= Solution
For the <elastic filament>, apply a variation $\eta(x)$. Two <integration by parts> operations give
$$
\delta\mathcal E=A\int_0^L h''\eta''dx
=A\int_0^L h^{(4)}\eta\,dx
+A[h''\eta'-h^{(3)}\eta]_0^L.
$$
At a free endpoint both $\eta$ and $\eta'$ are arbitrary. The <free-end bending boundary conditions> and the unloaded <Euler-Lagrange equation> are therefore
$$
\boxed{h^{(4)}=0\quad(0<x<L),\qquad
h''=h^{(3)}=0\quad\text{at }x=0,L.}
$$
Here $Ah''$ is the bending moment and $-Ah^{(3)}$ the transverse shear-force convention. The equilibrium shapes are affine, reflecting free translation and tilt.
On $L^2(0,L)$ define $K=A\partial_x^4$ on $H^4$ functions obeying those endpoints. The <self-adjoint endpoint conditions for filament bending> annul the boundary form
$$
\langle u,Kv\rangle-\langle Ku,v\rangle
=A[\overline u v^{(3)}-\overline{u'}v''+\overline{u''}v'-\overline{u^{(3)}}v]_0^L.
$$
This proves symmetry. To establish actual <self-adjointness>, the adjoint-domain boundary form must vanish for every free-end $u$. The endpoint values $u,u'$ can be varied independently, forcing $v''=v^{(3)}=0$ at both ends. The regular fourth-order expression then gives the same $H^4$ domain for the adjoint. Thus the domains coincide and $K$ is <self-adjoint>. Also $\langle h,Kh\rangle=A\int|h''|^2dx\geq0$.
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