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 (noniterative) 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]
I""od,ucvu= = O
(
j
in R
on a R
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 an j d S = O
(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
or surface. (2) Thc conductivity U is a real-valued strictly positive function in
curve
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.uVu=f
in0
on an,
Da,u = j
(
(1.4)
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.
-
For s 2 0, we have
To: H"(R) x H " c ' / z ( X l ) H"+'(n).
(1.5)
We note that Tois related to the resistive map R by taking f = 0 and restricting U
to the boundary. Thus
R = yT~l{CqxH.+.l.(an)
where y is the trace mapping
Y :U
++
Ulan
which maps
y :
where s
> 1/2.
Hyn)
-
HS-'/*(aR)
In the following, we write T for Tuwhen
U
= 1 identically.
902
E Somersalo et al
Proposition 1. The operator Tu admits a decomposition
+ TR
1
To= -T
U
-
where TR is a mapping H " ( n )x H s + 1 / 2 ( X 2 )
Proof. Let
U
= T,(f,j)and
UR
is smoother than
=U-
uo =
(1.6)
H " + 3 ( n ) s, 2 0.
T(f,j).
We want to show that the difference
1
-Uo
U
U , which
is itself in H s + 2 . TJ do this, we note that u R satisfies
V . oVuR = V . (uoVq)
( ua,uR = uaavl)
in
on
an
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 H s + 3 / 2 ( a S 2 ) .
This mapping properry together with (1.5) shows that u R is in
T operators, we can write (1.7) as
1
Tu = -T
U
(1.8)
In terms of the
1
+ TuBT = -T
+ TR,
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 H a + + ' / 2 ( a S 2to) H 5 + 5 / 2 ( a C L )so
, the principal part of the operator R is just
( l / u ) R a . 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
c,
= 3-,
1
.
Iml
Hence, the principal symbol of the operator Ro, denoted as n 1 ( R 0 )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-
E Somersalo er
904
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
1
Rm,n= lnlPm-*
+ (,;me,
RR
)'
(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
We integrate by parts and use the Schwarz inequality, obtaining
<
1
~m2 (eim8 , 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
u,(R)(z',(')
where
1
= --
1
lt'l dz';0)
[' 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 behaviour of the mapping R, the next step is to synthesize R on a subsurface infinitesimally 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 upwardpropagating 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
++
(1.16)
Ulan,
which have the mapping properties
R, : H3(Sn-') -+ H"+'(S"-')
s
n
1/2. Note that we have identified the boundary of Rv with the unit circle S' if
= 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). In the twodimensional case this polar form is
o
aTuaVu+ --aTu
+ -ar21
r
Oa,u = o
(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 ( Cra , u ) - R (fa,aa,u).
(1.19)
Finally, we use the boundary condition of (1.15) in (1.19), which gives
(i3,R)j
1
j
=;
+ ;RJ
.
+ -rz1R ( a , u a , R j ) .
(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 ~ ( H s ( S " - ' ) , H S t ' ( S " - ' ) ) ,s 2 3/2.
Moreover, the strong limit
1
lim - ( R ? - R,.-6) = D,R,
6-Ot
6
exists in L(H"(Sn-'), HS(Sn-')),and it satisfies
1
D,R, = U
whcre V,
1x1 = r.
Proof.
(R,
. oV,
n-1
+r R.. + R.VT
is the tangcntial part of the operator V . uV along the surface
First we prove the continuity claim. Let
- %6)j
(1.21)
uVT RP
= (R,j
-
U
be a solution of (1.15). We write
"11,1=,-6) + ("11z1=.-6 + Rr-6(+41+6
- (UIl+? - "11+6)
-
4&J
But a slight modification of theorem 3.1 in [30] shows that
IIUllzl=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
-1
6("llsl=r - 4 1 z I = , - 6 )
and similarly,
-
ar4151=T.
in
HS(Sn-')
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, AILE H s c ' / 2 ( { I z l
111 = P exists in H a ( S"-'), and
< l}),
so the trace to
= (%441,+v
+ -3 + V T . o V T R , j = 0
n-1.
r
so solving for the second-order radial derivative and substituting the result into equation (1.22) gives the claim.
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 t h e surfaces r, perpendicularly, so that (z', ..., z"-',t) forms a
with Si," = 6i,,g,,n. Here,
is the metric
new coordinate system in u,<,<,T,,
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
<
..
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 I
zI < 1). 'Ib get an explicit matrix formulation of the Riccati equaton
(1.16), we write the equation in the Fourier basis. Therefore, consider the twodimensional version of the equation,
1
1
1
+ -R,
+ rZR,a,ua,R,..
D,R, = U
T
If we define the operator W, = ( l / r ) R , . , we And that W , satisfies the equation
rD7WT=
U
+ w,asUa,w,..
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
elRBare the eigenfunctions of the operator W,, so
Wm,n
= 6,,,Wn.
Equation (1.23) for the diagonal elements becomes then
rdZU, = -1 - n2 u w 2, .
dr
U
(1.24)
Impedance imaging by layer sfripping
Furthermore, if we rescale the variable
T
by letting p
909
= 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
remains constant in an interval [po,pl]:In this case, equation (1.25) admits an
explicit solution. We have
U,
from which we find
Denoting by M the rational function M ( t ) = (1 - t ) / ( l
that M-' = M , we finally find that
+ t ) , and
using the fact
(1.26)
or, conversely,
(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 t h 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-
32-
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 . 2 5 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 ; ,
0Qi 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.
Impedance imaging by layer stripping
911
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
writing simply
p,
we get an approximation Z ( a , , , 0) for u ( a , , B ) by
< <
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 w m + ,
i.e.
<
+
repeating the step until an approximation for wm,-(al), - N
is obtained. Thus we have a new, synthesized data set
'D(N-l,al)={w,,,(al))
+ 1 < m , n < N - 1,
- 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
<
-
U =
-1
l
T
5(aN,B)dQ.
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.
912
E Somersalo et a1
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
-N
wm,n(%) = 6,,,wn
< 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
a( r ) =
{:
<
as r r ,
asr,<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.
difference of the voltages when j , is applied can be estimated as
11.
= 0. The
913
Impedance imaging by layer stripping
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 a i are used in the algorithm to determine where we discard the
higher modes no longer useful for t h e reconstruction, t h e 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 a N and a. = 1. Explicitly, we chose
a,=a+-
1-0
N-1
(N-1-n)
0<n<N
(2.6)
914
E Somersalo er ul
with a = 0.2. Heme, the last point where U is updated is a N - 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
aN-1=
(2.7)
a
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 c n / 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 nonconstant 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
I
R
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
10-3
10-4
10-5
10-6
5 . 2 5 x IO-’
1 . 5 2 x 10-2
3.91 x
3.88 x
3.88
x
io-2
Corrected
5 . 8 8 x 10-2
5.49 x
5.48
10-4
x 10-4
5 . 4 8 x 10-5
In further tests, we use rule (2.9), and we indicate only the size of a,henceforth
referred to as ‘the radius of the smallest detectable object’.
Before studying further the sensitivity of the method to the choice of a,we discuss
the effects of various other parameters appearing in the algorithm.
E Somersalo et al
916
1.0
(Dl
/
0.8
I
i
t
I
I bl
i
I
J
I
01
0.8
0.6
,
t
1-1
0.8 -
I
0
-
i
I
"
'
L
0.2
0.L
0.6
0.8
1.0
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
all cases the step size A was 0.01, and the radius of the smallesl detectable object a
was 0.3.
Impedance imaging by layer stripping
Table 2. This table is Similar to table 1, with the
L m norm of lhe reconstruction error.
Noise level
917
L1 n o m replaced by the companding
Arithmetic
Geometric
Corrected
10-3
10-4
10-5
1.22
10-6
3 . 8 1 x 10-2
1 . 4 3 x IO-'
1 . 4 3 x 10-2
1.43 x l o r 3
1.01 x 10-1
9.24 x
9 . 1 9 x 10-4
9 . 1 9 x 10-5
'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
00
=I
2pr:N
1-prfN
I <-
2r:N
1 - T:"
when no measurement error is present. Thus, the relative error decreases exponentially, 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]-').
1
0.9
-I
1
0.7
0.60.5
-
0.4
-
, ~ -
/
0.31.~
...... ..............i
~
0.1
Oo
a:i
012
0:3
0:4
015
0:6
0:7
ai
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 l o 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
4
8
16
32
64
Deviation
l/A
Deviation
1.63 x
1.48 Y
1.12 x
7.13 x
3.93 x
128
256
512
1024
2048
2.06 x
1.05 x
5.32 x
2.67 x l o r 3
1.34 x IO@
lo-’
IO-’
10-1
lo-’
10-2
deviatinnn of the. remnctrurtinns with d;tfe.re.nt
In table 3, we represent the
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 o l 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.
Impedance imaging by layer stripping
919
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 zeromean 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
0
I
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Figure 8. T h e set of reconslruclions l h a l 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
920
E Somersalo et a1
921
Impedance imaging by lnyer stripping
-
*.I
p
..> %
z 2 ;j
G.2
g l.F:
E
-5
ti :=
- e l
;?:
-2 5
S o $
.g; E%&m E:
m
2 .= o
.- .>_ c
3 s .-
,
a&g
-e $ ; ;:a
B U O j ;
U
2 .8S
2 -S
B
C
L'0 -
B IICO
E, b e l l
-
g.ss5
*d
p:zi
82%
B
U
-u-6
f $22
.-5
*-?-"
o
D
.+
2 ,
g
22.9
hclm;m
$dqX
.-,x
I1
2
.? L
5
g.l"
U
c *
8
'C
11
-
- 8
$ o
h.z
$5g
526..
6 g :8
U
.=SE
3 .5 0
-
s
E
a u.9
zl
.E g
U
9
u u g z
s5
8-52
E O - -
e g o . 5
- 5 MY)
o e a o
,8 0 B 0
2 5 : II
d $C
d-4
51
g
,u,ZO
5 II
E 5 b %
I
2 .L
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 experimental 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 evaluation, or medical diagnosis.
Acknowledgments
This work was done while Erkki Somcrsalo was visiting Rensselaer Polytechnic Institute; we thank the Academy of Finland, t h e Emil Aaltonen Foundation, and Rensselaer 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,
< < d,-l
T
O < k < K
where d, = 0 < ... < d , < d o = 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
for 0 6 k
923
< 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 RZwith the complex plane. Assume that
U is given as a function of a complex variable z as
a(.) =
U,
uo
in D ,
in D o = s 2 \ F
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
{
Au=O
u,a,u = e ,
[.Ir = [Ua”u], = 0
in D , U Do
on an
where e,(z) = zn and [ . Ir denotes t h e jump across the boundary r = aD,.
The idea is to map the disc
onto itself by a conformal mapping that renders
D , and S2 concentric. Such a mapping is
where
Under this mapping, the image of D , is then found to be f ( D l ) =
r:}, where
RI find the function
{
U , we
I lzl <
solve first the problem
Au* = 0
u,a,u*
{z E
in f ( D l ) U ( f l \ f ( o l ) )
= (0”
on
as2
[ ~ * l , ( r=i [oa,u*l,(r, = 0
and set U = U* o f . The boundary value function
condition
e , = o,a,u
= uoa,lfl(a,u*)
0
‘p,
is therefore determined by the
f = a , l f l ~ 0~f
924
E Somersalo et al
i.e.
Denoting by v i the Fourier coefkients of the function 1p, a t
and (2.3), we find the solution U * to be at the boundary
1
1
1
= 1 and using (2.2)
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
zero,
U
is smooth and bounded away from
Proof. We consider the integral
where u B 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
< c Il"ollL.(n,ll~URIlL.~").
IIV~RIILyn)
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
lUR(Z)I
< Clnl-3'2.
For large n, therefore,
U
behaves like u0, and this establishes (Bl).
(B5)
0
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