Choose $a=1$. The <power-transformed Taylor–Goldstein energy identity> reduces to
$$
\int_{-L}^L(U-c)^2(|q'|^2+k^2|q|^2)dz=\int_{-L}^LN^2|q|^2dz.
$$
The right side is real. For an unstable mode, $c_i>0$, so taking the <imaginary part> gives
$$
\int_{-L}^L(U-c_r)(|q'|^2+k^2|q|^2)dz=0.
$$
Consequently the <phase velocity> is a weighted average:
$$
\boxed{c_r=\frac{\int_{-L}^LU(|q'|^2+k^2|q|^2)dz}{\int_{-L}^L(|q'|^2+k^2|q|^2)dz},\qquad U_{\min}<c_r<U_{\max}.}
$$
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>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 331 / 1 / b / iii / Solution

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> gives
$$
c_r=\frac{\int U(|q'|^2+k^2|q|^2)dz}{\int(|q'|^2+k^2|q|^2)dz}.
$$
Thus 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>.

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