Barotropic Instability
A necessary condition for instability is that $d^{2}U/dy^{2}$ must change sign somewhere in the flow. This condition is called Rayleigh's inflection point criterion. In terms of mean flow vorticity, $\bar{\zeta} = - dU/dy$, the criterion states that $d\bar{\zeta}/dy$ must change sign somewhere in the flow. That analysis is extended here to a rotating earth to find that the criterion requires that $d(\bar{\zeta} + f)/dy$ must change sign somewhere within the flow
Consider a horizontal wind profile or current $U(y)$ in a medium of uniform density. In the absence of horizontal density gradients only the barotropic mode is allowed, and $U(y)$ does not vary with depth. The vorticity equation is $$ \left( \frac{\partial}{\partial t} + \mathbf{u} \cdot \nabla \right) (\zeta + f) = 0 $$ Let the total flow be decomposed into a background flow plus a disturbance $$ u = U(y) + u', \qquad v = v' $$ $$ \Rightarrow \zeta = \bar{\zeta} + \zeta' = -\frac{dU}{dy} + \underbrace{\left( \frac{\partial v'}{\partial x} - \frac{\partial u'}{\partial y} \right)}_{u' = \partial \psi / \partial y, v' = - \partial \psi / \partial x} = -\frac{dU}{dy} - \underbrace{\nabla^{2}\psi}_{\text{$\psi$ is the stream function for the disturbance}} $$
Substituting these relationships into \(\boxed{\left( \frac{\partial}{\partial t} + \mathbf{u} \cdot \nabla \right) (\zeta + f) = 0 }\) and linearizing \(\rightarrow\) the perturbation vorticity equation $$ \frac{\partial}{\partial t} (\nabla^{2}\psi) + U \frac{\partial}{\partial x} (\nabla^{2}\psi) + \left( \beta - \frac{d^{2}U}{dy^{2}} \right) \frac{\partial \psi}{\partial x} = 0 $$
Because the coefficients of the perturbation vorticity equation are independent of $x$ and $t$, its solutions can be of the form $$ \psi = \hat{\psi}(y) \exp \{ i k (x - ct) \} $$ The phase speed $c = c_{r} + i c_{i}$ may be complex and solutions are unstable when $c_{i} > 0$
The perturbation vorticity equation then becomes $$ (U - c)\left[\frac{d^{2}}{dy^{2}} - k^{2}\right] \hat{\psi} + \left[\beta - \frac{d^{2}U}{dy^{2}}\right] \hat{\psi} = 0 $$