m}^{n} J_{m} (k r) e^{i m θ}, 0 < r < b

with

f_{m}^{n} = \frac{{(- 1)}^{n}}{π i} \int_{- \infty}^{\infty + π i} A (τ) e^{2 k b sinh τ} e^{- (m + n) τ} d τ

Hence

ϕ_{n} (r, θ) = e^{- β b} \sum_{m} {(- 1)}^{m} V_{m}^{n} (r) e^{i m θ}

where

V_{m}^{n} (r) = {(- 1)}^{n} I_{m - n} (β r) H_{n}^{(1)} (k r) + \sum_{s} f_{s}^{n} I_{m - s} (β r) J_{s} (k r)

16

Thus, Eq. and orthogonality of

{e^{i m θ}}

give

\sum_{n} c_{n} V_{m}^{n^{'}} (a) = - U_{m}^{'} (a), all m

17

which is a linear system of algebraic equations for the coefficients

c_{n}

⁠.

Far-Field Behavior of Ψn

We should expect cylindrical waves in the homogeneous half-space. These arise from the far-field behavior of

Ψ_{n} (x, y)

⁠, for

x < 0

⁠. Thus, put

x = - R cos Θ, y = R sin Θ, | Θ | < π / 2

Then, making the substitution

k cosh τ = k_{0} cosh s

in Eq. gives

Δ = - k_{0} sinh s

and

Ψ_{n} = \frac{1}{π i} \int_{- \infty}^{\infty + π i} B_{n} (k_{0} cosh s; β) e^{k_{0} R sinh (s - i Θ)} d s, | Θ | < π / 2

18

where

B_{n} (ζ; β) = \frac{2 {(ζ^{2} - k_{0}^{2})}^{1 / 2} exp {- b {(ζ^{2} - k^{2})}^{1 / 2}}}{{(ζ^{2} - k^{2})}^{1 / 2} + β + {(ζ^{2} - k_{0}^{2})}^{1 / 2}} {(\frac{ζ + {(ζ^{2} - k^{2})}^{1 / 2}}{(- k)})}^{n}

the square roots being defined to have non-negative real parts.

The formula for

Ψ_{n}

⁠, Eq. , is convenient for estimating

Ψ_{n}

when

k_{0} R ⪢ 1

⁠, as we can use the saddle-point method ((16), Chap. 8). There is one relevant saddle point at

s = s_{0}

where

s_{0} = i (\frac{1}{2} π + Θ)

⁠. As

cosh s_{0} = - sin Θ

and

sinh (s_{0} + i Θ) = i

⁠, the standard argument gives

Ψ_{n} \sim \frac{1}{π i} B_{n} (- k_{0} sin Θ; β) e^{i k_{0} R} \int exp {\frac{1}{2} i k_{0} R {(s - s_{0})}^{2}} d s

19

\sim \sqrt{\frac{2}{π k_{0} R}} e^{i (k_{0} R - π / 4)} B_{n} (- k_{0} sin Θ; β) as R \to \infty

20

where the contour of integration in Eq. passes through the saddle point.

When $β = 0$ ⁠, we obtain $B_{n} (- k_{0} sin Θ; 0) = i^{n} e^{i k_{0} b cos Θ} e^{- i n Θ}$ ⁠. Then, Eq. 20 agrees with the known far-field expansion of $H_{n}^{(1)} (k r) e^{i n θ}$ ⁠, when one takes into account that $θ \sim π - Θ$ and $r \sim R + b cos Θ$ as $R \to \infty$ ⁠.

Near-Field Behavior of Φn

As the expression for

Φ_{n}

⁠, Eq. , is similar to Eq. , it is reasonable to ask if

Φ_{n}

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 + r^{'} cos θ^{'}

⁠,

y = r^{'} sin θ^{'}

⁠. Then, calculations similar to those described in the Appendix 0 show that

H_{n}^{(1)} (k r^{'}) e^{i n (π - θ^{'})} = \frac{{(- 1)}^{n}}{π i} \int_{- \infty}^{\infty + π i} e^{k x sinh τ} e^{- i k y cosh τ + k b sinh τ - n τ} d τ

21

for

| θ^{'} | < π / 2

⁠. The integral on the right-hand side of Eq. should be compared with the integral defining

Φ_{n}

⁠, Eq. . For them to be equal, the function

A (τ)

⁠, defined by Eq. , 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

Φ_{n}

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. , with

π - θ^{'}

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 $(d / d r) V_{m}^{n} (r)$ at $r = a$ ⁠, where $V_{m}^{n}$ 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. ), we have the integral representation

H_{n}^{(1)} (k r) = \frac{1}{π i} \int_{- \infty}^{\infty + π i} e^{k r sinh w - n w} d w

A1

The integration is along any contour in the complex

w

-plane, starting at

w = - \infty

and ending at

w = π i + \infty

⁠. When

w = ξ + i η

⁠, where

ξ

and

η

are real,

| e^{k r sinh w} | = e^{k r sinh ξ cos η}

⁠. Thus, we can generalize Eq. to

H_{n}^{(1)} (k r) = \frac{1}{π i} \int_{- \infty + i η_{1}}^{\infty + i η_{2}} e^{k r sinh w - n w} d w

A2

where the constants

η_{1}

and

η_{2}

must satisfy

- \frac{1}{2} π < η_{1} < \frac{1}{2} π and \frac{1}{2} π < η_{2} < \frac{3}{2} π

In other words, we have some flexibility in our choice of contour, flexibility that we shall exploit shortly.

Put

w = τ + i (θ - π)

⁠. Then Eq. becomes

H_{n}^{(1)} (k r) e^{i n θ} = \frac{{(- 1)}^{n}}{π i} \int_{- \infty + i β_{1}}^{\infty + i β_{2}} e^{- k r (sinh τ cos θ + i cosh τ sin θ)} e^{- n τ} d τ

A3

where the constants

β_{1}

and

β_{2}

must satisfy

- \frac{1}{2} π < β_{1} + θ - π < \frac{1}{2} π and \frac{1}{2} π < β_{2} + θ - π < \frac{3}{2} π

In particular, the choices

β_{1} = 0

and

β_{2} = π

show that

H_{n}^{(1)} (k r) e^{i n θ} = \frac{{(- 1)}^{n}}{π i} \int_{- \infty}^{\infty + π i} e^{- k r (\sinh τ cos θ + i cosh τ sin θ)} e^{- n τ} d τ for \frac{1}{2} π < θ < \frac{3}{2} π

A4

This is the integral representation that we use in Sec. 4A.

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by American Society of Mechanical Engineers

Scattering by a Cavity in an Exponentially Graded Half-Space

Introduction

Formulation

Incident Field

Scattering by a Buried Cavity

Multipole Functions

Imposing the Boundary Condition

Far-Field Behavior of Ψn

Near-Field Behavior of Φn

Discussion

Appendix: Integral Representations

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Scattering by a Cavity in an Exponentially Graded Half-Space

Introduction

Formulation

Incident Field

Scattering by a Buried Cavity

Multipole Functions

Imposing the Boundary Condition

Far-Field Behavior of Ψn

Near-Field Behavior of Φn

Discussion

Appendix: Integral Representations

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Cited By

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