Workbook on Aspects of Dynamical Meteorology A Self Discovery

Workbook on Aspects of Dynamical Meteorology

A Self Discovery Mathematical Journey for Inquisitive Minds

First Edition

Jan D. Gertenbach



First Edition



First Edition

Jan D. GertenbachSA Weather BureauPrivate Bag X097Pretoria, 0001South Africa

c©2001 by the Author.All Rights Reserved

No part of this publication may be reproduced or transmitted in any form or byany means, electronic or mechanical, without permission from the author.

ISBN 0-620-27229-5

This book was typeset in AMSTEX.Photographs were obtained from http://www.photolib.noaa.gov

Contents

1. Fundamental Mathematical Aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1. Vector algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.1. Vector space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.2. Basis and co-ordinate system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.3. Scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.4. Vector product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.5. Orthogonal vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2. Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3. Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3.1. Scalar valued functions of one variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3.2. Vector valued functions of one variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.3. Scalar valued functions of several variables . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.4. Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3.5. The total derivative and temperature advection . . . . . . . . . . . . . . . . . . . . 91.3.6. Vector valued functions of several variables . . . . . . . . . . . . . . . . . . . . . . . 111.3.7. Integration of knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4. Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4.1. Fundamental theorems of the Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4.2. Transformation of integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4.3. Interchanging total differentiation and integration . . . . . . . . . . . . . . . . . 14

1.5. More examples with meteorological applications . . . . . . . . . . . . . . . . . . . . . . 161.5.1. Motion of a particle in a circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.5.2. Circulation and vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5.3. Divergence and convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5.4. Advection and the material derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.5.5. Integration of knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.6. Further vector algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.6.1. Vector product (in R3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.6.2. The scalar triple product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.6.3. The vector triple product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.6.4. Integration of knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2. The momentum equation in rotating spherical co-ordinates . . . . . . 252.1. The inertial reference frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.1.1. Spherical co-ordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.2. The rotating frame of reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.2.1. Spherical co-ordinates in the rotating frame of reference . . . . . . . . . . . 262.2.2. The motion of a particle (parcel of air) in the rotating frame . . . . . . 27

2.3. Differentiation of an arbitrary vector A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.3.1. Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.4. Velocity and acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.4.1. The velocity vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.4.2. The acceleration vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3. Balance laws in physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

iv

3.1. One dimensional derivation of the mass conservation principle . . . . . . . . . 343.1.1. Eulerian approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.1.2. The constitutive equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.1.3. Lagrangian approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.2. Three dimensional derivation of the mass conservation principle . . . . . . . 353.2.1. Eulerian approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.2.2. The continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.2.3. Lagrangian approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.3. Abstract balance law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.3.1. Eulerian derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.3.2. Lagrangian derivation of the thermodynamic equation . . . . . . . . . . . . 40

3.4. Transformation of conservation equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.4.1. A choice: atmospheric pressure or height as independent variable . 403.4.2. Pressure and height as inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.4.3. Isobaric Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.4.4. Transformation of the Continuity equation . . . . . . . . . . . . . . . . . . . . . . . . 433.4.5. A Generalized Vertical Coordinate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4. The quasi-geostrophic approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.1. The geopotential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.2. The thickness of an atmospheric layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.3. The geopotential in isobaric co-ordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.4. The geostrophic wind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.5. The ageostrophic wind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494.6. The quasi-geostrophic prediction equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 494.7. The Q-vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.8. The quasi-geostrophic potential vorticity equation . . . . . . . . . . . . . . . . . . . . 53

5. Atmospheric modelling and simple numerical examples . . . . . . . . . . 555.1. Mathematical models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555.2. Atmospheric modelling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.3. Numerical models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.4. The quasi-geostrophic vorticity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595.5. A boundary value problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

v

PrefaceIn this Workbook1 the reader is invited to activate an eager inquisitive mind,

to go on a self discovery journey through the world of Dynamical Meteorology, bythe ready and efficient vehicle of Mathematics. The reader thus becomes a learner,an integrator of knowledge, capable of bringing together previous experiences andapplying new skills to problems. The attitude to the learning process should movefrom an information gathering process to an active outcome-based life-relevantexperience. It is the writer’s sincere hope that the problem-solving approach—to both mathematical theory and meteorological applications—will enhance thelearner’s intellectual development and self esteem. The writer believes that a soundself esteem and courage towards self discovery are vital prerequisites for success inMathematics and Dynamical Meteorology.

The reader is encouraged and challenged to develop a problem-solving atti-tude. As may be expected, some effort is needed to embark on such a journey.The satisfaction after success, the establishment of a long-term securely rootedfoundation, is worth it.

The learner is invited to develop a life-long learning attitude, grounded in life-long assessment of progress. Group work and evaluation by peers may be part ofthe learning process. Not only successes and correct answers are important—thelessons learnt from failures should not be forgotten.

For background theory, the reader is referred to the literature—no claim tocompleteness or self sufficiency is made. The purpose of the book is to help thereader discover through exercise.

The classroom, lecturer and library no longer remain the sole basis of infor-mation. The reader is encouraged to test, exercise and evaluate newly gainedexperience, using self developed computer programmes and graphical displays, aswell as the Internet and multi media information systems. The life-long learnershould use all appropriate senses to secure newly gained competence—competenceencompassing knowledge, skills and attitudes.

The levels of competence the learner should reach include the ability to dothings, to be able to demonstrate what was learnt and to reflect and apply to newproblems . The vision of the learner should be to

know that ,know why,understand ,know how ,know how and why not differently.

These aspects should constantly be kept in mind when doing the Workbook exer-cises.

1 The ambiguity in the subtitle of the book is intentional: the book is intended tobe both a discovery of the self through mathematical achievement and a discoveryof mathematics by self involvement and exercise.

vi

The workbook should be used in conjunction with other books on DynamicalMeteorology, e.g. the book by Holton (1992). Books on Calculus e.g. Apostol(1967) and (1969) can be used to revise and supplement fundamental Mathematics.The References include several works on Continuum Mechanics and the rationalfoundation thereof. The learner is challenged to first exercise self discovery, thento do self assessment against the literature and finally (if relevant) to participatein a group assessment (peer review).

Students frequently experience problems with interpreting and understandingwhat they have read. Moreover, knowledge gained (in other courses) but not usedtend to be forgotten and shelved very soon. This should not be the norm: inte-gration of knowledge, whereby the learners create their own cognitive structure,linking related topics from different parts of the text, is of utmost importance. Theauthor thus sets an example by frequently referring to previous statements of aspecific topic, e.g. the idea of an Eulerian or Lagrangian description of fluid flow(see Section 1.2, Example 1.3.5, Section 1.4.3, Section 1.5.2, Exercises 1.5.3.1 etc.).It is the author’s wish that the learners should constantly develop their abilityto, on the one hand, read accurately and, on the other, formulate their thoughtsprecisely.

I would like to thank my colleagues for their interest and suggestions for im-provement of the book.

I greatly appreciate Dr. A. P. Burger’s thorough reading of the manuscriptand valuable comments regarding content and preciseness.

Keywords to note:Workbook: Dynamical Meteorology; Learning: self discovering,involvement, multimedia, integration of knowledge, evaluation by peers and by self;Outcome based life-long assessment; Modelling of the atmosphere: mathematics,computing, graphical displays.

Orientation for a good study programme: the learner devise a curriculum,compile examination papers and memoranda, compile a portfolio containing suc-cesses, dead-ends and wrong efforts together with an evaluation why things did notwork.

vii

DuDt − 2Ωv sinφ+ 2Ωw cosφ− uv tanφ

a + uwa = − 1

ρ∂p∂x + Frx

DvDt + 2Ωu sinφ+

u2 tanφa + vw

a = − 1ρ∂p∂y + Fry

DwDt − 2Ωu cosφ− u2+v2

a = − 1ρ∂p∂z − g + Frz

∂ρ∂t +∇ · (ρU) = 0

cpD lnTDt −RD ln p

Dt − JT = 0

DVDt +∇pΦ+ fk × V = 0

∇p · V+ ∂ω∂p = 0

∂Φ∂p +

RTp = 0.

We see, measure,

and model

using the Greek alphabet

alpha α beta β gamma γdelta δ epsilon ε zeta ζeta η theta θ iota ιkappa κ lambda λ mu µnu ν xi ξ pi πrho ρ sigma σ tau τupsilon υ phi φ chi χpsi ψ omega ω

Alpha - Beta - Gamma ΓDelta ∆ Epsilon - Zeta -Eta - Theta Θ Iota -Kappa - Lambda Λ Mu -Nu - Xi Ξ Pi ΠRho - Sigma Σ Tau -Upsilon Υ Phi Φ Chi -Psi Ψ Omega Ω

and mathematical symbols

ddt

∂∂t ∂t ∂x ∇ ∫

V

∮ × → ∞ · · ·

Chapter 1

Fundamental Mathematical Aspects

A true story. The South African Weather Bureau (SAWB) uses mathematicalmodels for the prediction of the state of the atmosphere. Due to the complexityof the atmosphere and the consequential cost of model development, models thatwere developed elsewhere are used. An upgraded version of such a model, con-figured at an 80km horizontal resolution, was received during 1997 from overseas.The SAWB decided that an increase in horizontal resolution is vital. A researchproject, addressing amongst others the extent of the horizontal domain, the limita-tion of errors at the lateral boundaries, a feasible number of vertical layers and theoptimum configuration, keeping limited computer resources in mind, was approved.

The SAWB implemented the model with only limited help from overseas. Oneparticular issue was the determination of parameters related to the chosen gridresolution. A mathematical transformation is used to avoid the effect of the con-vergence of the meridians at the earth’s poles. The transformation was given inthe model documentation, but contained a typing error. After hours of dedicatedreasoning, marred by several dead-ends, a good sketch and sound vector calculuswere instrumental in obtaining the correct formula. Imagine the emotion when,after all this effort, the correct formula was found in older documentation of themodel! A computer programme for the calculation of the transformed grid waswritten, whereby a complete understanding of the workings of the transformationwas thought to be obtained. However, when a very large grid, was calculated,strange kinks occurred in the southeastern and southwestern corners. Investiga-tion of the computer source code revealed that a formula that differs from the one inthe manual was used. Further examination of the mathematical derivation gave anequivalent but different formula, and on using this no unexpected kinks occurred.

With this story the reader is motivated to pursue a problem based approachto the learning process. The identification of a problem will determine the toolsneeded for its solution. Which problems can be identified from this story? Thinkabout good up-to-date documentation and the background needed by an employee,responsible for the maintenance of a model, to be able to retrace the steps of theoriginal modeller and programmer.

2 Fundamental Mathematical Aspects

Introduction. The author’s aim is to help the learner obtain a complete under-standing of the mathematical tools available for making Dynamical Meteorologyeasier to understand and to provide an investment for the learner’s future. Chap-ter 1 is the launching site for the self discovery journey the learner is about toembark on. Instead of a mere mathematical introduction to Dynamical Meteorol-ogy, the learners are lead to enhance their mathematical ability through relevantmeteorological (or physical) examples and exercises. By early inclusion of topicslike geopotential, potential temperature, temperature advection, reference and cur-rent configurations, etc. learners are motivated to always strive for integration ofmeteorological knowledge and experience.

1.1. Vector algebra.

1.1.1. Vector space.Exercise 1.1.1. Revise the vector space concept. Use and expand on Fig. 1.1 tomotivate why the sums of the components of two vectors determine the sum vector.Repeat with the concept of scalar multiple of a vector (Fig. 1.2).

0 1 2 3 4 5 6 70

1

2

3

4

5

θ

0 1 2 3 4 5 60

1

2

3

4

5

Figure 1.1 Figure 1.2

1.1.2. Basis and co-ordinate system. With every basis e1, · · · , en of avector space X, a system of co-ordinates is associated as follows: to each vectorx ∈ X we assign the unique n-tuple of real numbers (x1, · · · , xn) such that

x = x1e1 + x2e2 + · · ·+ xnen = Σni=1xiei.

The numbers (x1, · · · , xn) are called co-ordinates or components of the vector xand depend on the basis vectors we have chosen. A basis determines a frame ofreference for the description of the physical (meteorological) variable. Whereas thephysical variable is co-ordinate free, it is convenient to choose a frame of referencefor manipulation and closer description.

1.1.3. Scalar product 3

Exercises 1.1.2.(a) Revise the following basic concepts: every vector space has a basis, and every

basis of a specific vector space has the same number of elements.(b) Check the following and use the result to show that the vector x = (1,−2, 3)

has different components relative to different frames of reference:

x = 1(1, 0, 0) + (−2)(0, 1, 0) + 3(0, 0, 1)= (−12)(1, 1, 1) + 18(1, 0, 0) + 5(−1, 2, 3).

1.1.3. Scalar product. Let x = (x1, x2, · · · , xn) and y = (y1, y2, · · · , yn) bevectors in Rn (n ≥ 2). Define the scalar product (or dot product) x · y and norm(length of a vector) ‖x‖ by

x · y = x1y1 + x2y2 + · · ·+ xnyn = Σni=1xiyi

‖x‖ = √x · x.

Exercises 1.1.3.(a) Revise the symmetric property

x · y = y · x

and the linear property

x · (y+ z) = x · y+ x · zx · λy = λ(x · y)

of the scalar product.(b) Prove that

x · y = ‖x‖ ‖y‖ cos θ

with θ the angle between the two vectors x and y (see Fig. 1.1). For simplicitylet x = (x, 0) and y = (y1, y2). Is this a severe restriction? Motivate youranswer. Revise the concept of projection of a vector along a given line.

(c) Spherical co-ordinates. Prove that the vectors

i = (− sinλ, cosλ, 0)j = (− sinφ cosλ,− sinφ sinλ, cosφ))k = (cosφ cosλ, cosφ sinλ, sinφ)

are linearly independent unit vectors. Next, expand and use Fig. 1.3 to en-lighten your findings.


Figure 1.3

1.1.4. Vector product (in R3). The following mnemonic form of the crossproduct is well known:

A × B =

∣∣∣∣∣∣∣i j k

Ax Ay Az

Bx By Bz

∣∣∣∣∣∣∣ .Exercises 1.1.4.(a) Revise the concept of cross product. Next, for simplicity, let x = (x, 0, 0)

and y = (y1, y2, 0). Prove that

x × y = (0, 0, ‖x‖ ‖y‖ sin θ)with θ the angle between the two vectors x and y. Show that the norm ofx × y equals the area of the parallelogram determined by x and y.

(b) Prove the following:

(1, 0, 0)× (0, 1, 0) = (0, 0, 1)(0, 1, 0)× (0, 0, 1) = (1, 0, 0)(0, 0, 1)× (1, 0, 0) = (0, 1, 0).

(c) Let i, j and k be as in Exercise 1.1.3(c). Prove the following:

i × j = k

j × k = i

k × i = j.

1.2. Functions 5

1.1.5. Orthogonal vectors. Definition: x ⊥ y if and only if x · y = 0.Exercises 1.1.5.(a) Show that

(a, b) ⊥ (−b, a)(1, 0, 0) ⊥ (0, 1, 0)(1, 0, 0) ⊥ (0, 0, 1)(0, 1, 0) ⊥ (0, 0, 1).

(b) Let a = (1, 2, 3), b = (3, 0,−1) and c = (5,−1,−1). Show that

a ⊥ b

a ⊥ c, butb ⊥ c.

(c) Integration of knowledge. Apply integration of knowledge to enhance yourunderstanding of vector algebra. Amongst others, write a computer programthat calculates vector sums, products of vectors with scalars, scalar productsand cross products. Test your program with mathematically worked out ex-amples. Use a computer graphics package to plot your results.

1.2. Functions. A function f , defined on domain D(f) in a region G ⊂ Rn, isdenoted by f : x → f(x). If the context is clear, the abbreviation, x → f(x), willbe used. The range of f is the set R(f) consisting of images of the elements ofD(f) under f . If D(f) = X , we will also write R(f) = f(X).

To describe the mechanics of bodies or the motion of fluids, we disregardthe microscopic structure and consider the body or fluid to be composed of a setof particles, distributed throughout some region of space. The set Ct in three-dimensional Euclidian space, associated with the particles of the body at a giveninstant of time, is called the configuration or state of the body at time t. Theco-ordinates (x1, x2, x3) ∈ Ct are known as spatial or Eulerian co-ordinates. Afunction f , defined on the configuration at time t is called a field function. We mayalso choose co-ordinates (X1, X2, X3) from a reference configuration C to describeindividual particles. These co-ordinates are called material (or Lagrangian) co-ordinates.Example 1.2. Atmospheric pressure in a vertical column may be considered tobe a function of height z → p(z). Conversely, the height z of an air parcel may beconsidered to depend on pressure p → z(p) in a region where p is monotone (i.e.either increasing or decreasing).Exercises 1.2.(a) Revise the following concepts: a function, the domain and range of a function,

the inverse of a function (if it exists) and continuity of a function.(b) In Example 1.2 the symbol z is used to indicate geometric height both as

independent variable and as a function of pressure, p. Similarly, the symbol pmay be used to indicate atmospheric pressure both as independent variable andas a function of height. A combination of the definitions of the two functions


gives a relationship p(z(p)) = p, showing that they are inverses of each other.However, p then becomes both independent and dependent variable in onesingle equation. To resolve this undesirable matter, define new functions pand z, for example, and make the above discussion precise and mathematicallysound.

(c) The geopotential. Define the geopotential Φ as Φ(z) =∫ z0g dz′. Use New-

ton’s law of universal gravitation to prove that, in the absence of centripetalacceleration,

Φ(z) = g0az

a+ z.

(d) Let Z = Φ(z)/g0 and a = 6000km. Calculate Z for z = 10, 20, · · ·150 km. Usea computer graphics package to plot your result.

(e) Use Taylor’s theorem to prove that for small z/a

Z ≈ z − a(za

)2

.

Calculate a(za

)2for z = 1, 10 and 100 km.

(f) Potential temperature Define the potential temperature θ as

θ = T

(psp

)R/cp

.

Let R = 287 J K−1 kg−1, cp = 1004 J K−1 kg−1 and ps = 1013.25 hPa. For θ =293 K, calculate T : p → T (p), in C, for pressure levels p = 1000, 900, · · · , 100hPa. Use a computer graphics package to plot your result. Repeat with θ = 303K and θ = 313 K. Plot your results on the same graph.

1.3. Differentiation.

1.3.1. Scalar valued functions of one variable. Revise the concept of thederivative of a function:

f ′(x) = limh→0

f(x+ h)− f(x)h

.

0 1 2 3-10123456

Figure 1.4. The derivative as slope of the tangent line

1.3.3. Scalar valued functions of several variables 7

Carefully note that the derivative is a function in its own right and that thenotation f ′ (or df

dx) refers to a single object, namely the derivative. Thus theargument (the element from the domain of f ′) has nothing in common with the xin the notation df

dx , see for example the use of the chain rule of differentiation inEquations (3.15) and (3.16) in Section 3.2.3 of Chapter 3.Exercise 1.3.1. Use the limit definition to determine the derivative function f ′ ofthe following functions:

f(x) = x

f(x) = x2

f(x) = ex.

1.3.2. Vector valued functions of one variable.Exercises 1.3.2. Let F(t) = (x(t), y(t)).(a) Explain how F can be used to describe a curve in the two dimensional plane.

Next, let x and y be differentiable. To investigate the differentiability of thevector valued function F, prove the following:

F(t+ h)− F(t)h

=(x(t+ h)− x(t)

h,y(t+ h)− y(t)

h

)

F(t+ h)− F(t)h

− (x′(t), y′(t)) = (x(t + h)− x(t)h

− x′(t),y(t+ h)− y(t)

h− y′(t))

so that

∥∥∥∥F(t+ h)− F(t)h

− (x′(t), y′(t))∥∥∥∥

=

√[x(t+ h)− x(t)

h− x′(t)

]2

+[y(t+ h)− y(t)

h− y′(t)

]2

→ 0 as h→ 0.

(b) Use the limit result above to motivate

F ′(t) = (x′(t), y′(t)).

(c) Let a be constant. The function t → F(t) defined by

F(t) = (x(t), y(t)) = (a cos θ(t), a sin θ(t))

describes circular motion with radius a around a fixed point. Calculate F ′(t).(d) Let x(t) = (x1(t), x2(t), x3(t)). Prove that

d

dt‖x(t)‖2 = 2x(t) · x′(t).


1.3.3. Scalar valued functions of several variables.Example 1.3.3. Heat conduction and Newton’s cooling law. Considerheat conduction in a material in the absence of motion. Consider the temperatureT (x, t) at position x and time t. If h is a fixed real number and n an arbitrary unitvector, then T (x + hn, t) gives the temperature on a sphere with radius h aboutthe point x. In general, the flux Φ of energy (rate of heat flow) at x in the directionn will depend on the direction n. We may imagine it to be proportional to thetemperature difference, T (x+hn, t)−T (x, t), between nearby points and inverselyproportional to the distance between them, i.e. proportional to T (x+hn,t)−T (x,t)

h .Assuming heat flow is from hot to cold, we look for a relation of the form Φ · n ∝−T (x+hn,t)−T (x,t)

h . For simplicity, we suppress the dependency of temperature ontime t, so that the flux becomes proportional to T (x+hn)−T (x)

h , which reminds ofthe difference quotients in Sections 1.3.1 or 1.3.2. We thus defineThe directional derivative. Let f : x → f(x) be a scalar valued function of thevariables x1, x2, · · · xk, where (x1, x2, · · · , xk) = x. Let n = (n1, n2, · · ·nk) be aunit vector denoting a fixed direction in space. The directional derivative at thepoint x in the direction of n is defined by

f ′(x;n) = limh→0

f(x+ hn)− f(x)h

,

see for example Apostol (1969), p. 252.If h→ 0, the difference quotient T (x+hn)−T (x)

h thus tends to T ′(x;n), so thatthe temperature flux in the direction n becomes Φ(x) · n = −κT ′(x;n), with κ aproportionality constant.Exercise 1.3.3. Let f(x, y) = x2+y2. Sketch a few curves on which f is constant.Next, let x = (x1, x2). Show that g(x) = x1

2+ x22 defines the same function, that

is, f = g. Use the limit definition to calculate f ′(x;n), for arbitrary directionsn = (n1, n2). In which direction is the derivative a maximum and in which aminimum? What happens in the other directions?

1.3.4. Partial derivatives. Define

∂f

∂x(x, y) = lim

h→0

f(x+ h, y)− f(x, y)h

∂f

∂y(x, y) = lim

h→0

f(x, y + h)− f(x, y)h

.

Notation 1.3.4. We also use the economical and sufficiently clear notation,

∂x =∂

∂x, ∂y =

∂

∂y, ∂p =

∂

∂p, etc.

for partial derivatives. In Cartesian co-ordinates x1, x2, · · ·xn we write

∂i =∂

∂xi, for i = 1, 2, · · ·n.

1.3.5. The total derivative and temperature advection 9

The del or nabla operator is denoted by

∇ = (∂1, ∂2, · · · , ∂n),or in three dimensions

∇ = ex∂

∂x+ ey

∂

∂y+ ez

∂

∂z,

with ex, ey and ez any three orthogonal unit vectors. Again, as mentioned at thebeginning of Section 1.3.1, note (and revise) the use of the chain rule of differenti-ation.

Remember that a vector represents a co-ordinate free quantity and that youhave the freedom to choose a frame of reference suitable for your application. Thegradient vector ∇f = (∂1f, ∂2f, · · · , ∂nf) or ∇f = ex

∂f∂x + ey

∂f∂y + ez

∂f∂z is a co-

ordinate dependent representation of a vector field. Its relation to the directionalderivative is given byTheorem 1.3.4. The directional derivative and partial differentiation.Once a frame of reference is chosen, the directional derivative can be representedby the gradient as follows:

f ′(x;n) = ∇f(x) · n,see Apostol (1969), p. 259.Exercises 1.3.4.(a) Let f(x, y) = x2 + y2 as in Exercise 1.3.3. Use differentiation to show that

∂f

∂x(x, y) = 2x

∂f

∂y(x, y) = 2y.

(b) Combine the result in (a) with Exercise 1.3.3 to show that

f ′((x, y);n) =(∂f

∂x(x, y),

∂f

∂y(x, y)

)· n.

(c) Use theorem 1.3.4 to prove

∂f

∂xi(x) = f ′(x; ei), for arbitrary i

and give an interpretation of partial derivatives in terms of the concept ofdirectional derivatives. Enlighten your answer with a sketch.

(d) Show that the flux vector in Example 1.3.3 satisfies

Φ(x) · n = −κ∇T (x) · n.Use the fact that n is arbitrary to show that

Φ(x) = −κ∇T (x)Φ(x, t) = −κ∇T (x, t).


1.3.5. The total derivative and temperature advection. A field functiondescribes a physical quantity from the viewpoint of a fixed observer while a materialdescription refers to an observer that moves with the fluid (Section 1.2). The totalderivative of a function of space and time reflects the possible influence of fluidmovement together with local changes with time. Heuristically, for an arbitrarychosen function f , it may be represented by:

Df

Dt=∂f

∂t+∂f

∂x

dx

dt+∂f

∂y

dy

dt+∂f

∂z

dz

dt

=∂f

∂t+∂f

∂xu+

∂f

∂yv +

∂f

∂zw,

see Holton (1992), p. 29. For a formal definition, let (x, t) → F (x, t) be anydifferentiable function, T (x, t) a temperature field and

U(x, t) = (u(x, t), v(x, t), w(x, t))

a three dimensional velocity field.Definitions 1.3.5.(a) The total derivative. The material , the total or the substantial derivative

of F is defined by

DF

Dt=∂F

∂t+U · ∇F = ∂F

∂t+

(u∂F

∂x+ v

∂F

∂y+ w

∂F

∂z

)

and represents the rate of change, following the motion, of the field variableF .

(b) Temperature advection. The temperature advection is defined as −U ·∇T .Note that definition 1.3.5(a) implies ∂T∂t =

DTDt − U · ∇T, so that the local time

rate of change of temperature, ∂T∂t , is the sum of the total derivative (following themotion) and the temperature advection.Example 1.3.5. Temperature advection. Let T (x, t) denote temperature in aspatial one dimensional region, at position x and at time t. T is a field functionand an Eulerian description (see Section 1.2) is used. Suppose the wind is blowingwith constant speed u and that the particle, initially at position x, is transporteddownstream a distance uδt while retaining its temperature. From Fig. 1.5 followsT (x+ uδt, t) = T (x, t− δt), and we find that

T (x+ uδt, t)− T (x, t)δt

=T (x, t− δt)− T (x, t)

δt.

In the limit δt→ 0, using L’Hospital’s rule, we deduce

u∂T

∂x(x, t) = −∂T

∂t(x, t).

1.3.6. Vector valued functions of several variables 11

T (x, t− δt) T (x+ uδt, t)

Figure 1.5

Exercises 1.3.5.(a) As may be expected, the material derivative of the temperature function T in

Example 1.3.5 vanishes. Prove this statement and discuss the local change intemperature caused by an easterly wind, blowing from a colder area towardsthe observer.

(b) Suppose T (x, t+ δt) = T (x− uδt, t). Interpret this statement and repeat theprevious argument to once more show that ∂T∂t (x, t) + u∂T∂x (x, t) = 0.

1.3.6. Vector valued functions of several variables.See for example Section 8.18 in Apostol (1969).Example 1.3.6. Velocity advection of a fluid in rigid body rotation. Theoperator U ·∇ = u ∂

∂x+v∂∂y +w

∂∂z in Definition 1.3.5 is simply a linear combination

of partial derivatives. It may thus be applied to a vector valued function. In twodimensions we define V = (u, v) and we may write V · ∇ = u ∂

∂x + v ∂∂y .

Using Notation 1.3.4, we put x = (x1, x2) and

V(x) = (−ωx2, ωx1) = (V1(x), V2(x)),

where ω represents a constant angular velocity. By direct differentiation

(V · ∇)V = (∂1V)V1 + (∂2V)V2

= (0, ω)V1 + (−ω, 0)V2.

Using V1 = −ωx2 and V2 = ωx1, the advection of velocity at x becomes

((V · ∇)V)(x) = −x2ω(0, ω) + x1ω(−ω, 0)= −ω2(x1, x2)

= −ω2x.

The general case. Let F(x) = (F1(x), F2(x), · · · , Fn(x)) and suppose each com-ponent Fi is differentiable.


Exercises 1.3.6.(a) Show that

F(x+ hn)− F(x)h

=

(F1(x+ hn)− F1(x)

h,F2(x+ hn)− F2(x)

h, · · · , Fn(x+ hn)− Fn(x)

h

).

(b) Complete the following:

∥∥∥∥F(x+ hn)− F(x)h

− (F1

′(x;n), · · · , Fn′(x;n))∥∥∥∥ =

√(F1(x+ hn)− F1(x)

h− F1

′(x;n))2

+ · · ·

→ 0 as h→ 0.

(c) Define the Jacobian matrix

A =

∇F1

∇F2...

∇Fn

=

∂1F1 ∂2F1 · · · ∂nF1

∂1F2 ∂2F2 · · · ∂nF2...

∂1Fn ∂2Fn · · · ∂nFn

.

Write A = DF(x) as on p. 270 of Apostol (1969) and show that the result of(b) for F ′(x;n) can be written in terms of the product of the Jacobian matrixDF(x) and the vector n. Compare the result with Theorem 1.3.4.

Remark 1.3.6. Note that relative to your frame of reference, the gradient vector∇F or the matrix DF forms part of a co-ordinate dependent representation of theco-ordinate free directional derivative F ′(x;n).

1.3.7. Integration of knowledge. Regarding the concept of functions and thederivatives of functions, make a summary of what was learnt, what mistakes andmisconceptions existed (and why) and other relevant ideas. Look for life-relevantaspects touched upon and applications to meteorology. Evaluate your work andput it in a portfolio for future reference.

1.4. Integration.

1.4.2. Transformation of integrals 13

1.4.1. Fundamental theorems of the Calculus. Revise the following:

F (x) =∫ x

a

f(x′)dx′ ⇒ F ′(x) = f(x) if f is continuous at x, x ∈ [a, b] (1.1)∫ b

a

F ′(x)dx = F (b)− F (a) if F is smooth enough. (1.2)

Exercises 1.4.1.(a) Differentiation of integrals. Define F (x) =

∫ xaf(x′)dx′. Use the chain rule of

differentiation ddt [F (r(t))] = F ′(r(t))r′(t) and (1.1) to show that

d

dt

∫ r(t)

a

f(x′)dx′ = f(r(t))r′(t). (1.3)

(b) Define g(t) =∫ ba f(x, t)dx. Use the limit definition of g

′(t) to show that

d

dt

∫ b

a

f(x, t)dx =∫ b

a

∂f

∂t(x, t)dx. (1.4)

You may assume the function f is such that you may interchange the order oflimit taking and integration.

(c) Define F (x, t) =∫ xa f(x

′, t)dx′. Use

d

dt(F (r(t), t)) =

∂F

∂x(r(t), t)r′(t) +

∂F

∂t(r(t), t)

in combination with (1.3) and (1.4) to show that

d

dt

∫ r(t)

a

f(x, t)dx = f(r(t), t)r′(t) +∫ r(t)

a

∂f

∂t(x, t)dx. (1.5)

(d) Use (1.5) to show that

d

dt

∫ b(t)

a(t)

f(x, t)dx = f(b(t), t)b′(t)− f(a(t), t)a′(t) +∫ b(t)

a(t)

∂f

∂t(x, t)dx. (1.6)

1.4.2. Transformation of integrals. Revise the general rule

∫ f(b)

f(a)

g(u)du =∫ b

a

g(f(x))f ′(x)dx. (1.7)

Example 1.4.2.


∫ t

0

1√1− x2

dx =∫ sin−1 t

0

cos θ√1− sin2 θ

dθ

=∫ sin−1 t

0

1dθ

= sin−1 t.

Exercises 1.4.2.(a) To test the result in Example 1.4.2, we need to prove

d

dt

∫ t

0

1√1− x2

dx =d

dtsin−1 t =

1√1− t2

.

Explain this statement and prove it.(b) Show that ∫ 1

0

2xex2dx = e− 1

by using the substitution x =√s.

(c) Show that ∫ x

0

11 + t2

dt = tan−1 x

by using the substitution t = tan θ.

1.4.3. Interchanging total differentiation and integration. Consider themotion of a part P of the reference configuration C in a Lagrangian descriptionof fluid flow. The fluid associated with the given set P of particles is called amaterial volume. As time progresses the material volume always consists of thesame particles and moves with the fluid. To describe such a material volume weneed a mapping from a reference configuration (e.g. the configuration at time t = 0)to the current configuration.

The reference and current configurations. Consider a one dimensional con-tinuum with reference configuration the interval (0, L). The transformation x =r(X), X ∈ (0, L) maps the interval (0, L) to the current configuration (0 , l) where0 = r(0) and l = r(L) (Fig. 1.6). Let [A,B] ⊂ (0, L) and put a = r(A) andb = r(B).

0 L

a x b

A X B

Figure 1.6

1.4.3. Interchanging total differentiation and integration 15

The following exercises show how total differentiation can arise from the differenti-ation of an integral defined on a material volume. For simplicity a one dimensionalapproach is followed. Let ρ denote fluid density and f : (x, t) → f(x, t) a fieldfunction. The main result of this Section is

d

dt

∫ r(B,t)

r(A,t)

ρ(x, t)f(x, t)dx =∫ r(B,t)

r(A,t)

ρ(x, t)Df

Dt(x, t)dx (1.8)

and it is proven in Exercises 1.4.3(d).Exercises 1.4.3. Suppose a function σ exists such that

∫ r(B)

r(A)

ρ(x)dx =∫ B

A

σ(X)dX for every [A,B] ⊂ (0, L).

(a) Use the transformation x = r(X) and the general rule (1.7) to prove that

ρ(r(X))r′(X) = σ(X).

(b) Use the transformation x = r(X) and the result ρ(r(X))r′(X) = σ(X) toprove that

∫ r(B)

r(A)

ρ(x)f(x)dx =∫ B

A

σ(X)f(r(X))dX.

(c) Let t be fixed and

∫ r(B,t)

r(A,t)

ρ(x, t)dx =∫ B

A

σ(X)dX for every A,B.

Consider the mapping r in Fig. 1.6 that maps the reference configuration [0, L]to the configuration [0 , l] at time t. We write x = r(X, t), X ∈ (0, L). Usereasoning similar to that in Exercise (b) to show that

ρ(r(X, t), t)∂r

∂X(X, t) = σ(X)

and that

∫ r(B,t)

r(A,t)

ρ(x, t)f(x, t)dx =∫ B

A

σ(X)f(r(X, t), t)dX.

(d) Prove (1.8) by completing the following:

d

dt

∫ r(B,t)

r(A,t)

ρ(x, t)f(x, t)dx =d

dt

∫ B

A

σ(X)f(r(X, t), t)dX

=∫ B

A

σ(X)(∂f

∂t(r(X, t), t) +

∂r

∂t(X, t)

∂f

∂x(r(X, t), t)

)dX.


Hint: replace σ(X) by the left hand side of the result in (c), replace ∂r∂t by the

velocity field v, given by

v(r(X, t), t) =∂r

∂t(X, t), (1.9)

use the definition of the material derivative DfDt

Df

Dt(x, t) =

∂f

∂t(x, t) + v(x, t)

∂f

∂x(x, t),

and transform back to the variable x, using x = r(X, t).

1.5. More examples with meteorological applications. To proceed, thestudent should have a thorough understanding of the concepts of vectors, func-tions and differentiation as described in the previous paragraphs and mathematicaltextbooks. To test his understanding and simultaneously to introduce new math-ematical, physical and meteorological concepts, the concepts of circular motion ofa particle, circulation and vorticity of a fluid, divergence and convergence, and ad-vection are presented. Chapter 1 ends with some useful vector algebraic results.The reader may expand on this by adding results on triple products, containingthe nabla operator and vector valued functions.

1.5.1. Motion of a particle in a circle. The motion of a particle in a circleis discussed to provide an example for introducing the concepts of velocity, accel-eration, circulation, divergence and advection. Moreover, it serves to prepare theway for the derivation of a formula for the acceleration of a small fluid element ina rotating frame of reference (Chapter 2). A solid body rotation in two dimensionscan be generated from a collection of circular particle motions, each with the sameangular velocity. To understand rotation of a fluid (or air mass) an understandingof solid body rotation is beneficial.

Consider the motion of a particle in a circle with radius a. Choose a frame ofreference and position vector

x(t) = (a cos θ(t), a sin θ(t)).

Example 1.5.1. Velocity and acceleration. Differentiation gives

x′(t) = aθ′(t)(− sin θ(t), cos θ(t))x′′(t) = aθ′′(t)(− sin θ(t), cos θ(t)) − aθ′(t)2(cos θ(t), sin θ(t)).

so that the definitions of the following vector valued functions are evident:• Position

x = (a cos θ, a sin θ)

• VelocityV = x′ = aω(− sin θ, cos θ)

1.5.1. Motion of a particle in a circle 17

• Acceleration

a = x′′ = aω′(− sin θ, cos θ)− aω2(cos θ, sin θ),

with θ the angle represented by the scalar valued function t → θ(t) and ω theangular velocity represented by the function t→ θ′(t).Definition 1.5.1. Define (see Fig. 1.7)

er = (cos θ, sin θ)eθ = (− sin θ, cos θ).

er

eθ

θ

Figure 1.7

Exercises 1.5.1. Co-ordinates or components with respect to a chosenframe of reference.(a) Check the differentiations in Example 1.5.1.(b) Show that

er · eθ = 0V · er = 0, the component in the direction of er

V · eθ = aω, the component in the direction of eθ

V = aωeθ

a · er = −aω2, the component in the direction of er

a · eθ = aω′, the component in the direction of eθ

a = aω′eθ − aω2er.

(c) Vector representation and the scalar product. Let x,V and a be as inExample 1.5.1. Simplify the scalar products in the following and interpret theresults:


x = (x · er)er + (x · eθ)eθV = (V · er)er + (V · eθ)eθa = (a · er)er + (a · eθ)eθV = (V · e1)e1 + (V · e2)e2.

Note that a vector has different components with respect to different bases. Com-pare, for example, the components of V in the second and fourth result above.Also, note the concept of projections (Exercise 1.1.3(b)). See Exercises 1.1.2 and1.6 (later on) for the concept of basis of a vector space.(d) Vector product, angular velocity vector. Define an angular velocity

vector ω = ω(0, 0, 1) = ωk. Let r = (x, 0) and U = (V, 0), with x and V asin Example 1.5.1. Show that

ω × r = aω(− sin θ, cos θ, 0)= (V, 0)= U.

Notation 1.5.1. As in Holton (1992), we use the symbol V to indicate a twodimensional velocity field and the symbol U to indicate a three dimensional velocityfield.

1.5.2. Circulation and vorticity. Let Ca be the circle with radius a and areaA = πa2. Consider an Eulerian description of the velocity V given in Exercise1.5.1(b) as V = aωeθ. Then Ca = x| ‖x‖ = a and the circulation C about theclosed contour Ca is (see Holton (1992), p. 88)

C =∮

V · dl

=∫Ca

V · dl

=∫ 2π

0

(aωeθ) · (adθ)eθ

=∫ 2π

0

(aωeθ · aeθ) dθ

=∫ 2π

0

a2ωdθ

= a2ω2π= 2ωA.

A mnemonic form of the rotation (curl) of a vector A is

1.5.2. Circulation and vorticity 19

curl A = ∇× A =

∣∣∣∣∣∣∣e1 e2 e3

∂1 ∂2 ∂3

A1 A2 A3

∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣i j k

∂x ∂y ∂z

Ax Ay Az

∣∣∣∣∣∣∣ .Exercise 1.5.2.1. Let U = (u(x, y), v(x, y), 0). Prove

curl U = (0, 0, ∂xv − ∂yu).

Exercises 1.5.2.2. In contrast with single particle motion, we now consider dif-ferent fluid particles, each rotating in a circle with its own radius r, around thesame fixed point. Consider circular motion, with radius r =

√x2 + y2 and velocity

U(x, y) = (−rω sin θ, rω cos θ, 0) = ω(−y, x, 0).Let ω be constant so that a rigid body motion ensues.(a) Compare the velocity field with V in Example 1.3.6 and discuss the direction

of the velocity vector.(b) Show that the vorticity vector ∇× U satisfies

∇× U = 2ωk

and that ∫∫A

∇× U · kdA = 2ωA =∮

U · dl.

1.5.2.3. Stokes’ theorem. Since ∇× U is constant, Stokes’ theorem∫∫A

∇× U · dA =∮

U · dlfollows trivially for this special case.Exercise 1.5.2.3. Stokes’ theorem on a rectangle. Consider the rectangleA = x| 0 < x1 < a, 0 < x2 < b, with boundary denoted by ∂A. Check each stepin the following:∫∫

A

curl U · dA =∫∫

A

curl U · kdA

=∫∫

A

(∂xv − ∂yu)dA

=∫ b

0

(∫ a

0

∂xvdx)dy −∫ a

0

(∫ b

0

∂yudy)dx

=∫ b

0

(v(a, y)− v(0, y))dy −∫ a

0

(u(x, b)− u(x, 0))dx

=∫∂A

U · dl.


Exercise 1.5.2.4. Polar co-ordinates for the rigid body motion. LetU(r, θ, z) = rωeθ and k = ez. For this example, a simplification of the generalrule in Calculus suffices, viz.

∇× A =1r

∣∣∣∣∣∣∣er reθ ez

∂r ∂θ ∂z

Ar rAθ Az

∣∣∣∣∣∣∣ .

Use this result to show that ∇×U = ∇× rωeθ = 2ωk and compare with Exercise1.5.2.2(b).

1.5.3. Divergence and convergence. Define the divergence of a vector A inCartesian co-ordinates by

∇ · A = ∂1A1 + ∂2A2 + ∂3A3

= ∂xAx + ∂yAy + ∂zAz .

Exercises 1.5.3.1.(a) Let U = (u(x, y), v(x, y), 0). Prove that ∇ · U = ∂xu+ ∂yv.(b) Let ω be constant. Show that the circular motion, with velocity field given by

U(x, y) = ω(−y, x, 0), is divergence free, i.e. ∇ · U = 0.(c) Explain how the vector field in (b) can be considered a field theoretical (Eule-

rian) description and find its relation to a material (Lagrangian) description.Hint: sketch the velocity vector at several points around the centre, definex(t) = a(cos(ωt), sin(ωt), 0) and find a relation between U(x(t)) and x′(t).

(d) The material derivative. Find a relation between x′′(t) and DUDt in Exercise

(c).(e) The geostrophic wind. Denote the two dimensional Laplace operator by

∇2 = ∂x2 + ∂y

2. Let f0 be a constant. Define the geostrophic wind by

Vg =(− 1f0∂yΦ

)i+

(1f0∂xΦ

)j.

Show that the geostrophic wind is divergence free:

∇ · Vg = 0

but has non zero vorticity:

k · ∇ × Vg =1f0

∇2Φ.

(f) Prove that Vg ⊥ ∇Φ and ∇Φ is perpendicular to all the curves where Φ =constant. What does this imply regarding the direction of Vg and surfaceswhere Φ = constant?

Exercises 1.5.3.2. Gauss’s theorem. Let B = (x1, x2, x3). Let V be a spherewith radius a and surface A.

1.5.4. Advection and the material derivative 21

(a) Show that ∇ · B = 3.(b) Since ∇ · B is constant, Gauss’s theorem

∫∫∫V

∇ · B dV =∫∫

A

B · ndA

follows trivially for this special case. Prove this statement.Exercise 1.5.3.3. Gauss’s theorem on a rectangular block. Consider theblock R = (x, y, z)|0 < x < a, 0 < y < b, 0 < z < c = (0, a) × (0, b)× (0, c). LetA denote the surface of R and n the outward directed normal vector. Check eachstep and complete the omitted parts in the following:

∫∫∫R

∇ · B dV =∫∫∫

R

(∂xBx + ∂yBy + ∂zBz)dxdy dz

=∫ c

0

(∫ b

0

(∫ a

0

∂xBx dx)dy)dz + · · ·

=∫ c

0

(∫ b

0

(Bx(a, y, z)− · · ·)dy)dz + · · ·

=∫ c

0

(∫ b

0

Bx(a, y, z)dy)dz −∫ c

0

(∫ b

0

Bx(0, y, z)dy)dz + · · ·

=∫ c

0

(∫ b

0

B(a, y, z) · n(a, y, z)dy)dz

−∫ c

0

(∫ b

0

B(0, y, z) · n(0, y, z)dy)dz + · · ·

=∫∫

A

B(x, y, z) · n(x, y, z)dA.

1.5.4. Advection and the material derivative. Let ω be a constant angularvelocity. To bring the rigid body fluid motion of Example 1.3.6 in line with theparticle motion of Section 1.5.1, put θ(t) = ωt. We thus aim to relate the veloc-ity field V, as defined in Example 1.3.6, to the particle motion described by thefollowing vector valued function r of a real variable t:

r : t → r(t) = (a cos(ωt), a sin(ωt)).

The velocity V(r(t)) experienced by the particle equals

V(a cos(ωt), a sin(ωt)) = (−ωa sin(ωt), ωa cos(ωt)) = aω(− sin(ωt), cos(ωt)).

The acceleration can be obtained by differentiation, as follows:

d

dt[V(r(t))] = aω2(− cos(ωt), − sin(ωt)) = −aω2er.


Substituting r(t) in the place of x in Example 1.3.6, we get the velocity advection((V · ∇)V)(r(t)) = −ω2r(t) = −aω2er,

which equals the derivative, following the motion. Thus the material derivativeequals the advection:

d

dt[V(r(t))] = ((V · ∇)V)(r(t)).

Note that in this example V is not explicitly a function of time t, so that ∂V∂t = 0and the above result conforms to the definition of the material derivative,

d

dt[V(r(t))] =

∂V

∂t(r(t)) + ((V · ∇)V)(r(t)) = DV

Dt(r(t)),

given in Definition 1.3.5.Exercise 1.5.4. A direct way of obtaining this result is to use the chain rule ofdifferentiation, to give

d

dt[V(r1(t), r2(t))] = (∂1V)r1′(t) + (∂2V)r2′(t)

= (∂1V)V1 + (∂2V)V2.

Make the derivation complete by proving amongst others that r1′(t) = V1 andr2

′(t) = V2.

1.5.5. Integration of knowledge. Develop your own integrated summary ofrelevant meteorological ideas, used as examples in Sections 1.4 an 1.5. Update yourportfolio of successes, mistakes and evaluations.

1.6. Further vector algebra. Define unit vectors in Rn

e1 = (1, 0, · · · , 0)e2 = (0, 1, · · · , 0)...

en = (0, 0, · · · , 1).Exercise 1.6. Show that B = e1, e2, · · · , en is a basis for Rn, by testing andinterpretation of the following:

x = x1e1 + x2e2 + · · ·xnen= Σni=1xiei

= Σni=1(x · ei)ei= (x · e1)e1 + (x · e2)e2 + · · ·+ (x · en)en.

Compare this result with Exercise 1.5.1(c).Remark 1.6. Let δij denote the Kronecker delta,

δij =

1, if i = j

0, if i = j.

Then ei · ej = δij and B is an orthonormal basis.

1.6.3. The vector triple product 23

1.6.1. Vector product (in R3).

x × y =

∣∣∣∣∣∣∣e1 e2 e3

x1 x2 x3

y1 y2 y3

∣∣∣∣∣∣∣= Σ3

i=1 ((x × y) · ei)ei= Σ3

i=1

(Σ3j=1Σ

3k=1εijkxjyk

)ei

with the permutation symbol εijk = 0 if any two indices are equal, otherwise |εijk| =1. A positive value, εijk = 1, is chosen if (i, j, k) ∈ (1, 2, 3); (2, 3, 1); (3, 1, 2), elsethe value is −1.1.6.2. The scalar triple product.Exercise 1.6.2. Combine the definitions for the scalar and cross product to showthat

a · b × c = Σ3i=1Σ

3j=1Σ

3k=1εijkaibjck.

a

b

c

b × c

Figure 1.8

Show that the volume of the parallelepiped in the sketch is a · b × c. (See Apostol(1967).)

1.6.3. The vector triple product. Combine the definitions for the crossproduct to show that

a × (b × c) = Σ3i=1

(Σ3j=1Σ

3k=1εijkajΣ

3l=1Σ

3m=1εklmblcm

)ei

= Σ3i=1

(Σ3j=1Σ

3k=1Σ

3l=1Σ

3m=1εkijεklmajblcm

)ei

= Σ3i=1

(Σ3j=1Σ

3l=1Σ

3m=1(δilδjm − δimδlj)ajblcm

)ei

= Σ3i=1Σ

3j=1 ((ajcj)bi − (ajbj)ci)ei

= (a · c)b − (a · b)c.


Exercise 1.6.3. Check each step in the previous derivation and make it morecomplete where necessary.

1.6.4. Integration of knowledge.Exercise 1.6.4. Expand the differentiations in expressions like

∇ · a × b

∇× (a × b)∇ · ca∇× ca, etc.

and compare with the literature.

Chapter 2

The momentum equation in rotating spherical co-ordinates

A true story about Coriolis. Persson (2000) writes: The early nineteenth cen-tury was a time of change, with the Industrial Revolution in full swing and Frenchindustry lagging behind the British. A radical and patriotic movement developedwithin l’Ecole Polytechnique to promote technical development by educating work-ers, craftsmen and engineers in ‘mechanique rationelle’ to make them understandthe functioning of machines in order to improve them. In his 1829 book Calcul del’effet des machines Coriolis presented mechanics in a way that could be used bythe industry. The book was also a milestone in the general development of physicssince it established for the first time the correct relation between potential and ki-netic energy, and showed that their sum remained constant in the absence of anyexternal force.

Persson (2000) also writes: Meteorological textbooks go to great lengths to im-press on the reader that the Coriolis force due to its inertial nature is ‘fictitious’,‘artificial’, a ‘pseudo force’ or even a ‘mental construct’. The centrifugal force,which of course is equally ‘fictitious’, is rarely talked about in this way. This mighteasily mislead an innocent reader into believing that some ‘fictitious’ forces are morefictitious than others.

Introduction. The purpose of Chapter 2 is to obtain the acceleration terms inthe momentum equations in the form (2.19) – (2.21) of Holton (1992), p. 37. Thebenefit to the student should be

a clear understanding of the movement of an air parcel in an inertial referenceframe• from the viewpoint of an observer on the rotating earth• in spherical co-ordinates (as in Holton (1992) paragraph 2.3, p.33)

by the mathematical approach of direct differentiation of the position andvelocity vectors in the inertial frame of reference.

2.1. The inertial reference frame. Suppose the movement of the earth aroundthe sun contributes in a short time period to a constant velocity of an air parcel

26 The momentum equation in rotating spherical co-ordinates

only—thus not contributing to the acceleration of the parcel. We thus choose ourinertial frame of reference to be a frame moving with the earth around the sun,but not rotating. We choose the origin of this inertial frame to be at the centre ofthe earth and the x–y plane to intersect the earth at the equator. A point in theinertial frame will be denoted by a Cartesian 3-tuple, e.g.

x = (x1, x2, x3). (2.1)

2.1.1. Spherical co-ordinates. The spherical co-ordinates, λa (longitude ininertial or absolute frame), φ (latitude) and r (radius) of the point x in (2.1) canbe obtained from

λa = arctan(x2/x1)

r =√x1

2 + x22 + x3

2

φ = arcsin(x3/r).

x1

x2

x3

λa

φ

Figure 2.1


x = r(cos φ cosλa, cosφ sinλa, sinφ). (2.2)

(b) Let x = (1.732, 1.732, 1.414). Calculate λa, φ, r.

2.2. The rotating frame of reference. Let λ denote the longitude of a parcelin the rotating frame. Then λa = Ωt+ λ and (2.2) becomes

x = r(cos φ cos(Ωt+ λ), cosφ sin(Ωt+ λ), sin φ). (2.3)

2.3. Differentiation of an arbitrary vector A 27

2.2.1. Spherical co-ordinates in the rotating frame of referenceExercises 2.2.1.(a) Let t be fixed in equation (2.3) and discuss the dependency of x on λ (lon-

gitude), φ (latitude) and r (radius) in the rotating frame. Hint: consider bigand small circles on the earth.

(b) Calculate ∂x∂λ ,∂x∂φ and

∂x∂r .

(c) Integration of knowledge. Interpret and expand Exercise 1.1.3(c) in thelight of new knowledge.

2.2.2. The motion of a particle (parcel of air) in the rotating frame. Lett → r(t) be a function describing the motion of the particle. Since its longitude,latitude and height z (z = r − a, a = earth radius) depend on time we may write

r(t) = r(t)(cosφ(t) cos(Ωt+ λ(t)), cosφ(t) sin(Ωt+ λ(t)), sin φ(t)) (2.4)

or suppressing the variable t

r = r(cosφ cosλa, cosφ sinλa, sinφ). (2.5)


dr

dt= (r cosφ)(Ω +

dλ

dt)i+ r

dφ

dtj+

dr

dtk (2.6)

where

i = (− sinλa, cosλa, 0) (2.7)j = (− sinφ cosλa,− sinφ sinλa, cosφ) (2.8)k = (cosφ cosλa, cosφ sinλa, sinφ) (2.9)

are orthogonal unit vectors rotating with the earth (see Exercise 1.1.3(c)). Notethat r = rk.(b) Define Ω = (0, 0,Ω). Let r be as in (2.5). Show that

Ω × r = (rΩcosφ)i. (2.10)

(c) Define

u = r cosφdλ

dt(2.11)

v = rdφ

dt(2.12)

w =dr

dt=dz

dt. (2.13)

Show that (2.6) implies

dr

dt= ui+ vj+ wk+Ω × r. (2.14)


2.3. Differentiation of an arbitrary vector A. We prefer to combine thereasoning of Holton (1992) Sections 2.1.1 and 2.2. Thus we choose Ax, Ay and Azto be the components of the vector A in the rotating frame of reference. We have

A = Axi+Ayj+Azk. (2.15)

(Note that Holton (1992) uses the notation A′x, A

′y and A

′z and i′, j′ and k′ instead.)

Exercises 2.3. Let λ, φ and r be functions of time t in (2.7) – (2.9). Let u, v andw be as in (2.11) – (2.13).(a) Show that

di

dt= (

u

rtanφ+Ωsinφ)j − (u

r+Ωcosφ)k (2.16)

dj

dt= −(u

rtanφ+Ωsinφ)i − v

rk (2.17)

dk

dt= (

u

r+Ωcosφ)i+

v

rj. (2.18)

(b) Show that

Ω = (Ω · i)i+ (Ω · j)j+ (Ω · k)k= (Ω cosφ)j+ (Ω sinφ)k. (2.19)

(c) Define the Coriolis parameter f by f = 2Ω sinφ. Show that Ω can be writtenin terms of horizontal and vertical components as follows:

Ω = Ωh +f

2k.

Explain your result by calculating the components from a sketch.(d) Show that

Ω × i = −(Ω cosφ)k+ (Ω sinφ)j (2.20)Ω × j = −(Ω sinφ)i (2.21)Ω × k = (Ω cosφ)i. (2.22)

(e) Use the results of (d) to show that (a) implies

di

dt= Ω × i+ (

u

rtanφ)j − u

rk (2.23)

dj

dt= Ω × j − (u

rtanφ)i − v

rk (2.24)

dk

dt= Ω × k+

u

ri+

v

rj. (2.25)

2.4. Velocity and acceleration 29

(f) Show that

dA

dt=dAxdt

i+dAydt

j+dAzdt

k+Ω × A

+1r(−Ayu tanφ+Azu)i

+1r(Axu tanφ+Azv)j

+1r(−Axu−Ayv)k. (2.26)

2.3.1. Notation. For convenience, but always keeping in mind the meaning ofeach term in (2.26) we define

DaA

Dt=dA

dt(2.27)

and

DA

Dt= (

dAxdt

− uAy tanφr

+uAzr)i

+ (dAydt

+uAx tanφ

r+vAzr)j

+ (dAzdt

− uAx + vAyr

)k (2.28)

so that (2.26) becomesDaA

Dt=DA

Dt+Ω × A. (2.29)

Exercise 2.3.1. Compare (2.28) with equation (2.14) in Holton (1992), Section2.3, p. 36.

2.4. Velocity and acceleration.

2.4.1. The velocity vector.Exercise 2.4.1. Apply (2.28) to the position vector r(t) to obtain

Dr

Dt= ui+ vj+ wk. (2.30)

Hint: Note that if A = r then Ax = 0, Ay = 0, Az = r in (2.28). Note also (2.13)which is (2.9) in Holton (1992), p.33.Definition 2.4.1. Define the (absolute) velocity vector as

Ua =Dar

Dt(2.31)


and the velocity vector as

U =Dr

Dt= ui+ vj+ wk (2.32)

so that (2.29) gives

Ua =Dar

Dt= U+Ω × r. (2.33)

Ω × r

Ω

Figure 2.2

2.4.2. The acceleration vector. We now apply (2.29) to the absolute velocityvector Ua to obtain

DaUaDt

=DaDt(U+Ω × r)

=DaU

Dt+Ω × Dar

Dt

=DU

Dt+Ω × U+Ω × (U+Ω × r) by use of (2.29) and (2.33)

=DU

Dt+ 2Ω × U+Ω × (Ω × r). (2.34)

Exercise 2.4.2.(a) Use (2.28) to prove the following

DU

Dt= (

du

dt− uv tanφ

r+uw

r)i

+ (dv

dt+u2 tanφ

r+vw

r)j

+ (dw

dt− u2 + v2

r)k. (2.35)

(b) Compare (2.34) with Holton (1992) (2.7) and (2.35) with Holton (1992) (2.14).

2.4. Velocity and acceleration 31

(c) Use (2.19) and (2.32) to show that

Ω × U = (wΩcosφ− vΩ sinφ)i+ (uΩ sinφ)j − (uΩcosφ)k. (2.36)

(d) Show that

Ω × (Ω × r) = (rΩ2 cosφ sin φ)j − (rΩ2 cos2 φ)k. (2.37)

(e) Substitute (2.35) – (2.37) in (2.34) and write your result in the form (2.19) –(2.21), Holton (1992), p.37.

(f) In (2.34) we have used DaΩDt = . Prove DΩ

Dt = , using the differentiationrule (2.28) and the expression (2.19) for Ω. Next, prove DaΩ

Dt = .(g) Let f be the Coriolis parameter as in Exercise 2.3(c). Show that

2Ω × U = 2Ω cosφ(wi − uk) + fk × V, (2.38)

withV = ui+ vj (2.39)

andU = V+ wk (2.40)

in accordance with Notation 1.5.1 for the two and three dimensional velocityfield respectively.

Chapter 3

Balance laws in physics

Cauchy wrote: The geometers who have investigated the equations of equi-librium or motion of thin plates or of surfaces, either elastic or inelastic, havedistinguished two kinds of forces, the one produced by dilatation or contraction, theother by the bending of these surfaces . . . It has seemed to me that these two kindsof forces could be reduced to a single one, which ought to be called always tensionor pressure, a force which acts upon each element of a section at will, not only in aflexible surface but also in a solid, whether elastic or inelastic, and which is of thesame kind as the hydrostatic pressure exerted by a fluid at rest upon the exteriorsurface of a body, except that the new pressure does not always remain perpendic-ular to the faces subject to it, nor is it the same in all directions at a given point.Cauchy, On the pressure or tension in a solid body, Exercises de Mathematiques,Seconde Annee (1827). (See Truesdell (1977), p. 118)

In many otherwise good textbooks a standing confusion reigns between threegroups of forces: 1. Internal and external forces. 2. Volume and surface forces—adistinction which the mechanics of points is altogether incapable of perceiving. 3.Applied forces and forces of reaction. Hamel, On the foundations of Mechanics,Mathematische Annalen 66, 350–397 (1909). (See Truesdell (1977), p. 118)

Introduction. The purpose of this chapter is to• give a thorough background of the physics and mathematics involved in thelaws of mass conservation, momentum balance and energy conservation

• show that these laws can be considered as special cases of a general or abstractbalance law

• show how the constitutive equations make the general law applicable to aspecial situation and specific material

• transform differential expressions to isobaric (pressure-based) co-ordinates.

The reader, familiar with Holton’s (1992) approach to conservation laws, may findit illuminating to know that these laws are treated from a fundamental point ofview in various books on continuum mechanics, e.g. Spencer (1980) and Truesdell

34 Balance laws in physics

(1977), as well as in other works, e.g. Serrin (1959) in Handbuch der Physik. Thischapter is based on ideas from Fung (1969), Serrin (1959), Atkin & Fox (1980)and Spencer (1980). Themes that will not be discussed in this chapter are theconservation of angular momentum, resulting in the symmetry property of thestress tensor; the Cauchy formula (T = τn), giving the stress vector T in termsof the stress tensor τ and the orientation, specified by n, of the surface on whichthe stress vector is desired; and the stress-strain-rate relationships (Fung, 1969).Instead, the interested reader is referred to the literature.

Note the symmetry between the Cauchy formula T = τn and the result F ′(x;n) =DF(x)n of Exercise 1.3.6(c): the stress vector is equal to the product of a matrixand a unit vector, and similarly, the directional derivative is equal to the productof the Jacobian matrix and a unit vector.

Let ρ be a density field (function) and note that it has dimension [ρ] = ML−3,where M denotes mass (kg) and L denotes length (m).

3.1. One dimensional derivation of the mass conservation principle. Toprepare the reader for the three dimensional situation and to clearly convey thebasic concepts, we first consider one dimensional flow in a pipe with constant crosssection A. We model the situation by considering a density function ρ : (x, t) →ρ(x, t), x ∈ [0, l], t ≥ 0. The interval [a, b] will always be an arbitrary subinterval of[0, l].

3.1.1. Eulerian approach. We will use the powerful tools of the Calculusto convert the balance of mass in an arbitrary interval [a, b] from a statement inintegral form to a differential equation. The mass in [a, b] is

∫ ba ρ(x, t)Adx and the

time rate of change of this mass is ddt

∫ baρ(x, t)Adx. The flow of mass (dimension

MT−1) is caused by inflow of material at a and outflow at b. If we denote the massflux by φ, where [φ] =MT−1L−2, we have the basic conservation law

d

dt

∫ b

a

ρ(x, t)Adx = Aφ(a, t)−Aφ(b, t). (3.1)

By (3.1) an increase (or convergence) of mass in [a, b] takes place if φ(a, t) > φ(b, t),i.e. more material enters at a than leaves at b. Decrease (or divergence) takes placeif φ(a, t) < φ(b, t). Dividing (3.1) by A and using the fundamental theorem of theCalculus (1.2), we obtain

d

dt

∫ b

a

ρ(x, t)dx = −∫ b

a

∂φ

∂x(x, t)dx.

Using (1.4) we get

∫ b

a

[∂ρ

∂t(x, t) +

∂φ

∂x(x, t)

]dx = 0 for every [a, b] ⊂ [0, l].

3.2.1. Eulerian approach 35

Since [a, b] is arbitrary, we may write the above balance law in the differentialequation form

∂ρ

∂t(x, t) +

∂φ

∂x(x, t) = 0 for every x ∈ [0, l]. (3.2)

3.1.2. The constitutive equation. The mass flux is the amount of mass thatflows out per unit time per unit area. A volume A(vδt), having mass ρA(vδt) flowsout during time δt. The constitutive equation for the mass flux is φ = ρA(vδt)

A(δt) = ρv.

The continuity equation. Substituting φ = ρv into (3.2) yields the continuityequation

∂ρ

∂t+∂(ρv)∂x

= 0. (3.3)

3.1.3. Lagrangian approach. We still consider fluid motion in a one dimen-sional continuum. It is convenient to consider the fluid at any time instant t to bea mapping from a reference configuration to its current configuration (see Section1.4.3). Thus, consider a mapping r : (X, t) → r(X, t) of the reference interval[0, L] to the configuration [0 , l] at time t. Let [A,B] be an arbitrary interval in thereference configuration, so that [r(A, t), r(B, t)] always contains the same material.In Section 1.4.3 such an interval was called a material volume. In (1.9) we definedthe fluid velocity by

v(r(X, t), t) =∂r

∂t(X, t).

It follows thatv(x, t) =

∂r

∂t(X, t) (3.4)

withx = r(X, t). (3.5)

Consider the conservation of mass, expressed by

d

dt

∫ r(B,t)

r(A,t)

ρ(x, t)dx = 0. (3.6)

Exercises 3.1.3(a) Use (1.6) to write the mass conservation principle (3.6) as

ρ(r(B, t), t)∂r

∂t(B, t)− ρ(r(A, t), t)

∂r

∂t(A, t) +

∫ r(B,t)

r(A,t)

∂ρ

∂t(x, t)dx = 0. (3.7)

(b) Use (3.7) and the above definition of the velocity v to prove

∫ r(B,t)

r(A,t)

∂

∂x[ρ(x, t)v(x, t)]dx +

∫ r(B,t)

r(A,t)

∂ρ

∂t(x, t)dx

= ρ(r(B, t), t)v(r(B, t), t) − ρ(r(A, t), t)v(r(A, t), t) +∫ r(B,t)

r(A,t)

∂ρ

∂t(x, t)dx

= 0.

Show that this again implies (3.3).


3.2. Three dimensional derivation of the mass conservation principle.

3.2.1. Eulerian approach. In Section 3.1.1 the fundamental theorem of theCalculus (1.2) was used to write the net flux over the end points of the interval [a, b]as an integral. In the three dimensional situation the divergence theorem (Gauss’divergence theorem) ∫

A

B · ndA =∫V

∇ · B dV (3.8)

serves a similar purpose. Let V be an arbitrary fixed volume in space. Let Vbe bounded by its surface A, having outward directed unit normal vector n at x.(Note that n : x → n(x) changes in direction as x moves on the surface A of V .)Let ρ be the fluid density and Φ the outward flux vector. The basic mass balancelaw is

d

dt

∫V

ρdV = −∫A

Φ · ndA. (3.9)

Exercises 3.2.1. Assume interchanging the time derivative with integration, asin (1.4), also holds for three dimensional integrals:

d

dt

∫V

f(x, t)dV =∫V

∂f

∂t(x, t)dV. (3.10)

(a) Discuss the direction and sign of the flux Φ and the integral of its normalcomponent, using an adaptation of the arguments in Section 3.1.1.

(b) Use (3.8) and (3.10) to write the basic mass balance law (3.9) as∫V

(∂ρ

∂t+∇ · Φ)dV = 0

and motivate the ensuing balance law

∂ρ

∂t+∇ · Φ = 0. (3.11)

Compare (3.11) with (3.2).(c) Let U be the fluid velocity field. Motivate the assumption Φ · n = ρU · n, by

considering flow out of an area element δA during a time interval δt. Then,use the fact that n is arbitrary (it was associated with the orientation, at thepoint x, of the surface area A of the arbitrary volume V ) together with theresult of Exercise 1.6 to prove that Φ = ρU.

3.2.2. The continuity equation. Substitution of Φ = ρU in (3.11) gives

∂ρ

∂t+∇ · (ρU) = 0. (3.12)

Exercise 3.2.2. Compare (3.12) with (3.3). Next, show that (3.12) implies

Dρ

Dt+ ρ∇ · U = 0, (3.13)

where the material derivative DDt is defined in Definition 1.3.5.

3.2.3. Lagrangian approach 37

3.2.3. Lagrangian approach. We first give an alternative derivation for theone dimensional case of Section 3.1.3. Put J(X, t) = ∂r

∂X (X, t). By Exercise 1.4.3(c)

∫ r(B,t)

r(A,t)

ρ(x, t)dx =∫ B

A

σ(X)dX

=∫ B

A

ρ(r(X, t), t)∂r

∂X(X, t)dX

=∫ B

A

ρ(r(X, t), t)J(X, t)dX. (3.14)

Since time differentiation of (3.14) requires knowledge of ∂∂t [ρ(r(X, t), t)] and

∂J∂t (X, t), consider

∂

∂t

[ρ(r(X, t), t)

]=∂ρ

∂t(r(X, t), t) +

∂ρ

∂x(r(X, t), t)

∂r

∂t(X, t)

=∂ρ

∂t(r(X, t), t) +

∂ρ

∂x(r(X, t), t)v(r(X, t), t)

=Dρ

Dt(r(X, t), t) (3.15)

and

∂J

∂t(X, t) =

∂

∂t

[∂r

∂X(X, t)

]

=∂2r

∂X∂t(X, t)

=∂

∂X[v(r(X, t), t)]

=∂v

∂x(r(X, t), t)

∂r

∂X(X, t)

=∂v

∂x(r(X, t), t)J(X, t)

= J(X, t)∂v

∂x(r(X, t), t). (3.16)

Thus, differentiation of (3.14) yields

d

dt

∫ r(B,t)

r(A,t)

ρ(x, t)dx =d

dt

∫ B

A

ρ(r(X, t), t)J(X, t)dX

=∫ B

A

∂

∂t

[ρ(r(X, t), t)J(X, t)

]dX


=∫ B

A

[J(X, t)

∂

∂t

ρ(r(X, t), t)

+ ρ(r(X, t), t)

∂J

∂t(X, t))

]dX

=∫ B

A

[J(X, t)

Dρ

Dt(r(X, t), t) + ρ(r(X, t), t)J(X, t)

∂v

∂x(r(X, t), t)

]dX

=∫ B

A

J(X, t)[Dρ

Dt(r(X, t), t) + ρ(r(X, t), t)

∂v

∂x(r(X, t), t)

]dX

=∫ r(B,t)

r(A,t)

(Dρ

Dt(x, t) + ρ(x, t)

∂v

∂x(x, t)

)dx. (3.17)

We are now ready to proceed with the three dimensional version of (3.16) and (3.17).Let f : x ∈ V0 → f(x) ∈ V be a one-to-one and continuously differentiable vectormapping from V0 to V . Then Df(x) is the Jacobian matrix defined in Exercise1.3.6(c). Let J be the Jacobian determinant: J(x) = detDf(x). The followingthree dimensional version of (1.7) is adapted from p. 408 in Apostol (1969):∫

V

g(u)du =∫V0

g(f(x))|J(x)|dx. (3.18)

Let χ : (X, t) → χ(X, t) be a mapping from the three dimensional reference config-uration C into the current configuration Ct. We choose f in (3.18) to be the functionχ(·, t) : X → χ(X, t), with t kept fixed. Thus f(X) = χ(X, t), for every X ∈ C.The derivative of the Jacobian determinant is proved by Serrin (1959) and Atkin& Fox (1980) to satisfy

DJ

Dt= J∇ · U. (3.19)

In continuum mechanics J > 0, since a body can not penetrate itself (Atkin & Fox(1980)).Exercise 3.2.3.(a) Compare (3.16) with (3.19) and compare (1.7) with (3.18).(b) Let V (t) = χ(C, t) be the material volume obtained when the reference con-

figuration C is mapped by χ to its current state. Use (3.18) to transform∫V (t)

ρ(u, t)du to an integral over the reference state C. Differentiate yourresult, using (3.10) to differentiate the integral over the reference state C. Use(3.15) and (3.19) to obtain∫

C(Dρ

Dt+ ρ∇ · U)J dV0 = 0, (3.20)

and thus ∫V (t)

(Dρ

Dt+ ρ∇ · U)dV = 0, (3.21)

so thatDρ

Dt+ ρ∇ · U = 0,

which again proves (3.13).

3.3.1. Eulerian derivation 39

3.3. Abstract balance law.

3.3.1. Eulerian derivation.(a) One dimensional case. Let f : (x, t) → f(x, t) be any quantity that is

conserved or for which a balance law is sought. Let F be the external forcing(or sources) per unit length and φ the flux of f . The balance of quantities inthe one dimensional interval (0, l) is

d

dt

∫ b

a

f(x, t)dx = φ(a, t)− φ(b, t) +∫ b

a

F (x, t)dx for every (a, b) ⊂ (0, l).(3.22)

The abstract balance law is described by the following partial differential equa-tion:

∂f

∂t(x, t) +

∂φ

∂x(x, t) = F for every x ∈ (0, l) and t > 0. (3.23)

(b) Three dimensional case. Let f : (x, t) → f(x, t) be a field variable thatis conserved (or for which a balance law is sought) in a region G ⊂ R3 andΦ : (x, t) → Φ(x, t) be the flux of f . Let F be the external forcing or sourcesper unit volume and S the external forcing or sources per unit area. Let V beany arbitrary subset of G. The integral form of the balance law for f in thethree dimensional region V is

d

dt

∫V

f(x, t)dV = −∫A

Φ(x, t) · ndA+∫V

F (x, t)dV +∫A

S · ndA (3.24)

for every V ⊂ G

and implies∂f

∂t+∇ · Φ = F +∇ · S. (3.25)

Exercises 3.3.1.(a) The heat equation. Let f/ρ = e = cvT be the internal energy per unit mass

(Holton (1992), Section 2.6, p. 51), Φ = −κ∇T , the heat flux (Example 1.3.3and Exercise 1.3.4(d)) and put k = κ

cvρ. Let F = 0 and S = . Show that

(3.25) implies∂T

∂t= k∇2T.

What is the one dimensional version (spatial) of this equation? Use (3.22) and(3.23).

(b) The momentum equations. Let f = ρUi be the momentum per unit volumein the i-th direction and Φ = fU the momentum flux. Let gi = g · ei bethe component of gravitational acceleration in the i-th direction. Put Fi =ρ(gi − 2Ω × U · ei) and S = −pei. Let V be any arbitrary set and A itsboundary. Show how the following momentum balance law can be consideredto be a generalization for fluid motion based on Newton’s second law:

d

dt

∫V

f(x, t)dV = −∫A

Φ(x, t)·ndA+∫V

ρ(gi−2Ω×U·ei)dV −∫A

pei ·ndA.


Show that the balance law implies

∂(ρUi)∂t

+∂p

∂xi+∇ · (ρUiU)− ρ(gi − 2Ω × U · ei) = 0.

Next, show that a combination with the continuity equation results in

DUiDt

= −1ρ

∂p

∂xi+ (g − 2Ω × U) · ei

and consequentlyDU

Dt= −1

ρ∇p+ g − 2Ω × U.

3.3.2. Lagrangian derivation of the thermodynamic equation. Choosef in (3.24) to be the total thermodynamic energy per unit volume, ρ[e+ 1

2‖U‖2].Let Φ = and let S = −pU, with −S the pressure flux. Let g · U be the rateat which body forces do work per unit mass, and let J denote the rate of heatingper unit mass owing to radiation, conduction, and latent heat release and combinethese two contributions into F = ρ(g · U+ J).Exercise 3.3.2. Formulate a three dimensional version of (1.8) and adapt (3.24)for the material description to show that

ρ(De

Dt+

D

Dt

12‖U‖2) = −∇ · (pU) + ρ(g · U+ J).

Note that the same symbol J is also used for the totally different conceptof the Jacobian determinant in Section 3.2.3.

3.4. Transformation of conservation equations. The pressure/height inter-dependency was already discussed in Section 1.2. The main result in this section isthe transformation of the continuity equation (3.13) from (x, z) co-ordinates to theso called isobaric co-ordinates (x, p). The transformed equation, (3.50) does notcontain density ρ and does not involve time derivatives. Holton (1992) sees the sim-plicity of (3.50) as one of the main advantages of the isobaric co-ordinate system. InChapter 4 density is also removed from the momentum equations, after introducingthe geopotential (Exercise 1.2(c)) and transforming to isobaric co-ordinates.

3.4.1. A choice: atmospheric pressure or height as independent variable.First consider atmospheric pressure, restricted to a vertical plane. Choose the x–zco-ordinate surface to coincide with this plane. Thus for each point (x, z) on thevertical plane there is defined a pressure p(x, z). Suppose p is constant on a curveC in this plane and that C can be described by a function z : x → z(x). In Holton(1992), p. 22 a part of z is sketched where it is assumed to be a linear increasingfunction. Along the curve C we have

3.4.3. Isobaric Coordinates 41

p(x, z(x)) = constant (3.26)d

dx[p(x, z(x))] = 0 (3.27)

∂p

∂x(x, z(x)) +

∂p

∂z(x, z(x))z′(x) = 0. (3.28)

Exercise 3.4.1. Compare (3.28) with(∂p

∂x

)z

+(∂p

∂z

)x

(∂z

∂x

)p

= 0 (3.29)

which is given on p. 22 of Holton (1992) and note carefully the subscripts z, x andp. Look how (3.29) is derived in Holton (1992) and note the curves along which z,x or p are being kept fixed.

Three dimensional situation. Next, consider a given point x = (x1, x2) on ahorizontal plane. Exercise 1.2(b) suggests the following possibilities:• Pressure as a function of three independent variables: during a vertical sound-ing a pressure, say p1, is measured at height z1, giving

p(x, z1) = p1. (3.30)

• Height as a function of three independent variables: while the vertical sound-ing, implied by (3.30), gives pressure p1 at height z1, it can equivalently besaid that the height z1 is reached when the pressure is p1. Thus

z(x, p1) = z1. (3.31)

3.4.2. Pressure and height as inverses.Example 3.4.2. A function notation for height and pressure. A combina-tion of (3.30) – (3.31) necessitates the definition of two symbols for pressure andtwo symbols for height: the pressure p at a height z and the functions p and zdescribing their interdependency. Thus (3.30) and (3.31) can be written as

p(x, z) = p (3.32)z(x, p) = z. (3.33)

Exercises 3.4.2.(a) Substitute (3.33) into (3.32), (thereby eliminating z) and write your result

down here, as equation (3.34):

p(x, . . . . . . . . . . . .) = p (3.34)

Differentiate your result (3.34) with respect to x1. Next, differentiate your(3.34) with respect to x2. Combine the result in vector form as

∇z p+(∂p

∂z

)x∇pz = 0. (3.35)

(b) Differentiate your equation (3.34) with respect to p to give∂p

∂z(x, z(x, p))

∂z

∂p(x, p) = 1. (3.36)


3.4.3. Isobaric Coordinates. For simplicity we only distinguish between p andp, z and z when necessary. Suppose the pressure p serves the role of independentvertical co-ordinate better than the vertical height z does. A function f : (x, z, t) →f(x, z, t) can be transformed into a function F : (x, p, t) → F (x, p, t) by using

f(x, z, t) = F (x, p(x, z, t), t). (3.37)

Examples 3.4.3.(a) Wind as a function of x and p. Let v : (x, z, t) → v(x, z, t) and V :

(x, p, t) → V(x, p, t) be field functions, respectively defined on (x, z) and (x, p)co-ordinates, describing the horizontal wind. The transformation (3.37) gives

v(x, z, t) = V(x, p(x, z, t), t). (3.38)

(b) The equation of state for dry air. If z is chosen as vertical coordinate wemay write

p(x, z) = ρ(x, z)RT (x, z).

If z is replaced by the left hand side of (3.33) then

p(x, z(x, p)) = ρ(x, z(x, p))RT (x, z(x, p)),

which impliesp = ρ(x, p)RT (x, p).

A barotropic atmosphere is one in which the density depends only on thepressure, ρ : p → ρ(p), so that the equation of state becomes

p = ρ(p)RT (x, p).

Exercises 3.4.3.(a) Prove that ∇pT = 0 in a barotropic atmosphere.(b) Prove the following by direct differentiation of (3.37):

∂f

∂t=∂F

∂t+∂F

∂p

∂p

∂t(3.39)

∇zf = ∇pF + ∂F

∂p∇zp (3.40)

∂f

∂z=∂F

∂p

∂p

∂z. (3.41)

(c) The material derivative in isobaric coordinates. Let w = DzDt and ω =

DpDt . Study the following carefully and compare with equation (3.3) on p. 59of Holton (1992):

3.4.4. Transformation of the Continuity equation 43

Df

Dt= (

∂

∂t+ v · ∇z + w

∂

∂z)f

= (∂

∂t+ V · ∇p)F + ∂F

∂p(∂p

∂t+ V · ∇zp+ w

∂p

∂z)

= (∂

∂t+ V · ∇p)F + ∂F

∂p

Dp

Dt

= (∂

∂t+ V · ∇p + ω

∂

∂p)F. (3.42)

(d) The w—ω relation. In Section 3.5 of Holton (1992) the hydrostatic approxi-mation is used together with scaling considerations to show that DpDt ≈ −ρgw.The hypsometric equation, Holton (1992), p. 20,

Z − Z0 =R < T >

g0ln

(p0

p

),

with < T > the layer mean temperature to be defined in (4.9), is also derivedusing the hydrostatic approximation. Assume p0 and R<T>

g0= H are constant

and that p = ρR < T >. Prove by differentiation with respect to time t that

ω = −ρg0w.

3.4.4. Transformation of the Continuity equation. Holton (1992) mentionsthat the isobaric form of the continuity equation can be obtain by a transformationfrom height co-ordinates to pressure co-ordinates. He then chooses to do the simplerdirect derivation, instead. The learner is now challenged: put away the Workbookand try to do the transformation yourself! However, if you need help, the followingexercises may put you back on track. The equations we need are

∇zp+(∂p

∂z

)x∇pz = 0 (3.43)

∂p

∂z= −ρg (3.44)

1ρ

Dρ

Dt+∇z · v+ ∂w

∂z= 0 (3.45)

andv(x, z(x, p, t), t) = V(x, p, t). (3.46)

Exercises 3.4.4.(a) Replace z in (3.38) with z(x, p), and use (3.34) to prove (3.46).(b) Let g be constant as indicated in the footnote on p. 60 of Holton (1992).

Complete the following argument and check every step. Write out in moredetail if necessary.


−gDρDt

=D

Dt

∂p

∂z

= (∂

∂t+ v · ∇z + w

∂

∂z)∂p

∂z

=∂

∂z(∂

∂t+ v · ∇z + w

∂

∂z)p− · · · · · ·

=∂

∂z

Dp

Dt+∂v

∂z· ∂p∂z

∇pz − · · ·

=∂ω

∂z+∂p

∂z(∂v

∂z· ∇pz − ∂w

∂z)

(c) Use (3.44) and (3.45) to rewrite the result of Exercise (b) as

∂ω

∂z− ρg(∇z · v+ ∂v

∂z· ∇pz) = 0. (3.47)

(d) Differentiate (3.46) with respect to xi to obtain

∂iv+∂v

∂z∂iz = ∂iV, i = 1 or 2. (3.48)

(e) Use v = (v1, v2) and V = (V1, V2) to write (3.48) in component form. Bychoosing appropriate values for i in the two resulting equations and adding,show that

∇z · v+ ∂v

∂z· ∇pz = ∇p · V. (3.49)

(f) Substitute (3.49) together with

∂ω

∂z=∂ω

∂p

∂p

∂z= −ρg ∂ω

∂p

in (3.47) to obtain the transformed continuity equation

∂ω

∂p+∇p · V = 0. (3.50)

3.4.5. A Generalized Vertical Coordinate. Consider the transformations

f(x, z, t) = F (x, s(x, z, t), t) (3.51)f(x, z(x, s, t), t) = F (x, s, t), (3.52)

where s is a single-valued monotonic function of height.Exercise 3.4.5. Prove the following and compare with p. 23 in Holton (1992):

∇zf = ∇sF + ∂F

∂s∇zs

∇zf + ∂f

∂z∇sz = ∇sF.

Chapter 4.

The quasi-geostrophic approximation

Gone with the wind. Zbigniew Sorbjan (1996) relates the following frighteningincidents in his book: In 1930, a deadly demonstration of the hail formation processwas provided by five glider pilots in Germany. The pilots soared into a thunderstormin the region of the Rhon Mountains, and were trapped in the cloud’s powerfuldrafts. After losing control of their planes, all five parachuted out. Powerful updraftsimmediately lifted them to regions of subfreezing temperatures. Then they fell downwith the downdraft, only to be raised up and frozen again. After a series of lifts andfalls they literally became human hailstones. When they finally reached the Earth,they were all frozen stiff an only one survived. Similarly in 1982, an Australianparachutist was trapped for half an hour in a cumulonimbus cloud. When he openedhis parachute, he was lifted by an updraft from a height of 2 km to 4 km. Theparachutist rescued himself by cutting out from his main parachute. As a result offree-falling, he was able to move through the cloud. About 500 m above the surface,he opened his reserve parachute and then landed safely on the ground.

Tropical cumulonimbus in the western Pacific.Photo obtained from http://www.photolib.noaa.gov.

Introduction. The purpose of this chapter is to briefly show how the conserva-tion laws for momentum, mass and thermodynamic energy can be combined andsimplified to become a vehicle for qualitative understanding of the movement and

46 The quasi-geostrophic approximation

development of atmospheric systems. Under certain restrictions the horizontal mo-mentum equation and the thermodynamic energy equation may be combined togive an equation describing the isobaric vertical velocity ω (the omega equation).Another equation, describing the geopotential tendency ∂φ

∂t , can be obtained sim-ilarly. The omega equation can be rewritten in such a form that the forcing ofthe vertical motion is expressed in terms of the divergence of a horizontal vectorforcing field (the Q-vector formulation). The geopotential tendency equation canbe rewritten in conservation form (the quasi-geostrophic potential vorticity equa-tion). For a qualitative description of how these equations may be used to explainthe development and movement of weather systems under the quasi-geostrophicapproximations, see Holton (1992).

After introduction of the concepts of the geopotential (Section 4.1) and thethickness of an atmospheric layer (Section 4.2), it is shown how atmospheric densitycan be eliminated from the pressure gradient force (Section 4.3), by use of theisobaric co-ordinate system (see Section 3.4 for a motivation and a discussion of thetransformation of conservation equations). The approximate balance on synopticscale between the horizontal pressure gradient force and the horizontal part of theCoriolis term leads to the definition of the geostrophic wind (Section 4.4) (see alsoExercise 1.5.3.1(e)). Following naturally is the definition of the ageostrophic wind(Section 4.5) as the difference between the horizontal velocity and the geostrophicwind. Next, assuming quasi-geostrophic approximations, the vorticity equation(Exercise 4.6(a)), the thermodynamic energy equation (Exercise 4.6(b)), the omegaequation (Exercise 4.6(c)) and the tendency equation (Exercise 4.6(d)) follow. InSection 4.7 the Q-vector formulation of the omega equation is derived in such away that it can clearly and easily be seen that purely geostrophic motion (Section4.7.3) will tend to destroy the thermal wind relationship. We close (Section 4.8)by showing how the quasi-geostrophic potential vorticity equation can be used tolink vertical motion with vorticity.

For the sake of simplicity, all the derivations in this chapter are done in Carte-sian co-ordinates, in accordance with the notation in Section 1.3.4. The moreadvanced learners may look at Burger & Riphagen (1990) for an exposition of thebasic equations in more general co-ordinates.

4.1. The geopotential. In Exercise 1.2(c) we defined the geopotential φ by

φ(z) =∫ z

0

g dz′, (4.1)

which conforms to Holton’s (1992) equation (1.8). Here we emphasize that φ is afunction of z only and that z′ is a dummy integration variable. By direct differen-tiation of (4.1), we obtain dφ

dz = g and a combination with the hydrostatic equation∂p∂z = −ρg gives

dφ

dz= g = −1

ρ

∂p

∂z= −RT

p

∂p

∂z. (4.2)

4.3. The geopotential in isobaric co-ordinates 47

Integration of (4.2) gives

φ(z2)− φ(z1) = −∫ p2

p1

1ρdp = −

∫ p2

p1

RT

pdp, (4.3)

where

p1 = p(z1) and p2 = p(z2). (4.4)

Exercise 4.1. Make the derivation mathematically more neat by meticulouslyusing the transformation rule (1.7) to show that

∫ z2

z1

1ρ(p(z))

∂p

∂z(z)dz =

∫ p(z2)

p(z1)

1ρ(p)

dp (4.5)

and that (4.3) follows.

Note that the dependency of p on x is suppressed in (4.3) – (4.5). Also note thatHolton (1992) uses the expression d ln p = 1

p dp.

4.2. The thickness of an atmospheric layer. As in Holton (1992), p. 20,we define

Z = φ(z)/g0Z1 = φ(z1)/g0Z2 = φ(z2)/g0

. (4.6)

The thickness of the atmospheric layer between the pressure surfaces p2 and p1, isgiven by

ZT = Z2 − Z1 =R

g0

∫ p1

p2

T

pdp (4.7)

=R

g0< T > ln

p1

p2(4.8)

where the mean temperature < T > is defined as

< T >=

∫ p1p2T d ln p∫ p1

p2d ln p

(4.9)

and p1 > p2.Exercise 4.2. Motivate (4.9) by showing that < T >= T1 if a constant tem-perature T = T1 is assumed. Apply integration of knowledge to enhance yourunderstanding of averaging and also revise Exercise 3.4.3(d).

48 The quasi-geostrophic approximation

4.3. The geopotential in isobaric co-ordinates. The right hand side of (4.3)leads us to define

Φ(x, p) = R

∫ p(x,0)

p

T (x, p′)p′

dp′. (4.10)

Note that by (4.3) and (4.5)

φ(z2)− φ(0) = −R∫ p(x,z2)

p(x,0)

T (x, p′)p′

dp′ = Φ(x, p(z2)).

so thatφ(z) = Φ(x, p(z)) for every z. (4.11)

A vector form of the momentum equations is given in Exercise 3.3.1(b) andin component form they are given in equations (2.19) – (2.21) of Holton (1992).Exercises 4.3 will now show that the pressure gradient force 1

ρ∇zp may be replacedby the geopotential gradient, ∇pΦ, which does not contain the atmospheric densityρ.Exercises 4.3.(a) Prove that

∂Φ∂p

= −RT

p

∇pΦ+ ∂Φ∂p

∇zp = 0

∇pΦ− 1ρ∇zp = 0.

(b) Show that a combination of (3.43), (3.44) and (4.11) again gives ∇pΦ = 1ρ∇zp.

Hint: use (4.11) to show that Φ(x, p) = φ(z(x, p)) and ∇pΦ = g∇pz.

4.4. The geostrophic wind. Referring to synoptic scaling of the horizontalmomentum equation, Holton (1992) uses the approximate balance between theCoriolis term and the pressure gradient force

−fv ≈ −1ρ

∂p

∂x(4.12)

fu ≈ −1ρ

∂p

∂y(4.13)

to define the variable-f geostrophic wind Vg by

Vg ≡ k × 1ρf

∇zp (4.14)

(see Section 2.4.1 in Holton (1992)).

4.6. The quasi-geostrophic prediction equations 49

A constant-f geostrophic wind is defined in Holton (1992) by use of a constantreference latitude value f0 for the Coriolis parameter f .Exercise 4.4. Assume that all the acceleration terms except the horizontal Cori-olis terms are negligibly small. Consider the approximate horizontal momentumequation, DV

Dt +fk×V = − 1ρ∇zp. Use the triple product rule given in Section 1.6.3

to show that (4.14) holds. The aim of this exercise is to obtain (4.14) by vectormanipulation without using the component form (4.12) – (4.13).

4.5. The ageostrophic wind. Define the ageostrophic wind Va = (ua, va) by

Va = V − Vg

so that V = Vg + Va.Exercise 4.5. Integration of knowledge. Revise the concepts of componentsof a vector, Exercise 1.5.1(b) and Exercise 1.5.3.1(e).

4.6. The quasi-geostrophic prediction equations. Using synoptic scalingconsiderations the total derivative following the geostrophic wind is defined by

DgDt

=∂

∂t+ Vg · ∇ =

∂

∂t+ ug

∂

∂x+ vg

∂

∂y.

The quasi-geostrophic momentum equation, (6.11) in Holton (1992), can be writtencomponent wise as

DgugDt

− f0va = βyvg

DgvgDt

+ f0ua = −βyug,

with f0 and β = ( dfdy )φ0 constants originating from the retention of only the firsttwo terms in a Taylor series expansion of the Coriolis parameter f about a referencelatitude φ0.Exercises 4.6.(a) Combine the above two approximate horizontal momentum equations to give

the quasi-geostrophic vorticity equation

∂ζg∂t

= −Vg · ∇(ζg + f) + f0∂ω

∂p.

Hint: use ζg =∂vg

∂x − ∂ug

∂y , assume ∇ ·Vg = 0 and write the answer in terms ofω by use of (3.50).

(b) Under adiabatic conditions (DsDt = 0), the thermodynamic energy equationcpD ln θDt = Ds

Dt , (see (2.46) in Holton (1992)) may be approximated for quasi-geostrophic conditions by

DgT

Dt− Spω = 0,

with Sp = −Tθ∂θ∂p . Prove this statement.

50 The quasi-geostrophic approximation.

(c) Assume ζg = 1f0∇2Φ by Exercise 1.5.3.1(e) and T = − p

R∂Φ∂p , by Exercise 4.3(a)

so that ∂ζg

∂t = 1f0∇2 ∂Φ

∂t and∂T∂t = − p

R∂∂p∂Φ∂t . Let σ =

Rp Sp be a constant.

Derive the omega equation[∇2 +

f20

σ

∂2

∂p2

]ω =

f0

σ

∂

∂p

[Vg · ∇

(1f0

∇2Φ + f

)]+1σ∇2

[Vg · ∇

(−∂Φ∂p

)].

Hint: eliminate the tendency ∂φ∂t from the quasi-geostrophic vorticity equationin (a) and the thermodynamic energy equation in (b).

(d) Assume ζg = 1f0∇2Φ, T = − p

R∂Φ∂p and σ =

Rp Sp as in (c). Put χ =

∂Φ∂t . Derive

the geopotential tendency equation[∇2 +

∂

∂p

(f20

σ

∂

∂p

)]χ = −f0Vg ·∇

(1f0

∇2Φ+ f

)+∂

∂p

[f20

σVg · ∇

(−∂Φ∂p

)].

Hint: eliminate ω from the quasi-geostrophic vorticity equation in (a) and thethermodynamic energy equation in (b).

4.7. The Q-vector. Using Cartesian co-ordinates, we follow the first part ofthe derivation in Holton’s (1992) Section 6.4.2 to derive two equations for Q ((4.31)and (4.33)) from which it can clearly and easily be seen that purely geostrophicmotion (Section (4.7.3)) will tend to destroy the thermal wind relationship.

Our derivation may be summarized as follows: the quasi-geostrophic equationson the f plane (Holton (1992), Section 6.4.2, p. 170-175)

DgugDt

− f0va = 0 (4.15)

DgvgDt

+ f0ua = 0 (4.16)

DgT

Dt− Spω = 0 (4.17)

may be combined with the thermal wind relationship (to be proven in Exercise 4.7)

∂Vg∂p

= − R

f0pk ×∇pT = − R

f0p(−∂T

∂yi+

∂T

∂xj), (4.18)

the fact that the constant-f geostrophic wind is divergence free

∇ · Vg = ∂ug∂x

+∂vg∂y

= 0, (4.19)

and the continuity equation

∇ · Va + ∂ω

∂p= ∇ · (Vg + Va) +

∂ω

∂p= ∇ · V+ ∂ω

∂p= 0 (4.20)

4.7. The Q-vector 51

to yield the so called Q-vector form of the omega equation:

σ∇2ω + f02 ∂

2ω

∂p2= −2∇ · Q, (4.21)

with

Q = (Q1, Q2) = −R

p

(∂Vg∂x

· ∇T, ∂Vg∂y

· ∇T). (4.22)

Exercise 4.7. Prove (4.18), (4.19) and (4.20). Does (4.19) hold for the variable-fgeostrophic wind? Hint for (4.18): write (4.14) in isobaric co-ordinates, and useExercises 4.3. For (4.19) refer to Exercise 1.5.3.1(e). For (4.20) revise Section 3.4.4.For the horizontal divergence of the variable-f geostrophic wind look at Holton’s(1992) Problem 3.19.

The Proof of (4.21) follows in Sections 4.7.1 to 4.7.4.

4.7.1. A differentiated form of the momentum equation (4.15).Differentiation (∂(4.15)∂p ) and manipulation gives

∂

∂p

[∂ug∂t

+ Vg · ∇ug]= f0

∂va∂p

DgDt

(∂ug∂p

)+∂Vg∂p

· ∇ug = f0∂va∂p

. (4.23)

Using (4.18), we may expand the second term in (4.23) as follows

∂Vg∂p

· ∇ug = − R

f0p

(−∂T

∂yi+

∂T

∂xj

)· ∇ug

=R

f0p

(−∂T

∂xj+

∂T

∂yi

)· ∇ug

=R

f0p

(−∂T

∂x

∂ug∂y

+∂T

∂y

∂ug∂x

)

=R

f0p

(−∂T

∂x

∂ug∂y

− ∂T

∂y

∂vg∂y

), using (4.19)

= − R

f0p∇T · ∂Vg

∂y

=1f0Q2 (4.24)

so thatf0∂Vg∂p

· ∇ug = Q2. (4.25)

Substitution of (4.25) in f0 times (4.23) gives

DgDt

(f0∂ug∂p

)+Q2 = f0

2 ∂va∂p

. (4.26)


Exercise 4.7.1. A differentiated form of the momentum equation (4.16).Use similar reasoning to prove that

DgDt

(∂vg∂p

)+∂Vg∂p

· ∇vg = −f0∂ua∂p

(4.27)

f0∂Vg∂p

· ∇vg = −Q1 (4.28)

DgDt

(f0∂vg∂p

)−Q1 = −f0

2 ∂ua∂p

. (4.29)

A vector combination of (4.29) and (4.26) is

(Q1, Q2)− f02 ∂

∂p(ua, va) = (Q1 − f0

2 ∂ua∂p

, Q2 − f02 ∂va∂p)

=(DgDt

(f0∂vg∂p

), −Dg

Dt

(f0∂ug∂p

))

=DgDt

(f0∂vg∂p

, −f0∂ug∂p

)

=DgDt

(−R

p

∂T

∂x, −R

p

∂T

∂y

)by (4.18)

= −DgDt

(R

p∇T

)(4.30)

which becomes

Q − f02 ∂Va∂p

= −DgDt

(R

p∇T

). (4.31)

4.7.2. A differentiated form of the energy equation (4.17). Assuming Spto be constant, another expression for the right hand side of (4.31) can be obtainedfrom (4.17) as follows: ∂(4.17)∂x implies

∂

∂x

[∂T

∂t+ Vg · ∇T

]= Sp

∂ω

∂x

DgDt

(∂T

∂x

)+∂Vg∂x

· ∇T = Sp∂ω

∂x(4.32)

and ∂(4.17)∂y implies

DgDt

(∂T

∂y

)+∂Vg∂y

· ∇T = Sp∂ω

∂y

which, combined with (4.32) and the definition of Q in (4.22), gives

DgDt(∇T )− p

R(Q1, Q2) = Sp∇ω

4.8. The quasi-geostrophic potential vorticity equation 53

orDgDt

(R

p∇T

)− Q = σ∇ω, σ =

RSpp

,

or

Q+ σ∇ω = DgDt

(R

p∇T

). (4.33)

4.7.3. Pure geostrophic motion. If Va = then ∂ω∂p = −∇ · Va = 0, from

(4.20). Thus ω is constant with height and since there is no vertical movement onthe surface, we have ω = 0. Then (4.31) and (4.33) may be written as

Q = −DgDt

(R

p∇T

)

Q =DgDt

(R

p∇T

),

which means that either Q = , or the thermal wind equation does not hold.

4.7.4. Ageostrophic motion. If Va = then (4.31) + (4.33) gives

2Q+ σ∇ω − f02 ∂Va∂p

=

2∇ · Q+ σ∇2ω − f02 ∂

∂p(∇ · Va) = 0

2∇ · Q+ σ∇2ω + f02 ∂

2ω

∂p2= 0,

where the last term follows from ∂ω∂p = −∇p ·Va, i.e. equation (4.20). Thus (4.21)

follows and dividing by σ we obtain

(∇2 +

f02

σ

∂2

∂p2

)ω = − 2

σ∇ · Q. (4.34)

Note the similarity between the partial differential operators on the left hand sideof the omega equation (Exercise 4.6(c)), the tendency equation (Exercise 4.6(d))and the one on the left hand side of equation (4.34).

4.7.5. Remark. Adams (1990) uses (4.24) and (4.28) as definition of the Q-vector and then derives (4.22).

4.8. The quasi-geostrophic potential vorticity equation. In Holton’s(1992) Section 6.3.2 the quasi-geostrophic potential vorticity q is defined by

q =1f0

∇2Φ+ f +∂

∂p

(f0

σ

∂Φ∂p

).


The tendency equation can be written as a conservation equation(∂

∂t+ Vg · ∇

) [1f0

∇2Φ+ f +∂

∂p

(f0

σ

∂Φ∂p

)]=Dgq

Dt= 0

for q. This conservation equation can be written in the form

DgDt(ζg + f)− Dg

Dt(f0

Sp

∂T

∂p) ≈ Dg

Dt

[1f0

∇2Φ+ f +∂

∂p

(f0

σ

∂Φ∂p

)]= 0 (4.35)

by use of the geostrophic vorticity ζg = 1f0∇2Φ from Exercise 1.5.3.1(e) and the

approximation f0Sp

∂T∂p ≈ ∂

∂p

(f0σ∂Φ∂p

)from Holton’s (1992) Section 6.3.2.

4.8.1. Vertical motion (sinking) accompanied by vorticity and geopoten-tial height changes. From (4.35) follows

DgDt(ζg + f) =

f0

Sp

DgDt

∂T

∂p

so that for f0 ≤ 0 (in the Southern Hemisphere) ζg + f increases (resp. decreases)with time if and only if ∂T∂p decreases (resp. increases) with time.Exercises 4.8.1.(a) Derive the quasi-geostrophic potential vorticity equation.(b) Ignoring density changes with time, show that ζg+f increases (resp. decreases)

with time if and only if ∂T∂z increases (resp. decreases) with time in the SouthernHemisphere.

(c) Consider the adiabatic sinking of a vertical atmospheric column depicted inFigure 4.1. Show that the geostrophic vorticity and the geopotential mustdecrease if f remains constant. You may consider a Southern Hemisphericsituation. Hint: use the sketch to determine whether ∂T∂z increases or decreaseswith time, use (b) and assume ∇2Φ = −cΦ with c a constant.

T

zadiabatic sinking

adiabatic sinking

Figure 4.1

Chapter 5

Atmospheric modelling and simple numerical examples

A true story. Zbigniew Sorbjan (1996) tells the following about Richardson’sforecasting plans: The first step toward mathematical weather forecasting was doneby Lewis Fry Richardson (1881–1953). Richardson believed that it might be possibleto solve the complex equations of atmospheric motion by working them out in stepby step computations. He described his ideas in his book Weather Prediction byNumerical Process. The first draft of it was prepared in May 1916, and revisedin 1916–1918, during World War I. During the battle of Champagne, in April1917, the working copy was sent to the rear lines and was lost. Fortunately, it wasrecovered a few months later under a heap of coal. Finally, the manuscript wasprinted in 1922.

To put his idea into practice, Richardson required data from about 2000 perma-nent weather stations collecting both surface and upper-air measurements aroundthe Earth. He also envisioned the globe divided up as a checkerboard into rows andcolumns. He thought that 32 individuals could compute a forecast at one column tokeep pace with the weather. If the column spacings were 200 km, he believed that2000 active columns would suffice to complete a forecast on the globe. Consequently,32 × 2000 = 64000 calculators would be needed to predict the weather for the entireglobe. Such an amount of computations would require a “forecast factory”.

Introduction. The purpose of Chapter 5 is to give the learner a glimpse ofthe world of numerical modelling by using a few examples from previous chaptersto introduce the new concepts. It will be fruitful to solve some problems on acomputer. First single equations in one independent variable need to be solved andthen systems of algebraic equations. It is not the intention of this chapter to beself contained and complete.

5.1. Mathematical models. The learner has encountered mathematical andmeteorological problems, has proposed solutions for them and practised integrationof knowledge. The process of mathematical modelling, which embraces mathemat-ical equations and their numerical solution, is summarised for the atmosphere inFig. 5.1.

56 Atmospheric modelling and simple numerical examples.

Du

Dt− 2Ωv sinφ+ 2Ωw cosφ− uv tanφ

a+uw

a= −1

ρ

∂p

∂x+ Frx

Dv

Dt+ 2Ωu sinφ+

u2 tanφa

+vw

a= −1

ρ

∂p

∂y+ Fry

Dw

Dt− 2Ωu cosφ− u2 + v2

a= −1

ρ

∂p

∂z− g + Frz

∂ρ

∂t+∇ · (ρU) = 0

cpD lnTDt

−RD ln pDt

− J

T= 0

DV

Dt+∇pΦ+ fk × V = 0

∇p · V+ ∂ω

∂p= 0

∂Φ∂p

+RT

p= 0.

15.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682839.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478446.4097 88.2764 21.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409748.798 92.8193 27.7548 50.1848 57.9137 50.1848 27.7549 92.8193 48.798646.4097 88.2764 11.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409739.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682815.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079415.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682839.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478446.4097 88.2764 21.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409748.798 92.8193 27.7548 50.1848 57.9137 50.1848 27.7549 92.8193 48.798646.4097 88.2764 11.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409739.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682815.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079415.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682839.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478446.4097 88.2764 21.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409748.798 92.8193 27.7548 50.1848 57.9137 50.1848 27.7549 92.8193 48.798646.4097 88.2764 11.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409739.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682815.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079415.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682839.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478446.4097 88.2764 21.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409748.798 92.8193 27.7548 50.1848 57.9137 50.1848 27.7549 92.8193 48.798646.4097 88.2764 11.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409739.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682815.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079415.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682839.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478446.4097 88.2764 21.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409748.798 92.8193 27.7548 50.1848 57.9137 50.1848 27.7549 92.8193 48.798646.4097 88.2764 11.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409739.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682815.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079415.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.079428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682839.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478446.4097 88.2764 21.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409748.798 92.8193 27.7548 50.1848 57.9137 50.1848 27.7549 92.8193 48.798646.4097 88.2764 11.5021 42.8343 50.1848 42.8343 21.5021 88.2764 46.409739.4784 75.0924 13.3558 21.5021 27.7548 21.5021 13.3559 75.0924 39.478428.6827 54.5578 75.0924 88.2764 92.8193 88.2764 75.0924 54.5578 28.682815.0794 28.6827 39.4784 46.4097 48.798 46.4097 39.4784 28.6827 15.0794

The Physical Atmosphere

A Mathematical model

Numerical Solution

Interpretation

Figure 5.1

5.2. Atmospheric modelling. We summarize the dynamical meteorolog-ical equations, using as independent variables x, y, p, t and dependent variablesu, v, ω,Φ, T .

DV

Dt= −∇pΦ− fk × V (5.1)

∂Φ∂p

= −RT

p(5.2)

DT

Dt− RT

pcpω − J

cp=

T

cp

(cpD lnTDt

−RD ln pDt

− J

T

)= 0 (5.3)

∇p · V+ ∂ω

∂p= 0. (5.4)

Boundary values must be specified at all times throughout the forecasting pe-riod. An approximate model topography will represent the real topography andthe model top will represent the upper atmospheric boundary. For a global modelthe lower and upper boundary values must be specified at all times throughoutthe forecast. For a Limited-Area Model (LAM) the lateral boundary values areneeded in addition. The lateral boundary values are normally obtained from aglobal model.

Initial values must be specified throughout the three dimensional model domain atthe start of the forecast.

5.3. Numerical models 57

5.3. Numerical models.

5.3.1. First order derivatives.Examples 5.3.1.(a) The geopotential. Consider Exercise 1.2(c). If we combine Newton’s gravity

law (allowing for centripetal acceleration) and the definition (4.1) of the geopo-tential, then a first order ordinary differential equation with lower boundaryvalue (at z = 0) arises:

Φ′(z) =g0

(1 + z/a)2

Φ(0) = 0.

(b) Time rate of change of temperature. In equation (5.3), DTDt =RTpcp

ω+ Jcp,

the temperature T depends on time t as well as position x. Consider the timerate of change first. The partial derivatives may be expanded to give

∂T

∂t= −U · ∇T + RT

pcpω +

J

cp. (5.5)

Approximation of first order derivatives. In view of Taylor’s theorem,

f(x+ h) = f(x) + hf ′(x) +h2

2f ′′(c)

it seems natural to look for approximations of the first order derivatives, based ona difference form, viz.

f ′(x) =f(x+ h)− f(x)

h− h

2f ′′(c). (5.6)

Since Φ is only a function of z in Example 5.3.1(a), the learner should have nodifficulty in calculating Φ for a discrete set of points zi, say (Exercise 5.3.1(a)). Forthe purpose of illustration consider now Example 5.3.1(b). We may write (5.5) inthe form

∂T

∂t= F (J,U, ω, T )

with F indicating the right hand side of (5.5). Using (5.6) to approximate the timederivatives, the finite difference scheme

T (t+ δt)− T (t)δt

≈ F (J,U, ω, T )

is obtained, implying

T (xi, yj , pk, tl+1) = T (xi, yj , pk, tl) + δtG(xi, yj , pk, tl), (5.7)

where G represents the value of F at the discrete points in space (xi, yj, pk) and aninstant of time tl at which all variables are known (p denotes a vertical co-ordinate).The temperature T may thus be calculated at a future instant of time tl+1.


This algorithm gives an explicit ‘marching’ ahead in time. By truncating theTaylor series, errors are introduced. The accuracy of the finite difference approx-imation for f ′(x), given by (5.6), is indicated by the symbol O(h). This symbolmeans that a positive number A exists such that the omitted part is less than orequal to Ah in absolute value. A higher order finite difference approximation for thefirst order derivatives can be obtained by subtracting the following two equations

f(x+ h) = f(x) + hf ′(x) +h2

2f ′′(x) +

h3

6f (3)(c1) (5.8)

f(x− h) = f(x)− hf ′(x) +h2

2f ′′(x)− h3

6f (3)(c2) (5.9)

to givef(x+ h)− f(x− h) = 2hf ′(x) +O(h3)

resulting in a central difference formulation

f ′(x) =f(x+ h)− f(x− h)

2h+O(h2). (5.10)

Likewise a forward difference and backward difference formulation may be obtainedfrom (5.8) and (5.9) respectively, but with accuracy O(h).Exercises 5.3.1.(a) Solve the problem in Example 5.3.1(a) numerically.(b) Assume numerical values R

pcpω = − 287

85000×1004 × 0.01 s−1 = −3 × 10−8 s−1,U = and J = 0 in (5.5). Solve (5.5), with T (t = 0) = 300 Kelvin, usinga forward finite difference scheme and compare with the analytical solution,T = 300e−0.00000003t Kelvin.

(c) Assume numerical values Rpcp

ω = −3 × 10−8 s−1 as in (b), U = (1, 0, 0) andJ/cp = 9× 10−6Ks−1 in (5.5). Solve (5.5), subject to T (x, t = 0) = 300+ sinxKelvin, using a forward finite difference scheme for the time derivative and acentral spatial derivative, as in (5.10) and compare with the analytical solution,T (x, t) = 300 + e−0.00000003t sin(x− t) Kelvin.

(d) Solve (5.5) as in (c), but using a backward spatial derivative, ∂T∂x ≈

T (x,t)−T (x−δx,t)δx and again compare with the analytical solution.

5.3.2. Higher order derivatives. A finite difference scheme for the secondorder derivatives can be obtained by adding equations like (5.8) and (5.9), involvingup to fourth order derivatives. The result

f(x+ h) + f(x− h) = 2f(x) + h2f ′′(x) +O(h4)

can be rewritten as

f ′′(x) =f(x+ h)− 2f(x) + f(x− h)

h2+O(h2). (5.11)

5.5. A boundary value problem 59

Example 5.3.2. Consider the oscillations of a parcel of air in a statically stableatmosphere. Equation (2.52) in Holton (1992) can be written as

f ′′(t) = −N2f(t),

where N represents the Brunt-Vaisala frequency and the function t → f(t) theparcel displacement. If we know the initial displacement f(0) and if we can ap-proximate the displacement at a small time step δt later on in terms of the previoustime steps (time t and time t− δt), then we may be able to obtain an approximatesolution. Alternatively, we may use an iterative method based on the discrete formf(t+h)−2f(t)+f(t−h)

h2 = −N2f(t) which implies

f(t) = 12(f(t+ h) + f(t− h) + h2N2f(t)).

Exercises 5.3.2.(a) Prove (5.11).(b) Use the approximation proposed in (5.11) in both the x- and y- directions

to give the following finite difference form of the horizontal Laplace operator(defined in Exercise 1.5.3.1(e))

∇2f ≈ fi,j+1 + fi,j−1 + fi+1,j + fi−1,j − 4fi,jh2

. (5.12)

5.4. The quasi-geostrophic vorticity equation. Let ζg = ∂xvg − ∂yug andassume

∂ζg∂t

= −Vg · ∇(ζg + f) + f0∂ω

∂p,

from Holton (1992), p.155, Section 6.2 and

ζg =1f0

∇2Φ, (5.13)

from Holton (1992), p.154, Section 6.2.2.Exercise 5.4. Compare the form of the vorticity equation above with that of thethermodynamic equation (5.3) and propose a ‘marching’ ahead in time similar to(5.7).

5.5. A boundary value problem. Let f0 = 10−4 s−1 and k = l = π2 × 10−6

m−1 as in Holton (1992), p.157. Let N = 2 × 106 m denote the east-west extentand M = 1× 106 m the north-south extent of the domain R = [−N,N ]× [−M,M ]of interest. Consider the boundary value problem

∇2Φ = −(k2 + l2) sin(kx) cos(ly) in RΦ(x,−M) = 50000 + 10f0M + 800 sin(kx) cos(lM)Φ(x,M) = 50000− 10f0M + 800 sin(kx) cos(lM)Φ(−N, y) = 50000− 10f0y − 800 sin(kN) cos(ly)Φ(N, y) = 50000− 10f0y + 800 sin(kN) cos(ly).

(5.14)


Note that the Partial Differential Equation (PDE) in (5.14) is of the form (5.13)with ζgf0 = −(k2 + l2) sin(kx) cos(ly).Exercises 5.5.(a) Use the discrete form (5.12) of the Laplace operator to write the Poisson

equation (5.13) in the finite difference form

Φi,j =14(Φi,j+1 +Φi,j−1 +Φi+1,j +Φi−1,j − h2f0ζgi,j ). (5.15)

Use the Gauss-Seidel method (5.15) to solve the boundary value problem(5.14).

(b) Write down the heat equation, obtained in Exercise 3.3.1(a), for the temper-ature function T : (x, t) → T (x, t). Find the numerical solution of the heatconduction problem, comprising of the heat equation in the spatial one dimen-sional interval [0, 1], boundary conditions T (0, t) = 0, T (1, t) = 0 for every tand the initial condition T (x, 0) = sin(πx) for every x ∈ (0, 1).

References 61

REFERENCES

Adams, M., (1990),Dynamical Meteorology, Bureau of Meteorology Training Cen-tre, GPO Box 1289K, Melbourne Victoria 3001, Australia.

Apostol, Tom M., (1967), Calculus Volume I, Second Edition, John Wiley &Sons, Inc., New York, Santa Barbara, London, Sydney Toronto.

Apostol, Tom M., (1969), Calculus Volume II, Second Edition, John Wiley &Sons, New York, London, Sydney Toronto.

Atkin, R.J. & Fox, N., (1980), An introduction to the theory of Elasticity,Longman, London and New York.

Burger, A.P. & Riphagen, H.A., (1990), The basic equations in meteorologicaldynamics - a reexamination of unsimplified forms for a general vertical co-ordinate, Contr. Atmos. Phys., 63, 151 – 164.

Fung, Y.C., (1969), A First Course In Continuum Mechanics, Prentice-Hall, INC.,Englewood, Cliffs, N.J.

Holton, James R., (1992), An introduction to Dynamic Meteorology, AcademicPress, INC, San Diego, California.

Persson, A., (2000), Back to basics: Coriolis: Part 2 – The Coriolis force accordingto Coriolis, Weather, 55, 182 – 188.

Serrin, James., (1959), Mathematical Principles of Classical Fluid Mechanics, inHandbuch der Physik Bd. VIII/1, Springer Verlag, Berlin, Gottingen, Heidel-berg.

Sorbjan, Zbigniew., (1996), Hands-On Meteorology Stories, Theories, and Sim-ple Experiments, Project Atmosphere, American Meteorological Society.

Spencer, A.J.M., (1980), Continuum mechanics, Longman, London and NewYork.

Truesdell, C., (1977), A first course in rational continuum mechanics, AcademicPress, New York, San Francisco, London.

The Author

Dr. Jan Gertenbach is a specialist scientist in the Directorate Research andTraining of the South African Weather Bureau. He started his career as lecturerin the Department of Applied Mathematics at the University of Pretoria, SouthAfrica, in August 1980. He pursued his interests in Applied Mathematics, FluidDynamics and Continuum Mechanics at the South African Council for Scientificand Industrial Research from 1987 to March 1994, from where he moved to hispresent position at the South African Weather Bureau. From 1994 to 1999 he alsolectured on Functional Analysis and Numerical Analysis in the post of Part-timeProfessor at the Medical University of Southern Africa.

Mail: SA Weather Bureau, Private Bag X097, Pretoria, 0001, South AfricaE-mail: [email protected]: +27 12 362 5218 (work) or +27 12 9986821 (home)Telephone +27 12 420 2170 (work) or +27 12 9986821 (home)

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