Thermocline problem
301
The O(1) system of equations in the boundary layer becomes
−˜ v
0 sin θ = −
1
cos θ
∂ ˜
p 0
∂φ
+
1
2
∂ 2 ˜
u 0
∂ξ 2 ,
(13.13a)
˜
u
0 sin θ = −
∂ ˜
p 0
∂θ
+
1
2
∂ 2 ˜
v 0
∂ξ 2 ,
(13.13b)
0=−
∂ ˜
p 0
∂ξ
,
(13.13c)
∂ ˜
u 0
∂φ
+
∂(˜ v 0 cos θ)
∂θ
+c o s θ
∂w 0
∂z
+
∂ ˜
w 0
∂ξ
=0,
(13.13d)
from which it follows that the pressure is again constant over the boundary layer,
i.e. ˜
p 0 = p 0 .
Let λ =
| sin θ| and define ¯
ξ = λξ. The equations (13.13a-b) are then
transformed into
−˜ v
0 = −v
0 +
1
2
∂ 2 ˜
u 0
∂ ¯
ξ 2 ,
(13.14a)
˜
u
0 = u
0 +
1
2
∂ 2 ˜
v 0
∂ ¯
ξ 2 ,
(13.14b)
where also (13.7) has been used. This is the same system of equations as for the
Ekman boundary layer on the β-plane. Hence, the solutions (5.49) can be copied
with ξ substituted by ¯
ξ.
Substitution of these solutions into (13.13) and use of (13.7d) gives
cos θ
∂ ˜
w 0
∂ξ
=
∂u 0
∂φ
cos λξ +
∂v 0
∂φ
sin λξ
e
−λξ
+
∂
∂θ
(v
0 cos λξ − u
0 sin λξ)e
−λξ cos θ
.
(13.15)
Through integration over the boundary layer and making use of the definite integrals
∞
0
e
−λξ
sin λξ
cos λξ
dξ =
1
2λ
1
1
,
(13.16a)
∞
0
ξe
−λξ
sin λξ
cos λξ
dξ =
1
2λ 2
1
0
,
(13.16b)
we find
lim
ξ→∞
˜
w
0 (φ, θ, ξ)= ˜
w
0 (φ, θ, 0) +
1
2λ
(
1
cos θ
(
∂u 0
∂φ
+
∂(v 0 cos θ)
∂θ
)+