The basic equations - Severe Convective Storms and Tornadoes: Observations and Dynamics - page 75

Differentiating (2.227) with respect to z, we find that

2 w

z 2

@=@ x ð@ u =@ z Þþ@=@ y ð@v=@ z Þ¼@ =@ ð 2 : 228 Þ

Then using (2.226) we find that the LHS of (2.228) must be zero at z ¼ 0 and 1, so
that

d 2 W dz 2 = ¼ 0atz ¼ 0 and 1 ð 2 : 229 Þ

Four boundary conditions down and two to go. The temperature of the bottom
plate and that of the top plate are held fixed, so that B is a constant on each
plate. If this is the case, then the plates must be perfectly conducting. So
r h B ¼ 0 tz ¼ 0 and 1

ð 2 : 230 Þ

We recall (2.215) and find that with (2.230)

2 2 w ¼ð@ 2

x 2

2

y 2

ð@=@ t r Þr =@ þ@ =@ Þ B ¼ 0 tz ¼ 0and1 ð 2 : 231 Þ

so that

d 4 W dz 4 = ¼ 0atz ¼ 0 and 1 ð 2 : 232 Þ

The six boundary conditions for the free-slip, perfectly conducting case are 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. A solution that satisfies the boundary
conditions is the following:

W ð z Þ¼ 1
n ¼ 1

A n sin ð n z Þ ð 2 : 233 Þ

Substituting (2.233) into (2.224) we find that

! i Þþ k 2 þ n 2

2

! i þ k 2 þ n 2

2

ð k 2 þ n 2

2

Þ¼ Ra k 2 ½ð! r þ i ½! r þ i ð 2 : 234 Þ

Since Ra is a real number, the imaginary part of the LHS must be zero. It follows
that

! r þð k 2 þ n 2

2

! i ½ 2 Þð 1 þÞ¼ 0 ð 2 : 235 Þ

This equation is satisfied if ! i ¼ 0, in which case there cannot be oscillating

! r could have any value, growing or decaying (i.e., unstable or

solutions and

stable), or if

! r þð k 2 þ n 2

2

2 Þð 1 þÞ¼ 0 ð 2 : 236 Þ

In the latter case, ! i could be anything, but

! r ¼ð k 2

þ n 2

2

Þð 1 þÞ=ð 2 Þ < 0 ð 2 : 237 Þ

because k 2 , n 2 ,

2 , and

0. So, in this case solutions could be oscillating,
but stable (decaying with time): unstable, oscillating solutions are not possible. We
now return to (2.234), but this time we equate the real part of the LHS to the real
part of the RHS. It follows that
ð! r þ k 2

are >

þ n 2

2

Þð! r þ k 2 þ n 2

2 2

i ¼ Ra k 2

=ð k 2 þ n 2

2

Þ! Þ ð 2 : 238 Þ

We now have two simultaneous equations in ! r and ! i ((2.235) and (2.238)). If

! i 6¼ 0, then we eliminate ! r by solving (2.236) for ! r and substituting it into