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For $G_{xx}=R$, zero bending moment and transverse force at the leading end give $G(x)={\int }_0^x(x-s)R(s)ds$. The two trailing-end free-end bending boundary conditions hold exactly when the displayed two integrals vanish. They express total transverse force and torque balance. Adding unknown rigid translation and yaw to a prescribed deformation supplies the two degrees of freedom of the swimming recoil correction. Its principal inertial matrix is the weighted moment matrix of $1$ and $x-x_c$ with weight $m_b+m$, hence is positive definite for a nondegenerate positive mass distribution.

A freely swimming body has neither an imposed end couple nor an imposed end transverse force. Its end bending moment is therefore zero, and the transverse force $S=G_x$ is also zero at both ends. These are the four free-end bending boundary conditions; they do not require the end displacement or slope to vanish.

Put $R=m_b{h}_{tt}+\mathcal{D}(m\mathcal{D}h)$, using the physically consistent sign. Integrating $G_{xx}=R$ with the two leading-end conditions givesThe two trailing-end conditions consequently requireThese free-end compatibility for a swimming beam conditions are total transverse force and torque balances. They explain why an arbitrary plausible prescribed $h$ does not suffice: two integration constants can satisfy only two end conditions unless the forcing already satisfies both compatibility integrals.

The missing freedom is the swimming recoil correction. An observed or prescribed deformation $h_d$ relative to the body must be supplemented by a lateral translation $Y(t)$ and a small yaw $\Theta (t)$:To identify $Y$ with the actual centre of mass displacement, first remove the body-mass-weighted mean from $h_d$. One may also remove its body-mass-weighted first moment about $x_c$ to define the yaw reference uniquely. This changes the decomposition of the observed motion, not the physical displacement.

For $\xi =x-x_c$, the recoil contribution to the forcing isInsert $R=R_d+R_{\mathrm{rec}}$ into the two compatibility integrals, using $1$ and $\xi$ as weights. This gives two coupled ordinary differential equations for the two recoil motions. Their principal acceleration matrix isFor a nondegenerate positive mass distribution it is a positive-definite matrix: its quadratic form at $(a,b)$ is ${\int }_0^L(m_b+m)(a+b\xi {)}^{2}dx>0$ unless $a=b=0$. Thus initial translation and yaw determine the recoil through these equations, and the corrected $h$ supplies the two compatibility conditions. The four free-end conditions determine the bending moment together with two rigid recoil motions; they do not overdetermine a physically complete swimming motion.

For the elastic filament, apply a variation $\eta (x)$. Two integration by parts operations giveAt a free endpoint both $\eta$ and ${\eta }^{\prime }$ are arbitrary. The free-end bending boundary conditions and the unloaded Euler-Lagrange equation are thereforeHere $A{h}^{\prime \prime }$ is the bending moment and $-A{h}^{(3)}$ 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 formThis proves symmetry. To establish actual self-adjointness, the adjoint-domain boundary form must vanish for every free-end $u$. The endpoint values $u,u^{\prime }$ can be varied independently, forcing $v^{\prime \prime }=v^{(3)}=0$ 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 $⟨h,Kh⟩=A\int |h^{\prime \prime }{|}^{2}dx\ge 0$.