= Solution
For an <Equatorial Kelvin wave>, set $\widehat v=0$. The zonal and meridional momentum equations give
$$
\widehat u=\frac{k}{\omega}\widehat\phi,
\qquad
\frac{d\widehat\phi}{dy}
=-\frac{\beta k}{\omega}y\widehat\phi,
$$
so
$$
\widehat\phi=\phi_0
\exp\left(-\frac{\beta k}{2\omega}y^2\right).
$$
Hydrostatic balance and buoyancy evolution give $\widehat\sigma=im\widehat\phi$ and $\widehat w=-\omega m\widehat\phi/N^2$. Continuity then gives
$$
\omega^2=\frac{N^2k^2}{m^2}.
$$
Because the mode must decay as $|y|\to\infty$, one needs $k/\omega>0$. With the stipulated $\omega>0$, the acceptable branch is therefore
$$
\boxed{\omega=\frac{Nk}{|m|},\qquad k>0.}
$$
The branch $\omega=-Nk/|m|$ would require $k<0$ when $\omega>0$, making the Gaussian exponent positive and the mode unbounded; it is not an equatorially trapped solution.
For real $k,m,\omega$,
$$
\overline{u'w'}
=\frac12\operatorname{Re}(\widehat u\widehat w^*)
=\boxed{-\frac12\frac{km}{N^2}|\widehat\phi|^2.}
$$
Upward group propagation has $m<0$, while a Kelvin wave has $k>0$, so $\overline{u'w'}>0$. Dissipation makes this eastward momentum flux decrease with height; therefore $-\partial_z\overline{u'w'}>0$ and the wave exerts an eastward force on the mean flow.
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