= Solution
Let $R=\sqrt{J^2+D^2}$ and $k_0=\arctan(D/J)$. Since $J\cos k+D\sin k=R\cos(k-k_0)$, the continuum minimum of the <dispersion relation> is
$$
\Delta=2S[J(1+u)-R],\qquad \boxed{u_c=\sqrt{1+(D/J)^2}-1.}
$$
For $D>0$, the <energy gap> closes at a positive $u_c$, at <wave vector> $k_0$. Below it, the quadratic expansion about the polarized state has a negative mode and is no longer a stable <ground state> approximation; higher-order terms and a different ordered reference state are needed.
Two qualifications are required by the actual assumptions. First, $D=0$ is permitted, in which case $u_c=0$; it is a concrete counterexample to the printed strict inequality $u_c>0$. Second, a finite ring only has discrete <momenta>. Its <finite-ring spin-wave threshold with antisymmetric exchange> is
$$
\boxed{u_c(L)=\max_{\ell=0,\ldots,L-1}\left[\cos\frac{2\pi\ell}{L}+\frac DJ\sin\frac{2\pi\ell}{L}\right]-1.}
$$
This approaches $u_c$ in the <thermodynamic limit> and equals it when $k_0$ is allowed. For example, $L=4$ and $0<D<J$ give $u_c(L)=0$, since the maximum of $1,D/J,-1,-D/J$ is one. Thus the continuum formula and its positive threshold require the large-ring interpretation. The two symmetry-related polarized <ground states> already have equal exact <energy>; the calculated <energy gap> refers to excitations about one chosen orientation.
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