Quadratic Hamilton–Poisson Systems in Three Dimensions: Equivalence, Stability, and Integration

Abstract

Quadratic Hamilton–Poisson systems on three-dimensional Lie–Poisson spaces are considered. The homogeneous (positive) semidefinite systems are classified up to linear isomorphism; an exhaustive and nonredundant list of 23 normal forms is exhibited. For each normal form, the stability nature of the equilibria is determined. Each normal form is explicitly integrated, with the exception of three families of systems. Based on the analysis of the normal forms, some simple invariants are identified.

Appendices

Appendix A: Taxonomy of Systems

Table 5 Taxonomy of 3D systems, semisimple algebras

Table 6 Taxonomy of 3D systems, solvable algebras

Appendix B: Selected Proofs for Stability and Integration

2.1 B.1 Stability and Integration of $\mathsf{P}(2)$

2.1.1 B.1.1 Stability of Equilibria

The equilibrium states are $\mathsf{e}_{1}^{\eta ,\mu }=(0,\eta , \mu )\neq 0$ and $\mathsf{e}_{2}^{\mu }=(\mu ,0,-\mu )$. The linearization of $\vec{H}$ at $\mathsf{e}_{1}^{\eta ,\mu }$ has eigenvalues $\{0,0,-2\eta \}$. Hence, the states $\mathsf{e}_{1} ^{\eta ,\mu }$, $\eta <0$ are spectrally unstable.

Consider the states $\mathsf{e}_{1}^{\eta ,\mu }$, $\eta >0$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1}C^{2}+\lambda_{2}F$, where $F(p)=p_{1}^{2}\geq 0$. Now $\vec{H}[F](p)=-4p_{1}^{2}p_{2}\leq 0$ for $p$ in a sufficiently small neighbourhood of $\mathsf{e}_{1}^{\eta ,\mu }$. We have $d H_{\lambda }(\mathsf{e} _{1}^{\eta ,\mu })= [{ 2 \mu \lambda_{0}\ \ 2 \eta \lambda_{0}\ \ 2 \mu ( \lambda_{0}+\lambda _{1} )} ]$ and

$$\begin{aligned} d^{2}H_{\lambda }\bigl(\mathsf{e}_{1}^{\eta ,\mu } \bigr)= \left[\textstyle\begin{array}{c@{\quad}c@{\quad}c} 2 \lambda_{0}+2 \lambda_{2} & 0 & 2 \lambda_{0} \\ 0 & 2 \lambda_{0} & 0 \\ 2 \lambda_{0} & 0 & 2 \lambda_{0}+2 \lambda_{1} \end{array}\displaystyle \right] . \end{aligned}$$

Suppose $\mu =0$ and let $\lambda_{0}=0$, $\lambda_{1}=\lambda _{2}=1$. Then $d H_{\lambda }(\mathsf{e}_{1}^{\eta ,\mu })=0$. As $d C^{2}= [{ 0\ \ 2 \eta \ \ 0 }]$, if follows that $d^{2}H_{\lambda }(\mathsf{e}_{1}^{\eta ,\mu })$ is positive definite when restricted to $W=\ker d C^{2}(\mathsf{e} _{1}^{\eta ,\mu })$. On the other hand suppose $\mu \neq 0$ and let $\lambda_{0}=\lambda_{1}=0$, $\lambda_{2}=1$. Then $d H_{ \lambda }(\mathsf{e}_{1}^{\eta ,\mu })=0$ and $d^{2}H_{\lambda }( \mathsf{e}_{1}^{\eta ,\mu })=\operatorname{diag}(2,0,0)$. Also, $d H(\mathsf{e}_{1}^{\eta ,\mu })= [{ 2 \mu \ \ 2\eta \ \ 2 \mu} ]$ and $d C^{2}(\mathsf{e}_{1}^{\eta ,\mu })= [{ 0\ \ 0\ \ \mu }]$. Hence $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\eta ,\mu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}_{1} ^{\eta ,\mu })\cap \ker d C^{2}(\mathsf{e}_{1}^{\eta ,\mu })$. Consequently, the states $\mathsf{e}_{1}^{\eta ,\mu }$, $\eta >0$ are weakly asymptotically stable.

Next, consider the states $\mathsf{e}_{1}^{0,\mu }$. We have that $p(t)=(\frac{-2 \mu }{1+4 \mu^{2}t^{2}},\frac{4\mu^{2}t}{1+4\mu^{2}t ^{2}},\mu )$ is a maximal integral curve (defined over ℝ). As $\lim_{t\to -\infty }p(t)=\mathsf{e}_{1}^{0,\mu }$, any neighbourhood of $\mathsf{e}_{1}^{0,\mu }$ contains a point $p(t_{0})$ for some $t_{0}<0$. Now $\|p(0)-\mathsf{e}_{1}^{0, \mu }\|=2|\mu |$. (Here $\|p\|$ denotes the norm $\|p\|=\sqrt{p _{1}^{2}+p_{2}^{2}+p_{3}^{2}}$.) Let $U=\{p : \|p\|<|\mu |\}$. Then for any neighbourhood $N\subset U$ there exists $t_{0}<0$ such that $p(t_{0})\in N$, but $p(0)\notin U$. Thus the states $\mathsf{e}_{1}^{0,\mu }$ are unstable.

Lastly, we consider the states $\mathsf{e}_{2}^{\mu }$, $\mu \neq 0$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1}C^{2}$. We have $d H_{\lambda }(\mathsf{e}_{2}^{\mu })= [{ 0\ \ 0\ \ -2 \mu \lambda_{1} }]$ and

$$\begin{aligned} d^{2}H_{\lambda }\bigl(\mathsf{e}_{2}^{\mu } \bigr)= \left[\textstyle\begin{array}{c@{\quad}c@{\quad}c} 2 \lambda_{0} & 0 & 2 \lambda_{0} \\ 0 & 2 \lambda_{0} & 0 \\ 2 \lambda_{0} & 0 & 2 \lambda_{0}+2 \lambda_{1} \end{array}\displaystyle \right] . \end{aligned}$$

Suppose $\mu =0$. Let $\lambda_{0}=\lambda_{1}=1$. Then $d H_{\lambda }(\mathsf{e}_{2}^{\mu })=0$ and $d^{2}H_{\lambda }(\mathsf{e}_{2}^{\mu })$ is positive definite. On the other hand, suppose $\mu \neq 0$. Let $\lambda_{0}=1$ and $\lambda_{1}=0$. We have $d C^{2}(\mathsf{e}_{2}^{\mu })= [{ 0\ \ 0\ \ -2 \mu }]$. Hence $d H_{\lambda }(\mathsf{e}_{2}^{\mu })=0$ and $d^{2}H_{\lambda }(\mathsf{e}_{2}^{\mu })$ is positive definite when restricted to $W=\ker d C^{2}(\mathsf{e}_{2}^{\mu })$. Thus the states $\mathsf{e}_{2}^{\mu }$ are stable.

2.1.2 B.1.2 Integration

Suppose $p(\cdot ):(-\varepsilon ,\varepsilon )\to (\mathfrak{g} _{2.1}\oplus \mathfrak{g}_{1})^{*}$ is a nonconstant integral curve of $\vec{H}$. Let $h_{0}=H(p(0))$ and $c_{0}=p_{3}(0)$. We claim the following.

• (a) If $c_{0}^{2}>h_{0}>0$, then there exists $t_{0}\in \mathbb{R}$ such that $p(t)=\bar{p}(t+t_{0})$ for $t\in (- \varepsilon ,\varepsilon )$, where

  $$\begin{aligned} \left\{ \begin{aligned} \bar{p}_{1}(t) &=-\frac{\delta^{2}}{ c_{0}-\sqrt{h_{0}}\cos (2 \delta t) } \\ \bar{p}_{2}(t)&=\frac{\delta \sqrt{h_{0}} \sin (2 \delta t) }{ c_{0}-\sqrt{h_{0}}\cos (2 \delta t)} \\ \bar{p}_{3}(t)&=c _{0} \end{aligned} \right . \quad \delta = \sqrt{c_{0}^{2}-h_{0}}. \end{aligned}$$

• (b) If $c_{0}^{2}=h_{0}>0$, then there exists $t_{0}\in \mathbb{R}$ such that $p(t)=\bar{p}(t+t_{0})$ for $t\in (- \varepsilon ,\varepsilon )$, where

  $$\begin{aligned} \left\{ \begin{aligned} \bar{p}_{1}(t) &=-\frac{2 c_{0}}{1+4 c_{0}^{2} t^{2}} \\ \bar{p}_{2}(t)&=\frac{4 c_{0}^{2} t}{1+4 c_{0}^{2} t^{2}} \\ \bar{p}_{3}(t)&=c_{0}. \end{aligned} \right . \end{aligned}$$

• (c) If $c_{0}^{2}< h_{0}$, then there exists $t_{0}\in \mathbb{R}$ and $\sigma \in \{-1,1\}$ such that $p(t)=\bar{p}(t+t _{0})$ for $t\in (-\varepsilon ,\varepsilon )$, where

  $$\begin{aligned} \left\{ \begin{aligned} \bar{p}_{1}(t) &=\frac{\sigma \operatorname{sgn}(c_{0}) \delta ^{2}}{\sigma |c_{0}|-\sqrt{h_{0}} \cosh (2 \delta t) } \\ \bar{p} _{2}(t)&=\frac{-\sqrt{h_{0} } \delta \sinh (2 \delta t) }{ \sigma |c_{0}|-\sqrt{h_{0}} \cosh (2 \delta t) } \\ \bar{p}_{3}(t)&=c _{0} \end{aligned} \right . \quad \delta = \sqrt{h_{0}-c_{0}^{2}}. \end{aligned}$$

Verification that, in each case, $\bar{p}(\cdot )$ is a maximal integral curve defined over ℝ is standard. Suppose $c_{0}^{2}>h_{0}>0$. We have $p_{3}(\cdot )=c_{0}$ and so $-\sqrt{h_{0}}-c_{0}\leq p_{1}(0)\leq \sqrt{h_{0}}-c_{0}$. Now $\bar{p}_{1}(0)=-\sqrt{h_{0}}-c_{0}$ and $\bar{p}_{1}(\tfrac{ \pi }{2\delta })=\sqrt{h_{0}}-c_{0}$. Thus there exists $t_{1} \in [0,\tfrac{\pi }{2\delta }]$ such that $p_{1}(0)=\bar{p}_{1}(t _{1})$. Then, as $H$ is a constant of motion, we get $p_{2}(0)= \pm \bar{p}_{2}(t_{1})$. However, $\bar{p}_{1}(-t_{1})=\bar{p}_{1}(t _{1})$, $\bar{p}_{2}(-t_{1})=-\bar{p}_{2}(t_{1})$ and $\bar{p} _{3}(-t_{1})=\bar{p}_{1}(t_{1})=c_{0}$. Thus, for $t_{0}=t_{1}$ or $t_{0}=-t_{1}$, we have that $\bar{p}(t_{0})=p(0)$. Consequently $t\mapsto p(t)$ and $t\mapsto \bar{p}(t+t_{0})$ solve the same Cauchy problem and therefore are identical.

Suppose $c_{0}^{2}=h_{0}>0$. Again, $-\sqrt{h_{0}}-c_{0}\leq p _{1}(0)\leq \sqrt{h_{0}}-c_{0}$, i.e., $-|c_{0}|-c_{0}\leq p_{1}(0) \leq |c_{0}|-c_{0}$. (If $c_{0}>0$, then $-2c_{0}\leq p_{1}(0) \leq 0$; if $c_{0}<0$, then $0\leq p_{1}(0)\leq -2c_{0}$.) Note that $p_{1}(0)\neq 0$, as $p(\cdot )$ is assumed nonconstant. Now $\bar{p}_{1}(0)=-2c_{0}$ and $\lim_{t\to \infty }\bar{p}_{1}(t)=0$. Thus there exists $t_{1} \in \mathbb{R}$ such that $\bar{p}_{1}(t_{1})=p_{1}(0)$. Likewise, for $t_{0}=t_{1}$ or $t_{0}=-t_{1}$, we get $p(0)=\bar{p}(t _{0})$. Hence $t\mapsto p(t)$ and $t\mapsto \bar{p}(t+t_{0})$ solve the same Cauchy problem and therefore are identical.

Lastly, suppose $c_{0}^{2}< h_{0}$. Let $\sigma = \operatorname{sgn}(p_{1}(0))\neq 0$. We have $-\sqrt{h_{0}}-c_{0} \leq p_{1}(0)\leq \sqrt{h_{0}}-c_{0}$. Also $\bar{p}_{1}(0)=- \sigma \operatorname{sgn}(c_{0}) \sqrt{h_{0}}-c_{0}$ and $\lim_{t\to \infty } \bar{p}_{1}(t)=0$. Thus there exists $t_{1}\in \mathbb{R}$ such that $\bar{p}_{1}(t_{1})=p_{1}(0)$. Consequently, either $\bar{p}(t+t_{1})=p(t)$ or $\bar{p}(t-t _{1})=p(t)$.

2.2 B.2 Stability of Equilibria for $\mathsf{P}(5)$

The equilibrium states are $\mathsf{e}^{\mu }=(0,0,\mu )$. The linearization of the system at $\mathsf{e}^{\mu }$ has eigenvalues $\{0,2\mu ,2\mu \}$. Hence, the states $\mathsf{e}^{\mu }$, $\mu >0$ are spectrally unstable. We consider the states $\mathsf{e}^{\mu }$, $\mu <0$. Let $H_{\lambda }=\lambda_{0}H+ \lambda_{1}F$, where $F(p)=p_{1}^{2}+p_{2}^{2}\geq 0$. Let $\lambda_{0}=0$ and $\lambda_{1}=1$. Now $\vec{H}[F](p)=4 (p _{1}^{2}+p_{2}^{2}) p_{3}\leq 0$ for $p$ in some neighbourhood of $\mathsf{e}^{\mu }$. We have $d H_{\lambda }(\mathsf{e}^{\mu })=0$ and $d^{2}H_{\lambda }(\mathsf{e}^{\mu })=\operatorname{diag}(2,2,0)$. As $d H(\mathsf{e}^{\mu })= [{ 0\ \ 0 \ \ 2\mu }]$, if follows that $d^{2}H_{\lambda }(\mathsf{e}^{\mu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}^{\mu })$. Hence the states $\mathsf{e}^{\mu }$, $\mu <0$ are weakly asymptotically stable. The state $\mathsf{e}^{0}$ is stable as $H^{-1}(0)=\{0 \}$. However, to prove weak asymptotic stability we consider $H_{\lambda }=\lambda_{0} H+\lambda_{1}F$, where $F(p)=(1+p_{3})(p _{1}^{2}+p_{2}^{2}+p_{3}^{2})$. Let $\lambda_{0}=0$ and $\lambda_{1}=1$. Now $F(p)\geq 0$ and $\vec{H}[F](p)=-2 ( p _{1}^{2}+p_{2}^{2} ) ( p_{1}^{2}+p_{2}^{2}+p_{3}^{2} ) \leq 0$ for $p$ in some neighbourhood of $\mathsf{e}^{0}$. We have $d H_{\lambda }(\mathsf{e}^{0})=0$ and $d^{2}H_{\lambda }( \mathsf{e}^{\mu })=\operatorname{diag}(2,2,2)$. Hence the state $\mathsf{e}^{0}$ is weakly asymptotically stable.

2.3 B.3 Stability of Equilibria for $\mathsf{Np}(2)$

We distinguish between the cases (a) $\delta =0$, (b) $0<\delta < \tfrac{1}{4}$, (c) $\delta =\tfrac{1}{4}$, and (d) $\delta > \tfrac{1}{4}$. We treat here only the cases (b) and (c); the other cases follow similarly.

(b) We have the following equilibrium states

$$\begin{gathered} \mathsf{e}_{1}^{\nu }=(0,0,\nu ),\qquad \mathsf{e}_{2}^{\nu }= \bigl(\nu , \tfrac{1}{2} (1+\sqrt{1-4 \delta })\nu ,0\bigr),\qquad \mathsf{e}_{3}^{ \mu }=\bigl(\mu ,\tfrac{1}{2} (1-\sqrt{1-4 \delta })\mu ,0\bigr). \end{gathered}$$

The linearization of $\vec{H}$ has eigenvalues $\{0, 2 \nu , 2 \nu \}$ at $\mathsf{e}_{1}^{\nu }$, whereas at $\mathsf{e}_{2} ^{\nu }$ it has eigenvalues $\{ 0,-2 (1-4 \delta )^{1/4} \nu ,2 (1-4 \delta )^{1/4} \nu \}$. Therefore the states $\mathsf{e} _{1}^{\nu }$, $\nu >0$ and $\mathsf{e}_{2}^{\nu }$, $\nu \neq 0$ are spectrally unstable.

We consider the states $\mathsf{e}_{1}^{\nu }$, $\nu <0$. Let $H_{\lambda }=\lambda_{0} H+\lambda_{1} F$, where $F(p)=p_{1}^{2}+p _{2}^{2}\geq 0$, $\lambda_{0}=0$, and $\lambda_{1}=1$. We have $\vec{H}[F](p)=4 (p_{1}^{2}-p_{1} p_{2}+p_{2}^{2}) p_{3}\leq 0$ for $p$ in some neighbourhood of $\mathsf{e}_{1}^{\nu }$. Furthermore, $d H_{\lambda }(\mathsf{e}_{1}^{\nu })=0$ and $d^{2} H_{ \lambda }(\mathsf{e}_{1}^{\nu })=\operatorname{diag}(2,2,0)$. Hence, as $d H(\mathsf{e}_{1}^{\nu })= [{ 0\ \ 0\ \ 2\nu}]$, we get that $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\nu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}_{1} ^{\nu })$. Thus the states $\mathsf{e}_{1}^{\nu }$, $\nu <0$ are weakly asymptotically stable.

We consider the states $\mathsf{e}_{3}^{\mu }$, $\mu \neq 0$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1}C$, where $C(p)=p_{1}\exp (\tfrac{p _{2}}{p_{1}})$, $\lambda_{0}=\frac{e^{\frac{1}{2}-\frac{1}{2} \sqrt{1-4 \delta }}}{1-\sqrt{1-4 \delta }}$, and $\lambda_{1}=- \mu$. We have $d H_{\lambda }(\mathsf{e}_{3}^{\mu })=0$ and

$$\begin{aligned} d^{2} H_{\lambda }\bigl(\mathsf{e}_{3}^{\mu } \bigr)= \left[\textstyle\begin{array}{c@{\quad}c@{\quad}c} e^{\frac{1}{2}-\frac{1}{2} \sqrt{1-4 \delta }} ( \sqrt{1-4 \delta }+\delta ) & \frac{1}{2} e^{\frac{1}{2}-\frac{1}{2} \sqrt{1-4 \delta }} ( 1-\sqrt{1-4 \delta } ) & 0 \\ \frac{1}{2} e^{\frac{1}{2}-\frac{1}{2} \sqrt{1-4 \delta }} ( 1-\sqrt{1-4 \delta } ) & \frac{e^{\frac{1}{2}- \frac{1}{2} \sqrt{1-4 \delta }} ( 1+\sqrt{1-4 \delta }-2 \delta ) }{2 \delta } & 0 \\ 0 & 0 & \frac{2 e^{\frac{1}{2}-\frac{1}{2} \sqrt{1-4 \delta }}}{1-\sqrt{1-4 \delta }} \end{array}\displaystyle \right] . \end{aligned}$$

A standard computation shows that $d^{2} H_{\lambda }(\mathsf{e} _{3}^{\mu })$ is positive definite. Therefore the states $\mathsf{e}_{3}^{\mu }$, $\mu \neq 0$ are stable. As $d H( \mathsf{e}_{3}^{0})=0$ and $d^{2} H(\mathsf{e}_{3}^{0})= \operatorname{diag}(2\delta ,2,2)$ is positive definite, it follows that $\mathsf{e}_{3}^{0}$ is stable.

(c) We have the following equilibrium states

$$\begin{aligned} \mathsf{e}_{1}^{\mu }=(0,0,\mu )\quad \text{and}\quad \mathsf{e}_{2} ^{\nu }=\bigl(\nu ,\tfrac{1}{2} \nu ,0 \bigr). \end{aligned}$$

The linearization of $\vec{H}$ has eigenvalues $\{0, 2 \mu , 2 \mu \}$ at $\mathsf{e}_{1}^{\mu }$. Therefore the states $\mathsf{e}_{1}^{\mu }$, $\mu >0$ are spectrally unstable.

We consider the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$. Let $H_{\lambda }=\lambda_{0} H+\lambda_{1} F$, where $F(p)=p_{1}^{2}+p _{2}^{2}\geq 0$, $\lambda_{0}=0$, and $\lambda_{1}=1$. We have $\vec{H}[F](p)=4 (p_{1}^{2}-p_{1} p_{2}+p_{2}^{2}) p_{3}\leq 0$ for $p$ in some neighbourhood of $\mathsf{e}_{1}^{\mu }$. Furthermore, $d H_{\lambda }(\mathsf{e}_{1}^{\mu })=0$ and $d^{2} H_{ \lambda }(\mathsf{e}_{1}^{\mu })=\operatorname{diag}(2,2,0)$. Hence, as $d H(\mathsf{e}_{1}^{\mu })= [{ 0\ \ 0\ \ 2\mu }]$, we get that $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\mu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}_{1} ^{\mu })$. Thus the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$ are weakly asymptotically stable.

Consider the state $\mathsf{e}_{1}^{0}$. As $d H(\mathsf{e}_{1} ^{0})=0$ and $d^{2} H(\mathsf{e}_{1}^{0})=\operatorname{diag}( \tfrac{1}{2},2,2)$ is positive definite, it follows that $\mathsf{e}_{3}^{0}$ is stable. To prove weak asymptotic stability, let $H_{\lambda }=\lambda_{0} H+\lambda_{1} F$, where $F(p)= ( 1+p _{3} ) H(p)$, $\lambda_{0}=0$, and $\lambda_{1}=1$. We have $F(p)\geq 0$ and $\vec{H}[F](p)=-2 ( \tfrac{1}{4} p_{1}^{2}-p _{1} p_{2}+p_{2}^{2} ) ( \tfrac{1}{4} p_{1}^{2}+p_{2}^{2}+p _{3}^{2} ) \leq 0$ for $p$ in some neighbourhood of $\mathsf{e}_{1}^{0}$. Moreover, $dH_{\lambda}(\mathsf{e}_{1}^{0})=0$ and $d^{2}H_{\lambda}(\mathsf{e}_{1}^{0})=\operatorname{diag}(\tfrac{1}{2},2,2)$. Thus the state $\mathsf{e}_{1}^{0}$ is weakly asymptotically stable.

Consider the states $\mathsf{e}_{2}^{\nu }$, $\nu \neq 0$. Let $F(p)=p_{1}^{2}+p_{2}^{2}$ and let $U$ be the open set given by $U=\{p :\sqrt{F(p)}>|\nu |\}$. As $\sqrt{F(\mathsf{e}_{2} ^{\nu })}=\tfrac{\sqrt{5}}{2} |\nu |$, it follows that $\mathsf{e}_{2}^{\nu }\in U$. We have $\dot{p}_{3}=-\frac{1}{2} ( p_{1}-2 p_{2} ) ^{2}\leq 0$. Therefore, $W_{\varepsilon }= \{p : p_{3}\leq \varepsilon \}$ is an invariant subset for $\varepsilon \in \mathbb{R}$ (in the sense that $p(t)\in W_{ \varepsilon }$ for $t>0$ whenever $p(0)\in W_{\varepsilon }$). Let $V=\{p : \sqrt{F(p)}>\frac{1}{2}|\nu |\}$. Note that $U\subset V$. Suppose $N\subset U$ is a neighbourhood of $\mathsf{e}_{2}^{\nu }$. There exists $\varepsilon <0$ such that $(\nu ,\tfrac{1}{2}\nu ,\varepsilon )\in N\cap V\cap W_{\varepsilon }$. Let $p(\cdot )$ be the integral curve satisfying $p(0)=( \nu ,\tfrac{1}{2}\nu ,\varepsilon )$. To prove instability, it suffices to show that there exists $t_{1}>0$ such that $\sqrt{F(p(t _{1}))}<|\nu |$ (and so $p(t_{1})\notin U$). If there exists $t_{1}>0$ such that $p(t_{1})\notin V$, then we are done. On the other hand, assume $p(t)\in V$ for $t>0$. As $W_{\varepsilon }$ is an invariant subset, it follows that $p(t)\in W_{\varepsilon }\cap V$ for $t>0$. We have

$$\begin{aligned} \vec{H}[F](p) &=4 \bigl( p_{1}^{2}-p_{1} p_{2}+p_{2}^{2} \bigr) p_{3}=4 \bigl( \bigl(\tfrac{1}{\sqrt{2}}p_{1}-\tfrac{1}{\sqrt{2}}p_{2} \bigr)^{2}+ \tfrac{1}{2}F(p) \bigr) p_{3}. \end{aligned}$$

Thus $\frac{d}{dt}F(p(t))<\tfrac{1}{2}\nu^{2}\varepsilon$. Accordingly, $F(p(t))\leq \tfrac{5}{4}\nu^{2}+\tfrac{1}{2}\nu^{2}\varepsilon t$. Hence, for $t_{1}=-\frac{11}{8 \varepsilon }>0$, we get $\sqrt{F(p(t _{1}))}\leq \tfrac{3}{4}|\nu |<|\nu |$.

2.4 B.4 Stability of Equilibria for $\mathsf{Np}(6)$

Let $\kappa_{\alpha }^{-}=-1+2 \alpha^{2}-2 \alpha \sqrt{\alpha ^{2}-1}$. We distinguish between the cases: (a) $\beta =0$, (b) ($0< \alpha <1$ and $0<\beta \leq 1$) or ($\alpha >1$ and $0< \beta <\kappa_{\alpha }^{-}$), (c) $\alpha >1$ and $\beta = \kappa_{\alpha }^{-}$, and (d) $\alpha >1$ and $\kappa_{\alpha }^{-}<\beta \leq 1$. We shall make use of the local Casimirs $C_{\pm }\in C^{\infty }(U_{\pm })$ as described in Sect. 3.1. We omit the proof for (d) as it is similar to (c).

(a) We have the following equilibrium states

$$\begin{aligned} \mathsf{e}_{1}^{\mu }=(0,0,\mu ),\qquad \mathsf{e}_{2}^{\nu }=( \alpha \nu ,\nu ,0),\quad \text{and}\quad \mathsf{e}_{3}^{\nu }=( \nu ,0,0). \end{aligned}$$

The linearization of $\vec{H}$ at $\mathsf{e}_{1}^{\mu }$ has eigenvalues $\{0, 2 (\alpha -1) \mu , 2 (\alpha +1) \mu \}$. Thus the states $\mathsf{e}_{1}^{\mu }\neq 0$ are spectrally unstable if $0<\alpha <1$, whereas the states $\mathsf{e}_{1}^{\mu }$, $\mu >0$ are spectrally unstable if $\alpha >1$. The linearization of $\vec{H}$ at $\mathsf{e}_{2}^{\nu }$ has eigenvalues $\{0, -2 \sqrt{ ( \alpha^{2}-1 ) \nu^{2}}, 2 \sqrt{ ( \alpha^{2}-1 ) \nu^{2}}\}$. Thus the states $\mathsf{e} _{2}^{\nu }$ are spectrally unstable if $\alpha >1$.

Consider the states $\mathsf{e}_{2}^{\nu }$, $\nu <0$ and suppose $0<\alpha <1$. (Note that $\mathsf{e}_{2}^{\nu }\in U_{+}$ as $(\alpha -1)\nu >0$.) Let $H_{\lambda }=\lambda_{0}H+\lambda_{1} C _{+}$. Setting $\lambda_{1}= -2^{\frac{2}{\alpha +1}} \nu ((\alpha -1) \nu )^{\frac{\alpha -1}{\alpha +1}} \lambda_{0}$, we get $d H_{\lambda }(\mathsf{e}_{2}^{\nu })=0$. Furthermore, we find that

$$\begin{aligned} d^{2} H_{\lambda }\bigl(\mathsf{e}_{2}^{\nu } \bigr)= \left[\textstyle\begin{array}{c@{\quad}c@{\quad}c} -\frac{2 \lambda_{0}}{\alpha^{2}-1} & \frac{2 \alpha \lambda_{0}}{ \alpha^{2}-1} & 0 \\ \frac{2 \alpha \lambda_{0}}{\alpha^{2}-1} & \frac{2 (\alpha -2) \alpha \lambda_{0}}{\alpha^{2}-1} & 0 \\ 0 & 0 & 2 \lambda_{0} \end{array}\displaystyle \right] . \end{aligned}$$

Setting $\lambda_{0}=1$, a simple calculation shows that $d^{2} H _{\lambda }(\mathsf{e}_{2}^{\nu })$ is positive definite. Hence the states $\mathsf{e}_{2}^{\nu }$, $\nu <0$ are stable whenever $0<\alpha <1$. A similar argument (involving $C_{-}$) shows that the states $\mathsf{e}_{2}^{\nu }$, $\nu >0$ are stable whenever $0<\alpha <1$.

Next, consider the states $\mathsf{e}_{3}^{\nu }$, $\nu \neq 0$. Let $H_{\lambda }=\lambda_{0} H+\lambda_{1}C_{\pm }$, with $\lambda _{0}=1$ and $\lambda_{1}=1$. (If $\nu >0$, then $\mathsf{e} _{3}^{\nu }\in U_{+}$; if $\nu <0$, then $\mathsf{e}_{3}^{\nu } \in U_{-}$.) We have $d H_{\lambda }(\mathsf{e}_{3}^{\nu })=0$ and $d^{2} H_{\lambda }(\mathsf{e}_{3}^{\nu })=\operatorname{diag}(0,1,1)$. Moreover, $W=\ker d C_{\pm }(\mathsf{e}_{3}^{\nu })=\{(\alpha x, x, y) : x,y\in \mathbb{R}\}$. Accordingly, $d^{2} H_{\lambda }( \mathsf{e}_{3}^{\nu })$ is positive definite when restricted to $W$. Thus the states $\mathsf{e}_{3}^{\nu }$, $\nu \neq 0$ are stable (for $\alpha >0$, $\alpha \neq 1$).

Consider the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$ and suppose $\alpha >1$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1} F$, with $F(p)=p_{1}^{2}+p_{2}^{2}\geq 0$, $\lambda_{0}=0$ and $\lambda_{1}=1$. We have $\vec{H}[F](p)=4 ( \alpha p_{1}^{2}-2 p_{1} p_{2}+\alpha p_{2}^{2} ) p_{3}\leq 0$ for $p$ in some neighbourhood of $\mathsf{e}_{1}^{\mu }$. Furthermore, $d H_{ \lambda }(\mathsf{e}_{1}^{\mu })=0$ and $d^{2} H_{\lambda }( \mathsf{e}_{1}^{\mu })=\operatorname{diag}(2,2,0)$. As $d H( \mathsf{e}_{1}^{\mu })= [{ 0\ \ 0\ \ 2\mu}]$, it follows that $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\mu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}_{1} ^{\mu })$. Thus the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$ are weakly asymptotically stable whenever $\alpha >1$.

(b) We have the following equilibrium states

$$\begin{aligned} \mathsf{e}_{1}^{\nu } =&(0,0,\nu ),\qquad \mathsf{e}_{2}^{\nu }= \biggl(\nu ,\frac{(1+ \beta )+\sqrt{1+\beta ( 2-4 \alpha^{2}+\beta ) } }{2 \alpha }\nu ,0\biggr),\quad\text{and} \\ \mathsf{e}_{3}^{\mu } =&\biggl(\mu , \frac{(1+\beta )-\sqrt{1+ \beta ( 2-4 \alpha^{2}+\beta ) } }{2 \alpha }\mu ,0\biggr). \end{aligned}$$

The linearization of $\vec{H}$ at $\mathsf{e}_{1}^{\nu }$ has eigenvalues $\{0,2 (\alpha -1) \nu ,2 (\alpha +1) \nu \}$. Hence the states $\mathsf{e}_{1}^{\nu }$, $\nu \neq 0$ are spectrally unstable when $0<\alpha <1$, whereas the states $\mathsf{e}_{1} ^{\nu }$, $\nu >0$ are spectrally unstable if $\alpha >1$. The linearization of $\vec{H}$ at $\mathsf{e}_{2}^{\nu }$ has eigenvalues $\lambda_{1}=0$ and

$$\begin{aligned} \lambda_{2,3}=\pm \frac{1}{\alpha }\sqrt{2} |\nu |\sqrt{(1+\beta ) \bigl( -1-\beta^{2}+*_{1}-2 \alpha^{2} *_{1}+\beta \bigl( -2+4 \alpha ^{2}+*_{1} \bigr) \bigr) } \end{aligned}$$

where $*_{1}=\sqrt{1+\beta ( 2-4 \alpha^{2}+\beta ) }$. It is easy to show that one of these eigenvalues have positive real part if $\alpha >1$ and $0<\beta <\kappa_{\alpha }^{-}$. Therefore, under these conditions, the states $\mathsf{e}_{2}^{\nu }$, $\nu \neq 0$ are spectrally unstable.

Consider the states $\mathsf{e}_{2}^{\nu }$, $\nu <0$ and suppose $0<\alpha <1$ and $0<\beta \leq 1$. Then $\mathsf{e}_{2}^{ \nu }\in U_{+}$, i.e., $C_{+}$ is defined in some neighbourhood of $\mathsf{e}_{2}^{\nu }$. (If $\nu >0$, then $\mathsf{e}_{2}^{ \nu }\in U_{-}$ and a similar argument involving $C_{-}$ yields the result.) Let $H_{\lambda }=\lambda_{0} H+\lambda_{1} C_{+}$. If we set

$$\begin{aligned} { \lambda_{0}=\frac{\lambda_{1}}{(1+\alpha ) \beta } 2^{\frac{-3+ \alpha }{1+\alpha }} ( -1+ \beta +*_{1} ) \biggl( -\frac{ ( 1-2 \alpha +\beta +*_{1} ) \nu }{\alpha } \biggr) ^{-\frac{2 \alpha }{1+ \alpha }}} \end{aligned}$$

where $*_{1}=\sqrt{1+\beta ( 2-4 \alpha^{2}+\beta ) }$, then $d H_{\lambda }(\mathsf{e}_{2}^{\nu })=0$. Moreover, if we set $\lambda_{1}=1$, a quite involved calculation shows that $d^{2} H _{\lambda }(\mathsf{e}_{2}^{\nu })$ is positive definite. Hence the states $\mathsf{e}_{2}^{\nu }$, $\nu <0$ are stable whenever $0<\alpha <1$ and $0<\beta \leq 1$. (Likewise, the states $\mathsf{e}_{2}^{\nu }$, $\nu >0$ are stable whenever $0<\alpha <1$ and $0<\beta \leq 1$.)

Consider the states $\mathsf{e}_{3}^{\mu }$, $\mu \neq 0$. Suppose $\mu >0$. Then $\mathsf{e}_{3}^{\mu }\in U_{+}$ i.e., $C_{+}$ is defined in a neighbourhood of $\mathsf{e}_{3}^{ \mu }$. Let $H_{\lambda }=\lambda_{0} H+\lambda_{1} C_{+}$. If we set

$$\begin{aligned} {\lambda_{0}=\frac{ 2^{\frac{-3+\alpha }{1+\alpha }}\lambda_{1}}{(1+ \alpha ) \beta } ( -1+\beta -*_{1} ) \biggl( \frac{ ( -1+2 \alpha -\beta +*_{1} ) \nu }{\alpha } \biggr) ^{-\frac{2 \alpha }{1+ \alpha }}} \end{aligned}$$

where $*_{1}=\sqrt{1+\beta ( 2-4 \alpha^{2}+\beta ) }$, then $d H_{\lambda }(\mathsf{e}_{3}^{\mu })=0$. Moreover, if we set $\lambda_{1}=1$, a quite involved computation shows that $d^{2} H _{\lambda }(\mathsf{e}_{2}^{\nu })$ is positive definite. Hence the states $\mathsf{e}_{3}^{\mu }$, $\mu >0$ are stable. (Likewise, the states $\mathsf{e}_{3}^{\mu }$, $\mu <0$ are stable.) As $dH(\mathsf{e}_{3}^{0})=0$ and $d^{2}H(\mathsf{e}_{3}^{0})=\operatorname{diag}(2\beta,2,2)$, it follows that the state $\mathsf{e}_{3}^{0}$ is stable.

Consider the states $\mathsf{e}_{1}^{\nu }$, $\nu <0$ and suppose $\alpha >1$ and $0<\beta <\kappa_{\alpha }^{-}$. Let $H_{ \lambda }=\lambda_{0}H+\lambda_{1} F$, with $F(p)=p_{1}^{2}+p_{2} ^{2}\geq 0$, $\lambda_{0}=0$ and $\lambda_{1}=1$. We have $\vec{H}[F](p)=4 ( \alpha p_{1}^{2}-2 p_{1} p_{2}+\alpha p_{2} ^{2} ) p_{3}\leq 0$ for $p$ in some neighbourhood of $\mathsf{e}_{1}^{\nu }$. Furthermore, $d H_{\lambda }(\mathsf{e} _{1}^{\nu })=0$ and $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\nu })= \operatorname{diag}(2,2,0)$. As $d H(\mathsf{e}_{1}^{\nu })= [{ 0\ \ 0\ \ 2\nu}]$, it follows that $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\nu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}_{1} ^{\nu })$. Thus the states $\mathsf{e}_{1}^{\nu }$, $\nu <0$ are weakly asymptotically stable whenever $\alpha >1$ and $0<\beta <\kappa_{\alpha }^{-}$.

(c) We have the following equilibrium states

$$\begin{gathered} \mathsf{e}_{1}^{\mu }=(0,0,\mu )\quad \text{and}\quad \mathsf{e}_{2} ^{\nu }=\bigl(\nu ,\bigl(\alpha -\sqrt{ \alpha^{2}-1} \bigr)\nu ,0\bigr). \end{gathered}$$

The linearization of $\vec{H}$ at $\mathsf{e}_{1}^{\mu }$ has eigenvalues $\{0,2 (\alpha -1) \mu ,2 (\alpha +1 ) \mu \}$. Therefore (as $\alpha >1$ by assumption) the states $\mathsf{e}_{1}^{ \mu }$, $\mu >0$ are spectrally unstable.

Consider the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1}F$, where $F(p)=p_{1}^{2}+p _{2}^{2}\geq 0$, $\lambda_{0}=0$, and $\lambda_{1}=1$. We have $\vec{H}[F](p)=4 ( \alpha p_{1}^{2}-2 p_{1} p_{2}+\alpha p_{2} ^{2} ) p_{3}\leq 0$ for $p$ in some neighbourhood of $\mathsf{e}_{1}^{\mu }$). Furthermore, we have that $d H_{\lambda }(\mathsf{e}_{1}^{\mu })=0$, $d^{2} H_{\lambda }(\mathsf{e}_{1} ^{\mu })=\operatorname{diag}(2,2,0)$, and $d H(\mathsf{e}_{1}^{ \mu })= [{ 0 \ \ 0\ \ 2\mu}]$. Therefore $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\mu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}_{1} ^{\mu })$. Thus the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$ are weakly asymptotically stable.

Consider the state $\mathsf{e}_{1}^{0}$. As $d H(\mathsf{e}_{1} ^{0})=0$ and $d^{2} H(\mathsf{e}_{1}^{0})=\operatorname{diag}(2 \kappa_{\alpha }^{-},2,2)$ is positive definite, it follows that the state $\mathsf{e}_{1}^{0}$ is stable. To prove weak asymptotic stability, let $H_{\lambda }=\lambda_{0} H+\lambda_{1} F$, where $F(p)= ( 1+p_{3} ) H(p)$, $\lambda_{0}=0$, and $\lambda _{1}=1$. We have $F(p)\geq 0$ and

$$\begin{aligned} \vec{H}[F](p)=-2 \bigl( \alpha \kappa_{\alpha }^{-} p_{1}^{2}-\bigl(1+ \kappa_{\alpha }^{-} \bigr) p_{1} p_{2}+\alpha p_{2}^{2} \bigr) H(p)\leq 0 \end{aligned}$$

for $p$ in some neighbourhood of $\mathsf{e}_{1}^{0}$. (The quadratic form $\alpha \kappa_{\alpha }^{-} p_{1}^{2}-(1+ \kappa_{\alpha }^{-}) p_{1} p_{2}+\alpha p_{2}^{2}$ is positive semidefinite; it has leading principle minors $\alpha \kappa_{ \alpha }^{-}$ and 0.) Moreover, $d H_{\lambda }(\mathsf{e} _{1}^{0})=0$ and $d^{2} H_{\lambda }(\mathsf{e}_{1}^{0})= \operatorname{diag}(2\kappa_{\alpha }^{-},2,2)$. Thus the state $\mathsf{e}_{1}^{0}$ is weakly asymptotically stable.

Consider the states $\mathsf{e}_{2}^{\nu }$, $\nu \neq 0$. Let $F(p)=p_{1}^{2}+p_{2}^{2}$ and let $U$ be the open set given by $\{p :\sqrt{F(p)}>\tfrac{1}{2}|\nu |\}$. As $\sqrt{F( \mathsf{e}_{2}^{\nu })}=\sqrt{1+ ( \alpha -\sqrt{\alpha^{2}-1} ) ^{2}} |\nu |$, it follows that $\mathsf{e}_{2}^{\nu }\in U$. We have $\dot{p}_{3}=-2(\alpha \kappa_{\alpha }^{-} p_{1}^{2}-(\kappa_{ \alpha }^{-}+1)p_{1}p_{2}+\alpha p_{2}^{2})\leq 0$. Therefore, $W_{\varepsilon }=\{p : p_{3}\leq \varepsilon \}$ is an invariant subset for $\varepsilon \in \mathbb{R}$ (in the sense that $p(t)\in W_{\varepsilon }$ for $t>0$ whenever $p(0)\in W _{\varepsilon }$). Let $V=\{p : \sqrt{F(p)}>\frac{1}{4}|\nu |\}$. Note that $U\subset V$. Suppose $N\subset U$ is a neighbourhood of $\mathsf{e}_{2}^{\nu }$. There exists $\varepsilon <0$ such that $(\nu ,(\alpha -\sqrt{\alpha^{2}-1} )\nu ,\varepsilon ) \in N\cap V\cap W_{\varepsilon }$. Let $p(\cdot )$ be the integral curve satisfying $p(0)=(\nu ,(\alpha -\sqrt{\alpha^{2}-1} )\nu , \varepsilon )$. To prove instability, it suffices to show that there exists $t_{1}>0$ such that $\sqrt{F(p(t_{1}))}<\tfrac{1}{2}| \nu |$ (and so $p(t_{1})\notin U$). If there exists $t_{1}>0$ such that $p(t_{1})\notin V$, then we are done. On the other hand, suppose $p(t)\in V$ for $t>0$. As $W_{\varepsilon }$ is an invariant subset, it follows that $p(t)\in W_{\varepsilon }\cap V$ for $t>0$. We have

$$\begin{aligned} \vec{H}[F](p) &=4 \bigl( \alpha p_{1}^{2}-2 p_{1} p_{2}+\alpha p_{2} ^{2} \bigr) p_{3} =4 \bigl( (p_{1}-p_{2})^{2}+( \alpha -1)F(p) \bigr) p _{3}. \end{aligned}$$

Thus $\frac{d}{dt}F(p(t))<\tfrac{1}{4}(\alpha -1)\nu^{2}\varepsilon$. Accordingly,

$$\begin{aligned} F\bigl(p(t)\bigr)\leq \bigl( 1+ \bigl( \alpha -\sqrt{\alpha^{2}-1} \bigr) ^{2} \bigr) \nu^{2}+\tfrac{1}{4}(\alpha -1) \nu^{2}\varepsilon t. \end{aligned}$$

Hence, for $t_{1}=\frac{ ( 1-18 \alpha^{2}+18 \alpha \sqrt{-1+ \alpha^{2}} ) \nu^{2}}{9 (-1+\alpha ) \varepsilon |\nu |}>0$, we get $\sqrt{F(p(t_{1}))}\leq \tfrac{1}{3}|\nu |<\tfrac{1}{2}| \nu |$.

Remark 12

The function $F(p)= ( 1+p_{3} ) ( \beta p_{1}^{2}+p_{2} ^{2}+p_{3}^{2} )$ suffices to show that the origin is weakly asymptotically stable whenever $4\alpha^{2}\beta \geq (1+\beta )^{2}$. This inequality holds for cases (c) and (d).

2.5 B.5 Stability of Equilibria for $\mathsf{Np}(8)$

Let $\kappa_{\alpha }^{-}=1+2 \alpha^{2}-2 \alpha \sqrt{\alpha ^{2}+1}$. We distinguish between the cases (a) $\beta =0$, (b) $0<\beta <\kappa_{\alpha }^{-}$, (c) $\beta =\kappa_{\alpha }^{-}$, and (d) $\kappa_{\alpha }^{-}<\beta \leq 1$. The proof for (d) is similar to that for (c) and hence omitted.

(a) We have the following equilibrium states

$$\begin{aligned} \mathsf{e}_{1}^{\mu }=(0,0,\mu ),\qquad \mathsf{e}_{2}^{\nu }=( \alpha \nu ,\nu ,0),\quad \text{and}\quad \mathsf{e}_{3}^{\nu }=( \nu ,0,0). \end{aligned}$$

Consider the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1}F$, where $F(p)=p_{1}^{2}+p _{2}^{2}\geq 0$, $\lambda_{0}=0$, and $\lambda_{1}=1$. We have $\vec{H}[F](p)=4 \alpha ( p_{1}^{2}+p_{2}^{2} ) p_{3} \leq 0$ for $p$ in some neighbourhood of $\mathsf{e}_{1}^{ \mu }$. Furthermore $d H_{\lambda }(\mathsf{e}_{1}^{\mu })=0$, $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\mu })=\operatorname{diag}(2,2,0)$, and $d H(\mathsf{e}_{2}^{\mu })= [{ 0 \ \ 0 \ \ 2\mu} ]$. Hence, as $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\mu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}_{1} ^{\mu })$, the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$ are weakly asymptotically stable.

The linearization of $\vec{H}$ at $\mathsf{e}_{1}^{\mu }$ has eigenvalues $\{0,2 (-i \mu +\alpha \mu ),2 (i \mu +\alpha \mu )\}$. Therefore, if $\mu >0$, then $\mathsf{e}_{1}^{\mu }$ is spectrally unstable. The linearization of $\vec{H}$ at $\mathsf{e}_{2}^{ \nu }$ has eigenvalues $\{0,-2 \sqrt{ ( 1+\alpha^{2} ) \nu^{2}},2 \sqrt{ ( 1+\alpha^{2} ) \nu^{2}}\}$. Thus the states $\mathsf{e}_{2}^{\nu }$, $\nu \neq 0$ are spectrally unstable.

Consider the states $\mathsf{e}_{3}^{\nu }$, $\nu \neq 0$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1} C$, with $\lambda_{0}=1$ and $\lambda_{1}=0$. (Note that $C$ is locally defined in some neighbourhood of $\mathsf{e}_{3}^{\nu }$.) We have $d H_{\lambda }(\mathsf{e}_{3}^{\nu })=0$ and $d^{2} H_{\lambda }(\mathsf{e} _{3}^{\nu })=\operatorname{diag}(0,2,2)$. Moreover $W=\ker d C( \mathsf{e}_{3}^{\nu })=\{(\alpha x,-x,y) : x,y\in \mathbb{R}\}$. Hence $d^{2} H_{\lambda }(\mathsf{e}_{3}^{\nu })$ is positive definite when restricted to $W$. Therefore the states $\mathsf{e} _{3}^{\nu }$, $\nu \neq 0$ are stable.

(b) We have the following equilibrium states

$$\begin{aligned} \mathsf{e}_{1}^{\mu } =&(0,0,\mu ),\qquad \mathsf{e}_{2}^{\nu }= \biggl(\nu ,\frac{1- \beta +\sqrt{1-2 \beta -4 \alpha^{2} \beta +\beta^{2}}}{2 \alpha } \nu ,0\biggr), \quad \text{and} \\ \mathsf{e}_{3}^{\nu } =&\biggl( \nu , \frac{1-\beta -\sqrt{1-2 \beta -4 \alpha^{2} \beta +\beta^{2}} }{2 \alpha }\nu ,0\biggr). \end{aligned}$$

Consider the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1}F$, where $F(p)=p_{1}^{2}+p _{2}^{2}\geq 0$, $\lambda_{0}=0$, and $\lambda_{1}=1$. We have $\vec{H}[F](p)=4 \alpha ( p_{1}^{2}+p_{2}^{2} ) p_{3} \leq 0$ for $p$ in some neighbourhood of $\mathsf{e}_{1}^{ \mu }$. Furthermore, $d H_{\lambda }(\mathsf{e}_{1}^{\mu })=0$, $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\mu })=\operatorname{diag}(2,2,0)$, and $d H(\mathsf{e}_{2}^{\mu })= [{ 0 \ \ 0 \ \ 2 \mu}]$. Hence, as $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\mu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}_{1} ^{\mu })$, the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$ are weakly asymptotically stable. As $d H(\mathsf{e}_{1}^{0})=0$ and $d^{2} H(\mathsf{e}_{1}^{0})=\operatorname{diag}(2 \beta , 2,2)$, it follows that $\mathsf{e}_{1}^{0}$ is stable.

The linearization of $\vec{H}$ at $\mathsf{e}_{1}^{\mu }$ has eigenvalues $\{0,2 (-i \mu +\alpha \mu ),2 (i \mu +\alpha \mu )\}$. Therefore, if $\mu >0$, then $\mathsf{e}_{1}^{\mu }$ is spectrally unstable. The linearization of $\vec{H}$ at $\mathsf{e}_{2}^{ \nu }$ has eigenvalues $\lambda_{1}=0$,

$$\begin{aligned} \lambda_{2,3}=\pm \frac{1}{\alpha }\sqrt{2} \sqrt{-(-1+\beta ) \bigl( 1+\beta^{2}-*_{1}-2 \alpha^{2} *_{1}+\beta \bigl( -2-4 \alpha ^{2}+*_{1} \bigr) \bigr) \nu^{2}} \end{aligned}$$

where $*_{1}=\sqrt{1+\beta ( -2-4 \alpha^{2}+\beta ) }$. A tedious computation shows that there is an eigenvalue with positive real part. Thus the states $\mathsf{e}_{2}^{\nu }$, $\nu \neq 0$ are spectrally unstable.

Consider the states $\mathsf{e}_{3}^{\nu }$, $\nu \neq 0$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1} C$, with $\lambda_{0}=1$ and

$$\begin{aligned} \lambda_{1}=\frac{2 \beta ( -\frac{\alpha +\alpha \beta +i \sqrt{1+ \beta ( -2-4 \alpha^{2}+\beta ) }}{(i+\alpha ) (-1+\beta )} ) ^{-i \alpha } }{1+\beta +\sqrt{1+\beta ( -2-4 \alpha^{2}+\beta ) }}. \end{aligned}$$

(Note that $C$ is locally defined in some neighbourhood of $\mathsf{e}_{3}^{\nu }$.) We have $d H_{\lambda }(\mathsf{e}_{3} ^{\nu })=0$. Moreover, a quite involved computation shows that $d^{2} H_{\lambda }(\mathsf{e}_{3}^{\nu })$ is positive definite when restricted to

$$\begin{aligned} W=\ker d C\bigl(\mathsf{e}_{3}^{\nu }\bigr)= \biggl\{ \biggl( -\frac{x}{2 \beta },\frac{x \alpha }{1-\beta -\sqrt{1-2 \beta -4 \alpha^{2} \beta +\beta^{2}}},y \biggr) : x,y\in \mathbb{R} \biggr\} . \end{aligned}$$

Therefore the states $\mathsf{e}_{3}^{\nu }$, $\nu \neq 0$ are stable.

(c) We have the following equilibrium states

$$\begin{aligned} \mathsf{e}_{1}^{\mu }=(0,0,\mu )\quad \text{and}\quad \mathsf{e}_{2}^{\nu }=\bigl( \nu , \bigl( -\alpha +\sqrt{1+ \alpha^{2}} \bigr) \nu ,0\bigr). \end{aligned}$$

Consider the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$. Let $H_{\lambda }=\lambda_{0}H+\lambda_{1}F$, where $F(p)=p_{1}^{2}+p _{2}^{2}\geq 0$, $\lambda_{0}=0$, and $\lambda_{1}=1$. We have $\vec{H}[F](p)=4 \alpha ( p_{1}^{2}+p_{2}^{2} ) p_{3} \leq 0$ for $p$ in some neighbourhood of $\mathsf{e}_{1}^{ \mu }$. Furthermore, $d H_{\lambda }(\mathsf{e}_{1}^{\mu })=0$, $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\mu })=\operatorname{diag}(2,2,0)$, and $d H(\mathsf{e}_{1}^{\mu })= [{ 0 \ \ 0 \ \ 2 \mu}]$. Hence, as $d^{2} H_{\lambda }(\mathsf{e}_{1}^{\mu })$ is positive definite when restricted to $W=\ker d H(\mathsf{e}_{1} ^{\mu })$, the states $\mathsf{e}_{1}^{\mu }$, $\mu <0$ are weakly asymptotically stable.

Consider the state $\mathsf{e}_{1}^{0}$. As $d H(\mathsf{e}_{1} ^{0})=0$ and $d^{2} H(\mathsf{e}_{1}^{0})=\operatorname{diag}(2 \kappa_{\alpha }^{-}, 2,2)$, it follows that $\mathsf{e}_{1}^{0}$ is stable. To prove weak asymptotic stability, let $H_{\lambda }= \lambda_{0} H+\lambda_{1} F$, where $F(p)= ( 1+p_{3} ) H(p)$, $\lambda_{0}=0$, and $\lambda_{1}=1$. We have $F(p)\geq 0$ and

$$\begin{aligned} \vec{H}[F](p)=-2 \bigl( \alpha \kappa_{\alpha }^{-} p_{1}^{2}+\bigl( \kappa_{\alpha }^{-} -1 \bigr) p_{1} p_{2} +\alpha p_{2}^{2} \bigr) H(p) \leq 0 \end{aligned}$$

for $p$ in some neighbourhood of $\mathsf{e}_{1}^{0}$. (The quadratic form $\alpha \kappa_{\alpha }^{-} p_{1}^{2}+(\kappa_{ \alpha }^{-} -1) p_{1} p_{2} +\alpha p_{2}^{2}$ is positive semidefinite; it has leading principle minors $\alpha \kappa_{ \alpha }^{-}$ and 0.) Moreover, $d H_{\lambda }(\mathsf{e} _{1}^{0})=0$ and $d^{2} H_{\lambda }(\mathsf{e}_{1}^{0})= \operatorname{diag}(2\kappa_{\alpha }^{-},2,2)$. Thus the state $\mathsf{e}_{1}^{0}$ is weakly asymptotically stable.

The linearization of $\vec{H}$ at $\mathsf{e}_{1}^{\mu }$ has eigenvalues $\{0,2 (-i \mu +\alpha \mu ),2 (i \mu +\alpha \mu )\}$. Therefore the states $\mathsf{e}_{1}^{\mu }$, $\mu >0$ are spectrally unstable.

Consider the states $\mathsf{e}_{2}^{\nu }$. Let $F(p)=p_{1}^{2}+p _{2}^{2}$ and let $U$ be the open set given by $\{p : \sqrt{F(p)}>\tfrac{1}{2}|\nu |\}$. As $\sqrt{F(\mathsf{e}_{2}^{ \nu })}=\sqrt{1+ ( \alpha -\sqrt{\alpha^{2}+1} ) ^{2}} | \nu |$, it follows that $\mathsf{e}_{2}^{\nu }\in U$. We have

$$\begin{aligned} \dot{p}_{3}=- \bigl( \sqrt{\tfrac{2}{\alpha }}\bigl(- \alpha^{2}+\sqrt{ \alpha^{2}+\alpha^{4}} \bigr)p_{1}-\sqrt{2 \alpha } p_{2} \bigr) ^{2} \leq 0. \end{aligned}$$

Therefore, $W_{\varepsilon }=\{p : p_{3}\leq \varepsilon \}$ is an invariant subset for $\varepsilon \in \mathbb{R}$ (in the sense that $p(t)\in W_{\varepsilon }$ for $t>0$ whenever $p(0) \in W_{\varepsilon }$). Let $V=\{p : \sqrt{F(p)}>\frac{1}{4}| \nu |\}$. Note that $U\subset V$. Suppose $N\subset U$ is a neighbourhood of $\mathsf{e}_{2}^{\nu }$. There exists $\varepsilon <0$ such that $(\nu ,(\alpha -\sqrt{\alpha^{2}-1} )\nu ,\varepsilon )\in N\cap V\cap W_{\varepsilon }$. Let $p(\cdot )$ be the integral curve satisfying $p(0)=(\nu , ( -\alpha +\sqrt{1+\alpha^{2}} ) \nu ,\varepsilon )$. To prove instability, it suffices to show that there exists $t_{1}>0$ such that $\sqrt{F(p(t_{1}))}< \tfrac{1}{2}|\nu |$ (and so $p(t_{1})\notin U$). If there exists $t_{1}>0$ such that $p(t_{1})\notin V$, then we are done. On the other hand, assume $p(t)\in V$ for $t>0$. As $W_{\varepsilon }$ is an invariant subset, it follows that $p(t)\in W_{\varepsilon }\cap V$ for $t>0$. We have $\vec{H}[F](p)=4 \alpha F(p) p_{3}$. Thus $\frac{d}{dt}F(p(t))<\tfrac{1}{4}\alpha \nu^{2}\varepsilon$. Accordingly,

$$\begin{aligned} F\bigl(p(t)\bigr)\leq \bigl( 1+ \bigl( \alpha -\sqrt{1+\alpha^{2}} \bigr) ^{2} \bigr) \nu^{2}+\tfrac{1}{4}\alpha \nu^{2}\varepsilon t. \end{aligned}$$

Hence, for $t_{1}=\frac{-68+72 \alpha ( -\alpha +\sqrt{1+ \alpha^{2}} ) }{9 \alpha \varepsilon }>0$, we get $\sqrt{F(p(t _{1}))}\leq \tfrac{1}{3}|\nu |<\tfrac{1}{2}|\nu |$.