= Solution
Take $a_1,a_2>0$ and $p_1+p_2=1$. Write $g^{(k)}=\nabla u$ in phase $k$ and $h^{(k)}=a_kg^{(k)}$. 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
$$
g^{(1)}_2=g^{(2)}_2=E_2,\qquad
g^{(1)}_3=g^{(2)}_3=E_3,\qquad
a_1g^{(1)}_1=a_2g^{(2)}_1=h_1.
$$
The mean <gradient> is $E=p_1g^{(1)}+p_2g^{(2)}$. Therefore
$$
E_1=h_1\left(\frac{p_1}{a_1}+\frac{p_2}{a_2}\right),\qquad
\langle h_2\rangle=(p_1a_1+p_2a_2)E_2,\qquad
\langle h_3\rangle=(p_1a_1+p_2a_2)E_3.
$$
By the definition of <effective conductivity>, $\langle h\rangle=a^*E$. Thus
$$
\boxed{
a^*=\operatorname{diag}\left(
\frac{a_1a_2}{p_1a_2+p_2a_1},\
p_1a_1+p_2a_2,\
p_1a_1+p_2a_2
\right).
}
$$
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.
Back to article page
