Adjustment
215
a. Show that in the part of the basin where the Sverdrup balance holds, the
bottom layer is motionless.
b. If one can measure the slope of the thermocline with an accuracy of 10%,
show that one cannot measure the sea surface height with a better accuracy.
(9.5) Adjustment: a numerical simulation
To study the adjustment of a motionless liquid layer to the Sverdrup-Stommel
steady solution, consider the adjustment problem defined by
∂
∂t
(∇
2 ψ − Λψ)+
∂ψ
∂x
= −H(t)siny − r∇
2 ψ
on the domain [0, 1] × [0,π], with the wind-stress field τ x (y)=−H(t)cosy
and τ y =0 . Here, H(t) is the Heaviside function and linear friction is added
(with coefficient r) for reasons which will become clear below. Kinematic
boundary conditions ψ =0hold on each lateral boundary. Furthermore, Λ=
0 for the barotropic case ψ = ˜
Ψ and Λ=F 1 + F 2 for the baroclinic case
(ψ = ¯
Ψ).
The equation above has solutions of the form
ψ(x, y, t)=−Φ(x, t)siny
and the equation for Φ becomes
Φ xxt − (Λ + 1)Φ t +Φ x + r(Φ xx − Φ) = H(t)
with boundary conditions Φ(0,t)=Φ ( 1 ,t)=0and initial condition
Φ(x, 0) = 0.
The aim of this exercise is to solve this equation numerically for given values
of the parameters r and Λ. Define the grid x i ,i =0 , ···,m, with x 0 =0
and x m =1, such that Δx =1/m. Use central discretization in space and an
implicit Crank-Nicholson in time, i.e.,
Φ xx ≈
Φ i+1 +Φ i−1 − 2Φ i
Δx
2
Φ x ≈
Φ i+1 − Φ i−1
2Δx
Φ t ≈
Φ n+1 − Φ n
Δt
where Δt = t n+1 − t n is the time step.
215
a. Show that in the part of the basin where the Sverdrup balance holds, the
bottom layer is motionless.
b. If one can measure the slope of the thermocline with an accuracy of 10%,
show that one cannot measure the sea surface height with a better accuracy.
(9.5) Adjustment: a numerical simulation
To study the adjustment of a motionless liquid layer to the Sverdrup-Stommel
steady solution, consider the adjustment problem defined by
∂
∂t
(∇
2 ψ − Λψ)+
∂ψ
∂x
= −H(t)siny − r∇
2 ψ
on the domain [0, 1] × [0,π], with the wind-stress field τ x (y)=−H(t)cosy
and τ y =0 . Here, H(t) is the Heaviside function and linear friction is added
(with coefficient r) for reasons which will become clear below. Kinematic
boundary conditions ψ =0hold on each lateral boundary. Furthermore, Λ=
0 for the barotropic case ψ = ˜
Ψ and Λ=F 1 + F 2 for the baroclinic case
(ψ = ¯
Ψ).
The equation above has solutions of the form
ψ(x, y, t)=−Φ(x, t)siny
and the equation for Φ becomes
Φ xxt − (Λ + 1)Φ t +Φ x + r(Φ xx − Φ) = H(t)
with boundary conditions Φ(0,t)=Φ ( 1 ,t)=0and initial condition
Φ(x, 0) = 0.
The aim of this exercise is to solve this equation numerically for given values
of the parameters r and Λ. Define the grid x i ,i =0 , ···,m, with x 0 =0
and x m =1, such that Δx =1/m. Use central discretization in space and an
implicit Crank-Nicholson in time, i.e.,
Φ xx ≈
Φ i+1 +Φ i−1 − 2Φ i
Δx
2
Φ x ≈
Φ i+1 − Φ i−1
2Δx
Φ t ≈
Φ n+1 − Φ n
Δt
where Δt = t n+1 − t n is the time step.
