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At a perfectly bonded elastic interface without an interfacial force, displacement and traction are continuous:Tangential derivatives of the common displacement trace are also continuous. Thus the jump of a piecewise constant displacement gradient tensor annihilates every tangential vector, and has the form $\alpha \otimes n$. Taking its symmetric part shows that the strain jump is $B\alpha$, whereThe adjoint formula follows from $\tau :B\alpha =(\tau n)\cdot \alpha$ for symmetric stress tensors. Tangential-tangential strain components are therefore common to all layers. Conversely, such gradient jumps can be realized by continuous piecewise affine displacement in the layered geometry.

Take positive elastic stiffness tensors with their usual minor and major symmetries, and a positive comparison elastic stiffness tensor $C^0$. A fictitious layer can have the same tangential displacement derivatives and the same traction as the physical layers. Hence ${\epsilon }^r={\epsilon }^0+B{\alpha }^r$. The common traction condition giveswhere $K^r=B^T{C}^{r}B$ is the positive acoustic tensor, with components $K_{ik}^r=C_{ijkl}^r{n}_j{n}_l$. Solving for ${\alpha }^r$ givesExpanding the two occurrences of $B$ gives exactly the symmetrization over $(ij)$ and $(kl)$ in the directional elastic strain Green operator specified in the paper.

For any two layers, the inverse identity for their acoustic tensors proves the required resolvent relation without interchanging noncommuting tensors:With $D_r=C^r-C^0$, this identity also gives ${\tilde{\Gamma }}^0{D}_r{\tilde{\Gamma }}^r={\tilde{\Gamma }}^0-{\tilde{\Gamma }}^r$. ThereforeThese finite-dimensional operators are inverses. Define $A_r=[I+{\tilde{\Gamma }}^0(C^r-C^0){]}^{-1}$. Then ${\epsilon }^r=A_r{\epsilon }^0$, and averaging strains and stresses givesEliminating the fictitious strain yieldsAll products have the displayed order; the elastic stiffness tensors need not commute.

It remains to justify the inverse and comparison independence. Consider two compatible equilibrated layer solutions with the same mean strain. Their difference has zero common tangential-tangential strain, so $\delta {\epsilon }^r=B{\beta }^r$, with ${\sum }_r{c}_r{\beta }^r=0$ because $B$ is injective and the mean difference is zero. Their traction difference is one common vector $g$. ThereforePositive elastic energy forces every $\delta {\epsilon }^r$ to vanish. In particular, if $⟨A⟩{\epsilon }^0=0$, all physical layer strains vanish. Their common tangential components then force ${\epsilon }^0=B{\beta }^0$, and their zero traction gives $K^0{\beta }^0=0$. Hence ${\epsilon }^0=0$, proving that $⟨A⟩$ is invertible. Finally, every positive choice of $C^0$ constructs the unique layer solution for the same $E$. Its averaged stress is unique, so the displayed effective stiffness is independent of the fictitious comparison material. This is the exact layered constitutive response, rather than a claim that piecewise constant fields meet arbitrary finite-body boundary data.

An elastic deformation is reversible: after removal of the deforming loads the material returns to its reference state, and its instantaneous stress is determined by its current deformation rather than a plastic or viscous history. In the infinitesimal approximation the displacement field $u_i$ gives the infinitesimal strain tensor $e_{ij}=(u_{i,j}+u_{j,i})/2$. The Cauchy stress tensor is defined by traction $t_i={\sigma }_{ij}{n}_j$ on a plane of unit normal $\mathbf{n}$; absence of body couples makes ${\sigma }_{ij}={\sigma }_{ji}$.

Let $\mathcal{E}(e)$ be the recoverable elastic energy per unit mass. The local mechanical power per unit volume isbecause the symmetric Cauchy stress tensor annihilates the antisymmetric part of the velocity gradient. Conservation of energy during the deformation gives $\rho \dot{\mathcal{E}}={\sigma }_{ij}{\dot{e}}_{ij}$. Since $\dot{\mathcal{E}}=(\partial \mathcal{E}/\partial e_{ij}){\dot{e}}_{ij}$ for arbitrary symmetric strain rates,Here the derivative is the gradient on the space of symmetric tensors, with the double-contraction inner product. This convention accounts for the two occurrences of each off-diagonal entry; treating six independent coordinates without their corresponding factors would give an erroneous factor of two in shear components.

Take a stress-free reference state of mass density ${\rho }_0$. The Taylor series of the elastic energy starts quadratically:Using $\rho ={\rho }_0+O(|e|)$ gives ${\sigma }_{ij}=c_{ijkl}{e}_{kl}+O(|e{|}^2)$, the relation of linear elasticity. If the reference state has initial stress, the first-order expansion also retains that constant stress. Symmetry of stress and strain gives the minor symmetries $c_{ijkl}=c_{jikl}=c_{ijlk}$, while equality of the mixed second derivatives of elastic energy gives the major symmetry $c_{ijkl}=c_{klij}$. Thus the elastic stiffness tensor is a symmetric map on a six-dimensional strain space: there are $6\cdot 7/2=21$ independent coefficients.

For twofold elastic symmetry, choose the symmetry axis as $x_3$. Rotation through $\pi$ is $Q=\mathrm{diag}(-1,-1,1)$, and transforms the infinitesimal strain tensor into $Qe{Q}^T$. Components $e_{11},e_{22},e_{33},e_{12}$ are even and $e_{23},e_{13}$ are odd. Invariance of elastic energy excludes every coupling between these parity classes. In the engineering strain ordering $(e_{11},e_{22},e_{33},2e_{23},2e_{13},2e_{12})$, with corresponding stress ordering $({\sigma }_{11},{\sigma }_{22},{\sigma }_{33},{\sigma }_{23},{\sigma }_{13},{\sigma }_{12})$, the symmetric stiffness matrix has the formIts symmetric even block has ten independent entries and its symmetric odd block has three. Hence a twofold axis leaves $13$ independent coefficients. No invariance under other rotation angles has been assumed.

Let $\Omega$ be the solid exterior to the cavity, and choose $\mathbf{n}$ outward from the solid, hence into the cavity on its boundary $S$. Subtraction of the incident solution from the total solution gives the scattered displacement field equationsThe traction-free cavity condition for the total field gives ${\sigma }_{ij}^s{n}_j=-{\sigma }_{ij}^0{n}_j$ on $S$. The scattered field has zero initial displacement and initial velocity before the pulse reaches the cavity and satisfies an outgoing radiation condition, excluding an additional incoming elastic wave. Equivalently one seeks the causal response to this prescribed boundary traction. Although the incident plane pulse extends across an infinite plane, the causal scattered field from the bounded cavity has finite propagation distance at each finite time.

To prove energy uniqueness for traction-driven elasticity, take the difference $\mathbf{w}$ of two solutions with the same initial conditions, body force, boundary traction and outgoing condition. It satisfiesFor a large enclosing ball $B_L$, multiply by ${\dot{w}}_i$ and integrate over $\Omega \cap B_L$. Integration by parts givesSymmetry of stress replaces the last derivative by ${\dot{e}}_{ij}(w)$, and major symmetry of the elastic stiffness tensor makes that integral the derivative of the strain energy. ThereforeThe cavity contribution is zero. On any fixed finite time interval choose $L$ beyond the causal support of the scattered difference; the outer contribution is then zero as well. This gives conservation of its nonnegative elastic energy. Positive mass density and positive strain energy imply that zero initial energy forces $\dot{\mathbf{w}}=0$ everywhere; the zero initial displacement field then gives $\mathbf{w}=0$. The same argument works for finite-energy outgoing solutions by taking the limit of enclosing surfaces and excluding incoming energy. Hence the causal scattered field ${\mathbf{u}}^s$ is unique. Specifying the incident field and cavity traction alone without the initial and outgoing conditions would not establish uniqueness.

Use the material-first nominal stress tensor convention $N_{\alpha i}$, whose reference-volume power is $N_{\alpha i}{\dot{A}}_{i\alpha }$. At a regular constraint point, $F_{,A}\ne 0$, all admissible deformation rates satisfy $F_{,A_{i\alpha }}{\dot{A}}_{i\alpha }=0$. Equality of mechanical power and the derivative of the strain energy density givesfor every vector in this eight-dimensional tangent hyperplane. Its orthogonal complement is the span of $F_{,A}$. Consequently the constraint reaction in hyperelastic stress isThe scalar Lagrange multiplier $q$ is not determined by the work identity: its contribution vanishes on every allowed rate. Regularity matters. For example, replacing a regular incompressibility constraint by $(\det A-1{)}^2=0$ would give zero gradient on the same admissible set and could not represent an arbitrary pressure reaction through that gradient.

For incompressibility use the regular function $F=J-1$, with $J=\det A$. Differentiating the alternating-tensor expression for $J$ givesThe first expression is the cofactor entry; contracting it with $A_{j\alpha }$ gives $J{\delta }_{ij}$, proving the inverse formula. On $J=1$ this yieldsWith this plus-sign convention $q$ is the negative of the usual pressure multiplier.

For an equibiaxially stretched sheet, incompressibility requires the principal stretches $(\lambda ,\lambda ,{\lambda }^{-2})$. The printed sheet expression's ${\lambda }^{-1/2}$ is inconsistent with this constraint and must be ${\lambda }^{-2}$; the balloon expression later on the same PDF page has the correct exponent. Let $W_j$ denote the derivative with respect to principal stretch ${\lambda }_j$, evaluated on this constrained path. With no applied thickness-direction traction,Isotropy gives $W_1=W_2$ at equal in-plane stretches. HenceThe equibiaxial nominal tension of an incompressible sheet is thereforeThe two equal in-plane nominal tensions do work $2N_{11}d\lambda$ per reference volume; the thickness-direction nominal stress does no work because it is zero. That is exactly the constrained energy differential, explaining energy conservation. As a check that the printed exponent is a genuine error, a neo-Hookean solid gives $N=\mu (\lambda -{\lambda }^{-5})$, zero at the undeformed state. The derivative with the erroneous exponent would instead give $3\mu /4$ at that state.

For the thin spherical balloon, use the leading reference shell volume $4\pi a_0^2{h}_0$. Its two tangential stretches are $\lambda =a/a_0$, while its thickness stretch is ${\lambda }^{-2}$ and its current thickness is $h=h_0{\lambda }^{-2}$. Define $\hat{W}(\lambda )=W(\lambda ,\lambda ,{\lambda }^{-2})$. Its total elastic energy at leading thin-shell order isDuring a quasistatic increment $da$, the enclosed volume changes by $d\mathcal{V}=4\pi a^{2}da$, and $d\lambda =da/a_0$. Equating energy change to work of the internal excess pressure givesThis is the inflation pressure of a thin incompressible elastic balloon. The approximation neglects through-thickness variation and curvature corrections beyond leading order, and requires $h/a\ll 1$. It also agrees with membrane force balance: $p=2h{\sigma }_t/a$, ${\sigma }_t=\lambda N$, and $N={\hat{W}}^{\prime }/2$. Here pressure is force per current area, while $N$ is force per reference area; confusing them would lose a factor of $\lambda$.

Retain $\beta =\sqrt{1-V^2/c^2}>0$, but normalize the complex potential as $F=\beta H$, so that $w=\mathrm{Re}F/\beta$. The Cauchy-Riemann equations give $w_x=\mathrm{Re}{F}^{\prime }/\beta$ and $w_y=-\mathrm{Im}{F}^{\prime }/\beta$. Since ${\partial }_{x_2}=\beta {\partial }_y$, the two stresses satisfyUsing the same unscaled $H$ from part (a) would instead put an additional $\beta$ on its derivative; the potential normalization matters.

Take the principal branch of $\sqrt{z}$, with its branch cut along the negative real axis and positive $\sqrt{x}$ for $x>0$. Write $Q=F^{\prime }$. For the antisymmetric crack-opening response, reflection gives $Q^{-}(s)=-\overline{Q^{+}(s)}$ on the faces. The prescribed equal face stress gives $\mathrm{Im}{Q}^{\pm }(s)=f(s)/\mu$. Multiplication by the square-root factor converts this face condition to an additive jump. In fact, with $G=\sqrt{z}Q$ and $r=\sqrt{-s}$,The Sokhotski–Plemelj theorem applied to this jump giveswhere $P$ is entire after imposing the usual finite-energy crack-tip regularity. This is the Hilbert problem for an antiplane crack, with the stress-potential phase chosen explicitly above.

For localized face loading with no additional remote loading, the far displacement is bounded and the weighted Cauchy transform decays. In particular, for compactly supported loading, $G\to 0$ at infinity and Liouville theorem sets $P=0$. The resulting Cauchy solution for a steadily moving semi-infinite antiplane crack isIntegrals and boundary values are understood for loadings with the corresponding convergence and regularity, or as limits of localized loading. The reflection used above is the pure antiplane opening response: reflection with a sign change of displacement leaves both prescribed face stresses unchanged, and the usual no-extra-load uniqueness condition eliminates an unforced symmetric addition.

One can check the face stresses directly. Put $I_{\mathrm{pv}}(s)=\mathrm{PV}{\int }_{-\infty }^0\sqrt{-t}f(t)/(s-t)dt$. The Cauchy boundary limits giveBoth have $-\mu \mathrm{Im}{Q}^{\pm }=-f(s)$, verifying the prescribed traction and its sign on both faces. The opposite real parts also give the required reflection. Ahead of the tip, the integral is real and $Q$ is purely imaginary, so the bonded line has zero tangential displacement gradient and the two halves match.

For $x=x_1-Vt>0$, take $z=x$ in the boxed analytic solution. The ahead-of-tip stress isNo $\beta$ appears: at fixed co-moving position and the specified co-moving face loading, this stress is independent of the speed. The displacement, however, still scales with $1/\beta$, so speed independence does not apply to the whole elastic field. If ${\int }_{-\infty }^0|f(s)|/\sqrt{-s}ds$ is finite, the near-tip stress is inverse-square-root with coefficient ${\pi }^{-1}\int f(s)/\sqrt{-s}ds$.

The far-field condition is a genuine uniqueness requirement. A homogeneous stress-intensity field of a semi-infinite antiplane crack has $Q_h=iC/\sqrt{z}$, for real $C$. It adds no face traction but changes the stress ahead by $-\mu C/\sqrt{x}$. Although its stress tends to zero at infinity, its primitive is $2iC\sqrt{z}$, which gives unbounded remote displacement and infinite total elastic energy. Thus only asking for stress to vanish at infinity would leave the homogeneous constant undetermined; excluding that added remote stress-intensity loading selects the displayed face-loaded solution. The ordinary inverse-square-root tip singularity itself has finite energy in a bounded neighbourhood and is not being excluded.

Use orthonormal cylindrical bases in the reference and deformed configurations. The radial coordinate remains $\rho$, the angular coordinate becomes $\theta =\Theta +\alpha {\xi }_3$, and the axial coordinate is unchanged. Removing the rotation of the cylindrical basis, the deformation gradient isIt is a local simple shear with unit determinant. The squared principal stretches are the eigenvalues ofOne principal stretch is one. The remaining squared principal stretches have product one and sum $2+s^2$. Consequently they can be written ${\lambda }^2$ and ${\lambda }^{-2}$, whereFor negative $s$ their ordering reverses, which is immaterial to the strain energy density of an isotropic material.

Let the reference length be $L$. By material isotropy, the total elastic energy and end rotation areFor a quasistatic change, incompressibility makes the constraint reaction do no work. The velocity on an end section is its rigid rotational velocity; its traction power is the resultant torque times the angular velocity. Thus conservation of energy, or equivalently the torque and axial force of a twisted incompressible cylinder work identity, gives $Md{\Theta }_{\mathrm{end}}=dE$ andThe radius $a$ is fixed in this derivative. This argument computes the required end torque along the prescribed deformation family; proving that a suitable end-only traction distribution sustains that family is unnecessary here.

For a neo-Hookean solid, the identity $\lambda -{\lambda }^{-1}=s$ gives ${\lambda }^2+{\lambda }^{-2}-2=s^2$. ThereforeThis is the signed torque in the twist direction; its magnitude is $\pi \mu a^4|\alpha |/2$. For the positive twist convention the signed value equals the magnitude.

Put $Z=\rho \beta =\sqrt{\rho \mu }>0$, the seismic impedance. In a uniform medium, define $w_{+}=v-\sigma /Z$ and $w_{-}=v+\sigma /Z$. The field equations giveThus $w_{+}=2F(x-\beta t)$ and $w_{-}=2G(x+\beta t)$, with arbitrary independent profiles $F,G$. Reconstruction givesThe two characteristic components therefore have $v_{+}=F$, ${\sigma }_{+}=-Z{v}_{+}$ and $v_{-}=G$, ${\sigma }_{-}=Z{v}_{-}$.

The kinetic plus elastic energy mass density is $e=\rho v^2/2+{\sigma }^2/(2\mu )$. Multiplying the equations by $v$ and $\sigma /\mu$ yields $e_t=(v\sigma {)}_x$. Hence the elastic-wave energy flux is $S=-v\sigma$. For a pure forward wave $S_{+}=Z{v}_{+}^2\ge 0$; for a pure backward wave $S_{-}=-Z{v}_{-}^2\le 0$. In the mixture the cross terms cancel, giving $S=Z(v_{+}^2-v_{-}^2)$. Strict sign holds for nonzero components.

If two harmonic solutions have the same body force and prescribed boundary traction, their difference has no body force and is traction-free. A nonzero difference would therefore be a free eigenfunction at that frequency. For any frequency outside the countable exceptional set found above, the difference must vanish, so the forced boundary-value solution, when it exists, is unique. At an eigenfrequency, a free eigenfunction can be added to any existing solution, and a resonant forcing may fail to admit a steady harmonic response. This distinguishes uniqueness from existence.

For the full time-dependent problem, let $\mathbf{w}$ be the difference between two fields with identical forces, tractions and initial displacement field and velocity. It obeys $\rho {\mathbf{w}}_{tt}=\nabla \cdot \mathbf{\sigma }(\mathbf{w})$, zero boundary traction, and zero initial data. Its energy uniqueness for traction-driven elasticity follows by direct integration:Multiply the equation by ${\mathbf{w}}_t$ and integrate by parts. Major symmetry turns the strain term into the derivative of elastic energy, giving $E^{\prime }(t)={\int }_{\partial \mathcal{D}}{\mathbf{w}}_t\cdot \mathbf{t}(\mathbf{w})dS=0$. Stable elasticity makes $E\ge 0$, and $E(0)=0$ gives ${\mathbf{w}}_t=0$. Thus $\mathbf{w}$ equals its initial value zero. Even rigid zero-strain motions cannot survive these initial conditions.

The PDF correctly specifies zero initial displacement field and zero initial velocity; the TeX duplicates displacement field in the second condition. The complete causal displacement field is unique, including when its Fourier representation contains resonant modal contributions fixed by the initial data.

Put $B=E{I}_2$, the filament bending modulus in the easier bending direction; $I_2$ is the relevant second moment of area. Take filament tension as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:The small transverse displacement induces a transverse tension force density $(T{X}_z{)}_z$, while bending produces $-B{X}_{zzzz}$. Their balance givesThe same result follows from elastic energy and gravitational energy. To quadratic order, local inextensibility shortens the height of the material point labelled $z$ by $\frac{1}{2}{\int }_0^z{X}_z(s{)}^{2}ds$. Integrating its weight and interchanging the integrals gives the second-order potential energyTwo integration by parts steps in its first variation giveThe clamped boundary conditions fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions areSince the filament tension vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the ordinary differential equation once: $B{X}_{zzz}-T{X}_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows thatThe other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)={\int }_0^{z}u(s)ds$ to impose the base position.

Set $x=L-z$, $\alpha =\lambda g/B$, and $\eta =(2/3)\sqrt{\alpha }{x}^{3/2}$. Then $d\eta /dx=3\eta /(2x)$ and $d^2\eta /d{x}^2=3\eta /(4x^2)$, so $u_{xx}+\alpha xu=0$ becomesSubstituting $u={\eta }^{1/3}F(\eta )$ givesThis is the Bessel differential equation of order $1/3$, whose two independent Bessel functions of the first kind giveThe tip is $x=\eta =0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:Since ${\eta }^2\propto x^3$ and ${\eta }^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requiresLet $j_{-1/3,1}$ be the first positive zero. The Bessel threshold for self-buckling of a vertical rod isTo identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac{1}{2}[B\int u_z^{2}dz-\lambda g\int (L-z)u^{2}dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum Rayleigh quotient $\int u_z^{2}dz/\int (L-z)u^{2}dz$; variation supplies the moment-free condition at the tip and the same Sturm-Liouville problem. The first zero gives the lowest eigenvalue, so the straight state is stable below this load, neutral at it, and unstable above it, with the easier $I_2$ direction buckling first.

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.