206
DYNAMICAL OCEANOGRAPHY
Note that the adjustment time scale depends on the latitude through L Di and β 0 .
At midlatitudes adjustment is relatively slow compared to more equatorial latitudes.
9.3.2. The spin-up problem
For convenience we choose d = π and consider the spin-up to the stationary
solution for the case f (t)=H(t). We choose τ p = τ β =1/(β 0 L),υ =1and can
separate the equation (9.34) by
Ψ n (x, y, t)=− sin ny Φ n (x, t),
(9.38)
for certain n. Through this separation, the boundary conditions for Ψ at y =0 , 1
are satisfied and we write Φ=Φ n . The equation for Φ becomes
Φ xxt − ˜
ΛΦ t +Φ x = H(t),
(9.39)
with ˜
Λ=Λ+n 2 and Φ=0for t =0 . Boundary conditions at the western and
eastern walls are kinematic and hence Φ=0.
Ex. 9.4
We see that for t →∞ , the Sverdrup solution Φ(x)=x − 1 is reached. In
addition, the homogeneous solutions of (9.39) are exactly the Rossby waves with
dispersion relation (note that β is absorbed into the time scale)
σ =
−k
k 2 + ˜
Λ
.
(9.40)
If there were no zonal boundaries, then the solution would be Φ I (t)=−t/ ˜
Λ.
This is also the initial response, because then gradients of Φ in x are still small.
However, Rossby waves will immediately be generated and these must eventually
take care of the zero mass flux at the zonal boundaries. Moreover, they must
collectively provide the Sverdrup flow in the limit t →∞.
The total solution of the problem (9.39) can be determined using Laplace transformation techniques but it turns out to be non transparent. Hence, we will present
a more qualitative analysis of the response by writing the solution as
Φ(x, t)=Φ
I (t)+Φ
L (x, t)+Φ
S (x, t),
(9.41)
where Φ I is the initial response, Φ L the response due to long Rossby waves and
Φ S the response due to short Rossby waves.
At the east coast long Rossby waves are generated and transport energy (cf.
section 9.2) westward. These waves satisfy the long-wave approximation (9.28),
i.e.,
∂Φ
∂x
L
− ˜
Λ
∂Φ
∂t
L
=0.
(9.42)
The solution of this equation is
Φ
L (x, t)=G(x + ˜
Λ
−1 t),
(9.43)
DYNAMICAL OCEANOGRAPHY
Note that the adjustment time scale depends on the latitude through L Di and β 0 .
At midlatitudes adjustment is relatively slow compared to more equatorial latitudes.
9.3.2. The spin-up problem
For convenience we choose d = π and consider the spin-up to the stationary
solution for the case f (t)=H(t). We choose τ p = τ β =1/(β 0 L),υ =1and can
separate the equation (9.34) by
Ψ n (x, y, t)=− sin ny Φ n (x, t),
(9.38)
for certain n. Through this separation, the boundary conditions for Ψ at y =0 , 1
are satisfied and we write Φ=Φ n . The equation for Φ becomes
Φ xxt − ˜
ΛΦ t +Φ x = H(t),
(9.39)
with ˜
Λ=Λ+n 2 and Φ=0for t =0 . Boundary conditions at the western and
eastern walls are kinematic and hence Φ=0.
Ex. 9.4
We see that for t →∞ , the Sverdrup solution Φ(x)=x − 1 is reached. In
addition, the homogeneous solutions of (9.39) are exactly the Rossby waves with
dispersion relation (note that β is absorbed into the time scale)
σ =
−k
k 2 + ˜
Λ
.
(9.40)
If there were no zonal boundaries, then the solution would be Φ I (t)=−t/ ˜
Λ.
This is also the initial response, because then gradients of Φ in x are still small.
However, Rossby waves will immediately be generated and these must eventually
take care of the zero mass flux at the zonal boundaries. Moreover, they must
collectively provide the Sverdrup flow in the limit t →∞.
The total solution of the problem (9.39) can be determined using Laplace transformation techniques but it turns out to be non transparent. Hence, we will present
a more qualitative analysis of the response by writing the solution as
Φ(x, t)=Φ
I (t)+Φ
L (x, t)+Φ
S (x, t),
(9.41)
where Φ I is the initial response, Φ L the response due to long Rossby waves and
Φ S the response due to short Rossby waves.
At the east coast long Rossby waves are generated and transport energy (cf.
section 9.2) westward. These waves satisfy the long-wave approximation (9.28),
i.e.,
∂Φ
∂x
L
− ˜
Λ
∂Φ
∂t
L
=0.
(9.42)
The solution of this equation is
Φ
L (x, t)=G(x + ˜
Λ
−1 t),
(9.43)