= Solution
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 <energy>
$$
E_{\rm bulk}(t)=\frac12\int_D\{\rho|w_t|^2+C_{ijkl}e_{ij}e_{kl}\}\,dV.
$$
The 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> gives
$$
\frac{dE_{\rm bulk}}{dt}=\int_S w_t\cdot t_w\,dS,
\qquad t_w=\sigma(w)n.
$$
All 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>,
$$
\int_Sw_t\cdot kw\,dS=\frac12\frac{d}{dt}\int_Sw\cdot kw\,dS.
$$
Hence the <elastic energy uniqueness with restoring boundary springs> identity is
$$
\boxed{\frac{d}{dt}\left(E_{\rm bulk}-\frac12\int_Sw\cdot kw\,dS\right)=0.}
$$
The 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 \b[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 is
$$
(I-nn^{\mathsf T})t=n\times F,\qquad u\cdot n=N.
$$
Thus 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 $(I-nn^{\mathsf T})t_w=0$ 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 $dE_{\rm bulk}/dt=0$. The same zero-initial-energy argument gives $w=0$. Therefore \b[the mixed boundary problem also has at most one solution].
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