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For spatially varying tension in filament bending, keep the derivative of the filament tension as well as the curvature term. The first variation isTherefore the Euler-Lagrange equation and fluctuation operator areFor real, sufficiently smooth $\sigma$, the boundary form is the bending boundary form minus $[\sigma (\overline{u}{v}^{\prime }-\overline{u^{\prime }}v){]}_0^L$. Because $\sigma$ vanishes at both ends, the four self-adjoint endpoint conditions for filament bending still apply. The natural endpoint force also reduces there to the bending shear term. Thus $K_{\sigma }$ is a self-adjoint fourth-order scalar differential operator on the same chosen domain, with compact resolvent.

Choose a real orthonormal basis ${\psi }_n$ of eigenfunctions, $K_{\sigma }{\psi }_n={\lambda }_n{\psi }_n$, and write $h={\sum }_n{b}_n{\psi }_n$. Using the endpoint conditions in integration by parts givesThe equipartition theorem now gives, on the strictly positive subspace,This is a formal modal construction; no explicit eigenfunctions are needed. Nonnegative filament tension makes the energy nonnegative. Any surviving zero-energy filament modes must again be fixed. If signed $\sigma$ permits compression, self-adjointness still holds but does not guarantee a canonical ensemble: sufficiently strong compression can create negative eigenvalues and Euler buckling of an elastic filament. For instance, on $A=L=1$, take $\sigma =-Cx(1-x)$ and the clamped trial function $h=x^2(1-x{)}^2$. Thenso the energy is negative when $C>252$, despite $\sigma (0)=\sigma (1)=0$. The equipartition theorem requires a stable positive quadratic energy, not merely a real modal spectrum.