The concept of stability is an important issue for any differential equation. The nonlinear stability of the equilibrium point of a dynamical system can be studied using the tools of mechanical geometry, so this is another good reason to find a Hamilton -Poisson realization. For more details, see [14]. We start this section with a short review of the most important notions.

#### Definition

An equilibrium state *x*_{e} is said to be **nonlinear stable** if for each neighborhood *U* of *x*_{e} in *D* there is a neighborhood *V* of *x*_{e} in *U* such that trajectory *x*(*t*) initially in *V* never leaves *U*.

This definition supposes well-defined dynamics and a specified topology. In terms of a norm ‖ ‖, nonlinear stability means that for each *ε* > 0 there is *δ* > 0 such that if
$$\begin{array}{}\u2225x(0)-{x}_{e}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u2225<\delta \end{array}$$

then
$$\begin{array}{}\u2225x(t)-{x}_{e}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u2225<\epsilon ,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\forall})\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t>0.\end{array}$$

It is clear that nonlinear stability implies spectral stability; the converse is not always true.

The equilibrium states of the dynamics Eq. (2) are
$$\begin{array}{}{e}_{1}^{M}=(M,0,0),\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{e}_{2}=\left(-\frac{(4-Ha)\alpha}{2Rey},0,0,\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}M\in \mathbf{R}.\end{array}$$

#### Proposition 1

The equilibrium states $\begin{array}{}{e}_{1}^{M}\end{array}$ = (*M*,0,0) are nonlinearly stable for any *M* ∈ **R**^{*}.

#### Proof

We will use energy-Casimir method, see [14] for details. Let
$$\begin{array}{}{\displaystyle \phantom{\rule{2em}{0ex}}{F}_{\phi}({f}_{1},{f}_{2},{f}_{3})=H({f}_{1},{f}_{2},{f}_{3})+\phi [C({f}_{1},{f}_{2},{f}_{3})]=}\\ \\ =\frac{1}{2}{f}_{2}^{2}-\frac{2}{3}\alpha Rey{f}_{1}^{3}-\frac{4-Ha}{2}{\alpha}^{2}{f}_{1}^{2}-{f}_{1}{f}_{3}-{f}_{3}-\alpha Rey{f}_{1}^{2}-\\ \\ \phantom{\rule{2em}{0ex}}-(4-Ha){\alpha}^{2}{f}_{1}+\phi ({f}_{3}+\alpha Rey{f}_{1}^{2}+(4-Ha){\alpha}^{2}{f}_{1})\end{array}$$

be the energy-Casimir function, where *φ* : **R** → **R** is a smooth real valued function.

Now, the first variation of *F*_{φ} is given by
$$\begin{array}{}{\displaystyle \delta {F}_{\phi}({f}_{1},{f}_{2},{f}_{3})={f}_{2}\delta {f}_{2}-{f}_{3}\delta {f}_{1}+[-{f}_{1}-1+}\\ \phantom{\rule{2em}{0ex}}+\dot{\phi}({f}_{3}+\alpha Rey{f}_{1}^{2}+(4-Ha){\alpha}^{2}{f}_{1})]\times \\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\times \left(2\alpha Rey{f}_{1}\delta {f}_{1}+(4-Ha){\alpha}^{2}\delta {f}_{1}+\delta {f}_{3}\right)\end{array}$$

so we obtain
$$\begin{array}{}{\displaystyle \delta {F}_{\phi}({e}_{1}^{M})=\left[-M-1+\dot{\phi}(\alpha Rey{M}^{2}+(4-Ha){\alpha}^{2}M)\right]\times}\\ \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\times \left(2\alpha ReyM\delta {f}_{1}+(4-Ha){\alpha}^{2}\delta {f}_{1}+\delta {f}_{3}\right)\end{array}$$

that is equals zero for any *M* ∈ **R**^{*} if and only if
$$\begin{array}{}\dot{\phi}\left(\alpha Rey{M}^{2}+(4-Ha){\alpha}^{2}M\right)=M+1.\end{array}$$(3)

The second variation of *F*_{φ} at the equilibrium of interest is given by
$$\begin{array}{}{\displaystyle {\delta}^{2}{F}_{\phi}({e}_{1}^{M})=\left[(2\alpha ReyM+(4-Ha){\alpha}^{2}{)}^{2}\cdot \ddot{\phi}(\alpha Rey{M}^{2}+\right.}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left.+(4-Ha){\alpha}^{2}M)-(2\alpha ReyM+(4-Ha){\alpha}^{2})\right]\cdot (\delta {f}_{1}{)}^{2}+\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+(\delta {f}_{2}{)}^{2}-2\left[1-(2\alpha ReyM+(4-Ha){\alpha}^{2})\times \right.\\ \phantom{\rule{1em}{0ex}}\left.\times \ddot{\phi}(\alpha Rey{M}^{2}+(4-Ha){\alpha}^{2}M)\right]\cdot \delta {f}_{1}\cdot \delta {f}_{3}+\\ \phantom{\rule{2em}{0ex}}+\ddot{\phi}(\alpha Rey{M}^{2}+(4-Ha){\alpha}^{2}M)\cdot (\delta {f}_{3}{)}^{2}.\end{array}$$

If we choose now *φ* such that the relation (3) is valid and
$$\begin{array}{}{\displaystyle (2\alpha ReyM+(4-Ha){\alpha}^{2})\cdot \ddot{\phi}(\alpha Rey{M}^{2}+(4-Ha){\alpha}^{2}M)-1>0,}\end{array}$$

then the second variation of *F*_{φ} at the equilibrium of interest is positive defined and so our equilibrium states $\begin{array}{}{e}_{1}^{M}\end{array}$ are nonlinearly stable.

#### Proposition 2

The equilibrium state *e*_{2} = $\begin{array}{}\left(-\frac{(4-Ha)\alpha}{2Rey},0,0,\right)\end{array}$ is nonlinearly stable for 0 ≤ *Ha* < 4.

#### Proof

We will use energy-Casimir method, see [14] for details. Let
$$\begin{array}{}{\displaystyle \phantom{\rule{2em}{0ex}}{F}_{\phi}({f}_{1},{f}_{2},{f}_{3})=H({f}_{1},{f}_{2},{f}_{3})+\phi [C({f}_{1},{f}_{2},{f}_{3})]=}\\ =\frac{1}{2}{f}_{2}^{2}-\frac{2}{3}\alpha Rey{f}_{1}^{3}-\frac{4-Ha}{2}{\alpha}^{2}{f}_{1}^{2}-{f}_{1}{f}_{3}-{f}_{3}-\alpha Rey{f}_{1}^{2}-\\ \phantom{\rule{2em}{0ex}}-(4-Ha){\alpha}^{2}{f}_{1}+\phi ({f}_{3}+\alpha Rey{f}_{1}^{2}+(4-Ha){\alpha}^{2}{f}_{1})\end{array}$$

be the energy-Casimir function, where *φ* : **R** → **R** is a smooth real valued function.

Now, the first variation of *F*_{φ} is given by
$$\begin{array}{}{\displaystyle \delta {F}_{\phi}({f}_{1},{f}_{2},{f}_{3})={f}_{2}\delta {f}_{2}-{f}_{3}\delta {f}_{1}+\left[-{f}_{1}-1+\right.}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left.+\dot{\phi}({f}_{3}+\alpha Rey{f}_{1}^{2}+(4-Ha){\alpha}^{2}{f}_{1})\right]\times \\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\times \left(2\alpha Rey{f}_{1}\delta {f}_{1}+(4-Ha){\alpha}^{2}\delta {f}_{1}+\delta {f}_{3}\right)\end{array}$$

so we obtain
$$\begin{array}{}{\displaystyle \delta {F}_{\phi}({e}_{2})=\left[\frac{(4-Ha){\alpha}^{2}}{2Rey}-1+\dot{\phi}\left(-\frac{(4-Ha{)}^{2}{\alpha}^{4}}{2Rey}\times \right.\right.}\\ \phantom{\rule{1em}{0ex}}\left.\left.\times (2-\alpha )\right)\right]\cdot \left((1-\alpha ){\alpha}^{2}(4-Ha)\delta {f}_{1}+\delta {f}_{3}\right)\end{array}$$

that is equals zero if and only if
$$\begin{array}{}{\displaystyle \dot{\phi}\left(-\frac{(4-Ha{)}^{2}{\alpha}^{4}}{2Rey}\cdot (2-\alpha )\right)=1-\frac{(4-Ha){\alpha}^{2}}{2Rey}.}\end{array}$$(4)

The second variation of *F*_{φ} at the equilibrium of interest is given by
$$\begin{array}{}{\displaystyle {\delta}^{2}{F}_{\phi}({e}_{2})=\left[(1-\alpha {)}^{2}{\alpha}^{4}(4-Ha{)}^{2}\cdot \ddot{\phi}\left(-\frac{(4-Ha{)}^{2}{\alpha}^{4}}{2Rey}\times \right.\right.}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left.\left.\times (2-\alpha )\right)+(\alpha -1){\alpha}^{2}(4-Ha)\right]\cdot (\delta {f}_{1}{)}^{2}+(\delta {f}_{2}{)}^{2}+\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+2\left[-1-(\alpha -1){\alpha}^{2}(4-Ha)\times \right.\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\left.\times \ddot{\phi}\left(-\frac{(4-Ha{)}^{2}{\alpha}^{4}}{2Rey}\cdot (2-\alpha )\right)\right]\cdot \delta {f}_{1}\cdot \delta {f}_{3}+\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}+\ddot{\phi}\left(-\frac{(4-Ha{)}^{2}{\alpha}^{4}}{2Rey}\cdot (2-\alpha )\right)\cdot (\delta {f}_{3}{)}^{2}.\end{array}$$

If we choose now *φ* such that the relation (4) is valid and
$$\begin{array}{}{\displaystyle (1-\alpha ){\alpha}^{2}(4-Ha)\cdot \ddot{\phi}(-\frac{(4-Ha{)}^{2}{\alpha}^{4}}{2Rey}\cdot (2-\alpha ))-1<0}\end{array}$$

for 4 − Ha > 0, then the second variation of *F*_{φ} at the equilibrium of interest is positive defined and so our equilibrium state *e*_{2} is nonlinearly stable. □

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