Calculus 4Course Notes for AMATH 231

J. Wainwright1 and G. TentiDepartment of Applied Mathematics

University of Waterloo

December 6, 2020

1©c J. Wainwright

Contents

Preface v

Acknowledgements vii

1 Curves & Vector Fields 11.1 Curves in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Curves as vector-valued functions . . . . . . . . . . . . . . . . . . . . 11.1.2 Reparametrization of a curve . . . . . . . . . . . . . . . . . . . . . . 71.1.3 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.1.4 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.5 Arclength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.2 Vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.2.1 Examples from physics . . . . . . . . . . . . . . . . . . . . . . . . . . 171.2.2 Field lines of a vector field . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Line Integrals & Green’s Theorem 272.1 Line integral of a scalar field . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.1 Motivation and definition . . . . . . . . . . . . . . . . . . . . . . . . 272.1.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2 Line integral of a vector field . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.2.1 Motivation and definition . . . . . . . . . . . . . . . . . . . . . . . . 342.2.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.2.3 Some technical matters . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.3 Path-independent line integrals . . . . . . . . . . . . . . . . . . . . . . . . . 412.3.1 First Fundamental Theorem for Line Integrals . . . . . . . . . . . . . 422.3.2 Second Fundamental Theorem for Line Integrals . . . . . . . . . . . . 452.3.3 Conservative (i.e. gradient) vector fields . . . . . . . . . . . . . . . . 46

2.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.4.1 The theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.4.2 Existence of a potential in R2 . . . . . . . . . . . . . . . . . . . . . . 53

2.5 Vorticity and circulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3 Surfaces & Surface Integrals 653.1 Parametrized surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.1.1 Surfaces as vector-valued functions . . . . . . . . . . . . . . . . . . . 65

i

3.1.2 The tangent plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693.1.3 Surface area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.1.4 Orientation of a surface . . . . . . . . . . . . . . . . . . . . . . . . . 73

3.2 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 743.2.1 Scalar fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 743.2.2 Vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 773.2.3 Properties of surface integrals . . . . . . . . . . . . . . . . . . . . . . 79

4 Gauss’ and Stokes’ Theorems 814.1 The vector differential operator ∇ . . . . . . . . . . . . . . . . . . . . . . . . 81

4.1.1 Divergence and curl of a vector field . . . . . . . . . . . . . . . . . . . 814.1.2 Identities involving ∇ . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.1.3 Expressing ∇ in curvilinear coordinates . . . . . . . . . . . . . . . . . 86

4.2 Gauss’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 934.2.1 The theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 934.2.2 Conservation laws and PDEs . . . . . . . . . . . . . . . . . . . . . . . 984.2.3 The Generalized Divergence Theorem . . . . . . . . . . . . . . . . . . 101

4.3 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1024.3.1 The theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1024.3.2 Faraday’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1074.3.3 The physical interpretation of ∇× F . . . . . . . . . . . . . . . . . . 109

4.4 The Potential Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1094.4.1 Irrotational vector fields . . . . . . . . . . . . . . . . . . . . . . . . . 1104.4.2 Divergence-free vector fields . . . . . . . . . . . . . . . . . . . . . . . 112

5 Fourier Series and Fourier Transforms 1175.1 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

5.1.1 Calculating Fourier coefficients . . . . . . . . . . . . . . . . . . . . . 1195.1.2 Pointwise convergence of a Fourier series . . . . . . . . . . . . . . . . 1235.1.3 Symmetry properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 1285.1.4 Complex form of the Fourier series . . . . . . . . . . . . . . . . . . . 133

5.2 Convergence of series of functions . . . . . . . . . . . . . . . . . . . . . . . . 1385.2.1 A deficiency of pointwise convergence . . . . . . . . . . . . . . . . . . 1395.2.2 The maximum norm and mean square norm . . . . . . . . . . . . . . 1415.2.3 Uniform and Mean Square Convergence . . . . . . . . . . . . . . . . . 1455.2.4 Termwise integration of series . . . . . . . . . . . . . . . . . . . . . . 151

5.3 A Second Look at Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . 1535.3.1 Uniform and mean square convergence . . . . . . . . . . . . . . . . . 1535.3.2 Integration of Fourier series . . . . . . . . . . . . . . . . . . . . . . . 155

5.4 The Fourier transform & Fourier integral . . . . . . . . . . . . . . . . . . . . 1575.4.1 The definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1575.4.2 Calculating Fourier transforms using the definition . . . . . . . . . . 1605.4.3 Properties of the Fourier transform . . . . . . . . . . . . . . . . . . . 1635.4.4 Parseval’s formula for a non-periodic function . . . . . . . . . . . . . 1675.4.5 Relation between the continuous spectrum and the discrete spectrum 168

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5.4.6 Things are simpler in the frequency domain . . . . . . . . . . . . . . 1705.5 Digitized signals – a glimpse . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

5.5.1 The sampling theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 1725.5.2 The discrete Fourier transform (DFT) . . . . . . . . . . . . . . . . . 174

References 177

Problem Sets 179

Sample Tests & Exams 221

iii

Preface

This course has two main themes:

i) the calculus of vector fields, and

ii) Fourier analysis.

The first theme, which includes the famous integral theorems associated with the namesof Green, Gauss and Stokes, represents the culmination of the traditional Calculus sequence.This material provides the mathematical foundation for continuum mechanics (AMATH361), fluid dynamics (AMATH 463), and electromagnetic theory (PHYS 252 & 253) and isof importance for partial differential equations (AMATH 353). The second theme, Fourieranalysis, is built on the remarkable idea that a variety of complicated functions can besynthesized from pure sine and cosine functions. This material provides the mathematicalfoundation for signal and image processing (course under development) and is of importancefor PDEs (AMATH 353) and quantum mechanics (AMATH 373).

v

Acknowledgements

Thanks are expressed to

Woei Chet Lim: for assistance with the figures.

Stanley Lipshitz: for many discussions and suggestions concerning the material in Chap-ter 5, and for comments on the remainder of the notes,

Bev Marshman: for detailed comments on the notes,

Ann Puncher: for excellent typesetting,Dan Wolczuk: for creating the index.

vii

Chapter 1

Curves & Vector Fields

The notion of a curve is of fundamental importance in Applied Mathematics courses becauseit arises in many physical and mathematical contexts. The path or trajectory of a particlemoving in space, e.g. a satellite orbiting the earth, is a curve in R3. A section of a power linesuspended between two pylons is another example of a curve in R3. Looking ahead in thiscourse, the field lines of a vector field F(x, y, z), such as magnetic flux lines, are a family ofcurves in R3 (more on this in Section 1.2). More generally, if one considers a physical systemwhose state at time t is a vector x(t) ∈ Rn, then the evolution in time of the system will bedescribed by a curve in Rn (the value of n will depend on the complexity of the system).

The first mathematical description of a curve that one encounters is the graph of afunction f : R→ R, described by an equation

y = f(x), a ≤ x ≤ b.

This way of describing curves has a number of limitations: it is only valid for curves in R2,and it doesn’t give a simple description of the motion of particles (e.g. a particle movingin a circle). So one introduces vector-valued functions and uses a parametric description ofcurves (MATH 138),

x = g(t), a ≤ t ≤ b,

which provides a sufficiently general description of curves for most purposes.In Section 1.1 we review vector-valued functions and their use in describing curves in R2

and R3, discussing the topic in greater depth than in MATH 138.

1.1 Curves in Rn

1.1.1 Curves as vector-valued functions

Let g : [a, b]→ Rn be a vector-valued function , i.e. a function which associates with each realnumber t ∈ [a, b] a unique vector g(t) ∈ Rn. We can express g(t) in terms of its componentsrelative to the standard basis in Rn:

g(t) = (g1(t), g2(t), . . . , gn(t)), (1.1)

1

where the gi, i = 1, . . . , n, are scalar functions called the component functions of g. We shallwork with functions g which are (at least) continuous, which means that the componentfunctions are continuous.

Consider a (continuous) vector-valued function g : [a, b] → Rn. As t runs from a to b,the function g determines an ordered succession of points g(t) in Rn, which we shall refer toas a curve C, in Rn. The curve C is described by the equations

x = g(t), a ≤ t ≤ b. (1.2)

We shall refer to the function g as a parametrization of C, and to t as a parameter on C. Wecan think of the curve C as the image of the interval [a, b] under the function g, representedpictorially as in Figure 1.1.

t

b

t

a

g

z

x

g(a)

g(t)C

g(b)

y

Figure 1.1: A vector-valued function g defines a curve C in R3.

One can think of the function g bending and stretching the line segment a ≤ t ≤ b so asto create the curve C. The ordering of points in the interval [a, b] determines an ordering ofpoints in C, called the orientation of C. The points g(a) and g(b) are called the endpoints ofC.

Example 1.1:The vector-valued function g : [0, a]→ R3 defined by

g(t) = r0 + tu, 0 ≤ t ≤ a, (1.3)

where r0 and u are given vectors in R3, determines a curve x = g(t), which is the straightline segment in R3, joining the points g(0) = r0 andg(a) = r0 + au, traversed in the direction of u.

2

tu

g(a)g(t)

u

g(0) = r0

Figure 1.2: The curve x = r0 + tu, 0 ≤ t ≤ a.

Comment:One can interpret an element a = (a1, a2, . . . , an) of Rn in two ways, either as the position

vector of a point in Rn (e.g. the vector r0 in Example 1.1), or as a vector attached to a pointin Rn (e.g. the vector u in Example 1.1). In the second interpretation the vector wouldrepresent some physical quantity such as the velocity or acceleration of a particle, or a forceacting on a particle.

Example 1.2:The vector-valued function g : [0, π]→ R2 defined by

g(t) = (b cos t, b sin t), 0 ≤ t ≤ π,

where b is a positive constant, determines a curve x = g(t) which is half the circle x2+y2 = b2

of radius b, traversed counterclockwise from (b, 0) to (−b, 0).

y

x = g(t)

g(0) = (b,0)g(π) = (−b,0)

x0

Figure 1.3: The curve x = (b cos t, b sin t), 0 ≤ t ≤ π.

Here is a more complicated example, which arises in many scientific contexts.

3

Example 1.3:Describe the curve C in R3 determined by the vector-valued function

g(t) =

(R cos t, R sin t,

h

2πt

), 0 ≤ t ≤ 2π, (1.4)

where R and h are positive constants.

Solution: The curve is described by the equation x = g(t), which in component form reads

x = R cos t, y = R sin t, z =h

2πt.

Sincex2 + y2 = R2,

and 0 ≤ t ≤ 2π, the curve rotates once around the cylinder of radius R, whose axis is thez-axis. Since z increases linearly with t, the curve rises uniformly as it moves around thecylinder, giving one revolution of a helix. The constant R is the radius of the helix, and theconstant h, the change in z during one revolution, is called the pitch of the helix.

h

R

z

y

g(2π) = (R,0,h)

g(0) = (R,0,0)x

Figure 1.4: A helix of radius R and pitch h.

In some situations one is given a curve C described geometrically, and it is necessary tofind a parametrization g(t) of C. Here’s an example.

Example 1.4:The cylinder x2 + z2 = R2 intersects the plane z = y in a closed curve (an ellipse).

We specify the orientation of the curve C by stating that the ellipse is traversed counter-clockwise when viewed from the positive y-direction. We complete the description of the

4

curve C by stating the the ellipse is traversed twice starting at the point (R, 0, 0). Find aparametrization for the curve C.

Solution: The equation of the cylinder, and Figure 1.5, suggest that we choose

x = R cos t, z = −R sin t,

with0 ≤ t ≤ 4π,

to give two revolutions. The equation of the plane, z = y, then determines y in terms of t,i.e.

y = −R sin t.

z

(R, 0)x

Figure 1.5: Projection of the ellipse into the xz-plane. The positive y-axis points into thepage.

Thus, a possible parametrization of C is

g(t) = (R cos t,−R sin t,−R sin t), 0 ≤ t ≤ 4π.

An important special case:

The graph of a scalar function f : [a, b] → R, given by y = f(x) is a curve C in R2.We take the orientation of C to be the direction of increasing x. It is easy to obtain aparametrization g(t) of this curve. Just let x = t, and then y is given by y = f(t), witha ≤ t ≤ b. Thus a parametrization of C is

g(t) = (t, f(t)), a ≤ t ≤ b. (1.5)

5

z

y

x

(R,0,0)

C

Figure 1.6: The curve C in Example 1.4.

A curve as the path of a particle:

Given a vector-valued function g : [a, b] → Rn, one can imagine the image point g(t)moving in Rn as t runs from a to b, thereby tracing out the curve C. When describing themotion of a particle, the parameter t will represent time and the curve C ∈ R3, (i.e. n = 3)will represent the path or trajectory of the particle in space during the time interval a ≤ t ≤ b.The orientation of the curve will give the direction of motion along the curve.

Comment:Given a vector-valued function g : [a, b] → Rn, it is important to make a distinction

between the range of g, i.e. the subset g(t)|a ≤ t ≤ b ⊂ Rn, and the curve C determinedby g, i.e. the ordered succession of points g(t) ∈ Rn, as t runs from a to b. For example,first consider g1 : [0, 2π]→ R2 defined by

g1(t) = (cos t, sin t).

The curve C1 given by x = g1(t), 0 ≤ t ≤ 2π, is the circle x2 + y2 = 1, traversed once ina counterclockwise direction,starting at the point (1, 0). Second, consider g2 : [0, 2π] → R2

defined byg2(t) = (cos 2t, sin 2t).

The curve C2 given by x = g2(t), 0 ≤ t ≤ 2π, is the circle x2 + y2 = 1 traversed twice incounterclockwise direction, starting at the point (1, 0).

The point is that g1 and g2 have the same range in R2, namely the points on the circlex2 + y2 = 1, but the curves C1 and C2 that they determine are not the same. For example, aparticle whose path is C1 will travel a distance of 2π, while for C2 the distance travelled willbe 4π.

Exercise 1.1:Describe the curve C determined by the vector-valued function

g(t) = (3 sin t, 4 cos t), 0 ≤ t ≤ π.

6

Exercise 1.2:A curve C in R3 is defined by the intersection of the plane x − y = 0 and the sphere

x2 + y2 + z2 = 2 (the intersection is a circle). The endpoints and orientation of the curveare as shown in Figure 1.7. Find a parametrization g(t) of the curve C.

z

x

y

(1,1,0)

C

Figure 1.7: The curve C in Exercise 1.2.

1.1.2 Reparametrization of a curve

An essential aspect of describing curves is that infinitely many different vector-valued func-tions describe the same curve C in Rn.

For example, consider g : [0, π]→ R2 defined by

g(t) = (cos t, sin t), 0 ≤ t ≤ π,

and also g :[0, π

2

]→ R2 defined by

g(τ) = (cos 2τ, sin 2τ), 0 ≤ τ ≤ π2,

where we use a different letter τ for the parameter in the second case. The point is that g andg are different functions, but determine the same curve in R2, namely the circle x2 + y2 = 1traversed halfway around counterclockwise from (1, 0) to (−1, 0). We say that g and g aredifferent parametrizations of the same curve C. Note that the two functions are related by

g(τ) = g(h(τ)),

whereh(τ) = 2τ.

In general, let h : [α, β] → [a, b] be a continuous, increasing (and hence one-to-one)function. Given g : [a, b]→ Rn, we define g : [α, β]→ Rn by

g(τ) = g(h(τ)), α ≤ τ ≤ β. (1.6)

7

As τ runs from α to β, t = h(τ) runs from a to b (see Figure 1.8). Thus by equation (1.6),g(τ) follows the same ordered succession of points in Rn as does g(t), and hence determinesthe same curve C in Rn.

!"

#

t

b

a

t = h (#)

Figure 1.8: The change of parameter function h.

We shall refer to g as a reparametrization of the curve C. the situation is representedpictorially in Figure 1.9.

z

y

x

a t b

g g

! " #

g (t) = g (h (")) = g (")

C

h

Figure 1.9: Two parametrizations g and g of a curve C.

In terms of the motion of a particle, reparametrizations have the following interpretation.Consider a particle that moves on a curve in 3-space between two given points. One can

8

imagine the particle speeding up or slowing down as it moves along the curve. Thus givenone curve, there are infinitely many ways a particle can move along the curve, and each onecorresponds to a different parametrization. We will shed more light on this matter when wediscuss tangent vectors and velocity vectors in the next subsection.

Comment:In these notes, when we write

“Consider a curve C in Rn . . .”

we shall mean that there is some vector-valued function g : [a, b] → Rn which gives aparametrization of this curve. At the same time, we know that there are infinitely manyother functions g that give reparametrizations of C. Which parametrization we use for doinga calculation will be a matter of convenience.

1.1.3 Limits

Given a vector-valued function g : [a, b]→ Rn, the Newton quotient at t is

1

∆t[g(t+ ∆t)− g(t)] (1.7)

for a non-zero change ∆t. The derivative of g at t, denoted g′(t), is defined to be the vectorthat is the limit of the Newton quotient as ∆t→ 0. So we need to think about the limit ofa vector-valued function.

Definition:Given a vector-valued function f defined in a neighbourhood of t0 and a constant vector

L, the statement limt→t0

f(t) = L means that

limt→t0‖ f(t)− L ‖= 0.

Comments:

1) Since ‖ f(t)− L ‖ is Euclidean distance between the vectors f(t) and L, the definitioncaptures the usual idea of “limit”, namely that the vector f(t) gets arbitrarily close tothe unique vector L as t approaches t0.

2) When doing calculations with vector-valued functions one works with components. Sowe need to express the basic concepts in terms of components. Let

f(t) = (f1(t), . . . , fn(t)), L = (L1, . . . , Ln).

Thenlimt→t0

f(t) = L ⇐⇒ limt→t0

fi(t) = Li, i = 1, 2, . . . , n. (1.8)

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This equivalence follows on noting that

‖ f(t)− L ‖=[(f1(t)− L1)2 + · · ·+ (fn(t)− Ln)2

]1/2,

which implies|fi(t)− Li| ≤ ‖ f(t)− L ‖, for i = 1, 2, . . . , n,

so that

limt→t0‖ f(t)− L ‖= 0 implies lim

t→t0fi(t) = Li, i = 1, . . . , n.

We can now talk about a vector-valued function being continuous.

Definition:f is continuous at t0 means that f(t0) is defined, lim

t→t0f(t) exists, and lim

t→t0f(t) = f(t0).

Comments:

i) In terms of component functions

f is continuous at t0 ⇔ fi is continuous at t0, i = 1, 2, . . . , n.

ii) The curves one encounters in physics are invariably defined by continuous functions. Ifthe function f was discontinuous at t0, say lim

t→t+0f(t) = L+ and lim

t→t−0f(t) = L− in terms

of one-sided limits, with L+ 6= L−, then the curve would have a “break” in it. Anobject traveling on such a trajectory would have to teleport at the break.

1.1.4 Derivatives

Definition:Given a vector-valued function g : [a, b] → Rn, the derivative of g at t, denoted g′(t), is

defined by

g′(t) = lim∆t→0

1

∆t[g(t+ ∆t)− g(t)] , (1.9)

provided this limit exists.

Comments:

i) If g′(t) exists we say that g is differentiable at t.

ii) In terms of component functions

g(t) = (g1(t), . . . , gn(t)),

the Newton quotient is given by

1

∆t[g(t+ ∆t)− g(t)] =

(g1(t+ ∆t)− g1(t)

∆t, . . . ,

gn(t+ ∆t)− gn(t)

∆t

).

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It follows from (1.8) and (1.9) that

g′(t) exists ⇐⇒ g′i(t) exists, i = 1, 2, . . . , n,

andg′(t) = (g′1(t), . . . , g′n(t)). (1.10)

In practice one calculates the derivative g′(t) using (1.10) and routine differentiation.

Physical interpretation of the derivative:

Consider a curve C in R3, x = g(t), which represents the path of a particle. We write

∆x = g(t+ ∆t)− g(t).

Then the Newton quotient (1.7) is(1

∆t

)[g(t+ ∆t)− g(t)] =

(1

∆t

)∆x.

Thus by (1.9), g′(t) represent the rate of change of position with respect to time t, i.e. g′(t)is the velocity v(t) of the particle. We write

v(t) = g′(t) or v(t) =dx

dt. (1.11)

Writing the position function g(t) in terms of its component functions,

g(t) = (x(t), y(t), z(t)),

equations (1.10) and (1.11) imply that

v(t) = (x′(t), y′(t), z′(t)), (1.12)

giving the components of the velocity vector. The magnitude of the velocity vector

‖ v(t) ‖=√x′(t)2 + y′(t)2 + z′(t)2 (1.13)

is of course the speed of the particle.If g′(t) is itself differentiable, the function g will have a second derivative denoted g′′(t),

andg′′(t) = (x′′(t), y′′(t), z′′(t))

in terms of the component functions. The second derivative represents the acceleration vectorof the particle, and we write

a(t) = g′′(t), or a(t) =d2x

dt2. (1.14)

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Geometrical interpretation of the derivative:

Given a curvex = g(t),

with g′(t0) 6= 000, then g′(t0) is a vector in the direction of the tangent line at the pointx = g(t0). This conclusion follows from Figure 1.10. Note that as ∆t → 0 the increment∆g tends to zero and the secant line approaches the position of the tangent line. Thisresult can also be confirmed on physical grounds: the velocity vector v(t0) = g′(t0) givesthe instantaneous direction of motion and hence must be tangent to the path of the particlewhen t = t0.

the secant line

the tangent line

0

!g = g(t0 + !t )" g(t

0)

g(t0 + !t )

# g (t0)

g( t0)

!g

Figure 1.10: The secant line and tangent line to a curve.

Equation of the tangent line:

Given a curve x = g(t) with g′(t0) 6= 000, the equation of the tangent line at x = g(t0) is

x = g(t0) + (t− t0)g′(t0), (1.15)

as follows from Figure 1.11. Referring to Figure 1.11, we note that

OQ = OP + PQ,

and PQ is a multiple of g′(t0) which we write as (t− t0)g′(t0). This choice of parameter onthe tangent line ensures that t = t0 gives the point of tangency.

In equation (1.15) we have defined a vector-valued function which we denote by Lt0(t):

Lt0(t) = g(t0) + (t− t0)g′(t0) (1.16)

12

Qtangent

line

P0

x = g(t )

! g (t0)

g( t0)

Figure 1.11: The tangent line to a curve x = g(t).

called the linearization of g at t0 Since the tangent line approximates the curve for t suffi-ciently close to t0, we obtain the linear approximation

g(t) ≈ Lt0(t),

or in full,g(t) ≈ g(t0) + (t− t0)g′(t0), (1.17)

for t sufficiently close to t0. We can write (1.17) more concisely in terms of the increments

∆x = g(t)− g(t0), ∆t = t− t0.

Rearranging (1.17) gives∆x ≈ (∆t)g′(t0), (1.18)

for ∆t sufficiently close to zero.

Terminology:

i) When we say that a vector-valued function g : [a, b] → Rn is of class C1 (or morebriefly, is C1) we mean that g has a derivative function g′ that is itself a continuousfunction: think of C1 as “C for continuous” and “1 for first derivative”. We shall usuallywork with C1 curves, i.e. curves described by C1 functions, although one can imaginesituations where the path of a particle would be a continuous but not a C1 curve.There is a subtle point concerning C1 curves. Based on experience with the graphs ofscalar functions one might expect that a C1 curve would look “smooth”. A C1 curvex = g(t) can, however, have cusps (i.e. sharp spikes, or corners). See Examples 1.5and 1.6 below. Physically, these occur when a particle stops, and changes its directionof motion.

13

ii) A function g : [a, b]→ Rn is piecewise C1 means that there is a partition of [a, b],

a = t0 < t1 < · · · < tN = b,

such that g, when restricted to each open interval (ti, ti+1), i = 0, 1, . . . , N−1, coincideswith a function that is C1 on the closed interval [ti, ti+1].

Example 1.5:Consider the curves defined by

i) x = g1(t) = (t |t|, t2), −1 ≤ t ≤ 1ii) x = g2(t) = (t, |t|), −1 ≤ t ≤ 1.

In both cases the component functions x(t), y(t) satisfy y = |x|, so the curve is a ∨, i.e. thecurve is not “smooth”. The point we wish to illustrate is that g1 is C1, while g2 is piecewiseC1 but not C1. The difference is that g′1 (0) = 000, while g′2(0) does not exist (convince yourselfof this). Physically, g1 represents the path of a particle that slows down and comes to restat time t = 0 and then changes direction and speeds up again, while g2 represents the pathof a particle that bounces off a hard surface without coming to rest (an idealized billiardball).

Example 1.6:The curve defined by

x = g(t) = b(t− sin t, 1− cos t), t ∈ R, (1.19)

where b is a constant, is called a cycloid.The derivative is

g′(t) = b(1− cos t, sin t),

so thatg′(2nπ) = 000, n = 0,±1,±2, . . . .

Note the properties of the component functions:

i) x′ ≥ 0, x is non-decreasing,

ii) y ≥ 0, y is periodic.

This curve is C1, but is not smooth. It is the path of a point on a uniformly rolling circle.(See Figure 1.12.)

Comment:A curve x = g(t), a ≤ t ≤ b, with g of class C1 and g′(t) 6= 000 for all t ∈ [a, b], will be

smooth, because it will have a unique non-zero tangent vector at each point. We shall referto such a curve as a smooth curve.

14

y

2b

0

x

2!b

Figure 1.12: The cycloid.

1.1.5 Arclength

We want to calculate the arclength s of a curve C in Rn given by

x = g(t), a ≤ t ≤ b. (1.20)

We assume that the curve is of class C1. If C represents the path of a particle then thearclength s represents the distance travelled by the particle in the time interval a ≤ t ≤ b.If C represents a hanging cable (e.g. a section of a power line) then s represents the actuallength of the cable.

A partition of [a, b],a = t0 < t1 < · · · < tN = b

defines N + 1 points on the curve (1.20), given by xi = g(ti), i = 0, 1, . . . , N . Joiningthese points in order with straight lines yields a polygonal arc that can be regarded asapproximating the curve C. See Figure 1.13 for the case N = 4.

x0

x1

x2

x3

x4

C

Figure 1.13: A polygonal arc approximating a curve C.

15

The ith increment∆xi = xi+1 − xi

represents the vector joining xi to xi+1. Its length will approximate the arclength ∆si of theith curve segment:

∆si ≈‖ ∆xi ‖ . (1.21)

Using the linear approximation (1.18),

∆xi ≈ (∆ti)g′(ti), (1.22)

x i

Δsi

xi+1

Δx

i

Figure 1.14: The length of the line segment approximates the arclength of the curve segment.

where∆ti = ti+1 − ti,

for ∆ti sufficiently small.It follows from (1.21) and (1.22) that

∆si ≈‖ g′(ti) ‖ ∆ti, i = 0, 1, . . . N − 1, (1.23)

for ∆ti sufficiently small. Summing over N curve segments gives an approximation for thetotal arclength s:

s ≈N−1∑i=0

‖ g′(ti) ‖ ∆ti.

The sum is a Riemann sum for the integral∫ b

a

‖ g′(t) ‖ dt,

16

and will thus approximate the integral with increasing accuracy as the partition becomesincreasingly fine. We thus expect that the arclength s of the curve x = g(t), a ≤ t ≤ b isgiven by

s =

∫ b

a

‖ g′(t) ‖ dt. (1.24)

Comments:

i) In practice we use equation (1.24) as the definition of arclength of a C1 curve. Thesteps leading to (1.21) do not constitute a proof because of the approximations made,but should be viewed as a heuristic justification of (1.24). It is important to understandthese steps, however, because they form the basis for the definition of the concept ofline integral in Chapter 2.

ii) If the curve C represents the path of a particle then the integrand ‖ g′(t) ‖ in (1.24) isthe speed of the particle, and so the formula (1.24) has a simple physical interpretation:

distance travelled is the integral of thespeed with respect to time.

1.2 Vector fields

So far we have considered vector-valued functions g : [a, b]→ Rn, which define curves. In thissection we consider a second type of vector-valued function of great importance in physicsand engineering applications, namely vector fields.

1.2.1 Examples from physics

Many physical quantities are characterized by giving a magnitude, i.e. a real number, ateach point of space. Such quantities are represented by real-valued functions f : R3 → R.Functions of this type, when used in a physical context, are called scalar fields. Someexamples are the following:

- the temperature T (x, y, z) at a point of a body,

- the pressure p(x, y, z) at a point of a fluid,

- the electric charge density σ(x, y, z) on the surface of a metallic object,

- the intensity of light I(x, y, z) falling on an object,

. . . and so on.

On the other hand there are many physical quantities that are characterized by givinga magnitude and a direction, i.e. a vector, at each point of space. Such quantities arerepresented by a function F : R3 → R3 that associates with each point (x, y, z) ∈ R3 a

17

unique vector F(x, y, z) ∈ R3. Functions of this type, when used in a physical context, arecalled vector fields.

Example 1.7:The gravitational field due to a spherical body of mass M , such as the earth, can be

represented by the vector field F defined by

F(x, y, z) = −GMr3

r, (1.25)

where r = (x, y, z), r =‖ r ‖=√x2 + y2 + z2, and G is the gravitational constant.

The vector F(x, y, z) represents the force exerted on a test body of unit mass placed atposition (x, y, z). The minus sign accounts for the fact that the test body will be attractedto the earth, i.e. F (x, y, z) points towards the origin.

x

z

y

F(x ,y, z)

r = (x , y, z)

Figure 1.15: The gravitational field of a spherical body.

Comment:In this example the domain of the vector field is the open set U = R3 − (0, 0, 0).

Example 1.8:Consider a distribution of fluid flowing in a steady state. Then at a given point (x, y, z)

the fluid has a uniquely defined velocity v(x, y, z) that is independent of the time. Thevector field v is called the velocity field of the fluid. See Figure 1.16.

Example 1.9:There is also a scalar field ρ(x, y, z) associated with a fluid, namely the mass density of

the fluid, which is independent of time for a steady state flow. One is interested in the rateat which mass is transferred by the fluid, and this rate is determined by another vector fieldformed from ρ and v, as follows.

Consider a plane surface element of area ∆S, with unit normal vector n, located atposition (x, y, z). In time ∆t, the fluid in a column of length ‖v ‖∆t will flow through thesurface element. Since the vertical height of the column is v ··· n∆t (see Figure 1.17) thevolume of the column, and hence the volume of fluid transported, is

∆V ≈ (v ··· n)∆S∆t.

18

(x , y,z )

v(x ,y ,z )

Figure 1.16: The velocity field of a fluid.

Figure 1.17: Flow of a fluid through a plane surface element.

It follows that mass of fluid transported across the surface element in time ∆t is

∆M ≈ ρ∆V ≈ ρv · n∆S∆t.

Thus, the mass transferred per unit time across the surface element is

(ρv) ··· n∆S. (1.26)

This quantity is called the mass flux across the surface element:

“flux” means “rate of flow”.

It is worth checking the physical dimensions of the mass flux (note that n, being a unitnormal, is dimensionless):

[(ρv) ··· n∆S] = (ML−3)(LT−1)(L2) = MT−1,

i.e. mass per unit time.

19

The mass flux (1.26) is determined by the vector field J defined by

J(x, y, z) = ρ(x, y, z)v(x, y, z), (1.27)

which has physical dimensions[ J ] = MT−1L−2,

i.e. mass per unit time per unit area. This vector field, which describes the transport ofmass by the fluid, is called the mass flux density of the fluid.

Comment:In (1.26) and the preceding equations, ρ and v are evaluated at the given point (x, y, z).

We are assuming that ‖ v ‖ ∆t and ∆S are sufficiently small that ρ and v can be regardedas constant throughout the cylinder in Figure 1.17.

Example 1.10:Any C1 scalar field u in R3 determines a vector field F in R3 according to

F(x, y, z) = ∇u(x, y, z), (1.28)

where ∇u is the gradient of the scalar field, defined by

∇u =

(∂u

∂x,∂u

∂y,∂u

∂z

). (1.29)

In other words, the gradient of a C1 scalar field is a vector field. This type of vector fieldwill arise frequently in the course.

The concept of a flux density vector field, as in Example 1.3, where we studied the massflux density, applies to any physical quantity that is transferred in space, e.g. heat energy,which is the next example.

Example 1.11:Consider a piece of material that is heated on one side and cooled on the other in such

a way that the temperature attains a steady state, i.e. the temperature u(x, y, z) dependson position (x, y, z) but not on time t. The temperature is a scalar field. From everydayexperience we know that heat will flow (i.e. energy will be transferred) from hot regions tocold regions. As with mass flow, heat flow can be described mathematically by a vector fieldj(x, y, z), which gives the rate of heat transfer through a small surface element at (x, y, z),called the heat flux density, i.e.

(j · n)∆S

is the rate of transfer of heat through a small plane surface element of area ∆S and havingunit normal n (compare with equation (1.26)).

Experimental work shows that the heat flux density may be represented by Fourier’s law

j(x, y, z) = −k∇u(x, y, z), (1.30)

20

where k > 0 is a constant. This law is plausible physically, since one expects that the larger isthe spatial temperature gradient, represented by the gradient ∇u, the larger will be the rateof heat transfer. The constant k, called the thermal conductivity, depends on the material,being bigger for a good conductor of heat than for a poor conductor (e.g. kcopper/kglass 1).

Having given some physical examples of vector fields we now introduce the terminologyformally.

Let U be an open subset of Rn. A vector field on U is a function F : U → Rn, whosedomain is U and whose range is in Rn, i.e. F assigns to each x ∈ U a unique vector F(x) ∈ Rn.

The vector fields we work with will usually be of class C1, which means that the ncomponent functions,

F(x) = (F1(x), . . . , Fn(x))

are C1 functions.We shall also assume that the domain U is a connected set, which means that U is not

the union of two or more disjoint sets.

1.2.2 Field lines of a vector field

One visualizes a vector field F on an open set U ⊂ R3 as a “field of vectors”, represented byarrows, attached to the points of U . The length of the vector at a point gives the strength ofthe field at the point, and the arrow gives the direction of the field.

It is also helpful to think of the family of curves in U with the property that at each pointP the tangent to the curve through P equals the vector field evaluated at P . These curves arecalled the field lines1 of the vector field. One thinks of a field line threading its way throughthe vector field, always following the direction of the vector field (see Figure 1.18).

Figure 1.18: Two field lines of a vector field.

1These curves are also called integral curves of the vector field.

21

N S

Figure 1.19: Field lines of a magnetic field.

In the case of the velocity field of a fluid the field lines are simply the paths of the fluidparticles (see Figures 1.16 and 1.18). In the case of a magnetic field one can do a simpleexperiment to visualize the field lines. Take a sheet of paper and sprinkle it with iron filings.Then put a bar magnet under the paper and shake the paper slightly. You will observe theiron filings arranging themselves in lines going from one magnetic pole to the other. Thestrength of the magnetic field is revealed by the density of packing of the filings. In this wayone obtains a picture of the field lines of the magnetic field.

We now consider the problem of determining the field lines of a given vector field F (x).

Let the curve x = g(t), assumed C1, be a field line. Its tangent vector is g′(t), and thusthe defining condition of a field line is written

g′(t) = F(g(t)). (1.31)

(see Figure 1.20). Equation (1.31) is a differential equation for the unknown vector-valuedfunction g(t). If you want to find the field line through a given point x0 then you shouldimpose the initial condition

g(t0) = x0. (1.32)

DEs such as (1.31) are studied in depth in the course AM 451. In general they can onlybe solved numerically using a computer. For this course, however, it will be enough toconcentrate on simple types that can be solved explicitly, as in the examples to follow.

Example 1.12:Find the field lines of the vector field

F(x, y) = (−y, x) (1.33)

22

x0

x = g(t )

′ g (t) = F(g(t))

Figure 1.20: The field line of a vector field F through a given point x0.

in R2, and sketch the field portrait.

Solution: A field line x = g(t) satisfies

g′(t) = F(g(t)).

In terms of components g(t) = (x(t), y(t)), this reads

(x′(t), y′(t)) = (−y(t), x(t)),

givingdx

dt= −y(t),

dy

dt= x(t). (1.34)

These two coupled DEs can be written as a single DE by using the chain rule:

dy

dt=dy

dx

dx

dt,

givingdy

dx= −x

y

by equations (1.34). Solving this separable DE yields∫y dy = −

∫x dx,

givingx2 + y2 = C,

where C is a constant.The conclusion is that any field line of the given vector field is a circle centred on the

origin. Equations (1.34) show that the circles are traversed counterclockwise, giving Figure1.21. We note that the field line through a given point (x0, y0) is the circle of radius

√x2

0 + y20

– specifying a point on the field line fixes the value of the constant C.

23

y

x0

Figure 1.21: The field lines x2 + y2 = C of the vector field F = (−y, x).

Exercise 1.3:The vector field in Example 1.12 can be interpreted physically as the velocity field of a

rigidly rotating disc with unit angular velocity. Verify that

θ′ ≡ dθ

dt= 1.

y

x

( ′ x , ′ y )= ( −y, x)

θ ( x, y) = (r cos θ, r sin θ )

Figure 1.22: Velocity of a point on a rotating disc.

Example 1.13:Find the field lines of the vector field

F(x, y) =

(−y

x2 + y2,

x

x2 + y2

)(1.35)

on the open set U = R2 − (0, 0), and sketch the “portrait”.

24

Solution: We have to solve the DEs

dx

dt=

−yx2 + y2

,dy

dt=

x

x2 + y2.

Proceeding as in Example 1.12, these equations lead to the same DE

dy

dx= −x

y,

givingx2 + y2 = C,

where C is a constant, as the field lines.

y

x

Figure 1.23: Field lines of the vector field (1.35).

Comment:The vector field (1.35) could represent the velocity field of a fluid swirling down a drain.

Note that for Example 1.12,‖ F(x, y) ‖=

√x2 + y2,

i.e. the speed equals the distance from the origin, while for Example 1.13,

‖ F(x, y) ‖= 1√x2 + y2

, (x, y) 6= (0, 0),

i.e. the speed equals the reciprocal of the distance from the origin. We have indicated thisdifference in Figures 1.21 and 1.23 by the size of the arrows. We note that Examples 1.12and 1.13 illustrate that different vector fields can have the same field lines.

Exercise 1.4:Find the field lines of the vector field F(x, y) = (x, 2y) in R2, and sketch the field portrait.

25

Gradient vector fields:

The field lines of a vector field F(x) = ∇u(x) in R2 that is the gradient of a scalar field canbe drawn without solving a DE. We know (Calculus 3) that the gradient ∇u of a scalar fieldu is orthogonal to the level curves u = constant of the scalar field. It follows that the fieldlines of the vector field F = ∇u are the orthogonal trajectories of the family of level curvesof u.

u= const

.

∇u

∇u

∇u

Figure 1.24: The field lines of F = ∇u intersect the level curves u = constant orthogonally.

26

Chapter 2

Line Integrals & Green’s Theorem

In this chapter we define two types of integral that are associated with a curve in Rn.

2.1 Line integral of a scalar field

2.1.1 Motivation and definition

Consider a nuclear fuel rod, with linear mass density ρ (i.e. the physical dimensions are[ρ] = ML−1) and length `. If ρ is a constant, the mass m of the rod is simply m = ρ`.Suppose that due to manufacturing defects, the density depends on position on the rod, sayρ = ρ(x), 0 ≤ x ≤ `. How does one calculate the mass of the rod? Well, one approximatesthe mass m as a Riemann sum of the density function on the interval 0 ≤ x ≤ `:

m ≈n∑i=1

ρ(xi)∆xi,

leading to the formula

m =

∫ `

0

ρ(x)dx. (2.1)

A similar but more difficult problem is to find the mass of a suspended wire (say, part ofa hydro line, see Figure 2.1) whose linear mass density depends on position. Evidently, wewill need some sort of integral along the curve C that represents the wire. This new type ofintegral, which we now introduce, is called the line integral of a scalar field.

Given a curve C in Rn, defined by x = g(t), a ≤ t ≤ b, where g is a class C1, and a scalarfield f continuous on C, the line integral of f along C is denoted by∫

C

f ds.

We first give a tentative definition, motivated by the problem of calculating the mass of thehanging wire. As in the calculation of arclength (Section 1.1.5) we introduce a partition of[a, b],

a = t0 < t1 < · · · < tN = b,

27

Figure 2.1: A curve C representing a hanging wire.

which defines N + 1 points xi = g(ti) on the curve C. Let ∆si be the arclength of the ith

curve segment. We make the tentative definition:∫C

f ds = lim N→∞|∆ti|→0

N−1∑i=0

f(xi)∆si, (2.2)

provided the limit exists (∆ti = ti − ti−1 as before).Thinking of the scalar field f as representing the linear density of the hanging wire, the

term f(xi)∆si approximates the mass of the ith segment of the wire, and so we expect thelimit of the sum to give the total mass of the wire.

Figure 2.2: The ith segment of the hanging wire.

Substituting xi = g(ti) and using the approximation ∆si ≈‖ g′(ti) ‖ ∆ti (see equation(1.23)), we can approximate the sum in (2.2) as

N−1∑i=0

f(xi)∆si ≈N−1∑i=0

f(g(ti)) ‖ g′(ti) ‖ ∆ti, (2.3)

28

for ∆ti sufficiently small. In the limit as N → +∞ and |∆ti| → 0 the right side of (2.3)

equals the Riemann integral

∫ b

a

f(g(t)) ‖ g′(t) ‖ dt, and we expect the approximation in

(2.3) to become increasingly accurate. Comparison of (2.2) and (2.3) then motivates thedefinition to follow.

Definition:Consider a curve C in Rn given by x = g(t), a ≤ t ≤ b, where g is of class C1, and a

scalar field f continuous on C. The line integral of f along C is defined by∫C

f ds =

∫ b

a

f(g(t)) ‖ g′(t) ‖ dt. (2.4)

Comment:We see that the line integral is defined in terms of an ordinary Riemann integral. The

formula (2.4) can be remembered easily as follows:

“f” is evaluated on the curve C giving “f(g(t))”, and the symbol “ds” remindsone of ∆s in (2.2), which is approximated as ∆s ≈‖ g′(t) ‖ ∆t, leading to“‖ g′(t) ‖ dt”.

Example 2.1:

Evaluate the line integral

∫C

f ds in R2, where the curve C is the upper semi-circle of

radius b joining (b, 0) and (−b, 0), and f is the scalar field defined by f(x, y) = x2 + 3y2.

Solution: We introduce the standard parametrization for the semi-circle,

x = g(t) = (b cos t, b sin t), 0 ≤ t ≤ π.

Theng′(t) = (−b sin t, b cos t),

and the magnitude is‖ g′(t) ‖=

√(−b sin t)2 + (b cos t)2 = b,

after simplifying. Evaluating the scalar field on the curve C gives

f(g(t)) = (b cos t)2 + 3(b sin t)2 = b2(3− 2 cos2 t),

after simplifying. By the definition (2.4) of the line integral,

29

∫C

f ds =

∫ π

0

f(g(t)) ‖ g′(t) ‖ dt

=

∫ π

0

b2(3− 2 cos2 t)b dt

= · · · = 2πb3. Aside: 2 cos2 t = 1 + cos 2t.

Exercise 2.1:

Evaluate the line integral

∫C

f ds in R3, where C is the helix x = g(t) = (R cos t, R sin t, t),

0 ≤ t ≤ 4π, and f is the scalar field f(x, y, z) = z.

An important consistency requirement:

Given a hanging wire represented by a curve C and a scalar field f representing the linear

mass density of the wire, we have seen that the line integral

∫C

f ds represents the mass of

the wire, which can be calculated using the definition (2.4). We know, however, that a curveC has infinitely many different parametrizations (Section 1.1.2). Clearly, the value of the lineintegral (the mass of the wire) should not depend on which parametrization we use. Considera different parametrization for the same curve C:

x = g(τ), α ≤ τ ≤ β,

with g(τ) = g(h(τ)) and t = h(τ) (see equation (1.6)). The definition (2.4) becomes∫C

f ds =

∫ β

α

f(g(τ)) ‖ g′(τ) ‖ dτ. (2.5)

The consistency requirement is that the Riemann integrals in (2.4) and (2.5) must be equal.The proposition to follow establishes the consistency.

Proposition 2.1:Consider a curve C given by

x = g(t), a ≤ t ≤ b,

with g of class C1. Under a change of parameter t = h(τ), with h of class C1 and h′(τ) > 0for τ ∈ [α, β], the curve is described by

x = g(τ), α ≤ τ ≤ β,

withg(τ) = g(h(τ)). (2.6)

30

Then ∫ b

a

f(g(t)) ‖ g′(t) ‖ dt =

∫ β

α

f(g(τ)) ‖ g′(τ) ‖ dτ. (2.7)

Proof: (outline)Differentiate (2.6) with respect to τ and use the Chain Rule to get

g′(τ) = h′(τ)g′(h(τ)).

Since h′(τ) > 0 it follows thatAside: See Problem Set 1, #20.

‖ g′(τ) ‖=‖ g′(h(τ)) ‖ h′(τ). (2.8)

Substitute (2.6) and (2.8) into the integral on the right in (2.7). Since t = h(τ) and a = h(α),b = h(β), the Change of Variable Theorem now implies the integral on the right equals theintegral on the left.

Exercise 2.2:Repeat Example 2.1 using a different parametrization of C, for example

x = g(τ) = (b cos 2τ, b sin 2τ), 0 ≤ τ ≤ π2,

and confirm that you get the same value for the line integral.

2.1.2 Applications

When working with a line integral in a physical context it is essential to keep in mind that

an integral is the limit of a sum1

(as in equation (2.2)). An important special case arises if f(x) = 1 for all x ∈ C. Then thedefinition (2.4) becomes ∫

C

ds =

∫ b

a

‖ g′(t) ‖ dt,

which by equation (1.24) equals the arclength of the curve C. In words, the line integral ofthe constant scalar field f(x) = 1 along a curve C equals the arclength of C, i.e. one thinks

of the line integral

∫C

ds as summing the elements of arclength along the curve C to give the

total arclength.We now give a glimpse of some other applications of the line integral of a scalar field

i) An “everyday” example:

The base of a vertical curved fence is a curve C in the xy-plane, and its height at position(x, y) is h(x, y). What is the total area of the fence?

1This statement applies to ANY integral.

31

Consideration of the preliminary definition (2.2) leads to the conclusion that the area Ais given by

A =

∫C

h ds.

AAAAAAAAAAAAAA

AAAAAAAAA

AAAAAA

AAAAAAAAAAAA

AAAAAAAAAAAAAA

AAAAAAAAA

h(x,y)

(x,y)

∆sC

Figure 2.3: A curved fence whose base is a plane curve C.

One thinks of the line integral as summing the product of height h and element of arclength∆s.

Exercise 2.3:The base of a vertical fence is given by x = g(t) = (b cos3 t, b sin3 t), 0 ≤ t ≤ π, where b

is a positive constant, and the height at position x = (x, y) is h(x, y) = b + 13y. Show that

the area of the fence is 175b2.

Reference: Marsden & Tromba, page 417, example 2.

ii) Line integral of a linear density function

It is helpful to think of the physical dimensions of quantities when interpreting a lineintegral

I =

∫C

f ds. (2.9)

In terms of the preliminary definition (2.2), we have

I = limN→∞

N−1∑i=0

f(x)∆si,

where ∆si is the arclength of the ith segment of C. Since2

[f(xi)∆si] = [f(xi)][∆si] = [f(xi)]L,

2We use the symbol [I] to denote the dimensions of a physical quantity I.

32

it follows that the dimensions of the integral I are

[I] = [f ]L. (2.10)

In the hanging wire problem, f represents the linear mass density, i.e. [f ] = ML−1, andthe line integral I in (2.9) represents the total mass of the wire. Equation (2.10) gives

[I] = ML−1L = M,

which is consistent with the interpretation of I.As another example, think of the curve C as representing a conductor (e.g. a copper

wire) which is charged, with f being the linear charge density, i.e. [f ] = [charge]L−1. Theline integral I will give the total charge on the conductor. Equation (2.10) implies

[I] = [charge]LL−1 = [charge],

which is again consistent with the interpretation of I.In general we can think of the scalar field f as representing the linear density of some

“physical stuff”, which is distributed on a curve C. Then the line integral I in (2.9) givesthe total amount of “physical stuff” on the curve. As before, equation (2.10) guaranteesdimensional consistency.

iii) Average value of a scalar field on a curve

For a continuous function f : [a, b]→ R, the average value of f over the interval [a, b] isdefined by

〈f〉 =

∫ baf(x)dx∫ badx

,

i.e. the average value is the integral of f over [a, b] divided by the length of the interval. Wecan generalize the concept of “average of a function” to the case of a scalar field f that iscontinuous on a curve C in Rn. In this case we define the average value by

〈f〉 =

∫Cf ds∫Cds

, (2.11)

i.e. the average value of f is the line integral of f along C divided by the arclength of C.

Comment:If we choose a partition so that the N curve segments of C are of equal length (i.e.

∆si = ∆s, i = 0, 1, . . . , N − 1), it follows from (2.11) and (2.2) that

〈f〉 ≈ 1

N

N−1∑i=0

f(xi),

i.e. the “continuous average of f” is approximated by the “discrete average” of N values ofthe scalar field on C.

33

Exercise 2.4:The steady state temperature of a circular metal plate of radius b centred on the origin

in the xy-plane is given by

u(x, y) =u0

b2(x2 − y2),

where u0 is a constant. Show that the average temperature along the diameter y = (tan θ)xis given by

〈u〉 = 13u0 cos 2θ.

2.2 Line integral of a vector field

2.2.1 Motivation and definition

Consider a particle moving along the x-axis from x = a to x = b under the action of a forceF. If F is constant the work done on the particle is simply

a bi

F

x

W = (F · i)(b− a),

where F · i is the component of F in the direction of motion (i is a unit vector in the x-direction). If F = F(x) then the work done can be calculated as an integral (the limit of aRiemann sum),

W =

∫ b

a

(F · i)dx.

A similar but more difficult problem is to calculate the work done by a force field F(x) actingon a particle moving in a complicated way in space. Evidently we will need some sort ofintegral along the curve C that represents the path of the particle in space. This new typeof integral is called the line integral of a vector field.

Given a curve C in Rn, defined by x = g(t), a ≤ t ≤ b, where g is of class C1, and avector field F continuous on C, the line integral of F along C is denoted by∫

C

F ··· dx.

We first give a tentative definition, motivated by the problem of calculating the workdone by a force field. The development parallels that of Section 2.1, 2.1.1 very closely. Weintroduce a partition of [a, b],

a = t0 < t1 < · · · < tN = b,

34

which defines N + 1 points xi = g(ti) on the curve C. Let

∆xi = xi+1 − xi

be the increment vector associated with the ith curve segment. We make the tentativedefinition: ∫

C

F ··· dx = limN→∞|∆ti|→0

N−1∑i=0

F(xi) ···∆xi, (2.12)

provided the limit exists. Thinking of the vector field as the force field acting on the movingparticle, the term F(xi)···∆xi approximates the work done on the particle while it moves fromxi to xi+1, and so we expect the limit of the sum to give the total work done on the particle.

∆xi

x i

F(xi) xi+1

C

Figure 2.4: The ith segment of the path of a particle and the increment vector ∆xi.

Substituting xi = g(ti) and using the approximation

∆xi ≈ (∆ti)g′(ti),

(see equation (1.22)), we can approximate the sum in (2.12) as

N−1∑i=0

F(xi) ···∆xi ≈N−1∑i=0

F(g(ti)) ··· g′(ti)∆ti (2.13)

for ∆ti sufficiently small. In the limit as N → ∞ and |∆ti| → 0 the right side of (2.13)

equals the Riemann integral

∫ b

a

F(g(t))···g′(t)dt, and we expect the approximation in (2.13) to

become increasingly accurate. Comparison of (2.13) and (2.12) then motivates the definitionto follow.

Definition:Consider a curve C in Rn given by x = g(t), a ≤ t ≤ b, where g is of class C1, and a

vector field F continuous on C. The line integral of F along C is defined by∫C

F·dx =

∫ b

a

F(g(t)) ··· g′(t) dt. (2.14)

35

Comment:The line integral of a vector field is defined in terms of an ordinary Riemann integral.

The formula (2.14) can be remembered easily, because F·dx reminds one of F(g(t)) ··· ∆x,which is approximated by F(g(t)) ··· g′(t)∆t.

Example 2.2:

Evaluate the line integral

∫C

F·dx, where C is the quarter circle of radius b, x2 + y2 = b2,

joining (b, 0) to (0, b), and F is the vector field defined by F(x, y) = (y, 0).

Solution: We use the usual parametrization for the circle:

x = g(t) = (b cos t, b sin t), 0 ≤ t ≤ π2.

Theng′(t) = (−b sin t, b cos t).

Evaluating the vector field on C gives

F(g(t)) = (b sin t, 0).

By the definition (2.14),

∫C

F·dx =

∫ π/2

0

(b sin t, 0) ··· (−b sin t, b cos t)dt

= −b2

∫ π/2

0

sin2 t dt

= −14πb2.

Aside: sin2 t = 12(1− cos 2t)

Comment:

The line integral

∫C

F·dx is the limit of the sum of scalar products F·∆x. One can thus

predict the sign of the line integral by referring to Figure 2.5, which shows that F·∆x < 0except at the point (b, 0), since the angle φ between F and ∆x satisfies π

2< φ ≤ π.

Exercise 2.5:Let C be the straight line joining (b, 0) to (0, b) and let F be the vector field F(x, y) =

(y, 0), as in Example 2.2. Show that

∫C

F·dx = −12b2.

Exercise 2.6:Compute the line integral of the vector field F(x) = (x, y, z) along the helix C defined by

x = g(t) = (cos t, sin t, t), 0 ≤ t ≤ 2π.

Answer: 2π2.

36

y

(0,b)

(b,0)x

F

F

FC ψ

Figure 2.5: The vector field F(x, y) = (y, 0) in Example 2.2.

2.2.2 Applications

So far we have considered the following physical interpretation of the line integral of a vectorfield along a curve C:

if F is a force field acting on a particle whose path is the curve C, then

∫C

F·dx

represents the work done by the force field on the particle.

In this course a major role is played by line integrals

∫C

F·dx, where C is a closed curve,

i.e. the end point coincides with the initial point. Line integrals of this type enter into two ofthe principal theorems, namely Green’s theorem (Section 2.4) and Stokes’ theorem (Section4.3). The physical interpretation of the line integral depends on the interpretation of thevector field, two important cases being where F is a velocity field (in fluid dynamics), andwhere F is an electric field (in electromagnetic theory).

C

37

2.2.3 Some technical matters

The consistency requirement:

The definition (2.14) of

∫C

F·dx is based on choosing a parametrization for the curve C.

However, the work done by a force field should be independent of the parametrization, andso in order to show that the definition is consistent, we should verify that it is independentof the parametrization. The situation is similar to that for the line integral of a scalar field,and a result analogous to Proposition 2.1 holds, namely∫ b

a

F(g(t)) ··· g′(t)dt =

∫ β

α

F(g(τ)) ··· g′(τ)dτ, (2.15)

where the notation is defined in Proposition 2.1.

Other notation for the line integral:

There is another notation for

∫C

F·dx that is popular in physics and engineering, namely

∫C

F·dx ≡∫C

F1dx+ F2dy + F3dz,

where F = (F1, F2, F3). In a purely formal sense, one expands the “scalar product” F·dxwith “dx = (dx, dy, dz)”. In terms of this alternate notation, the definition reads∫

C

F1dx+ F2dy + F3dz =

∫ b

a

[F1( )

dx

dt+ F2( )

dy

dt+ F3( )

dz

dt

]dt, (2.16)

where x = (x(t), y(t), z(t)) is a parametrization of C and Fi( ) = Fi(x(t), y(t), z(t)), i = 1, 2, 3.Mathematically, the quantity F1dx+F2dy+F3dz is called a differential form. The theory

of differential forms is more modern than that of vector fields, and is useful, for instance, ingeneralizing vector calculus, and in the subject of differential geometry. For our purposes,however, we do not need this more sophisticated approach.

Exercise 2.7:Evaluate the line integral

I =

∫C

cos z dx+ exdy + eydz,

where the curve C is given by x(t) = (1, t, et), 0 ≤ t ≤ 2.

Answer: 2e+ 12e4 − 1

2.

38

Properties of line integrals:

i) Linearity: ∫C

(F + G) ··· dx =

∫C

F·dx +

∫C

G·dx, (2.17)

∫C

(λF) ··· dx = λ

∫C

F·dx, (2.18)

where λ is a constant scalar.

These properties follow immediately from the definition (2.14) and the correspondingproperties of the Riemann integral.

ii) Additivity:

If C is a C1 curve that is the union of twocurves C1 and C2 joined end-to-end and consis-tently oriented (C = C1 ∪ C2), then∫

C

F·dx =

∫C1

F·dx +

∫C2

F·dx (2.19)

Line integral along a piecewise C1 curve:

Let C be a continuous curve which is piecewise of class C1 i.e. C = C1 ∪ · · · ∪ Cn, wherethe individual pieces Ci, i = 1, . . . , n are of class C1. Motivated by equation (2.19), we definethe line integral of a vector field F along C by∫

C

F·dx =

∫C1

F·dx + · · ·+∫Cn

F·dx. (2.20)

Each line integral on the right is, of course, defined as a Riemann integral by equation (2.14).

Example 2.3:Compute the line integral of the vector field F = (y,−2x) along the piecewise C1 curve

consisting of the two straight line segments joining (−b, b) to (0, 0) and (0, 0) to (2b, b), whereb is a positive constant.

Solution: For C1, x = g1(t) = (t,−t), with −b ≤ t ≤ 0, giving

g′1(t) = (1,−1), and F(g1(t)) = (−t,−2t).

By the definition (2.14), ∫C1

F·dx =

∫ 0

−b(−t,−2t) ··· (1,−1)dt

=

∫ 0

−bt dt = −1

2b2.

39

y

x(0,0)

(−b,b)

C1 C2

(2b,b)

For C2, x = g2(t) = (2t, t), 0 ≤ t ≤ b, and a similar calculation yields∫C2

F·dx = −b2.

Finally, by (2.20), ∫C

F·dx =

∫C1

F·dx +

∫C2

F·dx

= −12b2 − b2 = −3

2b2.

Exercise 2.8:Calculate the line integral of the vector field F = (y,−2x) along the piecewise smooth

curve consisting of two parabolic segments as drawn.

Answer: 2b2.

(0,0) (2b,0)

y

x

(b,b)

Reversal of orientation:

If C is a curve in Rn with a specific orientation, we denote by −C the curve that isobtained by reversing the orientation. Specifically if C is given by

x = g(t), a ≤ t ≤ b,

then −C is given byx = g(τ), a ≤ τ ≤ b,

whereg(τ) = g(a+ b− τ).

40

Observe that g(a) = g(b) and g(b) = g(a), which reverses the orientation.It is an important result that reversing the orientation of a curve changes the sign of the

line integral of a vector field along the curve.

Proposition 2.2:If C is a piecewise-smooth curve and F is a vector field continuous on C, then∫

−C

F·dx = −∫C

F·dx.

Proof:By definition of line integral∫−C

F·dx =

∫ b

τ=a

F(g(τ)) ··· g′(τ)dτ

= −∫ b

τ=a

F(g(a+ b− τ)) ··· g′(a+ b− τ)dτ (by the equation for g(τ))

= −∫ a

t=b

F(g(t)) ··· g′(t)(−1)dt (by the change of variable t = a+ b− τ)

= −∫ b

t=a

F(g(t)) ··· g′(t)dt (reverse the limits of integration)

= −∫C

F·dx. (by definition of line integral)

Comment:The change in sign is physically reasonable if one thinks in terms of work done by a force

field F; for example, if the height of an object above the earth’s surface is increased, thework done by the gravitational field is negative, whereas if an object falls, the work done ispositive.

2.3 Path-independent line integrals

Let F : U → Rn be a continuous vector field on the connected open set U ∈ Rn. Consider twopoints x1,x2 in U , and imagine all possible piecewise smooth curves in U joining x1 to x2.

In general, the value of the line integral

∫C

F·dx will depend on the particular curve C joining

x1 to x2. In physical terms, thinking of F as a force field, the work done on the particle asit moves from x1 to x2 in general depends on the path followed by the particle. There are,however, certain special vector fields (force fields) with the property that the line integral(the work done) depends only on the endpoints of the curve and not on the particular curvejoining the two points.

41

As an example, consider the vector field

E(x, y) = −kq(

x

(x2 + y2)3/2,

y

(x2 + y2)3/2

). (2.21)

As endpoints, consider x1 = (1, 0) and x2 = (0, 1). You will find that for any piecewise

smooth curve C joining x1 to x2,

∫C

E·dx has the same value, namely zero.

Exercise 2.9:Show that the line integral of the vector field (2.21) equals zero for each of the curves

C1 and C2 joining (1, 0) to (0, 1), where C1 is the quarter circle.

(0,1)

(0,0)

C2

(1,0)

C1

Definition:Let F be a continuous vector field on a connected open set in Rn. We say that the line

integral

∫C

F·dx is path-independent in U if, given any two points x1,x2 in U , the line integral

has the same value for all piecewise smooth curves in U that join x1 to x2.

Exercise 2.9 gives a hint that the line integral of the vector field (2.21) is path-independentin U = R2 − (0, 0). Of course we cannot prove that a line integral is path-independentby calculating its value all different curves joining different pairs of points because there areinfinitely many possibilities! So an important question is: how can we tell whether a givenline integral is path-independent or not? To find out, let us be guided by one of the mostimportant results in elementary calculus, the first Fundamental Theorem.

2.3.1 First Fundamental Theorem for Line Integrals

Recall that if f is continuous on an interval [a, b], then the new function g defined by

g(x) =

∫ x

a

f(t) dt, a ≤ x ≤ b, (2.22)

is such thatg′(x) = f(x). (2.23)

This result is the first Fundamental Theorem of Calculus (FTC).

42

Q: Can we extent this theorem to line integrals?

A: Yes, provided the line-integral is path-independent.

If the line integral

∫C

F·dx is path-independent and C joins x0 to x, we denote the line-

integral by ∫ x

x0

F·dx.

In this case we can define a new function – a scalar field φ – by

φ(x) =

∫ x

x0

F·dx,

in analogy with (2.22). We can now state the first Fundamental Theorem for line integrals.

Theorem 2.1:

Let U be a connected open subset of Rn, and let F : U → Rn be a continuous vector fieldwhose line integral is path independent in U .

If

φ(x) =

∫ x

x0

F·dx, (2.24)

where x0 is a specified point, then,

∇φ(x) = F(x) (2.25)

for all x ∈ U .

Proof:For simplicity we give the proof in R2. With φ defined by (2.24), we have to prove that

∂φ

∂x= F1,

∂φ

∂y= F2,

where F1 and F2 are the components of F.

The key idea is this: since the line integral is path-independent, we are free to make aspecial choice of the curve joining x0 = (x0, y0) to x = (x, y), i.e. to choose a “custom-designed” curve. Figure 2.6 shows the curve we need. Suitable parametrizations for C1 andC2 are

x = g1(t) = (x0, t), y0 ≤ t ≤ y,

x = g2(t) = (t, y), x0 ≤ t ≤ x.

43

y

x

(x,y)(x0,y)

(x0,y0)

C1

C2

Figure 2.6: A piecewise smooth curve C = C1 ∪ C2 joining (x0, y0) to (x, y).

Using these equations and the definition of line integral,

φ(x, y) =

∫C

F·dx =

∫C1

F·dx +

∫C2

F·dx

=

∫ y

y0

F2(x0, t)dt+

∫ x

x0

F1(t, y)dt.

It now follows from the first FTC for Riemann integrals (see equations (2.22) and (2.23))that

∂φ

∂x= 0 + F1(x, y),

since we are treating y as a constant.Similarly we get the result for ∂φ

∂yby choosing a different path (do it!).

Terminology:

The significance of Theorem 2.1 is this: any vector field F whose line integral is path-independent can be written as the gradient of a C1 scalar field. Such a vector field is calleda gradient field. The scalar field ψ is called a potential for F, for physical reasons that we’llsoon see. The level sets ψ(x) = C of the potential ψ are called equipotentials. In R2, wehave equipotential lines ψ(x, y) = C and in R3 we have equipotential surfaces ψ(x, y, z) = C.

As an example, we note that the vector field F(x, y) given by (2.21) (the electric fielddue to a point of charge q at the origin) is derivable from the potential

ψ(x, y) =kq√x2 + y2

(2.26)

i.e. E(x, y) = ∇ψ(x, y) (verify this!). The equipotential lines are given by ψ(x, y) = constant,i.e.

x2 + y2 = const.

44

2.3.2 Second Fundamental Theorem for Line Integrals

Continuing the train of thought from the previous subsection we ask

Q: In elementary calculus we learned that if G, g : [a, b]→ R are such that g is continuousand G′ = g, then ∫ b

a

g(x)dx = G(b)−G(a), (2.27)

(the second FTC). Is there a way to extend this result to line integrals?

A: Yes, provided the vector field F is a gradient field, i.e. F = ∇φ.

This generalization is the Second Fundamental Theorem for line integrals.

Theorem 2.2:

Let F : U → Rn be a continuous vector field on a connected open set U ⊂ Rn, and let x1,x2

be two points in U .If F = ∇φ, where φ : U → R is a C1 scalar field, and C is any curve in U joining x1 to

x2, then ∫C

F·dx = φ(x2)− φ(x1). (2.28)

Proof:Let C be given by

x = g(t), t1 ≤ t ≤ t2,

so thatx1 = g(t1), x2 = g(t2). (2.29)

By the hypothesis,∫C

F·dx =

∫C

(∇φ) ··· dx

=

∫ t2

t1

∇φ(g(t)) ··· g′(t)dt (by definition of line integral)

=

∫ t2

t1

d

dt[φ(g(t))]dt (by the Chain Rule)

= φ(g(t2))− φ(g(t1)) (by the second FTC)

= φ(x2)− φ(x1). (by (2.29))

45

Comment:The significance of Theorem 2.2 is two-fold:

i) if F is a gradient field, the line integral

∫C

F·dx only depends on the end points of the

curve C, and hence is path-independent,

ii) if the potential φ is known, then equation (2.28) gives the value of the line integralimmediately.

Exercise 2.10:We have seen that the vector field

E(x, y) = −kq(

x

(x2 + y2)3/2,

y

(x2 + y2)3/2

)on U = R2 − (0, 0) is a gradient field with potential

φ(x, y) =kq√x2 + y2

.

In exercise 9 you showed that the line integral of E along each of 2 curves joining (1, 0) to(0, 1) equalled zero. Use Theorem 2.2 to verify this result.

Looking ahead:

So far (Theorems 2.1 and 2.2) we have established that

∫C

F·dx is path-independent if and

only if F is a gradient field. Moreover, if F is a gradient field and we can find a potential φ,

then the line integral

∫C

F·dx can be quickly evaluated. Now we are faced with two problems:

i) if we are given a vector field F, how can we tell quickly whether it is a gradient field?

ii) if we know F is a gradient field, how do we find a potential φ?

Answering the first question requires the famous Green’s theorem, while the second ismore straightforward. But before dealing with these questions we first discuss the physicalsignificance of the Second Fundamental Theorem for line integrals.

2.3.3 Conservative (i.e. gradient) vector fields

Thinking of the vector field F in Theorem 2.2 as a force field, the line integral

∫C

F·dx equals

the work done by the force field on a particle as the particle moves along the curve C fromx1 to x2. The theorem asserts that if the force field is a gradient field, F = ∇φ, then thework done depends only on the potential at the end points x1 and x2.

46

In this physical context, it is customary, to define a scalar field V = −φ, so that

F = −∇V. (2.30)

Theorem 2.2 then has the form ∫C

F·dx = −V (x2) + V (x1). (2.31)

Since “work” is the same as “energy”, physicists call V (x) the potential energy of the particleat position x, when moving under the action of F.

Comment:The minus sign in equation (2.30) becomes appropriate when one thinks, for example

of the force field due to the earth’s gravitational field. The potential energy of a particleincreases if its distance from the earth’s centre increases i.e. ∇V points radially outwards,while the gravitational force field F acts radially inwards.

One can also relate the work done to the kinetic energy K of the particle, defined by

K = 12m ‖ v ‖2 . (2.32)

Describing the path C of the particle by x = r(t), t1 ≤ t ≤ t2, the work done can be written∫C

F·dx =

∫ t2

t1

F(r(t)) ··· r′(t)dt. (2.33)

But Newton’s second law tells us that

mr′′(t) = F(r(t)).

It follows that

F(r(t)) ··· r′(t) = mv′(t) ··· v(t) (since r′(t) = v(t))

= 12m[v(t) ··· v(t)]′ (property of the derivative)

=d

dtK(t) (by (2.32)).

Thus, by (2.33) and the FTCII, ∫C

F·dx = K(t2)−K(t1). (2.34)

Equation (2.34) and (2.31) gives

K(t1) + V (r(t1)) = K(t2) + V (r(t2)), (2.35)

for any two times t1 and t2. In words, for a gradient force field one can define a potentialenergy V of a particle in such a way that the sum of the potential energy and kinetic energyK is constant, i.e. conservation of energy holds.

47

It is for this reason that when vector fields are thought of a force fields, gradient fieldsare also called conservative fields.

Comment:In many applications in the real world, conservation of energy does not have the simple

form of (2.35), because, for example, of energy losses due to friction – think of the spaceshuttle re-entering the atmosphere. Dissipative, i.e. non-conservative forces have also to beconsidered. It is nevertheless important to be able to find out whether a given force field isconservative, and this is the problem we now consider.

2.4 Green’s Theorem

In this section we introduce Green’s theorem, and discuss a number of applications, includinghow to spot conservative/gradient vector fields in R2.

2.4.1 The theorem

We need some additional terminology related to curves.Consider a curve C in Rn given by

x = g(t), a ≤ t ≤ b,

with g continuous.

i) C is a closed curve means that g(a) = g(b).

ii) C is a simple closed curve means that g(a) = g(b) and g is a one-to-one function onthe interval a ≤ t < b. (Note the strict inequality t < b.) In geometric terms a simpleclosed curve has no self-intersections.

g(a) = g(b) g(a) = g(b)

Figure 2.7: A simple closed curve. A non-simple closed curve.

iii) In what follows we shall consider a bounded open subset D of R2, whose boundary,denoted by ∂D, is a simple closed curve. In this situation we assume that the curve∂D is oriented counter-clockwise, so that if you walk around the boundary, the regionD is on your left.

48

D

D∂

Figure 2.8: A bounded open subset D and its boundary ∂D oriented counter-clockwise.

Theorem 2.3 (Green’s theorem):

Let D be a bounded subset of R2 whose boundary ∂D is a piecewise C1 simple closed curveoriented counter-clockwise. If F = (F1, F2) is of class C1 on D ∪ ∂D then∫

∂D

F·dx =

∫∫D

(∂F2

∂x− ∂F1

∂y

)dx dy. (2.36)

Digression on iterated integrals:

If D is described by inequalities of the form

f(x) ≤ y ≤ g(x),

a ≤ x ≤ b,

and H(x, y) is continuous on D ∪ ∂D, then the double integral

∫∫D

H(x, y)dx dy can be

expressed as an iterated integral:

y

a x bx

(x,f(x))

(x,g(x))C2

C1

D

Figure 2.9: A region D and its boundary ∂D = C1 ∪ (−C2).

∫∫D

H(x, y)dx dy =

∫ b

x=a

[∫ g(x)

y=f(x)

H(x, y)dy

]dx. (2.37)

49

Proof of Green’s theorem:

We give a proof subject to the assumption that D can be described by inequalities of theform

f(x) ≤ y ≤ g(x), a ≤ x ≤ b, (2.38)

andh(y) ≤ x ≤ k(y), c ≤ y ≤ d, (2.39)

where f, g, h and k are C1 functions.It is sufficient to prove two special cases of (2.36), namely∫

∂D

(F1, 0) ··· dx =

∫∫D

−∂F1

∂ydx dy, (2.40)

and ∫∂D

(0, F2) ··· dx =

∫∫D

∂F2

∂xdx dy. (2.41)

The sum of (2.40) and (2.41) gives (2.36).We prove (2.40), using the inequalities (2.38). Consider the curves C1, C2 (see Figure 2.9)

given byx = (x, f(x)) and x = (x, g(x))

respectively, with a ≤ x ≤ b, i.e. we use x as parameter. The boundary ∂D is then theunion ∂D = C1 ∪ (−C2). By definition of the line integral,∫

∂D

(F1, 0) ··· dx =

∫C1

(F1, 0) ··· dx−∫C2

(F1, 0) ··· dx

(since

∫−C2

= −∫C2

)

=

∫ b

a

F1(x, f(x)), 0) ··· (1, f ′(x)dx−∫ b

a

F1(x, g(x)), 0) ··· (1, g′(x)dx

=

∫ b

a

[F1(x, f(x))− F1(x, g(x))] dx (2.42)

We now apply (2.37) to the right side of (2.40):∫∫D

−∂F1

∂ydx dy = −

∫ b

x=a

[∫ g(x)

y=f(x)

∂F1

∂ydy

]dx

= −∫ b

a

[F1(x, g(x))− F1(x, f(x))] dx,

(2.43)

by the FTCII. Equation (2.40) follows, on comparing (2.42) and (2.43). A similar argumentbased on (2.39) yields (2.41) (do it!), which completes the proof.

Example 2.4:Verify Green’s theorem for the vector field F(x) = (xy, 2xy) and the triangluar region D

with vertices (0, 0), (1, 0) and (0, 1).

50

Solution:

a) The boundary of D taken counterclockwise is the piecewise C1 curve

∂D = C1 ∪ C2 ∪ C3.

∫∂D

F · dx =

∫C1

F · dx +

∫C2

F · dx +

∫C3

F · dx. (2.44)

Observe that F = 000 on C1(y = 0) and on C3(x = 0), which implies that∫C1

F · dx = 0 =

∫C3

F · dx. (2.45)

A vector function for C2(x+ y = 1) is

x = g(t) = (1− t, t), 0 ≤ t ≤ 1.

It follows thatg′(t) = (−1, 1) and F(g(t)) = t(1− t)(1, 2),

after taking out a common factor. By definition of line integral,∫C2

F · dx =

∫ 1

0

F(g(t)) · g′(t)dt

=

∫ 1

0

t(1− t)(1, 2) · (−1, 1)dt

=

∫ 1

0

t(1− t)dt = · · · = 16

(2.46)

Substituting (2.45) and (2.46) in (2.44) gives∫∂D

F · dx = 16.

51

b) We calculate∂F2

∂x− ∂F1

∂y= −x+ 2y.

The right side of Green’s theorem is∫∫D

(∂F2

∂x− ∂F1

∂y

)dxdy =

∫∫D

(−x+ 2y)dxdy

=

1∫x=0

(∫ 1−x

y=0

(−x+ 2y)dy

)dx

= · · · = 16.

Green’s theorem is verified.

The limits of integration for the triangular region D.

Aside: D is defined by the inequalities

0 ≤ y ≤ 1− x,

0 ≤ x ≤ 1,

which give the limits of integration.

Green’s theorem can be used to express a given line integral as a double integral, orconversely, to express a given double integral as a line integral, provided you can choose asuitable vector field. Here is an example of the latter.

Example 2.5:Use a line integral to calculate the area enclosed by the ellipse

x2

a2+y2

b2= 1.

Solution: The area A of a plane region D can be expressed as a double integral:

A =

∫∫D

(1)dx dy.

The vector field F = (0, x) satisfies ∂F2

∂x− ∂F1

∂y= 1, so that Green’s theorem (2.36) gives

A =

∫∂D

(0, x) ··· dx,

where ∂D is the ellipse (a simple closed curve) oriented counter-clockwise. Using the standardparametrization,

x = (a cos t, b sin t), 0 ≤ t ≤ 2π,

52

the definition of line integral gives

A =

∫ 2π

0

(0, a cos t) ··· (−a sin t, b cos t)dt Aside: cos 2t = 2 cos2 t− 1

= · · · = π ab.

Exercise 2.11:Repeat example 2.5 using the vector field F = 1

2(−y, x).

2.4.2 Existence of a potential in R2

We now return to the question: how can you tell whether a given vector field F : U → R2 inR2 is derivable from a potential, i.e. is a conservative/gradient field?

So far we have proved the two fundamental theorems for line integrals (theorems 2.1 and2.2), which show that for continuous vector fields in Rn:

∫C

F·dx is path-independent ⇐⇒ F is a gradient field

in U ⊂ Rn in U ⊂ Rn (2.47)

We begin the final stage of the discussion by establishing that

“

∫C

F·dx is path-independent in U”

is equivalent to

“

∫C

F·dx = 0 for all simple closed curves in U”.

The reason for doing this is that Green’s theorem deals with line integrals around simpleclosed curves.

Proposition 2.3:Let U be an open subset of Rn. A continuous vector field F : U → Rn is path independent

in U if and only if

∫C

F·dx = 0 for every simple closed curve in U .

Proof:

1) Suppose

∫C

F·dx is path-independent in U . Let C be a simple closed curve in U .

Decompose C into C1 and C2 as in figure 2.10. Then∫C

=

∫C1

+

∫C2

=

∫C1

−∫−C2

= 0,

since the line integral is path independent.

53

C2

C1

− C2

C1

x1 x1

x2 x2

Figure 2.10: Decomposing a simple closed curve C into C1 and C2

2) Suppose

∫C

F·dx = 0 for every simple closed curve in U . Let x1,x2 be any two points

in U , and let C1 and C2 be two curves joining x1 to x2 which do not intersect eachother. Then C1 ∪ (−C2) is a simple closed curve C. It follows that∫

C1

−∫C2

=

∫C1

+

∫−C2

=

∫C

= 0.

C1 C

1

− C2

C2

x1 x1

x2 x2

Figure 2.11: Two curves C1, C2 joining x1 to x2 form a simple closed curve C = C1 ∪ (−C2).

If C1 and C2 intersect each other, introduce a third curve C3 that does not intersecteither C1 or C2. It follows as before that∫

C1

=

∫C3

and

∫C2

=

∫C3

.

We have thus shown that

∫C1

=

∫C2

for any two curves joining x1 to x2, i.e. the line

integral is path-independent.

54

x2

C2

C1

x1

C3

Figure 2.12: A third curve C3 avoids intersections.

Combining Proposition 2.3 with the result (2.47) gives the following:∫C

F·dx = 0 for every simple ⇐⇒ F is a gradient field

closed curve in U ⊂ Rn in U ⊂ Rn (2.48)

We now restrict our considerations to R2. Suppose that the C1 vector field F : U → R2

is a gradient field, i.e. F = ∇φ, or in component form,

(F1, F2) =

(∂φ

∂x,∂φ

∂y

).

It follows that∂F2

∂x− ∂F1

∂y=

∂2φ

∂x∂y− ∂2φ

∂y∂x= 0,

since φ is of class C2. This result, with (2.48), means that if F is a gradient field in R2, theformula in Green’s theorem,namely∫

∂D

F·dx =

∫∫D

(∂F2

∂x− ∂F1

∂y

)dx dy

is identically satisfied. This formula also suggests that if ∂F2

∂x− ∂F1

∂y= 0 in U , then

∫∂D

F·dx = 0

for any simple closed curve ∂D in U , so that by (2.48), F is a gradient field. The followingexample, however, shows that the situation is not as simple as this.

Example 2.6:Let U = R2 − (0, 0). Show that the vector field F : U → R2 defined by

F(x) =

(−y

x2 + y2,

x

x2 + y2

)(2.49)

satisfies∂F2

∂x− ∂F1

∂y= 0 in U ,

but that F is not a gradient field in U .

55

Solution: A simple calculation gives

∂F2

∂x=∂F1

∂y=−x2 + y2

(x2 + y2)2.

According to (2.48) we can show that F is not a gradient field by giving a simple closed curve

C in U such that

∫C

F·dx 6= 0.

Consider the unit circle C given by

x = (cos t, sin t), 0 ≤ t ≤ 2π.

A routine calculation gives

∫C

F·dx = 2π 6= 0. (do it!)

Comment:The essential point is that Green’s theorem cannot be applied to the vector field (2.49)

on the set D =

(x, y)∣∣ x2 + y2 ≤ 1

whose boundary is the circle C, x2 + y2 = 1, since F

is not C1 on D – it is not even defined at (0, 0). Another way of looking at the difficulty isthat the set U = R2−(0, 0) on which F is C1 has a “hole” in it – the point (0, 0) has beendeleted. So we need to introduce a “no holes” restriction on the set U on which F is C1.

Definition:A connected open set U ⊂ Rn is simply-connected means that every simple close curve in

U can be shrunk continuously to a point while remaining in U .

e.g. i) D =

(x, y)∣∣ 1 < x2 + y2 < 4

is not simply-connected in R2. The circle drawn cannot

be shrunk continuously to a point.

ii) D = R2 −

(x, 0)∣∣ x ≤ 0

is simply-connected in R2.

AAAAAAAAAAAAAAAAAAAA

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AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

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AAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAA

AAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAAA

iii) R3 minus a finite number of points is simply-connected.

iv) R3 minus an infinite line is not simply-connected.

We are now ready to state the theorem on detecting gradient vector fields in R2. Havingdone the preparatory work, the proof is short!

56

Theorem 2.4 (test for conservative fields)

If

i) U is a simply-connected open subset of R2,

ii) F is a C1 vector field that satisfies

∂F2

∂x− ∂F1

∂y= 0 (2.50)

in U , then there exists a single-valued C2 potential φ in U , i.e.

F = ∇φ in U .

Proof:Let C be any simple closed curve in U . Since U is simply-connected the interior D of C

belongs to U and thus (2.50) holds in D ∪ C. By Green’s theorem,∫C

F·dx = 0.

Since C is arbitrary it follows from (2.48) that F is a gradient field in U .

Example 2.7:Test the vector field

F(x) = (yexy, xexy + 2y) (2.51)

for being conservative, and if it is, find a potential φ.

Solution: A routine calculation shows that ∂F2

∂x− ∂F1

∂y= 0 on R2. Since R2 is simply-

connected, F is conservative by Theorem 2.4.To find a potential, we have to integrate the equations

∂φ

∂x= F1 = yexy,

∂φ

∂y= F2 = xexy + 2y. (2.52)

The first givesφ(x, y) = exy +K(y), (2.53)

where the “constant of integration” depends on y. Differentiate (2.53) with respect to y anduse the second equation in (2.52):

xexy + 2y = xexy +K ′(y),

giving K ′(y) = 2y, and henceK(y) = y2 + C,

57

where C is constant. By (2.53) the potential is

φ(x, y) = exy + y2 + C,

unique up to an additive constant C.

Exercise 2.12:Test whether the vector field F : U → R2 is conservative on the set U ⊂ R2, and if so,

find a potential φ:

i) F =(

1y,− x

y2

), U =

(x, y)

∣∣ y > 0

,

ii) F = (y cos(xy),−x cos(xy)), U = R2,

iii) F =(

xx2+y2 ,

yx2+y2

), U = R2 − (0, 0).

Answers:i) Yes; φ = x

y+ C ii) NO iii) Yes; φ = 1

2ln(x2 + y2) + C.

Comment:Exercise 2.12 iii) shows that even if U is not simply-connected the vector field may be

conservative.

2.5 Vorticity and circulation

Theorem 2.4 (test for conservative fields) shows that given a vector field F = (F1, F2) in R2,the quantity

∂F2

∂x− ∂F1

∂y

is of fundamental importance: if it is zero on a simply-connected set U ⊂ R2, then F is agradient/conservative field in U .

This quantity also plays an important role when the vector field is the velocity fieldv = (v1, v2) of a fluid flow in two dimensions3 and in this context it is called the vorticity ofthe fluid, denoted by Ω:

Ω =∂v2

∂x− ∂v1

∂y. (2.54)

We now describe the physical significance of the vorticity. Think of a small wooden paddlewheel that is carried along by the fluid. The question is: will the motion of the fluid causethe paddle wheel to rotate about its axis?

In order to illustrate the problem we consider two simple vector fields

u = (u, 0),

3By a fluid flow in two dimensions we mean a three dimensional flow which is “stratified”, i.e. the fluidvelocity is the same in each of a family of planes. Without loss of generality v(x) = (v1(x, y), v2(x, y), 0).

58

axis

Figure 2.13: A paddle wheel.

where u is a positive constant with [u] = LT−1, and

v = (αy, 0),

where α is a positive constant with [α] = T−1. It follows from figure 2.14 that the velocityfield u will not cause the paddle wheel to rotate, while figure 2.15 shows that the paddlewheel will rotate under the action of v.

a fluidflow line

y

x

Figure 2.14: The velocity field u = (u, 0) acting on a paddle wheel.

y

x

Figure 2.15: The velocity field v = (αy, 0) acting on a paddle wheel.

This difference between u and v can be characterized by considering the line integral ofthe velocity fields around the circle C of radius ρ that represents the circumference of thepaddle wheel. By (2.54) the vorticity for each velocity field is

Ωu =∂u2

∂x− ∂u1

∂y= 0, Ωv =

∂v2

∂x− ∂v1

∂y= −α,

59

and hence Green’s theorem applied to the paddle wheel disc gives∫C

u·dx = 0,

∫C

v·dx = −πρ2α.

The line integral of a velocity field v along a curve is in fact equal to the line integral of thetangential component v ·T along the curve, where T is the unit tangent vector to the curve.This result is seen as follows:∫

C

v·dx =

∫ b

a

v(g(t)) ··· g′(t)dt Aside: g′(t) = T ‖ g′(t) ‖

=

∫ b

a

v(g(t)) ···T ‖ g′(t) ‖ dt

=

∫C

(v ·T)ds.

(2.55)

If

∫C

v·dx < 0, i.e. v ·T is on balance negative along C, the paddle wheel will rotate in the

opposite direction to T. On the other hand, if

∫C

v·dx = 0, i.e. v ·T is on balance zero along

C, the paddle wheel will not rotate.

We now summarize the conclusion.

Let the curve C, oriented counter-clockwise as usual, be the circumference of a paddlewheel. If ∫

Cv·dx

< 0

= 0

> 0

,

then the fluid flow will cause the paddle wheel torotate clockwise

not rotate

rotate counter-clockwise.

The quantity

∫C

v·dx is called the circulation of the fluid around the simple closed curve C.

Green’s theorem leads to a relation between the vorticity Ω and the circulation

∫C

v·dx.

Let Dρ be the disc of radius ρ centred at x with boundary ∂Dρ oriented counter-clockwise.

60

By Green’s theorem ∫∂Dρ

v·dx =

∫∫Dρ

(∂v2

∂x− ∂v1

∂y

)dx dy

=

(∂v2

∂x− ∂v1

∂y

)(x)πρ2,

where x is some point in Dρ. The last step follows from the Mean Value Theorem forintegrals. Divide by the area πρ2 and let ρ→ 0 giving

(∂v2

∂x− ∂v1

∂y

)(x)︸ ︷︷ ︸

Ω(x)

= limρ→0

1

πρ2

∫∂Dρ

v·dx

. (2.56)

x

x ρ

∂Dρ

In words,

the vorticity at x equals the circulationper unit area at x.

For our purposes it is helpful to write an approximation,

Ω(x) ≈ 1

πρ2

∫∂Dρ

v·dx,

for ρ sufficiently close to 0.We thus obtain the desired physical interpretation of the vorticity:

Ω(x)

< 0

= 0

> 0

⇒ a paddle wheel at x will

rotate clockwise

not rotate

rotate counter-clockwise.

We conclude with two classic vector fields in R2 that illustrate vorticity and circulation.

Example 2.8:Consider the vector field

v = α(−y, x), (2.57)

61

where α is a positive constant with [α] = T−1. The flow lines are circles x2 + y2 = c2,traversed counter-clockwise (see example 1.12 in Chapter 1). The vorticity Ω is non-zero,

Ω =∂v2

∂x− ∂v1

∂y= 2α.

Since the domain of v is R2, Green’s theorem can be applied to any circle of radius ρ, giving∫ρ

v·dx = 2α(πρ2)

for the circulation. It follows that a paddle wheel will rotate counter-clockwise (see Figure2.16).

Example 2.9:Consider the vector field (a “vortex field”)

v =α

x2 + y2(−y, x), (x, y) 6= (0, 0). (2.58)

The flow lines are again circles x2 + y2 = c2 (see example 1.13 in Chapter 1). The vorticityΩ is, however, zero:

Ω =∂v2

∂x− ∂v1

∂y= 0,

(verify). In this case the domain of v is R2−(0, 0), which is not simply-connected. Green’stheorem can be applied to any circle that does not enclose or pass through the origin, giving∫

C

v·dx = 0

(since Ω = 0). Thus a paddle wheel will not rotate as it moves with the fluid (see Figure2.17). However, for a circle of radius b that encloses the origin the circulation is non-zero:∫

C

v·dx = 2παb.

(Verify using the definition of line integral.) Thus in this case, although the fluid is locallynon-rotating (i.e. the paddle wheel does not rotate), it does rotate globally, i.e. there aresimple closed curves with non-zero circulation.

62

Figure 2.16: The velocity field (2.57) causes the paddle wheel to rotate.

Figure 2.17: The velocity field (2.58) does not cause the paddle wheel to rotate.

63

Chapter 3

Surfaces & Surface Integrals

The notion of a surface arises in various physical and mathematical contexts, e.g. the wing orfuselage of an aircraft, an ocean wave at an instant of time, or a soap film formed by dipping aclosed wire loop into a soap solution. In a mathematical context, when deriving the equationthat governs heat transfer one considers an arbitrary finite chunk of the conducting mediumthat is bounded by a surface Σ, and one writes the heat flux through Σ as a surface integral.

3.1 Parametrized surfaces

3.1.1 Surfaces as vector-valued functions

The first mathematical description of a surface that one encounters is the graph of a functionf : R2 → R. The graph is a surface in R3, described by the equation

z = f(x, y), (x, y) ∈ D ⊂ R2.

This way of describing surfaces is somewhat limited: it cannot describe a surface that “foldsover” such as a sphere or a torus. So we think about vector-valued functions and generalizethe way they are used to describe curves by introducing two parameters u and v, and writing

x = g(u, v), (u, v) ∈ Duv. (3.1)

Here Duv is a subset of R2 (the uv-plane), x = (x, y, z) ∈ R3 (xyz-space), and g : Duv → R3

is a vector-valued function. Often Duv will be a rectangle in the uv-plane, but in general itwill be a bounded subset of R2, whose boundary is a simple closed curve. Figure 3.1 givesa schematic representation of the domain Duv and the surface Σ : as the point (u, v) movesthrough the set Duv the image point g(u, v) sweeps out the surface Σ in R3.

One thinks of the function g as a map from R2 → R3 that acts on the rectangle Duv,bending and stretching it so as to form the surface Σ. It is helpful to think of the surfaceΣ as being generated by two families of curves, namely the images of the two families ofstraight lines u = constant and v = constant. The two families of curves form a “grid” or“fish-net” that covers the surface. We think of u and v as being coordinates on the surfaceΣ and we refer to the curves on Σ that are defined by u = constant and v = constant ascoordinate curves which form a coordinate grid.

65

v (u,v)

u

Duv

g

z

x

y

g(u,v)

Σ

Figure 3.1: The vector-valued function g maps the domain Duv onto the surface Σ.

Example 3.1:The equation

x = g(u, v) = a + ue1 + ve2, (u, v) ∈ Duv, (3.2)

where Duv = (u, v) | −1 ≤ u, v ≤ 1, and e1, e2 are two linearly independent vectors inR3, describes a surface which is a piece of the plane through the point a, and containing thevectors e1 and e2. Referring to Figure 3.2, the vector x− a lies in the plane and hence is alinear combination of e1 and e2.

v (u,v)

u

(0,0)

g

x

z

x = g(u,v)

y

a = g(0,0)

e2

e1

n

Figure 3.2: Parametric representation of a plane.

One can obtain the equation of the plane in standard form

n1(x− a) + n2(y − b) + n3(z − c) = 0 (3.3)

by calculating a normal vector n. Since e1 and e2 lie in the plane the vector product e1× e2

is a vector normal to the plane:n = e1 × e2.

66

Then, taking the scalar product of equation (3.2) with n gives

n·(x− a) = 0,

which is the standard form (3.3).

Recall: The vector product a× b can be evaluated using a “symbolic determinant”:

a× b =

∣∣∣∣∣∣i j ka1 a2 a3

b1 b2 b3

∣∣∣∣∣∣ , (3.4)

which, when formally expanded by the first row, gives

a× b = (a2b3 − a3b2)i + (a3b1 − a1b3)j + (a1b2 − a2b1)k.

Exercise 3.1:A plane in R3 is given by

x = (1, 2, 3) + u(1, 1, 0) + v(1,−1, 1),

with (u, v) ∈ R2. Find the equation of the plane in standard form.

Answer: (x− 1)− (y − 2)− 2(z − 3) = 0.

Exercise 3.2:Find a vector-valued function to describe the plane x− 3y + z = 2.

Hint: Let u = x, v = y.

Answer: x = g(u, v) = (0, 0, 2) + u(1, 0,−1) + v(0, 1, 3).

Example 3.2:The vector-valued function g : Duv → R3 defined by

g(u, v) = (sinu cos v, sinu sin v, cosu), (3.5)

withDuv = (u, v) | 0 ≤ u ≤ π, 0 ≤ v ≤ 2π,

describes the surface of the unit sphere in R3. This can be verified by writing x = g(u, v)and verifying that

‖ x ‖2= x2 + y2 + z2 = 1.

The vector-valued function (3.5) is obtained from the formulas that relate spherical polarcoordinates to Cartesian coordinates:

x = r sin θ cosφ

y = r sin θ sinφ

z = r cos θ.

67

z

x

y

rθ

φ

Figure 3.3: Spherical polar coordinates.

circlesφ = const.

circlesθ = const.

fig3.4

Figure 3.4: The coordinate grid on the sphere created by the spherical polar angles θ and φ.

To obtain a unit sphere we choose r = 1, and then let u = θ, v = φ. The coordinate linesu = constant on the sphere are circles of constant latitude, and the coordinate lines v =constant are circles of constant longitude, as shown in Figure 3.4.

Comment:Consider the surface Σ defined by

z = f(x, y), (x, y) ∈ D ⊂ R2.

One can write a vector equation for Σ by introducing u = x and v = y as parameters. Thenz = f(u, v), and the vector equation for the surface is

x = g(u, v) = (u, v, f(u, v)), (u, v) ∈ D. (3.6)

Exercise 3.3:Give a vector-valued function to describe the cone x2 + y2 = z2 with 0 ≤ z ≤ 1.

Answer: x = g(u, v) = (u cos v, u sin v, u),with (u, v) ∈ Duv = (u, v) | 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π.

68

3.1.2 The tangent plane

Consider a surface in R3 given by

x = g(u, v), (u, v) ∈ Duv,

where g is of class C1. The coordinate curves on the surface, obtained by setting either u orv to be constant, are given by

x = g(u, v0), x = g(u0, v).

The vectors∂g

∂u(u0, v0) and

∂g

∂v(u0, v0) (3.7)

are tangent to these curves at the point g(u0, v0), and hence are tangent to the surface atthat point. We can thus regard these vectors, provided that they are linearly independent, asdefining the tangent plane of the surface at the point g(u0, v0).

zN

x

y

e2g(u0,v0)

e1

x = g(u, v0 )

x = g(u0 ,v)

Figure 3.5: The surface x = g(u, v), and tangent vectors e1 = ∂g∂u

(u0, v0) and e2 = ∂g∂v

(u0, v0).

Referring to Example 3.1, the equation of the tangent plane can thus be written by usingthe vectors (3.7) as e1 and e2, in the form

x = g(u0, v0) + (u− u0)∂g

∂u(u0, v0) + (v − v0)

∂g

∂v(u0, v0). (3.8)

Furthermore, the vectors (3.7) define a vector N that is normal to the surface,

N =∂g

∂u(u0, v0)× ∂g

∂v(u0, v0), (3.9)

where the vector product is defined by equation (3.4). Knowing N, the tangent plane at thepoint g(u0, v0) can also be written in the form

N ··· (x− g(u0, v0)) = 0. (3.10)

69

Comment:The requirement that the tangent vectors (3.7) be linearly independent is essential: if

they are linearly dependent (in particular if one or both is the zero vector), then the surfacemay not have a tangent plane at the point in question. A classic example is the cone

x = g(u, v) = (u cos v, u sin v, u),

with 0 ≤ v ≤ 2π and u ≥ 0. The function g is of class C1, but the surface does not have atangent plane at the point g(0, 0) = (0, 0, 0). The tangent vectors are

∂g

∂u= (cos v, sin v, 1),

∂g

∂v= (−u sin v, u cos v, 0),

and are linearly dependent when u = 0, since

∂g

∂v(0, v) = (0, 0, 0).

This requirement of linear independence is analogous to the requirement that g′(t) 6= 000for a curve: if g′(t0) = 000 the curve may have a cusp at g(t0).

For a vector-valued function g : Duv → R3 the linear approximation corresponds to usingthe tangent plane to approximate th surface x = g(u, v). Using (3.8) we have

g(u, v) ≈ g(u0, v0) + (u− u0)∂g

∂u(u0, v0) + (v − v0)

∂g

∂v(u0, v0),

for (u, v) sufficiently close to (u0, v0). Equivalently, introducing the increments

∆g = g(u, v)− g(u0, v0), ∆u = u− u0, ∆v = v − v0,

we obtain

∆g ≈ ∆u∂g

∂u(u0, v0) + ∆v

∂g

∂v(u0, v0), (3.11)

for ∆u, ∆v sufficiently close to zero.

Example 3.3:The surface S defined by

z = f(x, y), (x, y) ∈ D, (3.12)

can be described equivalently by

x = g(u, v) = (u, v, f(u, v)), (u, v) ∈ D

(see (3.6)). The tangent vectors (3.7) are given by

∂g

∂u=

(1, 0,

∂f

∂u

),

∂g

∂v=

(0, 1,

∂f

∂v

). (3.13)

70

The normal vector (3.9) is

N =

(−∂f∂u,−∂f

∂v, 1

), (3.14)

(verify, using (3.4)). As a check, one can also calculate a normal vector by writing (3.12) inthe form h(x, y, z) = 0, where

h(x, y, z) = z − f(x, y). (3.15)

We know that the gradient ∇h is orthogonal to the surface h(x, y, z) = 0. Using (3.15)

∇h =

(−∂f∂x,−∂f

∂y, 1

)in agreement with (3.14).

3.1.3 Surface area

We want to calculate the surface area of a surface Σ described by

x = g(u, v), (u, v) ∈ Duv,

where g is a C1 function. Consider a partition of Duv into small rectangles by means ofparallel lines u = constant and v = constant. These lines define a coordinate grid on thesurface Σ that decomposes Σ into small surface elements (see Figure 3.1). For ∆u and ∆vsufficiently small a surface element can be approximated as part of a plane, and the edgesof the surface element can be approximated by parallel vectors. Thus we can approximate asurface element as a small plane parallelogram.

v

u

z

g

x

y

magnify

Σ

∆v

∆u

Duv

g(u0,v0)

g(u0,v0)

g(u0 + ∆u, v0)

g(u0,v0 + ∆v)B

A

(u0,v0)

Figure 3.6: A surface element on a surface Σ.

Recall that the area of a parallelogram defined by two vectors A and B is ‖ A × B ‖,the magnitude of the vector product. This result follows from the formula

‖ A×B ‖=‖ A ‖ ‖ B ‖ sin θ,

where θ is the angle between the vectors.

71

In order to approximate the area ∆S of a surface element we need to approximate thevectors A and B that define the sides of the surface element in Figure 3.6. Using the linearapproximation (3.11) we obtain

A ≈ (∆u)∂g

∂u(u0, v0), B ≈ (∆v)

∂g

∂v(u0, v0).

Thus, simplifying ∆S ≈‖ A×B ‖ gives

∆S ≈∥∥∥∥∂g

∂u(u0, v0)× ∂g

∂v(u0, v0)

∥∥∥∥∆u∆v. (3.16)

To obtain the total area S we have to sum over all surface elements determined by thepartition of Duv and take the limit as N → ∞ and max(∆u), max(∆v) → 0. This processleads to a double integral over the set Duv in the uv-plane. With the above as motivationwe make the following.

Definition:The surface area of the surface by x = g(u, v), (u, v) ∈ Duv, where g is C1, is defined by

S =

∫∫Duv

∥∥∥∥∂g

∂u× ∂g

∂v

∥∥∥∥ du dv. (3.17)

Example 3.4:Calculate the surface area of a cone of radius b and height h.

Solution: The cone is given by

x2 + y2

b2=z2

h2, 0 ≤ z ≤ h.

A suitable vector-valued function is

x = g(u, v) = (bu cos v, bu sin v, hu), (u, v) ∈ Duv,

with Duv = [0, 1]× [0, 2π]. The tangent vectors are

∂g

∂u= (b cos v, b sin v, h),

∂g

∂v= (−bu sin v, bu cos v, 0),

giving

∂g

∂u× ∂g

∂v=

∣∣∣∣∣∣i j k

b cos v b sin v h−bu sin v bu cos v 0

∣∣∣∣∣∣ = (−bhu cos v,−bhu sin v, b2u),

and ∥∥∥∥∂g

∂u× ∂g

∂v

∥∥∥∥ =√b2 + h2 bu.

72

Thus by (3.17) the surface area is

S =

∫∫Duv

√b2 + h2 bu du dv

=√b2 + h2 b

1∫u=0

2π∫v=0

u dv

du

= · · · = πb√b2 + h2.

v

u10

D uv2π

Comment:This example is simply intended to show you how the “machinery” works. The surface

area of a cone can be calculated by simple geometry.

The most important part of this subsection is the approximation (3.16) for the area of asurface element, which we shall use when introducing surface integrals, the main goal of thischapter.

3.1.4 Orientation of a surface

Consider a surface Σ given by

x = g(u, v), (u, v) ∈ Duv,

where g is one-to-one and of class C1, and the normal vector

N =∂g

∂u× ∂g

∂v

is non-zero for all (u, v) ∈ Duv. Since g is one-to-one the surface does not intersect itself (seeFigure 3.7), and since g is of class C1, N varies continuously over Σ. In the sequel we shallneed to work with a unit normal n on Σ. There are two choices for n at each point, namely

n = ± 1

‖ N ‖N.

73

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAAAAAAAAAA

Figure 3.7: A surface which intersects itself.

Once we have made a choice of n we say that we have assigned an orientation to Σ. Anoriented surface has two distinct sides, a positive side which is the side on which n pointsand a negative side. You could paint the positive side red and the negative side blue andthe colours would not merge. A surface for which this cannot be done, i.e. a surface whichdoes not have two distinct sides, is said to be non-orientable. The existence of such surfacesis counter-intuitive, but there is a famous example called the Moebius band that can beconstructed easily using a piece of paper as shown in Figure 3.8.

B

A C

D B

A C

D

Figure 3.8: Identifying AB and CD creates a cylinder. Identifying BA and CD createsa Mobius band.

We shall assume that the surfaces we work with can be oriented. When working witha closed surface, e.g. the surface of a sphere or of a torus, the standard orientation is tochoose n to be the outward normal. On the other hand, when working with a surface Σwhose boundary ∂Σ is a piecewise smooth closed curve, we shall relate the orientation of Σto the orientation of ∂Σ as follows: when viewed from the side of Σ on which the normalvector points, the boundary ∂Σ is oriented counter-clockwise (see Figure 3.10).

3.2 Surface Integrals

3.2.1 Scalar fields

As motivation, think of a surface Σ coated with a thin film of silver of surface density ρ(mass per unit area), that varies over the surface. In order to calculate the total mass ofsilver on the surface we need to define a surface integral over Σ. This concept is analogousto the line integral of a scalar field, as defined in Section 2.1, but with surface area replacingarclength.

Consider a surface Σ given by

x = g(u, v), (u, v) ∈ Duv,

74

AAAAAAAAA

AAAAAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAA outward

normal

inwardnormal

Σ

Figure 3.9: The two normals on an orientable closed surface.

n

Σ

∂Σ

Figure 3.10: Orientation of the boundary of a surface.

and a scalar field f : R3 → R continuous on Σ. Consider a partition P of Σ into N surfaceelements as in Section 3.1.3. The surface integral of f over Σ is denoted by∫∫

Σ

f dS.

As a tentative definition we write∫∫Σ

f dS = lim|P|→0

∑P

f(x)∆S, (3.18)

where the summation is taken over all surface elements of the partition P , x is a point inthe surface element and ∆S is the area of the surface element. The symbol |P| denotes themaximum of the areas of the surface elements.

Using the approximation (3.16) we can write

f(x)∆S ≈ f(g(u, v))

∥∥∥∥∂g

∂u(u, v)× ∂g

∂v(u, v)

∥∥∥∥∆u∆v.

The sum (3.18) then becomes a Riemann sum for a double integral over the set Duv in theuv-plane. These considerations lead to the working definition below.

75

Definition:Consider a surface Σ given by x = g(u, v), (u, v) ∈ Duv, with g of class C1, and a scalar

field f continuous on Σ. The surface integral of f over Σ is defined by∫∫Σ

f dS =

∫∫Duv

f(g(u, v))

∥∥∥∥∂g

∂u× ∂g

∂v

∥∥∥∥ du dv. (3.19)

Comment:As regards physical interpretation, the scalar field f will usually represent the surface

density of some physical quantity and the surface integral

∫∫Σ

f dS gives the total amount of

the physical quantity on the surface Σ.

Example 3.5:

Evaluate the surface integral

∫∫Σ

f dS, where Σ is the surface of the sphere of radius b

centred on the origin and the scalar field f is given by f(x) = z2.

Solution: We use the standard parametrization (3.5) adapted to a sphere of radius b:

x = g(u, v) = b(sinu cos v, sinu sin v, cosu),

with (u, v) ∈ Duv = (u, v) | 0 ≤ u ≤ π, 0 ≤ v ≤ 2π. The tangent vectors are

∂g

∂u= b(cosu cos v, cosu sin v,− sinu),

∂g

∂v= b(− sinu sin v, sinu cos v, 0),

giving

∂g

∂u× ∂g

∂v= b2(sin2 u cos v, sin2 u sin v, sinu cosu)

= (b sinu)x.

It follows that ∥∥∥∥∂g

∂u× ∂g

∂v

∥∥∥∥ = (b sinu) ‖ x ‖= b2 sinu,

since sinu ≥ 0 on [0, π]. By definition of the surface integral (3.19)∫∫Σ

z2dS =

∫∫Duv

(b cosu)2(b2 sinu)du dv

= b4

2π∫v=0

π∫u=0

cos2 u sinu du dv

= · · · = 43πb4.

76

Exercise 3.4:

Evaluate the surface integral

∫∫Σ

f dS, where Σ is the cone z =√x2 + y2, 0 ≤ z ≤ 1,

and f(x) = z2.

Answer: π√2.

3.2.2 Vector fields

As motivation, think of a vector field F that is a flux density field, for example mass fluxdensity (example 3 in Section 1.6.1) or heat flux density (example 5 in Section 1.6.1). LetΣ be a (piece-wise) smooth oriented surface with unit normal n. Consider a surface elementof area ∆S on Σ. We have seen (Section 1.6.1) that the flux through the surface element isgiven by

(F · n)∆S. (3.20)

The total flux through Σ is obtained by summing over all surface elements. This processleads to the notion of the surface integral of the flux density field F over the surface Σ.

In general, consider an oriented surface Σ in R3 given by

x = g(u, v), (u, v) ∈ Duv,

with unit normal n. Let F be a vector field on R3 that is continuous on Σ. The surfaceintegral of F over Σ is denoted by ∫∫

Σ

F · n dS.

Consider a partition P of Σ into N surface elements, as in Section 3.1.3. Motivated by theform of the expression (3.20), we write the tentative definition of surface integral as follows:∫∫

Σ

F · n dS = lim|P|→0

∑P

F(x)·n ∆S, (3.21)

where the notation has the same meaning as in (3.18). The unit normal n is obtained bynormalizing1 the normal vector (3.9):

n =1∥∥∂g

∂u× ∂g

∂v

∥∥(∂g

∂u× ∂g

∂v

). (3.22)

Using (3.22) and the approximation (3.16) for ∆S i.e.

∆S ≈∥∥∥∥∂g

∂u× ∂g

∂v

∥∥∥∥∆u∆v,

1One has to verify that this normal vector does agree with the assigned orientation of Σ. If not, one hasto change the parametrization so as to reverse the direction of n.

77

we obtain

F(x)·n ∆S ≈ F(g(u, v)) ···(∂g

∂u× ∂g

∂v

)∆u∆v.

The sum (3.21) then becomes a Riemann sum for a double integral over the set Duv. Theseconsiderations lead to the working definition below.

Definition:Consider an oriented surface Σ given by x = g(u, v), (u, v) ∈ Duv, with g of class C1,

and a vector field F continuous on Σ. The surface integral of F over Σ is defined by∫∫Σ

F · ndS =

∫∫Duv

F(g(u, v)) ···(∂g

∂u× ∂g

∂v

)dudv. (3.23)

Comment:

If F is a flux density vector field, then the surface integral

∫∫Σ

F · ndS equals the total

flux of F through the surface Σ.

Example 3.6:The surface Σ is the piece of the plane y + z = 1 cut out by the cylinder x2 + y2 = a2,

oriented so that the unit normal n points upwards. The temperature in the vicinity of Σ isgiven by T (x) = x2 + y2 + z2. Calculate the total heat flux through Σ. Denote the thermalconductivity by k.

Solution: We parametrize Σ by writing

x = g(u, v) = (u, v, 1− v), u2 + v2 ≤ a2,

i.e. we choose x = u, y = v. The tangent vectors are

∂g

∂u= (1, 0, 0),

∂g

∂v= (0, 1,−1),

giving∂g

∂u× ∂g

∂v= (0, 1, 1),

which has the correct orientation. By equation (1.30) the heat flux density vector is

J = −k∇T = −2k(x, y, z).

The total heat flux J(Σ) through Σ is

J(Σ) =

∫∫Σ

J · ndS.

78

By definition of the surface integral

J(Σ) =

∫∫Duv

−(2k)(u, v, 1− v) ··· (0, 1, 1)dudv

=

∫∫Duv

(−2k)dudv.

Since Duv is the disc u2 + v2 ≤ a2 and the integrand is constant,

J(Σ) = −2πka2.

Comment:The temperature T in Example 3.6 increases radially outwards, and thus we expect

that heat will be transferred across Σ in the direction of −n. We thus expect that the heatflux across Σ will be negative.

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

z

n

0 Σ

Figure 3.11: The surface Σ in Example 3.6.

Exercise 3.5:

Evaluate the surface integral

∫∫Σ

F · ndS where Σ is the cylinder x2 +z2 = b2, 0 ≤ y ≤ h,

excluding the ends, and F is the vector field F(x) = (0, 0, z). The surface Σ is oriented sothat the normal is in the outward radial direction.

Answer: πb2h.

3.2.3 Properties of surface integrals

As with any integral, surface integrals possess the properties of linearity and additivity. Theformal statements are analogous to those for line integrals in Section 2.2.3. In addition, if

79

a surface Σ is piecewise C1 instead of being C1, i.e. the surface Σ is the union of severalsurfaces whose defining functions are C1, one can define the surface integral over Σ as thesum of the integrals over the separate pieces.

80

Chapter 4

Gauss’ and Stokes’ Theorems

In this chapter we first introduce the divergence and curl of a vector field, which we then usein the formulation of Gauss’ and Stokes’ theorems.

4.1 The vector differential operator ∇

4.1.1 Divergence and curl of a vector field

In elementary calculus it is useful to think of the process of differentiation as defining adifferential operator, denoted by d

dx, which acts on a C1 function f to give the derivative

function f ′:d

dx(f) = f ′.

When working with scalar and vector fields on R3 (i.e. functions of x, y and z) it is usefulto generalize the operator d

dx, and consider a vector differential operator which we denote by

∇:

∇ = i∂

∂x+ j

∂

∂y+ k

∂

∂z, (4.1)

or equivalently

∇ =

(∂

∂x,∂

∂y,∂

∂z

), (4.2)

relative to Cartesian coordinates.Firstly, ∇ can act on a scalar field f : R3 → R,

∇f = i∂

∂x(f) + j

∂

∂y(f) + k

∂

∂z(f),

which we rewrite as

∇f =∂f

∂xi +

∂f

∂yj +

∂f

∂zk, (4.3)

and recognize as the gradient of the scalar field f . Secondly, ∇ can act on a vector fieldF : R3 → R3 to give a scalar field ∇ ··· F:

∇ ··· F =

(∂

∂x,∂

∂y,∂

∂z

)··· (F1, F2, F3),

81

which, in analogy with the scalar product of two vectors, is written

∇·F =∂F1

∂x+∂F2

∂y+∂F3

∂z. (4.4)

This scalar field is called the divergence of F, and is also written div F. We note that Gauss’Theorem leads to a physical interpretation of ∇·F, which we shall discuss later.

Thirdly, in analogy with the vector product, ∇ can act on a vector field F on R3 to givea vector field ∇× F:

∇× F =

∣∣∣∣∣∣∣i j k

∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣∣ . (4.5)

Written out in full this gives

∇× F =

(∂F3

∂y− ∂F2

∂z

)i +

(∂F1

∂z− ∂F3

∂x

)j +

(∂F2

∂x− ∂F1

∂y

)k. (4.6)

This vector field is called the curl of F, and is also written curl F. On recalling that thevorticity scalar of a vector field on R2 is given by

Ω =∂F2

∂x− ∂F1

∂y,

(see equation (2.54)) we see that the curl is the generalization to three dimensions of thevorticity scalar in R2. Indeed in fluid mechanics, if v is the velocity field of a fluid thenw = ∇ × v is called the vorticity field of the fluid. We thus expect ∇ × F to describerotational properties of the vector field F. This interpretation will be clarified after wediscuss Stokes’ theorem.

Exercise 4.1:In this exercise, r = (x, y, z), r =‖ r ‖. By writing out components, verify that

i)∇r =1

rr, ii)∇ ··· r = 3, iii)∇× r = 000.

Exercise 4.2:Consider the vector field F defined by

F(x) = Ax,

where A is a 3×3 constant matrix with entries aij. The matrix A acts on the position vectorx written as a column vector. Show that

∇·F = a11 + a22 + a33,

∇× F = (a32 − a23, a13 − a31, a21 − a12).

As an example of the use of divergence and curl in physics, we now present Maxwell’sequations, the fundamentalequations that describe electromagnetic fields. Our purpose is not

82

to derive the equations or to study their physical implications – it is simply to present animportant example of the use of the divergence and curl.

An electromagnetic field is composed of an electric field E(t, x, y, z) and a magnetic fieldH(t, x, y, z), which are time-dependent vector fields on R3. With appropriate choice of units,Maxwell’s equations read

∂E

∂t= c∇×H− 4πJ (4.7)

∂H

∂t= −c∇× E (4.8)

∇·E = 4πε (4.9)

∇·H = 0, (4.10)

where ε is the charge density and j is the current vector. The constant c has the dimensionsof velocity, and is in fact the speed of light in a vacuum. Mathematically these equationsform a system of linear partial differential equations in 6 unknowns, namely the componentsof E and H.

One of the most remarkable predictions of Maxwell’s equations is that electromagneticfields can transport energy through space in the form of waves, called electromagnetic wavesor electromagnetic radiation (light, radio waves, X-rays are all examples, differing in fre-quency). See #6 in Problem Set 4.

4.1.2 Identities involving ∇The differential operator ∇ satisfies various identities which are used in deriving and workingwith the field equations of classical physics, e.g. Maxwell’s equations, or the equations offluid dynamics.

The first set of identities refers to the gradient.

G1 Sum of two scalar fields:∇(f + g) = ∇f +∇g.

G2 Product of two scalar fields:

∇(fg) = f∇g + g∇f.

G3 Scalar product of two vector fields:

∇(F ·G) = (F ··· ∇)G + (G·∇)F + F× (∇×G) + G× (∇× F).

The second set refers to the divergence

D1 Sum of two vector fields:

∇ ··· (F + G) = ∇·F +∇·G.

83

D2 Product of a scalar field and a vector field:

∇ ··· (fF) = f∇·F + (∇f)·F.

D3 Vector product of two vector fields:

∇ ··· (F×G) = G·(∇× F)− F·(∇×G).

The third set refers to the curl.

C1 Sum of two vector fields:

∇× (F + G) = ∇× F +∇×G.

C2 Product of a scalar field and a vector field:

∇× (fF) = f∇× F +∇f × F.

C3 Vector product of two vector fields:

∇× (F×G) = (∇·G)F− (∇·F)G + (G·∇)F− (F·∇)G.

The fourth set can be thought of as “zero identities”.

Z1 Curl of a gradient:∇× (∇f) = 000, (4.11)

for any C2 scalar field f ,

Z2 Divergence of a curl:∇ ··· (∇× F) = 0, (4.12)

for any C2 vector field F.

Note: In G3 and C3 there appear the terms (F · ∇)G and (G · ∇)F. In component formthe expression F · ∇ reads

F · ∇ = (F1, F2, F3) ·(∂

∂x,∂

∂y,∂

∂z

)= F1

∂

∂x+ F2

∂

∂y+ F3

∂

∂z.

Observe that F · ∇ is a scalar differential operator. For example

i · ∇ =∂

∂x,

84

where i = (1, 0, 0). Note that one cannot reverse the order of the symbols in the expressionF · ∇:

F · ∇︸ ︷︷ ︸a scalar

differential operator

6= ∇ · F︸ ︷︷ ︸divergence of F,

a scalar field

The scalar operator F · ∇ can act on a vector field G to give a vector field (F · ∇)G.

Exercise 4.3:Let A = (A1, A2, A3) be a constant vector field and let r = (x, y, z). Show that

(A · ∇)r = A.

One can also ask about the “curl of a curl”, i.e.

∇× (∇× F).

Unlike the other two expressions with repeated derivatives, namely Z1 and Z2, this one isnot identically zero, and in order to give an expression for it, we need to form a secondorder differential operator from ∇, called the Laplacian, and denoted by ∇2. This operatoris defined by taking the divergence of a gradient field:

∇2f = ∇ ··· (∇f) (4.13)

Writing this out in terms of partial derivatives, i.e.

∇2f =

(∂

∂x,∂

∂y,∂

∂z

)···(∂f

∂x,∂f

∂y,∂f

∂z

),

leads to

∇2f =∂2f

∂x2+∂2f

∂y2+∂2f

∂z2. (4.14)

One can formally write

∇2 = ∇ ··· ∇ =∂2

∂x2+

∂2

∂y2+

∂2

∂z2. (4.15)

The operator ∇2, which is a scalar differential operator, is called the Laplacian. Note thatthe Laplacian can act on a vector field F,

∇2F =∂2F

∂x2+∂2F

∂y2+∂2F

∂z2(4.16)

to give a vector field.We can now state the “curl of a curl” identity:

∇× (∇× F) = ∇(∇·F)−∇2F. (4.17)

In words: the curl of the curl of F equals the gradient of the divergence of F minus theLaplacian of F.

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The Laplacian operator ∇2 commutes with the vector operator ∇ in the following ways:

∇(∇2f) = ∇2(∇f), (4.18)

∇ · (∇2F) = ∇2(∇ · F), (4.19)

∇× (∇2F) = ∇2(∇× F), (4.20)

where f is a C3 scalar field and F is a C3 vector field. Note that in (4.18), ∇2f is a scalarfield while ∇f is a vector field.

Comment:Proving the identities listed in this section involves a straightforward but sometimes

lengthy calculation – one simply expresses each side of an identity in terms of components,thereby showing that they are equal. The “zero identities” Z1 and Z2 are essentially a

consequence of the equality of mixed second order partial derivatives e.g.∂2f

∂x∂y=

∂2f

∂y∂x.

See Problem Set 4, #3-7.

Exercise 4.4:Use exercise 4.1 and the chain rule to show that

∇f(r) =f ′(r)

rr,

where f : R→ R is a C1 function, r =‖ r ‖ and r = (x, y, z).

Exercise 4.5:Use exercises 4.1 and 4.4 and the identities involving ∇ to verify the following results:

i) ∇ · (f(r)r) = rf ′ + 3f ii) ∇ · (f(r)A) =f ′(r)

rr ·A

iii) ∇× (f(r)r) = 000 iv) ∇× (f(r)A) =f ′(r)

rr×A

v) ∇(A ·R) = A vi) ∇× (A× r) = 2A

vii) ∇2f(r) = f ′′(r) + 2f ′(r)r

Here f(r) is a C2 function of r =‖ r ‖, and A is a constant vector field in R3.

4.1.3 Expressing ∇ in curvilinear coordinates

In problems in which there is symmetry about a point it is usually helpful to introduce polarcoordinates (ρ, φ) in R2 or spherical coordinates (r, θ, φ) in R3. If there is symmetry about aline in R3, one thinks of cylindrical coordinates (ρ, φ, z). These three systems of coordinatesare examples of curvilinear coordinates. Let’s begin by reviewing the definitions of thesecoordinates.

86

Polar coordinates:

x = ρ cosφ

y = ρ sinφ,(4.21)

with ρ ≥ 0, 0 ≤ φ ≤ 2π.

Figure 4.1: Polar coordinates.

Spherical coordinates:

x = r sin θ cosφ

y = r sin θ sinφ

z = r cos θ,

(4.22)

with r ≥ 0, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π.

z

x

ry

θ

φ

Figure 4.2: Spherical coordinates.

Cylindrical coordinates:

x = ρ cosφ

y = ρ sinφ

z = z,

(4.23)

87

z

x

y

φρ

z

Figure 4.3: Cylindrical coordinates.

with ρ ≥ 0, 0 ≤ φ ≤ 2π, z ∈ R.Cartesian coordinates (x, y) in R2 define a coordinate grid consisting of two families of

orthogonal straight lines x = constant and y = constant. Polar coordinates (ρ, φ) definea coordinate grid consisting of concentric circles ρ = constant and radial half-lines φ =constant.

yy = const.

x

x = const.

θ = const.

r = const.

Figure 4.4: A Cartesian coordinate grid. A polar coordinate grid.

Coordinate systems in R3 define a coordinate grid consisting of three families of curves orlines. A coordinate system whose coordinate grid does not consist of families of parallellines is called a curvilinear coordinate system. If the families of curves intersect orthogonally,we call the coordinate system orthogonal. Polar, spherical and cylindrical coordinates areorthogonal.

Our goal is to learn how to write the gradient ∇f , the divergence ∇·F and the curl∇× F in terms of spherical and cylindrical coordinates in R3. In Cartesian coordinates theoperator ∇ has the form

∇ = i∂

∂x+ j

∂

∂y+ k

∂

∂z.

The key to writing this operator in a curvilinear coordinate grid is to construct an orthonor-mal basis e1, e2, e3), the analogue of the Cartesian basis i, j,k. Figure 4.5 shows theorthonormal basis eρ, eφ determined by polar coordinates ρ, φ in R2.

For simplicity we begin by introducing the orthonormal basis determined by an orthogonalcurvilinear coordinate system in R2. The generalization to R3 then becomes clear. A general

88

eφ

eρ

eφ eρ

Figure 4.5: The orthonormal basis associated with polar coordinates.

curvilinear coordinate system (v1, v2) in R2 is related to a Cartesian coordinate system by apair of equations

x = f(v1, v2)

y = g(v1, v2),(4.24)

(see for example (4.21)). It is convenient to write these equations in vector form

x = F(v1, v2), (4.25)

where x = (x, y), F = (f, g). The curves of the coordinate grid are then given by

x = F(v1, `), x = F(k, v2),

where k and ` are constants, and the tangent vectors are

∂x

∂v1

=

(∂x

∂v1

,∂y

∂v1

),

∂x

∂v2

=

(∂x

∂v2

,∂y

∂v2

). (4.26)

By assumption these vectors are orthogonal. We thus obtain an orthonormal basis by nor-malizing them

e1 =1

h1

∂x

∂v1

,

e2 =1

h2

∂x

∂v2

,

(4.27)

where

h1 =

∥∥∥∥ ∂x

∂v1

∥∥∥∥ =

√(∂x

∂v1

)2

+

(∂y

∂v1

)2

,

h2 =

∥∥∥∥ ∂x

∂v2

∥∥∥∥ =

√(∂x

∂v2

)2

+

(∂y

∂v2

)2

.

(4.28)

We can now express the gradient vector field ∇u of a given scalar field u relative to theorthonormal basis e1, e2:

∇u = (∇u·e1)e1 + (∇u·e2)e2. (4.29)

89

v1 = const.v2 = const.

e2 e1

Figure 4.6: The orthonormal basis.

Using the Chain Rule and equation (4.26) we obtain

∂u

∂v1

=∂u

∂x

∂x

∂v1

+∂u

∂y

∂y

∂v1

= ∇u ··· ∂x

∂v1

.

Dividing by h1 and using (4.27) gives

∇u ··· e1 =1

h1

∂u

∂v1

.

Similarly

∇u ··· e2 =1

h2

∂u

∂v2

.

It now follows from (4.29) that

∇u =1

h1

∂u

∂v1

e1 +1

h2

∂u

∂v2

e2.

We can write this equation in the form

∇u =

(1

h1

e1∂

∂v1

+1

h2

e2∂

∂v2

)u. (4.30)

Thus, relative to the curvilinear coordinate system (v1, v2), the operator ∇ has the form

∇ =1

h1

e1∂

∂v1

+1

h2

e2∂

∂v2

, (4.31)

where the basis vectors e1, e2 are given by (4.27) and the scale factors h1, h2 by (4.28).Equations (4.27) (4.31) generalizes to R3 in an obvious way:

ei =1

hi

∂x

∂vi, i = 1, 2, 3 (4.32)

∇ =1

h1

e1∂

∂v1

+1

h2

e2∂

∂v2

+1

h3

e3∂

∂v3

, (4.33)

where

hi =

∥∥∥∥ ∂x

∂vi

∥∥∥∥ (4.34)

We now give the expression for ∇ in each of the three classical coordinate systems.

90

Polar coordinates in R2:

The defining equations arex = ρ cosφ, y = ρ sinφ.

Using equations (4.27), (4.28) and (4.31) with ρ = v1, φ = v2 we obtain

eρ = (cosφ, sinφ), eφ = (− sinφ, cosφ), (4.35)

hρ = 1, hφ = ρ,

and

∇ = eρ∂

∂ρ+

1

ρeφ

∂

∂φ. (4.36)

Cylindrical coordinates in R3:

The defining equations are

x = ρ cosφ, y = ρ sinφ, z = z.

Using equations (4.32)-(4.34) we obtain with (v1, v2, v3) = (ρ, φ, z),

eρ = (cosφ, sinφ, 0), eφ = (− sinφ, cosφ, 0), ez = (0, 0, 1), (4.37)

hρ = 1, hφ = ρ, hz = 1,

and

∇ = eρ∂

∂ρ+

1

ρeφ

∂

∂φ+ ez

∂

∂z. (4.38)

Comment: It is important to note that an orthonormal basis (e1, e2, e3) determined by acurvilinear coordinate system (v1, v2, v3) differs from the Cartesian basis i, j,k in a signif-icant way – the vector fields (e1, e2, e3) vary from point to point (see for example (4.35) and(4.37).

Spherical coordinates in R3:

The defining equations are

x = r sin θ cosφ, y = r sin θ sinφ, z = r cos θ.

Using equations (4.32)-(4.34) with (v1, v2, v3) = (r, θ, φ), we obtain

er = (sin θ cosφ, sin θ sinφ, cos θ), hr = 1

eθ = (cos θ cosφ, cos θ sinφ,− sin θ), hθ = r

eφ = (− sinφ, cosφ, 0), hφ = r sin θ,

(4.39)

and

∇ = er∂

∂r+

1

reθ∂

∂θ+

1

r sin θeφ

∂

∂φ. (4.40)

91

We are now in a position to calculate the divergence ∇·F and the Laplacian ∇2f in thedifferent coordinate systems. We will illustrate the procedure by calculating ∇·F in polarcoordinates in R2.

Example 4.1: Calculate the divergence ∇ · F in polar coordinates in R2.

Solution: We write F in terms of the basis (4.35):

F = Fρeρ + Fφeφ.

Using (4.36), and taking into account that eρ and eφ are not constant vector fields,

∇·F = eρ∂

∂ρ··· (Fρeρ + Fφeφ) +

1

ρeφ

∂

∂φ··· (Fρeρ + Fφeρ)

= eρ ···(∂Fρ∂ρ

eρ + Fρ∂eρ∂ρ

)+ eρ ···

(∂Fφ∂ρ

eφ + Fφ∂eφ∂ρ

)+

1

ρeφ ···

(∂Fρ∂φ

eρ + Fρ∂eρ∂φ

)+

1

ρeφ ···

(∂Fφ∂φ

eφ + Fφ∂eφ∂φ

).

From (4.35) we obtain

∂eρ∂ρ

= 000,∂eρ∂φ

= eφ,

∂eφ∂ρ

= 000,∂eφ∂φ

= −eρ.

(4.41)

Using these results and the fact that

eρ·eρ = 1, eρ·eφ = 0, eφ·eφ = 1,

the expansion for ∇·F simplifies dramatically to give

∇·F =∂Fρ∂ρ

+1

ρFρ +

1

ρ

∂Fφ∂φ

. (4.42)

Comment: The surprise is the appearance of the term 1ρFρ, which is due to the fact that

the basis vectors are not constant. Equation (4.42) can be written more concisely as

∇·F =1

ρ

[∂

∂ρ(ρFρ) +

∂Fφ∂φ

]. (4.43)

Example 4.2: Calculate the Laplacian ∇2f in polar coordinates in R2.

Solution: We use (4.13),∇2f = ∇ · (∇f) (4.44)

92

Equation (4.36) implies that

∇f =∂f

∂ρeρ +

1

ρ

∂f

∂φeφ. (4.45)

We apply (4.43) with F = ∇f , i.e.

Fρ =∂f

∂ρ, Fφ =

1

ρ

∂f

∂φ, (4.46)

as follows from (4.45). Equation (4.44), in conjunction with (4.43) and (4.46) now gives

∇2f =1

ρ

∂

∂ρ

(ρ∂f

∂ρ

)+

1

ρ2

∂2f

∂φ2 (4.47)

In a similar fashion we can derive the expression for ∇·F in cylindrical coordinates:

∇·F =1

ρ

[∂

∂ρ(ρFρ) +

∂Fφ∂φ

]+∂Fz∂z

, (4.48)

and in spherical coordinates:

∇·F =1

r2

∂(r2Fr)

∂r+

1

r sin θ

(∂

∂θ(sin θFθ) +

∂Fφ∂φ

). (4.49)

The formulas for the Laplacian∇2f = ∇···(∇f) are also useful in applications. For cylindricalcoordinates it follows from (4.38) and (4.48) that

∇2f =1

ρ

∂

∂ρ

(ρ∂f

∂ρ

)+

1

ρ2

∂2f

∂φ2+∂2f

∂z2, (4.50)

and for spherical coordinates it follows from (4.40) and (4.49) that

∇2f =1

r2

∂

∂r

(r2∂f

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂f

∂θ

)+

1

r2 sin2 θ

∂2f

∂φ2. (4.51)

We leave the details as an exercise.

4.2 Gauss’ Theorem

In this section we state Gauss’ theorem, give a proof subject to a simplifying assumption onthe domain, and then discuss some applications in physics.

4.2.1 The theorem

Consider a bounded subset Ω of R3, whose boundary ∂Ω is a single piecewise smooth orientedclosed surface. For example, a solid sphere, a solid cube and a solid torus have this property.The region between two concentric spheres does not have this property because its boundaryis the union of two disjoint surfaces. We note that a closed surface is one which has no

93

boundary curve – it is the two-dimensional analogue of a closed curve, which has no boundarypoints (i.e. end points).

Gauss’s theorem – also known as the Divergence theorem – relates the surface integral ofa vector field over a surface ∂Ω to the triple integral of the divergence of F over the regionΩ enclosed by ∂Ω.

Theorem 4.1 (Gauss’ theorem):Let Ω be a bounded subset of R3 whose boundary ∂Ω is a single piecewise smooth oriented

closed surface. If the vector field F is of class C1 on Ω ∪ ∂Ω, then∫∫∫Ω

∇·FdV =

∫∫∂Ω

F · ndS,

where n is the unit outward normal to ∂Ω.In order to simplify the proof of the theorem we shall require Ω to be a special domain,

defined by the following restriction:

any line through an interior point of Ωintersects the boundary ∂Ω in two points.

A solid sphere and a solid cube are both special domains, but a solid torus is not.

F

n

nF

Ω

∂Ω

Figure 4.7: Gauss’ theorem.

Comment:If we project a special domain in the z-direction, as shown in Figure 4.8, we can describe

it by inequalities of the formf`(x, y) ≤ z ≤ fu(x, y),

where(x, y) ∈ Dxy.

It follows that a triple integral over Ω can be written in the form

94

z = fu (x,y)

Σu

Σl

z = fl (x,y)

Ω

(x,y,0)Dxy

∂Ω

Figure 4.8: A special domain projected in the z-direction.

∫∫∫Ω

h dV =

∫∫Dxy

(∫ fu(x,y)

z=f`(x,y)

h dz

)dx dy. (4.52)

In addition we can represent the boundary ∂Ω as the union of an upper surface

Σu : z = fu(x, y), (x, y) ∈ Dxy, (4.53)

and a lower surfaceΣ` : z = f`(x, y), (x, y) ∈ Dxy, (4.54)

i.e.∂Ω = Σu ∪ Σ`. (4.55)

This decomposition forms the basis of the proof of Gauss’ theorem. We shall also requirean expression for the surface integral of a vector field over a surface of the form Σu or Σ`,which assumes a particularly simple form, given in the proposition to follow.

Proposition 4.1:Consider a surface Σ given by z = f(x, y) with (x, y) ∈ Dxy, and with unit normal n

oriented in the positive z-direction. Then for any vector field of the form F = F3k,∫∫Σ

F · ndS =

∫∫Dxy

F3(x, y, f(x, y))dx dy. (4.56)

Proof:We use the parametrization

x = g(x, y) = (x, y, f(x, y)),

95

which leads to∂g

∂x× ∂g

∂y=

(−∂f∂x,−∂f

∂y, 1

)(exercise). By definition of the surface integral:∫∫

Σ

F · ndS =

∫∫Dxy

(0, 0, F3(x, y, f(x, y))) ···(−∂f∂x,−∂f

∂y, 1

)dx dy,

which simplifies to (4.56).

Proof of Gauss’ theorem for a special domain:

Consider a vector field of the form F = F3k. Decompose the boundary ∂Ω as in (4.55),noting that on Σu the outward normal is in the positive z-direction while on Σ` it is in thenegative z-direction. Thus by (4.56)∫∫

∂Ω

F · ndS =

∫∫Σu

F · ndS +

∫∫Σ`

F · ndS

=

∫∫Dxy

[F3(x, y, fu(x, y))− F3(x, y, f`(x, y))] dx dy.

On the other hand, using (4.52),∫∫∫Ω

∂F3

∂zdV =

∫∫Dxy

[∫ fu(x,y)

z=f`(x,y)

∂F3

∂zdz

]dx dy

=

∫∫Dxy

[F3(x, y, fu(x, y))− F3(x, y, f`(x, y))] dx dy,

after applying the second Fundamental Theorem of Calculus to the z-integral. We have thusshown that ∫∫∫

Ω

∂F3

∂zdV =

∫∫∂Ω

(F3 k)·n dV. (4.57)

By projecting in the y- and x-directions, one similarly obtains∫∫∫Ω

∂F2

∂ydV =

∫∫∂Ω

(F2 j)·n dV, (4.58)

∫∫∫Ω

∂F1

∂xdV =

∫∫Ω

(F1 i)·n dV. (4.59)

Summing (4.57)-(4.59) gives Gauss’ theorem.

Here is a simple application of Gauss’ theorem.

96

Example 4.1:Show that the flux of the vector field F = xi+yj+zk through any oriented closed surface

∂Ω equals 3V (Ω), where V (Ω) is the volume of the set Ω enclosed by ∂Ω.

Solution: Differentiation gives ∇·F = 3. Applying Gauss’ theorem to F gives∫∫∂Ω

F · ndS =

∫∫∫Ω

(3)dV = 3

∫∫∫Ω

(1)dV = 3V (Ω),

by the usual interpretation of the triple integral.

Physical interpretation of the divergence:

Gauss’ theorem leads directly to the physical interpretation of ∇·F at a given point x.Apply Gauss’ theorem to a sphere of radius ε centred at x:∫∫∫

Ω(ε)

∇·FdV =

∫∫∂Ω(ε)

F · n dS. (4.60)

∂Ω

c

x

Ω

By the Mean Value Theorem for integrals there exists a point c ∈ Ω such that∫∫∫Ω(ε)

(∇·F)dV = [∇·F(c)]V (ε),

(∇·F is continuous since F is of class C1), where V (ε) is the volume of the sphere. By (4.60),

∇·F(c) =1

V (ε)

∫∫∂Ω(ε)

F · n dS.

Taking the limit as ε→ 0+, so that c→ x, we get

∇·F(x) = limε→0+

1

V (ε)

∫∫∂Ω(ε)

F · n dS.

For ε sufficiently close to 0 we can write the approximation

∇·F(x) ≈ 1

V (ε)

∫∫∂Ω(ε)

F · n dS. (4.61)

97

Thus, the divergence ∇·F(x) equals the flux per unit volume at x of the vector field F.

Exercise 4.6:The figure shows four vector fields in R3 – the x-axis is out of the page and there is no

dependence on x. For each vector field estimate geometrically whether the flux through thesurface of a small cube is positive, negative or zero. Then equation (4.61) gives a predictionabout ∇·F. Verify your prediction by inventing a vector field for each picture and calculating∇·F.

z

z

z

z

y

y y

y

Figure 4.9: Four vector fields.

4.2.2 Conservation laws and PDEs

Gauss’ theorem plays a fundamental role in deriving the partial differential equations (PDEs)that govern a variety of physical phenomena. We will illustrate how the theorem is used byconsidering the law of conservation for a physical quantity described by a density scalar anda flux density vector field (e.g. mass or heat energy).

Let ψ(x, t) be a C1 scalar field that represents the density of some physical quantity Q(amount of Q per unit volume. It follows that the amount of Q in a bounded subset Ω ⊂ R3

is given by ∫∫∫Ω

ψdV.

Let j(x, t) be a C1 vector field that represents the flux density of the physical quantity Q(rate of flow of Q per unit area). It follows that the flux of Q across the boundary surface

98

∂Ω is ∫∫∂Ω

j · ndS,

where n is the unit outward normal.We now formulate the law of conservation in integral form for the physical quantity Q.

Let D ⊂ R3 be the domain of ψ and j, and let Ω be an arbitrary bounded subset, as in thestatement of Gauss’ theorem. If the quantity Q is neither created nor destroyed in D (i.e.no sources or sinks), then conservation of Q has the following form:

the rate at which the

amount of Q in Ω

increases

= −

the rate at which Q

leaves Ω across ∂Ω

.

The minus sign is needed because the amount of Q in Ω decreases if the flux across ∂Ω ispositive.

In terms of ψ and j the above conservation law reads

d

dt

∫∫∫Ω

ψdV = −∫∫∂Ω

j · n dS. (4.62)

Since ψ is of class C1 and Ω does not change with time,

d

dt

∫∫∫Ω

ψdV =

∫∫∫Ω

∂ψ

∂tdV.

In addition we can use Gauss’ theorem to write the surface integral as a triple integral,∫∫∂Ω

j · n dS =

∫∫∫Ω

∇·j dV.

Equation (4.62) then assumes the form∫∫∫Ω

(∂ψ

∂t+∇·j

)dV = 0. (4.63)

We now invoke a famous lemma from analysis (that applies to any type of integral).

Lemma 4.1:If the scalar field h is continuous on D ⊂ R3 and∫∫∫

Ω

hdV = 0

99

for any subset of Ω ⊂ D, thenh = 0 on D.

Proof:Suppose h(a) > 0 for some a ∈ D. Since h is continuous on D, h(x) > 0 for all x in some

neighbourhood N of a. Then

∫∫∫N

hdV > 0, contradicting the hypothesis. Thus h(x) = 0

for all x ∈ D.

Since the integral in equation (4.63) is evaluated over an arbitrary subset Ω ∈ D, thelemma implies that

∂ψ

∂t+∇·j = 0 (4.64)

in D. This equation is the differential form of the conservation law.We consider two special cases. In the first ψ is mass density of a fluid and j = ρv is the

mass flux density, where v is the velocity of the fluid. The conservation law (4.64) becomes

∂ρ

∂t+∇ ··· (ρv) = 0, (4.65)

which is called the equation of continuity for the fluid.In the second case, we consider a conducting medium whose temperature u(x, t) is a

function of position and time. The scalar field ψ is the heat energy density given by

ψ = cρu,

where c is the specific heat of the material (the amount of heat required to raise the tem-perature of unit mass by one degree), and ρ is the density. The vector field j is the heat fluxdensity, given by

j = −k∇u,

where k is the thermal conductivity (Fourier’s law). The conservation law (4.64) assumesthe form

∂

∂t(cρu) +∇ ··· (−k∇u) = 0. (4.66)

We assume that the medium is homogeneous so that k, c and ρ are independent of position,and we also assume that they do not change with time. Recalling the definition of theLaplacian, ∇2u = ∇ ··· (∇u) (see equation (4.13)), equation (4.66) assumes the form

∂u

∂t− a2∇2u = 0, (4.67)

where a2 = kcρ

is called the thermal diffusivity. Equation (4.67) is one of the fundamentalequations of mathematical physics, and is called the heat diffusion equation.

The heat diffusion equation also governs other diffusive processes. If a chemical is dis-solved in a solvent and the concentration is not uniform, then the chemical will diffuse (be

100

transported) from regions of high concentration to regions of law concentration. If C(x, t) isthe concentration, then the flux vector field will be given by Fick’s law:

j = −D∇C, (4.68)

where D is the diffusion coefficient. The conservation law (4.64), with ψ = C gives

∂C

∂t−D∇2C = 0, (4.69)

which has the same form as the heat diffusion equation.

4.2.3 The Generalized Divergence Theorem

Consider a subset Ω of R3 with boundary ∂Ω, as in Gauss’ theorem. Suppose that the vectorfield F is C1 on Ω ∪ ∂Ω except at one point a in the interior of Ω. In this situation Gauss’theorem cannot be applied. The theorem can, however, be generalized so as to apply in thiscase, as follows.

Surround the point a by a closed surface ∂H, lying entirely in Ω – a sufficiently smallsphere centred at a will do. Remove the region H inside ∂H from Ω. Then F is C1 on theresulting set Ω−H. In this situation Gauss’ theorem holds in the following modified form:

Ω

n

H

∂Ω

∂ H

na

Figure 4.10: A set Ω ⊂ R3 with a hole H enclosing the point a.

∫∫∂Ω

F · n dS =

∫∫∫Ω−H

∇·FdV +

∫∫∂H

F · n dS, (4.70)

where the normals on both ∂Ω and H are outward.Intuitively, you can think of the surface integral over ∂H as compensating for the fact

that you had to delete part of the original set Ω, thereby creating a hole H, in order toobtain a set Ω−H on which F is C1.

Equation (4.70) is proved by subdividing Ω−H into two regions on which the usual formof Gauss’ theorem can be applied. The details are left as a challenging exercise.

We now briefly discuss an important application of the generalized divergence theorem(4.70).

101

Gauss’ Law:

Let Ω be a bounded subset of R3 whose boundary ∂Ω is a single piecewise smooth orientedclosed surface. Suppose that (0, 0, 0) /∈ ∂Ω. Then

∫∫∂Ω

r · nr3

dS =

4π if (0, 0, 0) ∈ Ω

0 if (0, 0, 0) /∈ Ω.

where r = xi + yj + zk, r =‖ r ‖ and n is the unit outward normal on ∂Ω.

Proof:The proof depends on two facts, which we leave as exercises:

i) ∇ ···( r

r3

)= 0 for r 6= 0,

ii)

∫∫∂Hε

r · nr3

dS = 4π,

where ∂Hε is the sphere of radius ε centred at (0, 0, 0).

Apply Gauss’ theorem in one case, and the generalized form (4.70) in the other case.

Why is Gauss’ law important? Consider a point charge, such as an electron. The electric

force field around an electron placed at the origin is proportional tor · nr3

. Now assume we

integrate the electric field over the surface ∂Ω of a region Ω. By Gauss’ law, depending onwhether Ω contains the electron or not, the result is zero or it is proportional to the chargeof the electron. More generally, the integral of an electric field over ∂Ω can be used to countelectrons. Similarly, a point mass creates a gravitational force field of the same type aroundit. Therefore, the surface integral over the gravitational force field over Ω measures theamount of mass contained in Ω.

4.3 Stokes’ Theorem

Stokes’ theorem relates the integral of the curl of a vector field over a surface Σ to theline integral of the vector field around the boundary ∂Σ of Σ. The theorem is the naturalgeneralization in R3 of Green’s theorem.

4.3.1 The theorem

We motivate the form of the theorem by writing Green’s theorem in a vector form in R3.The formula in Green’s theorem is∫

C

F·dx =

∫∫D

(∂F2

∂x− ∂F1

∂y

)dxdy, (4.71)

102

AAAAAAAA

z

x

n = k

y

D = Σ

Figure 4.11: The plane region D viewed as a surface Σ in R3.

where F = (F1, F2). We can think of F as a vector field F = (F1, F2, 0) in R3 in which casewe can write

∂F2

∂x− ∂F1

∂y= (∇× F) ··· k, (4.72)

using the definition (4.6) of ∇ × F. We can also regard the plane region D as a surface Σin R3 whose unit normal n is n = k. It follows that a surface integral over Σ is simply adouble integral over D: ∫∫

Σ

(G · n)dS =

∫∫D

(G · k)dxdy,

for any vector field G. With these changes, (4.71) becomes∫C

F·dx =

∫∫Σ

(∇× F)·n dS,

which gives the form of Stokes’ theorem.

Theorem 4.2: (Stokes’ theorem)Let Σ be a piecewise C1 orientable surface whose boundary ∂Σ is a simple piecewise C1

closed curve. If F is of class C1 on some open set in R3 containing Σ ∪ ∂Σ, then∫∫Σ

(∇× F) ··· n dS =

∫∂Σ

F·dx, (4.73)

where ∂Σ is oriented counter-clockwise when viewed from the side on which the unit normaln points.

Proof: (outline)Suppose that Σ is given by

z = f(x, y), (x, y) ∈ Dxy.

103

∂Σ

n

Σ

Figure 4.12: An oriented surface Σ with piecewise C1 boundary ∂Σ.

Suppose that ∂Dxy is given by

h(t) = (x(t), y(t), 0), t1 ≤ t ≤ t2.

Then ∂Σ is given by

g(t) = (x(t), y(t), f(x(t), y(t))), t1 ≤ t ≤ t2.

∂D xy

∂Σ

n

Σ

D xy( z= 0)

Step 1: Using x, y as parameters on Σ, expand

∫∫Σ

(∇×F)·n dS in terms of components,

and write it as a double integral over Dxy.

Step 2: Show that

lineintegralin R3

↓∫∂Σ

F·dx =

lineintegralin R2

↓∫∂Dxy

(G1, G2) ··· (dx, dy),

104

where

G1(x, y) = F1(x, y, f(x, y)) + F3(x, y, f(x, y))∂f

∂x,

G2(x, y) = F2(x, y, f(x, y)) + F3(x, y, f(x, y))∂f

∂y.

Step 3: Apply Green’s theorem to

∫∂Dxy

(G1, G2) ··· (dx, dy), and compare the results to

Step 1.

Exercise 4.7: Do #14 in Problem Set 4. The common answer is π.

105

Corollary:If Σ1 and Σ2 are two oriented piecewise C1 surfaces with

∂Σ1 = ∂Σ2 = C,

where C is a simple closed curve, and A is a C1 vector field, then∫∫Σ1

(∇×A)·n1dS =

∫∫Σ2

(∇×A)·n2dS.

n1

Σ1

Σ2

C

n2

Figure 4.13: Two surfaces with the same boundary C.

Proof:

By Stokes’ theorem both surface integrals equal

∫C

A·dx.

Comment:Given a simple closed curve C we can imagine infinitely many surfaces Σ with ∂Σ = C.

The Corollary implies that the value of

∫∫Σ

(∇ × A)·ndS is independent of which surface

Σ we choose. One can thus talk about a surface integral being independent of surface, in

analogy with a line integral being independent of path. Moreover, since

∫∫Σ

(∇×A)·ndS is

independent of the surface and depends only on the simple closed curve C = ∂Σ, one cantalk about the flux of the vector field F = ∇×A through the simple closed curve C.

Example 4.2: Calculate the flux of the vector fieldF = ∇×A, where

A = (2z − y, x− z, y − x),

through the hemisphereΣ1 : x2 + y2 + z2 = a2,

with z − y ≥ 0.

106

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

Cz = y

Σ2

Σ1

n1

n2

Figure 4.14: A hemisphere Σ1 and a plane surface Σ2 with the same boundary C.

Solution: By the corollary we can replace the hemisphere Σ1 by the plane surface Σ2, i.e.the equatorial disc: ∫∫

Σ1

(∇×A)·n1dS =

∫∫Σ2

(∇×A)·n2dS.

The unit normal to the plane is n2 = 1√2(0,−1, 1), and the definition of ∇×A gives

∇×A = (2, 3, 2),

and hence(∇×A)·n2 = − 1√

2.

Thus ∫∫Σ1

(∇×A)·n1d =

∫∫Σ2

(− 1√

2

)dS

= − 1√2πa2,

since the disc Σ2 has radius a.

4.3.2 Faraday’s law

As an illustration of the use of Stokes’ theorem in deriving field equations in physics, weconsider Faraday’s law in the theory of electromagnetism, which states:

“The voltage change around a loop1 is proportional to the negative of the time rate of changeof the magnetic flux through the loop.”

The change in voltage ∆V across a curve segment ∆x is approximated by

∆V ≈ E·∆x.

1“loop” is synonymous with “simple closed curve”.

107

Thus, the change in voltage around the loop is∫C

E·dx.

H E

C

∆ x

magnetic field line

Figure 4.15: Magnetic field lines passing through a loop.

To write an expression for the magnetic flux through the loop we imagine a fixed surface Σwhose boundary is C, and calculate the flux of the magnetic field H through Σ:∫∫

Σ

H · n dS.

Note: In calculating the flux in this way, we are tacitly assuming that the flux integral isindependent of the surface Σ. That a magnetic field has this property is in fact a consequenceof the fact that the magnetic field satisfies ∇·H = 0. This surface-independence property ofH will be established in the next section.

Faraday’s law in integral form thus reads

d

dt

∫∫Σ

H · ndS

= −c∫C

E·dx, (4.74)

where c is a constant.We now use Stokes’ theorem to write the line integral in (4.74) as a surface integral. In

addition, since Σ does not change with time, we can take the t-derivative inside the surfaceintegral, giving ∫∫

Σ

(∂H

∂t+ c∇× E

)·n dS = 0.

Since Σ is an arbitrary surface in the domain U in which we are working, and the integrandis continuous by assumption, it follows that

∂H

∂t+ c∇× E = 000,

one of Maxwell’s equations (see Section 4.1.1).

108

4.3.3 The physical interpretation of ∇× F

In Section 2.5 (see equation (2.53)) we showed that(∂F2

∂x− ∂F1

∂y

)(x) = lim

ρ→0

1

πρ2

∫∂Dρ

F·dx, (4.75)

where F = (F1, F2), and Dρ is a disc of radius ρ centred at x. Thinking of F as a vector fieldin R3, F = (F1, F2, 0) we can use (4.72) to write (4.75) in the form

[(∇× F)·k] (x) = limρ→0

1

πρ2

∫∂Dρ

F·dx.

∂ Dρ

x

k

∂ Dρ

x

n

One can now use Stokes’ theorem and the Mean Value Theorem for Integrals to generalizethis result to a circle of radius ρ lying in an inclined plane in R3:

[(∇× F)·n] (x) = limρ→0

1

πρ2

∫∂Dρ

F·dx,

leading to the following interpretation:

The component of the curl ∇× F in the direction n, is the circulation per unitarea around a circle lying in the plane orthogonal to n. Moreover, the directionof ∇× F gives the direction of n for which the circulation is a maximum.

Thus, we see that the curl of a vector field on R3 is the appropriate generalization of thevorticity Ω(F) for a vector field in R2, i.e. ∇ × F describes the rotational aspects of thevector field F (see Section 2.5). In fluid dynamics, given a C1 velocity field v, the vectorfield

w = ∇× v

is called the vorticity field of the fluid. If w = 000 the fluid is said to be irrotational.

4.4 The Potential Theorems

There are two classes of vector fields that are special from a mathematical point of viewand important from a physical point of view. The first is the class of irrotational vectorfields and the second is the class of divergence-free vector fields. The mathematical bondbetween them is that, subject to a restriction on the domain, both classes are derivable froma potential, a scalar potential in the first case (F = ∇φ) and a vector potential in the secondcase (F = ∇×A).

109

4.4.1 Irrotational vector fields

Consider a gradient vector field (a.k.a. a conservative vector field) in R3

F = ∇φ, (4.76)

where φ is a C2 scalar field, called a scalar potential. We first summarize the principalproperties of gradient fields (already known in R2).

Proposition 4.1: If F = ∇φ, where φ is C2 in U ⊂ R3, then

i)

∫C

F·dx = φ(b) − φ(a), for any curve C in U joining a to b (i.e. the line integral is

path-independent in U),

ii)

∫C

F·dx = 0 for any simple closed curve in U ,

iii) ∇× F = 000 in U .

Proof:

i) is the second Fundamental Theorem for line integrals, valid in Rn.

ii) is an immediate consequence of i), by choosing b = a.

iii) is a consequence of the “zero identity” Z1.

The key question is: how does one determine whether a given vector field F is a gradientfield, i.e. has a scalar potential φ?

In view of Proposition 4.1 iii) one might conjecture that if F is irrotational (∇×F = 000),then F = ∇φ for some C2 scalar field. As in R2 (see Theorem 2.4) one needs a restrictionon the domain U ⊂ R3 in question. The following theorem generalizes Theorem 2.4.

Theorem 4.3 (scalar potential):

If F is of class C1 and ∇×F = 000 in U ⊂ R3, and U is simply-connected, then there existsa single-valued C2 scalar field φ such that F = ∇φ in U .

Proof:Let C be a simple closed curve in U . Since U is simply-connected it is plausible2 that

there exists an oriented C1 surface Σ in U such that C = ∂Σ. Apply Stokes’ theorem to Fon Σ and C: ∫

C

F·dx =

∫∫Σ

(∇× F)·n dS = 0.

2See the comment at the end of the proof.

110

Since

∫C

F·dx = 0 for any simple closed curve in U , it follows that the line integral is

independent of path in U and hence that there exists a potential function φ in U (seeProposition 2.3, Section 2.4.2).

Comment:It is difficult to prove that for any simple closed curve C in a simply-connected set U ⊂ R3,

there exists a surface Σ such that C = ∂Σ. We followed Marsden & Tromba (page 520) inrelying on geometrical intuition. Some authors (e.g. Corwin & Szczarba, page 343), definea simply-connected set U in R3 to be a set such that for any simple closed curve C thereis a surface Σ with C = ∂Σ, thereby side-stepping the problem. If one wants to avoid thisproblem completely, one can give a direct proof of Theorem 4.3 by defining a potential φas a line integral along a specific path and explicitly verifying that ∇φ = F (see Davis &Snider, page 206). We also refer to Flanigan & Kazdan for a simple proof of this theorem inRn (see the final conclusion Theorem 10.42, proved on page 574).

Here is a classical counter-example to show that the requirement that U be simply-connected is essential.

Example 4.3:Consider the vector field

F =

(−y

x2 + y2,

x

x2 + y2, 0

)on the subset

U = R3 − (x, y, z)|x = y = 0, z ∈ R.

Then F is C1 on U , and it follows from the definition (4.6) of ∇ × F that ∇ × F = 000 (doit!). Since a circle encircling the z-axis cannot be shrunk to a point in U the set U is notsimply-connected.3 Thus theorem 4.3 is not applicable, and leaves open the question as towhether a potential exists in U . This question can be answered using Proposition 4.1 ii). Astraight-forward calculation shows that for the circle C : x2 + y2 = b2, z = 0,∫

C

F·dx = 2π.

If F = ∇φ, this integral would be zero. Hence a potential φ does not exist in U .

Exercise 4.8:Show that the vector field

F = (y + z, z + x, x+ y)

is a gradient field on R3, and find a potential φ.

Answer: φ = xy + yz + zx.

3Equivalently, for such a circle, there is no surface Σ in U such that C = ∂Σ.

111

We now give a simple application of these ideas to fluid dynamics.

An irrotational and incompressible fluid:

Consider a vector field v(x, t) in R3 which represents the velocity of a fluid, and a scalarfield ρ(x, t) which represent its density. We have seen that conservation of mass leads to theequation of continuity (4.65):

∂ρ

∂t+∇ ··· (ρv) = 0. (4.77)

We assume that the fluid is incompressible i.e. that its density does not depend on time oron position (ρ = constant). Then (4.77) reduces to

∇·v = 0. (4.78)

We now assume that the motion is steady state i.e. v is independent of time, and irrotational,∇× v = 000, in R3. By Theorem 4.3 there exists a potential φ:

v = ∇φ. (4.79)

Substituting (4.79) in (4.78) gives ∇ ··· (∇φ) = 0, i.e.

∇2φ = 0,

where ∇2 is the Laplacian (see (4.13)). Thus, the velocity potential φ of an incompressiblefluid in steady state motion with zero vorticity satisfies Laplace’s equation ∇2φ = 0.

4.4.2 Divergence-free vector fields

Consider a vector field F defined by

F = ∇×A,

where A is a C2 vector field, called a vector potential for F. We first summarize the principalproperties of such vector fields.

Proposition 4.2:

If F = ∇×A, where A is C2 in U ⊂ R3, then

i)

∫∫Σ

F · ndS =

∫C

A·dx,

where Σ is any surface such that C = ∂Σ, (i.e. the surface integral is surface-independent in U).

ii)

∫∫Σ

F · ndS = 0 for any closed surface in U .

iii) ∇·F = 0.

112

Proof:i) is simply the statement of Stokes’ Theorem.ii) follows from Stokes’ Theorem by subdividing the closed surface Σ into two surfaces

Σ1 and Σ2, with common boundary C (see Figure 4.13).iii) is a consequence of the zero identity Z2 (see (4.12)).

The key question is: how does one determine whether a given vector field F has a vectorpotential A (F = ∇×A)?

In view of Proposition 4.2, it is natural to conjecture that if F is divergence-free (∇·F =0), then there will exist a vector potential A. This conjecture is in fact true, provided thatwe impose a restriction on the set U .

Definition:A subset U ⊂ R3 (or R2) is star-shaped means that there is a point P ∈ U such that for

any point Q ∈ U the line segment PQ is contained in U (see for example, Davis & Snider,page 159).

U1

Q1

Q2

PP

U2

Q

Figure 4.16: U1 is star-shaped, but U2 is not.

Comment:

i) The set U = R3 − (0, 0, 0) is not star-shaped, but is simply-connected. In fact if Uis star-shaped, then U is simply-connected.

ii) If U ⊂ R3 is star-shaped and Σ is a closed surface in U , then the interior of Σ lies inU .

Theorem 4.4 (vector potential):

If F is of class C1 and ∇·F = 0 in U ⊂ R3, and U is star-shaped, then there exists avector field A such that F = ∇×A in U .

Proof:We refer to Davis & Snider (pages 214-5).

113

Example 4.4:Any constant vector field B satisfies ∇·B = 0, and hence has a vector potential A in any

star-shaped set U . The identity∇× (B× r) = 2B, (4.80)

where B is a constant vector field and r = (x, y, z) (see Exercise 1 in Section 4.1.1) showsthat

A = 12B× r

is a vector potential4 for B i.e. B = ∇×A.

Exercise 4.9:Show that the vector field F = (y, z, x) is divergence-free in R3, and find a vector potential

A.

Answer: A = 12(z2, x2, y2)

Exercise 4.10:Show that any vector field of the form

F = (∇f)×B,

where B is a constant vector field and f is a C2 scalar field, is divergence-free on R3, andfind a vector potential A.

Answer: A = fB (see identities D3 and C2).

As a physical example of Theorem 4.4 we note that a magnetic field H is divergence-free(∇·H = 0), by Maxwell’s equations. Hence any magnetic field in a star-shaped set U ⊂ R3

has a vector potential, H = ∇ ×A. It follows that the magnetic flux through a surface Σbounded by a simple closed curve C = ∂Σ is independent of Σ (see the comment after thecorollary to Stokes’ theorem). Thus given a simple closed curve C, one can talk about themagnetic flux through C (see the discussion of Faraday’s law in Section 4.3.2).

Note: Because of the connection with magnetic fields, a vector field that satisfies ∇·H = 0is also called a solenoidal field.

We finally give a classical counter-example for Theorem 4.4, to show that the requirementthat U be star-shaped is essential.

Example 4.5:Consider the inverse square law vector field

F = − r

r3,

with r = (x, y, z) and r =‖ r ‖. Then F is C1 on the set U = R3−(0, 0, 0) and a standardcalculation (do it!) shows that ∇·F = 0. However, U is not star-shaped (missing a point!).

4Since (4.80) holds in any subset, A is a vector potential in any subset of R3.

114

Thus Theorem 4.4 is not applicable and leaves open the question of whether F has a vectorpotential A in U . This question can be answered using Proposition 4.2 ii). We have seenthat if Σ is any sphere centred on (0, 0, 0), then∫∫

Σ

F · n dS = −4π,

(see Gauss’ Law in Section 4.2.3). If F = ∇×A, then this surface integral would be zero.Hence a vector potential does not exist.

115

116

Chapter 5

Fourier Series and Fourier Transforms

In this Chapter we give an introduction to the branch of mathematics called Fourier analysis.The essential idea is that one can write a given function f(x) as a “sum” of sinusoidalfunctions sinωx and cosωx (or more concisely, eiωx, thinking of Euler’s formula). If thegiven function is periodic, the “sum” is an infinite series, the Fourier series of f , while iff is not periodic, the sum is an improper integral, the Fourier integral of f . The Fourierseries is obtained by calculating the Fourier coefficients of f , and the Fourier integral, bycalculating the Fourier transform of f . These are the key concepts that we shall discuss inwhat follows.

Historically, Fourier analysis first gained prominence in the early part of the nineteenthcentury, through the work of Joseph Fourier concerning the diffusion of heat.1 Earlier in thiscourse we showed that this process is described by a partial differential equation (PDE), theso-called diffusion equation (see equation (4.67)). Fourier showed that the solutions of thisPDE, subject to appropriate boundary and initial conditions, could be written as Fourierseries. This development exemplifies one area of application of Fourier analysis, namely thesolution of problems involving linear PDEs.2

In the twentieth century, Fourier analysis found application in a totally different area,namely signal processing. The idea is that the Fourier coefficients or the Fourier transformof a signal f(t), where t is time, represent the analysis of the signal into its constituent fre-quencies. This representation is of fundamental importance in connection with the problemof sampling a continuous signal in order to create a discretized version of it, which forms thetheoretical foundation for technological developments such as digital signal processing.

5.1 Fourier Series

In connection with his work on the diffusion of heat, Fourier argued that any3 function ofperiod 2π, even discontinuous ones, can be written as the sum of a series of sine and cosine

1Fourier’s book on this subject, “Theorie Analytique de la Chaleur”, was published in 1822, but he beganhis work in 1804. We refer to Korner 1988, pages 478-480, for a short biographical essay on Fourier.

2This topic forms a major part of AMATH 353.3We know now that the function has to satisfy certain restrictions.

117

functions of the form:

f(x) = 12a0 +

∞∑n=1

(an cosnx+ bn sinnx), (5.1)

where a0, an and bn are constants. The series (5.1), with the coefficients calculated in anappropriate way (see Section 5.1.1), is called the Fourier series of the function f .

Figure 5.1:

Partial sums S1, S3, S7 and S21 of the Fourier series

sinx+sin 3x

3+

sin 5x

5+ · · ·

of the function f(x) =

π4, 0 < x < π

−π4, −π < x < 0.

118

A remarkable fact about Fourier series is that the sum function f(x) is not necessarilycontinuous even though the terms in the series are continuous. This behaviour is in strongcontrast to the case of Taylor series, namely

f(x) =∞∑n=0

an(x− b)n,

for which the sum function has derivatives of all orders. Both Fourier series and Taylorseries, although differing significantly in their properties, provide an approximation of thegiven function by a finite sum of simpler terms, if one truncates the series after n terms.In Figure 5.1 we show a succession of partial sums for a Fourier series of a discontinuousfunction. Observe that the accuracy of the approximation increases as the number of termsincreases.

The goal of this section is to give the reader a working knowledge of Fourier series. Wefirst show how to find the Fourier coefficients of a given function, using symmetry to simplifythe calculations where possible. We then discuss (but do not prove) a convergence theoremwhich enables one to find the sum of a Fourier series.

5.1.1 Calculating Fourier coefficients

We begin by assuming that the given function f on the interval −π < x < π can be writtenas the sum of a trigonometric series

f(x) = 12a0 +

∞∑n=1

(an cosnx+ bn sinnx). (5.2)

Whereas the coefficients of a Taylor series are determined by the derivatives of the functionf , it turns out that the coefficients in a Fourier series can be expressed as integrals.

So we have to assume that the given function f is integrable i.e. that its Riemann integralexists on the interval −π ≤ x ≤ π. To proceed we need certain trigonometric integrals:

∫ π

−πcosmx cosnx dx =

0 if m 6= n

π if m = n 6= 0

2π if m = n = 0

(5.3)

∫ π

−πcosmx sin nx dx = 0, (5.4)

∫ π

−πsinmx sinnx dx =

0, if m 6= n

π, if m = n 6= 0,

(5.5)

119

where m and n are non-negative integers.These formulas can be verified by writing the products of the trigonometric functions as

sums, e.g.,

cos(m+ n)x = cosmx cosnx− sinmx sinnx

cos(m− n)x = cosmx cosnx+ sinmx sinnx,

givingcosmx cosnx = 1

2[cos(m+ n)x+ cos(m− n)x],

etc.

Exercise 5.1:Verify the formulae (5.3)-(5.5).

To find the coefficient a0 in (5.2) we integrate (5.2) from −π to π:∫ π

−πf(x)dx =

∫ π

−π

12a0dx+

∫ π

−π

∞∑n=1

(an cosnx+ bn sinnx)dx. (5.6)

We now make the assumption (which we shall justify in Section 5.3.2) that we can integratethe series term-by-term i.e.∫ π

−π

[∞∑1

( )

]dx =

∞∑1

[∫ π

−π( )dx

].

Since cosnx and sinnx are 2π periodic, the integral of the nth term of the series in (5.6)equals 0. Equation (5.6) thus gives ∫ π

−πf(x)dx = πa0, (5.7)

a formula for a0.We next multiply (5.2) by cosmx and integrate from −π to π. Again assuming we can

integrate the series term-by-term we get∫ π

−πf(x) cosmxdx =

∫ π

−π

12a0 cosmxdx

+∞∑n=1

[an

∫ π

−πcosmx cosnx dx+ bn

∫ π

−πcosmx sinnx dx

].

On using (5.3) and (5.4) this equation simplifies to∫ π

−πf(x) cosmxdx = πam, (5.8)

giving a formula for am,m = 1, 2, . . ..

120

Finally, we multiply (5.2) by sinnx and integrate from −π to π. Again assuming we canintegrate the series term-by-term we get∫ π

−πf(x) sinmxdx =

∫ π

−π

12a0 sinmxdx

+∞∑n=1

[an

∫ π

−πsinmx cosnx dx+ bn

∫ π

−πsinmx sinnx dx

].

On using (5.4) and (5.5) this equation simplifies to∫ π

−πf(x) sinmxdx = πbm, (5.9)

giving a formula for bm,m = 1, 2, . . . .We now summarize the results given by equations (5.7)-(5.9):

an =1

π

∫ π

−πf(x) cosnx dx, (5.10)

for n = 0, 1, 2, . . . , and

bn = 1π

∫ π

−πf(x) sinnx dx, (5.11)

for n = 1, 2, 3, . . .. Equations (5.10) and (5.11) are the formula for the Fourier coefficientsof an integrable function f .

Example 5.1: Calculate the Fourier series for thefunction

f(x) = 12(π− | x |), (5.12)

for −π < x < π.

y

xπ− π

π2

Figure 5.2: Graph of the function (5.12).

Solution: We begin by observing that f(x) is an even function. It thus follows immediatelyfrom (5.11) that

bn =1

π

∫ π

−πf(x) sin(nx)dx = 0,

121

since f(x) sin(nx) is an odd function4 and∫ π

−π( )dx = 0

for any odd function.By (5.10)

an =1

2π

∫ π

−π(π− | x |) cosnx dx.

Observe that (π− | x |) cosnx is an even function4 since both π− | x | and cosnx are botheven. For an even function ∫ π

−π( )dx = 2

∫ π

0

( )dx.

Thus the formula for an can be written

an = 1π

∫ π

0

(π − x) cosnx dx,

since | x |= x on the interval 0 ≤ x ≤ π. For n = 0 we obtain directly

a0 = 1π

∫ π

0

(π − x)dx = 12π.

For n > 0 we integrate by parts, obtaining

an = 1π

[(π − x) 1

nsinnx

∣∣∣∣π0

−∫ π

0

(−1) 1n

sinnx dx

]= − 1

πn2cosnx

∣∣∣∣π0

=1

πn2(1− cosnπ) =

0 if n is even

2πn2 if n is odd.

It is more convenient to relabel the coefficients by writing

a2n = 0, a2n−1 =2

π(2n− 1)2, n = 1, 2, . . . .

Thus the Fourier series of the function f(x) = 12(π− | x |) is

14π + 2

π

∞∑n=1

cos(2n− 1)x

(2n− 1)2. (5.13)

We discuss the convergence of this series in Section 5.1.2.

The following table gives some simple Fourier series and the functions from which theyarise. The functions we have chosen are either even, in which case the Fourier series isa cosine series, or odd, in which case the Fourier series is a sine series. The symmetrydetermines f(x) in the interval −π < x < 0.

4We discuss the use of symmetry to simplify the calculation of Fourier coefficients further in Section 5.1.3.

122

Table 5.1: Examples of Fourier series

f(x) 12a0 +

∞∑n=1

(an cosnx+ bn sinnx)

i) 12x, 0 < x < π

∞∑n=1

(−1)n+1 sinnx

n

ii) 12(π − x), 0 < x < π

∞∑n=1

sinnx

n

iii) 14π, 0 < x < π

∞∑n=1

sin(2n− 1)x

2n− 1

iv)

14π, 0 < x < 1

2π

−14π, 1

2π < x < π

∞∑n=1

(−1)n−1 cos(2n− 1)x

2n− 1

v) 112

(π2 − 3x2), 0 < x < π∞∑n=1

(−1)n−1 cosnx

n2

vi) 14(π − x)2 − 1

12π2, 0 < x < π

∞∑n=1

cosnx

n2

vii) 18π(π − 2x), 0 < x < π

∞∑n=1

cos(2n− 1)x

(2n− 1)2

viii)

14πx, 0 < x < 1

2π

14π(π − x), 1

2π < x < π

∞∑n=1

(−1)n−1 sin(2n− 1)x

(2n− 1)2

Exercise 5.2: Verify the Fourier series i)-iv) in Table 5.1. Hint: if on the right the expansionis an expansion in sines then this series represents a 2π-periodic function which is odd(because each of the sines in the series is 2π-periodic and odd). Therefore, the series extendsthe function on the left as an odd function to the interval [−π, 0]. Similarly, if the series onthe right is an expansion in cosines then this series represents a 2π-periodic function whichis even (because everyone of the cosines in the series is 2π-periodic and even). Therefore,the series extends the function on the left as an even function to the interval [−π, 0]. Inpractice, this means that to verify one of the entries in the table, check if the entry on theright contains sines or cosines and based on that do either a do a sine or a cosine expansionof the function on the left hand side.

5.1.2 Pointwise convergence of a Fourier series

We were led to the formulae (5.10)-(5.11) for the Fourier coefficients an, bn by a heuristicargument starting with the assumed series (5.2). We now regard (5.10) and (5.11) as the

123

definition of the Fourier coefficients of the given integrable function f . We are then facedwith the question: if we use these coefficients to form the Fourier series

12a0 +

∞∑n=1

(an cosnx+ bn sinnx),

does this series converge pointwise, and does its sum equal f(x)? The answer is that if fsatisfies certain conditions the series does converge for all x. Whether the sum of the seriesequals f(x) for −π < x < π depends on whether f is continuous at x.

In order to state the standard theorem on pointwise convergence of a Fourier series, weneed to introduce the notion of a function f being “piecewise C1” on an interval. Geomet-rically this means that the graph of f consists of a finite number of smooth sections. Forlater use we also define the related concept of “piecewise continuous”.

Definition 5.1:A function f : [a, b]→ R is piecewise continuous (respectively C1) means that there is a

partitiona = x0 < x1 < x2 < · · · < xn = b

such that f , when restricted to each open interval xi−1 < x < xi, coincides with a functionthat is continuous (respectively C1) on the closed interval xi−1 ≤ x ≤ xi.

Four examples are shown in Figure 5.3. The function H(x) is the Heaviside function:

H(x) =

0 if x < 0

1 if x ≥ 0.

a)

y

x

1

-1 1

b)

y

x

1

-1 1

c)

y

1

-1 1x

d)

y

x

1

-1 1

Figure 5.3:

a) f(x) = 1− | x | is continuous and piecewise C1.b) f(x) = 1− x2/3 is continuous but not piecewise C1.c) f(x) = H(x)(1− | x |) is piecewise continuous and piecewise C1.d) f(x) = H(x)(1− x2/3) is piecewise continuous but not piecewise C1.

124

If a function is piecewise continuous but not continuous on a finite interval, it will havea finite number of jump discontinuities, characterized by the fact that

limx→x+

0

f(x) 6= limx→x−0

f(x).

For convenience we introduce the notation

f(x+0 ) = lim

x→x+0

f(x), f(x−0 ) = limx→x−0

f(x). (5.14)

If a Fourier series converges pointwise on the interval −π ≤ x ≤ π, then the series willconverge pointwise for all x ∈ R and its sum will be a periodic function of period 2π (sinceeach term in the series is of period 2π). It is thus necessary to use period 2π functions whenstating theorems concerning convergence of Fourier series. This restriction does not limitthe scope of the theorems, since given any function f defined on the interval −π ≤ x ≤ πwe can introduce a period 2π extension of f : this is simply a function fp of period 2π thatequals the given function f on the open interval −π < x < π.

y = f ( x)π

−π

y = fp ( x)

x xπ−π

π

€

−π

€

π

€

2π

Figure 5.4: A period 2π extension of f(x) = x, −π < x < π.

If the given function f satisfies f(π) 6= f(−π), any period 2π extension fp will have jumpdiscontinuities at x = nπ. The value of fp(x) at x = nπ can be assigned arbitrarily, but it isnatural to choose the average,

fp(x) = 12[fp(x

+) + fp(x−)], (5.15)

in terms of the notation (5.14), as in Figure 5.4.5

We can now state the convergence theorem.

Theorem 5.1 (Pointwise convergence of a Fourier series):If fp has period 2π and is piecewise C1, then the Fourier series of fp converges pointwise

for all x, and

i) if fp is continuous at x, the sum is fp(x),

5If f satisfies f(π) = f(−π) then it will have a unique continuous period 2π extension fp.

125

ii) if fp is not continuous at x, the sum is 12[fp(x+) + fp(x−)].

Proof: See Churchill and Brown, pages 91-4, Theorem 1. The proof is lengthy.

Example 5.2: The Fourier series of f(x) = 12(π − |x|) on the interval −π ≤ x ≤ π is

14π + 2

π

∞∑n=1

cos(2n− 1)x

(2n− 1)2. (5.16)

(see Example 5.1).Sketch the graph of the period 2π function to which the series (5.16) converges pointwise

for all x ∈ R.

Solution: The period 2π extension fp of the given function f(x) is shown in Figure 5.5.Since f(−π) = f(π), the extension fp is continuous,6 as can be seen from the Figure. Sincefp is also piecewise C1 as can be seen by inspection, Theorem 5.1 implies that the Fourierseries (5.16) converges pointwise to fp for all x, and we can write

fp(x) = 14π + 2

π

∞∑n=1

cos(2n− 1)x

(2n− 1)2, for all x ∈ R. (5.17)

− πAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAy = f(x)

xπ

y = fp(x)

− π πAAAA

AAAA

− 3π 3π x

π2

Figure 5.5: The period 2π extension of f(x) = 12(π− | x |).

Example 5.3: The Fourier series of the function

f(x) = 12x, 0 < x < π,

is∞∑n=1

(−1)n+1 sinnx

n(5.18)

(see #i) in Table 5.1). Sketch the graph of the period 2π function to which the series (5.18)converges pointwise for all x ∈ R.

6Compare with Example 5.3 to follow, in which case f(−π) 6= f(π), leading to an extended function thatis discontinuous.

126

Solution: The sum of the series (5.18) on R is an odd function, since sin(nx), n = 1, 2, . . .is odd. So we first construct the odd extension of f(x) to 0 < x < π, shown in Figure 5.6(a).We then extend the function to have period 2π. Since f(−π) 6= f(π), the 2π − periodicextension fp has a jump discontinuity at x = kπ, k ∈ Z. At these points we define fp(x) by(5.15), leading to Figure 5.6(b). Since fp is piecewise C1 by inspection, Theorem 5.1 impliesthat the series (5.18) converges pointwise to fp(x) for all x ∈ R, and we can write

fp(x) =∞∑n=1

(−1)n+1 sinnx

n, for all x ∈ R. (5.19)

Example 5.4: Use the Fourier series (5.16) to sum∞∑n=1

1

(2n− 1)2.

Solution: We can choose x = 0 in (5.17), obtaining

fp(0) = 14π + 2

π

∞∑n=1

1

(2n− 1)2(5.20)

But fp(0) = f(0) = π2. Hence equation (5.20) gives

∞∑n=1

1

(2n− 1)2=π2

8.

Exercise 5.3: Use the Fourier series (5.18) to show that

∞∑n=1

(−1)n−1

2n− 1=π

4.

Hint: Choose x = π2.

Exercise 5.4: Sketch the graph of the period 2π function to which each of the Fourierseries in Table 5.1 converges pointwise for all x ∈ R. The previous example 5.3 gives theresult for series i).

The Gibbs phenomenon:

The graphs of selected partial sums of the series #iii) in Table 5.1 are shown in Figure5.1 (restricted to the interval 0 ≤ x ≤ 2π). The graphs illustrate a special feature ofthe convergence of Fourier series at a point of discontinuity, namely that the partial sums“overshoot” on either side of a jump discontinuity. One might expect that as the numberof terms in the partial sum increases the overshoot would tend to zero. The overshoot peakdoes become increasingly narrow as n increases and it moves successively closer to the pointof discontinuity, but its amplitude does not tend to zero, and in fact approaches a value of

127

y

x

−ππ

− π2

π2

y

x− 3π − π π 3π

(a) (b)

Figure 5.6: (a) shows the odd extension of f(x) = 12x, 0 < x < π. (b) shows the period 2π

extension fp of the function (a).

approximately 0.09H, where H is the jump in the function at the point of discontinuity. Itis important to note that the overshoot does not prevent the Fourier series from convergingpointwise at each point.

This behaviour, which is referred to as the Gibbs phenomenon, is illustrated schematicallyin Figure 5.7, for the function in # ii) in Table 5.1.

y

x

y

x

a)b)

≈ 0.09 H

graph of f

graph of Sn

H

Figure 5.7: The Gibbs phenomenon. The amplitude of the overshoot in Sn, shown in a),does not tend to 0 as n→∞, and is illustrated schematically in b).

5.1.3 Symmetry properties

As illustrated in Example 5.1, symmetry properties of the given function can simplify thecalculation of the Fourier coefficients. In this section we discuss symmetry properties in more

128

detail.We begin by briefly reviewing the properties of even and odd functions:

i) f is even means that f(−x) = f(x) for all x.

f is odd means that f(−x) = −f(x) for all x.

ii) The product of two even or two odd functions is even.

The product of an odd and an even function is odd.

iii) If f is even, then

∫ a

−af(x)dx = 2

∫ a

0

f(x)dx. If f is odd, then

∫ a

−af(x)dx = 0.

y

x

y

x

a) b)

AAAAAAAAAA

AAAAAA

AAAAAA

AAAA

AAAAAA

AAAAAAAAAAAA

f (-x)

-x

f (x)

x -x x

Figure 5.8: a) f is even, f(−x) = f(x). b) f is odd, f(−x) = −f(x).

Special cases of Fourier series:

i) Cosine series:

Suppose that f is an even function. Equations (5.10) and (5.11) for the Fourier coeffi-cients simplify to

an = 1π

∫ π

−πf(x) cosnx dx = 2

π

∫ π

0

f(x) cosnx dx,

bn = 1π

∫ π

−πf(x) sinnx dx = 0.

The Fourier series (5.2) then reduces to a cosine series:

12a0 +

∞∑n=1

an cosnx, (5.21)

with

an = 2π

∫ π

0

f(x) cosnx dx.

Thus any integrable function f on 0 < x < π has a cosine series (5.21). This cosineseries can be thought of as the full Fourier series for an even function feven on −π <x < π that coincides with f on 0 < x < π.

129

ii) Sine series:

Suppose f is an odd function. Equations (5.10) and (5.11) for the Fourier coefficientssimplify to

an = 1π

∫ π

−πf(x) cosnx dx = 0,

bn = 1π

∫ π

−πf(x) sinnx dx = 2

π

∫ π

0

f(x) sinnx dx.

The Fourier series (5.2) then reduces to a sine series:

∞∑n=1

bn sinnx, (5.22)

with

bn = 2π

∫ π

0

f(x) sinnx dx.

Thus any integrable function f on 0 < x < π has a sine series (5.22). This sine seriescan be thought of as the full Fourier series for an odd function fodd on −π < x < πthat coincides with f on 0 < x < π.

Symmetry about the line x = π2:

When working with cosine and sine series it is helpful to think of symmetry about theline x = π

2.

i) f is even with respect to the line x = π2

means

f(π − x) = f(x) for all x.

ii) f is odd with respect to the line x = π2

means

f(π − x) = −f(x) for all x.

Exercise 5.5:

Show that

i) cos 2nx and sin(2n− 1)x are even with respect to x = π2.

ii) cos(2n− 1)x and sin 2nx are odd with respect to x = π2.

Symmetry with respect to x = π2

and the cosine series:

130

AAAAAA

AAAAAA

a)y

0 x xππ −x

x = π2

AAAAAA

AAAA

x = π2

0 x

π−xπ x

y b)

Figure 5.9: a) f is even with respect to x = π2. b) f is odd with respect to x = π

2.

i) If f(π − x) = f(x) on 0 < x < π, then by (5.21),

a2n−1 = 2π

∫ π

0

f(x) cos (2n− 1)x dx = 0,

↑even

↑odd

and the cosine series is

12a0 +

∞∑n=1

a2n cos 2nx. (5.23)

ii) If f(π − x) = −f(x) on 0 < x < π, then

a2n = 2π

∫ π

0

f(x) cos 2nx dx = 0,

↑odd

↑even

and the cosine series is∞∑n=1

a2n−1 cos(2n− 1)x. (5.24)

Symmetry with respect to x = π2

and the sine series:

i) If f(π − x) = f(x) on 0 < x < π, then by (5.22),

b2n = 2π

∫ π

0

f(x) sin 2nx dx = 0,

↑even

↑odd

and the sine series is∞∑n=1

b2n−1 sin(2n− 1)x. (5.25)

131

ii) If f(π − x) = −f(x) on 0 < x < π, then

b2n−1 = 2π

∫ π

0

f(x) sin (2n− 1)x dx = 0,

↑odd

↑even

and the sine series is∞∑n=1

b2n sin 2nx. (5.26)

Example 5.5: Find the Fourier sine series of

f(x) = π4, 0 < x < π.

Use the fact that f is even with respect to x = π2.

(This series is # iii) in Table 5.1.)

Solution: Since we want the Fourier sine series, we first extend f to be an odd functionfodd on −π < x < π:

fodd(x) =

π4, 0 < x < π

−π4, −π < x < 0.

The Fourier sine coefficients are then giving by (5.11):

bn =1

π

∫ π

−πfodd(x) sin(nx)dx.

Since fodd(x) sin(nx) is even (the product of two odd functions), we can write

bn = 2π

∫ π

0

fodd(x) sin(nx)dx

= 2π

∫ π

0

(π4

)sin(nx)dx.

Next, since the given function is symmetric about x = π2, we have

b2n = 0, (since sin(2nx) is antisymmetric)

and

b2n−1 = 2π

∫ π

0

π4

sin(2n− 1)x dx

= 4π

∫ π2

0

π4

sin(2n− 1)x dx

= −cos(2n− 1)x

(2n− 1)

∣∣∣∣π20

=1

2n− 1.

132

The Fourier sine series is thus∞∑n=1

sin(2n− 1)x

2n− 1.

π−π

y = f p ( x)

x

y = f ( x)

xπ

xπ−π

y = f_(x)

AAAAAA

Figure 5.10: The odd, period 2π extension of f(x) = 12(π − x), 0 < x < π.

5.1.4 Complex form of the Fourier series

The general Fourier series (5.2) can be written more concisely by using complex numbers,i.e. by writing the nth term in terms of einx. This complex representation leads naturallyto the Fourier integral and Fourier transform, and is the most convenient representation forapplications to signal processing.

For a τ -periodic function f , the complex Fourier series is

f(t) =∞∑

n=−∞

cneinω0t, (5.27)

where

ω0 =2π

τ(5.28)

is the fundamental frequency. The complex Fourier coefficients cn, n = 0,±1,±2, . . . , aregiven by the formula

cn =1

τ

∫ τ2

− τ2

f(t)e−inω0tdt, (5.29)

and satisfyc−n = cn, (5.30)

where the overbar denotes complex conjugation. This condition ensures that the functionf(t) in (5.27) is real-valued. The complex basis functions

en(t) = einω0t (5.31)

satisfy the orthogonality condition

∫ τ/2

−τ/2en(t)em(t)dt =

0, if m 6= n

τ if m = n

(5.32)

133

(verify as an exercise).

134

Derivation of equation (5.29):

As in section 5.1.1, we use the fact that Fourier series can be integrated term-by-term(to be justified in Section 5.3.2). Using (5.31), write (5.27) as

f(t) =∞∑

n=−∞

cnen(t).

Multiply by em(t) and integrate from − τ2

to τ2:∫ τ

2

− τ2

f(t)em(t)dt =

∫ τ2

− τ2

(∞∑

n=−∞

cnen(t)em(t)

)dt

=∞∑

n=−∞

cn

∫ τ2

− τ2

en(t)em(t)dt (integrate term-by-term)

= τcm (by (5.32)).

Since em(t) = e−imω0t (see (5.31)), we obtain (5.29), with n replaced by m.

Relation between the real and complex forms

Write cn, n = 0, 1, 2, . . . in terms of real and imaginary parts:

cn = 12(an − ibn). (5.33)

Then by (5.30),c−n = 1

2(an + ibn). (5.34)

Setting n = 0 in these equations shows that

b0 = 0, c0 = 12a0. (5.35)

By Euler’s formula (eiθ = cos θ + i sin θ), the basis vectors (5.31) have the form

en(t) = cos(nω0t) + i sin(nω0t). (5.36)

We can write (5.27) as a series summed from 1 to ∞:

f(t) = 12a0 +

∞∑n=1

(cne

inω0t + c−ne−inω0t

), (5.37)

where we have made use of (5.35). It now follows, using (5.33), (5.34), (5.31) and (5.36)that

cneinω0t + c−ne

−inω0t = an cos(nω0t) + bn sin(nω0t)

(verify as an exercise). Thus (5.37) becomes the real form of the Fourier series of a τ -periodicfunction f(t):

f(t) = 12a0 +

∞∑n=1

[an cos(nω0t) + bn sin(nω0t)] , (5.38)

135

with ω0 = 2π/τ . If we specialize to τ = 2π, then ω0 = 1, and (5.38) agrees with (5.2), withx replaced by t.

Exercise 5.6: Choose τ = 2τ in (5.29). Show that (5.29) and (5.33) lead to equations(5.10) and (5.11) for the real Fourier coefficients an and bn.

The Fourier spectrum and Parseval’s theorem

We think of the Fourier series of a τ -periodic function f as the decomposition of a signalf(t) into its constituent frequencies or harmonics

ω0, 2ω0, 3ω0, . . .

where ω0 = 2πτ

is the fundamental frequency. The Fourier coefficients cn describe the relativeimportance of the various harmonics. We write

cn = |cn|eiψn

where |cn| is the amplitude and ψn is the phase of the nth harmonic. The Fourier coefficientscn essentially contain all information about the τ -periodic function f , and the set of numbers

cnn∈Z

is called the Fourier spectrum of f . In particular, the real numbers |cn| form the amplitudespectrum of f . The upshot is that we can represent a τ -periodic signal either directly as afunction of time t in the time domain, or through its spectrum in the frequency domain. Weillustrate this duality schematically below.

f(t) ←−−−−−−−−→ cnn∈Z Fourier spectrum

One can measure the strength of a signal in two ways, either in the time domain using theintegral

1

τ

∫ τ2

− τ2

[f(t)]2dt,

or in the frequency domain, using the series

∞∑n=−∞

|cn|2.

Parseval’s theorem states that these two quantities are equal.

Theorem 5.2 (Parseval’s formula for a τ -periodic function):If a τ -period function f has a complex Fourier series

f(t) =∞∑

n=−∞

cneinω0t, (5.39)

136

fig5.11

y = f (t )

−τ2

τ2

t0

c n

n−4 −3 −2 −1 0 1 2 3 4 L

L

a τ-periodic signalin the time domain

the amplitude spectrumin the frequency domain

Figure 5.11: The amplitude spectrum of a τ -periodic signal.

then1

τ

∫ τ2

− τ2

f(t)2dt =∞∑

n=−∞

|cn|2. (5.40)

Proof: Multiply (5.39) by f(t) and integrate from − τ2

to τ2:∫ τ

2

− τ2

f(t)2dt =

∫ τ2

− τ2

(∞∑

n=−∞

cnf(t)einω0t

)dt

=∞∑

n=−∞

(cn

∫ τ2

− τ2

f(t)einω0tdt

)(integrate term-by-term)

= τ∞∑

n=−∞

cncn (using the complex conjugate of (5.29))

= τ

∞∑n=−∞

|cn|2 (since zz = |z|2 for a complex number).

While the main significance of Parseval’s formula is in connection with signal analysis, itis also useful for evaluating the sums of series that are impossible to do by direct means.

Example 5.6: Prove that∞∑n=1

1

n2=π2

6. (5.41)

Solution: We have seen (Table 5.1, ii) that the Fourier series of

f(t) = 12(π − t), 0 < t < π,

137

is∞∑n=1

sinnt

n.

By (5.33)-(5.34) the complex Fourier coefficients are

cn = − i

2n, c−n =

i

2n, c0 = 0,

which implies

|cn| = |c−n| =1

2n.

We apply Parseval’s theorem (5.40) with τ = 2π, writing the 2-sided series as a 1-sidedseries:

1

2π

∫ π

−π[fodd(t)]2dt = 2

∞∑n=1

|cn|2,

where fodd is the odd extension of f to −π < t < π.Using the fact that fodd(t)2 is even we obtain

1

π

∫ π

0

14(π − t)2dt = 2

∞∑n=1

1

(2n)2.

Evaluating the integral gives the desired result.

Comment: The key to doing this problem is to pick a function whose Fourier coefficientshave a suitable form. There is no unique choice. For example, you could also use example i)in Table 3,

f(x) = 12x, −π < x < π.

whose Fourier series is∞∑n=1

(−1)n+1 sinnx

n.

Exercise 5.7: Use this series to verify (5.41).

5.2 Convergence of series of functions

In this section we discuss the convergence of series of functions in more depth. We have seenin Section 5.1.2 that if a series of functions converges pointwise to a function f ,

f(x) =∞∑n=1

an(x),

the sum function may be discontinuous even though the terms of the series are continuous.In Section 5.1.1, when calculating Fourier coefficients we encountered the need to integratea series of functions term-by-term. For these reasons it is necessary to discuss the followingquestions.

138

Q1: Under what conditions is the sum of a series of continuous functions itself continuous?

Q2: Under what conditions can a series of (integrable) functions be integrated term-by-term? i.e. under what conditions is it true that∫ b

a

∞∑n=1

an(x)dx =∞∑n=1

∫ b

a

an(x)dx. (5.42)

The key point as regards these questions is this: results that are valid for finite sumsof functions are not necessarily valid for infinite series of functions. In order to answer theabove questions, we have to introduce two new types of convergence for series of functions,namely

i) uniform convergence, and

ii) mean square convergence.

In studying the convergence of a series of functions∞∑n=1

an(x), one considers the Nth partial

sum

SN(x) =N∑n=1

an(x).

Convergence of the series is then defined in terms of convergence of the sequence SN ofpartial sums, i.e. does this sequence have a limit as N →∞ in some suitably defined sense.Thus the discussion in this section is initially formulated in terms of sequences of functions.

5.2.1 A deficiency of pointwise convergence

Any definition of convergence of a sequence of functions fn to a limit function f mustrequire that the difference | fn(x) − f(x) | becomes small in some sense as n → ∞. Sofar (in Section 5.1.2), we have worked with pointwise convergence. We now give the formaldefinition, and then illustrate the “deficiency” with an example.

Definition 5.2:A sequence of functions fn converges pointwise to f on [a, b] means that7

limn→∞

| fn(x)− f(x) |= 0, (5.43)

for all x ∈ [a, b].

7You could equivalently write limn→∞

fn(x) = f(x) for all x ∈ [a, b]. Equation (5.43) is preferable, however,

because it emphasizes that one is interested in the difference between fn(x) and f(x).

139

Definition 5.3:

A series of functions∞∑n=1

an converges pointwise to a function f on [a, b] means that the

sequence of partial sums, defined by

SN(x) =N∑n=1

an(x),

converges pointwise to f on [a, b]. The function f is called the sum of the series, and wewrite

f(x) =∞∑n=1

an(x) pointwise on [a, b].

The intention in defining convergence of a sequence of functions fn is that one wantsfn, for n sufficiently large, to approximate the limit function f with specified accuracy

f(x) ≈ fn(x) for n > N.

Heuristically, one wants the graph of fn to approximate the graph of f with specified accuracyover the whole interval. Surprisingly, the notion of pointwise convergence (5.30) does notguarantee this type of approximation, as we now show with a simple example.

Example 5.7: Consider the sequence of functions fn defined by

fn(x) =2nx

1 + n2x2, 0 ≤ x ≤ 1. (5.44)

Show thatlimn→∞

fn(x) = 0 for all x ∈ [0, 1], (5.45)

but thatmax0≤x≤1

|fn(x)| = 1, for all n ∈ N. (5.46)

Sketch the graphs of f2 and fn, for n large.

Solution: Consider x ∈ [0, 1], x 6= 0. Then

limn→∞

fn(x) = limn→∞

(1n

)( 2x

x2 + 1n2

)= (0)

(2x

)= 0.

Consider x = 0. Thenlimn→∞

fn(0) = limn→∞

(0) = 0.

We have thus established (5.45).Next, using first year calculus, it follows that fn has a maximum value of 1 at x = 1

n:

fn(

1n

)= 1, for all n ∈ N,

140

AAAAAAAAAA

0 1

1

y

x

fn

f2

f = 01n

Figure 5.12: The graphs of the functions f2, fn.

which establishes (5.46). The graphs are shown in Figure 5.11.

Discussion: The existence of the peaks may lead one to doubt (5.45) for x very close to 0.But the definition of pointwise convergence says: first pick an x-value and then let n→ +∞.So no matter how close to zero your x-value is, call it x∗, for n large enough the peaks willlie to the left of x∗, and so lim

n→∞fn(x∗) = 0.

The main conclusion from Example 5.5 is this: if a sequence fn converges pointwiseto f on an interval a ≤ x ≤ b, the graph of the function fn does not necessarily lie close tothe graph of the limit function f over the whole interval, no matter how large n is. We saycolloquially that “pointwise convergence permits the function fn to have spikes”, i.e. spike-like deviations from the graph of the limit function f . The situation can in fact be worsethan that depicted in Figure 5.11, since the height of the spikes can increase without boundas n → ∞. Consider, for example, gn(x) =

√nfn(x), where fn(x) is given by (5.44). The

function gn will have a spike of height√n. Finally, we note that the Gibbs phenomenon for

Fourier series, mentioned in Section 5.1.2, is reminiscent of the behaviour of fn in Example5.6.

5.2.2 The maximum norm and mean square norm

In order to define uniform and mean square convergence we have to introduce a concept ofdistance between two functions. Instead of focusing on individual values of functions we haveto simultaneously consider all values of the functions over the given interval, i.e. we have tothink of a function f as a single mathematical entity, as an element of a function space, anddefine a notion of distance in this space.

We begin by considering the space of all continuous functions on a finite interval [a, b],denoted by

C[a, b].

We recognize that C[a, b] is a vector space over the reals, with addition defined by

(f + g)(x) = f(x) + g(x) for all x ∈ [a, b], (5.47)

141

and multiplication by a scalar defined by

(λf)(x) = λf(x) for all x ∈ [a, b]. (5.48)

We now rely on the vector space Rn to provide an analogy. In Rn we introduce the Euclideannorm (i.e. the magnitude of a vector)

‖ x ‖2=√x2

1 + · · ·+ x2n, (5.49)

wherex = (x1, x2, . . . , xn),

and then define the distance d(x,y) between two points x and y by

d(x,y) =‖ x− y ‖2 . (5.50)

x - yx

yd (x,y)

Figure 5.13: Euclidean distance in Rn.

So we need to define a norm in C[a, b], which will give the magnitude of a continuousfunction. There are two “reasonable” ways to do this:

i) use the maximum value of |f(x)| over the interval, and

ii) use the mean value 〈|f |〉 of |f(x)| over the interval, defined by

〈|f |〉 =1

b− a

∫ b

a

|f(x)| dx.

Definition 5.4:The maximum norm ‖ f ‖∞ on the space C[a, b] is defined by

‖ f ‖∞= maxa≤x≤b

| f(x) | . (5.51)

For the norm based on the mean of the function it turns out to be more convenient toconsider the square root of the mean of the function squared,[∫ b

a

f(x)2dx

] 12

, (5.52)

142

b x

AAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAAA

y = | f(x)|

a

max | f(x)|

Figure 5.14: The Maximum value of | f(x) |.

y = | f (x)|

xba

f

Figure 5.15: The mean value of |f |.

omitting the normalizing factor (b − a). The reason for this preference will become clearlater, but for the moment we point out the analogy between the expression (5.52) and theEuclidean norm (5.49) in Rn, thinking of the integral as the limit of a sum.

Definition 5.5:The mean square norm ‖ f ‖2 on the space C[a, b] is defined by

‖ f ‖2=

[∫ b

a

f(x)2 dx

] 12

. (5.53)

The two norms are related by an important inequality, which we shall refer to as the basicinequality for function norms.

Proposition 5.1:

‖ f ‖2≤√b− a ‖ f ‖∞, (5.54)

for all functions f in C[a, b].

143

Proof:By definition of ‖ f ‖2:

‖ f ‖22 =

∫ b

a

f(x)2dx

≤ maxa≤x≤b

[f(x)]2 (b− a) (property of the integral)

=[

maxa≤x≤b

| f(x) |]2

(b− a)

= (b− a) ‖ f ‖2∞ (by definition of ‖ f ‖∞).

Technical digression:

i) Properties of norms

It should be verified that ‖ f ‖∞ and ‖ f ‖2, as defined by (5.39) and (5.42), do satisfythe properties of a norm on a vector space X over R:

N1 : ‖ f ‖≥ 0 for all f ∈ X; ‖ f ‖= 0 iff f = 0.

N2 : ‖ f + g ‖≤‖ f ‖ + ‖ g ‖, for all f, g ∈ X.(the triangle inequality)

N3 : ‖ λf ‖=| λ |‖ f ‖, for all f ∈ X, λ ∈ R.

We leave these details as an exercise.

ii) We have mentioned that the mean square norm ‖ f ‖2 on C[a, b] is analogous to theEuclidean norm ‖ x ‖2 on Rn. One can define another norm ‖ x ‖∞ on Rn, called themax norm, analogous to the max norm on C[a, b]:

‖ x ‖∞= max1≤i≤n

| xi |,

where x = (x1, x2, . . . , xn).

This norm has computational advantages over ‖ x ‖2, and is used in scientific compu-tation.

iii) Norms of piecewise C1 functions

When working with Fourier series we encounter piecewise continuous differentiablefunctions, i.e. functions that are piecewise in C1. The set of all piecewise continuousdifferentiable functions on an interval a ≤ x ≤ b forms a vector space, which we denoteby PC[a, b].

It is clear that the mean square norm ‖ f ‖2 can be defined for f ∈ PC[a, b] using(5.52), since the integral of a piecewise continuous differentiable function is defined asa sum of integrals over subintervals.

On the other hand, in defining the max norm (5.51) we are making use of the theoremaccording to which a function continuous on a finite closed interval attains a maximum

144

value on that interval. However, a piecewise continuous differentiable function doesnot necessarily attain a maximum value on a finite closed interval, as shown in Figure5.15, and for such a function

‖ f ‖∞= max | f(x) |

does not exist. But observe that in the example of Fig.5.16 below, we have thatf(x) < 1

2for all x and f(x) attains values arbitrarily close to 1

2. So one calls 1

2the

least upper bound or supremum of f on the interval 0 ≤ x ≤ 1 and we write

y

x0 1

12

12

Figure 5.16: A piecewise continuous differentiable function that does not attain a maximum.

sup0≤x≤1

| f(x) |= 12.

More generally, for any piecewise continuous differentiable function on an interval a ≤x ≤ b, | f(x) | is bounded above and hence has a least upper bound. So for functionsin PC[a, b] we define the max norm (also called the sup norm) by

‖ f ‖∞= supa≤x≤b

| f(x) | . (5.55)

More generally, (5.55) is defined for any bounded function on [a, b].

5.2.3 Uniform and Mean Square Convergence

We now return to one of our main goals in this section, to define uniform convergenceand mean square convergence (also called convergence in the mean). The definitions areformulated using the function norms ‖ ‖∞ and ‖ ‖2, as defined by equations (5.51) and(5.53), respectively.

Definition 5.6:A sequence fn in C[a, b] converges uniformly to f ∈ C[a, b] means that

limn→∞

‖ fn − f ‖∞= 0,

i.e.

limn→∞

[maxa≤x≤b

| fn(x)− f(x) |]

= 0. (5.56)

145

Definition 5.7:A sequence fn in C[a, b] converges in the mean to f ∈ C[a, b] means that

limn→∞

‖ fn − f ‖2= 0,

i.e.

limn→∞

[∫ b

a

(fn(x)− f(x))2 dx

] 12

= 0. (5.57)

Note: These definitions are also valid on the set PC[a, b] of piecewise continuous functions,with ‖ f ‖∞ defined by (5.55).

Discussion:In order to understand the difference between pointwise convergence (5.43) and uniform

convergence (5.56) it is helpful to think in terms of an ε-neighbourhood defined by a norm.

In R2, an ε-neighbourhood of a point x0 is defined byx ∈ R2| ‖ x− x0 ‖2< ε

,

and is a disc of radius ε, centred at x0.

AAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAA

AAAAAAAA

εx0

x

Figure 5.17: An ε-neighbourhood in R2.

In C[a, b] an ε-neighbourhood of a function f0, relative to the max norm, is similarlydefined by

f ∈ C[a, b]| ‖ f − f0 ‖∞< ε ,

and is the set of functions whose graphs lie within a strip of width 2ε, centred on the graphof f0.

Thus, if fn converges uniformly to f , given any ε > 0 it follows that for all n sufficientlylarge, the graph of fn will lie within the strip of width 2ε centred on the graph of f . In otherwords, for n sufficiently large, the graph of fn approximates the graph of f over the wholeinterval. This means that uniform convergence eliminates the possibility of “spikes” in thegraph of fn, as in example 5.6. On the other hand, convergence in the mean does noteliminate “spikes” in fn, since it only requires that the mean of | fn(x) − f(x) |2 over [a, b]tends to zero.

We note in passing that Figure 5.18 provides heuristic justification for the following result:

If the sequence fn converges uniformly to f on [a, b] and fn is continuous forall n, then f is continuous.

146

Figure 5.18: An ε-neighbourhood in C[a, b].

The idea is that if the graph of each fn has no “breaks” and convergence is uniform, thenthe graph of f is forced to have no “breaks”.

We now give an example in which we use the definition to prove that a sequence convergesuniformly.

Example 5.8: Consider the sequence fn in C[−1, 1] defined by

fn(x) =

√x2 +

1

n2. (5.58)

Find the function f to which the sequence fn converges pointwise. Prove that the sequencefn converges uniformly to f .

Solution: Observe that for any x ∈ [−1, 1],

limn→∞

fn(x) =√x2 =| x |,

i.e. fn converges pointwise tof(x) =| x | .

We now calculate ‖ fn − f ‖∞. We have

| fn(x)− f(x) | =√x2 +

1

n2−√x2

=1n2√

x2 + 1n2 +

√x2.

By inspection, the maximum occurs at x = 0 (when the denominator is a minimum). Thus

‖ fn − f ‖∞= max−1≤x≤1

| fn(x)− f(x) |= 1n.

147

x-1 1

y

f

fn , n large

Figure 5.19: The graph of fn, n large, approximates the graph of f .

Sincelimn→∞

‖ fn − f ‖∞= limn→∞

(1n

)= 0.

the sequence fn converges uniformly to f on the interval −1 ≤ x ≤ 1.

It is important to note that the three types of convergence in C[a, b] are not equivalent.From the discussion so far one expects that uniform convergence is the strongest of the threetypes. This expectation is confirmed by the proposition to follow.

Proposition 5.2:

If fn converges uniformly to f in C[a, b] (or in PC[a, b]), then

i) fn converges in the mean to f , and

ii) fn converges pointwise to f .

Proof:

i) By Proposition 5.1,‖ fn − f ‖2≤

√b− a ‖ fn − f ‖∞,

for all n. Since limn→∞

‖ fn − f ‖∞= 0 by hypothesis, the Squeeze theorem implies that

limn→∞

‖ fn − f ‖2= 0, i.e. fn converges to f in the mean.

ii) It follows from the definition (5.51) that

| fn(x)− f(x) | ≤ ‖ fn − f ‖∞ (5.59)

for all x ∈ [a, b]. Thus, since limn→∞

‖ fn−f ‖∞= 0 it follows that limn→∞

| fn(x)−f(x) |= 0

for all x ∈ [a, b], i.e. fn converges pointwise to f , by definition (see (5.43)).

148

Comment: There is no simple relation between convergence and the mean and pointwiseconvergence. This fact is illustrated by examples in Problem Set 5.

Having discussed uniform convergence and convergence in the mean for sequences of

functions, we can now define these concepts for a series∞∑n=1

an of continuous functions, i.e.

an ∈ C[a, b]. The definition is based on the sequence Sn of partial sums, defined by

Sn =n∑k=1

ak.

Definition 5.8:

A series∞∑k=1

ak in C[a, b] converges uniformly to f means that the sequence Sn of partial

sums converges uniformly to f , i.e.

limn→∞

∥∥∥∥∥f −n∑k=1

ak

∥∥∥∥∥∞

= 0. (5.60)

Definition 5.9:

A series∞∑k=1

ak in C[a, b] converges in the mean to f means that the sequence Sn of

partial sums converges in the mean to f , i.e.

limn→∞

∥∥∥∥∥f −n∑k=1

ak

∥∥∥∥∥2

= 0. (5.61)

Note: These definitions are also valid in the space of piecewise continuous functions, PC[a, b].

We now discuss a test for proving that a series of functions converges uniformly on aninterval.

Proposition 5.3 (Weierstrass M-test):

If H1 : | an(x) | ≤Mn for all x ∈ [a, b],

H2 :∞∑n=1

Mn converges,

then

i)∞∑n=1

an(x) converges absolutely for each x ∈ [a, b], with sum f(x), and

149

ii)∞∑n=1

an(x) converges uniformly to f(x) on [a, b].

Proof:

By H1 and H2,∞∑n=1

|an(x)| converges by the comparison test, for all x ∈ [a, b], i.e.∞∑n=1

an(x)

converges absolutely. Hence we can define a function f on [a, b] by

f(x) =∞∑n=1

an(x).

We also let

Sn(x) =n∑k=1

ak(x).

Then

| f(x)− Sn(x) | =

∣∣∣∣∣∞∑

k=n+1

ak(x)

∣∣∣∣∣ ,≤

∞∑k=n+1

|ak(x)| (since the series converges absolutely),

≤∞∑

k=n+1

Mk for all x ∈ [a, b] (by H1),

It follows that

‖ f − Sn ‖∞ ≤∞∑

k=n+1

Mk

< ε for n sufficiently large,

since∞∑k=1

Mk converges. Thus limn→∞

‖ f − Sn ‖∞= 0, and the series converges uniformly on

[a, b], by the definition.

Example 5.9: Prove that the Fourier series∞∑n=1

cosnx

n2converges uniformly on any finite

closed interval.

Solution: Observe that∣∣∣cosnx

n2

∣∣∣ ≤ 1

n2for all x, and the series

∞∑n=1

1

n2converges.

150

Thus by the Weierstrass M-test the result follows.

Continuity of the sum function

The continuity result for sequences stated in the discussion prior to Example 5.8 suggestsan analogous result for series.

Theorem 5.3: If the series∞∑k=1

ak(x) converges uniformly to f(x) on [a, b], and if each term

of the series is continuous, then the sum function is continuous.

Proof: See Taylor and Mann, 1983, pg. 620, Theorem III.

This result and Example 5.9 shows that the function

f(x) =∞∑n=1

cosnx

n2, −π ≤ x ≤ π,

is continuous. We find f(x) explicitly later (see Example 5.12).

5.2.4 Termwise integration of series

The theorem that we prove in this section states that if a series of functions converges in themean on some interval, then the series can be integrated term-by-term, and the integratedseries converges uniformly.

Theorem 5.4: If∞∑k=1

ak(t) converges in the mean on the interval a ≤ t ≤ b, then the

integrated series∞∑k=1

(∫ x

a

ak(t)dt

)converges uniformly on the interval a ≤ x ≤ b, and

∫ x

a

(∞∑k=1

ak(t)dt

)=∞∑k=1

(∫ x

a

ak(t)dt

).

The proof follows quickly from the following lemma, which states that if f and fn are closein the mean square norm, then their integrals are close in the max norm.

Lemma: If

g(x) =

∫ x

a

f(t)dt, gn(x) =

∫ x

a

fn(t)dt, (5.62)

for a ≤ x ≤ b, then‖ g − gn ‖∞≤

√b− a ‖ fn − f ‖2 .

151

Proof: For any x ∈ [a, b],

|g(x)− gn(x)| =∣∣∣∣∫ x

a

[f(t)− fn(t)]dt

∣∣∣∣ (property of the integral)

≤∫ x

a

|f(t)− fn(t)|dt (property of the integral)

≤∫ b

a

|f(t)− fn(t)|dt (since x ≤ b and the integrand is positive)

≤[∫ b

a

|f(t)− fn(t)|2dt] 1

2[∫ b

a

1dt

] 12

(by the Cauchy-Schwarz inequality8)

=√b− a ‖ f − fn ‖2 .

Since x is arbitrary, it follows that

‖ g − gn ‖∞≤√b− a ‖ f − fn ‖2 .

Proof of the theorem:

Choose

f(t) =∞∑k=1

ak(t), fn(t) =n∑k=1

ak(t)

in the lemma. Then by (5.62),

g(x) =

∫ x

a

(∞∑k=1

ak(t)

)dt (5.63)

and

gn(x) =

∫ x

a

(n∑k=1

ak(t)

)dt =

n∑k=1

(∫ x

a

ak(t)dt

), (5.64)

where we are integrating a finite sum. It follows from the Lemma and (5.64) that∥∥∥∥∥g(t)−n∑k=1

(∫ x

a

ak(t)dt

)∥∥∥∥∥∞

≤√b− a

∥∥∥∥∥f(t)−n∑k=1

ak(t)

∥∥∥∥∥2

. (5.65)

Since∞∑k=1

ak(t) converges in the mean to f(t) on the interval a ≤ t ≤ b,

limn→∞

∥∥∥∥∥f(t)−n∑k=1

ak(t)

∥∥∥∥∥2

= 0,

8The Cauchy-Schwarz inequality:

[∫ b

a

p(x)q(x)dx

]2≤

[∫ b

a

p(x)2dx

][∫ b

a

q(x)2dx

].

152

by definition (see (5.61)). It thus follows from (5.65) and the Squeeze Theorem that

limn→∞

∥∥∥∥∥g(x)−n∑k=1

(∫ x

a

ak(t)dt

)∥∥∥∥∥∞

= 0.

Hence, by definition of uniform convergence (see (5.61)),∞∑k=1

(∫ x

a

ak(t)dt

)converges uni-

formly to g(x), and

∞∑k=1

(∫ x

a

ak(t)dt

)= g(x) =

∫ x

a

(∞∑k=1

ak(t)

)dt,

by (5.63).

In Section 5.3.2 we shall apply this theorem to Fourier series.

5.3 A Second Look at Fourier Series

In this section we use the results from Section 5.2 to discuss the convergence of Fourier seriesin greater depth. We are then in a position to prove that Fourier series can be integratedtermwise.

5.3.1 Uniform and mean square convergence

We now give conditions that guarantee convergence in the mean and uniform convergencefor a Fourier series. For convenience we also repeat the conditions that guarantee pointwiseconvergence (Theorem 5.1).

Given an integrable function f defined on the interval −π ≤ x ≤ π, we construct theFourier series (5.2):

12a0 +

∞∑n=1

(an cosnx+ bn sinnx),

with an, bn given by (5.10) and (5.11). As in Section 5.1.2, we let fp denote the period 2πextension of f that satisfies

fp(x) = 12

[fp(x

+) + fp(x−)]

(5.66)

at any point of discontinuity of fp (see (5.15)). We now assume that fp satisfies one of thefollowing hypotheses9:

H1 : fp is piecewise continuous.

H2 : fp is piecewise C1.

H3 : fp is piecewise C1 and continuous.

9The weakest hypothesis is that the function be square integrable, i.e.

∫ π

−π[fp(x)]2dx < ∞. Such a

function describes a signal of finite energy. If fp is square integrable, then its Fourier series converges in themean to fp. The proof of this result is advanced (see J.D. Pryce, Basic methods of linear functional analysis,Theorem 13.5(ii), p. 195).

153

Theorem 5.5: (Convergence of Fourier series)

i) If fp satisfies H1, then the Fourier series of f converges in the mean to f on any finiteinterval,

ii) If fp satisfies H2, then the Fourier series of f converges pointwise to fp(x) for all x ∈ R,

iii) If fp satisfies H3, then the Fourier series of f converges uniformly to fp on any finiteinterval.

Proofs:

i) Davis, 1989, page 145,

ii) Churchill & Brown 1978, pages 91-4, Theorem 1,

iii) Davis, 1989, page 142, Theorem 1.

We illustrate these theorems with some examples.

Example 5.10: We have seen (Example 5.1) that the Fourier series of the function

f(x) = 12(π− | x |), −π < x < π

is

14π + 2

π

∞∑n=1

cos(2n− 1)x

(2n− 1)2.

The period 2π extension fp of f is shown in Figure 5.5. Since fp is continuous and piecewiseC1, Theorem 5.6 implies that the Fourier series of f converges uniformly to fp on any finiteclosed interval, and we write

fp(x) = 14π + 2

π

∞∑n=1

cos(2n− 1)x

(2n− 1)2uniformly.

This result is in agreement with Theorem 5.3 that the sum of a uniformly convergent seriesof continuous functions is a continuous function.

Example 5.11: We have seen (Example 5.3) that the Fourier series of the function

f(x) =

14π, 0 < x < π

−14π, −π < x < 0.

is∞∑n=1

sin(2n− 1)x

2n− 1.

154

Let us now consider the period 2π extension fp of f that satisfies

fp(x) = 12

[fp(x+) + fp(x−)]

at any jump discontinuity (see (5.66)). Since fp is piecewise continuous, Theorem 5.5, parti), implies that the Fourier series of f converges in the mean to fp. In addition, since fp ispiecewise C1, Theorem 5.5, part ii), implies that the Fourier series converges pointwise to fpon R. Hypothesis H3 of Theorem 5.6 is not satisfied. One cannot infer from Theorem 5.5,however, that the Fourier series does not converge uniformly. This conclusion does in factfollow from the contrapositive of Theorem 5.3 (the Continuity Theorem).

Exercise 5.8:Discuss the convergence properties on R of the Fourier cosine series of the function f

shown in Figure 5.20. Sketch the graph of the sum function. It is not necessary to find theseries.

y

xπ2

π2

π

Figure 5.20:

5.3.2 Integration of Fourier series

We can now establish a useful property of Fourier series, namely that term-wise integrationis permissible.

Theorem 5.6: The Fourier series of a period 2π piecewise continuous function can beintegrated term-by-term, over any finite interval.

Proof: Let fp be a period 2π piecewise continuous function. Theorem 5.5 implies thatthe Fourier series of fp converges in the mean to fp. We can thus apply Theorem 5.4 (theintegration theorem) to conclude that the Fourier series of fp can be integrated term-by-termover any finite interval.

For clarity we state the theorem explicitly:

If

fp(t) = 12a0 +

∞∑n=1

(an cosnt+ bn sinnt) in the mean,

155

then ∫ x

a

fp(t)dt =

∫ x

a

12a0dt+

∞∑n=1

∫ x

a

(an cosnt+ bn sinnt)dt, (5.67)

and the convergence is uniform.

Comment: Theorem 5.6 provides the justification for the derivation of the formulae forthe Fourier coefficients in Section 5.1 (see equation (5.6)), and for the proof of Parseval’stheorem in Section 5.1.4 (see equation (5.40)).

We now give an example to show how term-by-term integration can be used to derive anew Fourier series expansion from a known one.

Example 5.12: Given that

12(π − x) =

∞∑n=1

sinnx

n, 0 < x < π,

(see # ii), Table 5.1, show that

14(π − x)2 =

π2

12+∞∑n=1

cosnx

n2, (5.68)

uniformly on 0 < x < π.

Solution: We have

12(π − t) =

∞∑n=1

sinnt

n, 0 < t < π,

using t as the independent variable. The given series converges in the mean by Theorem5.5(i), since its odd period 2π extension is piecewise continuous. We can thus use Theorem5.6 to integrate term-by-term from 0 to x, x < π:∫ x

0

12(π − t)dt =

∫ x

0

(∞∑n=1

sinnt

n

)dt

=∞∑n=1

(∫ x

0

sinnt

ndt

)(the key step!, using Theorem 5.6)

Evaluating the integrals leads to

−14(π − t)2

∣∣∣∣x0

=∞∑n=1

(−cosnt

n2

∣∣∣∣x0

),

and hence

−14(π − x)2 + 1

4π2 = −

∞∑n=1

cosnx

n2+∞∑n=1

1

n2. (5.69)

156

In Section 5.1.4 we used Parseval’s theorem to show that∞∑n=1

1n2 = π2

6, (see Example 5.5)

which, when substituted in (5.69), leads to (5.68).

Exercise 5.9:Sketch the graph of the sum function on R of the Fourier series (5.68).

Exercise 5.10:Use the Fourier series (5.68) to show that

112x(π − x)(2π − x) =

∞∑n=1

sinnx

n3, 0 < x < π.

Discuss convergence of the series on R.

Exercise 5.11:Verify the Fourier series # vii) in Table 5.1 by integrating the series # iii) in the table.

5.4 The Fourier transform & Fourier integral

In this section we introduce the Fourier transform and the Fourier integral of a non-periodicfunction f(t). We begin with the Fourier coefficients and Fourier series for a τ -periodicfunction and then let τ → +∞.

5.4.1 The definition

Given a τ -periodic function f (a periodic signal), we have seen that the Fourier coefficientscn of f are defined by

cn =1

τ

∫ τ2

− τ2

f(t)e−inω0tdt, (5.70)

where ω0 = 2πτ

is the fundamental frequency (see (5.29)). The function f can be expressedin terms of the Fourier coefficients through the Fourier series

f(t) =∞∑

n=−∞

cneinω0t, (5.71)

(see (5.27)). We refer to (5.70) as the analysis equation (one is decomposing the periodicsignal into pure sinusoids), while (5.71) is the synthesis equation (one is reconstructing theoriginal signal from the pure sinusoids – the harmonics).

We now extend these ideas to non-periodic functions f , defined on the whole real line R.We restrict our considerations to functions that are square-integrable10 on R, i.e.∫ ∞

−∞f(t)2dt <∞, (5.72)

10Engineers refer to such a function as a finite-energy signal.

157

and satisfylimt→±∞

f(t) = 0. (5.73)

It turns out that the complex Fourier coefficients

cnn∈Z,

which constitute the discrete Fourier spectrum of f , have to be replaced by a complex-valuedfunction

F (ω), ω ∈ R.

This function, called the Fourier transform of f , is defined by

F (ω) =

∫ ∞−∞

f(t)e−iωtdt, (5.74)

and is said to constitute the continuous Fourier spectrum of f . In this situation the Fourierseries (5.71) is replaced by the Fourier integral, given by

f(t) =1

2π

∫ ∞−∞

F (ω)eiωtdω. (5.75)

Derivation of equation (5.75) from equation (5.74)

We give a heuristic derivation, by applying a limiting process to equations (5.70) and(5.71). Let f(t) be a function which is zero outside a finite interval. Construct the τ -periodicextension fτ (t) of f(t), with τ sufficiently large (see the figure).

− τ

2

− τ

2 τ2

τ2

− 3τ2

3τ2

f(t) fτ (t )

t t

Figure 5.21: A function f(t), zero outside a finite interval, and its τ -periodic extension fτ (t).

It follows that for all t ∈ R,limτ→∞

fτ (t) = f(t) (5.76)

(all humps except the central one get pushed out to infinity as τ →∞).We can represent fτ (t) by its complex Fourier series

fτ (t) =∞∑

n=−∞

cneinω0t, (5.77)

158

with

cn =1

τ

∫ τ2

− τ2

fτ (t)e−inω0tdt,

and

ω0 =2π

τ. (5.78)

By definition of fτ (t) (see the figure) we can write

cn =1

τ

∫ ∞−∞

f(t)e−inω0tdt. (5.79)

Comparing (5.74) and (5.79) yields

cn =1

τF (nω0). (5.80)

In the limit as τ →∞, the fundamental frequency ω0 tends to zero, by (5.78). We introducethe increment in the frequency domain:

∆ω =2π

τ. (5.81)

We now substitute (5.80) into (5.77), and express ω0 and τ in terms of ∆ω using (5.78) and(5.77). The result is

fτ (t) =1

2π

∞∑n=−∞

F (n∆ω)ein∆ωt∆ω.

We finally take the limit as τ →∞ and hence ∆ω → 0 by (5.81). On account of (5.76) weobtain

f(t) =1

2πlim

∆ω→0

∞∑n=−∞

F (n∆ω)ein∆ωt∆ω. (5.82)

We recognize the series as a Riemann sum for the improper integral∫ ∞−∞

F (ω)eiωtdω,

so that (5.82) becomes (5.75).We conclude this section by summarizing in the following table the four key formulae of

Fourier analysis .

159

τττ-periodic functions (ω0 = 2π/τ)(ω0 = 2π/τ)(ω0 = 2π/τ) square-integrable functions on RRR

Fourier coefficients Fourier transform

cn = 1τ

∫ τ2

− τ2

f(t)e−inω0tdt F (ω) =

∫ ∞−∞

f(t)e−iωtdt

Fourier series Fourier integral

f(t) =∞∑

n=−∞

cneinω0t f(t) = 1

2π

∫ ∞−∞

F (ω)eiωtdω

5.4.2 Calculating Fourier transforms using the definition

The rectangular window function, defined by

W (t) =

1, if |t| < 1

2

0, if |t| > 12

(5.83)

and the sinc function, defined by

sinc(x) =sinx

x, x 6= 0, (5.84)

with sinc(0) = 1, play an important role in Fourier analysis and in its application to signalprocessing.

Example 5.13: Calculate the Fourier transform of

f(t) = W (t).

Solution: By definition (5.74),

160

F (ω) =

∫ ∞−∞

W (t)e−iωtdt

=

∫ 12

− 12

e−iωtdt (by (5.83))

= − 1

iω(e−iωt)

∣∣∣t= 12

t=− 12

(by the FTC)

= − 1

iω(e−

12iω − e

12iω)

= − 1

iω

[(cos 1

2ω − i sin 1

2ω)−(cos 1

2ω + i sin 1

2ω)]

(by Euler’s formula)

=2

ωsin 1

2ω

= sinc(ω

2

)(by (5.84)).

To summarize,

the Fourier transform of the window function W (t) is sinc(

12ω).

We write this result symbolically as

F(W (t)) = sinc(

12ω), (5.85)

where we think of F as the operation of taking the Fourier transform.

The graphs of W (t) and its Fourier transform are shown in the figure.

Figure 5.22: The graph of W (t) and its Fourier transform.

161

Comment: If f(t) is even, the Fourier transform F (ω) is real, and the formula (5.74) canbe written in a real form, as follows:

F (ω) =

∫ ∞−∞

f(t)e−iωtdt

=

∫ ∞0

f(t)e−iωtdt+

∫ 0

−∞f(t)e−iωtdt

=

∫ ∞0

f(t)e−iωtdt−∫ 0

∞f(−t)eiωtdt (make the change of variable t = −t)

=

∫ ∞0

f(t)(eiωt + e−iωt)dt (use f(−t) = f(t) and relabel t as t)

= 2

∫ ∞0

f(t) cosωt dt (by Euler’s formula)

In summary, if f(t) is even, the Fourier transform of f can be written in the form

F (ω) = 2

∫ ∞0

f(t) cosωt dt. (5.86)

Exercise 5.12: Verify (5.85) using (5.86).

Similarly it follows that if f is odd, then

F(f(t)) = −2i

∫ ∞0

f(t) sinωt dt.

We now give a table containing a list of simple functions and their Fourier transforms.Two of the functions are the rectangular window functionW (t) (see (5.83)) and the triangularwindow function T (t):

W (t) =

1, if |t| < 1

2

0, if |t| > 12

T (t) =

1− |t|, if |t| < 1

0, if |t| > 1.

(5.87)

162

Table 5.2: Functions f(t) and their Fourier transforms F (ω)

f(t) F (ω)

W (t) sinc(

12ω)

sinc(t) πW(

12ω)

e−|t|2

1 + ω2

1

1 + t2πe−|ω|

e−12t2

√2πe−

12ω2

T (t)[sinc

(12ω)]2

5.4.3 Properties of the Fourier transform

In this section we develop some general properties of the Fourier transform operator F ,which maps a signal f(t) to its Fourier transform F (ω):

F(f(t)) = F (ω), (5.88)

where F (ω) is defined by

F (ω) =

∫ ∞−∞

f(t)e−iωtdt. (5.89)

These properties are needed when applying Fourier analysis (e.g., solving PDEs in AMATH353), and are also useful for calculating specific Fourier transforms.

• Linearity

A fundamental property of F is that it is a linear operator:

F(f + g) = F(f) + F(g)

F(cf) = cF(f),

where c is a constant.

• Scaling formula

We know the Fourier transform of the unit gate function W (t), defined by (5.83). Howcan we calculate the Fourier transform of the gate function W (at) of width 1/a? We can dothis without resorting to the definition, by using the dilation formula.

If F(f(t)) = F (ω)

then F(f(at)) =1

aF(ωa

),

(5.90)

163

where a > 0 is a constant.

Derivation: We use the definition (5.74) of the Fourier transform, and then make a changeof variable t = at:

F(f(at)) =

∫ ∞−∞

f(at)e−iωtdt

=1

a

∫ ∞−∞

f(t)e−i(ωa )tdt

=1

aF(ωa

)( replace ω by

ω

ain (5.74))

Comment: In performing a change of variable in an improper integral one should, strictlyspeaking, use the definition, as follows. With a > 0, make a change of variable t = at:∫ ∞

0

f(at)dt = limr→∞

∫ r

0

f(at)dt = limr→∞

1

a

∫ ar

0

f(t)dt =1

a

∫ ∞0

f(t)dt.

In these notes we will omit the intermediate steps.

Example 5.14: Calculate the Fourier transform of the gate function f(t) = W (at), a > 0,and sketch both graphs. This example illustrates the effect of a scaling in time.

Solution: By (5.85)F(W (t)) = sinc

(12ω).

The dilation formula (5.90) gives

F(W (at)) =1

asinc

( ω2a

). (5.91)

The graphs are plotted in Fig. 5.23. Thus, if 0 < a 1, the gate is wide and the sinc graphis sharply peaked, with rapid oscillations. On the other hand, if a 1, the gate is narrowand the sinc graph is very flat.

Exercise 5.13: Show that

i)

F(T (at)) =1

a

[sinc

( ω2a

)]2

, (5.92)

where T is the triangular window function defined by (5.87).

ii) F(e−a|t|) =2a

a2 + ω2, (see Table 5.2).

In both cases, sketch the graphs and indicate how their shape changes as the positiveconstant a changes.

164

f(t)

1

t − 1

2a 1

2a

ω

1

a

2aπ

F(ω)

time domain

frequency domain

Figure 5.23: The graph of W (t) and its Fourier transform.

• Shift formula (aka the frequency shifting property).

Suppose you know the Fourier transform of f(t):

F(f(t)) = F (ω).

Is there an easy way to calculate the Fourier transform of

g(t) = f(t) cos(ω0t),

i.e., the given signal is multiplied11 by a signal of frequency ω0?We can calculate such a Fourier transform without resorting to the definition, by using

the marvellous shift formula.

If F(f(t)) = F (ω),

then F(eiω0tf(t)) = F (ω − ω0),

(5.93)

i.e., multiplying the given signal f(t) by a sinusoid eiω0t in the time domain causes a trans-lation (aka a shift) in the frequency domain.

Derivation: By the definition (5.74) we have

F(f(t)) = F (ω) =

∫ ∞−∞

f(t)eiωtdt. (5.94)

11This process is called amplitude modulation of the “carrier” cos(ω0t) by the signal f(t).

165

Hence,

F(eiω0tf(t)) =

∫ ∞−∞

eiω0tf(t)e−iωtdt (replace f(t) by eiω0tf(t) in (5.94))

=

∫ ∞−∞

f(t)e−i(ω−ω0)tdt (rearrange)

= F (ω − ω0) (replace ω by ω − ω0 in (5.94))

Example 5.15: Calculate the Fourier transform of a gate-pulse modulated carrier cos(ω0t).

f(t) = W (at) cos(ω0t),

where W (t) is the unit gate function, and a > 0, ω0 > 0 are constants. Sketch both graphs.

Solution: We write f(t) in complex form by using Euler’s formula:

f(t) = 12W (at)

(eiω0t + e−iω0t

).

By (5.91)

F(W (at)) =1

asinc

( ω2a

).

By linearity and the shift formula (5.93),

F(W (at) cosω0t) = 12

[F(W (at)eiω0t) + F(W (at)e−iω0t)

]=

1

2a

[sinc

(ω − ω0

2a

)+ sinc

(ω + ω0

2a

)].

The graphs are shown in Figure 5.24.

Figure 5.24: The graph of f(t) = 12W (at) cos(W0t) and its Fourier transform.

Exercise 5.14: Calculate the Fourier transform of a triangular-pulse modulated carriercos(ω0t).

f(t) = T (at) cos(ω0t),

where T (t) is the triangular window function defined by (5.87). Sketch the graphs of f(t)and F (ω).

166

5.4.4 Parseval’s formula for a non-periodic function

As you might imagine, Parseval’s formula for a periodic function

1

τ

∫ τ2

− τ2

f(t)2dt =∞∑

n=−∞

|cn|2,

can be generalized to non-periodic functions.

• Parseval’s formula:

If F(f(t)) = F (ω),

then

∫ ∞−∞

f(t)2dt =1

2π

∫ ∞−∞|F (ω)|2dω.

(5.95)

Derivation: We have the Fourier integral (5.75):

f(t) =1

2π

∫ ∞−∞

F (ω)eiωtdω (5.96)

and the Fourier transform (5.74):

F (ω) =

∫ ∞−∞

f(t)e−iωtdt. (5.97)

Multiply (5.96) by f(t) and integrate from −∞ to ∞:∫ ∞−∞

f(t)2dt =1

2π

∫ ∞t=−∞

(∫ ∞ω=−∞

f(t)F (ω)eiωtdω

)dt

=1

2π

∫ ∞ω=−∞

F (ω)

(∫ ∞t=−∞

f(t)eiωtdt

)dω (interchange the order of integration,

valid subject to suitable restrictions on f and F )

=1

2π

∫ ∞ω=−∞

F (ω)F (ω)dω (by the complex conjugate of (5.97)

=1

2π

∫ ∞ω=−∞

|F (ω)|2dω.

Example 5.16: (an impossible integral!)

Show that ∫ ∞−∞

sin2 x

x2dx = π.

167

Solution: Recognizing the integrand as [sinc(x)]2, which is related to the Fourier transformof the gate function W (t), we think of using Parseval’s formula.

Choose a = 12

in (5.91):F(W (1

2t))

= 2sinc(ω).

Thus, by Parseval’s formula:

1

2π

∫ ∞−∞

[2sinc(ω)]2dω =

∫ ∞−∞

[W(

12t)]2

dt. (5.98)

On noting that

W(

12t)

=

1 if |t| < 1

0 if |t| > 1

we obtain ∫ ∞−∞

W(

12t)2dt =

∫ 1

−1

(1)dt = 2.

Thus (5.98) leads to ∫ ∞−∞

[sinc(ω)]2dω = π.

Exercise 5.15: Use the Fourier transform

F(e−|t|) =2

1 + ω2

to show that ∫ ∞−∞

1

(1 + x2)2dx =

π

2.

5.4.5 Relation between the continuous spectrum and the discretespectrum

A τ -periodic signal f(t) has a discrete Fourier spectrum determined by the complex Fouriercoefficients cnn∈Z, with discrete frequencies nω0, where ω0 = 2π/τ is the fundamentalfrequency (see Figure 5.11). A non-periodic finite-energy signal f(t) has a continuous Fourierspectrum determined by the complex Fourier transform F (ω), where the frequency ω assumesall real values (see figures 5.21 and 5.23). In this section we show that there is a simplerelation between the discrete spectrum and the continuous spectrum.

For fixed τ > 0 consider a signal f(t) that is non-zero only on a subinterval of the interval− τ

2≤ t ≤ τ

2(see Figure 5.25 a)). The Fourier transform of f(t), as given by equation (5.74),

becomes

F (ω) =

∫ τ2

− τ2

f(t)e−iωtdt. (5.99)

168

Let g(t) be the τ -periodic function such that

g(t) = f(t), if − τ

2≤ t ≤ τ

2, (5.100)

(see Figure 5.25 b)). The Fourier coefficients of g(t) are given by

cn =1

τ

∫ τ2

− τ2

g(t)e−inω0tdt, (5.101)

with ω0 = 2π/τ being the fundamental frequency (see (5.70)). We can substitute (5.100) in(5.101) and if we then compare (5.101) with (5.99), we obtain

τcn = F (nω0).

This equation gives the desired relation between the discrete spectrum and the continuousspectrum. Note that the continuous spectrum forms an “envelope” for the discrete spectrum(compare b) and d) in Figure 5.25). As τ → ∞, the discrete spectrum (d) becomes finerand finer (the spacing → 0) and approaches the continuous spectrum b).

Figure 5.25: A non-periodic signal a) and its continuous spectrum b), and the associatedperiodic signal c) and its discrete spectrum d), scaled by the period τ .

169

5.4.6 Things are simpler in the frequency domain

We give three brief illustrations of the claim made in the title of this section.

• Signal separation

Consider a signal of finite duration in the time domain consisting of the superposition oftwo high frequency components of widely differing frequencies:

f(t) = (A1 cosω1t+ A2 cosω2t)W (at),

with ω2 ω1. The graph of this function is complicated and it is not helpful to draw it. Onthe other hand, the Fourier transform of this signal is relatively simple. Using linearity andthe shift formula we obtain

F (ω) = F(f(t)) =1

2aA1

[sinc

(ω − ω1

2a

)+ sinc

(ω + ω1

2a

)]

+1

2aA2

[sinc

(ω − ω2

2a

)+ sinc

(ω + ω2

2a

)](exercise, similar to Example 5.15). The graph of F (ω) is shown in Figure 5.26.

Figure 5.26: Signal separation in the frequency domain.

We say the signals have been separated in the frequency domain.

• Differentiation

Differentiation in the time domain becomes multiplication by iω (i.e., a purely algebraicoperation) in the frequency domain.

Differentiation formula

If F(f(t)) = F (ω)

then F(f ′(t)) = iωF (ω),

(5.102)

170

provided that f is a suitably regular function on R.

Derivation: We apply the definition (5.74) and then integrate by parts:

F(f ′(t)) =

∫ ∞−∞

f ′(t)e−iωtdt

= f(t)e−iωt∣∣∣∞−∞

+iω

∫ ∞−∞

f(t)e−iωtdt

= 0 + iωF (ω). (by the definition (5.74)).

Note that the first term is zero since we require that limt→±∞

f(t) = 0 (see equation (5.73)).

Exercise 5.16: Use (5.102) and a result from Table 5.2 to find F(te−12t2).

• Response of a linear time-invariant system

Many signal processing devices can be modelled by using a so-called linear time-invariant(LTI) system, illustrated schematically below.

outputinputsystem

fin( t) fout (t ) L

“Linear” means that the output (aka the response) is determined from the input by theaction of a linear operator L:

fout(t) = Lfin(t). (5.103)

“Time-invariant” means that if the input is translated in time, then the output is translatedin time by the same amount.

One can show that if the input to an LTI system is a sinusoid eiωt then the output is asinusoid of the same frequency, but the amplitude and phase may change, depending on thefrequency:

L(eiωt) = α(ω)eiωt, (5.104)

where α(ω) is a complex function of frequency ω, called the system function.The key point is that the system function α(ω) determines the Fourier transform Fout(ω)

of the output fout(t) in a simple algebraic way in terms of the Fourier transform of the Fin(ω)of the input fin(t):

Fout(ω) = α(ω)Fin(ω). (5.105)

Derivation: Assuming that fin(t) has a Fourier integral representation

fin(t) =1

2π

∫ ∞−∞

Fin(ω)eiωtdω,

171

we obtain, using (5.103),

fout(t) = Lfin(t) =1

2πL

(∫ ∞−∞

Fin(ω)eiωtdω

)

=1

2π

∫ ∞−∞

Fin(ω)L(eiωt)dω (by linearity)

=1

2π

∫ ∞−∞

Fin(ω)α(ω)eiωtdω (by (5.104)).

But by definition the Fourier integral representation of fout(t) is

fout(t) =1

2π

∫ ∞−∞

Fout(ω)eiωtdω.

Comparing these two expressions yields (5.105).

5.5 Digitized signals – a glimpse

We end by briefly discussing two topics relating to digitized signals, that is, signals that areobtained when a given signal is sampled at discrete times. For example, digitized signals,sampled at a frequency of around 40,000 Hz, are recorded in computer files.

5.5.1 The sampling theorem

We first discuss the Sampling Theorem.12 one of the most important results in the analysisof digitized signals.

A signal f(t) is called band-limited if its Fourier transform F (ω) is zero except in a finiteinterval, that is, if

F (ω) = 0 for |ω| ≥ Ω. (5.106)

Then fB = Ω/2π is called the cut-off frequency. The sampling theorem essentially showsthat a band-limited signal f(t) with cut-off frequency Ω/2π can be reconstructed exactlyusing only the values of f(t) at the discrete sampling times t = 0, ± π

Ω, ±2π

Ω, . . . i.e. using

a time spacing ∆t = π/Ω, or equivalently, a sampling frequency fs = Ω/π. Thus, providedthe sampling frequency is twice the cut-off frequency (fs = 2fB), one can reconstruct thecontinuous signal f(t) exactly. This is another situation in which the sinc function

sinc(x) =sinx

x, x 6= 0; sinc(0) = 1,

comes into play.

12The origins of the Sampling Theorem can be traced to Cauchy and Borel in the nineteenth century,followed by E.T. Whittaker in 1915, motivated by interpolation theory. The theorem was rediscoveredby Nyquist in 1928 in the context of signal processing, while working at Bell Labs, and was subsequentlyincorporated into information theory by Claude Shannon in the late 1940’s.

172

Theorem 5.7 (Sampling theorem):If f(t) is a band-limited signal with cut-off frequency Ω/2π, then

f(t) =∞∑

n=−∞

f(nπ

Ω

)sinc(Ωt− nπ). (5.107)

Proof: Since f(t) is band-limited, i.e. (5.106) holds, equation (5.75) for the Fourier integralbecomes

f(t) =1

2π

∫ Ω

−Ω

F (ω)eiωtdω. (5.108)

We now take the unexpected step of writing the function F (ω) as a complex Fourier serieson the interval −Ω < ω < Ω. Replacing t by ω in (5.71), and choosing the period τ to be2Ω, we get

F (ω) =∞∑

n=−∞

cneinπω

Ω , −Ω < ω < Ω. (5.109)

The coefficients cn are given by (5.70), again with t replaced by ω, and τ = 2Ω:

cn =1

2Ω

∫ Ω

−Ω

F (ω)e−inπω

Ω dω. (5.110)

Comparing (5.108) with t = −nπΩ

and (5.110) leads to the conclusion that

cn =π

Ωf(−nπ

Ω

).

We now substitute this result in (5.85), obtaining

F (ω) =π

Ω

∞∑n=−∞

f(−nπ

Ω

)einπω

Ω

=π

Ω

∞∑n=−∞

f(nπ

Ω

)e−

inπωΩ , (5.111)

for −Ω < ω < Ω. In the last step we simply replaced n by −n. We finally obtain the desiredexpression for f(t) by substituting (5.111) in (5.108) and rearranging:

f(t) =1

2Ω

∞∑n=−∞

f(nπ

Ω

)∫ Ω

−Ω

e−inπω

Ω eiωtdω.

Evaluating the integral and using Euler’s identity gives

1

2Ω

∫ Ω

−Ω

e−inπω

Ω eiωtdω =1

2Ω

∫ Ω

−Ω

ei(Ωt−nπ)

Ωωdω

= sinc (Ωt− nπ)

(fill in the details as an exercise), leading to the desired result (5.107).

173

5.5.2 The discrete Fourier transform (DFT)

The problem of numerically computing the Fourier coefficients cn of a τ -periodic functionf(t) leads to another aspect of Fourier analysis, the discrete Fourier transform, as follows.It is convenient to write the formula for cn (see (5.70)) as an integral over the interval [0, τ ],instead of

[− τ

2, τ

2

]:

cn =1

τ

∫ τ

0

f(t)e−2πint/τdt. (5.112)

Given n, choose N > n, and consider a uniform partition of the interval [0, τ ] into Nsubintervals of length

∆t =τ

N,

and with left end points

tk = k( τN

), k = 0, 1, . . . , N − 1. (5.113)

The integral is then approximated by the Riemann sum associated with this partition, and(5.112) becomes

cn ≈1

N

N−1∑k=0

f

(kτ

N

)e−2πikn/N . (5.114)

This equation approximates the Fourier coefficients cn, n = 0, 1, . . . , N − 1 in terms of thevalues of f(t) at the N discrete points given by (5.113). We can think of the signal as havingbeen sampled at these discrete time values. We introduce the notation

f [k] = f

(kτ

N

), k = 0, 1, . . . , N − 1.

These N numbers then represent a digitized signal. Motivated by (5.114) we define thediscrete Fourier transform of f [k] (DFT) by

F [n] =1

N

N−1∑k=0

f [k]e−2πikn/N , (5.115)

n = 0, 1, . . . , N − 1. We can think of this formula as approximating the first N Fouriercoefficients cn of a τ -periodic signal f(t), or as giving the exact Fourier coefficients F [n] inthe frequency domain of a digitized signal f [k] in the time domain. It can be shown thatthe inverse of the DFT (5.115) is given by

f [n] =N−1∑k=0

F [k]e2πikn/N . (5.116)

In mathematical terms, the DFT (5.115) is a linear transformation from CN to CN , i.e.it maps a vector

f = (f [0], . . . , f [N − 1])

174

onto a vectorF = (F [0], . . . , F [N − 1]).

We can then write (5.115) in matrix form as

F =1

NFN f , (5.117)

where the N ×N matrix FN is given by

FN =

1 1 1 · · · 1

1 ω ω2 · · · ωN−1

1 ω2 ω4 · · · ω2N−2

......

......

1 ωN−1 ω2N−2 · · · ω(N−1)(N−1)

,

withω = e−2πi/N .

This matrix is called the Fourier matrix.In exploring large sets of data in many areas including digital signal analysis, it is nec-

essary to calculate the Fourier coefficients F [n], n = 0, 1, . . . , N − 1, as given by (5.115), forlarge values of N . Doing the matrix multiplication in (5.117) directly requires N2 opera-tions, which is time consuming for large N . The Fast Fourier Transform (FFT), introducedin 1965 by Cooley and Tukey, exploits the symmetry properties of FN to do the calculationusing O(N logN) operations, a significant saving. The FFT is one of the topics studied inthe numerical computation course AM 371/CM 271/CS 371.

Reference (a famous mathematical paper):Cooley, J.W. and Tukey, J.W. (1965) An algorithm for the machine computation of complexFourier series, Math. Comp. 19, 297-301.

175

References

1. Churchill, R.V. and Brown, J.W., 1978, Fourier Series and Boundary Value Problems,McGraw-Hill.

2. Colley, S.J., 2002, Vector Calculus, Second Edition, Prentice Hall.

3. Davis, H.F. and Snider, A.D., 1995, Introduction to Vector Analysis, Seventh Edition,Wm. C. Brown Communications, Inc.

4. Davis, H.F., 1989, Fourier Series and Orthogonal Functions, Dover Publications.

5. Flanigan, F.J. and Kazdan, J.L., 1990, Calculus 2, Linear and Nonlinear Functions,Second Edition, Springer-Verlag.

6. Korner, T.W., 1988, Fourier Analysis, Cambridge University Press.

7. Kammler, D.W., 2000, A First Course in Fourier Analysis, Prentice Hall.

8. Marsden, J.E. and Tromba, A.J., 1988, Vector Calculus, Third Edition, W.H. Freeman.

9. Taylor, A.E. and Mann, W.R., 1983, Advanced Calculus, Third Edition, Wiley.

177

Problem Sets

179

Problem Set 1:

Vector-valued functions

(a) Curves

1. The motion of a particle is given by x = g(t), where t is time and g : R→ R3 is givenby

g(t) = (t sin t, t cos t,√

3t), 0 ≤ t ≤ 4π.

(a) Show that the motion occurs along a quadric surface. Use this to sketch thetrajectory.

(b) Find the velocity, speed, and acceleration of the particle.

2. The path of a particle is given by x(t) = ti + tj + t2k.

(a) Sketch the path by first eliminating the time t.

(b) Suppose the particle flies off the path at t = 1 sec. Where will it be 2 secondslater assuming that it is subject to no further force?

3. The path of a particle is given by x(t) = (a cos t, b sin t, ct), where a, b, c > 0, and t istime.

(a) Show that the path lies on a quadric surface and sketch it.

(b) Show that the acceleration always points towards the z-axis.

4. Sketch the curves in R2 defined by x = g(t), where

(i) g(t) = (cos(t3), sin(t3)), 0 ≤ t ≤ (2π)13 .

(ii) g(t) = (cos3 t, sin3 t), 0 ≤ t ≤ 2π.

Find all t such that g′(t) = 0. If possible, find a parameterization

x = g(τ), τ1 ≤ τ ≤ τ2,

such that g is C1 and g′(τ) 6= 0 for all τ ∈ [τ1, τ2].

5. Find parametric equations for the curve of intersection of y + z = 1 and x2 + z2 = 4.

6. Find a vector-valued function g : R → R3 to describe the curve of intersection ofx2 + y2 + z2 = 6 and x+ y + z = 0.

7. Consider the vector-valued function g : R→ R3, defined by

g(t) = a + (sin t)b,

where a,b ∈ R3 are constants. Describe the curve x = g(t) in R3, and give a sketch.Interpret physically in terms of the motion of a particle.

181

8. The motion of a particle in the plane is described by x = (cos g(t), sin g(t)), whereg : R → R is a C1 function. Describe the motion, first assuming that g′(t) > 0 for allt, and then assuming that g′(t) changes sign at t = t1.

9. Prove that if a particle moves (in 3-space) with constant speed, then its accelerationvector is either zero or orthogonal to its velocity vector.

10. Prove that if a particle moves on the surface of a sphere centred at (0, 0, 0), then itsvelocity vector is orthogonal to its position vector. Is the converse true?

11. A curve joining two points P1 and P2 in R3 is given by

x = g(t), t1 ≤ t ≤ t2,

and also byx = g(τ), τ1 ≤ τ ≤ τ2,

whereτ = h(t), with h′(t) > 0,

and g, g, h are C1 functions.

a) Prove that ∫ t2

t1

‖ g′(t) ‖ dt =

∫ τ2

τ1

‖ g′(τ) ‖ dτ.

b) Interpret this result geometrically.

12. A disc rotates back and forth with angular velocity (cos t) rad/sec. An insect starting 1cm from the center of the disc at time t = 0 crawls outward at a rate of 2t cm/sec. Findthe position, velocity and speed of the insect after 2π seconds. Give the componentsof position and velocity relative to a fixed Cartesian coordinate system.

(b) Vector fields

13. Consider the vector field F : R2 → R2 given by F(x, y) = (x, x2). Find the field linesand sketch the field portrait.

14. The electric field E due to an electric charge q in the plane is given by

E(x, y) = kq

(x

(x2 + y2)3/2,

y

(x2 + y2)3/2

),

where k is a constant. Find the field lines and sketch the field portrait.

15. A fluid flow in R3 has velocity field v = (2x, z,−z2).

(a) Find the equation of the flow line which passes through an arbitrary point (x0, y0, z0)at time t = 0.

182

(b) A fluid particle is at position (e2, 0, 1) at time t = 0. Find its position at timet = 1.

16. Given the vector field F(x) = (−y, x,−z), find the equation of the field line throughan arbitrary point (x0, y0, z0). In particular, give a vector-valued function g(t) thatsatisfies

g(0) = (x0, y0, z0).

Sketch the exceptional field lines and some typical field lines of the vector field.

17. Suppose a vector field F on R3 is given by the gradient of a scalar field ψ, according toF = −∇ψ. Suppose also that x = g(t) is a field line of F. Show that h(t) = ψ(g(t))is a decreasing function of t.

18. Sketch the field lines of the vector field

F = B× r + λB,

where B is a constant vector field and λ is a constant. Consider the three cases λ > 0,λ = 0, λ < 0.

(c) The Chain Rule

19. Suppose that g : R → Rn is a vector-valued function and that f : Rn → R is a scalarfield. Define h : R→ R by

h(t) = f(g(t)).

Show that if f and g are of class C1, then

h′(t) = ∇f(g(t)) ··· g′(t).

20. Suppose that f is a vector-valued function on R and g is a real-valued function of onevariable. Define h : R→ Rn by

h(t) = f(g(t)).

Show that if f and g are of class C1, then

h′(t) = g′(t)f ′(g(t)).

21. Suppose that g is a scalar field on Rn and f is a real-valued function of one variable.Define h : Rn → R by

h(x) = f(g(x)).

Show that if f and g are of class C1, then

∇h(x) = f ′(g(x))∇g(x).

183

22. Let u = F(x) be a C1 map of the xy-plane into the uv-plane. Consider a smooth curvex = x(t) in the xy-plane. Suppose that F maps this curve into the curve u = u(t) inthe uv-plane. Show that the tangent vectors are related according to

u′(t) = DF(x(t))x′(t).

where DF is the derivative matrix of F.

Bonus questions

B1. Consider the family of curves

x = g(t) = (sin(t− α), sin t), 0 ≤ t ≤ 2π.

Describe how the curve changes as the constant α assumes values from 0 to 2π. Givea sketch.

Comment: These curves could be displayed on an oscilloscope by feeding in signalssin(t− α) and sin t to the horizontal and vertical inputs respectively.

B2. Can the Mean Value Theorem be generalized to vector-valued functions, i.e. is thefollowing statement true?

If g ∈ C1[a, b], then there exists a number c with a < c < b such that

g(b)− g(a) = (b− a)g′(c).

B3. Give a curve C : x = g(t), a ≤ t ≤ b in R2, where g ∈ C[a, b] such that C does not havefinite arclength.

B4. The dipole vector fieldThe electric potential φ` due to a pair of charges, one positiveand one negative, a distance ` apart as shown, is

φ`(x) =e√

x2 + y2 +(z − `

2

)2− e√

x2 + y2 +(z + `

2

)2.

We assume that the charge e and separation ` are related by

e` = µ,

where µ is a give constant. The potential for dipole of strength µ is obtained by letting`→ 0+ while keeping µ fixed.

184

i) By evaluating lim`→0+

φ`(x), show that the potential for a dipole of strength µ is

φ(x) =µz

r3,

where r =√x2 + y2 + z2.

ii) Describe the field lines of the dipole vector field F = −∇φ qualitatively, and givea sketch. It is helpful to show that each field line lies in a vertical plane throughthe z-axis.

B5. Sketch the field lines of the vector field F(x) = (sin y,− sinx) in R2. It is not necessaryto find the field lines explicitly.

B6. Consider the family of curves

x = g(t) = (sinωt, sin t), 0 ≤ t < +∞,

where ω ∈ R is given. Describe the curves qualitatively.

B7. Consider a charged particle of mass m and charge e moving in a constant electric fieldE = (0, E, 0) and constant magnetic field H = (0, 0, H). It is assumed that the speedv of the particle satisfies v/c 1, where c is the speed of light (i.e. non-relativisticmotion), and that E/H 1, i.e. the electric field is much weaker than the magneticfield. If the initial velocity is in the x-direction it can be shown that the path of theparticle is

x = g(t) =cE

ωH(ωt− ε sinωt, ε(1− cosωt), 0),

where ω = eH/mc, and ε is a dimensionless parameter that is determined by the initialvelocity:

g′(0) = cE

H(1− ε, 0, 0).

Sketch the path of the particle, and show how it depends qualitatively on the param-eter ε.

Reference: Landau L.D. and Lifshitz, E.M., The Classical Theory of Fields, FourthRevised English Edition, Section 22.

B8. Let m be the mass of a planet and M that of the sun, both idealized to be pointparticles in space. Newton’s law of gravitation states that the force between thesebodies is given by the vector field

F = −GmMr3

r,

where r is the position vector of the planet relative to the sun, r =‖ r ‖ and Gis Newton’s gravitational constant. The motion of the planet relative to the sun isdescribed by Newton’s second law:

md2r

dt2= F.

185

i) Show that the planet moves in a fixed plane through the centre of the sun.

Hint: Show that r and drdt

lie in a fixed plane.

ii) Let A(t) be the area swept out by the radius vector from time t = 0 to time t.Prove that that rate of change of area is constant i.e., the radius vector sweepsout equal areas in equal times. This result is known as Kepler’s second law.

Hint: Show that dAdt

= 12ρ2 dφ

dt, where ρ and φ are polar coordinates in the fixed

plane referred to in i).

B9. The force exerted on a particle of charge q and mass m moving in a magnetic field Bis

F = qv ×B.

i) Assuming that the motion is non-relativistic (v c) so that Newton’s Second Lawof Motion is valid, show that the kinetic energy 1

2mv2 of the particle is constant.

ii) Show that if the magnetic field is constant, then the component of the velocity ofthe particle in the direction of B is constant.

iii) Suppose that the magnetic field is constant. Choose coordinates so that B actsin the z-direction, i.e. B = Bk, where B is a constant. Let

R = r− (r · k)k,

i.e. R is the projection of the position of the particle into the xy-plane. Showthat

dR

dt= ω(R−R0)× k,

where R0 is a constant vector and

ω =qB

m.

Hence describe the motion of the charged particle.

Hint: Show that ‖ R−R0 ‖ is constant.

186

Problem Set 2:

Line Integrals and Green’s Theorem

1. Consider the scalar field f : R3 → R given by f(x, y, z) = 2x− y+ z−1, and the curve

C given by x = g(t) = (cos t, sin t, t). Compute

∫C

fds.

2. Evaluate the line integral

∫C

F · dx from (1, 0, 0) to (1, 0, 4), where F = xi− yj + zk,

(a) along the line segment joining (1, 0, 0) and (1, 0, 4)

(b) along the helix x(t) = (cos 2πt, sin 2πt, 4t).

3. Compute the line integral

∫C

F · dx, where F = (x + y)i + (y + z)j + (z + x)k, and C

is the intersection of the plane x+ y + z = 1 with the cylinder x2 + y2 = 1. Orient Cclockwise as viewed from above.

4. In each case calculate the work done by the force field F acting on a particle movingon the curve C.

(a) F = −y2i + x2j, C is the upper half of the ellipse x2

a2 + y2

b2= 1, from (a, 0) to

(−a, 0),

(b) F = yi + zj + xk, C is the curve x = (a cos t, a sin t, bt), 0 ≤ t ≤ 2π, and a, b arepositive constants.

(c) F = (y2 − z2)i + 2yzj− x2k, C is the curve of intersection of y = x2 and z = x3

joining the points (0, 0, 0) and (1, 1, 1).

(d) F = 2xi− 3yj + z2k, C is the line segment from (1, 0, 0) to (0, 1, π2).

5. A force field F : R3 → R3 is given by F(x) = (x3, y, z). Compute the work done by Fon a particle moving along the curve given by x = g(t) = (0, b cos t, b sin t), 0 ≤ t ≤ π.Explain the result geometrically.

6. Show by inspection that the velocity field

v(x) =(ye−2xy, xe−2xy

)is the gradient of a scalar field φ(x). Hence evaluate

∫C

v · dx along the curve C given

by x(t) = (t2 − t, t2 + t), 1 ≤ t ≤ 2.

7. (i) Consider the closed curve C, x(t) = (a cos t, b sin t), where a and b are positiveconstants, and the velocity field v = (−y, x). By sketching the velocity field atvarious points of C, predict whether the circulation of v around C is positive,negative or zero. Verify your prediction.

(ii) Repeat (i) with v = (x, y).

8. A piece of wire has the shape of the circle x2 + y2 = a2, and the mass density at (x, y)is |x|+ |y| (mass per unit length).

(i) Compute the mass of the wire.

(ii) Compute the average mass density of the wire.

9. Imagine a wire located at the intersection of x2 + y2 + z2 = 1 and x+ y+ z = 0, whosedensity depends on position according to φ(x) = x2 gm/unit length. Show that themass of the wire is 2

3π gm.

10. (i) Consider a continuous function f : R2 → R and a curve C parametrized in polarcoordinates by r = r(θ), θ1 ≤ θ ≤ θ2. Show that

∫C

fds =

∫ θ2

θ1

f(r cos θ, r sin θ)

√r2 +

(dr

dθ

)2

dθ.

(ii) Compute the arclength of the curve r = 1 + cos θ, 0 ≤ θ ≤ 2π.

11. Suppose that F : Rn → Rn is a continuous vector field that satisfies ‖ F(x) ‖≤ M forall x ∈ Rn, where M is a constant, and suppose that C is a C1 curve of length `. Provethat ∣∣∣∣∣∣

∫C

F···dx

∣∣∣∣∣∣ ≤M`.

It is necessary to begin by using the definition of line integral.

12. Decide whether Green’s theorem can be used to evaluate the line integral∫∂D

ln(x+√x2 + y2) dx ,

(see equation (2.16) for this notation) where ∂D is the boundary of the region D:

(i) D is the disc x2 + y2 ≤ 1, (ii) D is the disc (x− 2)2 + y2 ≤ 1,

(iii) D is the disc (x+ 2)2 + y2 ≤ 1.

13. Use Green’s theorem to evaluate the line integral

∫∂D

xy2dx+ 2x2y dy, where ∂D is the

ellipse 19x2 + 1

4y2 = 1, oriented suitably.

14. Show that the area of the region enclosed by the hypocycloid x2/3 +y2/3 = a2/3 is 38πa2.

15. Show that the area of the region enclosed by the curve x = (a sin 2t, a sin t),0 ≤ t ≤ 2π, is 8

3a2.

188

16. (a) The area of a region D ⊂ R2 bounded by a simple closed curve ∂D is given bythe line integral

A(D) = 12

∫∂D

(xdy − ydx).

Derive this formula using Green’s Theorem.

(b) A region D in the plane is defined in polar coordinates by r = f(θ), α ≤ θ ≤ β.Use (a) to show that

A(D) = 12

∫ β

α

f(θ)2dθ.

17. (a) Verify, by evaluating the line integral , that∫C

xdx+ xy dy = 0

for every circle C centered on the origin (see equation (2.16) for this notation).

(b) Explain this result geometrically by using Green’s theorem.

(c) Prove that

∫C

xdx+ xy dy is not independent of path in R2.

18. (a) Show that the vector field

F(x) = (sin xy + xy cosxy) i + x2 cosxy j

is conservative in R2, and find a potential function φ.

(b) Evaluate

∫C

F · dx,

(i) along the straight line from (1, 0) to (π, 1),

(ii) counterclockwise around the circle x2 + y2 = 1.

19. Determine whether the following vector fields are conservative or not; if they are, findthe corresponding potential:

(i) F(x) = (3x2y, x3) (ii) F(x) = (2xey + y, x2ey + x− 2y).

20. a) In Newton’s theory of gravity, the earth’s gravitational field is given by

g(r) = −GMr3

r,

where r = (x, y, z) is the position vector relative to the centre of the earth,r =‖ r ‖, M is the mass of the earth and G is the gravitational constant.

189

A rocket of mass m is launched from position (0, 0, R) and travels to (0, 0, R+ h)in a time T , along the straight line path C1. A similar rocket travels along thespiral path C2 given by

r = h(t) =

(R +

h

Tt

)(− sin

2πt

T, 0, cos

2πt

T

),

with 0 ≤ t ≤ T .

By evaluating the appropriate line integrals, show that the work done by thegravitational field along C1 equals the work done along C2.

Z

x

hC

1C

1

C2

b) Consider the gravitational field in a).

i) Verify that the field satisfies an inverse square law, i.e. ‖ g(x) ‖ is inverselyproportional to the square of the distance from the centre.

ii) Show that the gravitational field is conservative by finding a potential φ.

iii) Hence, verify the expression for the work done in a).

21. A “central” (or radial) force field in Rn has the form

F = f(r)r, r 6= 0 ,

where r = (x1, . . . , xn) and r = ‖r‖. Show that this force field is conservative inRn − 0 by finding a potential function φ.

22. Let f : R2 → R, g : R2 → R be of class C1 in D ⊂ R2. Prove that∫C

(f∇g) · dx = −∫C

(g∇f) · dx

for any piecewise smooth simple closed curve C in D.

23. Let D be a region bounded by a piecewise smooth simple closed curve C orientedcounterclockwise. Let f, g be of class C2 in an open set containing D. Prove that∫

C

f∇g · dx =

∫∫D

∂(f, g)

∂(x, y)dxdy .

190

24. Discuss whether or not the following sets are (i) connected (ii) simply-connected.

(a) R2 − (x, y) | y = 0, x ≤ 1,(b) R2 − (x, y) | y = 0, |x| ≤ 1,(c) (x, y) |x2 + y2 < 4 ∩ (x, y) |x2 + 4y2 > 4,(d) R3 − (x, y, z) | z = 0, x2 + y2 = 1.

25. Consider a frictional force which is constant in magnitude, and acts to oppose themotion of a particle. Show that the work done against friction is proportional to thelength of the path.

26. A particle moves according to Newton’s Second Law of Motion under the influence ofa conservative force field

F = −∇V,

where scalar field V is the potential of the field.

Prove that if the particle has constant speed, then it moves on an equipotential surface.

27. The force exerted on a particle of charge q moving with velocity v in a magnetic fieldB is

F = qv ×B.

Show that the work done by the magnetic field on the particle is zero.

Bonus questions

B1. The Generalized Green’s Theorem

Let ∂D be a simple closed curve in R2 enclosing a region D. Suppose that D containsn non-overlapping holes H1, . . . , Hn, with boundaries ∂H1, . . . , ∂Hn oriented as shown.Let F be a vector field which is C1 on D −

⋃ni=1Hi and on the boundary of this set.

Prove that ∫∂D

F • dx =

∫∫D−

⋃ni=1Hi

ω(F)dxdy +n∑i=1

∫∂Hi

F • dx

where

ω(F) =∂F2

∂x− ∂F1

∂y.

Comment: This theorem is useful when the vector field F is C1 on D except at afinite number of points, called singularities of F. One then excludes the singularitiesby deleting a neighbourhood of each singularity.

191

fig2ps2

∂H1

∂H2

D

€

∂ D

B2. An application of Green’s theorem to heat transfer

Consider a sheet of metal, with both faces insulated, represented by the plane regionD. The boundary ∂D of the sheet is kept at a prescribed temperature (for example,part of the boundary is in contact with ice and part with boiling water). The sheetis allowed to reach thermal equilibrium, which means that the temperature u(x, y) atposition (x, y) on the sheet no longer changes with time. Under these circumstances,it can be shown using Fourier’s law of heat transfer (see page 18) that the temperatureu(x, y) satisfies Laplace’s equation:

∂2u

∂x2+∂2u

∂y2= 0.

In order for this model to be physically reasonable, it should have the property thatif the temperature is constant all along the boundary, then, when the sheet reachesthermal equilibrium, the temperature at all points of the sheet must equal this constantvalue. Prove that the model satisfies this property.

Hint: Argue that the constant value can be chosen to be zero, and then show that∫∫D

‖ ∇u ‖2 dxdy = 0.

192

Problem Set 3:

Surfaces and Surface Integrals

1. Sketch each of the following surfaces x = g(u, v), where g : Duv → R3 is given below .Find the tangent plane at the indicated point.

i) g(u, v) = (cos v sinu, 1 + sin v sinu, cosu), Duv = [0, π]× [0, 2π]; at g(π4, π

3

)ii) g(u, v) = (sin v, u, cos v), Duv = [−1, 3]×

[−π

2, π

2

]; at g

(1, π

3

).

2. Sketch each surface x = g(u, v), where g : Duv → R3 is given below. Discuss whetherthe surface is smooth.

i) g(u, v) = (u, |u|, v), Duv =

(u, v)∣∣ |u| ≤ 1, 0 ≤ v ≤ 2

,

ii) g(u, v) =(u cos v, u sin v, ae−|u−b|

),

where Duv =

(u, v)∣∣u > 0, 0 ≤ v ≤ 2π

, and a, b are positive constants.

3. Sketch the “helicoid” x = g(u, v) = (u cos v, u sin v, v), (u, v) ∈ [0, 1] × [0, 2π], andfind its area.

4. Give the function g : R2 → R3 that describes the cone z =√x2 + y2, using as

parameters (u, v) = (r, θ),where r and θ are polar coordinates in R2. Then determinethe area of the portion of the cone lying:

(i) below z = 1,

(ii) inside the cylinder x2 + y2 = 2ay, a > 0.

5. Give a vector-valued function to describe the quadric surface x2 + y2 − z2 = 1, with|z| ≤ 3

4.

Hint: Use the identitycosh2 u− sinh2 u = 1.

for the hyperbolic functions. Recall that

coshu = 12(eu + e−u), sinh = 1

2(eu − e−u).

6. The surface of a torus is generated by revolving the circle

(y − b)2 + z2 = a2, x = 0,

where b > a > 0, about the z-axis.

(a) Show that the surface is given by x = g(u, v), where

g(u, v) = ((b+ a cosu) cos v, (b+ a cosu) sin v, a sinu),

and give the domain in the uv-plane.

(b) Calculate the surface area of the torus.

7. A surface of revolution is created by rotating a curve z = f(u), 0 ≤ a ≤ u ≤ b, aboutthe z-axis through an angle v, with 0 ≤ v ≤ 2π.

(a) Find a function g : R2 → R3 which describes the points x = g(u, v) on thissurface.

(b) Find an expression for the surface area of this surface.

8. Evaluate the surface integral

∫∫Σ

fdS where

(i) f(x, y, z) = xz and S is the part of the cylinder x2 + y2 = 1 between the planesz = 0 and z = x+ 2.

(ii) f(x, y, z) = x2 and S is the boundary of the region D in R3 inside the conez2 = x2 + y2 and between the planes z = 1 and z = 2.

9. a) A sphere of radius b centred at the origin has a silver coating of surface density

ρ(x, y, z) = k[1 + ε

(zb

)2]

mass per unit area, where k and ε are positive constants.

Calculate the total mass of silver.

b) Referring to part a), find the average density of silver. At what points on thesphere does the density equal its average value?

10. Using symmetry where possible, evaluate the surface integrals

(i)

∫∫Σ

xdS, (ii)

∫∫Σ

xzdS, (iii)

∫∫Σ

x2dS, (iv)

∫∫Σ

z2dS,

where S is the paraboloid y = x2 + z2, y ≤ 1.

11. Calculate the flux

∫∫Σ

F·n dS and circulation

∫∂Σ

F·dx for each vector field F, where

Σ is the triangular piece of the plane x + 2y + 3z = 6 with vertices on the coordinateaxes, and ∂Σ is the boundary of Σ, oriented counterclockwise as seen from (0, 0, 4).

(i) F = xi + yj + zk (ii) F = y2i + j + x2k.

12. Evaluate

∫∫Σ

r···n dS, where r = (x, y, z) and Σ is the unit sphere centred on the origin.

13. Suppose T (x, y, z) = x2 + y2 + z2 represents the temperature in a region of spacecontaining the origin of the coordinate system. Compute the heat flux across the unitsphere.

14. Compute the flux of the vector field F = (x3, y3, z3) across the unit sphere.

194

15. Compute the circulation of the vector field

F(x, y, z) = (x, x+ y, x+ y + z)

around the curve C defined by the equations

x2 + y2 = 1 and z = y.

16. Let S be the unit sphere centred on the origin. In each case, by sketching the vectorfield F, make a conjecture as to whether the flux of F across S in the outward directionis positive, negative or zero. Verify your conjecture by evaluating the surface integral

(i) F = (0, 0, z) (ii) F = (0, 0, z2) (iii) F = (−y, x, 0) .

17. Suppose that v = (0, y, 0) is the velocity field of a fluid and let S be the surface definedby y = x2 + z2, with y ≤ 1. Calculate the volume of fluid crossing S in unit time, inthe direction of increasing y.

18. Suppose that Σ is a smooth surface with area S(Σ) and suppose that F is a continuousvector field on R3 that satisfies ‖ F(x) ‖≤M for all x ∈ Σ. Prove that∣∣∣∣∣∣

∫∫Σ

F·n dS

∣∣∣∣∣∣ ≤MS(Σ).

It is necessary to begin by using the definition of surface integral.

Bonus questions

B1. Non-Euclidean geometry

Let Σ be a disc of ‘radius’ R (see diagram) on the surface of a sphere of radius b, where0 < R < πb. Show that the surface area of the disc is

S = πR2f(Rb

),

where f is a function which you should determine. Show that 0 < S < πR2 for0 < R < πb.

Comment: This result shows that the intrinsic geometry on a sphere is different fromthat on the plane. The intrinsic geometry of surfaces is discussed in detail in AM 333.

B2. Let x = g(u, v), (u, v) ∈ Duv and x = g(p, q), (p, q) ∈ Dpq be parametrizations of asurface Σ.

195

AAAAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAA

arclength R

i) What condition on the mapping (u, v) = h(p, q) that relates p, q and u, v willensure that the orientation of the surface is the same for both parametrizations?

ii) Verify that the definition of the surface integral of a vector field is independent ofthe parametrization.

B3. Let x = h(u), a ≤ u ≤ b, be a simple closed curve lying in a horizontal plane, and letb be a point not on the plane.

i) Describe the surface with parametrization

x = g(u, v) = b + vh(u),

anda ≤ u ≤ b, 0 ≤ v ≤ 1.

ii) Describe the surface with parametrization

x = g(u, v) = h(u) + vb,

anda ≤ u ≤ b, 0 ≤ v ≤ `.

B4. Describe the surface corresponding to the following parametrization:

x = g(u, v) = p(u) + vq(u),

where

p(u) = (cosu, sinu, 0),

q(u) =(cos u

2

)p(u) +

(sin u

2

)k,

and0 ≤ u ≤ 2π, −1

2≤ v ≤ 1

2.

196

B5. i) Find a parametrization for a trefoil knot in R3.

ii) Discuss whether the soap film surface bounded by the trefoil knot is orientable.

iii) The trefoil knot can be deformed continuously into the knot as shown. Discusswhether the soap film surface bounded by this knot is orientable.

197

Problem Set 4:

Gauss’ Theorem and Stokes’ Theorem

1. Let F = (x2 + y2 + z2)n(xi + yj + zk), and let Ω be the sphere of radius b centered onthe origin.

(i) Calculate the outward flux of F across the boundary ∂Ω.

(ii) Calculate the divergence of F, ∇ • F.

(iii) Referring to (i) and (ii), what is special about the value n = −32?

2. Verify the following properties of the divergence.

(i) Sum of two vector fields:

∇ · (F + G) = ∇ · F +∇ ·G .

(ii) Product of a scalar field and a vector field:

∇ · (fF) = f∇ · F + F · ∇f .

(iii) Vector product of two vector fields:

∇ · (F×G) = G · (∇× F)− F · (∇×G) .

3. Verify the following properties of the curl.

(i) Sum of two vector fields:

∇× (F + G) = ∇× F +∇×G .

(ii) Product of a scalar field and a vector field:

∇× (fF) = f∇× F +∇f × F .

(iii) Vector product of two vector fields:

∇× (F×G) = (∇ ·G)F− (∇ · F)G + (G · ∇)F− (F · ∇)G

4. Verify the following “zero identities”:

(i) ∇× (∇f) = 000, for any C2 scalar field f .

(ii) ∇ · (∇× F) = 0, for any C2 vector field F.

199

5. Let f, g be C2 scalar fields on R3.

(i) Verify that ∇× (f∇g) = ∇f ×∇g.

(ii) Simplify ∇× (f∇g + g∇f).

(iii) Verify that ∇ · (f∇g) = f∇2g +∇f · ∇g.

6. a) Verify the “curl of a curl” identity

∇× (∇× F) = ∇(∇·F)−∇2F.

where ∇2 is the Laplacian operator.

b) Use Maxwell’s equations (see Section 4.1.1) to show that if the charge densityε and current vector J are zero, the electric field vector E satisfies the “waveequation”

∂2E

∂t2− c2∇2E = 0.

The magnetic field H satisfies a similar equation.

7. Calculate the flux of the vector field F = (x3, y3, z3) outward across the unit spherex2 + y2 + z2 = 1 using Gauss’ Theorem.

8. Evaluate

∫Σ

∫F · n dS in the following cases. Use Gauss’ Theorem and/or symmetry

whenever convenient.

(a) F = (x2, y2, z2), Σ = ∂Ω, Ω = (x, y, z) :| x | ≤ 1, | y | ≤ 1, | z | ≤ 1,n is theoutward unit normal.

(b) F = (y, x+2, z), Σ is the part of the cylindrical surface x2 +y2 = 9, 0 ≤ z ≤ 4,x ≥ 0, y ≥ 0, n points away from the z-axis.

(c) F = ∇f, f(x, y, z) = xyz + 5, Σ is the sphere x2 + y2 + z2 = 9, n is the outwardunit normal.

9. Show that the volume V of the region enclosed by a closed piecewise smooth surfaceΣ can be expressed as

V = 13

∫∫Σ

r · n dS,

where n is the unit outward normal and r = (x, y, z).

10. The electric potential u(x) in R3 due to a point charge q at x0 is given by

u(x) =q

‖x− x0‖.

(i) Calculate the electric field E = −∇u, and draw the field lines.

200

(ii) Verify that∫∫Σ

E ·n dS = 4πq, where Σ is any piecewise smooth surface enclosing

the charge. Verify that the above integral is zero if Σ does not enclose the charge.

Hint: You are essentially being asked to prove Gauss’ Law. See Sec. 4.2.3 foran outline of the proof.

11. One of Maxwell’s equations reads

∇ ··· E = 4πε,

where ε is the charge density (see equation (4.9)). Let D being a region of R3 boundedby a piecewise smooth closed surface ∂D. Prove that the flux of the electric fieldthrough ∂D equals 4π times the total charge contained in D (i.e. the flux of theelectric field across ∂D essentially measures the charge contained within ∂D). Assumethat E is a C1 vector field in R3.

12. Consider a radial vector fieldF = f(r)r,

where r =‖ r ‖ and f is a C1 function of r, for r > 0. Let Ω be the solid sphere ofradius b centred on the origin.

i) Calculate the flux of F through the sphere ∂Ω of radius b.

ii) Assuming that f is C1 for r ≥ 0, calculate the integral of ∇···F over Ω, and verifythat Gauss’ theorem is satisfied.

iii) Show that if F satisfies ∇···F = 0, then Gauss’ theorem cannot be applied. Whatspecial property does the flux have in this case?

13. Verify Stokes’ Theorem for the vector field F = (z, x, y) and the surface Σ defined byz = 1− x2 − y2, z ≥ 0.

14. In each case, use Stokes’ Theorem to evaluate the circulation

∫C

F · dx of the vector

field F around the closed curve C.

(i) F = (x, x+y, , x+y+z), and C is the curve defined by x2 +y2 = 1, z = y, wherethe orientation of C is counterclockwise when viewed from the point (0, 0, 10).

(ii) F = (x sin y, −y sinx, (x + y)z2), and C is the union of the line segments suc-cessively joining the points (0, 0, 0), (π

2, 0, 0), (π

2, 0, 1), (0, 0, 1), (0, π

2, 1), (0, π

2, 0)

and (0, 0, 0).

15. Use Stokes’ Theorem to evaluate

∫∫Σ

∇× F · n dS, where

F = (y − z,−x − z, x + y), Σ is the portion of z = 9 − x2 − y2 with z ≥ 0 and n isthe upward pointing unit normal.

201

16. Determine whether the following vector fields F are conservative and if so, find a scalarpotential φ.

(i) F = xy(2yzi + 2xzj + xyk).

(ii) F = (y cosh z, x cosh z, xy sinh z).

17. Consider the vector field

F =

(−y

x2 + y2,

x

x2 + y2, z

), x2 + y2 6= 0 .

Verify that ∇ × F = 0 but that there does not exist a potential function on thedomain of F. Why does this not contradict the theorem on conservative vector fields?

18. Find the work done by the vector field

F = exy(z + xyz, x2z, x)

in moving a particle along the curve

x = g(t) = (t cos t, t sin t, t), 0 ≤ t ≤ 2π .

19. Any constant vector field F on R3 satisfies ∇ ·F = 0 and hence has a vector potential,i.e.F = ∇×A. Find a suitable vector field A. Is A uniquely determined?

20. A collinear vector field is a vector field of the form

F = λA,

where A is a constant vector field and λ is an arbitrary C1 scalar field. Find all collinearvector fields that satisfy

∇× F = 0 and ∇ · F = 0.

21. An axially symmetric vector field is a vector field of the form

F = g(ρ, z)(−y, x, 0),

where g is a C1 function and ρ =√x2 + y2.

i) Find all axially symmetric vectors that satisfy

∇× F = 000 and ∇ · F = 0.

ii) Find all axially symmetric vector fields with the property that the circulation∮C

F · dx

around a circle centred on the z-axis is independent of the radius of the circle.

202

iii) Show that any axially symmetric vector field has a vector potential of the form

A = f(ρ, z)k.

22. A radial vector field is a vector field of the form

F = f(r)r,

where f is a C1 function of the radial distance r =‖ r ‖, for r > 0.

i) Find all radial vector fields that satisfy

∇× F = 000 and ∇ · F = 0.

ii) Find all radial vector fields F = f(r)r with the property that the flux∫∫Σ

F · ndS

through a sphere Σ centred on the origin, is independent of the radius of thesphere.

23. The magnetic field due to a current I flowing along the z-axis is given by

B =2I

c(x2 + y2)(−y, x, 0), x2 + y2 > 0.

i) Show that the circulation of the magnetic field around any simple closed curvethat encircles the z-axis is given by∮

c

B · dx =2I

c.

Hint: First establish the result for any circle x2 + y2 = b2, z = h.

ii) Show that ∮C

B · dx = 0

for any simple closed curve that does not encircle the z-axis.

24. The wave equation in three dimensions is the partial differential equation

∂2u

∂t2− c2∇2u = 0,

where u = u(x, t).

203

i) Verify that

u =1

rf(r − ct),

where r =‖ x ‖=√x2 + y2 + z2 and f is an arbitrary C2 function, is a solution

of the wave equation. This solution is called a spherical wave, because the valueof u for fixed t is constant on spheres centred on the origin.

ii) Verify thatu = f(k · x− ct),

where k is a constant unit vector and f is an arbitrary C2 function, is a solutionof the wave equation. This solution is called a plane wave, because the value of ufor fixed t is constant on planes, the planes orthogonal to k.

25. Consider the vector field F = xyex + y2ey in R2 in terms of Cartesian coordinates.

i) Show that F = ρ2 sinφeρ in terms of polar coordinates.

ii) Calculate the divergence ∇ · F first using Cartesian coordinates and then usingpolar coordinates. Verify that the expressions are equal.

26. i) Starting with the expression (4.38) for ∇ in cylindrical coordinates, derive theexpression (4.48) for the divergence ∇ • F of a vector field.

ii) Hence derive the expression (4.50) for the Laplacian ∇2f of a scalar field.

27. i) Starting with the expression (4.40) for ∇ in spherical coordinates, derive the ex-pression (4.49) for the divergence ∇ • F of a vector field.

ii) Hence derive the expression (4.51) for the Laplacian ∇2f of a scalar field.

28. Elliptic coordinates (η, φ) in the plane are defined by the equation

(x, y) = (a cosh η cosφ, a sinh η sinφ),

where a is a positive constant.

i) Describe the two families of coordinate curves η = constant and φ = constant,and illustrate them with a sketch.

ii) Calculate the coordinate basis vector fields eη and eφ, and show that they aremutually orthogonal.

204

Bonus questions

B1. Acoustics (ref. W.A. Strauss, Partial Differential Equations)

The linearized equations for propagation of sound in a gas are

∂ρ

∂t+ ρ0∇ · v = 0,

∂v

∂t= − c

ρ0

∇ρ,

where ρ0 is the undisturbed density, a constant and c is also a constant. The unknownsare the perturbed density ρ of the gas and the velocity v of the disturbance.

i) Show that the vorticity ωωω = ∇× v is constant in time.

ii) Verify that the density ρ satisfies the wave equation

ρtt = c2∇2ρ.

iii) Verify that if ∇× v = 0, then the velocity v satisfies the wave equation

vtt = c2∇2v.

B2. Let r be the position vector of a planet or satellite moving in a plane orbit. Choosecoordinates so that the orbit lies in the xy-plane. Let er and eθ be the unit basis vectorassociated with polar coordinates in the plane. The planet’s position is described by

r = r(t) and θ = θ(t),

where t is time.

a) Starting withr = rer,

derive the following equations:

i) r× r′ = r2θ′ k

ii) r · r′′ = r[r′′ − r(θ′)2],

where ′ denotes differentiation with respect to time.

b) Newton’s law of gravitation and Newton’s second law of motion lead to

r′′ = −GMr3

r, (1)

where M is the mass of the sun and G is the gravitational constant (see B8 inProblem Set 1). Equation (1) is a vector DE for the position vector r(t) of theplanet at time t. One can solve it by using the results of a) to reduce it to a scalarDE as follows.

First show that r× r′ is constant. Then use a) to show that

r2θ′ = H

r′′ − r(θ′)2 = −GMr2

,

where H is a constant, the angular momentum per unit mass of the planet.

205

c) The orbit of the planet can be described by eliminating time t, thereby expressingr as a function of θ,

r = r(θ).

The resulting DE is much simplified if one introduces the new variable

u =1

r.

Show that a planetary orbit satisfies

d2u

dθ2+ u =

GM

H2.

Hint: Show thatdr

dt= −Hdu

dθ.

Hence show that planetary orbits are ellipses.

B3. The motion of an incompressible fluid with constant density ρ and constant viscosityµ is governed by the partial differential equation

ρ

[∂v

∂t+ (v · ∇)v

]= ρF−∇p+ µ∇2v ,

where v(x, t) is the velocity field, p(x, t) the pressure and F the external force acting(e.g. gravity). The vorticity field of the fluid is defined by

ω = ∇× v .

Derive the vorticity equation

∂ω

∂t+ (v · ∇)ω = (ω · ∇)v +

µ

ρ∇2ω +∇× F .

Hint: Use identity G3 to calculate (v · ∇)v.

B4. Let Ω be a bounded subset in R3, whose boundary is a simple piecewise smooth surface∂Ω. Let n be the unit outward normal on ∂Ω.

i) Prove that ∫∫∫Ω

∇f dV =

∫∫∂Ω

fndS,

for any C1 scalar field f .

Hint: Apply Gauss’ theorem to F = fC, where C is a constant vector field.

206

ii) Prove that ∫∫∫Ω

(∇× F)dV =

∫∫∂Ω

(n× F)dS,

for any C1 vector field F.

Comment: In the notes we defined two types of surface integrals:

i)

∫∫Σ

fdS, where f is a scalar field,

ii)

∫∫Σ

F · ndS, where F is a vector field.

Both of these integrals are scalar-valued. Question B2 uses vector-valued surface inte-grals of the form ∫∫

Σ

FdS.

As part of the solution you will need to define this concept.

B5. A body Ω with surface ∂Ω immersed in a fluid experiences a bouyancy force B givenby

B = −∫∫∂Ω

pndS.

Here the pressure p is a C1 scalar field satisfying

∇p = ρg,

where ρ(x) is the mass density of the fluid and g the constant acceleration of gravity.

Show that B equals the weight of the displaced fluid in magnitude, and acts in theopposite direction to the gravitational force.

B6. Let D be a bounded subset of R3 whose boundary is a piecewise smooth closed surface∂D, with outward unit normal n. Let f be a C1 scalar field and G be a C1 vector fieldon R3. Prove the generalized integration by parts formula:∫∫∫

D

∇f ·G dV =

∫∫∂D

fG · n dS −∫∫∫D

f∇ ·G dV .

207

Problem Set 5:

Fourier series and Fourier transforms

(a) Fourier Series

1. Consider the full-wave rectification function f defined by

f(x) =| sinx |, for − π < x < π.

a) Find the Fourier series of f .

b) Sketch the graph of the function to which the series converges pointwise on R.Use the pointwise convergence theorem to justify your answer.

c) By choosing a suitable value of x in the series that you have found, show that∞∑k=1

1

4k2 − 1= 1

2.

2. Let f(x) =

0 if − π < x < 0

x if 0 < x < π

.

(a) Find the Fourier series of f .

(b) Sketch the graph of function to which the series converges pointwise on R. Justifyyour answer.

(c) Show that∞∑m=1

1

(2m− 1)2=π2

8.

3. For each of the following functions defined on the interval (0, π):

(a) Find the Fourier sine series.

(b) Find the Fourier cosine series.

(c) Sketch the functions to which the Fourier sine and cosine series converge on R.

(i) f(x) = 2x−π (ii) f(x) =

1 if 0 < x < π/2

−1 if π/2 < x < π

(iii) f(x) = sin x.

4. (a) Show that the Fourier series for f(x) = x2, −π < x < π, is

π2

3− 4

∞∑n=1

(−1)n−1 cos(nx)

n2.

209

(b) Use the pointwise convergence theorem to find the pointwise sum function of theseries on R. Sketch its graph.

(c) Show that∞∑n=1

1

n2=π2

6.

(d) Show that∞∑n=1

(−1)n+1

n2=π2

12.

5. Let f be defined on 0 < x < π by

f(x) =

12ε, if

∣∣x− π2| < ε

0 otherwise.

a) Show that the Fourier sine series of f is

2

π

∞∑k=1

(−1)k−1 sinc [(2k − 1)ε] sin(2k − 1)x.

b) Sketch the graph of the function to which the series converges on R.

6. Consider the triangular window (or sawtooth pulse) function defined on the interval−1

2τ ≤ t ≤ 1

2τ by

Tε(t) =

1ε

(1− 1

ετ|t|), if |t| ≤ ετ

0, otherwise.

Show that the Fourier cosine coefficients of Tε are given by

an = 2 [sinc (nπε)]2 .

7. i) Show that the Fourier sine series of the function f defined on the interval0 < x < π by

f(x) =

x, 0 < x < π − ε

(π−ε)ε

(π − x), π − ε < x < π.

is

2∞∑1

(−1)n−1sinc (nε)sinnx

n.

ii) Sketch the graph of the function fp to which this series converges on R.

210

8. Let f be a C1 function on [−π, π]. Prove that the Fourier coefficients of f satisfy

| an | ≤K

n, | bn | ≤

L

n, n = 1, 2, · · · ,

for some constants K and L.

9. a) Derive the identity

12

+ cos θ + · · ·+ cosnθ =sin(n+ 1

2

)θ

2 sin θ2

.

Hint: Sum 1 + z + z2 + · · ·+ zn and let z = eiθ.

b) The Dirichlet kernel is defined by Dn(x) =sin(n+ 1

2)x

2π sin 12x

, −π ≤ x ≤ π. Give a

qualitative sketch of the graph for large n.

10. The Fourier series for the function f(x) = 12x, −π < x < π is

∞∑n=1

(−1)n−1 sinnx

n.

a) Sketch the graph of the function fp which is the pointwise sum of the series on R.

b) Use Parseval’s formula to show that∞∑n=1

1

n2=π2

6.

c) Use the integration theorem to show that

112

(π2 − 3x2) =∞∑n=1

(−1)n−1 cosnx

n2, −π < x < π

and

112x(π2 − x2) =

∞∑n=1

(−1)n−1 sinnx

n3, −π < x < π

d) Discuss whether the sum of each Fourier series in c) is piecewise continuous,continuous, piecewise C1 or C1 on R. Sketch the graphs of the sum functions.

11. Apply Parseval’s theorem to one of the series in #10 to show that

∞∑n=1

1

n4=π4

90.

12. Use Parseval’s formula and a suitable Fourier series to sum∞∑n=1

1

(2n− 1)2.

Hint: Look at Table 5.1.

211

13. Consider the function f defined by f(x) = b, 0 < x < L.

i) Extend f as an odd function and find its Fourier series (the sine series).

ii) Discuss convergence of the series on R, and sketch the graph of its sum function.

14. i) Prove that the series∞∑n=1

sinnx

n2

converges pointwise on R, and uniformly on any finite interval.. Call its sum f(x).

ii) Obtain the Fourier series of the function g defined by

g(x) =

∫ x

0

f(t)dt,

justifying your method by referring to appropriate theorems.

iii) What can you say about continuity and smoothness of the function g? Explain.

(b) Fourier transforms

15. Show that the Fourier transform of

f(t) = e−|t| is F (ω) =2

1 + ω2.

Sketch the graphs in the time domain and in the frequency domain.

16. Show thatF(T (t)) =

[sinc

(12ω)]2

,

where T (t) is the triangular gate function:

T (t) =

1− |t|, if |t| < 1

0, if |t| > 1.

Sketch the graphs of T (t) and its Fourier transform.

17. Find the Fourier transform of

f(t) =

1, −1 < t < 0

−1, 0 < t < 1

0, |t| > 1.

in two ways,

212

a) using the definition, and

b) using the fact that f(t) = T ′(t), where T (t) is the triangular gate function.

18. i) Knowing

∫ ∞−∞

e−u2du =

√π, evaluate

I(ω) =

∫ ∞−∞

e−12x2

cosωxdx.

Hint: Use Leibniz’s theorem to show that

I ′(ω) = −ωI(ω).

ii) Use i) to show that

F(e−12x2

) =√

2π e−12ω2

.

19. a) Find the Fourier transform F (ω) of the normal density

f(t) =1√2πσ

e−(t−µ)2

2σ2 ,

where σ, µ are real parameters with σ > 0.

b) Sketch the graphs of f(t) and of Re F (w) and Im F (w).

20. a) Find the Fourier transform F (ω) of

f(t) = cos(2πνt)e−π(t/α)2

,

where α and ν are positive real numbers.

b) Sketch the graphs of f(t) and F (ω).

21. a) Verify that ifg(t) = tf(t), then G(ω) = iF ′(ω),

where F (ω) = F(f(t)) and G(ω) = F(g(t)).

b) Find the Fourier transform of f(t) = te−t2

in two ways.

22. Show that if f is odd then

F(f(t)) = −2i

∫ ∞0

f(t) sinωt dt.

23. a) Show that if F(f(t)) = F (ω), then

F(f(t− c)) = e−icωF (ω).

213

b) Hence find the Fourier transform of

f(t) = 12

[W (t− c) +W (t+ c)] ,

where W (t) is the window function (5.83). Give a qualitative sketch of the graphsof f(t) and F(f(t)).

24. a) Show that if F(f(t)) = F (ω), then

F(F (t)) = 2πf(−ω).

b) Use a) and known results to show that

i) F(

1

1 + x2

)= πe−|ω|

ii) F(sinc (x)) = πW(

12ω).

25. Find the Fourier transform of

g(t) =

cos πt, if − 1

2< t < 1

2

0, otherwise

26. a) Verify that the Fourier transform of f(t) = e−tH(t) is F (ω) = 11+iω

, where

H(t) =

1 if t > 0

0 if t < 0

is the Heaviside step function.

b) Verify that if g(t) = f(−t), then the Fourier transforms are related by G(ω) =F (−ω).

c) Use a) and b) to verify that

F(e−|t|) =2

1 + ω2.

(c) Convergence

27. Consider the sequence fn of functions defined by

fn(x) =x2n

1 + x2n, for |x| ≤ 2.

214

i) Find the pointwise limit function f . Sketch the graph of f and a typical fn onthe same axes.

ii) Sketch the graph of fn−f and evaluate ‖ fn−f ‖∞. Does fn converge uniformlyto f on [−2, 2]?

28. Consider the sequence of functions fn defined by

fn(x) = nxe−12n2x2 − x, −1 ≤ x ≤ 1.

Let f denote the pointwise limit function.

i) Determine whether fn converges uniformly to f .

ii) Determine whether fn converges in the mean to f .

iii) Sketch graphs of f and a typical function fn on the same axes.

29. (a) Let fn(x) = (sin x)n, 0 ≤ x ≤ π. Find f(x) = limn→∞

fn(x). Does fn converge

uniformly to f? Does fn converge to f in the mean?

(b) Repeat (a) for fn(x) = (sinx)1n .

30. Consider the sequence fn on the interval 0 ≤ x ≤ 1 defined by

fn(x) =2nx

1 + n2x4.

a) Find the pointwise limit function f .

b) Evaluate

∫ 1

0

f(x)dx and limn→∞

∫ 1

0

fn(x)dx.

c) By referring to appropriate theorems, draw a conclusion about whether fni) converges in the mean to f and ii) converges uniformly to f .

Bonus questions

B1. Detective WorkJack calculates some Fourier coefficients and as a result, writes the formula

x

2=∞∑1

(−1)n+1 sinnx

n,

but forgets to specify the interval of x values. Jill does a similar calculation and arrivesat the formula

x

2=

2

π

∞∑1

(−1)n+1 sin(2n− 1)x

(2n− 1)2,

but likewise forgets to state the interval.

In each case, deduce the appropriate interval, and then sketch the graph of the functionto which the series converges on R.

215

B2. Fourier gives Laplace a headache

When Fourier first introduced, in 1807, the series that now bear his name, some of theleading mathematicians of the time, e.g., Laplace and Lagrange, were skeptical becausesome of the results were initially counterintuitive. Here’s an example (see, T. Korner,1988, page 478).

By calculating some Fourier coefficients, Fourier arrived at the amazing formula

cosx =8

π

∞∑n=1

(n

4n2 − 1sin 2nx

),

but forgot to write down the interval of validity. Laplace would not accept that thisexpansion was true because cos( ) is an even function, sin( ) is an odd function, anda sum of odd functions is odd.

i) Repeat Fourier’s calculation.

ii) Deduce the interval of validity of the above expansion, showing that Laplace’sconcerns were unwarranted.

iii) Sketch the graph of the sum function of the series, valid for all x ∈ R.

iv) By integrating, derive a cosine series for sinx, and sketch the graph of the sumfucntion of the series, valid for all x ∈ R.

B3. Frequency separation

a) Consider the rectangular window (or pulse) function Pε defined on the interval−1

2τ ≤ t ≤ 1

2τ by

Pε(t) =

1ε, if |t| ≤ 1

2ετ

0, otherwise,

where ε is a constant satisfying 0 < ε ≤ 1. Show that the Fourier cosine coeffi-cients of Pε are given by

an = 2sinc (nπε).

b) Consider the pulse modulated carrier cos(Nω0t), defined on the interval − τ2≤

t ≤ τ2

byf(t) = Pε(t) cos(Nω0t),

where ω0 = 2πτ

and N is a fixed positive integer (Nω0 is called the carrier frequencyof the signal). Show that the Fourier cosine coefficients of f are given by

an = sinc [(n+N)πε] + sinc [(n−N)πε] .

c) Suppose that Nπε 1. Give a qualitative sketch of the Fourier amplitudespectrum of f .

216

d) Consider a combined signal of the form

f(t) = A1Pε(t) cos(N1ω0t) + A2Pε(t) cos(N2ω0t),

where 1 N1πε N2πε. Give a qualitative sketch of the Fourier amplitudespectrum of f(t).

Comment: This result shows that a complicated signal in the time domain canhave a simple representation in the frequency domain. This behaviour is calledfrequency separation.

217

B4. The trapezoidal wave

δ

δ

− πt

π

€

0< δ <π2( )A trapezoidal wave −π π

1

y

t

€

δ =π2( )A saw-tooth wave

− π πt

y

1

€

(δ = 0)A square wave

i) Show that the Fourier series of the trapezoidal wave is

4

π

∞∑n=1

sinc [(2n− 1)δ]sin(2n− 1)t

(2n− 1).

ii) Hence find the Fourier series of the square wave and triangle wave by taking suit-able limits of the nth term.

Note: The goal is to do a complicated calculation with finesse.

B5. Prove that the trigonometric series

∞∑n=2

sinnx

lnn

converges pointwise for all x, but that it is not the Fourier series for any square inte-grable function.

218

B6. i) Show that the function f defined by

f(x) = − ln∣∣2 sin x

2

∣∣ , for |x| ≤ π, x 6= 0, and f(0) = 0,

is square integrable but not piecewise continuous on [−π, π].

ii) Find the Fourier series of f , and show that the series does not converge pointwiseat x = 0.

Comment: Since f is square integrable the series converges to f in the mean.

B7. It is known that the equilibrium temperature distribution u(x, y) in a plane sheet ofmetal satisfies Laplace’s equation:

∇2u ≡ ∂2u

∂x2+∂2u

∂y2= 0,

(see B4. in Problem Set 2).

i) For a sheet in the shape of a disc, it is convenient to use polar coordinates ρ andφ. Consider the series representation of the temperature function:

u(ρ, φ) = 12a0 +

∞∑n=1

ρn(an cosnφ+ bn sinnφ).

Show that this function satisfies Laplace’s equation ∇2u = 0 for 0 ≤ ρ ≤ b,0 ≤ φ ≤ 2π. Assume that term-by-term differentiation is permissible.

ii) Hence find the temperature of a metal disc of radius b, if the temperature on theboundary is 100 for 0 < φ < π and 0 for π < φ < 2π.

219

Sample Tests & Exams

221

Sample Exam 1

1. a) Explain the term “field line of a vector field” and give its defining equation.Without doing any calculations, give a qualitative sketch of the field lines of thevector field F = (x, x).

b) Suppose that f is a scalar field in R3 giving the temperature in space, and thatx = g(t) describes the path of a spacecraft. Let T (t) be the temperature of thespacecraft at time t. Express T (t) as a composite function, and write a formulafor its derivative.

c) Define the line integral of a scalar field φ : Rn → R along a curve C. Give onephysical interpretation.

d) Define the line integral of a vector field F : Rn → Rn along a curve C. Give onephysical interpretation.

2. a) State Green’s Theorem, giving all the hypotheses.

b) Explain how the area enclosed by a simple closed curve in the plane can becalculated by evaluating a line integral.

c) Verify Green’s theorem for the vector field F = (xy,−xy) and the region D lyingbetween y = x2 and y = x, by evaluating the line integral and the double integral.

3. a) Suppose that the line integral of F is path-independent in R2. Define a scalarfield ϕ : R2 → R by

ϕ(x, y) =

(x,y)∫C (0,0)

F • dx,

where C is any curve joining (0,0) to (x, y). Prove that

∇ϕ(x, y) = F(x, y), for all (x, y) ∈ R2.

It is sufficient to give the details for the second component of this vector equation.

b) Evaluate the line integral ∫C

sin ydx + x cos y dy

where C is the straight line joining(

2, π2

)and

(3,−π

2

). State any theorem you

use.

4. a) An inverse square force field in 3-space has the form

F(r) =k

r3r,

where k is a constant, r = (x, y, z) and r = ||r||. Show that the work done by Fin moving a particle from position r1 to position r2 along any path depends onlyon the distances r1 and r2.

223

b) Consider the linear vector field

F = (ax+ by, cx+ dy)

where a, b, c, and d are constants.

i) Let C be a simple closed curve enclosing a region of area A. Calculate thecirculation of F around C.

ii) For what values of a, b, c, and d will F be a conservative vector field? Justifyyour answer. Find a potential in this case.

c) Consider a vector field F in R2 that is C1 and has zero vorticity except at a finitenumber of singularities. Let C be a simple closed curve in the domain of F. Prove

that if

∮C

F • dx 6= 0, then C encloses a singularity.

d) Describe the curve

x = g(t) = (sin 2t, 2 sin t, 1 + cos 2t) , 0 ≤ t ≤ 2π,

by showing that it is the intersection of two quadric surfaces.

224

Sample Exam 2

1. a) State Green’s theorem, giving all the hypotheses.

b) Use Green’s theorem to evaluate the line integral∫

C

F • dx , where C is the

boundary of the triangle with vertices (1,0), (0,1) and (−1, 0), followed clockwise,and F = (cos(e−x), xy).

(c) Is the vector field F in b) a gradient field? Justify your answer.

2. a) Let C be a smooth curve x = g(t) , t1 ≤ t ≤ t2, joining two points a and b inRn. Let φ : Rn → R be a C1 scalar field.

(i) State the Chain Rule in vector form for the composite function h(t) = φ(g(t)).

(ii) Prove that if F(x) = ∇φ(x), then

b∫C a

F • dx = φ(b)− φ(a)

b) Show that the line integral

∫C

2x

ydx − x2

y2dy is path-independent in the subset

y > 0 of R2, and find its value if C joins a = (1, 2) to b = (−2, 4).

c) Let f and g be C1 scalar fields and let C be a simple closed curve. Prove that∮C

(f∇g) • dx = −∮

C

(g∇f) • dx

3. a) The path of a particle is the curve C, defined by

x = g(t) = (t− sin t, 1− cos t) , π ≤ t ≤ 5π.

(i) Determine at what points the particle is instantaneously at rest, and at whatpoints it is moving in the x-direction. Sketch the path.

(ii) Calculate the distance travelled by the particle.

Trig identity: cos t = 1− 2 sin2 12t

b) Invent a function x = g(t) to describe the curve of intersection of x2 + z2 = 4zand y + z = 2. Sketch the curve and indicate the orientation.

4. a) Verify that the curve x = g(t) = (tekt, ekt, e2t), where k is a constant, is a fieldline of the vector field F = (kx+ y, ky, 2z).

225

b) (i) Without doing any calculations sketch typical and exceptional field lines ofthe vector field

v = (sin y, 0), 0 ≤ y ≤ 2π.

(ii) Suppose that v is the surface velocity of water in a stream. Discuss whethera small floating object will rotate as it moves with the stream.

c) Consider a scalar field ψ = f(r), where r = ||r|| and r is the position vector inR3. Evaluate r×∇ψ.

d) The force F exerted on a particle of charge q and mass m moving with velocity vin a magnetic field B is

F = qv ×B.

Assuming that Newton’s Second Law of Motion is valid

(mdv

dt= F

), show that

the speed v of the particle is constant.

226

Sample Exam 3

1. a) i) Explain the term “field line of a vector field in Rn”, and give the definingequation.

ii) Find the field lines of the vector field F = (x,−y) in R2. Give a clearlylabelled sketch, indicating all exceptional field lines.

b) i) Define the line integral of a continuous vector field F along a C1 curve C givenby x = g(t), t1 ≤ t ≤ t2. Give one physical interpretation of the line integral.

ii) If C is a field line of the vector field F, show that

∫C

F • dx ≥ 0.

2. The surface Σ is the piece of the plane y+ z = b lying inside the cylinder x2 + y2 = b2,where b is a constant. The curve C is the curve of intersection of the plane with thecylinder, oriented counterclockwise when viewed from above. The vector field F isgiven by F = (kz, kx, 3y), where k is a constant.

i) Calculate the flux of the vector field ∇× F through Σ in the direction ofincreasing z.

ii) Calculate the circulation

∮C

F • dx of F around C.

iii) The quantities calculated in i) and ii) are equal as a result of a theorem. Give acomplete statement of this theorem.

3. a) Let C be a C1 curvex = g(t), t1 ≤ t ≤ t2,

joining two points a and b in R3. Let φ : R3 → R be a C1 scalar field. Prove thatif F(x) = ∇φ(x), then ∫

C

F • dx = φ(b)− φ(a).

b) Verify that ∇× (∇ψ) = 000, where ψ is a C2 scalar field on R3.

c) Give conditions that will guarantee that a C1 vector field F is a gradient field ona subset U ⊂ R3. Explain all terminology.

4. Consider the scalar field

u(x) =k

‖ x− a ‖n,

where k, n and a are constants.

i) Express u in terms of spherical coordinates r, θ, φ based on the point a, and thenexpress F = −∇u in terms of the basis vectors er, eθ and eφ.

ii) Hence, find n such that ∇ • F = 0.

227

iii) Calculate the flux of the vector field F in i) through the surface of the sphere ofradius b, centre a. Show that if ∇ • F = 0, then the flux is independent of theradius b.

5. a) Calculate the Fourier cosine series of the function f defined by

f(t) =

1, 0 < t <

π

2

−1,π

2< t < π.

Sketch the graph of the function fp to which this series converges pointwise on R.What is the sum of the series at the points of discontinuity?

b) i) Define in full the statement “the series∞∑n=1

an(t) converges in the mean on

the interval [a, b] to the function f(t).”

ii) Define in full the statement “the sequence of functions fn convergesuniformly to f on the interval [a, b]”.

6. a) i) Derive Parseval’s formula in complex form for a τ -periodic function f(t).

ii) In the case that f(t) is a 2π-periodic odd function, show that Parseval’sformula in i) can be written as

2

π

∫ π

0

f(t)2dt =∞∑n=1

b2n,

where the bn are the Fourier sine coefficients.

b) Consider the Fourier sine series

1 =4

π

∞∑n=1

sin(2n− 1)t

(2n− 1), 0 < t < π

i) Use Parseval’s formula to sum∞∑n=1

1

(2n− 1)2.

ii) By integrating the given series for 0 ≤ t ≤ x, find the sum of the Fourierseries

∞∑n=1

cos(2n− 1)x

(2n− 1)2

on the interval −π ≤ x ≤ π. Sketch the graph of the sum function.

7. The Fourier transform of f(t) is defined by

F(f(t)) = F (ω) =

∫ ∞−∞

f(t)e−iωtdt.

228

a) Show that if f is an even function then

F (ω) = 2

∫ ∞0

f(t) cos(ωt)dt.

b) Show that if F(f(t)) = F (ω), then F(eiω0tf(t)) = F (ω − ω0).

c) Find the Fourier transform F (ω) of f(t) = W (t) cos 100t where

W (t) =

1 if |t| < 1

2

0 if |t| > 12

.

You may assume that F(W (t)) = sinc(

12ω). Give a qualitative sketch of the

graphs of f(t) and F (ω).

8. a) Give a complete statement of Gauss’ theorem.

b) Let A be a constant vector. Calculate the flux of the vector field F = A × xthrough any simple closed piecewise C1 surface Σ.

c) Let Ω be a subset of R3 whose boundary ∂Ω is a smooth closed surface. Let f bea C2 function that satisfies

∇2f = 0 on Ω, f = 0 on ∂Ω.

Prove that f = 0 on Ω.

Hint: Integrate ∇ • (f∇f).

229

Sample Exam 4

1. a) Calculate the distance travelled by a particle moving on the curve C given by

x = g(t) = (sin t− t cos t, cos t+ t sin t), −2π ≤ t ≤ 2π.

b) Find the field lines of the vector field F = (1, 2x), and illustrate them with asketch. Without doing further calculations, sketch the field lines of the vectorfield G = (x, 2x2). Show typical field lines and any exceptional ones.

c) (i) If u(t) = f(g(t)), give a formula for u′(t).

(ii) If u(t) = f(g(t)), give a formula for u′(t).

2. i) Calculate the flux of the vector field

F = (x, y,−2z)

through the surface∑

given by

z = 1− x2 − y2, z ≥ 1− b2,

in the positive z-direction.

ii) Calculate the circulation of the vector field A = (yz,−xz, 0) around the curve

x2 + y2 = b2, z = 1− b2,

in the counter-clockwise direction.

iii) The answers in i) and ii) are equal. Explain why this is so by referring to a famoustheorem. Give a complete statement of this theorem.

3. a) Consider the radial vector field F = f(r)r, where r = ||r||.i) Calculate the flux of F through the sphere r = b.

ii) Suppose that F satisfies ∇ • F = 0. Draw a conclusion about the flux.

b) Give conditions that will guarantee that a C1 vector field F is a gradient field ona subset U ⊂ R3. Explain all terminology.

c) Calculate the work done by the force

F = ey−z(1, x,−x)

acting on a particle moving along the curve C given by

x = g(t) = (2− t, cos t, sin t), 0 ≤ t ≤ 2π.

230

4. a) In polar coordinates x = ρ cosφ, y = ρ sinφ the vector differential operator ∇ isgiven by

∇ = eρ∂

∂ρ+

1

ρeφ

∂

∂φ,

whereeρ = (cosφ, sinφ), eφ = (− sinφ, cosφ).

Calculate an expression for the Laplacian ∇2u in terms of the partial derivativesof u with respect to ρ and φ.

b) The integral form of the law of conservation of mass in a fluid flow is

d

dt

∫∫∫Ω

ρdV = −∫∫∂Ω

ρv • ndS,

where Ω is an arbitrary domain in R with piecewise smooth boundary ∂Ω. Themass density ρ and fluid velocity v are assumed to be C1 functions.Show that the ρ and v satisfy the partial differential equation

∂ρ

∂t+∇ • (ρv) = 0.

5. a) Calculate the Fourier sine series of the function

f(x) = 12(π − x), 0 < x < π.

b) Sketch the graph of the function fp that is the sum of the series derived in a), forall x ∈ R. What is the sum of the series at the points of discontinuity?

c) i) Given the Fourier series

112

(π2 − 3t2) =∞∑n=1

(−1)n−1 cosnt

n2, −π < t < π,

sum the series∞∑n=1

(−1)n−1 sinnx

n3.

ii) Sum the series

1− 1

33+

1

53− · · ·

6. a) Let en(t) = einw0t, w0 =2π

τ. Evaluate

∫ τ2

− τ2

en(t)em(t)dt, for n 6= m and for

n = m, where denotes complex conjugation.

231

b) The complex form of the Fourier series is

f(t) =∞∑

n=−∞

cneinw0t.

Use a) to derive the expression for the Fourier coefficients cn.

c) Use Parseval’s formula

1

τ

∫ τ2

− τ2

f(t)2dt =∞∑

n=−∞

|cn|2

and the Fourier series in 5 c) to sum∞∑n=1

1

n4.

7. a) Give the formula for the Fourier transform F (w) of a function f(t), and the for-mula for the Fourier integral.

b) Use a) to derive Parseval’s formula∫ ∞−∞

f(t)2dt =1

2π

∫ ∞−∞|F (w)|2dw.

c) If F(f(t)) = F (w), show that F(tf(t)) = iF ′(w).

d) Use the formula for the Fourier transform to show that if F(f(t)) = F (w),then

F(f(t) cosw0t) = 12[F (w − w0) + F (w + w0)].

8. a) The circulation of a constant vector field around any simple closed curve in R3 iszero. Explain why.

b) The flux of a constant vector field through any orientable closed surface in R3 iszero. Explain why.

c) A vector field is C1 and satisfies ∇ × F = 0 for all x ∈ R3, x /∈ `, where ` isan infinite line. Let C be a circle such that

∮c

F • dx 6= 0. Prove that C loopsaround `.

d) The base of a fence is described by the curve C : x = g(u), a ≤ u ≤ b, in thexy-plane. The height of the fence at position g(u) is given by z = h(u). Derive aformula for the surface area of the fence as an integral.

232

233

Index

ε-neighbourhood, 146

amplitude modulation, 165amplitude spectrum, 136analysis equation, 157arclength, 15

band-limited signal, 172boundary

of a solid in R3, 93, 101of a subset in R2, 48of a surface in R3, 74, 102

circulation of a fluid, 60class C1, 13closed curve, 48closed surface, 93complex Fourier coefficients, 133complex Fourier series, 133connected set, 21conservation law, 98

differential form, 100conservative field, 48, 53, 110

test for, 57convergence in the mean

sequence, 146series, 149

convergence of Fourier series, 154coordinate curves, 65coordinate grid, 65coordinates

curvilinear, 88cylindrical, 86, 91orthogonal, 88polar, 86, 91spherical, 86, 91

counter-clockwise orientation, 48curl, 82, 84

curvearclength, 15closed, 48counter-clockwise orientation, 48orientation, 2parameter on, 2parametrization, 2reparametrization, 8simple closed, 48smooth, 14

curvilinear coordinate system, 88cut-off frequency, 172cylindrical coordinates, 86, 91

differential form, 38differential operator, 81digitized signals, 172discrete Fourier transform, 174divergence, 82, 83divergence theorem, 94

generalized, 101

equation of continuity for a fluid, 100equipotentials, 44even function, 129

Faraday’s law, 107Fast Fourier Transform, 175field lines, 21fluid

circulation of, 60equation of continuity, 100irrotational, 109mass density, 18velocity field, 18vorticity, 82vorticity field, 109vorticity of, 58

234

flux density vector field, 20, 77heat, 20mass, 19

Fourier coefficients, 119–121, 157, 159complex, 133

Fourier integral, 158–159Fourier matrix, 175Fourier series, 117–159

complex, 133convergence, 154integration of, 155pointwise convergence, 125

Fourier spectrum, 136continuous, 168discrete, 168

Fourier transform, 158, 159dilation formula, 163discrete, 174Fast Fourier Transform, 175frequency shifting, 165linearity, 163scaling formula, 163shift formula, 165

Fourier’s Law, 100frequency domain, 136fundamental frequency, 136Fundamental Theorem of line integral, 43, 45

Gauss’ law, 102Gauss’ theorem, 94Gibbs phenomenon, 127gradien vector field, 53gradient of the scalar field, 20, 81, 83gradient vector field, 26, 44, 46, 110Green’s Theorem, 49

harmonics, 136heat diffusion equation, 100

jump discontinuities, 125

kinetic energy of a particle, 47

Laplacian, 85line integral

additivity, 39

along piecewise C1 curve, 39First Fundamental Theorem, 43linearity, 39of a linear density function, 32path-independent, 42scalar field, 29Second Fundamental Theorem, 45vector field, 35

linear approximationvector-value function on R2, 70vector-valued function on R, 13

linear time-invariant system, 171

mass density of a fluid, 18mass flux density, 19maximum norm, 142Maxwell’s equations, 82mean square norm, 143

non-orientable surface, 74norm, 142

basic inequality, 143maximum, 142mean square, 143of piecewise continuous functions, 144properties of, 144

odd function, 129orientation

curve, 2, 48reversal, 40surface, 73

orthogonal coordinate system, 88orthogonality condition, 133

Parseval’s formulanon-periodic function, 167periodic function, 136

path-independent, 46period 2π extension of f , 125piecewise C1, 14, 80, 124, 153piecewise continuous, 124, 153pointwise convergence

of a Fourier series, 125of a sequence of functions, 139

polar coordinates, 86, 91

235

polygonal arc, 15potential

existence of, 53, 110, 113scalar, 44, 110vector, 112

potential engery of a particle, 47

rectangular window function, 160reversal of orientation, 40

sampling frequency, 172Sampling theorem, 173scalar field, 17

average value on a curve, 33gradient of, 20, 81line integral of, 29surface integral of, 76

scalar potential, 44, 110existence of, 53, 110

signalband-limited, 172cut-off frequency, 172digitized, 172fundamental frequency, 136, 168harmonics, 136sampling frequency, 172

signal separation, 170simple closed curve, 48simply-connected, 56, 110sinc function, 160smooth curve, 14solenoidal field, 114special domain, 94spherical coordinates, 86, 91star-shaped set, 113Stokes’ theorem, 103surface

closed, 74, 93non-orientable, 74normal vector, 69orientation, 73tangent plane, 69vector function for, 65

surface area, 72surface integral

additivity, 79linearity, 79of a scalar field, 76of a vector field, 78

synthesis equation, 157system function, 171

termwise integration of series, 151time domain, 136

uniform convergencesequence, 145series, 149

vector differential operator, 81vector field, 17–21

field lines, 21lin integral of, 35surface integral of, 78

vector potential, 112existence of, 113

vector product, 67vector-valued function

component functions, 2continuous, 10derivative, 10for a curve, 1for a surface, 65limit, 9linear approximation, 13

velocity field, 18, 22vorticity field, 82, 109vorticity of a fluid, 58

Weierstrass M-test, 149

zero identities, 84

236