Etudes et évaluation de processus océaniques par des hiérarchies ...

238
64 CHAPTER 11. SOLUTION OF EXERCISES
Exercise 16:
(x r ,y r ) = R(cos(ωt), sin(ωt)), (u r ,v r ) = ωR(− sin(ωt), cos(ωt)) ∂ t (u r ,v r ) = −ω 2 R(cos(ωt), sin(ωt))
putting it together: ((−ω 2 R − fωR − f 2 /4R) cos(ωt), (−ω 2 R − fωR − f 2 /4R) sin(ωt)) = 0,
which is satisfied if and only if ω = −f/2
Exercise 17:
now the centrifugal force is balanced by the slope of the free surface and we have ((−ω 2 R − fωR) cos(ωt),
0 which is satisfied if and only if ω = −f (compare to previous exercise)!
Exercise 28:
tel-00545911, version 1 - 13 Dec 2010
The potential energy released per unit length (in the transverse direction):
E pot = 2 1 2 ρgη2 0
∫ ∞
0
(1 − (1 − exp(−x/a)))dx = 3 2 ρgη2 0a. (11.4)
The kinetic energy in the equilibrium (final) solution per unit length (in the transverse
direction):
E kin = 2 1 ∫ ∞
2 ρHg2 η0(fa) 2 −2 exp(−2x/a)dx = 1 2 ρgη2 0a. (11.5)
0
So energy is NOT conserved during the adjustment process. Indeed waves transport energy
from the region where the adjustment occurs to ±∞.
Exercise 30:
Calculus tells us that:
d
dt
( ) ζ + f
= 1
H + η
d
(ζ + f) −
ζ H + η dt
+ f d
(H + η) 2 dt η.
So take ∂ x of eq.(5.39) and substract ∂ y of eq. (5.38) to obtain:
After some algebra one obtains:
and eq. (5.40) gives
∂ t ∂ x v +∂ x (u∂ x v) + ∂ x (v∂ v v) + f∂ x u + g∂ xy η
−∂ t ∂ y u −∂ y (u∂ x u) − ∂ y (v∂ v u) + f∂ y v − g∂ xy η = 0.
d
dt (ζ + f) = −(ζ + f)(∂ xu + ∂ y v),
d
dt η = −(H + η)(∂ xu + ∂ y v),
putting this together gives the conservation of potential vorticity.