Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 331 / 2 / b

Set $y=\tanh z$ and $a=1-k$, so $dy/dz=1-y^2$ and $\hat{w}=(1-y^2{)}^{k/2}{y}^a$. Away from $y=0$, the logarithmic derivative givesFor $c=0$, ${\overline{U}}^{\prime \prime }/\overline{U}=-2(1-y^2)$ and $N^2/{\overline{U}}^2=J(1-y^2)/y^2$. The Taylor–Goldstein equation residual divided by $\hat{w}$ consequently simplifies toThus the neutral mode of the equal-width Hazel model requiresThe qualifier “solution” needs care at the critical level of an internal gravity wave $z=0$. For $0<k<1$, $\hat{w}\sim z^{1-k}$ and ${\hat{w}}^{\prime }\sim (1-k)z^{-k}$: 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 $\hat{u}=i{\hat{w}}^{\prime }/k$ and ${\int }_0^{ϵ}{z}^{-2k}dz$ then converges. At $k=1$, $J=0$ and $\hat{w}=\mathrm{sech}z$ is smooth after the removable ${\overline{U}}^{\prime \prime }/\overline{U}$ 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.

For the equal-width Hazel model with stable density stratification $J\ge 0$,The Miles–Howard theorem therefore excludes exponentially growing regular normal modes for $J\ge 1/4$. The candidate neutral modes of the equal-width Hazel model instead lie onThe parabola touches the sufficient-stability threshold at its maximum. There is no contradiction: neutrality has $c_i=0$, whereas the proof in part (a) assumes $c_i>0$. Moreover, the $k=1/2$ profile has a singular critical-layer derivative and logarithmically divergent horizontal kinetic energy; it is not a regular growing mode satisfying the proof's hypotheses.

For $J<1/4$, the local gradient Richardson number condition permits instability but does not establish it. Substitution of a neutral ansatz alone also does not determine on which side of the curve unstable eigenvalues lie. It identifies the formal neutral curve; concluding a full stability boundary requires additional continuation analysis of the eigenvalues. The nondecaying $k=0$ endpoint and the smooth $k=1,J=0$ endpoint have the distinct qualifications described above.