##### Citation preview

Mathematical Physics with Partial Differential Equations Second Edition

James Kirkwood

Professor of Mathematical Sciences, Sweet Briar College, Sweet Briar, VA, USA

Publisher: Katey Birtcher Acquisition Editor: Katey Birtcher Editorial Project Manager: Lindsay Lawrence Production Project Manager: Bharatwaj Varatharajan Designer: Victoria Pearson Typeset by TNQ Books and Journals

To Bessie, Katie, and Elizabeth. The lights of my life.

xiii

xiv

Preface

A couple of such examples are the fundamental solution of Laplace’s equation and the spectrum of the Laplacian. The major changes from the first edition are first that a chapter on generating functions has been added. This gives a nice way of considering solutions to LaGuerre equations, Hermite equations, Legendre’s equations, and Bessel equations. Second, in depth analyses for the rigid rotor, one-dimensional quantum mechanical oscillator, and the hydrogen atom are presented. Also, more esoteric coordinate systems have been moved to an appendix.

CHAPTER

Preliminaries

1

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 definitions, 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 function L : V/W for which v 1 Þ þ a2 Lðb v2Þ Lða1 vb1 þ a2 vb2 Þ ¼ a1 Lðb for all vb1 ; vb2 ˛ V and scalars a1 and a2. One of the most important linear operators for us will be L½y ¼ a0 ðxÞyðxÞ þ a1 ðxÞy0 ðxÞ þ a2 ðxÞy00 ðxÞ where a0(x), a1(x) and a2(x) are continuous functions. Definition: If L : V/V is a linear operator, then a nonzero vector vb is an eigenvector of L with eigenvalue l if Lðb v Þ ¼ lb v. Note that b 0 cannot be an eigenvector, but 0 can be an eigenvalue. Example: d For L ¼ , we have dx d Lðeax Þ ¼ ðeax Þ ¼ aeax dx so eax is an eigenvector of L with eigenvalue a. d2 An extremely important example is L ¼ 2 . Among its properties are dx d2 Lðsin nxÞ ¼ ðsin nxÞ ¼ n2 sin nx and dx2 Lðcos nxÞ ¼

d2 ðcos nxÞ ¼ n2 cos nx. dx2

1

2

CHAPTER 1 Preliminaries

We leave it as Exercise 1 to show that if vb is an eigenvector of L with eigenvalue l, then ab v is also an eigenvector of L with eigenvalue l. Definition: An inner product (also called a dot product) on a vector space V with scalar field or the complex number ) is a function F (which is the real number h ; i:V  V / F such that for all f, g, h e V and a e F haf ; gi ¼ ah f ; gi; h f ; agi ¼ ah f ; gi;

where x þ iy ¼ x  iy;

h f þ g; hi ¼ h f ; hi þ hg; hi; h f ; gi ¼ hg; f i; h f ; f i  0 with equality if and only if f ¼ 0. A vector space with an inner product is called an inner product space. If V ¼ ℝn , the usual inner product for ab ¼ ða1 ; .; an Þ; bb ¼ ðb1 ; .; bn Þ is E D ab; bb ¼ a1 b1 þ / þ an bn . If the vector space is ℂn , then we must modify the definition, because, for example, under this definition, if ab ¼ ði; iÞ, then h ab; ab i ¼ i2 þ i2 ¼ 2: Thus, on ℂn , for ab ¼ ða1 ; .; an Þ; bb ¼ ðb1 ; .; bn Þ, we define D E ab; bb ¼ a1 b1 þ / þ an bn . We use the notation h f ; f i ¼ k f k2 , which is interpreted as the square of the length of f, and k f  gk is the distance from f to g. We shall be working primarily with vector spaces consisting of functions that satisfy some property such as continuity or differentiability. In this setting, one usually defines the inner product using an integral. A common inner product is Z b f ðxÞgðxÞdx; h f ; gi ¼ a

where a or b may be finite or infinite. There might be a problem with some vector spaces in that h f ; f i ¼ 0 with f s 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 Z b f ðxÞwðxÞgðxÞdx h f ; giw ¼ a

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 ; gi ¼ h f ; Agi for all f ; g ε V. Self-adjoint operators are prominent in mathematical physics. One example 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 F ¼ ma. Example: d2 The operator 2 is self-adjoint on the inner product space dx V ¼ ff jf has a continuous second derivative and is periodic on ½a; bg; with inner product Z h f ; gi ¼

b

f ðxÞgðxÞdx.

a

We must show hf 00 ; gi ¼ h f ; g00 i; that is, Z

b a

Z

00

f ðxÞgðxÞdx ¼

b

a

f ðxÞg00 ðxÞdx.

To do this, we integrate by parts twice. Let u ¼ gðxÞ

du ¼ g0 ðxÞ

dv ¼ f 00 ðxÞ v ¼ f 0 ðxÞ so Z uv 

b Z  v du ¼ gðxÞf 0 ðxÞ  a

b a

f 0 ðxÞg0 ðxÞdx.

b  The periodicity of f and g forces gðxÞf 0 ðxÞ ¼ 0. Thus, a Z b Z b f 00 ðxÞgðxÞdx ¼  f 0 ðxÞg0 ðxÞdx: a

a

Integrating the integral on the right by parts with u ¼ g0 ðxÞ du ¼ g00 ðxÞ dv ¼ f 0 ðxÞ

v ¼ f ðxÞ

3

4

CHAPTER 1 Preliminaries

we have Z 

b

a

2

b Z  f 0 ðxÞg0 ðxÞdx ¼ 4f ðxÞg0 ðxÞ  a

a

b

3 f ðxÞg00 ðxÞdx5 ¼

Z

b a

f ðxÞg00 ðxÞdx.

Notice that if [a,b] is of length 2p, then fsinðnxÞ cosðnxÞjn ˛ Zg 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 1. The eigenvalues of L are real; 2. Eigenvectors of L with different eigenvalues are orthogonal; that is, their inner product is 0. Proof: 1. Suppose that f is an eigenvector of L with eigenvalue l. Then hLf ; f i ¼ hlf ; f i ¼ lh f ; f i and h f ; Lf i ¼ h f ; lf i ¼ lh f ; f i. Since L is self-adjoint, hLf ; f i ¼ h f ; Lf i so lh f ; f i ¼ lh f ; f i and since h f ; f is0, we have l ¼ l, so l is real. 2. Suppose Lf ¼ l1 f and Lg ¼ l2 g with l1 sl2 . Then hLf ; gi ¼ hl1 f ; gi ¼ l1 h f ; gi and hLf ; gi ¼ h f ; Lgi ¼ h f ; l2 gi ¼ l2 h f ; gi. So l1 h f ; gi ¼ l2 h f ; gi and thus h f ; gi ¼ 0 because l1 sl2 . Example: We have Z

2p 0

sinðnxÞ cosðmxÞ dx ¼ 0;

Z

2p

cosðnxÞ cosðmxÞ dx ¼ 0;

if msn;

0

Z

2p

sinðnxÞ sinðmxÞdx ¼ 0;

if msn;

0

for m and n integers. This is because sin (nx) and cos (mx) are eigenfunctions of the self-adjoint operator d2 =dx2 with the inner product defined above with different eigenvalues. We shall use the technique of the example above to prove the orthogonality 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 o n to a given basis. That is, if bb1 ; bb2 ; . is a basis for the vector space V, and if vb ε V, we want to find scalars a1, a2,. for which vb ¼ a1 bb1 þ a2 bb2 þ /: If the basis satisfies the characteristic below, then this is easy. Definition: o n If bb1 ; bb2 ; . is a set of vectors from an inner product space for which D E bbi ; bbj ¼ 0 if isj; o n then bb1 ; bb2 ; . is called an orthogonal set. If, in addition, D E bbi ; bbi ¼ 1 for all i; o n then bb1 ; bb2 ; . is called an orthonormal set. A basis that is an orthogonal (orthonormal) set is called an orthogonal (orthonormal) basis. Theorem: o n If bb1 ; bb2 ; . is an orthogonal basis for the inner product space V, and if vb ¼ a1 bb1 þ a2 bb2 þ /; then

D

E vb; bbk E ¼  2 . ak ¼ D b  bbk ; bbk bk  vb; bbk

E

D

5

6

CHAPTER 1 Preliminaries

Proof: We have

D

E D E vb; bbk ¼ a1 bb1 þ a2 bb2 þ /; bbk D E D E D E ¼ a1 bb1 ; bbk þ / þ ak bbk ; bbk þ / ¼ ak bbk ; bbk .

Thus

D

E vb; bbk E ¼  2 . ak ¼ D b  bbk ; bbk bk  o D n E Note that if bb1 ; bb2 ; . is an orthonormal basis, then ak ¼ vb; bbk . Definition: The constants {a1,a2,.} innthe theoremoabove are called the Fourier coefficients of vb with respect to the basis bb1 ; bb2 ; . . vb; bbk

E

D

Fourier coefficients are important because they provide the best approximation to a vector by a subset of an orthogonal basis in the sense of the following theorem. Theorem: o n Suppose vb is a vector in an inner product space V, and B ¼ bb1 ; bb2 ; . is an orthogonal basis for V. Let {c1, c2, .} be the Fourier coefficients of vb with respect to B. Then         n n X X     b b ci b i    vb  di b i   vb      i¼1 i¼1 for any numbers di. Equality holds if and only if ci ¼ di for every i ¼ 1,.,n. Proof: We assume the constants are real, and the basis is orthonormal to simplify the notation. We have  2 * +   n n n X X X   di bbi  ¼ vb  di bbi ; vb  di bbi  vb    i¼1 i¼1 i¼1 + (1) * n D n n E X X X b b b vb; di b i þ di b i ; di b i . ¼ hb v ; vbi  2 i¼1

Now,

*

n X i¼1

di bbi ;

n X i¼1

i¼1

+ di bbi

¼

n X i¼1

di2 ;

i¼1

1.1 Self-Adjoint Operators n o since bb1 ; bb2 ; . is an orthonormal basis, as we verify in Exercise 10. Also, E D vb; di bbi ¼ ci di . Thus, the right-hand side of Eq. (1) is + * n D n n n n E X X X X X b b b vb; di b i þ di b i ; di b i ¼ hb c i di þ di 2 v ; vbi  2 v ; vbi  2 hb i¼1 n X

¼ hb v ; vbi 

¼ hb v ; vbi 

i¼1 n X i¼1

n X

ci 2 þ

i¼1

i¼1

ci 2  2

i¼1

ci 2 þ

n X

n X

c i di þ

n X

i¼1

!

i¼1

i¼1

di 2

i¼1

ðci  di Þ2 .

i¼1

Following the first steps in the argument above, we get  2   n n n n X X X X   ci bbi  ¼ hb v ; vbi  2 ci ci þ ci 2 ¼ hb v ; vbi  ci 2 .  vb    i¼1 i¼1 i¼1 i¼1

(2)

Finally, v ; vbi  hb

n X

ci 2  hb v ; vbi 

i¼1

n X

ci 2 þ

i¼1

n X

ðci  di Þ2 ;

i¼1

with equality if and only if ci ¼ di for all i ¼ 1,.,n. Note that from Eq. (2), we have Bessel’s inequality n X

ci 2  hb v ; vbi ¼ kb v k2 .

i¼1

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 whose spring constant is k1. See Fig. 1.1.1. We assume that the surface is frictionless. If x1 is the

k1 m1

FIGURE 1.1.1

7

8

CHAPTER 1 Preliminaries

displacement of the spring from equilibrium, then, according to Hooke’s law, the spring creates a force Fb ¼  x1 k1 . Then d x1 Fb ¼ m1 2 ¼  x1 k1 . dt Now consider the coupled system shown in Fig. 1.1.2. We use the convention that force is positive if it pushes a body to the right. We suppose that both springs are under no tension if the masses are at points a and b. Suppose that the masses are at points x1 and x2. Force on mass m1: 2

1. Force due to spring 1: If x1 > a, then spring 1 is stretched an amount x1  a and pulls m1 to the left. If the spring constant of spring 1 is k1, then the force on m1 due to spring 1 is F1;1 ¼ k1 ðx1  a Þ: 2. Force due to spring 3: If x2  x1 < b  a, then spring 3 is compressed an amount (b  a)  (x2  x1). If the spring constant of spring 3 is k3 then spring 3 pushes the body m1 to the left with force F1;3 ¼ k3 ½ðb  aÞ  ðx2  x1 Þ ¼ k3 ½ðb  x2 Þ þ ðx1  aÞ. Thus, the total force on m1 is F1 ¼ F1;1 þ F1;3 ¼ k1 ðx1  aÞ  k3 ½ðb  x2 Þ þ ðx1  aÞ.

(3)

Force on mass m2: 1. Force due to spring 2: If x2 < b, then spring 2 is stretched an amount b  x2 and pulls m2 to the right. If the spring constant of spring 2 is k2, then the force on m2 due to spring 2 is F2;2 ¼ k2 ðb  x2 Þ.

k1 a

FIGURE 1.1.2

m1 x1

k3

m2 x2

k2 b

2. Force due to spring 3: If x2  x1 < b  a, then spring 3 is compressed an amount (b  a)  (x2  x1). If the spring constant of spring 3 is k3 then spring 3 pushes the body m2 to the right with force F2;3 ¼ k3 ½ðb  aÞ  ðx2  x1 Þ ¼ k3 ½ðb  x2 Þ þ ðx1  aÞ: Thus, the total force on m21 is F2 ¼ F2;2 þ F2;3 ¼ k2 ðb  x2 Þ þ k3 ½ðb  x2 Þ þ ðx1  aÞ.

(4)

If we let z1 ¼ x1  a and z2 ¼ x2  b, we get from Eq. (3) F1 ¼ k1 ðx1  aÞ  k3 ½ðb  x2 Þ þ ðx1  aÞ ¼ ðk1 þ k3 Þz1 þ k3 z2 and we get from Eq. (4) F2 ¼ k2 ðb  x2 Þ þ k3 ½ðb  x2 Þ þ ðx1  aÞ ¼ k3 z1  ðk1 þ k3 Þz2 . Using F ¼ ma ¼ kx, we get m1

d 2 z1 ¼ ðk1 þ k3 Þz1 þ k3 z2 dt2

m2

d 2 z2 ¼ k3 z1  ðk1 þ k3 Þz2 dt2

or   d 2 z1 k1 þ k3 k3 z1 þ z 2 ¼  2 dt m1 m1   d 2 z2 k3 k2 k3 z2 . ¼ z1  þ dt2 m2 m2 m2 Eqs. (5) and (6) can be written as the matrix equation 0 1 k1 þ k3 k3   B   C z  m1 m1 z1 d 2 z1 1 B C ¼ ¼ A B C 2 @ dt z2 z2 k3 k2 þ k3 A z2  m2 m2 where 0

k1 þ k3 B m 1 B A¼B @ k3 m2

k3 m1

1

C C C. k2 þ k3 A  m2

(5) (6)

9

10

CHAPTER 1 Preliminaries

Suppose that l is an eigenvalue for A, so that     z1 z1 ¼ l . A z2 z2 Then we would have d2 dt2



z1 z2



  z1 d 2 zi ¼ l so 2 ¼ lzi dt z2

and zi ðtÞ ¼ Bi e

pﬃﬃ lt

.

The eigenvalues of A are those values of l for which det (A  lI) ¼ 0. With the values we have, this would best be done with a computer algebra system (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 0 1 2k k  B C m C B m B C @ k 2k A  m m and

   2k  k   l       4k 3k2 k 3k m  m  lþ . detðA  lIÞ ¼   ¼ l2 þ l þ 2 ¼ l þ   m m m m k 2k     l  m m k 3k Thus, the eigenvalues for A are l ¼  and l ¼  . m 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 mi

d 2 z1 vV ¼ . 2 dt vzi

In the spring setting, V¼

X ki i

2

di2

where di is the distortion of the ith spring from equilibrium. In our problem k1 2 k2 2 k3 z1 þ z2 þ ðz1  z2 Þ2 . 2 2 2 k We now find the eigenvectors for the eigenvalues. For the eigenvalue l ¼  . m Suppose that V¼

0

2k B B m B @ k m

1 k   C  k z1 m C z1 ¼ . C m z2 2k A z2  m

Then 2k k k  z1 þ z2 ¼  z1 m m m k 2k k z1  z2 ¼  z2 m m m   1 k so z1 ¼ z2 and is an eigenvector for l ¼ . m 1 3k For the eigenvalue l ¼  . Suppose that m 0 1 2k k   B C  3k z1 m C z1 B m ¼ . B C @ k m z2 2k A z2  m m Then 2k k 3k  z 1 þ z2 ¼  z1 m m m k 2k 3k z1  z2 ¼  z2 m m m   1 3k so z1 ¼ z2 and is an eigenvector for l ¼ . m 1 Now the motion of the two masses is given by rﬃﬃﬃ ! rﬃﬃﬃﬃﬃ !     1 1 k 3k t þ C2 t exp i exp i zðtÞ ¼ C1 n n 1 1   z1 ðtÞ where zðtÞ ¼ . z2 ðtÞ The linear operators that we shall use throughout the text will be differential operators. A typical example of which is L½y ¼ y00 ðxÞ þ pðxÞy0 ðxÞ þ qðxÞ. We shall often use the Principle of Superposition, which states that if y1(x) and y2(x) are solutions to L½y ¼ y00 ðxÞ þ pðxÞy0 ðxÞ þ qðxÞ ¼ 0 then c1 y1 ðxÞ þ c2 y2 ðxÞ

11

12

CHAPTER 1 Preliminaries

is also a solution to L½y ¼ y00 ðxÞ þ pðxÞy0 ðxÞ þ qðxÞ ¼ 0 for any constants c1 and c2.

EXERCISES 1. Show that if vb is an eigenvector for A with eigenvalue l, then for any scalar a, the vector ab v is an eigenvector of A with eigenvalue l. 2. For, h ; i an inner product, show that h f ; g þ hi ¼ h f ; gi þ h f ; hi. 3. Show that L½y ¼ a0 ðxÞyðxÞ þ a1 ðxÞy0 ðxÞ þ a2 ðxÞy00 ðxÞ where a0(x), a1(x) and a2(x) are continuous functions, is a linear operator. 4. Show that the function Z b f ðxÞgðxÞdx h f ; gi ¼ a

where f (x) and g(x) are continuous functions on [a,b], is an inner product. 5. Show that the function Z b f ðxÞwðxÞgðxÞdx h f ; giw ¼ a

where f (x), w(x) and g(x) are continuous functions on [a,b] and w(x) > 0, is an inner product. What if w(x) < 0? 6. In ℝ2 or ℝ3 the angle q between the vectors ub and vb is determined by cos q ¼

ub\$b v : kb u kkb vk

 pﬃﬃ  pﬃﬃ pﬃﬃ pﬃﬃﬃﬃﬃ  a. Verify that 12; 23; 5 and 23; 25; 24 are points on the sphere of pﬃﬃﬃﬃﬃ radius 26 feet. b. Find the angle between ub and vb. c. What is the distance between ub and vb traveling along the surface of the sphere?

7. Let L½y ¼ 1  x2 y00 ðxÞ  2xy0 ðxÞ. Show that L is self-adjoint with the inner product Z 1 f ðxÞgðxÞdx. h f ; gi ¼ 1

8. Find the eigenvalue(s) and eigenfunction(s) for the following boundary value problems: d2 f ðxÞ ¼ lf ðxÞ; f ð0Þ ¼ f ðpÞ ¼ 0: a.  dx2 d 2 f ðxÞ ¼ lf ðxÞ; b. e dx2 c. 

f ð0Þ ¼ f ðLÞ ¼ 0:

d2 f ðxÞ ¼ lf ðxÞ; dx2

f ðp=2Þ ¼ f ðp=2Þ ¼ 0:

d2 f ðxÞ ¼ lf ðxÞ; f ðLÞ ¼ f ðLÞ ¼ 0: dx2 9. Show that if fb x 1 ; .; xbn g is a basis for the vector space V, then every vector in V can be written as a linear combination of xb1 ; .; xbn in exactly one way. 10. Show that if fb x 1 ; .; xbn g is an orthogonal basis for the vector space V, and d. 

vb ¼

n X

b¼ ai xbi and w

n X

i¼1

bi xbi

i¼1

then b ¼ v ; wi hb

n X n X

n X ai bj xbi ; xbj ¼ ai bi hb x i ; xbi i.

i¼1 j¼1

i¼1

What if fb x 1 ; .; xbn g is an orthonormal basis? 11. Suppose that fb x 1 ; .; xbn g is a basis for the vector space V and T:V / V is a linear transformation for which Tðb xÞ ¼ b 0 only if xb ¼ b 0. a. Show that T is a one-to-one function. x n Þg is a basis for V. b. Show that fTðb x 1 Þ; .; Tðb 12. For a function f (t), determine which of the following are linear transformations: Tðf Þ ¼ af 00 ðtÞ þ bf 0 ðtÞ Tðf Þ ¼ af 00 ðtÞ þ bf 0 ðtÞ þ 1 Tðf Þ ¼ et f 0 ðtÞ Tðf Þ ¼ ðf ðtÞÞ2 . 13. Suppose that fb x 1 ; xb2 g is an orthonormal basis for V and T:V / V is a linear transformation for which fTðb x 1 Þ; Tðb x 2 Þg is also an orthonormal basis for V. Show that for any vector xb, kTðb x Þk ¼ kb x k. (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 n  n matrix. The matrix depends on the choice of the basis. In particular, if fb x 1 ; .; xbn g is a basis for the vector space V and if Tðb x i Þ ¼ a1i xb1 þ / þ ani xbn

13

14

CHAPTER 1 Preliminaries

then

0

a11

B Tðb1 xb1 þ /bn xbn Þ ¼ @ «

1

an1 where

/ /

1 b1 xb1 CB C « A@ « A ¼ Ab x b b x n n ann a1n

10

a11 B A¼@ «

/ 1

1 a1n C « A and xb ¼ b1 xb1 þ /bn xbn .

an1

/

ann

0

Show that if T:V / V is a linear transformation and there is a basis of V, x i Þ ¼ li xbi , then the x 1 ; .; xbn g, consisting of eigenvectors of T, so that Tðb fb matrix of T with respect to this basis is the diagonal matrix 1 0 l1 / 0 C B @ « 1 « A. 0 / ln This is important because computations with diagonal matrices are particularly 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 kU xbk ¼ kb x k for every xb ˛V. a. Show that if U is an orthogonal transformation then hU xb; U ybi ¼ hb x ; ybi for every pair of vectors xb; yb ˛V. Hint: Use hUðb x þ ybÞ; Uðb x þ ybÞi ¼ hb x þ yb; xb þ yb i and expand both sides of the equation. b. What is the physical interpretation of part a in the case that V is ℝ2 or ℝ3. 16. Let V be the n þ 1 dimensional space of real polynomials in x. a. Show that {1,x,x2,.,xn} is a basis for V. d b. Let T:V / V be defined by T ¼ . Find the matrix of T with respect to this dx basis. c. What is the dimension of T(V)? Find a basis for T(V). 

 d. The kernel of a linear transformation T:V / V is x ˛V TðxÞ ¼ b 0 . Show that the kernel of T is a vector space. Find the kernel of T in the case V is the n þ 1 dimensional space of real polynomials in x of degree less than or d equal to n and T ¼ . dx 17. a. Show that {1, sin(nx), cos(nx) n ¼ 1,2,.;} is an orthogonal set of functions with respect to the inner product Z p f ðxÞgðxÞdx. h f ; gi ¼ p

1.2 Curvilinear Coordinates

b. Find a0, a1, a2, b1, b2 for ∞ ∞ P P i. ex ¼ a0 þ an cosðnxÞ þ bn sinðnxÞ. n¼1

ii. x þ

x2

¼ a0 þ

n¼1

∞ P n¼1

an cosðnxÞ þ

∞ P n¼1

bn sinðnxÞ.

1.2 CURVILINEAR COORDINATES Many problems have a symmetry associated with them and finding the solutions to such problemsdas well as interpreting the solutiondcan often be simplified 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 Appendix 3, 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 Fig. 1.2.1A we give a diagram of how cylindrical coordinates are defined and in Fig. 1.2.1B we do the same for spherical coordinates. We note that while the

Z P

O

z

θ

Y

y

x

r

X

FIGURE 1.2.1A

Z P

θ O

φ X

FIGURE 1.2.1B

z

r y

x P′

Y

15

16

CHAPTER 1 Preliminaries

convention we use for spherical coordinates is common, it is not universal. Some sources reverse the roles of q and 4. 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 group of variables u1,u2,u3 so that each of x,y,z is expressible in terms of u1,u2,u3; that is, we have x ¼ xðu1 ; u2 ; u3 Þ;

y ¼ yðu1 ; u2 ; u3 Þ;

z ¼ zðu1 ; u2 ; u3 Þ.

Cylindrical case: The variables in cylindrical coordinates are r,q and z. The relations are x ¼ r cos q; y ¼ r sin q; z ¼ z;

0  r < ∞; 0  q < 2p; ∞ < z < ∞. We write the vector b r ¼ xb i þ yb j þ z kb in terms of u1,u2,u3; that is, b b i þ yðu1 ; u2 ; u3 Þ b j þ zðu1 ; u2 ; u3 Þ k. r ¼ xðu1 ; u2 ; u3 Þ b

In cylindrical coordinates, this is b b r ¼ r cos qb i þ r sin q b j þ 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 ui ¼ constant and uj ¼ constant must meet at right angles. In the case of cylindrical coordinates, the surfaces r ¼ constant and q ¼ constant are shown in Fig. 1.2.2A, the surfaces r ¼ constant and z ¼ constant are shown in Fig. 1.2.2B, and the surfaces q ¼ constant and z ¼ constant are shown in Fig. 1.2.2C. Each pair does indeed meet at right angles. It is also possible to determine that the coordinates are orthogonal by analytical methods, as we now describe. v^ r will be tangent to the u1 curve, which is the In the general case the vector vu1 intersection of the u2 ¼ constant and u3 ¼ constant surfaces. Similar relations hold v^ r v^ r and . for vu2 vu3 z

r

y

x Surface of constant r

FIGURE 1.2.2A

1.2 Curvilinear Coordinates

z

y

x Surface of constant ␪

FIGURE 1.2.2B z

z y x Surface of constant z

FIGURE 1.2.2C

In cylindrical coordinates  vb r vb r v ¼ ¼ r cos q b i þ r sin q b j þ z kb ¼ cos qb i þ sin q b j. vu1 vr vr  vb r vb r v  ¼ ¼ r cos q b i þ r sin qb j þ z kb ¼ r sin q b i þ r cos q b j. vu2 vq vq  vb r vb r v b ¼ ¼ r cos q b i þ r sin q b j þ z kb ¼ k. vu3 vz vz We can show that a system of coordinates forms an orthogonal coordinate system v^ r are orthogonal; that is, by showing their inner prodby showing that the vectors vui uct is zero. In the cylindrical case   D E vb r vb r ; ¼ cos q b i þ sin q b j; r sin q b i þ r cos q b j vr vq ¼ r cos q sin q þ r cos q sin q ¼ 0  D E vb r vb r ; ¼ cos qb i þ sin q b j; kb ¼ 0 vr vz   D E vb r vb r b r sin q b ; ¼ k; i þ r cos q b j ¼ 0: vz vq 

17

18

CHAPTER 1 Preliminaries

SCALING FACTORS

v^ r are mutually vui orthogonal. We create an orthonormal system of vectors fb e 1 ; eb2 ; eb3 g by setting

We know that in an orthogonal coordinate system, the vectors

vb r vui  ebi ¼   vb .  r vu  i    v^ r  We define the scaling factors hi by hi ¼  vu , so that i vb r ¼ hi ebi . vui In the case of cylindrical coordinates, vb r vb r ¼ ¼ cos q b i þ sin q b j; vu1 vr so

vb r vb r ¼ ¼ r sin q b i þ r cos qb j; vu2 vq

vb r vb r ¼ ¼ kb vu3 vz

       vb  vb  vb r r r      h1 ¼ hr ¼   ¼ 1; h2 ¼ hq ¼   ¼ r; h3 ¼ hz ¼  vu  ¼ 1: vu1 vu2 3 Also, vb r vu i þ sin qb j; eb1 ¼ ebr ¼ 1 ¼ cos q b h1 vb r i þ r cos q b j vu2 r sin q b eb2 ¼ ebq ¼ ¼ ¼ sin q b i þ cos qb j; h2 r vb r vu3 b ¼ k. eb3 ¼ ebz ¼ h3 Back to the general case, we have db r¼

vb r vb r vb r du1 þ du2 þ du3 ¼ h1 eb1 du1 þ h2 eb2 du2 þ h3 eb3 du3 . vu1 vu2 vu3

For cylindrical coordinates, this is db r ¼ h1 eb1 du1 þ h2 eb2 du2 þ h3 eb3 du3 ¼ ebr dr þ rb e q dq þ ebz dz.

1.2 Curvilinear Coordinates

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 ofthe parallelepiped formed  volume   b Bb and Cb is  A\$ b Bb  Cb  (See Exercise 1). For the by three noncoplanar vectors A; Cartesian case, we compute an incremental volume element dV using b db b b r ¼ xb i þ yb j þ z k; r ¼ dx b i þ dy b j þ dz k. Then

     dV ¼ dx b i\$ dy b j  dz kb  ¼ dxdydz.

For the case db r ¼ h1 eb1 du1 þ h2 eb2 du2 þ h3 eb3 du3 . e 2  eb3 Þj dV ¼ jh1 du1 eb1 \$ðh2 du2 eb2  h3 du3 eb3 Þj ¼ h1 du1 h2 du2 h3 du3 jb e 1 \$ðb ¼ h1 h2 h3 du1 du2 du3 since jb e 2  eb3 Þj ¼ 1 because fb e 1 \$ðb e 1 ; eb2 ; eb3 g is an orthonormal system. Another way to do this computation is to use     vb r vb r vb r   dV ¼  \$ du1 du2 du3  vu1 vu2 vu3  and that   vx   vu  1     vx vb r vb r vb r ¼  \$  vu1 vu2 vu3  vu2   vx   vu3

vy vu1 vy vu2 vy vu3

 vz  vu1   vz  . vu2   vz  vu3 

(1)

The determinant in Eq. (1) is called the Jacobian of x,y,z with respect to u1, u2, u3 vðx; y; zÞ . So we have and is denoted vðu1 ; u2 ; u3 Þ    vðx; y; zÞ   du1 du2 du3 . dV ¼  vðu1 ; u2 ; u3 Þ We now compute dV for cylindrical coordinates. We demonstrate two methods. First, we use dV ¼ h1 h2 h3 du1 du2 du3 where u1 ¼ r, u2 ¼ q, u3 ¼ z so that du1 ¼ dr, du2 ¼ dq, u3 ¼ dz. We have previously found that h1 ¼ 1, h2 ¼ r, h3 ¼ 1 so

19

20

CHAPTER 1 Preliminaries

dV ¼ h1 h2 h3 du1 du2 du3 ¼ r dr dq dz. For the second method we compute the Jacobian. We have x ¼ r cos q;

vx vx vx vx vx vx ¼ cos q; ¼ r sin q; ¼0 ¼ ¼ ¼ vu1 vr vu2 vq vu3 vz

so

y ¼ r sin q;

so

z ¼ z;

vy vy vy vy vy vy ¼ sin q; ¼ r cos q; ¼0 ¼ ¼ ¼ vu1 vr vu2 vq vu3 vz

so

vz vz vz vz vz vz ¼ 0; ¼ ¼ 0; ¼ ¼ ¼ 1: vu1 vr vu2 vq vu3 vz

Thus   vx   vu  1   vx   vu  2   vx   vu3

vy vu1 vy vu2 vy vu3

 vz  vu1   cos q  vz   ¼  r sin q vu2   0   vz  vu3 

sin q r cos q 0

 0   0 ¼ r  1

and dV ¼ rdrdqdz. In multivariable calculus, one shows that if yb ¼ f ðb x Þ, then ZZZ ZZZ gðb y Þdy1 dy2 dy3 ¼ gðf ðb x ÞÞjdet Jðb x Þjdx1 dx2 dx3 V

V0

(2)

where Jðb x Þ is the matrix from which the Jacobian is formed. There are different ways that Eq. (2) is expressed in other sources. One other way is   ZZZ ZZZ  vðx; y; zÞ  dudvdw. f ðx; y; zÞdxdydz ¼ f ðxðu; v; wÞ; xðu; v; wÞ; xðu; v; wÞÞ vðu; v; wÞ V V0 Example: Evaluate

ZZ A

ex

2

þy2

dxdy

where A is the circle x2þy2  9, by changing to polar coordinates. In Cartesian coordinates, Z 3 Z y¼pﬃﬃﬃﬃﬃﬃﬃﬃ ZZ 9x2 2 2 x2 þy2 ex þy dxdy ¼ dxdy. pﬃﬃﬃﬃﬃﬃﬃﬃ e A

x¼3

y¼ 9x2

1.2 Curvilinear Coordinates

To convert to polar coordinates, we   vx    vðx; yÞ  vr ¼ x Þj ¼  jdetJðb vðr; qÞ  vy  vr

compute  vx    vq   cos q ¼ vy   sin q  vq

 r sin q  ¼ r. r cos q 

The region A in polar coordinates is 0  r  3, 0  q  2p. So, in this case, Eq. (2) says ZZ 2 2 ex þy dxdy A Z 3 Z y¼pﬃﬃﬃﬃﬃﬃﬃﬃ Z 3 Z 2p 9x2 2 x2 þy2 ¼ dxdy ¼ er rdqdr pﬃﬃﬃﬃﬃﬃﬃﬃ e x¼3 y¼ 9x2 Z 3 r2

r¼0

q¼0

e rdr ¼ p e9  1 .

¼ 2p

0

Example: We compute

ZZ

A

x2 þ y2 dxdy

where A is the region bounded by 1  xy  9, and the lines y ¼ x and y ¼ 4x. We seek a coordinate system (u,v) so that the transformed region of integration will be a rectangle au  b, c  v  d. The graph of the region A is shown in Fig. 1.2.3. y 10

y = 4x

8 6 y=x

4

xy = 9

2 xy = 1 2

FIGURE 1.2.3

4

x 6

8

21

22

CHAPTER 1 Preliminaries y If we let u ¼ xy then 1  u  9. If we let v ¼ , then 1  v  4. Now x  y pﬃﬃﬃﬃﬃ 2 uv ¼ ðxyÞ ¼ y ; so y ¼ uv x rﬃﬃﬃ u xy u ¼ y ¼ x2 ; so x ¼ . v v x We compute the Jacobian. We have rﬃﬃﬃ rﬃﬃﬃ rﬃﬃﬃ vx 1 vx 1 u vy 1 v vy 1 u ¼ pﬃﬃﬃﬃﬃ; ¼ ¼ ; ¼ ; . vu 2 uv vv 2v v vu 2 u vv 2 v Then   vx  vðx; yÞ  vu ¼ vðu; vÞ  vy  vu

 rﬃﬃﬃ    vx   p1ﬃﬃﬃﬃﬃ  1 u     2v v  1 vv   2 uv ¼ rﬃﬃﬃ rﬃﬃﬃ  ¼ 2v   vy   1 v 1 u      vv 2 u 2 v 

and ZZ

A

0

Z

2

x2 þ y dxdy ¼

4

@

v¼1

Z

9

1 vðx; yÞ duAdv þ uv  v vðu; vÞ

u

u¼1

1 9  2 Z 1 u 1 4 u u2  A @ þ vu du dv ¼ ¼ þ dv 2v 2 v¼1 2v2 2 u¼1 v¼1 u¼1 v 0

Z

4

Z

Z

4



¼ 1

9

 1 20 2 þ 1 dv ¼ 75: v

In the next section, we shall use the following forms of the change of variables equation: 1. If 41 exists and is differentiable for 4(a)  x  4(b), then Z b Z 4ðbÞ    0 f ðtÞh½4ðtÞdt ¼ f 41 ðxÞ hðxÞ 41 ðxÞ dx. 2.

RRR V

a

f ðx; y; zÞdxdydz ¼

4ðaÞ

RRR V0

f ðx1 ; x2 ; x3 Þ h1 h2 h3 dx1 dx2 dx3 .

THE GRADIENT Next we determine the gradient of a function f, denoted Vf . In Cartesian coordinates Vf ¼

vf b vf b vf b i þ j þ k. vx vy vz

1.2 Curvilinear Coordinates

To compute Vf in the general case, we set Vf ¼ Ab e 1 þ Bb e 2 þ Cb e 3 and write df in two different ways. First, df ¼ Vf \$db r and using that db r ¼ h1 eb1 du1 þ h2 eb2 du2 þ h3 eb3 du3 we get e 2 þ Cb e 3 Þ\$ðh1 eb1 du1 þ h2 eb2 du2 þ h3 eb3 du3 Þ df ¼ Vf \$db r ¼ ðAb e 1 þ Bb ¼ Ah1 du1 þ Bh2 du2 þ Ch3 du3 .

(3)

Second, df ¼

vf vf vf du1 þ du2 þ du3 . vu1 vu2 vu3

(4)

From Eqs. (3) and (4), we get Ah1 du1 þ Bh2 du2 þ Ch3 du3 ¼

vf vf vf du1 þ du2 þ du3 vu1 vu2 vu3

so A¼

1 vf 1 vf 1 vf ; B¼ ; C¼ . h1 vu1 h2 vu2 h3 vu3

Thus e 2 þ Cb e3 ¼ Vf ¼ Ab e 1 þ Bb

1 vf 1 vf 1 vf eb1 þ eb2 þ eb3 . h1 vu1 h2 vu2 h3 vu3

Accordingly, we write V as the operator V¼

1 v 1 v 1 v eb1 þ eb2 þ eb3 . h1 vu1 h2 vu2 h3 vu3

For the cylindrical coordinate case, we again have h1 ¼ 1; h2 ¼ r; h3 ¼ 1;

eb1 ¼ ebr ; eb2 ¼ ebq ; eb3 ¼ ebz

so in cylindrical coordinates, v 1 v v þ ebq þ ebz and vr r vq vz vf 1 vf vf vf 1 vf vf Vf ¼ ebr þ ebq þ ebz ¼ ebr þ ebq þ ebz . vr r vq vz vr r vq vz V ¼ ebr

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 Df ( some authors use V2 f ) in Cartesian coordinates is defined by Df ¼ V2 f ¼

v2 f v2 f v2 f þ þ . vx2 vy2 vz2

23

24

CHAPTER 1 Preliminaries

The notation V2 for the Laplacian is suggestive because V\$Vf gives the Laplacian of f. We have V¼

1 v 1 v 1 v þ eb2 þ eb3 ; and eb1 h1 vu1 h2 vu2 h3 vu3

Vf ¼

1 vf 1 vf 1 vf þ eb2 þ eb3 eb1 h1 vu1 h2 vu2 h3 vu3

v^ ei but computing V\$Vf is not as straightforward as it might seem. This is because is vuj not a simple expression. In fact, vb e i ebj vhj ¼ vuj hi vui

if isj and

X 1 vhi vb ei ebk . ¼ h vuk vui ksi k

We demonstrate the validity of X 1 vhi vb ei ¼ ebk vui h vuk ksi k for cylindrical coordinates with i ¼ 2. We have  v  r cos qb i þ r sin q b j þ z kb ¼ r sin q b i þ r cos qb j eb2 ¼ ebq ¼ vq so vb eq ¼ r cos q b i  r sin q b j. vq Also eb1 ¼ ebr ¼

 v r cos q b i þ r sin qb j þ z kb ¼ cos q b i þ sin q b j. vr

Now

      X 1 vh2 1 vh2 1 vh2 ebk ¼  eb1 þ eb3 h vuk h1 vu1 h3 vu3 ks2 k      vh2 vh2 ¼ eb1 þ eb3 vu1 vu3   vh2 ¼ 1. Also u3 ¼ z so since h1 ¼ h3 ¼ 1. Now h2 ¼ r and u1 ¼ r so vu1   vh2 ¼ 0. Thus vu3        vh2 vh2 i þ r sin q b j e 1 ¼  r cos q b  eb1 þ eb3 ¼ b vu1 vu3 

so the formula holds in this case.

1.2 Curvilinear Coordinates

.

In fact, it will be simpler to derive expressions for V2 f and V  F after we have studied the divergence theorem and Stokes’ theorem in Section 2.3. For now, we simply state the results:        1 v h2 h3 vf v h1 h3 vf v h1 h2 vf þ þ ð4Þ V2 f ¼ h1 h2 h3 vu1 h1 vu1 vu2 h2 vu2 vu3 h3 vu3    h eb h eb h eb  2 2 3 3  1 1   . 1  v v v  V F ¼ (5) h1 h2 h3  vu1 vu2 vu3     h 1 F 1 h 2 F2 h 3 F 3  .

where F ¼ F1 eb1 þ F2 eb2 þ F3 eb3 . For the case of cylindrical coordinates, we have h1 ¼ h3 ¼ 1; h2 ¼ r; so

eb1 ¼ ebr ; eb2 ¼ ebq ; eb3 ¼ ebz ;

u1 ¼ r; u2 ¼ q; u3 ¼ z;

       1 v vf v 1 vf v vf V f ¼ r þ þ r r vr vr vq r vq vz vz   1 v2 f vf 1 v2 f v2 f v2 f 1 vf 1 v2 f v2 f ¼ þ r þ þ . ¼ r 2þ þ þ r vr vr r vq2 vz2 vr 2 r vr r 2 vq2 vz2 2

We also have    eb rb e q ebz   r     . 1  v v v  V F ¼  r  vr vq vz       F1 rF2 F3         1 vF3 v vF3 vF1 v vF1 ¼ ðrF2 Þ   ðrF2 Þ ebr   rb eq þ ebz r vz vr vq vr vz vq       1 vF3 vF2 vF1 vF3 1 vF2 vF1 ¼ r   þ F2  ebr þ ebq þ ebz . r vq r vz vz vr vr vq

SPHERICAL COORDINATES One of the most commonly used three-dimensional coordinate systems is spherical coordinates. (In the examples above, we used cylindrical coordinates because the computations are simpler.)

25

26

CHAPTER 1 Preliminaries

The transformations are x ¼ r sin q cos 4 y ¼ r sin q sin 4 z ¼ r cos q. The ranges of the variables are 0  r 0 and d > 0, there is a number N(ε,d) so that if n  N(ε,d), then Z d fn ðxÞdx > 1  ε. d

Conditions 1 and 2 say that each fn(x) is a probability density function, and Condition 3 says that as n becomes large the area under the graph of fn(x) becomes concentrated near x ¼ 0. Fig. 1.3.1 illustrates the idea of an approximate identity.

15 f4 10 f3 f2

5 f1

−1

FIGURE 1.3.1

−0.5

0.5

1

29

30

CHAPTER 1 Preliminaries

In Exercise 1 we give several examples of approximate identities. The next theorem gives a crucial feature of approximate identities. Theorem: Suppose that {fn(x)} is an approximate identity at x ¼ 0 and g(x) is a bounded function that is continuous at x ¼ 0. Then Z ∞ fn ðxÞgðxÞdx ¼ gð0Þ. lim n/∞

∞

Proof: Let ε > 0 be given. We show that if g(x) is a bounded function that is continuous at x ¼ 0, then there is a positive integer N(g,ε) (emphasizing that N depends on both g(x) and ε) so that if n > N(g,ε), then Z   

∞ ∞

  fn ðxÞgðxÞdx  gð0Þ < ε.

The intuition of the proof is simple: For x very close to 0, g(x) is approximately g(0) because g(x) is continuous. Also, because {fn(x)} is an approximate identity at x ¼ 0, for d small Z d Z d Z ∞ fn ðxÞgðxÞdx z fn ðxÞgðxÞdx z gð0Þ fn ðxÞdx z gð0Þ\$1 ¼ gð0Þ. ∞

d

d

To make the argument rigorous, note that Z   

∞

 Z   fn ðxÞgðxÞdx  gð0Þ ¼ 

∞

Z  ¼  Now Z   

∞ ∞

Z fn ðxÞgðxÞdx  gð0Þ

∞

d ∞

Z þ

d

for any d > 0.

Z fn ðxÞ½gðxÞ  gð0Þdx þ   fn ðxÞ½gðxÞ  gð0Þdx;

d

d

∞ ∞

  fn ðxÞ½gðxÞ  gð0Þdx.

  fn ðxÞ½gðxÞ  gð0Þdx

Z  ¼ 

fn ðxÞ½gðxÞ  gð0Þdx

  fn ðxÞdx

1.3 Approximate Identities and the Dirac-d Function

Since g(x) is bounded, there is a number M > 0 so that jgðxÞj  M for all x, so Z   

d

∞

 Z  fn ðxÞ½gðxÞ  gð0Þdx  2M

d ∞

fn ðxÞdx;

 R∞ and likewise, for  d fn ðxÞ½gðxÞ  gð0Þdx. We can now choose d > 0 so that if jx  0j < d, then jgðxÞ  gð0Þj < 2ε . After d is chosen, we can find an N(d) so that if n > N(d), then Z ∞ Z d ε fn ðxÞdx þ fn ðxÞdx < . 8M ∞ d Thus, we have Z   

Z

d

fn ðxÞ½gðxÞ  gð0Þdx þ

∞

Z  Z þ

d ∞ d

d

d

d

Z d

Z jfn ðxÞjjgðxÞ  gð0Þjdx þ

  fn ðxÞ½gðxÞ  gð0Þdx

fn ðxÞ½gðxÞ  gð0Þdx þ

jfn ðxÞjjgðxÞ  gð0Þjdx

d

jfn ðxÞjjgðxÞ  gð0Þjdx < 2\$2M\$

ε ε þ 1\$ ¼ ε. 8M 2

We now heuristically describe the limit of an approximate identity. If for each n, fn(x) is a continuous function then, under some rather reasonable conditions, for each xs0; lim fn ðxÞ ¼ 0. With these conditions, it must be that lim fn ð0Þ ¼ ∞. This is n/∞

n/∞

the idea behind the Dirac-d function at x ¼ 0, which we denote d0(x). Definition: The Dirac-d function at x ¼ 0, denoted d0(x), is defined by the condition Z ∞ gðxÞd0 ðxÞdx ¼ gð0Þ ∞

for any function g(x) that is continuous at x ¼ 0. We define dx0 (x) ¼ d0(x  x0), so that Z ∞ gðxÞdx0 ðxÞdx ¼ gðx0 Þ ∞

for any function g(x) that is continuous at x0. The Dirac-d function is not actually a function, but is what is known as a “generalized function” or a “distribution.”

31

32

CHAPTER 1 Preliminaries

THE DIRAC-d FUNCTION IN PHYSICS The Dirac-d function was used by physicists before mathematicians created a structure that would make it mathematically rigorous. (This structure is called distribution theory.) We explain the ideas of the Dirac-d function by a physical example. Suppose we want to model an impulse I as being a large constant force being applied for a very brief period of time, such as a hammer blow. If f (t) is the force at time t, then in general Z ∞ f ðtÞdt. I¼ 0

We want f (t) to be a positive constant during a small period of time, and 0 otherwise. If [0,h] is the period for which f (t) is nonzero, then Z h I¼ kdt ¼ kh; 0

so

8 > < I 0  t  h; f ðtÞ ¼ h > : 0 otherwise.

(1)

The Heaviside function, H(t), is defined by  0 t0 Some sources define H(0) ¼ 1/2, but the difference will not affect our work. Suppose we take I ¼ 1 in Eq. (1). We want to know what happens as h / 0. As h / 0, f (t) must change as h changes if it is to be true that Z h f ðtÞdt ¼ 1: 0

So, for a given h, we denote the force function by fh. With these modifications we have 8 >

: 0 otherwise Note that HðtÞ  Hðt  hÞ ¼



1 if t > 0 and t  h < 0 i.e. 0 < t < h . 0 otherwise

1.3 Approximate Identities and the Dirac-d Function

Thus HðtÞ  Hðt  hÞ . (2) h Now we could attempt to take the limit in Eq. (2) as h / 0. While this is not appropriate mathematically, if we examine the right-hand side of Eq. (2) formally, it appears that we are looking for the derivative of the Heaviside function. We would have to establish a context in which this makes sense, but one could also call HðtÞ  Hðt  hÞ dðtÞ ¼ lim h/0 h fh ðtÞ ¼

the Dirac-d function. Thus, in some contexts, the Dirac-d function is said to be the derivative of the Heaviside function. We now establish the connection between the Dirac-d function and approximate identities. HðtÞ  Hðt  hÞ for h ¼ 1; 12; 14 and 18 are shown in Fig. 1.3.2. The graphs of h Note that the area under each graph is 1, the function is positive, and the area becomes more concentrated about t ¼ 0 as h / 0. Also 8

h=

1 8

h= 4

1 4

h= 2

1 2

h=1

1

1 8

FIGURE 1.3.2

1 4

1 2

1

33

34

CHAPTER 1 Preliminaries

Z b a

 HðtÞ  Hðt  hÞ dt ¼ 1 h

if [a,b] is any interval that contains [0,h]. Proceeding heuristically, if we take the limit as h / 0, we have Z b dðtÞdt ¼ 1 a

whenever [a,b] contains 0. Obviously, we have done some things that are not mathematically rigorous. Notice, however, that  Z b HðtÞ  Hðt  hÞ lim gðtÞdt ¼ gð0Þ h/0 a h if g(t) is a function that is continuous at t ¼ 0 and 0 ˛ (a,b). Thus, we could also define d(t) by the condition Z b dðtÞgðtÞdt ¼ gð0Þ (3) a

for every function g(t) that is continuous at t ¼ 0 and every interval (a,b) that contains 0. Recapping what we have done, d(t) or d0(t) is the function (actually the distribution) described by the condition of Eq. (3).

SOME CALCULUS FOR THE DIRAC-d FUNCTION We now demonstrate some formulas involving the Dirac-d function. One such formula is xd0 ðxÞ ¼ dðxÞ. In this and other formulas for the Dirac-d function, we mean equality in the weak sense. That is, in the sense that if g(x) is any suitably well-behaved function, then Z b Z b xd0 ðxÞgðxÞdx ¼  dðxÞgðxÞdx. a

a

We shall proceed formally rather than rigorously. We give a heuristic interpretation of Z ∞ d0 ðxÞgðxÞdx. ∞

Suppose we consider Z ∞

Z lim

∞ h/0

dðx þ hÞ  dðxÞ gðxÞdx. h ∞

1.3 Approximate Identities and the Dirac-d Function

Now 1 h

Z

1 1 dðx þ hÞgðxÞdx ¼ gðhÞ and h h ∞

so

Z

∞

Z

1 dðxÞgðxÞdx ¼ gð0Þ; h ∞

dðx þ hÞ  dðxÞ gðhÞ  gð0Þ gðxÞdx ¼ . h h

Also lim

h/0

Thus, we define

Z

gðhÞ  gð0Þ ¼ g0 ð0Þ. h

∞

d0 ðxÞgðxÞdx ¼ g0 ð0Þ:

With this convention, one also writes Z Z ∞ d0 ðxÞgðxÞdx ¼  ∞

∞

g0 ðxÞdðxÞdx.

Example: We show xd0 ðxÞ ¼ dðxÞ by showing that if g(x) is a differentiable function that is continuous at x ¼ 0 then Z ∞ d gðxÞ xd0 ðxÞdx ¼  gðxÞxjx¼0 ¼ ½gðxÞ þ g0 ðxÞxjx¼0 ¼ gðxÞ dx ∞ Z ∞ ¼ gðxÞ dðxÞdx. ∞

Example: Problems involving a change of variables. We want to compute d( f (x)) where f (x) is some suitably well-behaved function. To begin, we recall a change of variables formula of integration. If 41 exists and is differentiable for 4(a)  x  4(b), then Z b Z 4ðbÞ    0 f ðtÞh½4ðtÞdt ¼ f 41 ðxÞ hðxÞ 41 ðxÞ dx. (4) a

4ðaÞ

Taking h(x) ¼ d(x) in Eq. (4), we have Z b Z 4ðbÞ    0 f ðtÞd½4ðtÞdt ¼ f 41 ðxÞ dðxÞ 41 ðxÞ dx. a

4ðaÞ

35

36

CHAPTER 1 Preliminaries

Suppose 4(t) is monotone increasing for a  t  b. Now

0 41 ðtÞ ¼

1 40 ð41 ðtÞÞ

.

Suppose also that 4(c) ¼ 0 for some c between a and b. Then Z 4ðbÞ    0   0  f 41 ðxÞ dðxÞ 41 ðxÞ dx ¼ f 41 ð4ðcÞÞ 41 ðxÞ 

x¼4ðcÞ

4ðaÞ

¼

f ðcÞ . 40 ðcÞ

If 4(t) is monotone decreasing for a  t  b, we get the same result except for a change in sign (Exercise 7). This yields the following result: If 4(t) is monotone and 4(c) ¼ 0 and 40 (c) s 0, then dð4ðtÞÞ ¼

dðt  cÞ . j40 ðcÞj

(5)

Example: By (5), to compute d(3t  6), we have 4(t)  6, so 4(t) ¼ 0 if t ¼ 1/2. Then  ¼ 3t 

c ¼ 12 and j40 ð1=2Þj ¼ 3 so dð3t  6Þ ¼ 13 d t  12 .

Example: We next compute d[(ta) (tb)] where a < b. In this case, the function 4(t) ¼ (ta) (tb) is not monotone, but is decreasing for t < ða þ bÞ=2 and increasing for t > ða þ bÞ=2. Also 4(a) ¼ 0 and 4(b) ¼ 0. Now 4ðtÞ ¼ t2  ða þ bÞt þ ab;

so 40 ðtÞ ¼ 2t  ða þ bÞ.

Thus j40 ðaÞj ¼ ja  bj and j40 ðbÞj ¼ jb  aj so Z

ðaþbÞ=2

∞

and thus

f ðtÞdð4ðtÞÞdt ¼ Z

∞

f ðaÞ ja  bj

Z and

f ðtÞdð4ðtÞÞdt ¼

ðaþbÞ=2

f ðtÞdð4ðtÞÞdt ¼

1 ½f ðaÞ þ f ðbÞ. ja  bj

Hence, d½ðt  aÞðt  bÞ ¼

1 ðdðt  aÞÞ þ ðdðt  bÞÞ. ja  bj

This example can be generalized to show that if Y f ðxÞ ¼ ðx  xi Þ i

f ðbÞ jb  aj

1.3 Approximate Identities and the Dirac-d Function

and f 0 (xi) s 0 then d½f ðxÞ ¼

X dðx  xi Þ jf 0 ðxi Þj

i

.

THE DIRAC-d FUNCTION IN CURVILINEAR COORDINATES The representation of the Dirac-d function in higher dimensions for Cartesian coordinates is fairly obvious. In three dimensions, if xb ¼ ðx; y; zÞ

xb1 ¼ ðx1 ; y1 ; z1 Þ

then dðb x  xb1 Þ ¼ dðx  x1 Þ dðy  y1 Þdðz  z 1 Þ so that



ZZZ V

f ðb x Þdðb x  xb1 Þdb x¼

f ðb x1Þ 0

if xb1 ˛V . if xb1 ;V

The case of orthogonal curvilinear coordinates is not as straight forward. In spherical coordinates, one might expect that dðb rb r 1 Þ ¼ dðr  r1 Þ dðq  q1 Þdð4  4 1 Þ; but this is not the case. What we are doing in 3space in evaluating.

RRR V

f ðb x Þdðb x

x is evaluating over an increment of volume dV. The reason for the simplicity in xb1 Þdb Cartesian coordinates, is that in that case dV ¼ dxdydz. In curvilinear coordinates, recall from Section 1.2 that converting from Cartesian coordinates to an orthogonal curvilinear coordinate system (x1,x2,x3) we must introduce the scaling factors "   2  2 #12 vx 2 vy vz þ þ . hi ¼ vxi vxi vxi For example, in spherical coordinates x ¼ r sin q cos 4;

y ¼ r sin q sin 4;

z ¼ cos q

and h1 ¼ 1, h2 ¼ r, h3 ¼ r sin q. We have 1 dxdydz h1 h 2 h 3

dx1 dx2 dx3 ¼ so

ZZZ

ZZZ f ðx; y; zÞdxdydz ¼ V

so that in spherical coordinates

V0

f ðx1 ; x2 ; x3 Þ h1 h2 h3 dx1 dx2 dx3 ;

37

38

CHAPTER 1 Preliminaries

ZZZ

ZZZ f ðx; y; zÞdxdydz ¼

V

f ðr; q; 4Þr 2 sin q drdqd4. V0

Thus,



ZZZ dðb x 0 Þdxdydz ¼ V

ZZZ

if xb1 ˛V if xb1 ;V

1 0 

dðb r 0 Þr sin q drdqd4 ¼ 2

V0

1 0

if b r 0 ˛V . if b r 0 ;V

We then have the correspondence in spherical coordinates, dðb rb r 1Þ ¼

1 dðr  r1 Þ dðq  q1 Þdð4  41 Þ r2 sin q

and, in the general case, dðb rb r 1 Þ ¼ dððx1 ; x2 ; x3 Þ  ðx11 ; x21 ; x31 ÞÞ ¼

dðx1  x11 Þ dðx2  x21 Þ dðx3  x31 Þ . h1 h2 h3

If there is symmetry in a problem, the integral with respect to the symmetrical variable must be projected (integrated) out. For example, in spherical coordinates, if there is symmetry with respect to 4, we have Z 2p r 2 sin qd4 ¼ 2pr 2 sin q 4¼0

so we would have 1 dðr  r1 Þ dðq  q1 Þ. 2pr2 sin q We note that in spherical coordinates, it is often useful to replace the variable q with cosq, in which case dðb rb r 1Þ ¼

dðb rb r 1Þ ¼

1 dðr  r1 Þ dðq  q1 Þdð4  41 Þ r2

as we show in Exercise 6.

EXERCISES 1. Show that the following sequences of functions define an approximate identity at x ¼ 0. 8 >

: 0 otherwise n . b. fn ðxÞ ¼ ½pð1 þ n2 x2 Þ

1.3 Approximate Identities and the Dirac-d Function   2 2 n n x . exp  2 2p (

n c n 1  x2 jxj < 1 d. fn ðxÞ ¼ 0 otherwise

n R1 where cn is chosen so that 1 cn 1  x2 dx ¼ 1. 

c. fn ðxÞ ¼

2. Suppose that {fn(x)} is an approximate identity. Show that for any positive integer k, {kfn(kx)} is also an approximate identity. 3. Prove the following properties for the Dirac-d function: a. xn dðnÞ ðxÞ ¼ ð1Þn n!dðxÞ. 1 dðxÞ. b. dðaxÞ ¼ jaj

c. sin x d0 ðxÞ ¼ dðxÞ. d. cos x d0 (x) ¼ d0 (x). 4. a. Show that in cylindrical coordinates, 1 dðb rb r 1 Þ ¼ dðr  r1 Þdðq  q1 Þdðz  z1 Þ. r b. Find the Dirac-d function in the case there is symmetry about q. 5. Find the Dirac-d function in spherical coordinates when there is symmetry with respect to both the q and 4 variables. 6. Find d(x2  x  12). 7. Evaluate Z ∞

ð3x  4Þd x2  4 dx. ∞

8. Show that in spherical coordinates, if we replace the variable q with cosq, then dðb rb r 1 Þ ¼ r12 dðr  r1 Þ dðq  q1 Þdð4  41 Þ. 9. Show that if f (x), f 0 (x),..,f (n) (x) are continuous at x ¼ 0, then Z ∞ f ðxÞdðnÞ ðxÞdx ¼ ð1Þn f ðnÞ ð0Þ. ∞

10. Let xb ¼ ðx1 ; x2 ; x3 Þ denote an arbitrary point in ℝ3 and bx ¼ ðx1 ; x2 ; x3 Þdenote an arbitrary point in an orthogonal curvilinear coordinate system. Show that   d b xb x0  dðb x  xb0 Þ ¼   vðx1 ; x2 ; x3 Þ vðx ; x ; x Þ 1 2 3 vðx1 ; x2 ; x3 Þ is the Jacobian. where vðx1 ; x2 ; x3 Þ

39

40

CHAPTER 1 Preliminaries

11. The convolution of the functions f and g, denoted f * g, is defined by Z ∞ ðf  gÞðtÞ ¼ f ðxÞgðx  tÞdx. ∞

What is d * g? 12. We can show

 2 pﬃﬃﬃ x exp  2 dx ¼ b p. b ∞

Z

What is  2 1 x lim pﬃﬃﬃ exp  2 ? bY0 b p b

1.4 THE ISSUE OF CONVERGENCE Many functions in mathematics and physics are expressed as a series of functions, often because the desired function is the superposition of several components. One example is that the sound of a note on a musical instrument is the result of a fundamental frequency and many harmonics. (The reason middle C on a trumpet sounds different from the same note on a violin is that the relative amplitudes of the harmonics are different.) When dealing with series, we must be careful that processes such as differentiation and integration behave with infinite sums as they do with finite sums. Whether they do depend on the way the series converges. In this section, we discuss different types of convergence of series of functions, how this impacts mathematical operations on such series and interchanging the order of limits. We also discuss the representation of a function by its Taylor series.

SERIES OF REAL NUMBERS Series are inextricably linked with sequences, and we begin with some basic ideas of sequences of numbers. This section is intended as a review and will primarily consist of a list of facts. A more complete discussion is available in many sources, including Kirkwood, An Introduction to Analysis, Second Edition. Intuitively, a sequence is an infinite list of numbers. A sequence converges to the number L provided the numbers in the list can be made arbitrarily close to L by going sufficiently far out in the list. More precisely: Definition: The sequence {an} converges to L provided given ε > 0, there is a number N(ε) such that if n > N(ε), then jan  Lj < ε. If a sequence does not converge, then it is said to diverge.

1.4 The Issue of Convergence

Definition: The sequence {an} is a Cauchy sequence provided given ε > 0, there is a number N(ε) such that if n,m > N(ε), then jan  am j < ε. Theorem: A sequence {an} converges if and only if it is a Cauchy sequence. Intuitively, a series is the sum of an infinite number of numbers. A series is usually represented by ∞ X

an ¼ a1 þ a2 þ /;

n¼1

P or simply an. P Associated with each series an is a sequence of partial sums {Sn}, defined by S k ¼ a 1 þ a2 þ / þ a k . Definition:P The series an converges P to L if the associated sequence of partial sums {Sn} converges to L. The series an diverges if {Sn} diverges. As was mentioned above, we shall want to know if a series of functions converges in a manner that permits certain manipulations. A frequently used test (the Weierstrass M-test) relies on knowing that a related series of numbers converges. We now list some of the most common results pertaining to convergence of series of numbers. Theorem: P If an converges, then lim an ¼ 0. n/∞

then P It is the contrapositive of this theorem that is most useful; i.e., if lim an s0 P an diverges. There are many examples ofPseries for which lim an ¼ 0 and an 1 diverges.) diverges. (One example: We shall see that n Theorem: P P If the P series an converges to L and the series bn converges to M, then the series (aan þ bbn) converges to aL þ bM. A series of the form a þ ar þ ar 2 þ ar 3 þ / is called a geometric series. Theorem: The geometric series a þ ar þ ar 2 þ ar 3 þ /

ðas0Þ a converges to if jrj < 1 and diverges if jrj  1: 1r Theorem (Comparison tests) Suppose 0  an  bn.

41

42

CHAPTER 1 Preliminaries P P 1. If Pbn converges, thenP an converges. 2. If an diverges, then bn diverges. Theorem: 1. Ratio test: If

P

an is a series of positive terms, and anþ1 r ¼ lim ; n/∞ an

P then an converges P if r < 1 and diverges if r > 1. The test is inconclusive if r ¼ 1. 2. Root test: If an is a series of positive terms, and 1

r ¼ lim ðan Þn ; then

P

n/∞

an converges if r < 1 and diverges if r > 1. The test is inconclusive if r ¼ 1.

Theorem (Integral test): is a monotone decreasing function Suppose a1  a2  a3  /  0, and f (x)P R ∞ with f (n) ¼ an for each positive integer n. Then an converges if and only if 1 f ðxÞdx converges. Theorem ( p-series test): P1 converges if p > 1 and diverges if p  1. The series np Theorem: The series ! X ak nk þ ak1 nk1 þ / þ a0 bj nj þ bj1 nj1 þ / þ b0 ! P nk converges if and only if converges. (This assumes that nj bjnj þ bj  1nj  1 þ / þ b0 s 0 for any n, and the coefficients are bounded.)

CONVERGENCE VERSUS ABSOLUTE CONVERGENCE The series we have considered so far, with the exception of geometric series, have been made up of positive terms. Another type of series that often appears is alternating series. An alternating series is the one in which each term differs in sign from its predecessor. Such series are typically represented as X X ð1Þnþ1 an ð1Þn an or where an > 0. It may be that a series where not all terms have the same sign converges, but if we change the series so that all terms do have the same sign, then the altered series diverges. Such series give rise to different notions of convergence called absolute and conditional convergence.

1.4 The Issue of Convergence

Definition: P P P A series P an is said to converge jan j converges. If an conP absolutely if verges, but jan j diverges, then an is conditionally convergent. Theorem: P P If jan j converges, then an converges. Theorem (Alternating series test) Suppose a1  a2  a3  /  0; lim an ¼ 0;

n/∞

then

X

ð 1Þn an and

X

ð 1Þnþ1 an

converge.

SERIES OF FUNCTIONS The idea with series of functions is to add infinitely many functions. The approach to developing a theory for series of functions follows that of series of numbers, in that we begin with sequences of functions. Our setting is for each positive integer n we have a function fn(x) defined on a set E. We want to know if there is a “limit function” to which the sequence converges. The first question we need to address is what is meant by a limit function. For the limit function f (x) that we consider, we assume that for each x ˛ E, the sequence of numbers {fn(x)} converges, and we define the limit function f (x) by f ðxÞ ¼ lim fn ðxÞ. n/∞

We shall see that this idea requires more precision and, as a consequence, we distinguish between pointwise convergence and uniform convergence. So suppose that for each x ˛ E, the sequence of numbers {fn(x)} converges, and we define f (x) as above. Definition: We say that the sequence of functions { fn(x)} converges pointwise to f (x) on E if given e > 0 and x ˛ E, there is a number N(x,e), so that if n > N(x,e), then jfn ðxÞ  f ðxÞj < e. Note that N(x,ε) depends on both e and x in pointwise convergence. Definition: We say that the sequence of functions {fn(x)} converges uniformly to f (x) on E if given e > 0 there is a number N(e), so that if n > N(e), then jfn ðxÞ  f ðxÞj < e

for every x ˛E.

43

44

CHAPTER 1 Preliminaries

In uniform convergence, N(ε) depends only on ε. That is, with ε fixed, there is a number N(ε) that works for every x ˛ E. Example: The sequence of functions {fn(x)} defined by fn(x) ¼ xn on E ¼ [0,1] converges pointwise, but not uniformly, to  0 0x 0 be given. We show that there is a d(ε) > 0 so that if jx  cj < dðεÞ, then jf ðxÞ  f ðcÞj < ε. We have jf ðxÞ  f ðcÞj ¼ jf ðxÞ  fn ðxÞ þ fn ðxÞ  fn ðcÞ þ fn ðcÞ  f ðcÞj  jf ðxÞ  fn ðxÞj þ jfn ðxÞ  fn ðcÞj þ jfn ðcÞ  f ðcÞj. Since {fn(x)} converges uniformly to f (x), there is a number N(ε) so that if n > N(ε), then jfn ðxÞ  f ðxÞj < 3ε for every x. So we have ε ε jfn ðxÞ  f ðxÞj < and jfn ðcÞ  f ðcÞj < ; 3 3 whenever n > N(ε). Now choose a fixed N1 > N(ε). Since fN1 ðxÞ is continuous at x ¼ c, there is a number d(ε)>0 so that if jx  cj < dðεÞ, then j fN1 ðxÞ  fN1 ðcÞj < 3ε. Thus if jx  cj < dðεÞ, then jf ðxÞ  f ðcÞj < ε so f (x) is continuous at x ¼ c. Theorem: Suppose that each of the functions fn(x) is integrable on [a,b] and {fn(x)} converges uniformly to f (x) on [a,b]. Then f (x) is integrable on [a,b] and 0 1 Z b Z b Z b  lim fn ðxÞ dxh fn ðxÞdxA ¼ f ðxÞdx. lim @ n/∞

a

a

n/∞

a

We do not give a proof, which may be found in Kirkwood (1995). The difficult part of the proof is showing that f (x) is integrable on [a,b]. We note that uniform convergence does not ensure convergence of improper integrals as the example 8 >

: 0 otherwise shows. The next example shows that uniform convergence does not ensure that derivatives behave as we might hope. Example: Let sin nx fn ðxÞ ¼ pﬃﬃﬃ . n 1 Then jfn ðxÞj  pﬃﬃﬃ ; so {fn(x)} converges uniformly to 0. But n pﬃﬃﬃ fn0 ðxÞ ¼ ncos nx;

 and so fn0 ðxÞ does not converge.

1.4 The Issue of Convergence

This provides an example where     fn ðxÞ  fn ðcÞ fn ðxÞ  fn ðcÞ s lim lim . lim lim n/∞ x/c x/c n/∞ xc xc The situation for derivatives is addressed by the next theorem. Theorem: Let {fn(x)} be a sequence of functions that are differentiable on an open interval containing [a,b]. Suppose that 1. there isa point x0 ˛ [a,b] where {fn(x0)} converges, and 2. fn0 ðxÞ converges uniformly on [a,b]. Then 1. {fn(x)} converges uniformly to a function f (x) on [a,b] and 2. f 0 ðxÞ ¼ lim fn0 ðxÞ on (a,b). n/∞

The proof of this theorem is omitted, and we refer the reader to Kirkwood, An Introduction to Analysis, Second Edition. The next two theorems give criteria for when a sequence of functions converge uniformly. Theorem: The sequence of functions {fn(x)} converges uniformly to f (x) on E if and only if lim Mn ¼ 0, where

n/∞

Mn ¼ sup jfn ðxÞ  f ðxÞj. x ˛E

The proof follows directly from the definition as is left as Exercise 19. Example: We show {xenx} converges uniformly to 0 on [0,∞). We have d xenx ¼ enx  xnenx ¼ enx ð1  nxÞ. dx d xenx ¼ 0 if 1  nx ¼ 0 or x ¼ 1. One can check that f (x) ¼ xenx attains So dx n n its maximum on [0,∞) at x ¼ 1n. Thus

1 n 1n 1 Mn ¼ e ¼ e1 and lim Mn ¼ 0: n/∞ n n Theorem (Cauchy criterion for uniform convergence of sequences of functions): A sequence of functions {fn(x)} converges uniformly on E if and only if it is uniformly Cauchy on E; that is, if and only if given ε > 0, there is a number N(ε) so that if m,n > N(ε), then jfn ðxÞ  fm ðxÞj < ε for every x ˛ E.

47

48

CHAPTER 1 Preliminaries

Proof: Suppose {fn(x)} is uniformly Cauchy on E. We show that {fn(x)} is uniformly convergent. For any x0 ˛ E, {fn(x0)} is a Cauchy sequence of real numbers, and therefore converges. For each x ˛ E define f ðxÞ ¼ lim fn ðxÞ. n/∞

Now {fn(x)} converges pointwise to f (x), but we must show that the convergence is uniform. Let ε > 0 be given. There is a number N(ε) so that if m,n > N(ε), then ε jfn ðxÞ  fm ðxÞj < 2 for every x ˛E. Now suppose n > NðεÞ, and choose x0 ˛E. We shall show that jfn ð x0 Þ  f ð x0 Þj < ε. We have jfn ð x0 Þ  f ð x0 Þj  jfn ð x0 Þ  fm ð x0 Þj þ jfm ð x0 Þ  f ð x0 Þj for any m: Since f fm ð x0 Þg/ f ð x0 Þ, there is a number Mðε; x0 Þ (that depends on ε and x0) so that if m > Mðε; x0 Þ, then jfm ð x0 Þ  f ð x0 Þj < 2ε. Choose m > maxfNðεÞ; Mðε; x0 Þ g. Then ε ε jfn ð x0 Þ  f ð x0 Þj  jfn ð x0 Þ  fm ð x0 Þj þ jfm ð x0 Þ  f ð x0 Þj < þ ¼ ε 2 2 if n > N(ε). The proof that a uniformly convergent sequence of functions is uniformly Cauchy is left as Exercise 20. The work in this section that will be most important for us later deals with series of functions. The connection between series and sequences of functions is thePsame as with numbersdpartial sums. In particular, we say the series of functions fn(x) converges pointwise (uniformly) to f (x) on E if and only if the sequence of functions {Sn(x)} defined by Sn ðxÞ ¼ f1 ðxÞ þ f2 ðxÞ þ / þ fn ðxÞ converges pointwise (uniformly) to f (x) on E. We give two criteria to determine uniform convergence of series of functions. Theorem (Cauchy criterion P for uniform convergence of series of functions): The series of functions fn(x) converges uniformly on E if and only if given ε > 0, there is a number N(ε) so that if m > n > N(ε), then    m  X   fi ðxÞ < ε for every x˛E.    i¼nþ1 Proof: The theorem follows by noting that

1.4 The Issue of Convergence   m  X    fi ðxÞ; jSm ðxÞ  Sn ðxÞj ¼   i¼nþ1 

P where {Sn(x)} is the sequence of partial sums associated with the series fn(x). Perhaps the most frequently used test to show uniform convergence of series of functions is the Weierstrass M-test. TheoremP (Weierstrass M-test): Suppose fn(x) is a series of functions defined on a set E. Let Mn ¼ sup jfn ðxÞj. x ˛E

P

P If the series of numbers Mn converges, then fn(x) converges uniformly on E. Proof: We have, for any m and n with m > n     X m m m X X   fi ðxÞ  Mi for every x ˛E. jfi ðxÞj    i¼nþ1  i¼nþ1 i¼nþ1 By the Cauchy criterion for convergence of series real numbers, givenε > 0 there is a number N(ε) so that if m > n > N(ε), then m X

Mi < ε.

i¼nþ1

Thus,

   X m    fi ðxÞ < ε   i¼nþ1 

for every x˛E

P if m > n > N(ε), so fn(x) converges uniformly on E. We note that there are series of functions that converge uniformly for which the Weierstrass M-test fails. OneP example is developed in Exercise 22. Uniform convergence of fn(x) on [a,b] allows us to conclude 1 0 ! Z b X Z b ∞ ∞ X @ fn ðxÞdxA ¼ fn ðxÞ dx. n¼1

a

a

n¼1

POWER SERIES A series of the form a0 þ a1 ðx  cÞ þ a2 ðx  cÞ2 þ a3 ðx  cÞ3 þ / ¼

∞ X

an ðx  cÞn

n¼0

is called a power series. If we let z ¼ x  c, we can express the series as

P

anzn.

49

50

CHAPTER 1 Preliminaries

A major result about power series is the following theorem. Theorem: P For the power series anzn there is an extended real number R, 0  R  ∞ for which the series 1. converges absolutely if jzj < R 2. diverges if jzj > R. 3. If 0 < R < ∞, then the power series converges uniformly on [eRþε, Rε] for any ε > 0. P n If R P¼ 0,n then the series anz converges only for x ¼ 0. If R ¼ ∞, then the series anz converges absolutely for all values of z, and uniformly on any bounded interval. The number R in the theorem is called the radius of convergence of the power series. P n It can be calculated in either of the following ways: For the power series anz , let   l ¼ lim jan j1=n n/∞

or l ¼ lim

n/∞

janþ1 j jan j

(if both limits exist, they are equal) then 1 R¼ . l Examples: P xn . Here an ¼ 1=n!, so Consider the series n! 1 1 1 janþ1 j ðn þ 1Þ! ¼ 0 and R ¼ ¼ ∞. ¼ lim l ¼ lim ¼ lim 1 n/∞ jan j n/∞ n/∞ n þ 1 l n! Thus, this series converges absolutely for all values of x. Consider the series X n3 xn 5n

 Here, an ¼ n3 5n so 

l ¼ lim jan j n/∞

1=n



¼ lim

n/∞

.

3 1=n n ð5n Þ1=n

3 1=n n 1 ¼ lim ¼ n/∞ 5 5

1.4 The Issue of Convergence

since lim ðn Þ1=n ¼ 1:

n/∞

Thus, R ¼ 1l ¼ 5.

TAYLOR SERIES In Taylor series, we begin with a function f (x) that has derivatives of all orders at x ¼ c, and then construct a polynomial of degree n, called the Taylor polynomial of degree n, that provides “the best approximation” to f (x) at x ¼ c. To determine this polynomial, we must define what we mean by “the best approximation.” In the case of Taylor polynomials, if Pn(x) is the Taylor polynomial of degree n for f (x), we require that Pn ðcÞ ¼ f ðcÞ; P0n ðcÞ ¼ f 0 ðcÞ; P00n ðcÞ ¼ f 00 ðcÞ; .; Pn ðnÞ ðcÞ ¼ f ðnÞ ðcÞ. We now determine a formula for the constants ak in the Taylor polynomial for f (x) at x ¼ c. Suppose that f (x) that has derivatives of all orders at at x ¼ c, and let Pn ðxÞ ¼ a0 þ a1 ðx  cÞ þ a2 ðx  cÞ2 þ / þ an ðx  cÞn . The first condition we impose is f (c) ¼ Pn(c). Now Pn ðcÞ ¼ a0 ¼ f ðcÞ;

so a0 ¼ f ðcÞ.

Next, we require f 0 ðcÞ ¼ P0n ðcÞ We have P0n ðxÞ ¼ a1 þ 2a2 ðx  cÞ þ 3a3 ðx  cÞ2 þ 4a4 ðx  cÞ3 þ / þ nan ðx  cÞn1 so P0n ðcÞ ¼ a1 ¼ f 0 ðcÞ;

so a1 ¼ f 0 ðcÞ.

We seek to determine a formula for ak. Continuing, we have P00n ðxÞ ¼ 2a2 þ 2\$3a3 ðx  cÞ þ 4\$3a4 ðx  cÞ2 þ / þ nðn  1Þan ðx  cÞn2 so P00n ðcÞ ¼ 2a2 ¼ f 00 ðcÞ;

and a2 ¼

f 00 ðcÞ . 2

Next, n3 P000 n ðxÞ ¼ 2\$3a3 þ 4\$3\$2a4 ðx  cÞ þ / þ nðn  1Þðn  2Þan ðx  cÞ

so 000 P000 n ðcÞ ¼ 2\$3a3 ¼ f ðcÞ;

and a3 ¼

f 000 ðcÞ f 000 ðcÞ ¼ . 2\$3 3!

51

52

CHAPTER 1 Preliminaries

A pattern begins to emerge (that can be proved using induction) that ak ¼

f ðkÞ ðcÞ . k!

Definition: Let f (x) be a function that has derivatives of all orders at x ¼ c. The Taylor series for f (x) at x ¼ c is the power series ∞ X

an ðx  cÞn

where an ¼

n¼0

f ðnÞ ðcÞ . n!

If c ¼ 0, the series is called the Maclaurin series. If there is an open interval about x ¼ c for which f ðxÞ ¼

∞ ðnÞ X f ðcÞ ðx  cÞn n! n¼0

then f (x) is said to be analytic at x ¼ c. The most frequently used Taylor series (Maclaurin series) are those for ex, sin x and cos x. We demonstrate the computations for ex and sin x. Example: We compute the Maclaurin series for ex. We have f ðxÞ ¼ ex

f ð0Þ ¼ e0 ¼ 1

a0 ¼ f ð0Þ ¼ 1

f 0 ðxÞ ¼ ex

f 0 ð0Þ ¼ e0 ¼ 1 a1 ¼

f 0 ð0Þ ¼1 1!

f 00 ðxÞ ¼ ex

f 00 ð0Þ ¼ e0 ¼ 1 a2 ¼

f 00 ð0Þ 1 ¼ 2! 2!

f 000 ð0Þ ¼ e0 ¼ 1

f 000 ð0Þ 1 ¼ 3! 3!

f 000 ðxÞ ¼ ex

a3 ¼

and f ðkÞ ð0Þ 1 ¼ . k! k! x Thus, the Maclaurin series for e is given by ak ¼

∞ X 1

xk . k! n¼0

(1)

The series in (1) converges for all values of x as we showed above. We would like to say that ex ¼

∞ X 1 n¼0

k!

xk ¼ 1 þ x þ

x2 þ/ 2!

1.4 The Issue of Convergence

for all values of x, and, in fact, this is the case but we do not prove it. (For a discussion of how this can be done, see Kirkwood, 1995.) For the problems that we shall consider, if the Taylor series of a function converges, then it converges to the function. One use of Taylor series is to replace a function by the first few terms of its Taylor series. This often gives a sufficiently accurate approximation of the function by a simple expression. For example, some applications replace ex when x is close to 2 0 by 1 þ x or 1 þ x þ x2!. To get a feel for the accuracy of the approximation, we let x ¼ 0.02, and observe that ð0:02Þ2 ¼ 1:0202 2! and note that to 6 decimal places of accuracy, e0.02 ¼ 1.020201. Example: We compute the Maclaurin series for sin x. We have 1 þ 0:02 ¼ 1:02; and 1 þ 0:02 þ

f ðxÞ ¼ sin x f ð0Þ ¼ sin 0 ¼ 0 f 0 ðxÞ ¼ cos x

f 0 ð0Þ ¼ cos 0 ¼ 1

a0 ¼ f ð0Þ ¼ 0 a1 ¼

f 00 ðxÞ ¼ sin x f 00 ð0Þ ¼ sin 0 ¼ 0

f 0 ð0Þ ¼1 1!

a2 ¼

f 00 ð0Þ ¼0 2!

f 0ð0Þ 1 ¼ . 3! 3! By taking enough terms of the series, it is not hard to convince oneself that the Maclaurin series for sin x is f 000 ðxÞ ¼ cos x

f 000 ð0Þ ¼ cos 0 ¼ 1 a3 ¼

x3 x5 x7 þ  þ /. 3! 5! 7! In Exercise 2 we show that the Maclaurin series for cos x is x

1

x2 x4 x6 þ  þ /. 2! 4! 6!

Example: Bessel functions are fundamental in solving differential equations where a particular type of symmetry occurs. One way to define the Bessel function J0(x) is Z 2p 1 J0 ðxÞ ¼ eix sin q dq. 2p 0 We shall find a power series representation for J0(x). We use the Maclaurin series expansion for eix sinq, which is eix sin q ¼

∞ X ðix sin qÞn n¼0

n!

.

53

54

CHAPTER 1 Preliminaries

Now

∞ P

  ðix sin qÞn  xn    n!  n! xn n!

converges for every x. So Z 2p Z 2pX ∞ ∞ Z 2p 1 1 ðix sin qÞn 1 X ðix sin qÞn dq ¼ dq eix sin q dq ¼ J0 ðxÞ ¼ 2p 0 2p 0 n¼0 2p n¼0 0 n! n! Z ∞ 1 X ðixÞn 2p ¼ ðsin qÞn dq. 2p n¼0 n! 0

and

n¼0

R 2p If n is odd, then 0 ðsin qÞn dq ¼ 0, so Z ∞ Z 2p ∞ 1 X ðix sin qÞn 1 X ðixÞ2n 2p dq ¼ ðsin qÞ2n dq. 2p n¼0 0 2p n¼0 ð2nÞ! 0 n! We show in Exercise 17 that Z 2p ð2n  1Þ!2p . ðsin qÞ2n dq ¼ 2n1 2 ðn  1Þ!n! 0 Thus J0 ðxÞ ¼

∞ ∞ X 1 X ðixÞ2n ð2n  1Þ!2p ð1Þn x 2n ¼ . 2p n¼0 ð2nÞ! 22n1 ðn  1Þ!n! n¼0 ðn!Þ2 2

Another notion of convergence that will be important to us is L2[a,b] convergence. The setting for L2[a,b] convergence is a vector space with inner product typically defined by an integral. We say that {fn} converges to f in the L2 sense provided lim kfn  f k ¼ 0

n/∞

where Z ðkfn  f2 kÞ2 ¼ hfn  f ; fn  f i ¼

a

b

jfn ðxÞ  f ðxÞj2 dx.

This sense of convergence is important in many areas of mathematics, including Fourier analysis. There are some connections among the notions of convergence that we have seen. These include the following: (1) Uniform convergence implies pointwise convergence. (2) Uniform convergence implies L2 convergence if the interval of integration is finite. (3) L2 convergence does not imply pointwise nor uniform convergence. (4) Pointwise convergence does not imply L2 convergence.

1.4 The Issue of Convergence

There will be many instances in which we want to “differentiate under the integral.” More precisely, we want to claim 0 1  Z b Z b d@ v f ðt; xÞ dx. f ðt; xÞdxA ¼ dt a a vt This is not always legitimate as the example 8 3 > < t et2 =x x > 0; f ðt; xÞ ¼ x2 > : 0 x¼0 shows. (See Gelbaum and Olmsted, Counterexamples in Analysis.) In most of the cases that we shall encounter, the function f (t,x) will be sufficiently well behaved so that the equation 1 0  Z b Z d@ b v A f ðt; xÞdx ¼ f ðt; xÞ dx dt a a vt is valid. The next theorem gives a condition that enables us to move the derivative under the integral. Theorem: Suppose that for each t ˛ [c,d], Z b f ðt; xÞdx FðtÞ ¼ a

vf vt

exists. If is continuous on {(x,t)ja  x  b, c  t  d}, then t ˛ (c,d), and  Z b dFðtÞ v ¼ f ðt; xÞ dx. dt a vt

dF dt

exists for each

That is, 0 d@ dt

EXERCISES

Z

b a

1 f ðt; xÞdxA ¼

 v f ðt; xÞ dx. vt

Z b a

1 1. Suppose a sequence {an} has the property that janþ1  an j < . Does this mean n 1 that {an} is a Cauchy sequence? What if janþ1  an j < e ? n 2. Find the Maclaurin series for f (x) ¼ cos x.

55

56

CHAPTER 1 Preliminaries

3. Find the Maclaurin series for f (x) ¼ ln (1x). What is the radius of convergence of the power series? 4. Use the Maclaurin series for sin x, cos x and ex to find the Maclaurin expression for the following functions: a. sinx x b. x2 cosð3xÞ ex  1  x . c. x2 P 5. If (1)nan is an alternating P series for which jan j  janþ1 j and lim an ¼ 0 n/∞ and L is the number to which (1)nan converges, then    X k   n ð1Þ an  L < akþ1 .    n¼0 Use this fact and a CAS to calculate the following expressions to three decimal place accuracy: R 1 sin x a. 0 x R2 2 b. 0 ex dx. 6. Find the Maclaurin series for a. sinh x b. cosh x. 7. Show that the following series of functions converge uniformly on the given interval P ½sinðnxÞ on ð∞; ∞Þ a. n2 h  P nk on 2; ∞ b. xn P 1 c. for a > 0 on ½eM; M. n1þa P 8. Show that if the power seriesP anxn has radius of convergence p R,ﬃﬃﬃthen for any positive integer k the series anxkn has radius of convergence k R. 9. FindPthe radius of convergence for the series a. nkxn for k an integer P ðx þ 2Þn b. ln n 1\$2\$3/n\$x2n c. . 1\$3\$5/ð2n  1Þ 10. Use the first four terms of the Taylor series for sin x about c ¼ p6 to estimate sin 32 . Note that 32 will have to be converted to radians. 11. Find the first four terms of the Taylor series for the following functions about the given point. a. cos x about x ¼ p4 b. ln x about x ¼ 2

1.4 The Issue of Convergence

c. x1=2 about x ¼ 4 d. x2 about x ¼ 3. 12. The integral

Z

ε 0

x3 dx ex  1

where ε is small, arises in heat capacity in solids. a. Use Maclaurin series to show ex

x3 ¼ 1

x2 . x x2 1þ þ þ/ 2 6

b. Write x2 x x2 1 2 ¼ x ð1 þ yÞ where y ¼ 1 þ þ þ /. 2 6 x x2 1þ þ þ/ 2 6 Use geometric series to show that for jyj 0. Use integration by parts to show that Gðx þ 1Þ ¼ xGðxÞ x > 0. Show that Gð1Þ ¼ 1. Use induction to show that GðnÞ ¼ ðn  1Þ! for every positive integer n. (Recall that 0! ¼ 1.) e. Show that lim GðxÞ ¼ ∞.

a. b. c. d.

xY0

The gamma function is used to extend the idea of factorial to positive non-integer values of x. This is used in several applications, including quantum mechanics. Rp 4. Determine whether 0 sinx x dx converges.

67

CHAPTER

Vector Calculus

2

2.1 VECTOR INTEGRATION In this section we present results from vector analysis that pertains to integration. The presentation is somewhat brief, and for a more complete explanation we recommend a standard text in vector analysis such as Marsden and Tromba, Vector Calculus, Third Edition. A superb reference for much of what we do in Section 2.2 is Div, Grad, Curl, and All That by H. M. Schey. Definition: If f : U4Rn /R, the gradient of f, denoted Vf, in Cartesian coordinates is   vf vf . ; .; Vf ¼ vx1 vxn If Rn ¼ R3 ; as is often the case, then   vf vf vf vf b vf b vf b ; ; ¼ iþ j þ k. Vf ¼ vx vy vz vx vy vz We showed in Section 1.3 that in cylindrical coordinates Vf ¼

vf 1 vf vf ebr þ ebq þ ebz vr r vq vz

and in spherical coordinates Vf ¼

vf 1 vf 1 vf ebr þ ebq þ eb4 . vr r vq r sin q v4

Example: Let f (x, y, z) ¼ exy2 þ 4xysin z. Then   b i þ ð2ex y þ 4x sin zÞ b j þ ð4xy cos zÞ k. Vf ðx; y; zÞ ¼ ex y2 þ 4y sin z b If f (x, y, z) ¼ constant, then the graph of the equation is a three-dimensional surface. If (x0, y0, z0) is a point on the surface, and the surface is smooth at that point, then a normal (perpendicular) vector to the surface at that point is Vf (x0, y0, z0).

69

70

CHAPTER 2 Vector Calculus

Example: Let z ¼ x2y þ 3y (or x2y þ 3y  z ¼ 0). Then   b j  k. Vf ðx; y; zÞ ¼ 2xy b i þ x2 þ 3 b b So a vector normal to the surface at (1, 2, 8) is 4b i þ 4b j  k. Definition: If f : U4Rn /R, the directional derivative of f at xb in the direction of the unit vector vb is Vf ðb x Þ\$ vb. If vb ¼ ðv1 ; .; vn Þ then Vf ðb x Þ\$ vb ¼

vf vf v1 þ / þ vn . vx1 vxn

We now describe some different kinds of integrals that are important in vector analysis. These are (in the order in which we discuss them) as follows: 1. 2. 3. 4.

Path integrals, where we integrate a real-valued function along a curve. Line integrals, where we integrate a vector-valued function along a curve. Integration of a real-valued function over a surface. Integration of a vector-valued function over a surface.

PATH INTEGRALS A path integral is very similar to a Riemann integral on [a, b]. In constructing the Riemann integral of f (x) on [a, b] we divide [a, b] into n subintervals by choosing x0, x1, . , xn, with a ¼ x0 < x1 < . < xn ¼ b. This creates subintervals [x0, x1], [x1, x2], . , [xn1, xn] whose union is [a, b]. In each subinterval we choose a value of x. Let xi denote the value chosen in the subinterval [xi1, xi]. We then form the n P f ðxi ÞDxi , where Dxi ¼ xi  xi1 is the width of the ith subinterval. Riemann sum, i¼1

If f (x) is sufficiently well-behaved, then lim

maxDxi /0

n X i¼1

f ðxi ÞDxi

Rb exists, and the limit is denoted a f ðxÞdx. In a path integral and a line integral, a path is used as the quantity over which we integrate instead of an interval on the x-axis. Intuitively, a path is a curve in space that has a direction associated with it. The mathematical definition is as follows. Definition: A path in Rn is a continuous function s : ½a; b/Rn . The end points of the path are s(a) and s(b). The image of s is the curve of s. The path s is differentiable if each of the components of s is differentiable. We often have t as the variable, and think of s(t) as being the position of a particle at time t.

2.1 Vector Integration

Example: The function s(t) ¼ (cos t, sin t), 0  t  2p is a path whose curve is a circle of radius 1 centered at the origin. The motion of the particle that describes the curve begins at (0,1) and travels counterclockwise around the circle. Example: The function s(t) ¼ (t, t2), 1  t  2, is a path whose curve is a parabola that begins at (1, 1) and ends at (2, 4). If s(t) is a path, then the velocity vector is s0 (t). If s(t) ¼ (x(t), y(t), z(t)), then 0 s (t) ¼ (x0 (t), y0 (t), z0 (t)). The speed of s(t) is qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ks0 ðtÞk ¼ ðx0 ðtÞÞ2 þ ðy0 ðtÞÞ2 þ ðz0 ðtÞÞ2 . We note that s0 (t) is tangent to s(t), and the length of s(t) is Z b ks0 ðtÞkdt. t¼a

We often use ds to denote an infinitesimal element of arc length. If this is the case and s(t) ¼ (x(t), y(t), z(t)) then sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  2  2  2ﬃ dx dy dz dt. þ þ ds ¼ dt dt dt We shall construct Riemann sums for path integrals (where we integrate a realvalued function) and line integrals (where we integrate a vector-valued function). In both cases, the construction involves splitting the path into small pieces and forming a Riemann sum, with the length of the pieces of the path taking the role of Dx in the Riemann integral. We now investigate how these lengths are computed. Let s : ½a; b/Rn . Choose t0, t1, t2, . , tn with a ¼ t0 < t1 < t2 < . < tn ¼ b. Then s(ti) is a point on the path s, and the role of Dxi ¼ xiþ1  xi in the Riemann sum is now played by the length between s(tiþ1) and s(ti). See Fig. 2.1.1.

σ (b)

σ (t2) σ (t1)

σ (a)

FIGURE 2.1.1

σ (t3)

71

72

CHAPTER 2 Vector Calculus

If s is a path in R3 , and s(t) ¼ (x(t), y(t), z(t)) then sðtiþ1 Þ  sðti Þ ¼ ðxðtiþ1 Þ  xðti Þ; yðtiþ1 Þ  yðti Þ; zðtiþ1 Þ  zðti ÞÞ and ksðtiþ1 Þ  sðti Þk ¼

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðxðtiþ1 Þ  xðti ÞÞ2 þ ðyðtiþ1 Þ  yðti ÞÞ2 þ ðzðtiþ1 Þ  zðti ÞÞ2 .

If s(t) is differentiable, then so are the component functions, and by the mean value theorem xðtiþ1 Þ  xðti Þ ¼ x0 ðti Þ or xðtiþ1 Þ  xðti Þ ¼ x0 ðti Þðtiþ1  ti Þhx0 ðti ÞDti tiþ1  ti where ti is between ti and tiþ1. We have the analogous relations for the other two components. Thus qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ksðtiþ1 Þ  sðti Þk z ½x0 ðti Þ2 þ ½y0 ðti Þ2 þ ½z0 ðti Þ2 Dti ¼ ks0 ððti ÞÞkDti . This method is applicable to curves in dimensions other than three. We continue to work in three dimensions. Suppose f : R3 /R and s : ½a; b/R3 is a path. To construct a Riemann sum for the path integral we choose t0, t1, t2, . , tn with a ¼ t0 < t1 < t2 < . < tn ¼ b as above and from each subinterval [ti1, ti] choose a point ti . Form the Riemann sum n X    f s ti ksðtiþ1 Þ  sðti Þk. i¼1

Notice how this compares with an ordinary Riemann sum. Since in the limit as max (tiþ1  ti) / 0, we have ksðtiþ1 Þ  sðti Þk/ks0 ðti Þkdt, and we have Z Z b n    X  f s ti ksðtiþ1 Þ  sðti Þkh f ds ¼ f ðsðtÞÞks0 ðtÞkdt. lim maxðtiþ1 ti Þ/0

s

i¼1

t¼a

Recapping, we have the following: Definition: Suppose that f : Rn /R and s : ½a; b/Rn is differentiable. The path integral of f along s is Z Z b f ds ¼ f ðsðtÞÞks0 ðtÞkdt. s

t¼a

If f : R /R, and s(t) ¼ (x(t), y(t), z(t)) then sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  2  2  2ﬃ Z Z b dx dy dz f ds ¼ f ðxðtÞ; yðtÞ; zðtÞÞ þ þ dt. dt dt dt a s 3

2.1 Vector Integration

Example: Let f (x, y, z) ¼ x2 þ y2 þ 4z and s(t) ¼ (cos t, sin t, t) 0  t  2. Then qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ 0 0 s (t) ¼ (sin t, cos t, 1) so ks ðtÞk ¼ ðsin tÞ2 þ ðcos tÞ2 þ 1 ¼ 2 and f (s(t))¼(cos t)2 þ (sin t)2þ4t ¼ 1 þ 4t. Thus, sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ   2   2  2 Z Z b dx dy dz dt fds ¼ f ðxðtÞ; yðtÞ; zðtÞÞ þ þ dt dt dt s a Z

2

¼ 0

pﬃﬃﬃ pﬃﬃﬃ pﬃﬃﬃ 2 ð1 þ 4tÞ 2dt ¼ 2 t þ 2t2 0 ¼ 10 2.

LINE INTEGRALS In line integrals, we use the idea of a vector field. Definition: . A vector field on Rn is.a function F : Rn /Rn . For most of our work, F : R3 /R3 . An intuitive way to visualize a vector field is a flowing fluid. At each point, the fluid has a particular velocity. This velocity we can represent as an arrow. Thus each point in space has an arrow associated with it. It may also be visualized as that at each point there is an associated force, as shown in Fig. 2.1.2. . . i þ F2 ðx; y; zÞ b jþ In the case, F : R3 /R3 we often write F ðx; y; zÞ ¼ F1 ðx; y; zÞ b 3 b F3 ðx; y; zÞ k where Fi ðx; y; zÞ : R /R. . We say that the vector field F is continuous (differentiable) if each component function is continuous (differentiable). The idea of a line integral is well-illustrated by computing work. The work done in moving an object along the x-axis from x ¼ a to x ¼ b by a force F (x) that acts Rb only in the horizontal direction is a FðxÞdx. Suppose that we are in higher z

F(x) x y x

FIGURE 2.1.2

73

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CHAPTER 2 Vector Calculus

t(x, y, z) F(x, y, z)

(x, y, z)

FIGURE 2.1.3

dimensions and we move an object along a path s from s(a) to s(b), and the object is .

acted on by a vector field F ðx; y; zÞ. Consider what happens along an infinitesimal part of the path. See Fig..2.1.3. The only portion of F ðx; y; zÞ that contributes to the motion of the body is that .

t ðx; y; zÞ is a unit vector that is tangent part of F ðx; y; zÞ that is tangent to the path. If b to the path at (x, y, z), then the force that contributes to work at that point is

.

t ðx; y; zÞ. The body moves in almost a straight line along the path from F ðx; y; zÞ\$b (x, y, z) to (x þ Dx, y þ Dy, z þ Dz). If we denote the distance between those points as Ds, and DW is the work done in moving between those points, then .

t ðx; y; zÞDs DW z F ðx; y; zÞ\$b so that in the limit as Ds / 0 we get Z . W¼ F ðx; y; zÞ\$b t ðx; y; zÞds. .

t ðx; y; zÞDs as a quantity that we can integrate. We We now . express F ðx; y; zÞ\$b can express F ðx; y; zÞ in Cartesian coordinates as .

i þ F2 ðx; y; zÞ b j þ F3 ðx; y; zÞ kb F ðx; y; zÞ ¼ F1 ðx; y; zÞ b

or .

b F ðsðtÞÞ ¼ F1 ðsðtÞÞ b i þ F2 ðsðtÞÞ b j þ F3 ðsðtÞÞ k.

We noted earlier in the section on path integrals that s0 (t) is tangent to s(t), so s0 ðtÞ ks0 ðtÞk is a unit vector tangent to s(t). We also noted that Ds z ks0 (t)kDt. Thus, we have

2.1 Vector Integration Z W¼

b . t¼a

F ðsðtÞÞ\$

s0 ðtÞ ks0 ðtÞk dt ¼ ks0 ðtÞk

Z

b .

t¼a

F ðsðtÞÞ\$s0 ðtÞdt.

Definition: . . Let F be a vector field on Rn and s : ½a; b : /Rn . The line integral of F along R . s, denoted s F \$db s , is defined by Z Z b . . F \$db s¼ F ðsðtÞÞ\$s0 ðtÞdt. a

s .

i þ F2 ðx; y; zÞ b j þ F3 ðx; y; zÞ kb and We often have F ðx; y; zÞ ¼ F1 ðx; y; zÞ b s(t) ¼ (x(t), y(t), z(t)) and write Z Z . F \$db s ¼ F1 dx þ F2 dy þ F3 dz. s

s

Example: . i þ xz b j þ xyz kb and s(t) ¼ (t,t2,4t) 0  t  2. Then Let F ðx; y; zÞ ¼ x2 y b x ¼ t; y ¼ t2 ; z ¼ 4t dx ¼ dt; dy ¼ 2tdt; dz ¼ 4dt so Z

. s

Z

F \$db s¼

2

  t2 t2 1 þ tð4tÞ2t þ t t2 ð4tÞ4 dt

0

  2   17 25 t5 8t4 16t5  ¼ þ þ ¼ þ 2 24 .  5 4 5 0 5 Example: Compute

Z s

xy2 dx þ x2 dy

along the path y ¼ x2, 0  x  3. We have dy ¼ 2x dx so 3 Z Z 3h i  2 2 x6 x4  36 34 2 2 2 x x xy dx þ x dy ¼ þ x ð2xÞ dx ¼ þ 2  ¼ þ . 6 4 0 6 2 s 0 Example: Compute

Z s

xy2 dx þ x2 dy

from (0, 0) to (3, 9) along the path s from (0, 0) to (3, 0) and then from (3, 0) to (3, 9).

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CHAPTER 2 Vector Calculus

Let s1be the path from (0, 0) to (3, 0) and s2 be the path from (3, 0) to (3, 9). Then Z Z Z xy2 dx þ x2 dy ¼ xy2 dx þ x2 dy þ xy2 dx þ x2 dy s

s1

s2

and Z

Z s1

2

s2

9

32 dy ¼ 81  27 ¼ 54:

3

Z s

Now compute

0dx ¼ 0

0

since y ¼ 0 and dy ¼ 0 on s1, and Z Z xy2 dx þ x2 dy ¼ Thus,

3

xy dx þ x dy ¼ 2

xy2 dx þ x2 dy ¼ 54: Z s

xy2 dx þ x2 dy

from (0, 0) to (3, 9) along the path y ¼ 3x. Then 3 Z Z 3h i x4  xð3xÞ2 þ x3 3 dx ¼ 12  ¼ 243: xy2 dx þ x2 dy ¼ 4 0 0 s Notice that the answers to the last two examples are different, even though we are integrating the same function between the same end points.

SURFACES The next two integrals we discuss involve integrating over a surface rather than a curve. Like line integrals that extended integrating a function over an interval to integrating over a curve, surface integrals extend the idea of integrating over a planar region to integrating over a surface. In developing line integrals it was fundamental that we develop an approximation for an increment of the pathda quantity we denoted Ds. Our first task with creating surface integrals is to develop an approximation for an increment of the surface. We denote this incremental element DS. The simplest situationdthe one that we now considerdis when the surface can be expressed z ¼ f (x, y). The cases where y ¼ g(x, z) and x ¼ h(y, z) are conceptually identical.

2.1 Vector Integration

z N

γ

k

v

Q u

ΔP

y0

0

y0 + Δy

y

x0 ΔA

x0 + Δx

Δy

x

Δx

FIGURE 2.1.4

Suppose that D is a region in the x,y plane and f (x, y) has continuous partial derivatives. Divide D into small rectangles whose dimensions are Dx and Dy. We consider the particular rectangle whose corners are (x0, y0), (x0 þ Dx, y0), (x0 þ Dx, y0 þ Dy) and (x0, y0 þ Dy). See Fig. 2.1.4. Denote this rectangle DA. If we project DA onto the surface z ¼ f (x, y), we get a portion of the surface that we denote DS. To estimate the area of DS, choose a point p on DS and construct the plane tangent to the surface at that point. It is notationally convenient to choose p ¼ (x0, y0, f (x0, y0)). We project DA onto this plane and get a planar region we denote DP. We compute the area of DP, and this will be our estimate for the area of DS. The sides of DP are the vectors ub ¼ Dx b iþ

vf ðx0 ; y0 Þ b Dx k vx

vf ðx0 ; y0 Þ b Dy k. vb ¼ Dy b jþ vy The area of DP ¼ kb u  vbk. Now     b b   b j k   i  

 vf ðx0 ; y0 Þ  vf ðx0 ; y0 Þ b vf ðx0 ; y0 Þ b b   Dx 0 Dx iþ jk ub  vb ¼   ¼ DxDy vx   vx vy   vf ðx0 ; y0 Þ   Dy   0 Dy   vy

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CHAPTER 2 Vector Calculus

so

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2  2  vf ðx0 ; y0 Þ vf ðx0 ; y0 Þ þ þ 1. DP ¼ kb u  vbk ¼ DxDy vx vy

When we develop integrals over surfaces, this will be our incremental surface element if we can write the surface as z ¼ f (x, y).

PARAMETERIZED SURFACES In situations where the surface is not conveniently expressed as z ¼ f (x, y), such as when cylindrical or spherical coordinates are advantageous, we parameterize the surface with different variables. Definition: A parameterized surface is a function F : D3R2 /R3 ; where D is a connected set. The surface corresponding to F is the image of F, F(D). If F(u, v) ¼ (x(u, v), y(u, v), z(u, v)) then S ¼ F(D) is continuous (differentiable) if each of the coordinate functions is continuous (differentiable). Example: The function F(q, 4) ¼ (rsin qcos 4, rsin qsin 4, rcos q) 0  q  p, 0 4 < 2p is a parameterization of the surface of a sphere of radius r. To compute the element DS in parameterized surfaces, we follow the same idea of finding the tangent plane at a point, and estimating DS by the area of the increment of the tangent plane. As before, we need two nonparallel vectors that are tangent to the surface to create the tangent plane. We now describe how to do this. If F(u,v) ¼ (x(u, v), y(u, v), z(u, v)) is a surface and F(u0, v0) is a point on the surface, then vx vy vz i þ ðu0 ; v0 Þ b j þ ðu0 ; v0 Þ kb Tbu ðu0 ; v0 Þ ¼ ðu0 ; v0 Þ b vu vu vu is a vector tangent to the surface at the point F(u0,v0) parallel to the curve F(t,v0) on the surface. Likewise, vx vy vz i þ ðu0 ; v0 Þ b j þ ðu0 ; v0 Þ kb Tbv ðu0 ; v0 Þ ¼ ðu0 ; v0 Þ b vv vv vv is a vector tangent to the surface at the point F(u0, v0) parallel to the curve F(u0, t) on the surface. The vector Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ is normal to the surface F(u, v) at the point F(u0, v0). The surface is said to be smooth at F(u0, v0) if 0. Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þsb Following what we did in the case where the surface is described by z ¼ f (x, y), our estimate for DS is Tbu  Tbv DuDv. Example: Consider the parameterization of the sphere of radius r, F(q, 4) ¼ (rsin qcos 4,rsin qsin 4,rcos q) 0  q  p, 0 4 < 2p. Then

2.1 Vector Integration

x ¼ r sin q cos 4; y ¼ r sin q sin 4; z ¼ r cos q so Tbq ¼ r cos q cos 4b i þ r cos q sin 4 b j  r sin q kb i þ r sin q cos 4 b j Tb4 ¼ r sin q sin 4b and

    b b b   i j k    b b T q  T 4 ¼  r cos q cos 4 r cos q sin 4 r sin q     r sin q sin 4 r sin q cos 4  0 h i i þ ðsin qÞ2 sin 4 b j þ sin q cos q kb . ¼ r2 ðsin qÞ2 cos 4 b

Note that

h i Tbq  Tb4 ¼ r2 sin q sin q cos 4 b i þ sin q sin 4 b j þ cos q kb

 ¼ r2 sin q x b i þ yb j þ z kb ¼ r2 sin qb r

so that the normal to the tangent plane to a sphere is radially directed, as we would expect. Also Tbq  Tb4 ¼ r 2 sin q. Thus the estimate for DS is r2sin qDqD4. Fact: If F(D) is a surface S, then the area of S is ZZ Tbu  Tbv dudv. D

Example: We compute the surface area S of a sphere. We have Z p Z 2p Z ZZ  2  2 b b T u  T v dudv ¼ r sin qd4 dq ¼ 2pr S¼ D

q¼0

4¼0

p

sin qdq

0

¼ 2pr2 cos qjp0 ¼ 4pr 2 .

INTEGRALS OF SCALAR FUNCTIONS OVER SURFACES Suppose that f : R3 /R and S is a surface in R3 . We begin by giving an intuitive R description of what S f dS should mean. The idea is simpler than it may appear. All we are doing is picking a point on the surface, evaluating the function at that

79

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CHAPTER 2 Vector Calculus

point and then multiplying that number by a small amount of the surface area close to that point. We then sum all the pieces. As with most integrals, we can go back to Riemann sums. We approximate the surface S by covering S with nonoverlapping planar segments, DPi, each of which is tangent to S at one point. We described how these planar segments are determined above. Let kDPki ¼ diameter of DPi ¼ supfjxi  yi j jxi ; yi ˛DPi g so that if kPi k/0, then every dimension of Pi must go to 0. For each planar segment Pi, let (xi, yi, zi) denote the point on the surface S at which Pi is tangent to S. Form the product f (xi, yi, zi)DPi where DPi is the area of Pi. In the work above, we showed how to calculate DPi in the special instances that we consider. Again referring to our work above, f (xi, yi, zi)DPi is an estimate for f (xi, yi, zi)DS. Depending on how we describe the surface, the Riemann sum that we form is 0sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1  2  2 X vf ðx0 ; y0 Þ vf ðx0 ; y0 Þ @ þ þ 1 ADxDy f ðxi ; yi ; zi Þ vx vy or

X

X f ðxi ; yi ; zi Þ Tbu  Tbv DuDv ¼ f ðFððui ; vi ÞÞÞ Tbu  Tbv DuDv.

If the function f is reasonably well behaved, then the limit of these Riemann sums exists as kDPi k/0, and we define ZZ X f ðxi ; yi ; zi Þ fdS ¼ lim Dx/0;Dy/0

S

0sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1  2  2 vf ðx0 ; y0 Þ vf ðx0 ; y0 Þ @  þ þ 1 ADxDy. vx vy

(In this case, z is a function of x and y so the integrand is actually a function of x and y.) Likewise, ZZ X f ðFððui ; vi ÞÞÞ Tbu  Tbv DuDv. fdS ¼ lim S

Du/0;Dv/0

Example: We compute the surface area of a cone, which is determined by

by computing

RR

x2 þ y 2 ¼ z 2 ; S 1dS.

0z3

We parameterize the surface by the variables u,v with

z ¼ u; x ¼ u cos v; y ¼ u sin v;

0  u  3; 0  v < 2p.

2.1 Vector Integration

(A more natural choice for the names of the parameterizing variables is z and q, but we are following the notation we have established.) Then Fðu; vÞ ¼ ðu cos v; u sin v; uÞ vF ¼ cos v b i þ sin v b j þ kb Tbu ¼ vu vF Tbv ¼ ¼ u sin v b i þ u cos v b j vv     b b  i j kb    Tbu  Tbv ¼  cos v sin v 1     u sin v u cos v 0    ¼ u cos v b i  u sin v b j þ u cos2 v þ u sin2 v kb qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ Tbu  Tbv ¼ ðu cos vÞ2 þ ðu sin vÞ2 þ u2 ¼ 2u. So ZZ S¼

Z

S

1dS ¼

0 3

@

Z

u¼0

1

pﬃﬃﬃ Z 2udvAdu ¼ 2 2p

2p pﬃﬃﬃ v¼0

3

u du ¼

0

pﬃﬃﬃ 2 3 pﬃﬃﬃ 2pu 0 ¼ 9 2p.

Example:  RR  Compute the surface integral S 5xy3 þ 4xy dS where S is the surface determined by z ¼ x2þ3y, 0  x  y  1. We have

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ  2  2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ vf vf þ þ 1 ¼ 4x2 þ 10 z ¼ f ðx; yÞ ¼ x2 þ 3y; so dS ¼ vx vy

and ZZ

 S



5xy3 þ 4xy dS ¼

Z

0 1

y¼0

@

Z

y x¼0



1 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  5xy3 þ 4xy 4x2 þ 10dxAdy.

We leave it to show Z

0 1 y¼0

@

Z

y x¼0



1 pﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  49 14 485 10 3 2 A  5xy þ 4xy 4x þ 10dx dy ¼ 15 168

as an exercise using a computer algebra system (CAS).

81

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CHAPTER 2 Vector Calculus

SURFACE INTEGRALS OF VECTOR FUNCTIONS A major reason why we study vector functions over a surface is to measure fluxdan idea that we now describe. Flux is an important concept in electricity and magnetism. In fact, much of what we shall study in the later part of this chapter was developed as part of the effort to give a mathematical formulation to the principles of electricity and magnetism. The culmination of this effort was Maxwell’s equations. An intuitive way to visualize flux is the passage of a fluid through a membrane. We want to measure the amount of fluid that passes through one unit of area of the membrane in one unit of time. Fig. 2.1.5 illustrates how this depends on two factors: (1) the velocity of the fluid and (2) the orientation of the surface with respect to the direction of flow of the fluid. The more closely the surface is to being perpendicular . to the flow of the fluid, the more fluid will pass through the surface. Thus, if F is the vector field that describes the flow of the fluid, and nb is the unit vector normal to the . n would be the appropriate mathematical quantity to combine these surface, then F \$b two conditions. . If S is a given two-sided surface and F is the vector field describing the motion of the fluid, we determine how much fluid passes through, dS, a small amount of the surface in one unit of time. (All of our surfaces are two sided. An example of a surface that is one sided is a Mobius strip.) To compute this volume of fluid, suppose that F is a parameterization of S, and F:D / S. Partition D into small rectangles, and consider one of the rectangles, Dij. Suppose the length of Dij is Du and the width is Dv. Let Sij ¼ F(Dij) and suppose that (u0, v0) is a point in Dij. Then Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ is a vector that is normal to Sij at F(u0, v0), and Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ nb ¼ Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ is a unit vector that is normal to Sij at F(u0, v0). In the case that the surface can be described by z ¼ f (x, y), then vf ðx0 ; y0 Þ b vf ðx0 ; y0 Þ b b iþ jk vx ﬃ vx . nb ¼ sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  2  2 vf ðx0 ; y0 Þ vf ðx0 ; y0 Þ þ þ1 vx vy

θ

FIGURE 2.1.5

S

2.1 Vector Integration

F•(Tu × Tv)

v

z | |Tu × Tv| | F

T u × Tv

Φ (D) = S

Tu

Tv y

u

x

FIGURE 2.1.6 .

Let q denote the angle that nb makes with F ðFðu0 ; v0 ÞÞ. See Fig. 2.1.6. The parallelogram with sides Tbu ðu0 ; v0 ÞDu and Tbv ðu0 ; v0 ÞDv has approximately the same area as Sij. The volume of fluid that passes through Sij in one unit of time is approximately . F ð Fðu0 ; v0 ÞÞ cos q Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ DuDv. But

. F ð Fðu0 ; v0 ÞÞ cos q Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ . ¼ F ð Fðu0 ; v0 ÞÞ\$b n Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ DuDv

. Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ Tb ðu ; v Þ  Tbv ðu0 ; v0 Þ DuDv ¼ F ð Fðu0 ; v0 ÞÞ\$ Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ u 0 0   . ¼ F ð Fðu0 ; v0 ÞÞ\$ Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ DuDv. Also, this approximation becomes exact as Du / 0, Dv / 0. Thus X.   F ð Fðu0 ; v0 ÞÞ\$ Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ DuDv i;j

is a Riemann sum that approximates the amount of fluid that flows through the surface S in one unit of time, and the exact value is given by ZZ   . F ð Fðu ; v ÞÞ\$ Tbu ðu0 ; v0 Þ  Tbv ðu0 ; v0 Þ dudv: (2) D

83

84

CHAPTER 2 Vector Calculus R. R . It is common to write expression (2) as s F \$d Sb or S F \$b n dS. In the case the equation of the surface is z ¼ f (x, y) the integral is   vf ðx; yÞ b vf ðx; yÞ b b ZZ iþ jk . vx vy ﬃ dxdy F ðx; y; zÞ\$sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ     2 D vf ðx; yÞ vf ðx; yÞ 2 þ þ1 vx vx .

keeping in mind that F ðx; y; zÞ can be expressed in terms of x and y. The process of evaluating the integral of a vector function over a surface can be broken down into steps. Step 1. If the surface is described in parametric form, compute the tangent vectors Tu and Tv as before; that is, vx vy vz Tbu ¼ b iþ b j þ kb vu vu vu vx vy vz b iþ b j þ k. Tbv ¼ b vv vv vv Step 2. Form a normal vector. If the surface is described in parametric form, a normal vector is Tbu  Tbv . If the surface is described z ¼ f (x, y), a normal vector is vf ðx; yÞ b vf ðx; yÞ b b iþ j  k. vx vy Step 3. Normalize the vector found in Step 2. If surface is defined parametrically, the vector is Tbu  Tbv and then a unit normal vector is Tbu  Tbv . nb ¼ Tbu  Tbv If the surface is defined by z ¼ f (x, y), normalize the vector in Step 2 to get   vf ðx; yÞ b vf ðx; yÞ b b iþ jk vx vy ﬃ. nb ¼ sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ     vf ðx; yÞ 2 vf ðx; yÞ 2 þ þ1 vx vx

2.1 Vector Integration

Step 4. Take the dot product of the unit normal vector with the vector field. This is either Tbu  Tbv F ð Fðu ; v ÞÞ\$ Tbu  Tbv

.

in the parametric case or

  vf ðx; yÞ b vf ðx; yÞ b b iþ jk . vx vy ﬃ F ðx; y; zÞ\$sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ     vf ðx; yÞ 2 vf ðx; yÞ 2 þ þ1 vx vx

in the case z ¼ f (x, y). Step 5. Integrate the quantity in Step 4 as a scalar function over a surface using dS as the infinitesimal element. For example, integrating over x and y, dS ¼ dxdy; integrating over the surface of a sphere of radius R, dS ¼ R2sin qdq d4. Example: b y; zÞ ¼ x2 y b i þ 2yz b j þ 4x kb and RRS .be the surface defined by Let Fðx; z ¼ 1  x  2y in the first octet. We compute S F \$dS. We have b b y; zÞ ¼ x2 y b i þ 2yz b j þ 4x kb ¼ x2 y b i þ 2yð1  x  2y Þ b j þ 4x k. Fðx; Also, z ¼ f ðx; yÞ ¼ 1  x  2y; so

  vf ðx; yÞ ^ vf ðx; yÞ ^ ^ iþ jk 

vx vy i^ 2j^ k^ 1ﬃﬃ b ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ ¼ p i þ 2b j þ kb . ¼p 2 2 2  2  6 ð1Þ þð2Þ þ1 vf ðx; yÞ vf ðx; yÞ þ þ1 vx vx Then

  vf ðx; yÞ b vf ðx; yÞ b b iþ jk . vx vy ﬃ F ðx; y; zÞ\$sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2    vf ðx; yÞ vf ðx; yÞ 2 þ þ1 vx vx  h i 1 i þ 2b j þ kb ¼ x2 y b i þ 2yð1  x  2y Þ b j þ 4x kb \$ pﬃﬃﬃ b 6  1  ¼ pﬃﬃﬃ x2 y þ 4y  4xy  8y2 þ 4x : 6

85

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CHAPTER 2 Vector Calculus

z

1

1

y

½

x

FIGURE 2.1.7A

The surface is shown in Fig. 2.1.7A, and the region of integration for x and y is shown in Fig. 2.1.7B. We have 1 0 Z 12y Z 1 ZZ 2  C . 1  B pﬃﬃﬃ x2 y þ 4y  4xy  8y2 þ 4x dxAdy. F \$dS ¼ @ 6 S y¼0 x¼0 Using a CAS 1 0   Z 1 Z 12y 2   1 2 1 91 C B 2 p ﬃﬃ ﬃ p ﬃﬃ ﬃ y þ 4y  4xy  8y þ 4x dx dy ¼ x . A @ 6 6 240 x¼0 y¼0

y

½

x 1

FIGURE 2.1.7B

2.1 Vector Integration

Example: . i þ yb j þ z kb out of the upper half of the unit We compute the flux of F ðx; y; zÞ ¼ x b sphere. Parameterizing with spherical coordinates with r ¼ 1, we have x ¼ cos 4 sin q; y ¼ sin 4 sin q; z ¼ cos q. Now, vx vy vz iþ b j þ kb ¼ cos 4 cos q b i þ sin 4 cos q b j  sin q kb Tbq ¼ b vq vq vq vx b vy b vz b Tb4 ¼ iþ jþ k ¼ sin 4 sin q b i þ cos 4 sin q Jb v4 v4 v4 so

  b b  i j  Tbq  Tb4 ¼  cos 4 cos q sin 4 cos q   sin 4 sin q cos 4 sin q

  kb   sin q   0 

b ¼ cos 4sin2 q b i þ sin 4sin2 q b j þ sin q cos q k. Then Tbq  Tb4 2 ¼ cos2 4sin4 q þ sin2 4sin4 q þ sin2 qcos2 q ¼ sin4 q þ sin2 qcos2 q ¼ sin2 q so Tbq  Tb4 b ¼ cos 4 sin q b i þ sin 4 sin qb j þ cos q k: Tbq  Tb4 Now .

i þ sin 4 sin q b j þ cos q kb F ðFðq; 4ÞÞ ¼ cos 4 sin q b

so Tbq  Tb4 ¼ cos2 4 sin2 q þ sin2 4 sin2 q þ cos2 q ¼ 1: F ðFðq; 4ÞÞ\$ Tbq  Tb4

.

Thus the flux over the upper hemisphere is Z Z p Z 2p 2 sin qd4dq ¼ 2p q¼0

4¼0

p 2

0

sin qdq ¼ 2p.

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CHAPTER 2 Vector Calculus

EXERCISES

1. If Vf ðb x Þs b 0, show that Vf ðb x Þ points in the direction in which f is increasing most rapidly. 2. Find the directional derivative of f ðx; y; zÞ ¼ xyz3  4x at the point b ð1; 3; 2Þ in the direction 4 b i  2b j þ k. 3. Find Vf for the following functions at the given point. Find the direction in which f is increasing most rapidly. a. f ðx; y; zÞ ¼ xey þ sin z at (3, 2, 0). b. f (r, q, z) ¼ zrcos q at (5, p/4, 2).  c. f ðr; q; 4Þ ¼ r sin q þ r cos 4 at 3; p6 ; p3 . 4. Let S be the surface consisting of the points (x, y, z) for which f (x, y, z) ¼ k where k is a constant. The equation of the plane tangent to S at the point (x0, y0, z0) is Vf ðx0 ; y0 ; z0 Þ\$ðx  x0 ; y  y0 ; z  z0 Þ ¼ 0: Find the equation of the plane tangent to the following surfaces at the given point. a. xy2 þ 3z ¼ 7 at ð1; 2; 1Þ. b. e3x þ 4yz ¼ 8 at ð0; 1; 2Þ. 5. Find the length of the curve s(t) ¼ (cos t, sin t, t3/2) 0  t  1.

Path Integrals

R 6. Evaluate the following path integrals s f ðx; y; zÞds. a. f ðx; y; zÞ ¼ x2 þ y2 þ z2 ; sðtÞ ¼ ðcos t; sin t; t Þ; 0  t  4: b. f ðx; y; zÞ ¼ x2 þ y þ 3z; sðtÞ ¼ ð2t; 5t; 9Þ; 1  t  5: pﬃﬃﬃ   c. f ðx; y; zÞ ¼ xy; sðtÞ ¼ 2t; 3t; 3 t , 0  t  4.

Line Integrals 7. Compute the following line integrals. R a. s x3 y dx þ xz dy þ y2 dz where s(t) ¼ (t2, t3, 4), 0  t  2.   R b. s sin zdx þ cos z dy þ 4dz where sðtÞ ¼ cos3 t; sin3 t; t ; 0  t  3p 2. R c. s yz dx  zx dy þ xy dz where s(t) consists of the line segments from (2, 0, 0) to (0, 2, 0) to (0, 0, 2).  R  2 d. x þ y2 þ z2 ds sðtÞ ¼ cos t b i þ sin t b j þ 2k from ð1; 0; 0Þ to s

ð1; 0; 8pÞ. R e. x2 y dx þ x dy from ð0; 0Þ to ð1; 1Þ along the path y ¼ x. R f. x2 y dx þ x dy from ð0; 0Þto ð1; 1Þ along the path y ¼ x2 . R g. s xydx þ ðx þ yÞdy sðtÞ ¼ 1 þ 2t ; t3 , 2  t  4. R b where 8. Find s F\$ds 2 b b j ; x ¼ t3 ; y ¼ t; 0  t  1: a. F ¼ xy i  xy b

2.1 Vector Integration

i þ y2 b j; s is the unit circle; from 0 to 2p. b. Fb ¼ x2 b 2 2 b sðtÞ ¼ t2 b b 0  t  2: c. Fb ¼ x y z b i þ 2yz b j  xy k; i þ tb j  t3 k; 9. Find the work done going on upper half of the unit circle from (1, 0) to (1, 0) when the force is .

y b x b iþ 2 j. x2 þ y2 x þ y2

10. Calculate the work done by going from (0, 0) to (1, 1) along two different paths . . where F ¼ ðx þ yÞ b i þ ðx  yÞ b j. What does your answer tell you about F ? 11. Gravitational force is given by  pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 i þ yb j þ z kb where r ¼ x2 þ y 2 þ z 2 . Fb ¼ 3 x b r R b where s is any path for which Compute the work done s F\$ds ks(initial point)k ¼ R1 and ks(final point)k ¼ R2.

Integrals of Scalar Functions Over Surfaces

RR 12. Evaluate S f ðx; y; zÞdS where f (x, y, z) ¼ 3x2, S is the surface determined by x2 þ y2 ¼ 4, 2  z  2. RR 13. Evaluate S f ðx; y; zÞdS where f (x, y, z) ¼ 3x2cos y, S is the surface determined by z ¼ x2, 0  x  2, 0  y  p/2. pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ RR 14. Evaluate S f ðx; y; zÞdS where f ðx; y; zÞ ¼ x2 þ y2 ; S is the surface determined by z ¼ 9  x2  y2, z  0. 2 2 15. Find the surface  paraboloid z ¼ 4x þ 4 y , 0  z  16. RR  2 area of the 16. Evaluate S x y þ 3xyz dS where, S is the plane in the first octet 2x þ 4 y þ z ¼ 8. 17. Let S be a sphere of radius R. Evaluate ZZ 1 h i dS 2 S ðx  x0 Þ þ ðy  y0 Þ2 þ ðz  z0 Þ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x0 2 þ y0 2 þ z0 2 < R pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ b. in the case x0 2 þ y0 2 þ z0 2 > R.

a. in the case

18. Parameterize the surface of a right circular cone of radius R and height h and find the area of the surface of the cone.

Integrals of Vector Functions Over Surfaces .

19. Find the flux of the vector field F ðx; y; zÞ ¼ x b iþy b j þ z kb through the upper 2 2 2 half of the hemisphere x þ y þ z ¼ 4.

89

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CHAPTER 2 Vector Calculus

.

20. Find the flux of the vector field F ðx; y; zÞ ¼ 5 kb through the surface z ¼ 9  x2  y2, z  0. . j  z kb where S is the surface 21. Find the flux of the vector field F ðx; y; zÞ ¼ y b 2 2 2 2 consisting of two pieces y ¼ x þ z and x þ z  1. . j þ 5z kb and S is the plane 22. Find the flux of the vector field F ðx; y; zÞ ¼ 2y b 2x þ y þ 3z ¼ 6 in the first octet. . i þ 2b j þ xz kb and S is the surface 23. Find the flux of the vector field F ðx; y; zÞ ¼ 3z2 b 2 y ¼ x 0  x  1, 0  z  2. . 24. Find the flux of the vector field F ðx; y; zÞ ¼ 2y b j  z kb and S is the surface 2 2 bounded by the paraboloid y ¼ 4x þ 4z and the plane y ¼ 1. . 25. Find the flux of the vector field F ðx; y; zÞ ¼ x b i þ yb j þ z kb and S is the plane z ¼ 4  2x  y in the first octet. . i þ xb j þ z kb and S is the surface 26. Find the flux of the vector field F ðx; y; zÞ ¼ y b 2 2 z ¼ 4  x  y , z  0.

2.2 THE DIVERGENCE AND CURL The divergence and curl are two of the most important operators in vector calculus. One way of presenting them is to define them in terms of mathematical formulas. We choose instead to analyze the phenomena from which they arise, and then derive the associated formulas. Both the phenomena describe the action of a vector field at a point. The derivation involves analyzing what happens within a small volume, dividing by the volume to determine the effect per unit volume, and then taking the limit as the volume goes to 0.

DIVERGENCE .

.

Let F be a vector field and S be an enclosed surface. Then the flux of F through S is ZZ . F \$b n dS S

where nb is the outward pointing unit vector to the surface of S. Let V be the volume

.

of the region enclosed by S. Let p ¼ (x0, y0, z0) be a point in R3 . The divergence of F .

at p, denoted div F ðpÞ, is .

div F ðpÞ ¼

lim

kVk/0; p˛V

1 V

ZZ

. S

F \$b n dS

where the limit exists. . . Thus, div F ðpÞ is the flux of F at p. The formula used to calculate the divergence depends on the coordinate system used. We develop the formulas for Cartesian, cylindrical, and spherical coordinates.

2.2 The Divergence and Curl

CARTESIAN COORDINATE CASE Choose the point p ¼ (x0, y0, z0) and enclose p in the center of a rectangular parallelepiped. See Fig. 2.2.1. Let Dx, Dy and Dz denote the lengths of the sides of the parallelepiped, and let .

b i þ Fy ðx; y; zÞ b j þ Fz ðx; y; zÞ k. F ðx; y; zÞ ¼ Fx ðx; y; zÞ b

This notation will help keep the component functions straight, but we must remember that for the rest of the section, the subscript does not mean a partial derivative. We assume for the remainder of this chapter that each component function is differentiable. Our parallelepiped has six faces, and we compute ZZ . F \$b n dS S

by computing the integral over each of the faces and then adding the results. Let S1 and S2 denote the faces that are parallel to the y, z plane. See Fig. 2.2.1.  . . . .  n ¼ F \$b i ¼ Fx ðx; y; zÞ and on S2 ; F \$b n ¼ F \$ b i ¼ Fx ðx; y; zÞ. On S1 ; F \$b (Recall that nb is the unit normal vector that points outward from the surface.) The x coordinate of any point on S1 is x0 þ Dx 2 . If Dy and Dz are small, then

S2 S5

Δz S4

S1

, y + Δy , z – Δz (x0 – Δx 2 0 2 0 2)

(x0, y0, z0) Δx

S3

Δy

(x0 + Δx2 , y0 + Δy2 , z0 – Δz2 ) S6

FIGURE 2.2.1

91

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CHAPTER 2 Vector Calculus ZZ



 Z Z  Dx Dx F \$b n dS ¼ Fx x0 þ ; y; z dS z Fx x0 þ ; y0 ; z0 dydz 2 2 S1 S1 S1   Dx ¼ Fx x0 þ ; y0 ; z0 DyDz. 2 ZZ

.

The x coordinate of any point on S2 is x0  Dx 2 . If Dy and Dz are small, then   ZZ ZZ . Dx F \$b n dS ¼  Fx x0  ; y; z dS 2 S2 S2  Z Z   Dx Dx z  Fx x0  ; y0 ; z0 dydz ¼ Fx x0  ; y0 ; z0 DyDz. 2 2 S2 Thus ZZ

    Dx Dx F \$b n dS z Fx x0 þ ; y0 ; z0  Fx x0  ; y0 ; z0 DyDz 2 2 S1 þS2

    Dx Dx Fx x 0 þ ; y 0 ; z 0  F x x 0  ; y 0 ; z 0 2 2 DxDyDz. ¼ Dx Now DxDyDz ¼ V, so

    Dx Dx ZZ F x x 0 þ ; y 0 ; z 0  Fx x 0  ; y 0 ; z 0 . 1 2 2 : F \$b n dS z Dx V S1 þS2 .

In the limit as Dx / 0 we have ZZ . 1 vFx ðx0 ; y0 ; z0 Þ lim F \$b n dS ¼ . Dx/0 V vx S1 þS2 Similarly, if S3 and S4 are the faces that are parallel to the x, z plane as shown in Fig. 2.2.1 then ZZ . vFy ðx0 ; y0 ; z0 Þ 1 F \$b n dS ¼ lim Dy/0 V vy S3 þS4 and if S5 and S6 are the faces that are parallel to the x, y plane as shown in Fig. 2.2.1 then ZZ . 1 vFz ðx0 ; y0 ; z0 Þ F \$b n dS ¼ lim . Dz/0 V vz S5 þS6

2.2 The Divergence and Curl

Thus, lim

kVk/0

1 V

ZZ S

.

F \$b n dS ¼ lim

kVk/0

1 V

ZZ S1 þ/þS6

.

F \$b n dS

ZZ ZZ ZZ . . . 1 1 1 F \$b n dS þ lim F \$b n dS þ lim F \$b n dS V V kVk/0 V kVk/0 kVk/0 S1 þS2 S3 þS4 S5 þS6 ZZ ZZ ZZ . . . 1 1 1 F \$b n dS þ lim F \$b n dS þ lim F \$b n dS ¼ lim Dx/0 V Dy/0 V Dz/0 V S1 þS2 S3 þS4 S’5 þS6 ¼ lim

¼ and

. vFx ðx0 ; y0 ; z0 Þ vFy ðx0 ; y0 ; z0 Þ vFz ðx0 ; y0 ; z0 Þ þ þ hdiv F vx vx vz



 v v v . b i þb j þ kb \$F vx vx vx    v bv v b b \$ Fx ðx; y; zÞ b ¼ i þj þk i þ Fy ðx; y; zÞ b j þ Fz ðx; y; zÞ kb vx vx vx ¼

(1)

vFx ðx; y; zÞ vFy ðx; y; zÞ vFz ðx; y; zÞ þ þ . vy vx vz

Many texts will give the right-hand side of Eq. (1) as the definition of the divergence of a vector field in Cartesian coordinates. It is common to refer to v þb v þ kb v as the “del operator.” b i vx j vx vx Example: The divergence of .

i þ 3yz b j þ xy kb F ðx; y; zÞ ¼ x2 ey sin z b

is  v v 2 y v x e sin z þ ð3yzÞ þ ðxyÞ ¼ 2xey sin z þ 3z. vx vy vz

CYLINDRICAL COORDINATE CASE In integrating in cylindrical coordinates, the incremental element of volume is shown in Fig. 2.2.2. The volume of the incremental piece is DV ¼ rdqdrdz. We have the faces of the incremental element and will proceed by grouping the faces in pairs, as we did in the case of Cartesian coordinates. We describe the vector field as .

e r þ Fq ðr; q; zÞb e q þ Fz ðr; q; zÞb ez. F ðr; q; zÞ ¼ Fr ðr; q; zÞb

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CHAPTER 2 Vector Calculus

z

S1

( r0 + Δr2 , θ 0 + Δ2θ , z0 + Δz2 )

S4

(r0, θ 0, z0) Δz

Δθ

S3

Δr

S2 x

y

( r0 + Δr2 , θ 0 – Δ2θ , z0 – Δz2 )

FIGURE 2.2.2

Consider an incremental volume element centered at p ¼ (r0, q0, z0). We compute   ZZ ZZ . . Dz F \$b n dS ¼ F \$b e z dS z Fz r0 ; q0 ; z0 þ rðDqÞðDrÞ 2 S1 S1   ZZ ZZ . . Dz F \$b n dS ¼  F \$b e z dS z  Fz r0 ; q0 ; z0  rðDqÞðDrÞ 2 S2 S2 so ZZ

  

 Dz Dz  Fz r0 ; q0 ; z0  rðDqÞðDrÞ F \$b n dS z Fz r0 ; q0 ; z0 þ 2 2 S1þ S2  

  Dz Dz  Fz r0 ; q0 ; z0  Fz r0 ; q0 ; z0 þ 2 2 rðDqÞðDrÞðDzÞ ¼ Dz .

z Also, ZZ

vFz ðr0 ; q0 ; z0 Þ DV. vz

  Dr Dr ðDqÞðDzÞ r0 þ F \$b n dS ¼ F \$b e r dS z Fr r0 þ ; q0 ; z0 2 2 S3 S3    ZZ ZZ . . Dr Dr r0  F \$b n dS ¼  F \$b e r dS z  Fr r0  ; q0 ; z0 ðDqÞðDzÞ. 2 2 S4 S4 .

ZZ

.



2.2 The Divergence and Curl

Thus ZZ

  

 Dr Dr F \$b n dS z Fr r0 þ ; q0 ; z0  Fr r0  ; q0 ; z0 r0 ðDqÞðDzÞ 2 2 S3 þS4 .

þFr ðr0 ; q0 ; z0 ÞðDrÞðDqÞðDzÞ

    Dr Dr Fr r0 þ ; q0 ; z0  Fr r0  ; q0 ; z0 2 2 r0 ðDqÞðDrÞðDzÞ ¼ Dr 1 vFðr0 ; q0 ; z0 Þr 1 DV þ Fr ðr0 ; q0 ; z0 ÞDV. þ Fr ðr0 ; q0 ; z0 ÞðDrÞrðDqÞðDzÞ z r r vr For the remaining two faces, we have   ZZ ZZ . . Dq F \$b n dS ¼ F \$b e q dS z Fq r0 ; q0 þ ; z0 ðDrÞðDzÞ 2 S5 S5   ZZ ZZ . . Dq F \$b n dS ¼  F \$b e q dS z  Fq r0 ; q0  ; z0 ðDrÞðDzÞ 2 S6 S6 so

   Dq Dq F \$b n dS z Fq r0 ; q0 þ ; z0  Fq r0 ; q0  ; z0 ðDrÞðDzÞ 2 2 S5 þS6

    Dq Dq Fq r0 ; q0 þ ; z0  Fq r0 ; q0  ; z0 1 2 2 ¼ r Dq 1 vFq ðr0 ; q0 ; z0 Þ ðDrÞrðDqÞðDzÞ z DV. r vq Putting the faces together, we have ZZ ZZ . . vFz ðr0 ; q0 ; z0 Þ vFðr0 ; q0 ; z0 Þr DV þ DV F \$b n dS ¼ F \$b n dS z vz vr S S1 þ/þS6 ZZ

.



1 1 vFq ðr0 ; q0 ; z0 Þ DV þ Fr ðr0 ; q0 ; z0 ÞDV þ r r vq so that 1 DV

ZZ

.

1 F \$b n dS z DV S

ZZ S1 þ/þS6

.

F \$b n dS z

vFz ðr0 ; q0 ; z0 Þ vFr ðr0 ; q0 ; z0 Þr þ vr vz

1 1 vFq ðr0 ; q0 ; z0 Þ þ Fr ðr0 ; q0 ; z0 Þ þ . r r vq

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CHAPTER 2 Vector Calculus

As long as the component functions are continuously differentiable, we can take the limit as kDVk / 0, and we have in cylindrical coordinates ZZ . . 1 vFz ðr0 ; q0 ; z0 Þ vFr ðr0 ; q0 ; z0 Þ þ F \$b n dS ¼ div F ðr0 ; q0 ; z0 Þ ¼ lim vz vr kDVk/0 DV S 1 1 vFq ðr0 ; q0 ; z0 Þ þ Fr ðr0 ; q0 ; z0 Þ þ r r vq ¼

1 v 1 vFq vFz ðrFr Þ þ þ . r vr r vq vz

Thus, we say .

div F ¼

1 v 1 vFq vFz þ . ðrFr Þ þ vz r vr r vq

Example: We compute the divergence of .

F ðr; q; zÞ ¼ 5r 2 ez sin qb e r þ r tan qz3 ebq þ zb ez.

We have .

div F ¼ ¼

 v  1 v  1 v  2 z r 5r e sin q þ r tan qz3 þ ðzÞ r vr r vq vz  1 2 z 15r e sin q þ sec2 qz3 þ 1 ¼ 15r ez sin q þ sec2 qz3 þ 1: r

SPHERICAL COORDINATE CASE In integrating in spherical coordinates, the incremental element of volume is shown in Fig. 2.2.3. The volume of the incremental piece is DV ¼ r2sin qd4dqdr. We proceed as in the two previous cases. Consider an incremental volume . e r þ Fq ðr; q; 4;Þb eqþ element centered at (r0, q0, 40). Let F ðr; q; 4;Þ ¼ Fr ðr; q; 4;Þb e 4 . We first compute F4 ðr; q; 4;Þb ZZ ZZ ZZ ZZ . . . . F \$b n dS þ F \$b n dS ¼ F \$b e q dS  F \$b e q dS S1

S2

S1

S2

 Z Z  Z Z Dq Dq dS  Fq r0 ; q0  ; 40 dS z Fq r0 ; q0 þ ; 40 2 2 S1 S2     Dq Dq DrD4 z Fq r0 ; q0 þ ; 40 r sin q þ 2 2     Dq Dq Fq r0 ; q0  ; 40 r sin q  DrD4. 2 2

2.2 The Divergence and Curl

z S2 S4

S3

Δr

( r0 + Δr2 , θ 0 + Δ2θ , φ 0 + Δ2φ ) S1

Δθ

y Δφ

x

FIGURE 2.2.3

We

expand

f (x þ Dx) z f (x) þ

and likewise,

Thus



sin q þ Dq 2

using

the

Taylor

f 0 (x)Dx,

so that   Dq Dq z sin q þ cos q sin q þ 2 2

  Dq Dq z sin q  cos q . sin q  2 2 

   Dq Dq DrD4 Fq r0 ; q0 þ ; 40 r sin q þ 2 2  

Dq Dq z Fq r0 ; q0 þ ; 40 r sin q þ cos q DrD4 2 2     Dq Dq Fq r0 ; q0  ; 40 r sin q  DrD4 2 2  

Dq Dq DrD4 z Fq r0 ; q0  ; 40 r sin q  cos q 2 2

approximation

97

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CHAPTER 2 Vector Calculus

and so     Dq Dq DrD4 Fq r0 ; q0 þ ; 40 r sin q þ 2 2     Dq Dq Fq r0 ; q0  ; 40 r sin q  DrD4 2 2     Dq Dq z Fq r0 ; q0 þ ; 40 r sin qDrD4  Fq r0 ; q0  ; 40 r sin qDrD4 2 2   Dq Dq þFq r0 ; q0 þ ; 40 r cos q DrD4 2 2   Dq Dq þFq r0 ; q0  ; 40 r cos q DrD4 2 2     Dq Dq Fq r0 ; q0 þ ; 40  Fq r0 ; q0  ; 40 2 2 r sin qDrD4Dq z Dq þFq ðr0 ; q0 ; 40 Þr cos qDqDrD4     Dq Dq F q r 0 ; q 0 þ ; 40  F q r 0 ; q 0  ; 4 0  2  1 2 2 r sin qDrD4Dq ¼ Dq r   1 þ Fq ðr0 ; q0 ; 40 Þcos q r2 sin qDrD4Dq r sin q     " F r ; q þ Dq; 4  F r ; q  Dq; 4 # 0 0 q 1 q 0 0 1 2 0 2 0 ¼ Fq ðr0 ; q0 ; 40 Þcos q DV þ r r sin q Dq

1 vFq ðr0 ; q0 ; 40 Þ 1 þ Fq ðr0 ; q0 ; 40 Þcos q DV. z r vq r sin q .

In our approximations we have assumed that each component of F is continuously differentiable. The term 1 vFq 1 Fq cos q þ r vq r sin q is often written in the more compact form

1 v ðsin qFq Þ . r sin q vq

2.2 The Divergence and Curl

We next compute ZZ ZZ ZZ ZZ . . . . F \$b n dS þ F \$b n dS ¼ F \$b e 4 dS  F \$b e 4 dS S3

S4

S3

S4

4 4 z F4 r0 ; q0 ; 40 þ rDrDq  F4 r0 ; q0 ; 40  rDrDq 2 2

4 4 F4 r 0 ; q 0 ; 4 0 þ  F4 r 0 ; q 0 ; 4 0  1 2 2 2 r sin qDrDqD4 ¼ D4 r sin q z

1 vF4 ðr0 ; q0 ; 40 Þ DV. r sin q v4

Finally, ZZ ZZ ZZ ZZ . . . . F \$b n dS þ F \$b n dS ¼ F \$b e r dS  F \$b e r dS S5



S6

S5

S6

    Dr Dr Dr r0 þ z F r r 0 þ ; q 0 ; 40 sin qD4 r0 þ Dq 2 2 2      Dr Dr Dr r0  sin qD4 r0  Dq  Fr r0  ; q0 ; 40 2 2 2     Dr Dr 2 2 ¼ Fr r0 þ ; q0 ; 40 sin qD4Dq r0 þ r0 Dr þ 4 2     Dr Dr 2  Fr r0  ; q0 ; 40 sin qD4Dq r02  r0 Dr þ 2 4   

 Dr Dr z Fr r0 þ ; q0 ; 40  Fr r0  ; q0 ; 40 r02 sin qD4Dq 2 2 þ 2Fr ðr0 ; q0 ; 40 Þr0 Dr sin qD4Dq

    Dr Dr Fr r0 þ ; q0 ; 40  Fr r0  ; q0 ; 40 2 2 Drr02 sin qD4Dq ¼ Dr 2 vFr ðr0 ; q0 ; 40 Þ DV Fr ðr0 ; q0 ; 40 Þr02 Dr sin qD4Dq z r0 vr

2 vFr ðr0 ; q0 ; 40 Þ 2 þ Fr ðr0 ; q0 ; 40 ÞDV ¼ þ Fr ðr0 ; q0 ; 40 Þ DV. r0 vr r0 þ

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CHAPTER 2 Vector Calculus

The expression vFr 2 þ Fr r vr is often written in the final formula for the divergence as

1 2 vFr . 2rF þ r r vr r2 Thus,

ZZ S

and so

1 2 vFr 2rF þ r DV r vr r2 S1 þ/þS6

1 vF4 1 v DV þ þ ðsin qFq Þ DV r sin q v4 r sin q vq ZZ

.

F \$b n dS ¼

.

F \$b n dS z

1 2 vFr 2rF þ r r vr kDVk/0 r2 S

1 vF4 1 v ðsin qFq Þ . þ þ r sin q v4 r sin q vq lim

1 DV

ZZ

.

F \$b n dS ¼

Finally, we remark that it is important to remember that the divergence is a property of a vector field and not a formula. Example: We compute the divergence of .

e r þ er cos q sin 4b e q þ cos q cos 4b e4. F ðr; q; 4Þ ¼ r 2 sin q cos 4b

We have Fr ¼ r 2 sin q cos 4; Fq ¼ er cos q sin 4; F4 ¼ cos q cos 4 so that vFr ¼ 2r sin q cos 4; vr   v v ðsin qFq Þ ¼ sin qer cos q sin 4 ¼ er sin 4  sin2 q þ cos2 q ; vq vq vF4 ¼ cos q sin 4. v4

2.2 The Divergence and Curl

Then

  . 1 vFr 1 vF4 1 v ðsin qFq Þ þ þ div F ¼ 2 2rFr þ r 2 r sin q v4 r sin q vq vr r ¼

 1  2 1 ðcos q sin 4Þ 2r r sin q cos 4 þ r 2 ð2r sin q cos 4Þ þ 2 r sin q r

þ

  1 r e sin 4  sin2 q þ cos2 q r sin q

¼ 4r sin q cos 4 

  1 1 r cos q sin 4 þ e sin 4  sin2 q þ cos2 q . r sin q r sin q

THE CURL The line integral of a vector field over a path gives the tendency of the vector field to follow that path. This is often called the circulation of the vector field along the path. A positive (negative) circulation indicates that we move with (against) the direction of the vector field. To compute the curl of a vector field at a point we compute the circulation of a vector field per unit area in an infinitesimally small circle about a particular point in a particular plane. This is equal to the component of the curl of the vector field in the direction normal to plane. We then say that we have found the tendency of a vector field to rotate or “curl” about the point. Said another way, the curl of a vector field measures the tendency of a vector field to cause rotation.

THE CURL IN CARTESIAN COORDINATES Let the vector field be denoted .

b F ðx; y; zÞ ¼ Fx ðx; y; zÞ b i þ Fy ðx; y; zÞ b j þ Fz ðx; y; zÞ k.

Choose (x0, y0, z0). We form three rectangular loops centered around (x0, y0, z0), and compute the circulation around each. The first loop is parallel to the x,y b The coordinates of the corners of plane (so that it is normal to the k).  vector    

Dy Dy Dy Dx Dx the loop are x0  Dx 2 ; y0  2 ; z0 ; x0 þ 2 ; y 0  2 ; z0 ; x 0 þ 2 ; y0 þ 2 ; z0 ;   Dy x0  Dx . See Fig. 2.2.4. ; y þ ; z 0 0 2 2

101

102

CHAPTER 2 Vector Calculus

z C4 Δz

C1

(x0, y0, z0) C3

C2

y Δx Δy x

, y + Δy , z – Δz (x0 + Δx 2 0 2 0 2)

FIGURE 2.2.4

In this discussion bt will be either bi, bj, or b k, whichever is parallel to the path. We .

compute the path integral of F around the loop. On C1 and C3 the only contribution . i. Thus of F is due to Fx b  Dy F \$b t ds ¼ Fx x; y0  ; z0 dx Dx 2 C1  x0  2  Dy we approximate Fx on C1 by Fx x0 ; y0  2 ; z0 . So   Z . Dy b F \$ t ds z Fx x0 ; y0  ; z0 Dx. 2 C1 Z

.

Z

x0 þDx 2



Because the direction of the path C3, we have    Z x0 Dx  Z 2 . Dy Dy b F \$ t ds z Fx x; y0 þ ; z0 dx z  Fx x0 ; y0 þ ; z0 Dx 2 2 C3 x0 þDx 2 so

  

 Dy Dy F \$b t ds z Fx x0 ; y0  ; z0  Fx x0 ; y0 þ ; z0 Dx 2 2 C1 þC3   

 Dy Dy  Fx x0 ; y0 þ ; z0  Fx x0 ; y0  ; z0 2 2 DxDy ¼ Dy

Z

.

z

vFx ðx0 ; y0 ; z0 Þ vFx ðx0 ; y0 ; z0 Þ DxDy ¼  DA vy vy

where DA ¼ DxDy is the area of the loop.

2.2 The Divergence and Curl

On C2 we have Z Z . F \$b t ds ¼

y0 Dy 2

C2

and on C4 Z Z . b F \$ t ds ¼

y0 Dy 2

y0 þDy 2

C4

so

y0 þDy 2

    Dx Dx Fy x0 þ ; y ; z0 dy z Fy x0 þ ; y0 ; z0 Dy 2 2

    Dx Dx Fy x0  ; y ; z0 dy z  Fy x0  ; y0 ; z0 Dy 2 2

    Dx Dx F \$b t ds z Fy x0 þ ; y0 ; z0 Dy  Fy x0  ; y0 ; z0 Dy 2 2 C2 þC4

    Dx Dx Fy x0 þ ; y0 ; z0  Fy x0  ; y0 ; z0 2 2 DxDy ¼ Dx

Z

.

vFy ðx0 ; y0 ; z0 Þ DA. vx Thus the circulation around this loop is   Z . vFy ðx0 ; y0 ; z0 Þ vFx ðx0 ; y0 ; z0 Þ b  F \$ t ds z DA. vx vy C1 þ/þC4 z

This is the circulation normal to the kb vector. We divide by DA and take the limit as kDAk/0 to get Z . vFy ðx0 ; y0 ; z0 Þ vFx ðx0 ; y0 ; z0 Þ 1  F \$b t ds ¼ lim vx vy kDAk/0 kDAk C1 þ/þC4 is the circulation about (x0, y0, z0) in the x,y plane. We repeat this procedure for the loop parallel to the x,z plane, so it will be perpendicular to the b j vector. Our explanation will be more brief. The loop perpendicular to the b j vector is shown in Fig. 2.2.5. In this case we have    Z x0 þDx  Z 2 . Dz Dz dx z Fx x0 ; y0 ; z0  Dx F \$b t ds ¼ Fx x; y0 ; z0  2 2 C1 x0 Dx 2 Z

. C3

F \$b t ds ¼

Z

x0 Dx 2 x0 þDx 2

    Dz Dz dx z  Fx x0 ; y0 ; z0 þ Dx Fx x; y0 ; z0 þ 2 2

103

104

CHAPTER 2 Vector Calculus

z C3

C4 Δz

C2

(x, y, z)

y C1

Δx

Δy x

FIGURE 2.2.5

so

Z

  

 Dz Dz  Fx x0 ; y0 ; z0 þ Dx F \$b t ds z Fx x0 ; y0 ; z0  2 2 C1 þC3   

 Dz Dz  Fx x 0 ; y 0 ; z0   F x x 0 ; y0 ; z0 þ 2 2 DxDz ¼ Dz .

z

vFx ðx0 ; y0 ; z0 Þ vFx ðx0 ; y0 ; z0 Þ DxDz ¼  DA vz vz

where DA ¼ DxDz is the area of the loop. We also have    Z z0 þDz  Z 2 . Dx Dx F \$b t ds ¼ Fz x0  ; y0 ; z dz z Fz x0  ; y0 ; z0 Dz 2 2 C2 z0 Dz 2 Z C4

so

Z

.

F \$b t ds ¼

Z

z0 Dz 2

z0 þDz 2

    Dx Dx Fz x0  ; y0 ; z0 dz z  Fz x0 þ ; y0 ; z0 Dx 2 2

    Dx Dx F \$b t ds z Fz x0  ; y0 ; z0 Dz  Fz x0 þ ; y0 ; z0 Dz 2 2 C2 þC4

    Dx Dx Fz x0 þ ; y0 ; z0  Fz x0  ; y0 ; z0 2 2 ðDxÞðDzÞ ¼ Dx .

z

vFz ðx0 ; y0 ; z0 Þ DA. vx

2.2 The Divergence and Curl

Thus the circulation around this loop is   Z . vFx ðx0 ; y0 ; z0 Þ vFz ðx0 ; y0 ; z0 Þ DA. F \$b t ds z  vz vx C1 þ/þC4 We divide by DA and take the limit as kDAk/0 to get lim

kDAk/0

1 DA

Z

.

C1 þ/þC4

F \$b t ds ¼

vFx ðx0 ; y0 ; z0 Þ vFz ðx0 ; y0 ; z0 Þ  vz vx

is the circulation about (x0, y0, z0) in the x,z plane, which is the circulation normal to the b j vector. The loop perpendicular to the b i vector is shown in Fig. 2.2.6. We leave it as an exercise to show lim

kDAk/0

1 DA

Z

.

C1 þ/þC4

F \$b t ds ¼

vFz ðx0 ; y0 ; z0 Þ vFy ðx0 ; y0 ; z0 Þ  vz vy

is the circulation about (x0, y0, z0) in the y,z plane, which is the circulation normal to the b i vector. . . b denoted V  F , in Cartesian coordinates is i þ Fy b j þ Fz k, The curl of F ¼ Fx b defined to be       . vFy vFx b vFz vFy b vFx vFz b  iþ  jþ  k. V F ¼ vy vz vz vx vx vy

z C3 C2 C4

C1

y Δx Δy

x

FIGURE 2.2.6

Δz

(x, y, z)

105

106

CHAPTER 2 Vector Calculus

The formula defies memorization, but there is an easy way to do the computation, namely     b b  b i j k     . v v  v V F ¼ .  vx vy vz       F x Fy Fz  Example: The curl of .

F ðx; y; zÞ ¼ x2 ey sin z b i þ 3yz b j þ xy kb

is

  b  i   . v  V F ¼  vx   2 y  x e sin z

b j v vy 3yz

  kb     v  b j  x2 ey sin z k. i þ x2 ey cos z  y b  ¼ ðx  3yÞ b vz    xy 

THE CURL IN CYLINDRICAL COORDINATES In cylindrical coordinates we denote the vector field .

F ðr; q; zÞ ¼ Fr ðr; q; zÞb e r þ Fq ðr; q; zÞb e q þ Fz ðr; q; zÞb ez.

Choose a point (r0, q0, z0) an construct a basic increment of volume in cylindrical coordinates centered at (r0, q0, z0). As in the Cartesian coordinate case, we construct three loops around the incremental piece, each centered at (r0, q0, z0) and each perpendicular to one of the vectors ebr ; ebq ; ebz . We compute the circulation around each loop. We first consider the loop perpendicular to ebz . The coordinates of the corners of   

Dq; z Dr ; q  Dq; z Dr ; q þ Dq; z ; r ; r the loop are r0  Dr ; q  þ þ 0 0 0 0 0 0 0 0 2 2 2 2 2 2

 Dq and r0  Dr 2 ; q0 þ 2 ; z0 . See Fig. 2.2.7. The path integral on C1    Z r0 þDr  2 Dq Dq Fr r ; q0  ; z0 dr z Fr r0 ; q0  ; z0 Dr 2 2 r0 Dr 2     Dq; z . where we have approximated Fr r ; q0 þ Dq on C r ; z by F ; q þ r 0 0 1 2 0 2 0

2.2 The Divergence and Curl

z

(r −

Δr Δθ ,θ+ , z) 2 2

êz Δθ

(r −

Δr Δθ ,θ − , z) 2 2

(r +

Δr Δθ ,θ− , z) 2 2

Δr

C4 C1 C2

(r +

Δr Δθ ,θ+ , z) 2 2

C3

y

x

FIGURE 2.2.7

The path integral on C3 is Z

r0 Dr 2

r0 þDr 2

    Dq Dq Fr r ; q0 þ ; z0 dr z  Fr r0 ; q0 þ ; z0 Dr 2 2

so the sum of the two integrals is approximately

    Dq Dq Fr r0 ; q0  ; z0  Fr r0 ; q0 þ ; z0 Dr 2 2

    Dq Dq Fr r0 ; q0 þ ; z0  Fr r0 ; q0  ; z0 2 2 DqDr. ¼ Dq The area enclosed by the loop, denoted DA, is approximately DA z r0 ðDqÞðDrÞ so

    Dq Dq Fr r0 ; q0 þ ; z0  Fr r0 ; q0  ; z0 1 1 vFr ðr0 ; q0 ; z0 Þ 2 2  DA. r0 DqDr z  Dq r0 r0 vq

107

108

CHAPTER 2 Vector Calculus

The path integral on the path C3 is Z

Dq    Dr 2 F r þ Dr ; q ; z dq r0 þ 0 Dq q 0 2 2 q0  2    Dr Dr z Fq r0 þ ; q0 ; z0 r0 þ Dq 2 2 q0 þ

and the path integral on the path C4 is Dq    Dr 2 F r  Dr ; q ; z r  dq 0 0 Dq q 0 2 2 q0 þ 2    Dr Dr r0  Dq: z  Fq r0  ; q0 ; z0 2 2

Z

q0 

Summing the integrals on C3 and C4 gives       Dr Dr Dr Dr r0 þ r0  Dq  Fq r0  ; q0 ; z0 Dq F q r 0 þ ; q0 ; z 0 2 2 2 2   

 Dr Dr z Fq r0 þ ; q0 ; z0  Fq r0  ; q0 ; z0 r0 Dq þ Fq ðr0 ; q0 ; z0 ÞDrDq 2 2

    Dr Dr Fq r0 þ ; q0 ; z0  Fq r0  ; q0 ; z0 1 2 2 r0 DqDr þ Fq ðr0 ; q0 ; z0 Þr0 DrDq ¼ Dr r0 z

vFq ðr0 ; q0 ; z0 Þ 1 DA þ Fq ðr0 ; q0 ; z0 ÞDA. vr r0

Combining the integrals over the four paths gives 1 vFr ðr0 ; q0 ; z0 Þ vFq ðr0 ; q0 ; z0 Þ DA þ DA þ Fq ðr0 ; q0 ; z0 ÞDA.  r0 vq vr We next consider the loop perpendicular to ebq . The coordinates of the corners of

   Dz Dr Dz Dr Dz the loop are r0  Dr 2 ; q0 ; z0  2 ; r0 þ 2 ; q0 ; z0  2 ; r0 þ 2 ; q0 ; z0 þ 2 

Dz . The area of the loop is DA ¼ (Dr)\$(Dz). See Fig. 2.2.8. ; q ; z þ and r0  Dr 0 0 2 2

 The path integral on C1 is approximately Fr r0 ; q0 ; z0  Dz 2 Dr, the path

 integral on C3 is approximately Fr r0 ; q0 ; z0 þ Dz 2 Dr so the combined value is

2.2 The Divergence and Curl

z

(r0 – Δr, θ , z0 + Δz ) 2 2 (r0 + Δr, θ , z0 + Δz ) 2 2

C3

C2

êθ (r0 – Δr, θ , z0 – Δz ) 2 2

C4 C1

Δz

Δr

y

(r0 + Δr, θ , z0 – Δz ) 2 2 x

FIGURE 2.2.8

   Dz Dz Dr  Fr r0 ; q0 ; z0  Dr Fr r0 ; q0 ; z0 þ 2 2

    Dz Dz Fr r 0 ; q 0 ; z 0 þ  Fr r0 ; q0 ; z0  2 2 DrDz ¼ Dz 

z

vFr ðr0 ; q0 ; z0 Þ vFr ðr0 ; q0 ; z0 Þ DrDz ¼ DA. vz vz

 The path integral on C2 is approximately Fz r0 þ Dr 2 ; q0 ; z0 Dz; the path inte  gral on C4 is approximately Fz r0  Dr ; q ; z 0 0 Dz so the combined value is 2     Dr Dr Fz r0 þ ; q0 ; z0 Dz þ Fz r0  ; q0 ; z0 Dz 2 2   

 Dr Dr  Fz r0 þ ; q0 ; z0  Fz r0  ; q0 ; z0 2 2 ¼ DrDz Dr vFz ðr0 ; q0 ; z0 Þ DA. vr Combining the integrals on the four paths gives the integral around the loop. This z

is vFr ðr0 ; q0 ; z0 Þ vFz ðr0 ; q0 ; z0 Þ DA  DA. vz vr

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CHAPTER 2 Vector Calculus

Δz (r0, θ + Δθ, z0+ ) 2 2

Δz Δ (r0, θ – θ, z0+ ) 2 2 z

C3 C2 Δz Δθ

êr

C4

y

C1 Δ Δz (r0, θ – θ, z0– ) 2 2 x

(r0, θ +

Δθ Δz ,z – ) 2 0 2

FIGURE 2.2.9

We next consider the loop perpendicular to ebr . The coordinates of the corners of the   

Dz ; Dq; z þ Dz ; Dq; z þ Dz and r r loop are r0 ; q0  Dq ; z  ; q  ; q þ 0 0 0 0 0 0 0 2 2 2 2 2 2 

Dz r0 ; q0 þ Dq 2 ; z0  2 . See Fig. 2.2.9. The length of one side of the loop is r0Dq, and the length of the other side is Dz so the area of the loop is DA ¼ r0DqDz. Note that the length of C1 is r0Dq so the path integral on C1 is approx  imately Fq r0 ; q0 ; z0  Dz 2 r0 Dq, the path integral on C3 is approximately

 Fq r0 ; q0 ; z0 þ Dz 2 r0 Dq, so the combined value is     Dz Dz r0 Dq  Fq r0 ; q0 ; z0 þ r0 Dq Fq r0 ; q0 ; z0  2 2

    Dz Dz Fq r0 ; q0 ; z0   Fq r0 ; q0 ; z0 þ 2 2 ¼ r0 DzDq Dz

    Dz Dz Fq r0 ; q0 ; z0 þ  Fq r0 ; q0 ; z0  2 2 ¼ r0 DzDq Dz z ¼

vFq ðr0 ; q0 ; z0 Þ r0 DzDq vz

vFq ðr0 ; q0 ; z0 Þ DA. vz

2.2 The Divergence and Curl

 Dz, the path inteThe path integral on C2 is approximately Fz r0 ; q0 þ Dq ; z 0 2

 gral on C4 is approximately Fz r0 ; q0  Dq 2 ; z0 Dz so the combined value is     Dq Dq Fz r0 ; q0 þ ; z0 Dz  Fz r0 ; q0  ; z0 Dz 2 2

    Dq Dq Fz r0 ; q0 þ ; z0  Fz r0 ; q0  ; z0 1 2 2 ¼ r0 DqDz Dq r0 z

1 vFz ðr0 ; q0 ; z0 Þ DA. r0 vq

Thus, the path integral around the loop is 1 vFz ðr0 ; q0 ; z0 Þ vFq ðr0 ; q0 ; z0 Þ DA  DA. r0 vq vz .

Finally, we have that the curl of F ðr; q; zÞ ¼ Fr ðr; q; zÞb e r þ Fq ðr; q; zÞb eqþ e z in cylindrical coordinates is Fz ðr; q; zÞb    

. 1 vFz vFq vFr vFz 1 v 1 vFr ðrFq Þ    V F ¼ ebr þ ebq þ ebz . r vq r vr r vq vz vz vr Example: We compute the curl of .

e r þ r tan qz3 ebq þ zb ez. F ðr; q; zÞ ¼ 5r 2 ez sin qb

We have Fz ¼ z; Fq ¼ rðtan qÞz3 so Fr ¼ 5r 2 ez sin q so Thus,

so

vFz ¼0 vq

and

vFz ¼0 vr

vFq v ðrFq Þ ¼ 2rz3 tan q ¼ 3z2 r tan q and vr vz vFr vFr ¼ 5r2 ez sin q and ¼ 5r2 ez cos q. vz vq

    1 vFz vFq vFr vFz 2  ¼ 3z r tan q;  ¼ 5r2 ez sin q; r vq vz vz vr 1 v 1 vFr ¼ 2 z3 tan q  5r ez cos q ðrFq Þ  r vr r vq

111

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CHAPTER 2 Vector Calculus

so

   

1 vFz vFq vFr vFz 1 v 1 vFr ðrFq Þ    ebr þ ebq þ ebz r vq r vr r vq vz vz vr   ¼ 3z2 r tan qb e r þ 5r 2 ez sin qb e q þ 2 z3 tan q  5r ez cos q ebz .

.

V F ¼

THE CURL IN SPHERICAL COORDINATES

.

In spherical coordinates, we define the vector field to be F ¼ Fr ebr þ Fq ebq þ F4 eb4 . We construct an incremental element of volume in spherical coordinates about the point (r0, q0, 40) as shown in Fig. 2.2.10. We construct paths about (r0, q0, 40) and eb4 . perpendicular to the vectors ebr , ebq , . We compute the path integral of F around the path that is perpendicular to ebr . A diagram of the path is shown in Fig. 2.2.11. 2 The area enclosed by the path is DA  0(D4)r0(Dq) ¼ r0sinq0(D4) (Dq).

z r0sinq We have the length of C1 ¼ r0 sin q0 þ Dq 2 D4; the length of C2 ¼ r0Dq; the 

Dq length of C3 ¼ r0 sin q0  2 D4; the length of C4 ¼ r0Dq. 

From Taylor’s theorem, we have f (x þ Dx) z f (x) þ f0 (x)Dx, so sin q0 þ Dq 2 z 

Dq Dq sin q0 þ cos q0 Dq 2 and sin q0  2 z sin q0  cos q0 2 .

z

(r0, θ 0, φ 0) Δr

Δθ y Δφ x

FIGURE 2.2.10

2.2 The Divergence and Curl

z (r0 , θ 0 –

(r0 , θ0 –

Δφ Δθ ,φ – ) 2 0 2

C3

φ Δθ ,φ +Δ ) 2 0 2

êr C2

(r0 , θ 0 +

Δφ Δθ ,φ – ) 2 0 2

(r0 , θ0 +

C4 C1

Δθ Δφ ,φ + ) 2 0 2

Δθ y Δφ

x

FIGURE 2.2.11

Thus,

    Dq Dq b F \$ t ds z F4 r0 ; q0 þ ; 40 r0 sin q0 þ D4 2 2 C1  

Dq Dq cos q0 D4 z F4 r0 ; q0 þ ; 40 r0 sin q0 þ 2 2   Dq ¼ F4 r0 ; q0 þ ; 40 r0 sin q0 D4 2   Dq Dq þF4 r0 ; q0 þ ; 40 r0 cos q0 D4 2 2       Z . Dq Dq b F \$ t ds z  F4 r0 ; q0  ; 40 r0 sin q0  D4 2 2 C3    Dq z  F4 r0 ; q0  ; 40 r0 sin q0 D4 2    Dq Dq F4 r0 ; q0  ; 40 r0 cos q0 D4 . 2 2 Z

.

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CHAPTER 2 Vector Calculus

So

Z

  Dq F \$b t ds z F4 r0 ; q0 þ ; 40 r0 sin q0 D4 2 C1 þC3   Dq Dq þF4 r0 ; q0 þ ; 40 r0 cos q0 D4 2 2    Dq  F4 r0 ; q0  ; 40 r0 sin q0 D4 2    Dq Dq F4 r0 ; q0  ; 40 r0 cos q0 D4 2 2 

 Dq ¼ F4 r0 ; q0 þ ; 40 r0 sin q0 D4 2   Dq F4 r0 ; q0  ; 40 r0 sin q0 D4 2   Dq þF4 r0 ; q0 þ ; 40 r0 cos q0 DqD4 2     Dq Dq F4 r0 ; q0 þ ; 40  F4 r0 ; q0  ; 40 2 2 r0 sin q0 DqD4 ¼ Dq   Dq þF4 r0 ; q0 þ ; 40 r0 cos q0 DqD4 2 .

vF4 ðr0 ; q0 ; 40 Þ r0 sin q0 D4Dq þ F4 ðr0 ; q0 ; 40 Þr0 cos q0 DqD4 vq 1 vF4 ðr0 ; q0 ; 40 Þ 2 1 r0 sin q0 D4Dq þ ¼ F4 ðr0 ; q0 ; 40 Þr02 sin q0 cos q0 D4Dq r0 vq r0 sin q0

1 vF4 ðr0 ; q0 ; 40 Þ 1 þ ¼ F4 ðr0 ; q0 ; 40 Þcos q0 DA. r0 vq r0 sin q0 z

Note that  1 vF4 1 1 v  þ cos q F4 ¼ sin qF4 ; r vq r sin q r sin q vq and we shall use the more compact form of this expression in the final formula. Next,   Z . Dq F \$b t ds z  Fq r0 ; q0 þ ; 40 r0 Dq 2 C2

2.2 The Divergence and Curl

and

so

Z

  Dq F \$b t ds z Fq r0 ; q0  ; 40 r0 Dq 2 C4 .

  

 Dq Dq b F \$ t ds z  Fq r0 ; q0 þ ; 40  Fq r0 ; q0  ; 40 r0 Dq 2 2 C2 þC4     "F r ; q þ Dq; 4  F r ; q  Dq; 4 # 0 0 0 0 q q 1 2 0 2 0 r02 sin q0 D4Dq ¼ D4 r0 sin q0

Z

.

z Thus, we have Z C1 þC2 þC3 þC4

1 vFq ðr0 ; q0 ; 40 Þ DA. r0 sin q0 v4 

.

F \$b t ds z

1 vF4 ðr0 ; q0 ; 40 Þ 1 F4 ðr0 ; q0 ; 40 Þ þ r0 vq r0 sin q0  1 vFq ðr0 ; q0 ; 40 Þ  DA. r0 sin q0 v4

Dividing by DA and taking the limit as kDAk/0, we get the rotational compo. nent of F in the direction of ebr is Z . 1 1 vF4 1 1 vFq F4  þ lim F \$b t ds ¼ r vq r sin q r sin q v4 kDAk/0 kDAk C1 þC2 þC3 þC4 ¼

 1 v  1 vFq . sin qF4  r sin q vq r sin q v4 .

We next compute the path integral of F around the path that is perpendicular to eb4 . A diagram of the path is shown in Fig. 2.2.12.   The area thepath is DA z r0(Dq)\$(Dr). The length of C1 is r0  Dr 2 Dq , and the  length of C3 is r0 þ Dr 2 Dq. The length of C2 is Dr, as is length of C4. Now    Z . Dr Dr Dq r0  F \$b t ds z  Fq r0  ; q0 ; 40 2 2 C1     Dr Dr Dr ¼ Fq r0  ; q0 ; 40 r0 Dq þ Fq r0  ; q0 ; 40 Dq 2 2 2    Z . Dr Dr Dq r0 þ F \$b t ds z Fq r0 þ ; q0 ; 40 2 2 C3     Dr Dq Dr ¼ Fq r0 þ ; q0 ; 40 r0 Dq þ Fq r0 ; q0 þ ; 40 Dq 2 2 2

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CHAPTER 2 Vector Calculus

z (r0 +

C2

Δθ Δr ,θ – ) 2 0 2 C3 êφ

(r0 –

Δθ Δr ,θ – ) 2 0 2

(r0 +

Δθ

C1

Δθ Δr ,θ + ) 2 0 2 y

C4 x

(r0 –

Δr Δθ ,θ + ) 2 0 2

FIGURE 2.2.12

so

Z

  

 Dr Dr F \$b t ds z Fq r0 þ ; q0 ; 40  Fq r0  ; q0 ; 40 r0 Dq 2 2 C1 þC3 .

þ Fq ðr0 ; q0 ; 40 ÞDrDq

    Dr Dr Fq r0 þ ; q0 ; 40  Fq r0  ; q0 ; 40 2 2 ðDrÞr0 Dq ¼ Dr þ z Also,

1 Fq ðr0 ; q0 ; 40 ÞðDrÞr0 ðDqÞ r0

vFq ðr0 ; q0 ; 40 Þ 1 DA þ Fq ðr0 ; q0 ; 40 ÞDA. vr r0

  Dq b F \$ t ds z Fr r0 ; q0  ; 40 Dr 2 C2

Z

so

.

  Dq b F \$ t ds z  Fr r0 ; q0 þ ; 40 Dr 2 C4

Z

.

    Dq Dq F \$b t ds z Fr r0 ; q0  ; 40 Dr  Fr r0 ; q0 þ ; 40 Dr 2 2 C2 þC4

    Dq Dq Fr r0 ; q0 þ ; 40  Fr r0 ; q0  ; 40 2 2 ðDqÞðDrÞ ¼ Dq 1 vFr ðr0 ; q0 ; 40 Þ 1 vFr ðr0 ; q0 ; 40 Þ z r0 ðDqÞðDrÞ ¼  DA. r0 vq r0 vq

Z

.

2.2 The Divergence and Curl

Dividing by DA and taking the limit as kDAk/0, we get the rotational compo. nent of F in the direction of eb4 is lim

kDAk/0

1 kDAk

Z

. C1 þC2 þC3 þC4

F \$b t ds ¼

vFq 1 vFq 1 vFr 1 v 1 vFr ðrFq Þ  þ  ¼ . r vr r vq r vr r vq vr .

We next compute the path integral of F around the path that is perpendicular to ebq . A diagram of the path is shown in Fig. 2.2.13. (r0 , –

Δφ Δr ,θ ,φ + ) 2 0 0 2

C1

(r0 , –

Δφ Δr ,θ ,φ – ) 2 0 0 2

C4

(r0, θ 0 , φ 0 ) C2

(r0 , +

Δφ Δr ,θ ,φ + ) 2 0 0 2

C3

(r0 , +

Δφ Δr ,θ ,φ – ) 2 0 0 2

FIGURE 2.2.13

The length of C1 is   TheDr area the path is DA z r0sin q0(D4)(Dr). sin q r0  2 sin q0 ðD4Þ and the length of C3 is r0 þ Dr 0 ðD4Þ. The length of C2 2 is Dr, as is length of C4. Now    Z . Dr Dr sin q0 ðD4Þ r0  F \$b t ds z F4 r0  ; q0 ; 40 2 2 C1   Dr ¼ F4 r0  ; q0 ; 40 r0 sin q0 ðD4Þ 2   Dr Dr  F4 r0  ; q0 ; 40 sin q0 ðD4Þ 2 2 and

   Dr Dr sin q0 ðD4Þ r0 þ F \$b t ds z  F4 r0 þ ; q0 ; 40 2 2 C3   Dr ¼ F4 r0 þ ; q0 ; 40 r0 sin q0 ðD4Þ 2   Dq Dr  F4 r0 ; q0 þ ; 40 sin q0 ðD4Þ 2 2

Z

.

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CHAPTER 2 Vector Calculus

so Z

  

 Dr Dr F \$b t ds z  F4 r0 þ ; q0 ; 40  F4 r0  ; q0 ; 40 r0 sin q0 ðD4Þ 2 2 C1 þC3 .

 F4 ðr0 ; q0 ; 40 ÞDr sin q0 ðD4Þ

    Dr Dr F4 r0 þ ; q0 ; 40  F4 r0  ; q0 ; 40 2 2 ðDrÞr0 sin q0 ðD4Þ ¼ Dr  z Also,

so

1 F4 ðr0 ; q0 ; 40 Þr0 ðDrÞsin q0 ðD4Þ r0

vF4 ðr0 ; q0 ; 40 Þ 1 DA  F4 ðr0 ; q0 ; 40 ÞDA. vr r0   D4 b F \$ t ds z Fr r0 ; q0 ; 40 þ Dr 2 C2   Z . D4 Dr F \$b t ds z  Fr r0 ; q0 ; 40  2 C4 Z

.

    D4 D4 Dr  Fr r0 ; q0 ; 40  Dr F \$b t ds z Fr r0 ; q0 ; 40 þ 2 2 C2 þC4

    D4 D4 Fr r0 ; q0 ; 40 þ  Fr r0 ; q0 ; 40  2 2 ðD4ÞðDrÞ ¼ D4

Z

.

z

1 vFr ðr0 ; q0 ; 40 Þ r0 sin q0 ðD4ÞðDrÞ r0 sin q0 v4

z

1 vFr ðr0 ; q0 ; 40 Þ DA: r0 sin q0 v4

Dividing by DA and taking the limit as kDAk/0, we get the rotational compo. nent of F in the direction of ebq is Z . 1 vF4 1 1 vFr A  F4 þ lim F \$b t ds ¼  r r sin q v4 vr kDAk/0 kDAk C1 þC2 þC3 þC4 ¼

 1 vFr 1 v   rF4 . r sin q v4 r vr

2.2 The Divergence and Curl

.

Thus, in spherical coordinates, if F ¼ Fr ebr þ Fq ebq þ F4 eb4 , then       . 1 v  1 vFq 1 vFr 1 v   sin qF4  rF4 ebq ebr þ V F ¼ r sin q vq r sin q v4 r sin q v4 r vr   1 v 1 vFr þ ðrFq Þ  eb4 . r vr r vq Example: We compute the curl of .

e r þ er cos q sin 4b e q þ cos q cos 4b e4: F ðr; q; 4ÞÞ ¼ r 2 sin q cos 4b

We have Fr ¼ r 2 sin q cos 4; Fq ¼ er cos q sin 4; F4 ¼ cos q cos 4 so    v  v sin qF4 ¼ sin q cos q cos 4 ¼  sin2 q þ cos2 q cos 4 vq vq vFq ¼ er cos q cos 4; v4 vFr ¼ r 2 cos q cos 4; vq .

v ðrFq Þ ¼ ðrer þ er Þcos q sin 4 vr

 vFr v ¼ r 2 sin q sin 4; rF4 ¼ cos q cos 4. v4 vr

Thus, the curl of F is       1 v  1 vFq 1 vFr 1 v   sin qF4  rF4 ebq ebr þ r sin q vq r sin q v4 r sin q v4 r vr   1 v 1 vFr ðrFq Þ  þ eb4 r vr r vq

 1  1  sin2 q þ cos2 q cos 4  er cos q cos 4 ebr ¼ r sin q r sin q

  1 1 2 þ r sin q sin 4  cos q cos 4 ebq r sin q r

1 1 þ ðrer þ er Þcos q sin 4  r 2 cos q cos 4 eb4 . r r

EXERCISES 1. Compute the gradient of a. f (r, q, z) ¼ z2ersin q þ 2rz. b. f (r, q, 4) ¼ rsin 4cos q  tan 4. c. f (r, q, z) ¼ r2 þ rzsec q.

119

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CHAPTER 2 Vector Calculus

d. f (r, q, 4) ¼ r q 4. e. f (r, q, 4) ¼ ln(r4)þsin q. f. f (r, q, z) ¼ z/rcos q. 2. Compute the curl and divergence of . 1 e a. . F ðr; q; 4Þ ¼ r tan q b4 . b. . F ðr; q; 4Þ ¼ cos qb e r þ sin qb e q þ r 2 tan 4b e4. c. . F ðr; q; zÞ ¼ rb eq. d. . F ðr; q; 4Þ ¼ sin qb e q þ r4b e4. e. . F ðr; q; zÞ ¼ r 2 tan qb e r þ zb e q þ er z sin qb ez. f. F ðr; q; zÞ ¼ ln rb e r  cos qb e q þ 4b ez. 3. . a. In cylindrical coordinates, let F ¼ Fr ebr þ Fq ebq þ Fz ebz . Show that the curl of can be computed using the formula   1 1   eb ebz   r ebq r r    . v v . V  F ¼  v  vr vq vz      Fr rFq Fz  .

b. In spherical coordinates, let F ¼ Fr ebr þ Fq ebq þ F4 eb4 . Show that the curl of can be computed using the formula     1 1 1   b b b e e e r 4 q   2 r sin q r   r sin q   . .  v v v V F ¼    vr vq v4      Fr rFq r sin qF4 

2.3 GREEN’S THEOREM, THE DIVERGENCE THEOREM, AND STOKES’ THEOREM In this section, our main focus is the three major theorems of integral vector calculusdGreen’s theorem, the divergence theorem (Gauss’ theorem), and Stokes’ theorem. In a sense, each of these is an extension of the fundamental theorem of calculus. We begin with Green’s theorem, which relates the line integral over the boundary of a region to the area enclosed by the region. (This relates to the fundamental theorem of calculus, Z b f 0 ðxÞdx ¼ f ðbÞ  f ðaÞ a

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

in that the left-hand side is an expression over a region [a, b] and the right-hand side is an expression over the boundary of the region.) We assume that the boundary of the region is a simple closed curve with positive orientation. A simple closed curve is the one that does not intersect itself. A positively oriented curve is a curve that is traversed in the counterclockwise direction, so that the region lies on the left of the direction of transversal. See Fig. 2.3.1.

FIGURE 2.3.1

Recall that to compute

Z C

Pdx þ Qdy

we parameterize C by a function s(t) ¼ (x(t), y(t)), a  t  b and compute Z b

dx dy PðxðtÞ; yðtÞÞ þ QðxðtÞ; yðtÞÞ dt. dt dt t¼a We assume that s0 (t) is piecewise continuous. Theorem (Green’s theorem): Let C be a simple closed curve in R2 that is positively oriented and piecewise smooth. Let D be the region that is enclosed by C. Suppose that P : R2 /R and Q : R2 /R have continuous partial derivatives on an open set containing D. Then  Z ZZ  vQðx; yÞ vPðx; yÞ Pðx; yÞdx þ Qðx; yÞdy ¼  dxdy. vx vy C D We present a major idea in the proof for a simplified region. Consider the region shown in Fig. 2.3.2. We show that in this case  Z ZZ  vPðx; yÞ dydx ¼  Pðx; yÞdx. vy D C

121

122

CHAPTER 2 Vector Calculus

Referring to Fig. 2.3.2, we have

C2

y = φ 2 (x)

C1

C3 D C4

y = φ 1 (x)

a

a

b

b

FIGURE 2.3.2

ZZ  D

 vPðx; yÞ dydx vy Z ¼

x¼a

Z ¼

Z

b

a

b

  vPðx; yÞ dydx vy y¼41 ðxÞ 42 ðxÞ

½Pðx; 42 ðxÞÞ  Pðx; 41 ðxÞÞdx.

We also have Z Z Z Z Z Pðx; yÞdx ¼ Pðx; yÞdx þ Pðx; yÞdx þ Pðx; yÞdx þ Pðx; yÞdx. C

C1

C2

C3

C4

Now dx ¼ 0 on C1 and C3, so Z Z Z Pðx; yÞdx ¼ Pðx; yÞdx þ Pðx; yÞdx. C

C2

C4

Also Z C2

Z Pðx; yÞdx ¼

a

x¼b

Z Pðx; 42 ðxÞÞdx ¼ 

b x¼a

Pðx; 42 ðxÞÞdx

and Z

Z C4

Pðx; yÞdx ¼

b

x¼a

Pðx; 41 ðxÞÞdx

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

so Z

Z C

Pðx; yÞdx ¼  2 ¼ 4

Thus ZZ  D

Z

b

x¼a

Z

Pðx; 42 ðxÞÞdx þ

b x¼a

b x¼a

Pðx; 41 ðxÞÞdx 3

Pðx; 42 ðxÞÞ  Pðx; 41 ðxÞÞdx5.

 Z Z b vPðx; yÞ ½Pðx; 42 ðxÞÞ  Pðx; 41 ðxÞÞdx ¼  Pðx; yÞdx. dydx ¼ vy C a

In Exercise 2 we show Z C

ZZ Qðx; yÞdy ¼

vQðx; yÞ dxdy vx D

for a region of a particular type. Putting these results together gives Green’s theorem. Vector form of Green’s theorem: . If F is a vector field and D is a region in the xy plane with boundary C as mentioned above, then  Z ZZ  . . b V  F \$ kdxdy. F \$db r¼ C

D

This relates to the first statement of Green’s theorem by taking i þ Qðx; yÞ b j. F ¼ Pðx; yÞ b The next example illustrates that it is often  easier to compute  the line integral RR vQðx; yÞ vPðx; yÞ dxdy rather than  vy around a simple closed curve using D vx R C Pðx; yÞdx þ Qðx; R yÞdy. On the other hand, areas can sometimes be more easily computed using C Pðx; yÞdx þ Qðx; yÞdy. Example: We verify Green’s theorem in the case P(x, y) ¼ xy, Q(x, y) ¼ y, and D is the region x2 þ y2  1. To compute Z Pdx þ Qdy .

C

we parameterize C by s(q) ¼ (cos q, sin q) 0  q  2p, so that x ¼ cos q; y ¼ sinq;

dx dy ¼ sin q; ¼ cos q; Pðx; yÞ ¼ cos q sin q; Qðx; yÞ ¼ sin q. dq dq

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CHAPTER 2 Vector Calculus

Then, Z

Z C

2p 

Pdx þ Qdy ¼ 0

To compute

2p  sin3 q sin2 q  sin2 q cos q þ sin q cos q dq ¼ ¼ 0: þ 3 2 0  ZZ  vQðx; yÞ vPðx; yÞ  dxdy vx vy D

note that vQðx; yÞ ¼ 0; vx

vPðx; yÞ ¼x vy

so that in polar coordinates 1 0  ZZ  Z 2p Z 1 Z vQðx; yÞ vPðx; yÞ 1 2p 2 A @ cosqr dr dq ¼  cosqdq ¼ 0:  dxdy ¼  vx vy 3 0 D q¼0 r¼0 An Application: If we take P ¼ y and Q ¼ x, then  ZZ ZZ  vQðx; yÞ vPðx; yÞ  dxdy ¼ 2 dx dy ¼ 2ðArea of DÞ vx vy D D so the area of D can be computed according to its line integral using Z ZZ ydx þ xdy ¼ 2 dx dy ¼ 2ðArea of DÞ C

D

or 1 Area of D ¼ 2

Z C

ydx þ xdy.

We use this idea to compute the area of the ellipse x2 y2 þ ¼ 1: a2 b2 We parameterize the boundary using x ¼ a cos q;

y ¼ b sin q;

0  q  2p

so that dx ¼ a sin q; dq

dy ¼ b cos q dq

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

and 1 2

Z

Z 1 2p ydx þ xdy ¼ ½ðb sin qÞða sin qÞ þ ða cos qÞðb cos qÞdq 2 0 C Z 2p  2 1 1 sin q þ cos2 q dq ¼ abð2pÞ ¼ pab. ¼ ab 2 2 0

Green’s theorem can be extended to regions with “holes” such as the one shown in Fig. 2.3.3A. Note that each part of the boundary has been assigned an orientation. The idea is to divide the region into parts as shown in Fig. 2.3.3B, and compute the integral of each piece. What makes it work is that parts of the integrals cancel each other as Fig. 2.3.3B shows.

(A)

C1

C2

(B)

FIGURE 2.3.3A AND B

The divergence (Gauss’) theorem: We give the statement of the divergence theorem in two and three dimensions but give the proof only in the two dimensional case. The intuition of the theorem is better described in the three-dimensional case, and this is where we begin. Recall . that in Cartesian coordinates, the divergence of a vector field in three dii þ F2 ðx; y; zÞ b j þ F3 ðx; y; zÞ kb is given by mensions F ðx; y; zÞ ¼ F1 ðx; y; zÞ b      . . v b v b v b i þ j þ k \$ F1 ðx; y; zÞ b i þ F2 ðx; y; zÞ b j þ F3 ðx; y; zÞ kb div F ¼ V\$ F ¼ vx vy vz ¼

vF1 vF2 vF3 þ þ . vx vy vz .

Suppose the flow of a fluid in a region is described by the vector field F . The amount of fluid that flows through a membrane in time t depends on the area of the surface, the orientation of the surface with the direction of the flow, and the velocity of the flow. See Fig. 2.3.4.

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CHAPTER 2 Vector Calculus

Surface S

θ

FIGURE 2.3.4

If the surface is perpendicular to the direction of flow, then the amount of flow through the surface in time t is velocity  area  time and if the normal to the surface makes an angle q with the direction of flow, then the amount of flow through the surface in time t is velocity  area  time  cos q. If nb is the unit vector normal to the surface, then the amount of fluid that crosses one unit of area of the surface in one unit of time is .

velocity  cos q ¼ F \$b n.

(1)

Now consider an infinitesimal element of volume DV. For specificity, suppose this is a parallelepiped DxDyDz and the fluid flows through six faces. See Fig. 2.3.5. Consider the two surfaces parallel to the yz plane that we have labeled S1 and S2, and suppose the direction of flow is as indicated by the arrows. The net outflow per unit time through surfaces S1 and S2 is the outflow per unit area per unit time through

S2 z

(x, y, z + Δz) (x, y + Δy, z + Δz) (x + Δx, y + Δy, z + Δz)

(x + Δx, y, z + Δz)

Δz (x, y, z) (x, y + Δy, z) y Δx

x

FIGURE 2.3.5

(x + Δx, y, z)

S1

Δy

(x + Δx, y + Δy, z)

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

surface S2 minus the inflow per unit area per unit time through surface S1 multiplied by the area of each surface, which is DyDz. In terms of a formula, this is ½F1 ðx þ Dx; y; zÞ  F1 ðx; y; zÞ DxDyDz Dx vF1 DV. z vx Similar relations hold for the other pairs of faces. Summing, we get that the net outflow through the volume DV is     . vF1 vF2 vF3 þ þ DV ¼ V\$ F dV. vx vy vz ½F1 ðx þ Dx; y; zÞ  F1 ðx; y; zÞDyDz ¼

The outflow through the total volume V is  ZZZ  . V\$ F dV. V

But from Eq. (1), we get that this outflow is also ZZ . F \$b n dS. S

The divergence theorem is the statement that these two quantities are the same. Theorem (divergence theorem in two dimensions): Suppose that D is a region in R2 , and C is a piecewise smooth simple closed curve that encloses the region D. Let nb be the unit normal vector that points outward from C (so that nb changes from point to point along C). Suppose that Fb is a continuously differentiable vector field on an open set that contains CWD. Then ZZ Z   b n ds ¼ F\$b V\$ Fb dA. C

D

Remark: Notice that the expression on the right is in a sense the integral of the derivative over a region and the expression on the left is the integral of the function over the boundary. Proof: Let s(t) ¼ (x(t), y(t)) be a parameterization of C that is positively oriented. Then ðy0 ðtÞ; x0 ðtÞÞ nb ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ½x0 ðtÞ2 þ ½y0 ðtÞ2 is a unit normal vector that points outward from C, as we show in Exercise 3. Let b yÞ ¼ FðxðtÞ; b Fðx; yðtÞÞ ¼ QðxðtÞ; yðtÞÞ b i  PðxðtÞ; yðtÞÞ b j. (We make this unusual looking choice of functions so that at the end of the computations it will be apparent how we are using Green’s theorem in the proof.)

127

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CHAPTER 2 Vector Calculus

Now

so

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ds ¼ ½x0 ðtÞ2 þ ½y0 ðtÞ2 dt

9 8 > =qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 2 b qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ½x0 ðtÞ þ ½y0 ðtÞ dt F\$b n ds ¼ > > 2 2 0 0 ; : ½x ðtÞ þ ½y ðtÞ ¼ ½QðxðtÞ; yðtÞÞy0 ðtÞ þ PðxðtÞ; yðtÞÞx0 ðtÞdt. Thus

Z C

b n ds ¼ F\$b

By Green’s theorem Z C

Pdx þ Qdy ¼

Z C

Qdy þ Pdx.

 ZZ  vQ vP  dx dy. vy D vx

b For our choice of F, V\$ Fb ¼ so

ZZ



D

 V\$ Fb dxdy ¼

vQ vP  vx vy

 ZZ  Z Z vQ vP b n ds.  dx dy ¼ Qdy þ Pdx ¼ F\$b vy D vx C C

Example: RR   R b n ds in the case We evaluate D V\$ Fb dxdy and C F\$b b yÞ ¼ x b Fðx; i þ yb j and C : x2 þ y2 ¼ 16:  RR  RR We have V\$ Fb ¼ 2, so D V\$ Fb dxdy ¼ 2 D 1dxdy ¼ 2ð16pÞ ¼ 32p; since the area of the Rcircle enclosed by C is 16p. b n ds we parameterize C using x ¼ 4cos q, y ¼ 4 sinq, 0  q  2p. To find C F\$b Then 4 cos q b i þ 4 sin qb j i þ sin q b j and ds ¼ 4dq. nb ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ cos q b 2 ð4 sin q Þ þ ð4 cos qÞ2 So

  b n ds ¼ 4 cos q b F\$b i þ 4 sin q b j \$ cos q b i þ sin q b j 4dq ¼ 16dq

and thus Z C

b n ds ¼ F\$b

Z

2p 0

16dq ¼ 32p.

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

Theorem (divergence theorem in three dimensions): Suppose that V is a closed bounded region in R3 with piecewise smooth boundary S, and Fb ðx; y; zÞ is a continuously differentiable vector field on an open set containing V. Then ZZ ZZZ   b n dS V\$ Fb dV ¼ F\$b V

S

where nb is the unit vector that points outward from the surface S. The next example highlights RRR the idea that it is often easier RR to compute the flux ðV\$FÞdV rather than through a closed surface using U vU F\$dS, which is an important application of the divergence theorem. Example: Let U be the cylinder whose base radius is a, which is centered at the origin of the b We first x,y plane,RRR and whose height is h. Let F be the vector field F ¼ xi þ y b j þ z k. compute U ðV\$FÞdV. We have V\$F ¼ so that ZZZ

vx vy vz þ þ ¼3 vx vy vz

ZZZ U

ðV\$FÞdV ¼

To compute

U

ð3ÞdV ¼ 3\$volume of the cylinder ¼ 3pa2 h. ZZ

ZZ vU

F\$dS ¼

vU

ðF\$nÞdS;

we must consider three surfaces. Let vU1 ¼ top of the cylinder; vU2 ¼ bottom of the cylinder; vU3 ¼ side of the cylinder. On vU1, z ¼ h, so F ¼ x b i þ yb j þ h kb and the outward pointing unit normal is b n ¼ k. Thus ZZ ZZ ðF\$nÞdS ¼ hdS ¼ h\$Area of the top ¼ h\$pa2 . vU1

vU1

j þ 0 kb and the outward pointing unit normal is On vU2, z ¼ 0, so F ¼ xi þ y b b Thus n ¼  k: ZZ ZZ ðF\$nÞdS ¼ 0dS ¼ 0: vU2

vU2

On vU3, the curved surface, x þ y ¼ a , the vector   V x2 þ y2 ¼ 2x b i þ 2y b j 2

2

2

129

130

CHAPTER 2 Vector Calculus

is normal to the surface, so 2x b i þ 2y b j 2x b i þ 2y b j pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ ¼ nb 2 2 2a 4x þ 4y is a unit vector normal to the surface. Then !! ZZ ZZ

 2x b b i þ 2y j xi þ y b j þ z kb \$ ðF\$nÞdS ¼ dS 2a vU3 vU3 ZZ

2x2 þ 2y2 dS ¼ 2a vU3

¼

ZZ

2a2 dS ¼ a\$area of vU3 ¼ a\$2pah vU3 2a

¼ 2pa2 h. Thus ZZ vU

ðF\$nÞdS ZZ ¼ ZZ þ

vU1

vU2

ðF\$nÞdS ZZ

ðF\$nÞdS þ

vU3

ðF\$nÞdS ¼ h\$pa2 þ 0 þ 2pa2 h

¼ 3pa2 h. Example: Evaluate

ZZ S

x3 dydz þ x2 ydxdz þ x2 zdxdy

over the surface of the cylinder bounded by x2 þ y2 ¼ 16, z ¼ 0, z ¼ 3. Solution: We use the divergence theorem with b j þ x2 z k. i þ x2 y b F ¼ x3 b Then V\$F ¼ 3x2 þ x2 þ x2 ¼ 5x2 so that

ZZ S

ZZZ x3 dydz þ x2 ydxdz þ x2 zdxdy ¼

V

5x2 dxdydz.

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

Converting to cylindrical coordinates, Z 2p Z ZZZ 2 5x dxdydz ¼ V

44 ¼ 5\$3\$ 4

Z

q¼0 2p

q¼0

4

Z

r¼0

3 z¼0

5r 2 cos2 qr dzdrdq

cos2 qdq ¼ 960p.

Example: Gauss’ law for inverseesquare fields . An inverseesquare field F is a vector field for which . . c F ¼ c b and F ðb rÞ ¼ r. 2 rk r k3 kb kb In R3 c

.

F ðx; y; zÞ ¼

ðx2 þ y2 þ z2 Þ

3 2

 xb i þ yb j þ z kb . .

We show in Exercise 30 that for such a vector field div F ðx; y; zÞ ¼ 0. Gauss’ law for inverseesquare fields says that if s. is a closed orientable surface that surrounds the origin, then the outward flux of F across s is ZZ . F \$b n dS ¼ 4pc. s

Let G be the region enclosed . by s. To prove the result, we cannot use the divergence theorem directly because F is not continuous at the origin. We circumvent this problem by constructing a sphere of radius ε about the origin where ε is small enough so that the sphere is contained in G. See Fig. 2.3.6. Let sε denote the surface of this sphere.

σ G

σε

FIGURE 2.3.6

131

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CHAPTER 2 Vector Calculus

Let Hε denote the solid that lies between s and sε. This is a solid with a hole, but the divergence theorem applies to such regions by a proof identical in spirit to the proof that Green’s theorem applies to a region with holes. Thus, we have ZZ ZZ ZZZ . . . div F dV ¼ F \$b n dS þ F \$b n dS Hε

.

s

but since div F ¼ 0, we have ZZ s

ZZ

.

F \$b n dS ¼ 

. sε

F \$b n dS.

Now ZZ sε

  ZZ b r r k2 kb b r \$  dS. dS ¼ c rk kb r k4 r k3 sε kb sε kb

ZZ

.

F \$b n dS ¼

c

On sε ; kb r k ¼ ε, so ZZ ZZ c c r k2 kb dS ¼  1dS ¼  2 \$4pε2 ¼ 4pc c 4 2 ε ε rk sε kb sε since the surface area of a sphere of radius ε is 4pε2. Thus ZZ . F \$b n dS ¼ 4pc. s

We return to the problem introduced in Section 1.2 of finding the Laplacian in a general orthogonal coordinate system. We denote the coordinates by (u1, u2, u3) and the scaling factors by h1, h2, h3. Consider an infinitesimal cube with sides parallel to the curvilinear coordinate axes as shown in Fig. 2.3.7. The lengths of the edges of the cube are h1du1, h2du2 and h3du3. The gradient of the function J in the u1 direction is h11 vJ vu1 . The divergence of a . vector field F ¼ F1 eb1 þ F2 eb2 þ F3 eb3 at a point p is the flux through the six faces of the infinitesimal cube centered at p divided by the volume of the cube. We consider the flux through the pair of parallel faces labeled Face 1 and Face 2. As in the proof of the divergence theorem, the net flow across the two faces is v ðh2 h3 F1 Þdu1 du2 du3 . vu1 h2 du2

Face 1 h3 du3

Face 2

FIGURE 2.3.7

h1 du1

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

The flow across the other pairs of faces has analogous expressions. According to the divergence theorem,   . v v ðh2 h3 F1 Þdu1 du2 du3 þ ðh1 h3 F2 Þdu1 du2 du3 V\$ F ðvolumeÞ ¼ vu1 vu2 þ so



and so

v ðh1 h2 F3 Þdu1 du2 du3 vu3

 V\$ F h1 du1 h2 du2 h3 du3 .

  . V\$ F ¼

¼

v v ðh2 h3 F1 Þdu1 du2 du3 þ ðh1 h3 F2 Þdu1 du2 du3 vu1 vu2

þ

v ðh1 h2 F3 Þdu1 du2 du3 vu3

  1 v v v ðh2 h3 F1 Þ þ ðh1 h3 F2 Þ þ ðh1 h2 F3 Þ . h1 h2 h3 vu1 vu2 vu3

(2)

.

To determine an expression for the Laplacian, we let F ¼ Vf in Eq. (2) to get   V\$Vf ¼ Df or V2 f        1 v vf v vf v vf ¼ h2 h 3 þ h 1 h3 þ h1 h 2 . h1 h2 h3 vu1 vu1 vu2 vu2 vu3 vu3 Stokes’ theorem: Stokes’ theorem is a generalization of Green’s theorem to R3 . In Stokes’ theorem we relate an integral over a surface to a line integral over the boundary of the surface. We assume that the surface is two sided (a Mobius strip is an example of a one-sided surface) that consists of a finite number of pieces, each of which has a normal vector at each point. This allows us to consider surfaces such as cubes that do not have normal vectors at the edges. We assume an orientation of the surface, which means we agree on the direction of the unit normal vector. This is necessary because there will be two unit normal vectors at each point p on the surface, nbðpÞ and b n ðpÞ. This induces an orientation for the boundary of the surface. Theorem (Stokes’ theorem): . Let S be an oriented surface, and let vS denote the oriented boundary of S. If F is a continuously differentiable vector field on S, then  ZZ  Z . . F \$ds. V  F dS ¼ S

vS

133

134

CHAPTER 2 Vector Calculus

S Sij ∂s

FIGURE 2.3.8

We give a sketch of the central idea in the proof of Stokes’ theorem, which is simply Green’s theorem. We begin by dividing the surface S into small squares, and focus on one square, that we call Sij. See Fig. 2.3.8. . b so that i þ Fy b j þ Fz kb be a vector field and b Let F ¼ Fx b s ¼ xb i þ yb j þ z k, . b b b d s ¼ dx i þ dy j þ dz k. We compute I . . F\$ds vSij

by which we mean the line integral around the line integral around the boundary of Sij. We choose the axes so that Sij is perpendicular to the z-axis. Then .

.

F \$ d s ¼ Fx dx þ Fy dy þ Fz dz:

Consider Sij as shown in Fig. 2.3.9. We have Z Z . . F\$ds ¼ C1

x1 þDx

x¼x1

Fx ðx; y1 Þdx

and Z C3

.

.

F\$ds ¼

FIGURE 2.3.9

Z

Z

x1 x¼x1 þDx

Fx ðx; y1 þ DyÞdx ¼ 

(x1, y1 + Δy)

C3

C4

Sij

(x1, y1)

C1

x1 þDx

x¼x1

Fx ðx; y1 þ DyÞdx

(x1 + Δx, y1 + Δy)

C2

(x1 + Δx, y1)

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

so Z C1

.

Z

.

F \$d s þ

.

.

Z

Z

x1 þDx

x1 þDx

F \$d s ¼ Fx ðx; y1 Þdx  Fx ðx; y1 þ DyÞdx C3 x¼x1 Z x1 þDx x¼x1 ¼ ½Fx ðx; y1 Þ  Fx ðx; y1 þ DyÞdx x¼x1 Z x1 þDx ¼ ½Fx ðx; y1 þ DyÞ  Fx ðx; y1 Þdx. x¼x1

Also Z

.

Z

.

F\$ds ¼

C2

y1 þDy

y¼y1

Fy ðx1 þ Dx; yÞdy

and Z C4

.

Z

.

F \$d s ¼

Z

y1

y¼y1 þDy

Fy ðx1 ; yÞdy ¼ 

y1 þDy y¼y1

Fy ðx1 ; yÞdy

so Z

. C2

Z

.

F \$d s þ

.

C4

Z

.

F \$d s ¼

y1 þDy

y¼y1

By Green’s theorem we have Z Z I . . . . F \$d s ¼ F \$d s þ vSij

C1

.

C3

½Fy ðx1 þ Dx; yÞ  Fy ðx1 ; yÞdy.

.

Z

F \$d s þ

.

C2

.

Z

F \$d s þ

 ZZ  vFy vFx ¼ Fx dx þ Fy dy ¼  dxdy. vy vSij Sij vx I

. C4

.

F \$d s

Recall that vFy vFx  ; ðV  F Þ\$ kb ¼ vx vy .

so that in this special case I vSij

.

.

F\$ds ¼

ZZ S

.

ðV  F Þ\$ kb dS.

In the general case, the square will not be in the xy plane but if we replace kb by the unit normal vector nb, we get I ZZ . . . F\$ds ¼ ðV  F Þ\$b n dS. vSij

S

135

136

CHAPTER 2 Vector Calculus

Sij

FIGURE 2.3.10

For a second way to get the same result, note that Z Z y1 þDy Z . . vFy . . DxDy F\$ds þ F \$d s ¼ ½Fy ðx1 þ Dx; yÞ  Fy ðx1 ; yÞdy z vx y¼y1 C2 C4 and Z

. C1

.

F \$d s þ

Z

. C3

.

Z

F\$ds ¼ 

x1 þDx

x¼x1

½Fx ðx; y1 þ DyÞ  Fx ðx; y1 Þdx z

vFx DxDy. vy

Now we put the squares together. Fig. 2.3.10 describes the idea. For adjoining squares, the contributions at the common boundary cancel one another because they are in opposite directions. Thus, when we sum up all squares the only pieces that do not cancel are those on the boundary of S. Thus we get  Z ZZ  . . F \$ds. V  F dS ¼ S

vS

Example: . i þ 4x b j þ ð2x þ 2zÞ kb where S is We verify Stokes’ theorem in the case F ¼ 3y b 2 2 the upper half of the hemisphere x þ y þ z2 ¼ 1. Now vS is x2 þ y2 ¼ 1, so we let sðtÞ ¼ ðcos t; sin t; 0Þ and then on vS .

s0 ðtÞ ¼ ð  sin t; cos t; 0Þ

F ðsðtÞÞ ¼ ð3 sin t; 4 cos t; 2 cos tÞ .

.

F ðsðtÞÞ\$s0 ðtÞ ¼ 3 sin2 t þ 4 cos2 t

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

We then have Z

Z

. vS

F \$ds ¼

2p 

  3 sin2 t þ 4 cos2 t dt ¼ p.

0

We next compute

ZZ  S

We have

   b  i  . v V F ¼  vx    3y

 V  F dS. .

     v v  b j þ k.  ¼ 2 b vy vz   4x 2x þ 2z  b j

kb

We parameterize S with spherical coordinates x ¼ sin q cos 4;

y ¼ sin q sin 4;

z ¼ cos q

so that Tq ¼ cos q cos 4 b i þ cos q sin 4 b j  sin q kb i þ sin q cos 4 b j T4 ¼ sin q sin 4b and Tq  T4 ¼ sin2 q cos 4 b i þ sin2 q sin 4 b j þ sin q cos q kb is normal to the surface. Then       . . V  F \$b n ¼ V  F \$ Tq  T4 ¼ 2 sin2 q sin 4 þ sin q cos q and ZZ  S

 Z V  F \$b n dS ¼ .

0 p=2 q¼0

B @

Z

2p  4¼0

1  C  2 sin2 q sin 4 þ sin q cos q d4Adq ¼ p.

Note that we could have computed the normal vector using T4  Tq instead of Tq  T4, in which case we would have  ZZ  . V  F \$b n dS ¼ p. S

To make the signs of the integrals agree in Stokes’ theorem, we must follow the “right-hand rule” that determines the outward pointing normal for a surface that is not closed according to the direction of the line integral of the boundary. If the signs are opposite, you should have taken the cross product in the reverse order.

137

138

CHAPTER 2 Vector Calculus

The following theorem gives results that will be important in the solution of partial differential equations. Theorem (Green’s identities): 1. (Green’s first identity) If f and g are scalar functions with continuous partial derivatives in a region R, and if V is a region within R with surface S, then ZZ ZZZ   2 vg f dS f V g þ Vf \$Vg dV ¼ V S vn vg ¼ Vg\$b n the directional derivative of g in the direction that is outward vn normal to the surface S. 2. (Green’s second identity) If f and g are scalar functions with continuous partial derivatives in a region R, and if V is a region within R with surface S, then  ZZ  ZZZ  2  vg vf g dS. f f V g  gV2 f dV ¼ vn vn V S

where

We leave the proof of the theorem to Exercises 12 and 13. An application of Stokes’ theorem: We can use Stokes’ theorem to show that Ampere’s law is equivalent to one of b and Maxwell’s equations. Consider a group of wires that are carrying current I, the wires are enclosed by a closed loop C, as shown in Fig. 2.3.11. b Ampere’s law says Suppose the current induces a magnetic field B. I b b B\$d l ¼ Ib. C

Let Jb denote the current density, which is the current crossing a unit area that is b n dS is the current across the surface element dS, where perpendicular to Jb. Then J\$b nb is a unit vector parallel to Ib. Then ZZ Jb\$b n dS S

C

I

FIGURE 2.3.11

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

is the current across any surface S that is bounded by C. So we have ZZ I b b b n dS. B\$d l¼ J\$b C

By Stokes’ theorem

I C

so

ZZ S

S

b b B\$d l¼

ZZ

 S

Jb\$b n dS ¼

 V  Bb \$b n dS

ZZ

 S

 V  Bb \$b n dS.

Since this holds for any surface S, we have V  Bb ¼ Jb, which is one of Maxwell’s equations. Note that we could have begun with V  Bb ¼ Jb and arrived at Ampere’s law by reversing the steps. An application of the divergence theorem: . Coulomb’s law says that the electric field E at the point r due to a point charge q located at the origin is . q E¼ ebr ; 4pεr 2 where ebr is the outward pointing unit normal vector in the direction from the origin to r. . The electric displacement D is defined by q ebr . 4pr 2 If nb is the outward pointing normal to the sphere centered at the origin at the point r, then .

.

D ¼ εE ¼

q . 4pr 2 If S is a sphere of radius jrj centered at the origin, and vS is the boundary of S, then .

n¼ D \$b

q dS ¼ q. (3) 4pr2 Gauss’ law says that Eq. (3) holds for any surface that has only the charge q in its interior. If there is more than one charge in S, then X . %vS D \$b n dS ¼ qi . .

n dS ¼ %vS %vS D \$b

Now suppose that at each point x interior to S we have a charge density r(x). Then in the infinitesimal volume dV centered at x the quantity of charge in dV is approximately r(x)dV

139

140

CHAPTER 2 Vector Calculus

Then

%vS.D\$bn dS ¼ But by the divergence theorem ZZZ rðxÞdV ¼ S

ZZZ S

rðxÞdV.

%vS D\$bn dS ¼ .

ZZZ

.

.

S

V\$D dV.

Since this is true for any surface, we have V\$D ¼ r, which is another of Maxwell’s equations. Note that we could have begun with this Maxwell’s equation and derived Gauss’ law.

CONSERVATIVE FIELDS

.

In this section we suppose that F is an R3 vector field. One way to create a vector field is by taking the gradient of a function. Such vector fields are called gradient fields or conservative fields. These vector fields are prominent in physics because if f represents a potential, such as gravitational or electrical potential, then Vf represents a force. The next theorem characterizes conservative fields. Definition: A vector field satisfying the conditions in the theorem below is said to be conservative. Theorem: . Let F be a differentiable R3 vector field. The following conditions are equivalent: 1. 2. 3. 4.

.

There.is a function f for which F ¼ Vf . V  F ¼ 0. R . For any oriented simple closed curve C, C F \$db s ¼ 0. If C1 and C2 are two simple oriented curves that have the same end points, then Z

.

C1

Z

F \$db s¼

.

C2

F \$db s.

Proof: ð1Þ0ð2Þ. In Exercise 15 we show that for any function f with continuous second partial derivatives, V  Vf ¼ 0. ð2Þ0ð3Þ. Suppose that C is an oriented simple closed curve, and S is a onesided surface whose boundary is C. Let s be a path that represents C. Then by Stokes’ theorem Z C

.

F \$db s¼

Z s

.

F \$db s¼

 Z  . V  F \$dS. S

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

C1

C2

−C2

FIGURE 2.3.12 .

But V  F ¼ 0, so

  R . . V  F \$dS ¼ C F \$db s ¼ 0. S

R

ð3Þ0ð4Þ. Suppose C1 and C2 are two simple oriented curves that have the same end points. Let C2 denote the curve with the same image, but opposite orientation as C2. See Fig. 2.3.12. Then C ¼ C1 WðC2 Þ is a closed curve. The curve C may not be simple, but if it is, then Z Z Z Z . . . . F \$db s¼ F \$db s¼ F \$db sþ F \$db s 0¼ C C1 C2 Z C1 WðC2 Þ Z . . ¼ F \$db s F \$db s¼0 C1

so

C2

Z C1

.

F \$db s¼

Z

.

C2

F \$db s.

If C ¼ C1 WðC2 Þ is not a simple closed curve, a more elaborate proof, that is beyond the scope of this.text, is required. b i þ F2 ðx; y; zÞ b j þ F3 ðx; y; zÞ k. For ð4Þ0ð1Þ. Let F ðx; y; zÞ ¼ F1 ðx; y; zÞ b 3 ðx0 ; y0 ; z0 Þ˛R , let s be a path from (0,0,0) to (x0, y0, z0) and define Z . f ðx0 ; y0 ; z0 Þ ¼ F \$ds. s

Since we assume (4) holds, the definition of f (x0, y0, z0) is independent of the path. We shall show that vf vf vf ¼ F1 ; ¼ F2 ; ¼ F3 . vx vy vz

vf ¼ F3. Let s be the path from (0, 0, 0) to (x0, y0, z0) given by vz s ¼ s1 þ s2 þ s3 where We first show

s1 is the path along the x-axis from (0, 0, 0) to (x0, 0, 0) s2 is the path parallel to the y-axis from (x0, 0, 0) to (x0, y0, 0) s3 is the path parallel to the z-axis from (x0, y0, 0) to (x0, y0, z0). See Fig. 2.3.13.

141

142

CHAPTER 2 Vector Calculus

z

(x0, y0, z0)

σ3 y

σ1 (x0, 0, 0)

(x0, y0, 0)

σ2

x

FIGURE 2.3.13

Then

Z

Z

. s

F \$ds ¼

s1

Z

.

F \$ds1 þ

.

s2

F \$ds2 þ

Z s3

.

F \$ds3 .

Now

  i þ F2 ðx; y; zÞ b j þ F3 ðx; y; zÞ kb \$ dx b i þ 0b j þ 0 kb F \$ds1 ¼ F1 ðx; y; zÞ b

.

¼ F1 ðx; y; zÞdx and. similarly, . F \$ds2 ¼ F2 ðx; y; zÞdy and F \$ds3 ¼ F3 ðx; y; zÞdz. Note that on s1, y ¼ 0 and z ¼ 0 so Z Z x . F \$ds1 ¼ F1 ðt; 0; 0Þdt; s1

on s2, x ¼ x0 and z ¼ 0, so Z s2

t¼0

Z

.

F \$ds2 ¼

and on s3, x ¼ x0 and y ¼ y0, so Z Z . F \$ds3 ¼ s3

Thus,

Z

f ðx0 ; y0 ; z0 Þ ¼ ¼

Zs

t¼0

z t¼0

F2 ðx0 ; t; 0Þdt;

F3 ðx0 ; y0 ; tÞdt.

.

F \$ds

Z . F \$ds2 þ F \$ds3 s2 s3 Z y0 Z F1 ðt; 0; 0Þdt þ F2 ðx0 ; t; 0Þdt þ

.

F \$ds1 þ

s1 x0

Z ¼

y

t¼0

Z

.

t¼0

z0 t¼0

F3 ðx0 ; y0 ; tÞdt

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

so f ðx0 ; y0 ; z0 þ DzÞ Z Z x0 F1 ðt; 0; 0Þdt þ ¼ t¼0

y0 t¼0

Z F2 ðx0 ; t; 0Þdt þ

z0 þDz t¼0

F3 ðx0 ; y0 ; tÞdt.

Then f ðx0 ; y0 ; z0 þ DzÞ  f ðx0 ; y0 ; z0 Þ Z z0 Z z0 þDz F3 ðx0 ; y0 ; tÞdt  F3 ðx0 ; y0 ; tÞdt ¼ t¼0 Z t¼0 z 0 þDz   ¼ F3 ðx0 ; y0 ; tÞdt ¼ F3 x0 ; y0 ; z Dz t¼z0

for some z˛½z0 ; z0 þ Dz . So   f ðx0 ; y0 ; z0 þ DzÞ  f ðx0 ; y0 ; z0 Þ ¼ F3 x0 ; y0 ; z Dz and vf f ðx; y; z þ DzÞ  f ðx; y; zÞ ¼ lim ¼ F3 ðx; y; zÞ. vz Dz/0 Dz vf To show that ¼ F2 ðx; y; zÞ, let s be the path from (0, 0, 0) to (x0, y0, z0) given by vy s ¼ s1 þ s2 þ s3 where s1 is the path along the z-axis from (0, 0, 0) to (0, 0, z0). s2 is the path parallel to the x-axis from (0, 0, z0) to (x0, 0, z0). s3 is the path parallel to the z-axis from (x0, 0, z0) to (x0, y0, z0). See Fig. 2.3.14. Similar to the analysis above, we get Z z0 Z F3 ð0; 0; tÞdt þ f ðx0 ; y0 ; z0 Þ ¼ t¼0

x0

t¼0

Z F1 ðt; 0; z0 Þdt þ

y0 t¼0

F3 ðx0 ; t; z0 Þdt

z (0, 0, z0)

τ2

τ3

(x0, 0, z0)

(x0, y0, z0)

τ1 y

x

FIGURE 2.3.14

143

144

CHAPTER 2 Vector Calculus

and Z f ðx0 ; y0 þ Dy; zÞ ¼

z0

t¼0

Z F3 ð0; 0; tÞdt þ

x0

t¼0

Z F1 ðt; 0; z0 Þdt þ

y0 þDy t¼0

F3 ðx0 ; t; z0 Þdt

so Z f ðx0 ; y0 þ Dy; z0 Þ  f ðx0 ; y0 ; z0 Þ ¼ From this, we can arrive at

vf ¼ F2 ðx; y; zÞ. vy

y0 þDy

t¼y0

F3 ðx0 ; t; z0 Þdt.

vf ¼ F1 ðx; y; zÞ. In Exercise 29, we show vx Example: Show that the vector field .

i þ x2 z b j þ x2 y kb F ðx; y; zÞ ¼ ð2xyz þ sin xÞ b .

is conservative, and find a function f (x, y, z) for which F . ¼ Vf . . We show that F is conservative by showing that V  F ¼ 0. We have     b b  i j kb          . v v v   V F ¼ vx vy vz         2 2  2xyz þ sin x x z x y 

v 2  v 2  b v 2  v x y  x z i x y  ð2xyz þ sin xÞ b j vy vz vx vz

v 2  v x z  ð2xyz þ sin xÞ kb þ vx vy   i  ð2xy  2xyÞ b j þ ð2xz  2xzÞ kb ¼ b 0. ¼ x2  x 2 b

¼

.

vf b vf b vf b We seek a function f for which Vf ¼ vx i þ vy j þ vz k ¼ F . If we have

vf b vf b vf b . b i þ F2 b j þ F3 k. i þ j þ k ¼ F ¼ F1 b vx vy vz Then we must have vf vf vf ¼ F1 ; ¼ F2 ; ¼ F3 . vx vy vz

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

In our example, this is vf vf vf ¼ 2xyz þ sin x; ¼ x2 z; ¼ x2 y. vx vy vz vf We begin our analysis with ¼ x2 z. From this, we get vy Z Z  2  vf dy ¼ x z dy. f ðx; y; zÞ ¼ vy Here, we must be careful in integrating with respect to y because the constant of v gðx; zÞ ¼ 0. Thus, integration could be a function of x or z because vy Z  2  f ðx; y; zÞ ¼ x z dy ¼ x2 yz þ gðx; zÞ. vf ¼ 2xyz þ sin x: We have vx vf v vf ¼ 2yz þ gðx; zÞ and ¼ F1 ¼ 2xyz þ sin x. vx vx vx v gðx; zÞ ¼ sin x, and so gðx; zÞ ¼ cos x þ hðzÞ. Thus, we write Thus vx

Next we use the condition

f ðx; y; zÞ ¼ x2 yz  cos x þ hðzÞ. vf Finally, we use the condition ¼ x2 y. We have vz vf d vf ¼ x2 y þ hðzÞ and ¼ F3 ¼ x2 y. vz dz vz Thus

d dz hðzÞ

¼ 0, so h is a constant. We conclude f ðx; y; zÞ ¼ x2 yz  cos x þ C.

Theorem: . . Suppose that F is a conservative vector field for which F ¼ Vf . Then Z ðx1 ; y1 ; z1 Þ Z ðx1 ; y1 ; z1 Þ . F \$db r¼ Vf \$db r ¼ f ðx1 ; y1 ; z1 Þ  f ðx0 ; y0 ; z0 Þ. ðx0 ; y0 ; z0 Þ

ðx0 ; y0 ; z0 Þ

Example: . The work done by the force field F ðx; y; zÞ ¼ ð2xyz þ sin xÞ b i þ x2 z b j þ x2 y kb in moving a particle from (10, 1, 4) to (10, 3, 4) is Z ð0; 3; 4Þ h i ð10; 3; 4Þ j þ x2 y kb \$ db r ¼ x2 yz  cos xð10; 1; 4Þ ð2xyz þ sin xÞ b i þ x2 z b ð0; 1; 4Þ

¼ 1200 þ 400

  j þ x2 y kb ¼ V x2 yz  cos x . since ð2xyz þ sin xÞ b i þ x2 z b

145

146

CHAPTER 2 Vector Calculus

y=d

ψ1(y)

ψ2(y)

y=c

FIGURE 2.3.15

EXERCISES 1. Find a unit normal vector to the following surfaces at the point (x0, y0): a. z ¼ 9  x2  y2. 2 þ z2ﬃ ¼ 1. b. x2 pyﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ c. z ¼ 1  x2 : 2. Derive an expression for a unit normal vector to the surface y ¼ g(x, z). 3. Show that for the region shown in Fig. 2.3.15 ZZ I vQðx; yÞ dxdy. Qðx; yÞdy ¼ vx C S 4. Show that in R3 if Sε ¼ {xjkxk < ε} and f : R3 /R is bounded, then ZZZ lim kxk2 f ðxÞdS ¼ 0: εY0

5. Show that ðy0 ðtÞ; x0 ðtÞÞ nb ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðx0 ðtÞÞ2 þ ðy0 ðtÞÞ2 is a unit vector normal to s0 (t) where s(t) ¼ (x(t), y(t)). 6. If i þ u2 b j þ u3 kb and ub ¼ u1 b

nb ¼ cos a b i þ cos b b j þ cos g kb

where a, b, and g are the angles that nb makes with the positive x, y, and z axes respectively, then ZZ ZZ ðu1 cos a þ u2 cos b þ u3 cos gÞdS ¼ ðu1 dydz þ u2 dxdz þ u3 dxdyÞ. S

S

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

Problems on Green’s theorem 7. Verify Green’s theorem for the following examples. Each boundary C is assumed to be positively oriented. H a. C x2 y dx þ xy3 dy where C is the rectangle whose vertices are (0, 0), (2, 0), (2, 2), and (0, 2). b. H i. H C y3 dx  x3 dy where C is the circle x2 þ y2 ¼ 1. 3 x3 dy where the region is between x2 þ y2 ¼ 1 and x2 þ y2 ¼ 4. Hii. 2C y dx  2 c. C y dx þ x dy where the region is between y ¼ x and y ¼ x2. 8. Evaluate using Green’s theorem: H a. HC 2xy dx þ ðx þ yÞdy where the region is between y ¼ 0 and y ¼ 1  x2. b. HC ex cos y dx  ex sin ydy where C is the circle x2 þ y2 ¼ 1. c. C ð2y þ 3xÞdx þ ðy  4xÞdy where C is the circle x2 þ y2 ¼ 9. 9. Suppose that f (x,y) has continuous second partial derivatives and v2 f v2 f þ ¼ 0: vx2 vy2 Show that

Z

vf vf dx  dy ¼ 0 vx vD vy

where D is any circle and vD is the boundary of D.

Problems on Stokes’ Theorem 10. Verify Stokes’ theorem in the following cases: pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ b S : z ¼ 9  x2  y2 ; z  0. a. Fb ¼ y b i þ 2x b j þ k; b S : upper half of the sphere x2 þ y2 þ z2 ¼ 16. b. Fb ¼ 2y b i þ 3x b j  z k; b b b c. F ¼ z i þ x j; S is the square 0  x  4; 0  y  4; z ¼ 2. d. Fb ¼ y2 b i þ x2 b j; S : x2 þ y2  25; z ¼ 0. b S : upper half of the sphere b b b e. F ¼ y i þ 2x j þ ðx þ zÞ k; 2 2 2 x þ y þ z ¼ 1. 11. (Green’s first identity) Show that if f and g are scalar functions with continuous partial derivatives in a region R, and if V is a region within R with surface S, then ZZ ZZZ  2  vg f V g þ Vf \$Vg dV ¼ f dS V S vn vg ¼ Vg\$n; the directional derivative of g in the direction that is outward vn normal to the surface S.

where

147

148

CHAPTER 2 Vector Calculus

Use this result to show ZZZ

ZZ V

jVf j2 dV ¼

vf dS. S vn

12. (Green’s second identity) Show that if f and g are scalar functions with continuous partial derivatives in a region R, and if V is a region within R with surface S, then  ZZ  ZZZ  2  vg vf f f V g  gV2 f dV ¼ g dS. vn vn V S 13. The flux of a vector field Fb is the flow rate of a quantity through a unit area. If S is a surface and nb is the outward directed unit normal to the surface, then the rate of flow through the surface S is given by ZZ ZZ b b n dSh b S. F\$b F\$d S

S

Suppose we have a material whose density is r that flows with velocity v kb through a hemisphere whose equation is pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ z ¼ R2  x2  y2 . a. Show that the unit normal to the surface is  1 b b nb ¼ x i þ y j þ z kb . R  b n ¼ r vz R. b. Show that F\$b c. Show that the flow rate through S is rvpR2. 14. Show that if S is any closed surface and Fb is a vector field, then ZZ   curl Fb \$b n dS ¼ 0: S

15. Show that for a differentiable function f (x, y, z), V  Vf ¼ 0. 16. Verify that Fb is a conservative force, and find a function f for which Fb ¼ Vf in the following cases;  2 2   2 b b iþ a. Fb ¼ 2xz3 þ 6y b  2ð6x  2yzÞ j þ 3x z  y k. b j þ 4y k. b. Fb ¼ ð2xy þ 3Þ b i þ x  4z b H b r in the following cases: 17. Use Stokes’ theorem to evaluate C F\$db b C : x2 þ y2 ¼ 4 in the z ¼ 1 plane. a. Fb ¼ y b i þ 2x b j þ 2z k; b C : x2 þ y2 ¼ 9 in the z ¼ 0 plane. i þ 3b j þ 2z k; b. Fb ¼ x3 y4 b b C is the boundary of the triangle b b c. F ¼ ð2x  yÞ i þ ð4x þ 3zÞ b j þ ð2y  zÞ k, whose vertices are (0, 0, 2), (0, 2, 0), (2, 0, 0).

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

18. Faraday’s law states that if a current I passes through a loop C encloses that encloses a region S, then the electromotive force around the loop is the negative of the rate of change of the magnetic flux through the surface. As an equation, this is stated ZZ I b Ib ¼  v b n dS E\$d B\$b vt S C where nb is the normal vector to S. Use Stokes’ theorem to show that Faraday’s law implies v Bb V  Eb ¼  vt which is one of the Maxwell’s equations.

 19. The magnetic field Bb due to a current loop is given by Bb ¼ curl Ab where y x b b Ab ¼  3 i þ 3 j. 2 2 2 2 2 ðx þ y þ z Þ ðx þ y2 þ z2 Þ2 For C a circle of radius a centered at (0, 0, a) that is parallel to the xy-plane, use Stokes’ theorem to calculate the flux of Bb through C for a large. 20. A solenoid is a wound spiral of wire (see Fig. 2.3.16). A magnetic field Bb is created when current flows through the wire. We orient the solenoid so that the axis lies along the z-axis. If the solenoid is infinitely long, then Bb is zero outside the solenoid, and Bb ¼ b kb inside the solenoid, where b is a constant that depends on the current strength and the spacing of the turns of the coil. Suppose that the radius

of the coil is a. b Show that B ¼ curl Ab where 8 1 2 y b x b > > if r > a < 2a b  2 i þ 2 j r r Ab ¼

 > > : 1 b  yb i þ xb j if r < a 2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ where r ¼ x2 þ y2 . 21. Show that the flux of Bb through a circle in the xy-plane of radius r < a is pr2b.

a

FIGURE 2.3.16

149

150

CHAPTER 2 Vector Calculus

Problems on Divergence Theorem 22. Verify the divergence theorem in the following cases: i þ 2xy b j þ z2 kb and S is the surface of the cylinder bounded by a. Fb ¼ y2 b 2 2 x þ y ¼ 9 and the planes z ¼ 2 and z ¼ 3. b. Fb ¼ x b i þ yb j þ z kb and S is the surface of the cube bounded by the planes x ¼ 0 and x ¼ 1, y ¼ 0 and y ¼ 1, and z ¼ 0, and z ¼ 1. 23. Evaluate the following integrals using the divergence theorem: R a. ðx þ zÞdydz þ ðy þ zÞdxdz þ ðx þ yÞdxdy where S is the surface of problem R 10(a). b. x2 dydz þ y2 dxdz þ z2 dxdy where S is the surface of problem 10(b). 24. A function f (x, y) is harmonic on S if v2 f v2 f þ ¼0 vx2 vy2 on S. Show that if f (x, y) is harmonic on S, then ZZ vf dS ¼ 0: S vn 25. By letting Fb ¼ Vf in .

div F ðpÞ ¼

lim

V/0; p˛V

show that V2 f ¼

lim

V/0; p˛V

1 V

1 V

ZZ

ZZ

. S

F \$b n dS

vf dS. vn S

Problems on Conservative Field 26. Show that the gravitational field GMm ebr r2 is conservative. To convert to Cartesian coordinates .

F ðx; y; zÞ ¼ 

ebr ¼

! x y z pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ; pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ; pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ . x2 þ y2 þ z2 x2 þ y2 þ z2 x2 þ y2 þ z2

    . i þ x3 þ x 2 y b j þ 2 kb is 27. Show that the vector field F ðx; y; zÞ ¼ xy2 þ 3x2 y b .

conservative, and find a function f (x, y, z) for which F ¼ Vf .

2.3 Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem

.

j þ 2x kb is conserva28. Show that the vector field F ðx; y; zÞ ¼ ð6xy . þ 2zÞ b i þ 3x2 b tive, and find a function f (x, y, z) for which F ¼ Vf . 29. Show that in the proof of the theorem of the equivalence of conditions for a vf ¼ F1 ðx; y; zÞ. vector field, vx 30. Show that the divergence of the vector field

 . c xb i þ yb j þ z kb F ðx; y; zÞ ¼ 3 ðx2 þ y2 þ z2 Þ2 is 0.

151

CHAPTER

Green’s Functions

3

3.1 INTRODUCTION In this chapter we discuss how to solve a nonhomogeneous linear second-order ordinary differential equation with given boundary conditions by presenting the solution in terms of an integral. More precisely, we find a function G(x,t), for which the solution to L½y ¼ a0 ðxÞy00 ðxÞ þ a1 ðxÞy0 ðxÞ þ a0 ðxÞyðxÞ ¼ f ðxÞ yð0Þ ¼ 0; yð1Þ ¼ 0

0 0 be given. Consider Z tþε Z tþε 00 0 ½a0 ðxÞy ðxÞ þ a1 ðxÞy ðxÞ þ a2 ðxÞyðxÞdx ¼ dðx  tÞdx ¼ 1: tε

tε

Since the functions ai(x) are continuous, ai(x) z ai(t) if x ε ½t  ε; t þ ε and ε is small. In that case, Z tþε ½a0 ðxÞy00 ðxÞ þ a1 ðxÞy0 ðxÞ þ a2 ðxÞyðxÞdx tε

Z z a0 ðtÞ

tþε

tε

00

y ðxÞdx þ a1 ðtÞ

Z

tþε

tε

0

y ðxÞdx þ a2 ðtÞ

Z

tþε tε

yðxÞdx ¼ 1:

155

156

CHAPTER 3 Green’s Functions

Now Z

tþε tε

y0 ðxÞdx ¼ yðt þ εÞ  yðt  εÞ z 0

since y0 (x) and y(x) are continuous and ε is small. Likewise, Z tþε yðxÞdx z 0. tε

Also Z a0 ðtÞ

tþε tε

y00 ðxÞdx ¼ a0 ðtÞ½y0 ðt þ εÞ  y0 ðt  εÞ.

Taking the limit as ε/0 in Z Z tþε ½a0 ðxÞy00 ðxÞ þ a1 ðxÞy0 ðxÞ þ a2 ðxÞyðxÞdx ¼ tε

tþε

tε

dðx  tÞdx ¼ 1

we get

  dGðx; tÞ dGðx; tÞ  lim ¼ 1: lim a0 ðtÞ½y ðt þ εÞ  y ðt  εÞ ¼ a0 ðtÞ lim ε/0 tYx t[x dx dx 0

0

This gives the jump condition on G(x,t); namely lim xYt

dGðx; tÞ dGðx; tÞ 1 .  lim ¼ x[t dx dx a0 ðtÞ

A final condition is that G(x,t) be symmetric; i.e., G(x,t) ¼ G(t,x). Recapping, there are five conditions that determine G(x,t) for the differential equation given by Eq. (1): 1. Gð0; tÞ ¼ 0. 2. Gð1; tÞ ¼ 0. 3. lim Gðx; tÞ ¼ lim Gðx; tÞ. x[t

4.

xYt dGðx;tÞ lim dx  lim dGðx;tÞ dx xYt x[t

¼ a01ðtÞ.

5. Gðx; tÞ ¼ Gðt; xÞ. There is an intuitive way to see how the solution to L[y] ¼ d(x  t) yields the solution Z 1 f ðxÞGðx; tÞdx yðtÞ ¼  0

using the example y00 ðxÞ ¼ f ðxÞ 0 < x < 1;

yð0Þ ¼ yð1Þ ¼ 0:

(2)

3.2 Construction of Green’s Function Using the Dirac-Delta Function

The explanation we give follows that in Stakgold, Green’s Functions and Boundary-Value Problems. A physical interpretation of Eq. (2) is we have a wire stretched between the points x ¼ 0 and x ¼ 1. At a point x between x ¼ 0 and x ¼ 1 a vertical pressure f (x) is applied. (We assume that f (x) is a continuous function.) The function y(x) is the deflection of the string at the point x. Choose a positive integer N and divide the interval [0,1] into N equal parts. The length of each part is then 1/N. Let xk denote the center of the kth subinterval and let Dx denote the length of each interval. We assume the deflection of the string due to the continuous pressure f (x) is approximately equal to the deflection of the string due to the sum of forces of magnitude f (xk)Dx each concentrated at the point xk. The defection at x ¼ t due to the single force f (xk)Dx concentrated at xk is Gðt; xk Þf ðxk ÞDx and, by the principle of superposition, the deflection due to the sum of the concentrated forces is N X Gðt; xk Þf ðxk ÞDx: (3) k¼1

The limit of expression (3) as N / ∞ (equivalently, as Dx / 0) is Z 1 Gðt; xÞf ðxÞdx. 0

Note: For our examples we use the boundary conditions y(0) ¼ 0, y(1) ¼ 0. But analogous ideas work if we use boundary conditions such as y0 (0) ¼ 0, y0 (1) ¼ 0 as we do in some of the exercises. Next, we give two examples that demonstrate the computations involved in finding the Green’s function. Example: We find the Green’s function for y00 ðxÞ ¼ f ðxÞ;

yð0Þ ¼ 0;

yð1Þ ¼ 0

using the Dirac-delta function. Two linearly independent solutions to the associated homogeneous equation y00 ðxÞ ¼ 0 are y1 ðxÞ ¼ 1 and y2 ðxÞ ¼ x. Thus

 Gðx; tÞ ¼

c1 ðtÞ1 þ c2 ðtÞx

0x : t  tx

0x 0 be given. We must show there is a number N(ε) such that if n > N(ε), then jsn ðxÞ  f ðxÞj < 3 for every x˛½ep; p. Since f (x) is continuous on [ep,p], it is bounded and uniformly continuous on [ep,p]. Thus there is a number M such that j f (x)j  M if x ˛½ep; p and a d > 0 0 such that if jx  x00 j < d, then j f (x0 )  f (x00 )j < ε for x0 , x00 ˛½ep; p. Now Z 1 p f ðx  tÞDk ðtÞdt Sk ðxÞ ¼ p p so S0 ðxÞ þ S1 ðxÞ þ / þ Sn ðxÞ nþ1 

Z p 1 D0 ðtÞ þ D1 ðtÞ þ / þ Dn ðtÞ dt ¼ f ðx  tÞ p p nþ1 Z 1 p ¼ f ðx  tÞKn ðtÞdt. p p

sn ðxÞ ¼

Since 1 p

Z

p p

Kn ðtÞdt ¼ 1;

then f ðxÞ ¼

1 p

Z

p p

f ðxÞKn ðtÞdt.

Thus

   Z p  Z 1  1 p  ½ f ðx  tÞ  f ðxÞKn ðtÞdt  jsn ðxÞ  f ðxÞj ¼  j f ðx  tÞ  f ðxÞjKn ðtÞdt p p  p p 2 Z d Z 1 4 d ¼ j f ðx  tÞ  f ðxÞjKn ðtÞdt þ j f ðx  tÞ  f ðxÞjKn ðtÞdt p p d Z þ

d

Z þ

p

d

d

3

1 2 0 Z p Z d 1 j f ðx  tÞ  f ðxÞjKn ðtÞdt5  42M @ Kn ðtÞdt þ Kn ðtÞdtA p p d 3 j f ðx  tÞ  f ðxÞjKn ðtÞ dt5

for 0 < d < p.

4.3 Methods of Convergence of Fourier Series

Now choose d so that if jtj < d, then j f ðx  tÞ  f ðxÞj < ε=2: This will make Z Z d ε d ε Kn ðtÞdt < for every n. j f ðx  tÞ  f ðxÞjKn ðtÞdt < 2 2 d d Since KN ðxÞ 

1 ðN þ 1Þð1  cos dÞ

for 0 < d  jxj  p;

after d has been chosen, there is a number N(ε) such that if n > N(ε), then Kn ðxÞ < εp 8M if d  jxj  p. Thus if n > N(ε), then jsn ðxÞ  f ðxÞj 2 0 1 3 Z Z p Z d 1 4 @ d  2M Kn ðtÞdt þ Kn ðtÞdtA þ j f ðx  tÞ  f ðxÞjKn ðtÞ dt5 p p d d  3 i 1 3 p 3 p 3  1 h  3p 3p þ þ þ 2M þ ¼ ¼3 p 8M 8M p 4 4 2 2 Corollary: If f (x) is piecewise continuous, then
0, there is a continuously differentiable function fε(x) for which Z h f ðxÞ  fε ðxÞ; f ðxÞ  fε ðxÞi ¼

p

ε j f ðxÞ  fε ðxÞj2 dx < . 4 p

(See Rudin’s “Real and Complex Analysis,” for instance.) Since fε(x) is continuously differentiable, there is a number N(ε) so that if n > N(ε) then   ε Sn ðxÞ  fε ðxÞ; Sεn ðxÞ  fε ðxÞ ¼

Z

  ε S ðxÞ  fε ðxÞ2 dx < ε n 4 p p

where Sεn ðxÞ is the nth partial sum of the Fourier expansion of fε(x). Now, h f ðxÞ  SN ðxÞ; f ðxÞ  SN ðxÞi  Z p Z p  2 ε   f ðxÞ  Sn ðxÞ dx ¼ ¼ p

p



 f ðxÞ  Sεn ðxÞ; f ðxÞ  Sεn ðxÞ

   f ðxÞ  fε ðxÞ þ fε ðxÞ  Sε ðxÞ2 dx. n

We now bound   

  f ðxÞ  fε ðxÞ þ fε ðxÞ  Sε ðxÞ2 ¼ ð f ðxÞ  fε ðxÞÞ þ fε ðxÞ  Sε ðxÞ 2 . n n Note that ða þ bÞ2 ¼ a2 þ 2ab þ b2 and 2ab  a2 þ b2 Thus ða þ bÞ2  2a2 þ 2b2



 since ða  bÞ2  0 .

4.3 Methods of Convergence of Fourier Series

and so

Z

p p

   f ðxÞ  fε ðxÞ þ fε ðxÞ  Sε ðxÞ2 dx n

0

Z

 2@

p p

Z ð f ðxÞ  fε ðxÞÞ2 dx þ

p

p

2

1

fε ðxÞ  Sεn ðxÞ dxA

ε ¼ ε if n > NðεÞ. 4 4 Fourier Series on Arbitrary Intervals: If instead of having our functions being periodic of period 2p they are periodic of period 2L, the formulas for Fourier coefficients for f (x) would change to Z npx 1 L f ðxÞcos dx n ¼ 0; 1; 2; . an ¼ L L L Z npx 1 L f ðxÞsin bn ¼ dx n ¼ 1; 2; . L L L

1 > > sin N þ x > > 2 > > x ; if xs2pm; m an integer > < 2 sin : ¼ 2 > > > > > > 1 > > :N þ if x ¼ 2pm 2

DN ðxÞ ¼

2. Show that A cos t þ B sin t ¼

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ A2 þ B2 sinðt þ 4Þ

where 4 ¼ tan1 ðA=BÞ. This yields another way to express a Fourier series.

205

206

CHAPTER 4 Fourier Series

3. Let KN ðtÞ ¼

R

D0 ðtÞ þ D1 ðtÞ þ / þ DN ðtÞ . Nþ1

Show that a.

p 1 p p KN ðtÞdt

¼ 1:

1 b. KN ðxÞ ¼ 2ðNþ1Þ

c. KN ðxÞ 

sin2 ½ðNþ1Þx=2   sin2

1 ðNþ1Þð1cos dÞ

x 2

 0:

for 0 < d  jxj  p.

d. lim KN ðxÞ ¼ Nþ1 2 . x/0

4. Show that

Z

p

p

Z f ðuÞDN ðx  uÞdu ¼

p p

f ðx  tÞDN ðtÞdt.

5. Suppose that f (x) is piecewise continuous and periodic of period 2L. Mimic the case for a function that is periodic of period 2p to show Z npx 1 L f ðxÞcos an ¼ dx n ¼ 0; 1; 2; . L L L Z npx 1 L f ðxÞsin dx n ¼ 1; 2; . bn ¼ L L L and the formula the Fourier series for f (x) would change to ∞ npx npx a0 X an cos þ bn sin . f ðxÞw þ L L 2 n¼1

6. a. Find the Fourier coefficients for f ðxÞ ¼ x. ∞ P b. Use the answer to part (a) and Parseval’s theorem to show n¼1 7. Find the Fourier series for x 0xp f ðxÞ ¼ : 0 p  x < 0

1 n2

¼ p6 . 2

4.4 The Exponential Form of Fourier Series

8. Find the Fourier series for f ðxÞ ¼

x 0xp : x p  x < 0

Use this to show ∞ X

1

n¼1

ð2n  1Þ2

¼

p2 . 8

Note the result is same as problem 4, Section 3.1, but the technique is different. 9. Find the Fourier series for f ðxÞ ¼

cos x

0xp

0

p  x < 0

:

10. Use a CAS to plot the functions in problems 6e9 and the first n terms of the Fourier series for n ¼ 3, 10, and 25. 11. Find the Fourier series for f (x) ¼ x2, p < x < p and use it along with Parseval’s theorem to show ∞ X 1 p4 . ¼ n4 90 n¼1

4.4 THE EXPONENTIAL FORM OF FOURIER SERIES In later sections, we shall sometimes find it convenient to express the Fourier series of a function in exponential form; that is, f ðxÞ ¼

∞ X

cn einx .

n¼∞

This representation is possible because of Euler’s formula einx ¼ cosðnxÞ þ i sinðnxÞ: We will determine a formula for the coefficients cn and a relationship between cn and the an and bn we defined previously. We have 0 1 ! Z p Z p Z p X ∞ ∞ X f ðxÞeikx dx ¼ cn einx eikx dx ¼ cn @ eiðnþkÞx dxA: p

p

n¼∞

n¼∞

p

207

208

CHAPTER 4 Fourier Series

Now Z

p p

Thus Z

p

eiðnþkÞx dx ¼

Z f ðxÞe dx ¼

0

if n þ ks0; i.e.; unless n ¼ k

2p

if n þ k ¼ 0; i.e.; if n ¼ k

∞ X

p

ikx

p

and so ck ¼

1 2p

p

Z

p p

! cn e

Z e dx ¼

inx

p

ikx

p

n¼∞

f ðxÞeikx dx or ck ¼

1 2p

Z

p

p

.

ck dx ¼ 2pck

f ðxÞeikx dx.

Example: Find the exponential form of the Fourier series for 0 0xp f ðxÞ ¼ : x p  x < 0 We have for k > 0 ck ¼

1 2p

Z

p p

f ðxÞeikx dx ¼

1 2p

Z

p p

xeikx dx.

Using integration by parts or a CAS  Z 0 cosðpkÞ sinðpkÞ 1 p cosðpkÞ sinðpkÞ þ  . xeikx dx ¼   p þ i k2 k k2 k k2 p For k an integer, this is

8 > > > >
k k k ip 2 > > > þ 2 : k k Thus,

8 > > > >
1 p 2 > > > : 2p  i k þ 2 k ck

1 ¼ 2p

Z

p

if k is odd .

if k > 0 is even if k > 0 is odd

1 f ðxÞe dx ¼ 2p p ikx

if k is even

Z

0 p

xeikx dx.

4.4 The Exponential Form of Fourier Series

Now Z 0

 cosðpkÞ sinðpkÞ 1 p cosðpkÞ sinðpkÞ xe dx ¼  p þ 2þi  þ . k2 k k k k2 p ikx

For k an integer, this is

 cosðpkÞ 1 p cosðpkÞ þ 2þi   k2 k k 8  > i > p >  > < k cosðpkÞ 1 p cosðpkÞ þ 2þi ¼   > k2 k k ip 2 > > > þ : k k2

Thus,

8 > > > >
1 p 2 > > > : 2p  i k þ 2 k

if k is even if k is odd :

if k is even if k is odd.

We now relate cn and the an and bn we defined previously. We have an cosðnxÞ ¼

an inx bn inx e þ einx ; bn sinðnxÞ ¼ e  einx . 2 2i

So ∞ X

an cosðnxÞ þ bn sinðnxÞ ¼ a0 þ ða1 cosðxÞ þ b1 sinðxÞÞ

n¼0

ða2 cosð2xÞ þ b2 sinð2xÞÞ þ / ¼ a0 þ

a2 i2x a1 ix e þ eix þ e þ ei2x þ / 2 2

b1 ix b2 i2x e  eix þ e  ei2x 2i 2i   a1 b1 a2 b2 ¼ a0 þ eix þ þ e2ix þ þ/ 2 2i 2 2i   b1 b2 ix a1 2ix a2 þe  þe  þ/ 2 2i 2 2i þ

∞ X n¼∞

cn einx ¼ c0 þ c1 ei1x þ c2 ei2x þ / þ c1 eix þ c2 e2ix þ /.

209

210

CHAPTER 4 Fourier Series

If ∞ X

an cosðnxÞ þ bn sinðnxÞ ¼

∞ X

cn einx

n¼∞

n¼0

then we must have     a1 b1 a1 b2 a 1 b1 a1 b2 þ ; c2 ¼ þ ; c1 ¼  ; c2 ¼  : c 0 ¼ a0 ; c 1 ¼ 2 2i 2 2i 2 2i 2 2i So cj ¼

  aj b j aj b j þ ; cj ¼  . 2 2i 2 2i

From these equations, we also have   aj bj aj bj þ þ  ¼ aj cj þ cj ¼ 2 2i 2 2i and cj  cj

  aj bj aj bj 1 þ   ¼ bj ¼ i 2 2i 2 2i

so bj ¼ iðcj  cj Þ.

Note: if trigonometric series is as above, then c0 ¼ a0, but we often give the first term as a0/2 in which case c0 ¼ a0/2.

EXERCISES 1. Find the exponential form of the Fourier series for the Dirac delta function, given that the Fourier series for d(x) is ∞ 1 1X cos nx. þ 2p p n¼1

2. Find the exponential form of the Fourier series for 1 0xp f ðxÞ ¼ . 1 p  x < 0 3. Find the exponential form of the Fourier series for x 0xp f ðxÞ ¼ : 0 p  x < 0

4.4 The Exponential Form of Fourier Series

4. Find the exponential form of the Fourier series for x 0xp f ðxÞ ¼ : x p  x < 0 5. Find the exponential form of the Fourier series for cos x 0  x  p f ðxÞ ¼ : 0 p  x < 0 6. Find the exponential form of the Fourier series for f (x) ¼ jsin xj. 7. Find ak and bk in terms of ck and ck. 8. This exercise develops the Poisson integral formula in polar coordinates. We suppose that u(r,q) is a solution to Du ¼ 0; uð1; qÞ ¼ f ðqÞ, which is Laplace’s equation on a circle of radius 1. Suppose that 2 Z p Z p ∞ X 1 n4 f ð4Þd4 þ r f ð4Þcos nq cos n4 d4 uðr; qÞ ¼ 2p p p n¼1 Z þ

p

p

a. Show that uðr; qÞ ¼

3 f ð4Þsin nq sin n4 d45.

1 p

Rp

"

p f ð4Þ

1þ 2

∞ P

 r n cos½nðq  4Þ d4.

n¼1

 ∞

P b. Show that r 2 þ 1  2r cos q r n cos nq ¼ r cos q  r 2 .

c. Show that r 2 þ 1  2r cos q

"

n¼1 1þ 2

∞ P n¼1

#

rn

cos nq ¼ 12 1  r 2 .

d. Replace q by (q  4) and conclude Z 1  r2 p f ð4Þ d4. uðr; qÞ ¼ 2  2r cosðq  4Þ 2p 1 þ r p

211

212

CHAPTER 4 Fourier Series

In the case the radius of the circle is R, the formula changes to Z R2  r 2 p f ð4Þ d4. uðr; qÞ ¼ 2 þ r 2  2rR cosðq  4Þ R 2p p Either of these formulas is the Poisson integral formula.

4.5 FOURIER SINE AND COSINE SERIES Suppose that f (x) is a function defined on (0,p]. We can extend f (x) to be an even function on [ep,p], which we denote by f1(x), by defining 8 if 0 < x  p > < f ðxÞ f1 ðxÞ ¼ a0 if x ¼ 0 . > : f ðxÞ if  p  x < 0 We can then extend f1(x) to be periodic of period 2p on (∞,∞) by defining f1(x þ 2p) ¼ f1(x). Likewise, we can extend f (x) to be an odd function on [ep,p], which we denote by f2(x), by defining 8 if 0 < x  p > < f ðxÞ if x ¼ 0 . f2 ðxÞ ¼ 0 > : f ðxÞ if  p  x < 0 We can then extend f2(x) to be periodic of period 2p on (∞,∞) by defining f2(x þ 2p) ¼ f2(x). If either f1(x) or f2(x) can be expressed as a Fourier series, then both can be expressed as a Fourier series. Suppose that this is the case. Then f1(x) being an even function can be written in its Fourier series f1 ðxÞ ¼

∞ a0 X þ an cos nx 2 n¼1

and f2(x) being an odd function can be written in its Fourier series f2 ðxÞ ¼

∞ X

bn sin nx.

n¼1

The point of this argument is that if f (x) is a suitably well-behaved function on (0,p], then it can be expressed as either a Fourier cosine series or a Fourier sine series. We now determine a formula for the Fourier coefficients. Consider

4.5 Fourier Sine and Cosine Series

f2 ðxÞ ¼

∞ X

bn sin nx.

n¼1

Then 1 bn ¼ p

Z

p p

f2 ðxÞsin nx dx.

Since both f2(x) and sin nx are odd functions, their product is an even function, so Z Z 1 p 2 p bn ¼ f2 ðxÞsin nx dx ¼ f2 ðxÞsin nx dx. p p p 0 ∞ P bn sin nx where Thus we can expand f (x) on (0,p] into a sine series n¼1 Z p 2 bn ¼ f ðxÞsin nx dx. p 0 If instead of the interval being (0,p] it is (0,L], the formula for bn will change to Z npx 2 L f ðxÞsin bn ¼ dx. L 0 L Similarly, we can expand f (x) on (0,p] into a cosine series ∞ a0 X þ an cos nx 2 n¼1

where an ¼

2 p

Z

p

f ðxÞcos nx dx

0

if the interval is (0,p] and 2 an ¼ L

Z

L 0

f ðxÞcos

npx dx L

if the interval is (0,L]. Example: Let f (x) ¼ sin x on (0,p]. We find the cosine series for f (x). We have Z Z 2 p 2 p f ðxÞcos nx dx ¼ sin x cos nx dx. an ¼ p 0 p 0 Now sinðx þ nxÞ ¼ sin x cos nx  sin nx cos x sinðx  nxÞ ¼ sin x cos nx þ sin nx cos x

213

214

CHAPTER 4 Fourier Series

so 1 sin x cos nx ¼ ½sinðx þ nxÞ þ sinðx  nxÞ 2 and

Z

2 an ¼ p

p 0

1 sin x cos nxdx ¼ p

Z

p

½sinðx þ nxÞ þ sinðx  nxÞdx.

0

If n ¼ 1 then 1 a1 ¼ p Otherwise,

Z

p 0

1 ½sinðx þ xÞ þ sinðx  xÞdx ¼ p

Z

p

½sinð2xÞdx ¼ 0:

0

3 2 Z p Z 14 p sinð1 þ nÞxdx þ sinð1  nÞxdx5. an ¼ p 0 0

Now, if n s 1 Z

p

sinð1 þ nÞxdx ¼ 

0

and

Z

p

1 1 cosð1 þ nÞxjpx¼0 ¼  ½cosð1 þ nÞp  1 nþ1 nþ1

sinð1  nÞxdx ¼ 

0

1 1 cosð1  nÞxjpx¼0 ¼ ½cosð1  nÞp  1. 1n n1

Now

cosð1 þ nÞp  1 ¼ cosð1  nÞp  1 ¼

so

Z

¼

Thus

8 > >
> :0

2 0

if n is even . if n is odd

3 2 Z p Z 14 p 2 1 þ ð1Þn an ¼ sinð1 þ nÞxdx þ sinð1  nÞxdx5 ¼ , : p p 1  n2 0 0

4.6 Double Fourier Series

Hence, the cosine series for f (x) ¼ sin x on (0,p] is ∞ ∞ ∞ a0 X 2 2X 1 þ ð1Þn 2 4X cosð2 nxÞ þ . an cos nx ¼ þ cos nx ¼  2 n¼1 p p n¼2 1  n2 p p n¼1 4n2  1

EXERCISES 1. Let f (x) ¼ cos x on (0,p]. Find the sine series for f (x). 2. Find the sine and cosine series for the following functions on (0,p]. a. f (x) ¼ 1. b. f (x) ¼ x. c. f (x) ¼ x2.

4.6 DOUBLE FOURIER SERIES In later applications we shall want to express a function of two variables f (x,y), 0  x  p, 0  y  p as a double series of sine functions. That is, we want to find constants Bmn so that ! ∞ ∞ X X Bmn sin my sin nx. f ðx; yÞ ¼ n¼1

m¼1

We show how the numbers Bmn are determined. Fix y ˛ ½0; p. Then f ðx; yÞ is a function of x. If f (x,y) is continuous in both x and y, then f ðx; yÞ ¼

∞ X

bn ðyÞsin nx

(1)

n¼1

where the notation bn ðyÞ is used to emphasize that the constants depend on the fixed value of y that we have chosen. We know that bn ðyÞ is computed according to Z 2 p bn ðyÞ ¼ f ðx; yÞsin nx. p 0 Now f ðx; yÞ ¼

∞ ∞ X X n¼1

! Bmn sin my sin nx.

m¼1

From Eqs. (1) and (2) we get bn ðyÞ ¼

∞ X m¼1

! Bmn sin my

(2)

215

216

CHAPTER 4 Fourier Series

so 2 p

Z

p

f ðx; yÞsin nx ¼

0

∞ X

! Bmn sin my

m¼1

for each y˛½0; p. Now fix n and define a function of y, denoted Fn(y), by Z 2 p f ðx; yÞsin nxdx Fn ðyÞ ¼ p 0 so that ∞ X

Fn ðyÞ ¼

(3)

! Bmn sin my .

(4)

m¼1

From Eq. (4) we get Bmn

Z

2 ¼ p 2 ¼ p ¼

p 0

Z

4 p2

p

Fn ðyÞsin my dy

2

Z

0

42 p

Z

pZ p 0

p

3 f ðx; yÞsin nx dx5sin my dy

0

f ðx; yÞsin nx sin my dxdy.

0

EXERCISES 1. Find the coefficients for the double Fourier sine series for the following functions for 0  x, y  p a. f (x,y) ¼ x þ y. b. f (x,y) ¼ xy. c. f (x,y) ¼ x2y2.

CHAPTER

Three Important Equations

5

5.1 INTRODUCTION Partial differential equations (PDEs) are extremely important in both mathematics and physics. A major purpose of this text is to give an introduction to some of the simplest and most important PDEs in both disciplines, and techniques for their solution. Accordingly, we focus on three equations: 1. the heat equation vuðt; xbÞ ¼ kDuðt; xbÞ vt 2. the wave equation v2 uðt; xbÞ ¼ a2 Duðt; xbÞ vt2 3. Laplace’s equation Duðx; yÞ ¼ 0: It would appear that we are severely limiting ourselves by examining only three equations. However, these encompass any PDE of the form v2 uðt; xÞ v2 uðt; xÞ v2 uðt; xÞ þC þB ¼0 (1) 2 vt vxvt vx2 because by a linear transformation of variables, any equation of the form of Eq. (1) can be transformed into one of the three equations. We show at the end of the section that the transformation is exactly the same as transforming an equation of the form A

Ax2 þ Bxy þ Cy2 ¼ 0

(2)

into the standard form of a parabola, hyperbola, or an ellipse by a rotation of axes according to the sign of B2  4AC. Following the nomenclature of the geometrical Mathematical Physics with Partial Differential Equations. https://doi.org/10.1016/B978-0-12-814759-7.00005-3 Copyright © 2018 Elsevier Inc. All rights reserved.

217

218

CHAPTER 5 Three Important Equations

figures, if B2  4AC < 0 the PDE is said to be parabolic, if B2  4AC ¼ 0 the equation is elliptic, and if B2  4AC > 0 the equation is hyperbolic. Thus the heat equation is the prototypical parabolic equation, the wave equation is the prototypical wave equation, and Laplace’s equation is the prototypical elliptical equation. We shall see that for the heat equation, the initial conditions diffuse in time whereas in the wave equation initial conditions are propagated, changing position but not shape. The Laplacian appears in each of these equations, and we begin by exploring the physical significance of that operator. Consider the heat equation. Here, u(t,x,y,z) is the temperature of a solid homogenous body at the point (x,y,z) at time t. Then vuðt; x; y; zÞ is the rate of change of the temperature at the point (x,y,z). The tempervt ature at (x,y,z) will undergo a change if and only if the temperature in the immediate vicinity at (x,y,z) is different than at (x,y,z). We consider the case in one space dimension. Suppose we want to compare the value of a function f (x) with the average value of the function at x þ h and x  h; i.e., we compare f (x) with ½ f ðx þ hÞ þ f ðx  hÞ=2. How does f 00 (x) enter in? By the definition of the second derivative f 00 ðxÞ ¼ lim

h/0

f 0 ðx þ hÞ  f 0 ðxÞ f 0 ðxÞ  f 0 ðx  hÞ ¼ lim : h/0 h h

(3)

We use the second limit. Approximate f 0 (x) and f 0 (xh) using f 0 ðxÞ ¼ lim

h/0

f ðx þ hÞ  f ðxÞ h

and f 0 ðx  hÞ ¼ lim

h/0

f ðxÞ  f ðx  hÞ . h

Substitute the approximations into Eq. (3) to get f ðx þ hÞ  f ðxÞ f ðxÞ  f ðx  hÞ  f ðx þ hÞ  2f ðxÞ þ f ðx  hÞ h h ¼ f ðxÞ z h h2   2 1 ¼ 2 ½ f ðx þ hÞ þ f ðx  hÞ  f ðxÞ . h 2 00

Thus f 00 (x) is a measure of the difference between the value of a function at a point and the average of the values of the function in the immediate vicinity.

5.2 LAPLACE’S EQUATION The simplest second-order PDE is Laplace’s equation, which in two variables is Duðx; yÞ ¼ uxx þ uyy ¼ 0:

5.2 Laplace’s Equation

Solutions to Laplace’s equation are called harmonic functions, which play a key role in complex analysis. We review some facts from complex analysis that will be important for us. An analytic function f (x,y) is of the form f ðx; yÞ ¼ uðx; yÞ þ ivðx; yÞ where u(x,y) and v(x,y) are harmonic functions. It is not true that any combination of two harmonic functions is an analytic function. For f (x,y) to be analytic, v(x,y) must be a harmonic conjugate of (x,y). This means u(x,y) and v(x,y) must satisfy the CauchyeRiemann equations, which are vu vv vu vv ¼ and ¼ . vx vy vy vx A domain in the complex plane is a connected open set. A simply connected domain is a domain with no holes. If D is a simply connected domain and u(x,y) is a harmonic function, then the function v(z) is defined by Z z ux dy  uy dx vðzÞ ¼ z0

for any path in D that connects z0 and z is a harmonic conjugate of u(x,y) and f ðx; yÞ ¼ uðx; yÞ þ ivðx; yÞ is analytic in D. From complex analysis, we have the following result:

MAXIMUM MODULUS PRINCIPLE If f (z) is analytic and not constant on a domain D, then j f ðzÞj does not attain its maximum on D. That is, the maximum of j f ðzÞj on the closure of D is attained on the boundary of D. Corollary: If f (z) is analytic and not constant on a domain D and if f (z) s 0 on D, then j f ðzÞj does not attain its minimum on D. Proof: 1 . Then g(z) is analytic and jgðzÞj does not attain its maximum on D. Let gðzÞ ¼ f ðzÞ Corollary: If f ðx; yÞ ¼ uðx; yÞ þ ivðx; yÞ is analytic and not constant on a domain D, then u(x,y) does not attain its maximum on D. Proof: We have

        f ðzÞ   uðx;yÞþivðx;yÞ   uðx;yÞ  ivðx;yÞ  e  ¼ e  ¼ e e  ¼ euðx;yÞ

219

220

CHAPTER 5 Three Important Equations

since

   iq  e  ¼ 1 and euðx;yÞ  ¼ euðx;yÞ

and if f (z) is analytic, then e f(z) is analytic.

EXERCISES 1. Determine whether the following functions are harmonic: a. f ðx; yÞ ¼ x2 y  y2 x. b. f ðx; yÞ ¼ x3 y  y3 x. c. f ðx; yÞ ¼ ex sin y. d. f ðx; yÞ ¼ sin x  cos y.   e. f ðx; yÞ ¼ ln x2 þ y2  f. f ðx; yÞ ¼ arctan yx . 2. Show that u(r,q) ¼ ln r, r > 0,p0ﬃﬃ < q < 2p is harmonic. 3. Determine whether uðr; qÞ ¼ r eiq=2 ; r > 0; 0 < q < 2p is harmonic.

5.3 DERIVATION OF THE HEAT EQUATION IN ONE DIMENSION Before presenting the heat equation, we review the concept of heat. Energy transfer that takes place because of temperature difference is called heat flow. The energy transferred in this way is called heat. Thus heat refers to the transfer of energy, not the amount of energy contained within a system. An example of a unit of heat is the calorie. One calorie is the amount of heat required to raise 1 g of water from 14.5 C to 15.5 C. The quantity of heat H required to raise a body of mass m from T1 to T2 is approximately proportional to DT ¼ T1  T2, and is proportional to m. It is also dependent on the material of the body. This quality is called the specific heat of the materiald typically denoted c. Thus we have H ¼ mcDT. The amount of heat required for an infinitesimal change in temperature dT is denoted dH. To summarize, heat is energy in transit, and dH does not represent the change in the amount of heat in the body inasmuch as “the amount of heat in a body” has no meaning. When a quantity of heat dH is transferred in time dt, then the rate of energy transfer is dH dt . We now derive the heat equation in one dimension. Suppose that we have a rod of length L. While the derivation will be for the case that the rod is one-dimensional, it

5.3 Derivation of the Heat Equation in One Dimension

is advantageous to visualize the rod as having a cross-sectional area of one square unit. Let u(x,t) ¼ the temperature of the rod at the point x (0  x  L) at time t(t  0). qbðx; tÞ ¼ the heat flow at point x at time t (a vector quantity) r ¼ the density of the material (assumed to be constant) c ¼ the specific heat of the material DQ ¼ change in internal energy Du ¼ change in temperature. For nb, a unit vector, qb\$b n , is the heat flux in the direction of nb. The two laws that we use in our derivation of the heat equation are conservation of energy and Fourier’s law. Fourier’s law is qbðx; tÞ ¼ kVuðx; tÞ which, in one dimension, is vuðx; tÞ b i. (1) vx The number k is a constant of the material called the thermal conductivity. The reason for the negative sign is that heat flows from higher to lower temperatures and vuðx; tÞ is positive if u(x,t) is increasing as x increases. vx Consider the small segment ½x; x þ Dx. (See Fig. 5.2.1.) The amount of heat energy passing through the segment at point x in time Dt is approximately qbðx; tÞ\$ b iðDtÞ, and the amount of heat energy passing through the segment at point x þ Dx in time Dt is approximately qbðx þ Dx; tÞ\$ b iðDtÞ. Depending on the direction of heat flow and in the absence of a heat source or a heat sink, one of these quantities will represent an addition to the internal energy of the segment, and the other will represent a removal of internal energy from the segment. For definiteness, suppose energy is being added at x and removed at x þ Dx. Then change in internal energy is h i (2) DQ ¼ qbðx; tÞ\$ b i  qbðx þ Dx; tÞ\$ b i Dt. qbðx; tÞ ¼ k

When the amount of energy DQ is added to a body of mass m and specific heat c, the temperature of the body rises according to DQ ¼ mcDu.

(3)

qˆ (x, t)

qˆ (x + Δx, t) x

FIGURE 5.2.1

x + Δx

221

222

CHAPTER 5 Three Important Equations

Thus from Eqs. (2) and (3) we get h i qbðx; tÞ\$ b i  qbðx þ Dx; tÞ\$ b i Dt ¼ mcDu or Du . qbðx; tÞ\$ b i  qbðx þ Dx; tÞ\$ b i ¼ mc Dt Taking the limit as Dt / 0 gives qbðx; tÞ\$ b i  qbðx þ Dx; tÞ\$ b i ¼ mc

vu . vt

So, from Eqs. (1) and (2) we get

vuðx; tÞ vuðx þ Dx; tÞ vu vu b b qbðx; tÞ\$ i  qbðx þ Dx; tÞ\$ i ¼ k  ¼ mc ¼ rðDxÞc vx vx vt vt or

k

vuðx; tÞ vuðx þ Dx; tÞ  vu vx vx ¼ rc . Dx vt

Taking the limit as Dx / 0 gives k

v2 uðx; tÞ vu ¼ rc . vx2 vt

This is often written

where a2 ¼

k . rc

v2 uðx; tÞ 1 vu ¼ 2 ; 2 vx a vt

EXERCISES 1. Show that if there is a heat source f (x), then the heat equation becomes 1 vu v2 uðx; tÞ þ f ðxÞ. ¼ a2 vt vx2

5.4 Derivation of the Wave Equation in One Dimension

5.4 DERIVATION OF THE WAVE EQUATION IN ONE DIMENSION Suppose we have a string stretched along the x-axis between x ¼ 0 and x ¼ L with tension T. The string is fixed at the endpoints. We distort the string in the vertical direction by plucking it, which will cause the string to vibrate in the vertical direction. We assume there is no damping and no external forces, and we want to find the equation that governs the dynamics of motion. Consider Fig. 5.3.1 below that depicts the forces on a small element of string in the interval ½x; x þ Dx. We let u(x,t) ¼ the vertical displacement of the string from the x-axis at point x at time t q(x,t) ¼ the angle the string makes with a horizontal line at point x at time t T(x,t) ¼ the tension in the string at point x at time t r(x) ¼ the mass density of the string at point x. We assume that the string is perfectly flexible, so that the forces that parts of the string exert on one another are tangential to the string. This means that T(x,t) is tangent to the string at (x,t) so vTðx; tÞ . vx We assume that there is no net force in the horizontal direction so that the string vibrates vertically. This means tan½qðx; tÞ ¼ the slope of the tangent line ¼

Tðx; tÞ cos ½qðx; tÞ ¼ Tðx þ Dx; tÞ cos ½qðx þ Dx; tÞ; i.e., the horizontal force to the left is equal to the horizontal force to the right. T(x + Δx, t)

θ (x + Δx, t)

Δu

θ (x, t) Δx T(x, t)

u(x, t)

x

FIGURE 5.3.1

223

224

CHAPTER 5 Three Important Equations

Analysis of forces in the u (vertical) direction: The net force in the vertical direction is not zero, and we apply Newton’s law, Force ¼ mass  acceleration. The mass of the string between x and x þ Dx is approximately qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rðxÞ ðDxÞ2 þ ðDuÞ2 . v2 uðx; tÞ The acceleration is the second derivative with respect to time, which is vt2 at the point (x,t). The force in the u-direction is the sum of T(x,t) sin [q(x,t)] and Tðx þ Dx; tÞ sin ½qðx þ Dx; tÞ, which, because of the direction of T(x,t) and Tðx þ Dx; tÞ, is Tðx þ Dx; tÞ sin ½qðx þ Dx; tÞ  Tðx; tÞ sin ½qðx; tÞ. Thus the governing equation in the u-direction is qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 v uðx; tÞ rðxÞ ðDxÞ2 þ ðDuÞ2 vt2 ¼ Tðx þ Dx; tÞ sin ½qðx þ Dx; tÞ  Tðx; tÞ sin ½qðx; tÞ.

(1)

This is the equation that connects the tension and the vertical displacement from equilibrium. Divide Eq. (1) by Dx to get sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  2 2 Du v uðx; tÞ Tðx þ Dx; tÞ sin ðx þ Dx; tÞ  Tðx; tÞ sin ½qðx; tÞ . ¼ rðxÞ 1 þ Dx vt2 Dx (2) We now work with the right-hand side of Eq. (2). Recall that Tðx; tÞ cos ½qðx; tÞ ¼ Tðx þ Dx; tÞ cos ½qðx þ Dx; tÞ so the right-hand side of Eq. (1) is equal to 0

1 Tðx þ Dx; tÞ sin ½qðx þ Dx; tÞ Tðx; tÞ sin ½qðx; tÞ BTðx þ Dx; tÞ cos ½qðx þ Dx; tÞ  Tðx; tÞ cos ½qðx; tÞC B C B CTðx; tÞ cos ½qðx; tÞ @ A Dx

¼

tan ½qðx þ Dx; tÞ  tan ½qðx; tÞ Tðx; tÞ cos ½qðx; tÞ. Dx

Now tan ½qðx; tÞ ¼

vTðx; tÞ vx

(3)

5.4 Derivation of the Wave Equation in One Dimension

so the right-hand side of Eq. (3) can be written 1 0 vTðx þ Dx; tÞ vTðx; tÞ  B vx vx C CTðx; tÞ cos ½qðx; tÞ. B A @ Dx Thus Eq. (2) can be written sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  2 2 Du v uðx; tÞ rðxÞ 1 þ Dx vt2 0 1 vTðx þ Dx; tÞ vTðx; tÞ  B vx vx C CTðx; tÞ cos ½qðx; tÞ. ¼B @ A Dx Taking the limit as Dx / 0 in Eq. (4) gives sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  2ﬃ 2 vu v uðx; tÞ v2 Tðx; tÞ rðxÞ 1 þ ¼ Tðx; tÞ cos ½qðx; tÞ. vt2 vx2 vx

(4)

(5)

We want to express the right-hand side of Eq. (5) in terms of u(x,t). To do this, we refer back to Fig. 5.3.1 and note that vTðx; tÞ Du vu v2 Tðx; tÞ v2 u ¼ 2. ¼ tan ½qðx; tÞ ¼ lim ¼ and Dx/0 Dx vx vx vx2 vx So we have sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  2ﬃ 2 vu v uðx; tÞ v2 u ¼ 2 Tðx; tÞ cos ½qðx; tÞ. rðxÞ 1 þ vx vt2 vx

(6)

Eq. (6) is not solvable, so all derivations of the equation known as the wave equation make simplifying assumptions. (Actually, we have ignored some factors already, including elasticity.) For a more complete derivation that includes these factors, see Weinberger, A First Course in Partial Differential Equations. We assume that the mass density of the string is constant and replace r(x) by r and assume that q is small so that Tðx; tÞ cos ½qðx; tÞ z Tðx; tÞ and tan ½qðx; tÞ z 0 and thus

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  2ﬃ vu z 1: 1þ vx

225

226

CHAPTER 5 Three Important Equations

If we assume that T(x,t) is constant, T(x,t) ¼ T, with these assumptions and approximations, Eq. (6) becomes r

v2 uðx; tÞ v2 uðx; tÞ ¼ T vt2 vx2

or 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c vt2 vx2

(7) T where c2 ¼ . r Eq. (7) is the wave equation with no external forces. If there were external forces present whose sum was F(x,t), then we would modify Eq. (7) to qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 v uðx; tÞ rðxÞ ðDxÞ2 þ ðDuÞ2 vt2 ¼ Tðx þ Dx; tÞ sin ½qðx þ Dx; tÞ  Tðx; tÞ sin ½qðx; tÞ þ Fðx; tÞ and continue the analysis as we did. The result would be the equation r

v2 uðx; tÞ v2 uðx; tÞ ¼T þ Fðx; tÞ. 2 vt vx2

EXERCISES 1. Complete the derivation of the wave equation in the case there is an external force F(x,t). 2. Show that a solution to v2 uðx; hÞ ¼0 vxvh is uðx; hÞ ¼ f ðxÞ þ gðhÞ. (We use this in the next section.) 3. Show that a solution to v2 uðx; tÞ 1 v2 uðx; tÞ ¼ 2 2 vx v vx2 is u(x,t) ¼ f (x  vt) þ g(x þ vt) where f and g have continuous second derivatives with respect to x and t.

5.5 AN EXPLICIT SOLUTION OF THE WAVE EQUATION The wave equation in one dimension v2 uðx; tÞ v2 uðx; tÞ  c2 ¼0 2 vt vx2

5.5 An Explicit Solution of the Wave Equation

can be solved explicitly by making a change of variables that will convert the equation into the form v2 uðx; hÞ ¼ 0: vxvh We demonstrate that the change of variables x ¼ x þ ct h ¼ x  ct accomplishes the desired transformation. To do this, we use the chain rule several times. We have vu vu vx vu vh vu vu ¼ \$ þ \$ ¼ \$1 þ \$1: vx vx vx vh vx vx vh We also have     v2 u v vu v vu vu ¼ ¼ þ . vx2 vx vx vx vx vh Now       v vu v vu vx v vu vh v2 u v2 u v2 u v2 u ¼ \$ þ \$ ¼ 2 \$1 þ \$1 ¼ 2 þ vx vx vx vx vx vh vx vx vx vxvh vxvh vx and       v vu v vu vx v vu vh v2 u v2 u ¼ \$ þ \$ ¼ þ 2. vx vh vx vh vx vh vh vx vxvh vh Thus v2 u v2 u v2 u v2 u þ ¼ þ 2 . vx2 vx2 vxvh vh2 We also have

  vu vu vx vu vh vu vu vu vu ¼ \$ þ \$ ¼ \$c þ \$ðcÞ ¼ c  vt vx vt vh vt vx vh vx vh

and   v2 u v vu vu ¼c  . vt2 vt vx vh Now       v vu v vu vx v vu vh v2 u v2 u ¼ \$ þ \$ ¼ 2 \$c þ \$ðcÞ vt vx vx vx vt vh vx vt vxvh vx

227

228

CHAPTER 5 Three Important Equations

and       v vu v vu vx v vu vh v2 u v2 u ¼ \$ þ \$ ¼ \$c þ 2 \$ðcÞ. vt vh vx vh vt vh vh vt vxvh vh Thus

2 v2 u v u v2 u v2 u v2 u ¼ c \$c þ \$ðcÞ \$ðcÞ  \$c þ vxvh vxvh vt2 vx2 vh2  2  v u v2 u v2 u þ ¼ c2  2 vxvh vh2 vx2 and we have 0¼

2 v2 u 2v u  c vt2 vx2

 2  v u v2 u v2 u v2 u v2 u v2 u þ  ¼c 2  2 vxvh vh2 vx2 vxvh vh2 vx2 v2 u . ¼ 4c2 vxvh 2

We now solve v2 u ¼ 0: vxvh

Integrating

v2 u with respect to x gives vxvh Z 2 Z vu v u ¼ dx ¼ 0 dx ¼ JðhÞ vh vxvh

because d ðJðhÞÞ ¼ 0: dx

Integrating

vu with respect to h gives vh Z uðx; hÞ ¼ JðhÞdh ¼ pðhÞ þ qðxÞ

where pðhÞ is an antiderivative of JðhÞ that involves only h: Thus uðx; tÞ ¼ pðx þ ctÞ þ qðx  ctÞ for any twice differentiable functions p and q. This is d’Alembert’s formula. If we have appropriate initial conditions, then we can specify p and q as we now show.

5.5 An Explicit Solution of the Wave Equation

Suppose that uðx; 0Þ ¼ f ðxÞ and ut ðx; 0Þ ¼ gðxÞ. So uðx; 0Þ ¼ pðx þ 0Þ þ qðx  0Þ ¼ pðxÞ þ qðxÞ ¼ f ðxÞ

(1)

and ut ðx; tÞ ¼ cp0 ðx þ ctÞ  cq0 ðx  ctÞ so ut ðx; 0Þ ¼ cp0 ðx þ 0Þ  cq0 ðx  0Þ ¼ cp0 ðxÞ  cq0 ðxÞ ¼ gðxÞ.

(2)

If we differentiate Eq. (1) and multiply by c, we get cp0 ðxÞ þ cq0 ðxÞ ¼ cf 0 ðxÞ.

(3)

Adding Eqs. (2) and (3) gives 2cp0 ðxÞ ¼ gðxÞ þ cf 0 ðxÞ so that p0 ðxÞ ¼ Integrating, we get 1 pðxÞ ¼ 2c

Z

1 1 gðxÞ þ f 0 ðxÞ. 2c 2 x

0

1 gðsÞds þ K þ f ðxÞ. 2

Then

2 qðxÞ ¼ f ðxÞ  pðxÞ ¼ f ðxÞ  4 1 1 ¼ f ðxÞ  2 2c

Z

x

1 2c

3

Z

x 0

1 gðsÞds þ K þ f ðxÞ5 2

gðsÞds  K.

0

So pðx þ ctÞ ¼

1 2c

Z

xþct 0

1 gðsÞds þ f ðx þ ctÞ þ K 2

and 1 1 qðx  ctÞ ¼ f ðx  ctÞ  2 2c 1 1 ¼ f ðx  ctÞ þ 2 2c

Z

xct

gðsÞds  K

0

Z

0 xct

gðsÞds  K.

229

230

CHAPTER 5 Three Important Equations

Finally, we have 1 1 uðx; tÞ ¼ pðx þ ctÞ þ qðx  ctÞ ¼ ½ f ðx þ ctÞ þ f ðx  ctÞ þ 2 2c

Z

xþct

gðsÞds. xct

(4) Eq. (4) is also called d’Alembert’s formula. We shall give the solution for the heat equation later in the text, but we give a cartoon of graphs below showing how the solutions of the wave equation and the heat equation evolve in time for the same initial condition of a Gaussian distribution. In Fig. 5.4.1, we show the evolution of the wave equation, and in Fig. 5.4.2 we show the evolution of the heat equation. Notice that for the wave equation, the distribution splits into two equal parts, each as the same shape but one half the size as the original distribution. The wave equation is the prototypical example of a hyperbolic PDE, and the solutions of such equations behave in this manner. For the heat equation, the distribution of the initial condition diffuses in time. (The heat equation is also called the diffusion equation.) The heat equation is the prototypical example of a parabolic PDE, and the solutions of such equations behave in this manner. Graphs of the solution to the wave equation as it evolves from its initial condition through t¼1, 2 and 3. The graphs of the solution to the heat equation for t ¼ 0.2, 0.5 and 1.

y

y

x

x +t = 1

+t = 0

y

y

x +t = 2

FIGURE 5.4.1

+t = 3

x

5.5 An Explicit Solution of the Wave Equation

y

5 4

+t

= .2

3 2 1 x

–4

–3

–2

–1

–1

1

2

3

1

2

3

1

2

3

–2 y

5 4 3

+t

= .5

2 1 x

–4

–3

–2

–1

–1 –2 y

5 4 3

+t

=1

2 1 x

–4

–3

–2

–1

–1 –2

FIGURE 5.4.2

231

232

CHAPTER 5 Three Important Equations

EXERCISES 1. 2. 3. 4.

Solve the wave equation on [0,p] if u(x,0) ¼ 0 and ut(x,0) ¼ sin x. Solve the wave equation on [0,p] if u(x,0) ¼ sin x and ut(x,0) ¼ 0. Solve the wave equation on [0,1] if u(x,0) ¼ 1x2 and ut(x,0) ¼ 0. Solve the wave equation on [0,1] if 8 1 > > < x; 0 < x < 2 ; ut ðx; 0Þ ¼ sin px. uðx; 0Þ ¼ > > : 1  x; 1  x < 1 2

5. Solve the wave equation on [0,1] if u(x,0) ¼ 0 and ut(x,0) ¼ 1. 6. Consider the PDE utt ðx; tÞ þ 2aut ðx; tÞ þ a2 uðx; tÞ ¼ a2 uxx ðx; tÞ t > 0; ∞ < x < ∞; a > 0

(5)

which models voltage in a power line. a. Show that if y(x,t) satisfies ytt(x,t) ¼ a2yxx(x,t), then u(x,t) ¼ eaty(x,t) satisfies Eq. (5). b. Find the solution for Eq. (5) with initial conditions u(x,0) ¼ 0 and ut(x,0) ¼ sin x. c. Find the solution for Eq. (5) with initial conditions u(x,0) ¼ sin x and ut(x,0) ¼ 0. 7. Solve the wave equation on ∞ < x < ∞ if 1 ; 1 . ut ðx; 0Þ ¼ 1þx a. uðx; 0Þ ¼ 1þx 2 2 b. uðx; 0Þ ¼ ex ; 2

ut ðx; 0Þ ¼ xex . 2

5.6 CONVERTING SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS TO STANDARD FORM Next we demonstrate how to convert a second-order PDE of the form A

v2 uðx; yÞ v2 uðx; yÞ v2 uðx; yÞ þC þB ¼0 2 vx vxvy vx2

into one of the three equations we studied earlier by a linear change of variables. The process is computationally similar to rotating axes in the plane to convert an equation of the form Ax2 þ Bxy þ Cy2 þ Dx þ Ey þ F ¼ 0 into a parabola, ellipse, or a hyperbola in standard form. We now review that process. Start with a vector vb. We find the coordinates of vb with respect to two sets of axes. The first set of axes, (x,y), is in standard position, and the second set of axes, (x0 ,y0 ), is

5.6 Converting Second-Order Partial Differential Equations

(x, y)

y

P

y′

(x′, y ′) x′ Q′

α θ O

Q

x

FIGURE 5.5.1

obtained by rotating the first set of axes counterclockwise through an angle q. (See Fig. 5.5.1.) Suppose in the (x0 ,y0 ) system, the vector vb makes an angle of a with the x0 axis. Then in the (x,y) system, the vector vb makes an angle of a þ q with the x axis. Then x0 ¼ jb v j cos a; x ¼ jb v j cosða þ qÞ;

y0 ¼ jb v j sin a y ¼ jb v j sinða þ qÞ.

So x ¼ jb v j cos ða þ qÞ ¼ jb v j½cos a cos q  sin a sin q ¼ x0 cos q  y0 sin q y ¼ jb v j sin ða þ qÞ ¼ jb v j½sin a cos q þ sin q cos a ¼ x0 sin q þ y0 cos q. We want to find A0 ,., F0 so that Ax2 þ Bxy þ Cy2 þ Dx þ Ey þ F ¼ A0 x02 þ B0 x0 y0 þ C0 y02 þ D0 x0 þ E0 y0 þ F 0 . Now Ax2 þ Bxy þ Cy2 þ Dx þ Ey þ F ¼ Aðx0 cos q  y0 sin qÞ

2

þ Bðx0 cos q  y0 sin qÞðx0 sin q þ y0 cos qÞ þ / þ F. If one expands the right-hand side of Eq. (1) and sets the result equal to A0 x02 þ B0 x0 y0 þ C 0 y02 þ D0 x0 þ E0 y0 þ F 0

(1)

233

234

CHAPTER 5 Three Important Equations

then one finds that A0 ¼ A cos2 q þ B sin q cos q þ C sin2 q   B0 ¼ B cos2 q  sin2 q þ 2ðC  AÞ sin q cos q C 0 ¼ A sin2 q  B sin q cos q þ C cos2 q D0 ¼ D cos q þ E sin q E0 ¼ D sin q þ E cos q F 0 ¼ F. We want to choose q so there is no x0 y0 term; i.e., we want B0 ¼ 0. We use the identities cos2 q  sin2 q ¼ cosð2qÞ and 2 sin q cos q ¼ sinð2qÞ to get B0 ¼ B cosð2qÞ þ ðC  AÞ sinð2qÞ so that B0 ¼ 0 if B cos(2q) ¼ (AC) sin(2q) or sinð2qÞ B ¼ tanð2qÞ ¼ cosð2qÞ AC so q¼

  1 B tan1 . 2 AC

Example: We convert the equation x2 þ 3xy þ y2 ¼ 7 to standard form. We have A ¼ C ¼ 1 and B ¼ 3. Thus 2q ¼ 90 so q ¼ 45 and x0  y0 x ¼ x0 cos q  y0 sin q ¼ pﬃﬃﬃ ; 2

x0 þ y0 y ¼ x0 sin q þ y0 cos q ¼ pﬃﬃﬃ 2

so x2 þ 3xy þ y2 ¼

  0   0   0  0 x  y0 2 x  y0 x þ y0 x þ y0 2 pﬃﬃﬃ pﬃﬃﬃ pﬃﬃﬃ þ þ 3 pﬃﬃﬃ 2 2 2 2

5x02  y02 ¼7 2 which is the equation of a hyperbola. Repeating what we said earlier, recall that if B2  4AC > 0 the graph is a hyperbola, if B2  4AC ¼ 0 the graph is an ellipse, if B2  4AC < 0 the graph is a parabola. We follow the same nomenclature with second-order PDEs, designating them as hyperbolic, elliptic, or parabolic according to the sign of B2  4AC. ¼

5.6 Converting Second-Order Partial Differential Equations

We now demonstrate by an example that the same change of variables used to rotate the axes to convert a second-degree equation to standard form works to convert a second-order PDE to standard form. Example: We convert the equation v2 uðx; yÞ v2 uðx; yÞ v2 uðx; yÞ þ 3 ¼7 þ vx2 vxvy vx2 to A

v2 uðx0 ; y0 Þ v2 uðx0 ; y0 Þ þB ¼C 02 vx vy02

using the change of variables x0  y0 x ¼ pﬃﬃﬃ ; 2

x0 þ y0 y ¼ pﬃﬃﬃ . 2

Recall the chain rule for partial derivatives: If u is a function of x and y and x and y are each functions of x0 and y0, then vu vu vx vu vy vu vu vx vu vy ¼ þ and ¼ þ . vx0 vx vx0 vy vx0 vy0 vx vy0 vy vy0 We have vx 1 vx 1 vy 1 vy 1 ¼ pﬃﬃﬃ; ¼ pﬃﬃﬃ; ¼ pﬃﬃﬃ; ¼ pﬃﬃﬃ vx0 2 vy0 2 vx0 2 vy0 2 so vu vu 1 vu 1 1 pﬃﬃﬃ þ pﬃﬃﬃ ¼ pﬃﬃﬃ ðux þ uy Þ and ¼ vx0 vx 2 vy 2 2   vu vu 1 vu 1 1 pﬃﬃﬃ þ pﬃﬃﬃ ¼ pﬃﬃﬃ ðuy  ux Þ. 0 ¼ vx vy 2 vy 2 2 Then     v2 u v vu v 1 ¼ 0 pﬃﬃﬃ ðux þ uy Þ ¼ vx vx02 vx0 vx0 2 1 v vx 1 v vy ¼ pﬃﬃﬃ ðux þ uy Þ 0 þ pﬃﬃﬃ ðux þ uy Þ 0 vx vx 2 vx 2 vy 1 1 1 1 1 1 ¼ pﬃﬃﬃ ðuxx þ uxy Þ pﬃﬃﬃ þ pﬃﬃﬃ ðuyx þ uyy Þ pﬃﬃﬃ ¼ uxx þ uxy þ uyy . 2 2 2 2 2 2

235

236

CHAPTER 5 Three Important Equations

Likewise,     v2 u v vu v 1 pﬃﬃﬃ ðuy  ux Þ ¼ 0 0 ¼ vy0 vy02 vy vy 2

1 v vx v vy ðuy  ux Þ 0 þ ðuy  ux Þ 0 ¼ pﬃﬃﬃ vy vy vy 2 vx      1 1 1 ¼ pﬃﬃﬃ ½uxy  uxx  pﬃﬃﬃ þ ½uyy  uyx  pﬃﬃﬃ 2 2 2 ¼

1 1 uxx  uxy þ uyy . 2 2

We now show v2 u v2 u  v u v u v u vx02 vy02 . þ 3 ¼ þ 2 vx2 vxvy vy2 2

2

2

5

We have 5

v2 u v2 u      1 1 1 1 vx02 vy02 5 1 uxx þ uxy þ uyy  uxx  uxy þ uyy ¼ 2 2 2 2 2 2 2       5 1 5 1 5 1 ¼ uxx  þ uxy þ þ uyy  ¼ uxx þ 3uxy þ uyy . 4 4 4 4 4 4

EXERCISES 1. Classify the PDEs below as parabolic, hyperbolic, or elliptic and convert them to standard form. a. b. c. d.

uxx  2uxy þ 9uyy  ux þ 5uy ¼ 0: uxx þ 4uxy  9uyy  3uy ¼ 0: uxx  uxy ¼ 0: uxx  6uxy þ 4uyy þ ux ¼ 0:

CHAPTER

SturmeLiouville Theory

6

6.1 INTRODUCTION Earlier, we studied Fourier series for trigonometric functions. One reason this is important is because the ordinary differential equation y00 ðxÞ þ lyðxÞ ¼ 0 with periodic boundary conditions arises when solving certain partial differential equations using separation of variables. Our approach was to recognize the problem of solving the differential equation as an eigenfunction/eigenvalue problem, the solutions to the equation being the eigenfunctions. It was necessary to determine when the Fourier expansion of a function in terms of the eigenfunctions converges to the function. In that problem, we considered pointwise convergence, uniform convergence, and L2 convergence. For the equation above, the eigenfunctions are sines and cosines. We shall see that other ordinary differential equations arise when the partial differential equations that we shall study are solved by separation of variables, including Bessel’s equation and Legendre’s equation. Each of these is a Sturme Liouville differential equation. In this chapter, we present the problem of solving a SturmeLiouville differential equation as an eigenfunction/eigenvalue problem and find conditions on a function to ensure that the Fourier expansion of the function in terms of eigenfunctions converges to the function. We shall consider only uniform convergence and L2 convergence. A differential equation of the form 0

½rðxÞy0 ðxÞ  qðxÞyðxÞ þ lrðxÞyðxÞ ¼ 0 for

a 0. Using integration by parts, with du ¼ y0 ðxÞ

u ¼ yðxÞ dv ¼ ½rðxÞy0 ðxÞ we have Z

1

0

v ¼ ½rðxÞy0 ðxÞ

0

yðxÞ½rðxÞy0 ðxÞ dx ¼ ½yðxÞrðxÞy0 ðxÞj

0

x¼1  x¼0

Since y(0) ¼ y(1) ¼ 0, we have Z Z 1 0 0 yðxÞ½rðxÞy ðxÞ dx ¼  0

1

Z

0

Z

1

yðxÞrðxÞyðxÞdx 0

rðxÞ½y0 ðxÞ dx.

0

rðxÞ½y0 ðxÞ dx.

Now, lhy; yir ¼ l

1

2

2

6.3 Completeness of Eigenfunctions for SturmeLiouville Equations

and Z

o 1 n 0 ½rðxÞy0 ðxÞ  qðxÞyðxÞ dx rðxÞ 0 Z 1 n o 0 ¼ yðxÞ ½rðxÞy0 ðxÞ  qðxÞyðxÞ dx 1

hy; Lyir ¼ 

yðxÞrðxÞ

0

Z

1

¼

rðxÞ½y0 ðxÞ dx þ 2

Z

1

yðxÞqðxÞyðxÞdx.

0

0

Since lhy; yir ¼ hy; Lyir we have Z l

1

Z

1

yðxÞrðxÞyðxÞdx ¼

0

0

2

Z

1

rðxÞ½y ðxÞ dx þ

0

qðxÞ½yðxÞ2 dx

0

so Z l¼

1

rðxÞ½y0 ðxÞ dx þ 2

0

Z

hy; yir

1

qðxÞ½yðxÞ2 dx

0

.

Now r(x) is positive and q(x) is nonnegative, and y0 ðxÞ cannot be zero everywhere. This is because if y0 ðxÞ ¼ 0 everywhere, then y(x) would be constant, the boundary condition y(0) ¼ 0 would mean that y(x) would be the zero function. This is impossible for an eigenvector. Thus l > 0. This completes Step 1. Step 2. Recall that 0

½rðxÞy0 ðxÞ  qðxÞyðxÞ þ lrðxÞyðxÞ ¼ FðxÞ; yð0Þ ¼ yð1Þ ¼ 0

(1)

has a unique solution if and only if 0

½rðxÞy0 ðxÞ  qðxÞyðxÞ þ lrðxÞyðxÞ ¼ 0; yð0Þ ¼ yð1Þ ¼ 0 has no solution other than y(x) ¼ 0. Since 0 is not an eigenvalue for Eq. (1), there is a Green’s function for 0

½rðxÞy0 ðxÞ  qðxÞyðxÞ ¼ FðxÞ; yð0Þ ¼ yð1Þ ¼ 0:

(2)

Let Gðx; xÞ be the Green’s function for Eq. (2) and take F (x) ¼ lr(x)u(x) to get that, if u(x) is an eigenfunction of Eq. (1), with eigenvalue l, then Z uðxÞ ¼ l

1

Gðx; xÞrðxÞuðxÞdx. 0

243

244

CHAPTER 6 SturmeLiouville Theory

Fix x ¼ x. Consider Gðx; xÞ. For any function f ðxÞ, we can write the Fourier expansion of f ðxÞ in terms of the eigenfunctions of an operator (even if we do not know the expansion converges to the function). If the eigenfunctions are u1, u2, . and the inner product is h; ir , the Fourier expansion is ∞ hu ðxÞ; f ðxÞi X n r n¼1

hun ðxÞ; un ðxÞir

un ðxÞ.

We are not claiming at this point that the Fourier series of the function is equal to the function. We do this for f ðxÞ ¼ Gðx; xÞ to get R ∞ hu ðxÞ; Gðx; xÞi X n r n¼1

hun ðxÞ; un ðxÞir

un ðxÞ ¼

Bessel’s inequality states that if any N N X

c2n hun ðxÞ;

P

1 0 un ðxÞGðx; xÞrðxÞdx

∞ X

hun ðxÞ; un ðxÞir

n¼1

 un ðxÞ.

cn un is the Fourier expansion for f ðxÞ, then for Z

un ðxÞir 

1

½ f ðxÞ2 rðxÞdx.

0

n¼1

For this problem, Z f ðxÞ ¼ Gðx; xÞ and

cn ¼

1 0

un ðxÞGðx; xÞrðxÞdx

Z

1

½un ðxÞ2 rðxÞdx

0

so we have, for any N Z N X

1 0

!2 un ðxÞGðx; xÞrðxÞdx

Z

1

n¼1

Z

1



½Gðx; xÞ2 rðxÞdx.

0

½un ðxÞ2 rðxÞdx

0

But Z

1

un ðxÞGðx; xÞrðxÞdx ¼

0

un ðxÞ ln

so we have N X ½un ðxÞ2 n¼1

l2n

Z

1

,Z

1 0

1

 2

½un ðxÞ rðxÞdx

0

½Gðx; xÞ2 rðxÞdx.

(3)

6.3 Completeness of Eigenfunctions for SturmeLiouville Equations

Now multiply both sides of (3) by rðxÞ then integrate with respect to x from 0 to 1 to get Z 1 N X 1 1 ½un ðxÞ2 rðxÞdx Z 1 2 l x¼0 2 n¼1 ½un ðxÞ rðxÞdx n 0

Z 

1

Z

1

2

½Gðx; xÞ rðxÞdx rðxÞdx.

x¼0

x¼0

So

Z 1 Z 1 N X 1 2  ½Gðx; xÞ rðxÞdx rðxÞdx. 2 0 0 n¼1 ln

(4)

Since Gðx; xÞ is continuous on [0, 1]  [0, 1], it is bounded, and so the right-hand side of Eq. (4) is finite. Thus if there are infinitely many eigenvalues, then the ∞ P 1 must converge and lim ð1=l Þ ¼ 0: series n l2 n¼1

n/∞

n

This completes Step 2. Step 3. The proof of this step is beyond the scope of the text. A salient part of Step 3 is described in Step 4. A proof of Step 3 may be found in Courant and Hilbert, Vol. I. Step 4. In this part of the proof, we need a few facts to make the argument flow more smoothly. First, if f (x) is continuously differentiable with f (0) ¼ f (1) ¼ 0, then Z 1 Z 1  0 0 0 rðxÞf ðxÞun ðxÞdx ¼ f ðxÞ  rðxÞu0n ðxÞ dx. 0

0

This is because if we integrate by parts with  0 w ¼ rðxÞu0n ðxÞ dw ¼ rðxÞu0n ðxÞ dv ¼ f 0 ðxÞ

v ¼ f ðxÞ

we get Z 0

1

x¼1 rðxÞf 0 ðxÞu0n ðxÞdx ¼ rðxÞu0n ðxÞf ðxÞx¼0  Z ¼ 0

because f (0) ¼ f (1) ¼ 0.

1



Z 0

0

1

 0 f ðxÞ rðxÞu0n ðxÞ dx

f ðxÞ  rðxÞu0n ðxÞ dx

245

246

CHAPTER 6 SturmeLiouville Theory

From this we get Z

1

0

 rðxÞf 0 ðxÞu0n ðxÞ þ qðxÞf ðxÞun ðxÞ dx ¼ Z

1

¼

Z

1

f ðxÞln rðxÞun ðxÞdx ¼ ln

0

Z

1

f ðxÞ



0

 rðxÞu0n ðxÞ þ qðxÞun ðxÞ dx

0

Z

1

f ðxÞrðxÞun ðxÞdx ¼ ln cn

0

rðxÞ½un ðxÞ2 dx.

0

(5) The last step follows because Z

1

cn ¼ Z0

f ðxÞrðxÞun ðxÞdx

1

; rðxÞ½un ðxÞ2 dx

0

so Z

1

Z

1

f ðxÞrðxÞun ðxÞdx ¼ cn

0

rðxÞ½un ðxÞ2 dx.

0

A second fact is that if we replace f (x) by um(x) in Z

1 0

rðxÞf

0

ðxÞu0n ðxÞ þ

 qðxÞf ðxÞun ðxÞ dx ¼

Z

1

f ðxÞln rðxÞun ðxÞdx

0

we get Z

1

 rðxÞu0m ðxÞu0n ðxÞ þ qðxÞum ðxÞun ðxÞ dx ¼ 0 8 Z 1 >

: 0 if msn

Z

1

um ðxÞln rðxÞun ðxÞdx

0

(6)

We now apply these equalities. We continue to assume that f (x) is a function for ∞ P which f (0) ¼ f (1) ¼ 0 and whose Fourier expansion is cn un with n¼1

Z

1

cn ¼ Z0 0

f ðxÞrðxÞun ðxÞdx

1

. 2

rðxÞ½un ðxÞ dx

6.3 Completeness of Eigenfunctions for SturmeLiouville Equations

We want to establish a bound on " Z 1

rðxÞ f ðxÞ 

0

N X

#2 c n un

dx

n¼1

so that the bound will go to 0 as. N/∞. A key part of the proof of Step 3 is to show that the minimum of the set  8Z 9 i  1h > > 2 2  > > 0 > > rðxÞðf ðxÞÞ þ qðxÞðfðxÞÞ dx < =  2 0 fðxÞeC ½0; 1 and fð0Þ ¼ fð1Þ ¼ 0 Z 1  > > > > > >  rðxÞðfðxÞÞ2 dx : ;  0

is actually achieved, that the minimum value of this set is l1, and the minimizing function is a multiple of. u1. One then proceeds to take the minimum of the set above, except restricting f(x) to be in the orthogonal complement of u1. That is, we take the minimum of the set  8Z i  1h > 2 2 > 0 > rðxÞðf ðxÞÞ þ qðxÞðfðxÞÞ dx <  0  fðxÞeC 2 ½0; 1 and Z 1  > > 2  > rðxÞðfðxÞÞ dx :  0

Z fð0Þ ¼ fð1Þ ¼ 0

1

and 0

9 =

rðxÞu1 ðxÞf ðxÞdx ¼ 0 . ;

The minimum value of this set is l2 and the minimizing function is a multiple of u2. One then continues recursively to show that lk is the minimum value of the set  8Z i  1h > 2 2 > 0 > rðxÞðf ðxÞÞ þ qðxÞðfðxÞÞ dx <  0 fðxÞeC 2 ½0; 1 and Z 1  > >  > rðxÞðfðxÞÞ2 dx :  0

Z

1

fð0Þ ¼ fð1Þ ¼ 0 and

rðxÞui ðxÞfðxÞdx ¼ 0

0

and the minimizing function is a multiple of uk.

for

i ¼ 1; 2; .; k  1

9 = ;

247

248

CHAPTER 6 SturmeLiouville Theory

Thus for any function f(x) for which Z 1 rðxÞui ðxÞfðxÞdx ¼ 0 for i ¼ 1; 2; .; k  1, we have 0

Z 1h i 2 rðxÞðf0 ðxÞÞ þ qðxÞðfðxÞÞ2 dx lk  0 . Z 1 rðxÞðfðxÞÞ2 dx

(7)

0

We show in Exercise 1 that Z

1

"

k1 X

rðxÞ f ðxÞ 

0

# cn un ðxÞ ui ðxÞdx ¼ 0

i  k  1.

for

n¼1

In Eq. (7) we take k1 X

fðxÞ ¼ f ðxÞ 

cn un ðxÞ

n¼1

to get Z 1(

"

k1 X

0

rðxÞ f ðxÞ 

0

lk 

Z

#2

n¼1 1

"

cn u0n ðxÞ

þ qðxÞ f ðxÞ 

"

rðxÞ f ðxÞ 

0

k1 X

#2

k1 X

#2 ) cn un ðxÞ

dx

n¼1

cn un ðxÞ dx

n¼1

so that Z

1

lk 

Z 1(

" rðxÞ f ðxÞ 

0

" 0

rðxÞ f ðxÞ 

0

k1 X

#2 cn u0n ðxÞ

k1 X n¼1

#2 cn un ðxÞ dx "

þ qðxÞ f ðxÞ 

n¼1

k1 X

#2 ) cn un ðxÞ

dx

n¼1

and thus Z

1  lk

8 Z 1< 0

:

1

" rðxÞ f ðxÞ 

0

" rðxÞ f 0 ðxÞ 

k1 X

#2 cn un ðxÞ dx

n¼1 k1 X n¼1

#2 cn u0n ðxÞ

#2 9 = dx. þ qðxÞ f ðxÞ  cn un ðxÞ ; n¼1 "

k1 X

(8)

6.3 Completeness of Eigenfunctions for SturmeLiouville Equations

Now Z 1(

" 0

rðxÞ f ðxÞ 

0

k1 X

#2 cn u0n ðxÞ

Z

1

rðxÞ½ f 0 ðxÞ dx  2 2

0 k1 X k1 X

cn 0 n¼1 k1 X k1 X

0

Z

1

cn cm

2

cn

0 n¼1 k1 X k1 X

k1 X

#2 ) cn un ðxÞ

dx

n¼1

Z

1

cn 0

rðxÞf 0 ðxÞu0n ðxÞdx

rðxÞu0n ðxÞu0m ðxÞdx

Z

1

þ

qðxÞ½ f ðxÞ2 dx

0

qðxÞf ðxÞun ðxÞdx

m¼1 n¼1 Z 1 k1 X 

þ

1

cn cm

2

k1 X n¼1

Z

m¼1 n¼1 Z 1 k1 X

þ

þ qðxÞ f ðxÞ 

n¼1

¼ þ

"

qðxÞun ðxÞum ðxÞdx ¼

Z 1n

0

o 2 rðxÞ½ f 0 ðxÞ þ qðxÞ½ f ðxÞ2 dx

0

 rðxÞf 0 ðxÞu0n ðxÞ þ qðxÞf ðxÞun ðxÞ dx Z

1

cn cm

m¼1 n¼1

0

 rðxÞu0n ðxÞu0m ðxÞ þ qðxÞun ðxÞum ðxÞ dx. (9)

By Eq. (5) Z 0

 rðxÞf 0 ðxÞu0n ðxÞ þ qðxÞf ðxÞun ðxÞ ¼ ln

1

Z

1

f ðxÞrðxÞun ðxÞdx

0

Z

1

¼ ln cn

rðxÞ½un ðxÞ2 dx

0

and Z 0

1

rðxÞu0m ðxÞu0n ðxÞ þ

 qðxÞum ðxÞun ðxÞ dx ¼

Z

1

um ðxÞln rðxÞun ðxÞdx

0

¼

8 Z > > < ln > > :0

1

½un ðxÞ2 rðxÞdx if

m¼n

if

msn:

0

249

250

CHAPTER 6 SturmeLiouville Theory

Thus the right-hand side of Eq. (9) is Z 1n Z k1 o X 2 cn l n rðxÞ½ f 0 ðxÞ þ qðxÞ½ f ðxÞ2 dx  2 0

þ

k1 X n¼1



k1 X

1

½un ðxÞ2 rðxÞdx ¼

0

Z c2n ln

Z 1n

f ðxÞrðxÞun ðxÞdx

0

n¼1

Z c2n ln

1

o 2 rðxÞ½ f 0 ðxÞ þ qðxÞ½ f ðxÞ2 dx

0 1

½un ðxÞ2 rðxÞdx.

0

n¼1

So from Eq. (8) we have Z

1

" rðxÞ f ðxÞ 

0

k1 X

#2 cn un ðxÞ

dx

n¼1

9 8 Z 1 Z 1n k1 = < o X 1 2  rðxÞ½ f 0 ðxÞ þ qðxÞ½ f ðxÞ2 dx  c2n ln ½un ðxÞ2 rðxÞdx . ; lk : 0 0 n¼1 Since lim ð1=lk Þ ¼ 0, we have k/∞ + * k1 k1 X X cn un ðxÞ; f ðxÞ  cn un ðxÞ lim f ðxÞ  k/∞

Z

n¼1

¼ lim

1

k/∞ 0

n¼1

"

rðxÞ f ðxÞ 

k1 X

#2

r

cn un ðxÞ dx ¼ 0.

n¼1

Thus if Z

1

rðxÞ½ f ðxÞ2 dx < ∞

0

then the Fourier series of f (x) converges to f (x) in the L2 sense (i.e., in the mean). Said another way, the eigenfunctions for the SturmeLiouville problem are complete.

EXERCISES 1. Show that Z

1 0

" rðxÞ f ðxÞ 

k1 X

# cn un ðxÞ ui ðxÞdx ¼ 0 for

i  k  1:

n¼1

2. Verify that the eigenvalues and eigenfunctions of y00 ðxÞ þ lyðxÞ ¼ 0; yð0Þ ¼ 0; y0 ð1Þ þ yð1Þ ¼ 0 pﬃﬃﬃ pﬃﬃﬃ  pﬃﬃﬃﬃﬃ  are tan l ¼  l; yn ðxÞ ¼ sin x ln .

6.4 Uniform Convergence of Fourier Series

3. Find the eigenvalues and eigenfunctions for y00 ðxÞ þ lyðxÞ ¼ 0 with the following boundary conditions: a. yð0Þ ¼ 0; y0 ð1Þ ¼ 0: b. y0 ð0Þ ¼ 0; yð1Þ ¼ 0: c. yð0Þ ¼ 0; y0 ð1Þ  yð1Þ ¼ 0:

6.4 UNIFORM CONVERGENCE OF FOURIER SERIES We want to determine conditions on f (x) that will ensure that the Fourier series of f (x) converges uniformly to f (x). In this case, we mean Fourier series of eigenfunctions of an eigenvalue problem. One example of these we have seen previouslydthe sine and cosine functions. P Suppose that cn un ðxÞ is the Fourier series of f (x). Since each function un(x) is an eigenfunction of L, in our setting, it is differentiable and thus continuous. We P u ðxÞ is uniformly Cauchy. shall determine conditions on f (x) that ensure c n n P That will mean that cn un ðxÞ converges uniformly, and the limit is a continuous function that we call g(x). It will follow that f (x) ¼ g(x). We have shown that if G(x, y) is the Green’s function for the second-order linear differential operator L where the interval is 0 < x < 1, and if {fn} is the set of normalized eigenfunctions for L and lm is the eigenvalue for fm and no eigenvalue is 0, then Gðx; yÞ ¼

X fn ðxÞfn ðyÞ ln

n

.

If the inner product is Z h f ðxÞ; gðxÞir ¼

1

f ðxÞrðxÞgðxÞdx

0

we have for {un}, the set of eigenfunctions that are not necessarily normalized Gðx; yÞ ¼

X un ðxÞun ðyÞ n

ln hun ; un ir

¼

X n

un ðxÞun ðyÞ

Z

1

ln

. 2

rðwÞ½un ðwÞ dw

0

Also recall Schwarz’s inequality states that for ai, bi, i ¼ 1,.,n !1=2 !1=2 n n n X X X 2 2 ai bi  ai bi i¼1

i¼1

i¼1

so that n X i¼1

!2 ai bi



n X i¼1

! a2i

n X i¼1

! b2i .

251

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CHAPTER 6 SturmeLiouville Theory

Fix x. Let

sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Z 1 ui ðxÞ ﬃ. rðxÞ½un ðxÞ2 dx bi ¼ sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ai ¼ ci li Z 1

0

li

rðxÞ½ui ðxÞ2 dx

0

Then

Z

ai bi ¼ ci ui ðxÞ;

a2i

¼

c2i li

1 0

rðxÞ½ui ðxÞ2 dx; b2i ¼

½ui ðxÞ2

Z

1

li

rðxÞ½ui ðxÞ2 dx

0

and so n X

0 ! Z 1 n n BX X B c2i li rðxÞ½ui ðxÞ2 dx B @ i¼1 0 i¼1

!2 ci ui ðxÞ



i¼1

1 C C C. 1 2 A rðxÞ½ui ðxÞ dx ½ui ðxÞ2

Z li 0

Now Gðx; xÞ ¼

X i

½ui ðxÞ2

Z

1

li

rðxÞ½ui ðxÞ2 dx

0

so n X

!2 ci ui ðxÞ

i¼1

0 @

n X

1

Z c2i li

i¼1

1

rðxÞ½ui ðxÞ2 dxAGðx; xÞ.

0

We want to invoke the Cauchy criterion for uniform convergence, so we change the values over which we sum to get, for positive integers M and N with M < N, 1 !2 0 Z 1 N N X X ci ui ðxÞ  @ c2i li rðxÞ½ui ðxÞ2 dxAGðx; xÞ. i¼Mþ1

i¼Mþ1

0

PN Now Gðx; xÞ is uniformly bounded, so i¼Mþ1 ci ui ðxÞ will be uniformly P∞ 2 R 1 2 Cauchy if c l rðxÞ½u ðxÞ dx converges. We now derive conditions for i i¼1 i i 0 which that is the case. P Suppose fðxÞ ¼ cn un ðxÞ. We want to find the kth Fourier coefficient in the expansion of o 1 n 0 ½rðxÞf0 ðxÞ  qðxÞfðxÞ . rðxÞ That is, we compute Z 1 o 1 n 0 ak ¼ ½rðxÞf0 ðxÞ  qðxÞfðxÞ rðxÞuk ðxÞdx 0 rðxÞ Z 1n o 0 ¼ ½rðxÞf0 ðxÞ  qðxÞfðxÞ uk ðxÞdx. 0

6.4 Uniform Convergence of Fourier Series

We integrate

R1 0

½rðxÞf0 ðxÞ0 uk ðxÞdx by parts with dw ¼ uk 0 ðxÞ

w ¼ uk ðxÞ dv ¼ ½rðxÞf0 ðxÞ

0

v ¼ rðxÞf0 ðxÞ

to get Z

1 0

0

Z

1

½rðxÞf ðxÞ uk ðxÞdx ¼ rðxÞf  rðxÞf0 ðxÞuk 0 ðxÞdx 0 Z 1 rðxÞf0 ðxÞuk 0 ðxÞdx: ¼

We integrate

0

R1 0

0

ðxÞuk ðxÞjx¼1 x¼0

0

 0 fðxÞ rðxÞuk 0 ðxÞ dx by parts with w ¼ fðxÞ dw ¼ f0 ðxÞ 0  v ¼ rðxÞuk 0 ðxÞ dv ¼ rðxÞuk 0 ðxÞ

to get Z

1 0

Z 1  0 fðxÞ rðxÞuk 0 ðxÞ dx ¼ fðxÞ rðxÞuk 0 ðxÞjx¼1  rðxÞuk 0 ðxÞ f0 ðxÞdx x¼0 0 Z 1 0 0 rðxÞuk ðxÞf ðxÞdx. ¼ 0

Thus Z

1

0

½rðxÞf0 ðxÞ uk ðxÞdx ¼

Z

0

1

 0 fðxÞ rðxÞuk 0 ðxÞ dx

0

and so Z 1n Z o 0 ½rðxÞf0 ðxÞ  qðxÞfðxÞ uk ðxÞdx ¼ 0

1

fðxÞ

n

o 0 rðxÞuk 0 ðxÞ  qðxÞuk ðxÞ dx:

0

Now 

0 rðxÞuk 0 ðxÞ  qðxÞuk ðxÞ ¼ lk rðxÞuk ðxÞ

so Z

1 0

Z n o 0 fðxÞ rðxÞuk 0 ðxÞ  qðxÞuk ðxÞ dx ¼ lk

1

fðxÞrðxÞuk ðxÞdx ¼ lk ck .

0

Thus we have the Fourier expansion o X 1 n 0 lk ck uk . ½rðxÞf0 ðxÞ  qðxÞfðxÞ w  rðxÞ

253

254

CHAPTER 6 SturmeLiouville Theory

Note that *

+ Z 1 o  1 0 0 1 n 0 ¼ f ðxÞrðxÞ ½rðxÞf 0 ðxÞ  qðxÞf ðxÞ dx f ; ðrf Þ  qf r rðxÞ 0 r Z 1 n o 0 f ðxÞ ½rðxÞf 0 ðxÞ  qðxÞf ðxÞ dx ¼ Z0 1 Z 1 dx ¼ f ðxÞ½rðxÞf 0 ðxÞ  qðxÞ½f ðxÞ2 dx. 0

0

We have seen that integrating by parts gives Z

1

Z n o 0 f ðxÞ ½rðxÞf 0 ðxÞ dx ¼ 

0

1

rðxÞ½ f 0 ðxÞ dx; 2

0

so *

 1 0 f ; ðrf 0 Þ  qf r

+ ¼

 Z 1

  2 rðxÞ½ f 0 ðxÞ þ qðxÞ½ f ðxÞ2 dx .

0

r

We also have E D X X lk ck h f ðxÞ; uk ðxÞir f ðxÞ;  lk ck uk ðxÞ ¼  r

¼

X

Z

1

l k ck

f ðxÞrðxÞuk ðxÞdx.

0

Now Z

1

ck ¼ Z0

f ðxÞrðxÞuk ðxÞdx

1

rðxÞ½uk ðxÞ2 dx

0

so Z

1

Z

1

f ðxÞrðxÞuk ðxÞdx ¼ ck

0

rðxÞ½uk ðxÞ2 dx

0

and thus 

X

Z

1

l k ck

f ðxÞrðxÞuk ðxÞdx ¼ 

X

Z

0

¼

X

1

lk ck ck 0

Z lk c2k

rðxÞ½uk ðxÞ2 dx

1 0

rðxÞ½uk ðxÞ2 dx.

6.4 Uniform Convergence of Fourier Series

Since the vectors {un} form a complete set, and since o X 1 n 0 lk ck uk ; ½rðxÞf0 ðxÞ  qðxÞfðxÞ w  rðxÞ if f (x) is a function that has a continuous second derivative and f (0) ¼ f (1) ¼ 0,we have  E o D X 1 n 0 f ðxÞ; ¼ f ðxÞ;  lk ck uk . ½rðxÞf0 ðxÞ  qðxÞfðxÞ r rðxÞ r That is,  Z 1    o 1 n 2 0 2 0 0  rðxÞ½ f ðxÞ þ qðxÞ½ f ðxÞ dx ¼ f ðxÞ; ½rðxÞf ðxÞ  qðxÞfðxÞ rðxÞ 0 r E D X ¼ f ðxÞ;  l k c k uk ¼ Thus Z 1n 0

X

Z lk c2k

1

r

rðxÞ½uk ðxÞ2 dx.

0

Z o X 2 rðxÞ½ f 0 ðxÞ þ qðxÞ½ f ðxÞ2 dx ¼ lk c2k

1

rðxÞ½uk ðxÞ2 dx;

0

 R 1 so if 0 rðxÞ½ f 0 ðxÞ2 þ qðxÞ½ f ðxÞ2 dx is finite, then the series P 2 R1 PN 2 lk ck 0 rðxÞ½uk ðxÞ2 dx converges. This means we can make i¼Mþ1 ci li R1 2 0 rðxÞ½ui ðxÞ dx as small as we like, by making M sufficiently large. Summarizing, we have If f (x) has a continuous second derivative and f (0) ¼ f (1) ¼ 0 and Z 1n o 2 rðxÞ½ f 0 ðxÞ þ qðxÞ½ f ðxÞ2 dx 0

is finite, then the Fourier series of f (x) with respect to the eigenfunctions of L converges uniformly to f (x).

255

CHAPTER

Using Generating Functions to Solve Specialized Differential Equations

7

7.1 INTRODUCTION Our focus for the rest of the text will be solving Laplace’s equation, the wave equation, and the heat equation using different techniques and in different scenarios. Some of these will involve curvilinear coordinates, and in curvilinear coordinates certain specialized differential equations will arise. In this chapter, we describe the solutions to these specialized equations through generating functions. We recommend investigating these specific equations when the need arises. There are several types of generating functions, and their uses in mathematics abound. We will use generating functions to give explicit formulae for solutions to some differential equations that occur prominently in mathematical physics. In addition to providing solutions, the search for these functions will uncover relationships among them that may not be as transparent with other techniques. The fundamental idea of a generating function is that we express a function as a power series, and the coefficients of the power series form a sequence. Examples: The function f ðxÞ ¼

1 1x

can be expressed as f ðxÞ ¼ 1 þ x þ x2 þ x3 þ /: Each coefficient of xn is 1, so f (x) generates the sequence {1,1,1,.}. With generating functions, we are proceeding formally and will not be concerned about convergence of the power series. The function x f ðxÞ ¼ 1  x  x2 Mathematical Physics with Partial Differential Equations. https://doi.org/10.1016/B978-0-12-814759-7.00007-7 Copyright © 2018 Elsevier Inc. All rights reserved.

257

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CHAPTER 7 Using Generating Functions

can be expressed as f ðxÞ ¼ x þ x2 þ 2x3 þ 3x4 þ 5x5 þ / so f (x) generates the sequence {1,1,2,3,5,.}, which is the Fibonacci sequence. In the setting of this chapter, a generating function G(x,t) is a function of two variables, and we will express the function as a power series in t. That is, we will have Gðx; tÞ ¼ P0 ðxÞ þ P1 ðxÞt þ P2 ðxÞt2 þ P3 ðxÞt3 þ / ¼

∞ X

Pn ðxÞtn .

n¼0

We will be given the function G(x,t) and be asked to determine the functions Pn(x) and show that each function Pn(x) satisfies a certain differential equation. Four important differential equations in physics are Laguerre’s equation, Hermite’s equation, Legendre’s equation, and Bessel’s equation. Each of these has a solution that is a polynomial of order n. For Laguerre’s equation, Hermite’s equation, and Legendre’s equation n is any nonnegative integer, and for Bessel’s equation, any integer n has an associated solution. One way to obtain an expression for these polynomials is through generating functions. The idea is that we hypothesize a generating function G(x,t), expand G(x,t) in a Maclaurin series, and group the terms in powers of t. We then have an expression of the form Gðx; tÞ ¼ g0 ðxÞ þ g1 ðxÞt þ g2 ðxÞt2 þ g3 ðxÞt3 þ /. The function gi(x) will turn out to be the ith Hermite polynomial in the case of the generator for Hermite polynomials and, similarly, for the other cases. Thus, we have a conceptually simple method of finding these functions. To justify the claim that these are indeed the desired functions, we must show that they satisfy the appropriate differential equation. To do this, one computes v v Gðx; tÞ and Gðx; tÞ vx vt and from these, obtains recurrence relations. With some manipulations, the functions gi(x) are shown to satisfy the pertinent differential equations, thus justifying their names as Hermite polynomials, Legendre polynomials, Laguerre polynomials, and Bessel functions. This technique is also helpful in determining relationships among different orders of the functions that might not be transparent. The major difficulty is finding the generating function for a particular equation, and there is not an established algorithm or strategy for doing this. The generating functions that we will use are exponential functions that use the Maclaurin expansion e f ðx;tÞ ¼ 1 þ f ðx; tÞ þ

ð f ðx; tÞÞ2 ð f ðx; tÞÞ3 þ þ /. 2! 3!

7.2 Generating Function for Laguerre Polynomials

7.2 GENERATING FUNCTION FOR LAGUERRE POLYNOMIALS Laguerre’s equation is a second-order differential equation of the form xy00 þ ð1  xÞy0 þ ny ¼ 0 where n is a positive integer. Here, we develop the solutions to Laguerre’s equation from a generating function. The generating function for Laguerre polynomials is   X ∞ 1 xt Ln ðxÞ n exp  ¼ t Gðx; tÞ ¼ 1t 1t n! n¼0 where Ln(x) will be shown to be polynomials that satisfy Laguerre’s equation for the value n. We describe vG=vt in two ways, and set the expressions equal to one another. This will yield a recursion relation. Doing the same thing for vG=vx gives a second recursion relation. Manipulating the two relations will enable us to demonstrate that Ln ðxÞ is a solution for Laguerre’s equation. We will now determine the two recurrence relations. The first is obtained by noting ð1  tÞ2

vG ¼ ð1  t  xÞG: vt

This says ð1  tÞ2

∞ ∞ X X Ln ðxÞ n1 Ln ðxÞ n nt t : ¼ ð1  t  xÞ n! n! n¼0 n¼0

Expanding gives ∞ ∞ ∞ X X Ln ðxÞ n1 Ln ðxÞ n X Ln ðxÞ nþ1 2 ¼ nt nt þ nt n! n! n! n¼0 n¼0 n¼0 ∞ ∞ ∞ X X Ln ðxÞ n X Ln ðxÞ nþ1 Ln ðxÞ n x t  t t : n! n! n! n¼0 n¼0 n¼0

Reindexing yields ∞ ∞  X Ln ðxÞ n X Lnþ1 ðxÞ Ln ðxÞ Ln1 ðxÞ Ln ðxÞ x  t ¼  2n þ nðn  1Þ  n! n! n! n! n! n¼0 n¼0  Ln1 ðxÞ n t : þ n!

259

260

CHAPTER 7 Using Generating Functions

This lead to the first recurrence relation ðn þ 1ÞLnþ1 ðxÞ ¼ ð2n þ 1  xÞLn ðxÞ  n2 Ln1 ðxÞ. Next, we use the fact that vG t ¼ G vx 1  t so ð1  tÞ

vG ¼ tG: vx

Now ∞ ∞ vG v X Ln ðxÞ n X L’n ðxÞ n ¼ t ¼ t vx vx n¼0 n! n! n¼0

so ð1  tÞ

∞ ∞ ∞ X vG L’n ðxÞ n X L’n ðxÞ n X L’n ðxÞ nþ1 ¼ ð1  tÞ t ¼ t  t : vx n! n! n! n¼0 n¼0 n¼0

Also tG ¼ t

∞ ∞ X X Ln ðxÞ n Ln ðxÞ nþ1 t ¼ t : n! n! n¼0 n¼0

Now ∞ ∞ ∞ ∞ X Ln1 ðxÞ n X nLn1 ðxÞ n X nLn1 ðxÞ n Ln ðxÞ nþ1 X t ¼ t ¼ t ¼ t ¼ ðn  1Þ! n! n! n! n¼0 n¼1 n¼1 n¼0 ∞ ∞ ∞ ∞ X L’n1 ðxÞ n X nL’n1 ðxÞ n X nL’n1 ðxÞ n L’n ðxÞ nþ1 X t ¼ t ¼ t t ¼ ðn  1Þ! n! n! n! n¼0 n¼1 n¼1 n¼0

so ð1  tÞ

∞ ∞ ∞ ∞ nL’n1 ðxÞ n vG X L’n ðxÞ n X L’n ðxÞ nþ1 X L’n ðxÞ n X t ¼ ¼ t  t t  n! vx n! n! n! n¼0 n¼0 n¼0 n¼0

and tG ¼ t

∞ ∞ X X nLn1 ðxÞ n Ln ðxÞ n t : t ¼ n! n! n¼0 n¼0

(1)

7.2 Generating Function for Laguerre Polynomials

Thus we have ∞ ∞ ∞ X X nL’n1 ðxÞ n nLn1 ðxÞ n L’n ðxÞ n X t ¼ t t  n! n! n! n¼0 n¼0 n¼0

and so L’n ðxÞ  nL’n1 ðxÞ ¼ nLn1 ðxÞ

(2)

which is the second recurrence relation. We now manipulate these relations to show that Ln ðxÞ is a solution to Laguerre’s equation. The problem now is to manipulate the recurrence relations to show that Ln(x) is a solution to Laguerre’s equation. In Eq. (2), replace n by n þ 1 to get L0nþ1 ðxÞ ¼ ðn þ 1ÞL0n ðxÞ  ðn þ 1ÞLn ðxÞ.

(3)

Differentiate Eq. (3) to get L00nþ1 ðxÞ ¼ ðn þ 1ÞL00n ðxÞ  ðn þ 1ÞL0n ðxÞ. Replace n by n þ 1 to get L00nþ2 ðxÞ ¼ ðn þ 2ÞL00nþ1 ðxÞ  ðn þ 2ÞL0nþ1 ðxÞ. Use Eq. (3) to get

  L00nþ2 ðxÞ ¼ ðn þ 2ÞL00nþ1 ðxÞ  ðn þ 2Þ ðn þ 1ÞL0n ðxÞ  ðn þ 1ÞLn ðxÞ   ¼ ðn þ 1Þðn þ 2Þ L00nþ1 ðxÞ  2L0n ðxÞ þ Ln ðxÞ .

The first recurrence relation is Lnþ1 ðxÞ þ ðx  2n  1ÞLn ðxÞ þ n2 Ln1 ðxÞ ¼ 0. Differentiating (using the product rule on the second term) gives L0nþ1 ðxÞ þ ðx  2n  1ÞL0n ðxÞ þ Ln ðxÞ þ n2 L0n1 ðxÞ ¼ 0. Differentiating a second time gives L00nþ1 ðxÞ þ ðx  2n  1ÞL00n ðxÞ þ L0n ðxÞ þ L0n ðxÞ þ n2 L00n1 ðxÞ ¼ 0. Replace n by n þ 1 to get L00nþ2 ðxÞ þ xL00nþ1 ðxÞ þ 2L0nþ1 ðxÞ  ð2n þ 2ÞL00nþ1 ðxÞ þ ðn þ 1Þ2 L00n ðxÞ ¼ 0. Substituting L0nþ1 ðxÞ ¼ ðn þ 1ÞL0n ðxÞ  ðn þ 1ÞLn ðxÞ L00nþ1 ðxÞ ¼ ðn þ 1ÞL00n ðxÞ  ðn þ 1ÞL0n ðxÞ

(4)

261

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CHAPTER 7 Using Generating Functions

  L00nþ2 ðxÞ ¼ ðn þ 1Þðn þ 2Þ L00nþ1 ðxÞ  2L0n ðxÞ þ Ln ðxÞ into Eq. (4) gives     ðn þ 1Þðn þ 2Þ L00nþ1 ðxÞ  2L0n ðxÞ þ Ln ðxÞ þ x ðn þ 1ÞL00n ðxÞ  ðn þ 1ÞL0n ðxÞ   þ 2 ðn þ 1ÞL0n ðxÞ  ðn þ 1ÞLn ðxÞ    ð2n þ 2Þ ðn þ 1ÞL00n ðxÞ  ðn þ 1ÞL0n ðxÞ þ ðn þ 1Þ2 L00n ðxÞ ¼ 0. Collecting terms, we have xL00n ðxÞ þ ð1  xÞL0n ðxÞ þ nLn ðxÞ ¼ 0. Thus, Ln(x) is a solution to Laguerre’s equation.

EXERCISES 1. Compute the first four Laguerre polynomials using the generating function. 2. Compute the first four Laguerre polynomials using Rodrigues formula Ln ðxÞ ¼ ex

dn  n x x e . dxn

3. Using a computer algebra system (CAS) evaluate Z

ex L2 ðxÞL3 ðxÞdx and

0

Z

ex L3 ðxÞL3 ðxÞdx.

0

7.3 HERMITE’S DIFFERENTIAL EQUATION Hermite’s differential equation is y00 ðxÞ  2xy0 ðxÞ þ 2nyðxÞ ¼ 0: For Hermite polynomials, we will show that the generating function is 2

Gðx; tÞ ¼ e2txt . Expanding in a power series Gðx; tÞ ¼

 n ∞ ∞ n X X 2tx  t2 t ð2x  tÞn ¼ . n! n! n¼0 n¼0

Generating functions are convenient because they provide an easy way to compute the coefficients of a power series. Similar to Maclaurin series, if Gðx; tÞ ¼ a0 ðxÞ þ a1 ðxÞt þ a2 ðxÞt2 þ a3 ðxÞt3 þ /

7.3 Hermite’s Differential Equation

then vk Gðx; tÞjt¼0 ¼ k!ak ðxÞ. vtk In our case, this means vk Hk ðxÞ ¼ Hk ðxÞ Gðx; tÞjt¼0 ¼ k! k! vtk where Hk(x) is the kth Hermite polynomial. We now compute the recurrence equations. Since 2

Gðx; tÞ ¼ e2txt ¼

∞ X Hn ðxÞ n t n! n¼0

we have ∞ ∞ X v 2 Hn ðxÞ n X 2tnþ1 Gðx; tÞ ¼ 2te2txt ¼ 2t t ¼ Hn ðxÞ vx n! n! n¼0 n¼0

and we also have ∞ ∞ n v v X Hn ðxÞ n X t 0 Gðx; tÞ ¼ t ¼ Hn ðxÞ: vx vx n¼0 n! n! n¼0 v Thus, we have computed Gðx; tÞ in two different ways. In the first method we vx differentiated the generating function, and in the second we differentiated the power series. We get the first recurrence relation by equating the coefficients of tn in the two power series expansions. Typically, we will need to reindex some of the series. We have ∞ ∞ n X X 2tnþ1 t 0 Hn ðxÞ ¼ Hn ðxÞ n! n! n¼0 n¼0

Reindexing the term on the left gives ∞ ∞ X X 2tnþ1 2tn Hn ðxÞ ¼ Hn1 ðxÞ: n! ðn  1Þ! n¼0 n¼1

Also, d d H0 ðxÞ ¼ ð1Þ ¼ 0 dx dx so ∞ n ∞ n X X t 0 t 0 Hn ðxÞ ¼ H ðxÞ n! n! n n¼0 n¼1

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and we have ∞ X n¼1

∞ n X 2tn t 0 Hn1 ðxÞ ¼ H ðxÞ: ðn  1Þ! n! n n¼1

Equating the coefficients of tn gives, 2 1 Hn1 ðxÞ ¼ Hn0 ðxÞ ðn  1Þ! n! so 2nHn1 ðxÞ ¼ Hn0 ðxÞ.

(1)

This is the first recurrence equation that we will use later. To get the second recurrence relation, we do a similar procedure, but using

v Gðx; tÞ. vt We have

∞ ∞ X v 2 Hn ðxÞ n X ð2x  2tÞHn ðxÞ n Gðx; tÞ ¼ ð2x  2tÞe2txt ¼ ð2x  2tÞ t ¼ t vt n! n! n¼0 n¼0

and ∞ ∞ ∞ v v X Hn ðxÞ n X Hn ðxÞ n1 X Hn ðxÞ n1 Gðx; tÞ ¼ t ¼ t t n ¼ n vt vt n¼0 n! n! n! n¼0 n¼1

¼

∞ ∞ X Hn ðxÞ n1 X Hnþ1 ðxÞ n t t : ¼ ðn  1Þ! n! n¼1 n¼0

Thus, ∞ ∞ X ð2x  2tÞHn ðxÞ n X Hnþ1 ðxÞ n t ¼ t : n! n! n¼0 n¼0

Now ∞ ∞ ∞ X ð2x  2tÞHn ðxÞ n X 2tHn ðxÞ n X 2xHn ðxÞ nþ1 t ¼ t þ t n! n! n! n¼0 n¼0 n¼0

and ∞ ∞ ∞ ∞ X 2tHn ðxÞ n X 2Hn ðxÞ nþ1 X 2Hn1 ðxÞ n X 2nHn1 ðxÞ n t ¼ t t ¼ t . ¼ n! n! ðn  1Þ! n! n¼0 n¼0 n¼1 n¼1

7.3 Hermite’s Differential Equation

Thus, ∞ ∞ ∞ X 2nHn1 ðxÞ n X 2xHn ðxÞ n X Hnþ1 ðxÞ n t þ t ¼ t . n! n! n! n¼1 n¼0 n¼0

When n ¼ 0 2xHn ðxÞ n t ¼ 2xH0 ðxÞ ¼ 2xð1Þ ¼ 2x n! and Hnþ1 ðxÞ n t ¼ H1 ðxÞ ¼ 2x n! so ∞ ∞ ∞ X 2nHn1 ðxÞ n X 2xHn ðxÞ n X Hnþ1 ðxÞ n t þ t ¼ t n! n! n! n¼1 n¼1 n¼1

and so 2nHn1 ðxÞ þ 2xHn ðxÞ ¼ Hnþ1 ðxÞ n ¼ 1; 2; .

(2)

This is the second recurrence relation. The problem now is to manipulate the recurrence relations to show that Hn(x) solves Hermite’s equation. Differentiating Eq. (2) gives 0 0 ðxÞ ¼ 2Hn ðxÞ þ 2xHn0 ðxÞ  2nHn1 ðxÞ. Hnþ1

We also found in Eq. (1) that Hn(x) solves the differential equation 2nHn1 ðxÞ ¼ Hn0 ðxÞ

ð1Þ

so 0 ðxÞ: Hn00 ðxÞ ¼ 2nHn1

Replacing n by n þ 1 Eq. (1) gives 0 Hnþ1 ðxÞ ¼ 2ðn þ 1ÞHn ðxÞ.

Thus 0 0 ðxÞ ¼ 2Hn ðxÞ þ 2xHn0 ðxÞ  2nHn1 ðxÞ Hnþ1

is equivalent to 2ðn þ 1ÞHn ðxÞ ¼ 2Hn ðxÞ þ 2xHn0 ðxÞ  Hn00 ðxÞ  0 0 since Hn00 ðxÞ ¼ 2nHn1 ðxÞ and Hnþ1 ðxÞ ¼ 2ðn þ 1ÞHn ðxÞ

(3)

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or 2nHn ðxÞ ¼ 2xHn0 ðxÞ  Hn00 ðxÞ so Hn00 ðxÞ  2xHn0 ðxÞ þ 2nHn ðxÞ ¼ 0 which is Hermite’s equation. Thus, we have justified the claim that Hn(x) is an nth degree polynomial that satisfies Hermite’s equation and so is a Hermite polynomial.

EXERCISES 1. Use the generating function for Hermite polynomials to find Hn(x) for n ¼ 1,2,3 and show Hn ðxÞ ¼ ð1Þn ex

2

dn x2 e dxn

for n ¼ 1; 2; 3:

2. Using a CAS find Z

∞

e

x2

Z H2 ðxÞH3 ðxÞdx

and

∞

ex H3 ðxÞH3 ðxÞdx: 2

7.4 GENERATING FUNCTION FOR LEGENDRE’S EQUATION The differential equation  1  x2 y00 ðxÞ  2xy0 ðxÞ þ nðn þ 1ÞyðxÞ ¼ 0 is Legendre’s equation. We will show that the function 1 Gðx; tÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1  2tx þ t2 is the generating function for Legendre polynomials. Denoting the Legendre polynomial of degree n by Pn(x), we have Gðx; tÞ ¼

∞ X n¼0

Pn ðxÞtn .

7.4 Generating Function for Legendre’s Equation

We will get two recursion relations by computing v v Gðx; tÞ and Gðx; tÞ. vt vx Letting 1 Gðx; tÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1  2tx þ t2 gives v t t 1 ¼ Gðx; tÞ ¼ 2Þ 3=2 1=2 2 vx ð1  2xt þ t ð1  2xt þ t Þ ð1  2xt þ t2 Þ ∞ X t t Gðx; tÞ ¼ ¼ Pn ðxÞtn ð1  2xt þ t2 Þ ð1  2xt þ t2 Þ n¼0 and expressing Gðx; tÞ ¼

∞ X

Pn ðxÞtn

n¼0

gives ∞ X v Gðx; tÞ ¼ P0n ðxÞtn : vx n¼0

Thus, we have ∞ X t Pn ðxÞtn ð1  2xt þ t2 Þ n¼0

¼

∞ X

P0n ðxÞtn :

n¼0

So t

∞ X

∞ X  Pn ðxÞtn ¼ 1  2xt þ t2 P0n ðxÞtn :

n¼0

n¼0

Expanding gives ∞ X

Pn ðxÞtnþ1 ¼

∞ X

n¼0

P0n ðxÞtn 

n¼0

∞ X

2xP0n ðxÞtnþ1 þ

n¼0

∞ X

P0n ðxÞtnþ2 :

n¼0

Reindex to get ∞ X n¼0

Pn ðxÞtnþ1 ¼

∞ X

P0nþ1 ðxÞtnþ1 

n¼1

∞ X

2xP0n ðxÞtnþ1 þ

n¼0

∞ X

P0n1 ðxÞtnþ1 :

n¼1

So for n  1 we have Pn ðxÞ ¼ P0nþ1 ðxÞ  2xP0n ðxÞ þ P0n1 which is our first recurrence equation.

(1)

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Also v xt xt 1 Gðx; tÞ ¼ 3 ¼ 1 2 vt ð1  2xt þ t2 Þ2 ð1  2xt þ t Þ ð1  2xt þ t2 Þ2 ∞ X xt xt Gðx; tÞ ¼ ¼ Pn ðxÞtn 2 2 ð1  2xt þ t Þ ð1  2xt þ t Þ n¼0 and ∞ ∞ X v v X Pn ðxÞtn ¼ nPn ðxÞtn1 Gðx; tÞ ¼ vt vt n¼0 n¼0

so ∞ ∞ X X xt n P ðxÞt ¼ nPn ðxÞtn1 n ð1  2xt þ t2 Þ n¼0 n¼0

or ðx  tÞ

∞ X

∞ X  Pn ðxÞtn ¼ 1  2xt þ t2 nPn ðxÞtn1 :

n¼0

n¼0

Expanding gives ∞ X

xPn ðxÞtn 

n¼0

∞ X

Pn ðxÞtnþ1 ¼

n¼0

∞ X

nPn ðxÞtn1 

n¼0

þ

∞ X

∞ X

2xnPn ðxÞtn

n¼0

nPn ðxÞtnþ1 :

n¼0

Reindexing, we have ∞ X n¼0

xPn ðxÞtn 

∞ X n¼1

Pn1 ðxÞtn ¼

∞ X

ðn þ 1ÞPnþ1 ðxÞtn 

n¼1 ∞ X

þ

∞ X

2xnPn ðxÞtn

n¼0

ðn  1ÞPn1 ðxÞtn :

n¼0

So xPn ðxÞ  Pn1 ðxÞ ¼ ðn þ 1ÞPnþ1 ðxÞ  2xnPn ðxÞ þ ðn  1ÞPn1 ðxÞ or ðn þ 1ÞPnþ1 ðxÞ  ð2xn þ xÞPn ðxÞ þ nPn1 ðxÞ ¼ 0 which is the second recurrence equation.

(2)

7.4 Generating Function for Legendre’s Equation

Our goal is to demonstrate  1  x2 P00n ðxÞ þ 2xP0n ðxÞ þ nðn þ 1ÞPn ðxÞ ¼ 0: This will be done by manipulating Pn ðxÞ þ 2xP0n ðxÞ ¼ P0n1 ðxÞ þ P0nþ1 ðxÞ

ð3Þ

Pn ðxÞ þ 2xP0n ðxÞ ¼ P0n1 ðxÞ þ P0nþ1 ðxÞ:

ð4Þ

and

We have ð2n þ 1ÞxPn ðxÞ ¼ ðn þ 1ÞPnþ1 ðxÞ þ nPn1 ðxÞ

(5)

P’nþ1 ðxÞ þ P’n1 ðxÞ ¼ 2xP’n ðxÞ þ Pn ðxÞ.

(6)

Adding 2  Eq. (5) to (2n + 1)  Eq. (6) gives ð2n þ 1ÞPn ðxÞ ¼ P’nþ1 ðxÞ  P’n1 ðxÞ.

(7)

Computing [Eq. (6) + Eq. (7)]/2 gives P’nþ1 ðxÞ ¼ ðn þ 1ÞPn ðxÞ þ xP’n ðxÞ.

(8)

Similarly, [Eq. (6)  Eq. (7)]/2 gives P’n1 ðxÞ ¼ nPn ðxÞ þ xP’n ðxÞ.

(9)

In Eq. (7), replace n by n  1 to get ½2ðn  1Þ þ 1P’n1 ðxÞ ¼ P’n ðxÞ  P’n2 ðxÞ or ð2n  1ÞP’n1 ðxÞ ¼ P’n ðxÞ  P’n2 ðxÞ.

(10)

Multiply Eq. (9) by x to get xP’n1 ðxÞ ¼ nxPn ðxÞ þ x2 P’n ðxÞ.

(11)

Add Eq. (10) to Eq. (11) to get  1  x2 P’n ðxÞ ¼ nPn1 ðxÞ  nxPn ðxÞ.

(12)

Differentiate Eq. (12) to get  1  x2 P’’n ðxÞ  2xP’n ðxÞ ¼ nP’n1 ðxÞ  nPn ðxÞ  nxP’n ðxÞ.

(13)

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CHAPTER 7 Using Generating Functions

Multiply Eq. (9) by n to get nP’n1 ðxÞ ¼ n2 Pn ðxÞ þ nxP’n ðxÞ:

(14)

Adding Eq. (13) to Eq. (14) gives  1  x2 P’’n ðxÞ  2xP’n ðxÞ þ nðn þ 1ÞPn ðxÞ ¼ 0: Thus, Pn(x) is a polynomial of degree n, which satisfies Legendre’s equation and is a Legendre polynomial.

EXERCISES 1. Use the generating function to find P2(x), P3(x), P4(x). 2. Use a CAS to evaluate R1 a. 1 P3 ðxÞP4 ðxÞdx: R1 b. 1 P3 ðxÞP3 ðxÞdx: R1 c. 1 P4 ðxÞP4 ðxÞdx: 3. Show that P3 ðxÞ ¼

3 1 d3  2 x 1 : 23 3! dx3

4. Show that 1 P2 ðcos qÞ ¼ ð1 þ 3 cosð2qÞÞ 4

7.5 GENERATOR FOR BESSEL FUNCTIONS OF THE FIRST KIND Bessel’s equation is

We will show that

 x2 y00 ðxÞ þ xy0 ðxÞ þ x2  n2 yðxÞ ¼ 0: 

x t  t1 Gðx; tÞ ¼ exp 2



is a generator for Bessel functions of the first kind. We will show that Jn(x) satisfies Bessel’s equation.

7.5 Generator for Bessel Functions of the First Kind

We will get two recurrence relations by taking the partial derivatives of the generating function. We have     x vGðx; tÞ v x x x ¼ exp t  t1 ¼ exp t  t1 þ 2 vt vt 2 2 2 2t      ∞    x 1 x x 1 X 1 1 þ 2 exp tt 1þ 2 ¼ ¼ Jn ðxÞtn 2 t 2 2 t n¼∞ ∞  h i X x Jn ðxÞtn þ Jn ðxÞtn2 ¼ n¼∞ 2 ∞   ∞   X X x x n Jn ðxÞt þ Jnþ2 ðxÞtn ¼ n¼∞ 2 n¼∞ 2 We also have ∞ ∞ ∞ X X vGðx; tÞ v X Jn ðxÞtn ¼ nJn ðxÞtn1 ¼ ðn þ 1ÞJnþ1 ðxÞtn : ¼ vt vt n¼∞ n¼∞ n¼∞

Thus,

  x ½Jn ðxÞ þ Jnþ2  ¼ ðn þ 1ÞJnþ1 ðxÞ 2

or Jn1 ðxÞ þ Jnþ1 ¼

2n Jn ðxÞ x

which is the first recurrence relation. Also,      1  vGðx; tÞ v x x ¼ exp t  t1 t  t1 ¼ exp t  t1 vx vx 2 2 2   ∞ X 1 Jn ðxÞtn ¼ t  t1 2 n¼∞ #  " X ∞ ∞ X 1 nþ1 n1 ¼ Jn ðxÞt  Jn ðxÞt 2 n¼∞ n¼∞ #  " X ∞ ∞ X 1 n n Jn1 ðxÞt  Jnþ1 ðxÞt : ¼ 2 n¼∞ n¼∞ We also have ∞ ∞ X vGðx; tÞ v X ¼ Jn ðxÞtn ¼ Jn0 ðxÞtn . vx vx n¼∞ n¼∞

(1)

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CHAPTER 7 Using Generating Functions

Thus,

  1 ½Jn1 ðxÞ  Jnþ1 ðxÞ ¼ Jn0 ðxÞ 2

or Jn1 ðxÞ  Jnþ1 ðxÞ ¼ 2Jn0 ðxÞ

(2)

which is the second recurrence relation. Adding Eqs. (1) and (2) gives Jn1 ðxÞ þ Jnþ1 ¼

2n Jn ðxÞ x

ð1Þ

Jn1 ðxÞ  Jnþ1 ðxÞ ¼ 2 Jn0 ðxÞ

ð2Þ

n Jn1 ðxÞ ¼ Jn ðxÞ þ Jn0 ðxÞ x or xJn0 ðxÞ ¼ xJn1 ðxÞ  nJn ðxÞ

(3)

and so nxJn0 ðxÞ ¼ nxJn1 ðxÞ  n2 Jn ðxÞ: Thus, nxJn0 ðxÞ  nxJn1 ðxÞ þ n2 Jn ðxÞ ¼ 0:

(3 0)

Differentiating Eq. (3) gives 0 xJn00 ðxÞ þ Jn0 ðxÞ ¼ xJn1 ðxÞ þ Jn1 ðxÞ  nJn0 ðxÞ

or 0 ðxÞ ¼ 0: xJn00 ðxÞ þ ðn þ 1ÞJn0 ðxÞ  Jn1 ðxÞ  xJn1

(4)

Multiplying Eq. (4) by x gives 0 ðxÞ ¼ 0: x2 Jn00 ðxÞ þ ðn þ 1ÞxJn0 ðxÞ  xJn1 ðxÞ  x2 Jn1

Eq. (40 )  Eq. (30 ) gives  2 00  0 x Jn ðxÞ þ ðn þ 1ÞxJn0 ðxÞ  xJn1 ðxÞ  x2 Jn1 ðxÞ   Eq. ð30 Þ  nxJn0 ðxÞ  nxJn1 ðxÞ þ n2 Jn ðxÞ

(4 0)

Eq. ð40 Þ

0 ¼ x2 Jn00 ðxÞ þ xJn0 ðxÞ  n2 Jn ðxÞ þ ðn  1ÞxJn1 ðxÞ  x2 Jn1 ðxÞ ¼ 0:

(5)

7.5 Generator for Bessel Functions of the First Kind

We will show in Exercise 1 that 0 ðxÞ ¼ x2 Jn ðxÞ. ðn  1ÞxJn1 ðxÞ  x2 Jn1

Knowing this, we have 0 x2 Jn00 ðxÞ þ xJn0 ðxÞ  n2 Jn ðxÞ þ ðn  1ÞxJn1 ðxÞ  x2 Jn1 ðxÞ

¼ x2 Jn00 ðxÞ þ xJn0 ðxÞ  n2 Jn ðxÞ þ x2 Jn ðxÞ  ¼ x2 Jn00 ðxÞ þ xJn0 ðxÞ þ x2  n2 Jn ðxÞ ¼ 0 which is Bessel’s equation.

EXERCISE 1 Show that 0 ðn  1ÞxJn1 ðxÞ  x2 Jn1 ðxÞ ¼ x2 Jn ðxÞ.

(6)

273

CHAPTER

Separation of Variables in Cartesian Coordinates

8

8.1 INTRODUCTION For the remainder of the text, we concentrate on solving partial differential equations that involve the Laplacian. We analyze the three prototypical equationsdthe heat equation, the wave equation, and the Laplace’s equationdin significant detail. We shall consider four techniques of solving partial differential equations; separation of variables, the Fourier transform, the Laplace transform, and Green’s functions. In this chapter we solve each of these equations in Cartesian coordinates by separation of variables. The idea of separation of variables is to assume that the solution to the partial differential equation, u(a, b, g), can be written as uða; b; gÞ ¼ f ðaÞgðbÞhðgÞ and determine an ordinary differential equation (ODE) that each of f (a), g(b), and h(g) must satisfy. Each of the ODEs is then solved, and the solutions are “pasted together” to give the solution to the partial differential equation. The validity of the solution should be verified because we began with the assumption that the variables could be separated. In Section 8.2, we consider the case of Laplace’s equation in two variables. We shall see that in all of the examples of this chapter the resulting ODEs are familiar and elementary to solve. In Section 8.3, we analyze Laplace’s equation in three variables. In Section 8.4, we give a detailed description of the heat equation in one space dimension, and in Section 8.5 we study the wave equation in one space dimension.

8.2 SOLVING LAPLACE’S EQUATION ON A RECTANGLE We begin with Laplace’s equation in two variables Duðx; yÞ ¼

v2 u v2 u þ ¼0 vx2 vy2

275

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

and hypothesize that uðx; yÞ ¼ XðxÞ YðyÞ. Then v2 u ¼ X 00 ðxÞ YðyÞ vx2

and

v2 u ¼ XðxÞ Y 00 ðyÞ vy2

so X 00 ðxÞ YðyÞ þ XðxÞ Y 00 ðyÞ ¼ 0: Hence X 00 ðxÞ YðyÞ XðxÞ Y 00 ðyÞ X 00 ðxÞ Y 00 ðyÞ þ ¼ þ ¼0 XðxÞ YðyÞ XðxÞ YðyÞ XðxÞ YðyÞ and thus X 00 ðxÞ Y 00 ðyÞ ¼ . XðxÞ YðyÞ

(1)

The left-hand side of Eq. (1) is a function only of x and the right hand side a function only of y, so the common value must be a constant, which we denote l. So we have XðxÞ ¼ lXðxÞ and

YðyÞ ¼ lYðyÞ

or X 00 ðxÞ  lXðxÞ ¼ 0

(2a)

Y 00 ðyÞ þ lYðyÞ ¼ 0:

(2b)

There are three cases: l ¼ 0, l > 0, and l < 0. If l ¼ 0, then X 00 ðxÞ ¼ 0 and X(x) ¼ Ax þ B and likewise, Y(y) ¼ Cy þ D. The more interesting cases that allow for nontrivial boundary conditions are l > 0 and l < 0. Suppose that l > 0. Then X 00 ðxÞ þ lXðxÞ ¼ 0 has the solution pﬃﬃﬃ pﬃﬃﬃ XðxÞ ¼ A cos lx þ B sin lx and Y 00 ðyÞ  lYðyÞ ¼ 0 has the solution pﬃﬃﬃ pﬃﬃﬃ YðyÞ ¼ C cosh ly þ D sinh ly. The case l < 0 is nearly identical and is left as Exercise 1. We continue to consider the case l > 0 and assign boundary conditions to the rectangle 0  x  a, 0  y  b. See Fig. 8.2.1. We assign the boundary conditions uðx; 0Þ ¼ f1 ðxÞ; 0  x  a;

uðx; bÞ ¼ f2 ðxÞ; 0  x  a;

uð0; yÞ ¼ g1 ðyÞ; 0  y  b;

uða; yÞ ¼ g2 ðyÞ; 0  y  b.

8.2 Solving Laplace’s Equation on a Rectangle

f2 (x)

(0, b)

(a, b)

g1 (y)

g2 (y) f1 (x) (a, 0)

FIGURE 8.2.1

The simplest way to solve the problem is to consider four boundary value problems Du(x, y) ¼ 0 with the boundary values on three of the sides being zero and the value of the given function on the fourth side. Solve each of the four problems, and sum the solutions. The result will be Du(x, y) ¼ 0, and all four boundary conditions will be satisfied. One such boundary value problem, which we now consider, will be Duðx; yÞ ¼ 0; uðx; 0Þ ¼ f1 ðxÞ; 0  x  a; uð0; yÞ ¼ 0; 0  y  b; We have

uðx; bÞ ¼ 0; 0  x  a; uða; yÞ ¼ 0; 0  y  b.

pﬃﬃﬃ pﬃﬃﬃ XðxÞ ¼ A cos lx þ B sin lx; Xð0Þ ¼ 0; XðaÞ ¼ 0:

Now Xð0Þ ¼ A; so A ¼ 0;

pﬃﬃﬃ XðaÞ ¼ B sin la ¼ 0:

To avoid having only the trivial solution, we must have integer n

pﬃﬃﬃ la ¼ np, so for each

n 2 p2 a2 is an eigenvalue for the boundary value problem ln ¼

X 00 ðxÞ þ lXðxÞ ¼ 0; Xð0Þ ¼ 0; XðaÞ ¼ 0

277

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

and Xn ðxÞ ¼ sin

npx a

is the corresponding eigenfunction. Now consider Y 00 ðyÞ  ln YðyÞ ¼ 0; We have determined that ln ¼ Y 00 ðyÞ 

n2 p 2 a2

YðbÞ ¼ 0:

so Eq. (3) is

n 2 p2 YðyÞ ¼ 0; a2

YðbÞ ¼ 0:

The solution to Y 00 ðyÞ 

n 2 p2 YðyÞ ¼ 0 a2

is

npy npy þ D sinh a a and the boundary condition Yn(b) ¼ 0 gives the solution     npb npb þ D sinh ¼0 C cosh a a Yn ðyÞ ¼ C cosh

so

  npb   C cosh npb a   ¼ C coth D¼ npb a sinh a

and

npy a a    npy npy npb sinh . ¼ Cn cosh  coth a a a

Yn ðyÞ ¼ Cn cosh

npy

þ Dn sinh

A more convenient way to write the solution is   npðb  yÞ Yn ðyÞ ¼ Fn sinh a as the eigenfunction for Y 00 ðyÞ þ as we verify in Exercise 2.

n2 p2 YðyÞ ¼ 0; a2

YðbÞ ¼ 0

(3)

8.2 Solving Laplace’s Equation on a Rectangle

Thus, we have un ðx; yÞ ¼ Xn ðxÞYn ðyÞ ¼ sin

  npy npy npx npb cosh sinh  coth a a a a

or sin

  npx npðb  yÞ ; sinh a a

and so uðx; yÞ ¼

∞ X

cn un ðx; yÞ

n¼1

¼

∞ X

cn sin

n¼1

  npx npy npy npb cosh sinh  coth a a a a

(4)

  npx npðb  yÞ . cn sin ¼ sinh a a n¼1 ∞ X

We now determine the constants cn so that f1 ðxÞ ¼ uðx; 0Þ ¼

∞ X

cn un ðx; 0Þ ¼

n¼1 ∞ X

npx

∞ X

cn sin

n¼1



  npx npðb  0Þ sinh a a



npb . sinh a a n¼1

Letting dn ¼ cn sinh npb a , we get ¼

cn sin

f1 ðxÞ ¼ uðx; 0Þ ¼

∞ X n¼1

dn sin

npx a

which is the Fourier expansion of f1(x) in a sine series. The coefficients are given by Z npx 2 a f1 ðxÞsin dn ¼ dx. a 0 a Thus

cn ¼

d n  ¼ npb sinh a

2 a

Z

a 0

npx f1 ðxÞsin dx a   npb sinh a

279

280

CHAPTER 8 Separation of Variables in Cartesian Coordinates

and uðx; yÞ ¼

∞ X

cn un ðx; yÞ ¼

n¼1

∞ X

cn sin

n¼1

npx a

 npðb  yÞ sinh a 

2 Z npx 3 2 a   f1 ðxÞsin dx7  ∞ 6 X npx npðb  yÞ a 6a 0 7   sin . ¼ sinh 6 7 npb 4 5 a a n¼1 sinh a In Exercise 3 we show the solution Duðx; yÞ ¼ 0; uðx; 0Þ ¼ 0; 0 < x < a; uð0; yÞ ¼ g1 ðyÞ; 0 < y < b; is uðx; yÞ ¼

(" Z ∞ X 2 n¼1

b

b 0

uða; yÞ ¼ 0; 0 < x < a; uða; yÞ ¼ 0; 0 < y < b

# npy npx dy cosh g1 ðyÞsin b b

" Z # ) 2 b npy npa npx npy g1 ðyÞsin  dy coth sinh sin . b 0 b b b b Example: We consider the case where a ¼ b ¼ 1, f1(x) ¼ x2, g1(y) ¼ y. Two cases are pertinent to this example. For the first case, Duðx; yÞ ¼ 0; uðx; 0Þ ¼ x2 ; 0 < x < 1; uð0; yÞ ¼ 0; 0 < y < 1;

uðx; 1Þ ¼ 0; 0 < x < 1; uð1; yÞ ¼ 0; 0 < y < 1

the solution is

2 Z 3 2 1 2 npx   x sin dx7  ∞ 6 X npx npð1  yÞ 1 61 0 7   : uðx; yÞ ¼ sinh 6 7sin np1 4 5 1 1 n¼1 sinh 1

For the second case, Duðx; yÞ ¼ 0; uðx; 0Þ ¼ 0; 0 < x < 1;

uðx; 1Þ ¼ 0; 0 < x < 1;

uð0; yÞ ¼ y; 0 < y < 1;

uð1; yÞ ¼ 0; 0 < y < 1

8.2 Solving Laplace’s Equation on a Rectangle

the solution is 82 3 Z 1 ∞ < X npy 5 npx 42 y sin uðx; yÞ ¼ dy cosh : 1 1 1 0 n¼1 2 4

2 1

9 npy 5 np1 npx= npy dy coth sinh sin y sin 1 1 1 ; 1 3

Z

1 0

and we would add the solutions.

EXERCISES 1. Describe the solutions to Eqs. (2a) and (2b) in the case l < 0. 2. Verify that   npðb  yÞ Yn ðyÞ ¼ sinh a is a solution to Y 00 ðyÞ þ

n 2 p2 YðyÞ ¼ 0; a2

YðbÞ ¼ 0:

3. Show that the solution to Duðx; yÞ ¼ 0; uðx; 0Þ ¼ 0; 0 < x < a; uð0; yÞ ¼ g1 ðyÞ; 0 < y < b;

uða; yÞ ¼ 0; 0 < x < a; uða; yÞ ¼ 0; 0 < y < b

is 82 3 Z b ∞ < X 2 npy npx 4 uðx; yÞ ¼ g1 ðyÞsin dy5cosh : b 0 b b n¼1 9 2 3 Z b 2 npy 5 npa npx= npy sin g1 ðyÞsin 4 dy coth sinh . b 0 b b b ; b

281

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

4. Find the solution to Duðx; yÞ ¼ 0; uðx; 0Þ ¼ 0; 0 < x < a;

uðx; bÞ ¼ f2 ðxÞ; 0 < x < a;

uð0; yÞ ¼ 0; 0 < y < b;

uða; yÞ ¼ 0; 0 < y < b.

5. Find the solution to Duðx; yÞ ¼ 0; uðx; 0Þ ¼ 0; 0 < x < a; uð0; yÞ ¼ 0; 0 < y < b;

uðx; bÞ ¼ 0; 0 < x < a; uða; yÞ ¼ g2 ðyÞ; 0 < y < b.

6. Find the solution to Duðx; yÞ ¼ 0; uðx; 0Þ ¼ 0; 0 < x < p;

uðx; pÞ ¼ sin x; 0 < x < p;

uð0; yÞ ¼ 0; 0 < y < p;

uðp; yÞ ¼ 0; 0 < y < p.

7. Find the solution to Duðx; yÞ ¼ 0; uðx; 0Þ ¼ 0; 0 < x < p; uð0; yÞ ¼ cos y; 0 < y < p;

uðx; pÞ ¼ 0; 0 < x < p; uðp; yÞ ¼ 0; 0 < y < p.

8. Find the solution to Duðx; yÞ ¼ 0; uðx; 0Þ ¼ sin x  sin 4 x; 0 < x < p;

uðx; pÞ ¼ 0; 0 < x < p;

uð0; yÞ ¼ 0; uðp; yÞ ¼ 0: 9. Find the solution to Duðx; yÞ ¼ 0; uðx; 0Þ ¼ 0; 0 < x < p;

uðx; pÞ ¼ p; 0 < x < p;

uð0; yÞ ¼ y; uðp; yÞ ¼ y.

10. Find the solution to Laplace’s equation on the semiinfinite strip 0 < x < 1, 0 < y < ∞ given by Duðx; yÞ ¼ 0;

8.3 Laplace’s Equation on a Cube

283

uðx; 0Þ ¼ f ðxÞ; 0 < x < 1; uð0; yÞ ¼ 0; 0 < y < ∞;

uð1; yÞ ¼ 0; 0 < y < ∞;

lim uðx; yÞ ¼ 0:

y/∞

11. Use separation of variables to write the ODEs necessary to solve the following problems: a. Du(x, y) ¼ u(x, y) b. Du(x, y) þ ux(x, y) ¼ u(x, y) c. Du(x, y) ¼ ux(x, y) ¼ uy(x, y). 12. We shall see that the steady state for the heat equation in two dimensions is given by Txx ðx; yÞ þ Tyy ðx; yÞ ¼ 0 Find the steady-state temperature distribution for the square [0, p]  [0, p] if the boundary conditions are Tx ð0; yÞ ¼ Tx ðp; yÞ ¼ 0 ðwhich is the case for insulated edgesÞ

and

Tðx; 0Þ ¼ 0; Tðx; pÞ ¼ T0 sin x.

8.3 LAPLACE’S EQUATION ON A CUBE We now consider Laplace’s equation on a cube. We solve the problem on a cube whose edges are each of length p instead of an arbitrary parallelepiped to make the computations somewhat less cumbersome. We model the solution on the twodimensional case. Major differences are that the boundary consists of six faces rather than four edges, and the boundary conditions are functions of two variables. Accordingly, we will need to solve boundary value problems of the type Duðx; y; zÞ ¼

v2 u v2 u v2 u þ þ ¼ 0; vx2 vy2 vz2

uðx; y; zÞ ¼ 0

if

0 < x < p; 0 < y < p; 0 < z < p;

0 < z < p;

x ¼ 0; x ¼ p; y ¼ 0; y ¼ p; z ¼ p; uðx; y; 0Þ ¼ f ðx; yÞ.

The approach is identical in spirit to the two-dimensional case. We hypothesize that u(x, y, z) ¼ X(x)Y(y)Z(z) so that v2 u ¼ X 00 ðxÞYðyÞZðzÞ; vx2

v2 u ¼ XðxÞY 00 ðyÞZðzÞ; vy2

v2 u ¼ XðxÞYðyÞZ 00 ðzÞ. vz2

284

CHAPTER 8 Separation of Variables in Cartesian Coordinates

Then v2 u v2 u v2 u þ þ ¼ X 00 ðxÞYðyÞZðzÞ þ XðxÞY 00 ðyÞZðzÞ þ XðxÞYðyÞZ 00 ðzÞ ¼ 0: vx2 vy2 vz2 Dividing by X(x)Y(y)Z(z) gives X 00 ðxÞ Y 00 ðyÞ Z 00 ðzÞ þ þ ¼0 XðxÞ YðyÞ ZðzÞ so X 00 ðxÞ Y 00 ðyÞ Z 00 ðzÞ þ ¼ . XðxÞ YðyÞ ZðzÞ

(1)

The left-hand side of Eq. (1) is a function of x and y, and the right-hand side is a function of z so each must be a constant that we denote a. Thus X 00 ðxÞ Y 00 ðyÞ þ ¼ a and XðxÞ YðyÞ

Z 00 ðzÞ ¼ a. ZðzÞ

Since X 00 ðxÞ Y 00 ðyÞ þ ¼ a; XðxÞ YðyÞ then X 00 ðxÞ Y 00 ðyÞ ¼a . XðxÞ YðyÞ

(2) 00

ðxÞ

Then, reasoning as before, each side of Eq. (2) must be a constant. Let b ¼ XXðxÞ . We then have X 00 ðxÞ ¼ b; XðxÞ so X 00 ðxÞ  bXðxÞ ¼ 0: Also, Y 00 ðyÞ X 00 ðxÞ ¼a ¼ab YðyÞ XðxÞ so Y 00 ðyÞ  ða  bÞYðyÞ ¼ 0: Collecting the equations and noting the boundary conditions, we have X 00 ðxÞ  bXðxÞ ¼ 0;

Xð0Þ ¼ XðpÞ ¼ 0

(3)

8.3 Laplace’s Equation on a Cube

Y 00 ðyÞ  ða  bÞYðyÞ ¼ 0; Z 00 ðzÞ þ aZðzÞ ¼ 0;

Yð0Þ ¼ YðpÞ ¼ 0

(4)

ZðpÞ ¼ 0:

(5)

We determine appropriate values for the constants. The equation X 00 ðxÞ  bXðxÞ ¼ 0;

Xð0Þ ¼ XðpÞ ¼ 0

because of the boundary conditions must have b > 0. Experience with the twodimensional case prompts us to take b ¼ n2 . We thus have Eq. (3) is X 00 ðxÞ þ n2 XðxÞ ¼ 0;

Xð0Þ ¼ XðpÞ ¼ 0:

We saw in the two-dimensional case, the solution for this equation is Xn ðxÞ ¼ sin nx. Likewise, for the equation Y 00 ðyÞ  ða  bÞYðyÞ ¼ 0;

Yð0Þ ¼ YðpÞ ¼ 0

we must have e(ab) > 0, and we take ab ¼  m2. We then have for Eq. (4) Y 00 ðyÞ þ m2 YðyÞ ¼ 0;

Yð0Þ ¼ YðpÞ ¼ 0

for which the solution is Ym ðyÞ ¼ sin my. Note that a ¼ b  m2 ¼ m2  n2 so the equation Z 00 ðzÞ þ aZðzÞ ¼ 0; is

ZðpÞ ¼ 0

Z 00 ðzÞ  m2 þ n2 ZðzÞ ¼ 0;

ZðpÞ ¼ 0

and we saw in the two-dimensional case the solution for this equation is pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Zmn ðzÞ ¼ sinh m2 þ n2 ðp  zÞ. Thus, the solution to Duðx; y; zÞ ¼

v2 u v2 u v2 u þ þ ¼ 0; 0 < x < p; 0 < y < p; 0 < z < p; 0 < z < p; vx2 vy2 vz2

uðx; y; zÞ ¼ 0

if

x ¼ 0; x ¼ p; y ¼ 0; y ¼ p; z ¼ p

is uðx; y; zÞ ¼

∞ X ∞ X m¼1 n¼1

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ amn sin nx sin my sinh m2 þ n2 ðp  zÞ.

285

286

CHAPTER 8 Separation of Variables in Cartesian Coordinates

We now determine the constants amn so that u(x, y, 0) ¼ f (x, y). We have uðx; y; 0Þ ¼

∞ X ∞ X

amn sin nx sin my sinh

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ m2 þ n2 ðp  0Þ

m¼1 n¼1

¼

∞ X ∞ X

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ amn sin nx sin my sinh p m2 þ n2 .

m¼1 n¼1

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Let cmn ¼ amn sinh p m2 þ n2 . Then f ðx; yÞ ¼ uðx; y; 0Þ ¼

∞ X ∞ X

cmn sin nx sin my.

m¼1 n¼1

To satisfy the boundary condition u(x, y, 0) ¼ f (x, y) we must choose cmn

4 ¼ 2 p

Z p Z 0

p 0

Then

amn ¼

 f ðx; yÞsin nx dx sin my dy.

cmn pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ sinh p m2 þ n2

4 p2

Z p Z

 f ðx; yÞsin nx dx sin my dy 0 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ . sinh p m2 þ n2

0

p

Thus, uðx; y; zÞ ¼

∞ X ∞ X

amn sin nx sin my sinh

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ m2 þ n2 ðp  zÞ

m¼1 n¼1

2 4 Z p Z 2 ∞ X ∞ 6 X 6p 6 ¼ 6 m¼1 n¼1 4

0

p 0

 sin nx sin my sinh

 f ðx; yÞsin nx dx sin my dy

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ sinh p m2 þ n2

3 7 7 7 7 5

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ m2 þ n2 ðp  zÞ.

EXERCISES 1. Solve the initial value problem Duðx; y; zÞ ¼

v2 u v2 u v2 u þ þ ¼ 0; vx2 vy2 vz2

0 < x < p; 0 < y < p; 0 < z < p;

0 < z < p;

8.4 Solving the Wave Equation in One Dimension by Separation of Variables

uðx; y; zÞ ¼ 0 if

287

x ¼ 0; x ¼ p; y ¼ 0; y ¼ p; z ¼ p; uðx; y; 0Þ ¼ f ðx; yÞ

for the following values of f (x, y). a. f (x, y) ¼ xy. b. f (x, y) ¼ y2. 2. Use separation of variables to solve the initial value problem Duðx; y; zÞ ¼

v2 u v2 u v 2 u þ þ ¼ 0; vx2 vy2 vz2

0 < x < p; 0 < y < p; 0 < z < p;

vu ¼ 0; x ¼ 0; x ¼ p; vx

0 < z < p;

u ¼ 0; y ¼ 0; y ¼ p; z ¼ p.

3. Use separation of variables to solve v2 u v2 u v 2 u þ þ ¼ u; 0 < x < p; 0 < y < p; 0 < z < p; vx2 vy2 vz2 u¼0 if

if

z ¼ p;

x ¼ p; y ¼ 0; z ¼ 0;

uy ¼ 0

if

y ¼ p;

0 < z < p; uz ¼ 0

ux ðx; y; 0Þ ¼ f ðx; yÞ.

8.4 SOLVING THE WAVE EQUATION IN ONE DIMENSION BY SEPARATION OF VARIABLES Consider the wave equation utt ðx; tÞ  c2 uxx ðx; tÞ ¼ 0;

uð0; tÞ ¼ uðL; tÞ ¼ 0;

uðx; 0Þ ¼ f ðxÞ; ut ðx; 0Þ ¼ gðxÞ.

Note that in the wave equation, we need two initial conditions because of the utt(x, t) term and two boundary conditions because of the uxx(x, t) term. Suppose that uðx; tÞ ¼ XðxÞTðtÞ. Then utt ðx; tÞ ¼ XðxÞT 00 ðtÞ and

uxx ðx; tÞ ¼ X 00 ðxÞTðtÞ.

Thus utt ðx; tÞ  c2 uxx ðx; tÞ ¼ XðxÞT 00 ðtÞ  c2 X 00 ðxÞTðtÞ ¼ 0; or XðxÞT 00 ðtÞ ¼ c2 X 00 ðxÞTðtÞ.

288

CHAPTER 8 Separation of Variables in Cartesian Coordinates

Dividing by X(x) T(t) gives T 00 ðtÞ X 00 ðxÞ ¼ c2 . TðtÞ XðxÞ

(1)

The left-hand side of Eq. (1) is a function only of t, and the right-hand side is a function only of x, so the common value is a constant that we denote a2. (We show the constant must be negative in Exercise 1.) So T 00 ðtÞ ¼ a2 TðtÞ

or

T 00 ðtÞ þ a2 TðtÞ ¼ 0

the solution for which is TðtÞ ¼ A sin at þ B cos at. Also, c2

X 00 ðxÞ ¼ a2 XðxÞ

or X 00 ðxÞ þ

a2 XðxÞ ¼ 0 c2

the solution for which is ax ax þ E cos . c c The boundary values will determine the value of  a2. Since XðxÞ ¼ D sin

uð0; tÞ ¼ Xð0ÞTðtÞ ¼ 0 we must have Xð0Þ ¼ E ¼ 0 so XðxÞ ¼ D sin

ax . c

Then uðL; tÞ ¼ XðLÞTðtÞ ¼ 0

forces

XðLÞ ¼ D sin

To avoid having only the trivial solution, we must have an L ¼ np c where n is an integer. Thus an ¼ where n is an integer.

npc L

aL ¼ 0: c

8.4 Solving the Wave Equation in One Dimension by Separation of Variables

Many derivations of the wave equation in one dimension will substitute c ¼ 1/y. We now do that so as to conform to the more common expression of the solution. Thus, we have an ¼

np yL

where n is an integer. So Tn ðtÞ ¼ An sin

npt npt þ Bn cos yL yL

and npx L

Xn ðxÞ ¼ Dn sin and thus, by superposition, uðx; tÞ ¼

X

un ðx; tÞ ¼

X

n

Xn ðxÞTn ðtÞ ¼

X

n

n

  npx npt npt Dn sin An sin þ Bn cos . L yL yL

We use the initial conditions to determine the constants. We have X npx . Bn Dn sin f ðxÞ ¼ uðx; 0Þ ¼ L n Let bn ¼ BnDn. Then f ðxÞ ¼ uðx; 0Þ ¼

X

bn sin

n

and bn is determined by bn ¼ We also have ut ðx; tÞ ¼

Xnp n

yL

2 L

 An cos

Z

L

f ðxÞsin

0

npx L

npx dx. L

     npx npt np npt  Dn sin Bn sin yL yL yL L

so gðxÞ ¼ ut ðx; 0Þ ¼

npx Xnp An Dn sin . yL L n

Let gn ¼

np An D n yL

so

An D n ¼

gn yL . np

289

290

CHAPTER 8 Separation of Variables in Cartesian Coordinates

Then gðxÞ ¼

X

gn sin

n

and so

npx L

and

2 gn ¼ L

Z

L

gðxÞsin 0

npx dx L

  npx npt npt An sin þ Bn cos L yL yL n npx npt npx npt X An Dn sin þ Bn Dn sin ¼ sin cos L yL L yL n npx npt Xgn yL npx npt sin þ bn sin ¼ sin cos np L yL L yL n

uðx; tÞ ¼

X

Dn sin

2 0 1 npx npt X  yL  2 Z L npx 4 @ A gðxÞsin ¼ dx sin sin np L 0 L L yL n 0 2 þ@ L

Z

L 0

1 3   npx A npx npt 5 f ðxÞsin dx sin . cos L L yL

We should still check that the series converges, and the series is a solution. Example: Suppose that we initially distort the string by lifting it at the center of the interval by lifting it by the amount a. See Fig. 8.4.1. The boundary conditions are then uð0; tÞ ¼ uðL; tÞ ¼ 0 and one initial condition (because the string is at rest immediately before it is released) is ut ðx; 0Þ ¼ 0:

(L / 2, α )

0

FIGURE 8.4.1

L

8.4 Solving the Wave Equation in One Dimension by Separation of Variables

The other initial condition, u(x, 0), is the equation of the graph in Fig. 8.4.1. In Exercise 2 we show that this is 8 2ax L > > 0x < L 2 . (2) u ðx; 0Þ ¼ > > : 2a ðL  xÞ L  x  L L 2 Note that u(x, 0) is f (x) in the derivation above. Thus, in the formula we derived "Z # Z Z L L=2 2 L npx 2 2ax npx 2a npx f ðxÞsin bn ¼ dx ¼ sin dx þ ðL  xÞsin dx . L 0 L L 0 L L L L=2 L One can check with a computer algebra system that the value of this integral is np 8a bn ¼ 2 2 sin . n p 2 Since g(x) ¼ ut(x, 0) ¼ 0, we have Z 2 L npx dx ¼ 0: gðxÞsin gn ¼ L 0 L Substituting into npx npt Xgn yL npx npt uðx; tÞ ¼ sin þ bn sin sin cos np L yL L yL n gives ∞  np npx npt X 8a . sin sin cos n2 p 2 2 L yL n¼1   Note that when n is even, sin np 2 ¼ 0.

uðx; tÞ ¼

Example: We compute the kinetic and potential energy of a vibrating string. Let u(x, t) denote the vertical distance of the string at point x at time t. Kinetic energy K is computed according to 1 K ¼ mv2 . 2 Divide the interval [0, L] into n equal subintervals of length Dx by inserting x0, x1, x2,.,xn with 0 ¼ x0 < x1 < x2 < / < xn ¼ L and Dx ¼ xixi1. The mass of the string between xi1 and xi is rDx and the velocity at xi is

vuðx; tÞ

. vt x¼xi ;

291

292

CHAPTER 8 Separation of Variables in Cartesian Coordinates

Thus the kinetic energy of the string is approximately given by the Riemann sum

 2 n X 1 vuðx; tÞ

rDx 2 vt x¼xi ; i¼1 so, in the limit as Dx / 0 we get the exact value of the kinetic energy is  Z  r L vuðx; tÞ 2 K¼ dx. 2 0 vt To compute the potential energy, we appeal to Hooke’s law. According to Hooke’s the force a spring exerts is proportional to the distance the spring is distorted from equilibrium. If the spring constant is k and the distance the spring is distorted is x, we have f ¼ kx. The work performed in moving a unit mass from a displaced distance a to a displaced distance b is Z b Z b

1 f ðxÞdx ¼ kx dx ¼  k b2  a2 W¼ 2 a a which is the difference in potential energy between the points. If T is the tension of the nonstretched string then we have that the potential energy is approximately  n n  1 T X T X uðxi ; tÞ  uðxi1 ; tÞ 2 ðuðxi ; tÞ  uðxi1 ; tÞÞ2 ¼ Dx. Uz 2 Dx i¼1 2 i¼1 Dx Taking the limit as Dx / 0, we get Z   T L vu 2 U¼ dx. 2 0 vx

EXERCISES 1. In Eq. (1), we derived T 00 ðtÞ X 00 ðxÞ ¼ c2 . TðtÞ XðxÞ 00

ðxÞ Since T 00 ðtÞ=TðtÞ is a function only of t and c2 XXðxÞ is a function only of x, it must be that this is a constant. Show that this constant is negative. 2. Show that the equation given for u(x, 0) in the example is valid. 3. Find the solution for the wave equation if we have the same conditions as in the example, except instead of plucking the string at the point x ¼ L/2 we pluck it at the point x ¼ L/3.

8.4 Solving the Wave Equation in One Dimension by Separation of Variables

4. Solve the wave equation for the following initial conditions on 0 < x < L: a. u(x, 0) ¼ 0, ut(x, 0) ¼ 2. b. u(x, 0) ¼ sin px/L, ut(x, 0) ¼ 0. c. u(x, 0) ¼ x(Lx), ut(x, 0) ¼ 2. d. u(x, 0) ¼ sin px/L, ut(x, 0) ¼ sin px/L. e. u(x, 0) ¼ 0, ut(x, 0) ¼ x. 5. Verify that the wave equation c2 uxx  utt ¼ hðx; tÞ; with initial conditions uðx; 0Þ ¼ f ðxÞ; ut ðx; 0Þ ¼ gðxÞ has as its solution Z xþct f ðx þ ctÞ þ f ðx  ctÞ 1 gðqÞdq þ uðx; tÞ ¼ 2 2c xct # Z t " Z xþcðtsÞ 1 þ hðq; sÞdq ds. 2c 0 xcðtsÞ 6. Consider an infinitely long string that is released from rest with displacement 2 f ðxÞ ¼ ex . Show that a solution to the wave equation with these initial conditions is i 2 2 1h uðx; tÞ ¼ eðxctÞ þ eðxþctÞ . 2 7. Show that energy is conserved in a vibrating string. To do this, show the time derivative of the energy we derived for the vibrating string in the last example is zero. Take T ¼ rc2. 8. Show that the solution to utt ¼ c2 uxx  g; 0 < x < p; t > 0; uð0; tÞ ¼ 0; uðp; tÞ ¼ 0; uðx; 0Þ ¼ 0; ut ðx; 0Þ ¼ 0 is

# " ∞ 4g X sinð2n  1Þx 1 cosð2n  1Þct  xðp  xÞ . uðx; tÞ ¼ 2 c n¼1 ð2n  1Þ3 8

This models a string that is initially at rest and at equilibrium that when released falls due to the force of gravity. 9. In this exercise we solve utt ðx; tÞ ¼ uxx ðx; tÞ  2aut ðx; tÞ;

uð0; tÞ ¼ uðp; tÞ ¼ 0;

uðx; 0Þ ¼ 0; ut ðx; 0Þ ¼ b.

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

a. Show that separation of variables leads to the ODEs X 00 ðxÞ þ lXðxÞ ¼ 0; Xð0Þ ¼ 0; XðpÞ ¼ 0 T 00 ðtÞ þ 2aT 0 ðtÞ þ lTðtÞ ¼ 0; Tð0Þ ¼ 0: b. Show that ln ¼ n2 and that the solutions to the resulting ODEs are Xn ðxÞ ¼ An sin nx pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  Tn ðtÞ ¼ Bn eat sin n2  a2 t . c. Conclude that uðx; tÞ ¼

∞ X

Cn eat sin

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  n2  a2 t sin nx.

n¼1

d. Use the initial condition ut(x, 0) ¼ b to show Cn ¼

2b 1  ð1Þn \$ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ . p n n 2  a2

10. The telegraph equation is utt ðx; tÞ þ aut ðx; tÞ ¼ buxx ðx; tÞ  cuðx; tÞ   where a, b, c > 0. Show that the substitution uðx; tÞ ¼ exp at2 yðx; tÞ yields the equation   1 2 a  c yðx; tÞ. ytt ðx; tÞ ¼ byxx ðx; tÞ þ 4

8.5 SOLVING THE WAVE EQUATION IN TWO DIMENSIONS IN CARTESIAN COORDINATES BY SEPARATION OF VARIABLES In this section we solve the wave equation   2 v u v2 u 1 v2 u ¼ þ Duðx; y; tÞ ¼ vx2 vx2 c2 vt2

0 < x < a; 0 < y < b.

(1)

8.5 Solving the Wave Equation in Two Dimensions in Cartesian Coordinates

The method combines the ideas of what we did in solving Laplace’s equation in two dimensions and what we did for the wave equation in one dimension. As we did with Laplace’s equation, we take a ¼ b ¼ p to make the computations less cumbersome. Our boundary conditions will be uðx; 0Þ ¼ uðx; pÞ ¼ uð0; yÞ ¼ uðp; yÞ ¼ 0 for all t and the initial conditions will be uðx; y; 0Þ ¼ f ðx; yÞ; ut ðx; y; 0Þ ¼ gðx; yÞ. We hypothesize uðx; y; tÞ ¼ XðxÞYðyÞTðtÞ. Then Eq. (1) is X 00 ðxÞYðyÞTðtÞ þ XðxÞY 00 ðyÞTðtÞ ¼

1 XðxÞYðyÞT 00 ðtÞ. c2

Dividing by X(x) Y(y) T(t) gives   00 X ðxÞ Y 00 ðyÞ 1 T 00 ðtÞ þ ¼ 2 . XðxÞ YðyÞ c TðtÞ

(2)

The left-hand side of Eq. (2) is a function of x and y, and the right-hand side is a function of t, so each must be a constant l. We show in Exercise 1 that this constant must be negative, so we let  a2 ¼ l. We then have 1 T 00 ðtÞ ¼ a2 c2 TðtÞ

so T 00 ðtÞ þ c2 a2 TðtÞ ¼ 0:

We also have   00 X ðxÞ Y 00 ðyÞ þ ¼ a2 XðxÞ YðyÞ

so

X 00 ðxÞ Y 00 ðyÞ ¼ a2  . XðxÞ YðyÞ

(3)

(4)

The left-hand side of Eq. (4) is a function of x, and the right-hand side is a function of y, so each must be a constant. As in the case of Laplace’s equation on a square, we must have X 00 ðxÞ ¼ m2 XðxÞ

and

XðxÞ þ m2 XðxÞ ¼ 0

and

Y 00 ðyÞ ¼ n2 YðyÞ

where m2 þ n2 ¼ a2. Thus we have YðyÞ þ n2 YðyÞ ¼ 0:

We also have the boundary conditions Xð0Þ ¼ XðpÞ ¼ 0

and

Yð0Þ ¼ YðpÞ ¼ 0:

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

As we have seen on several occasions, the solutions are Xm ðxÞ ¼ sin mx; Yn ðyÞ ¼ sin ny. Returning to

T 00 ðtÞ þ c2 a2 TðtÞ ¼ T 00 ðtÞ þ c2 m2 þ n2 TðtÞ ¼ 0

we have Tmn ðtÞ ¼ amn cos

h pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  i h pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  i c m2 þ n2 t þ bmn sin c m2 þ n2 t .

Thus, ∞ X

uðx; y; tÞ ¼

um:n ðx; y; tÞ ¼

m;n¼1

∞ n X

amn cos

h pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  i c m 2 þ n2 t

m;n¼1

þbmn

h pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ  io sin c m2 þ n2 t sin mx sin ny.

We use the initial conditions to determine amn and bmn. We have f ðx; yÞ ¼ uðx; y; 0Þ ¼

∞ X

amn sin mx cos ny

m;n¼1

so amn

4 ¼ 2 p

Z p Z 0

p

 f ðx; yÞsin ny dy sin mx dx.

0

We also have ut ðx; y; 0Þ ¼

∞ n h pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ io X sin mx sin ny. bmn c m2 þ n2 m;n¼1

Letting cmn

h pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ i ¼ bmn c m2 þ n2 we have ∞ X

gðx; yÞ ¼ ut ðx; y; 0Þ ¼

cmn sin mx sin ny

m;n¼1

and cmn ¼

4 p2

Z p Z 0

 gðx; yÞsin ny dy sin mx dx

0

so cmn bmn ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ c m2 þ n 2

p

4 p2

Z p Z 0

0

p

 gðx; yÞsin ny dy sin mx dx pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ . c m 2 þ n2

Substituting the values of amn and bmn into Eq. (5) gives the solution.

(5)

8.6 Solving the Heat Equation in One Dimension Using Separation

Example: Solve the wave equation on the rectangle 0  x  p, 0  y  p given that f ðx; yÞ ¼ uðx; y; 0Þ ¼ xðp  xÞyðp  yÞ

and

gðx; yÞ ¼ ut ðx; y; 0Þ ¼ 0:

Since g(x, y) ¼ 0, then bmn ¼ 0. Also  Z p Z p 4 f ðx; yÞsin ny dy sin mx dx amn ¼ 2 p 0 0  Z p Z p 4 ¼ 2 xðp  xÞyðp  yÞsin ny dy sin mx dx. p 0 0

EXERCISES Solve the wave equation on the rectangle 0 < x < p, 0 < y < p given the following initial data (do not attempt to simplify the expressions for amn and bmn): f (x, y) ¼ u(x, y, 0) ¼ 1  x2  y2; g(x, y) ¼ ut(x, y, 0) ¼ sinx cosy. f (x, y) ¼ u(x, y, 0) ¼ 1  x siny; g(x, y) ¼ ut(x, y, 0) ¼ x  2y. f (x, y) ¼ u(x, y, 0) ¼ 1  x  y2; g(x, y) ¼ ut(x, y, 0) ¼ sinx. f (x, y) ¼ u(x, y, 0) ¼ sinx siny; g(x, y) ¼ ut(x, y, 0) ¼ 0. (It is possible to simplify this problem with reasonable effort.) 5. Solve utt(x, y, t) ¼ uxx(x, y, t) þ uyy(x, y, t), 0  x  p, 0  y p, with boundary conditions u(x, 0, t) ¼ u(x, p, t) ¼ u(0, y, t) ¼ u(p, y, t) ¼ 0 and initial conditions u(x, y, 0) ¼ x(x  p)y(y  p), ut(x, y, 0) ¼ 0.

1. 2. 3. 4.

8.6 SOLVING THE HEAT EQUATION IN ONE DIMENSION USING SEPARATION OF VARIABLES NO HEAT SOURCE In one dimension the heat equation with no heat source is v2 uðx; tÞ 1 vuðx; tÞ . (1) ¼ 2 vx2 a vt Note: By using the change of variables s ¼ at, the 1/a2 term in Eq. (1) can be  eliminated. In some examples we make this substitution.  v2 uðx; tÞ To solve the equation,we need two boundary  conditions because of vx2 vuðx; tÞ . For our initial considerations we let and one initial condition because of vt these be uð0; tÞ ¼ 0; uð0; LÞ ¼ 0; uðx; 0Þ ¼ f ðxÞ.

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

To apply separation of variables, we assume that the solution can be written as the product of functions of a single variable. In this problem we assume uðx; tÞ ¼ XðxÞTðtÞ.

(2)

From Eqs. (1) and (2), we get X 00 ðxÞTðtÞ ¼

1 XðxÞT 0 ðtÞ. a2

(3)

Dividing Eq. (3) by X(x) T(t) gives X 00 ðxÞ 1 T 0 ðtÞ ¼ 2 . XðxÞ a TðtÞ

(4)

Eq. (4) again illustrates the salient point of the technique of separation of variables. We have one side of the equation as a function of one set of variables (in this case, x) and the other side as a totally different set of variables (in this case, t). Thus, each side must be the same constant. Suppose X 00 ðxÞ ¼ b and XðxÞ

1 T 0 ðtÞ ¼ b. a2 TðtÞ

Then we must solve X 00 ðxÞ  bXðxÞ ¼ 0

T 0 ðtÞ ¼ a2 b. TðtÞ

and

The solution to the second equation is lnðTðtÞÞ ¼ a2 bt þ C

or TðtÞ ¼ Tð0Þea bt . 2

We now determine the sign of b. If b ¼ 0, then T(t) is constant, which is impossible unless u(x, 0) ¼ f (x) ¼ 0. This yields the trivial solution u(x, t) ¼ 0. If b > 0 then T(t) grows exponentially, which is impossible with no heat source. Thus, b < 0, and we set b ¼  l2. We must now solve X 00 ðxÞ þ l2 XðxÞ ¼ 0

(5a)

T 0 ðtÞ þ a2 l2 TðtÞ ¼ 0:

(5b)

and

The solutions to Eqs. (5a) and (5b) are XðxÞ ¼ a1 sinðlxÞ þ a2 cosðlxÞ

(6a)

and TðtÞ ¼ a3 ea

l t

2 2

(6b)

8.6 Solving the Heat Equation in One Dimension Using Separation

respectively. We use the boundary conditions and initial condition to determine a1, a2, and a3. We have uð0; tÞ ¼ Xð0ÞTðtÞ ¼ 0 uðL; tÞ ¼ XðLÞTðtÞ ¼ 0: Since T(t) ¼ 0 yields the trivial solution, we must have X(0) ¼ X(L) ¼ 0. In Eq. (6a), when x ¼ 0 we have Xð0Þ ¼ a1 sinðl0Þ þ a2 cosðl0Þ ¼ a2 ¼ 0: So XðxÞ ¼ a1 sinðlxÞ and thus when x ¼ L we have XðLÞ ¼ a1 sinðlLÞ ¼ 0: One possibility is that a1 ¼ 0, but that gives the trivial solution u(x, t) ¼ 0. The only other possibility is lL ¼ np where n is an integer. Thus, we must have ln ¼ np/L where n is an integer, and so npx Xn ðxÞ ¼ sin . L Thus    anp2 a2 l2n t ¼ exp  t Tn ðtÞ ¼ e L and for any positive integer n we have

npx  anp2  t un ðx; tÞ ¼ Xn ðxÞ Tn ðtÞ ¼ sin exp  L L

is a solution to v2 uðx; tÞ 1 vuðx; tÞ ¼ 2 vx2 a vt By superposition ∞ X

uð0; tÞ ¼ uðL; tÞ ¼ 0:

cn un ðx; tÞ

n¼1

is also a solution to Eq. (7). We use the initial condition u(x, 0) ¼ f (x) to determine the cn’s. Now uðx; 0Þ ¼

∞ X

cn un ðx; 0Þ

n¼1 ∞ npx  anp2  X npx cn sin 0 ¼ cn sin ¼ exp  ¼ f ðxÞ. L L L n¼1 n¼1 ∞ X

(7)

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

We need to express f (x) as a sine series. We have f ðxÞ ¼

∞ X

cn sin

n¼1

npx L

so 2 cn ¼ L

Z

L 0

f ðxÞsin

npx L

dx.

Thus, uðx; tÞ ¼

∞ X n¼1

cn sin

npx  anp2  t exp  L L

" Z # ∞ npx npx  anp2  X 2 L f ðxÞsin t ¼ dx sin exp  L 0 L L L n¼1 is the solution to the equation v2 uðx; tÞ 1 vuðx; tÞ uð0; tÞ ¼ uðL; tÞ ¼ 0 uðx; 0Þ ¼ f ðxÞ ¼ 2 vx2 a vt if indeed u(x, t) can be written as X(x) T(t). We leave it as Exercise 1 to show the solution we have asserted is valid.

THE INITIAL CONDITION IS THE DIRAC-DELTA FUNCTION We now consider the equation v2 uðx; tÞ 1 vuðx; tÞ ¼ 2 uðp; tÞ ¼ uðp; tÞ ¼ 0 uðx; 0Þ ¼ d0 ðxÞ. vx2 a vt This equation would be appropriate to model a sudden release of pollution where the boundary of the region is absorbing. (The heat equation is also models diffusion.) To do this problem, we use the Fourier expansion of the Dirac-delta function, dx0 ðxÞ. We note that Z 1 p 1 dx ðxÞcosðnxÞdx ¼ cosðnx0 Þ an ¼ p p 0 p Z 1 p 1 bn ¼ dx0 ðxÞsinðnxÞdx ¼ sinðnx0 Þ p p p so dx0 ðxÞ ¼

∞ 1 1X þ ½cosðnx0 ÞcosðnxÞ þ sinðnx0 ÞsinðnxÞ. 2p p n¼1

8.6 Solving the Heat Equation in One Dimension Using Separation

Now cosðnx0 ÞcosðnxÞ þ sinðnx0 ÞsinðnxÞ ¼ cosðnx  nx0 Þ ¼ cos½nðx  x0 Þ so the Fourier expansion of dx0 ðxÞ is ∞ 1 1X þ cos½nðx  x0 Þ 2p p n¼1

and the Fourier expansion of d0(x) is ∞ 1 1X cosðnxÞ. þ 2p p n¼1

As in the previous example, we assume uðx; tÞ ¼ XðxÞTðtÞ. Since we know T(0) ¼ 1, we have from previous examples, u(x, 0) ¼ X(x). But uðx; 0Þ ¼ d0 ðxÞ ¼

∞ 1 1X cosðnxÞ. þ 2p p n¼1

As before, Tn ðtÞ ¼ ea

2 2

n t

so that uðx; tÞ ¼

∞ X

Xn ðxÞTn ðtÞ ¼ X0 ðxÞT0 ðtÞ þ

n¼0

¼

∞ X

Xn ðxÞTn ðtÞ

n¼1

∞ 2 2 1 1X cosðnxÞea n t þ 2p p n¼1

where we use the fact that the coefficients have been determined by the initial condition. We note that another approach to the problem is to use the fundamental solution, which we derive in Chapter 11, to get  2 1 x . uðx; tÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp  kt 4pkt

EXERCISES 1. Solve the heat equation vuðx; tÞ v2 uðx; tÞ ¼ a2 vt vx2

0 < x < p; t > 0

301

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

for the following initial and boundary conditions: a. u(0, t) ¼ u(p, t) ¼ 0; u(x, 0) ¼ 1cos x. b. u(0, t) ¼ u(p, t) ¼ 0; u(x, 0) ¼ sin2(x). c. u(0, t) ¼ u(p, t) ¼ 0; u(x, 0) ¼ x(px). d. ux(0, t) ¼ ux(p, t) ¼ 0; u(x, 0) ¼ x(px). 2. Solve vuðx; tÞ v2 uðx; tÞ  uðx; tÞ 0 < x < p; t > 0 ¼ vt vx2 uð0; tÞ ¼ 0;

vuðp; tÞ ¼ 0; vx

uðx; 0Þ ¼ sin x.

3. Solve the heat equation vuðx; tÞ v2 uðx; tÞ ¼ vt vx2

0 < x < p; t > 0

vuð0; tÞ vuðp; tÞ ¼ 0; ¼ 0; vx vx

uðx; 0Þ ¼ x.

4. Solve the heat equation vuðx; tÞ v2 uðx; tÞ ¼ vt vx2

 p < x < p; t > 0;

vuðp; tÞ vuðp; tÞ ¼ ; vx vx

uðp; tÞ ¼ uðp; tÞ;

uðx; 0Þ ¼ cd0 ðxÞ:

5. In this exercise we solve by separation of variables the equation vuðx; tÞ v2 uðx; tÞ þ auðx; tÞ ¼ 0  vt vx2 uð0; tÞ ¼ 0;

0 < x < p; t > 0;

vuðp; tÞ ¼ 0; uðx; 0Þ ¼ xðp  xÞ. vx

a. Show that separation of variables yields the equations X 00 ðxÞ þ l2 XðxÞ ¼ 0 T 0 ðtÞ þ a þ l2 TðtÞ ¼ 0: b. Show that the eigenvalues are ln ¼ n þ 12. c. Show that

8.7 Steady State of the Heat Equation

Tn ðtÞ ¼ e

½ð2n1Þ2 þ 4at=4

  1 Xn ðxÞ ¼ sin n  x 2

;

and thus uðx; tÞ ¼

∞ X

cn e½ð2n1Þ

n¼1

2

þ 4at=4

  1 sin n  x. 2

d. Let k ¼ n  1=2 so that uðx; 0Þ ¼

∞ X

bk sin kx ¼ xðp  xÞ.

k¼1

Find bk and show that " #   ∞ X 32 8 cosðnpÞ ½ð2n1Þ2 þ4at=4 1 þ sin n  x. uðx; tÞ ¼ e 3 2 ð2n  1Þ2 n¼1 pð2n  1Þ

8.7 STEADY STATE OF THE HEAT EQUATION The heat equation has a steady (equilibrium) state exactly when vuðx; tÞ ¼ 0: vt If we have a rod of length L, then having a steady state means for each value of x, 0  x  L, there is a number (temperature) Tx, for which lim uðx; tÞ ¼ Tx . t/∞ One of the simplest cases is constant boundary conditions lim

t/∞

uð0; tÞ ¼ T0 ; uðL; tÞ ¼ TL . In the event that the heat equation has a heat source, g(x), which depends only on x, the heat equation is vuðx; tÞ v2 uðx; tÞ ¼ þ gðxÞ vt vx2 where we have rescaled the model so that a2 ¼ 1. We consider the heat equation with a source vuðx; tÞ v2 uðx; tÞ þ gðxÞ; uð0; tÞ ¼ T0 ; uðL; tÞ ¼ TL ; uðx; 0Þ ¼ f ðxÞ. ¼ vt vx2 (If g(x) > 0 then g(x) is said to be a heat source; if g(x) < 0 then g(x) is said to be a heat sink.) If there is a steady state, then

303

304

CHAPTER 8 Separation of Variables in Cartesian Coordinates " # vuðx; tÞ v2 uðx; tÞ ¼ 0 ¼ lim lim þ gðxÞ . t/∞ t/∞ vt vx2 vuðx; tÞ v2 uðx; tÞ must exist. ¼ 0; then lim t/∞ t/∞ vt vx2 Let uS(x) be the steady state; i.e., uS ðxÞ ¼ lim uðx; tÞ ¼ Tx . Then uS(x) must t/∞ satisfy Note that if lim

d2 uS ðxÞ þ gðxÞ ¼ 0; uS ð0Þ ¼ T0 ; uS ðLÞ ¼ TL . dx2 From u(x, t) we subtract the steady state uS(x) to obtain vðx; tÞ ¼ uðx; tÞ  uS ðxÞ. We show that v(x, t) satisfies the heat equation. We have vvðx; tÞ vuðx; tÞ v2 vðx; tÞ v2 uðx; tÞ d2 uS ðxÞ v2 uðx; tÞ ¼  ¼ þ gðxÞ ¼ ; vt vt vx2 vx2 dx2 vx2 so vvðx; tÞ vuðx; tÞ v2 uðx; tÞ v2 vðx; tÞ þ gðxÞ ¼ ¼ ¼ vt vt vx2 vx2 and vð0; tÞ ¼ uð0; tÞ  uS ð0Þ ¼ T0  T0 ¼ 0 vðL; tÞ ¼ uðL; tÞ  uS ðLÞ ¼ TL  TL ¼ 0 vðx; 0Þ ¼ f ðxÞ  uS ðxÞ 0  x  L. We have seen how to solve vvðx; tÞ v2 vðx; tÞ ; vð0; tÞ ¼ 0; vð0; tÞ ¼ 0; vðx; 0Þ ¼ f ðxÞ  uS ðxÞ ¼ vt vx2

0xL

in the previous section. The function uS(x) must satisfy d2 uS ðxÞ ¼ gðxÞ; uS ð0Þ ¼ T0 ; uS ðLÞ ¼ TL . dx2 In the next two examples, we apply this method to solve forms of the heat equation. Example: Suppose there is no heat source, but the body reaches a steady state. In this problem, we continue to rescale so that a2 ¼ 1. Find the steady state and the solution for the heat equation that is given by vuðx; tÞ v2 uðx; tÞ ¼ uð0; tÞ ¼ T1 ; uðL; tÞ ¼ T2 ; uðx; 0Þ ¼ f ðxÞ. vt vx2

8.7 Steady State of the Heat Equation

Since we are given the temperature of the body approaches a steady state, we also have vuðx; tÞ ¼ 0: vt Because u(0, t) ¼ T1 and u(L, t) ¼ T2, we must have lim

t/∞

uS ð0Þ ¼ lim uð0; tÞ ¼ T1 t/∞

and

uS ðLÞ ¼ lim uðL; tÞ ¼ T2 . t/∞

Now lim

t/∞

v2 uðx; tÞ vuðx; tÞ ¼0 ¼ lim 2 t/∞ vx vt

and lim

t/∞

v2 uðx; tÞ d2 uS ðxÞ ¼ vx2 dx2

so d 2 uS ðxÞ ¼ 0: dx2 Thus uS ðxÞ ¼ Ax þ B. We have T1 ¼ uS(0) ¼ B and T2 ¼ uS(L) ¼ AL þ B ¼ AL þ T1, so A ¼ (T2T1)/L. Thus, uS ðxÞ ¼

ðT2  T1 Þ x þ T1 . L

If, as before, we let vðx; tÞ ¼ uðx; tÞ  uS ðxÞ then we have seen vvðx; tÞ v2 vðx; tÞ ¼ ; vð0; tÞ ¼ 0; vð0; tÞ ¼ 0; vðx; 0Þ ¼ f ðxÞ  uS ðxÞ vt vx2 As in the previous section, (although here we have set a2 ¼ 1)  2 2 2  ∞ X a n p t npx sin cn exp vðx; tÞ ¼ . L L n¼1 To determine cn we must use vðx; 0Þ ¼ f ðxÞ  uS ðxÞ ¼ f ðxÞ 



  T2  T1 x þ T1 . L

0  x  L.

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

Thus 2 cn ¼ L

Z L 0

   T2  T1 npx x sin f ðxÞ  T1 þ dx. L L

Thus, vðx; tÞ ¼

∞ X

 cn exp

n¼1

  a2 n2 p2 t npx sin L L

" Z  #    ∞ X 2 L T2  T1 npx f ðxÞ  T1 þ x sin ¼ dx L 0 L L n¼1   a2 n2 p2 t npx sin  exp . L L 

Next we consider the case where there is a constant heat source and an equilibrium state. Example: Consider the equation vuðx; tÞ v2 uðx; tÞ þ b; uð0; tÞ ¼ T1 ; uð0; LÞ ¼ T2 ; uðx; 0Þ ¼ f ðxÞ ¼ vt vx2 where b is a constant. We first find the equilibrium state uS(x) that satisfies the equation d 2 uS ðxÞ þb¼0 dx2

uS ð0Þ ¼ T1 ; uS ðLÞ ¼ T2 .

Then d2 uS ðxÞ ¼ b dx2 so b uS ðxÞ ¼  x2 þ Ax þ B. 2 Since uS(0) ¼ T1, we have B ¼ T1 and b uS ðLÞ ¼ T2 ¼  L2 þ AL þ T1 2 so b AL ¼ T2  T1 þ L2 2

and

T2  T1 bL þ . 2 L

(1)

8.7 Steady State of the Heat Equation

Therefore

  b T2  T1 bL x þ T1 . þ uS ðxÞ ¼  x2 þ 2 2 L

We let vðx; tÞ ¼ uðx; tÞ  uS ðxÞ. If we can find v(x, t), then we know u(x, t). Again vvðx; tÞ v2 vðx; tÞ ¼ ; vð0; tÞ ¼ 0; vð0; tÞ ¼ 0; vðx; 0Þ ¼ f ðxÞ  uS ðxÞ vt vx2

0  x  L.

We proceed as before to find   a2 n2 p2 t npx sin cn exp . vðx; tÞ ¼ L L n¼1 ∞ X

To determine cn we must use



   b 2 T2  T1 bL þ x þ T1 . vðx; 0Þ ¼ f ðxÞ  uS ðxÞ ¼ f ðxÞ   x þ L 2 2 

Thus cn ¼

2 L

Z L 0

     b T2  T1 bL npx x þ T1 sin þ f ðxÞ   x2 þ dx. 2 2 L L

Thus,  2 2 2  a n p t npx sin cn exp vðx; tÞ ¼ L L n¼1 " Z      ∞ X 2 L b 2 T2  T1 bL x þ T1 þ f ðxÞ   x þ ¼ L 0 2 2 L n¼1 #  2 2 2  npx a n p t npx  sin sin dx exp . L L L ∞ X

Finally, we have uðx; tÞ ¼ vðx; tÞ þ uS ðxÞ " Z      ∞ X 2 L b 2 T2  T1 bL x þ T1 þ f ðxÞ   x þ ¼ L 0 2 2 L n¼1 #  2 2 2  npx a n p t npx sin  sin dx exp L L L     b T2  T1 bL þ x þ T1 : þ  x2 þ L 2 2

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EXERCISES 1. Find the steady state and the solution to the heat equation vuðx; tÞ v2 uðx; tÞ ¼ þ gðxÞ vt vx2

0 < x < L; t > 0; uð0; tÞ ¼ T0 ; uðL; tÞ ¼ TL ;

uðx; 0Þ ¼ f ðxÞ for the following functions g(x). a. g(x) ¼ 2. b. g(x) ¼ sin x. c. g(x) ¼ x2. d. g(x) ¼ 3ex. e. g(x) ¼ 1 þ 3x. px 2. Find the solutions of problem 1 for f ðxÞ ¼ T0 þ ðTL  T0 Þsin . 2L 3. Solve the heat equation vuðx; tÞ v2 uðx; tÞ ¼ 0 < x < p; uð0; tÞ ¼ T0 ; uðp; tÞ ¼ TL ; uðx; 0Þ ¼ 1  x vt vx2 assuming there is an equilibrium state. 4. Solve the heat equation vuðx; tÞ v2 uðx; tÞ ¼ ; 0 < x < 1; vt vx2 uðx; 0Þ ¼ ð1  xÞT0 þ xT1

uð0; tÞ ¼ T0 ; uð1; tÞ ¼ T1 ;

assuming there is an equilibrium state. 5. In this problem we solve the heat equation vuðx; tÞ v2 uðx; tÞ þ f ðx; tÞ; 0 < x < p; t > 0; ¼ vt vx2 uðx; 0Þ ¼ gðxÞ.

uð0; tÞ ¼ 0; uðp; tÞ ¼ 0;

a. Show that the solution to vuðx; tÞ v2 uðx; tÞ ; 0 < x < p; t > 0; ¼ vt vx2

uð0; tÞ ¼ 0; uðp; tÞ ¼ 0; uðx; 0Þ ¼ gðxÞ

is of the form ∞ X

An en t sinðnxÞ

n¼1

where the An’s are determined using g(x).

2

8.7 Steady State of the Heat Equation

b. Assume that f (x, t) can be expanded in a sine series f ðx; tÞ ¼

∞ X

fn ðtÞsinðnxÞ.

n¼1

Show that fn ðtÞ ¼

Z

2 p

p

f ðx; tÞsinðnxÞdx

0

and thus f ðx; tÞ ¼

∞  Z X 2

p

n¼1

p

 f ðx; tÞsinðnxÞdx sinðnxÞ.

0

c. Let uðx; tÞ ¼

∞ X

Bn ðtÞsinðnxÞ.

n¼1

Show u(x, 0) ¼ u(x, p) ¼ 0. Also show that u(x, t) as defined above gives the equation

vuðx; tÞ v2 uðx; tÞ þ f ðx; tÞ with ¼ vt vx2

∞ ∞ ∞ X X X d n2 Bn ðtÞsinðnxÞ þ fn ðtÞsinðnxÞ. Bn ðtÞsinðnxÞ ¼  dt n¼1 n¼1 n¼1

This means d Bn ðtÞ þ n2 Bn ðtÞ ¼ fn ðtÞ. dt

(2)

d. Show that the solution to Eq. (2) is R 2 fn ðtÞen t dt þ Cn Bn ðtÞ ¼ 2 en t where the Cn’s are to be determined. P e. Use that uðx; 0Þ ¼ gðxÞ ¼ ∞ n¼1 Bn ð0ÞsinðnxÞ to conclude Z 2 p gðxÞsinðnxÞdx. Bn ð0Þ ¼ p 0 f. Show Cn ¼ Bn(0) and thus Bn ðtÞ ¼ en

2

Z

t

t 0

fn ðsÞen s ds þ Bn ð0Þen t . 2

2

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

g. Show that uðx; tÞ ¼

∞ X

Bn ð0Þe

n2 t

sinðnxÞ þ

n¼1

∞ Z X n¼1

t

e

n2 ðtsÞ

 fn ðsÞds sinðnxÞ.

0

6. Use the result of problem 5 to solve vuðx; tÞ v2 uðx; tÞ þ xqðtÞ; ¼ vt vx2 uðp; tÞ ¼ 0; uðx; 0Þ ¼ 0:

0 < x < p; t > 0;

uð0; tÞ ¼ 0;

It may be helpful to use the Fourier expansion x¼2

∞ X ð1Þnþ1 sin nx. n n¼1

8.8 CHECKING THE VALIDITY OF THE SOLUTION In this chapter we have assumed that the solution to the initial value problem could be written as the product of functions of one variable. With that assumption, we found the solution. Along the way, we have cautioned that the result should be verified. In this section we show one method for checking the solution. Some facts about series of functions that we shall use are 1. The M-test. PWeierstrass 2. If an xn is a power series with radius of convergence R, then P a. an xn converges uniformly and absolutely on the interval [eR þ ε, R  ε] for any ε > 0. P P d an xn ¼ nan xn1 . b. dx P P c. The series an nk xn has the same radius of convergence as an xn for any positive integer k. Consider the heat equation 1 ut ðx; tÞ; uð0; tÞ ¼ uðp; tÞ ¼ 0; uðx; 0Þ ¼ f ðxÞ. a2 We have found that the solution using separation of variables is uxx ðx; tÞ ¼

uðx; tÞ ¼

∞ X n¼1

un ðx; tÞ

8.8 Checking the Validity of the Solution

where un ðx; tÞ ¼ Now

 Z p  2 2 f ðxÞsin nx dx sin nxeðanÞ t . p 0

Z p Z

2 p

2



f ðxÞsin nx dx j f ðxÞjdx.

p

p 0 0

If f (x) is piecewise continuous, then it is bounded. Thus there is a number M for which Z 2 p j f ðxÞjdx  M. p 0 So for every positive integer n, we have

2 2

jun ðx; tÞj  M sin nxeðanÞ t ¼ MeðanÞ t . Note that d2 ðun ðx; tÞÞ ¼ n2 un ðx; tÞ and dx2 so that

2

d 2 ðanÞ2 t

dx2 ðun ðx; tÞÞ  n Me

and

d ðun ðx; tÞÞ ¼ ðanÞ2 un ðx; tÞ dt

d

2

ðun ðx; tÞÞ  n2 a2 MeðanÞ t .

dt

Since for any ε > 0 the series of numbers ∞ X

n2 en

2

ε

n¼1

converges, each of the series ∞ X n¼1

un ðx; tÞ;

∞ X v2 ðun ðx; tÞÞ vx2 n¼1

and

∞ X v ðun ðx; tÞÞ vt n¼1

converges uniformly for t  ε, where ε is an arbitrary positive number. Thus, if we let uðx; tÞ ¼

∞ X

un ðx; tÞ

n¼1

then ∞ X v2 v2 uðx; tÞ ¼ un ðx; tÞ vx2 vx2 n¼1

and

∞ X v v uðx; tÞ ¼ un ðx; tÞ vt vt n¼1

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CHAPTER 8 Separation of Variables in Cartesian Coordinates

if t  ε. Since v2 1 v un ðx; tÞ  2 un ðx; tÞ ¼ 0 and vx2 a vt we have v2 1 v uðx; tÞ  2 uðx; tÞ ¼ 0 a vt vx2 Finally, we must show

un ð0; tÞ ¼ un ðp; tÞ ¼ 0

uð0; tÞ ¼ uðp; tÞ ¼ 0

and

for every n

for t  ε.

lim uðx; tÞ ¼ f ðxÞ. tY0

To proceed, we assume that f (x) is continuous on [0, p] and f (0) ¼ f (p) ¼ 0 and that Z p 2 ½ f 0 ðxÞ dx < ∞. 0

The last assumption ensures that the Fourier series for f (x) converges uniformly to f (x). Let sN(x, t) be the Nth partial sum of ∞ X

bn eðanÞ t sin nx 2

n¼1

i.e., sN ðx; tÞ ¼

N X

bn eðanÞ t sin nx. 2

n¼1

Then sN ðx; 0Þ ¼

N X

bn sin nx.

n¼1

Since the Fourier series of f (x) converges uniformly to f (x), given ε > 0 there is a number N(ε) so that if n > N(ε) then

X

ε N

bn sin nx  f ðxÞ < jsN ðx; 0Þ  f ðxÞj ¼

n¼1

2 so jsN ðx; 0Þ  sM ðx; 0Þj ¼ jsN ðx; 0Þ  f ðxÞ þ f ðxÞ  sM ðx; 0Þj  jsN ðx; 0Þ  f ðxÞj ε ε þj f ðxÞ  sM ðx; 0Þj < þ ¼ ε 2 2

8.8 Checking the Validity of the Solution

if M, N > N(ε). Thus fsn ðx; 0Þg is a uniformly Cauchy sequence of continuous functions that converges uniformly to a continuous function, and this function must be f (x). A fact that we do not prove (see Weinberger, pp. 58e60) is that in our setting we must then have jsN ðx; tÞ  sM ðx; tÞj  ε

if

0  x  p; t  0

if

M; N > NðεÞ.

This shows lim tY0 uðx; tÞ ¼ f ðxÞ. We note that in verifying the solution for the heat equation we were aided substantially by the fact that we had an exponentially decreasing factor. This is not true with the wave equation. To prove validity of the solution for the wave equation, we need some properties of the wave equation that we have not yet developed. A proof of the validity of the solution of the wave equation may be found in Brown and Churchill, pp. 338e341.

313

CHAPTER

Solving Partial Differential Equations in Cylindrical Coordinates Using Separation of Variables

9

9.1 INTRODUCTION In Chapter 8, we solved Laplace’s equation, the wave equation, and the heat equation in Cartesian coordinates using separation of variables. In this chapter, we solve the same equations in polar or cylindrical coordinates. In Cartesian coordinates, the ordinary differential equations (ODEs) that arose were simple to solve. We shall see that, in cylindrical and spherical coordinates, not all the ODEs are as agreeable. The solutions to these more difficult ODEs go by the names Bessel functions and Legendre polynomials. In cylindrical coordinates, we need only Bessel functions. We begin this section by showing how these equations arise.

AN EXAMPLE WHERE BESSEL FUNCTIONS ARISE Consider the heat equation ut ¼ KDu. The reason the equations that arise from separation of variables in cylindrical coordinates are not as simple as in Cartesian coordinates is the form of the Laplacian. In cylindrical coordinates, the Laplacian is given by 1 1 Du ¼ urr þ ur þ 2 uqq þ uzz . r r It will simplify our computations and still allow us to demonstrate how Bessel functions arise if we assume that u is a function of r, q and t, but not a function of z. We suppose uðr; q; tÞ ¼ RðrÞQðqÞTðtÞ so that ut ¼ KDu Mathematical Physics with Partial Differential Equations. https://doi.org/10.1016/B978-0-12-814759-7.00009-0 Copyright © 2018 Elsevier Inc. All rights reserved.

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CHAPTER 9 Solving Partial Differential Equations

can be expressed as   1 0 1 00 00 RðrÞQðqÞT ðtÞ ¼ K R ðrÞQðqÞTðtÞ þ R ðrÞQðqÞTðtÞ þ 2 RðrÞQ ðqÞTðtÞ : r r 0

Dividing by KR(r)Q(q)T(t) gives 1 T 0 ðtÞ R00 ðrÞ 1 R0 ðrÞ 1 Q00 ðqÞ ¼ þ þ : K TðtÞ RðrÞ r RðrÞ r 2 QðqÞ

(1)

The left-hand side of Eq. (1) is a function only of t, and the right-hand side is a function of r and q, so it must be that each is a constant. In Exercise 1, we show that this is a negative number that we denote el. Thus 1 T 0 ðtÞ ¼ el or T 0 ðtÞ þ lKTðtÞ ¼ 0: K TðtÞ

(2)

Also, R00 ðrÞ 1 R0 ðrÞ 1 Q00 ðqÞ þ þ ¼ el: RðrÞ r RðrÞ r 2 QðqÞ So R00 ðrÞ 1 R0 ðrÞ 1 Q00 ðqÞ þ þ l¼ 2 RðrÞ r RðrÞ r QðqÞ or r2

  00 R ðrÞ 1 R0 ðrÞ Q00 ðqÞ þ þ l ¼ . RðrÞ r RðrÞ QðqÞ

(3)

The left-hand side of Eq. (3) is a function of r and the right-hand side is a function of q, so each must be a constant that we denote m. Thus we have Q00 ðqÞ þ m QðqÞ ¼ 0 and

 r2

(4)

 R00 ðrÞ 1 R0 ðrÞ R00 ðrÞ 1 R0 ðrÞ m þ þ l ¼ m or þ þ l¼ 2 RðrÞ r RðrÞ RðrÞ r RðrÞ r

so  1 0 m R ðrÞ þ l  2 RðrÞ ¼ 0: (5) r r Thus, to solve the heat equation in polar coordinates, we need to solve Eqs. (2), (4) and (5). Of these, only Eq. (5) requires additional attention. It is (like) a Bessel equation, and we construct its solution in the next section. This equation will arise R00 ðrÞ þ

9.1 Introduction

when we use the Laplacian in polar or cylindrical coordinates in the wave equation or the heat equation. In Laplace’s equation, we shall see the equation is of the form   1 0 n2 00 2 R ðrÞ þ R ðrÞ  m þ 2 RðrÞ ¼ 0 r r and will have to be handled differently (although very similarly). Had we assumed that the function u also depended on z and that uðr; q; z; tÞ ¼ RðrÞQðqÞTðtÞZðzÞ; Eq. (5) would still have been the only complicated ODE that would have arisen (see Exercise 3.) Eq. (5) is a “Bessel-like” equation. We next define a Bessel equation, demonstrate one solution to such equations, and then make a transformation that will enable us to solve the equation above. (Because this is a second-order differential equation, there are two solutions, but one is unbounded at r ¼ 0. Because of physical considerations, this will be an inadmissible solution for our problems.) A Bessel equation is an equation of the form  x2 y00 ðxÞ þ xy0 ðxÞ þ x2  n2 yðxÞ ¼ 0; 0  x < ∞. The method of solution that we use is power series: Step 1: Hypothesize a solution of the form y¼

∞ X

an xnþa .

n¼0

For the solution to be bounded at x ¼ 0, we require that a  0. Step 2: Differentiate and collect terms. We have y0 ¼

∞ X

an ðn þ aÞxnþa1

n¼0

so xy0 ¼

∞ X

an ðn þ aÞxnþa

n¼0

y00 ¼

∞ X

an ðn þ aÞðn þ a  1Þxnþa2

n¼0

so x2 y00 ¼

∞ X n¼0

an ðn þ aÞðn þ a  1Þxnþa

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CHAPTER 9 Solving Partial Differential Equations

then ∞

 X

an ðn þ aÞðn þ a  1Þxnþa þ an ðn þ aÞxnþa x2 y00 þ xy0 þ x2  n2 ¼ n¼0

þ



an xnþaþ2  n2 an xnþa ¼ 0:

This can be written as

a0 aða  1Þ þ a  n2 xa þ a1 ða þ 1Þa þ ða þ 1Þ  n2 xaþ1 ∞

X

an ððn þ aÞÞððn þ aÞ  1Þ þ ðn þ aÞ  n2 þ an2 xaþn þ n¼2

h i ¼ a0 a2  n2 xa þ a1 ða þ 1Þ2  n2 ∞ nh i o X ðn þ aÞ2  n2 an þ an2 xaþn ¼ 0: þ 

n¼2

The coefficient of each power of x must be 0. The coefficient of xa must be 0 and this gives the indicial equation, from which we determine the value of a. If a0 s 0, we have a2  n2 ¼ 0 so a2 ¼ n2. Also, h i h i a1 ða þ 1Þ2  n2 ¼ a1 ða þ 1Þ2  a2 ¼ a1 ½2a þ 1 ¼ 0: If the solution is bounded, then a is nonnegative and a1 ¼ 0. Step 3: The recurrence relation is

an ðn þ aÞððn þ aÞ  1Þ þ ðn þ aÞ  n2 þ an2 ¼ 0 or

h i

an ðn þ aÞððn þ aÞ  1Þ þ ðn þ aÞ  n2 ¼ an ðn þ aÞ2  n2 þ an2 ¼ 0:

Substituting n for a gives h i an2 . an ðn þ nÞ2  n2 ¼ an nðn þ 2nÞ ¼ an2 or an ¼ nðn þ 2nÞ Because a1 ¼ 0, then ak ¼ 0 for every odd integer k. Step 4: We now determine a general description of a2k. We have a0 a2 ¼  2ð2 þ 2nÞ a4 ¼ 

a2 ð1Þ ð1Þa0 ¼ 4ð4 þ 2nÞ 4ð4 þ 2nÞ 2ð2 þ 2nÞ

9.1 Introduction

a4 a0 a6 ¼  ¼ 3 6ð6 þ 2nÞ 2 ð1\$2\$3Þð3 þ nÞð2 þ nÞð1 þ nÞ and a2k ¼

ð1Þk a0 . 22k ðk!Þðk þ nÞðk  1 þ nÞ/ð1 þ nÞ

Thus one of the solutions of Bessel’s equation is # " ∞ k 2k X ð1Þ x y1 ðxÞ ¼ a0 xn 1 þ . 22k ðk!Þðk þ nÞðk  1 þ nÞ/ð1 þ nÞ k¼1 This is a solution for any value of a0. (Notice that we have not imposed any boundary conditions.) This is Bessel’s function of the first kind of order n, denoted Jn(x). If we let a0 ¼ 1=n!2n . We can express Jn(x) as  2kþn k x ∞ ∞ ð1Þ k 2kþn X X ð1Þ x 2 ¼ . (6) Jn ðxÞ ¼ 2kþn k!ðk þ nÞ! 2 k!ðk þ nÞ! n¼0 k¼0 By the ratio test, this series converges for all values of x, and in Exercise 5 we show that it is a solution for x > 0. In the case that the equation is of the form   1 n2 R00 ðrÞ þ R0 ðrÞ  m2 þ 2 RðrÞ ¼ 0; r r as will be the case for Laplace’s equation, the solution is a modified Bessel function (sometimes called a Bessel function with imaginary argument) denoted by In(x), which is defined by " # ∞ X xn x2n 1þ . In ðxÞ ¼ n 2 n! 22n n!ð1 þ nÞ/ðn þ nÞ n¼1 (See Pinsky, 1998, p. 187.) The modified Bessel’s function is obtained by replacing x with ix in Bessel’s equation. Bessel’s equation is a second-order equation, so there will be two linearly independent solutions. For our purposes, we shall not be concerned with the second solution because it diverges at x ¼ 0. The discussion and derivation of the second solution can be found in many differential equation texts, including Boyce and DiPrima (2008), but the second solution is of the form y2 ðxÞ ¼ y1 ðxÞlnjxj þ jxjn

∞ X n¼1

bn x n .

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CHAPTER 9 Solving Partial Differential Equations

The two equations given at the beginning of this section are of the form  ðd  1Þ 0 m y ðrÞ þ l  2 yðrÞ ¼ 0: ð1Þ r r In cylindrical coordinates, d ¼ 2 and m will typically be m2. In spherical coordinates, d ¼ 3 and m will typically be k(k þ 1). In our later work we consider this to be an eigenvectoreeigenvalue problem. y00 ðrÞ þ

EXERCISES 1. Show that the constant that arises in Eq. (1) is negative. 2. Repeat the separation of variables argument in the case that uðr; q; z; tÞ ¼ RðrÞQðqÞTðtÞZðzÞ. 3. Repeat the separation of variables argument for the wave equation. 4. Show that Jn(x) that arises in Eq. (6) converges for x  0 and solves  x2 y00 ðxÞ þ xy0 ðxÞ þ x2  n2 yðxÞ ¼ 0; 0  x < ∞. 5. Give a bounded solution for the following equations:  a. x2 y00 ðxÞ þ xy0 ðxÞ þ x2  4 yðxÞ ¼ 0: b. x2 y00 ðxÞ þ xy0 ðxÞ þ x2 yðxÞ  ¼ 0:  00 0 c. y ðxÞ þ ð1=xÞy ðxÞ þ 1  92 yðxÞ ¼ 0: x

9.2 THE SOLUTION TO BESSEL’S EQUATION IN CYLINDRICAL COORDINATES In our analysis of the heat equation in cylindrical coordinates using separation of variables, we arrived at an equation of the form d2 RðrÞ dRðrÞ 2 þr þ lr  m RðrÞ ¼ 0: 2 dr dr For our applications, m ¼ n2 where n is an integer, so we want to solve r2

r2

d2 RðrÞ dRðrÞ 2 þ lr  n2 RðrÞ ¼ 0: þr 2 dr dr

(1)

9.2 The Solution to Bessel’s Equation in Cylindrical Coordinates

Eq. (1) is not a Bessel equation pﬃﬃﬃ but can be transformed into a Bessel equation by the change of variables x ¼ r l, as we now demonstrate. We have dRðrÞ dRðxÞ dx dRðxÞ pﬃﬃﬃ ¼ ¼ l and dr dx dr dx d 2 RðrÞ d2 RðxÞ ¼ l dr 2 dx2 so r2

d2 RðrÞ dRðrÞ 2 þ lr  n2 RðrÞ ¼ 0 þr 2 dr dr

is transformed to     2  2 2 x d RðxÞ x pﬃﬃﬃ dRðxÞ x 2 pﬃﬃﬃ l l þ pﬃﬃﬃ þ l  n RðxÞ dx l dx2 l l

(2) RðxÞ dRðxÞ 2 2 ¼x þ x  n RðxÞ ¼ 0: þx dx dx2 Eq. (2) is a Bessel equation that has two solutions. The solution that is bounded at x ¼ 0 is typically the only one of interest to us. That solution is given by  pﬃﬃﬃ  RðxÞ ¼ Jn ðxÞ ¼ Jn r l . 2d

2

Any Bessel function of the first type has infinitely many positive roots. The graphs of several Bessel functions of the first type are shown in Fig. 9.2.1.

y 1 0.8 0.06

J0(x) J1(x)

J2(x)

J3(x)

0.4

J4(x)

J5(x)

0.2 x –0.2 –0.4

FIGURE 9.2.1

2

4

6

8

10

321

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CHAPTER 9 Solving Partial Differential Equations

Let {xn,m} denote the positive roots of Jn(x); i.e., Jn(xn,m) ¼ 0. In many equations involving a cylinder, one of the boundary conditions is that the value of R(r) at the surface of the cylinder is constant. Suppose that the radius of the cylinder is a and R(a) ¼ 0. (If R(a) is a different constant, this boundary condition can be obtained by rescaling the temperature. We could also rescale the length to take a ¼ 1, which is common.) Thus, in cylindrical coordinates, an initial value problem of interest is d2 RðxÞ dRðxÞ 2 þ x  n2 RðxÞ ¼ 0; RðaÞ ¼ 0: (3) þx 2 dx dx pﬃﬃﬃ The solution to Eq. (3) that is continuous at x ¼ 0 is R(x) ¼ Jn(x) ¼ Jn(r l) and the boundary condition requires that Jn(a) ¼ 0. Since we have Jn(xn,m) ¼ 0, if we let  r Rm ðrÞ ¼ Jn xn;m a then   a Rm ðaÞ ¼ Jn xn;m ¼ Jn xn;m ¼ 0: a Thus,  r Rm ðrÞ ¼ Jn xn;m a is an eigenfunction for the initial value problem given by Eq. (3). The eigenvalue for x 2 n;m as we verify in Exercise 1. Rm(r) is a In Exercise 2 we show that x2

d2 RðxÞ dRðxÞ 2 þx þ x  n2 RðxÞ ¼ 0; RðaÞ ¼ 0 dx2 dx is a SturmeLiouville problem with weight function w(x) ¼ x. By the SturmeLiouville theory, the eigenfunctions are complete. That is, if f (x) is a function on [0, a] for which Z a x½ f ðxÞ2 dx < ∞; x2

0

then f ðxÞ ¼ where

 x bm Jn xn;m a m¼0 ∞ X

 x x f ðxÞx J dx n n;m 0 a . bm ¼ R h  i 2 x a dx 0 x Jn xn;m a Ra

9.3 Solving Laplace’s Equation in Cylindrical Coordinates

The final points in this section are reiterated with some intuitive explanation in Section 9.5 when we discuss the heat equation on a disk.

EXERCISES 1. Show that

 x Jn xn;m a

is an eigenfunction for x2

d2 RðxÞ dRðxÞ 2 þ x  n2 RðxÞ ¼ 0; þx 2 dx dx

RðaÞ ¼ 0; RðxÞ is bounded at x ¼ 0

with eigenvalue x

 n;m 2 a

.

2. Show that d2 RðxÞ dRðxÞ 2 þ x  n2 RðxÞ ¼ 0; RðaÞ ¼ 0 þx 2 dx dx is a SturmeLiouville problem with weight function w(x) ¼ x. x2

9.3 SOLVING LAPLACE’S EQUATION IN CYLINDRICAL COORDINATES USING SEPARATION OF VARIABLES In cylindrical coordinates, Laplace’s equation is 1 1 Duðr; q; zÞ ¼ urr þ ur þ 2 uqq þ uzz ¼ 0: r r We solve the boundary value problem 1 1 urr þ ur þ 2 uqq þ uzz ¼ 0; r r uðr; q; 0Þ ¼ uðr; q; bÞ ¼ 0;

0 < r < a; 0 < z < b; uða; q; zÞ ¼ f ðq; zÞ.

We hypothesize that uðr; q; zÞ ¼ RðrÞQðqÞZðzÞ so 1 Duðr; q; zÞ ¼ R00 ðrÞQðqÞZðzÞ þ R0 ðrÞQðqÞZðzÞ r 1 þ 2 RðrÞQ00 ðqÞZðzÞ þ RðrÞQðqÞZ 00 ðzÞ ¼ 0: r

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CHAPTER 9 Solving Partial Differential Equations

Dividing by R(r) Q(q) Z(z) gives R00 ðrÞ 1 R0 ðrÞ 1 Q00 ðqÞ Z 00 ðzÞ þ þ þ ¼0 RðrÞ r RðrÞ r 2 QðqÞ ZðzÞ or R00 ðrÞ 1 R0 ðrÞ 1 Q00 ðqÞ Z 00 ðzÞ þ þ 2 ¼ . RðrÞ r RðrÞ r QðqÞ ZðzÞ

(1)

The left-hand side of Eq. (1) is a function of r and q, and the right-hand side is a function of z, so each is a positive constant that we call C. Thus we have 

Z 00 ðzÞ ¼C ZðzÞ

so Z 00 ðzÞ þ CZðzÞ ¼ 0: Then ZðzÞ ¼ A cos

pﬃﬃﬃﬃ pﬃﬃﬃﬃ C z þ B sin C z

and Zð0Þ ¼ 0; so A ¼ 0;

ZðbÞ ¼ 0

so np2 pﬃﬃﬃﬃ np C¼ and C ¼ . b b Thus Zn ðzÞ ¼ sin

npz . b

We also have R00 ðrÞ 1 R0 ðrÞ 1 Q00 ðqÞ þ þ ¼C RðrÞ r RðrÞ r 2 QðqÞ so

 r2

R00 ðrÞ 1 R0 ðrÞ þ C RðrÞ r RðrÞ



and there is a constant D with Q00 ðqÞ ¼D  QðqÞ

¼

Q00 ðqÞ QðqÞ

9.3 Solving Laplace’s Equation in Cylindrical Coordinates

so Q00 ðqÞ þ DQðqÞ ¼ 0 and r2

  00 R ðrÞ 1 R0 ðrÞ þ  C ¼ D. RðrÞ r RðrÞ

(2)

The periodicity conditions Q(p) ¼ Q(p) and Q0 (p) ¼ Q0 (p) require that D > 0. We let D ¼ m2 and so we have   00 1 R0 ðrÞ R00 ðrÞ 1 R0 ðrÞ D 2 R ðrÞ þ  C ¼ D or þ C 2 ¼0 r RðrÞ r RðrÞ RðrÞ r RðrÞ r so      1 0 D 1 0 np 2 m2 þ 2 RðrÞ ¼ 0: RðrÞ þ R ðrÞ  C þ 2 RðrÞ ¼ RðrÞ þ R ðrÞ  r r r b r It will simplify the notation if we let b ¼ p, and we now make that substitution. We then have   1 m2 R00 ðrÞ þ R0 ðrÞ  n2 þ 2 RðrÞ ¼ 0: r r Recapping, there are three equations we must solve: Z 00 ðzÞ þ n2 ZðzÞ ¼ 0;

Zð0Þ ¼ ZðpÞ ¼ 0

Q00 ðqÞ þ m2 QðqÞ ¼ 0;

QðpÞ ¼ QðpÞ and Q0 ðpÞ ¼ Q0 ðpÞ   1 0 m2 00 2 R ðrÞ þ R ðrÞ  n þ 2 RðrÞ ¼ 0: r r

(3)

Eq. (3) has no boundary condition. Instead, we require that the solution is bounded at r ¼ 0. Note that to solve Eq. (3), we will need to use a modified Bessel function. We have shown that Zn ðzÞ ¼ an sin nz. The equation Q00 ðqÞ þ m2 QðqÞ ¼ 0 has only the continuity conditions Q(p) ¼ Q(p) and Q0 (p) ¼ Q0 (p) so Qm ðqÞ ¼ bm cos mq þ cm sin mq.

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CHAPTER 9 Solving Partial Differential Equations

The equation   1 0 m2 2 R ðrÞ þ R ðrÞ  n þ 2 RðrÞ ¼ 0 r r 00

has AmnIm(nr) as its solution. We choose Amn ¼ 1/Im(na) so that Rmn ðrÞ ¼ Amn Im ðnrÞ ¼

Im ðnrÞ Im ðnaÞ

and so Rmn ðaÞ ¼

Im ðnaÞ ¼ 1: Im ðnaÞ

Thus, we have umn ðr; q; zÞ ¼ Rmn ðrÞQm ðqÞZn ðzÞ ¼ Amn Im ðnrÞðbm cos mq þ cm sin mqÞan sin nz ¼

1 Im ðnrÞðbm cos mq þ cm sin mqÞan sin nz. Im ðnaÞ

The constants can be combined to write unm ðr; q; zÞ ¼

1 Im ðnrÞsin nzðdnm cos mq þ enm sin mqÞ. Im ðnaÞ

Thus, uðr; q; zÞ ¼

∞ 1X 1 I0 ðnrÞ dn0 2 n¼1 I0 ðnaÞ

þ

∞ X

1 Im ðnrÞsin nzðdnm cos mq þ enm sin mqÞ. ðnaÞ I m; n¼1 m

We apply the boundary condition u(a,q, z) ¼ f (q, z) to get uða; q; zÞ ¼ f ðq; zÞ ¼ þ ¼

∞ 1X 1 I0 ðnaÞ dn0 2 n¼1 I0 ðnaÞ

∞ X

1 Im ðnaÞsin nzðdnm cos mq þ enm sin mqÞ I ðnaÞ m; n¼1 m

∞ ∞ X 1X dn0 þ sin nzðdnm cos mq þ enm sin mqÞ. 2 n¼1 m; n¼1

9.3 Solving Laplace’s Equation in Cylindrical Coordinates

Then dnm

1 2 ¼ pp

Z

2

p

4

Z

3 2p q¼0

z¼0

f ðq; zÞcos mqdq5sin nzdz

and enm

1 2 ¼ pp

Z

p z¼0

2 4

Z

3 2p q¼0

f ðq; zÞsin mqdq5sin nzdz.

We note that the zcoordinate can be problematic. Some sources, such as Pinsky, Partial Differential Equations and Boundary-Value Problems with Applications, Third Edition, do the problem in polar coordinates. If no boundary conditions on Z(z) are given, the sign of C cannot be assigned. We have followed the problem and solution as given in Weinberger, A First Course in Partial Differential Equations.

EXERCISES 1. Solve Du ¼ 0; 0 < r < 1; 0 < z < p; uðr; q; 0Þ ¼ uðr; q; pÞ ¼ 0; uð1; q; zÞ ¼ zð1  zÞ. 2. Solve Du ¼ 0;

0 < r < 1; 0 < z < p; uðr; q; 0Þ ¼ uðr; q; pÞ ¼ 0;

uð1; q; zÞ ¼ sin q. 3. a. Solve Laplace’s equation on a cylinder in the case that the solution is independent of z and q. b. Use the result in part a to solve Laplace’s equation in the cylinder r1 0. As usual, we hypothesize uðr; q; tÞ ¼ RðrÞQðqÞTðtÞ so that Eq. (1) becomes 1 1 RðrÞQðqÞT 0 ðtÞ ¼ R00 ðrÞQðqÞTðtÞ þ R0 ðrÞQðqÞTðtÞ þ 2 RðrÞQ00 ðqÞTðtÞ r r

9.5 The Heat Equation on a Disk

and dividing by R(r) Q(q) T(t) gives T 0 ðtÞ R00 ðrÞ 1 R0 ðrÞ 1 Q00 ðqÞ ¼ þ þ . TðtÞ RðrÞ r RðrÞ r2 QðqÞ

(2)

The left-hand side of Eq. (2) is a function of t, and the right-hand side is a function of r and q, so each must be a constant that we designate k. We now determine the sign of k. We have T 0 ðtÞ ¼k TðtÞ so TðtÞ ¼ ekt . As there are no heat sources, u(r,q,t) cannot grow unboundedly. Thus k must be negative. We set k ¼ l2. Thus we have T 0 ðtÞ ¼ l2 TðtÞ. Setting R00 ðrÞ 1 R0 ðrÞ 1 Q00 ðqÞ þ þ ¼ l2 RðrÞ r RðrÞ r2 QðqÞ and rearranging gives r2

R00 ðrÞ R0 ðrÞ Q00 ðqÞ þr þ l2 r 2 ¼  . RðrÞ RðrÞ QðqÞ

Again, this means each side of the equation must be a constant that we denote j. We now determine the sign of j. We have Q00 ðqÞ  ¼j QðqÞ so QðqÞ ¼ jQðqÞ

and

QðqÞ þ jQðqÞ ¼ 0:

The periodicity conditions Q(p) ¼ Q(p), and Q0 (p) ¼ Q0 (p) force j  0. We set j ¼ m2. We thus have three equations T 0 ðtÞ ¼ l2 TðtÞ Q00 ðqÞ þ m2 QðqÞ ¼ 0 r2

R00 ðrÞ R0 ðrÞ þr þ l2 r 2 ¼ m2 . RðrÞ RðrÞ

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CHAPTER 9 Solving Partial Differential Equations

The last equation can be rewritten as  r 2 R00 ðrÞ þ rR0 ðrÞ þ l2 r 2  m2 RðrÞ ¼ 0: We want to analyze a Bessel equation of order 0. If we assume that u(r,q,t) is independent of q, then m2 ¼ 0 and we get r 2 R00 ðrÞ þ rR0 ðrÞ þ l2 r2 RðrÞ ¼ 0: This is a Bessel equation of order 0. It has two solutions, but only J0(r) is bounded at r ¼ 0 so that is the only admissible solution. Thus we have r 2 R00 ðrÞ þ rR0 ðrÞ ¼  l2 r 2 RðrÞ or 1 R00 ðrÞ þ R0 ðrÞ ¼ l2 RðrÞ. r

(3)

If we take 1 L½R ¼ R00 ðrÞ þ R0 ðrÞ r then Eq. (3) can be recognized as an eigenvalue/eigenvector problem L½R ¼ l2 RðrÞ for which J0(r) is an eigenfunction. We have imposed the boundary condition u(a,q,t) ¼ 0, t > 0, so we must have J0(a) ¼ 0. We want to emphasize the similarities between this problem and the familiar problem L½Q ¼ Q00 . This is an eigenvalue/eigenvector problem with two eigenfunctions, sin aq and cos aq. If we add the boundary condition Q(a) ¼ 0, then sin aq is the only admis   2 np x are the eigenfunctions and  np sible eigenfunction, and a ¼ np . Thus, sin a a a are the eigenvalues. Returning to the Bessel function case, we know that J0(r) (like sin q) has infinitely many values where the function is 0. Previously, we have designated these as x01, x02, x03, .. Then x  0n r J0 a are the eigenfunctions that satisfy the boundary conditions. Thus, the numbers x 2 0n  a

9.5 The Heat Equation on a Disk

are the eigenvalues for L½R ¼ l2 RðrÞ;

RðaÞ ¼ 0:

Going back to the problem of Fourier series, recall that if {f1(x), f2(x), .} is a complete orthogonal set of functions and f (x) is a suitably well-behaved function, then f ðxÞ ¼

∞ X

an fn ðxÞ

n¼1

where hf ; fn i . hfn ; fn i n  o The SturmeLiouville theory tells us that J0 xa0n r is a complete orthogonal an ¼

set of functions. The inner product for this problem is Z a f ðxÞxgðxÞdx hf ; gir ¼ 0

as we verified in Exercise 2, Section 6.2. In our particular problem, we set the initial condition to be u(r,q,0) ¼ f (r,q). Since we later assumed independence of q, we amend this so that the initial condition is u(r,q,0) ¼ f (r). Because of the independence of q, our solution is uðr; tÞ ¼

∞ X

an Rn ðrÞTðtÞ ¼

n¼1

∞ X

an J 0

x

0n

a

n¼1

 r ekt

with f ðrÞ ¼ uðr; 0Þ ¼

∞ X n¼1

so that

an J0

x

0n

a

 r

x  0n r dr f ðrÞrJ0 a x  x  . an ¼ R 0n 0n a 0 J0 a r rJ0 a r dr Ra 0

Thus we have 3 2 R x  0n a ∞   r dr f ðrÞrJ X 0 0 7 x0n 6 r . uðr; tÞ ¼ 4R a x0n  ax0n  5J0 a n¼1 0 J0 a r rJ0 a r dr

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CHAPTER 9 Solving Partial Differential Equations

EXERCISES 1. Describe (do not work out the calculus) the solution to the heat equation on a disk  of radius a with boundary condition u(a,q,t) ¼ 0 if uðr; 0Þ ¼ sin pr a . 2. Describe (do not work out the calculus) the solution to the heat equation on a disk of radius a with boundary condition u(a,q,t) ¼ 0 if u(r,0) ¼ a  r. 3. Describe (do not work out the calculus) the solution to the heat equation on a disk  of radius a with boundary condition u(a,q,t) ¼ 0 if uðr; 0Þ ¼ 1  cos pr a .

CHAPTER

Solving Partial Differential Equations in Spherical Coordinates Using Separation of Variables

10

10.1 AN EXAMPLE WHERE LEGENDRE EQUATIONS ARISE In Chapter 9 we studied solving partial differential equations (PDEs) in which the Laplacian appeared in cylindrical coordinates using separation of variables. We saw that among the differential equations that arose was a Bessel (or, at least a “Bessel-like”) equation. In this chapter, we follow a similar approach except we work in spherical coordinates. We shall see that in addition to a Bessel equation we encounter a differential equation called Legendre’s equation. Legendre equations arise when solving a PDE in spherical coordinates that uses the Laplacian. We demonstrate this with the wave equation. In spherical coordinates, the Laplacian is 1 v 2  1 v 1 r ur þ 2 ðsin q uq Þ þ 2 2 uww r 2 vr r sin q vq r sin q so the wave equation in spherical coordinates is Duðr; q; wÞ ¼



 1 v 2  1 v 1 utt ¼ KDu ¼ K 2 r ur þ 2 ðsin q uq Þ þ 2 2 uww . r vr r sin q vq r sin q We note that v ðsin quq Þ ¼ cos q uq þ sin q uqq vq and  2 1 v 2  1 r ur ¼ 2 2rur þ r 2 urr ¼ ur þ urr 2 r vr r r

337

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CHAPTER 10 Solving Partial Differential Equations in Spherical

so 1 v 2  1 v 1 r ur þ 2 ðsin quq Þ þ 2 2 uww 2 vr r r sin q vq r sin q   2 1 1 ur þ urr þ 2 ðcos quq þ sin quqq Þ þ 2 2 uww ¼ r r sin q r sin q     2 1 uww . ur þ urr þ 2 cot quq þ uqq þ 2 ¼ r r sin q

Duðr; q; wÞ ¼

(1)

We assume uðr; q; w; tÞ ¼ RðrÞQðqÞFðwÞTðtÞ . Then Eq. (1) can be written as 2 Duðr; q; wÞ ¼ ðR0 QFT þ R00 QFTÞ r   1 1 0 00 00 þ 2 cot qRQ FT þ RQ FT þ 2 RQF T r sin q and

  Du 2 R0 R00 1 Q0 Q00 1 F00 þ þ 2 cot q þ þ 2 . ¼ R r Q Q sin q F u r R

Also 1 utt 1 T 00 ¼ . K u K T We rewrite the wave equation as Du 1 T 00 ¼ u K T or

  2 R0 R00 1 Q0 Q00 1 F00 1 T 00 þ þ 2 cot q þ þ 2 ¼ . r R K T R r Q Q sin q F

(2)

In what follows, we argue that Eq. (2) can be solved by solving four ordinary differential equations (ODEs), one for each of the functions T, Q, F, and R. The algebra is somewhat tedious, but we are doing what we have done before. That is, we show, one step at a time, that independence of an expression of certain variables forces the expression to be constant. The expression on the right-hand side of Eq. (2) depends only on t, and the expression on the left-hand side is independent of t. Therefore each side is constant. Thus we have 1 T 00 ¼ l or T 00 þ lKT ¼ 0 K T where l > 0. (As before, the sign of the constants will be justified in the exercises.)

(3)

10.1 An Example Where Legendre Equations Arise

Multiply Eq. (2) by r2 to get   R0 R00 Q0 Q00 1 F00 1 T 00 þ 2 ¼ r2 ¼ lr 2 2r þ r 2 þ cot q þ R R Q Q sin q F K T so

  00 Q0 Q00 1 F00 R0 2R 2 cot q þ þ ¼  2r þ r þ lr . Q Q sin2 q F R R

(4)

The left-hand side of Eq. (4) is independent of r and the right-hand side depends only on r, so each side must be a constant. We set Q0 Q00 1 F00 þ þ 2 ¼ m. Q Q sin q F Multiply Eq. (5) by sin2 q to get cot q

sin2 q cot q

(5)

Q0 Q00 F00 þ sin2 q þ ¼ m sin2 q Q Q F

so Q0 Q00 F00 þ sin2 q þ m sin2 q ¼  . (6) Q Q F The terms on the left-hand side of Eq. (6) depend only on q, and the right-hand side depends only on 4, so F00 =F is another constant. We set sin2 q cot q

F00 ¼ n F

(7)

so F00 þ nF ¼ 0: Thus Eq. (6) is sin2 q cot q  Multiply by Q sin2 q to get

Q0 Q00 þ sin2 q þ m sin2 q ¼ n. Q Q

cot q Q0 þ Q00 þ mQ ¼ n so

 cot q Q þ Q þ m  0

00

Q sin2 q

 n Q ¼ 0: sin2 q

(8)

Finally, we determine the ODE that R(r) must satisfy. We began with Eq. (2)   2 R0 R00 1 Q0 Q00 1 F00 1 T 00 þ þ 2 cot q þ þ 2 ¼ R r Q Q sin q F r R K T

339

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CHAPTER 10 Solving Partial Differential Equations in Spherical

and have set 1 T 00 F00 ¼ l and ¼ n. K T F The equation F00 ¼ n F can be written F00 þ nF ¼ 0. Since F is periodic, n > 0, so we take n ¼ m2. So Eq. (2) could now be written     1 2 0 1 n 1 00 R þ R þ 2  m þ 2 2 ðnÞ þ l ¼ 0 R r r sin2 q r sin q or

  2 0 m

00 R þ R þ l  2 R ¼ 0: r r

If we look back at what we have done, we see that to solve the wave equation in spherical coordinates using separation of variables, we need to solve the following ODEs: T 00 þ lKT ¼ 0

(9)

F00 þ m2 F ¼ 0   n 0 00 Q¼0 cot q Q þ Q þ m  2 sin q   2 0 m

00 R þ R þ l  2 R ¼ 0: r r

(10) (11) (12)

The first two of these equations are familiar. The fourth is a “Bessel-like” equation. It is similar to what we considered in Chapter 9, but different enough so that we will give it a separate analysis. The third equation can be converted into a “Legendre-like” equation that we shall study. In the next sections, we discuss the solutions of these equations.

10.2 THE SOLUTION TO BESSEL’S EQUATION IN SPHERICAL COORDINATES In spherical coordinates, the “Bessel-like” equation that must be solved when analyzing equations that involve the Laplacian using separation of variables is d2 RðrÞ 2 dRðrÞ h mi þ l  þ RðrÞ ¼ 0 dr 2 r dr r2

10.2 The Solution to Bessel’s Equation in Spherical Coordinates

or d2 RðrÞ dRðrÞ 2 þ lr  m RðrÞ ¼ 0: þ 2r dr 2 dr In practice, it is most common that m ¼ n(n þ 1) and l ¼ k2 where n and k are positive integers. Thus we seek a solution to r2

  d2 RðrÞ 2 dRðrÞ nðn þ 1Þ 2 RðrÞ ¼ 0 þ þ k  dr 2 r dr r2 or d2 RðrÞ dRðrÞ 2 2 þ k r  nðn þ 1Þ RðrÞ ¼ 0 þ 2r (1) 2 dr dr that is bounded at r ¼ 0. We convert Eq. (1) to a Bessel equation by first making a change of variables and then transforming a function. The change of variables is x ¼ kr. Then r2

dRðrÞ dRðxÞ dx dRðxÞ d 2 RðrÞ d2 RðxÞ 2 ¼ ¼ k and ¼ k dr dx dr dx dr 2 dx2 so r

r

dRðrÞ x dRðxÞ dRðxÞ ¼ k¼x and dr k dx dx 2d

2

RðrÞ ¼ dr 2

 2 2 2 x d RðxÞ 2 2 d RðxÞ k ¼ x . k dx2 dx2

Thus r2

d2 RðrÞ dRðrÞ 2 2 þ 2r þ r k  nðn þ 1Þ RðrÞ 2 dr dr ¼ x2

d2 RðxÞ dRðxÞ 2 þ x  nðn þ 1Þ RðxÞ ¼ 0: þ 2x 2 dx dx

Now let YðxÞ ¼

pﬃﬃﬃ 1 xRðxÞ or RðxÞ ¼ x2 YðxÞ.

Then 1 dYðxÞ dRðxÞ 1 3 ¼ x 2  x2 YðxÞ dx dx 2

341

342

CHAPTER 10 Solving Partial Differential Equations in Spherical 2 3 1 2 3 3 5  d YðxÞ d RðxÞ dYðxÞ 1  1 6  dYðxÞ 3  7 ¼x 2  x 2  4x 2  x 2 YðxÞ5 2 2 dx 2 2 dx 2 dx dx 2

1 3 5  d 2 YðxÞ  dYðxÞ 3  2 2 YðxÞ. ¼x 2  x þ x dx 4 dx2 So d2 RðxÞ dRðxÞ 2 þ x  nðn þ 1Þ RðxÞ þ 2x 2 dx dx 2 3 2 3 1 2 3 5 1 3  dYðxÞ 3  6  d YðxÞ 7 6  dYðxÞ 1 2 7 þ x 2 YðxÞ5 þ 2x4x 2  x YðxÞ5 ¼ x2 4 x 2 x 2 2 dx 4 dx 2 dx

x2

1  þ x2  nðn þ 1Þ x 2 YðxÞ ¼ 0: Multiplying by x1=2 gives x2

 2    d YðxÞ 3 2 dYðxÞ 1 1 1 dYðxÞ  x YðxÞ þ 2x YðxÞ þ x  x dx 4 dx 2 dx2 2 þ x  nðn þ 1Þ YðxÞ ¼ 0

or   YðxÞ dYðxÞ 3 2 þ ðx þ 2xÞ x þ  1 þ x  nðn þ 1Þ YðxÞ dx 4 dx2    2 dYðxÞ 1 2 d YðxÞ 2 2 ¼x þx þ x  n þnþ YðxÞ dx 4 dx2 "   # 2 dYðxÞ 1 2 2 d YðxÞ 2 þx þ x  nþ YðxÞ ¼ 0: ¼x dx 2 dx2 2d

2

The equation x

2d

2 YðxÞ

dx2

"  #  dYðxÞ 1 2 2 þ x  nþ YðxÞ ¼ 0 þx dx 2

(2)

10.2 The Solution to Bessel’s Equation in Spherical Coordinates

is a Bessel equation of half-integer order. The requirement that the solution be bounded at x ¼ 0 means the solution is of the form YðxÞ ¼ AJnþ1 ðxÞ. 2

Now Jnþ1 ðxÞ Jnþ1 ðkrÞ YðxÞ RðxÞ ¼ pﬃﬃﬃ ¼ A p2ﬃﬃﬃ and RðrÞ ¼ A p2ﬃﬃﬃﬃﬃ . x x kr pﬃﬃﬃﬃﬃﬃﬃﬃ It is common to take the constant A to be p=2 and denote the spherical solution of the first kind (the solution that is bounded when r ¼ 0) by jn ðrÞ ¼

p 1=2 Jnþ1 ðrÞ p2ﬃﬃ 2 r

so that RðrÞ ¼

p 1=2 Jnþ1 ðkrÞ p2ﬃﬃﬃﬃﬃ hjn ðkrÞ. 2 kr

Next we find the solution to   2 kðk þ 1Þ yðrÞ þ y0 ðrÞ þ l  yðrÞ ¼ 0 r r2 or

r 2 yðrÞ þ 2ry0 ðrÞ þ lr2  kðk þ 1Þ yðrÞ ¼ 0: We consider the case l > 0. We let yðrÞ ¼

∞ X

an r nþa

n¼0

and proceed as we did in Chapter 9. We have y0 ðrÞ ¼

∞ X

an ðn þ aÞr nþa1

n¼0

y00 ðrÞ ¼

∞ X

an ðn þ aÞðn þ a  1Þrnþa2

n¼0

so ry0 ðrÞ ¼

∞ X n¼0

an ðn þ aÞr nþa

r 2 y00 ðrÞ ¼

∞ X n¼0

an ðn þ aÞðn þ a  1Þr nþa .

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Then

r 2 yðrÞ þ 2ry0 ðrÞ þ lr 2  kðk þ 1Þ yðrÞ ¼

∞ X

an ½ðn þ aÞðn þ a  1Þ þ 2ðn þ aÞ  kðk þ 1Þrnþa þ l

n¼0

¼ ¼

∞ X

∞ X

an rnþaþ2

n¼0

an ½ðn þ aÞðn þ a þ 1Þ  kðk þ 1Þrnþa þ l

∞ X

n¼0

n¼0

∞ X

∞ X

an ½ðn þ aÞðn þ a þ 1Þ  kðk þ 1Þrnþa þ l

n¼0

an r nþaþ2 an2 r nþa

n¼2 a

¼ a0 ½ðaÞða þ 1Þ  kðk þ 1Þr þ a1 ½ð1 þ aÞð2 þ aÞ  kðk þ 1Þraþ1 þ

∞ X

fan ½ðn þ aÞðn þ a þ 1Þ  kðk þ 1Þ þ lan2 gr nþa ¼ 0:

n¼2

Each coefficient must be 0, so a0 ½ðaÞða þ 1Þ  kðk þ 1Þ ¼ 0, and if a0 s 0, then a ¼ k. If a ¼ k, since a1 ½ð1 þ aÞð2 þ aÞ  kðk þ 1Þ ¼ 0, then a1 ¼ 0. Since an ½ðn þ aÞðn þ a þ 1Þ  kðk þ 1Þ þ lan2 ¼ an ½ðn þ kÞðn þ k þ 1Þ  kðk þ 1Þ þ lan2 ¼ 0 we have the recurrence relation an ¼

lan2 lan2 ¼ . ðn þ kÞðn þ k þ 1Þ  kðk þ 1Þ nðn þ 2k þ 1Þ

If a1 ¼ 0, then a2nþ1 ¼ 0 for the very positive integer n. By induction, one can show # " ∞ X ðlÞn r 2n k yðrÞ ¼ a0 r 1 þ 2n n!ð2k þ 3Þð2k þ 5Þ/ð2k þ 2n þ 1Þ n¼1 # " ∞ X ðlÞn r 2n k     .  ¼ a0 r 1 þ 3 5 1 n¼1 22n n! k þ kþ / kþnþ 2 2 2 Let l ¼ k þ 1=2. Then the right-hand side of Eq. (3) is # " ∞ X ðlÞn r2n l12 a0 r . 1þ 22n n!ðl þ 1Þðl þ 2Þ/ðl þ nÞ n¼1

(3)

10.3 Legendre’s Equation and Its Solutions

If we set l ¼ 1 and denote the resulting expression by jk(r), we have a0 a0 jk ðrÞ ¼ pﬃﬃ Jl ðrÞ ¼ pﬃﬃ Jkþ1=2 ðrÞ. r r The function jk(r) is called pﬃﬃﬃ the kth spherical Bessel function. We next show that jk l r satisfies 2 m y00 ðrÞ þ y0 ðrÞ  2 yðrÞ ¼ lyðrÞ. r r pﬃﬃﬃ pﬃﬃﬃ  pﬃﬃﬃ d d yðzÞ ¼ d y We let z ¼ lr so that dr l r ¼ l dz yðzÞ. dr Also d2 pﬃﬃﬃ

d2 y l r ¼ l 2 yðzÞ. 2 dr dz Then d 2 pﬃﬃﬃ 2 d pﬃﬃﬃ m pﬃﬃﬃ

y lr þ y lr  2y lr r dr dr 2 r pﬃﬃﬃ 2 d 2 l d m ¼ l 2 yðzÞ þ yðzÞ  2 yðzÞ r dz r dz " # d2 2 d m yðzÞ  pﬃﬃﬃ 2 yðzÞ ¼ l 2 yðzÞ þ pﬃﬃﬃ dz lr dz lr   2 d 2 d m ¼ l 2 yðzÞ þ yðzÞ  2 yðzÞ ¼ lyðzÞ z dz dz z since d2 2 d m yðzÞ  2 yðzÞ ¼ yðzÞ. yðzÞ þ z dz z dz2 What we have done is shown that Bessel functionsdthe solutions to Bessel’s equationdare eigenfunctions for a particular linear operator. In the next section we show that Legendre polynomialsdthe solutions to Legendre’s equationdare also eigenfunctions of a linear operator. These problems are one case of a general theory, where the differential operator is self-adjoint so that eigenfunctions belonging to different eigenvalues are orthogonal with respect to an appropriate weight function.

10.3 LEGENDRE’S EQUATION AND ITS SOLUTIONS We have seen that in solving a PDE that uses the Laplacian by separation of variables in spherical coordinates, it is necessary to solve   n Q ¼ 0: (1) cot q Q0 þ Q00 þ m  2 sin q

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Our procedure will be similar to what we did in solving “Bessel-like” equations. We begin by solving Legendre’s equation. With a change of variables, Eq. (1) can be transformed to a “Legendre-like” equation. We then solve the “Legendre-like” equation. Legendre’s equation is   (2) 1  x2 y00 ðxÞ  2xy0 ðxÞ þ myðxÞ ¼ 0  1 < x < 1: We solve the equation by power series; that is, we assume yðxÞ ¼

∞ X

an x n

n¼0

and determine a recurrence relation for the an 0 s. We have y0 ðxÞ ¼

∞ X

an nxn1 and y00 ðxÞ ¼

n¼0

so



∞ X

an nðn  1Þxn2

n¼0

 1  x2 y00 ðxÞ  2xy0 ðxÞ þ myðxÞ ∞ ∞ ∞ ∞ X X X X an nðn  1Þxn2  an nðn  1Þxn  2 an nxn þ m an xn ¼ 0: ¼ n¼0

n¼0

n¼0

n¼0

(3) Now ∞ X

an nðn  1Þxn2 ¼

n¼0

∞ X

an nðn  1Þxn2 ¼

n¼2

∞ X

anþ2 ðn þ 2Þðn þ 1Þxn

n¼0

so Eq. (3) can be written as ∞ X

anþ2 ðn þ 2Þðn þ 1Þxn 

n¼0

∞ X

an ½nðn  1Þ þ 2n  mxn

n¼0

¼

∞ X

anþ2 ðn þ 2Þðn þ 1Þxn 

n¼0

∞ X

an n2 þ n  m xn ¼ 0:

n¼0

Thus we have 

 anþ2 ðn þ 2Þðn þ 1Þ ¼ an n þ n  m or anþ2 ¼ 2

So we have two linearly independent solutions y0 ðxÞ ¼ a0 þ a2 x2 þ a4 x4 þ / y1 ðxÞ ¼ a1 x þ a3 x3 þ a5 x5 þ /.



 n2 þ n  m an . ðn þ 2Þðn þ 1Þ

10.3 Legendre’s Equation and Its Solutions

In many problems, the value of m is k(k þ 1), where k is a positive integer. We show that if this is the case, then there is a solution to Legendre’s equation that is a polynomial of degree k. First, note that in the case m ¼ k(k þ 1) we have  2  n þ n  k2  k an anþ2 ¼ ðn þ 2Þðn þ 1Þ so



akþ2

 k2 þ k  k 2  k ak ¼ 0 ¼ ðk þ 2Þðk þ 1Þ

and if a0 ¼ 0, then akþ2 ¼ akþ4 ¼ akþ6 ¼ . ¼ 0. To get the polynomial solution when k is even, take a1 ¼ 0 and a0 s 0. The solution is yðxÞ ¼ a0 þ a2 x2 þ a4 x4 þ / þ ak xk . To get the polynomial solution when k is odd, take a0 ¼ 0 and a1 s 0. The solution is yðxÞ ¼ a1 x þ a3 x3 þ / þ ak xk . In these cases, the solution is called the Legendre polynomial of degree k. We have seen how the differential equation   n cot q Q0 þ Q00 þ m  2 Q¼0 sin q arises in problems that involve the Laplacian which have spherical symmetry. We now convert this to a Legendre equation. We let x ¼ cos q. Then Q0 ¼

dQ dQ dx dQ ¼ ¼ sin q . dq dx dq dx

To compute Q00 , we have Q00 ¼

        d dQ d dQ d dQ dQ d ¼  sin q ¼ sin q þ ðsin qÞ. dq dq dq dx dq dx dx dq

Now     d dQ d dQ dx d 2 Q ¼ ¼ 2 ðsin qÞ dq dx dx dx dq dx

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CHAPTER 10 Solving Partial Differential Equations in Spherical

so

    d dQ dQ d þ ðsin qÞ dq dx dx dq   d2 Q dQ ¼ ðsin qÞ 2 ðsin qÞ þ ðcos qÞ dx dx

Q00 ¼ sin q

¼ sin2 q Thus

d2 Q dQ .  cos q 2 dx dx

  n

dQ Q ¼ cot q  sin q dx sin2 q     d2 Q dQ n Q þ m 2 þ sin2 q 2  cos q dx dx sin q   2 dQ dQ n 2 d Q Q ¼ cos q þ sin q 2  cos q þ m 2 dx dx dx sin q   2 dQ n 2 d Q ¼ sin q 2  2 cos q Q ¼ 0: þ m 2 dx dx sin q

cot q Q þ Q þ m  0

00

Now use cos q ¼ x;

sin2 q ¼ 1  x2

to get 

1x

2

  dQ n Q ¼ 0:  2x þ m dx2 dx 1  x2

 d2 Q

(4)

This is not exactly of the form of Legendre’s equation, but if we take m ¼ l(l þ 1) and n ¼ m2, then Eq. (4) is     d2 Q dQ m2 Q ¼ 0: (5)  2x 1  x2 þ lðl þ 1Þ  dx2 dx 1  x2 Eq. (5) often is the equation that needs to be solved in physical problems. In such problems, m and l are integers with m  l. The solution is called an associated Legendre function, which we study in the next section.

EXERCISES 1. Legendre’s polynomial of degree n, denoted Pn(x), is a solution (there are two) to the differential equation   1  x2 y00 ðxÞ  2xy0 ðxÞ þ nðn þ 1ÞyðxÞ ¼ 0; 1 < x < 1 where n is a nonnegative integer.

10.4 Associated Legendre Functions

a. Verify that P0(x) ¼ 1 and P1(x) ¼ x are Legendre polynomials. b. Given that Legendre polynomials satisfy the recursion relation ðn þ 1ÞPnþ1 ðxÞ  ð2n þ 1ÞxPn ðxÞ þ nPn1 ðxÞ ¼ 0;

n  1;

find P2(x), P3(x), and P4(x). 2. Rodrigues’ formula can be used to generate Legendre polynomials. This formula is n ð1Þn d n  1  x2 . n n 2 n! dx Verify that Rodrigues’ formula is valid for n ¼ 0, 1, 2, 3. 3. Verify for n ¼ 0, 1, 2, 3 that Z 1 2 ½Pn ðxÞ2 dx ¼ . 2n þ 1 1 Pn ðxÞ ¼

(Thus Legendre’s polynomials are not normalized.) 4. If f ðxÞ ¼

∞ X

an Pn ðxÞ

n¼0

show that ak ¼

  Z 1 1 f ðxÞPk ðxÞdx . ð2k þ 1Þ 2 1

5. Use the result of Exercise 4 to find the first two nonzero terms for the expansion of f (x) ¼ x2 in terms of Legendre polynomials.

10.4 ASSOCIATED LEGENDRE FUNCTIONS We relate the solution of 

   m2 yðxÞ ¼ 0 1  x2 y00 ðxÞ  2xy0 ðxÞ þ lðl þ 1Þ  1  x2

to Legendre polynomials using the substitution  m yðxÞ ¼ 1  x2 2 uðxÞ where u(x) is a solution to   1  x2 u00 ðxÞ  2xðm þ 1Þu0 ðxÞ þ ½lðl þ 1Þ þ mðm þ 1ÞuðxÞ ¼ 0:

(1)

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CHAPTER 10 Solving Partial Differential Equations in Spherical

With this substitution, we have m m1   y0 ¼ 1  x2 2 u0  mx 1  x2 2 u so that m  m1  2xy0 ¼ 2x 1  x2 2 u0 þ 2mx2 1  x2 2 u. Also   m m1 m1 m1    d  . y00 ¼ 1  x2 2 u00  mx 1  x2 2 u0  mx 1  x2 2 u0 þ mu x 1  x2 2 dx Now

2



d 4 x 1x dx

m 2 1 2

3

2 3

 m m1 m2   5 ¼ 1  x2 2 þ 4x  1 1  x2 2 ð 2xÞ5 2 

 m  m1 m2  1 1  x2 2 . ¼ 1  x2 2  2x2 2 Thus    m m1 m1 y00 ¼ 1  x2 2 u00  2mx 1  x2 2 u0  mu 1  x2 2

 m m2 þ2mux2  1 1  x2 2 2 so 

 mþ1 m m    1  x2 y00 ¼ 1  x2 2 u00  2mx 1  x2 2 u0  mu 1  x2 2

 m m1  1 1  x2 2 . þ2mux2 2

Since  m1 m  2xy0 ¼ 2mx2 1  x2 2 u  2x 1  x2 2 u0 we have 

 mþ1 m   1  x2 y00  2xy0 ¼ 1  x2 2 u00  2xðm þ 1Þ 1  x2 2 u0 3 2

 m m1 m m1      þ42mx2  1 1  x2 2  m 1  x2 2 þ 2mx2 1  x2 2 5u. 2

10.4 Associated Legendre Functions

Thus the coefficient of the u term in 

   m2 y 1  x2 y00  2xy0 þ lðl þ 1Þ  1  x2  mþ1 m  ¼ 1  x2 2 u00  2xðm þ 1Þ 1  x2 2 u0 2

 m  m  1 m  m1  1 1  x2 2  m 1  x2 2 þ 2mx2 1  x2 2 þ42mx2 2 3   m   m2 1  x2 2 5 u ¼ 0 þ lðl þ 1Þ  1  x2

is 2mx2

 m m1 m  m1   1 1  x2 2  m 1  x2 2 þ 2mx2 1  x2 2 2    m m2 2 2 1  x þ lðl þ 1Þ  1  x2

m i  m  1 h m   1 þ 2mx2  m2 þ 1  x2 2 ½lðl þ 1Þ  m 2mx2 ¼ 1  x2 2 2 m  1     m m2 x2  m2 þ 1  x2 2 ½lðl þ 1Þ  m ¼ 1  x2 2  m  m ¼ m2 1  x2 2 þ 1  x2 2 ½lðl þ 1Þ  m m  ¼ 1  x2 2 ½lðl þ 1Þ  mðm þ 1Þ.

Hence 

   m2 y 1  x2 y  2xy0 þ lðl þ 1Þ  1  x2  m þ 1 00 m m   ¼ 1  x2 2 u  2xðm þ 1Þ 1  x2 2 u0 þ 1  x2 2 ½lðl þ 1Þ  mðm þ 1Þu m     ¼ 1  x2 2 1  x2 u00  2xðm þ 1Þu0 þ ½lðl þ 1Þ  mðm þ 1Þu .

So 

  m2 y¼0 1  x y  2xy þ lðl þ 1Þ  1  x2 2



00

0

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CHAPTER 10 Solving Partial Differential Equations in Spherical

if and only if 

 1  x2 u00  2xðm þ 1Þu0 þ ½lðl þ 1Þ  mðm þ 1Þu ¼ 0:

(2)

Note that if m ¼ 0, then Eq. (2) is Legendre’s equation. Recall that Legendre’s equation has two linearly independent solutions, and if l is a positive integer, then one of the solutions is a polynomial that we denote Pl(x). We now demonstrate that Pl 0 ðxÞ is a solution to Eq. (2) when m ¼ 1. Differentiating Eq. (2) gives   1  x2 u000  2xu00  2ðm þ 1Þu0  2xðm þ 1Þu00 þ ½lðl þ 1Þ  mðm þ 1Þu0   ¼ 1  x2 u000  2xðm þ 2Þxu00  ½lðl þ 1Þ  mðm þ 1Þu0 ¼ 0: This means that if Pl(x) is a solution to Eq. (2) for m ¼ 0, that is, if   1  x2 Pl 00 ðxÞ  2xPl 0 ðxÞ þ lðl þ 1ÞPl ðxÞ ¼ 0 then



 1  x2 Pl 000 ðxÞ  2xPl 00 ðxÞ þ ½lðl þ 1Þ  1ð1 þ 1ÞPl 0 ðxÞ ¼ 0:

That is, Pl 0 ðxÞ solves Eq. (2) when m ¼ 1. One can follow this procedure to show d m P ðxÞ is a solution to Eq. (2) for any positive integer m. This inductively that dx m l means  m d m y ¼ 1  x2 2 m Pl ðxÞ dx is a solution to Eq. (1). These solutions are denoted Pl m ðxÞ and  m d m Pl m ðxÞ ¼ 1  x2 2 m Pl ðxÞ dx are called associated Legendre functions. We return to the original problem of solving n

cot q Q0 þ Q00 þ m  2 Q ¼ 0 sin q in the case that m ¼ l(l þ 1) and n ¼ m2 where l and m are integers with m  l. Thus we want to solve   m2 0 00 cot q Q þ Q þ lðl þ 1Þ  2 Q ¼ 0: (3) sin q With the substitution x ¼ cos q, Eq. (3) became    2  dQ m2 2 d Q þ lðl þ 1Þ  Q ¼ 0:  2x 1x dx2 dx 1  x2

(4)

10.5 Laplace’s Equation in Spherical Coordinates

We have shown that the solution to Eq. (4) is m dm  Pl m ðxÞ ¼ 1  x2 2 m Pl ðxÞ dx and so the solution to Eq. (3) is Pl m ðcos qÞ.

EXERCISES 1. Use the formula  m dm Pl m ðxÞ ¼ 1  x2 2 m Pl ðxÞ dx to compute Pl m ðcos qÞ for a. l ¼ 0; m ¼ 0: b. l ¼ 0; m ¼ 1: c. l ¼ 1; m ¼ 1: d. l ¼ 2; m ¼ 1: e. l ¼ 3; m ¼ 3:

10.5 LAPLACE’S EQUATION IN SPHERICAL COORDINATES In this section, we solve the boundary value problem Duðr; q; wÞ ¼ 0;

0 < r < a;

uða; q; wÞ ¼ f ðq; wÞ

using separation of variables. The approach is the one we have been using; namely, we hypothesize that uðr; q; wÞ ¼ RðrÞQðQÞFð4Þ and find an ODE of each R, Q, and F. We solve each of the ODEs, and each solution will involve arbitrary constants. We then use the boundary condition uða; q; wÞ ¼ f ðq; wÞ to determine the constants. We have already done most of the preliminary work, but we now repeat it. In Section 10.1, we found that in spherical coordinates   2 1 uww

ur þ urr þ 2 cot quq þ uqq þ 2 . Duðr; q; wÞ ¼ r r sin q Following what we did in Section 9.1, we get   Du 2 R0 R00 1 Q0 Q00 1 F00 ¼ þ þ þ ¼ 0: cot q þ u r R R r2 Q Q sin2 q F

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CHAPTER 10 Solving Partial Differential Equations in Spherical

We then get   2 R0 R00 1 Q0 Q00 1 F00 þ ¼  2 cot q þ þ 2 r R r R Q Q sin q F so

  0 2 R R00 Q0 Q00 1 F00 þ ¼ cot q þ þ 2 . r R Q Q sin q F r R 2

(1)

We repeat the familiar argument that the left-hand side of Eq. (1) is a function of r and the right-hand side is a function of q and 4, so each is a constant (that will turn out to be negative) that we denote m. We then have cot q

Q0 Q00 1 F00 þ þ 2 ¼ m Q Q sin q F

and

  0 2 R R00 þ ¼ m. r R r R 2

(2)

(3)

Multiplying Eq. (2) by sin2 q gives sin2 q cot q

Q0 Q00 F00 þ sin2 q þ ¼ m sin2 q Q Q F

so Q0 Q00 F00 þ sin2 q þ m sin2 q ¼  . (4) Q Q F The left-hand side of Eq. (4) is a function of q and the right-hand side a 4, so each is a constant that we denote n. We now show that n > 0. We have sin2 q cot q



F00 ¼F F

so F þ nF ¼ 0: The periodicity of F forces n > 0. We let m2 ¼ n. Thus we have F00 þ m2 F ¼ 0 and so Fm ð4Þ ¼ am cosðm4Þ þ bm sinðm4Þ.

(5)

10.5 Laplace’s Equation in Spherical Coordinates

We also have Q0 Q00 þ m sin2 q ¼ m2 sin2 q cot q þ sin2 q Q Q  and multiplying by Q sin2 q gives cot qQ0 þ Q00 þ mQ ¼

m2 Q sin2 q

or   m2 Q00 þ cot qQ0 þ m  2 Q ¼ 0: sin q In problems that concern us m ¼ l(l þ 1), so Eq. (6) becomes   m2 00 0 Q ¼ 0: Q þ cot qQ þ lðl þ 1Þ  2 sin q

(6)

(7)

In Section 10.4, we found that the solution to Eq. (7) is Pl m ðcos qÞ. Eq. (3) states   0 2 R R00 þ ¼m r2 r R R so

  1 2 m R00 þ R0 ¼ 2 R r r

or 2 m R00 þ R0  2 R ¼ 0: r r If m ¼ l(l þ 1), then we have 2 lðl þ 1Þ R00 þ R0  ¼0 r r2 or r2 R00 þ 2rR0  lðl þ 1ÞR ¼ 0: As we show in Exercise 1, the solution to Eq. (8) is RðrÞ ¼ ar l .  If we want R(a) ¼ 1, we take a ¼ 1 al .

(8)

355

356

CHAPTER 10 Solving Partial Differential Equations in Spherical

Thus Duðr; q; wÞ ¼ 0;

0 0 with the boundary condition uðx; 0Þ ¼ f ðxÞ and suppose that uðx; yÞ is bounded. We take the Fourier transform with respect to x to get Z ∞ 1 F fuðx; yÞg ¼ pﬃﬃﬃﬃﬃﬃ uðx; yÞeikx dxhUðk; yÞ. 2p ∞ Then F fuxx ðx; yÞg ¼ k2 Uðk; yÞ;

F fuyy ðx; yÞg ¼

d2 Uðk; yÞ; dy2

F fuðx; 0Þg ¼ Uðk; 0Þ ¼ F ff ðxÞg ¼ FðkÞ. Taking the Fourier transform of Laplace’s equation and the boundary condition above, we get the ordinary differential equation in y, d2 Uðk; yÞ  k2 Uðk; yÞ ¼ 0; dy2

Uðk; 0Þ ¼ FðkÞ.

The solution to this equation is Uðk; yÞ ¼ Aejkjy þ Bejkjy . If we assume that U(k,y) is bounded as y / ∞, then Uðk; yÞ ¼ Bejkjy . The initial condition gives B ¼ U(k,0) ¼ F (k), so Uðk; yÞ ¼ FðkÞejkjy . If we take the inverse Fourier transform of U(k,y), we obtain the solution u(x,y). In taking the inverse transform of FðkÞejkjy , we are taking the inverse of the product of two transforms, and this can be accomplished by using the convolution theorem. To apply the convolution theorem to this problem, we need to find   F 1 ejkjy .

11.5 Solving Partial Differential Equations Using the Fourier Transform

Now

Z ∞ n o 1 F 1 ejkjy ¼ pﬃﬃﬃﬃﬃﬃ ejkjy eikx dk 2p k¼∞ Z 0 1 eky ½cosðkxÞ þ i sinðkxÞdk ¼ pﬃﬃﬃﬃﬃﬃ 2p ∞ Z ∞ 1 þ pﬃﬃﬃﬃﬃﬃ eky ½cosðkxÞ þ i sinðkxÞdk. 2p 0

In

Z

0 ∞

eky ½cosðkxÞ þ i sinðkxÞdk

let m ¼ k. Then Z Z 0 ky e ½cosðkxÞ þ i sinðkxÞdk ¼ ∞

¼

0

emy ½cosðmxÞ þ i sinðmxÞðdmÞ

∞ Z ∞

emy ½cosðmxÞ  i sinðmxÞdm:

0

Thus

Z

ejkjy eikx dk ¼ 2

k¼∞

eky cosðkxÞdk.

0

Integrating by parts twice gives Z ∞ eky cosðkxÞdk ¼ 0

Thus

Z

y x2 þ y2

if y > 0:

n o 1 2y : F 1 ejkjy ¼ pﬃﬃﬃﬃﬃﬃ 2 2p x þ y2

If we let gbðkÞ ¼ ejkjy , we have 1 2y gðxÞ ¼ pﬃﬃﬃﬃﬃﬃ 2 2p x þ y2 and gðx  zÞ ¼

2y ðx  zÞ2 þ y2

.

So uðx; yÞ ¼ F 1 fUðk; yÞg ¼ F 1 fFðkÞGðkÞg ¼ ðf  gÞðxÞ Z 1 ∞ y f ðzÞ ¼ dz. p ∞ ðx  zÞ2 þ y2 This is one form of Poisson’s integral formula.

389

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CHAPTER 11 The Fourier Transform

In Section 12.3 we use Poisson’s integral in polar coordinates, which we developed in Exercise 8 of Section 4.3. This is the solution to the boundary value problem on a circle 1 1 Vuðr; qÞ ¼ urr þ ur þ 2 u qq ¼ 0 for r < R; uðR; qÞ ¼ f ðqÞ r r which is given by Z R2  r 2 p f ð4Þ uðr; qÞ ¼ d4. 2 2 2p p R þ r  2rR cosðq  4Þ

EXERCISES Solve the following PDEs using the Fourier transform. 1. ut ðx; tÞ ¼ Kuxx ðx; tÞ; t  0: Assume

1 ; uðx; 0Þ ¼ 1þx 2

 ∞ < x < ∞;

lim uðx; tÞ ¼ 0:

x/∞

2. ut ðx; tÞ ¼ Kuxx ðx; tÞ; uðx; 0Þ ¼ cex ; t  0: Assume lim uðx; tÞ ¼ 0: 2

 ∞ < x < ∞;

x/∞

3. Use Euler’s formula eiq ¼ cos q þ i sin q to show cosðcxtÞ ¼

eicxt þ eicxt eicxt  eicxt and sinðcxtÞ ¼ . 2 2i

4. Solve the wave equation utt ðx; tÞ  c2 uxx ðx; tÞ ¼ 0; uðx; 0Þ ¼

1 ; 1 þ x2

t > 0;

ut ðx; 0Þ ¼ 0:

5. In digital processing, the function sincðxÞ ¼

∞ 0 we have Z ∞  pﬃﬃﬃ  1 p 1 ðA þ cÞ f ðxÞ ¼ pﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ exp  c jx  yj f ðxÞdx c 2p ∞ if f (x) is sufficiently well behaved. We let m2 ¼ c and consider u00 ðxÞ þ m2 uðxÞ ¼ f ðxÞ

(1)

where f (x) is a rapidly decreasing, infinitely differentiable function. Proceeding formally, we rewrite Eq. (1) as

d2  2 þ m2 uðxÞ ¼ f ðxÞ dx and uðxÞ ¼



d2 þ m2 dx2

1 f ðxÞ.

Applying the Fourier transform to Eq. (1), we get b k2 Ub ðkÞ þ m2 Ub ðkÞ ¼ FðkÞ so  1 b Ub ðkÞ ¼ k2 þ m2 FðkÞ.  2 1 b If there is a function g(x) for which GðkÞ ¼ k þ m2 ; then    1 b FðkÞ b b FðkÞ ¼ GðkÞ ¼ gd  f ðkÞ Ub ðkÞ ¼ k2 þ m2 and uðxÞ ¼ ðg  f ÞðxÞ. To find g(x), we take the inverse Fourier transform of (k2 þ m2)1. That is, Z ∞ 1 eikx dk. (2) gðxÞ ¼ pﬃﬃﬃﬃﬃﬃ 2 2p ∞ k þ m2

11.6 The Spectrum of the Negative Laplacian in One Dimension

CR mi

−R

R

FIGURE 11.6.1

We shall show that Z

∞

k2

eikx p dk ¼ emjxj 2 m þm

for m > 0:

We do this by using the residue theorem for two cases. The first case is for x > 0. Consider the contour shown in Fig. 11.6.1. The only pole of f ðkÞ ¼

eikx eikx ¼ k2 þ m2 ðk  miÞðk þ miÞ

that occurs within the contour is a simple pole at k ¼ mi. The residue of f (k) at k ¼ mi is lim ðk  miÞ

k/mi

eikx eiðmiÞx emx ¼ ¼ . ðk  miÞðk þ miÞ ðmi þ miÞ 2mi

Thus Z R

mx Z Z eikx e eikx p mx eikx e  dk ¼ 2pi dk ¼  dk. 2 2 2 2 2 2 m 2mi CR k þ m CR k þ m R k þ m

 ikxIf k is a point on CR, then, because the imaginary part of k > 0 and x > 0, we have e  < 1: Also for k a point on CR,    1  1   . k 2 þ m 2   ðR  mÞ2 Since the length of CR is pR, we Z  Z   eikx    k2 þ m2 dk  CR

have  ikx   e  1   pR k2 þ m2 dk  ðR  mÞ2 CR

393

394

CHAPTER 11 The Fourier Transform

−R

R

mi

CR

FIGURE 11.6.2

and so Z    eikx  ¼ 0: dk lim   2 2 R/∞ CR k þ m Thus for x > 0 we have Z ∞

eikx p dk ¼ emx 2 2 m ∞ k þ m

for m > 0:

The second case is for x < 0. Consider the contour in Fig. 11.6.2. In this case, the only pole of f ðkÞ ¼

k2

eikx eikx ¼ 2 þm ðk  miÞðk þ miÞ

that occurs within the contour is a simple pole at k ¼ mi. The residue of f (k) at k ¼ mi is lim ðk þ miÞ

k/mi

eikx eiðmiÞx emx ¼ ¼ . ðk  miÞðk þ miÞ ðmi  miÞ 2mi

Thus for x < 0 " #

Z Z R Z R eikx eikx emx eikx dk ¼  dk ¼  2pi dk  2 2 2 2 2mi k2 þ m2 CR k þ m R k þ m R

Z Z emx eikx p mx eikx þ dk ¼ þ dk. e 2 2 2 2 2mi m CR k þ m CR k þ m

¼ 2pi

 ikxIf k is a point on CR, then, because the imaginary part of k < 0 and x < 0, we have e  < 1: Also for k is a point on CR,    1  1   . k 2 þ m 2   ðR  mÞ2

11.7 The Fourier Transform in Three Dimensions

Since the length of CR is pR, we  Z Z   eikx    k2 þ m2 dk  CR

again have  ikx   e    k2 þ m2 dk 

CR

1 ðR  mÞ2

pR

and so Z    eikx  dk ¼ 0: lim R/∞ CR k2 þ m2 Thus for x < 0 we have Z ∞ ∞

k2

eikx p dk ¼ emx þ m2 m

From the two cases we conclude Z ∞ eikx p dk ¼ emjxj 2 þ m2 m k ∞ Thus 1 uðxÞ ¼ fd  gðxÞ ¼ pﬃﬃﬃﬃﬃﬃ 2p

Z

for m > 0:

for m > 0:

1 p gðx  yÞf ðyÞdy ¼ pﬃﬃﬃﬃﬃﬃ 2p m ∞

Z

∞ ∞

emjxyj f ðyÞdy.

So we have formally written

1 Z d2 1 p ∞ mjxyj 2  2þm f ðxÞ ¼ pﬃﬃﬃﬃﬃﬃ e f ðyÞdy; dx 2p m ∞

(3)

where m  > 0. 1 d 2 þ m2 is given by Eq. (3), and at least part of the spectrum of Thus  dx 2  2  d dx2 is the positive real axis. In fact, the positive real axis is the entire spectrum of D. This is because  1  d 2  m2 is defined and bounded, and because  d 2 dx2 is self-adjoint,  dx 2 the spectrum is a subset of the real numbers.

11.7 THE FOURIER TRANSFORM IN THREE DIMENSIONS The Fourier transform in three dimensions in rectangular coordinates is given by ZZZ   1 b b F k ¼ pﬃﬃﬃﬃﬃﬃ 3 f ðx; y; zÞeiðkx xþky yþkz zÞ dxdydz R3 2p

395

396

CHAPTER 11 The Fourier Transform

where kb ¼ ðkx ; ky ; kz Þ. The inverse Fourier transform is determined by ZZZ   1 Fb kb eiðkx xþky yþkz zÞ dkx dky dkz f ðx; y; zÞ ¼ pﬃﬃﬃﬃﬃﬃ 3 R3 2p Some of the most important applications in three dimensions occur when f (x,y,z) pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ depends only on r ¼ jb r j ¼ x2 þ y2 þ z2 . In these instances, it is often advantageous to use spherical coordinates. Recall that in spherical coordinates dV ¼ r 2 sin qdrdqd4; so that Z ∞ Z p Z 2p   1 ^ f ðrÞeik\$^r r 2 sin q d4 dq dr. Fb kb ¼ pﬃﬃﬃﬃﬃﬃ 3 r¼0 q¼0 4¼0 2p b b r ¼ kr cos q where We   take k so that it is parallel to the z-axis and then k\$b  b k ¼  k . Then Z ∞ Z p Z 2p   1 f ðrÞeikr cos q r 2 sin q d4 dq dr. Fb kb ¼ pﬃﬃﬃﬃﬃﬃ 3 r¼0 q¼0 4¼0 2p Now Z

p q¼0

Z

2p 4¼0

! d4 eikr cos q sin q dq ¼ 2p

Z

p q¼0

eikr cos q sin q dq

 2p ikr cos q p 2p  ikrð1Þ ikrð1Þ ¼  e e e q¼0 ikr ikr 2p 4p sinðkrÞ ¼ 2i sinðkrÞ ¼ . ikr kr

¼

Thus 1 b FðkÞ ¼ pﬃﬃﬃﬃﬃﬃ 3 2p Similarly,

Z

∞ r¼0

f ðrÞ r

2 4p

sinðkrÞ dr ¼ kr

rﬃﬃﬃ Z 2 ∞ sinðkrÞ dr. f ðrÞ r p r¼0 k

rﬃﬃﬃ Z 2 ∞ b k sinðkrÞ f ðrÞ ¼ FðkÞ dk. p r¼0 r

We consider the example x Þ ¼ f ðrÞ Duðb x Þ þ m2 uðb in three dimensions. As in the one-dimensional case, we take the Fourier transform and find  1 b FðkÞ. Ub ðkÞ ¼ k2 þ m2

11.7 The Fourier Transform in Three Dimensions

We then take the inverse Fourier transform of (k2 þ m2)1; that is, we compute rﬃﬃﬃ Z 2 ∞ k sinðkrÞ gðrÞ ¼ dk. p 0 rðk2 þ m2 Þ To compute the integral, we again use the residue theorem, but the estimates are more delicate than in the one-dimensional case. We begin by computing Z ∞ x sin x dx 2 þ m2 x ∞ using the residue theorem. Let f ðzÞ ¼ We compute

z2

z . þ m2

Z f ðzÞeiz dz C

where C is the contour shown in Fig. 11.7.1. The function f (z)eiz has a pole of order 1 at z ¼ mi. The residue of f (z)eiz at z ¼ mi is lim ðz  miÞ

z/mi

zeiz mieimi em ¼ ¼ ðz  miÞðz þ miÞ mi þ mi 2

so Z

m Z xeix e  dx ¼ 2pi f ðzÞeiz dz. 2 þ m2 x 2 R CR R

Thus Z

  x sin x dx ¼ Im piem  Im 2 2 R x þ m R

Z f ðzÞeiz dz. CR

CR

mi

−R

FIGURE 11.7.1

R

397

398

CHAPTER 11 The Fourier Transform

Consider

 Z    f ðzÞeiz dz.   CR

If z is a point on CR, then j f ðzÞj 

R ðR  mÞ2

and the length of CR is pR. So we can say Z     f ðzÞeiz dz    CR

R ðR  mÞ2

pR

but lim

R

R/∞

ðR  mÞ2

pRs0:

Thus more is needed to conclude that Z    iz   lim  f ðzÞe dz ¼ 0: R/∞

CR

The pertinent result is Jordan’s lemma. (See Brown and Churchill, p. 272.) It states that if f (z) is analytic at all points in the upper half plane that are exterior to a circle jzj ¼ R0 and CR denotes a semicircle z ¼ Reiq ; 0  q  p; where R > R0 and j f ðzÞj  MR for z on CR and lim MR ¼ 0

R/∞

then for every positive real number a, Z lim f ðzÞeiaz dz ¼ 0: R/∞ CR

Thus Z ∞

x sin x dx ¼ lim 2 2 R/∞ ∞ x þ m

Z

R

x sin x dx 2 2 R x þ m

  ¼ Im piem  Im lim

Z

R/∞ CR

  f ðzÞeiz dz ¼ Im piem ¼ pem .

11.7 The Fourier Transform in Three Dimensions

Now consider rﬃﬃﬃ Z rﬃﬃﬃ Z ∞ 2 ∞ k sinðkrÞ 2 1 rk sinðkrÞ  2  dk ¼  rdk  2 2 3 p ∞ r k þ m p r ∞ k þ m2 rﬃﬃﬃ 2 Z ∞ 2 r rk sinðkrÞ   rdk ¼ p r 3 ∞ r2 k2 þ m2 rﬃﬃﬃ Z 2 1 ∞ rk sinðkrÞ  rdk ¼ p r ∞ ðr2 k2 þ r 2 m2 rﬃﬃﬃ Z pﬃﬃﬃﬃﬃﬃ 1 2 1 ∞ u sinðuÞ  du ¼ 2p emr . ¼ 2 2 2 p r ∞ ðu þ r m r Since

k sinðkrÞ rðk2 þm2 Þ

is an even function in k, and r  rﬃﬃﬃ Z ∞ k sinðkrÞ 2 dk ¼ 2 þ m2 Þ rðk p 0

0 1 mr e . r

Returning to  1 b Ub ðkÞ ¼ k2 þ m2 FðkÞ we now know 1 uðb x Þ ¼ ðg  f Þðb x Þ ¼ pﬃﬃﬃﬃﬃﬃ3 2p

ZZZ

1 x y^jb

ybj

EXERCISES 1. In three dimensions, find the Fourier transform of f ðrÞ ¼

1 ð4prÞ

3=2

er

2

=4

.

y Þdb y. emj^x^yj f ðb

399

CHAPTER

The Laplace Transform

12

12.1 INTRODUCTION A second integral transform that plays a prominent role in the solution of differential equations is the Laplace transform. If f ðxÞ is a piecewise continuous function for which f ðxÞ ¼ 0

if x < 0

then the Laplace transform of f ðxÞ; denoted ðLf ÞðsÞ; is defined to be Z ∞ ðLf ÞðsÞ ¼ esx f ðxÞdx 0

for functions for which the integral converges. We note a relationship between the Laplace transform and the Fourier transform. We have Z ∞ 1 f ðxÞeixs ds ðF f ÞðsÞ ¼ pﬃﬃﬃﬃﬃﬃ 2p ∞ so 1 ðF f ÞðisÞ ¼ pﬃﬃﬃﬃﬃﬃ 2p

Z

∞ ∞

f ðxÞe

ixðisÞ

1 dx ¼ pﬃﬃﬃﬃﬃﬃ 2p

Z

∞ ∞

f ðxÞexs dx.

If f ðxÞ ¼ 0 for x < 0; then Z ∞ Z ∞ 1 1 1 xs p ﬃﬃﬃﬃﬃ ﬃ p ﬃﬃﬃﬃﬃ ﬃ ðF f ÞðisÞ ¼ f ðxÞe dx ¼ f ðxÞexs dx ¼ pﬃﬃﬃﬃﬃﬃ ðLf ÞðsÞ: (1) 2p ∞ 2p 0 2p (This is one place where our definition of the Fourier transform makes things a little messier. There is a definition of the Fourier transform for which ðF f ÞðisÞ ¼ ðLf ÞðsÞ.)

401

402

CHAPTER 12 The Laplace Transform

EXERCISES 1. Show that L½tn  ¼

n! snþ1

s > 0:

2. Use the Maclaurin series for sin t and assume that the series can be integrated term by term to show Lfsin tg ¼

1 ; s2 þ 1

s > 1:

3. Use the Maclaurin series for

8 > < sin t; ts0 t f ðtÞ ¼ > : 1; t¼0

and assume that the series can be integrated term by term to show Lff ðtÞg ¼ tan1 ð1=sÞ; tan1

s > 1:

3 x  x3

5

x is þ x5  / x < 1: Note that the Taylor series for 4. Use the Maclaurin series expansion for the order zero Bessel function J0 ðtÞ ¼

∞ X ð1Þn t2n n¼0

22n ðn!Þ2

and assume that the series can be integrated term by term to show  1=2 LfJ0 ðtÞg ¼ s2 þ 1 ; s > 1:

12.2 PROPERTIES OF THE LAPLACE TRANSFORM The relationship between the Laplace and Fourier transforms suggests that certain properties of the two transforms are shared. For example, the Laplace transform can be viewed as a method to decompose a function. Also, as we now show, the analogous result about the Laplace transform of the convolution of two functions is the product of the Laplace transforms. Theorem: Let FðsÞ and GðsÞ denote the Laplace transforms of f ðxÞ and gðxÞ, respectively. Then ðLð f  gÞÞðsÞ ¼ FðsÞGðsÞ.

12.2 Properties of the Laplace Transform

Proof: We have Z ðLð f  gÞÞðsÞ ¼

ð f  gÞðtÞest dt ¼

Z

t¼0

0 @

1

Z

t¼0

t

f ðzÞgðt  zÞdzAest dt:

(1)

z¼0

Rt The reason the limits of integration are as they are in z¼0 f ðzÞgðt  zÞdz is because f ðzÞ ¼ 0 if z < 0; and gðt  zÞ ¼ 0 if z > t. This is why the convolution of functions f ðxÞ and gðxÞ that have a Laplace transform is usually given by Z t f ðzÞgðx  zÞdz ð f  gÞðxÞ ¼ z¼0

rather than

Z ð f  gÞðxÞ ¼

f ðzÞgðx  zÞdz:

z¼∞

We change the order of integration in the double integral in Eq. (1). The region of integration is shown in Fig. 12.2.1. z

z=t

t

FIGURE 12.2.1

Changing the order of integration gives 0 1 Z ∞ Z t Z @ f ðzÞgðt  zÞdzAest dt ¼ t¼0

z¼0

0 @

Z

z¼0

Z ¼

Z

∞ t¼z

1 f ðzÞgðt  zÞest dtAdz

t¼z

0 f ðzÞ@

z¼0

In the integral

gðt  zÞest dt

Z

∞ t¼z

403

404

CHAPTER 12 The Laplace Transform

we make the change of variables u ¼ t  z: When t ¼ z; then u ¼ 0: When t ¼ ∞; then u ¼ ∞. Also; dt ¼ du and est ¼ esðuþzÞ ¼ esu esz : Thus Z ∞ Z ∞ Z ∞ gðt  zÞest dt ¼ gðuÞesu esz du ¼ esz gðuÞesu du t¼z

u¼0

so Z

0 f ðzÞ@

z¼0

Z

u¼0

t¼z

0 ¼@

Z

10 f ðzÞesz dzA@

z¼0

Z

1 gðuÞesu duA ¼ FðsÞGðsÞ:

u¼0

Corollary: With the notation of the theorem above L1 ðFðsÞGðsÞÞ ¼ ðf  gÞðxÞ: As with the Fourier transform, it is the corollary that is useful in solving differential equations. Some important results that we shall use in solving differential equations with the Laplace transform are the following theorem and its corollaries. Theorem: Suppose that the Laplace transform of f ðxÞ and f 0 ðxÞ exist and f ðxÞ is continuous. Then ðLf 0 ÞðsÞ ¼ sðLf ÞðsÞ  f ð0Þ: Proof: Integrating by parts with u ¼ esx and dv ¼ f 0 ; we have Z ∞ Z ∞ ∞ ðLf 0 ÞðsÞ ¼ esx f 0 ðxÞdx ¼ esx f ðxÞx¼0 þ s esx f ðxÞdx ¼ sðLf ÞðsÞ  f ð0Þ: 0

0

Corollary: Suppose that the Laplace transform of f ðxÞ; f 0 ðxÞ, and f 00 ðxÞ exist and f ðxÞ and 0 f ðxÞ are continuous. Then ðLf 00 ÞðsÞ ¼ s2 ðLf ÞðsÞ  sf ð0Þ  f 0 ð0Þ: Proof: By the theorem, we have ðLf 00 ÞðsÞ ¼ sðLf 0 ÞðsÞ  f 0 ð0Þ ¼ s½sðLf ÞðsÞ  f ð0Þ  f 0 ð0Þ ¼ s2 ðLf ÞðsÞ  sf ð0Þ  f 0 ð0Þ:

12.2 Properties of the Laplace Transform

Continuing the idea of the above corollary, we have the following result. Corollary: Suppose the Laplace transform of f ðxÞ; f 0 ðxÞ; . ; f ðnÞ ðxÞ exist and f ðxÞ; f 0 ðxÞ; . ; f ðn1Þ ðxÞ are continuous. Then   Lf ðnÞ ðsÞ ¼ sn ðLf ÞðsÞ  sn1 f ð0Þ  /  sf ðn2Þ ð0Þ  f ðn1Þ ð0Þ: Table 12.2.1 gives some formulas for the Laplace transform. Several of these are proven in the exercises. In the table, FðsÞ is the Laplace transform of f ðxÞ: Table 12.2.1 Some Properties and Formulas for the Laplace Transform LðafðxÞ þ bgðxÞÞ ¼ aFðsÞ þ bGðsÞ d FðsÞ LðxfðxÞÞ ¼ ds d FðsÞ L½x n fðxÞ ¼ ð1Þn ds n  1 bs=a F as L½fðax  bÞ ¼ a e n

a > 0; b  0

¼ Fðs  aÞ hR i x L 0 fðtÞdt ¼ 1s FðsÞ

L½eax fðxÞ

L½ðf  gÞ ¼ FðsÞGðsÞ  L f ðnÞ ðxÞ ¼ sn FðsÞ  sn1 fð0Þ  sn2 f 0 ð0Þ  /  f ðn1Þ ð0Þ ( If uðxÞ ¼

1 if x  0

as

then L½uðxÞ ¼ 1s and L½uðx  aÞ ¼ e s

0 if x < 0

L½uðx  aÞfðx  aÞ ¼ eas FðsÞ s>0

L½x ¼ s12 L½x n 

¼ snþ1

L½eax 

s>0

n!

¼

1 sa

L½sinðaxÞ ¼ s2 þa a2

s>0

L½cosðaxÞ ¼ s2 þs a2 L½x sinðaÞ ¼ 22ax2 2 ðs þ a Þ

s>0 s > jaj

s2  a2 ðs2 þ a2 Þ2

L½x cosðaxÞ ¼  L½fðaxÞ¼ 1a F as L½eax x n  ¼

n! ðsaÞnþ1

s > jaj

a>0 s>a

L½eax sinðbxÞ ¼

b ðsaÞ2 þb2 L½eax cosðbxÞ ¼ sa ðsaÞ2 þb2 as L½dðs  aÞ ¼ e

L½ðxÞn fðxÞ ¼ F ðnÞ ðsÞ

s>a s>a

s > 0; then

405

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CHAPTER 12 The Laplace Transform

EXERCISES 1. Show that d FðsÞ. a. Lðxf ðxÞÞ ¼ ds a b. L½sinðaxÞ ¼ s2 þa s > 0. 2

c. L½x cosðaxÞ ¼

s2 a2 ðs2 þa2 Þ2

s > jaj:

2. If FðsÞ ¼ Lff ðtÞg, show  that if FðsÞ exists for s > b  0; then a. Lff ðatÞg ¼ 1a F

s a

;

s > ab:

Lfeat f ðtÞg

¼ Fðs  aÞ; s > a þ b:  o c. L 1a ebt=a f at ¼ Fðas þ bÞ; a > 0:

b.

n

3. Find the Laplace transform for the Dirac-d function. 4. FindR the Laplace transform of the functions below: t a. 0 ðt  sÞsin sds: Rt b. 0 cosðt  sÞes ds: Rt c. 0 ðt  sÞsinð4sÞds: 5. Use the convolution theorem to find the inverse Laplace transform of the given functions: 1 : a. 2 2 s ðs þ 1Þ 2 b. 2 2 : s ðs þ 4Þ s c. : 2 2 ðs þ 4Þ

12.3 SOLVING DIFFERENTIAL EQUATIONS USING THE LAPLACE TRANSFORM The Laplace transform can be applied to solve both ordinary and partial differential equations. We first demonstrate how the Laplace transform can be used to solve ordinary differential equations. The major steps of the process are as follows: 1. Apply the Laplace transform to the ordinary differential equation. We shall see that in the case where the coefficients are constants, this converts the ordinary differential equation to an algebraic equation. The point of the process is that the algebraic equation is easier to solve. 2. Solve the resulting algebraic equation. 3. Find the function whose Laplace transform is that of the solution found in Step (2). The result is the solution to the differential equation. Like with solving differential equations with the Fourier transform, it is the third step, where we find the “inverse Laplace Transform” of a function that is the most

12.3 Solving Differential Equations Using the Laplace Transform

difficult. There is a transform that accomplishes this, called the Mellin transform, but it is beyond the scope of this text. We give two examples of how to use the Laplace transform to solve ordinary differential equations. Example: Solve the differential equation y00 ðtÞ þ 6y0 ðtÞ þ 5yðtÞ ¼ f ðtÞ;

yð0Þ ¼ 1; y0 ð0Þ ¼ 0

using the Laplace transform. This example demonstrates how the Laplace transform converts an ordinary differential equation with constant coefficients into an algebraic equation and use of the convolution theorem. Solution: Letting LyðtÞ ¼ YðsÞ; we have Ly00 ðtÞ ¼ s2 YðsÞ  syð0Þ  y0 ð0Þ ¼ s2 YðsÞ  s Ly0 ðtÞ ¼ sYðsÞ  yð0Þ ¼ sYðsÞ  1 and we let FðsÞ ¼ Lf ðtÞ: Thus taking the Laplace transform of the equation yields s2 YðsÞ  s þ 6½sYðsÞ  1 þ 5YðsÞ ¼ FðsÞ or



 s2 þ 6s þ 5 YðsÞ  s  6 ¼ FðsÞ:

So YðsÞ ¼

sþ6 FðsÞ sþ6 FðsÞ þ ¼ þ : ðs2 þ 6s þ 5Þ ðs2 þ 6s þ 5Þ ðs þ 5Þðs þ 1Þ ðs þ 5Þðs þ 1Þ

Taking the inverse Laplace transform gives

sþ6 FðsÞ 1 1 þ yðtÞ ¼ L ðYðsÞÞ ¼ L ðs þ 5Þðs þ 1Þ ðs þ 5Þðs þ 1Þ

sþ6 FðsÞ þ L1 : ¼ L1 ðs þ 5Þðs þ 1Þ ðs þ 5Þðs þ 1Þ Using a CAS to decompose the first expression on the right into partial fractions yields sþ6 5 1 ¼  : ðs þ 5Þðs þ 1Þ 4ðs þ 1Þ 4ðs þ 5Þ

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CHAPTER 12 The Laplace Transform

Now L1

s ðs þ 5Þðs þ 1Þ

5 1  4ðs þ 1Þ 4ðs þ 5Þ

5 1 1 1 1 1 5 1  L ¼ et  e5t : ¼ L 4 sþ1 4 sþ5 4 4 ¼ L1

We use the convolution theorem to get

FðsÞ 1 ¼ ð f  gÞðtÞ L ðs þ 5Þðs þ 1Þ where gðtÞ ¼ L1

1 ðs þ 5Þðs þ 1Þ

So Z ð f  gÞðtÞ ¼

t

¼ L1

 1 1 1  ¼ et  e5t : 4ðs þ 1Þ 4ðs þ 5Þ 4

Z

t

f ðuÞgðt  uÞdu ¼

0

0

Finally, 5 1 yðtÞ ¼ et  e5t þ 4 4

Z

t 0

f ðuÞ

  1 ðtuÞ  e5ðtuÞ du: e 4

  1 ðtuÞ 5ðtuÞ du: f ðuÞ e e 4

The next example gives a linear equation where the coefficients are not constants. The main point of the example is to demonstrate a use of the relation Lðty0 Þ ¼ 

d  0 L y : ds

Example: Solve the equation y00 ðtÞ þ ty0 ðtÞ þ yðtÞ ¼ 0; yð0Þ ¼ 1; y0 ð0Þ ¼ 0: Solution: Letting LyðtÞ ¼ YðsÞ; we have Ly00 ðtÞ ¼ s2 YðsÞ  syð0Þ  y0 ð0Þ ¼ s2 YðsÞ  s Ly0 ðtÞ ¼ sYðsÞ  yð0Þ ¼ sYðsÞ  1: Then d  d L½ty0 ðtÞ ¼  L y0 ¼  ½sYðsÞ  1 ¼ YðsÞ  sY 0 ðsÞ: ds ds Taking the Laplace transform of the differential equation thus yields 2 s YðsÞ  s þ ½ YðsÞ  sY 0 ðsÞ þ ½YðsÞ ¼ 0

12.3 Solving Differential Equations Using the Laplace Transform

or s2 YðsÞ  sY 0 ðsÞ  s ¼ 0 or Y 0 ðsÞ  s YðsÞ ¼ 1: In this case, taking the Laplace transform has reduced the order of the differential equation from two to one. The resulting differential equation is one that is familiar in a first differential equations course. It is solved by multiplying by an integrating fac2 tor, which in this case is es =2 ; to give es

2

=2 0

Y ðsÞ  es

2

=2

s YðsÞ ¼ es =2 . 2

But  2 2 d  s2 =2 YðsÞ ¼ es =2 Y 0 ðsÞ  es =2 s YðsÞ e ds so we have 

es

Then es =2 YðsÞ  C ¼ 2

Z

2

=2

YðsÞ

∞

0

¼ es

2

=2

:

Z 0 2 ez =2 YðzÞ dz ¼

s

ez =2 dz 2

s

where C ¼ lim ez =2 YðzÞ ¼ 0: 2

z/∞

Integrating and simplifying gives YðsÞ ¼ es

2 2

=2 4

Z

3 ez

2

=2

z5:

s

Now,

2 YðsÞ ¼ 4

Z

3

e

z =2 s =2 2

e

s

2

dz5 ¼

Z

eðz s 2

2

Þ=2

dz:

s

We can find yðtÞ by making the change of variables u ¼ z  s. Then du ¼ dz; z2  s2 ¼ ðz  sÞðz þ sÞ ¼ uðu þ 2sÞ ¼ u2 þ 2su; if z ¼ s; then u ¼ 0; if z ¼ ∞; then u ¼ ∞. So Z ∞ Z ∞ Z ∞ 2 2 2 2 eðz s Þ=2 dz ¼ eðu þ2suÞ=2 du ¼ esu eu =2 du: YðsÞ ¼ s

0

0

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CHAPTER 12 The Laplace Transform

We recognize that the expression on the right is the Laplace transform of et Thus the solution to the differential equation is yðtÞ ¼ et

2

=2

2

=2 :

:

Example: In this example we show how to find the Laplace transform of a function with a jump discontinuity. Let  0 t2

Then f ðtÞ ¼ t þ uc ðtÞðt  2Þ3 : So n o   Lff ðtÞg ¼ Lftg þ L uc ðtÞðt  2Þ3 ¼ Lftg þ e2s L t2 : Consider the initial value problem y00 ðtÞ þ 2y0 ðtÞ  8yðtÞ ¼ 3d0 ðt  2Þ;

yð0Þ ¼ 2; y0 ð0Þ ¼ 1:

Taking the Laplace transform, we have 2 s YðsÞ  syð0Þ  y0 ð0Þ þ 2½sYðsÞ  yð0Þ  8YðsÞ ¼ s2 YðsÞ  2s þ 1 þ 2sYðsÞ  4  8YðsÞ ¼ 3e2s so

  YðsÞ s2 þ 2s  8 ¼ 3e2s þ 2s þ 3: Thus YðsÞ ¼ e2s

3 2s þ 3 þ : ðs þ 4Þðs  2Þ ðs þ 4Þðs  2Þ

Using 3 1 1 1 1 2s þ 3 5 1 7 1 ¼  and ¼ þ ðs þ 4Þðs  2Þ 2 s  2 2 s þ 4 ðs þ 4Þðs  2Þ 6 ðs þ 4Þ 6 ðs  2Þ we have YðsÞ ¼ e2s Y1 ðsÞ þ Y2 ðsÞ where Y1 ðsÞ ¼

1 1 1 1 5 1 7 1  and Y2 ðsÞ ¼ þ : 2 ðs  2Þ 2 ðs þ 4Þ 6 ðs þ 4Þ 6 ðs  2Þ

12.4 Solving the Heat Equation Using the Laplace Transform

So

yðtÞ ¼ u2 ðtÞy1 ðt  2Þ þ y2 ðtÞ ¼ u2 ðtÞ

1 2ðt2Þ 1 4ðt2Þ 5 7 e þ e4t þ e2t :  e 2 2 6 6

EXERCISES Use the Laplace transform to solve the following initial value problems: 1. 2. 3. 4. 5. 6.

y00 ðxÞ þ yðxÞ ¼ 0; yð0Þ ¼ 2; y0 ð0Þ ¼ 3: y00 ðxÞ þ 7y0 ðxÞ þ 12yðxÞ ¼ 0; yð0Þ ¼ 3; y0 ð0Þ ¼ 2: y00 ðxÞ þ 4y0 ðxÞ þ 3yðxÞ ¼ x; yð0Þ ¼ 0; y0 ð0Þ ¼ 4: y00 ðxÞ þ 5y0 ðxÞ þ 6yðxÞ ¼ ex ; yð0Þ ¼ y0 ð0Þ ¼ 0: y00 ðxÞ  4y0 ðxÞ þ 3yðxÞ ¼ sin x; yð0Þ ¼ 2; y0 ð0Þ ¼ 0: For c > 0 define  0 t 0 uð0; tÞ ¼ gðtÞ

411

412

CHAPTER 12 The Laplace Transform

and vuðx; tÞ v2 uðx; tÞ ¼ 0;  vt vx2

0 < x < ∞; t > 0

uðx; 0Þ ¼ f ðxÞ;

uð0; tÞ ¼ gðtÞ

where gðtÞ ¼ 0 if t < 0: Since uðx; tÞ is a function of two variables, it might seem that we could apply the Laplace transform with respect to either variable. However, the Laplace transform is defined only for functions that are zero when the variable is negative. Thus we could apply the Laplace transform with respect to either variable in the second problem, but only with respect to t in the first problem. Let uðx; tÞ be a function for which uðx; tÞ ¼ 0 if x < 0 or t < 0: Then we can take the Laplace transform of uðx; tÞ with respect to either variable. Let Lt denote the Laplace transform with respect to t and Lx denote the Laplace transform with respect to x: That is, Z ∞ Lt ðuðx; tÞÞ ¼ Uðx; sÞ ¼ uðx; tÞest dt Z

0

Lx ðuðx; tÞÞ ¼ Uðs; tÞ ¼

uðx; tÞesx dx:

0

If we take the Laplace transform with respect to t; then Z ∞

vuðx; tÞ vuðx; tÞ st ¼ e dt ¼ s Uðx; sÞ  uðx; 0Þ ¼ s Uðx; sÞ  f ðxÞ Lt vt vt 0 and

2 Z ∞ 2 Z ∞ v uðx; tÞ v uðx; tÞ st v2 v2 Uðx; sÞ st ¼ e dt ¼ uðx; tÞe dt ¼ : Lt vx2 0 vx2 vx2 vx2 0 Analogous equations hold if one takes the Laplace transform with respect to x: The boundary condition transforms to Z ∞ gðtÞest dt ¼ GðsÞ: Lt ðuð0; tÞÞ ¼ 0

Thus taking the Laplace transform with respect to t of the given equation yields v2 Uðx; sÞ ¼ 0 Uð0; sÞ ¼ GðsÞ: vx2 In the next example, we solve such a problem for given functions f ðxÞ and gðtÞ: As one would expect because of the relationship between the Fourier transform and the Laplace transform given by Eq. (1) in Section 12.1, the process of solving partial s Uðx; sÞ  f ðxÞ 

12.4 Solving the Heat Equation Using the Laplace Transform

differential equations with the two techniques is virtually identical. We recall that the major steps with the Fourier transform were as follows: 1. Convert the PDE to an ODE using the transform. 2. Solve the ODE. 3. Convert the solution back to the original space-time variables. We reiterate that the Fourier transform when applied to the space ðxÞ variable is valid for all problems, but to apply the Laplace transform to the x variable, the function must be zero when x is negative. Example: We solve the heat equation vuðx; tÞ v2 uðx; tÞ  ¼ 0; vt vx2 uðx; 0Þ ¼ 1;

uð0; tÞ ¼

0 < x < ∞; t > 0 

1 0

0 0 then L½f ðtÞ ¼ p1ﬃs ek s : Hint: Note that

=4t ;

=4t :

2. Use the Laplace transform to solve the heat equation vuðx; tÞ v2 uðx; tÞ  ¼ 0; vt vx2 0 < x < ∞; t > 0:

uðx; 0Þ ¼ 0; uð0; tÞ ¼ 4;

lim uðx; tÞ ¼ 0;

x/∞

3. Use the Laplace transform to solve the heat equation vuðx; tÞ v2 uðx; tÞ ¼ 0;  vt vx2 0 < x < ∞; t > 0:

uðx; 0Þ ¼ 0; u ð0; tÞ ¼ sin t;

lim uðx; tÞ ¼ 0;

x/∞

4. Use the Laplace transform to solve the heat equation vuðx; tÞ v2 uðx; tÞ ¼ 0;  vt vx2 0 < x < ∞; t > 0:

uðx; 0Þ ¼ 0; uð0; tÞ ¼ f ðtÞ;

lim uðx; tÞ ¼ 0;

x/∞

5. a. Use the Laplace transform to solve the heat equation vuðx; tÞ v2 uðx; tÞ ¼ 0;  vt vx2 0 < x < ∞; t > 0:

uðx; 0Þ ¼ 0; ux ð0; tÞ ¼ f ðtÞ;

b. Show that for t > 0

Z

t

uð0; tÞ ¼ 0

f ðsÞ pﬃﬃﬃ ds: pðt  sÞ

6. The flux at x ¼ 0 is 

v uðx; tÞjx¼0 : vx

lim uðx; tÞ ¼ 0;

x/∞

417

418

CHAPTER 12 The Laplace Transform

Let vuðx; tÞ v2 uðx; tÞ ¼ 0;  vt vx2 0 < x < ∞; t > 0:

uðx; 0Þ ¼ 0; uð0; tÞ ¼ f ðtÞ;

lim uðx; tÞ ¼ 0;

x/∞

h i pﬃ v uðx; tÞ ¼  v Uðx; sÞ ¼ We have shown that Uðx; sÞ ¼ fbðsÞex s so that L  vx vx pﬃﬃ b xpﬃs pﬃﬃ b v Uðx; sÞj v s f ðsÞe so that vx x¼0 ¼ s f ðsÞ; and vx uðx; tÞjx¼0 hpﬃﬃ i h i ¼ L1 s fbðsÞ ¼ L1 p1ﬃs s fbðsÞ . Use

h i 1 1 L1 pﬃﬃ ¼ pﬃﬃﬃﬃﬃ and L1 s fbðsÞ ¼ f 0 ðtÞ s pt h i 1 pﬃﬃ b s f ðsÞ . to find L

12.5 THE WAVE EQUATION AND THE LAPLACE TRANSFORM We apply the Laplace transform to solve the wave equation 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c ; ∞ < x < ∞; t > 0; v2 vx2 vuðx; 0Þ uðx; 0Þ ¼ f ðxÞ; ¼ gðxÞ: vt The steps in the solution are as follows:

(1)

1. Take the Laplace transform with respect to t to change the partial differential equation to an ordinary differential equation. 2. Solve the ordinary differential equation using a variation of parameters. 3. Take the inverse Laplace transform of the ordinary differential equation to obtain the solution to the original partial differential equation. Let Uðx; sÞ denote the Laplace transform of uðx; tÞ with respect to t: That is, Z ∞ est uðx; tÞdt: Uðx; sÞ ¼ L½uðx; tÞ ¼ 0

Recall that L

v2 uðx; tÞ vuðx; 0Þ ¼ s2 Uðx; sÞ  s uðx; 0Þ  2 vt vt

so that taking the Laplace transform of Eq. (1) gives s2 Uðx; sÞ  s uðx; 0Þ 

vuðx; 0Þ v2 Uðx; sÞ ¼ c2 vt vx2

12.5 The Wave Equation and the Laplace Transform

or s2 Uðx; sÞ  s f ðxÞ  gðxÞ ¼ c2

v2 Uðx; sÞ vx2

so s f ðxÞ  gðxÞ ¼ c2

v2 Uðx; sÞ  s2 Uðx; sÞ vx2

or v2 Uðx; sÞ s2 s 1  2 Uðx; sÞ ¼  2 f ðxÞ  2 gðxÞ: (2) c vx2 c c We fix s and then regard Eq. (2) as an ordinary differential equation with respect to the variable x: Two linearly independent solutions of the associated homogeneous equation d 2 Uðx; sÞ s2  2 Uðx; sÞ ¼ 0 dx2 c are sx

sx

y1 ðx; sÞ ¼ e c and y2 ðx; sÞ ¼ e c : We find a particular solution to Eq. (2) using Z xh y1 ðx; sÞy2 ðx; sÞ  y1 ðx; sÞy2 ðx; sÞ i yp ðx; sÞ ¼ hðx; sÞdx 0 0 0 y1 ðx; sÞy2 ðx; sÞ  y1 ðx; sÞy2 ðx; sÞ where hðx; sÞ ¼ 

s 1 f ðxÞ  2 gðxÞ: 2 c c

Now s

y1 ðx; sÞy2 0 ðx; sÞ  y1 0 ðx; sÞy2 ðx; sÞ ¼ e c e c

sx

s

 s  sx 2s sx  e c  ec ¼ c c c

and y1 ðx; sÞy2 ðx; sÞ  y1 ðx; sÞy2 ðx; sÞ ¼ e

sðxxÞ c

e

sðxxÞ c

:

419

420

CHAPTER 12 The Laplace Transform

Thus the general solution to Eq. (2) is Uðx; sÞhyðx; sÞ Z

h sðx  xÞ

x

¼ AðsÞy1 ðx; sÞ þ BðsÞy2 ðx; sÞ þ 0

c e 2s

c

i

sðx  xÞ c e hðx; sÞdx: (3)

Substituting sx

sx

y1 ðx; sÞ ¼ e c and y2 ðx; sÞ ¼ e c Eq. (3) may be written Uðx; sÞhyðx; sÞ

2 3 Z x sx sx Z x sx sx sx sx c e c e c ðhðx; sÞÞdx  e c e c ðhðx; sÞÞdx5 ¼ AðsÞe c þ BðsÞe c þ 4 2s 0 0 3 2 2 3 sx Z x sx Z x sx sx c c ¼ e c 4AðsÞ  e c ðhðx; sÞÞdx5 þ e c 4BðsÞ þ c ðhðx; sÞÞdx5: 2s 0 2s 0 e If

lim Uðx; sÞ is finite, then we must have Z x sx c lim AðsÞ  e c ðhðx; sÞÞdx ¼ 0 x/∞ 2s 0

x/∞

so AðsÞ ¼

c 2s

Z

∞

0

sx c e c ðhðx; sÞÞdx ¼  2s

Z

0

∞

sx

e c ðhðx; sÞÞdx:

If lim Uðx; sÞ is finite, then we must have x/∞ Z x sx c lim BðsÞ þ e c ðhðx; sÞÞdx ¼ 0 x/∞ 2s 0 so BðsÞ ¼ 

c 2s

Z 0

sx

e c ðhðx; sÞÞdx:

12.5 The Wave Equation and the Laplace Transform

Thus Uðx; sÞhyðx; sÞ 2 3 sx Z x sx c ¼ e c 4AðsÞ  e c ðhðx; sÞÞdx5 2s 0 2 3 sx Z x sx c þ e c 4BðsÞ þ e c ðhðx; sÞÞdx5 2s 0 2 3 sx Z 0 sx Z x sx c c e c ðhðx; sÞÞdx  e c ðhðx; sÞÞdx5 ¼e c 4 2s ∞ 2s 0 2 3 sx Z ∞ sx Z x sx c c e c ðhðx; sÞÞdx þ e c ðhðx; sÞÞdx5 þec4 2s 0 2s 0 2 3 sx Z 0 sx Z x sx c e c ðhðx; sÞÞdx þ e c ðhðx; sÞÞdx5 ¼ e c 4 2s ∞ 0 2 3 sx Z ∞ sx Z x sx c e c ðhðx; sÞÞdx  e c ðhðx; sÞÞdx5  ec4 2s 0 0 8 9 2 3 sx Z x sx sx Z ∞ sx = c< e c 4 ¼ e c ðhðx; sÞÞdx5 þ e c e c ðhðx; sÞÞdx ; 2s : ∞ x 2 3 sðx  xÞ Z Z ∞ sðx  xÞ c4 x ¼ e c hðx; sÞdx þ e c ðhðx; sÞÞdx5 2s ∞ x

¼

c 2s

Z

∞

e

sjx  xj c hðx; sÞdx:

Now hðx; sÞ ¼ 

s 1 f ðxÞ  2 gðxÞ 2 c c

421

422

CHAPTER 12 The Laplace Transform

so c Uðx; sÞ ¼  2s 1 ¼ 2c

Z

Z

∞

e

sjx  xj

Z ∞ sjx  xj c s 1 c c e hðx; sÞdx ¼   2 f ðxÞ  2 gðxÞ dx 2s ∞ c c

sjx  xj sjx  xj Z ∞ 1 1 c c e e f ðxÞdx þ gðxÞdx: 2c ∞ s ∞ ∞

We now find L1 ½Uðx; sÞ We first note that 3 3 2 2

Z ∞ Z ∞ Z ∞ 1 1 4 f ðxÞd t  jx  xj dx5 ¼ f ðxÞd t  jx  xj dx5est dt: L4 c c ∞ 0 ∞ Interchanging the order of integration, this is 2 3

Z ∞ Z ∞ Z ∞ sjxxj 1 st 4 5 e d t  jx  xj dt f ðxÞdx ¼ e c f ðxÞx: c ∞ 0 ∞ Since

2

3

Z ∞ sjxxj 1 L4 f ðxÞd t  jx  xj dx5 ¼ e c f ðxÞdx c ∞ ∞

then

Z

2 L

1 4

Z

3

∞ ∞

e

sjxxj c

f ðxÞdx5 ¼

1 f ðxÞd t  jx  xj dx: c ∞

Z

(5)

If HðtÞ is the Heaviside function, then 3 2

Z ∞ 1 gðxÞH t  jx  xj dx5 L4 c ∞ Z ¼ 0

3

1 4 gðxÞH t  jx  xj dx5est dt: c ∞ 2

Z

Interchanging the order of integration, this is 3 2

Z ∞ Z ∞ 1 4 est H t  jx  xj dt5gðxÞ dx: c ∞ 0

(6)

12.5 The Wave Equation and the Laplace Transform

Now

Z

∞ 0

423

Z ∞ 1 1 sjxxj est H t  jx  xj dt ¼ est dt ¼ e c 1 c s jxxj c

so Eq. (6) is

Z

1 sjxxj e c gðxÞ dx: ∞ s

So, since

then

3

Z ∞ 1 1 sjxxj e c gðxÞ dx L4 gðxÞH t  jx  xj dx5 ¼ c ∞ ∞ s 2

Z

3

Z ∞ sjxxj 1 1 1 4 5 c e L gðxÞ dx ¼ gðxÞH t  jx  xj dx: c ∞ s ∞ 2

Z

(7)

Thus 2 1 uðx; tÞ ¼ L1 ½Uðx; sÞ ¼ L1 4 2c

Z

3 sjx  xj sjx  xj Z ∞ 1 1 c c e e f ðxÞdx þ gðxÞdx5 2c ∞ s ∞ ∞

8 2 2 3 39 sjx  xj sjx  xj Z Z ∞ = 1 < 1 4 ∞ 1 c c e L e f ðxÞdx5 þ L1 4 gðxÞdx5 ¼ ; 2c : ∞ ∞ s 2 3

Z Z ∞ 14 ∞ 1 1 f ðxÞd t  jx  xj dx þ gðxÞH t  jx  xj dx5: ¼ 2c c c ∞ ∞ (8) Consider the first integral on the right-hand side of Eq. (8). We have

Z ∞ Z 1 1 1 ∞ 1 dx : f ðxÞd t  jx  xj dx ¼ f ðxÞd t  jx  xj 2c ∞ c 2 ∞ c c 1 1 Let z ¼ xc. Then dz ¼ dx c ; x ¼ cz; f ðxÞ ¼ f ðczÞ and t  c jx  xj ¼ t  c jx  czj:

Now t  1c jx  czj ¼ 0 if x  cz ¼ ct or x  cz ¼ t; which occurs when z¼

x  ct x þ ct or z ¼ : c c

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CHAPTER 12 The Laplace Transform

Then

Z Z 1 ∞ 1 dx 1 ∞ 1 ¼ f ðxÞd t  jx  xj f ðczÞd t  jx  czjdz 2 ∞ c c 2 ∞ c

1 x  ct x þ ct 1 ¼ f c þf c ¼ ½ f ðx  ctÞ þ f ðx þ ctÞ: 2 c c 2 Next, consider the second integral in the right-hand side of Eq. (8). Note that 8 1 >

> < 0 if t  jx  xj < 0 1 c : H t  jx  xj ¼ > c > : 1 if t  1 jx  xj > 0 c Now, t  1c jx  xj > 0 if and only if jx  xj < ct; that is, if and only if ct < x  x < ct So, 1 2c

Z

or

x  ct < x < x þ ct:

Z xþct 1 1 gðxÞH t  jx  xj dx ¼ gðxÞdx: c 2c xct ∞ ∞

Thus 1 1 uðx; tÞ ¼ ½ f ðx þ ctÞ þ f ðx  ctÞ þ 2 2c

Z

xþct

gðxÞdx xct

which is d’Alembert’s formula.

EXERCISES 1. Use the Laplace transform to solve the wave equation v2 uðx; tÞ v2 uðx; tÞ ¼ c2 ; 2 vt vx2

0 < x < ∞; t > 0; uðx; 0Þ ¼ ut ðx; 0Þ ¼ 0; uð0; tÞ

¼ f ðtÞ; lim uðx; tÞ ¼ 0: x/∞

2. Use the Laplace transform to solve the wave equation 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c ; 0 < x < p; t > 0; vt2 vx2 uðx; 0Þ ¼ sin x; ut ðx; 0Þ ¼ 0; uð0; tÞ ¼ uðp; tÞ ¼ 0:

12.5 The Wave Equation and the Laplace Transform

3. Use d’Alembert’s formula to solve the wave equation 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c ; ∞ < x < ∞; t > 0; lim uðx; tÞ ¼ 0: x/∞ vt2 vx2 for the following initial conditions: 2 a. uðx; 0Þ ¼ ex ; ut ðx; 0Þ ¼ sin x: 1 ; ut ðx; 0Þ ¼ 0: b. uðx; 0Þ ¼ 1þx 2 c. uðx; 0Þ ¼ cos x; ut ðx; 0Þ ¼ sin x:

Check the validity of your answer by substituting the solution into the wave equation.

425

CHAPTER

Solving PDEs With Green’s Functions

13

13.1 SOLVING THE HEAT EQUATION USING GREEN’S FUNCTION We construct Green’s function for the heat equation using the Dirac-d function, following the method of Section 3.1. The first form of the problem we consider is Duxx ðx; tÞ ¼ ut ðx; tÞ

 ∞ < x < ∞; t > 0

uðx; 0Þ ¼ dðx  x0 Þ:

(1)

We solve this problem using the Fourier transform. Recall that if the Fourier transform of u(x,t) with respect to x is denoted by U(k,t), then Z ∞ 1 Uðk; tÞ ¼ pﬃﬃﬃﬃﬃﬃ uðx; tÞeikx dx 2p ∞ Z ∞ 1 uðx; tÞ ¼ pﬃﬃﬃﬃﬃﬃ Uðk; tÞeikx dk 2p ∞  Z ∞ n vn uðx; tÞ 1 v Uðk; tÞ kx p ﬃﬃﬃﬃﬃ ﬃ e dk ¼ vtn vtn 2p ∞ and vn uðx; tÞ 1 ¼ pﬃﬃﬃﬃﬃﬃ vxn 2p

Z

From Eq. (1), we have 1 0 ¼ ut ðx; tÞ  Duxx ðx; tÞ ¼ pﬃﬃﬃﬃﬃﬃ 2p

∞ ∞

Z

Uðk; tÞðikÞn eikx dk:

∞

∞

 vUðk; tÞ 2 þ Dk Uðk; tÞ eikx dk: vt

Thus, we need to solve the ordinary differential equation dUðk; tÞ þ Dk2 Uðk; tÞ ¼ 0: dt

(2)

427

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CHAPTER 13 Solving PDEs With Green’s Functions

The initial condition for Eq. (2) is Uðk; 0Þ ¼ F ðuðx; 0ÞÞ ¼ F ðdðx  x0 ÞÞ Z ∞ 1 1 ¼ pﬃﬃﬃﬃﬃﬃ dðx  x0 Þeikx dx ¼ pﬃﬃﬃﬃﬃﬃ eikx0 : 2p ∞ 2p The solution to dUðk; tÞ þ Dk2 Uðk; tÞ ¼ 0; dt

1 Uðk; 0Þ ¼ pﬃﬃﬃﬃﬃﬃ eikx0 2p

is 2 1 Uðk; tÞ ¼ pﬃﬃﬃﬃﬃﬃ eikx0 ek Dt : 2p

To find u(x,t) we take the inverse Fourier transform of U(k,t), which is Z ∞ 2 1 1 pﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃ eikx0 ek Dt eikx dk: 2p ∞ 2p In Section 1.5, we showed that " # Z ∞ 2 2 1 1 ðx  x Þ 0 uðx; tÞ ¼ eikx0 ek Dt eikx dk ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp . 2p ∞ 4Dt 4pDt Thus the Green’s function is

" # 1 ðx  sÞ2 Gðx; t; sÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp . 4Dt 4pDt

When we studied deriving Green’s functions from the Dirac-d function earlier, we gave a procedure to solve a differential equation with initial condition u(x,0) ¼ f (x), where f (x) is a piecewise continuous function, once we knew the solution for the initial condition being a Dirac-d function. We now review that procedure. Suppose that f (x) is a piecewise continuous function. Divide the x-axis into small subintervals of width Dx, and construct a step function X f ðxk ÞjDk ðxÞ f  ðxÞ ¼ k

where f (xk) is the value of f (x) in the center of the kth interval, and  1 if x is in the kth interval . jDk ðxÞ ¼ 0 otherwise

13.1 Solving the Heat Equation Using Green’s Function

P By superposition and the fact that the effect of f*(x) is the same as k f ðxk ÞDxdðx  xk Þ, we have that u(x,t) is approximated by # " X 1 ðx  xk Þ2 f ðxk Þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp Dx: 4Dt 4pDt k Taking the limit as Dx / 0, we have # " Z ∞ 1 ðx  sÞ2 uðx; tÞ ¼ f ðsÞ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp ds: 4Dt 4pDt ∞

GREEN’S FUNCTION FOR THE NONHOMOGENEOUS HEAT EQUATION Consider the equation ut ðx; tÞ ¼ Duxx ðx; tÞ þ f ðx; tÞ

 ∞ < x < ∞; t > 0; uðx; 0Þ ¼ 0:

(3)

We find the Green’s function for Eq. (3) by solving ut ðx; tÞ  Duxx ðx; tÞ ¼ dðx  yÞdðt  sÞ

 ∞ < x < ∞; t > 0; uðx; 0Þ ¼ 0: (4)

Taking the Fourier transform of Eq. (4) with respect to x gives dUðk; tÞ þ k2 DUðk; tÞ dt Z ∞ dðx  yÞdðt  sÞeikx dx ¼ ∞

¼ dðt  sÞ

Z

∞

dðx  yÞeikx dx ¼ dðt  sÞeiky :

The initial condition is F ðuðx; 0ÞÞ ¼ Uðk; 0Þ ¼ F ð0Þ ¼ 0: Thus, we seek to solve dUðk; tÞ þ k2 DUðk; tÞ ¼ dðt  sÞeiky ; Uðk; 0Þ ¼ 0: (5) dt One method of solving an equation such as Eq. (5) is to multiply by an integrating factor (see Edwards and Penny [2008] for example) which for this problem 2 is ek Dt . This gives 2 2 dUðk; tÞ þ ek Dt k2 DUðk; tÞ ¼ ek Dt dðt  sÞeiky : dt The point of the integrating factor is  2 2 dUðk; tÞ d  k2 Dt ek Dt þ ek Dt k2 DUðk; tÞ ¼ e Uðk; tÞ dt dt

ek

2

Dt

429

430

CHAPTER 13 Solving PDEs With Green’s Functions

so that  2 d  k2 Dt e Uðk; tÞ ¼ ek Dt dðt  sÞeiky dt and e

k2 Dt

Z

t

Uðk; tÞ ¼

ek

2

Dt

dðt  sÞeiky dt ¼ ek

2

Ds iky

e

for s < t:

0

Thus

( Uðk; tÞ ¼

  exp k2 Dt exp k2 Ds expðikyÞ

for s < t

0

for s > t

:

To find the Green’s function, we apply the inverse Fourier transform to U(k,t). Thus, we have Z ∞ 2 2 1 Gðx; t; s; yÞ ¼ pﬃﬃﬃﬃﬃﬃ eikx ek Dt ek Ds eiky dk 2p ∞ Z ∞ 2 1 ek DðtsÞikðyxÞ dk for s < t: ¼ pﬃﬃﬃﬃﬃﬃ 2p ∞ Since 1 2p we have 1 2p and thus

Z

∞ ∞

ek

2

Z

2 2 1 eikx eatk dk ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ex =4at 4pat ∞

# " 2 1 ðy  xÞ DðtsÞikðyxÞ dk ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp 4Dðt  sÞ 4pDðt  sÞ

8 " # 2 > 1 ðy  xÞ > < pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp for s < t 4Dðt  sÞ : Gðx; t; s; yÞ ¼ 4pDðt  sÞ > > : 0 for s > t

By superposition, the solution to ut ðx; tÞ ¼ Duxx ðx; tÞ þ f ðx; tÞ is

Z uðx; tÞ ¼

Z

s¼0

Z ¼

t s¼0

 ∞ < x < ∞; t > 0; uðx; 0Þ ¼ 0

Gðx; t; s; yÞf ðy; sÞdyds

y¼∞

Z

" # 1 ðy  xÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðy; sÞdyds: 4Dðt  sÞ 4pDðt  sÞ y¼∞ ∞

13.2 The Method of Images

EXERCISES In the following exercises it may be difficult to find the inverse Fourier transform that gives the solution. If that is the case, find the function that is to be inverted and give the solution as expressed by the Green’s function. 1. Solve ut ðx; tÞ ¼ Duxx ðx; tÞ þ ejxj sin t; t > 0; ∞ < x < ∞; uðx; 0Þ ¼ 0: 2. Solve Duxx ðx; tÞ ¼ ut ðx; tÞ

 ∞ < x < ∞; t > 0; uðx; 0Þ ¼ ejxj :

3. Solve Duxx ðx; tÞ ¼ ut ðx; tÞ

 ∞ < x < ∞; t > 0; uðx; 0Þ ¼

1 : 1 þ x2

4. Solve ut ðx; tÞ ¼ Duxx ðx; tÞ þ tet ; t > 0; ∞ < x < ∞; uðx; 0Þ ¼ 0:

13.2 THE METHOD OF IMAGES The method of images is a technique for solving heat equationetype problems on a bounded interval or semiinfinite interval. It uses imaginary heat sources or sinks at points outside the interval to obtain the desired boundary conditions.

METHOD OF IMAGES FOR A SEMIINFINITE INTERVAL Example: Consider the heat equation ut ðx; tÞ ¼ Duxx ðx; tÞ;

0 < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ dðx0  xÞ; uð0; tÞ ¼ 0: We present two methods to find the Green’s function.

431

432

CHAPTER 13 Solving PDEs With Green’s Functions

Method 1 In the first section of this chapter we saw that the Green’s function for ut ðx; tÞ ¼ Duxx ðx; tÞ;

∞ < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ dðx0  xÞ; uð0; tÞ ¼ 0 is

# " 1 ðx  x0 Þ2 Gðx; t; x0 Þ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp : 4Dt 4pDt

The problem that we now consider is on a semiinfinite interval and has a boundary condition. We construct a second Green’s function where an added Dirac-d function balances the given Dirac-d function at x ¼ 0. That is, we construct the Green’s function for ut ðx; tÞ ¼ Duxx ðx; tÞ;

∞ < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ dðx0  xÞ;

uð0; tÞ ¼ 0:

(1)

One way to visualize this is to consider the following example from electrostatics: To balance the effect at x ¼ 0 of a positive unit charge placed at x ¼ x0, place a negative unit charge at x ¼ x0. The Green’s function for Eq. (1) is ½x  ðx0 Þ2 ðx þ x0 Þ2 ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ exp p4Dt ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ : GI ðx; t; x0 Þ ¼  Gðx; t; x0 Þ ¼ exp p4Dt 4pDt 4pDt Then 1 ðx  x0 Þ2 1 ðx þ x0 Þ2 Gðx; t; x0 Þ þ GI ðx; t; x0 Þ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp  pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp 4Dt 4Dt 4pDt 4pDt solves ut ðx; tÞ ¼ Duxx ðx; tÞ;

∞ < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ dðx0  xÞ  dð x0  xÞ;

uð0; tÞ ¼ 0:

Now consider the problem ut ðx; tÞ ¼ Duxx ðx; tÞ;

0 < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ f ðxÞ; uð0; tÞ ¼ 0: The solution to ut ðx; tÞ ¼ Duxx ðx; tÞ;

∞ < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ feðxÞ; uð0; tÞ ¼ 0

(2)

13.2 The Method of Images

where feðxÞ ¼ is given by

Z

uðx; tÞ ¼ ¼

∞ Z ∞



f ðxÞ

if x > 0

0

if x  0

½Gðx; t; sÞ þ GI ðx; t; sÞfeðsÞds

½Gðx; t; sÞ þ GI ðx; t; sÞfeðsÞds

0

Z

¼ 0

Z

¼ 0

½Gðx; t; sÞ þ GI ðx; t; sÞf ðsÞds "

# 1 ðx  sÞ2 1 ðx þ sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp  pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðsÞds: 4Dt 4Dt 4pDt 4pDt

This is the same as the solution to ut ðx; tÞ ¼ Duxx ðx; tÞ;

0 < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ f ðxÞ;

uð0; tÞ ¼ 0:

(3)

Thus the Green’s function for Eq. (3) is 1 ðx  sÞ2 1 ðx þ sÞ2 e t; sÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ exp Gðx;  pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp : 4Dt 4Dt 4pDt 4pDt

Method 2 In this method we use the fact that we know the solution to ut ðx; tÞ ¼ Duxx ðx; tÞ;

∞ < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ gðxÞ to construct the solution to ut ðx; tÞ ¼ Duxx ðx; tÞ;

0 < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ f ðxÞ; uð0; tÞ ¼ 0: We are doing something similar to what we did in the first method. We are “balancing” the function u(x,0) ¼ f (x), 0 < x < ∞, by extending the function to ∞ < x < ∞, so that it is an odd function. We let  f ðxÞ if x > 0 gðxÞ ¼ : f ðxÞ if x < 0

433

434

CHAPTER 13 Solving PDEs With Green’s Functions

Then

" # 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp gðsÞds uðx; tÞ ¼ 4Dt ∞ 4pDt " " # # Z 0 Z ∞ 1 ðx  sÞ2 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp ¼ gðsÞds þ gðsÞds 4Dt 4Dt 4pDt ∞ 4pDt 0 Z

" " # # Z ∞ 1 ðx  sÞ2 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðsÞds þ f ðsÞds: ¼ 4Dt 4Dt 4pDt ∞ 4pDt 0 Z

In

0

" # 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ð sÞds  4Dt ∞ 4pDt Z

0

we make the change of variables w ¼ s. Then ds ¼ dw; if s ¼ ∞; then w ¼ ∞; if s ¼ 0; then w ¼ 0, so " " # # Z 0 Z 0 1 ðx  sÞ2 1 ðx  wÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp  f ðsÞds ¼  f ðwÞðdwÞ 4Dt 4Dt 4pDt ∞ 4pDt ∞ # " Z ∞ 1 ðx  wÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp ¼ f ðwÞdw: 4Dt 4pDt 0 Thus, we have " " # # Z 0 Z ∞ 1 ðx  sÞ2 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp uðx; tÞ ¼  f ðsÞds þ f ðsÞds 4Dt 4Dt 4pDt ∞ 4pDt 0 # # " " Z ∞ Z ∞ 1 ðx  wÞ2 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðwÞdw þ f ðsÞds ¼ 4Dt 4Dt 4pDt 4pDt 0 0 " " # ## Z ∞" 1 ðx  sÞ2 1 ðx þ sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp  pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðsÞds: ¼ 4Dt 4Dt 4pDt 4pDt 0 which is what we concluded in Method 1. We can use a modification of the ideas in Method 2 to solve the heat equation with a Neumann boundary condition at x ¼ 0. Now we want to solve ut ðx; tÞ ¼ Duxx ðx; tÞ;

0 < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ f ðxÞ; ux ð0; tÞ ¼ 0: The difference in the approach to the problems is that now we want to balance the derivative of the function u(x,0) ¼ f (x), 0 < x < ∞, by extending the function f (x) to ∞ < x < ∞. We do this by extending f (x) to be an even function. For if f ðxÞ ¼ f ðxÞ then f 0 ðxÞ ¼ f 0 ðxÞ ¼ f 0 ðxÞ:

13.2 The Method of Images

Accordingly, we define ( gðxÞ ¼

f ðxÞ f ðxÞ

if x > 0 : if x < 0

If we repeat the ideas of Method 2, we need to evaluate " # Z ∞ 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp gðsÞds uðx; tÞ ¼ 4Dt ∞ 4pDt # # " " Z ∞ Z 0 1 ðx  sÞ2 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp gðsÞds þ gðsÞds ¼ 4Dt 4Dt 4pDt ∞ 4pDt 0 # # " " Z ∞ 1 ðx  sÞ2 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðsÞds þ f ðsÞds: ¼ 4Dt 4Dt 4pDt ∞ 4pDt 0 Z

0

We again make the change of variables w ¼ s in " # Z 0 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðsÞds 4Dt ∞ 4pDt to obtain Z 0

1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðsÞds ¼ 4Dt ∞ 4pDt ¼

Z

0

Z

∞ ∞ 0

1 ðx þ wÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðwÞðdwÞ 4Dt 4pDt 1 ðx þ wÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðwÞdw 4Dt 4pDt

so that Z

Z ∞ 1 ðx  sÞ2 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp uðx; tÞ ¼ f ðsÞds þ f ðsÞds 4Dt 4Dt 4pDt ∞ 4pDt 0 Z ∞ Z ∞ 1 ðx þ wÞ2 1 ðx  sÞ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp f ðwÞdw þ f ðsÞds ¼ 4Dt 4Dt 4pDt 4pDt 0 0 # Z ∞" 1 ðx þ sÞ2 ðx  sÞ2 þ exp ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ f ðsÞds: exp 4Dt 4Dt 4pDt 0 0

Thus, for the Neumann boundary condition we have " # 1 ðx þ sÞ2 ðx  sÞ2 þ exp Gðx; t; x0 Þ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp : 4Dt 4Dt 4pDt

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CHAPTER 13 Solving PDEs With Green’s Functions

METHOD OF IMAGES FOR A BOUNDED INTERVAL Now we consider the heat equation ut ðx; tÞ ¼ Duxx ðx; tÞ; 0 < t < ∞;

L  x < L;

uðL; tÞ ¼ uðL; tÞ ¼ 0;

uðx; 0Þ ¼ dðx0  xÞ:

We want to do something similar to the semiinfinite case, but now the “balancing” is more involved. Our approach will be similar to Method 1 for the semiinfinite interval. We again use an example from electrostatics to develop some intuition. Consider a charge q to be placed at x ¼ 0 on the interval [L, L]. We want to place charges eq or q outside the interval so that the potential at x ¼ L and x ¼ L will be zero. We proceed in steps: Step 1. We make the potential at x ¼ L be 0 by balancing the charge q at x ¼ 0 with a charge of q at x ¼ 2L as shown in Fig. 13.2.1A. –q

q −L

0

L

2L

FIGURE 13.2.1A

Step 2. We make the potential at x ¼ L be 0 by adding two charges; one to balance the charge at x ¼ 0 and another to balance the charge at x ¼ 2L. To balance the charge q at x ¼ 0 we add a charge of q at x ¼ 2L. To balance the charge q at 2L we add a charge q at x ¼ 4L. We now have the charges as shown in Fig. 13.2.1B. L –q

q −4L

L

−3L

−2L 3L

−q

q −L

0

L

2L

3L

FIGURE 13.2.1B

Step 3. We return to make the potential at x ¼ L be 0. To do this, we must balance the charges we added in Step 2. In Fig. 13.2.1C we show only the charges we added in Step 2.

13.2 The Method of Images

q

–q

–4L

–2L

FIGURE 13.2.1C

To balance the charge of eq at x ¼ 2L we add the charge of q at x ¼ 4L. To balance the charge of q at x ¼ 4L we add the charge of q at x ¼ 6L. See Fig. 13.2.1D. 5L

5L

q

–q

q

–q

–4L

–2L

4L

6L

L 3L

3L

FIGURE 13.2.1D

The charges to this point are shown in Fig. 13.2.1E. q

–q

q

–q

q

q

–4L

–2L

0

2L

4L

6L

FIGURE 13.2.1E

We can perhaps see a pattern emerging that we are adding charges eq at x ¼ 2L, 6L, 10L,. and charges q at x ¼ 4L, 8L, 12L,.. We can now find Green’s function for ut ðx; tÞ ¼ Duxx ðx; tÞ; L  x < L; 0 < t < ∞; uðL; tÞ ¼ uðL; tÞ ¼ 0; uðx; 0Þ ¼ dð0  xÞ: (Conceptually, this is the problem we just solved.) We get "

! ! 1 x2 ðx  2LÞ2 ðx þ 2LÞ2  exp  Gðx; t; 0Þ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp   exp  4Dt 4Dt 4Dt 4pDt ! ! # ðx  4LÞ2 ðx þ 4LÞ2 þ exp  / : þ exp  4Dt 4Dt

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CHAPTER 13 Solving PDEs With Green’s Functions

For the last example of using the method of images, we demonstrate how to balance a charge of q at x0 ˛ (L, L) so that the potential at x ¼ L and x ¼ L is zero. Consider Fig. 13.2.2A. a q

–L

x0

L

FIGURE 13.2.2A

We let a ¼ L  x0 ¼ the distance from x0 to L: We follow the same ideas as in the previous example. We first make the potential at x ¼ L be zero by putting a charge of q at x ¼ L þ a (see Fig. 13.2.2B). q

–q

x0

–L

L+a

a

a

FIGURE 13.2.2B

Figs. 13.2.2Ce2F illustrate how we proceed. In Fig. 13.2.2C we show how the charges after the first step are configured with respect to eL. In Fig. 13.2.2D we show how to add charges so that the potential at x ¼ L will be zero. a

2L – a

a

q –L

–q

x0

L

L+a

FIGURE 13.2.2C

2L + a 2L – a

2a q

–q

–3L – a

–3L + a

FIGURE 13.2.2D

2L+ a 2L – a

2a q

–L

x0

–q L

L+a

13.2 The Method of Images

4L – a –q

q

L 4L + a

FIGURE 13.2.2E

4L – a q

–q

L 4L + a

FIGURE 13.2.2F

In Fig. 13.2.2E we show the distances the new charges are from L, and in Fig. 13.2.2F we show how to place additional charges so that the potential at x ¼ L will be zero. Finally, Fig. 13.2.2G shows the positioning of the charges so far. q

–q

–3L – a

–3L + a

q –L

x0

L

–q

q

–q

L+a

5L – a

5L + a

= L–a

FIGURE 13.2.2G

We now determine a pattern for the placement of the charges. We have placed charges q at x ¼ 3L  a; L  a;

and

5L  a:

Since a ¼ L  x0, these points are x ¼ 3L  ðL  x0 Þ; L  ðL  x0 Þ; and 5L  ðL  x0 Þ ¼ 4L þ x0 ; x0 ; 4L þ x0 : It appears that it is correct to place a charge q at x ¼ 4nL þ x0 where n is an integer.

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CHAPTER 13 Solving PDEs With Green’s Functions

We have placed charges q at x ¼ 3L þ a; L þ a; and 5L þ a ¼ 3L þ ðL  x0 Þ; L þ ðL  x0 Þ; and 5L þ ðL  x0 Þ ¼ 2L  x0 ; 2L  x0 ; 6L  x0 : It appears that a charge of q should be placed at x ¼ 4 nL þ 2Lx0 where n is an integer. We can now find the Green’s function for ut ðx; tÞ ¼ Duxx ðx; tÞ;

L  x < L; 0 < t < ∞;

uðL; tÞ ¼ uðL; tÞ ¼ 0; uðx; 0Þ ¼ cdðx0  xÞ: Following the analysis for the electrostatic potential, we get # ( " ∞ . i h X c Gðx; t; x0 Þ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ exp  ðx  x0 þ 4nLÞ2 4Dt 4pDt n¼∞ " #) ∞ h . i X 2  exp  ðx  x0 þ 2L þ 4nLÞ 4Dt : n¼∞

Thus the solution to ut ðx; tÞ ¼ Duxx ðx; tÞ; L < x < L; 0 < t < ∞; uðL; tÞ ¼ uðL; tÞ ¼ 0; uðx; 0Þ ¼ f ðxÞ is 8 8" # Z L ∞ < < h . i X c 1 2 f ðuÞ exp  ðx  x0 þ 4nLÞ 4Dt uðx; tÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ : 4pDt n¼∞:2L L #9 9 " . i = = h du :  exp  ðx  x0 þ 2L þ 4nLÞ2 4Dt ; ; We have found the Green’s function for several forms of the heat equation. There are a few factors that can complicate a particular form of the heat equation, including boundary terms, initial conditions, and existence of a heat source or sink. When complicating factors are present, it is sometimes advantageous to separate the problem into pieces, each of which contains one of the complicating factors. The solutions to each of the pieces are added together to give the solution to the original problem. We give two examples of this method.

13.2 The Method of Images

Example: Consider ut ðx; tÞ ¼ Duxx ðx; tÞ þ f ðx; tÞ;

∞ < x < ∞; 0 < t < ∞; uðx; 0Þ ¼ gðxÞ:

Suppose that v(x,t) solves vt ðx; tÞ ¼ Dvxx ðx; tÞ þ f ðx; tÞ;

∞ < x < ∞; 0 < t < ∞; vðx; 0Þ ¼ 0

and w(x,t) solves wt ðx; tÞ ¼ Dwxx ðx; tÞ

 ∞ < x < ∞; 0 < t < ∞; wðx; 0Þ ¼ gðxÞ:

(Note that we have found v(x,t) and w(x,t) earlier.) Let u(x,t) ¼ v(x,t) þ w(x,t). Then ut ðx; tÞ ¼ vt ðx; tÞ þ wt ðx; tÞ ¼ Dvxx ðx; tÞ þ f ðx; tÞ þ Dwxx ðx; tÞ ¼ Dðvxx ðx; tÞ þ wxx ðx; tÞÞ þ f ðx; tÞ ¼ Duxx ðx; tÞ þ f ðx; tÞ;

∞ < x < ∞;

0 < t < ∞;

uðx; 0Þ ¼ vðx; 0Þ þ wðx; 0Þ ¼ 0 þ gðxÞ ¼ gðxÞ: Example: Suppose ut ðx; tÞ ¼ Duxx ðx; tÞ þ f ðx; tÞ; 0 < x < ∞; 0 < t < ∞; uðx; 0Þ ¼ gðxÞ; uð0; tÞ ¼ hðtÞ: Suppose p(x,t) solves pt ðx; tÞ ¼ Dpxx ðx; tÞ þ f ðx; tÞ; 0 < x < ∞; 0 < t < ∞; pðx; 0Þ ¼ 0; pð0; tÞ ¼ 0; and q(x,t) solves qt ðx; tÞ ¼ Dqxx ðx; tÞ;

0 < x < ∞; 0 < t < ∞; qðx; 0Þ ¼ gðxÞ; qð0; tÞ ¼ 0

and r(x,t) solves rt ðx; tÞ ¼ Drxx ðx; tÞ;

0 < x < ∞; 0 < t < ∞; rðx; 0Þ ¼ 0; rð0; tÞ ¼ hðtÞ:

We have already found the solutions for p(x,t), q(x,t), and r(x,t). Let uðx; tÞ ¼ pðx; tÞ þ qðx; tÞ þ rðx; tÞ: Then ut ðx; tÞ ¼ pt ðx; tÞ þ qt ðx; tÞ þ rt ðx; tÞ ¼ Dpxx ðx; tÞ þ f ðx; tÞ þ Dqxx ðx; tÞ þ Drxx ðx; tÞ ¼ Dðpxx ðx; tÞ þ qxx ðx; tÞ þ rxx ðx; tÞÞ þ f ðx; tÞ ¼ Duxx ðx; tÞ þ f ðx; tÞ;

0 < x < ∞;

uðx; 0Þ ¼ pðx; 0Þ þ qðx; 0Þ þ rðx; 0Þ ¼ 0 þ gðxÞ þ 0 ¼ gðxÞ uð0; tÞ ¼ pð0; tÞ þ qð0; tÞ þ rð0; tÞ ¼ 0 þ 0 þ hðtÞ ¼ hðtÞ:

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CHAPTER 13 Solving PDEs With Green’s Functions

EXERCISES 1. Solve the heat equation ut ðx; tÞ  Duxx ðx; tÞ ¼ 0; 1 < x < 1; t > 0; uð1; tÞ ¼ uð1; tÞ ¼ 0; uðx; 0Þ ¼ f ðxÞ using the method of images. 2. Solve ut ðx; tÞ  Duxx ðx; tÞ ¼ 0; 0 < x < ∞; t > 0; uðx; 0Þ ¼ 0; ux ð0; tÞ ¼ cos t lim uðx; tÞ ¼ 0: x/∞

If you do this using the Laplace transform, ( pﬃ ) Dx s x2 D2 1 1 e pﬃﬃ L ¼ pﬃﬃﬃﬃﬃ e 4t s pt is helpful. 3. Solve ut ðx; tÞ  Duxx ðx; tÞ ¼ 0; L < x < L; t > 0; uðx; 0Þ ¼ sin x; uðL; tÞ ¼ uðL; tÞ ¼ 0: 4. Solve ut ðx; tÞ ¼ Duxx ðx; tÞ þ et ;

0 < x < ∞; 0 < t < ∞;

uðx; 0Þ ¼ sin x; uð0; tÞ ¼

1 : 1þt

13.3 GREEN’S FUNCTION FOR THE WAVE EQUATION Consider the wave equation in one dimension. Let G(x,t;x0,t0) denote the deflection of a string initially at rest when a unit force is applied at the point x0 at the time t0. Then G(x,t;x0,t0) satisfies the equation v2 Gðx; t; x0 ; t0 Þ v2 Gðx; t; x0 ; t0 Þ ¼ c2 þ dðx  x0 Þdðt  t0 Þ; 2 vt vx2 Gðx; t; x0 ; t0 Þ ¼ 0 for t < t0 :

t  t0 ;

13.3 Green’s Function for the Wave Equation

Applying the Fourier transform with respect to x gives b t; x0 ; t0 Þ v2 Gðk; b t; x0 ; t0 Þ þ p1ﬃﬃﬃﬃﬃﬃ eikx0 dðt  t0 Þ ¼ c2 ðikÞ2 Gðk; 2 vt 2p where b t; x0 ; t0 Þ ¼ p1ﬃﬃﬃﬃﬃﬃ Gðk; 2p

Z

∞ ∞

(1)

Gðx; t; x0 ; t0 Þeikx dx:

Also, b t; x0 ; t0 Þ ¼ 0 for t < t0 : Gðk; Solving for (1) and (2) gives b t; x0 ; t0 Þ ¼ Gðk;



0 aeikct þ beikct

for t < t0 for t > t0

(2)

(3)

where a ¼ aðk; x0 ; t0 Þ and b ¼ bðk; x0 ; t0 Þ: For ε > 0 we have Z t0 þε 2 b Z t0 þε v Gðk; t; x0 ; t0 Þ b t; x0 ; t0 Þdt dt þ c2 k2 Gðk; vt2 t0 ε t0 ε Z 1 ikx0 t0 þε ¼ pﬃﬃﬃﬃﬃﬃe dðt  t0 Þdt: 2p t0 ε b are continuous, Since G and G Z t0 þε b t; x0 ; t0 Þdt ¼ 0: c2 k2 Gðk; lim εY0

Also Z

t0 ε

b t; x0 ; t0 Þ b t þ ε; x0 ; t0 Þ v Gðk; b t  ε; x0 ; t0 Þ v Gðk; v Gðk;  dt ¼ 2 vt vt vt

t0 þε 2

t0 ε

and 1 pﬃﬃﬃﬃﬃﬃeikx0 2p

Z

t0 þε

t0 ε

1 dðt  t0 Þdt ¼ pﬃﬃﬃﬃﬃﬃeikx0 : 2p

(4)

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CHAPTER 13 Solving PDEs With Green’s Functions

Thus from Eq. (4), we have # " b t þ ε; x0 ; t0 Þ v Gðk; b t  ε; x0 ; t0 Þ v Gðk; 1  ¼ pﬃﬃﬃﬃﬃﬃ eikx0 : lim εY0 vt vt 2p

(5)

We use these conditions to solve for a ¼ a(k,x0,t0) and b ¼ b(k,x0,t0) in Eq. (3). b t; x0 ; t0 Þ is continuous in t, Since Gðk; b t; x0 ; t0 Þ ¼ lim Gðk; b t; x0 ; t0 Þ: lim Gðk; t[t0

tYt0

Then aeikct0 þ beikct0 ¼ 0 so b ¼ ae2ikct0 : Differentiating Eq. (3) gives  b t; x0 ; t0 Þ 0 v Gðk; ¼ vt ikcaeickt  ikcbeickt

for t < t0 : for t > t0

(6)

Substituting b ¼ ae2ikct0 into Eq. (6) gives 

 ikc aeickt   ae2ikct0 eickt ¼ ikca eickt þ e2ikct0 ikct : As t / t0

 ikca eickt þ e2ikct0 ikct / 2ikcaeikct0 :

b is p1ﬃﬃﬃﬃﬃ eikx0 , so by Eq. (5) says that the jump condition on the derivative of G 2p Eq. (6), we have 1 2ikcaeikct0 ¼ pﬃﬃﬃﬃﬃﬃ eikx0 ; 2p so eikx0 eikct0 a ¼ pﬃﬃﬃﬃﬃﬃ 2p2ikc and thus eikx0 eikct0 2ikct0 eikx0 eikct0 e : ¼  pﬃﬃﬃﬃﬃﬃ b ¼ ae2ikct0 ¼  pﬃﬃﬃﬃﬃﬃ 2p2ikc 2p2ikc

13.3 Green’s Function for the Wave Equation

Thus b t; x0 ; 0 Þ ¼ Gðk;

¼

¼

8 t0

8 > 0 > > >
> > pﬃﬃﬃﬃﬃﬃ e þ > : 2p2ikc

! eikx0 eikct0 ikct e  pﬃﬃﬃﬃﬃﬃ 2p2ikc

8 0 > > >
t0

for t < t0

" # ikcðtt0 Þ  eikcðtt0 Þ > ikx0 e > > pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ :e 2p2ikc

for t > t0

:

(7) Now eiq  eiq ¼ ðcos q þ i sin qÞ  ðcos q  i sin qÞ ¼ 2i sin q so sin q ¼

eiq  eiq : 2i

Thus, Eq. (7) can be expressed 8 >

: pﬃﬃﬃﬃﬃﬃ sin½kcðt  t0 Þ 2pkc

for t < t0 for t > t0

:

b t; x0 ; t0 Þ gives We now show that taking the inverse Fourier transform of Gðk; 8 > < 1 if jx  x j < cðt  t Þ 0 0 : Gðx; t; x0 ; t0 Þ ¼ 2 > : 0 if jx  x j > cðt  t Þ 0 0 We note that jx  x0j < c(t  t0) if and only if x0  x ε (c(t  t0), c(t  t0)) if and only if x0 ε (xc(t  t0), x þ c(t  t0)).

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CHAPTER 13 Solving PDEs With Green’s Functions

To prove the claim, we compute for t > t0 Z ∞ 1 eikx0 pﬃﬃﬃﬃﬃﬃ sin½kcðt  t0 Þdk eikx kc 2p ∞ Z 1 1 ∞ ikx ikx0 eikcðtt0 Þ  eikcðtt0 Þ ¼ pﬃﬃﬃﬃﬃﬃ e e dk ikc 2p 2 ∞ 0 1 Z ∞ Z xþcðtt0 Þ 1 1 B C ¼ pﬃﬃﬃﬃﬃﬃ eikx0 @ eikx dxA dk: 2c 2p ∞ xcðtt0 Þ

(8)

Reversing the limits of integration, we have 0 1 Z xþcðtt0 Þ Z ∞ 1 1 B C pﬃﬃﬃﬃﬃﬃ eikx0 @ eikx dxAdk 2p 2c ∞ xcðtt0 Þ 1 1 ¼ pﬃﬃﬃﬃﬃﬃ 2p 2c Now 1 pﬃﬃﬃﬃﬃﬃ 2p

Z

∞ ∞

xþcðtt0 Þ

0 @

Z

xcðtt0 Þ

∞ ∞

 eikx0 eikx dk ¼ F 1 eikx0 ¼ F 1 ðF ðdðx0 ÞÞÞ ¼ dðx0 Þ:

Thus 1 1 pﬃﬃﬃﬃﬃﬃ 2p 2c

Z

Z

0 ∞ ∞

B eikx0 @

Z

1 xþcðtt0 Þ xcðtt0 Þ

1 1 ¼ pﬃﬃﬃﬃﬃﬃ 2p 2c

Z

xþcðtt0 Þ

xcðtt0 Þ

0 @

Z

∞ ∞

Z xþcðtt0 Þ 1 ðdðx0 ÞÞ dx 2c xcðtt0 Þ 8 1 > > if x0 ε ðx  cðt  t0 Þ; x þ cðt  t0 ÞÞ < ¼ 2c : > > : 0 otherwise

¼

We note that x0 ε(x  c(t  t0), x þ c(t  t0)) if and only if x0  x ε (c(t  t0), c(t  t0)) if and only if jx  x0j < c(t  t0).

13.3 Green’s Function for the Wave Equation

Thus the solution to 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c þ f ðx; tÞ; vt2 vx2

uðx; 0Þ ¼ 0; is

Z uðx; tÞ ¼

Z

∞ x0 ¼∞

∞ t0 ¼0



where U is the region

∞ < x < ∞; t > 0;

vuðx; 0Þ ¼0 vt

Gðx; t; x0 ; t0 Þf ðx0 ; t0 Þdt0 dx0 ¼

1 2c

 ðx0 ; t0 Þ j 0 < t0 < t  1c jx  x0 j .

ZZ U

f ðx0 ; t0 Þdt0 dx0

Example: Solve 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c ; ∞ < x < ∞; t > 0; vt2 vx2

uðx; 0Þ ¼ 0;

vuðx; 0Þ ¼ gðxÞ: vt

Solution: In this problem there is no forcing term, but there is an initial velocity. We show that this is equivalent to having a forcing term that acts only at t ¼ 0. We replace the given equation by 2 v2 uðx; tÞ vuðx; 0Þ 2 v uðx; tÞ ¼ 0: ¼ c þ gðxÞdðt  0Þ; uðx; 0Þ ¼ 0; vt2 vx2 vt If this replacement is valid, then the solution will be Z ∞ Z ∞ uðx; tÞ ¼ Gðx; t; x0 ; t0 Þgðx0 Þdðt0  0Þdt0 dx0

Z ¼

x0 ¼∞ ∞ x0 ¼∞

t0 ¼0

Gðx; t; x0 ; t0 Þgðx0 Þdx0 :

We now demonstrate that this is the case. Suppose G(x,t;x0,t0) satisfies 2 v2 Gðx; t; x0 ; t0 Þ 2 v Gðx; t; x0 ; t0 Þ ¼ c þ dðx  x0 Þdðt  t0 Þ: vt2 vx2

We show

Z uðx; tÞ ¼

∞

Gðx; t; x0 ; t0 Þgðx0 Þdx0

satisfies 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c ; uðx; 0Þ ¼ 0; vt2 vx2

vuðx; 0Þ ¼ gðxÞ: vt

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CHAPTER 13 Solving PDEs With Green’s Functions

We have  Z ∞ 2 2 2 v2 uðx; tÞ v Gðx; t; x0 ; 0Þ 2 v uðx; tÞ 2 v Gðx; t; x0 ; 0Þ gðx0 Þdx0 c ¼ c vt2 vx2 vt2 vx2 ∞ Z ∞ ¼ dðt  0Þ dðx  x0 Þgðx0 Þdx0 ¼ dðt  0ÞgðxÞ ¼ 0 if t > 0: ∞

Also

Z uðx; 0Þ ¼

Now

∞

Gðx; 0; x0 ; 0Þgðx0 Þdx0 :

8 >

:0

if jx  x0 j < cðt  t0 Þ if jx  x0 j > cðt  t0 Þ

so Gðx; 0; x0 ; 0Þ ¼ 0 ¼ uðx; 0Þ: Now

8 > < 1 if jx  x j < cðt þ DtÞ 0 Gðx; t þ Dt; x0 ; 0Þ ¼ 2c > : 0 otherwise

so Gðx; t þ Dt; x0 ; 0Þ  Gðx; t; x0 ; 0Þ ¼ 0 unless jx  x0 j < cDt; that is, unless cDt < x0  x < cDt or x  cDt < x0 < x þ cDt: So uðx; DtÞ  uðx; 0Þ ¼ Dt

Z



∞ ∞

1 ¼ Dt

Z

 Gðx; t þ Dt; x0 ; 0Þ  Gðx; t; x0 ; 0Þ gðx0 Þdx0 Dt

xþcDt

xcDt

1 1 gðx0 Þdx0 z gðxÞ2cDt/gðxÞ as Dt/0: 2c 2cDt

Thus, vuðx; 0Þ ¼ gðxÞ: vt

13.3 Green’s Function for the Wave Equation

Example: Show that 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c vt2 vx2

has the solution uðx; tÞ ¼ 

Z

 ∞ < x < ∞; t > 0; uðx; 0Þ ¼ f ðxÞ;

vuðx; 0Þ ¼0 vt

vGðx; t; x0 ; 0Þ f ðx þ ctÞ þ f ðx  ctÞ : f ðx0 Þdx0 ¼ vt 2 0 ∞

It is easy to check that uðx; tÞ ¼

f ðx þ ctÞ þ f ðx  ctÞ 2

satisfies the equation. To show that the integral form of the solution is valid, we express G(x,t;x0,t0) in terms of the Heaviside function. Recall that the Heaviside function, H(x), is defined by  0 if x < 0 HðxÞ ¼ : 1 if x > 0 Also, the derivative of the Heaviside function is the Dirac-d function. We have 8 > < 1 if jx  x j < cðt  t Þ 0 0 : Gðx; t; x0 ; t0 Þ ¼ 2c > : 0 if jx  x j > cðt  t Þ 0 0 Note that H(a)  H(b) ¼ 1 if and only if a > 0 and b < 0. Now b < a < b if and only if jaj < b, so jx  x0 j < cðt  t0 Þ if and only if cðt  t0 Þ < x  x0 < cðt  t0 Þ which is true if and only if ðx  x0 Þ þ cðt  t0 Þ > 0 and ðx  x0 Þ  cðt  t0 Þ < 0: Thus Hððx  x0 Þ þ cðt  t0 ÞÞ  Hððx  x0 Þ  cðt  t0 ÞÞ ¼ 1 if and only if jx  x0 j < cðt  t0 Þ:

449

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CHAPTER 13 Solving PDEs With Green’s Functions

So we have 1 ½Hððx  x0 Þ þ cðt  t0 ÞÞ  Hððx  x0 Þ  cðt  t0 ÞÞ: 2c Keeping in mind that the derivative of the Heaviside function is the Dirac-d function, we have Gðx; t; x0 ; t0 Þ ¼

vGðx; t; x0 ; t0 Þ 1 ¼ ½fd½ðx  x0 Þ þ cðt  t0 ÞðcÞ  d½ðx  x0 Þ  cðt  t0 ÞðcÞg vt0 2c ¼

c ½f  d½ðx  x0 Þ þ cðt  t0 Þ  d½ðx  x0 Þ  cðt  t0 Þg 2c

1 ¼  fd½ðx  x0 Þ þ cðt  t0 Þ  d½ðx  x0 Þ  cðt  t0 Þg: 2 Thus, vGðx; t; x0 ; 0Þ 1 ¼  fd½ðx  x0 Þ þ ct  d½ðx  x0 Þ  ctg vt0 2 and so

Z

¼ ¼

∞ ∞

8 Z 1< 2:

vGðx; t; x0 ; 0Þ f ðx0 Þdx0 vt0

∞ ∞

Z d½ðx  x0 Þ þ ct f ðx0 Þdx0 þ

∞

9 =

d½ðx  x0 Þ  ct f ðx0 Þdx0 : ;

Now ðx  x0 Þ þ ct ¼ 0 if x0 ¼ x þ ct and ðx  x0 Þ  ct ¼ 0 if x0 ¼ x  ct so

8 Z 1< 2:

∞ ∞

Z d½ðx  x0 Þ þ ct f ðx0 Þdx0 þ

∞

d½ðx  x0 Þ  ct f ðx0 Þdx0

f ðx þ ctÞ þ f ðx  ctÞ : 2 Combining the examples above, we have the solution to ¼

2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c þ Qðx; tÞ; vt2 vx2

uðx; 0Þ ¼ f ðxÞ;

∞ < x < ∞; t > 0;

vuðx; 0Þ ¼ gðxÞ vt

9 = ;

13.4 Green’s Function and Poisson’s Equation

is ZZ

f ðx þ ctÞ þ f ðx  ctÞ 1 þ 2 2c U   where U is the region ðx0 ; t0 Þ j 0 < t0 < t  1c jx  x0 j . uðx; tÞ ¼

1 2c

Z

xþct

Qðx0 ; t0 Þdx0 dt0 þ

gðx0 Þdx0

xct

EXERCISES 1. Solve 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c ; vt2 vx2

uðx; 0Þ ¼ 0;

∞ < x < ∞; t > 0; vuðx; 0Þ ¼ ex : vt

2. Solve 2 v2 uðx; tÞ 2 v uðx; tÞ ¼ c þ et sin x; vt2 vx2

uðx; 0Þ ¼

1 ; 1 þ x2

∞ < x < ∞; t > 0;

vuðx; 0Þ ¼ cos x: vt

3. Solve v2 uðx; tÞ v2 uðx; tÞ ¼ c2 ; 2 vt vx2

∞ < x < ∞; t > 0; uðx; 0Þ ¼ ex ; 2

vuðx; 0Þ ¼ 0: vt

13.4 GREEN’S FUNCTION AND POISSON’S EQUATION In this section we find the solution to Laplace’s equation and Poisson’s equation in R2 using Green’s function. A boundary value problem for Laplace’s equation on a domain D is Du ¼ 0 on D; u ¼ f on vD: Poisson’s equation is the nonhomogeneous form of Laplace’s equation. A boundary value problem for Poisson’s equation is Du ¼ g on D; u ¼ f on vD:

451

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CHAPTER 13 Solving PDEs With Green’s Functions

The pattern that we follow is conceptually the same as what one does in ordinary differential equation to find the solution to a nonhomogeneous problem. Namely, we find the solution to the homogeneous problem and a particular solution. One then gets the general solution by adding the two. To determine the solution to Poisson’s equation, we find the superposition of uf and ug where uf is the solution to Laplace’s equation Duf ¼ 0 on D; uf ¼ f on vD and ug is the solution to Dug ¼ g on D; ug ¼ 0 on vD: We consider the problem on the disk Duðr; qÞ ¼ 0; uðR; qÞ ¼ f ðqÞ: We have previously noted (in Exercise 8, Section 4.3 and at the end of Section 10.5) that the solution is given by the Poisson integral formula Z R 1 R2  r 2 uðr; qÞ ¼ f ðsÞ 2 ds: R  2rR cosðq  sÞ þ r 2 2p 0 The second part of the problem is a boundary value problem with homogeneous boundary conditions, and we find the solution using Green’s function. We find Green’s function for Poisson’s equation inside a circle of radius R centered at the origin. We want to find Gðb x ; xb0 Þ so that x  xb0 Þ DGðb x ; xb0 Þ ¼ dðb inside the circle and satisfies the boundary condition Gðb x ; xb0 Þ ¼ 0 if jb x j ¼ R: To achieve the boundary condition we use the method of images. We follow the ideas of the previous section and hypothesize that we balance a charge q within the circle at point xb0 by a charge eq at point outside the circle at point xb0 . We situate the point xb0 so that xb0 and xb0 lie on the same radial line from the origin. See Fig. 13.4.1. x^*0 x^

θ x^0

FIGURE 13.4.1

13.4 Green’s Function and Poisson’s Equation

We must determine the distance from the origin to xb0 so that Gðb x ; xb0 Þ ¼ 0 if jb x j ¼ R: We have Gðb x ; xb0 Þ ¼ ¼

 1 1  lnjb ln xb  xb0  þ C x  xb0 j  2p 2p 1 jb 1 x  xb0 j2 ln  ln e4pC 2   4p  xb  xb  4p 0

¼

1 x  xb0 j2 jb ln   4p e4pC  xb  xb 2 0

¼

1 x  xb0 j2 jb 4pC : ln   where D ¼ e 4p D xb  xb 2 0

For jb x j ¼ R, we want Gðb x ; xb0 Þ ¼ 0,which will be true if and only if x  xb0 j2 jb  2 ¼ 1; D xb  xb  0

that is, if and only if  2 x  xb0 j2 ¼ D xb  xb0  : jb x 0 . We Since xb0 and xb0 are on the same radial line from the origin, then xb0 ¼ ab must find D and a. Now jb x  xb0 j2 ¼ hb x  xb0 ; xb  xb0 i ¼ hb x ; xbi þ hb x 0 ; xb0 i  2hb x ; xb0 i ¼ jb x 0 j2  2jb x jjb x j2 þ jb x 0 jcos q where q is the angle between xb and xb0 . Likewise,   2     xb  xb 2 ¼ jb x j2 þ  xb0   2jb x j xb0 cos q: 0 x j ¼ R. We let jb x 0 j ¼ r, so that   If xb is on the boundary of the circle, then jb  xb  ¼ ar. Thus, we have 0  R2 þ r 2  2rR cos q ¼ D R2 þ a2 r 2  2arR cos q (1) and Eq. (1) must hold for all angles q. Letting q ¼ p/2 we have  R2 þ r 2 ¼ D R2 þ a2 r 2 :

453

454

CHAPTER 13 Solving PDEs With Green’s Functions

Then

  R2 þ r 2  2rR cos q ¼ D R2 þ a2 r 2  2rR cos q ¼ D R2 þ a2 r 2  2arR cos q  ¼ D R2 þ a2 r 2  D2arR cos q so 2rR cos q ¼ D2arR cos q and thus aD ¼ 1: One can use the quadratic formula applied to   1 2 R þ a2 r 2 or a2 r2  a R2 þ r 2 þ R2 ¼ 0 R2 þ r 2 ¼ D R2 þ a2 r 2 ¼ a to get a¼

R2 1 r2 and thus D ¼ ¼ 2 : 2 a R r

Finally, we have Gðb x ; xb0 Þ ¼

1 1 R2 x  xb0 j2 jb ¼ ln  ln  2 4p D xb  xb  4p r 2 0

¼

1 R jb x  xb0 j2 x  xb0 j jb  ln    ¼  xb  xb 2 2p r  xb  xb0  0

1 R jb x  xb0 j   ln 2p jb x 0 j  xb  xb0 

where xb0 ¼

R2 x 0 j2 jb

xb0 :

We thus have the following result: Theorem: The solution to Dug ¼ g on D; ug ¼ 0 on vD where D is a disc of radius R is Z uðPÞ ¼

1 R jP  Qj ln gðQÞdSQ 0 jQj¼R 2p jQj jP  Q j

where Q0 ¼

R2 jQj2

Q.

13.4 Green’s Function and Poisson’s Equation

We thus have the following result: Theorem: The solution to Du ¼ g on D; u ¼ f on vD where D is the disc of radius R is given by Z p 1 R2  r 2 uðr; qÞ ¼ f ðqÞ 2 dq 2p p R  2Rr cosðq  q0 Þ þ r 2 Z 1 R jP  Qj ln gðQÞdSQ þ 2p jQj jP  Q0 j jQj¼R where P ¼ ðr; qÞ: We note that Z

1 R jP  Qj ln gðQÞdSQ 0 jQj¼R 2p jQj jP  Q j

Z 1 1 1 ¼  ln gðQÞdSQ : ln 2p jQj¼R jP  Qj jP  Q0 j

We note also that if P ¼ (r0,q0) and Q ¼ (r,q) then jP  Qj ¼ r 2 þ r02   2 4 2 2rr0 cosðq  q0 Þ and Q0 ¼ Rr ; q so that jP  Q0 j ¼ Rr2 þ r02  2 Rr cosðq  q0 Þ. Thus, u(r,q) can be expressed as Z p 1 R2  r 2 f ðqÞ 2 dq uðr; qÞ ¼ 2p p R  2Rr cosðq  q0 Þ þ r2 2 1 þ 2p

Z

p6 6

1 6ln 6 2 2 þ r  2rr r p 4 0 cosðq  q0 Þ 0

0 B B  lnB 4 @R

1 2 R 2 cosðq  q0 Þ þ r  2 0 r r2

!

13 C7 C7 gðRqÞdq: C7 A7 5

455

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CHAPTER 13 Solving PDEs With Green’s Functions

EXERCISES 1. Let D be the disc of radius. 2. Find an expression for the solution to Du ¼ g on D,u ¼ f on vD for the following functions: a. f ¼ r sin q; g ¼ cos q. b. f ¼ 3; g ¼ sin2 q. cos q c. f ¼ ; g ¼ 1. 1 þ r2

Appendix 1 COMPUTING THE LAPLACIAN WITH THE CHAIN RULE Here we compute the Laplacian in polar and spherical coordinates using the chain rule. The computations are tedious and are presented only because this is the approach of many texts. Suppose that we are in the (x,y,z) coordinate system, and we want to convert to the (a,b,c) coordinate system where a, b, and c are functions of x, y, and z. The chain rule says that vf vf va vf vb vf vc ¼ \$ þ \$ þ \$ . (1) vx va vx vb vx vc vx vf Note that vx is expressed in terms of x, y, and z, and the right-hand side of Eq. (7) vf

vf

is in terms of a, b, and c. Also vy and vz are expressed in a similar manner. We shall apply the idea of Eq. (1) several times to find v2 f v2 f v2 f þ þ vx2 vy2 vz2 in cylindrical and spherical coordinates.

CYLINDRICAL COORDINATES Cylindrical coordinates are related to rectangular coordinates by x ¼ r cos q;

y ¼ r sin q;

or 1

z ¼ z;

  y ; x

r ¼ x þ y ; q ¼ tan z ¼ z: 1  2 Thus r ¼ x2 þ y2 , so   12 x r cos q vr ¼ ¼ ¼ x x2 þ y2 ¼ cos q; vx r r   vq 1 1 y r sin q sin q ¼ ; ¼ 2 ¼  2 \$y 2 ¼ 2 2 vx x x þy r r y 1þ x 2

2

2

457

458

Appendix 1

vz ¼ 0: vx Similarly, vr ¼ sin q; vy

vq cos q ¼ ; vy r

vz ¼ 0; vy

and vr ¼ 0; vz

vq ¼ 0; vz

vz ¼ 1: vz

So   vf vf vr vf vq vf vz vf vf sin q ¼ \$ þ \$ þ \$ ¼ cos q þ  : vx vr vx vq vx vz vx vr vq r Also vf vf vf cos q ¼ sin q þ vy vr vq r

v2 f

vf vf ¼ : vz vz

We next compute vx2 . We have       v2 f v vf v vf v vf sin q ¼ cos q  : ¼ vx2 vx vx vx vr vx vq r Now         v vf v vf vr v vf vq v vf vz cos q ¼ cos q þ cos q þ cos q : vx vr vr vr vx vq vr vx vz vr vx Consider   v vf v2 f vf v v2 f cos q ¼ 2 cos q þ \$ cos q ¼ 2 cos q: vr vr vr vr vr vr So   v vf vr v2 f 2 cos q: cos q ¼ vr vr vx vr 2 Now   v vf v2 f vf cos q ¼ cos q  sin q vq vr vrvq vr

Appendix 1

so    2   v vf vq v f vf sin q cos q ¼ cos q  sin q  vq vr vx vrvq vr r ¼

v2 f cos q sin q vf sin2 q þ : vrvq r vr r

Also vz ¼ 0. vx Thus       v vf v vf vr v vf cos q ¼ cos q þ cos q vx vr vr vr vx vq vr     v vf vr v vf cos q þ cos q ¼ vr vr vx vq vr  2 v f 2 v2 f cos q sin q vf þ cos q  ¼ vrvq r vr vr 2

  vq v vf vz þ cos q vx vz vr vx vq vx  sin2 q : r (2)

Similarly,         v vf sin q v vf sin q vr v vf sin q vq v vf sin q vz ¼ þ þ vx vq r vr vq r vx vq vq r vx vz vq r vx     v vf sin q vr v vf sin q vq ¼ þ : vr vq r vx vq vq r vx We have   v vf sin q v2 f sin q vf sin q ¼ \$  ; vr vq r vrvq r vq r 2 so   v vf sin q vr v2 f sin q cos q vf sin q cos q : ¼ \$  vr vq r vx vrvq r v r2 Likewise,   v vf sin q v2 f sin q vf cos q ¼ 2\$ þ \$ ; vq vq r r vq r vq

459

460

Appendix 1

so    2   v vf sin q vq v f sin q vf cos q sin q ¼ þ \$  \$ vq vq r vx vq r r vq2 r ¼

v2 f sin2 q vf sin q cos q  : vq vq2 r 2 r2

Thus   v vf sin q v2 f sin q cos q vf sin q cos q ¼ \$  vx vq r vrvq r vq r2

(3) v2 f sin2 q vf sin q cos q  2  : vq vq r2 r2 2 v f To obtain vx2 , we subtract expression (3) from expression (2) to get     v2 f v vf v vf sin q ¼ cos q  vx vq r vx2 vx vr  2  v f 2 v2 f cos q sin q vf sin2 q cos q  ¼ þ vrvq r vr r vr2   2 v f sin q cos q vf sin q cos q v2 f sin2 q vf sin q cos q (4) \$   2   vrvq r vq vq r2 vq r2 r2 ¼

v2 f 2 v2 f cos q sin q v2 f sin2 q cos q  2 þ 2 2 vrvq r vr vq r2 þ2

vf sin q cos q vf sin2 q : þ vq vr r r2 v2 f

Likewise, we must calculate vy2 : We have vf vf vf cos q ¼ sin þ ; vy vr vq r so     v2 f v vf v vf cos q sin q þ : ¼ vy2 vy vr vy vq r Now       v vf v vf vr v vf vq sin q ¼ sin q þ sin q : vy vr vr vr vy vq vr vy

Appendix 1

We have   v vf vr v2 f sin q\$sin q; sin q ¼ vr vr vy vr 2 and    2  v vf vq v f vf cos q sin q ¼ sin q þ cos q : vq vr vy vqvr vr r So   v vf v2 f v2 f sin q cos q vf cos2 q sin q ¼ 2 sin2 q þ þ : vy vr vr vqvr r vr r

(5)

Also       v vf cos q v vf cos q vr v vf cos q vq ¼ þ : vy vq r vr vq r vy vq vq r vy Now     2 v vf cos q vr v f cos q vf cos q sin q ¼  vr vq r vy vrvq r vq r2 ¼

v2 f cos q sin q vf cos q sin q ;  vrvq r vq r2

(6)

and    2  v vf cos q vq v f cos q vf sin q cos q ¼  vq vq r vy vq r r vq2 r

(7)

v2 f cos2 q vf cos q sin q  : ¼ 2 vq vq r2 r2 Adding (6) and (7) gives     2 v vf cos q v f cos q sin q vf cos q sin q ¼  vy vq r vrvq r vq r2  2  v f cos2 q vf cos q sin q þ  vq vq2 r 2 r2 ¼

v2 f cos q sin q vf cos q sin q v2 f cos2 q þ 2 : 2 vrvq r vq r2 vq r2

(8)

461

462

Appendix 1

Adding (5) and (8) yields     v2 f v vf v vf cos q sin q þ ¼ vy vq r vy2 vy vr   2 v f 2 v2 f sin q cos q vf cos2 q þ sin q þ ¼ vqvr r vr r vr 2  2  v f cos q sin q vf cos q sin q v2 f cos2 q þ 2 : þ 2 vrvq r vq r2 vq r2 v2 f

v2 f

Adding the expressions for vx 2 and vy2 gives  2 v2 f v2 f v f 2 v2 f cos q sin q v2 f sin2 q þ ¼ cos q  2 þ 2 2 2 2 vrvq r vx vy vr vq r2  vf sin q cos q vf sin2 q þ2 þ vq vr r r2  2  v f 2 v2 f sin q cos q vf cos2 q sin q þ þ þ vqvr r vr r vr 2   2 v f cos q sin q vf cos q sin q v2 f cos2 q 2 þ þ vrvq r vq r2 vq2 r 2 ¼

v2 f 2 v2 f 2 vf sin2 q cos q þ sin q þ vr r vr2 vr 2 þ

vf cos2 q v2 f sin2 q v2 f cos2 q þ 2 vr r vq2 r 2 vq r2

v2 f 1 vf 1 v2 f þ þ : vr2 r vr r 2 vq2 Thus the Laplacian in cylindrical coordinates is given by ¼

Df ¼

v2 f 1 vf 1 v2 f v2 f þ þ 2 2 þ 2: 2 vr r vr r vq vz

(9)

Appendix 2 THE LAPLACIAN IN SPHERICAL COORDINATES To compute the Laplacian in spherical coordinates, we use the variables r, q, f,where q and f are as shown in Fig. A.1 and r2 ¼ x2 þ y2 þ z2. We have x ¼ r sin q cos f; y ¼ r sin q sin f; z ¼ r cos q. We shall adapt some of our computations from cylindrical coordinates. In doing so we must be careful, because the standard representation of r in spherical coordinates is not the same r as in cylindrical coordinates. If we take r in spherical coordinates to be r2 ¼ x2 þ y2 ;

z (r, θ , φ )

θ

r

y

φ x

FIGURE A.1

463

464

Appendix 2

then r corresponds to r in cylindrical coordinates. Also f in spherical coordinates was q in cylindrical coordinates. We shall also use the subscript notation for partial derivatives. Adapting what we found in cylindrical coordinates to spherical coordinate notation, and using u to stand for the function instead of f, we have 1 1 uxx þ uyy ¼ urr þ ur þ 2 uff : r r

(10)

We need to convert from the variable r to the standard spherical coordinates of r, q, and f. A key observation is that z and r in spherical coordinates are obtained from r and q by the same functions that give x and y from r and f. Namely, x ¼ r cos f

and

y ¼ r sin f;

z ¼ r cos q

and

r ¼ r sin q;

and

The last relationship follows from r2 ¼ x2 þ y2 ¼ ðr sin q cos fÞ2 þ ðr sin q sin fÞ2  ¼ ðr sin qÞ2 cos2 f þ sin2 f ¼ ðr sin qÞ2 so r ¼ r sin q: This correspondence means that since we have in cylindrical coordinates 1 1 uxx þ uyy ¼ urr þ ur þ 2 uff ; r r we have in spherical coordinates 1 1 uzz þ urr ¼ urr þ ur þ 2 uqq : r r We want to compute uxx þ uyy þ uzz : Adding Eqs. (10) and (11) gives

(11)

1 1 1 1 uxx þ uyy þ uzz þ urr ¼ urr þ ur þ 2 uff þ urr þ ur þ 2 uqq r r r r so 1 1 1 1 uxx þ uyy þ uzz ¼ ur þ 2 uff þ urr þ ur þ 2 uqq : r r r r

(12)

Appendix 2

To convert Eq. (12) to the desired form, we must eliminate the variable r. If we can express ur in terms of r, q, and f, the rest will be easy. The chain rule gives vu vu vr vu vq vu vf ¼ \$ þ \$ þ \$ : vr vr vr vq vr vf vr 1 r Now r ¼ r2 þ z2 2 ; q ¼ tan1 ; f ¼ f; so z vr r r ¼ ; 1 ¼ vr ðr2 þ z2 Þ2 r vq ¼ vr

1 1 z z r cos q cos q ¼ ¼ ¼ ; r2 \$ ¼ 2 z z þ r2 r 2 r2 r 1þ z vf ¼ 0: vr

Thus vu vu r vu cos q vu ¼ \$ þ \$ þ \$0: vr vr r vq r vf Or, converting to subscript notation, r cos q : ur ¼ ur þ uq r r Substituting into Eq. (12) gives 1 1 1 1 uxx þ uyy þ uzz ¼ ur þ 2 uff þ urr þ ur þ 2 uqq r r r r   1 r cos q 1 1 1 u r þ uq þ 2 uff þ urr þ ur þ 2 uqq ¼ r r r r r r ¼

ur 1 cos q 1 1 1 þ uq þ 2 uff þ urr þ ur þ 2 uqq r r r r r r

2 1 cos q 1 1 þ 2 uff þ urr þ 2 uqq : ¼ ur þ uq r r r r r Substituting r sin q for r gives 2 1 cos q 1 1 þ uxx þ uyy þ uzz ¼ ur þ uq u þ 2 uqq þ urr ; 2 ff r r sin q r r ðr sin qÞ which may be rearranged to yield

  2 1 1 Du ¼ urr þ ur þ 2 uqq þ cot quq þ 2 uff : r r sin q

We note that other sources may express this formula in a slightly different way.

465

Appendix 3 SOME OTHER COORDINATE SYSTEMS ELLIPTIC CYLINDRICAL COORDINATES In elliptic cylindrical coordinates the transformations are x ¼ a cosh u sin v y ¼ a sinh u sin v z¼z 0  u < N; 0  v < 2p; N < z < N where a is a positive constant. We take u1 ¼ u; u2 ¼ v; u3 ¼ z. We have b b r ¼ xb i þ yb j þ z kb ¼ a cosh u cos v b i þ a sinh u sin v b j þ z k; so

vb r vb r ¼ ¼ a sinh u cos v b i þ a cosh u sin v b j vu1 vu vb r vb r ¼ a cosh u sin v b i þ a sinh u cos v b j ¼ vu2 vv vb r vb r b ¼ k: ¼ vu3 vz

1. We demonstrate that the system is orthogonal. We have 

 vb r vb r ; ¼ a2 sinh u cos v cosh u sin v þ a2 cosh u sin v sinh u cos v vu vv ¼ 0:

Clearly, 

 vb r vb r ; ¼0 vu vz

 and

 vb r vb r ; ¼ 0: vv vz

467

468

Appendix 3

2. We compute the scaling factors. We have h i1=2 h1 ¼ ða sinh u cos vÞ2 þ ða cosh u sin vÞ2  1=2 ¼ a sinh2 u cos2 v þ cosh2 u sin2 v    1=2 ¼ a sinh2 u cos2 v þ 1 þ sinh2 u sin2 v since cosh2 u ¼ 1 þ sinh2 u. Then   1=2  a sinh2 u cos2 v þ 1 þ sinh2 u sin2 v    1=2 ¼ a sinh2 u cos2 v þ sin2 v þ sin2 v  1=2 ¼ a sinh2 u þ sin2 v : Thus

 1=2 h1 ¼ hu ¼ a sinh2 u þ sin2 v :

Also

h i1=2 h2 ¼ hv ¼ ða cosh u sin vÞ2 þ ða sinh u cos vÞ2  1=2 ¼ a cosh2 u sin2 v þ sinh2 u cos2 v  1=2 ¼ a sinh2 u þ sin2 v :

Finally, h3 ¼ hz ¼ 1. 3. The orthonormal basis fb e 1 ; eb2 ; eb3 g is eb1 ¼ ebu ¼ eb2 ¼ ebv ¼

sinh u cos v 1=2

b iþ

1=2

b iþ

½sinh2 u þ sin2 v cosh u sin v ½sinh u þ sin v 2

2

cosh u sin v ½sinh2 u þ sin2 v

1=2

sinh u cos v 1=2

½sinh2 u þ sin2 v

b j b j

b eb3 ¼ ebk ¼ k: 4. We have

  dV ¼ h1 h2 h3 dudvdz ¼ a2 sinh2 u þ sin2 v dudvdz:

5. Now, Vf ¼

1 vf 1 vf 1 vf þ eb2 þ eb3 eb1 h1 vu1 h2 vu2 h3 vu3

1 vf 1 vf vf ¼  1=2 vu ebu þ  1=2 vv ebv þ vz ebz : 2 2 2 2 a sinh u þ sin v a sinh u þ sin v

Appendix 3

6. Next



1 v h2 h3 vf v h1 h3 vf v h1 h2 vf þ þ h1 h2 h3 vu1 h1 vu1 vu2 h2 vu2 vu3 h3 vu3 "   1=2 1 v sinh2 u þ sin2 v vf  ¼ 2   2 2 1=2 vu a sinh u þ sin v vu sinh2 u þ sin2 v

V2 f ¼

# 1=2   vf v sinh2 u þ sin2 v vf v 2 2 2 þ a sinh u þ sin v þ  vv sinh2 u þ sin2 v1=2 vv vz vz  2   v2 f 1 v f v2 f 2 2 2  ¼ 2 : þ þ a sinh u þ sin v a sinh2 u þ sin2 v vu2 vv2 vz2 7. Finally,

h eb 1 1 . 1 v V F ¼ h1 h2 h3 vu1 h 1 F1

h2 eb2 v vu2

h3 eb3 v vu3 h 3 F3

h 2 F2  1=2  1=2 ebu a sinh2 u þ sin2 v ebv ebz a sinh2 u þ sin2 v 1 v v v  ¼ 2 vu vv vz a sinh2 u þ sin2 v     a sinh2 u þ sin2 v 1=2 F1 a sinh2 þ sin2 v 1=2 F2 F3

  1=2 vF2  1=2 1 vF3 2 2  ¼ 2  a sinh a sinh2 u þ sin2 v ebu u þ sin v 2 2 vv vz a sinh u þ sin v   1=2 vF1  1=2 vF3 2 2  a sinh u þ sin v a sinh2 u þ sin2 v ebv  vu vz    1=2 vF2 1=2 þ a sinh2 u þ sin2 v þ a sinh2 u þ sin2 v sinh u cosh u F2 vu    1=2 vF1 1=2  a sinh2 u þ sin2 v a sinh2 u þ sin2 v sin v cos v ebz : vv

469

470

Appendix 3

We present some additional examples. 1. Parabolic cylindrical coordinates (x; h; z) whose transformation equations are x ¼ xh y¼

 1 2 h  x2 2 z ¼ z:

The ranges of the variables are N < x < N; 0  h < N, and N < z < N. It can be shown that qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hx ¼ hv ¼ h2 þ x2 ; hz ¼ 1;   dV ¼ h2 þ x2 dh dx dz; 2

1 v f v2 f v2 f 2 þ þ : Df ¼ V f ¼ 2 vz2 h þ x2 vh2 vh2 One example where these coordinates could be used is to describe an electric field around a semiinfinite conducting plate. 2. Parabolic coordinates (x; h; 4) whose transformation equations are x ¼ xh cos 4 y ¼ xh sin 4  1 2 h  x2 : 2 The ranges of the variables are 0 < x < N; 0  h < N, are 0  4 < 2p. It can be shown that qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hx ¼ hh ¼ h2 þ x2 ; hz ¼ xh;   dV ¼ xh h2 þ x2 dx dh dz; 

1 1 v vf 1 v vf 1 v2 f 2 : Df ¼ V f ¼ 2 x þ h þ vx h vh vh h þ x2 x vx x2 h2 v42 z¼

3. Bipolar coordinates (x; h; z) whose transformation equations are x¼

a sinh h cosh h  cos x

a sin x cosh h  cos x z ¼ z:

Appendix 3

where 0  x < 2p; N < h < Nand N < z < N. It can be shown that a ; hz ¼ 1; hx ¼ hh ¼ cosh h  cos x dV ¼

a2

dx dh dz; ðcosh h  cos xÞ2

ðcosh h  cos xÞ2 v2 f v2 f v2 f 2 þ þ : Df ¼ V f ¼ a2 vh 2 vx 2 vz2 An example where these coordinates could be used is in describing the electric field around two parallel cylindrical cylinders. 4. Prolate spheroidal coordinates (u; v; fÞ whose transformation equations are x ¼ a sinh u sin v cos f y ¼ a sinh u sin v sin f z ¼ a cosh u cos v: The ranges of the variables are 0  u < N; 0  v  p, are 0  f < 2p. It can be shown that  1=2 hu ¼ hv ¼ a sinh2 u þ sin2 v ; hf ¼ a sinh u sin v;   dV ¼ a3 sinh2 u þ sin2 v sinh u sin v; Df ¼ V2 f

 2 1 v f v2 f 1 vf 1 vf ¼ 2 þ þ þ a ðsinh2 u þ sin2 vÞ vu2 vv2 tanh u vu tan v vv þ2

1 v2 f : 2 vf2 sinh u sin v 2

Prolate spheroidal coordinates is a coordinate system that results from rotating an ellipse about the axis on which the foci are located. Oblate spheroidal coordinates is a coordinate system that results from rotating an ellipse about the axis that separates the foci. Both coordinates are sometimes used to solve partial differential equations when the boundary conditions are defined on particular shapes. Prolate spheroidal coordinates could be used to describe the electric field generated by two electrode tips. 5. Oblate spheroidal coordinates (u; v; fÞ whose transformation equations are x ¼ a cosh u cos v cos f y ¼ a cosh u cos v sin f z ¼ a sinh u sin v:

471

472

Appendix 3

The ranges of the variables are 0  u < N; p2  v  p2 , are 0  f < 2p. It can be shown that  1=2 ; hf ¼ cosh u cos v; hu ¼ hv ¼ a sinh2 u þ sin2 v   dV ¼ a3 cosh u cos v sinh2 u þ sin2 v du dv df; Df ¼ V2 f

"

1 v vf ¼ 2 a cosh u cos v vu a ðsinh2 u þ sin2 vÞcosh u cos v vu #  

a2 sinh2 u þ sin2 v v2 f v vf : þ a cosh u cos v þ a cosh u cos v vv vv vf2

Oblate coordinates can be used in diffusion problems such as the scattering of sound through a circular hole or flow of liquid through a hole.

Bibliography Apostol, T.M., 1974. Mathematical Analysis, second ed. Addison eWesley Publishing Company, Reading MA. Arfken, G., 1970. Mathematical Methods for Physicists, second ed. Academic Press, New York. Boas, M.L., 1966. Mathematical Methods in the Physical Sciences. John Wiley & Sons, Inc., New York. Boyce, W., DiPrima, R., 2008. Elementary Differential Equations and Boundary Value Problems. John Wiley & Sons, Inc., New York. Brown, J.W., Churchill, R.V., 2008. Fourier Series and Boundary Value Problems. McGrawHill Book Company, New York. Courant, R., Hilbert, D., 1989. Methods of Mathematical Physics, Vol. 1, p. c1937. New York. Edwards, C., Penny, D., 1994. Elementary Differential Equations with Applications, third ed. Prentice Hall, Englewood Cliffs, NJ. Edwards, C., Penny, D., 2008. Elementary Differential Equations. Gelbaum, B., Olmsted, J., 1964. Counterexamples in Analysis. Holden Day, San Francisco. Growers, T. (Ed.), 2008. The Princeton Companion to Mathematics. Princeton University Press, Princeton, NJ. Hoskins, R.F., 1979. Generalized Functions. Halsted Press, New York. Kirkwood, J.R., 1995. An Introduction to Analysis, second ed. Waveland Press, Inc., Prospect Heights, IL. Kreysig, E., 1967. Advanced Engineering Mathematics, second ed. John Wiley & Sons, Inc., New York. Marsden, J.E., Tromba, A.J., 1988. Vector Calculus, third ed. W.H. Freeman and Company, New York. McQuarrie, D., 2003. Mathematical Methods for Scientists and Engineers. University Science Books, Sausalito, CA. Morse, P.M., Feshbach, H., 1953. Methods of Theoretical Physics. McGraw-Hill Book Company, New York. Park, D., 1964. Introduction to the Quantum Theory. McGraw-Hill Book Company, New York. Pinsky, M.A., 1998. Partial Differential Equations and Boundary e Value Problems with Applications, third ed. Waveland Press, Inc, Prospect Heights, IL. Rogawski, J., 2008. Calculus. W.H. Freeman and Company, New York. Rudin, W., 1964. Principles of Mathematical Analysis. McGraw-Hill Book Company, New York. Rudin, W., 1973. Functional Analysis. McGraw-Hill Book Company, New York. Stakgold, I., 1967ae68. Boundary Problems in Mathematical Physics. Macmillan, New York. Stakgold, I., 1967be68. Green’s Functions and Boundary Value Problems. Macmillan, New York. Schey, H.M., 1973. Div, Grad, Curl and All that; an Informal Text on Vector Calculus. Norton, New York. Weinberger, H.F., 1965. A First Course in Partial Differential Equations. John Wiley and Sons, New York.

473

Index ‘Note: Page numbers followed by “f ” indicate figures and “t” indicate tables.’

A Absolute convergence, 42e43 Alternating series test, 43 Ampere’s law, 138 Analytic function, 52, 219 Approximate identities, 29e31, 29f and Dirac-d function, 29e40 examples of, 38e40 Associated Legendre function, 349e353

B Bessel equation, 159, 164, 317, 319 cylindrical coordinates, 320e323, 321f generating function for, 258 of half-integer order, 342e343 order 0, 334 second solution, 319 in spherical coordinates, 340e345 Bessel function, 53, 315e320 cylindrical coordinates, 320e323, 321f of first kind of order, 319 modified (with imaginary argument), 319 Bessel’s inequality, 7, 191, 199, 242e250

C Calculus for Dirac-d function, 34e37 Cartesian coordinates, 15, 23, 69, 125 curl in, 101e106, 102f, 104fe105f cylindrical case, 16 general case, 16 separation of variables, 275e313 Cauchy criterion for uniform convergence of sequences of functions, 47 for uniform convergence of series of functions, 48 Cauchy integral formula, 62e64 Cauchy sequence, 41 Cesaro sum, 200 Change of variables (integration), 28e29 Circulation, 101 Comparison test, 41e42 Completeness of eigenfunctions, 204, 241e251 Conservative fields, 140e145 exercises, 150e151 Continuity of limit function, 43 Convergence, 40e59, 187 absolute, 42e43

conditional, 42e43 functions, 43e49, 44t pointwise, 43, 187 uniform, 43e44, 187 L2 convergence, 187 numbers, 40e42 Convolution of functions, 40 Fourier transform, 380 Laplace transform, 402e406, 403f Convolution theorem, 380 Coordinates Cartesian coordinates, 15 curvilinear, 15e29 cylindrical, 15e16, 15f orthogonal, 16, 16fe17f spherical, 15e16, 15f, 25e26 Cosine series and Fourier series, 212e215 Coulomb’s law, 139 Coupled spring system, 7e8 Curl, 90e120 Cartesian coordinates, 101e106, 102f, 104fe105f cylindrical coordinates, 106e112, 107f, 109fe110f spherical coordinates, 112e119 Curvilinear coordinates, 15e29, 101e106 Curvilinear systems, 26 Cylindrical coordinates, 15e16, 15fe16f Bessel’s equation in, 320e323, 321f curl in, 106e112, 107f, 109fe110f

D Del operator, 93 Dirac-d function, 32e34, 33f, 300e301 calculus for, 34e37 in curvilinear coordinates, 37e38 in Fourier series, 300e301 Green’s function, construction, 154e162 Directional derivative, 70 Dirichlet kernel, 194 Divergence, 90e120 Cartesian coordinate case, 91e93, 91f cylindrical coordinate case, 93e96, 94f spherical coordinate case, 96e101, 97f

475

476

Index

Divergence theorem, 120e151, 179 application of, 138e139 exercises, 147e149 for inverse square fields, 131 in three dimensions, 129 in two dimensions, 127 Dot product. See Inner product Double Fourier series, 215e216 Drum head problem, 328e332 D’Alembert’s formula, 230, 424

E Eigenfunctions completeness of, 241e251 Green’ function from, 166e169 Eigenvalue, 1 Eigenvector, 1 Elliptic equation, 217e218 Equilibrium (steady) state of the heat equation, 303e310

F Fibonacci sequence, 257e258 Fejer kernel, 200 Fejer’s theorem, 201 Flux, 82 Fourier coefficients, 5e12 Fourier inversion theorem, 379e380 Fourier series, 187e216 on arbitrary intervals, 205 convergence methods, 194e207, 195f definitions, 188e193 double, 215e216 eigenfunctions, 188 exponential form of, 207e212 pointwise convergence, 187, 194 sine and cosine series, 212e215 uniform convergence, 187, 194, 251e255 Fourier transform, 375e399 as decomposition, 376e377 from Fourier series, 377e379 fundamental solution of, 385e390 of Gaussian distribution, 59, 379e383, 379fe380f heat equation, solving, 384 Laplace’s equation, solving, 388 negative Laplacian spectrum, in one dimension, 391e395, 393fe394f partial differential equations, solving, 383e391 properties of, 379e383, 379fe380f

in three dimensions, 395e399, 397f wave equation, solving, 386e387 Fourier’s law, 221 Fredholm alternative, 171e178, 174f, 176f

G Gamma function, 67 Gaussian functions, 379e383, 379fe380f Gauss’s law for inverse square fields, 131, 139 Gauss’ theorem. See Divergence theorem Generating functions, 257e273 for Bessel functions of the first kind, 270e273 for Hermite’s differential equation, 262e266 for Laguerre’s equation, 259e262 for Legendre’s equation, 266e270 Geometric series, 41 Gradient, 22e23 Cartesian coordinates, 22 cylindrical coordinates, 23 Green’ function, 153e185 for Bessel’s equation, 159, 164 and Dirac-d function, 154e162 from eigenfunctions, 166e169 general boundary conditions, 169e171 for heat equation, 427e431 for the Laplacian, 178e185 for Poisson’s equation, 451e456, 452f for wave equation, 442e451 using variation of parameters, 162e166 Green’s identities, 138, 147e148 Green’s theorem, 120e151 exercises, 147

H Hamiltonian operator, 3 Heat equation, 217 derivation, in one dimension, 220e222, 221f on disk, 332e336 Fourier transform, 384 fundamental solution of, 385e390 Green’s function, 427e431 Laplace transform, 411e418 method of images, 431e442 nonhomogeneous, 429e430 in one dimension, 297e303 with no heat source, 297e300 steady state of, 303e310 Heaviside function, 32, 422, 449 Hermite equation generating function for, 258, 262e266 of order n, 365

Index

Hermite polynomial of degree n, 365 Hydrogen atom, 368e373 Hyperbolic equation, 217e218

I Increments of volume, 19 Inner product, 2 Integrals, 69e90 line, 73e76, 73fe74f parameterized surfaces, 78e79 path, 70e73 of scalar functions over surfaces, 79e81, 89 surface, 76e78, 77f, 82e87, 82fe83f, 89e90 Integral test, 42 Integration formulas, 59e67 Inverse square field, 131

J Jacobian, 19e20 Jordan’s lemma, 398 Jump condition on Green’s function, 156

K Kernel, 14

L Laguerre’s equation, 258 generating function for, 259e262 Laplace transform, 401e425 of convolution of functions, 402e406, 403f heat equation, 411e418 properties and formulas for, 355t solving differential equations, 406e411 using Fourier transform to solve, 388 wave equation, 418e425 Laplace’s equation, 27, 217e220 on cube, 283e287 in cylindrical coordinates, 323e328 on rectangle, 275e283 spherical coordinates, 337, 353e356 Laplacian, 23e25, 132, 218 cylindrical coordinates, 24 Green’s function, 178e185 spherical coordinates, 25e26 Legendre equation, 337e340, 345e349 associated Legendre function, 349e353 generating function for, 258, 266e270 Legendre polynomial, 347 Linear function. See Linear operator Linear operator, 1 self-adjoint, 1e15 Line integrals, 73e76, 73fe74f exercises, 88e89

M Maclaurin series, 52, 258 Maximum modulus principle, 219e220 Method of images, 431e442 on bounded interval, 436e441, 436fe439f on semi-infinite interval, 431e435

N Newton’s law, 3, 224

O One dimension quantum mechanical oscillator, 360e367 Ordinary differential equations (ODEs), 26, 187, 237, 315, 406 Orthogonal basis, 5 coordinates, 16, 16fe17f set, 5 transformation, 14 vectors, 13 Orthogonal coordinate system, 17e18 Orthonormal basis, 5 set, 5

P Parabolic equation, 217e218 Parameterized surfaces, 78e79 Parseval’s formula, 203 Parseval’s theorem, 381e382 Partial differential equations (PDEs) second-order, converting to standard form, 232e236, 233f solving, 383e391 Path integrals, 70e73 exercises, 88 Poisson’s equation, 451e456, 452f Poisson’s integral formula, 211e212, 389 Polar coordinates, 332 Power series, 49e51 Power series expansion, 60e62 Principle of Superposition, 11e12 p-series test, 42

R Radius of convergence, 50 Ratio test, 42 Real numbers, series of, 40e42 Resolvent, 392 Riemann-Lebesgue lemma, 191 Rigid rotor, 356e360 Root test, 42

477

478

Index

S Scaling factors, 18 Schrodinger’s equation for hydrogen atom, 368 for rigid rotor, 358 Schwarz inequality, 198e205 Second-order PDEs to standard form, converting, 232e236, 233f Self-adjoint operator, 1e15, 188e193, 238e241 Separation of variables in Cartesian coordinates, 275e313 heat equation, 297e310 Laplace’s equation, 275e283, 277f wave equation, 287e294 in cylindrical coordinates, 315e336 Bessel functions, 315e320 Bessel’s equation, 320e323, 321f heat equation, on disk, 332e336 Laplace’s equation, 323e328 wave equation, on disk, 328e332 Sequences functions, 43 numbers, 40 Series cosine series, 212e215 of functions, 43e49, 44t power series, 49e51 of real numbers, 40e42 sine series, 212e215 Taylor series, 51e55 Spectrum, 391e395, 393fe394f Spherical coordinates, 15e16, 15f, 25e26 Bessel’s equation in, 340e345 curl in, 112e119 Laplace’s equation in, 353e356 Steady state (heat equation), 303e310 Stokes’ theorem, 120e151 application of, 138e139 exercises, 147e149 Stone’s theorem, 3 Sturm-Liouville theory, 237e255, 322 eigenfunctions completeness for, 241e251

self-adjoint property of, 238e241 uniform convergence of Fourier series, 251e255 Surface, 76e78, 77f parameterized, 78e79 Surface integral, 76e78, 82e87, 83f, 89e90

T Taylor polynomial, 51 Taylor series, 51e55 Telegraph equation, 294 Trigonometric polynomials, 189

U Uniform convergence, 43e44, 47e48 of Fourier series, 251e255

V Variation of parameters, 162e166 Vector calculus, 69e151 integrals, 69e90 line, 73e76, 73fe74f parameterized surfaces, 78e79 path, 70e73 of scalar functions over surfaces, 79e81 surface, 76e78, 77f, 82e87, 83f Vector field, 73 Volume integrals, 19e22

W Wave equation, 217, 328e332 on disk, 328e332 explicit solution of, 226e232, 230fe231f Fourier transform, 386e387 Green’s function, 442e451 Laplace transform, 388, 418e425 in one dimension derivation, 223e226, 223f separation of variables, 287e294 spherical coordinates, 328e332 in two dimension, 294e297 Weierstrass M-test, 49, 310 Weight function, 2, 239, 322