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

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 areThe mean gradient is $E=p_1{g}^{(1)}+p_2{g}^{(2)}$. ThereforeBy the definition of effective conductivity, $⟨h⟩=a^{\ast }E$. ThusThe 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.

Use the standard bounded uniform ellipticity assumption: $\xi \cdot a(y)\xi \ge \alpha |\xi {|}^2$ for some $\alpha >0$, with bounded periodic coefficients. It holds for the usual finite set of positive-conductivity phases. Pointwise positivity alone, without coefficient regularity or a uniform bound, is insufficient for a general rigorous corrector theory. The calculation below is the requested formal bulk expansion; for nonsymmetric uniformly elliptic coefficients the flux derivation still works.

Set $y=x/ϵ$. In a two-scale expansion for periodic conductivity, differentiation becomesThe leading, order ${ϵ}^{-2}$, equation isFor fixed $x$, multiply by $u_0$ minus its cell mean and integrate over the periodic cell. Opposite-face terms cancel, givingHence $u_0$ is independent of $y$.

The order ${ϵ}^{-1}$ equation is nowFor each coordinate direction, choose a periodic zero-mean corrector $w_j$ satisfying the periodic conductivity cell problemThe right side has zero cell mean. In weak form,for every periodic test function $\varphi$. On the space of zero-mean periodic functions, the Poincaré inequality and uniform ellipticity give coercivity, so the corrector exists and is unique by the Lax-Milgram theorem. Discontinuous phase coefficients are handled by this weak formulation.

The order ${ϵ}^{-1}$ solution iswhere the last term is independent of $y$ and does not affect leading flux. Thus the leading microscopic flux isAt order one the equation readsAverage over $Y$. The second term integrates to zero by periodicity, and $|Y|=1$. ThereforeThe effective tensor is constant, so the last equation is exactly $-a_{ij}^{\ast }{\partial }_{x_i}{\partial }_{x_j}{u}_0=f$. The macroscopic boundary condition is $u_0=0$ on $\partial \Omega$. Higher bulk correctors usually do not individually vanish on the physical boundary; a boundary corrector in periodic homogenization handles this without changing the leading homogenized equation.

For symmetric $a$, a useful check on the effective conductivity isThe extra term involving $\nabla w_i$ vanishes by the weak cell equation. This formula proves symmetry. With $w_{\xi }={\xi }_j{w}_j$, it also gives $\xi \cdot a^{\ast }\xi \ge \alpha {\int }_Y|\xi +\nabla w_{\xi }{|}^2\ge \alpha |\xi {|}^2$, confirming that the effective tensor is positive-definite.

In one dimension the periodic conductivity cell problem reduces toHence $a(y)(1+w^{\prime })=C$ is constant. Periodicity requires ${\int }_0^1{w}^{\prime }dy=0$, soThe effective flux average is $a^{\ast }={\int }_0^{1}a(1+w^{\prime })dy=C$, givingOne can exhibit the corrector explicitly: $w^{\prime }=a^{\ast }/a_1-1$ in phase 1 and $w^{\prime }=a^{\ast }/a_2-1$ in phase 2. Its integrated change over a period is zero, so a continuous piecewise affine periodic $w$ exists, and an additive constant fixes its mean.

This is precisely the normal component of the laminate conductivity in part (i). In one dimension there is no tangential direction: every heat-flow path crosses the phases in sequence, so only the harmonic-mean response remains.