The basic equations - Severe Convective Storms and Tornadoes: Observations and Dynamics - page 79

So,

p 0 ¼ 2

H 2 O U =ð U = Þ ð 2 : 257 Þ

In other words, the square root of the Taylor number represents the relative
importance of the acceleration due to the Coriolis force to viscous acceleration.
As in the case without rotation, we seek solutions of the form

ð 1 ð 1

e ! t e i ð k x þ k y Þ dk x dk y w ¼ W ð z Þ Re ð 2 : 258 Þ

1 1

The characteristic value equation for the Rayleigh problem including rotation is

! i Þþ k 2 d 2 dz 2 ! i þ k 2 d 2 dz 2

2

ð k 2 d 2 dz 2 ½ð! r þ i = ½! r þ i = = Þ W

! i Þþ k 2 d 2 dz 2 k 2 W þ T 0 ½ð! r þ i ! i Þþ k 2 d 2 dz 2 d 2 W dz 2 ¼ Ra ½ð! r þ i = = =

ð 2 : 259 Þ

As in the case without rotation, the simplest boundary conditions mathematically
are the free-slip boundary conditions. Six of the eight boundary conditions for the
free-slip, perfectly conducting case are, as for the non-rotating case, that W as
well as the second and fourth derivatives of W with respect to z all vanish at
z ¼ 0 and 1, at the top and bottom plates. We need two additional boundary
conditions.
To get them, let us consider the characteristic value equation (2.259) evaluated
when ! i and ! r are each zero (i.e., at the onset of instability when there are no

oscillations):

ð k 2 d 2 dz 2 Þ½ð k 2 d 2 dz 2

3 W Ra c k 2 W T 0 d 2 W

dz 2

= = Þ = ¼ 0 ð 2 : 260 Þ

Recall that at the onset of instability Ra ¼ Ra c , the critical Rayleigh number.
We integrate (2.260) twice with respect to z and find that

½ð k 2 d 2

dz 2

3

Ra c k 2 T 0 d 2

dz 2

= Þ = W þ F ¼ 0 ð 2 :

261 Þ

where ð k 2 d 2 dz 2 Þ F ¼ 0. At z ¼ 0 and 1, W ¼ 0 and d 2 W dz 2 = = ¼ 0, so F ¼ 0.

The remaining two boundary conditions are therefore that

F ¼ 0atz ¼ 0 and 1 ð 2 : 262 Þ

which means that

½ð k 2 d 2 dz 2

3

Ra c k 2 T 0 d 2 dz 2 = Þ = W ¼ 0 ð 2 : 263 Þ

at z ¼ 0and1.
For the free-slip boundary conditions, then, as for the non-rotating case
W ð z Þ¼ 1
n ¼ 1

A n sin ð n z Þ ð 2 : 264 Þ

Substituting (2.264) into the reduced version of the characteristic value equation
(2.263), we find that

ð k 2 þ n 2

2 3 2

Ra c k 2 Þ þ T 0 ð n Þ ¼ 0 ð 2 : 265 Þ