= Solution
Let $V=U-c$ and suppose a growing <normal mode> has $\operatorname{Im}c=c_i>0$, taking $\alpha>0$ without loss of generality. Since $V$ never vanishes on the real channel, choose a continuous square-root branch and write $\phi=V^{1/2}H$. Direct differentiation gives
$$
\phi'=V^{1/2}H'+\frac{U'}{2V^{1/2}}H,
$$
$$
\phi''=V^{1/2}H''+\frac{U'}{V^{1/2}}H'+\left(\frac{U''}{2V^{1/2}}-\frac{U'^2}{4V^{3/2}}\right)H.
$$
Substitution into the <Taylor–Goldstein equation> and division by $V^{1/2}$ gives
$$
VH''+U'H'-\alpha^2VH-\frac12U''H+\frac{J-U'^2/4}{V}H=0.
$$
Thus the required transformation is
$$
\boxed{D(V DH)-\left(\alpha^2V+\frac12U''+\frac{U'^2/4-J}{V}\right)H=0.}
$$
The impermeable walls give $\phi(\pm1)=0$, hence $H(\pm1)=0$. Multiply by $\overline H$, integrate across the channel, and use <integration by parts>. With the boundary term zero, the resulting <power-transformed Taylor–Goldstein energy identity> is
$$
\int_{-1}^1\left[V\left(|H'|^2+\alpha^2|H|^2\right)+\frac12U''|H|^2+\frac{U'^2/4-J}{V}|H|^2\right]dy=0.
$$
Both $U''$ and $J$ are real. Since $\operatorname{Im}V=-c_i$ and $\operatorname{Im}(1/V)=c_i/|V|^2$, its <imaginary part>, divided by $-c_i$, is
$$
\int_{-1}^1\left[|H'|^2+\alpha^2|H|^2+\frac{J-U'^2/4}{|U-c|^2}|H|^2\right]dy=0.
$$
If $J\ge U'^2/4$ everywhere, the integrand is nonnegative and the first two terms have positive integral for any nonzero mode. This contradicts the identity. Therefore instability requires
$$
\boxed{J(y)<\frac14U'(y)^2\quad\text{somewhere in the channel}.}
$$
This is the necessary instability condition of the <Miles–Howard theorem>. Where $U'\ne0$, it says that the <gradient Richardson number> $J/U'^2$ must be below $1/4$ somewhere. It is not a sufficient condition for instability, and the conclusion concerns exponential normal-mode growth, not exclusion of the algebraic <lift-up effect> in part (a). The algebra uses the given nondimensional $J$; in dimensional notation its counterpart is the squared <buoyancy frequency>.
Back to article page
