inverse problems l(1985)

Inverse Problems l(1985) 133-172. Printed in Great Britain

Some inverse scattering problems of geophysics

A G Ramm Mathematics Department. Cardwell Hall. Kansas State University. Manhattan. K S 66506, USA

Received 28 December 1984

Abstract. Several problems arising in geophysics are considered. The problems consist of finding an unknown coefficient in a partial differential equation from the knowledge of the solution to this equation on a certain surface. The unknown coefficient is usually the refraction index of a medium, the scattering potential, or the conductivity profile. Analytical inversion algorithms are given.

1. Introduction

There are many problems in geophysics, optics, nondestructive testing and elsewhere in which a finite inhomogeneity in a homogeneous space is to be recovered from the measurements of a field scattered by the inhomogeneity. The field can be measured far away from the inhomogeneity or in some region which is not too far from the inhomogeneity. Several problems of this type i r e solved in this paper. The problems are formulated in $5 2-5. Section 2 deals with recovery of the refraction index (velocity profile) from the knowledge of the field on the plane for all positions of the source and receiver. Other problems of a similar nature are also considered. Section 3 deals with the two-parameter inversion: both the density and velocity are recovered from the knowledge of the field for all positions of the source and receiver on the plane and for two distinct frequencies. In Q 2 the theory is exact, in 5 3 it is based on the first Born approximation. In Q 4 the three-dimensional inverse problem of the theory of scattering by a potential is solved in the Born approximation. Section 5 deals with a model problem of induction logging theory.

The results in this paper were announced by Ramm (1983a, b). The presentation in Q 2 is based partly on the work of Martin and Ramm (1985), in Q 2.5 an idea of Blackledge (1 983) is used, Q 3 is the work of Ramm and Weglein (1984) and Q 5 is the work of Ramm (1 985).

2. Inverse scattering for geophysical problems

2.1

Let

v2U+W*tz(X)U=-J(X--y) in R 3

n(x)= 1 if 1x1 ) R n(x) = 1 + v(x) if 1x1 < R

n(x) >0, v(x)=O if 1x1 >R or x3 > 0.

0266-561 1/85/020133 + 40 $02.25 @ 1985 The Institute of Physics 133

134 A G Ramm

The function U is the outgoing (i.e. U satisfies the radiation condition) scalar, e.g. acoustic, field generated by a point source located at the point y in the homogeneous space with a compactly supported inhomogeneity v(x) located in the lower half-space R I = { x : s 3 < O } . Let P= { x : x 3 =O}, BR = { x : 1x1 G R } . The inhomogeneity is not assumed small in any sense. No symmetry assumptions of the type v = v ( x 3 ) or u = u ( p , x 3 ) , where p=(x: + x:)”~, are made. The only assumptions about v(x) are

u(x) = 0, 1x1 > R or x3 > 0. v(x) E L ‘(R 3) . (2.4)

The inverse problem we are concerned with consists of finding v(x) from measurements of the field u(x. JJ, w ) for all positions x, 2’ E P of the receiver and source and for small U.

The notion of the smallness of o will be specified later. This is a model seismic exploration problem in which the refraction coefficient n(x) = c-’(x), where c(x) is the wave velocity, and it is assumed that the measured field is the acoustic scalar field. One could assume that c = c(x. U) but in this case only c(x, 0) is recovered, Our method is exact and analytic. The Born approximation is not used. The assumption v(x) = 0 if 1x1 2 R can be relaxed but fast decay of c(x) at infinity is necessary for our method. For example, one can assume that

but further relaxation of the assumptions on the rate of decay of z! would require some new ideas.

The requirement that u(x,J’, o) is measured for w+O does not restrict the resolution ability of our method as will be clear later. Therefore the usual concept, according to which the size of the details that can be resolved by testing an object with waves of length 1 is of order 1, is not applicable to our scheme. The reason is that the data we use differ from the usual data.

There are quite a few practical examples in which the condition 0-10 is satisfied. Currently the wavelength 1 used in seismic prospecting is in the range 50-400 m. If the characteristic dimension a of the inhomogeneity v(x) satisfies the condition a < 1, then the condition w + 0 is satisfied.

Practical examples include (1) metallic ore lodes (a - 30 m), (2) caves (a - 20-80 m), (3) old mine structures (a N 50 m), (4) submarines (a - 50-100 m), (5) pipelines (a - 3 m). The last problem requires some modifications in the theory because the inhomogeneity is infinite in one direction (see Ramm 1983a).

2.2

The method we use for solving the inverse problem is simple. One starts with the integral equation

u ( x , ~ , U)=g(x,Jl, U)+ w2 g(x , z , o)v(z)u(z,y, w ) d z = g + o 2 B ~ (2.6)

where g= exp(iolx-yl)/4~lx-yj and the integral is taken actually over R? because of (2.4). Equation (2.6) is equivalent to (2.1) which can be checked by differentiation, and the solution to (2.6) is an outgoing wave.

Under assumption (2.4) the linear integral operator B in (2.6) is compact in the space Cy of functions with a fixed singularity at the point y. The functions are of the form U = C/jx-y/ +h(x,y) where h(x,y) is continuous, and the norm in this space is given by the equation l/ull= /Cl + maxXER3 !h(x, y)l.

j

Some inverse scattering problems of geophysics 135

Lemma 1. For sufficiently small o equation (2.6) is uniquely solvable in Cy, its solution is analytic in w in the neighbourhood of the point o = 0 and can be obtained by iterations

U,+ =g + w ~ B u , U0 = g. (2.7) Prooj The conclusion of lemma 1 is an immediate consequence of the Banach fixed point theorem.

Let us rewrite (2.6) as

U-g 1 6n2 - = Bg + 0(o2) o2

and, using lemma 1, pass to the limit U - 0 in (2.8). The result is

where

(2.10) U-g 1 6n2 lim -, f ( x , y ) m-0 w2

Let

x1 =(x1,x2,0) Y ' =(Yl ,Y2 ,0) .

Setting x=xl andy=y ' in (2.9) yields

(2.1 1)

(2.12)

This is our basic equation for finding c(z) from the data f ( x ' , y ').

Remark 1. One can measure the data on surfaces other than P. For example, on a sphere SR, with radius R 1 > R or on a cylinder Cx, ={x:x; + x i =R:}, R1 >R. The corresponding inverse problems can be useful also. See 0 2.5 below.

2.3.

Let us solve equation (2.12). Taking the Fourier transform in x1 and y ' and using the formula

(2.13)

136 A G Ramm

one obtains

where

and

--m

Let

p=A+,u PI =A, +PI P2=A2 +P2

P3 = 111 P4 = IPI.

Then (2.14) can be written as

(2.15)

(2.16)

(2.17)

1 0 1 0 0 1 0 1

J = A11A1-l 12111-1 0 0 c 0 PIIPI - I iu2lPl - I

so that Jf 0 ifp and A are linearly independent. Let

P3 =P4 = t 4 . (2.20)

Equation (2.20) determines a three-dimensional manifold in the four-dimensional space of variablesp1,p2,p3,p4. Equation (2.18) reduces to

jr exp(-qO( 0 dzl dzz ~ X P ( ~ P * Z')V(ZI, 22, -0 d C = 7 1 2 ~ 7 ( ~ ~ , ~ 2 , -m 1 (2.2 1)

where

dP19P23 4)=q2F(Pl3P2, (2.22)

andc=Jz31=-z3.

transform and one-dimensional inverse Laplace transform:

~ Z I , ZZ, ~ 3 ) = a dpl dp2 exp(-ip z') - j dq ~ P I , P ~ , q) exp(-qzd (2.23)

where p 6 z' =plz l +p2z2 and c > 0 is an arbitrary constant. The Mellin integral, which gives the inversion of the Laplace transform, does not depend on c > 0. Computational

Equation (2 1) can be solved analytically by taking the two-dimensional inverse Fourier

P). 1 -c+im

2711 c-im -m


aspects of the problem will be discussed in 0 2.6. Let us summarise the result that we have just proved.

Theorem 1. Equation (2.12) is uniquely solvable by formula (2.23) in which q is given in (2.22) and F ( p l , p 2 , p j , p4) is the function (2.16) in the variables (2.17).

Remark 2. The above argument does not give any explicit characterisation of the functions f ( x ' , y ' ) for which equation (2.12) is solvable. Equation (2.12) is an integral equation of the first kind since the support of v(z) is a compact set. Therefore the range of the operator Mp :L2(BR)+C(Px P) is not closed; small perturbations off(xl,y ') in the norm of C(P x P) may lead to an equation which has no solution in L2(BR) , and therefore the problem of solving equation (2.12) is ill-posed. An implicit characterisation of admissible data, that is of the functions f ( x l , y l ) for which equation (2.12) is solvable in L2(BR) can be given as follows: f ( x ' , y ' ) are admissible data iff formula (2.23) gives a function U which satisfies conditions (2.4).

If the data are not measured on all of P but only on a finite domain, e.g. on a disc Q,, of radius R I , then one can still apply the inversion formula (2.23),7@, p) will be the Fourier transform of the data collected on 8,, , but the integral (2.23) will not necessarily vanish outside BR.

One could try a numerical method to solve equation (2.12). Since this is an equation of the first kind some regularisation is needed.

2.4

If the data are measured with a random error 7 uniformly distributed on the interval (-6, 6), 6 > 0, then one can define the right-hand side of equation (2.12) as the solution of a linear regression problem. One measures U = u(x, y, oj) at several small frequencies wj,

uj = g + W;b + vj l < j < n (2.24)

and estimates b from these measurements. This b is then taken as the right-hand side of equation (2.12). The least-squares method gives

(2.25)

where =g(x, y, wj), n is the number of various wj used and b = b(x ', y I).

2.5

Consider the case when the lower half-space, which contains the inhomogeneity, has different properties from the upper half-space. Let p l , a , and p2, a2(x) be the density and velocity in the upper (x3 > 0) and lower (x3 < 0) half-spaces respectively. Let us assume that there is a compactly supported inhomogeneity in the lower half-space R! , that p 1 =constant, a, =constant, p2 =constant, a2(x)= a2 + 2;(x), where a2 = constant and v (x ) = 0 if 1x1 > R or xj > 0. The Green function for the problem without an inhomogeneity, i.e. for 2; = 0, satisfies the equations

v 2 (f l i+T(fl i=-8(x-y) i= 1,2 ai

-=- 3471 a(fl2 P2472 atxj=O. ax3 ax,

(2.26)

(2.27)

138 A G Ramm

Conditions (2.27) mean that the normal displacement and the pressure are continuous across the interface. p1 (q2) in (2.26)-(2.27) denotes the value of the Green function in Rt (R!).

Ewing er a1 (1957, pp 95-6) have calculated the Green function:

1 exp(w3 -w3)JO(kP)k dk x3 < (2.28) 472=-

where y3>0 (i.e. the source is in R : ) , ~ ~ = ( k ~ - - w ~ c r ; ~ ) " ~ , p= lx ' - y ' l = [ (x, -y1)' + (x2

6Vl +U2 271 1 6 = p 2 p ; ' . Takey, = O to obtain

P2 =- x3 < 0.

If U-+ 0 then vi -+ k and (2.29) takes the form

(2.29)

(2.30)

Therefore the basic integral equation (2.12) remains essentially the same:

where 6=p2/p1 and f ( x ' , y ' ) is given in (2.10) with p2 in place ofg.

2.6. Inversion of the data given on a sphere

Consider equation (2.6) in the Born approximation at an arbitrary positive frequency w. This equation can be written as

x, y E sR1 (2.32)

where the right-hand side is the datum

f(x, Y ) = (U -g)o - 2 (2.33)

and RI > R , so that the data are collected on the sphere S R I , = j B R . Let us use the well known formulae (Gradshteyn and Ryzhik 1965)

m

g(x, z, U)= c ~;lClfe(z)lClr(x) 1=0

/=O m=- / (2.34)

W m i

exp(iwn - z)= c a I @ ( z ) ~ = 4 7 c 1 i ' j i ( w l z l ) Y / m ( z o ) ~ (2.35) i=o i=O m = - i


where Y, are the normalised spherical harmonics, the regular waves

are the outgoing waves, and @? are

= ( ~ / 2 r ) l i 2 J r + I , * W hi'' (r) = (J2r ) 1'2Hjy (r)

(x) = $')(IxI) Yrm(xo) if'(x)=jr(IxI)Yim(xO)

21+ 1 (1-m)! 477 ( I+m)! - - (e, v).

Substitute (2.34) into (2.32) to obtain

(2.36')

1 U(Z)+~'(Z)@;?(Z) dz = U//,. (2.37)

Multiply (2.37) by alal,Yl(n')Y,,(-n), where n and n' are arbitrary unit vectors, and use (2.35) to obtain

-__

-- a J U(Z) exp[-iw(n-n') * Z] d z = 1 arar,Uir,Yi(n')YI~(-n). (2.38) /,/'=O

Denote the right-hand side of (2.38) by F(w(n - n')) and set w(n - n') =p. Then

f ~ ( z ) exp(-ip z ) dz =F( p) IPI < 2w. = BR

This equation is solved in $ 4 (see equation (4.9)). Let us summarise the result.

(2.39)

Theorem 2. Equation (2.32) has at most one solution U E L2(BR) . If it is solvable then the solution can be computed as follows: (1) compute numbers Arc by formula (2.36'), (2) compute the right-hand side of (2.38), (3) solve equation (2.39) for a(z) by the method

In a similar way one can solve the inverse problem in the case when x and y run

If w=O then the above scheme should be modified because the right-hand side of

If w = 0 then the numbers

Of $ 4 .

through a cylindrical surface, x: + xt = R:: R > R.

(2.39) is known only at a point p = 0 if w =O.

[ t.(z)lr~'+''Y~(zO)Y~,(zO) dz=t..i, (2.40)

are uniquely determined by the data f ( x , y ) known for x, y E S R I . These numbers define v(z) uniquely.

140 A G Ramm

Lemma 2. Equation (2.32) for w = 0 has at most one solution.

Proof. Suppose f(x,y)=O, x, y E S,, , Then the numbers u&=O, V I , l‘>O. We want to prove that in this case v(z) = 0. Let

m u(z)= 1 ur.(r)~,.(z - O<r=jzi<R. (2.4 1)

Clearly, it is sufficient to prove that ui(r)=O, V I > 0. It follows from (2.40) and (2.41) that if qi, = 0 then

I”=O

where

b(I, l‘, 1”)=Js2 Y,(zo)Yp(zo)Yp(zo) dzo.

Let us consider for simplicity the two-dimensional analogue of (2.42)-(2.43):

V 1, I‘

where I , l‘, l”=O, 21, 2 2 , . . . ,and

It follows from (2.44) and (2.45) that

(2.42)

(2.43)

(2.44)

(2.45)

(2.46)

Let I” be fixed. Then 1 + 111 + /l’j runs through all integers starting from 1 + 11”1 when 1 and I‘ run through the set 0, 2 1, 22, and 1 + l‘=l“ is fixed. Therefore

dr I-9vi. (r) = 0 q> 1 + I1”l. -0

(2.47)

From Miintz’s theorem (Akhiezer 1956) it follows that uiln(r) = 0 for all I”. Thus u(z) = 0. In the three-dimensional case one uses the formula (Lubarskii 1957, 0 54)

Kim, (~)yi,m,(w)= 2 ~ ( [ i , 1 2 , /1>(1112mIm2/llm’)~1,m,(w) m’ = m , + m2

where (I1l2mlm2/lm) are the Clebsch-Gordan coefficients and p ( I 1 , 1 2 , I ) = [(21, + 1)(212 + 1)/(21+ l)]’~2(l,lzOO[10). Multiply the above equation by F i jm j (u ) and integrate over S 2 to obtain

11 +12

1’=111 -121


where the formula Is2 Ypmjylm dw=JIldmm" was used. It follows now from (2.43') and (2.42) that

O= 2 5 j.R drr2+11+12 vi3 m3 ( r l~( l i 12 13)(1112 m 1m2 I l m )

where we wrote I , , 1, and l3 for I , i t and I" . Multiply this equation by (l1l2ml, m21LM), sum over m, and m2, and use the orthogonality conditions (Lubarskii 1957 5 54)

g = O m3=-/3 0

to obtain

(2.43)

The coefficient p( l l , I , , L ) vanishes unless I I + 1, + L is even and 11, - 121 <L <1, + 1,. Take 1, = I I +L. Then the above conditions are met, and therefore p ( l , , I , + L, L)#O. Thus (2.48) yields

O = i R drr2+211+L ULM(~) l ,=O, 1,2, ....

As above, Miintz's theorem yields vLM(r)=O. Since LM: L =0, 1, 2 , . . . , -L <M<L, are arbitrary, v(z) = 0.

Remark 3. The moment problem which consists of finding a function f(r) from the knowledge of its moments f,,

f(r)rm dr=fm m=mo, mo + 1, mo + 2, . . .

has been studied extensively (see, e.g., Akhiezer 1965).

2.7. Computational aspects of the inversion scheme based on equation (2.21)

function (a) If one inverts the Laplace transform using the Mellin contour Re q=c, then the

c + i m

i( P,, ~ 2 , z3 ) = (274 - j" P(PI>P2,4) exp(-qz3) dq (2.49)

is computed for some values p I , p 2 , and z3 < 0. If q is a real number then, according to (2.171, (p1 i <p3 +p4 < q, Ip2 1 < q. In particular, lp, 1 < c and lp2J < c. Therefore, for a fixed c > 0, one obtains the values of the Fourier transform of v(zl, z2, z3) in a disc on the real two-dimensional space of the variablesp, andp, :

JJ exp(ipz1)v(z~,z2,z3)dr, dz2=n2i(pl,p2,z3) bli2 + /p2t2<c2. (2.50)

Of course, the function @(pl ,p2, z3) can be computed for any desired real point ( p I , p 2 ) by the Mellin integral provided that the parameter c in this integral is chosen so that lp,/ <c and lpzj < c. Therefore, the function i ( p l , p 2 , z3) can be computed for all realp, andp,.

e-im

-m

142 A G Ramm

However, it is of interest practically to invert the Fourier transform of a function L' with a known compact support (say L' = O for / z / > R) if the Fourier transform is known on a compact set only, for example on a disc lp112 + lp2j2 <c2. This problem is solved in 0 4.

(b) Let us briefly review some methods for inverting the Laplace transform (see Bellman et a1 1966, Krylov and Skoblya 1969, Davies and Martin 1979).

1. Let

F(p) = L f = exp(-pt)f(t) d t p = G + id (2.5 1)

where I f 1 <c( l + t)-', U > 1, c= constant > 0. Then F ( p ) is analytic in the half-plane G > 0 and continuous up to the line G = 0.

Lm If F(G), G > uo > 0, is given, where G~ is an arbitrary fixed number, then

(2.52)

This is a well known formula (e.g., Davies and Martin 1979). Although higher order derivatives have appeared in (2.52) this formula can be used in practice, e.g. in the case when F ( p ) is a rational function.

Example ,n+ 1

= lim = e -'. n-m (n + tin+'

1 F ( p ) = -

P + 1

If we can approximate F(p) by a rational function P,(p)Q;'(p) so that

IF(P)-P~(P)Q;~(P)I + IPI)-' U > 1, G > G o , (2.53)

then

where f , ( t ) = L -'(P,(p)Q;'(p)). The difficulty is that the simple stability estimate (2.54) holds if (2.53) holds in the complex half-plane. It would be interesting to obtain an estimate for If(t) -f, (t)l if an estimate of the type (2.53) is known on the real axis o > G~ only.

2. If F(p) is known in a complex half-plane G > oo > 0 then the Mellin inversion formula can be used:

(2.55)

Integral (2.55) can be easily reduced to the Fourier transform

1 5 0 f ( t ) = 1 exp(at)F(uo + 2) d ; ~ exp(u,t), (2.56)

which is convenient for numerical computation. There are a number of quadrature formulae and tables for numerical computation of (2.55) and the FFT routine for numerical computation of (2.56).

- m

Some incerse scattering problems of geophysics 143

3. Inversion methods which use the values of F ( p ) on the real axis only are of particular interest. Formula (2.52) is very sensitive to the errors in F ( p ) because it requires high-order derivatives of F(p) . Therefore the following methods, which are more stable towards the errors in F(p) , are of interest (Krylov and Skoblya 1969).

Suppose that f is expanded in the Fourier-Laguerre series

33 exp(-it)Ln(t) n ! f(f)= An

n=O

where L,(t) are the Laguerre polynomials

(2.57)

t n - 2 +. . .+(-I)%! (2.58) n2(n - 1)2

L,(t)=(-l)n t"--tt"-' + ( ; ' 2!

Then

1 A , =- exp(-it)l,(t)f(t) dt.

Let -33

am= 1 exp(-ft)t"f dt. J O

Then

a, = F"'(f)(- 1)"

so that

and, by (2.57), (2.61) and (2.62),

(2.59)

(2.60)

(2.6 1)

(2.62)

(2.63)

The derivatives of F are present in this formula, but if the series (2.57) converges rapidly, so that only a few first terms are needed for approximating f(& then (2.53) might goik better as an inversion algorithm than (2.52).

4. One can use other expansions off(t) for inverting the Laplace transform. If exp(-t)=x then (2.5 1) becomes

jol xP-'q(x) d u = F ( p ) &Y)= f(-ln x).

Let p = 1.2,3, . . . . Then

(2.64)

- 1 1 .u"p(x) dX=F, m=O, 1, 2 , . . . . (2.65)

This is a well known moments problem. Let P,(x) be the Legendre polynomials and ~n(.u)=Pn(2.u- 1) be the polynomials defined on the interval [0, 11.

- 0

144 A G Ramm

Let m

f ( ~ ) = 1 ( 2 ~ + l)CnFn(x) n=O

where 1

c, = j f(X)Fn(x) h. 0

Let n

F ~ ( Y ) = 1 p n m Y m . m=O

It follows from (2.68) and (2.65) that n

Cn= C PnmFm. m=O

The inversion formula is (2.66) with C, given by (2.69).

(2.66)

(2.67)

(2.68)

(2.69)

5. One can expand f in a series of Chebyshev polynomials or a Fourier series (Krylov and Skoblya 1969). In the first case one sets exp(--t)=i(I +cos 8) andf(t)+p(O), O<8(n. In the second case one sets exp(-of)=cos 8, 0 > 0 , f ( t ) = f ( - u - ~ In cos O)=p(8), 0<8<$n.

6. Finally, one can consider equation (2.5 1) with p = o > 0 as an equation of the first kind and solve this equation for f ( t ) numerically. This is an ill-posed problem and therefore one should use some regularisation techniques (iterative methods, variational methods, singular value decomposition or the method of quasisolutions (e.g., Ivanov et a1 1978, Tikhonov and Arsenin 1977, Ramm 1980, 1981)).

2.8

Consider the two-dimensional analogue of the problem (2.1)-(2.3). The data now are the values of the field u(x, y , k) on the line L = {x :x2 = 0) for all positions of x andy on L. The assumptions about v(x) are the same as in Q 2.1. The governing equation is equation (2.6) with

g(x, ,U, w) =$iH$"(wr,) rxY=/x--Yl (2.70)

where H','' is the Hankel function. The basic difference between the three-dimensional and two-dimensional problems is that the function (2.70) does not have a finite limit as w+O. One has

g(x, y ) = a(w) + go(x, y ) + O(w2r2 log l/w) asw-0 (2.71)

where

(2.72)

and y=0.5772.. . is Euler's constant.


where

(2.73)

(2.74)

The field if w is sufficiently small. It follows from (2.71)-(2.74) that

is the scattered field. Equation (2.73) is uniquely solvable in C ( R 2 ) by iteration

i ( x , y ) = w2 [a2 I U dz + a ( go (x, z)v(z) d z + go(y, z)v(z) dz J’

Therefore

so-= lim (wa)-2@(x,y, U)= v(z)dz=vo 0-0 i’

(2.75)

(2.76)

and

(2.78)

The quantities$, j = O , 1,2, can be measured, and they are our data. In particular,fo gives the number uo. I f x = y then equation (2.77) becomes

X€L, 9 = { x : v ( x ) # O } . (2.79)

This equation is not unique!y solvabie in general. indeed, the right-hand side of (2.79) is the potential of the mass distribution v(z). This potential is given on the line L. These data and u0 determine the potential uniquely in the half-plane x2 > 0. By the unique continuation property the potential is uniquely determined in 52 = R 2 \ 9 . However, there are many distributions U which produce the zero potential in R. For example, take the origin in 9 and let the disk 9,, belong to 9. Choose r2 and r3 such that 0 < r3 < r2 < rl . Define ~ = [ n ( d - - r ; ) ] - ~ if r3<r<r2 , v = - - [ ~ ( r ~ - r ~ ) ] - ~ if r2<r<r1, V = O if r > r , or r<r3 , r=Ix[. Then the left-hand side of (2.79) vanishes for r > r l r but vfO. Under some special assumptions on v(z) equation (2.79) can be uniquely solvable. For example, if v = v ( x 2 ) for a < x l <b and -R <x2 < 0 and is zero otherwise, then taking the Fourier transform of (2.79) yields

146 A G Ramm

where m

F1(d)=(27r)-' 1, fi(x, x ) exp(i1x) dx. (2.81)

Inverting the Laplace transform (2.80) yields u(z2). Thus, under the above assumption about U(Z) = u(z2) an explicit analytical inversion procedure is given in terms of the back- scattered data measured on the line x2 = 0.

If no special assumptions on U are made then one can use equation (2.78) to recover U:

X , Y E L (2.82)

f ( X 3 Y ) = 4712f2k Y) .

This equation is similar to (2.9). Taking the Fourier transform in x=(x, 0) and y = ( y , 0) yields

where

(2.84)

and

Integral (2.85) can be calculated as follows. Notice that

(274-l im exp(ilx) 1n{[(x-zZ1)2 +z:]'/~} d x -m

=exp(ilzl)(2n)-' exp(i1y) In[(y2 + z:)'/~] dy E and

m n Iny=(271)-' [ lnlyl exp(iy) dy=-(2l1l)-',

.-m

Indeed,

y - ' exp(ily) dy=(2n)-'i71 sgn A = $ sgn d)

(2.8 5 a)

(2.8 5 6)

and (2.85b) follows. Formula (2.85) follows from (2.85a) and (2.856) if one takes into account that integral (2.85) is a harmonic function of(z, , z2). Another way to derive (2.85) is to differentiate the integral on the right-hand side of (2.85) with respect to z2, calculate the resulting standard integral and integrate over z2. Integral (2.853) serves as an initial condition at z2 = 0.


Let

p=,’.+p and 4= 111 + IPI. (2.86)

As in 5 2.3 one can prove that equation (2.83) has at most one solution U E L2. The Jacobian of the mapping (2.86), {A, p } --$ {p, q } is J = sgn p - sgn A # 0 if i p < 0. Take p > 0 and d < 0. Then (2.83) becomes

I’ v(z) exp(ipz, + 422) dz= lP2-q2IF(P, 4) z 2 < 0 (2.87)

where

F(P, 4 ) = F N p - q ) , i(P+4)1. (2.88)

If d < 0 and p > 0 then the corresponding point (p, q) lies inside the wedge 0 < q < CO,

Equation (2.87) can be written as -4 < P < 4.

(2.89)

where

d P , 4)= lP2 -q2JF(P, 4). (2.90)

Equation (2.89) is analogous to (2.21), and one can repeat the analysis given in $ 5 2.3 and 2.6.

2.9. Consider the one-dimensional case

The problem now is to find v(x), x E R ’ , from the measurements of u(x, y , w ) for x>O and y >O. We assume that v(x) E L z , v(x) = 0 if x < -R or x > - E where E and R are positive constants, E < R. The Green function is given by

1 1 (2.9 1)

1 g=- exp(iwlx-yj)=-+g, =-+go + O(wr,) a5 w+O

2w 2w 2w

where go = -ilx-yl, rxv = Ix-y/ . The integral equation (2.6) becomes

exp(iwlx-zl)v(z)u(z,y> dz.

Let u-g=@. Then

(2.92)

(2.93)

where

@o(x,y)=-a ~exp(icolx-zl + Iy-zl)v(z) dz. (2.94)

Equation (2.93) is uniquely solvable in C ( R ’ ) by iteration if jut dz < 1, in particular, for sufficiently small w. Thus, 3 is analytic in w in the neighbourhood of the point w=O. The function i+b is the scattered field, it is the datum.

148 A G Ramm

It follows from (2.93) that

Let

(2.95)

(2.96)

and m

@ 0 = 4 c @oJiWy’ j = O

i+kOj=+ ~ ( I x - z ~ + iy-zi)jv(z) dz. (2.97)

Substitute (2.96) and (2.97) into (2.93) and compare the coefficients in front of (iw)’ to obtain the recurrence formula

J .

Cj(x,y)=t 2 j- I x - ~ ~ ~ ~ ( z ) C , ( z , y ) dz-$@oj j = O , 1, 2 , . . . . (2.98) m + n + l = j m!

I f j=O then (2.98) gives

CO (x, y ) = - + @m = - 4 U g . (2.99)

I f j = 1, then

c,(x,y)=-!v; +Iv,-$(x+y)v0. (2.100)

I f j = 2, then

c, (x , y ) = -$U; + i U 0 U l -$ U;(x + y ) - 4 U& - $Uo (x + y), f 4v l(X + y ) - f C 2 . (2.101)

The coefficients Cj can be considered as data. Equations (2.99)-(2.100) determine uo , U, and v2 recurrently. In principle all the moments U, (see (2.95)) can be determined. It is a classical prob!em, Eamburger’s moment problem (Akhiezer i965), to find U(Z) from the knowledge of its moments U, :

0 lR U(Z)Z, dz= U, n=O, 1, 2,. . . . (2.102)

We had a similar problem in 5 2.5, equation (2.48). Since the system {z‘], n=O, 1, 2 , . . . , is closed in L2(-R, 0), the moment problem (2.102) has at most one solution. Necessary and sufficient conditions are known for the solvability of problem (2.102) (Akhiezer 1956). A numerical solution similar to the T-matrix approach studied by Ramm (1982a, b) and Kristensson et a1 (1983) can be suggested. Namely, take a basis (qi(z)} of L2(-R, 0), look for an approximate solution in the form

m

(2.103)


substitute (2.103) into (2.102) and use the first m equations (2.102) to obtain a linear system for the coefficients a y ) :

(2.104)

We will not go into further detail because it is probably not so easy to measure higher order coefficients Cj(x, y ) , j > 2, and therefore find the higher order moments.

It is worth mentioning that an inversion procedure within the Born approximation can be constructed easily. In the Born approximation one has @= y?o, so that

1 exp(iwlx - zl + I y - zl)v(z) dz =f(x, y ) x, Y>O (2.105)

where

f(x, Y , 0) = -4y?(x, Y ) (2.106)

is the data. Set x = y in (2.105) to obtain

exp(-2iwz)v(z) dz = exp(-2iwx)f(x, x, w) =f(x, w).

Let us fix x>O arbitrarily and consider the data f(x, w)=q(w) as a function of w. Then one can Fourier-invert the data q(w) and compute u(z). Since v(z) is compactly supported the Fourier inversion can be done also in the case when the data ~ ( w ) are given on a finite interval of frequencies, wo < w < w,, This is done in Q 4 below.

In conclusion, it is worth mentioning that if the incident field is a plane wave and the measured quantity is the reflection coefficient which is known at all frequencies, then the one-dimensional problem can be reduced by the Liouville transform to the problem of inverse scattering by a potential. This problem has been studied in the literature (e.g., Chadan and Sabatier 1977).

(2.107) C

2.10. Inversion of the data given on a cylinder

Consider equation(2.32)withx,yECR,,whereCR, = { x : p = R l } , R 1 > R , p = ( x : + ~ : ) ” ~ . We have

where a = ( k 2 and Hn(r)=Hil)(r) is the Hankel function, px=py=R1, px= (x: +

150 A G Ramm

Equation (2.32) can be written as

-i(Lxg +n~3) ]Hn(aR, )Hn , (a ’R , ) dz v(z) exp[i(n + n’)q, .1

where a’=(k2 and

Heref(yx, x,, q,.,y3) is the data, the right-hand side of the equation (2.32). From (2.1 10) it follows that

Multiply (2.1 11) by in+” exp[-i(nq + dq’)], sum over n, n’ and use (2.109) to get

dz v(z) exp[ik(m + m’) z] =F[k(m + m’)] k > O (2.1 12) I S R

where

m = (sin 8 cos q, sin B sin q, cos 8) m‘ =(sin 0‘ cos q’, sin 8’ sin p’, cos 0’)

Equation (2.1 12) is of the type (4.9) and can be solved as in 9 4.

homogeneous equation (2.9) with x. y E CR, has only the trivial solution. If

A = k c o s B a = k s i n B 1,’ = k cos 6‘ a’ = k sin 6’.

Equation (2.9) with x, y E CR, is equation (2.32) with o = 0. Let us prove that the

w(x,y)= r dz v(z)lx-zi-‘iy-zl-’=O x . Y E C R j J lz l<R

then for any fixed y , jyl > R, and for 1x1 > R the function w(x, y ) solves the problem

V:w(x.y)=O if Ixl>R and I j l>R (2.1 13)

R and y E CRI , because (2.1 13) has only the trivial solution by the

w(x,y)=O if YEC,, and XECR, ; W ( C O , Y ) = O .

Thus w(x, y)= 0 if 1x1 maximum principle.


Fix x, 1x1 > R, and notice that w(x, y ) as a function ofy solves the problem

Vjw(x,y)=O if Ixl>R and iyl>R

~ ( x , y)=O if Y E CR, and Ixl>R; w(x, co)=O.

This problem has only the trivial solution w(x, y ) = 0, y : +yi > R:, 1x1 > R. Therefore, by the unique continuation property for solutions to Laplace’s equation, w(x, y ) = 0 if 1x1 >R and IyI >R. In particular, w(x,y)=O if x , y ~ P, where P is a plane which does not intersect the support of t’. This implies that U = 0 as is proved in theorem 1.

3. Two-parameter inversion

3.1

In 0 2, one unknown function n(x) was found from the seismic data. In this section we use the data at two frequencies to find the density of the inhomogeneity and the velocity. Consider the equation

where x, y E R 3 , p(x) is the density and k(x) is the bulk modulus. Let us assume that

1 1 a2(x) - 1 1 al(x) k(x) - kr P r P P r Pr (3.2)

where k, and pr are positive constants, and aj(x), j = 1, 2, are functions with coapact support, i.e. aj(x) = 0 if 1x1 > R , where R > 0 is an arbitrarily large fixed number. We assume that U , E L 2 ( B R ) , BR = {x: 1x1 < R ) and a2 E H’(BR).

Equation (3.1) can be written as

V 2 f-j- G-w’a1G-V * (azVG)=-p,d(x-y) (3.3) 3 where c2 =k,p; ’. Let us choose units in which

c= 1 & = l . (3.4)

This is possible: take x= ax’, y=ay’, o=Pw’ where a and ,8 are constants. Then (3.3) takes the form

(a”V” + , L ? 2 ~ ‘ 2 / ~ 2 ) G - / ? 2 ~ 2 ~ , G - a - 2 V ‘ . ~~V’G)=-p;~ad(x’ -y’ ) . (3.5)

(3.6)

Let a =pr , ,L?= cp;’ and define a; = c2al and U$ = u2. Then (3.5) takes the form

(V’2 + C O ’ ~ ) G - ~ ~ U ; ( X ) G - V ’ * (a$VG)=-d(x’-y’).

Therefore one can study the equation

(V2 +w2)G-o2a1(x)G-V - a*VG)=-d(x-y). (3.7)

This equation can be written as the integral equation (which is analogous to equation (2.6))

G=g-w2 j‘galG d z - [gVz - (a2(z)VzG) dz (3.8)

152 A G Ramm

where = l R 3 , dz = dz, dz2 dz3 and

g= exp(iwlx -y1)/4zlx-yl. (3.9)

In operator form equation (3.8) can be written as

G=g- w2gq G -gV - (a2VG) =g-gVG (3.10)

where the potential Vis given by

V=w2al -I v - a 2 v (3.1 1)

Notice that G=g-GVg so that GVg=gVG. Let us define the T matrix

T= V- VGV. (3.12)

The T matrix here is not to be confused with the Tmatrix of Ramm (1982a, b, 1983~). It is easy to check that

T= V- VgT and

Tg= VG.

(3.13)

(3.14)

The operator Vis an integral operator in impulse space with kernel

( 2 ~ ) - ~ ? ( k , p ) = ( 2 n ) - ~ dxdy V(x,y)exp(ik. x - p .y) JJ where V(x,y) is the kernel of V. This kernel may be a distribution. For V defined in (3.1 1) we have V(x,y)=w2a,(x)6(x-y) + V a2(y)V6(x-y), so that

?(k?p)= W 2 6 1 ( k -p) - k * p62(k-p) (3.15)

where E(k) = exp(ik - x)a(x) dx. The scattered field G, = G - g according to equations (3.10) and (3.14) can be written as

G, =-gVG=-gTg. (3.16)

Let us keep the variables x and y (the positions of the receiver and the source) on the plane x3 = O and Fourier transform (3.16) in R=(xl, x2) andj=(y l ,y2) . Let i; x = k l x , + k 9 2 , d i = dx, dx2, etc. Then

- I

G,(k,j,w)=Jexp(i&.P-ij .j)G,(R,j,w)dRd?

= - J’ exp(iE e . ? - i j - j)gTg d.? d j

J exp(iwlz’-Ei +iEi) = - i dz’ dz” T(z’, z”) d i

4 4 z’ - il


- - dq exp(ililp cos q) 477 lo

where (see Erdelyi et a1 1954, formula (8.6.21))

(3.19)

(3.20)

The vectors i and are real valued but k , and p 3 may be complex numbers. If V has compact support (and this was our assumption) then T has compact support as follows from (3.12). Therefore the Fourier transform of T is meaningful for complex p 3 and k3.

154 A G Ramm

From (3.17) and (3.19) one obtains

1 G, = - ?’ 1 dz’ dz” T(z’, z”) exp(ii - 2’ + ik3 Iz; j - i j - 2‘‘ + ip3 1z41). 4k3 P3

(3.21)

Let us assume that the support of V belongs to the half-space z3 < 0. Then jz3/ = - z 3 and (3.21) takes the form

1 -- y(k‘,p) - 4k3P3

lkl = IPI = w

(3.22)

k ‘=( i , -k3) lk’l= Ikl

and

y(k, p) --= dz‘ dz”T(z’, 2”) exp(ik a z’ -p - 2”). (3.23)

The basic problem can be formulated as follows. Given the scattering data G(1, x3 = 0,j,y3 =0, w) for all .? and j and two distinct frequencies oj, j = 1,2, or small CO, find

Formulae (3.12), (3.15) and (3.22) are basic for our first inversion scheme with the data given at two distinct frequencies. Our second inversion scheme is based on the method given in 0 2.

The inversion schemes can be generalised to include the dissipative terms (corresponding to first derivative in time with a coefficient depending on x but not on t i n the time-dependent problem). Our method is based on the Born approximation. Since the potential (3.1 1) contains second derivatives it does not become small even for w+O.

uj (x ) , j= 1,2.

3.2

Let us describe the first inversion scheme. Formula (3.15) shows that under the assumption

Ikl = ( P I =U, (3.24)

which means that the Tmatrix in (3.22) is known on shell, ?can be written as

(3.25)

where k o = kjkl-’ and are unit vectors which do not depend on W . Since the functions a, and a2 have compact support their Fourier transforms 5, and 2i2 are entire functions of the three complex variables k,, k2 and k3, which decay a t infinity in the real space in R 3 .

In the Born approximation T= V because one neglects the quadratic terms in V in (3.14), and in this approximation (3.22) becomes

.. ViL -\&,p,-COW(k?p) -\ - w‘=Zi1(k-p)-k3 p0Zi2(k-p)

- 1 , CO2 G, =- V(k’, p) = - Wk’, PI.

4k3 P3 4k3P3 Thus

4 G . , k 3 p 3 ~ - ~ = W(k’,p)

(3.26)

(3.27)

Some inverse scattering problems of geophysics

Let

k ’ - p = q k‘+p=s

k’=i(S + q )

k‘ ’ P=:(ls12 - 1412)

k’O * p o =(lsl2- 1q12)(402)-1

1qI2 = 2 0 2 ( 1 -k’O PO) j ~ / ~ = l k ’ 1 ~ + j p 1 ~ + 2 k ’ . p = 2 w ~ ( l +k’O - P O ) W > O

lsl2-Iql2=402-2/q[2.

P =f(s - 4).

Then

s . q=o

Therefore

Let us take two arbitrary frequencies U, and w2 # wi and solve the two equations

155

(3.28)

(3.29)

(3.30)

(3.3 1)

(3.32)

(3.33)

(3.34)

(3.35)

(3.36)

for 2, and &(q). The functions Wj, j = 1, 2, are the data according to (3.27). One obtains

(3.37)

(3.38)

Taking the inverse Fourier transform of the functions (3.37) and (3.38) one obtains a2(x) and al (x) . It is clear from (3.34) that the following conclusion holds.

Conclusion. One cannotjind both functions a , and a2 if the data are given at only one frequency. Indeed, it follows from (3.34) that for a fixed frequency the data depends on q only, and there is no parameter to vary in order to find both 2 , and E 2 . The conclusion that one cannot recover both functions aj , j = 1, 2 , from the data given at one frequency is not at all obvious: at first glance one can think that the two conditions Ikl= IpI = w leave four degrees of freedom in the six-dimensional space Ri x R j , which should be enough to determine two functions Gj, j=. 1, 2, of three variables. This argument, however, is not valid as one saw above. The reason is that the function W( k’, p ) has a very special structure as a function of two vectors k’ and p . Our reasoning applies to the theory in the Born approximation. It does not imply that the conclusion holds in exact theory. The exact theory unfortunately is out of reach.

Another important point concerns the numerical problem of inverting the Fourier transforms of Z j , j = 1, 2. Let W , < w2. It follows from (3.32) that the real-valued vectors q run through the ball Bzw = {q : 141 < 2wJ when k’O and po run through the unit sphere S 2 , so that the number k” - p o runs through the interval [-I, I] . Therefore, for real-valued vectors q formulae (3.37) and (3.38) determine the Fourier transforms Z j , j = I ? 2, in the ball B2w, only. Since aj(x) are compactly supported the functions 6j(q) are entire analytic

156 A G Ramm

functions of q. Therefore if one knows these functions in a ball BZw,, one can uniquely determine these functions everywhere in R 3 by analytic continuation. However, the basic numerical problem is as follows.

Problem. Given the Fourier transform C(q) of a function a(x), a(x) = 0 if 1x1 > R, in the ball BZo = {q, 141 < 2 4 , compute a(x) in the ball BR = {x : 1x1 < R } with a prescribed accuracy.

This problem will be treated in Q 4.

3.3

Let us describe the second inversion scheme. If CO-+ 0 the limit of equation (3.8) is

The scattered field G, = G -go in the Born approximation for x and y on the plane x 3 = 0 is of the form

(3.40)

This equation can be solved analytically by the method given in 0 2. One takes the Fourier transform in 1 a n d j of equation (3.40), uses formula (2.13) and obtains

(3.41)

Here we integrated once by parts, used the equality Iz3/ = - z 3 and set p=d+,u, q= 121 + /P I , (p1,p2), p3 = 111, p4 = 1 ~ 1 . These p and q should not be confused with p and q in 0 3.2. They are the same as in Q 2.

Define

m

= dp exdip 3 lm d5 ~ z G , -0 exp(-qf) (3.42)

as in Q 2. Here C = - Z ~ . The function F(p, q) is the datum. Taking the inverse Fourier and Laplace transforms of F(p, q) one finds az (z) = a z ( i , -5):

-a


If uz (z) is found then equation (3.8) in the Born approximation and for x = (a, 0), y = (j, 0) can be written as

-16z2 lim -- w - 0 o2

where

(3.44)

Q = -J g(a, z)V, - (a2(z)V,g(z9 3) dz (3.45)

and a2(z) is given by (3.43).

First, one computes az by formula (3.43), then al by solving equation (3.44). Equation (3.44) was solved in 5 2. Let us summarise the second inversion scheme.

3.4

In this section we give sufficient conditions for convergence of the Born series. These conditions can be considered as conditions under which the Born approximation is valid. Error estimates will be obtained.

Consider the equation

u=f+m2T1u+T2u (3.46)

where

T1u= galudz s TZU = g V . ( u ~ ( z ) V U ) dz. s

The assumptions about aj(x) are

(3.47)

(3.48)

The Born series for equation (3.47) is of the form u=C:==, (w2T, + T2)y It converges in some space H if ilmzTl + Tzll < 1 where IlTIl is the norm of a linear operator Tacting in this space. Let H=Hz(BR) be the Sobolev space, BR = { x : 1x1 <R}. Since the functions aj(x) vanish outside BR the values of the functions f are determined entirely by the values off on BR. The estimate

IlgfIl2,R <c(R)llf IlO,R (3.49)

is known (see, e.g., Kantorovich and Akilov 1980). Here I J u I ~ ~ , ~ = I I u I I ~ " ( ~ ~ ) . If w+O the inequality ~ ~ 0 2 T l + T2ii < 1 holds if ilT,Il<c and llT211 < 1. Here and below we denote various constants by c. One has, using (3.49),

(3.50) liTlf IIZ,R < c R l l a l f i l O , R <cRcII/f IIO,R <cilflIZ,R. Thus IlT,(l<cif lall<cl. Actually, l l T l ~ ~ < c i f l ~ a l ~ l o , R <cl. Furthermore

i i T 2 f 112,R <cRIIV * (a2Vf )IlO,R G C R * c211fllZ,R. (3.51)

Therefore IITzll < 1 if cRcz < 1. Let us summarise the result.

158 A G Ramm

Proposition 1. If (3.48) holds and q=wzc,cR + czcR < 1 then the Born series X,"=o (w2T, + Tz)"f converges in Hi and is majorised by the series 9".

3.5

Consider the inverse scattering problem for equation (3.7) with the scattering amplitude as the data. If the incident field is the plane wave uo=exp(ioso . x), then the scattering amplitude is given by

f ( s , , s, o ) = - ~ j'exp(-iws -y)[al(y)u(y)w2 + V . (a2(y)Vu)] dy. (3.52)

In the Born approximation U = u0, and, integrating the second term in (52) by parts,

4n d

one obtains for the scattering amplitude

f=fB(so, s, U)=-- exp[iw(so-s) ~yl(al(y)-so . sady)) dy. (3.53) w 2 477 i'

Consider D=-47iw-tfB as the data. Assume that the data are known at u = w j , j = 1, 2, w, < wz. Then the Fourier transforms g j of aj(y), j = 1, 2, can be found from (3.53). Indeed, set

w(so-s)=wp, where [pI2=2-2y and y = s so= l-iIp12.

Then

Dl = 51 (%P) - (1 - f IP lZ)~Z(~1P) (3.54)

Dz = 4 (WZP) - (1 - f lPl 2)5z (WZP).

Set q=w,p. Then (3.54) and (3.55) can be written as

(3.55)

From (3.56) and (3.57) one finds I;,.(q) and, by applying the Fourier inversion, aj(y). j = 1. 2. Again the problem formulated at the end of 0 3.2 arises.

4. An inversion formula in scattering theory

4.1

In this section we solve a three-dimensional inverse scattering problem in the Born approximation, and the problem in 0 3.2.

Let

(Vz +k'-q(x))~=O in R 3

U = exp(iks e x) + U

1, N __. f(s, s', k) exp(i kr)

r as r= 1x1 -+ CO, xr-' = s'. (4.3)


The integral equation for U is

Therefore

In the Born approximation (U = exp(iks . x)) one has

1 - f =.& =-- j exp[ik(s-s’) . y]q(y) dy.

471

Let us assume that

d x ) E L 2 ( B ~ ) q(x) = 0 if 1x1 > R.

Then one obtains for 9(y) the integral equation

exp[-ik(s’-s) . y]9 (y ) dy=F[k(s’-s)] ?YG

(4.8)

where F[k(s‘-s)] =-4zfB. In (4.8) k > 0 is fixed and s and s’ run through the unit sphere S2. Let p=k(s‘-s). When s and s‘ run through S 2 the vector p runs through the ball B2k = ( p : 1p/ < 2 k J . Equation (4.8) can be written as

This equation is solvable in the class of functions 9 satisfying conditions (4.7) iff F ( ~ ) E W R 3 where W, is the set of entire functions of exponential type <R square integrable over R 3 , that is of entire functions such that

This is a variant of the Paley-Wiener theorem. If

G(p) = 1 exp(-ip y)9 dy = 0 for Ipl > 2ko

(4.10)

(4.11)

then one can find 9(.1’)~ the solution of equation (4.9), by taking the inverse Fourier transform of the function F(p) . But (4.1 1) is equivalent to the assumption that q(j.1 is an entire function of class W2,, which is not the case in applications where assumptions (4.7) are more realistic. The idea of what follows is to approximate 9(s) satisfying conditions (4.7) by an entire function 9,,&) E W2b with arbitrary accuracy on BR :/lq,v-qllR<~(N)-tO, N - t x . Here ( 1 . / I R is the norm in L 2 ( B , ) or C(BR depending on whether 9 E L 2 ( B R ) or C(BR) . The function qN(x) will be explicitly constructed analyticaify from the data.

160 A G Ramm

4.2

The data in (4.8) are given on Sz x S2 and it is useful to have a formula which transforms the three-dimensional Fourier integral into an integral over Sz x Sz. Such a formula in R d , d 2 2 , is known (Devaney 1982, d= 2; Beylkin 1983, d 2 2 ) :

(2n)-d F ( p ) exp(ip - x) dp jIpl<2k

’ Is-s’J exp[ik(s-s’) x]F[k(s-s‘)] ds ds‘ (4.12)

(4 - -s/~2)(d-3)/2

where ds is the element of the area of the unit sphere S d - ’ . The identity (4.12) holds for any FE L2(&), Sd-’ is the unit sphere in Rd, wd is the area of Sd-’ , cod= 2nd”/r(d/2), where T(x) is the gamma function. For d= 3 one has

k3 =4 Is-s’I exp[ik(s-s’) 0 x]F[k(s-s’)] dsds‘.

( 2 4 s2 s2

For d= 2 one has

( 2 7 ~ ) - ~ 1 F ( p ) exp(ip - x) dp - lp1<2k

x exp[ik(s-s‘) + x]F[k(s-s’)] ds ds‘.

If (4.1 1) holds then (4.8) is solvable by the formula

q(x) = ( W W 3 [ exdip - x)F(p) dp J

-- - 1s -s’I exp[ik(s - s’) . x]F[k(s - s’)] ds ds‘

where formula (4.13) was used. Before we pass to the formula for qN let us prove the following two lemmas.

Lemma I . Equation (4.8) has no more than one solution.

Lemma 2. Formula (4.12) holds.

Proof of lemma 1. Equation (4.8) is equivalent to (4.9). If F ( p ) = 0, IpI < 2k then

(4.13)

(4.14)

(4.15)

exp(-ip * Y M Y ) dY = 0 p E R3. (4.16)

Indeed, the left-hand side of (4.9) is an entire function ofp. If this function vanishes in the ball B2, it vanishes everywhere. From (4.16) one concludes that q=O.


Proof of lernrna 2. In this proof we use the identity for the spherical means (see John 1955, p 8 1). Let us define the spherical mean for a continuous function F a t the point x:

Z(x, r) = )‘ F(x + rs) ds. Wd Sd-I

Clearly, if F is continuous, then

Z(x, 0) = F(x) .

Let us define the iterated spherical mean

(4.17)

(4.18)

M(x. A, p) = -7j- 1’ i’ F(x+As+ps’ ) dsds’. (4.19) wd Sd-1 Sd-1

One has M(x, 1, p) = M(x? p, I.), M(x, i, 0) = M(x, 0, A) =I(,u, A), M(x, 0,O) = F(x). The identity which we need is of the form (1, ,U > 0):

, y+1

x 1 [ ( r + p - A)(r + p + Ib)(r-,u + A)@ - r + p)](d-3)’2rZ(,~. r) dr. (4.20)

Set I. = p = k in (4.20) to obtain

(4.2 1)

Let h,(p) = Jp1(4k2 - lp12)-‘d-3)’2P(p) exp(ip J ) . where P ( p ) is a continuous function vanishing outside B2k. Taking the spherical mean of h,(p) yields

1 r I,(O, r ) = - P(rs) exp(irs y ) ds.

Wd (4k2 - r2)(d-3)’2 j S d - 1

In formula (4.21), the iterated spherical mean of h,(p) is -2k I dr rd - ’ jsd- * I ~ ( v s ) exp(irs y ) ds Wd- 1 M(0, k, k) =

4 k 2 d - 4 2 d - 3 - 0

Let

(4.22)

(4.23)

= I dy exp(-ip . y)F(y). (4.24)

Let us write M(0 , k, k) for the function h,(p) using definition (4.19) and then equate the “R3

162 A G Ramm

expression to (4.23). This will lead to (4.15). Take F ( p ) = h,(p) in (4.19) to obtain

M(03 k, k ) = - [k(s + s‘)l

x exp[ik(s + s’) e y ] ds ds‘.

Compare (4.25) and (4.23) to obtain

Ld-l ( 4 - ( s + ~ ‘ I ~ ) - ( ~ - ~ ) / ~

(4.25)

x (s + s‘/>[k(s + s’)] exp[ik(s + s’) - y ] ds ds’. (4.26)

If one changes s‘ to -s‘ in (4.26) then one obtains (4.12).

4.3

Let us proceed to the construction of qN(x) which approximates the solution q(y) of equation (4.8). We assume that the data F[k(s’-s)] are exact, so that equation (4.8) is solvable. The calculations are given for the general case of d-dimensional space.

Let us multiply (4.8) by exp[ik(s’-s) a x]h,[k(s’-s)] and integrate over Sd-’ x S d - ’ to obtain

I dy q(y) exp[ik(s’-sj. (x-y)]hN[k(s’-s)l ds ds‘ Sd-1 sd-1

= L d - l i d - I

lr14R

F[k(s’-s)]hN[k(s’-s)] exp[ik(s’-s) . x] ds ds’. (4.27)

Here hN(p) is some function which will be specified later. Our problem will be solved if the sequence defined by the formula

exp[ik(s’-s) x]hN[k( s ’ - s ) ] ds ds’ (4.28) 6N(x)’ I d - I J i d - I

is a delta sequence in L2(BR) or C(BR) in the sense that

(4.29)

(4.30)

and the norm 1 ) J/R in (4.29) is the L2(BR) (C(BR)) norm if q(x) E L2(BR) (q(x) E C(BR)). One can estimate &(N) if some additional information about q(x) is known, e.g. bounds

Let us choose h, such that the sequence (4.28) becomes a delta sequence in the sense 141 < a, Pql< b.

defined above. It follows from (4.12) that

exp[ik(s’-s) x]h,[k(s’-s)] ds ds‘ I d - I L d - I

(4.3 1)


Let us denote

Then, comparing (4.28), (4.3 1 ) and (4.32) yields

(4.33)

Therefore we need to construct a delta sequence in L 2 ( B x ) of functions B,E W2k Such a sequence was constructed by Ramm (1980):

(4.34)

Here lBkI is the volume of the ball of radius k , lBkl = kdnd i2 / r [ i (d+ 2)] and

(2k)d12(2N + d)d/2(2n)d12 I’ d p e x p ( e ) = 2 N + d IxJd’2 Jd /2 (g) (4.35) IPI < 2k

where J,(x) is a Bessel function. Thus

(4.36) Jdl(2 [2klxl/(2N f d ) ] ) [2kx/(2N+ d)ld12

where T(x) is the gamma function. In particular, for d= 3 one has

Notice that the sequence (4.34) is a delta sequence on L2(BR) but not on L2(Rd). One cannot construct a delta sequence &NE W2, on L 2 ( R d ) because the Fourier transform &(p) of a function a,,, E w2k should have its support in the ball B2k while the support of the Fourier transform of a delta sequence on L 2 ( R d ) cannot stay within a fixed compact as N + CO. It follows from (4.32) and (4.33) that

(4.38)

and

hN(p)= lp1(4k2 - lp12)-(d-3)/2A ,(p)k2d- 4( 8ZdWd- 1) - (4.39)

It follows from (4.27), (4.28) and (4.30) that

q,dx)=[ 1 F[k(s’-s)]h,[k(s’-s)] exp[ik(s’-s) x] ds ds’. (4.40)

Let us formulate the result.

~ d - I . s d - I

Theorem 1. If equation (4.8) is solvable then formula (4.40) gives an approximate analytical solution to equation (4.8) in the sense that equation (4.29) holds. The sequence h N ( x ) in (4.40) is given by formulae (4.39), (4.38) and (4.36).

164 A G Ramm

Remarkl. If IslGa and Pql G b (4.4 1)

then, in the norm of C ( B R ) , the estimate for E(N) in (4.29) is of the form

E ( N ) ( T [ f ( d + 1)](I'(fd))-1bRN-''2 + O(N- ' ) N-+ Co. (4.42)

This is proved by Ramm (1 980, p 21 1).

4.4

Let d= 3. Suppose that the data are given with some error q, so that F,(kls'-sl) is known such that IF-F,] < 7.

If one applies the inversion formula (4.40) to the noisy data F , then one obtains a function qN,, such that

1iqN,,-411R(11qN,-qNIIR + i l q N - q i i R < < b N + d N ) (4.43)

where the norm is taken in C ( B R ) ; &(AV), given by (4.29), is the error of the inversion formula (4.40) in the absence of noise, and

(4.44)

where, for d=3, from (4.39) one hash&)= l p l k 2 A N ( p ) ( 1 6 ~ 4 ) - 1 , A N ( p ) is given by (4.38) and JA, is given by (4.37). For a fixed q > 0 one can find N = N ( q ) for which

a(N: q) = qbN + E ( N ) = min. (4.45)

Such an N(q) does exist because c(N)-+O and bN-+ CO as N-+ CO. For this N(7) the error a(q) = a(N(q), 17) -+ 0 as q+ 0. Therefore

l l q ~ ~ l ( , ) , r i ( x ) - q ( x ) I I R -0 as q-+O (4.46)

so that qN(,),rl(x) is a stable approximation of q ( x ) on B R .

The following lemma is useful for computing b,..

Lemma 3.

1,,_, j s d - l .i,l d 2k

- 1

O d - I h[k(s'-s)] ds ds'= 2 d - ~ k Z d - 4 dp h(p))pj-'(4k2- l ~ l ~ ) - ( ~ - ~ ) " . (4.47)

?roc$ The proof is simiiar to the proof of lemma 2 and therefore is left to the reader. From (4.44) and (4.47) it follows that

2n * 1 . bN=F Jlp,d2k dP lhN(P)llPl - I =s lAN(P)l dP. (4.48)

k I d 2 k

A rough estimate for b N , which follows from (4.48), is

(4.49)

Another estimate for bN follows directly from (4.44):

b N G (47d2 max Ih" (4.50) I ~ l d 2 k

Estimate (4.50) can also be obtained from the rough estimate of the first integral in (4.48).

Some inuerse scattering problems of geophysics 165

5. A model inverse problem of induction logging

5.1

Consider a medium whose properties we want to find from the knowledge of the field on the ray xj < 0. One can consider a borehole-a cylindrical hole in the medium; along the x3 axis the source and receiver can be moved. They can either be kept at a fixed distance d, or their positions can vary arbitrarily. The problem is to recover the properties of the medium from the measurements recorded by the receiver.

In induction logging the medium is characterised by the dielectric and magnetic parameters E = 1, p=constant, and by the conductivity U = G ~ + U@), where uo= constant and v ( X ) is the inhomogeneous part of U. The displacement currents 2 9 / a t are negligible. The vector potential A satisfies the equation (V2 + k 2 ) A =-j, k2=iwpo(X) and j is the source current. Let X=(p, q, [) be the cylindrical coordinates of the point x. In induction logging it is always assumed that j is a current in a circular loop of wire. The loop is perpendicular to the xg axis, it has a small radius a, and its centre is at a point (0, 0, [). The vector potential A under these assumptions and under the additional assumption u = v(p, z ) has only one non-zero component, A,, in the cylindrical coordinates, V . A =0, E= iwp l , H = V x A .

We consider a model scalar acoustic problem in which the field U satisfies the equation

V’U + k 2 ( X ) U = - S ( X - Y ) X, Y € R 3 .

k2(X)=iWG(X), u(X)=uO + u ( X ) , oo=constant>O, w>O, v(X)=O if x3 > O or 1x1 > R , where R > 0 is a fixed (arbitrarily large) number, u E L2, Im k ( X ) > 0, ko=(iwuo)1’2. The integral equation for U is

u(X, Y, w)=g(X , U) + i o g(X, Z)v(Z)u(Z, Y ) d Z (5.1)

where exp(ikolX- Yi)

g= 4nIX-YI . If w is sufficiently small then the integral equation is uniquely solvable by iteration and U is analytic in the variablep=w’” in a neighbourhood of the pointp=O.

Therefore the following limit exists:

- ~ ( 2 ) d Z f ( X , Y ) = lim 16n2 -=

W - 0

The function f ( X , Y ) is the datum. Let X=(O, 0, x), Y=(O, 0, y) . Under various assumptions on v(Z) we give inversion algorithms for recovering v(Z) from the dataf(x, 1’) measured on the x3 axis.

5.2. Inversion in the one-dimensional case

Assume that v(Z)=v(z , , z2 , z3)=c(z3) , lz31<R, r<R, and v = O otherwise. Here r=(z: +z;)”*. Let X=(O, O,x), Y=(O,O,JJ), z 3 = ( , y=x-d, d=constant. Then (5 .2) takes the form

-m

166 A G Ramm

where

@(x) = (271) - ' f ( x , x - d )

Equation (5.3) is of convolution type and can be solved by Fourier's transform. If the data are noisy, i.e. $* is known instead of @, I$J - @I < 6, then a regularising algorithm should be used for the numerical solution of (5.3). For example, one can solve (5.3) with $8 on the right-hand side by the formula

CO

v8( i )= i-, d l exp(iA()$*(A)K-'(A) exp(--a(S)IlI)(2~)-'

where a(6)+0 and 6a-'(6)+0 (see Ramm 1980, p 207 for details and for an optimal choice of a(@). One can also solve the Wiener-Hopf-type equation

(5.3')

which models the case when measurements are taken strictly inside the borehole.

Born approximation. The function analogous to h(x- <) in (5.4) will be The same method is valid for inversion of the data a t a fixed frequency w > 0 in the

h , ( x - [ ) = d r r exp(iwp)a-' i where

cr=[(5-x)' + r 2 ] 1 / 2 [ ( ( - ~ + d ) 2 + r 2 ] 1 / 2 = a l u 2 ,B=a,+a2.

5.3. The two-dimensional case

5.3.1. The results of this section show that the data in the Born approximation

F(X , Y)= d Z ( 5 . 5 )

measured for all X, Y belonging to the 5 axis and at a fixed frequency w > 0 do determine v(r, z) in somecases. Assume that v=e(r,z)=Oifr>.!? cr lzl>R.

Let us use the spheroidal coordinates as used by Ramm (1966):

z,=Ist+;(x3 +y, ) z2=1(~2-1)1'2(1 --12)1/2cos B + + ( . ~ ~ + ~ ~ )

z l =l( s2 - 1)''*(1 - t 2 ) l i 2 sin e+;(xl +yl)

where I=ilX- YI, IX-ZI + IZ- YI = 21s, 1 <s < m, (X-Z/ - IZ- Yl=2/t. -1 <t< 1 and 0 is the angle between the plane spanned by X-Z and Y - Z and a fixed plane containing X and Y. The Jacobian of the transformation Z+(s, t , 0) is J = l 3 ( s 2 - 1 2 ) . In our case X and Y belong to the 5 axis. As the fixed plane we take the plane ~ 1 ~ 3 , z3=5, so that 0 is the angular variable in the cylindrical coordinate system. In the new variables (5.5) takes the form

dsexp(2ikols) 1 dt e [ l ( s 2 - 1)1'2(1 - t 2 ) 1 / 2 . 1 s t + ~ ( . u + ~ . ) ] = F ( . u . ~ . ) (5.6) ' - 1


where x and y denote the [ components of the vectors X and Y, q(x-y) = I if x>y. From (5.6) it is clear that under our assumption v = c(r, z ) the data depend on m = f ( x + y ) and

5.3.2. Let us assume that F is given for all frequencies w, i.e. for all ko, ko= [ ( l + ~)/&]C#~CO~’~. Then the function

/ = ; Ix - y 1.

p(s, l ) = f dtv[l(sz- 1) ’ /2 (1 -t2)’/’, ls t+m] - 1

can be found from (5.6) by taking the inverse Fourier transform in ko. Therefore one obtains the equation for u(r, [)

-1 J c [ l ( s 2 - l)1/2(l-t2)1’2, Is t+m] dt=p(s, 1,m) s > l 120 (5.7) - I

wherep(s, I , m) is known.

given by the equation t2(s2 - l ) - ’ + (17- m)2s-2 = 1 2 . Equation (5.7) can be written as Let r= /(sz - 1)1’2( 1 - t 2 ) l i 2 , v = 1st + m. Consider the ellipse E(s, I ) on the plane (5, q)

Here d S is the arc length of the ellipse,

and a(& v ) is the function [( 1 - t z ) / (s2 - t 2)] ‘ l2 /1 expressed in the variables <,q, s2 + t = cz1-2 + ( ~ - m ) ~ l - ’ + 1, sZ t2 = ( ~ - m ) ~ l - * . Problem (5.8) is an integral geometry problem. If w is found from (5.8) then v is found since a(5, 7 ) is known. It seems that equation (5.8) was not studied in the literature. A similar equation in which the ellipses had one focus at the origin and the other running through one of the coordinate axes was studied by Lavrentiev er a1 (1970, 1980), but their results are not applicable to eqnation (5.8) because the families of ellipses are different.

5.3.3. Consider the inversion scheme for the case when v(r, z)= v(z). If L‘ = u(z) does not depend on r then (5.7) takes the form

I” d t L>(/.st + m ) =p(s, 1, m) s > l , l>O. - 1

Then s = m// and m(t + 1)= 7. Then (5.9) can be written as

(5.9)

c(r) d s = mp(m/l, I , m) = h(m, I ) l>O$ m>l. (5.10) -0

From (5.10) one finds by differentiation that

m > 1. (5.1 1) 1 dh(m, /)

~‘(2m)=- __ 2 dm

168 A G Ramm

Thus, the inversion scheme in this subsection can be summarised as follows. First, take the inverse Fourier transform of the data F(x, y , ko) in the variable ko (see equation (5.6)) to obtain (5 .7 ) . Then, take s=ml-' in (5 .7 ) and find v(z) by formula (5.1 1). In this scheme the data are assumed to be three-dimensional (a function of 1, m and ko), while the conductivity is assumed to be a function of z only, i.e. a function of one variable. The differentiation in (5.1 1) makes the inversion an ill-posed problem. Ramm (198 1, 1984) has given an algorithm for stable differentiation of noisy data. There is considerable freedom in the choice of the method for solving (5.9). For example, one could set m = 1 and s= 1 and solve (5.9) as in Q 5.3.6 below.

5.3.4. Consider the two-dimensional inhomogeneity v(r, z ) and assume that the dependence on r is linear near the borehole, that is u(r, z)= v(0, z ) + (av(0, z)/Zr)r= v(z) + w(z)r. Then (5 .7) can be written as

!", v(1st + m) dt + 1 w(1st + m)(l - t2)'/' dt(s2 - 1)'/'

=P(s, 1, m) s > l , 12.0. (5.12)

Take s= 1 in (5.12) and solve (5.12) for U. (See Q 5.3.6 below.) If v(z) is known then (5.12) can be written as an equation for w:

w(lst + m)( 1 - t')'/' dt=pl(s, 1, m) l' p l = p(s,l,m)- v(lst+m)dt ( ~ ~ - l ) - ' / ~ l - ' . (5.13)

Equation (5.13) can be solved for w. Notice that the functionp,(s, I , m) is continuous as s + l becauseofthechoiceofv. One has,ifm=I,p,(l, l )=pl ( l , I , I ) ,

( 1:' 1 r' w(lr)[ 1 -(t- 1)']'/' dr=p,( l , 1) l>O. (5.14)

JO

Take the Mellin transform of (5.14) to obtain

+(u)a(u) = b(u) $(U) = b(u)a - '(U) (5.15)

where iV(u)=j," w(x)xU-' dx, b(u)=j,(l , l ) , &)=Si r - ' [1- (7- 1)2]1/2 dt, and the formula ~ ( X T ) = + ( U ) ~ - ~ was used. To find w(z ) take the inverse Mellin transform of a - (u)b( U).

5.3.5. A general idea f o r inversion. Let A =Ao + iwp gvA d Z be equation (5.1) corresponding to the source which produces potential A,, in the absence of the inhomogeneity U. A-A,, =A' is the scattered field. In the Born approximation one has A'=iwp gvAo dZ. One can try to find& such that this equation can be solved for U.

5.3.6. A way to solve the equation

z(lt + m) dt =p( 1, I , m) 1>0


is to take m = i, set l(t + 1) = 7, write the equation as

- 0 ~ v ( ~ ) d 7 = ~ ( l , l , l ) ~ h ( l )

and find

v(z) = - - 1 dh(/) 2 dl

421) = - -

5.3.7. Assume U =u(r). Then (5.7) becomes

v { l [ ( ~ ~ - l ) ( l - t ~ ) ] ~ / ~ } dt=p(s,l) ~ ) 1 , l>O (5.16)

(p does not depend on m). This equation can be solved analytically even in the case when s (or I ) is fixed. Indeed, let l(s2 - 1)lI2 =z. Then (5.16) becomes

1' ~ [ z ( i - t ~ ) ' / ~ ] dt=f(z)

Let (1 -t2)li2=y, zy=x. Then(5.17) becomes

f(z)=O.Sp[s, z(s2- 1)-'12] s > 1 iffixed. (5.17)

1 v(x)x(z2 - x ~ ) ' - ' / ~ dx=zf(z) z a o .

This equation can be solved by the formula

1 2 d 2 xlf(x)dx u(z)=- - - z d z l ( z 2 -x2)1/2'

(5.18)

(5.19)

To derive (5.19) multiply (5.18) by 2 ~ ( y ~ - z ' ) - ' ~ ~ , integrate over z in the range (O,y), and use the formula i,' 2z dz/[(y2 -z2)(z2 -x2)]

5.3.8. Assume u(r, z ) = vo(z) + vl(z)h(ro - r) where uo(z), ul(z) are to be found, ro > 0 is to be found, h(x) = I for x)O, h(x)= 0 for x < 0. This model is used in geophysics. Then (5.7) becomes

= n. This is a variant of Abel's equation.

I t.o(Ist+m)dt+ ul(Isl+km)h[ro-i(s2- 1)1/2(l-t2)1i2] dt=p(s, 1,m). (5.20)

Set s= 1 and find uo + v1 as in 5 5.3.6. Take l(s2 - 1)lI2 large, then the second integral in (5.20) is small and (5.20) reduces to (5.9) with U = uo. Solve for uo. Since v o + v 1 and u 0 are found both functions uo and u1 are found. Equation (5.20) can be written as

J - 1 il

.1

vl(lst + m) dt=pl(s, 1, m)=p(s, 1, m)- \ vo(lst + m) dt (5.21) s 1>lf > [ 1 - 7 ; l - ~ ( 5 2 - 1 ) - ~ 1 ~ ~ 2 ' - 1

wherepl(s, I , m)is known. Let l - 2 ( s 2 - l)- '=P. Then (5.21) becomes

V[m + IS(I -rip)1/2] - V(m-ls(1 -r,2P)1/2] =p2(s, p, m) (5.22)

where p2 = - [pi(& I , m)is - V(m + Is) + V(m - Is)] and Vis an antiderivative of u l , V' = v 1 .

170 A G Ramm

Choose Is( 1 -rip)’/‘ and p - r i 2 . Then Taylor’s formula and (5.22) yield

1. This is possible: Is = l(1 + Z-2p-1)1/2 and one can take I small

2k(1 - r i , 8 ) ” 2 ~ l ( m ) = p 2 ( ~ , P, m) P ( S 2 - l)-’=p. (5.23)

Therefore

Let us give an illustrative example of one of the inversion procedures. In this example all the computations are analytical.

Example. Assume that v ( X ) = 1 if r ,< 1, - 1 < xj < 0 v ( X ) = 0 otherwise, p = o9 = 1, and d=O, so that the positions of the source and the receiver are the same. The inversion procedure from 4 5.2 is based on equation (5.3). Under our assumptions this equation is

Let &y) = JTw @(x) exp(ixy) dx. Taking the Fourier transform of (5.25) one obtains

(5.26)

where formula (1.5.10) of Erdelyi et a1 (1954) was used. Therefore

i.(A =9$(v)[2n2(1 - exP(-lYi))l-’. (5.27)

In particular, under the above assumptions about 2; one has

1 - exp(-iy)

iY &Y) = 71 2ny-’( 1 - exp(-y)).

From (5.27) and the last formula one finds

1 - exp (-iy) :y ;(A= .

Therefore the inhomogeneity is recovered from the exact data analytically. Suppose now that the data are noisy: @6 is given instead of @, - @11,<6, where 6 > 0 is known, and ll@li=(J:W I @ i Z dx)’/2. Let

(5.28)

We will show that under an appropriate choice of a(6) the function u6(x) gives a stable approximation to the solution v(x) (corresponding to the exact data) in the sense that I ~ u ~ ( x ) - v(x)ll+O as d+O. The regularisation parameter a(6) will be chosen optimally.


Let us assume that the unknown solution v(x) satisfies the inequality [-= lC12(l + y 2 ) b dy<M2, where b and M are given positive constants. This assumption means that the solution lies in a certain compact set in L 2 . Assume for simplicity that b > 4. It will be clear soon how this assumption is used. Parseval’s equality yields

m

~ c ( b a - ~ / ~ + M a ) c = constant > 0.

The estimate I / ( 1 - exp(-alyl))/( 1 +y2)b’2/1 < ca holds if b > i. Let us find a = a(6) such that 6a-3/2 +Ma=min. One has a ( 6 ) = ( ~ M / 6 ) - 2 ’ 5 = ~ 1 ~ 2 / s , cI = ( $ M ) - 2 / 5 . For this optimal a(6) the error of the approximate solution is ljua - vII < constant x 6’15, as 6- 0.

Coen and Yu (1981), Minerbo (1983) and Oristaglio (1983) give many references on the work done in induction logging theory.

Acknowledgments

The author thanks his colleagues Drs J Blackledge, P Martin, G Minerbo, M Oristaglio, 0 L Weaver and A Weglein for discussions and collaboration.

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Bel!man R, Kalaba R and Lockett J 1966 Numerical Inuersion of the Laplace Transform (New York: Elsevier) Beylkin G 1983 The fundamental identity for iterated spherical means and the inversion formula for diffraction

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