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Take the difference $w$ of two solutions with the same body force, initial displacement and initial velocity. It satisfies the homogeneous linear elasticity equations. With $e=e(w)$ the infinitesimal strain tensor, define the bulk energyThe usual stability assumptions make the elastic quadratic form nonnegative and $\rho >0$. Multiplying momentum balance by $w_t$, using integration by parts and using symmetry of the stress tensor givesAll interior terms cancel: the strain term is exactly the time derivative of elastic energy.

For the restoring boundary law, $t_w=kw$. The prescribed support displacement cancels in the difference. Since $k$ is a time-independent symmetric matrix,Hence the elastic energy uniqueness with restoring boundary springs identity isThe boundary contribution is nonnegative because $k$ is negative semidefinite. The initial difference and its velocity vanish, so the conserved energy is zero. Its kinetic term implies $w_t=0$ everywhere; the zero initial displacement then gives $w=0$. This proves uniqueness without needing to exclude rigid displacements separately.

Physically, $-k$ is a nonnegative boundary spring-stiffness matrix, and the prescribed vector $U$ is the moving support position. The traction restores the displacement towards that support; null directions of $k$ have no spring force. The interpretation and conserved-energy proof use a fixed spring matrix, as indicated by the given boundary law.

On the mixed part $S_1$, write $t=\sigma n$. The condition isThus the tangential traction and normal displacement are prescribed. The normal traction is a constraint reaction; tangential displacement is free. Only the tangential component of $F$ matters. On $S_2$, the entire displacement is prescribed.

For the difference of two such solutions, $w\cdot n=0$ and $(I-n{n}^{\mathsf{T}})t_w=0$ on $S_1$. Consequently $w_t$ is tangential and $t_w$ normal, so $w_t\cdot t_w=0$. On $S_2$, $w_t=0$. The boundary power vanishes everywhere, giving $d{E}_{\mathrm{bulk}}/dt=0$. The same zero-initial-energy argument gives $w=0$. Therefore the mixed boundary problem also has at most one solution.