Some Universal Properties of the Green’s Functions Associated with the Wave Equation in Bounded Partially-Homogeneous Domains and Their Use in Acoustic Tomography ()
1. Introduction
Let
denote a bounded region composed of n sub-domains having different (constant) constitutive parameters, say ej and mj (
) as shown in Figure 1 below. The constitutive parameters of the surrounding space, say
, will be denoted by
and
. Let a function
, defined for
and
, satisfy the wave equation
(1)
in the sense of distribution. Here
stands for the density of the source which will be assumed located inside V while
and
denote 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
. Therefore it is not necessary to write down here these boundary and initial conditions explicitly (see for ex. (14b)). To determine
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
is Fourier transformable with respect to
. The transform of each quantity are denoted by a hat on that quantity. For example one has
(2a)
and, inversely
(2b)
Thus the transform of (1) is as follows:
(3)
The radiation conditions satisfied by
are
(4a)
and
(4b)
where we put
(4c)
The outgoing Green’s function associated with (3)-(4c) is the function
determined uniquely through the following relations:
(5a)
(5b)
(5c)
Here
stands for any point while
is the Dirac distribution concentrated at
. The sub-indices
in
and
, which appear in (5a), means that the derivatives are to be computed with respect to the components of
. Notice that because of the radiation condition (5c), the wave associated with G1 propagates in the direction
. This Green’s function permits one to transport the data known at the source point
to the observation point
. Indeed, as we will see in Section 2.1, the solution to the problem (3)-(4c) (i.e. the direct problem) is as follows:
(6)
As to the ingoing Green’s function
, it satisfies
(7a)
in the sense of distribution under the radiation conditions
(7b)
and
(7c)
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
propagates in the direction
and serves to transport the measured data at
to the source point
.
Observe that the Green’s functions defined above differ from the classical ones by the factor
(see [1] ). In Section 2 we will prove that these Green’s functions have the universal properties stated in the following theorems:
Theorem-1. For all
the Green’s function
is symmetric with respect to
and
:
(8)
Theorem-2. For
the Green’s functions
and
are interrelated with the following functional relations:
(9a)
(9b)
(9c)
Here
means the complex conjugate. The relations in (9a) and (9c) are also valid for complex
.
Theorem-3. For all
and
one has
(10a)
(10b)
(10c)
Here
stands for any sequence such that
as
.
Theorem-4. For all
and
one has
(11a)
(11b)
(11c)
Here
stands for any sequence such that
as
.
Theorem-5. For all
,
,
and
one has
(12a)
(12b)
Here
stands for any sequence such that
as
.
Theorem-6. For all
,
and
one has
(13)
Here
stands for any sequence such that
as
and
2. 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.
2.1. Proof of the Formula (6)
Multiply first (1) with
and (5a) with
, 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
, then we write
Now let us make
. Because of the radiation conditions satisfied by
and
, the surface integral tends to zero and yields (6).
2.2. Proof of Theorem-1
Now consider the function
defined as follows:
One can easily check that
satisfies (3) as well as the radiation conditions (4a, b). Therefore it is equal to
If we compare this new expression of
with (6), then we write also
Since
is arbitrary, from the latter one gets (8).
2.3. Proof of Theorem-2
Assume
and make the change
in the differential equation and the radiation conditions satisfied by
. One obviously obtains the equations satisfied by
. From this one concludes (9a).
If under the same assumption
one considers the complex conjugates of the equations satisfied by
, one sees that the resulting equations are nothing but those satisfied by
. 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
also.
2.4. Proof of Theorem-3
Consider the Equation (1) for the case when
with
for
. In this case one obviously has
for
. Thus from (2a) and (2b) written for
one gets
(14a)
Assume first that
which yields
. In this case from (1) one gets
(14b)
A comparison of (14a) with (14b) yields
(14c)
with
From (14c) one concludes that when
the sequence
defines a generalized function (= distribution) [2] , namely:
The latter proves (10a).
To prove (10b), let us define a polar coordinate system
with origin at the point
and consider the distribution
. If
denotes any test function, then we write
(15a)
where we put
Now it is worthwhile to remark that the transformation to the polar coordinates causes to a confusion in (15a) because the distribution
is defined on the space of the test functions defined for
while the function
is known only for
. To overcome the difficulty, we extend the functions
defined in
into
as an even function (which is continuous at
!!). If the extended function is denoted by
, then we write
which yields
(15b)
This formula defines the extension of
into
to be used in
and
.
From (15a) and (15b) one gets
and
(15c)
which reduces (10a) to (10b).
Finally, let us consider the distribution
and write
with
.
By virtue of (15b) we write also
This shows that
(15d)
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
(16a)
and
(16b)
where
. To this end one has to observe that in Â3
2.5. Proof of Theorem-4
By considering (9a) we get directly
Thus from (10a), (10b) and (10c) one gets directly (11a), (11b) and (11c).
2.6. Proof of Theorem-5
Since the function
is symmetrical with respect to
and
, as a function of
, it satisfies the Helmholtz equation
(17a)
under the radiation conditions
(17b)
Here we put
. Now let us define a polar co-ordinate system
whose origin is at the point
and
. In this system (17a) and (17b) become
(18a)
and
(18b)
(18c)
Now for each
let us make the substitutions
and
which reduce (18 a-c) to
(19a)
and
(19b)
(19c)
Notice that (19a) is satisfied outside the region V, and
may depend on
(and
) through the boundary conditions on
also.
It is obvious that (19a) and (19b, c) define the radiating (=outgoing) solution to the Helmholtz equation. When
, one has
, which reduces the expression of
to the so-called far field pattern of
[3] :
(20)
Here
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
then we conclude that
and the explicit expression of (20) is
as
. This yields
(22)
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
because
is finite. It states merely that when computing the integral appearing on the left-hand side of (21), for
can be replaced by (22).
2.7. Proof of Theorem-6
Let us now define two polar coordinates
and
whose origins are located at the points
and
, respectively. If we make the substitutions
and
(23a)
(23b)
then from (22) we write
(24a)
and
(24b)
Now let us define the distribution defined through the sequence [2]
By virtue of (24a, b) we have
This proves (13).
Remark. (24c) can easily be checked in the simplest case of homogeneous space for which
and
are given by (16a, b).
3. 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
is exposed to a light (or thermal) pulse, excited at a certain time, say
, then the energy stored in various regions creates a pressure wave
which propagates towards
. The initial value of the pressure function, say
, 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
on a closed surface S lying in
in a certain time interval (0, T) and then inserts them into an integral on S to obtain
in V (see Figure 2). That means that the acoustic tomography consists in an inverse initial-value problem. The pressure function
satisfies the Equation (1) with
, namely [4] :
(25)
Figure 2. Various parameters connected with the tomography problem.
where
and
. It is worthwhile to remark that in realistic problems both the source density
and the constitutive parameters
(i.e. the geometrical shapes of the regions
) are not known beforehand. Remark also that all of the previous theoretical works concern the simplest case where the space is assumed to be homogeneous and the aim consists in the determination of the source density
(inverse source problem) [4] - [12] . The basic principles and different implementations along with a long reference list are given in [4] and [5] .
Let
and
be the Green’s functions associated with the configuration shown in Figure 1. The pressure
measured at a point
is connected to the source density
through the outgoing Green’s function
as given in Formula (6), where
.
Now, by using the ingoing Green’s function
we transport the data known at the point
to the point
through the following function
(26)
Here
stands for any sequence such that
as
,
is the unit outward normal vector to the surface S at the point
and
means the derivative in this direction (see Figure 2). First we will show that when
one has
(27)
To prove (27), let us replace
in (26) by (6) and write
where we put
(28)
(28) can also be written as follows (in what follows
shows the domain bounded by S):
(29)
where we put
and
Now reconsider the equations satisfied by
and
(see (5a) and (7a)) to write
Now let us add the left hand-side again to both sides to obtain a more symmetrical expression
By using the latter in (29) one obtains
where
and
(30)
with
.
In what follows we will show that if
is convex (see Sec.4 below), then
(31)
which means that the contribution of
to
is naught when
. As to the contribution of Pn1, one has
.
This equation defines a distribution (generalized function), say
, which permits us to write [2]
(32a)
and
On the other hand, from Theorem-4 we already know that
,
which reduces (32a) to
(32b)
The second equality in (32b) is an obvious issue of the distributional validity of (25).
(32b) is the basic formula of the present work. It can be used in photo-acous- tic and thermo-acoustic tomography problems connected with non-homoge- neous media. To obtain a more explicit expression of
, let us insert the expression of
given by (26) into (27), and make the change of variable
. If we use also (12b), then we write successively
Here
(see Figure 2). It is obvious that when
the contribution of the part taking place in
tends to zero if the contribution of the constant
is finite. Therefore, by omitting the terms in
and writing inversely
we write also
Now observe that the integral on
gives the inverse Fourier transform at
. Thus we obtain finally
where
(see Figure 2).
(33) gives the solution to the tomography problem in question. Since its right- hand 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
and
, (33) is reduced to the known formula obtained in [12] .
From (33) one concludes that in order to get a chart of the sensitive points, one has to know the values of
at all
at the times
. Maximum values of these t, say T, for all
and all
is obviously
. Therefore, (33) can be stated as the following theorem.
Theorem. Let the values of
be known on the boundary S of a convex bounded domain DS involving V for all
where
with
. Then the initial values configuration
is given by the universal Formula (33) while the density
of the source appearing in (25) can be determined through the Formula (32b).
4. Proof of (31)
One can easily check that the integrand in (30) is equal to
where
while
. Thus, by using the Gaus-Ostrogradski theorem one can write also
or
When
the inner integral tends to the distribution given by (13) and yields
Now it is important to observe that the function in the bracket is a function of the form
and
This shows that (34) can also be written as follows:
(35)
By a theorem due to the author, the integral taking place in (35) is equal to naught whenever the domain DS is convex (see Lemma in [12] ). This proves (31).
5. Conclusions and Concluding Remarks
From the results obtained above one concludes that the solution to the inverse initial-value problem (i.e. the tomography problem!) connected with spaces involving partially homogeneous bounded regions is given by (33). It is extremely important to observe that it is a universal formula which does not depend on the configuration. The geometrical shape and constitutive parameters effect only the data measured on S. Since the shapes and number of the sub-regions does not appear in (33), we can conjecture that it is also valid for any non-homogeneous medium.
Furthermore, since the Formula (28) considered above is symmetrical with respect to the Green’s functions
and
, we could replace
given by (26) by
In this case, instead of (33) one gets more simply
;
and
.