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A normal mode of a uniformly bending elastic filament fixed in position and slope at both ends satisfies $W^{\prime \prime \prime \prime }=k^{4}W$. With $\beta =kL>0$, the allowed wavenumbers solve $\cos \beta \cosh \beta =1$. The first root is $4.730040745$ and ${\beta }_n=(n+\frac{1}{2})\pi +O(e^{-(n+1/2)\pi })$ for $n\ge 1$. The apparent root at zero is not an eigenfunction: the zero-eigenvalue cubic polynomial satisfying all four clamped boundary conditions is identically zero. This spectrum differs from that of a filament clamped only at one end.

Elastic energy is the recoverable energy stored by deforming an elastic material. In the small-deflection model of an elastic filament, the bending contribution is a positive quadratic functional of the curvatures, such as $\frac{1}{2}\int (A{y}^{\prime \prime 2}+B{z}^{\prime \prime 2})dx$. Work by applied loads must also be included when minimizing the total potential energy.

Take $A>0$ and use primes for spatial derivatives. Two integration by parts operations in the first variation of the elastic filament's bending energy giveThus the Euler-Lagrange equation is $A{h}^{\prime \prime \prime \prime }=0$, and the fluctuation differential operator is $K=A{d}^4/d{x}^4$. In the L2 space inner product, its boundary form isThe conjugate endpoint trace pairs are $(h,h^{\prime \prime \prime })$ and $(h^{\prime },h^{\prime \prime })$. Requiring one member of each pair to vanish gives the four standard self-adjoint endpoint conditions for filament bending, applied at both ends:

Each pair annihilates the boundary form for all $u,v$ in the domain. Conversely, the remaining two endpoint traces can be chosen freely: requiring the boundary form to vanish against every such $u$ forces an adjoint-domain function $v$ to satisfy the same two conditions. This proves self-adjointness, rather than just formal symmetry, on the corresponding fourth-order Sobolev space domain.

The count four concerns these elementary homogeneous choices. Identical-end boundary conditions do not restrict all self-adjoint operators to these four possibilities. For example, $h=0$, $h^{\prime \prime }=b{h}^{\prime }$ at both ends, with any fixed real $b\ne 0$, also annihilates the boundary form: the remaining expression is $-\overline{u^{\prime }}b{v}^{\prime }+b\overline{u^{\prime }}{v}^{\prime }=0$. This Robin boundary condition supplies a continuous family beyond the four listed pairs.

Choose real eigenfunctions with ${\int }_0^L{W}_n{W}_{m}dx={\delta }_{nm}$. For positive modes, self-adjointness and integration by parts diagonalize the energy:At temperature $T$, the canonical ensemble is a product of centered Gaussian distributions. The equipartition theorem, with Boltzmann constant $k_B$, givesThe resulting thermal covariance of an elastic filament isFor an unnormalized eigenfunction of squared L2 norm $N_n$, divide its summand by $N_n$. This normalization factor cannot be absorbed silently into the modal variance.

For clamped boundary conditions at both ends, the inverse of $d^4/d{x}^4$ has Green function, for $x\le y$,It is cubic on each side of $y$, satisfies the four clamped conditions, has continuous first two derivatives, and has unit jump in its third derivative. Thus ${\partial }_x^{4}G=\delta (x-y)$, and its eigenfunction expansion is the sum above. In particular,These finite variances require removal of every zero-energy filament mode. An unconstrained free-free elastic filament can translate and tilt at no energy cost; a torqued-torqued elastic filament can translate. Their unrestricted Boltzmann distributions are not normalizable, so the full displacement variance is undefined. Fix those rigid degrees of freedom before applying the positive-mode formula.

For one small transverse displacement, the elastic energy of the elastic filament iswhere $A$ is the filament bending modulus. Its first variation isThe clamped boundary conditions remove the left boundary terms. A force-free and torque-free tip permits independent $\delta h(L)$ and $\delta h^{\prime }(L)$, so the natural boundary conditions for a free endpoint are $h^{\prime \prime }(L)=h^{\prime \prime \prime }(L)=0$: zero bending moment and transverse shear.

There is an important distinction between an energy minimum and a normal mode. The unloaded higher-order Euler-Lagrange equation is $h^{\prime \prime \prime \prime }=0$, whose only solution with these four boundary conditions is $h=0$. A general fluctuating shape is a sum of modes, not one of the stated sinusoidal/hyperbolic functions. To obtain the clamped--free bending modes, extremize the bending Rayleigh quotient, or equivalently $U[W]-\frac{1}{2}A{k}^4\int W^{2}dx$ with a fixed norm. Its Euler-Lagrange equation isThis is the constrained variational characterization of bending modes. Its four characteristic roots are $\pm k,\pm ik$. Clamping gives $F=-D$ and $E=-B$, henceThe free-tip conditions reduce toIts determinant is $-2(1+\cos q\cosh q)$, so nontrivial modes requireNo zero mode exists, since a cubic satisfying the homogeneous clamp/free conditions is zero. The clamped-free bending spectrum starts with $q_1\in (\pi /2,2)$; bisection or Newton iteration gives $q_1\simeq 1.875104$, so $k_1\simeq 1.875104/L$. Choosing $D=N(\sin q_n+\sinh q_n)$ and $B=-N(\cos q_n+\cosh q_n)$ givesThe second boundary condition follows from the root equation, and $N$ is arbitrary until a normalization is chosen.

The bending operator with these boundary conditions is positive and self-adjoint. Twice integrating by parts givesSymmetry implies orthogonality when $n\ne m$. Expand $h(x)={\sum }_n{a}_n{W}_n(x)$ and use $I_n=L^{-1}{\int }_0^L{W}_n^{2}dx$. Thenby the equipartition theorem. The supplied endpoint identity therefore yields the thermal bending fluctuations of a clamped filament:To evaluate the sum, apply a tip force $f$ and minimize $U-fh(L)$. The modified free-end conditions are $h^{\prime \prime }(L)=0$ and $-A{h}^{\prime \prime \prime }(L)=f$, and $h^{\prime \prime \prime \prime }=0$ in the interior. Integration gives $h(x)=f{x}^2(3L-x)/(6A)$, so the tip-force compliance of a cantilever is $L^3/(3A)$. On the other hand, minimizing the modal energy minus $f{\sum }_n{a}_n{W}_n(L)$ gives $a_n=f{W}_n(L)/(AL{k}_n^4{I}_n)$ and henceComparison evaluates the fourth inverse-power sum of the cantilever spectrum without truncating the modes:The boxed variance is for the specified single transverse direction. An independent equilibrium check follows by differentiating the Gaussian partition function with respect to $f$: $\partial ⟨h(L)⟩/\partial f=⟨h(L{)}^2⟩/(k_{B}T)$ at zero load, reproducing the same result from the static compliance. Keeping only the first normal mode gives about $0.32356k_{B}T{L}^3/A$, roughly $97.1\%$ of the exact variance. With two independent equivalent transverse directions, their summed variance is twice the boxed result. The small-slope model requires $k_{B}TL/A\ll 1$, or $L$ small compared with the usual three-dimensional persistence length $A/(k_{B}T)$.

For the elastic filament operator $K=A{\partial }_x^4$, two integration by parts operations give boundary form $A[\overline{u}{v}^{\prime \prime \prime }-\overline{u^{\prime }}{v}^{\prime \prime }+\overline{u^{\prime \prime }}{v}^{\prime }-\overline{u^{\prime \prime \prime }}v]$. Four elementary homogeneous choices at each endpoint are $h=h^{\prime }=0$ (clamped), $h=h^{\prime \prime }=0$ (hinged), $h^{\prime \prime }=h^{\prime \prime \prime }=0$ (free), and $h^{\prime }=h^{\prime \prime \prime }=0$ (slope-constrained and transverse-force-free). Each defines a self-adjoint operator on a finite interval when imposed at both ends. More general real Robin boundary conditions also give self-adjointness; the four choices are not an exhaustive classification of all boundary subspaces.

Adding $\frac{1}{2}\int \sigma (x)(h^{\prime }{)}^{2}dx$ to the quadratic bending energy of an elastic filament contributes $-(\sigma h^{\prime }{)}^{\prime }$ to its variational derivative. The endpoint term is $A{h}^{\prime \prime }{\eta }^{\prime }+(\sigma h^{\prime }-A{h}^{\prime \prime \prime })\eta$. If the real filament tension vanishes at both ends, the usual self-adjoint endpoint conditions for filament bending remain valid. A complete eigenfunction expansion diagonalizes the energy even when the eigenfunctions have no explicit formula. Thermal equipartition theorem arguments require positive eigenvalues after fixing any kernel; strong compression can violate this requirement.

For a clamped elastic filament of filament bending modulus $A$, a transverse tip force $f$ gives $h(x)=f{x}^2(3L-x)/(6A)$ in the small-slope model. Thus $h(L)/f=L^3/(3A)$. This differs from the effective tip stiffness under a distributed load. The zero-load Gaussian response reproduces the thermal bending fluctuations of a clamped filament, $⟨h(L{)}^2⟩=k_{B}T{L}^3/(3A)$ for one transverse coordinate.

A zero-energy mode lies in the kernel of the quadratic elastic filament operator. For pure bending, zero curvature makes the displacement affine. Free-free endpoints therefore leave translation and tilt modes, while slope-constrained, force-free endpoints leave only translation. Both clamped-clamped and hinged-hinged endpoints remove these modes. An unrestricted modal amplitude has constant Boltzmann distribution weight, so no normalizable canonical ensemble or finite displacement variance exists. Fixing the rigid degrees of freedom allows the equipartition theorem on the positive complement.