Estimation of the internal pressure gradient
Jarle Berntsen1and Lie-Yauw Oey2
1University of Bergen, Norway and2Princeton University, USA
Workshop on Modeling the Ocean 2009 Taipei 24. February
Outline of the talk
Background and history The standard POM approach The Green’s Theorem approach The Seamount case
The Northwest Atlantic Discussion
Background
The independent variables(x,y,z,t)are transformed to (x∗,y∗, σ,t∗), where
x∗ = x y∗ =y σ = z−η
H+η t∗ =t.
The x -component of the internal pressure gradient term
− 1 ρ0
∂p
∂x z
= − gD
ρ0 Z 0
σ
∂ρ
∂x − σ D
∂D
∂x
∂ρ
∂σ
dσ.
History
Haney - 1991
Mellor, Ezer and Oey (1994) - Internal pressure errors die out prognostically
The seamount Case - Beckmann and Haidvogel - 1993 Eight eddies around the seamount
The errors grow prognostically Mellor, Oey and Ezer - 1998
Sigma Errors of the First and Second Kind Shchepetkin and McWilliams 2003 (SM03)
Berntsen (2002) + SM03 - Sigma Errors of the Second Kind may grow in time
Errors for real oceanic basins?
Methods to reduce the IPG errors
Subtraction ofρref(z)
Extra smoothing of the topography
Take the IPG term in z-coordinates (Stelling and van Kester 1994)
Higher order (McCalpin (1994) and papers by Chu and Fan) Rotated method - Thiem and Berntsen (2006)
Alternative mathematical formulations of the IPG term (SM03)
Hydrostatic Inconsistency
The scheme is Hydrostatic Consistent (Haney91) if
σ δσ
δH H
< 1.
This criterion does NOT need to be satisfied.
From MEO94
E δxb
δx
= H 4
δxH δx
∂2b
∂z2 (
(δσ)2−σ2 δxH
H 2)
.
The POM approach
− 1 ρ0
∂p
∂x z
= − gD
ρ0
Z 0
σ
∂ρ
∂x − σ D
∂D
∂x
∂ρ
∂σ
dσ.
Using integration by parts this may be written
− 1 ρ0
∂p
∂x z
= − gD
ρ0
"
Z 0
σ
∂ρ
∂x + 1 D
∂D
∂xρ
dσ+σρ D
∂D
∂x
# .
The Green’s theorem approach
Shchepetkin and McWilliams 2003 transform the area integrals around U-points into line integrals using the Green’s theorem
Z
A
Z ∂ρ
∂x z
dxdz = I
ρdz. (1)
The line integrals may be computed using splines. The Cubic-H method appear to be a very good alternative. H for Harmonic averaging.
Open Questions - Work Plan
In SM03 - POM 2nd order results are compared to SM03 higher order results
Is the progress reported due to the Greens’ theorem approach or to higher order?
Is the progress due to a better integration of the horizontal or vertical integrals?
To address this, we try to compare results using the two approaches and the same order of accuracy in both (2nd order POM to 2nd order SM03 etc.)
Methods for the horizontal approximations
POM - 2nd order method, 4th order method (McCalpin), Cubic-H (spline), Cubic-A (spline), 4th order compact differencing, 6th order method (Chu and Fan)
SM03 - 2nd order method, 4th order method, Cubic-H (spline), Cubic-A (spline), 6th order method (Berntsen and Oey)
The results are robust to the vertical methods 2nd order trapezoidal rule as in the POM suffice
The seamount case
Beckmann and Haidvogel (1993) DX = 6 km
10 and 40 equidistantσ-layers
The experiments are run for 180 days AM = 100 m2s−1
The Burger number is 3.0
The linearized POM model as in MOE98
Accelerations from initial IPG errors
−50 0 50
−50
−40
−30
−20
−10 0 10 20 30 40 50
ci=2E−7 m s−2 Distance [km]
Distance [km]
Vertical integral of∂u∂t =−ρ10∂p∂x
−50 0 50
−50
−40
−30
−20
−10 0 10 20 30 40 50
ci=2E−11 s−2 Distance [km]
Distance [km]
Vertical integral of∂ω∂t =
∂2v
∂x∂t − ∂
2u
∂y∂t
Correlation between initial IPG errors and E
kinafter 10 days
10−7 10−8
10−7 10−6 10−5 10−4 10−3
2
4
6 4′
Norm(∂ u/ ∂ t) [m s−2] Ekin [m2s−2]
Ekin(10 days) as function of the
∂u 1 ∂p
10−13 10−12
10−8 10−7 10−6 10−5 10−4 10−3
2
4
6 4′
Norm(∂ ω/ ∂ t) [s−2] Ekin [m2s−2]
Ekin(10 days) as function of the
∂ω ∂2v ∂2u
Time series of v
maxand E
kin0 30 60 90 120 150 180
0 0.05 0.1 0.15 0.2 0.25
Time [days]
Vmax [m/s]
POM−2nd SM03−2nd POM−4th SM03−4th POM−6th SM03−6th
vmax
0 30 60 90 120 150 180
10−8 10−7 10−6 10−5 10−4 10−3
Time [days]
Ekin [m2/s2]
POM−2nd SM03−2nd POM−4th SM03−4th POM−6th SM03−6th
Ekin
Comparisons of 2nd order, 4th order, and 6th order results for the two approaches. 40 equidistantσ-layers. The results for the SM03 methods are dashed. Thick solid line for the 2nd order POM.
4th order Compact Differencing
0 30 60 90 120 150 180
0 0.05 0.1 0.15 0.2 0.25
Time [days]
Vmax [m/s]
POM−2nd POM−4th POM−6th POM−4th−CD
vmax
0 30 60 90 120 150 180
10−8 10−7 10−6 10−5 10−4 10−3
Time [days]
Ekin [m2/s2]
POM−2nd POM−4th POM−6th POM−4th−CD
Ekin
Results for the 4th order compact differencing scheme (dashed lines). Results of 2nd order, 4th order, and 6th order POM
Cubic-H and Cubic-A Spline methods
0 30 60 90 120 150 180
0 0.05 0.1 0.15 0.2 0.25
Time [days]
Vmax [m/s]
POM−2nd SM03−4th SM03−CubicH SM03−CubicA
vmax
0 30 60 90 120 150 180
10−8 10−7 10−6 10−5 10−4 10−3
Time [days]
Ekin [m2/s2]
POM−2nd SM03−4th SM03−CubicH SM03−CubicA
Ekin
Results for the SM03 Cubic-H method (dashed lines) and SM03 Cubic-A method (dotted line) compared to results for the 2nd order POM method (thick solid line) and 4th order SM03 method (thin solid line).
Discussion - The seamount case
Error growth linked to the vortices In the vertical: Trapezoidal rule suffice
Small effects of the special treatment of the upper 1/2-cell suggested in SM03
Errors are reduced when introducing higher order methods (4th or 6th)
4th order compact differencing slightly better than 4th order McCalpin
The results for the POM methods and the SM03 methods generally similar, if they have the same order of accuracy No evidence so far that Cubic-H (or Cubic-A) is better than straightforward 4th order methods
The Northwest Atlantic POM model
Horizontal grid size∼20 km
Stratification as in the seamount case No extra filtering to smooth H
24σ-layers - finer resolution near the bottom and the surface Results are robust to the vertical method
Time series vmax and Ekin
Horizontal plots of erroneous streamfunctions and speeds
Streamfunctions after 110 days - 1
Ψ(110 days)-2nd order POM Ψ(110 days)-4th order McCalpin
Streamfunctions after 110 days - 2
Ψ(110 days)-4th order McCalpin Ψ(110 days)-4th order Compact Differencing
Streamfunctions after 110 days -3
Ψ(110 days)-4th order McCalpin Ψ(110 days)-The Cubic-H method
Time series 2nd, 4th, and 6th order
0 30 60 90 120
0 0.05 0.1 0.15 0.2 0.25 0.3
Time [days]
Vmax [m/s]
POM−2nd SM03−2nd POM−4th SM03−4th POM−6th SM03−6th
vmax
0 30 60 90 120
10−8 10−7 10−6 10−5 10−4 10−3
Time [days]
Ekin [m2/s2]
POM−2nd SM03−2nd POM−4th SM03−4th POM−6th SM03−6th
Ekin
Comparisons of 2nd order, 4th order, and 6th order results for the two approaches. The results for the SM03 methods are dashed. Thick solid line for the 2nd order POM. The errors for the 6th order methods are larger than the errors in the 4th order
Time series Compact Differencing
0 30 60 90 120
0 0.05 0.1 0.15 0.2 0.25 0.3
Time [days]
Vmax [m/s]
POM−2nd POM−4th POM−4th−CD
vmax
0 30 60 90 120
10−8 10−7 10−6 10−5 10−4 10−3
Time [days]
Ekin [m2/s2]
POM−2nd POM−4th POM−4th−CD
Ekin
Comparisons of 2nd order, 4th order-POM, 4th order compact results (dashed lines) for the POM methods.
Time series Cubic-H (SM03)
0 30 60 90 120
0 0.05 0.1 0.15 0.2 0.25 0.3
Time [days]
Vmax [m/s]
POM−2nd POM−4th SM03−4th SM03−CubicH
vmax
0 30 60 90 120
10−8 10−7 10−6 10−5 10−4 10−3
Time [days]
Ekin [m2/s2]
POM−2nd POM−4th SM03−4th SM03−CubicH
Ekin
Comparisons of 2nd order POM, 4th order POM, 4th order SM03 (dashed lines), and Cubic-H (dotted lines)
Discussion - The Northwest Atlantic case
4th order (McCalpin and Compact Differencing) and SM03-CubicH method very promising
Is the good performance due to the unfiltered bottom topography?
For this more realistic case 6th did not further reduce the errors
Overall Discussion
In the toolbox: Standard 2nd order POM
Test on the simple case withρ=ρ(z)if theσ-errors are acceptable
If not: Test 4th order POM and/or CubicH(SM03) For smooth problems, 6th order methods may help
No clear evidence so far that the Green’s theorem approach is better than the standard POM approach
Many variants are coded up and can be made available to the POM users
Is further improvement of the methods possible?
New Class of methods to explore
We may integrate over the horizontal surrounding cell according to (Chen et al. 2007 FVCOM)
− g ρ0
"
Z 1/2
−1/2
Z 1/2
−1/2
"
Z 0
σ
D∂ρ
∂x +∂D
∂xρ
dσ+σρ∂D
∂x
# dxdy
# .
The integral may be reduced to a two dimensional integral, in y andσ, using the Green’s theorem
Z 1/2
−1/2
Z 0
σ
D(1/2,y, σ)ρ(1/2,y, σ)−D(−1/2,y, σ)ρ(−1/2,y, σ)dσdy .
THANKS
This is not the end of the story Only the end of the talk
