Linearization of the torque-free Euler system at a principal-axis spin, and the oscillatory/exponential dichotomy

Expanding with [L1], $(\mathrm{\Omega }{e}_3+\delta )\times I(\mathrm{\Omega }{e}_3+\delta )=\mathrm{\Omega }{e}_3\times I\delta +\mathrm{\Omega }{I}_3\delta \times e_3+\delta \times I\delta$, and $e_3\times I\delta =I_1{\delta }_1{e}_2-I_2{\delta }_2{e}_1$ while $\delta \times e_3={\delta }_2{e}_1-{\delta }_1{e}_2$; collecting components gives $(\mathrm{\Omega }{e}_3+\delta )\times I(\mathrm{\Omega }{e}_3+\delta )=\mathrm{\Omega }(I_3-I_2){\delta }_2{e}_1+\mathrm{\Omega }(I_1-I_3){\delta }_1{e}_2+\delta \times I\delta$. Applying $-I^{-1}$ and [L1] yields $F(\mathrm{\Omega }{e}_3+\delta )=A\delta -I^{-1}(\delta \times I\delta )$ with $A$ as displayed, and $F(\mathrm{\Omega }{e}_3)=A\cdot 0=0$.

Let $\delta$ solve ${\delta }^{\mathrm{\prime }}=A\delta$. The third equation gives ${\delta }_3^{\mathrm{\prime }}=0$, so ${\delta }_3$ is constant; the first two equations read ${\delta }_1^{\mathrm{\prime }}=c{\delta }_2$, ${\delta }_2^{\mathrm{\prime }}=d{\delta }_1$ with $c=\mathrm{\Omega }(I_2-I_3)\mathrm{/}{I}_1\mathrm{\ne }0$ and $d=\mathrm{\Omega }(I_3-I_1)\mathrm{/}{I}_2\mathrm{\ne }0$ by distinctness and $\mathrm{\Omega }\mathrm{\ne }0$, so ${\delta }_1^{\mathrm{\prime }\mathrm{\prime }}=cd{\delta }_1=\lambda {\delta }_1$ and ${\delta }_2={\delta }_1^{\mathrm{\prime }}\mathrm{/}c$; conversely every solution of $x^{\mathrm{\prime }\mathrm{\prime }}=\lambda x$ produces the solution $\delta =(x,x^{\mathrm{\prime }}\mathrm{/}c,\text{const})$.

The remainder satisfies $\parallel I^{-1}(\delta \times I\delta ){\parallel }_2\le \parallel I^{-1}{\parallel }_F\parallel \delta {\parallel }_2\parallel I\delta {\parallel }_2\le \parallel I^{-1}{\parallel }_F\parallel I{\parallel }_F\parallel \delta {\parallel }_2^2$ by [L1] and two applications of [L2]; hence $R(\delta ):=-I^{-1}(\delta \times I\delta )$ obeys the bound of clause 1 with $C=\parallel I^{-1}{\parallel }_F\parallel I{\parallel }_F$, and $A$ is the total derivative of $F$ at $\mathrm{\Omega }{e}_3$ by [L3].

Suppose $\lambda <0$ and put $\kappa =\sqrt{-\lambda }$. The function $x(t):=x(0)\cos (\kappa t)+(x^{\mathrm{\prime }}(0)\mathrm{/}\kappa )\sin (\kappa t)$ satisfies $x^{\mathrm{\prime }\mathrm{\prime }}=-{\kappa }^{2}x=\lambda x$ by the derivative formulas and chain rule [L5]; the pair $(x,x^{\mathrm{\prime }})$ solves the first-order system $(x,x^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=(x^{\mathrm{\prime }},\lambda x)$, which is locally Lipschitz, so by uniqueness [L4] it is the solution with the given initial data. Taking $x={\delta }_1$ and using ${\delta }_2={\delta }_1^{\mathrm{\prime }}\mathrm{/}c$, ${\delta }_1^{\mathrm{\prime }}(0)=c{\delta }_2(0)$, gives ${\delta }_1(t)={\delta }_1(0)\cos (\kappa t)+(c{\delta }_2(0)\mathrm{/}\kappa )\sin (\kappa t)$ and ${\delta }_2(t)={\delta }_2(0)\cos (\kappa t)-(\kappa \mathrm{/}c){\delta }_1(0)\sin (\kappa t)$; substituting $c=\mathrm{\Omega }(I_2-I_3)\mathrm{/}{I}_1$, ${\kappa }^2=-{\mathrm{\Omega }}^2(I_2-I_3)(I_3-I_1)\mathrm{/}(I_1{I}_2)$ and $\kappa \mathrm{/}c=-\mathrm{\Omega }(I_3-I_1)\mathrm{/}(I_2\kappa )$ gives the two displayed formulas. Boundedness follows from $\mid \sin \mid ,\mid \cos \mid \le 1$ [L5].

Suppose $\lambda >0$ and put $\kappa =\sqrt{\lambda }$. The same argument with $\cosh ,\sinh$ in place of $\cos ,\sin$ verifies the two displayed formulas: $x(t)=x(0)\cosh (\kappa t)+(x^{\mathrm{\prime }}(0)\mathrm{/}\kappa )\sinh (\kappa t)$ satisfies $x^{\mathrm{\prime }\mathrm{\prime }}={\kappa }^{2}x$ by [L5], and uniqueness [L4] identifies it as the solution.

Let $\lambda >0$ and set $A_0:={\delta }_1(0)$, $B_0:=c{\delta }_2(0)\mathrm{/}\kappa$, so that ${\delta }_1(t)=A_0\cosh (\kappa t)+B_0\sinh (\kappa t)=\frac{1}{2}(A_0+B_0)e^{\kappa t}+\frac{1}{2}(A_0-B_0)e^{-\kappa t}$ by [L6]. If $A_0+B_0\mathrm{\ne }0$ then $\mid {\delta }_1(t)\mid \ge \frac{1}{2}\mid A_0+B_0\mid e^{\kappa t}-\frac{1}{2}\mid A_0-B_0\mid e^{-\kappa t}\to +\mathrm{\infty }$ as $t\to +\mathrm{\infty }$, because $e^{\kappa t}\to +\mathrm{\infty }$ and $e^{-\kappa t}\to 0$ [L6]; since $A_0+B_0={\delta }_1(0)+c{\delta }_2(0)\mathrm{/}\kappa$, this is the growth criterion of clause 2, and the choice ${\delta }_2(0)=0\mathrm{\ne }{\delta }_1(0)$ satisfies it, so the system has unbounded solutions and clause 3 follows.

Finally the sign of $\lambda$ is the sign of $(I_2-I_3)(I_3-I_1)$ because ${\mathrm{\Omega }}^2\mathrm{/}(I_1{I}_2)>0$; this vanishes exactly when $I_3$ equals $I_1$ or $I_2$, which distinctness excludes, so $\lambda \mathrm{\ne }0$, and $\lambda <0$ exactly for extreme, $\lambda >0$ exactly for intermediate $I_3$, completing the sign criterion. In the intermediate case, for every $\delta >0$ choose $0<\eta <\min \{\delta ,1\}$ and initial data $\eta e_1$. Step 3.1 gives an unbounded solution with initial norm $\eta$; by continuity it reaches norm $1$ at some $t>0$. Thus the fixed $\epsilon =1$ violates the stability condition for every $\delta$, proving that the zero solution is not Lyapunov stable (Equilibria, Lyapunov stability and asymptotic stability of an autonomous first-order system) and completing clause 3.