in the neighborhood of the following equilibria $$\label{eq4_5}
   \alpha_1 = a, \quad \alpha_2=\alpha_3=0, \quad \beta_1=\beta_3=0, \quad \beta_2 = \pm b,\quad M_1=M_2=M_3=0.$$

In variables we have $K=\xi_1 \xi_2$, where $$\label{eq4_6}
  \begin{array}{l}
    \xi_1=w_1^2 + {\varepsilon}_0 x_1 - \rmi {\varepsilon}_1 (x_1 w_3 - 2 z_1 w_1) - {\varepsilon}_1 ^2 r^2, \\
    \xi_2=w_2^2 + {\varepsilon}_0 x_2 + \rmi {\varepsilon}_1 (x_2 w_3 - 2 z_2 w_2) - {\varepsilon}_1 ^2 r^2.
  \end{array}$$ The system is equivalent to $$\label{eq4_7}
  \xi_1=0, \qquad \xi_2=0.$$ To find possible singularities of ${\mathcal{M}}_1$ we apply equations to the function with undefined multipliers $f=\mu_1 \xi_1+\mu_2 \xi_2$ ($\mu_2=\overline{\mu_1}$, $\mu_1 \mu_2 \ne 0$). The only solution is $$w_1=w_2=w_3=0, \quad x_1=x_2=\frac{{\varepsilon}_1 ^2 r^2}{{\varepsilon}_0 }, \quad y_1=y_2=\frac{{\varepsilon}_0 }{{\varepsilon}_1 ^2}, \quad z_1=z_2=0,$$ and equations reduce to . According to the choice of the sign in we come to the corresponding point .

It follows from that the Poisson bracket of $Z_1$ and $Z_2$ is a partial integral on ${\mathcal{M}}_1$. In the classical case such integral was found by D.N.Goryachev, and in the Bogoyavlensky case it has the form [@Bogo] $$F_{\rm B}= \frac{1}{4} (M_1^2 + M_2^2) M_ 3+ \alpha_3 M_1 + \beta_3 M_2.$$

Recall one general fact from the symplectic geometry.

\[lem1\] Suppose a submanifold ${\mathcal{M}}$ of a symplectic 