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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-80/1/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 80 / 1 / i / Solution by

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

Take

a_{1}, a_{2} > 0

and

p_{1} + p_{2} = 1

. Write

g^{(k)} = \nabla u

in phase

k

and

h^{(k)} = a_{k} g^{(k)}

. Here

h

is the positive conductivity flux; physical heat flux is

- h

. Continuity of temperature on a planar interface implies continuity of its tangential derivatives, while conservation gives continuity of normal heat flux. With interface normal

e_{1}

, these conditions are

g_{2}^{(1)} = g_{2}^{(2)} = E_{2}, g_{3}^{(1)} = g_{3}^{(2)} = E_{3}, a_{1} g_{1}^{(1)} = a_{2} g_{1}^{(2)} = h_{1} .

(1)

The mean gradient is

E = p_{1} g^{(1)} + p_{2} g^{(2)}

. Therefore

E_{1} = h_{1} (\frac{p _{1}}{a _{1}} + \frac{p _{2}}{a _{2}}), ⟨ h_{2} ⟩ = (p_{1} a_{1} + p_{2} a_{2}) E_{2}, ⟨ h_{3} ⟩ = (p_{1} a_{1} + p_{2} a_{2}) E_{3} .

(2)

By the definition of effective conductivity,

⟨ h ⟩ = a^{*} E

. Thus

a^{*} = diag (\frac{a _{1} a _{2}}{p _{1} a _{2} + p _{2} a _{1}}, p_{1} a_{1} + p_{2} a_{2}, p_{1} a_{1} + p_{2} a_{2}) .

(3)

The laminate conductivity is a weighted harmonic mean across the layers and a weighted arithmetic mean along them. This also explains why isotropic constituents can yield an anisotropic thermal conductivity tensor: flux must cross each resistance in sequence in the normal direction, whereas tangential transport occurs in parallel.

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