= Solution
Substitution of $e^{i(kx-\omega t)}$ gives
$$
\boxed{\omega(k)=-\frac{\beta k}{k^2+\ell^2},\qquad c(k)=\frac\omega k=-\frac\beta{k^2+\ell^2},\qquad
c_g(k)=\omega'(k)=\frac{\beta(k^2-\ell^2)}{(k^2+\ell^2)^2}.}
$$
The <phase velocity> is negative for all nonzero <wavenumbers>, so crests travel left. The <group velocity> is negative for $|k|<\ell$, zero at $|k|=\ell$, and positive for $|k|>\ell$. Its minimum is $-\beta/\ell^2$ at zero and maximum $\beta/(8\ell^2)$ at $|k|=\sqrt3\ell$; it tends to zero from above at infinity. The odd <dispersion relation> has extrema at $k=\pm\ell$, with $\omega(\ell)=-\beta/(2\ell)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-4-rossby.png]
{title=Rossby dispersion, phase velocity and group velocity in dimensionless variables}
{height=360}
The Fourier solution is $\varphi(x,t)=\int_{\mathbb R}A(k)e^{i[kx-\omega(k)t]}\,dk$. The stationary-phase statements require enough amplitude regularity and decay, for example $A$ smooth and rapidly decreasing; realness and evenness alone do not guarantee a convergent integral or a stationary-phase expansion. For $x=Vt$, set $\Phi_V(k)=kV-\omega(k)$ and solve $\omega'(k)=V$ implicitly.
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