Thermocline problem
315
Because z 2 (φ E ,θ)=0and z 3 (φ E ,θ) has to be constant in the meridional direction (since u 2 =0at φ = φ E ) it follows that z 3 (φ E ,θ)=H 2 . This determines
the thickness h through the Ekman vertical velocity ˆ
w E and hence h 1 (φ, θ) and
h 2 (φ, θ) as well.
◮
Example 13.2: The shadow zone
Just as in Example 13.1, let the Ekman vertical velocity be given by
ˆ
w E (φ, θ)=
α
2
θ − θ 0
sin
2 θ
,
(13.80)
with α>0. The solution (13.79) in the domain S(θ) then becomes
h
2 (φ, θ)=
α(θ 0 − θ)(φ E − φ)+γ 2 H 2
2
γ 1 (1 −
f
f2 ) 2 + γ 2
,
(13.81)
and the layer thicknesses are
h 1 = h(1 −
sin θ
sin θ 2
),
(13.82a)
h 2 = h
sin θ
sin θ 2
.
(13.82b)
The solution matches the solution in the domain N (θ) at θ = θ 2 , for which (see
Example 13.1)
h 1 =0 ,
(13.83a)
γ 2 (h
2
2 (φ, θ) − H
2
2 )=α(θ 0 − θ 2 )(φ E − φ).
(13.83b)
Now consider the streamlines that, in layer 2, intersect the curves θ = θ 2 at a
certain longitude ˜
φ (Fig. 13.5). Because z 3 = −h, curves of constant pressure h
are also streamlines in layer 2 (note p 2 − p 3 = γ 2 h). For the streamline through
φ = ˜
φ we find
h
2 (φ, θ)=
α(θ 0 − θ)(φ E − φ)+γ 2 H 2
2
γ 1 (1 −
f
f2 ) 2 + γ 2
=
= h
2 ( ˜
φ, θ 2 )=
α(θ 0 − θ 2 )(φ E − ˜
φ)+γ 2 H 2
2
γ 2
.
(13.84)
The streamline that connects to the eastern boundary in layer 2 ( ˜
φ = φ E ) is
determined in the domain S(θ) through the relation
α(θ 0 − θ)(φ E − φ)=H
2
2 γ 1 (1 −
f
f 2
)
2 .
(13.85)