AOSC 652: Week 6, Day 1 - atmos.umd.edurjs/class/fall2016/week_06/AOSC652_2016_1003.pdfWhy might you...

Copyright © 2016 University of Maryland.This material may not be reproduced or redistributed, in whole or in part, without written permission from Ross Salawitch. 3 Oct 2016

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Numerical Integration

Week 6, Day 1

3 Oct 2016

Analysis Methods in Atmospheric and Oceanic Science

AOSC 652


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AOSC 652: Analysis Methods in AOSC

OH Column: Units of 1012 molecules cm−2 OR1012 cm−2

How is Trop OH column from a model such as theGoddard “Global Modeling Initiative” (GMI)

Chemical Transport Model (CTM) found ???

Trop: Troposphere


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τ CH4 : CH4 Lifetime with respect to loss by reaction w/ tropospheric OH

How found ?!?


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GMIGEOS-Chem

DECCAM-Chem


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Air Quality Model Satellite Data



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22 March

NASA Aura MLS ClO490 K pot’l temp

~18 km22 March

2010 2011

ppb ppb


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2 October 2011


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2 October 2011

{ }HIGH

LOW

*3

θ*

3θ

3

3

O (θ) is measure

Ozone Deficit = Factor O (θ) O (θ) θ

where

O (θ) is profile withou and t any

d prof

chemi

ile (

cal l

RED

oss (BLUE)

)

d× −∫


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AOSC 652: Analysis Methods in AOSCNumerical Integration

Why else might you need to compute an integral ?

Calculate carbon emissions to compare with change in atmospheric CO2

CO2 emission, 1959 to 2014 = 323.6 Gt C


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Why else might you need to compute an integral ?

Calculate carbon emissions to compare with change in atmospheric CO2

Change in atmospheric CO2, 1959 to 2014 = 82.64 ppm82.64 ppm × 2.18 Gt C / ppm CO2 = 180.16 Gt C

CO2 emission, 1959 to 2014 = 323.6 Gt C

Legacy of Charles Keeling, Scripps Institution of Oceanography, La Jolla, CAhttp://www.esrl.noaa.gov/gmd/ccgg/trends/co2_data_mlo.html

About half of the CO2 released by combustion offossil fuels stays in the atmosphere:

rest taken up by land biosphere and oceans

http://www.esrl.noaa.gov/gmd/ccgg/trends/co2_data_mlo.html


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Why might you need to compute an integral ?

Heat Balance in Ocean:

( ) /

0

0 −

− − −

=

∫∫ OCEAN WARMING

t

SOLAR IR RADIATION EVAPORATION SENSIBLE HEAT LOSS GAIN

t

Q Q Q Q dt

Q dt


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Heat Balance in Ocean:

( ) /

0

0 −

− − −

=

∫∫ OCEAN WARMING

t

SOLAR IR RADIATION EVAPORATION SENSIBLE HEAT LOSS GAIN

t

Q Q Q Q dt

Q dt

This heat then affects temperature in a particular ocean layer via:

T Z

OCEAN WARMING p

tc dzQ dt ρ− ∆= ∫∫ 00


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Also known as “quadrature”


Find pressure based on the mass of the overlying atmosphere:

Find dynamic height D of ocean based on specific volume α:

( ) dz

p z g zρ∞

= ∫

2

1

dp

p

D pα= ∫


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Trapezoidal Rule:

1

11

11

( ) + ( )( ) ( ) 2

NNx i i

i ix i

f x f xf x dx x x−

++

=

≈ −Σ∫


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Trapezoidal Rule:

1

11

11

( ) + ( )( ) ( ) 2

NNx i i

i ix i


++

=

≈ −Σ∫

Analysis Methods for Engineers, Ayyub and McCuen


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Trapezoidal Rule:

1

11

11

( ) + ( )( ) ( ) 2

NNx i i

i ix i


++

=

≈ −Σ∫

Base of trapezoid



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Trapezoidal Rule:

1

11

11

( ) + ( )( ) ( ) 2

NNx i i

i ix i


++

=

≈ −Σ∫Avg height of end points



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Trapezoidal Rule:

1

11

11

( ) + ( )( ) ( ) 2

NNx i i

i ix i


++

=

≈ −Σ∫


Case where integration using the trapezoidal rule should work well:


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Trapezoidal Rule:

1

11

11

( ) + ( )( ) ( ) 2

NNx i i

i ix i


++

=

≈ −Σ∫


Case where integration using the trapezoidal rule may not work well:


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Trapezoidal Rule:

1

11

11

( ) + ( )( ) ( ) 2

NNx i i

i ix i


++

=

≈ −Σ∫

Error order d2f/dx2: i.e., second derivative of function evaluated at some placein the interval

Hence, trapezoidal rule is exact for any function whose second derivativeis identically zero.


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Simpson’s Rule:

1

22 2 1

1, 3, 5

( ) + 4 ( ) ( )( ) 2 3

NNx i i i i i

x i

x x f x f x f xf x dx−

+ + +

=

− +≈ Σ∫

What is the basis of this formula?


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Simpson’s Rule:

1

22 2 1

1, 3, 5

( ) + 4 ( ) ( )( ) 2 3

NNx i i i i i

x i


+ + +

=

− +≈ Σ∫

What is the basis of this formula? each point fit by “quadratic” Lagrange polynomials



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Simpson’s Rule:

1

22 2 1

1, 3, 5

( ) + 4 ( ) ( )( ) 2 3

NNx i i i i i

x i


+ + +

=

− +≈ Σ∫

What is the basis of this formula? each point fit by “quadratic” Lagrange polynomials


See pages 187 to 190 of Numerical Analysis, Burden and Faires,for a derivation of Simpson’s Rule

in terms of Quadratic Lagrange Polynomials


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Simpson’s Rule:

1

22 2 1

1, 3, 5

( ) + 4 ( ) ( )( ) 2 3

NNx i i i i i

x i


+ + +

=

− +≈ Σ∫

Error order d4f/dx4: i.e., fourth derivative of function evaluated at some placein the interval

Hence, Simpson’s rule is exact for what order polynomial?


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Boole’s Rule (aka as Bode’s rule):

1

44 4 3 2 1

1, 5, 11, ...

( )

14 ( ) 64 ( ) 24 ( ) + 64 ( ) 14 ( ) 2 45

−+ + + + +

=

≈

− + + +

∫

Σ

Nx

x

Ni i i i i i i

i

f x dx

x x f x f x f x f x f x

Error order d6f(x)/dx6: i.e., sixth derivative of function evaluated at some placein the interval

Hence, Boole’s rule is exact for what order polynomial?


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Gaussian Quadrature:

y1, y2, … yn nodes are not uniformly spacedc1, c2, … cn Gauss coefficients are determined once n is specified

Nice descriptions of theory at http://en.wikipedia.org/wiki/Gaussian_quadratureand http://www.efunda.com/math/num_integration/num_int_gauss.cfm

( )n1

1i=1

1 ( ) = ( ( )) ( ( )) 2

b

i ia

dxf x dx f g y dy b a c f g ydy−

≈ − ∑∫ ∫

Note:

produces exact result for polynomials of degree 3 or less

1

1

3 3( ) 3 3−

−≈ +

∫ f y dy f f

http://en.wikipedia.org/wiki/Gaussian_quadrature

http://www.efunda.com/math/num_integration/num_int_gauss.cfm


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Numerical Integration: Derivation of Simpson’s rule, page 1

R. L. Burden and J. D. Faires, Numerical Analysis, 8th edition


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AOSC 652: Week 6, Day 1 - atmos.umd.edurjs/class/fall2016/week_06/AOSC652_2016_1003.pdfWhy might you...

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