An inverse problem seeks to measure a physical quantity2. Ohm’s law, equation (3.1) defines conductivity pointwise in correspondence with pointwise values of the potential gradient and current density. However, in practise, we cannot measure the gradient of a potential at a point, nor the density of a current flux: instead, scientists and engineers resort to measuring integrated quantities, such as the difference in potential between two points, and the current flux which is an integral of current density over an area. For practical purposes, conductivity will turn out to be an integral average of the pointwiseσappearing in Ohm’s Law. The idea of reporting integral quantities is not specific to inverse problems: it is central to discrete differential geometry, an idea presented in [34, 73].
2We might argue that every scientific measurement provides data for an inverse problem.
Starting simply, in one dimension, what can we measure about conductivity? By corre- spondence to our problem inR2, we are given the one-dimensional rod 0≤x≤1 with con- ductivityσ(x). We measure the potentialsu(0), u(1), and the boundary currentsσ du
dx x=0,1
. The potential obeys the conservation law
− d dx
σdu
dx
= 0, 0< x <1, (3.5)
so we immediately note that consistent with our physical intuition, the currentσdu
dx =f = constant throughout the rod. Furthermore, since only the first derivative of u appears in the conservation law, we can add a constant to u(0), u(1) without changing the solution, and only the difference u(1)−u(0) is the relevant measurement of potential. Finally, the equation is linear: In this context, the only unique data we can collect about this rod is the potential differenceg=u(1)−u(0) when the current is equal tof, and when, for example, we doublef,gwill double also. Hence, the Neumann-to-Dirichlet map in one dimension is
Λσ(f) =ag (3.6)
for some constant a. This valueais parameterised by σ and is the only information about σ that we can hope to obtain from studying Λσ.
It is helpful for what comes next to know what that constant a actually is. Multiply- ing (3.5) byu and integrating by parts, we have
Z 1 0
σ du
dx 2
= uσdu dx
1 0
(3.7)
σdu dx
2Z 1 0
1
σ =f g (3.8)
f /g= Z 1
0
1 σ
−1
=σ
¯. (3.9)
We have used the fact that the current,f =σdu
dx, is constant throughout the domain. The interpretation is that we identify the harmonic mean of σ
¯ as the effective conductivity of the domain. In one dimension, the harmonic mean is the only quantity we can measure about conductivity based only on the Neumann-to-Dirichlet map3.
3This is really a consequence of the coincidence of the gradient and divergence in one dimension: since
From an experimentalist’s point of view, even if we are able to probe the material at its interior points, with only a finite number of measurements, all we can deduce is the harmonic mean of σ between our sample points — we can know integral values, but not pointwise values. The harmonic mean is the same conductivity we compute when we homogenise by the RVE method, the asymptotic homogenisation, and, most importantly, by metric-based up-scaling. The message is that the averaged conductivities we measure using the Neumann-to-Dirichlet map are also the homogenised conductivities. In fact, the only experiment we can do to determine conductivity is measure a Neumann-to-Dirichlet map. And, this is not just true in one dimension.
When the conductivity is homogenised by metric-based up-scaling, the homogenised conductivity admits piecewise linear solutions. In one dimension, this means that the ho- mogenisation solution has constant gradient, so since the current Qdu
dx is constant, the homogenised conductivity is constant also. However, Q homogenised from σ is almost never constant in more than one dimension, but Q’s admission of solutions with constant gradient can tell us precisely what we can measure about homogenisation of conductivity using the Neumann-to-Dirichlet map.
Consider our domain to be the triangleT =ijk. As usual, (ϕi) form the piecewise linear basis over the triangle. To each test potentialgi =ϕi|∂T, we associate a test currentfi|∂T. In keeping with our restriction to integrated quantities, the information available to us are the fi integrated over edges of the triangle. For example, for test potentiali and edge ij, we have
fi(ij)= Z
ij
fi (3.10)
= Z
ij
(Q∇ϕi)Tdˆn (3.11)
= Z
∂T
(ϕi+ϕj)(Q∇ϕi)Tdˆn (3.12)
= Z
T
∇(ϕi+ϕj)TQ∇ϕi (3.13)
=− Z
T
∇ϕTkQ∇ϕi =qik. (3.14)
We have used the divergence theorem and the fact that∇ϕi+∇ϕj+∇ϕk= 0 on a triangle.
div(σ∇u) =∇(σ∇u), x∈R, we knowσ∇u= constant.
Similar calculations show that the three independent elements qij of the stiffness matrix, namely
qij =− Z
T
∇ϕTi Q∇ϕj, (3.15)
together completely determine the relationship between these test potentials and integrated currents. In some sense this is obvious: the stiffness matrix for an element determines the relationship between the solution u and the sources R
Tf ϕj over the element. However, we use the piecewise linear basis functions as test potentials because we know that they are exact solutions to the homogenised equation usingQat every interior point of the triangle.
Of course, the uniqueness results in the inverse problems literature say we can do far more in dimensions two and greater, stating that given Λσ, we can uniquely specify the corresponding isotropic conductivity everywhere in the domain [13, 50, 54, 82]. However, even in the broader context of an entire triangulated domain, deducing the elements of the stiffness matrix, the (qij) for the piecewise linear basis, is the correct thing to do, and we should interpret these edge-wise values as stiffness matrix elements for the homogenised conductivity. Further, since Q is divergence-free, we are justified in parameterising its stiffness matrix by s(x), just as we did for the forward problem.