Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 66 / 1

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.

For each of the four standard pairs, integration by parts givesConsequently the eigenvalues are nonnegative. For a positive eigenvalue ${\mu }_n=A{k}_n^4$, the eigenvalue equation is $W_n^{\prime \prime \prime \prime }=k_n^4{W}_n$. Its four characteristic roots are $\pm k_n,\pm i{k}_n$, soHere $C_j$ are coefficients, avoiding a collision with the filament bending modulus $A$. The regular finite-interval self-adjoint operator has compact resolvent; applying the spectral theorem for compact self-adjoint operators to a shifted inverse supplies a complete orthonormal basis of eigenfunctions.

For clamped boundary conditions at zero, a clamped--clamped bending mode can be writtenAt $L$, writing $\beta =kL$, the two remaining boundary conditions areThe determinant is $2(1-\cos \beta \cosh \beta )$. Hence the positive wavenumbers obeyEquivalently, intersect $\cos \beta$ with $\mathrm{sech}\beta$. Numerical root bracketing givesThe entire sequence has the useful large-$n$ descriptionIndeed, put $\beta =(n+\frac{1}{2})\pi +\delta$ in $\cos \beta =\mathrm{sech}\beta$ and use $\cos \beta =(-1{)}^{n+1}\delta +O({\delta }^3)$ and $\mathrm{sech}\beta =2e^{-\beta }+O(e^{-3\beta })$.

The apparent root $\beta =0$ in the determinant equation is spurious for the clamped problem: at zero eigenvalue, $W$ is a cubic polynomial, and its four clamped conditions force $W=0$. For other endpoint choices, zero-energy filament modes must be treated separately from the trigonometric formula. The free-free kernel consists of affine functions, the torqued-torqued kernel consists of constants, and the hinged-hinged kernel is trivial. Including those kernels is necessary for a complete eigenfunction expansion.

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 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.