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VII Inverse Problem

[] Up to now the direct problem was discussed which consists in finding \varepsilon(\omega) for given \mu(\phi) and \lambda(\phi) from the nonlinear eq. (3.2). [] The inverse problem is to determine \alpha(\phi) or \mu(\phi) from a knowledge of \varepsilon(\omega). [] This is most important for applications such as well logging. [] Particularly important is the problem of determining \alpha(\phi) from \varepsilon(\omega) and \mu(\phi) in view of the fact that the local porosity distribution \mu(\phi) can be observed much more easily than the local percolation probabilities.

[] Consider therefore briefly the problem of determining \lambda(\phi) from eq. (3.2) given \varepsilon(\omega) and \mu(\phi). [] A general theoretical discussion can be given based on the observation that eq. (3.2) is now linear. [] It can be written as

\int _{0}^{1}K(\omega;\phi)\lambda(\phi)\,\mathrm{d}\phi=f(\omega), (7.1)


\displaystyle K(\omega;\phi) \displaystyle=\mu(\phi)\left(\frac{\varepsilon _{C}(\omega;\phi)-\varepsilon(\omega)}{\varepsilon _{C}(\omega;\phi)+2\varepsilon(\omega)}-\frac{\varepsilon _{B}(\omega;\phi)-\varepsilon(\omega)}{\varepsilon _{B}(\omega;\phi)+2\varepsilon(\omega)}\right), (7.2)
\displaystyle f(\omega) \displaystyle=-\int _{0}^{1}\frac{\varepsilon _{B}(\omega;\phi)-\varepsilon(\omega)}{\varepsilon _{B}(\omega;\phi)+2\varepsilon(\omega)}\mu(\phi)\,\mathrm{d}\phi. (7.3)

[] Equation (7.1) is a linear Fredholm integral equation of the first kind. [] Because the kernel K(\omega;\phi) is not symmetric, define

\displaystyle K_{1}(\omega,\omega^{\prime}) \displaystyle=\int _{0}^{1}K(\omega,\phi)K(\omega^{\prime},\phi)\,\mathrm{d}\phi,
\displaystyle K_{2}(\phi,\phi^{\prime}) \displaystyle=\int _{0}^{1}K(\omega,\phi)K(\phi,\phi^{\prime})\,\mathrm{d}\omega.

[] General results can be employed to solve eq. (7.1) if f(\omega) is continuous and such that \int _{0}^{1}f(\omega)^{2}\,\mathrm{d}\omega exists, and if \int _{0}^{1}\int _{0}^{1}K_{i}(\omega;\phi)\,\mathrm{d}\omega\mathrm{d}\phi exists and K_{i}(\omega;\phi) is piecewise continuous in 0\leq\omega\leq 1, 0\leq\phi\leq 1. [] The K_{i} are symmetric and have eigenvalues \lambda^{2}_{n}. [] The normal modes called \Omega _{{ni}} and \Phi _{{ni}} can be chosen orthonormal and satisfy

\displaystyle\lambda _{n}\int _{0}^{1}K(\omega;\phi)\Omega _{{ni}}(\phi)\,\mathrm{d}\phi=\Phi _{{ni}}(\omega),
\displaystyle\lambda _{n}\int _{0}^{1}K(\omega;\phi)\Phi _{{ni}}(\omega)\,\mathrm{d}\omega=\Omega _{{ni}}(\phi).

[] These modes are used to solve eq. (7.1). [] If (7.1) has any solution, then the inhomogeneity f(\omega) can be written as

f(\omega)=\sum _{{n,i}}\Phi _{{ni}}(\omega)\int _{0}^{1}f(\omega^{\prime})\Phi _{{ni}}(\omega^{\prime})\,\mathrm{d}\omega^{\prime}. (7.4)

[] It is assumed that the two sets \Omega _{{ni}} and \Phi _{{ni}} have been made orthonormal. [] Then the solution to eq. (7.1) is given as

\lambda(\phi)=\sum _{{i,n}}\lambda _{n}\Omega _{{ni}}(\phi)\int _{0}^{1}f(\omega^{\prime})\Phi _{{ni}}(\omega^{\prime})\,\mathrm{d}\omega^{\prime}. (7.5)

[] The eigenvalues \lambda _{n}^{2} are the solutions of D(\lambda^{2})=0, where

\displaystyle D(\lambda^{2})=\sum _{{n=0}}^{\infty}\kappa _{n}\lambda^{{2n}},
\displaystyle\kappa _{0}=1,\qquad K_{i}^{{(0)}}(x,y)=0,
\displaystyle\kappa _{n}=-\frac{1}{n}\int _{0}^{1}K_{i}^{{(n)}}(x,x)\,\mathrm{d}x,
\displaystyle K_{i}^{{(n)}}(x,y)=\int _{0}^{1}K_{i}(x,t)K_{i}^{{(n-1)}}(t,y)\,\mathrm{d}t+\kappa _{{n-1}}K_{i}(x,y).

[] These brief remarks about the inverse problem are intended to outline the general characteristics of the problem. [] A more detailed discussion must await the availability of experimentally observed local porosity distibutions.[27]