= Solution
The <Dirichlet uniqueness for the Poisson equation> follows from a positive integral. Suppose $T_a,T_b$ satisfy the same data, and put $w=T_a-T_b$. Then $w$ is a <harmonic function> in $V$ and vanishes on $S$. <Green's first identity>, applied with both arguments $w$, gives
$$
\int_V|\nabla w|^2dV=\int_Sw\,\partial_nw\,dS-\int_Vw\Delta w\,dV=0.
$$
Thus $\nabla w=0$, and $w$ is constant on the connected body. Its zero boundary value makes that constant zero. This proves \b[at most one temperature field]; it does not require an independent existence construction. As usual, the argument assumes a bounded body and sufficient regularity for <Green's first identity>, or the corresponding finite-energy weak formulation.
For the composite body, let $w_-$ and $w_+$ be the differences in $V_1$ and $V_2$. They are <harmonic functions>, have a common interface value $w$, and obey
$$
\alpha\partial_nw_+=\beta\partial_nw_-,\qquad w_+|_{S_2}=0.
$$
The <unit normal vector> $n$ points out of $V_1$; the outward normal of $V_2$ on its inner boundary is $-n$. Keeping that sign in <Green's first identity> gives
$$
E_1:=\int_{V_1}|\nabla w_-|^2dV=\int_{S_1}w\,\partial_nw_-\,dS,
\qquad
E_2:=\int_{V_2}|\nabla w_+|^2dV=-\int_{S_1}w\,\partial_nw_+\,dS.
$$
The outer boundary term in $E_2$ is zero. Weight these equations by the positive coefficients that match the flux condition:
$$
\boxed{\beta E_1+\alpha E_2=\int_{S_1}w\left(\beta\partial_nw_--\alpha\partial_nw_+\right)dS=0.}
$$
Both <Dirichlet energies> must vanish. Hence both differences are constants; the outer data set the constant in $V_2$ to zero, and interface continuity sets the constant in $V_1$ to zero. This establishes <uniqueness for a harmonic transmission problem>. Physically $\beta$ and $\alpha$ play the roles of the inner and outer <thermal conductivities>; continuity of temperature and normal heat flux is what cancels the interface contribution.
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