 Solution

Let $D=\overline{U}-c$ and $F=\hat{w}/D^{1/2}$. For $c_i>0$, $D$ never vanishes, so an analytic branch of a square root exists along the real flow domain. Take real smooth velocity and buoyancy frequency coefficients, nonzero $k$ (chosen positive for the growing-wave convention), a nontrivial regular normal mode, and boundary decay strong enough to remove the integration-by-parts term.

Substitution into the Taylor–Goldstein equation gives the half-power case of the power-transformed Taylor–Goldstein energy identity:For completeness, differentiating $\hat{w}=D^{1/2}F$ yields ${\hat{w}}^{\prime \prime }=D^{1/2}{F}^{\prime \prime }+{\overline{U}}^{\prime }{D}^{-1/2}{F}^{\prime }+[\frac{1}{2}{\overline{U}}^{\prime \prime }{D}^{-1/2}-\frac{1}{4}({\overline{U}}^{\prime }{)}^2{D}^{-3/2}]F$, which explains the coefficient $1/4$.

Multiply by $F^{\ast }$ and integrate over $z$. The boundary term $[F^{\ast }D{F}^{\prime }]$ is zero for the given homogeneous endpoint conditions, or for sufficiently decaying finite-energy modes at infinity. ThereforeSince $\mathrm{Im}D=-c_i$, $\mathrm{Im}(1/D)=c_i/|D{|}^2$, and $S,{\overline{U}}^{\prime \prime }$ are real, taking the imaginary part and dividing by $-c_i$ givesThe first two terms have strictly positive integral for a nonzero mode. Thus $S$ cannot be nonnegative everywhere:This proves the Miles–Howard theorem in contrapositive form. Where ${\overline{U}}^{\prime }\ne 0$, the corresponding gradient Richardson number must fall below $1/4$ somewhere. Failure of this sufficient-stability criterion does not prove instability. The derivation uses $c_i>0$; it cannot be applied unchanged to a neutral singular critical layer.