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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.

The compressive tip force is limited by Euler buckling of an elastic filament. With unsupported length $\ell$ and filament bending modulus $B=EI$,For a hollow microtubule, use $I=\pi (r_o^4-r_i^4)/4$ rather than the solid-cylinder value. For a clamped seed and a moment-free tip free to slide along the wall, linearized angle balance is $B{\theta }^{\prime \prime }+F\theta =0$, with $\theta (0)=0$ and ${\theta }^{\prime }(\ell )=0$. Its first nonzero solution has $\sqrt{F/B}\ell =\pi /2$, hence $F_c={\pi }^{2}B/(4{\ell }^2)$. A different tip constraint changes this coefficient.

Increasing the unsupported length decreases that threshold. For growth against a wall at fixed projected gap $d$, polymerization-driven cantilever postbuckling gives a more direct quantitative statement. The inextensible elastica obeys $B{\theta }^{\prime \prime }+F\sin \theta =0$. Multiplying by ${\theta }^{\prime }$ and using the zero tip moment givesPut $k=\sin ({\theta }_t/2)$ and substitute $\sin (\theta /2)=k\sin \varphi$. Integrating arc length and its horizontal projection giveswhere the complete elliptic integrals are $K(k)={\int }_0^{\pi /2}(1-k^2{\sin }^2\varphi {)}^{-1/2}d\varphi$ and $E(k)={\int }_0^{\pi /2}(1-k^2{\sin }^2\varphi {)}^{1/2}d\varphi$. Using $K=(\pi /2)(1+k^2/4+\cdots )$ and $E=(\pi /2)(1-k^2/4+\cdots )$ gives $\ell /d=1+k^2+O(k^4)$, and thereforeThus the initial postbuckling force decreases as more contour is inserted. This result describes the weakly buckled branch before overhang, further wall contact, or altered tip constraints invalidate the model. It is a mechanical force law; a chemical force limit from polymerization can impose a smaller force.

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.

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.