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is constant on the connected body. Its zero boundary value makes that constant zero. This proves 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

α \partial_{n} w_{+} = β \partial_{n} w_{-}, w_{+} ∣_{S_{2}} = 0.

(2)

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_{-} ∣^{2} d V = \int_{S_{1}} w \partial_{n} w_{-} d S, E_{2} := \int_{V_{2}} ∣\nabla w_{+} ∣^{2} d V = - \int_{S_{1}} w \partial_{n} w_{+} d S .

(3)

The outer boundary term in

E_{2}

is zero. Weight these equations by the positive coefficients that match the flux condition:

β E_{1} + α E_{2} = \int_{S_{1}} w (β \partial_{n} w_{-} - α \partial_{n} w_{+}) d S = 0.

(4)

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

β

and

α

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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