= Solution
Set $y=\tanh z$ and $a=1-k$, so $dy/dz=1-y^2$ and $\widehat w=(1-y^2)^{k/2}y^a$. Away from $y=0$, the <logarithmic derivative> gives
$$
\frac{\widehat w'}{\widehat w}=\frac ay-y,\qquad
\frac{\widehat w''}{\widehat w}=\frac{a(a-1)}{y^2}-a-1+2y^2.
$$
For $c=0$, $\overline U''/\overline U=-2(1-y^2)$ and $N^2/\overline U^2=J(1-y^2)/y^2$. The <Taylor–Goldstein equation> residual divided by $\widehat w$ consequently simplifies to
$$
[J-k(1-k)]\left(\frac1{y^2}-1\right).
$$
Thus the <neutral mode of the equal-width Hazel model> requires
$$
\boxed{J=k(1-k).}
$$
The qualifier “solution” needs care at the <critical level of an internal gravity wave> $z=0$. For $0<k<1$, $\widehat w\sim z^{1-k}$ and $\widehat w'\sim(1-k)z^{-k}$: it is a solution separately on either side, not a classical smooth <eigenfunction> across the <critical level of an internal gravity wave>. A branch for negative $\tanh z$ is also needed. For example, a limit from $c_i>0$ assigns $(\tanh z)^{1-k}=|\tanh z|^{1-k}e^{-i\pi(1-k)}$ on $z<0$.
The <critical-layer regularity of a neutral Hazel mode> further gives local finite horizontal <kinetic energy> only for $0<k<1/2$, since <incompressible flow> gives $\widehat u=i\widehat w'/k$ and $\int_0^\epsilon z^{-2k}dz$ then converges. At $k=1$, $J=0$ and $\widehat w=\operatorname{sech}z$ is smooth after the removable $\overline U''/\overline U$ quotient is continued. At $k=0$, $J=0$ and the formal $\tanh z$ profile does not decay at infinity, so it fails the remote endpoint condition. These qualifications prevent interpreting the whole closed parameter interval as a family of classical decaying modes.
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