Cm Math Rigid — Examples

First the finite-atomic integration formula is recorded: if $s={\sum }_j{a}_j{\chi }_{E_j}$ is a nonnegative simple function, then $\int sd\nu ={\sum }_j{a}_j\nu (E_j)={\sum }_j{a}_j{\sum }_{k:x_k\in E_j}{c}_k={\sum }_k{c}_{k}s(x_k)$, and the identity $\nu (E_j)={\sum }_k{c}_k{\delta }_{x_k}(E_j)$ is [L1]; taking the supremum over simple $0\le s\le f$ in the definition of the nonnegative integral gives $\int fd\nu \ge {\sum }_k{c}_{k}f(x_k)$ from the minorant $s_0:={\sum }_{k}f(x_k){\chi }_{\{x_k\}}\le f$, and $\int fd\nu \le {\sum }_k{c}_{k}f(x_k)$ from $\int sd\nu ={\sum }_k{c}_{k}s(x_k)\le {\sum }_k{c}_{k}f(x_k)$ for every simple $0\le s\le f$; hence $\int fd\nu ={\sum }_k{c}_{k}f(x_k)$, and the real case follows by $f=f^{+}-f^{-}$.

With the formula of step 1.1, $I_{jk}={\sum }_l{c}_l(\parallel x_l{\parallel }_2^2{\delta }_{jk}-(x_l{)}_j(x_l{)}_k)$ for arbitrary data, by [L1]; in particular $I$ is the sum over the atoms of $c_l(\parallel x_l{\parallel }_2^2{I}_3-x_l{x}_l^{\mathsf{T}})$ [L6].

Case (ii): $\mu (\nu )={\sum }_k{x}_k=0$ and ${\sum }_k(x_k{)}_1^2=\frac{3}{2}$, ${\sum }_k(x_k{)}_2^2=\frac{3}{2}$, ${\sum }_k(x_k{)}_3^2=0$, ${\sum }_k(x_k{)}_1(x_k{)}_2=0$, ${\sum }_k(x_k{)}_1(x_k{)}_3={\sum }_k(x_k{)}_2(x_k{)}_3=0$ by [L5]; hence $I_{11}={\sum }_k((x_k{)}_2^2+(x_k{)}_3^2)=\frac{3}{2}$, $I_{22}={\sum }_k((x_k{)}_1^2+(x_k{)}_3^2)=\frac{3}{2}$, $I_{33}={\sum }_k((x_k{)}_1^2+(x_k{)}_2^2)=3$ and the off-diagonal entries vanish, so $I=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\frac{3}{2},\frac{3}{2},3)$ by steps 1.1 and 2.1.

Case (i): $I=(1\cdot (I_3-e_3{e}_3^{\mathsf{T}}))+(1\cdot (I_3-e_3{e}_3^{\mathsf{T}}))=2(I_3-e_3{e}_3^{\mathsf{T}})=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(2,2,0)$ by step 2.1 and [L6], and the moment vanishes by symmetry; for $\omega =e_3$ the line $\mathbb{R}{e}_3$ contains both atoms, so $\nu ({\mathbb{R}}^3\setminus \mathbb{R}{e}_3)=0$, so $I$ is not positive definite by [L3], matching the zero eigenvalue.

Case (ii) checked against the quadratic identity: for $\omega =e_1$, ${\sum }_k\parallel e_1\times x_k{\parallel }_2^2={\sum }_k(1-(x_k{)}_1^2)=3-\frac{3}{2}=\frac{3}{2}={\omega }^{\mathsf{T}}I\omega$, and for $\omega =e_3$, ${\sum }_k\parallel e_3\times x_k{\parallel }_2^2={\sum }_k((x_k{)}_1^2+(x_k{)}_2^2)=3$; both agree with $I=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\frac{3}{2},\frac{3}{2},3)$, and no line through the origin contains all three $x_k$, so for every $\omega \mathrm{\ne }0$ at most two of the three points satisfy $\omega \times x_k=0$, at least one atom of weight $1$ lies off $\mathbb{R}\omega$, so $\nu ({\mathbb{R}}^3\setminus \mathbb{R}\omega )\ge 1>0$; by [L3] and [L4] this is exactly positive definiteness of $I$, matching the positive eigenvalues.

Case (iii): $I=(I_3-e_1{e}_1^{\mathsf{T}})+(4I_3-4e_1{e}_1^{\mathsf{T}})=5I_3-5e_1{e}_1^{\mathsf{T}}=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(0,5,5)$ by steps 2.1 and [L6]; the support lies on the line $\mathbb{R}{e}_1$, which is the zero set of $\omega \mapsto \omega \times \rho$ for $\omega =e_1$, and the complement of that line has $\nu$-measure $0$, so $I$ is not positive definite by [L3], matching the zero eigenvalue; the first moment is $\mu (\nu )=e_1+2e_1=3e_1\mathrm{\ne }0$, so this distribution is not centred and its inertia matrix is computed about the origin of coordinates, not about its centre of mass. This last observation is a bookkeeping check, not a claim about any translated inertia matrix.

Certificates at $(1,2,3)$: $c=\frac{1}{2}(1-\frac{1}{2})=\frac{1}{4}$ and $c_3=\frac{1}{2}(\frac{1}{2}-\frac{1}{3})=\frac{1}{12}$; the instability data are $a=\sqrt{(1-\frac{1}{2})\mathrm{/}(\frac{1}{2}-\frac{1}{3})}=\sqrt{3}$, $1+a^2=4$, ${\delta }_0=\frac{1}{4}$, $c_0=\sqrt{3}\cdot \frac{1}{6}\cdot \frac{\sqrt{3}}{2}=\frac{1}{4}$ and ${\epsilon }_0=\frac{1}{2}$ for $L=1$.

Stability certificate check about $e_3$ with $L=1$ and initial data $l(0)=\frac{1}{10}{e}_1+\frac{3\sqrt{11}}{10}{e}_3$ on the unit sphere, where $\frac{3\sqrt{11}}{10}=\sqrt{\frac{99}{100}}$: $D_3(0)=\frac{1}{2}(1\mathrm{/}{I}_1-1\mathrm{/}{I}_3)l_1(0{)}^2=\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{1}{100}=\frac{1}{300}$ (the $l_2$-term of $D_3$ vanishes at $t=0$), so the bound $l_1(t{)}^2+l_2(t{)}^2\le D_3(0)\mathrm{/}{c}_3=\frac{12}{300}=\frac{1}{25}$ holds at all times by [L2]; moreover $C(0)=\frac{1}{100}+\frac{99}{100}=1$, so $l_3(t{)}^2=1-(l_1^2+l_2^2)\ge \frac{24}{25}$, and since $l_3$ is continuous with $l_3(0)=\frac{3\sqrt{11}}{10}>0$ we have $l_3(t)\ge \sqrt{\frac{24}{25}}\ge \frac{97}{100}$, whence $\mathrm{\mid }{l}_3(t)-1\mathrm{\mid }=\frac{l_1(t{)}^2+l_2(t{)}^2}{1+l_3(t)}\le \frac{4}{197}<\frac{21}{1000}$; hence $\parallel l(t)-e_3{\parallel }_2^2\le \frac{1}{25}+(\frac{21}{1000}{)}^2=\frac{40441}{{10}^6}<\frac{41}{1000}$ for all $t\ge 0$: the certificate inequality of [L2] is verified at explicit data, $e_3$ being the extreme axis of largest moment.

Instability check about $e_2$ with $L=1$ and $\eta =\frac{1}{100}$: with $a=\sqrt{3}$ the initial point is $l^{\eta }=(\frac{1}{100},\frac{\sqrt{2499}}{50},-\frac{\sqrt{3}}{100})$, with $C(l^{\eta })=\frac{1}{{10}^4}+\frac{2499}{2500}+\frac{3}{{10}^4}=1$ and $E(l^{\eta })=\frac{1}{2}(\frac{1}{{10}^4}+\frac{2499}{5000}+\frac{1}{{10}^4})=\frac{1}{4}=\frac{L^2}{2I_2}$ exactly, so it lies on the invariant level set of [L1] and [L3] requires no further hypothesis; the escape time is bounded by $c_0^{-1}\log ({\delta }_0\mathrm{/}\eta )=4\log 25$, and there is an escape time no later than this bound at which the solution is at distance at least ${\epsilon }_0=\frac{1}{2}$ from $e_2$.

Linearization check, first case: keep $(I_1,I_2,I_3)=(1,2,3)$ and take $\mathrm{\Omega }=2$, so $\lambda =4\cdot (2-3)(3-1)\mathrm{/}(1\cdot 2)=-4$ and $\kappa =2$; with data $(1,0,0)$ the formulas of [L4] give ${\delta }_1=\cos 2t$, ${\delta }_2=\sin 2t$, and indeed ${\delta }_1^{\mathrm{\prime }}=-2\sin 2t=\mathrm{\Omega }(I_2-I_3){\delta }_2\mathrm{/}{I}_1=-2\sin 2t$ and ${\delta }_2^{\mathrm{\prime }}=2\cos 2t=\mathrm{\Omega }(I_3-I_1){\delta }_1\mathrm{/}{I}_2=2\cos 2t$: the solution is bounded, in the oscillatory branch, and $e_3$ is an extreme moment for these data.

Linearization check, second case: take $(I_1,I_2,I_3)=(1,3,2)$, so $I_3=2$ is the intermediate moment, and $\mathrm{\Omega }=2$: then $\lambda =4\cdot (3-2)(2-1)\mathrm{/}(1\cdot 3)=\frac{4}{3}>0$, $\kappa =\frac{2}{\sqrt{3}}$, and the data $(1,0,0)$ give ${\delta }_1(t)=\cosh (\kappa t)$ and ${\delta }_2(t)=\frac{\mathrm{\Omega }(I_3-I_1)}{I_2\kappa }\sinh (\kappa t)=\frac{2\cdot 1}{3\kappa }\sinh (\kappa t)$, both unbounded as $t\to +\mathrm{\infty }$ because $\cosh (\kappa t)\ge \frac{1}{2}{e}^{\kappa t}$ by [L5]; this exhibits the exponential branch of the dichotomy and the linearized instability of the intermediate axis.

Value of the parametrization: $\mathrm{\Theta }(0,\frac{\pi }{2},0)=R_z(0)R_y(\frac{\pi }{2})R_z(0)=R_y(\frac{\pi }{2})=\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ -1& 0& 0\end{array}\right)$ by [L4] and [L5]; its columns $(0,0,-1),(0,1,0),(1,0,0)$ are orthonormal and its determinant is $1$ (expansion along the second row), so the value lies in $SO(3)$, as the parametrization guarantees.

Angular-velocity check: take $\phi (t)=t$, $\theta (t)=\frac{\pi }{2}$, $\psi (t)=0$, so $R(t)=R_z(t)R_y(\frac{\pi }{2})$. Then $R^{\mathrm{\prime }}(t)=R_z^{\mathrm{\prime }}(t)R_y(\frac{\pi }{2})=R_z(t)[e_3{]}_{\times }{R}_y(\frac{\pi }{2})$ by [L1], hence $R(t{)}^{\mathsf{T}}{R}^{\mathrm{\prime }}(t)=R_y(\frac{\pi }{2}{)}^{\mathsf{T}}[e_3{]}_{\times }{R}_y(\frac{\pi }{2})=[R_y(\frac{\pi }{2}{)}^{\mathsf{T}}{e}_3{]}_{\times }=[-e_1{]}_{\times }$ by the conjugation identity; so $\omega =-e_1$. The formula of [L2] with ${\phi }^{\mathrm{\prime }}=1$, ${\theta }^{\mathrm{\prime }}={\psi }^{\mathrm{\prime }}=0$, $\cos \psi =1$, $\sin \psi =0$ gives $(-\sin \frac{\pi }{2},0,\cos \frac{\pi }{2})=(-1,0,0)$, in agreement.

Steady precession: put $\theta (t)\equiv \frac{\pi }{3}$, $\phi (t)=2t$, $\psi (t)=\frac{1}{2}t$. Then $b=C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta )=\frac{1}{2}+2\cdot \frac{1}{2}=\frac{3}{2}$ and $a=A{\phi }^{\mathrm{\prime }}{\sin }^2\theta +b\cos \theta =2\cdot \frac{3}{4}+\frac{3}{2}\cdot \frac{1}{2}=\frac{9}{4}$ are constants, so the first and third system equations hold identically; and the second equation reads $\frac{d}{dt}(A{\theta }^{\mathrm{\prime }})=0=A{\phi }^{\mathrm{\prime }2}\sin \theta \cos \theta -C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta ){\phi }^{\mathrm{\prime }}\sin \theta +mgd\sin \theta$, the right side being $4\cdot \frac{\sqrt{3}}{2}\cdot \frac{1}{2}-\frac{3}{2}\cdot 2\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\sqrt{3}-\frac{3\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=0$; hence this is a heavy-top trajectory.

The reduction clauses at this trajectory: ${\phi }^{\mathrm{\prime }}=\frac{a-b\cos \theta }{A{\sin }^2\theta }=\frac{\frac{9}{4}-\frac{3}{4}}{\frac{3}{4}}=2$ and ${\psi }^{\mathrm{\prime }}=\frac{b}{C}-\frac{(a-b\cos \theta )\cos \theta }{A{\sin }^2\theta }=\frac{3}{2}-\frac{\frac{3}{2}\cdot \frac{1}{2}}{\frac{3}{4}}=\frac{1}{2}$, recovering the chosen rates; and $U^{\mathrm{\prime }}(\theta )={\phi }^{\mathrm{\prime }}\sin \theta (b-A{\phi }^{\mathrm{\prime }}\cos \theta )-mgd\sin \theta =2\cdot \frac{\sqrt{3}}{2}\cdot (\frac{3}{2}-1)-\frac{\sqrt{3}}{2}=0$, consistent with the constant $\theta$. The reduced energy evaluates to $\frac{1}{2}A{\theta }^{\mathrm{\prime }2}+\frac{(a-b\cos \theta {)}^2}{2A{\sin }^2\theta }+\frac{b^2}{2C}+mgd\cos \theta =0+\frac{9\mathrm{/}4}{3\mathrm{/}2}+\frac{9\mathrm{/}8}{+}\frac{1}{2}=\frac{3}{2}+\frac{9}{8}+\frac{1}{2}=\frac{25}{8}$, and directly as $\frac{1}{2}A{\phi }^{\mathrm{\prime }2}{\sin }^2\theta +\frac{1}{2}C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos \theta {)}^2+mgd\cos \theta =\frac{1}{2}\cdot 4\cdot \frac{3}{4}+\frac{1}{2}\cdot \frac{9}{4}+\frac{1}{2}=\frac{3}{2}+\frac{9}{8}+\frac{1}{2}=\frac{25}{8}$; both agree, as the reduction theorem requires.