Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 9D / Solution

The quadratic first integrals of the torque-free Euler equations for a torque-free rigid body areDifferentiation and substitution of all three Euler equations make each derivative vanish: the coefficients respectively sum to $(I_2-I_3)+(I_3-I_1)+(I_1-I_2)=0$ and $I_1(I_2-I_3)+I_2(I_3-I_1)+I_3(I_1-I_2)=0$. At either endpoint these constants are $2E=I_2{\Omega }^2$, $L^2=I_2^2{\Omega }^2$, proving $L^2=2E{I}_2$.

Eliminating the other squared components gives, with $s={\omega }_2/\Omega$,Hence $|s|\le 1$ and the second Euler equation gives $ds/d\tau =\pm (1-s^2)$, where $\tau =\Omega t\sqrt{(I_2-I_1)(I_3-I_2)/(I_1{I}_3)}$ increases with $t$. The specified orbit goes from $s=1$ to $s=-1$, so its sign must be negative. The sign in the original PDF's reduced equation is therefore inconsistent with its endpoint conditions. The corrected heteroclinic satisfiesIntegrating by separation gives this family for $|s|<1$; the integration constant is a time translation. The printed positive-sign equation instead has the increasing solutions $s=\tanh (\tau -{\tau }_0)$ and cannot meet the stated endpoints. For completeness its other real solutions are $s=\pm 1$ and the branches $s=\coth (\tau -{\tau }_0)$ on intervals avoiding their pole. These latter branches are excluded by the conserved-energy constraint $|s|\le 1$. The scalar heteroclinic is unique up to translation. For the full angular-velocity vector there are two sign-related branches: simultaneously reversing ${\omega }_1$ and ${\omega }_3$ preserves all three Euler equations and both endpoints. Full-vector uniqueness therefore also requires choosing one of those branches.