Some Universal Properties of the Green ’ s Functions Associated with the Wave Equation in Bounded Partially-Homogeneous Domains and Their Use in Acoustic Tomography

Direct and inverse scattering problems connected with the wave equation in non-homogeneous bounded domains constitute challenging actual subjects for both mathematicians and engineers. Among them one can mention, for example, inverse source problems in seismology, nondestructive archeological probing, mine prospecting, inverse initial-value problems in acoustic tomography, etc. In spite of its crucial importance, almost all of the available rigorous investigations concern the case of unbounded simple domains such as layered planar or cylindrical or spherical structures. The main reason for the lack of the works related to non-homogeneous bounded structures is the extreme complexity of the explicit expressions of the Green’s functions. The aim of the present work consists in discovering some universal properties of the Green’s functions in question, which reduce enormously the difficulties arising in various applications. The universality mentioned here means that the properties are not depend on the geometrical and physical properties of the configuration. To this end one considers first the case when the domain is partially-homogeneous. Then the results are generalized to the most general case. To show the importance of the universal properties in question, they are applied to an inverse initial-value problem connected with photo-acoustic tomography.


Introduction
V ⊂ R denote a bounded region composed of n sub-domains having dif- ferent (constant) constitutive parameters, say ε j and µ j ( 1, , j n =  ) as shown in Figure 1 below.The constitutive parameters of the surrounding space, say 0 V , will be denoted by 0 ε and 0 µ .Let a function ( ) , p x t , defined for x ∈ ℜ and t ∈ ℜ , satisfy the wave equation ( ) ( ) in the sense of distribution.Here ( ) 0 , p x t stands for the density of the source which will be assumed located inside V while ( ) ( ) the above-mentioned constitutive parameters: ( ) , .
The validity of (1) in the sense of distribution involves in itself both the boundary conditions satisfied on the interfaces between the adjacent sub-regions and the initial conditions, if any, at a certain time, say 0 t = .Therefore it is not necessary to write down here these boundary and initial conditions explicitly (see for ex.(14b)).To determine ( ) , p x t uniquely, one has to add to (1) the so- called radiation conditions also.They will be clarified later on (see (4a, b) below).
As is well known, in a direct propagation problem the so-called outgoing Green's function plays important role in transporting the data known in the source region towards the observation points.This is a wave propagating towards infinity.But in an inverse source problem, the data observed (measured) on a surface S, surrounding the source region, is transported inversely towards the source region.This transportation is accomplished through a Green's function which propagates in the inverse direction (ingoing Green's function).Figure 1.A partially-homogeneous bounded domain.
Therefore, in inverse source problems one has to use both of these Green's functions together.Now our aim is to reveal some universal properties of these functions.Here the universality means that the properties in question do not depend on the geometrical shapes and constitutive parameters of the sub-domains.
In what follows we will assume that ( ) , p x t is Fourier transformable with respect to t .The transform of each quantity are denoted by a hat on that quantity.For example one has and, inversely Thus the transform of ( 1) is as follows: ( ) The radiation conditions satisfied by ( ) ( ) where we put The outgoing Green's function associated with (3)-(4c) is the function , , G x w η determined uniquely through the following relations: Here 3 η ∈ R stands for any point while ( ) The sub-indices x in div x and grad x , which appear in (5a), means that the derivatives are to be computed with respect to the components of x .Notice that because of the radiation condition (5c), the wave as- sociated with G 1 propagates in the direction x η → .This Green's function permits one to transport the data known at the source point η to the observa- tion point x .Indeed, as we will see in Section 2.1, the solution to the problem (3)-(4c) (i.e. the direct problem) is as follows: ( ) ( ) ( ) As to the ingoing Green's function in the sense of distribution under the radiation conditions and ( ) Notice that (7a) and (7b) are quite identical to (5a) and (5b) while (7c) differs from (5c) only with the sign of (i).Because of this difference, the wave associated with ( )

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, , G x η ω propagates in the direction x η → and serves to transport the measured data at x to the source point η .
Observe that the Green's functions defined above differ from the classical ones by the factor ( ) x ε (see [1]).In Section 2 we will prove that these Green's functions have the universal properties stated in the following theorems: , , G x η ω is symme- tric with respect to x and η : , , , , .
Here ( ) * means the complex conjugate.The relations in (9a) and (9c) are also valid for complex ω .

Proofs of the Theorems
Now we will give the proofs of the above-mentioned theorems in turn.But, first of all we prove the Formula (6) which will be used later on in the proofs of the theorems.

Proof of the Formula (6)
Multiply first (1) with 1 G and (5a) with p  , and subtract side by side to get If we integrate both sides in a sphere of radius R and transform the first side to the surface integral on x R = , then we write Now let us make R → ∞ .Because of the radiation conditions satisfied by 1 G and p  , the surface integral tends to zero and yields (6).

Proof of Theorem-1
Now consider the function ( ) ( ) ( ) One can easily check that 1 p  satisfies (3) as well as the radiation conditions (4a, b).Therefore it is equal to ( )


If we compare this new expression of ( ) , p x ω  with ( 6), then we write also is arbitrary, from the latter one gets (8).

Proof of Theorem-2
Assume ω ∈ R and make the change If under the same assumption ω ∈ R one considers the complex conjugates of the equations satisfied by 1 G , one sees that the resulting equations are noth- ing but those satisfied by 2 G .This proves (9b).
As to (9c), it is a direct consequence of (9a) and Theorem-1.The relations in (9a) and (9c) are obviously valid for complex ω also.

Proof of Theorem-3
Consider the Equation (1) for the case when ( ) ( ) ( ) In this case one obviously has ( ) , 0 p x t ≡ for 0 t < .Thus from (2a) and (2b) written for . In this case from (1) one gets A comparison of (14a) with (14b) yields .
To prove (10b), let us define a polar coordinate system ( ) , , R θ ϕ with origin at the point η and consider the distribution , , x x x x φ φ ≡ denotes any test function, then we write where we put This formula defines the extension of ( ) From (15a) and (15b) one gets which reduces (10a) to (10b).
Finally, let us consider the distribution By virtue of (15b) we write also which reduces (10a) to (10c).
Remark.All the results obtained above can easily be checked in the simplest case of homogeneous space for which one has where R x η = − .To this end one has to observe that in ℜ 3 ( )

Proof of Theorem-4
By considering (9a) we get directly

Proof of Theorem-5
Since the function , , G x η ω is symmetrical with respect to x and η , as a function of 0 V η ∈ , it satisfies the Helmholtz equation under the radiation conditions ( ) ( ) Here we put 0 0 0 k ω ε µ = .Now let us define a polar co-ordinate system ( ) , , R θ ϕ whose origin is at the point x and R x η = − .In this system (17a) and (17b) become Now for each n Ω let us make the substitutions [ ] .
Notice that (19a) is satisfied outside the region V, and 1n G may depend on n Ω (and α ) through the boundary conditions on V ∂ also.G to the so-called far field pattern of 1n G  [3]: Here A ∞ stands for the scattering amplitude.It does not depend on r.By using (20), we write successively If we compare (21) with (10c) written as follows ( ) ( ) and the explicit expression of (20) is ( ) as n → ∞ .This yields ( ) The latter proves (12a).The proof of (12b) is a direct result of (12a) and (9a).
It is important to remark here that ( 22) is not the far-field expression of 1 G because R is finite.It states merely that when computing the integral appearing on the left-hand side of (21), for n → ∞ 1 G can be replaced by (22).

Proof of Theorem-6
Let us now define two polar coordinates ( ) , , R θ ϕ whose origins are located at the points x and y , respectively.If we make the substi- tutions then from (22) we write and Now let us define the distribution defined through the sequence [2] ( ) ( ) , , , , Remark.(24c) can easily be checked in the simplest case of homogeneous space for which 1 G and 2 G are given by (16a, b).

Application. Inverse Initial-Value Problem of Acoustic Tomography in a Non-Homogeneous Domain
Reconsider the domain shown in Figure 1 above.Such a domain can model, for example, small-seized animals, women's breasts or other biological domains where different regions are the bone, skin, muscle, fat, cancerous tissues, epileptic tumor etc.If V is exposed to a light (or thermal) pulse, excited at a certain time, say 0 t = , then the energy stored in various regions creates a pressure wave ( ) , p x t which propagates towards 0 V .The initial value of the pressure function, say ( ) , 0 p x + , gives an idea about the configuration.The so-called photo-acoustic and thermo-acoustic tomographies are based on this phenomenon.To this end one first measures the pressure intensity ( ) , p x t on a closed surface S lying in 0 V in a certain time interval (0, T) and then inserts them into an integral on S to obtain ( ) , 0 p x + in V (see Figure 2).That means that the acoustic tomography consists in an inverse initial-value problem.

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conditions satisfied by 1 G .One obviously obtains the equa- tions satisfied by 2G .From this one concludes (9a).
Now it is worthwhile to remark that the transformation to the polar coordinates causes to a confusion in (15a) because the distribution ( ) R δ is defined on the space of the test functions defined for
It is obvious that (19a) and (19b, c) define the radiating (=outgoing) solution to the Helmholtz equation.When n → ∞ , one has n n r R Ω → ∞ ⇒ = Ω → ∞ , which reduces the expression of 1n By virtue of (24a, b) we have

Figure 2 .
Figure 2. Various parameters connected with the tomography problem.
Now observe that the integral on ω gives the inverse Fourier transform at gives the solution to the tomography problem in question.Since its righthand side does not explicitly involve any information about the configuration of the domain V, it is valid for all non-homogeneous bounded domains.In the case of homogeneous space, where ( ) 0 x ε ε ≡ and ( ) 0 x µ µ ≡ , (33) is reduced to the