= Solution
For <spatially varying tension in filament bending>, keep the derivative of the <filament tension> as well as the curvature term. The first variation is
$$
\delta\mathcal E=\int_0^L\{Ah''''-(\sigma h')'\}\eta\,dx+
[Ah''\eta'+(\sigma h'-Ah''')\eta]_0^L.
$$
Therefore \b[the <Euler-Lagrange equation> and fluctuation operator are]
$$
\boxed{Ah''''-(\sigma h')'=Ah''''-\sigma h''-\sigma'h'=0,\qquad
K_\sigma=A\partial_x^4-\partial_x(\sigma\partial_x).}
$$
For real, sufficiently smooth $\sigma$, the boundary form is the bending boundary form minus $[\sigma(\overline u v'-\overline{u'}v)]_0^L$. 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 $h=\sum_nb_n\psi_n$. Using the endpoint conditions in <integration by parts> gives
$$
\int_0^L[A\psi_n''\psi_m''+\sigma\psi_n'\psi_m']\,dx=\lambda_n\delta_{nm},\qquad
\mathcal E=\frac12\sum_n\lambda_nb_n^2.
$$
The <equipartition theorem> now gives, on the strictly positive subspace,
$$
\boxed{\langle b_nb_m\rangle=\frac{k_BT}{\lambda_n}\delta_{nm},\qquad
\operatorname{Var}(h(x))=k_BT\sum_{n:\,\lambda_n>0}\frac{\psi_n(x)^2}{\lambda_n}.}
$$
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 $\sigma=-C x(1-x)$ and the clamped trial function $h=x^2(1-x)^2$. Then
$$
\int_0^1(h'')^2\,dx=\frac45,\qquad
\int_0^1 x(1-x)(h')^2\,dx=\frac1{315},
$$
so 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.
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