Past exam of the elasticity and plasticity course of the University of Oxford / 2011 / 3 / c

Set $\ell =\sqrt{EI/F}$. The Euler elastica equation becomes ${\theta }^{\prime \prime }={\ell }^{-2}\sin \theta$. Multiplying by ${\theta }^{\prime }$ gives the first integralStraight asymptotic ends require ${\theta }^{\prime }\to 0$ and a horizontal tangent, so $C=F$. This is the separatrix of a simple pendulum with its angle shifted by $\pi$. A nonconstant solution moves from $\theta =0$ to $2\pi$, with the same physical tangent at both ends, taking infinite arc length to approach either limit. One explicit representative isUsing $\cos \theta =1-2{\mathrm{sech}}^2(s/\ell )$ and $\sin \theta =-2\mathrm{sech}(s/\ell )\tanh (s/\ell )$ and integrating $r^{\prime }=(\cos \theta ,\sin \theta )$ gives the tensile elastica loopThis solution is nontrivial, tends to horizontal straight tails, and has finite excess bending energy. It is also necessarily self-intersecting. Indeed $y$ is even and $x$ is odd. For $z>0$, let $f(z)=z-2\tanh z$. Near zero $f(z)<0$, while $f(2)>0$, so there is a root $z_{\ast }\in (0,2)$. At the two distinct parameters $s=\pm \ell z_{\ast }$, both points have $x=0$ and the same $y>0$. Numerically $z_{\ast }\simeq 1.91501$.

Any nonconstant smooth solution with the same straight asymptotic conditions has this first integral. Between consecutive horizontal orientations, separating variables gives the displayed solution up to translation, reflection, and reversal. It cannot pass through a straight equilibrium at a finite parameter, by uniqueness of the ordinary differential equation. Thus all such nontrivial branches have the same loop crossing. This is an ideal zero-thickness rod solution; a physical rod with self-contact excluded cannot realize the crossing without adding contact forces.