= Solution
Choose the unitary <discrete Fourier transform>
$$
a_j=\frac1{\sqrt L}\sum_ke^{ikj}b_k,\qquad b_k=\frac1{\sqrt L}\sum_je^{-ikj}a_j,\qquad k=\frac{2\pi\ell}{L},\quad \ell=0,\ldots,L-1.
$$
The finite-sum orthogonality identity gives $[b_k,b_q^\dagger]=\delta_{kq}$. It also gives $\sum_ja_{j+1}^\dagger a_j=\sum_ke^{-ik}b_k^\dagger b_k$. Substituting $A,B$ from <2/c/solution> therefore gives the <dispersion relation>
$$
\boxed{H_{\rm LSW}=\sum_k\epsilon(k)b_k^\dagger b_k,\qquad\epsilon(k)=2S[J(1+u)-J\cos k-D\sin k].}
$$
Using the opposite sign in the <Fourier transform> reverses $k$ and the sine term together, without changing the spectrum.
For $u\gg1$ every $\epsilon(k)>0$, so the <ground state> of this quadratic <Hamiltonian> is the bosonic vacuum, $b_k|{\rm GS}\rangle=0$ for every $k$. In the original <spin> variables it is the fully $+z$ polarized product state. Its <energy> is zero after the constant subtraction. Repeating the construction about the fully $-z$ polarized state describes the other easy-axis sector of the original <Hamiltonian>.
A one-<magnon> excitation $b_k^\dagger|{\rm GS}\rangle$ has <energy> $\epsilon(k)$ above the reference <ground state>. A normalized many-<magnon> <Fock state> with occupations $n_k$ has excitation <energy> $\sum_kn_k\epsilon(k)$, within the dilute regime of the <linear spin-wave approximation>. The lowest excitation in this sector has the minimum allowed $\epsilon(k)$.
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