 Solution

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.