})+B_{\alpha\lambda}({\bf p })v_{\lambda\gamma}({\bf p})\right]=\omega_{{\bf p}\gamma}u_{\alpha\gamma}({\bf p}), \nonumber \\
\sum_{\lambda=x,y}\left[A^{*}_{\alpha\lambda}({\bf p })v_{\lambda\gamma}({\bf p })+B^{*}_{\alpha\lambda}({\bf p })u_{\lambda\gamma}({\bf p})\right]= -\omega_{{\bf p}\gamma}v_{\alpha\gamma}({\bf p}). \label{eq:4.17}\end{gathered}$$ This system of homogeneous liner equations has non-zero solution when the corresponding determinant turns to zero. Therefore, taking into account Eqs. (\[eq:4.11\]), one finds the equation for $\omega_{{\bf p}x}$: $$(A^{2}-4B^{2}-2A{\cal A}+{\cal A}^{2}-\omega_{x}^{2})((A+{\cal A})^{2}-\omega_{x}^{2})=0,$$ Next, taking into account Eqs. (\[eq:4.12\]), one obtains two different excitation spectra, $$\label{eq:4.18}
\omega_{{\bf p}x}^{(\rm II)}=\varepsilon_{\bf p}+2h+2n_{0}(J({\bf p})-J(0))=\omega_{{\bf p}z}^{\rm (I)}+h$$ and $$\label{eq:4.19}
\omega_{{\bf p}x}^{({\rm III})}=\left[\varepsilon_{\bf p}^{2}+2\varepsilon_{\bf p}\left(n_{0}U({\bf p})+{4\over 3}n_{0}J({\bf p })\right)\right]^{1/2}.$$ Note that $\omega_{{\bf p}y}$ satisfies exactly the same equation as $\omega_{{\bf p}x}$ and, consequently, the corresponding spectra are identical or degenerate [@Ohmi]. Therefore, the ferromagnetic phase of a weakly interacting Bose gas with condensate is characterized by three types of excitations with the dispersion laws given by Eqs. (\[eq:4.13\]), (\[eq:4.18\]), (\[eq:4.19\]) and any of $\omega_{{\bf p}x}$ can be related to operators $a^{\dagger}_{{\bf p}y}$, $a_{{\bf p}y}$ in the Hamiltonian determined by Eq. (\[eq:4.15\]).

The spectrum given by Eq. (\[eq:4.19\]) is independent of magnetic field and represents the gapless Bogoliubov mode modified by the interaction of internal degrees of freedom. At small momenta, it represents the phonon excitations, $$\omega_{{\bf p}x}^{\rm (II)}\approx cp, \quad c=\left[{n_{0}\over m}\left(U(0)+{4\over 3}J(0)\right)\right]^{1/2},$$ where $c$ 