Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 80 (Macroscopic Behaviour of Microscopic Structures in Fluid and Solid Media)

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.

Within each constant-conductivity region the temperature is harmonic. For $u=v(r)\cos \theta$, the spherical Laplacian givesTrying $v=r^s$ yields $s(s-1)+2s-2=(s-1)(s+2)=0$. Thus $v=Cr+D{r}^{-2}$. Boundedness at the origin removes the interior $r^{-2}$ term, while the prescribed far gradient fixes the exterior growing term:Continuity of temperature and normal flux at $r=r_1$ giveSolving these two equations yields $B=-r_1^3(a_1-a_0)/(a_1+2a_0)$ and $C_{\mathrm{in}}=3a_0/(a_1+2a_0)$. ConsequentlyPositive conductivities make the denominator nonzero. The matched solution is unique in the usual finite-energy decaying-perturbation class: the difference of two solutions has zero far data and interface jumps, and an energy integration gives zero gradient.

For the polarization field of a conductivity inclusion, use the sign convention consistent with the perturbation equation:ThusFor a general incident constant gradient $E$, rotational symmetry and linearity replace $e_3$ by $E$. The polarization vanishes outside the sphere because its local conductivity equals the reference value. The temperature perturbation outside is dipolar; this is the spherical conductivity inclusion response used in the remaining parts.

Let $E=⟨\nabla u⟩$ and retain only the independent response of each well-separated small sphere. In the leading dilute spherical-inclusion conductivity approximation,Averaging gives $⟨P⟩\simeq p\beta E$, where $p={\sum }_j|B_j|/|\Omega |$. Since $a\nabla u=a_0\nabla u+P$,The spherical response holds for every $E$, soThis retains the order-$p$ term and neglects interactions. A precise dilute regime can be obtained by shrinking the spheres while holding separated centres fixed. Volume fraction alone is not enough if the inclusions instead remain in close clusters whose collective polarization persists at leading order.

Set $E=e_3$ and $\beta =3a_0(a_1-a_0)/(a_1+2a_0)$. The incident field relevant to each sphere is its exciting field of a conductivity inclusion: the macroscopic field plus the regular field due to other inclusions and periodic images, excluding its own singular dipole. In the first interaction approximation writeThe interior response from part (i) then gives the improved polarizationAveraging, using $⟨{\chi }_{B_j}⟩=p_j$, givesTherefore the dipole interaction correction to effective conductivity isThe factors $p_j$ cancel against the denominator in the given field correction. Linearity permits repeating the calculation for three independent incident gradients, recovering the tensor rather than just its action on $e_3$. This is a second-reflection estimate; further dipole reflections and higher multipoles have been omitted.

A literal pointwise reading of the field printed “on the surface” needs qualification. Even the isolated-sphere solution in part (i) has, on a concentric sphere of radius $R$,For $E=e_3$, its values at $n=e_3$ and $n=e_1$ are respectively $2k{r}_1^{3}E/R^3$ and $-k{r}_1^{3}E/R^3$, where $k=(a_1-a_0)/(a_1+2a_0)$. They differ whenever there is nonzero contrast. Thus a single constant matrix cannot give the total pointwise field everywhere on that sphere.

The consistent local-field interpretation is a spherical average, or equivalently the regular exciting field at the centre. Indeed $⟨n\otimes n{⟩}_S=I/3$, so the angular mean of the self-dipole is zero. The other-inclusion field is harmonic inside the surrounding sphere, and the mean value property gives its central value. With ${\Lambda }_j$ encoding this average regular correction, the calculation above is justified. If “on the surface” is instead insisted upon as a constant total pointwise field, that premise is false and the isolated-sphere values above are a counterexample. The distinction between dipole fields and local exciting fields is also treated in sections 10.2–10.3 of www.math.utah.edu/~milton/TheoryCompositesNOPRINT.pdf .

Let $k=(a_1-a_0)/(a_1+2a_0)$. Far from all the small spheres, their leading dipole contributions have the same direction and their leading $|x{|}^{-2}$ potential terms add:Translations of their centres change only higher far-field multipoles. Since the outer body is a sphere of radius $R_0$,Replacing the whole body by an isotropic effective sphere of conductivity $a^{\ast }$ gives its leading dipole coefficient $R_0^3(a^{\ast }-a_0)/(a^{\ast }+2a_0)$. Matching coefficients yieldsSolving this equation gives the Maxwell approximation for conductivity:For positive phase conductivities and $0\le p\le 1$, 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, $\beta =3a_{0}k$, so its dilute expansion isThe isotropic interaction relation supplied in part (iii) givesThus the Maxwell approximation and the interaction correction agree through second order in volume fraction.

Use symmetric positive conductivity, for example $\alpha I⪯a(x)⪯MI$ with $\alpha >0$. Let $v$ have the same Dirichlet boundary data as $u$, and write $v=u+\varphi$ with $\varphi \in H_0^1(\Omega )$. Symmetry givesThe middle integral vanishes by the weak field equation; equivalently integrate by parts, use $\varphi =0$ on the boundary and $\nabla \cdot (a\nabla u)=0$. HenceThereforeEquality forces $\nabla \varphi =0$, and its zero boundary trace gives $\varphi =0$, proving uniqueness. This is the Dirichlet principle for the conductivity equation.

The positivity is essential to the physical minimum-energy interpretation; symmetry by itself would give only stationarity. For example, take $a=-I$, $\lambda =0$ and $u=0$. The field equation holds, but every nonzero smooth compactly supported $v$ has $J(v)=-{\int }_{\Omega }|\nabla v{|}^2<0$, and scaling $v$ makes the energy unbounded below. Thus if the printed word “conductivity” were not understood to include positivity, the stated minimum claim would require that additional hypothesis.

Choose the affine trial function $v(x)=\lambda \cdot x$. It has the required trace and constant gradient, so the Dirichlet principle givesFor the macroscopic isotropic effective medium, the corresponding affine-boundary energy is $|\Omega |a^{\ast }|\lambda {|}^2$. Dividing by $|\Omega ||\lambda {|}^2$ for a nonzero imposed gradient yields the Voigt boundMore generally the same calculation is the tensor inequality in the Loewner order $a^{\ast }⪯⟨a⟩$. For a finite unresolved specimen the calculation bounds its apparent affine-boundary response; identification with the bulk effective conductivity uses the macroscopic scale separation.

Write $\Delta =a_1-a_2<0$, and let $q=t\Delta \lambda {\chi }_1$. The inverse contrast in the Hashin-Shtrikman conductivity variational principle is interpreted only in phase 1: the admissible $q$ vanishes in phase 2, where $a-a_{2}I=0$. Thus

For the nonlocal term define $K(x,x^{\prime })={\nabla }_x{\nabla }_{x^{\prime }}g(x,x^{\prime })$. The given Green representation impliesThe two-point phase correlation function therefore givesBecause the unweighted double integral of $K$ is zero, subtract $p_1^2$ inside the bracket. Set $h(r)=p_{11}(r)-p_1^2$; it vanishes for $r>\rho$, so only the microscopic near-diagonal kernel contributes. Its contact part isAt coincident points ${\chi }_1^2={\chi }_1$, hence $h(0)=p_1-p_1^2=p_1{p}_2$. The ordinary dipole part has zero angular integral by the stipulated spherical-shell identity. Since $h$ is radial, its weighting preserves this cancellation:Consequently, in the macroscopic bulk regime implicit in the supplied local Green approximation,andThis is the dipole Green tensor angular cancellation: the constant background correlation disappears and the connected contact value supplies $p_1{p}_2$.

Insert all three averaged terms in the variational upper inequality and normalize by $|\Omega ||\lambda {|}^2$. For the bulk effective conductivity one obtainsThe kernel in the question has an approximation sign. At fixed finite size, spheres around points near $\partial \Omega$ are truncated and the radial cancellation is not an exact finite-body identity. Their contribution and the regular boundary part disappear in the stated microscopic-to-macroscopic limit. The calculation derives the intended homogenized bound, rather than silently treating that Green approximation as exact for arbitrary finite $\rho$.

Let $C=1+\Delta p_2/(3a_2)$. Since $a_1>0$, $\Delta >-a_2$, so $C>0$. Also $\Delta p_1<0$ when both phases are present. Completing the square in the trial upper bound givesThus the tightest upper bound is obtained by minimizing at $t=1/C$, not maximizing. It isPure-phase cases follow directly or by continuity. These Hashin-Shtrikman bounds for conductivity are volume-fraction bounds on the statistically isotropic homogenized material; the full variational framework is discussed in Chapter 23 of www.math.utah.edu/~milton/TheoryCompositesNOPRINT.pdf .

For the lower variational inequality choose reference conductivity $a_1$ and a trial polarization $q$ that vanishes in phase 1. Let $v_q\in H_0^1(\Omega )$ solveThe Hashin-Shtrikman conductivity variational principle isIt holds for every admissible trial $q$; the exact principle takes the supremum of the right side. Writing the last integral only over phase 2 avoids applying an inverse to the zero contrast in the reference phase.

With $D=a_2-a_1>0$, the isotropic phase trial $q=tD\lambda {\chi }_2$ gives the lower expressionHere the quadratic is concave, so maximize it at $t=[1+D{p}_1/(3a_1){]}^{-1}$. The lower member of the Hashin-Shtrikman conductivity bounds isTogether with part (iii), this gives the required upper and lower interval. All denominators are positive for positive phase conductivities.