What's the Matter with the Volume Integral Equat ion?



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What's the Matter with the Volume Integral Equat ion? Jukka Liukkonen Geological Survey of Finland Geophysical Research P. 0. Box 96 FIN-02151 Espoo Finland E-mail: Jukka. LiukkonenQgsf. f i Abstract February 20, 2002 Numerous authors have reported that a numerical solution of the volume integral equation for an electric field behaves poorly when there is a large contrast in conductivity between the background space and the anomalous body. This paper is devoted to the examination of this problem. A proposal of how to avoid the known numerical problems is introduced. For the sake of simplicity the background space is assumed to be homogeneous. 1 Dipoles Denote the complex permittivity by y := E + ia/w and the fundamental solution of the scalar Helmholtz operator by 4-H Let Q be a constant vector. The electric dipole (GE, GE) with the dipole moment Q located at y E R3 in a homogeneous space is defined by It is the unique solution to =E -H - V x GE(., Y) - ~WPOGE (*, Y) = 0, =E - V X G:(-, y) + iw)")'ge(., 9) = - i ~ ~ ~ 6 ~ f that satisfies the Silver-Muller radiation conditions (see, e.g., [4] for the radiation conditions). -E -H Likewise, the magnetic dipole (GH, GH) in a homogeneous space is defined by =E It is the unique solution to GH ($7 Y). M = ~WPOV~ -H GH(x,y).M = Vz x Vz x (4(~- x ($(x - Y)M)~ Y)M) -6y(x)M. =E 3 V x GH(., y) - iwpogh(., y) = iwpobyi, -H =E - V x GH (., Y) + ~WYOGH (*, Y) = ö,

2 VOLUME INTEGRAL EQUATION satisfying the radiation conditions. The corresponding dipoles for a space containing a bounded inhomogeneity are denoted =e =h =e =h by (G,, G, ) and (G,, G,). They satisfy the radiation conditions and the equations It is assumed that the source point y has a neighbourhood where the materia1 functions are equal to the constants po, 70 of the background space. Dimensions or SI units of some of the quantities are listed below: Quantity Explanation Unit 9-t. GE, G, 9-h GE, G, =E =e GH, G, =H =h GH, G, (3 M Electric response of an electric dipole moment Magnetic response of an electric dipole moment Electric response of a magnetic dipole moment Magnetic response of a magnetic dipole moment Electric dipole moment Magnetic dipole moment 3D Dirac's delta distribution AV-1 VA-1 Vm 2 Am 2 2 Volume Integral Equation Assume that St c R3 is an Open domain with a piecewise smooth boundary and that the closure supp(y - yo) of the inhomogeneous region is totally included in St (in connection with geophysical applications it is customary to start with the presumption that p = po everywhere). Then the volume integral equation for the electric field of a magnetic dipole located at x 4 is where the field point y varies within 0. Multiply this from the left by a dimensionless constant projection vector? and from the right by M to obtain

3 PROBLEMS AND SOLUTIONS - Here X(y) := ~ h(y, x) - is the unknown vector field to be solved in R. Moreover =H GE (x, Y) - M = -~wyovz x (4(x - Y)G), =E GE(x,y).M = Vz x Vz x (~(X-~)M) -6y(x)M which leads to the integral equation + d M - Z(y) = iw,u0v - GE(X,Y) -M + V, x V, x (g(r - y)m). X(Z) dr (3) or after multiplication by yo Y(Y)M + =H - Z(y) = iwpov GE (x, y) - M + f Since the equation (4) is equivalent to Y(Y)M + =H - Z(y) = iwpov - GE (x, y) M + This equation is essentially (6.4.7) in [2]. Note that 3 Problerns and Solutions If the host conductivity 00 and/or the frequency w are low with respect to the anomaly serious numerical problems occur when trying to solve (7). In this case the wavenumber k defined by 2 2 k = w ~ O Y O = w ~ + iwp0a0 ~ ~ E ~ almost vanishes. Kim and Song consider an electric Hertzian dipole in [3] but the phenomenon is the same with a magnetic dipole. They report two sources of numerical problems:

REFERENCES 1 1 1. In (7) the ratio of the coefficients - and 1 in (-VV - +1) increases without k 2 k 2 limit as k tends to 0. When solving the integral equation numerically the integral corresponding to 1 is practically omitted by the computer program (see also [5]). 2. When discretizing the unknown electric field 2 one uses elements such that they do not satisfy the continuity conditions on element boundaries. This is the reason why the model does not adequately admit induction currents in the conducting anomalous body. On the other hand, the strong primary field, according to continuity, forces currents to flow on the surface of the body. The author's proposal is that the first problem could be avoided by using (4) instead of (7). In the following the awkward term is tracked as the calculation proceeds: Put this into (6): Note how the problematic term was annihilated! The second problem could be avoided by using Whitney 1-elements or edge elements in the discretization of the unknown electric field. These elements have the property of tangential continuity at the element boundaries (see [l]). The discretized integral equation is a matrix equation for the coefficients of the element functions. Acknowledgments The author thanks Geophysicist Matti Oksama and Geophysicist Ilkka Suppala for valuable discussions. References [l] Alain Bossavit: Whitney forms: a class of finite elements for three-dimensional computations in electromagnetism. IEE Proceedings 135 (1988) 493-500.

REFERENCES [2] Lauri Eskola: Geophysical Interpretation using Integral Equations. Chapman & Hall, London 1992. [3] Hee Joon Kim and Yoonho Song: A no-charge integral equation formulation for three-dimensional electromagnetic modeling. Butsuri-Tansa 54 (2001) 103-107. [4] Jukka Liukkonen: Uniqueness of electromagnetic inversion by local surface measurements. Inverse Problems 15 (1999) 265-280. [5] Jussi Rajala and Jukka Sarvas: Electromagnetic Scattering from a Plate-Like, Non- Uniform Conductor. Acta Polytechnica Scandinavica, Mathematics and Computer Science Series No. 43, Helsinki 1985.