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The elastic energy $U=A\int (W^{\prime \prime }{)}^{2}dx/2$ minimized without forcing gives $W^{\prime \prime \prime \prime }=0$. With fixed $\int W^{2}dx$, stationarity of $U-A{k}^4\int W^{2}dx/2$ gives $W^{\prime \prime \prime \prime }=k^{4}W$: a bending eigenmode. A general admissible shape expands in these normal modes. The natural boundary conditions for a free endpoint are $W^{\prime \prime }(L)=W^{\prime \prime \prime }(L)=0$, while the clamp imposes $W(0)=W^{\prime }(0)=0$.

An elastic filament is a slender elastic body modeled by a centerline and bending rigidities. Its elastic energy penalizes curvature. In a small-deflection representation $(x,y(x),z(x))$, the bending energy is quadratic in $y^{\prime \prime }$ and $z^{\prime \prime }$. Compression can cause Euler buckling of an elastic filament.

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 $A/(k_{B}T)$.

The arclength element is $ds=\sqrt{1+y^{\prime 2}+z^{\prime 2}}dx$. For uniformly small slopes, the Taylor expansion $\sqrt{1+s}=1+s/2+O(s^2)$ givesThus to quadratic order in the slopes the excess length is $\boxed{\ell -L\simeq \frac{1}{2}{\int }_0^L(y^{\prime 2}+z^{\prime 2})dx}$. This is the small-slope approximation used in the subsequent elastic energy.

Vary $y$ by $y+\epsilon \eta$, where the clamped endpoint conditions require $\eta ={\eta }^{\prime }=0$ at both endpoints. The first variation of the quadratic elastic energy isBoth integrations by parts have zero boundary terms. Since the variation is arbitrary in the interior, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. Varying $z$ independently gives the second equation:The clamped endpoint values accompany these equations. If $A$ or $B$ depends on $x$, it must remain inside the derivatives.

With unit constant bending coefficients and unit length, the quadratic elastic energy isTo find the first loss of positivity, put $v=y^{\prime }$. The clamped conditions give $v(0)=v(1)=0$ and ${\int }_0^{1}vdx=0$. Extend $v$ periodically with period one. The sharp periodic Wirtinger inequality gives ${\int }_0^1{v}^{\prime 2}dx\ge 4{\pi }^2{\int }_0^1{v}^{2}dx$. For completeness, expand the mean-zero periodic $v$ in its Fourier series ${\sum }_{n\ge 1}(a_n\cos 2\pi nx+b_n\sin 2\pi nx)$. Parseval identity giveswhich proves the inequality and its equality condition. The same holds for $z^{\prime }$. ConsequentlyFor $F<4{\pi }^2$, the straight filament is the unique minimum. Equality in the periodic Wirtinger inequality has $v=a\cos 2\pi x+b\sin 2\pi x$; the endpoint value $v(0)=0$ eliminates $a$. Integration and the position boundary conditions therefore give the first buckling load and modes:At this load every such pair has zero quadratic elastic energy and solves the Euler-Lagrange equations. For $F>F_c$ these modes give negative energy, signaling instability of the straight configuration. The quadratic small-slope model does not select a finite post-buckling amplitude; nonlinear terms would be needed for that. This is Euler buckling of an elastic filament.