1 Ekman Flow: Assume that the Coriolis force and friction force are the dominant terms in the horizontal component equations of motion. Also assume that f and z A are constants. With the first two assumptions, here are the relevant horizontal equations of motion: 2 2 z dz u d A v f 0 2 2 z dz v d A u f 0 Employing a trick similar to one used to solve for the Inertial Flow, we multiply the bottom equation by i , add the two equations, obtain the sum, multiply the left-hand term by 2 i and the right-hand term by -1. 2 2 z 2 dz ) iv u ( d A ) v iu ( f i 0 or 2 2 z dz ) iv u ( d A ) iv u ( if 0 Rearranging: ) iv u ( A if dz ) iv u ( d 0 z 2 2
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1
Ekman Flow:
Assume that the Coriolis force and friction force are the dominant terms in the horizontal component equations
of motion. Also assume that f and zA are constants. With the first two assumptions, here are the relevant
horizontal equations of motion:
2
2
zdz
udAvf0
2
2
zdz
vdAuf0
Employing a trick similar to one used to solve for the Inertial Flow, we multiply the bottom equation by i , add
the two equations, obtain the sum, multiply the left-hand term by 2i and the right-hand term by -1.
2
2
z2
dz
)ivu(dA)viu(fi0
or
2
2
zdz
)ivu(dA)ivu(if0
Rearranging:
)ivu(A
if
dz
)ivu(d0
z2
2
2
Here we contrast the mathematics of Inertial and Ekman Flow:
Inertial Flow: Ekman Flow:
)iuv(ifdt
)iuv(d0
)ivu(
A
if
dz
)ivu(d0
z2
2
First order ordinary differential equation Second order ordinary differential equation
Independent variable is “ t ” Independent variable is “ z ”