Layer stripping: a direct numerical method for …eaton.math.rpi.edu/Faculty/Cheney/papers/EITLSreprint.pdfInverse Problems 7 (1991) 899-926.Printed in the UK Layer stripping: a direct

Inverse Problems 7 (1991) 899-926. Printed in the UK

Layer stripping: a direct numerical method for impedance imaging

Er!& Somersalot, Margaret Cheneyi, David IsaacsonT and Eli Isaacson) t Department of Mathematics, University of Helsinki, Hallituskatu IS, M)lW Helsinki, Finland t Department of Mathematical Sciences, Rensselaer Polytechnic Institute. 'Roy, NY 12180, USA 3 Deparlment of Mathematics, University of Wyoming, Laramie, WY 82071, USA

Received 29 July 1991

AbstracL An impedance imaging problem is to find the electrical conductivity and permittivity distributions inside a body from measurements made an the boundary. The following experiment is considered here: a set of electric currenu are applied to the surface of the body and the resulting voltages are measured on that surface. The present paper describes the performance of a direct numerical method for approximating the conductivity in the interior. The algorithm proceeds via two steps: first the conductivity is found near the bounding surface of the body from the data having the highest available spatial frequency; next the boundary data on an interior surface are synthesized using a nonlinear differential equation of Riccati type. The process is then repeated, and an estimate of the conductivity is found, layer by layer. This paper establishes the theoretical basis for the algorithm and reports on numerical tesu.

Introduction

The impedance imaging problem is to form an image of the electromagnetic properties of an inaccessible region of space by applying patterns of electric currents to the bounding surface of the region and measuring the resulting voltages on that surface. Passible applications range from non-destructive materials testing 114) geophysical prospecting [13,33], and process control [42] to medical imaging [1,24].

In recent years, a large number of papers have been written about the impedance imaging problem both from the more theoretical [2,29,31,34,38] as well as from the computational and practical [1,7,15,16,21,26,28,32,36,44] points of view. The list of references above is far from being complete.

The impedance imaging problem is nonlinear and ill-posed. Nevertheless, solutions based on linearizing approximations have produced useful images [I, 7,14,24]. On the other hand, constructive solutions that take the full nonlinearity into account tend to involve either quite sophisticated analytical methods with restrictions on the dimensionality of the space 1311 or several iteration steps (e.g. quasi-Newton methods [12,14,19,25,44]).

In the present article, we give the details of an impedance imaging algorithm described in [5,6,8,9]. This algorithm has the advantage of being a direct (non- iterative) method, yet it takes the full nonlinearity into account. Apart from that,

02665611/91/060899+28~3.50 @ 1991 IOP Publishing Ltd 899

900 E Somersaio et ai

the method has the extra advantage of being conceptually very transparent The idea behind the algorithm can be described in a few words as follows. Assume first that we apply a current density with very rapid spatial variation to the surface of the body. Since most of the corresponding current does not penetrate very deeply into the body but rather is affected mainly by the conductivity near the boundary, the corresponding voltage measuremen& can be used to estimate the conductivity in a thin layer at the boundary. Once the conductivity in this thin layer is approximately known, we compute the outcome of the same kind of experiment if the known boundary layer were stripped away. We then determine the conductivity on this next layer, strip it away, and so on through the body. In this way an estimate of the conductivity of the body is found, layer by layer.

The paper is divided into two parts. In section 1, we discuss the theoretical background necessary to understand the algorithm. Section 2 describes the algorithm itself in a simple geometry, and contains a discussion of some of the numerical tests that we have performed.

1. Theoretical background of the algorithm

The inversion method is based on two separate steps, namely (1) the reconstruction of the conductivity at the boundary, and (2) the synthesis of the data on an interior surface. We dedicate one subsection to each one of these steps.

i.i. Reconslruclion at lhe boundO!y

We start this subscction with a mathematical formulation of the inverse problem under consideration.

We denote the body by R C R", n = 2 or 3, and the conductivity by U. Tne electric potential U satisfies [22]

in R on a R I " " cvu = O

( od,u = j

where a, denotes the exterior normal derivative at the boundary and j is the Current density applied to the boundary BSI. I n order for a solution of (1.1) to exist, j must satisfy the conservation of charge condition

j j d S = O an

(1.2)

Solutions of (1.1) are determined only up to an additive constant; choosing this constant corresponds to specifying the ground or reference potential. We do this by means of the condition

We make the following assumptions. (1) The body R consists of an open, connected, simply connected, bounded set whose boundary aSI consists of a single smooth curve - or surface. (2) Thc conductivity U is a real-valued strictly positive function in R. Tb avoid extra complications, we assume that U is smooth. This assump?ion can

Impednnce imaging by layer stripping 901

be relaxed: e.g. U E C3(n) would be enough, and most of the results apply even for less smooth functions. (3) The current density j is a function in H"(aR) for s 2 P/2, i.e. j is in the Lz-based Sobolev space with smoothness index s. It follows that the electric potential U is in Hs+3/2(CL); see e.g. [17,18].

We assume that we can apply any current density j to the boundary and measure the corresponding voltage U at every point on the boundary. In other words, we know the map

R : j H ulan

which we call the resistive map. Thus R is a map between the following spaces:

R : HS(an) -t H"'(8Q). (In fact, both the domain of R and its range are the closed subspaces of the Sobolev spaces where the integral constraints (1.2)-(1.3) are valid.) We denote by Ro the resistive map in the case when D is identically one.

The inverse boundary value problem that we seek to solve numerically is to reconstruct U from the partial knowledge of the operator R on an.

We work with the mapping R instead of its inverse, the Dirichlet-to-Neumann mapping, because measuring the voltages corresponding to known current densities is in practice a more stable procedure than the reverse [20].

We next establish a formula for reconstructing D on the boundary ail from the resistive map. This reconstruction problem can be solved uniquely in a constructive manner [29,34,37,39]. We present here a simple calculation that gives a formula for U.

We consider the Neumann problem V . u V u = f i n 0

(1.4) ( Da,u = j on an, where. U satisfies (1.3) and where f and j are subject to the integral constraint

This condition is required to guarantee the existence of the solution. We denote by T, the map

TAf,j) = U.

To : H"(R) x H " c ' / z ( X l ) - H"+'(n).

R = yT~l{CqxH.+.l.(an)

Y : U ++ Ulan

y : H y n ) - HS-'/*(aR)

For s 2 0, we have

(1.5) We note that To is related to the resistive map R by taking f = 0 and restricting U to the boundary. Thus

where y is the trace mapping

which maps

where s > 1/2. In the following, we write T for Tu when U = 1 identically.

902 E Somersalo et al

Proposition 1. The operator Tu admits a decomposition

1 To = -T + TR (1.6) U

where TR is a mapping H " ( n ) x H s + 1 / 2 ( X 2 ) - H " + 3 ( n ) , s 2 0.

Proof. Let U = T,(f,j) and uo = T(f,j). We want to show that the difference

1 U R = U - -Uo

U

is smoother than U, which is itself in H s + 2 . TJ do this, we note that uR satisfies

V . oVuR = V . (uoVq) in on an ( ua,uR = uaavl)

where q = log U. We can therefore write uR in terms of the composition of maps

uR = TuBu0 = T , B T ( f , j ) (1.7)

where B denotes the mapping

BP = (v ' (PVj.,J), Planav,v)

The map B maps between the spaces

E : H"+'(n) -3 Hst'(Q) x Hs+3/2 (aS2) . (1.8)

This mapping properry together with (1.5) shows that uR is in T operators, we can write (1.7) as

In terms of the

1 1 Tu = -T + Tu BT = -T + TR,

U U

We have shown that the operator TR has the claimed mapping properties. 0

We convert (1.6) into information about R by taking f = 0 and restricting to the hnnndaq. T%US we hzve

R = yTul{ojxw+~yan)

= 'I U a n R a + RR

where R, is defined to be the last term in the second line of (1.9). Clearly, R, maps Ha++' /2 (aS2) to H5+5/2 (aCL) , so the principal part of the operator R is just ( l /u)Ra. In the following example, the explicit form of Ra is given in a simple geometly.

Impedance iniaging by layer stripping 903

Example 1 (the unit disc). In this example, we show how expansion (1.9) can be used to recover U on the boundary in the case when 0 is the unit disc in R2. The recovey formula given below in proposition 2 is the one we used in the numerical tests described in section 2.

In this case, the operator Ra appearing in (1.9) can be found explicitly. It is the resistive map corresponding to the boundary value problem

(1.10)

where a, is the derivative in the radial direction. This problem is separable in the polar coordinates ( T-, Q), so we can write the solution as

m=-m

where the prime on the sum indicates that the m = 0 term is omitted. This expansion can be used in the boundary condition of (1.10) to obtain

If we expand j in its Fourier series, with j , being the mth Fourier coeficient of j , then we have

1 . c , = --3, Iml

Hence, the principal symbol of the operator Ro, denoted as n1(R0), is

1 u l ( R 0 ) ( Q , n ) = -. In1

From (1.9) we then obtain the principal symbol of R:

(1.11)

The following proposition shows that we can recover o from the matrix elements of R in the Fourier basis, which we denote by

R,,,n = (eimB, Re'"' ).

More specifically, we recover the Fourier coefficients

-ins-

904 E Somersalo er a1

Proposition 2. Let R be the unit disc, and assume that U is smooth. Then

lim 14Rn+h,,, = P k , (1.12) 14-m

Hence, the high-frequency limit of R gives the function l/a at the boundary.

Proof. From (1.9) and the defintion of R,,,, we have

Rm,n = - + (,;me, R R )' (1.13)

'Ib estimate the decay propcrties of the second term on the right side of (1.13), we use proposition 1. 'Ib do this, we consider the expression

1 lnlPm-*

We integrate by parts and use the Schwarz inequality, obtaining

1 ~m 2 (e im8 , RReina)I < -IIe".IIH-c.+,/.)(s,) IIRd"'lIH.+u'(S~) ?.T

where S' denotes the unit circle and s is non-negative. Since we know from proposition 1 that RR is a bounded operator mapping H"+'l2(S1) to Ha+5/2(S1), this estimate becomes

where C is independent of n and m. In particular, we have

(1.14)

Finally, we use this in '(1.13) and let m = n + IC. We then see that the last term Of 0 (1.13) decays like In)' for large n; thus we obtain (1.12).

A different, morc clcmen:aly- proof is given ill appendix B.

Remark 1. In a more complex geometry than that of the previous example, the principal part of R must be calculated in local coordinates. If p , is a point on a R and U is a coordinate neighbourhood of p , such that in the local coordinates (z1,2', ..., ~ " ) = ( 2 ' ; 1 " ) , t h e b o u n d a r y i s g i v e n b y a R n U = ( ( z ' ; ~ " ) 12% = O } , then the principal symbol of R is

1 1 u,(R)(z',(') = -- lt'l dz'; 0)

where [' is the dual coordinate of 2'. For a related calculation, see e.g. [34,39].

Impedance imaging by layer stripping 905

1.2. Subsurface synthesis of data in the interior

Once the conductivity is determined at the boundary from the high-frequency be- haviour of the mapping R, the next step is to synthesize R on a subsurface infinites- imally close to the boundary. We do this by means of a differential equation that R satisfies. This differential equation, which is of the Riccati type, is discussed below.

Riccati equations have also been obtained for scattering data arising in hyperbolic problems [3,10,11,40]. The derivation of the Riccati equations for such time-dependent problems is based on the idea of splitting the wave into upward- propagating and downward-propagating components. In the present case, however, the differential equation governing the field is elliptic, and a different approach is needed. The approach taken in [43] is to convert the elliptic problem to a hyperbolic one and then use wave-splitting, but this approach requires analytic continuation and symmetry assumptions on the medium. Our approach is based on the ideas of [4], in which a Riccati-type equation was found for the Schrodinger Green's function.

For the sake of clarity, we derive the Riccati equation for R only in a spherical domain, thus avoiding the use of local coordinate systems. Therefore, let Q he either a unit ball in R3 or a unit disc in R2. If r is the radial coordinate in R, we denote by R, the domain {z : 121 < r) , 0 < r < 1, and R l = Q. For each r , we can now solve the boundary value problem

(1.15)

where j satisfies the conservation of charge constraint (1.2). As before, the solution U is unique if we impose the condition (1.3) to specify a choice of ground potential. This again enables us to define the family of operators,

R, : j ++ Ulan, (1.16)

which have the mapping properties

R, : H3(Sn-') -+ H"+'(S"- ')

s 1/2. Note that we have identified the boundary of Rv with the unit circle S' if n = 2 and with the unit sphere S2 if n = 3. The original R is now denoted by RI.

To understand how R, changes with r, we simply differentiate (1.16) with respect to r . This gives us

a,u = ( a m j + R(a,.j). (1.17)

The last term of (1.17) contains the expression

a,j = avuaYu which arises in the polar form of the differential equation (1.15). dimensional case this polar form is

In the two-

o 1 aTuaVu + --aTu + -a Oa,u = o r r2 (1.18)

906 E Somersaio et a1

where a, is the angular derivative. Wc therefore use (1.18) in (1.17), obtaining

8,u = ( a , R ) j - R(Ca,u) r - R (fa,aa,u). (1.19)

Finally, we use the boundary condition of (1.15) in (1.19), which gives

j 1 . 1 rz (i3,R)j = ; + ;RJ + -R(a ,ua ,Rj ) . (1.20)

In the above calculation, we have assumed that R , is differentiable. The following theorem makes precise the sense in which this derivative exists, and gives the form of the Riccati equation in higher dimensions.

Theorem 1. The operators R,, 0 < r < 1, form a left-continuous family with respect to r in the strong operator topology of ~(Hs(S"-') ,HSt'(S"-')) , s 2 3/2. Moreover, the strong limit

1 6 - O t 6 lim - (R? - R,.-6) = D,R,

exists in L(H"(Sn-'), HS(Sn-')), and it satisfies

1 n - 1 U r D,R, = - + - R.. + R.VT uVT RP (1.21)

whcre V, . oV, is the tangcntial part of the operator V . uV along the surface 1x1 = r.

Proof. First we prove the continuity claim. Let U be a solution of (1.15). We write

(R, - %6)j = ( R , j - "11,1=,-6) + ("11z1=.-6 - - - (UIl+? - "11+6) + R r - 6 ( + 4 1 + 6 - 4&J

But a slight modification of theorem 3.1 in [30] shows that

I I U l l z l = r UI,=I=~-611H'+L(S"-1) - 0

as 6 tends to zero. Since the family {R,,-6) is bounded for 6 small, the continuity claim follows.

To show the differentiability, we have to show that

i n HS(Sn-') 1 - 6("llsl=r - 4 1 z I = , - 6 ) - ar4151=T.

and similarly,

Impedance imaging by layer srripping 907

These limits follow again by the continuity of the trace operator with respect to r. Indeed, if we consider the case r = I , we have

The second limit follows in the same way, by replacing U with aa,u in the above argument.

Finally, to get the differential equation for R,, we note that

by the above proof. On the other hand, AIL E H s c ' / 2 ( { I z l < l}), so the trace to 111 = P exists in H a ( S"-'), and

n - 1 . = (%441,+v + - 3+VT.oVTR, j=0 r

so solving for the second-order radial derivative and substituting the result into equa- 0

Remark 2. It is not hard to see how this result above could be generalized to an arbitrary geometry. Let p, E a52 and assume that U is a neighbourhood of p , having local coordinates zl, ..., z". We may assume that 852 n U = {(z'; z") I Z" = 0}, where z' = (z', ..., ."-I) . Assume further that we have a one-parameter family of surfaces T,, 0 6 t < 1, that foliates U and is such that To = ai2 n U. Furthermore, we assume that the trajectory t Y p ( z ' ; t ) E U of each boundary point (z'; 0) E as2 n U intersects the surfaces r, perpendicularly, so that (z', ..., z"- ' , t) forms a new coordinate system in u,<,<,T,, with Si," = 6i,,g,,n. Here, is the metric tensor, i.e. ds2 = gi,jdzidzJ (figure 1) and 6 denotes the Kronecker delta. If we define R, as the resistive mapping on the boundary anl obtained by replacing To by rr, a straightfonvard modification of the proof of theorem 1 shows that RI satisfies

tion (1.22) gives the claim.

..

where r:,; is the Riemannian connection in the local coordinate system (z'; t). We skip the details here.

908 E Soniersalo et nl

Flgurr 1. h c a l deformation of lhe domain in arbitrary geometry.

Remark 2, together with remark 1, gives the generalization of the layer-stripping algorithm to arbitrary geometry.

From now on, we confine ourselves to the case of a two-dimensional disc, R = {z E R2 I IzI < 1). 'Ib get an explicit matrix formulation of the Riccati equaton (1.16), we write the equation in the Fourier basis. Therefore, consider the two- dimensional version of the equation,

1 1 1 U T

D,R, = - + -R, + rZR,a,ua,R,..

If we define the operator W, = (l /r)R,. , we And that W, satisfies the equation

rD7WT = + w,asUa,w,.. U

Let ( w m , n ( r ) ) be the Fourier representation of the operator W,, i.e.

We will refer to the w , , , ~ as the Fourier modes of W . Denoting by p n ( r ) and a,(.) the Fourier coefficients of 1/0(r ,S) and u(r,O), we find that the matrix elements w, , ,~ satisfy the equations

This formula is the one that will be employed in the actual implementation of the aigorithm iater in section 2 of this work.

We close this subsection by considering the Riccati equation in a special case that illuminates the synthesis procedure.

Erample 2. When the conductivity distribution U is rotationally invariant, i.e. 0 = U(.) , equation (1.23) takes an especially simple form. In this case, the exponentials elRB are the eigenfunctions of the operator W,, so

Wm,n = 6,,,Wn.

Equation (1.23) for the diagonal elements becomes then

rdZU, = - 1 2 2 - n uw,. d r U

(1.24)

Impedance imaging by layer sfripping 909

Furthermore, if we rescale the variable T by letting p = rn, and define

Y,(P) = ""JP"") = I1",(T)

u,(p) = u(p1'") = U ( T )

we get simply

(1.25)

Tb get an idea of the instability of the solutions y,, consider the simple case where U, remains constant in an interval [po,pl]: In this case, equation (1.25) admits an explicit solution. We have

from which we find

Denoting by M the rational function M ( t ) = (1 - t ) / ( l + t ) , and using the fact that M-' = M , we finally find that

or, conversely,

(1.26)

(1.27)

In particular, assume that U and hence U, is constant over the whole interval [0,1]. Then the solution of the equation (1.25) is simply

1 Y, = 0'

Assume, however, that we want to find a numerical approximation y; to y,, starting with a boundary value y;( 1) that contains measurement errors. We may write

Yi(1) = ( 1 + E ) I . U

Assuming that we keep the value of U fured in the equation (1.25), the solution y:, is

It is easy to see that r reaches a critical value T;, where y;(r) either becomes negative or infinite, depending on the sign of E , at

(1.28)

(see figure 2).

From this simple example we see that in general, the numerical determination of the n th mode becomes less and less stable as 1111 increases. This principle can be used as a guideline in designing the numerical algorithm.

910 E Somersalo et a1

4 -

3 -

2 -

Figure 2. Solutions of lhe one-dimensional Riccati equation (1.24) with n = 1 and v = 1. In the diverging solulion, the error in the boundaly data is E = 0.25, whereas the ermr e = -0 .25 yields the solution passing lhrough the r-axis.

2. The algorithm: description and performance

In this section, we explain a numerical algorithm we used to approximate the conductivity c r ( ~ - ~ 0 ) frnm the rpsistivvp map E. !E suhscrtirrn 2.1 >ve desrrihp the z!g~:khz, and in subsection 2.2, we discuss the choice of various parameters and the algorithm’s performance with these parameters. Appendix A gives details of the generation of synthetic test data.

Throughout this section, R is the unit disc.

2.1. Description of the algorithm

We start with the numerical data consisting of the matrix elemenls of the operator W, at the boundaly in the Fourier basis (see formula (1.23)). If N is the number of the highest Fourier mode available, the initial data set is

DD(1,N) = {w,,,,,,(I) I - N < n,m < N )

the numbers tun:,, including possible numerical or measurement errors. In D( 1, N), the first variable 1 refers to the value of P.

The reconstruction of the conductivity proceeds through the following steps: Step 1. Choose mode-dropping radii. Choose first a decreasing sequence of radii a; , 0 Q i < N + 1, i.e.

0 = < a,,, < ... < a. = 1

and divide the disc R into rings R;,

These rings form the radial subdivisions of the mesh on which we display the reconstruction.


Step 2. Reconsfnrction at the boundafy. Estimate the resistivity p = l / a at the outer boundary a,, = 1 of Q, by using the reconstruction formula of proposition 2:

Using both n = N and n = -N, we obtain an approximation p ( a,,, e) for p( ao, 0) by writing

= @(ao, 8 ) . Having this approximation for p, we get an approximation Z ( a , , , 0) for u(a,,B) by writing simply

and calculate the Fourier coelficients U,,, -N < n < N, of 5 ( a o , B ) for later use. We retain the Fourier coefficients of ,5 and 5, and assume that p, = an = 0 if In1 > N . Step 3. Synthesis ofdata on subsurface. Assume that n( r , 0) = 5( a,,, e ) and p( r, 6') = 6(ao, 0) in the whole ring R,, i.e. for a, < r < a,,. Choose a step size A < a. - a l , and propagate the matrix elemens w,,,, - N + 1 < m, n < N - 1 through the ring Q, by applying, for example, the Euler method to the differential equation for wm+, i.e.

repeating the step until an approximation for wm,-(al) , - N + 1 < m, n < N - 1, is obtained. Thus we have a new, synthesized data set

'D(N- l ,a l )={w, , , (a l ) ) - N + 1 < m , n , < N - 1 ) .

Repeat. Now we go back to the beginning of step 2, replacing N by N - 1 and a. by al everywhere, and repeating the procedure we find an approximation for a everywhere in a N < r < 1. Finally, in the central mesh element R, we set Z equal to the average

- l T U = -1 5 ( a N , B ) d Q .

2n - * Note that at the nth radial subdivision, we drop the zk)n)th modes from the

calculation. This is a method of regularizing, because (a) using fewer Fourier modes in the approximation for p restricts p to a smoother class of functions, and (b) the Riccati equation is more stable for the lower Fourier modes.


2.2. Numerical tests and discussion

In our tests of the algorithm, we were careful not to commit any of the 'inverse crimes' described by Kress [27]; in particular, in most of our tests, the conductivity distributions are not smooth; the discontinuities were chosen so that they do not coincide with the mesh elements; and the starting data were generated by a method different from that used in the reconstruction algorithm (see appendix A).

In our tests, we started with the simplest case, namely the one in which o is rotationally invariant, i.e. U = U(.). The number N of Fourier modes in the data set was chosen to be N = 16. This number was chosen for compatibility with the device described in [32], which uses 32 electrodes.

In the rotationally invariant case, we have a diagonal data matrix

wm,n(%) = 6,,,wn - N < m,n < N

as explained in subsection 2.1. Furthermore, the steps 2 and 3 simplify considerably: we have

and the differential equations are uncoupled. Evidently, the first problem to solve is to decide how to choose the radii a i ,

i = 1 , 2 , . . . , N, where the higher modes are dropped. A natural, but naive, choice would be equally spaced points between the centre and the edge. However, a better choice can be made by considering the size ol the smallest object that can be detected with a given mode and a given measurement precision. To see this, consider a conductivity distribution 01 the form

as r < r, {:: a s r , < r < ] a( r ) =

If we apply the current density j n ( 0 ) = cine at the boundary r = 1, the corresponding voltage at r = 1 is

where

(see [21]). Let u0 be the solution in the case of a homogenous body, i.e. 11. = 0. The difference of the voltages when j , is applied can be estimated as


Assume that the measurement precision at the frequency n is E,. Clearly, then the smallest size T, of an object at the centre of the body that still contributes a measurable difference to the boundary voltage is obtained by solving the equation

for T,, yielding

Objects smaller than this radius T , are unobsenable with this mode. In particular, if we assume that E = l n l u o ~ , / 2 = constant, we get

(Note the close relationship with the formula (1.28) of T;. )

Since the radii ai are used in the algorithm to determine where we discard the higher modes no longer useful for the reconstruction, the formulae (2.4) and (2.5) should he used as guidelines for this choice.

Figures 3 and 4 represent the effects of choice of the radii a,. We have assumed that the body has a constant conductivity U = 0.5. The step size A appearing in the synthesis step 3 was kept fixed at A = 0.01.

Figure 3. Reconstructions of a constant conduclivily wilh noisy data. The relative noise level is 0.0001%, and the radius of lhe smallest deleclable object is a = 0.2. The propagalion step Size is A = 0.01. In this figure, we used formulae (2.6) ('arithmetic') and (2.8) ('geometric') to choosc the radii where the Fourier modes are dropped.

For the first reconstruction in figure 3, labelled as 'arithmetic', we choose equally spaced lattice points a, between the values aN and a. = 1. Explicitly, we chose

(2.6) 1 - 0 N - 1 a , = a + - ( N - 1 - n ) 0 < n < N

914 E Somersalo er ul

with a = 0.2. Heme, the last point where U is updated is aN-l = a. The reconstruction of (Y was obtained by applying the layer-stripping algorithm to data produced as described in appendix A with random error added to each Fourier mode w,,. The noise for each mode was assumed to be independent and normally distributed, with variance 0.0001 times the value of the largest of the data points w,. We see that with even this low noise level, the reconstructed conductivity deviates greatly from the true profile.

Tb compare the choice (2.6) to differently chosen lattice points, we produced another reconstruction from the same data set hut we used

This choice is suggested hy formula (2.5). Tb obtain figures comparable to those with the ‘arithmetic’ mesh, we picked

a N - 1 = a (2.7)

i.e.

a,, = OiZ/(N+1-n) 1 < n N (2.8)

with o( = 0.2. The reconstruction with this choice is depicted also in figure 3, labelled as ‘geometric’. The reconstruction is clearly much more stable than with the previous cnoice. Hence, we discard the arirhmeric choice and proceei witn the geometric one.

For the next comparison of different choices of the lattice points, we assumed that in fcrmula (2.4), the noise levels E , are equal for all modes, so that cn/2 = E .

The approximate identity

then suggests that one should use the choice

a , = ( ( N + 1 - n)uN)l’(N+l-n).

In order that this choice satisfy (2.7), we choose a N = a 2 / 2 , and hence

Again, the reconstruction is made with the same data as in the previous cases, and the outcome is depicted in figure 4, labelled as ‘corrected’. For comparison, we have plotted the reconstruction with the previous choice in the same picture. (Note that figures 3 and 4 are plotted on very different scales!) In this example, the new choice (2.9) seems to improve the stability almost as much as did the ‘geometric’ choice compared to the ‘arithmetic’ one.

We also compared the ‘geometric’ (2.8) and ‘corrected’ (2.9) choices for a non- constant conductivity distribution. Figure 5(u) shows a reconstructed conductivity distribution with the ‘geometric’ choice; figure 5 ( b ) shows the same thing with the ‘corrected’ choice. We see that the ‘geometric’ choice produces a somewhat better

Impedance imaging by layer slripping 915

R I

0.41 0 0.1 0.2 0.3 0.4 0.5 0.6 0.1 0.8 0.9 1

Figure 4, All the parameters are as in figure 4, and formulae (2.8) (‘geomelric’) and (2.9) (Lcorrectedl) were used.

reconsauction. The situation changes, however, when we add noise. Figures 5(c) and 5(e) show ‘geometric’ reconstructions from data with different samples of 0.05% added noise; figure 5(d) shows the ‘corrected’ reconstructions from the same data. We see that the ‘geometric’ one is much more sensitive to noise than is the ‘corrected’ one.

To get a better idea of the tolerance of different choices to additive noise, we ran the reconstruction algorithm with different noise levels and computed the average L2 and L” norms of the reconstruction errors with a large number of data samples containing independent additive noise. The outcomes of these tests are represented in tables 1 and 2, indicating that each improvement in the choice of lattice points improved the noise tolerance of the method almost by a decade. Thus, we conclude that formula (2.9) is the best of the three.

Table 1. This table gives the L2 deviations of a large sample of reconstructions with noisy data. The true conductivity was assumed to be o = 0.5 throughout lhe body, and the step size A was 0.01. The noise was additive, normally distribuled wilh zero mean and variance equal to lhe noise level given in the first column times the maximum of the noiseless dalapainls. The radius of the Smallest detectable object was set lo 0.25. The stars in some of the columns indicate lhal reconstmction failed for some of lhe data sets with the indicated noise Iwel.

Noise level Arithmetic Geometric Corrected

10-3 5.88 x 10-2 10-4 3.91 x io-2 5.49 x 10-5 5 .25 x IO-’ 3.88 x 5.48 x 10-4 10-6 1 .52 x 10-2 3.88 x 1 0 - 4 5.48 x 10-5

In further tests, we use rule (2.9), and we indicate only the size of a, henceforth

Before studying further the sensitivity of the method to the choice of a, we discuss referred to as ‘the radius of the smallest detectable object’.

the effects of various other parameters appearing in the algorithm.


0.8

1.0 ( D l I

-

i 0.8 / t J

0 1 I

0.8

0.6

I" , 1 - 1

I 0 0.2 0.L 0 . 6 0.8 1.0

I bl I i I

t -i - L 'i 0 0.2 0 . 4 0.6 0.8 1.0

Figure 5. A comparison of the 'gcomctric' and 'corrected' choices far radii at which modes are dropped. ( n ) shows the 1Ne conductivity distribution and lhe 'geometric' reconstruction from noiseless data. (b) shows the 'corrected' reconstruction from the Same dsta. (c) shows the 'geometric' reconstruction from data with 0.05% added noise. (d ) shows the corresponding 'corrected' reconstruction. (e) shows the 'geomelric' n C O n s t N C - lion in which a different seed was used in lhe random number generator that generated the 0.05% added noise. (f) shows the 'corrected' reconstruction from the Same data. In al l cases the step size A was 0.01, and the radius of the smallesl detectable object a was 0.3.


Table 2. This table is Similar to table 1, with the L1 n o m replaced by the companding Lm norm of lhe reconstruction error.

Noise level Arithmetic Geometric Corrected

10-3 1.01 x 10-1

10-5 1.22 1 .43 x 10-2 9 . 1 9 x 10-4 10-6 3.81 x 10-2 1.43 x l o r 3 9 . 1 9 x 10-5

10-4 1.43 x IO-' 9.24 x

0.6

0.5

0.4

0.31.~

'Ib get an idea of the sensitivity of the method to the choice of N, the number of Fourier modes available, consider again the simple model (2.1), in which the conductivity is constant in a neighbourhood of the boundary T = 1. At the boundary, the relative error in estimating the conductivity is

I%( I,@) - uol 2pr:N 2r:N <- = I 1 - p r f N I 1 - T:"

when no measurement error is present. Thus, the relative error decreases exponen- tially, as the number of modes is increased. However, when the conductivity varies on the boundary, the error due to truncating the Fourier series of U could dominate the O(lnl-2) error from (1.12); if U is discontinuous on the boundary, the error could be as large as O( In]-').

0 0

1

- , ~ -

- / -

i ...... ~ ..............

0.9 1 -I

0.6

0.5

0.4

0.31.~

0.7

- , ~ -

- / -

i ...... ~ ..............

0.1

Oo a:i 012 0:3 0:4 015 0:6 0:7 a i 0:9 I

Figure 6. Reconstructions of a three-ring conductivily distribution from noiselm boundary dala with different propagation step size. The reconstruction platted with a solid line is obtained by choosing the slep size A equal to the width of the ring n,, whereas the dashed one uses the exact solution (1.26) of the Riccati equation lo propagate the data through each ring, i.e. A = 0.

Next, we tested the sensitivity of the algorithm to the choice of the step length A used in step 3 to solve the Riccati equation by Euler's method. For the reconstructions in figure 6, we started with data containing no noise other than that due to the numerical round-off errors. The true conductivity profile is plotted in figure 6, as a solid line. For the reconstruction plotted as a solid line with circles, we chose the

918 E Somersalo e! a1

step A be the same as the width of the corresponding ring. Thus, in the synthesis step 3, we made just one big leap from the outer boundary of Q, to the inner one. The reconstruction plotted with a dotted line was obtained by using formula (1.26), which provides an exact solution to the Riccati equation. These two reconstructions represent the extremes of large and small step sizes.

These two reconstructions also illustrate the general feature of the method that decreasing the step sue in the propagation procedure improves the dynamical range of the reconstruction. A larger step size can thus be seen as one possible regularization parameter. On the other hand, if wider dynamical range is needed, it may be desirable to replace the Euler method by higher-order schemes.

Table 3. This table gives the maximum deviation of the reconstruction from that with zero step size as a function of the step size. The true conductivity profile as well as the reconstructions with zero and large step sizes are seen in figure 5.

I /A Deviation l / A Deviation

4 1.63 x lo-’ 128 2.06 x 8 1.48 Y IO-’ 256 1.05 x

16 1 . 1 2 x 10-1 512 5.32 x 32 7 .13 x lo-’ 1024 2.67 x l o r 3 64 3.93 x 10-2 2048 1.34 x IO@

In table 3, we represent the deviatinnn of the. remnctrurtinns with d;tfe.re.nt step sizes from that with zero step size. The initial data was the same noiseless data used in figure 6. It may not be surprising to find that the use of a first-order method in the propagation step yields a first-order reconstruction method, i.e. doubling the step size doubles the L2 deviation ol the reconstruction from that obtained by an ‘exact’ propagation.

IO’ 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

Figure 7. The wade-all curve, representing the L2-variance of the recanstruction versus the radius of the smallcst dcteclable object a. The variancc is computed from 20 reconstructions with independent normally distributed noise added to the boundary data. The noise was normally distribulcd, independent at each Fourier frequency, with variance 0.001.


The previous examples give already some idea of the robustness of the algorithm. Tb get a better idea of how the reconstruction depends on the radius of the smallest detectable object a and on the nOke level of the data, we made the following test We chose the same rotationally invariant conductivity profile as in the reconstructions in figure 6, and computed the corresponding matrix elements of W . From this data, we produced a sample of 20 data sets by adding independent normally distributed zero- mean random noise vectors with variance 0.001, With each of these 20 data vectors, we then ran the reconstruction program with different choices of a in formula (2.9). For each value of a, we then computed the mean of these reconstructions, and the mean square deviation of the reconstructions from this mean profile. In each reconstruction, the step size was zero, i.e. the exact solution to the Riccati equation with constant d was used. Figure 7 shows how the mean square deviation varies as a function of a; we see that the bigger we choose a, the less the reconstruction deviates from the expected value. Tb get an idea of the deviation of the reconstructed profiles, we have plotted the 20 reconstructions when a = 0.29 in figure 8.

- 1 I

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 8. The set of reconslruclions lhal were used lo campule the variance a1 a = 0.29 in the previous figure.

Finally, we tested the method in the more realistic non-rotationally invariant case. The method we used to produce synthetic data for the reconstruction algorithm is explained in appendix A.

Figure 9 shows a series of reconstructions, where the true conductivity consists of a circular inhomogeneity against a constant background. In all of the reconstructions, the true values of the conductivity of the background is d = 0.2, while the conductivity of the inhomogeneity is 0 = 0.4. The radius of the circular inhomogeneity is T = 0.2, and the locations of the midpoints are 0, 0.1, 0.2, 0.3, 0.4, 0.5, and 0.6. The propagation step size was A = 0.01 in all of thc reconstructions. We chose the radius of the smallest detectable object to be a = 0.05 in the first four reconstructions, a = 0.08 in the fifth one, and a = 0.1 in the last two cases. The need to increase the size of ct when the inhomogeneity approaches the boundary of the body is due to the fact that the forward solver described in appendix A becomes more unstable when the inhomogeneity gets closer to the boundary, and consequently the precision


Impedance imaging by lnyer stripping 921

*.I p .> - % z 2 ;j G . 2 g l . F : E - 5 ti := - e l ;?: - 2 5 S o $ ; E m .g % : m & E 2 .= o .- .>_ c 3 s - ,

B U O j ; - :a e $ ; ; U

2 . 8 S S B C 2 - L ' -

B I I C O E, b e l l g . s s 5 i *d - p:z- B 8 2 % U

- u - 6 f $ 2 2 .- 5 o *-?-" D

2 , .+ g 22.9 h c l m ; m $ d q X .- ,x I1 2 11 .? L 5 g.l" - U - 8 c * $ o 8 'C h.z U $ 5 g 5 2 6 . . 6 g :8

.E - g U 9 u u g z s 5 - 8 - 5 2 E O - - e g o . 5 - 5 M Y ) o e a o ,8 0 B 0

25: II d $ C 51 d - 4 g ,u ,ZO I 5 2 II .L

.-

a & g .-

0 -

. = S E E 3 .5 0 s a - u.9 z l

E 5 b %

922 E Somersalo er a1

of the data decreases. 'Ib make the reconstructions comparable with each other, the same grey scale is used in all of them.

3. Conclusions

We have presented a preliminary study of a layer-stripping method. This study has demonstrated that at least on a limited class of data sets, the method is capable of making useful reconstructions.

We have not presented here a thorough numerical or analytical study of this method, nor have we described all of the difficulties that need to be overcomc in order to make this into a practical, robust mcthcd.

For example, in order t o reconstruct conductivity images that come from experi- mental data [24,32], more realistic models of the electrode-body interface should be incorporated into the algorithm [45,46].

Formally, the algorithm described here could be extended to many fixed-frequency and fixed-energy inverse problems. For example, it could be used to solve inverse spectral, inverse scattering, and inverse boundary value problems for many systems of equations arising in mathematical physics. It remains to be seen whether it and its generalizations will prove useful in geophysical prospecting, non-destructive eval- uation, or medical diagnosis.

Acknowledgments

This work was done while Erkki Somcrsalo was visiting Rensselaer Polytechnic Insti- tute; we thank the Academy of Finland, the Emil Aaltonen Foundation, and Rens- selaer for making this visit possible. This work was supported in part hy NIH (GM 39388, GM 42935). NSF (BCS-8706340), ONR (N-0001449-J-1129), GE and RPI. We are grateful to John Sylvester for sharing with us his ideas on a related approach to this problem. We would also like to thank Ed Coffey, whose initial computations gave us insight into the instability of the method. We are grateful to Jon Newell, Ed Coffey, Peter Edic, and John Goble for their help in producing figure 9. Finally, we would like to thank all the members of the impedance imaging group at RPI, whose work continues to inspire and enlighten us.

Appendix A

In this appendix, we explain the numerical algorithms that were used to generate the data 'D( 1, N) used in the numerical tests.

Consider first the rotationally invariant case, when W is diagonal. We assume that U(.) is a piecewise constant function,

U(.) = U, as d, < T < d,-l O < k < K

where d, = 0 < ... < d , < do = 1. Thcn thc valucs of the diagonal elemens w , ( l ) , - N < n < N , are obtained recursively by formula (1.27) as

Impedance imaging by lnyer stripping 923

for 0 6 k < K - 2, 1 6 n < N, and

Finally, set w-,,(l) = w , ( l ) . ’lb generate the data for the non-rotationally invariant examples, we used the

above formulae together with several below that we obtained by conformal mapping. ’RI derive the needed formulae, we identify RZ with the complex plane. Assume that U is given as a function of a complex variable z as

U , in D , uo in D o = s 2 \ F a(.) =

where D , is a disc non-concentric with s2, i.e. D , = { z I Iz - zll < r , } for some zl E 0, and 0 < T , < 1 - Izl/. We shall asume that zl is real and positive. The boundary value problem that we seek to solve is

A u = O in D , U Do u,a,u = e , on an { [.Ir = [Ua”u], = 0

where e,(z) = zn and [ . I r denotes the jump across the boundary r = aD,.

D , and S2 concentric. Such a mapping is The idea is to map the disc onto itself by a conformal mapping that renders

where

Under this mapping, the image of D , is then found to be f ( D l ) = { z E r:}, where

I lzl <

RI find the function U, we solve first the problem

Au* = 0 in f (D l )U( f l \ f (o l ) ) u,a,u* = (0” on as2 { [~*l,(ri = [oa,u*l,(r, = 0

and set U = U* of. The boundary value function ‘p, is therefore determined by the condition

e , = o,a,u = uoa,lfl(a,u*) 0 f = a , l f l ~ ~ 0 f


i.e.

Denoting by v i the Fourier coefkients of the function 1p, a t 111 = 1 and using (2.2) and (2.3), we find the solution U* to be at the boundary

and consequently

This is the outcome of R acting on e'"@; the Fourier modes R,,,+ are then obtained by a further Fourier transformation.

Appendix B

Here we give an elementary proof of proposition 2.

Proposition E l . When il is the unit disk and U is smooth and bounded away from zero,

~~

Proof. We consider the integral

where uB and U,, are as defined in the proof of proposition 1. Applying Green's theorem to (82); we ohtain

The two boundary integrals in (B3) cancel because of the boundary condition satisfied by uR (see proof of proposition 1). TI the remaining integral on the right side of (B3), we apply Schwarz's inequality; in the remaining integral on the left side we bound U below by U,,. Thus we obtain

2 IIV~RIILyn) < c Il"ollL.(n,ll~URIlL.~").

Impedance imaging by lnyer stripping 92.5

We divide both sides by llVunll and bound the left side below by the PoincarB inequality. We thus obtain

where 21 denotes the average of uR over R. The right side of (B4) can be computed explicitly, since uo = dnlein8/lnl; this right side is bounded above by a constant times If we integrate (B4) over BQ, the term involving uR vanishes by (1.3). This shows that U itself is bounded by a constant times lnl-3/2; therefore we have

l U R ( Z ) I < Clnl-3'2. (B5)

For large n, therefore, U behaves like u0, and this establishes (Bl). 0

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