1 No friction

The figure shows the cyclonic wind (azimuthal velocity, v) in the planetary boundary layer (the lowest kilometers) of a polar low. The wind varies from very strong close to the center, to zero at some radius and negative (anticyclonic) at larger radius. The variation is due to conservation of rotation (angular momentum, M). In the figure, M is controlled by setting the latitude and the radius of zero cyclonic wind.

Motivation:

We assume that the polar low is in a steady state and axis-symmetric. We treat the boundary layer as a vertically homogeneous slab. This means that there is no variation in time, nor in the vertical and the azimuthal directions. The only variation is in the radial direction. With our assumptions the momentum equation for the azimuthal direction becomes
(1) u(dv)/(dr) =  − u((v)/(r) + f) + Fa, 
where u is the velocity in the radial direction, r is the radial co-ordinate with zero at the cyclone centre, v is the velocity in the azimuthal direction, f is the Coriolis parameter and Fa is the friction in the azimuthal direction. For a frictionless flow Fa =  0. Dividing equation (1) with u we get
(2) (dv)/(dr) =  − ((v)/(r) + f), 
with the solution
(3) v = (M)/(r) − (f)/(2)r, 
where M is an integration constant. Note that we have no information about the radial velocity. We see that M is identical to the angular momentum defined as
(4) M ≡ rv + (f)/(2)r2.
In other words, equation (3) tells us that the variation of azimuthal velocity along the radius is caused by the conservation of angular momentum. The figure shows the azimuthal velocity calculated with equation (3). Since M is conserved, we may also define M as
(5) M ≡ (f)/(2)r20, 
where r0 is the radius of zero azimuthal velocity. In the figure, M is set by setting r0 and the latitude, yielding f.

2 Adding friction

In the frictionless case the cyclonic wind is unrelated to the radial wind. Adding friction changes that. The friction leads to a continous loss of angular momentum. To maintain a steady state there must be a compensating radial transport of angular momentum. We assume that the radial wind is negative (directed towards the center) and hence the gradient of angular momentum must be positive.
Conservation of volume requires a lower radial velocity at larger radius, which means longer time for the friction to act. This leads to a vannishing gradient of the cyclonic wind at large radius and the wind never becomes anticyclonic.

Motivation:

Let us assume that the frictional acceleration in the azimuthal direction can be calculated as
(6) Fa =  − (CD)/(h)Uv, 
where CD is a dimensionless drag coefficient, h is the height of the boundary layer and U = (u2 + v2) is the magnitude of the (horizontal) wind. With this parametrisation equation (1) becomes
(7) u(dv)/(dr) =  − u((v)/(r) + f) − (CD)/(h)Uv.
We want to divide equation (7) with u, but first we define a parameter, A,
(8) A ≡ (urh)/(CD).
A is the radial volumeflow, scaled with 2πρCD. Note that convergence towards the cyclone center corresponds to negative values of u and A. Let us assume that A is constant for all radii. Now, dividing equation (7) with u, using equation (8), yields
(9) (dv)/(dr) =  − (v)/(r) − f − (r)/(A)Uv.
The analytical solution for v of this equation eludes me. However, at large radius it is reasonable to assume that both the gradient of the wind and the curvature of the wind are small. The asymptote for v at large radii can hence be calculated from the balance of only the two last terms on the right hand side of equation (9). We further assume that the radial wind is small compared to the cyclonic wind (U = |v|). We get
(10) v = (f|A|)/(r)(1)/(2), 
where v should have the opposite sign to A. We can find v by using this asymptote and integrating equation (9) numerically. [A]  [A] The result at smaller radius is actually not very sensitive to the starting value of v. Test this by typing ’v’ in the figure and move the pointer for v at r =  500 km. Return to standard method by typing ’V’. At small radii we could guess that the friction term is little and that the solution is identical to equation (3). The angular momentum (M0) of the small radius asymptote appears to be [B]  [B] My conviction rest upon dimensional arguments and on numerical testing. I am sure there is a more rigid proof...
(11) M0 = (fA2)/(e1)(1)/(3), 
(12) A = fr30((e1)/(2)), 
where we have used equation (5). Note that the squareroot expression is approximately unity, so Afr30 is a good approximation. We find that in a boundary layer with friction the azimuthal velocity is determined by the scaled volumeflow, A. The figure shows the asymptotes together with the result of numerically integrating equation (9). The figure shows true values of u, using CD =  10-3and h =  2 km.

3 Adding convection

We have this far ignored the question of convection. Implicitly we have assumed that any radial massflow is compensated by convection (updraft) at the very center. However, a finite updraft velocity requires a finite updraft area. We will refer to this area as the eye-wall. The updraft in the eye-wall interupts the radial flow. The figure explores different assumptions about the updraft velocity and the position of the eye-wall. A common feature is that the convergence may reach close to the center, resulting in very high wind speed.
To increase the realism, we can let the uppdraft in the eye-wall be compensated by a downdraft on the outside. A radius of balanced convection (inside that radius) correspond with zero radial velocity (and hence zero radial volume transport) and also zero azimuthal velocity. This radius becomes the outer limit of the cyclone. However, introducing downdraft also introduces the question of the properties this downdraft brings to our boundary layer. Ignoring that question (which we will do for now) is the same as assuming zero vertical gradient.

Motivation:

If volume is to be conserved in the boundary layer, there must be a balance between the radial and vertical massflows. The radial flow A(r1) at radius r1 is thus the integrated vertical velocity from the center to that radius,
(13) A(r1) =  − r10rW dr, 
where W is the vertical velocity scaled by CD ⁄ ρ. The figure shows true vertical velocity, using CD =  10-3 and ρ =  1.0 kg/m3.
If all convection is localised to the eye-wall, then A =  0 inside the eye and A = A(rew) outside of the eye, where rew is the outer limit of the eye-wall. The figure explores the implication of moving the eye-wall. If the width of the eye-wall is constant, then a more central eye-wall yields a smaller eye-wall area. For a given A, a more central eye-wall thus requires either a higher mean uppdraft velocity or a wider eye-wall. In either case, the maximum updraft in the eye-wall will increase dramatically towards the center.
From equation (10) we realize that if A is constant, then the azimuthal velocity will tend to zero very slowly with larger radius. If we expect the azimuthal velocity to become zero at some finite radius, then A must be zero at that radius. From equation (13) we see that this means that the convection, integrated from the center to the radius of zero azimuthal velocity, must be zero. The figure shows a case where the updraft in the eye-wall is balanced by downdraft over the area outside of the eye-wall, with A decreasing linearly from its maximum at r = rew to zero at r =  500 km, which becomes the outer limit of the cyclone. We see that the azimuthal velocity decreaces from its maximum in the eye-wall to zero at r =  500 km.

4 Adding density (unfinished!!)

In the previous sections we have ignored the cause of the updraft. Now, let us assume that the cause of the updraft is that the air has become less dens than the ambient atmosphere. We assume that the density depend on the pressure and the temperature. We find the the pressure from the momentum equation for the radial velocity,
(14) u(du)/(dr) =  − (1)/(ρ)(dp)/(dr) + v((v)/(r) + f) + Fr.
(15) Fr =  − (CD)/(h)|u|u.
(16) cpu(dT)/(dr) = (u)/(ρ)(dp)/(dr) + J, 
(17) J =  − cp(CHE)/(h) U ΔT, 
(18) cp(dT)/(dr) =  − u(du)/(dr) + v((v)/(r) + f) − (r)/(A)|u|u2 − (r)/(A) U cpΔT.
(19) (dT)/(dr) = (1)/(cpρ)(dp)/(dr) − (r)/(Q) U ΔT.