Mathematical physics with partial differential equations james kirkwood

Solving Partial Differential Equations in Cylindrical8-2 Solving Laplace’s Equation in Cylindrical Coordinates 9.. Solving Partial Differential Equations in Spherical 9-2 The Solution to

Differential Equations

www.TheSolutionManual.com

Mathematical Physics

with Partial Differential

Equations

James R Kirkwood

Sweet Briar College

AMSTERDAM • BOSTON • HEIDELBERG • LONDON NEW YORK • OXFORD • PARIS • SAN DIEGO SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO

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12 13 14 15 10 9 8 7 6 5 4 3 2 1

Integrals of Scalar Functions Over Surfaces 83

4-4 Fourier Sine and Cosine Series 208

7-4 Solving the Wave Equation in Two Dimensions in Cartesian

8 Solving Partial Differential Equations in Cylindrical

8-2 Solving Laplace’s Equation in Cylindrical Coordinates

9 Solving Partial Differential Equations in Spherical

9-2 The Solution to Bessel’s Equation in

11-2 Solving Differential Equations Using the Laplace Transform 356

www.TheSolutionManual.com

The major purposes of this book are to present partial differential equations

(PDEs) and vector analysis at an introductory level As such, it could be

con-sidered a beginning text in mathematical physics It is also designed to provide

a bridge from undergraduate mathematics to the first graduate mathematics

course in physics, applied mathematics, or engineering In these disciplines, it

is not unusual for such a graduate course to cover topics from linear algebra,

ordinary and partial differential equations, advanced calculus, vector analysis,

complex analysis, and probability and statistics at a highly accelerated pace

In this text we study in detail, but at an introductory level, a reduced list

of topics important to the disciplines above In partial differential equations,

we consider Green’s functions, the Fourier and Laplace transforms, and how

these are used to solve PDEs We also study using separation of variables to

solve PDEs in great detail Our approach is to examine the three prototypical

second-order PDEs—Laplace’s equation, the heat equation, and the wave

equation—and solve each equation with each method The premise is that in

doing so, the reader will become adept at each method and comfortable with

each equation

The other prominent area of the text is vector analysis While the usual

topics are discussed, an emphasis is placed on understanding concepts rather

than formulas For example, we view the curl and gradient as properties of a

vector field rather than simply as equations A significant—but optional—

portion of this area deals with curvilinear coordinates to reinforce the idea of

conversion of coordinate systems

Reasonable prerequisites for the course are a course in multivariable

cal-culus, familiarity with ordinary differential equations including the ability to

solve a second-order boundary problem with constant coefficients, and some

experience with linear algebra

In dealing with ordinary differential equations, we emphasize the linear

operator approach That is, we consider the problem as being an eigenvalue/

eigenvector problem for a self-adjoint operator In addition to eliminating

some tedious computations regarding orthogonality, this serves as a unifying

theme and an introduction to more advanced mathematics

The level of the text generally lies between that of the classic encyclopedic

texts of Boas and Kreysig and the newer text by McQuarrie, and the partial

differential equations books of Weinberg and Pinsky Topics such as Fourier

series are developed in a mathematically rigorous manner The section on

completeness of eigenfunctions of a Sturm-Liouville problem is considerably

xi

more advanced than the rest of the text, and can be omitted if one wishes to

merely accept the result

The text can be used as a self-contained reference as well as an introductory

text There was a concerted effort to avoid situations where filling in details

of an argument would be a challenge This is done in part so that the text

could serve as a source for students in subsequent courses who felt “I know I’m

supposed to know how to derive this, but I don’t.” A couple of such examples

are the fundamental solution of Laplace’s equation and the spectrum of the

Laplacian

I want to give special thanks to Patricia Osborn of Elsevier Publishing

whose encouragement prompted me to turn a collection of disjointed notes

into what I hope is a readable and cohesive text, and also to Gene Wayne of

Boston University who provided valuable suggestions

James Radford Kirkwood

1-1 SELF-ADJOINT OPERATORS

The purpose of this text is to study some of the important equations and techniques

of mathematical physics It is a fortuitous fact that many of the most important

such equations are linear, and we can apply the well-developed theory of linear

operators We assume knowledge of basic linear algebra but review some

defini-tions, theorems, and examples that will be important to us

Definition: A linear operator (or linear function), from a vector space V to a

vector space W, is a functionL : V-W for which

Lða1 ^v11 a2 ^v2Þ5 a1Lð^v1Þ1 a2Lð^v2Þfor all ^v1; ^v2AVand scalars a1and a2

One of the most important linear operators for us will be

L½y 5 a0ðxÞyðxÞ1 a1ðxÞy0ðxÞ1 a2ðxÞyvðxÞ;

where a0(x), a1(x), and a2(x) are continuous functions

Definition: If L : V-V is a linear operator, then a nonzero vector ^v is an

eigenvector ofL with eigenvalue λ if Lð^vÞ 5 λ^v

Note that ^0 cannot be an eigenvector, but 0 can be an eigenvalue

Example: For L 5 d

dx, we haveLðeax

Þ5 d

dxðe

ax

Þ5 aeax;

so eaxis an eigenvector ofL with eigenvalue a

An extremely important example is forL 5 d2

dx2

Lðsin nxÞ 5 d2

dx2ðsin nxÞ5 2n2sin nx; and

1

Mathematical Physics with Partial Differential Equations

© 2013 Elsevier Inc All rights reserved.

Lðcos nxÞ 5 d2

dx2ðcos nxÞ5 2 n2cos nx:

We leave it as exercise 1 to show that if ^v is an eigenvector of L with

eigenvalueλ, then a^v is also an eigenvector of L with eigenvalue λ

Definition: An inner product (also called a dot product) on a vector space V

with scalar field F (which is the real numbersℝ or the complex numbers ℂ)

is a function h , i : V3 V-F such that for all f, g, h A V, and a A F,

ha f; gi 5 ah f ; gi;

hf; agi 5 ahf ; gi; where x 1 iy 5 x 2 iy;

hf1 g; hi 5 h f ; hi 1 hg; hi;

hf; gi 5 hg; f i;

and h f, fi$ 0 with equality if and only if f 5 0

A vector space with an inner product is called an inner product space

If V5 ℝn, the usual inner product for ^a 5 ða1; ; anÞ; ^b5 ðb1; ; bnÞis

^a; ^b5 a1b11 ? 1 anbn:

If the vector space isℂn, then we must modify the definition, because, for

example, under this definition, if ^a 5 ði; iÞ, then

h^a; ^ai 5 i21 i25 22:

Thus, onℂn, for ^a 5 ða1; ; anÞ; ^b5 ðb1; ; bnÞ, we define

h^a; ^bi 5 a1b11 ? 1 anbn:

We use the notation h f, fi5 :f:2, which is interpreted as the square of

the length of f, and:f 2 g: is the distance from f to g

We will be working primarily with vector spaces consisting of functions

that satisfy some property such as continuity or differentiability In this

set-ting, one usually defines the inner product using an integral A common

inner product is

hf; gi 5

ðb a

f ðxÞgðxÞdx;where a or b may be finite or infinite There might be a problem with some

vector spaces in that h f, fi5 0 with f 6¼ 0 This problem can be overcome by

a minor modification of the vector space or by restricting the functions to

being continuous, and will not affect our work We leave it as exercise 4 to

show that the function defined above is an inner product

On some occasions it will be advantageous to modify the inner product

above with a weight function w(x) If w(x)$ 0 on [a, b], then

hf; giw5

ðb a

f ðxÞwðxÞgðxÞdx

is also an inner product as we show in exercise 5

Definition: A linear operator A on the inner product space V is self-adjoint if

hAf, gi5 h f, Agi for all f, g A V

Self-adjoint operators are prominent in mathematical physics One

exam-ple is the Hamiltonian operator It is a fact (Stone’s Theorem) that energy is

conserved if and only if the Hamiltonian is self-adjoint Another example is

shown below Part of the significance of this example is due to Newton’s

law F5 ma

Example: The operator d 2

dx 2is self-adjoint on the inner product space

V5 ff jf has a continuous second derivative and is periodic on ½a; bg;

with inner product

hf; gi 5

ðb a

f ðxÞgvðxÞdx:

To do this, we integrate by parts twice Consider the integral on the left Let

u5 gðxÞ; du5 g0ðxÞ;

dv5 f vðxÞ; v 5 f0ðxÞ;so

fvðxÞgðxÞdx 5 2

ðb a

f0ðxÞg0ðxÞdx:Integrating the integral on the right by parts with

u5 g0ðxÞ; du 5 gvðxÞ;

dv5 f0ðxÞ; v 5 f ðxÞ;

f ðxÞgvðxÞdx

5

ðb a

f ðxÞgvðxÞdx:

Notice that if [a, b] is of length 2π, then {sin (nx) cos (nx) j n A ℤ} is a

subset of V

We next prove two important facts about self-adjoint operators

Theorem: If L : V-V is a self-adjoint operator, then

a The eigenvalues of L are real

b The eigenvectors of L with different eigenvalues are orthogonal; that is,

their inner product is 0

Proof: (a) Suppose that f is an eigenvector of L with eigenvalue λ Then

hLf ; f i 5 hλf ; f i 5 λhf ; f iand

hf; Lf i 5 hf ; λf i 5 λhf ; f i:

SinceL is self-adjoint,

hLf ; f i 5 hf ; Lf i so λhf ; f i 5 λhf ; f iand since h f, fi 6¼ 0, we haveλ 5 λ , so λ is real

(b) Suppose

Lf 5 λ1f and Lg 5 λ2g with λ16¼λ2:Then

hLf ; gi 5 hλ1f; gi 5 λ1hf; giand

hLf ; gi 5 hf ; Lgi 5 h f ; λ2gi5 λ2hf; gi:

So

λ1hf; gi 5 λ2hf; giand thus h f; gi 5 0 because λ16¼λ2:

ð2 π 0

sin ðnxÞ sin ðmxÞ dx5 0; if m 6¼ n

for m and n integers This is because sin (nx) and cos (mx) are eigenfunctions

of the self-adjoint operator d 2

dx 2with the inner product defined above with ferent eigenvalues

dif-We will use the technique of the example above to prove the

orthogonal-ity of functions, such as Bessel functions and Legendre polynomials, without

having to resort to tedious calculations

Fourier Coefficients

We now describe how to determine the representation of a given vector with

respect to a given basis That is, if f ^b1; ^b2; g is a basis for the vector space

V, and if ^vAV, we want to find scalars a1, a2, for which

^v 5 a1 ^b11 a2 ^b21 ?:

If the basis satisfies the characteristic below, then this is easy

Definition: If f ^b1; ^b2; g is a set of vectors from an inner product space for

which

h ^bi; ^bji5 0 if i 6¼ j;

then f ^b1; ^b2; g is called an orthogonal set If, in addition,

h ^bi; ^bii5 1 for all i;

then f ^b1; ^b2; g is called an orthonormal set A basis that is an orthogonal

(orthonormal) set is called an orthogonal (orthonormal) basis

Theorem: If f ^b1; ^b2; g is an orthogonal basis for the inner product space

h^v; ^bki5 ha1 ^b11 a2 ^b21 ? 1 a1^bk1 ?; ^bki

5 a1h ^b1; ^bki1 ? 1 akh ^bk; ^bki1 ? 5 akh ^bk; ^bki:Thus,

ak5 h^v; ^bki

h ^bk; ^bki5 h^v; ^bki

: ^bk:2:Note that if f ^b ; ^b ; g is an orthonormal basis, then a 5 h^v; ^b i

Definition: The constants {a1, a2, } in the theorem above are called the

Fourier coefficients of ^v with respect to the basis f ^b1; ^b2; g

Fourier coefficients are important because they provide the best

approxi-mation to a vector by a subset of an orthogonal basis in the sense of the

fol-lowing theorem

Theorem: Suppose ^v is a vector in an inner product space V, and

ℬ 5 f ^b1; ^b2; g is an orthogonal basis for V Let {c1, c2, .} be the Fourier

coefficients of ^v with respect to ℬ Then

Proof: We assume the constants are real, and the basis is orthonormal to

simplify the notation We have

with equality if and only if ci5 difor all i5 1, , n.

Note that from equation(2)we have Bessel’s inequality,

Xn

i 51

ci2# h^v; ^vi 5 :^v:2

:

Example: In this example, we demonstrate an application of eigenvalues

and eigenfunctions (eigenvectors) to solve a problem in mechanics

Suppose that we have a body of mass m1 attached to a spring of which

the spring constant is k1 See Figure 1-1-1 We assume that the surface is

frictionless If x1 is the displacement of the spring from equilibrium, then,

according to Hooke’s law, the spring creates a force ^F5 2x1k1 Then,

^F 5 m1

d2x1

dt2 5 2x1k1:Now consider the coupled system shown in Figure 1-1-2 We use the

convention that force is positive if it pushes a body to the right We suppose

that no spring is under no tension if the masses are at points a and b

Suppose that the masses are at points x1and x2

Force on mass m1:

a Force due to spring 1: If x1 a, then spring 1 is stretched an amount

x12 a and pulls m1to the left If the spring constant of spring 1 is k1,

then the force on m1due to spring 1 is

F1;15 2k1ðx12 aÞ:

b Force due to spring 3: If x22 x1, b 2 a, then spring 3 is compressed an

amount (b2 a) 2 (x22 x1) If the spring constant of spring 3 is k3 then

spring 3 pushes the body m1to the left with force

F1;35 2k3½ðb2 aÞ 2 ðx22 x1Þ5 2k3½ðb2 x2Þ1 ðx12 aÞ:

Thus, the total force on m1is

F15 F1 ;11 F1 ;35 2 k1ðx12 aÞ 2 k3½ðb2 x2Þ1 ðx12 aÞ: ð3Þ

Force on mass m2:

a Force due to spring 2: If x2, b, then spring 2 is stretched an amount

b2 x2 and pulls m2to the right If the spring constant of spring 2 is k2,

then the force on m2due to spring 2 is

F2 ;25 k2ðb2 x2Þ:

b Force due to spring 3: If x22 x1, b 2 a, then spring 3 is compressed an

amount (b2 a) 2 (x22 x1) If the spring constant of spring 3 is k3, then

spring 3 pushes the body m2to the right with force

1CC:

Suppose thatλ is an eigenvalue for A, so that

Then we would have

t:The eigenvalues of A are those values of λ for which det(A 2 λI) 5 0

With the values we have, this would best be done with a CAS; however, if

we set the value of each mass to be m, and the value of each spring constant

to be k, then the matrix A is

22km

kmk

m

0BB

1CCand

detðA2λIÞ5

22k

mk

:

Thus, the eigenvalues for A areλ 5 2k

m and5 23k

m.Before continuing, we note there is an alternate method to calculate the

equations of motion If V is the potential energy of the system, then

kmk

m

0BB

1CC

m, suppose that

22km

kmk

m

0BB

1CC

mz122k

mz25 23k

mz2;

zðtÞ5 C1

11

 

ei

ffiffik m

p

t1 C2

121

ei

ffiffiffi3k m

p

t;where zðtÞ5 z1ðtÞ

z2ðtÞ

.The linear operators that we will use in the text will usually be differen-

tial operators A typical example of which is

L½y5 yvðxÞ 1 pðxÞy0ðxÞ1 qðxÞ:

We will often use the Principle of Superposition, which states that if

y1(x) and y2(x) are solutions to

L½y5 yvðxÞ 1 pðxÞy0ðxÞ1 qðxÞ 5 0;

1 Show that if ^v is an eigenvector for A with eigenvalue λ, then for any

scalar a, the vector a^v is an eigenvector of A with eigenvalue λ

2 For h , i an inner product, show that h f, g 1 hi 5 h f, gi 1 h f, hi

3 Show that the following is a linear operator,

L½y5 a0ðxÞyðxÞ1 a1ðxÞy0ðxÞ1 a2ðxÞyvðxÞ;

where a0(x), a1(x), and a2(x) are continuous functions

4 Show that the following function is an inner product,

hf; gi 5

ðb a

f ðxÞgðxÞdx;where f(x) and g(x) are continuous functions on [a, b]

5 Show that the following function is an inner product,

hf; giw5

ðb a

f ðxÞwðxÞgðxÞdx;where f(x), w(x), and g(x) are continuous functions on [a, b] and w(x) 0

What if w(x), 0?

6 In ℝ2orℝ3the angleθ between the vectors ^u and ^v is determined by

cosθ 5 ^uU^v: ^u::^v::

b Find the angle between ^u and ^v

c What is the distance between ^u and ^v traveling along the surface of

9 Show that if f^x1; ; ^xngis a basis for the vector space V, then every vector

in V can be written as a linear combination of^x1; ; ^xnin exactly one way

10 Show that if f^x1; ; ^xngis an orthogonal basis for the vector space V, and

What if f^x1; ; ^xngis an orthonormal basis?

11 Suppose that f^x1; ; ^xngis a basis for the vector space V and T : V- V

is a linear transformation for which Tð^xÞ 5 ^0 if and only if ^x 5 ^0

a Show that T is a one-to-one function

b Show that fTð^x Þ; ; Tð^x Þgis a basis for V

12 For a function f(t), determine which of the following are linear

13 Suppose that f^x1; ^x2gis an orthonormal basis for V and T : V-V is a

lin-ear transformation for which fTð^x1Þ; Tð^x2Þgis also an orthonormal basis

for V Show that for any vector ^x; :Tð^xÞ: 5 :^x: (This is also true for

the case of any finite basis.)

14 Recall that if T : V-V is a linear transformation and V is an

n-dimensional vector space, then T can be represented as multiplication

by an n3 n matrix The matrix depends on the choice of the basis In

particular, if f^x1; ; ^xngis a basis for the vector space V and if

Tð^xiÞ5 a1i^x11 ? 1 ani^xn;then

Show that if T : V-V is a linear transformation and there is a basis of V,

f^x1; ; ^xng, consisting of eigenvectors of T, so that Tð^xiÞ5 λi^xi, then

the matrix of T with respect to this basis is the diagonal matrix

This is important because computations with diagonal matrices are

par-ticularly simple

15 A linear transformation U : V-V is called an orthogonal transformation

(or a unitary transformation if the field is the complex numbers rather

than the real numbers) if:U ^x: 5 :^x: for every ^x ε V

a Show that if U is an orthogonal transformation, then hU^x; U ^yi 5 h^x; ^yi

for every pair of vectors ^x; ^y ε V Hint: Use hUð^x 1 ^yÞ; Uð^x 1 ^yÞi 5

h^x 1 ^y; ^x 1 ^yi and expand both sides of the equation

b What is the physical interpretation of part (a.) in the case that V is

c What is the dimension of T(V)? Find a basis for T(V)

17 The kernel of a linear transformation T : V-V is fxAVjTðxÞ 5 ^0g

a Show that the kernel of T is a vector space

b Find the kernel of T in the case where V is the n 1 1-dimensional

space of real polynomials in x of degree less than or equal to n and

T5 d

dx

1-2 CURVILINEAR COORDINATES

Many problems have a symmetry associated with them, and finding the

solu-tions to such problems, as well as interpreting the solution, can often be

simpli-fied if we work in a coordinate system that takes advantage of the symmetry In

this section we describe the methods of transforming some important functions

to other coordinate systems The most common coordinate systems besides

Cartesian coordinates are cylindrical and spherical coordinates, but the methods

we develop are applicable to other systems as well In our discussion, we

include some of the less common systems The less common systems will not

be used in later sections but are included to reinforce the techniques of the

transformations

In Figure 1-2-1a we give a diagram of how cylindrical coordinates are

defined, and in Figure 1-2-1b we do the same for spherical coordinates

We note that while the convention we use for spherical coordinates

is common, it is not universal Some sources reverse the roles of θ

Y y

θ

FIGURE 1-2-1a

Our approach will be to describe the general case of converting from

Cartesian coordinates (x, y, z) to a system of coordinates (u1, u2, u3)

After making a statement that holds in the general case, to visualize that

statement, we demonstrate how the statement applies to cylindrical

coordinates

General Case: We start with Cartesian coordinates (x, y, z) and select

the group of variables u1, u2, u3so that each of x, y, z is expressible in terms

of u1, u2, u3; that is, we have

x5 xðu1; u2; u3Þ; y 5 yðu1; u2; u3Þ; z 5 zðu1; u2; u3Þ:

Cylindrical Case: The variables in cylindrical coordinates are r, θ, and z

The relations are

x5 r cos θ; y 5 r sin θ; z 5 z; 0 # r , N 0 # θ , 2π; 2N , z , N:

We write the vector ^r 5 x^i1 y^j1 z ^k in terms of u1, u2, u3; that is,

^r 5 xðu1; u2; u3Þ ^i1 yðu1; u2; u3Þ ^j1 zðu1; u2; u3Þ ^k:

In cylindrical coordinates, this is

^r 5 r cos θ ^i1 r sin θ ^j1 z ^k:

For some of our relations to be viable, the coordinates (u1, u2, u3) must

be orthogonal This means that the pairs of surfaces ui5 constant and

uj5 constant must meet at right angles In the case of cylindrical

coordi-nates, the surface r5 constant is shown in Figure 1-2-2a, the surface

θ 5 constant is shown in Figure 1-2-2b, and the surface z5 constant is

shown inFigure 1-2-2c Each pair does indeed meet at right angles It is also

possible to determine that the coordinates are orthogonal by analytical

meth-ods, as we now describe

In the general case, the vector @^r

@u 1 will be tangent to the u1curve, which

is the intersection of the u25 constant and u35 constant surfaces Similar

relations hold for @^r

P′

Z

Y y

FIGURE 1-2-1b

FIGURE 1-2-2a

z

y x

u Surface of constant u

We can show that a system of coordinates forms an orthogonal

coordi-nate system by showing that the vectors @^r

@u i are orthogonal; that is, by ing their inner product is zero In the cylindrical case,

5 hcos θ ^i1 sin θ ^j; 2 r sin θ ^i1 r cos θ ^ji

5 2r cos θ sin θ 1 r cos θ sin θ 5 0;

mutu-setting

^ei5 @u@^ri:@^r

h2 5 2r sin θ ^i1 r cos θ ^j

r 5 2sin θ ^i1 cos θ ^j;

^e35 ^ez5 @u@^r3

h 5 ^k:

Back to the general case, we have

d^r 5 h1^e1du11 h2^e2du21 h3^e3du35 ^erdr1 r ^eθdθ 1 ^ezdz:

Volume Integrals

We now describe how to convert volume integrals to other coordinate

systems

General Case: Our aim is to determine an expression for an incremental

volume element dV in a general coordinate system The volume of the

paral-lelepiped formed by three non-coplanar vectors ^A; ^B, and ^C is j ^AUð ^B 3 ^CÞj

(See exercise 1.) For the Cartesian case, we compute an incremental volume

since^e1Uð^e23 ^e3Þ51 because f^e1; ^e2; ^e3gis an orthonormal system

Another way to do this computation is to use

The determinant in equation (1) is called the Jacobian of x, y, z with

respect to u1, u2, u3, and is denoted @ðx;y;zÞ

We now compute dV for cylindrical coordinates We demonstrate two

methods First, we use

dV5 h1h2h3du1du2du3;where u15 r, u25 θ, u35 z so that du15 dr, du25 dθ, u35 dz We have pre-

viously found that h15 1, h25 r, h35 1, so

There are different ways that equation(2) is expressed in other sources

One other way is

Example: Evaluate ðð

A

ex21y 2dxdy;where A is the circle x21 y2# 9, by changing to polar coordinates

In Cartesian coordinates,

ðð

A

ex21y 2dxdy5

The region A in polar coordinates is 0 # r # 3, 0 # θ # 2π So, in this

case, equation(2)says

ðð

A

ex21y 2dxdy

er2rdθdr

5 2π

ð3 0

We seek a coordinate system (u, v) so that the transformed region of

inte-gration will be a rectangle a# u # b, c # v # d The graph of the region A

v5xyy x

5 x2; so x 5

ffiffiffiuv

r:

We compute the Jacobian We have

ffiffiffiuv

r

; @y

@u5

12

ffiffiffivu

r

; @y

@v5

12

ffiffiffiuv

r:

2 ffiffiffiffiffiuv

2v

ffiffiffiuvs

12

ffiffiffivu

s12

ffiffiffiuvs

ð4

v 51

u22v2 1 u22

The Gradient

Next, we determine the gradient of a function f, denoted rf In Cartesian

coordinates,

rf5@f@x^i1 @@yf ^j1 @@zf ^k:

To compute rf in the general case, we set rf5 A^e11 B^e21 C ^e3 and

write df in two different ways

First, df5 rf Ud ^r and using that d ^r 5 h1^e1du11 h2^e2du21 h3^e3du3 we

The Laplacian

The final function that we consider in this section is the Laplacian, one of the

most important operators in mathematics and physics The Laplacian of the

function f, denoted Δf (some authors use r2f ) in Cartesian coordinates, is

for cylindrical coordinates with i5 2 We have

^e25 ^eθ5 @θ@ ðr cosθ ^i1 r sin θ ^j1 z ^kÞ 5 2r sin θ ^i1 r cos θ ^j

since h15 h35 1 Now h25 r and u15 r so @h 2

5 2 ^e15 2 ðr cos θ^i1 r sin θ^jÞ;

so the formula holds in this case

In fact, it will be simpler to derive expressions for r2f and r3 F, after

we have studied the Divergence Theorem and Stokes’ Theorem in Section 2-3

For now, we simply state the results:

Spherical Coordinates

One of the most commonly used coordinate systems is spherical coordinates

(In the previous examples, we used cylindrical coordinates because the

com-putations are simpler.)

The transformations are

Other Curvilinear Systems

The most common curvilinear systems are cylindrical and spherical

coordi-nates, but there are several lesser-known systems, some of which we

dis-cuss here We will not use these examples in the sequel but present these

results to develop familiarity and facility in the computations The list we

give is not complete Other examples are given in Morse and Feshbach

(1953) and Arfken (1970)

The first example is elliptic cylindrical coordinates We give the

transfor-mations and

G Demonstrate that the system is orthogonal

G Compute the scaling factors

G Determine the orthonormal basis