Reduction of the heavy symmetric top system: conserved momenta, one-dimensional $\theta$ equation and conserved reduced energy

By the first and third equations of [L1], the functions ${\mathrm{\partial }}_{{\phi }^{\mathrm{\prime }}}L=A{\phi }^{\mathrm{\prime }}{\sin }^2\theta +C\cos \theta ({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta )$ and ${\mathrm{\partial }}_{{\psi }^{\mathrm{\prime }}}L=C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta )$ are constant on $I$; the second of these is $b$, and the first equals $A{\phi }^{\mathrm{\prime }}{\sin }^2\theta +b\cos \theta =a$. Hence $a$ and $b$ are constants by A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$; consequently two such functions with the same derivative differ by a constant applied to their continuous differentiable restrictions to $I$, which is clause 1.

Define $E(t):={\phi }^{\mathrm{\prime }}{\mathrm{\partial }}_{{\phi }^{\mathrm{\prime }}}L+{\theta }^{\mathrm{\prime }}{\mathrm{\partial }}_{{\theta }^{\mathrm{\prime }}}L+{\psi }^{\mathrm{\prime }}{\mathrm{\partial }}_{{\psi }^{\mathrm{\prime }}}L-L$ along the trajectory. By [L4] the derivative of $t\mapsto L(q(t),q^{\mathrm{\prime }}(t))$ is ${\sum }_i{\mathrm{\partial }}_{q_i}L{q}_i^{\mathrm{\prime }}+{\sum }_i{\mathrm{\partial }}_{q_i^{\mathrm{\prime }}}L{q}_i^{\mathrm{\prime }\mathrm{\prime }}$, and substituting the three system equations $\frac{d}{dt}{\mathrm{\partial }}_{q_i^{\mathrm{\prime }}}L={\mathrm{\partial }}_{q_i}L$ of [L1] turns this into $\frac{d}{dt}\left({\sum }_i{q}_i^{\mathrm{\prime }}{\mathrm{\partial }}_{q_i^{\mathrm{\prime }}}L\right)$; hence $E^{\mathrm{\prime }}=0$. Since $E$ is continuous on the interval $I$ and differentiable at its interior points, A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$; consequently two such functions with the same derivative differ by a constant makes it constant on $I$.

Compute $U^{\mathrm{\prime }}$: writing $N=(a-b\cos \theta {)}^2$ and $D=2A{\sin }^2\theta$, the quotient rule [L3] and [L2] give $\frac{d}{d\theta }\frac{N}{D}=\frac{N^{\mathrm{\prime }}D-N{D}^{\mathrm{\prime }}}{D^2}=\frac{4Ab{\sin }^3\theta (a-b\cos \theta )-4A\sin \theta \cos \theta (a-b\cos \theta {)}^2}{4A^2{\sin }^4\theta }$, which after factoring $(a-b\cos \theta )\mathrm{/}(A{\sin }^3\theta )$ is $\frac{a-b\cos \theta }{A{\sin }^3\theta }(b{\sin }^2\theta -\cos \theta (a-b\cos \theta ))$; and the remaining terms contribute $-mgd\sin \theta$, so $U^{\mathrm{\prime }}(\theta )=\frac{a-b\cos \theta }{A{\sin }^3\theta }(b{\sin }^2\theta -\cos \theta (a-b\cos \theta ))-mgd\sin \theta$.

Solving the two defining identities: $b=C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta )$ gives ${\psi }^{\mathrm{\prime }}=b\mathrm{/}C-{\phi }^{\mathrm{\prime }}\cos \theta$, and $a=A{\phi }^{\mathrm{\prime }}{\sin }^2\theta +b\cos \theta$ gives ${\phi }^{\mathrm{\prime }}=(a-b\cos \theta )\mathrm{/}(A{\sin }^2\theta )$ because $\sin \theta >0$ by [L2]; substituting the second into the first gives the displayed formula for ${\psi }^{\mathrm{\prime }}$. This is clause 2.

Evaluating $E$ with the partial derivatives of [L1] and $L$ of the definition gives $E=\frac{1}{2}A{\theta }^{\mathrm{\prime }2}+\frac{1}{2}A{\phi }^{\mathrm{\prime }2}{\sin }^2\theta +\frac{1}{2}C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta {)}^2+mgd\cos \theta ,$ the cross terms cancelling after collecting ${\phi }^{\mathrm{\prime }}(A{\phi }^{\mathrm{\prime }}{\sin }^2\theta +C\cos \theta ({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta ))+{\psi }^{\mathrm{\prime }}C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta )-\frac{1}{2}C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta {)}^2=\frac{1}{2}C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta {)}^2$ and the analogous axial term; substituting ${\phi }^{\mathrm{\prime }}=(a-b\cos \theta )\mathrm{/}(A{\sin }^2\theta )$ and ${\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta =b\mathrm{/}C$ from steps 1.1 and 2.1 gives the displayed formula for $E$. Since $E$ is constant by step 1.2, this is clause 3.

Substituting $a-b\cos \theta =A{\phi }^{\mathrm{\prime }}{\sin }^2\theta$ from step 2.1 into the first term of step 1.3 gives $U^{\mathrm{\prime }}(\theta )={\phi }^{\mathrm{\prime }}\sin \theta (b-A{\phi }^{\mathrm{\prime }}\cos \theta )-mgd\sin \theta$, hence $-U^{\mathrm{\prime }}(\theta )=A{\phi }^{\mathrm{\prime }2}\sin \theta \cos \theta -b{\phi }^{\mathrm{\prime }}\sin \theta +mgd\sin \theta$; by step 1.1 and the identity ${\mathrm{\partial }}_{\theta }L=A{\phi }^{\mathrm{\prime }2}\sin \theta \cos \theta -C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta ){\phi }^{\mathrm{\prime }}\sin \theta +mgd\sin \theta$ of [L1] with $C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta )=b$, the right-hand side equals ${\mathrm{\partial }}_{\theta }L$. Since the second system equation reads $\frac{d}{dt}(A{\theta }^{\mathrm{\prime }})={\mathrm{\partial }}_{\theta }L$ and $\frac{d}{dt}(A{\theta }^{\mathrm{\prime }})=A{\theta }^{\mathrm{\prime }\mathrm{\prime }}$, we obtain $A{\theta }^{\mathrm{\prime }\mathrm{\prime }}=-U^{\mathrm{\prime }}(\theta )$ on $I$, which is clause 4.