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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-80/2/iv/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 80 / 2 / iv / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Let

k = (a_{1} - a_{0}) / (a_{1} + 2 a_{0})

. Far from all the small spheres, their leading dipole contributions have the same direction and their leading

∣ x ∣^{- 2}

potential terms add:

u (x) - x_{3} ≃ - k (\sum_{j} r_{j}^{3}) \frac{x _{3}}{∣ x ∣ ^{3}} .

Translations of their centres change only higher far-field multipoles. Since the outer body is a sphere of radius

R_{0}

,

p = \frac{\sum _{j} r _{j}^{3}}{R _{0}^{3}} .

Replacing the whole body by an isotropic effective sphere of conductivity

a^{*}

gives its leading dipole coefficient

R_{0}^{3} (a^{*} - a_{0}) / (a^{*} + 2 a_{0})

. Matching coefficients yields

\frac{a ^{*} - a _{0}}{a ^{*} + 2 a _{0}} = p k .

Solving this equation gives the Maxwell approximation for conductivity:

a^{*} = a_{0} \frac{1 + 2 p k}{1 - p k} = a_{0} + \frac{3 p a _{0} ( a _{1} - a _{0} )}{3 a _{0} + ( 1 - p ) ( a _{1} - a _{0} )} .

For positive phase conductivities and

0 \leq p \leq 1

, the denominator is positive. This rational approximation is obtained by dipole matching; its extrapolation beyond the dilute regime is a closure assumption, not an exact formula for arbitrary inclusion arrangements.

For the consistency check,

β = 3 a_{0} k

, so its dilute expansion is

a^{*} = a_{0} + \frac{pβ}{1 - pβ / ( 3 a _{0} )} = a_{0} + pβ + \frac{p ^{2} β ^{2}}{3 a _{0}} + O (p^{3}) .

The isotropic interaction relation supplied in part (iii) gives

a^{*} ≃ a_{0} I + pβ I - β^{2} (- \frac{p ^{2}}{3 a _{0}} I) = (a_{0} + pβ + \frac{p ^{2} β ^{2}}{3 a _{0}}) I .

Thus the Maxwell approximation and the interaction correction agree through second order in volume fraction.

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