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A smooth inviscid stratified parallel flow with gradient Richardson number at least $1/4$ everywhere has no exponentially growing two-dimensional normal modes. Put $a=1/2$ in the power-transformed Taylor–Goldstein energy identity and take its imaginary part:The integral is positive for a nonzero mode when the numerator is nonnegative, forcing $c_i=0$. This is a modal stability theorem, not a prohibition on transient growth. Maslowe's review discusses the theorem and the role of critical layers.

Choose $a=1$. The power-transformed Taylor–Goldstein energy identity reduces toThe right side is real. For an unstable mode, $c_i>0$, so taking the imaginary part givesConsequently the phase velocity is a weighted average:The weights are nonnegative and not identically zero. For a nonconstant shear profile, equality at an extremum would force $q$ to vanish on an interval where $U$ differs from that extremum; uniqueness for the regular ordinary differential equation would then give the zero mode. This is the phase-speed bound for unstable stratified shear modes.

Choose $a=1/2$ in the power-transformed Taylor–Goldstein energy identity and let $Q=N^2-(U^{\prime }{)}^2/4$. The result isBecause $\mathrm{Im}V=-c_i$ and $\mathrm{Im}(1/V)=c_i/|V{|}^2$, its imaginary part givesIf $4N^2>(U^{\prime }{)}^2$ everywhere, the integral is strictly positive for every nonzero eigenfunction, so a mode with $c_i\ne 0$ is impossible. Thus the flow is neutrally stable to these inviscid normal modes. This is the Miles–Howard theorem, expressed as a lower bound $\mathrm{Ri}>1/4$ on the gradient Richardson number. It excludes exponential modal growth; it does not by itself exclude transient growth.

For a possibly unstable normal mode, $c_i\ne 0$ ensures $V=U-c$ never vanishes, so choose one continuous branch of $V^a$. Substitute $\hat{w}=V^{a}q$ into the Taylor–Goldstein equation and collect terms:Multiplication by $V^{2a}$ puts its first two terms in divergence form:Multiply by the complex conjugate $\overline{q}$ and apply integration by parts. Since $q(\pm L)=0$, the boundary term vanishes and the power-transformed Taylor–Goldstein energy identity isThe weight is $V^{2a}$, not $|V{|}^{2a}$: preserving its complex phase is essential for the subsequent stability proofs.

For an unstable Taylor–Goldstein equation mode in a finite channel, take $a=1$ in the power-transformed Taylor–Goldstein energy identity. Its imaginary part givesThus the real phase velocity lies strictly between the extremes of a nonconstant smooth shear profile. Equality would force the regular eigenfunction to vanish on an interval, hence everywhere by uniqueness for its ordinary differential equation.