Euler's equations $I{\omega }^{\mathrm{\prime }}+\omega \times I\omega =R^{\mathsf{T}}\tau$ under an adopted torque balance, and the torque-free invariants and principal-axis equations

By [L1] the angular velocity $\omega$ is differentiable, hence so is $l=I\omega$; the product $Rl$ satisfies $(Rl{)}^{\mathrm{\prime }}=R^{\mathrm{\prime }}l+R{l}^{\mathrm{\prime }}$ by [L3], and $R^{\mathrm{\prime }}l=R[\omega {]}_{\times }l=R(\omega \times l)$ by [L1] and [L3].

By the balance hypothesis (B) and step 1.1, $\tau =R(\omega \times l+l^{\mathrm{\prime }})$. Multiplying by $R^{\mathsf{T}}$ and using [L2] and linearity of matrix-vector multiplication gives $R^{\mathsf{T}}\tau =\omega \times l+l^{\mathrm{\prime }}$. Since $l=I\omega$ with $I$ a constant matrix, $l^{\mathrm{\prime }}=I{\omega }^{\mathrm{\prime }}$ by the finite-sum and scalar rules of [L7], and $\omega \times l=\omega \times I\omega$; therefore $I{\omega }^{\mathrm{\prime }}+\omega \times I\omega =R^{\mathsf{T}}\tau$, which is clause 1.

Assume $\tau =0$. Then step 2.1 gives $I{\omega }^{\mathrm{\prime }}=-\omega \times I\omega$; multiplying by $I^{-1}$ from [L4] gives ${\omega }^{\mathrm{\prime }}=-I^{-1}(\omega \times I\omega )$, hence $l^{\mathrm{\prime }}=I{\omega }^{\mathrm{\prime }}=-(\omega \times I\omega )=(I\omega )\times \omega =l\times (I^{-1}l)$, using antisymmetry of the cross product [L5] and $I^{-1}l=\omega$ [L4]. This is the first part of clause 2.

The two functions of clause 2 are constant: the product rule and symmetry of $I^{-1}$ in [L7] give $\frac{d}{dt}⟨l,l⟩=2⟨l^{\mathrm{\prime }},l⟩=2⟨l\times (I^{-1}l),l⟩$ and $\frac{d}{dt}\frac{1}{2}⟨l,I^{-1}l⟩=⟨l^{\mathrm{\prime }},I^{-1}l⟩=⟨l\times (I^{-1}l),I^{-1}l⟩$. Both inner products vanish because a cross product is orthogonal to each of its factors [L5]. Also $\frac{d}{dt}\frac{1}{2}⟨l,I^{-1}l⟩=\frac{d}{dt}\frac{1}{2}{\omega }^{\mathsf{T}}I\omega$ because $I^{-1}l=\omega$ and $I$ is symmetric [L4], [L6]. These functions are continuous on the solution interval $J$ and differentiable at every interior point, so their zero derivatives make them constant 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.

For the eigenbasis, $l={\sum }_j{l}_j{e}_j$ gives $I^{-1}l={\sum }_j(l_j\mathrm{/}{I}_j)e_j$ by [L4]; by orthonormality $⟨l,l⟩={\sum }_j{l}_j^2$ and $\frac{1}{2}⟨l,I^{-1}l⟩={\sum }_j{l}_j^2\mathrm{/}(2I_j)$, and these are constant by step 4.1.

Let $Q$ have columns $(e_1,e_2,e_3)$ in the standard basis. The positive orientation and orthonormality give $Q\in SO(3)$; [L2] gives $Q(u\times v)=(Qu)\times (Qv)$ from the skew/cross correspondence. Thus the standard cross-product coordinate formula applies in this principal basis. Coordinate-wise, $l^{\mathrm{\prime }}=l\times (I^{-1}l)$ reads $l_1^{\mathrm{\prime }}=l_2(l_3\mathrm{/}{I}_3)-l_3(l_2\mathrm{/}{I}_2)=(1\mathrm{/}{I}_3-1\mathrm{/}{I}_2)l_2{l}_3$, $l_2^{\mathrm{\prime }}=l_3(l_1\mathrm{/}{I}_1)-l_1(l_3\mathrm{/}{I}_3)=(1\mathrm{/}{I}_1-1\mathrm{/}{I}_3)l_1{l}_3$ and $l_3^{\mathrm{\prime }}=l_1(l_2\mathrm{/}{I}_2)-l_2(l_1\mathrm{/}{I}_1)=(1\mathrm{/}{I}_2-1\mathrm{/}{I}_1)l_1{l}_2$, using the coordinate formula for the cross product [L5] and step 5.1. An arbitrary orthonormal eigenbasis can be made positively oriented by negating one eigenvector; neither its eigenvalues nor the coordinate-free motion/invariant claims change. This is clause 3.