METHODS FOR CALCULATING FRÃCHET DERIVATIVES AND ...

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~520 P. R. McGILLIVRAY AND D. W. OLDENBURG201.260.80hEv5anQ1 O0h>v0.63>i.- c.- >.- cm0.44 C0.3 1200N IdAOX (rn)FIG. 5. Absolute sensitivities computed for the conductive prism model shown in Fig. 4 usingthe adjoint-equation approach.dOO N0.22SUMMARYThe efficient solution of the non-linear inverse problem requires that the dependenceof the data on changes in the model be easily quantified. The most convenient wayof quantifying this dependence is to derive the analytic Fréchet derivative using oneof the three procedures described in this paper. In most cases, however, it is notpossible to derive an analytic expression for either the forward responses or theFréchet derivatives, so one must resort to a model parameterization to simplify theproblem. The dependence of the data on changes in the model is then given by a setof partial derivatives or sensitivities. These sensitivities can be computed from theirfinite-difference approximation, although this requires the solution of a largenumber of forward problems. When dealing with boundary-value problems we canderive a new set of equations whose solution yields either the sensitivities themselvesor quantities from which the sensitivities can be calculated. The similarity of thesenew equations and the original forward problem allows the sensitivities to be computedefficiently if a numerical solution based on a matrix factorization is used.ACKNOWLEDGEMENTSThe authors wish to thank S. Dosso and U. Haussman for helpful discussions concerningthe series-expansion approach to the calculation of Fréchet derivatives. Weare also grateful to the two anonymous referees who provided many constructivesuggestions for this paper.
CALCULATING FRÉCHET DERIVATIVES 521APPENDIX AForward responses for the ID resistivity problemLet the earth be represented by a sequence of M, layers of constant conductivityover a half-space. Within any of the layers or the half-space, the governing differentialequation for the 1D resistivity problem reduces to-_ d2hdz2Â2k = O.The general solution to (Al) for the kth layer can then be written ash(Â, .) = Uke-"Zk-l-Z) + Dked(Zk-i-z) >('42)and for the half-space, ask(1, z) = D, + 1Making use of the continuity relationshipk(Â, z:) - k(Â, zk) = O,('43)and the conservation of charge relationship-- bk+lOkk(Â, z:) --- k(Â, z;) =1 dz 1 dz2n for k = k,O otherwise,one can solve for the coefficients U and D for the kth layer in terms of those for the(k + 1)th layer. This leads to the propagator matrix expressionwhereandThe source vector s k is given byfor k = k.
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