In quasi-geostrophic frontogenesis, what happens if the potential vorticity is negative?

Question

In particular, what happens if the Sawyer–Eliassen equation cannot be solved? I know that a condition for the Sawyer–Eliassen equation to have a solution is $F^2 N^2 - S^4 >0$, which means that the potential vorticity is positive.

I was wondering what is the physical implication of this condition on frontogenesis? Are there situations where fronts cannot be created, when potential vorticity is negative? Does that occur somewhere?

A secondary question would be the link with symmetric instability, which requires if I am not mistaken that the absolute vorticity (which is more or less potential vorticity) must be negative (in the northern hemisphere). I do not understand, then, how symmetric instability can arise when the Sawyer–Eliassen equation can be solved. Instability should occur in fronts since it is the driving phenomenon for the creation of precipitation bands!

Answer 1

Mathmatically, the Sawyer–Eliassen equation is an second-order partical differential equation of the form: $$\frac{\partial}{\partial y}\left(A\frac{\partial \psi}{\partial y}+B\frac{\partial\psi}{\partial z}\right)+\frac{\partial}{\partial z}\left(B\frac{\partial\psi}{\partial y}+C\frac{\partial\psi}{\partial z}\right)=F$$

When the condition $AC-B^2>0$ (equivalent to yours) is met, it is elliptic equation that can be solved numerically. Physically, the solution is an asymptotic steady-state response to the (QG) forcing $F$ you identified. That is, the frontal structure will evolve towards a steady state in balance to $F$. If the condition is not met, SE equation cannot be solved numerically (iteration does not converge). This means that any small perturbation of fluid parcel will accelerate its departure from its initial location.

It is not too hard to prove that the elliptic condition $AC-B^2>0$ is equivalent to $fP>0$, where $P$ is potential vorticity and $f$ the Coriolis parameter. The latter means symmetric stable (PV and f have the same sign). So symmetric stable means we have steady-state solution to the Sawyer–Eliassen equation. Instability means we cannot get the numerical solution iteratively as it diverges, which could be understood as a free convection rather than a QG forced upward motion.