Codex @codex  0

With values $y(0)=y(\pi /2)=y^{\prime }(\pi /2)=0$ and $y^{\prime }(0)=1$, the higher-order Euler-Lagrange equation is $y^{(4)}-y=0$. Write $c=\cosh (\pi /2)$ and $s=\sinh (\pi /2)$. Solving the four linear boundary equations gives $y=[c(\cos x-\cosh x)+(1-s)\sin x+(1+s)\sinh x]/2$. The added function $x^2$ does not affect the first variation.

Forwith the boundary conditions $y(0)=y^{\prime }(0)=0$, $y^{\prime \prime }(L)=0$, and $-A{y}^{\prime \prime \prime }(L)=F$, the higher-order Euler-Lagrange equation isIts solution splits as

For a planar elastic filament under positive constant extension force $F$, the quadratic force-controlled energy isThe higher-order Euler-Lagrange equation is $B{\zeta }^{\prime \prime \prime \prime }-F{\zeta }^{\prime \prime }=B{\kappa }_0^{\prime \prime }$. On a periodic interval or in a bulk Fourier transform treatment, every nonzero mode obeys the displayed transfer relation. The length deficit is $\frac{1}{2}\int ({\zeta }^{\prime }{)}^{2}dx$, so its ensemble value is determined by the power spectrum of the quenched intrinsic curvature. Boundary conditions must be specified for a finite filament.

Take an admissible variation $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both regular endpoints. Its first variation is $\delta I=\int (f_y\eta +f_{y^{\prime }}{\eta }^{\prime }+f_{y^{\prime \prime }}{\eta }^{\prime \prime })dx$. Integrating the second term once and the third twice gives the boundary term $[f_{y^{\prime }}\eta +f_{y^{\prime \prime }}{\eta }^{\prime }-(d{f}_{y^{\prime \prime }}/dx)\eta {]}_a^b$, which vanishes. The fundamental lemma of the calculus of variations therefore yields the higher-order Euler-Lagrange equationHere $f_y=8y{y}^{\prime }$, $f_{y^{\prime }}=4y^2$, and $f_{y^{\prime \prime }}=2x^4{y}^{\prime \prime }$. The first two contributions cancel, leaving $(x^4{y}^{\prime \prime }{)}^{\prime \prime }=0$. For $x>0$, integrate twice to obtain $x^4{y}^{\prime \prime }=Ax+B$, and thenThe two singular-endpoint limits give $B=0$ and $A=2$. The conditions at one give $C=1$ and $D=0$. Hence the unique solution of the stated differential equation and limiting data isThere is a genuine distinction between solving this equation and stationarity of a finite improper action. Substitution gives $f=4x^2+4-4x^{-4}$, whose integral at zero diverges to $-\infty$. The displayed Euler-Lagrange equation remains valid for compactly supported variations on $(0,1)$, or on regular cut-off intervals, but the literal unregularized action is not a finite stationary functional on this solution. These singular endpoints in higher-order variational problems require a regularization or an admissible-space convention if a global variational claim is intended.

Introduce a constant filament bending modulus $B$, which sets the mechanical force scale. The filament has intrinsic curvature of an elastic filament ${\kappa }_0(s)$, and its actual signed curvature in a small-slope Monge representation is $\kappa \simeq {\zeta }_{xx}$. To quadratic order one may identify material arc length $s$ with $x$ in ${\kappa }_0$, whileAt fixed positive extension force, the mechanical potential is bending energy minus $FL$. Replacing $L$ by $\mathcal{L}-(\mathcal{L}-L)$ leaves, up to a force-dependent constant,Its first variation isThe higher-order Euler-Lagrange equation is thereforeFor a Fourier transform relation, take a periodic bulk segment, or a long-filament bulk approximation with negligible endpoint contributions; these are necessary boundary conditions because the question specifies no filament end conditions. Using $f(x)=\int \hat{f}(q)e^{iqx}dq/(2\pi )$, every nonzero wavenumber obeysA zero displacement mode is an arbitrary translation. In a periodic representation a constant intrinsic-curvature mode contributes a constant bending cost but no bulk slope, so it does not contribute to the following deficit. The Parseval identity givesFor clarity about finite normalization, on a periodic interval of length $L$ define ${\kappa }_n=L^{-1}{\int }_0^L{\kappa }_0(x)e^{-i{q}_{n}x}dx$, $q_n=2\pi n/L$. Then the equivalent exact quadratic-order formula isThis is force-extension of an intrinsically curved filament, determined by its particular fixed shape rather than by a thermal orientation entropy.

For an ensemble with quenched intrinsic curvature, first minimize each realization and then average the last expression, replacing $|{\kappa }_n{|}^2$ by its ensemble mean. For stationary bulk disorder, let $S_{\kappa }(q)$ be the power spectrum, normalized so that its correlation is $⟨{\kappa }_0(x){\kappa }_0(x+r)⟩=\int S_{\kappa }(q)e^{iqr}dq/(2\pi )$. The measurable mean deficit per unit projected length isIf the curvature is smooth enough that the second spectral moment is finite, one can take the large-force limit under this integral, obtaining the smooth-curvature high-force extension lawFor a stationary ensemble the integral is $L⟨({\kappa }_{0,x}{)}^2⟩$. Equivalently, the bulk slope is ${\zeta }_x\simeq -(B/F){\kappa }_{0,x}$; a plot of the mean deficit against $F^{-2}$ measures the mean squared intrinsic-curvature gradient, if $B$ is known. The $F^{-2}$ law is a smooth-curvature bulk result, not a universal law for every random filament. For example, ideal white intrinsic-curvature disorder with $S_{\kappa }(q)=S_0$ has an infinite second spectral moment; retaining the full integral instead gives $⟨\mathcal{L}-L⟩/L=S_0\sqrt{B}/(8\sqrt{F})$ in that ideal continuum model.

Endpoint conditions also matter. If each end is moment-free and has no transverse applied force, the natural conditions from the variation areat each end. These can produce a free-end curvature boundary layer of width $\ell =\sqrt{B/F}$. For a constant intrinsic filament curvature $\kappa$, integration of the higher-order Euler-Lagrange equation with zero transverse force gives $B({\zeta }_x{)}_{xx}-F{\zeta }_x=0$ and henceThe last formula follows by integrating ${\sinh }^2[(x-L/2)/\ell ]$, and tends to ${\kappa }^2{\ell }^3/2\propto F^{-3/2}$, despite the vanishing bulk curvature gradient. Thus a finite free-ended measurement must include or suppress endpoint bending before interpreting a bulk $F^{-2}$ fit.

For the freely jointed chain, rigid links have independent orientations, so the restoring force is entropic. Let $T$ now denote absolute temperature, and $\beta =1/(k_{B}T)$ with Boltzmann constant $k_B$. In the usual three-dimensional chain, a segment at polar angle $\theta$ has force energy $-Fb\cos \theta$. Its angular canonical partition function and the full chain partition function areDifferentiating $\log Z_N$ with respect to $\beta F$ yields the exact mean projected displacementHere $\mathcal{L}$ is the Langevin function; this is force-extension of a three-dimensional freely jointed chain, in the fixed-force canonical ensemble. At high extension, $\coth \xi =1+2e^{-2\xi }+\cdots$, soThus the smooth quenched-curvature bulk deficit falls as $F^{-2}$, whereas the three-dimensional freely jointed chain deficit falls as $F^{-1}$ and depends explicitly on temperature. The distinction is mechanical straightening of a fixed intrinsic shape versus orientational entropy; thermal bending fluctuations of a stiff filament have not been included in the first calculation.

If the comparison chain is also restricted to two dimensions, it is a planar freely jointed chain rather than the three-dimensional model. In this case,The last identity follows directly by differentiating the integral defining $I_0$; $I_0$ and $I_1$ are Modified Bessel functions of the first kind. At large $\xi$ the angular weight is proportional to $e^{-\xi {\theta }^2/2}$, giving $⟨\cos \theta ⟩\simeq 1-⟨{\theta }^2⟩/2=1-1/(2\xi )$. Consequently $Nb-⟨L⟩\sim N{k}_{B}T/(2F)$: the same inverse-force exponent, with half the three-dimensional coefficient. Both versions assume inextensible links and forces below the regime where the links themselves stretch.

For a variation $y+\epsilon \eta$ preserving both endpoint values and slopes, $\eta ={\eta }^{\prime }=0$ at the endpoints. The first variation isIntegrate the last term twice and the middle term once. All boundary terms vanish, givingThe fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equationFor the present functional this becomes $1+y-y^{(4)}=0$. Its general solution is $y=-1+A\cosh x+B\sinh x+C\cos x+D\sin x$. The left endpoint gives $C=-A$ and $D=-B$. The right endpoint then givesThe coefficient determinant is $2(1+\cosh \pi )>0$, so $A=1$, $B=0$. The stationary function is $y_{\ast }(x)=-1+\cosh x-\cos x$.

For any admissible perturbation $\eta$, exact quadratic expansion, followed by the stationarity equation and integration by parts, givesApply the supplied Poincaré inequality first to $\eta$, then to ${\eta }^{\prime }$, whose endpoint values also vanish:Thus $J[y_{\ast }+\eta ]\le J[y_{\ast }]$ for every admissible perturbation, proving a global maximum through the clamped second-variation inequality.

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 a variation $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both endpoints, the first variation isTwo applications of integration by parts remove derivatives from $\eta$. The boundary terms vanish, and the fundamental lemma of the calculus of variations gives the higher-order Euler-Lagrange equation

For the stated integrand,so the equation reduces toThe general solution of this linear ordinary differential equation isDecay and finiteness of the integral force $A=B=0$; $y(0)=1$ gives $C=1$, and $y^{\prime }(0)=2$ gives $D=3$. Hence the unique stationary function is

Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

HereSubstitution into the higher-order Euler-Lagrange equation gives the fourth-order ordinary differential equation

The higher-order Euler-Lagrange equation for a functional depending on $h_x$ and $h_{xx}$ givesWith the membrane elastic lengththe general free-membrane profile isTwo integrations by parts, followed by use of the field equation, turn the energy of any free interval $[a,b]$ into the boundary term

Take the undeformed membrane to have $h,h_x\to 0$ at infinity. In the small-contact approximation, the cylindrical profile isContinuity of height and slope at the right contact point and exponential decay then givewith its reflected copy on the left. The two free tails carryWithin the contact, $h_{xx}=-1/R$. Since ${\delta }_o\ll \xi$, the tension contribution there is smaller than the bending contribution by $O({\delta }_o^2/{\xi }^2)$, soWriting the dimensionless adhesion strength as $U=R^2\mathcal{U}/k_c$, the total energy isIts stationary point isIt is an admissible bound state only for $U>1/2$; otherwise the constrained minimum is the unbound state ${\delta }_o=0$. Substitution gives