= Solution
Put $q=\delta\theta$, $p=\delta\phi$, so $\theta=\pi/2-q$ and $\phi=p$. The <quadratic Lagrangian> is obtained by taking
$$
\mathcal L_2=sq\dot p-
\frac{\rho_s}{2}\left[(\nabla q)^2+(\nabla p)^2\right].
$$
The two real fluctuations form a coordinate and its <canonical momentum> rather than two independent modes. Their <Euler-Lagrange equations> combine to give
$$
\ddot q+\left(\frac{\rho_s}{s}\right)^2\nabla^4q=0,
$$
and similarly for $p$. The <ferromagnetic magnon> therefore has the quadratic <dispersion relation>
$$
\omega_k=\frac{\rho_s}{s}k^2.
$$
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