= Solution
Write $\theta=\pi/2-\vartheta$ and $\phi=\varphi$. To quadratic order,
$$
\cos\theta=\vartheta+O(\vartheta^3),
\qquad
\sin^2\theta=1+O(\vartheta^2).
$$
After dropping the total derivative $S\varphi_t$, the quadratic Lagrangian density is
$$
\boxed{
\mathcal L_2
=S\vartheta\varphi_t
-u(\vartheta_x^2+\varphi_x^2)
-K\vartheta^2}.
$$
Its Euler--Lagrange equations are
$$
S\varphi_t+2u\vartheta_{xx}-2K\vartheta=0,
\qquad
-S\vartheta_t+2u\varphi_{xx}=0.
$$
For modes proportional to $e^{i(kx-\omega t)}$, nontrivial amplitudes require
$$
\boxed{
\omega^2(k)
=\frac{4uk^2(uk^2+K)}{S^2}}.
$$
Thus
$$
\boxed{
\omega\sim\frac{2\sqrt{uK}}S|k|}
\quad(k\to0),
\qquad
\boxed{
\omega\sim\frac{2u}{S}k^2}
\quad(|k|\to\infty).
$$
At short wavelength the anisotropy is negligible and the quadratic dispersion is that of a conventional <ferromagnetic magnon>. At long wavelength, <easy-plane anisotropy> makes the out-of-plane fluctuation the conjugate density of the in-plane phase; eliminating it produces the linear Goldstone sound mode characteristic of a <superfluid>.
Solved by gpt-5.6-sol high.
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