series solutions of odes. special functions · series solutions of odes. special functions ......

167

C H A P T E R 5

Series Solutions of ODEs.Special Functions

In the previous chapters, we have seen that linear ODEs with constant coefficients can besolved by algebraic methods, and that their solutions are elementary functions known fromcalculus. For ODEs with variable coefficients the situation is more complicated, and theirsolutions may be nonelementary functions. Legendre’s, Bessel’s, and the hypergeometricequations are important ODEs of this kind. Since these ODEs and their solutions, theLegendre polynomials, Bessel functions, and hypergeometric functions, play an importantrole in engineering modeling, we shall consider the two standard methods for solvingsuch ODEs.

The first method is called the power series method because it gives solutions in theform of a power series .

The second method is called the Frobenius method and generalizes the first; it givessolutions in power series, multiplied by a logarithmic term or a fractional power ,in cases such as Bessel’s equation, in which the first method is not general enough.

All those more advanced solutions and various other functions not appearing in calculusare known as higher functions or special functions, which has become a technical term.Each of these functions is important enough to give it a name and investigate its propertiesand relations to other functions in great detail (take a look into Refs. [GenRef1],[GenRef10], or [All] in App. 1). Your CAS knows practically all functions you will everneed in industry or research labs, but it is up to you to find your way through this vastterrain of formulas. The present chapter may give you some help in this task.

COMMENT. You can study this chapter directly after Chap. 2 because it needs nomaterial from Chaps. 3 or 4.

Prerequisite: Chap. 2.Section that may be omitted in a shorter course: 5.5.References and Answers to Problems: App. 1 Part A, and App. 2.

5.1 Power Series MethodThe power series method is the standard method for solving linear ODEs with variablecoefficients. It gives solutions in the form of power series. These series can be usedfor computing values, graphing curves, proving formulas, and exploring properties ofsolutions, as we shall see. In this section we begin by explaining the idea of the powerseries method.

x rln x

a0 � a1x � a2 x2 � a3 x3 � Á

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168 CHAP. 5 Series Solutions of ODEs. Special Functions

From calculus we remember that a power series (in powers of ) is an infiniteseries of the form

(1)

Here, x is a variable. are constants, called the coefficients of the series. is a constant, called the center of the series. In particular, if , we obtain a power

series in powers of x

(2)

We shall assume that all variables and constants are real.We note that the term “power series” usually refers to a series of the form (1) [or (2)]

but does not include series of negative or fractional powers of x. We use m as thesummation letter, reserving n as a standard notation in the Legendre and Bessel equationsfor integer values of the parameter.

E X A M P L E 1 Familiar Power Series are the Maclaurin series

Idea and Technique of the Power Series MethodThe idea of the power series method for solving linear ODEs seems natural, once weknow that the most important ODEs in applied mathematics have solutions of this form.We explain the idea by an ODE that can readily be solved otherwise.

E X A M P L E 2 Power Series Solution. Solve .

Solution. In the first step we insert

(2) y � a0 � a1x � a2 x2 � a3 x3 � Á � a�

m�0

am xm

yr � y � 0

� sin x � a�

m�0

(�1)mx2m�1

(2m � 1)!� x �

x3

3!�

x5

5!� � Á .

cos x � a�

m�0

(�1)mx2m

(2m)!� 1 �

x2

2!�

x4

4!� � Á

ex � a�

m�0

xm

m!� 1 � x �

x2

2!�

x3

3!� Á

1

1 � x� a

�

m�0

xm � 1 � x � x2 � Á

( ƒ x ƒ � 1, geometric series)

a�

m�0

am xm � a0 � a1x � a2 x2 � a3 x3 � Á .

x0 � 0x0

a0, a1, a2, Á

a�

m�0

am(x � x0)m � a0 � a1(x � x0) � a2(x � x0)2 � Á .

x � x0

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Ketiga fungsi ini diturunkan menggunakan prinsip deret MacLaurin

SEC. 5.1 Power Series Method 169

and the series obtained by termwise differentiation

(3)

into the ODE:

Then we collect like powers of x, finding

Equating the coefficient of each power of x to zero, we have

Solving these equations, we may express in terms of , which remains arbitrary:

With these values of the coefficients, the series solution becomes the familiar general solution

Test your comprehension by solving by power series. You should get the result

We now describe the method in general and justify it after the next example. For a givenODE

(4)

we first represent p(x) and q(x) by power series in powers of x (or of if solutionsin powers of are wanted). Often p(x) and q(x) are polynomials, and then nothingneeds to be done in this first step. Next we assume a solution in the form of a power series(2) with unknown coefficients and insert it as well as (3) and

(5)

into the ODE. Then we collect like powers of x and equate the sum of the coefficients ofeach occurring power of x to zero, starting with the constant terms, then taking the termscontaining x, then the terms in , and so on. This gives equations from which we candetermine the unknown coefficients of (3) successively.

E X A M P L E 3 A Special Legendre Equation. The ODE

occurs in models exhibiting spherical symmetry. Solve it.

(1 � x2)ys � 2xyr � 2y � 0

x2

ys � 2a2 � 3 # 2a3 x � 4 # 3a4 x2 � Á � a

�

m�2

m(m � 1)am xm�2

x � x0

x � x0

ys � p(x)yr � q(x)y � 0

�y � a0 cos x � a1 sin x.ys � y � 0

y � a0 � a0 x �a0

2! x2 �

a0

3! x3 � Á � a0 a1 � x �

x2

2!�

x3

3!b � a0ex.

a1 � a0, a2 �a1

2�

a0

2! , a3 �

a2

3�

a0

3! , Á .

a0a1, a2, Á

a1 � a0 � 0, 2a2 � a1 � 0, 3a3 � a2 � 0, Á .

(a1 � a0) � (2a2 � a1)x � (3a3 � a2)x2 � Á � 0.

(a1 � 2a2 x � 3a3 x2 � Á ) � (a0 � a1x � a2 x2 � Á ) � 0.

yr � a1 � 2a2 x � 3a3 x2 � Á � a�

m�1

mam xm�1

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Solution. Substitute (2), (3), and (5) into the ODE. gives two series, one for and one forIn the term use (3) and in 2y use (2). Write like powers of x vertically aligned. This gives

Add terms of like powers of x. For each power equate the sum obtained to zero. Denote these sumsby (constant terms), (first power of x), and so on:

Sum Power Equations

This gives the solution

and remain arbitrary. Hence, this is a general solution that consists of two solutions: x and. These two solutions are members of families of functions called Legendre polynomials

and Legendre functions ; here we have and . Theminus is by convention. The index 1 is called the order of these two functions and here the order is 1. More onLegendre polynomials in the next section.

Theory of the Power Series MethodThe nth partial sum of (1) is

(6)

where If we omit the terms of from (1), the remaining expression is

(7)

This expression is called the remainder of (1) after the term .For example, in the case of the geometric series

we have

s0 � 1, R0 � x � x2 � x3 � Á ,

s1 � 1 � x, R1 � x2 � x3 � x4 � Á ,

s2 � 1 � x � x2, R2 � x3 � x4 � x5 � Á , etc.

1 � x � x2 � Á � xn � Á

an(x � x0)n

Rn(x) � an�1(x � x0)n�1 � an�2(x � x0)n�2 � Á .

snn � 0, 1, Á .

sn(x) � a0 � a1(x � x0) � a2(x � x0)2 � Á � an(x � x0)n

�

1 � x2 � 13 x4 � 1

5 x6 � Á � �Q1(x)x � P1(x)Qn(x)Pn(x)1 � x2 � 1

3 x4 � 15 x6 � Á

a1a0

y � a1x � a0(1 � x2 � 13 x4 � 1

5 x6 � Á ).

30a6 � 18a4, a6 � 1830 a4 � 18

30 (�13)a0 � �1

5 a0.[x4][4]

a5 � 0 since a3 � 0[x3][3]

12a4 � 4a2, a4 � 412

a2 � �13

a0[x2][2]

a3 � 0[x][1]

a2 � �a0[x0][0]

[1][0]x0, x, x2, Á

2y � 2a0 � 2a1x � 2a2x2 � 2a3 x3 � 2a4 x4 � Á .

�2xyr � � 2a1x � 4a2 x2 � 6a3 x3 � 8a4 x4 � Á

�x2ys � � 2a2 x2 � 6a3 x3 � 12a4 x4 � Á

ys � 2a2 � 6a3 x � 12a4 x2 � 20a5 x

3 � 30a6 x4 � Á

�2xyr�x2ys.ys(1 � x2)ys

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In this way we have now associated with (1) the sequence of the partial sums. If for some this sequence converges, say,

then the series (1) is called convergent at , the number is called the valueor sum of (1) at , and we write

Then we have for every n,

(8)

If that sequence diverges at , the series (1) is called divergent at .In the case of convergence, for any positive there is an N (depending on ) such that,

by (8)

(9)

Geometrically, this means that all with lie between and (Fig. 104). Practically, this means that in the case of convergence we can approximate thesum of (1) at by as accurately as we please, by taking n large enough.sn(x1)x1s(x1)

s(x1) � Ps(x1) � Pn � Nsn(x1)

for all n � N.ƒ Rn(x1) ƒ � ƒ s(x1) � sn(x1) ƒ � P

PP

x � x1x � x1

s(x1) � sn(x1) � Rn(x1).

s(x1) � a�

m�0

am(x1 � x0)m.

x1

s(x1)x � x1

limn:�

sn(x1) � s(x1),

x � x1s0(x), s1(x), s2(x), Á

∈ ∈

s(x1) – ε s(x

1)∈ s(x

1) + ε∈

Fig. 104. Inequality (9)

R R

x0

– R x0

x0 + R

ConvergenceDivergence Divergence

Fig. 105. Convergence interval (10) of a power series with center x0

Where does a power series converge? Now if we choose in (1), the series reducesto the single term because the other terms are zero. Hence the series converges at .In some cases this may be the only value of x for which (1) converges. If there are othervalues of x for which the series converges, these values form an interval, the convergenceinterval. This interval may be finite, as in Fig. 105, with midpoint . Then the series (1)converges for all x in the interior of the interval, that is, for all x for which

(10)

and diverges for . The interval may also be infinite, that is, the series mayconverge for all x.

ƒ x � x0 ƒ � R

ƒ x � x0 ƒ � R

x0

x0a0

x � x0

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The quantity R in Fig. 105 is called the radius of convergence (because for a complexpower series it is the radius of disk of convergence). If the series converges for all x, weset (and ).

The radius of convergence can be determined from the coefficients of the series bymeans of each of the formulas

(11)

provided these limits exist and are not zero. [If these limits are infinite, then (1) convergesonly at the center .]

E X A M P L E 4 Convergence Radius , 1, 0

For all three series let

Convergence for all is the best possible case, convergence in some finite interval the usual, andconvergence only at the center is useless.

When do power series solutions exist? Answer: if p, q, r in the ODEs

(12)

have power series representations (Taylor series). More precisely, a function is calledanalytic at a point if it can be represented by a power series in powers of with positive radius of convergence. Using this concept, we can state the following basictheorem, in which the ODE (12) is in standard form, that is, it begins with the Ifyour ODE begins with, say, , divide it first by and then apply the theorem tothe resulting new ODE.

T H E O R E M 1 Existence of Power Series Solutions

If p, q, and r in (12) are analytic at then every solution of (12) is analyticat and can thus be represented by a power series in powers of withradius of convergence .

The proof of this theorem requires advanced complex analysis and can be found in Ref.[A11] listed in App. 1.

We mention that the radius of convergence R in Theorem 1 is at least equal to the distancefrom the point to the point (or points) closest to at which one of the functionsp, q, r, as functions of a complex variable, is not analytic. (Note that that point may notlie on the x-axis but somewhere in the complex plane.)

x0x � x0

R � 0x � x0x � x0

x � x0,

h(x)h(x)ysys.

x � x0x � x0

f (x)

ys � p(x)yr � q(x)y � r(x)

�(R � 0)x (R � �)

a�

m�0

m!xm � 1 � x � 2x2 � Á , ` am�1

am` �

(m � 1)!

m!� m � 1 : �, R � 0.

1

1 � x� a

�

m�0

xm � 1 � x � x2 � Á , ` am�1

am

` �1

1� 1, R � 1

ex � a�

m�0

xm

m!� 1 � x �

x2

2!� Á , ` am�1

am` �

1>(m � 1)!

1>m!�

1

m � 1 : 0, R � �

m : �

R � �

x0

^ limm:�

` am�1

am`^ lim

m:�2m

ƒ am ƒ (b) R � 1(a) R � 1

1>R � 0R � �

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Further Theory: Operations on Power SeriesIn the power series method we differentiate, add, and multiply power series, and we obtaincoefficient recursions (as, for instance, in Example 3) by equating the sum of thecoefficients of each occurring power of x to zero. These four operations are permissiblein the sense explained in what follows. Proofs can be found in Sec. 15.3.

1. Termwise Differentiation. A power series may be differentiated term by term. Moreprecisely: if

converges for , where , then the series obtained by differentiating termby term also converges for those x and represents the derivative of y for those x:

Similarly for the second and further derivatives.

2. Termwise Addition. Two power series may be added term by term. More precisely:if the series

(13)

have positive radii of convergence and their sums are and g(x), then the series

converges and represents for each x that lies in the interior of the convergenceinterval common to each of the two given series.

3. Termwise Multiplication. Two power series may be multiplied term by term. Moreprecisely: Suppose that the series (13) have positive radii of convergence and let andg(x) be their sums. Then the series obtained by multiplying each term of the first seriesby each term of the second series and collecting like powers of , that is,

converges and represents for each x in the interior of the convergence interval ofeach of the two given series.

f (x)g(x)

� a�

m�0

(a0bm � a1bm�1 � Á � amb0)(x � x0)m

a0b0 � (a0b1 � a1b0)(x � x0) � (a0b2 � a1b1 � a2b0)(x � x0)2 � Á

x � x0

f (x)

f (x) � g(x)

a�

m�0

(am � bm)(x � x0)m

f (x)

a�

m�0

am(x � x0)m and a�

m�0

bm(x � x0)m

( ƒ x � x0 ƒ � R).yr(x) � a�

m�1

mam(x � x0)m�1

yrR � 0ƒ x � x0 ƒ � R

y(x) � a�

m�0

am(x � x0)m

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4. Vanishing of All Coefficients (“Identity Theorem for Power Series.”) If a powerseries has a positive radius of convergent convergence and a sum that is identically zerothroughout its interval of convergence, then each coefficient of the series must be zero.

1. WRITING AND LITERATURE PROJECT. PowerSeries in Calculus. (a) Write a review (2–3 pages) onpower series in calculus. Use your own formulations andexamples—do not just copy from textbooks. No proofs.(b) Collect and arrange Maclaurin series in a systematiclist that you can use for your work.

2–5 REVIEW: RADIUS OF CONVERGENCEDetermine the radius of convergence. Show the details ofyour work.

2.

3.

4.

5.

6–9 SERIES SOLUTIONS BY HANDApply the power series method. Do this by hand, not by aCAS, to get a feel for the method, e.g., why a series mayterminate, or has even powers only, etc. Show the details.

6.

7.

8.

9.

10–14 SERIES SOLUTIONSFind a power series solution in powers of x. Show the details.

10.

11.

12.

13.

14. ys � 4xyr � (4x2 � 2)y � 0

ys � (1 � x2)y � 0

(1 � x2)ys � 2xyr � 2y � 0

ys � yr � x2y � 0

ys � yr � xy � 0

ys � y � 0

xyr � 3y � k (� const)

yr � �2xy

(1 � x)yr � y

a�

m�0

a23b

m

x2m

a�

m�0

x2m�1

(2m � 1)!

a�

m�0

(�1)m

km x2m

a�

m�0

(m � 1)mxm

15. Shifting summation indices is often convenient ornecessary in the power series method. Shift the indexso that the power under the summation sign is .Check by writing the first few terms explicity.

16–19 CAS PROBLEMS. IVPsSolve the initial value problem by a power series. Graphthe partial sums of the powers up to and including . Findthe value of the sum s (5 digits) at .

16.

17.

18.

19.

20. CAS Experiment. Information from Graphs ofPartial Sums. In numerics we use partial sums ofpower series. To get a feel for the accuracy for variousx, experiment with . Graph partial sums of theMaclaurin series of an increasing number of terms,describing qualitatively the “breakaway points” ofthese graphs from the graph of . Consider otherMaclaurin series of your choice.

sin x

sin x

(x � 2)yr � xy, y(0) � 4, x1 � 2

x1 � 0.5yr(0) � 1.875,y(0) � 0,(1 � x2)ys � 2xyr � 30y � 0,

x � 0.5yr(0) � 1,y(0) � 1,ys � 3xyr � 2y � 0,

yr � 4y � 1, y(0) � 1.25, x1 � 0.2

x1

x5

a�

s�2

s(s � 1)

s2 � 1 xs�1, a

�

p�1

p2

( p � 1)! xp�4

xm

P R O B L E M S E T 5 . 1

–0.5

0.5

0

1

1.5

1 2 3 4 5 6

–1

–1.5

x

Fig. 106. CAS Experiment 20. and partial sums s3, s5, s7

sin x

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SEC. 5.2 Legendre’s Equation. Legendre Polynomials 175Pn(x)

5.2 Legendre’s Equation. Legendre Polynomials

Legendre’s differential equation1

(1) (n constant)

is one of the most important ODEs in physics. It arises in numerous problems, particularlyin boundary value problems for spheres (take a quick look at Example 1 in Sec. 12.10).

The equation involves a parameter n, whose value depends on the physical orengineering problem. So (1) is actually a whole family of ODEs. For we solved itin Example 3 of Sec. 5.1 (look back at it). Any solution of (1) is called a Legendre function.The study of these and other “higher” functions not occurring in calculus is called thetheory of special functions. Further special functions will occur in the next sections.

Dividing (1) by , we obtain the standard form needed in Theorem 1 of Sec. 5.1and we see that the coefficients and of the new equationare analytic at , so that we may apply the power series method. Substituting

(2)

and its derivatives into (1), and denoting the constant simply by k, we obtain

.

By writing the first expression as two separate series we have the equation

It may help you to write out the first few terms of each series explicitly, as in Example 3of Sec. 5.1; or you may continue as follows. To obtain the same general power in allfour series, set (thus ) in the first series and simply write s insteadof m in the other three series. This gives

.a�

s�0

(s � 2)(s � 1)as�2 xs � a

�

s�2

s(s � 1)as xs � a

�

s�1

2sas xs � a

�

s�0

kas xs � 0

m � s � 2m � 2 � sxs

a�

m�2

m(m � 1)am xm�2 � a�

m�2

m(m � 1)am xm � a�

m�1

2mam xm � a�

m�0

kam xm � 0.

(1 � x2) a�

m�2

m(m � 1)am xm�2 � 2x a�

m�1

mam xm�1 � k a�

m�0

am xm � 0

n(n � 1)

y � a�

m�0

am xm

x � 0n(n � 1)>(1 � x2)�2x>(1 � x2)

1 � x2

n � 1

(1 � x2)ys � 2xyr � n(n � 1)y � 0

Pn(x)

1ADRIEN-MARIE LEGENDRE (1752–1833), French mathematician, who became a professor in Paris in1775 and made important contributions to special functions, elliptic integrals, number theory, and the calculusof variations. His book Éléments de géométrie (1794) became very famous and had 12 editions in less than30 years.

Formulas on Legendre functions may be found in Refs. [GenRef1] and [GenRef10].

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(Note that in the first series the summation begins with .) Since this equation withthe right side 0 must be an identity in x if (2) is to be a solution of (1), the sum of thecoefficients of each power of x on the left must be zero. Now occurs in the first andfourth series only, and gives [remember that ]

(3a) .

occurs in the first, third, and fourth series and gives

(3b) .

The higher powers occur in all four series and give

(3c)

The expression in the brackets can be written , as you mayreadily verify. Solving (3a) for and (3b) for as well as (3c) for , we obtain thegeneral formula

(4) .

This is called a recurrence relation or recursion formula. (Its derivation you may verifywith your CAS.) It gives each coefficient in terms of the second one preceding it, exceptfor and , which are left as arbitrary constants. We find successively

and so on. By inserting these expressions for the coefficients into (2) we obtain

(5)

where

(6)

(7) y2(x) � x �(n � 1)(n � 2)

3! x3 �

(n � 3)(n � 1)(n � 2)(n � 4)

5! x5 � � Á .

y1(x) � 1 �n(n � 1)

2! x2 �

(n � 2)n(n � 1)(n � 3)

4! x4 � � Á

y(x) � a0y1(x) � a1y2(x)

�(n � 3)(n � 1)(n � 2)(n � 4)

5! a1 �

(n � 2)n(n � 1)(n � 3)

4! a0

a5 � �

(n � 3)(n � 4)

5 # 4 a3 a4 � �

(n � 2)(n � 3)

4 # 3 a2

a3 � �

(n � 1)(n � 2)

3! a1 a2 � �

n(n � 1)

2! a0

a1a0

(s � 0, 1, Á )as�2 � �

(n � s)(n � s � 1)

(s � 2)(s � 1) as

as�2a3a2

(n � s)(n � s � 1)[ Á ]

(s � 2)(s � 1)as�2 � [�s(s � 1) � 2s � n(n � 1)]as � 0.

x2, x3, Á

3 # 2a3 � [�2 � n(n � 1)]a1 � 0

x1

2 # 1a2 � n(n � 1)a0 � 0

k � n(n � 1)x0

s � 0

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These series converge for (see Prob. 4; or they may terminate, see below). Since(6) contains even powers of x only, while (7) contains odd powers of x only, the ratio

is not a constant, so that and are not proportional and are thus linearlyindependent solutions. Hence (5) is a general solution of (1) on the interval

Note that are the points at which , so that the coefficients of thestandardized ODE are no longer analytic. So it should not surprise you that we do not geta longer convergence interval of (6) and (7), unless these series terminate after finitelymany powers. In that case, the series become polynomials.

Polynomial Solutions. Legendre Polynomials The reduction of power series to polynomials is a great advantage because then we havesolutions for all x, without convergence restrictions. For special functions arising assolutions of ODEs this happens quite frequently, leading to various important families ofpolynomials; see Refs. [GenRef1], [GenRef10] in App. 1. For Legendre’s equation thishappens when the parameter n is a nonnegative integer because then the right side of (4)is zero for , so that . Hence if n is even, reduces to a polynomial of degree n. If n is odd, the same is true for . Thesepolynomials, multiplied by some constants, are called Legendre polynomials and aredenoted by . The standard choice of such constants is done as follows. We choosethe coefficient of the highest power as

(8) (n a positive integer)

(and ). Then we calculate the other coefficients from (4), solved for interms of , that is,

(9)

The choice (8) makes for every n (see Fig. 107); this motivates (8). From (9)with and (8) we obtain

Using in the numerator and andin the denominator, we obtain

cancels, so that we get

an�2 � �

(2n � 2)!

2n(n � 1)! (n � 2)! .

n(n � 1)2n(2n � 1)

an�2 � �

n(n � 1)2n(2n � 1)(2n � 2)!

2(2n � 1)2nn(n � 1)! n(n � 1)(n � 2)! .

n! � n(n � 1)(n � 2)!n! � n(n � 1)!(2n)! � 2n(2n � 1)(2n � 2)!

an�2 � �

n(n � 1)

2(2n � 1) an � �

n(n � 1)

2(2n � 1)#

(2n)!

2n(n!)2

s � n � 2pn(1) � 1

(s � n � 2).as � �

(s � 2)(s � 1)

(n � s)(n � s � 1) as�2

as�2

asan � 1 if n � 0

an �(2n)!

2n(n!)2 �1 # 3 # 5 Á (2n � 1)

n!

xnan

Pn(x)

y2(x)y1(x)an�2 � 0, an�4 � 0, an�6 � 0, Ás � n

Pn(x)

1 � x2 � 0x � 1�1 � x � 1.

y2y1y1>y2

ƒ x ƒ � 1

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Similarly,

and so on, and in general, when ,

(10)

The resulting solution of Legendre’s differential equation (1) is called the Legendrepolynomial of degree n and is denoted by .

From (10) we obtain

(11)

where , whichever is an integer. The first few of these functionsare (Fig. 107)

and so on. You may now program (11) on your CAS and calculate as needed.Pn(x)

P0(x) � 1, P1(x) � x

P2(x) � 12 (3x2 � 1), P3(x) � 1

2 (5x3 � 3x)

P4(x) � 18 (35x4 � 30x2 � 3), P5(x) � 1

8 (63x5 � 70x3 � 15x)

(11�)

M � n>2 or (n � 1)>2

�(2n)!

2n(n!)2 xn �

(2n � 2)!

2n1! (n � 1)! (n � 2)!xn�2 � � Á

Pn(x) � a

M

m�0

(�1)m (2n � 2m)!

2nm! (n � m)! (n � 2m)! xn�2m

Pn(x)

an�2m � (�1)m (2n � 2m)!

2nm! (n � m)! (n � 2m)! .

n � 2m 0

�(2n � 4)!

2n2! (n � 2)! (n � 4)!

an�4 � �

(n � 2)(n � 3)

4(2n � 3) an�2

–1

–1 x

Pn(x) P0

P1

P4

P3

P2

1

1

Fig. 107. Legendre polynomials

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The Legendre polynomials are orthogonal on the interval , a basicproperty to be defined and used in making up “Fourier–Legendre series” in the chapteron Fourier series (see Secs. 11.5–11.6).

�1 � x � 1Pn(x)

1–5 LEGENDRE POLYNOMIALS ANDFUNCTIONS

1. Legendre functions for Show that (6) withgives and (7) gives (use

)

Verify this by solving (1) with , setting and separating variables.

2. Legendre functions for Show that (7) withgives and (6) gives

3. Special n. Derive from (11).

4. Legendre’s ODE. Verify that the polynomials in satisfy (1).

5. Obtain and .

6–9 CAS PROBLEMS6. Graph on common axes. For what x

(approximately) and is ?

7. From what n on will your CAS no longer producefaithful graphs of ? Why?

8. Graph , and some further Legendrefunctions.

9. Substitute into Legen-dre’s equation and obtain the coefficient recursion (4).

10. TEAM PROJECT. Generating Functions. Generatingfunctions play a significant role in modern appliedmathematics (see [GenRef5]). The idea is simple. If wewant to study a certain sequence and can find afunction

,

we may obtain properties of from those of G,which “generates” this sequence and is called agenerating function of the sequence.

( fn(x))

G(u, x) � a�

n�0

fn(x)un

( fn(x))

asxs � as�1x s�1 � as�2x s�2

Q0(x), Q1(x)

Pn(x)

ƒ Pn(x) ƒ � 12n � 2, Á , 10

P2(x), Á , P10(x)

P7P6

(11r)(11r)

� 1 �12

x ln 1 � x1 � x

.

y1 � 1 � x2 �13

x4 �15

x6 � Á

y2(x) � P1(x) � xn � 1n � 1.

z � yrn � 0

y2(x) � x �13

x3 �15

x5 � Á �12

ln 1 � x1 � x

.

x � 12 x2 � 1

3 x3 � Á

ln (1 � x) �P0(x) � 1n � 0n � 0.

(a) Legendre polynomials. Show that

(12)

is a generating function of the Legendre polynomials.Hint: Start from the binomial expansion of then set , multiply the powers of out, collect all the terms involving , and verify thatthe sum of these terms is .

(b) Potential theory. Let and be two points inspace (Fig. 108, ). Using (12), show that

This formula has applications in potential theory. (is the electrostatic potential at due to a charge Qlocated at . And the series expresses in terms ofthe distances of and from any origin O and theangle between the segments and .)OA2OA1u

A2A1

1>rA1

A2

Q>r

�1r2 a

�

m�0

Pm(cos u) ar1

r2b

m

.

1r �

1

2r12 � r2

2 � 2r1r2 cos u

r2 � 0A2A1

Pn(x)unun

2xu � u2v � 2xu � u21>11 � v,

G(u, x) �1

21 � 2xu � u2� a

�

n�0

Pn(x)un


r2

r

A2

θA

1r1

0

Fig. 108. Team Project 10

(c) Further applications of (12). Show that, and

.

11–15 FURTHER FORMULAS11. ODE. Find a solution of

, by reduction to the Legendreequation.

12. Rodrigues’s formula (13)2 Applying the binomialtheorem to , differentiating it n times termby term, and comparing the result with (11), show that

(13) Pn(x) �1

2nn!

dn

dxn [(x2 � 1)n].

(x2 � 1) n

a � 0n(n � 1)y � 0,(a2 � x2)ys � 2xyr �

P2n(0) � (�1) n # 1 # 3 Á (2n � 1)>[2 # 4 Á (2n)]Pn(1) � 1, Pn(�1) � (�1) n, P2n�1(0) � 0

2OLINDE RODRIGUES (1794–1851), French mathematician and economist.

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15. Associated Legendre functions are needed, e.g.,in quantum physics. They are defined by

(15)

and are solutions of the ODE

(16)

where . Find , and and verify that they satisfy (16).P4

2(x)P22(x) P2

1(x),P1

1(x),q(x) � n(n � 1) � k2>(1 � x2)

(1 � x2)ys � 2xyr � q(x)y � 0

Pnk(x) � (1 � x2)k>2

dkpn(x)

dxk

Pnk

(x)13. Rodrigues’s formula. Obtain from (13).

14. Bonnet’s recursion.3 Differentiating (13) withrespect to u, using (13) in the resulting formula, andcomparing coefficients of , obtain the Bonnetrecursion.

(14)

where . This formula is useful for com-putations, the loss of significant digits being small(except near zeros). Try (14) out for a few computationsof your own choice.

n � 1, 2, Á

(n � 1)Pn�1(x) � (2n � 1)xPn(x) � npn�1(x),

un

(11r)

3OSSIAN BONNET (1819–1892), French mathematician, whose main work was in differential geometry.4GEORG FROBENIUS (1849–1917), German mathematician, professor at ETH Zurich and University of Berlin,

student of Karl Weierstrass (see footnote, Sect. 15.5). He is also known for his work on matrices and in group theory.In this theorem we may replace x by x � x0 with any number x0. The condition a0 � 0 is no restriction; it

simply means that we factor out the highest possible power of x.The singular point of (1) at x � 0 is often called a regular singular point, a term confusing to the student,

which we shall not use.

5.3 Extended Power Series Method: Frobenius Method

Several second-order ODEs of considerable practical importance—the famous Besselequation among them—have coefficients that are not analytic (definition in Sec. 5.1), butare “not too bad,” so that these ODEs can still be solved by series (power series times alogarithm or times a fractional power of x, etc.). Indeed, the following theorem permitsan extension of the power series method. The new method is called the Frobeniusmethod.4 Both methods, that is, the power series method and the Frobenius method, havegained in significance due to the use of software in actual calculations.

T H E O R E M 1 Frobenius Method

Let and be any functions that are analytic at . Then the ODE

(1)

has at least one solution that can be represented in the form

(2)

where the exponent r may be any (real or complex) number (and r is chosen so that).

The ODE (1) also has a second solution (such that these two solutions are linearlyindependent) that may be similar to (2) (with a different r and different coefficients)or may contain a logarithmic term. (Details in Theorem 2 below.)

a0 � 0

(a0 � 0)y(x) � xra

�

m�0

am xm � xr(a0 � a1x � a2 x2 � Á )

ys �b(x)

x yr �

c(x)

x2 y � 0

x � 0c(x)b(x)

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For example, Bessel’s equation (to be discussed in the next section)

(v a parameter)

is of the form (1) with and analytic at , so that the theoremapplies. This ODE could not be handled in full generality by the power series method.

Similarly, the so-called hypergeometric differential equation (see Problem Set 5.3) alsorequires the Frobenius method.

The point is that in (2) we have a power series times a single power of x whose exponentr is not restricted to be a nonnegative integer. (The latter restriction would make the wholeexpression a power series, by definition; see Sec. 5.1.)

The proof of the theorem requires advanced methods of complex analysis and can befound in Ref. [A11] listed in App. 1.

Regular and Singular Points. The following terms are practical and commonly used.A regular point of the ODE

is a point at which the coefficients p and q are analytic. Similarly, a regular point ofthe ODE

is an at which are analytic and (so what we can divide by and getthe previous standard form). Then the power series method can be applied. If is not aregular point, it is called a singular point.

Indicial Equation, Indicating the Form of SolutionsWe shall now explain the Frobenius method for solving (1). Multiplication of (1) by gives the more convenient form

We first expand and in power series,

or we do nothing if and are polynomials. Then we differentiate (2) term by term,finding

(2*)

� xr�23r(r � 1)a0 � (r � 1)ra1x � Á 4.

ys(x) � a�

m�0

(m � r)(m � r � 1)am xm�r�2

yr(x) � a�

m�0

(m � r)am xm�r�1 � xr�13ra0 � (r � 1)a1x � Á 4

c(x)b(x)

b(x) � b0 � b1x � b2 x2 � Á , c(x) � c0 � c1x � c2 x2 � Á

c(x)b(x)

x2ys� xb(x)yr � c(x)y � 0.(1r)

x2

x0

h~

h~

(x0) � 0h~

, p~, q~x0

h~

(x)ys � p~(x)yr(x) � q~(x)y � 0

x0

ys � p(x)yr � q(x)y � 0

x � 0c(x) � x2 � v2b(x) � 1

ys �1x

yr � ax2 � v2

x2 b y � 0

SEC. 5.3 Extended Power Series Method: Frobenius Method 181

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By inserting all these series into we obtain

(3).

We now equate the sum of the coefficients of each power to zero. Thisyields a system of equations involving the unknown coefficients . The smallest poweris and the corresponding equation is

.

Since by assumption , the expression in the brackets must be zero. Thisgives

(4) .

This important quadratic equation is called the indicial equation of the ODE (1). Its roleis as follows.

The Frobenius method yields a basis of solutions. One of the two solutions will alwaysbe of the form (2), where r is a root of (4). The other solution will be of a form indicatedby the indicial equation. There are three cases:

Case 1. Distinct roots not differing by an integer .

Case 2. A double root.

Case 3. Roots differing by an integer .

Cases 1 and 2 are not unexpected because of the Euler–Cauchy equation (Sec. 2.5), thesimplest ODE of the form (1). Case 1 includes complex conjugate roots and because Im is imaginary, so it cannot be a real integer. Theform of a basis will be given in Theorem 2 (which is proved in App. 4), without a generaltheory of convergence, but convergence of the occurring series can be tested in eachindividual case as usual. Note that in Case 2 we must have a logarithm, whereas in Case 3we may or may not.

T H E O R E M 2 Frobenius Method. Basis of Solutions. Three Cases

Suppose that the ODE (1) satisfies the assumptions in Theorem 1. Let and bethe roots of the indicial equation (4). Then we have the following three cases.

Case 1. Distinct Roots Not Differing by an Integer. A basis is

(5)

and

(6)

with coefficients obtained successively from (3) with and , respectively.r � r2r � r1

y2(x) � x r2(A0 � A1x � A2 x2 � Á )

y1(x) � xr1(a0 � a1x � a2 x2 � Á )

r2r1

r1r1 � r2 � r1 � r1 � 2ir2 � r1r1

1, 2, 3, Á

1, 2, 3, Á

r (r � 1) � b0r � c0 � 0

[ Á ]a0 � 0

[r (r � 1) � b0r � c0]a0 � 0

xram

x r, x r�1, x r�2, Á

� (c0 � c1x � Á ) x r(a0 � a1x � Á ) � 0

xr[r(r � 1)a0 � Á ] � (b0 � b1x � Á ) x r(ra0 � Á )

(1r)

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Case 2. Double Root A basis is

(7)

(of the same general form as before) and

(8) .

Case 3. Roots Differing by an Integer. A basis is

(9)

(of the same general form as before) and

(10)

where the roots are so denoted that and k may turn out to be zero.

Typical ApplicationsTechnically, the Frobenius method is similar to the power series method, once the rootsof the indicial equation have been determined. However, (5)–(10) merely indicate thegeneral form of a basis, and a second solution can often be obtained more rapidly byreduction of order (Sec. 2.1).

E X A M P L E 1 Euler–Cauchy Equation, Illustrating Cases 1 and 2 and Case 3 without a Logarithm

For the Euler–Cauchy equation (Sec. 2.5)

( constant)

substitution of gives the auxiliary equation

which is the indicial equation [and is a very special form of (2) ]. For different roots we get a basis, and for a double root r we get a basis . Accordingly, for this simple ODE, Case 3

plays no extra role.

E X A M P L E 2 Illustration of Case 2 (Double Root)

Solve the ODE

(11) .

(This is a special hypergeometric equation, as we shall see in the problem set.)

Solution. Writing (11) in the standard form (1), we see that it satisfies the assumptions in Theorem 1. [Whatare and in (11)?] By inserting (2) and its derivatives into (11) we obtain

(12)

.� 3 a�

m�0

(m � r)am xm�r � a�

m�0

(m � r)am xm�r�1 � a�

m�0

am xm�r � 0

a�

m�0

(m � r)(m � r � 1)am xm�r � a�

m�0

(m � r)(m � r � 1)am xm�r�1

(2*)c(x)b(x)

x(x � 1)ys � (3x � 1)yr � y � 0

�x r, x r ln xy1 � xr1, y2 � xr2

r1, r2!y � xr

r(r � 1) � b0r � c0 � 0,

y � xr

b0, c0x2ys � b0 xyr � c0y � 0

r1 � r2 � 0

y2(x) � ky1(x) ln x � x r2(A0 � A1x � A2 x2 � Á ),

y1(x) � x r1(a0 � a1x � a2 x2 � Á)

(x � 0)y2(x) � y1(x) ln x � x r(A1x � A2 x2 � Á)

[r � 12 (1 � b0)]y1(x) � x r(a0 � a1x � a2 x2 � Á)

r1 � r2 � r.

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The smallest power is , occurring in the second and the fourth series; by equating the sum of its coefficientsto zero we have

.

Hence this indicial equation has the double root .

First Solution. We insert this value into (12) and equate the sum of the coefficients of the powerto zero, obtaining

thus . Hence , and by choosing we obtain the solution

.

Second Solution. We get a second independent solution by the method of reduction of order (Sec. 2.1),substituting and its derivatives into the equation. This leads to (9), Sec. 2.1, which we shall use inthis example, instead of starting reduction of order from scratch (as we shall do in the next example). In (9) ofSec. 2.1 we have , the coefficient of in (11) in standard form. By partial fractions,

Hence (9), Sec. 2.1, becomes

,

and are shown in Fig. 109. These functions are linearly independent and thus form a basis on the interval (as well as on ). �1 � x � �0 � x � 1

y2y1

y2 � uy1 �ln x

1 � x .u � ln xur � U � y1

�2e��p dx �(x � 1)2

(x � 1)2x�

1

x ,

��p dx � �� 3x � 1x(x � 1)

dx � ��a 2x � 1

�1xb dx � �2 ln (x � 1) � ln x.

yrp � (3x � 1)>(x2 � x)

y2 � uy1

y2

( ƒ x ƒ � 1)y1(x) � a�

m�0

xm �1

1 � x

a0 � 1a0 � a1 � a2 � Áas�1 � as

s(s � 1)as � (s � 1)sas�1 � 3sas � (s � 1)as�1 � as � 0

xsr � 0

r � 0

[�r (r � 1) � r]a0 � 0, thus r 2 � 0

xr�1

4

3

2

–1

–2

–2 2 6

–3

–4

0

1

x

y

y2

y1

4

Fig. 109. Solutions in Example 2

E X A M P L E 3 Case 3, Second Solution with Logarithmic Term

Solve the ODE

(13) .

Solution. Substituting (2) and into (13), we have

.(x2 � x) a�

m�0

(m � r)(m � r � 1)am xm�r�2 � x a�

m�0

(m � r)am xm�r�1 � a�

m�0

am xm�r � 0

(2*)

(x2 � x)ys � xyr � y � 0

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We now take , x, and x inside the summations and collect all terms with power and simplify algebraically,

.

In the first series we set and in the second , thus . Then

(14) .

The lowest power is (take in the second series) and gives the indicial equation

.

The roots are and . They differ by an integer. This is Case 3.

First Solution. From (14) with we have

.

This gives the recurrence relation

.

Hence successively. Taking , we get as a first solution .

Second Solution. Applying reduction of order (Sec. 2.1), we substitute andinto the ODE, obtaining

.

xu drops out. Division by x and simplification give

.

From this, using partial fractions and integrating (taking the integration constant zero), we get

Taking exponents and integrating (again taking the integration constant zero), we obtain

and are linearly independent, and has a logarithmic term. Hence and constitute a basis of solutionsfor positive x.

The Frobenius method solves the hypergeometric equation, whose solutions includemany known functions as special cases (see the problem set). In the next section we usethe method for solving Bessel’s equation.

�y2y1y2y2y1

ur �x � 1

x2 �1

x�

1

x2, u � ln x �1

x , y2 � xu � x ln x � 1.

ln ur � ln 2 x � 1

x2 2 .usur

� �

x � 2

x2 � x� �

2

x�

1

1 � x ,

(x2 � x)us � (x � 2)ur � 0

(x2 � x)(xus � 2ur) � x(xur � u) � xu � 0

ys2 � xus � 2ury2 � y1u � xu, yr2 � xur � u

y1 � xr1a0 � xa0 � 1a1 � 0, a2 � 0, Á

(s � 0, 1, Á )as�1 �s2

(s � 2)(s � 1) as

a�

s�0

3s2as � (s � 2)(s � 1)as�14xs�1 � 0

r � r1 � 1

r2 � 0r1 � 1

r(r � 1) � 0

s � �1xr�1

a�

s�0

(s � r � 1) 2as xs�r � a

�

s��1

(s � r � 1)(s � r)as�1x s�r � 0

s � m � 1m � s � 1m � s

a�

m�0

(m � r � 1)2 am xm�r � a�

m�0

(m � r)(m � r � 1)am xm�r�1 � 0

xm�rx2

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5CARL FRIEDRICH GAUSS (1777–1855), great German mathematician. He already made the first of his greatdiscoveries as a student at Helmstedt and Göttingen. In 1807 he became a professor and director of the Observatoryat Göttingen. His work was of basic importance in algebra, number theory, differential equations, differentialgeometry, non-Euclidean geometry, complex analysis, numeric analysis, astronomy, geodesy, electromagnetism,and theoretical mechanics. He also paved the way for a general and systematic use of complex numbers.

1. WRITING PROJECT. Power Series Method andFrobenius Method. Write a report of 2–3 pagesexplaining the difference between the two methods. Noproofs. Give simple examples of your own.

2–13 FROBENIUS METHOD Find a basis of solutions by the Frobenius method. Try toidentify the series as expansions of known functions. Showthe details of your work.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14. TEAM PROJECT. Hypergeometric Equation, Series,and Function. Gauss’s hypergeometric ODE5 is

(15)

Here, a, b, c are constants. This ODE is of the form, where are polyno-

mials of degree 2, 1, 0, respectively. These polynomialsare written so that the series solution takes a most prac-tical form, namely,

(16)

.

This series is called the hypergeometric series. Its sumis called the hypergeometric function and is

denoted by F(a, b, c; x). Here, . Bychoosing specific values of a, b, c we can obtain anincredibly large number of special functions as solutions

c � 0, �1, �2, Á

y1(x)

�a(a � 1)(a � 2)b(b � 1)(b � 2)

3! c(c � 1)(c � 2) x3 � Á

y1(x) � 1 �ab

1! c x �

a(a � 1)b(b � 1)2! c(c � 1)

x2

p0 p1,p2,p2 ys � p1yr � p0y � 0

x(1 � x)ys� [c � (a � b � 1)x]yr� aby � 0.

xys � (1 � 2x)yr � (x � 1)y � 0

x2ys � 6xyr � (4x2 � 6)y � 0

xys � (2 � 2x)yr � (x � 2)y � 0

xys � 2yr � 4xy � 0

2x(x � 1)ys � (x � 1)yr � y � 0

xys � yr � xy � 0

ys � (x � 1)y � 0

xys � 2x3yr � (x2 � 2)y � 0

xys � (2x � 1)yr � (x � 1)y � 0

xys � y � 0

xys � 2yr � xy � 0

(x � 2)2ys � (x � 2)yr � y � 0

of (15) [see the small sample of elementary functionsin part (c)]. This accounts for the importance of (15).

(a) Hypergeometric series and function. Show thatthe indicial equation of (15) has the roots and

. Show that for the Frobeniusmethod gives (16). Motivate the name for (16) byshowing that

(b) Convergence. For what a or b will (16) reduce toa polynomial? Show that for any other a, b, c( ) the series (16) converges when

.

(c) Special cases. Show that

Find more such relations from the literature on specialfunctions, for instance, from [GenRef1] in App. 1.

(d) Second solution. Show that for theFrobenius method yields the following solution (where

:

(17)

Show that

.

(e) On the generality of the hypergeometric equation.Show that

(18) (t 2 � At � B)# #y � (Ct � D)y

#� Ky � 0

y2(x) � x1�cF(a � c � 1, b � c � 1, 2 � c; x)

� Á b .

x2�(a � c � 1)(a � c � 2)(b � c � 1)(b � c � 2)

2! (�c � 2)(�c � 3)

y2(x) � x1�c a1 �(a � c � 1)(b � c � 1)

1! (�c � 2) x

c � 2, 3, 4, Á)

r2 � 1 � c

ln 1 � x1 � x

� 2xF(12 , 1, 32 ; x2).

ln (1 � x) � xF(1, 1, 2; �x),

arcsin x � xF(12 , 12 , 32 ; x2),

arctan x � xF(12 , 1, 32 ; �x2)

(1 � x)n � 1 � nxF (1 � n, 1, 2; x),

(1 � x)n � F (�n, b, b; �x),

ƒ x ƒ � 1c � 0, �1, �2, Á

F (1, 1, 1; x) � F(1, b, b; x) � F (a, 1, a; x) �1

1 � x .

r1 � 0r2 � 1 � cr1 � 0


c05.qxd 10/28/10 1:33 PM Page 186

SEC. 5.4 Bessel’s Equation. Bessel Functions J� (x) 187

with , etc., constant A, B, C, D, K, and , can be reduced to

the hypergeometric equation with independent variable

and parameters related by . From this you see that (15)

is a “normalized form” of the more general (18) andthat various cases of (18) can thus be solved in termsof hypergeometric functions.

K � abC � a � b � 1,Ct1 � D � �c(t2 � t1),

x �t � t1

t2 � t1

At � B � (t � t1)(t � t2), t1 � t2

t 2 �y#

� dy>dt 15–20 HYPERGEOMETRIC ODEFind a general solution in terms of hypergeometricfunctions.

15.

16.

17.

18.

19.

20. 3t(1 � t)y# #

� ty#

� y � 0

2(t 2 � 5t � 6)y# #

� (2t � 3)y#

� 8y � 0

4(t 2 � 3t � 2)y# #

� 2y#

� y � 0

4x(1 � x)ys � yr � 8y � 0

x(1 � x)ys � (12 � 2x)yr � 2y � 0

2x(1 � x)ys � (1 � 6x)yr � 2y � 0

5.4 Bessel’s Equation. Bessel Functions One of the most important ODEs in applied mathematics in Bessel’s equation,6

(1)

where the parameter (nu) is a given real number which is positive or zero. Bessel’sequation often appears if a problem shows cylindrical symmetry, for example, as themembranes in Sec.12.9. The equation satisfies the assumptions of Theorem 1. To see this,divide (1) by to get the standard form . Hence, accordingto the Frobenius theory, it has a solution of the form

(2) .

Substituting (2) and its first and second derivatives into Bessel’s equation, we obtain

We equate the sum of the coefficients of to zero. Note that this power corresponds to in the first, second, and fourth series, and to in the thirdseries. Hence for and , the third series does not contribute since .m 0s � 1s � 0

m � s � 2m � sxs�rx s�r

� a�

m�0

am xm�r�2 � �2a�

m�0

am xm�r � 0.

a�

m�0

(m � r)(m � r � 1)am xm�r � a�

m�0

(m � r)am xm�r

(a0 � 0)y(x) � a�

m�0

am xm�r

ys � yr>x � (1 � �2>x2)y � 0x2

�

x2ys � xyr � (x2 � �2)y � 0

J�(x)

6FRIEDRICH WILHELM BESSEL (1784–1846), German astronomer and mathematician, studied astronomyon his own in his spare time as an apprentice of a trade company and finally became director of the new KönigsbergObservatory.

Formulas on Bessel functions are contained in Ref. [GenRef10] and the standard treatise [A13].

c05.qxd 10/28/10 1:33 PM Page 187

For all four series contribute, so that we get a general formula for all these s.We find

(a)

(3) (b)

(c) .

From (3a) we obtain the indicial equation by dropping ,

(4) .

The roots are and .

Coefficient Recursion for For , Eq. (3b) reduces to Hence since . Substituting in (3c) and combining the three termscontaining gives simply

(5)

Since and , it follows from (5) that . Hence we have todeal only with even-numbered coefficients with . For , Eq. (5) becomes

.

Solving for gives the recursion formula

(6) , .

From (6) we can now determine successively. This gives

and so on, and in general

(7) .

Bessel Functions for Integer Integer values of v are denoted by n. This is standard. For the relation (7) becomes

(8) .m � 1, 2, Áa2m �(�1)ma0

22mm! (n � 1)(n � 2) Á (n � m) ,

� � n

� � nJn(x)

m � 1, 2, Áa2m �(�1)ma0

22mm! (� � 1)(� � 2) Á (� � m) ,

a4 � �

a2

222(v � 2)�

a0

242! (� � 1)(� � 2)

a2 � �

a0

22(� � 1)

a2, a4, Á

m � 1, 2, Áa2m � � 1

22m(� � m) a2m�2

a2m

(2m � 2�)2ma2m � a2m�2 � 0

s � 2ms � 2mas

a3 � 0, a5 � 0, Á� 0a1 � 0

(s � 2�)sas � as�2 � 0.

as

r � �� 0a1 � 0(2� � 1)a1 � 0.r � �r � r1 � v.

r2 � ��r1 � � ( 0)

(r � �)(r � �) � 0

a0

(s � 2, 3, Á )(s � r)(s � r � 1)as � (s � r)as � as�2 � �2as � 0

(s � 1)(r � 1)ra1 � (r � 1)a1 � �2a1 � 0

(s � 0)r(r � 1)a0 � ra0 � �2a0 � 0

s � 2, 3, Á


c05.qxd 10/28/10 1:33 PM Page 188

is still arbitrary, so that the series (2) with these coefficients would contain this arbitraryfactor . This would be a highly impractical situation for developing formulas orcomputing values of this new function. Accordingly, we have to make a choice. The choice

would be possible. A simpler series (2) could be obtained if we could absorb thegrowing product into a factorial function Whatshould be our choice? Our choice should be

(9)

because then in (8), so that (8) simply becomes

(10) .

By inserting these coefficients into (2) and remembering that we obtaina particular solution of Bessel’s equation that is denoted by :

(11) .

is called the Bessel function of the first kind of order n. The series (11) convergesfor all x, as the ratio test shows. Hence is defined for all x. The series convergesvery rapidly because of the factorials in the denominator.

E X A M P L E 1 Bessel Functions and

For we obtain from (11) the Bessel function of order 0

(12)

which looks similar to a cosine (Fig. 110). For we obtain the Bessel function of order 1

(13) ,

which looks similar to a sine (Fig. 110). But the zeros of these functions are not completely regularly spaced(see also Table A1 in App. 5) and the height of the “waves” decreases with increasing x. Heuristically, in (1) in standard form [(1) divided by ] is zero (if ) or small in absolute value for large x, and so is

, so that then Bessel’s equation comes close to , the equation of ; also actsas a “damping term,” in part responsible for the decrease in height. One can show that for large x,

(14)

where is read “asymptotically equal” and means that for fixed n the quotient of the two sides approaches 1as .x : �

�

Jn(x) � B

2px

cos ax �

np

2�

p

4b

yr>xcos x and sin xys � y � 0yr>xn � 0x2

n2>x2

J1(x) � a�

m�0

(�1)mx2m�1

22m�1m! (m � 1)!�

x

2�

x3

231! 2!�

x5

252! 3!�

x7

273! 4!� � Á

n � 1

J0(x) � a�

m�0

(�1)mx2m

22m(m!)2� 1 �

x2

22(1!)2�

x4

24(2!)2�

x6

26(3!)2� � Á

n � 0

J1(x)J0(x)

Jn(x)Jn(x)

(n 0)Jn(x) � xna�

m�0

(�1) mx2m

22m�nm! (n � m)!

Jn(x)c1 � 0, c3 � 0, Á

m � 1, 2, Áa2m �(�1)m

22m�nm! (n � m)! ,

n! (n � 1) Á (n � m) � (n � m)!

a0 �1

2nn!

(n � m)!(n � 1)(n � 2) Á (n � m)a0 � 1

a0

a0


c05.qxd 10/28/10 1:33 PM Page 189

Formula (14) is surprisingly accurate even for smaller . For instance, it will give you good startingvalues in a computer program for the basic task of computing zeros. For example, for the first three zeros of you obtain the values 2.356 (2.405 exact to 3 decimals, error 0.049), 5.498 (5.520, error 0.022), 8.639 (8.654,error 0.015), etc. �

J0

x (�0)


0

1

0.5

x105

J0

J1

Fig. 110. Bessel functions of the first kind J0 and J1

Bessel Functions for any . Gamma FunctionWe now proceed from integer to any . We had in (9). So wehave to extend the factorial function to any . For this we choose

(15)

with the gamma function defined by

(16) .

(CAUTION! Note the convention on the left but in the integral.) Integrationby parts gives

.

This is the basic functional relation of the gamma function

(17) .

Now from (16) with and then by (17) we obtain

and then and in general

(18) .(n � 0, 1, Á ) (n � 1) � n!

(2) � 1 # (1) � 1!, (3) � 2 (1) � 2!

(1) � ��

0

e�t dt � �e�t `0

�

� 0 � (�1) � 1

� � 0

(� � 1) � � (�)

(� � 1) � �e�tt � `�

0� ��

�

0

e�tt ��1 dt � 0 � � (�)

�� 1

(� � �1) (� � 1) � ��

0e�tt � dt

(� � 1)

a0 �1

2� (� � 1)

� 0n!a0 � 1>(2nn!)� 0� � n

� 0J�(x)

c05.qxd 10/28/10 1:33 PM Page 190

Hence the gamma function generalizes the factorial function to arbitrary positive .Thus (15) with agrees with (9).

Furthermore, from (7) with given by (15) we first have

.

Now (17) gives and so on,so that

.

Hence because of our (standard!) choice (15) of the coefficients (7) are simply

(19) .

With these coefficients and we get from (2) a particular solution of (1), denotedby and given by

(20) .

is called the Bessel function of the first kind of order �. The series (20) convergesfor all x, as one can verify by the ratio test.

Discovery of Properties from SeriesBessel functions are a model case for showing how to discover properties and relations offunctions from series by which they are defined. Bessel functions satisfy an incredibly largenumber of relationships—look at Ref. [A13] in App. 1; also, find out what your CAS knows.In Theorem 3 we shall discuss four formulas that are backbones in applications and theory.

T H E O R E M 1 Derivatives, Recursions

The derivative of with respect to x can be expressed by or (x) bythe formulas

(21)(a)

(b) .

Furthermore, and its derivative satisfy the recurrence relations

(21)(c)

(d) J��1(x) � J��1(x) � 2Jr�(x).

J��1(x) � J��1(x) �2�x J�(x)

J�(x)

[x��J�(x)]r � �x��J��1(x)

[x�J�(x)]r � x�J��1(x)

J��1J��1(x)J�(x)

J�(x)

J�(x) � x�a�

m�0

(�1)mx2m

22m��m! (� � m � 1)

J�(x)r � r1 � �

a2m �(�1)m

22m��m! (� � m � 1)

a0

(� � 1)(� � 2) Á (� � m) (� � 1) � (� � m � 1)

(� � 1) (� � 1) � (� � 2), (� � 2) (� � 2) � (� � 3)

a2m �(�1)m

22mm! (� � 1)(� � 2) Á (� � m)2� (� � 1)

a0

� � n�


c05.qxd 10/28/10 1:33 PM Page 191

P R O O F (a) We multiply (20) by and take under the summation sign. Then we have

We now differentiate this, cancel a factor 2, pull out, and use the functionalrelationship [see (17)]. Then (20) with insteadof shows that we obtain the right side of (21a). Indeed,

(b) Similarly, we multiply (20) by , so that in (20) cancels. Then we differentiate,cancel 2m, and use . This gives, with ,

Equation (20) with instead of and s instead of m shows that the expression onthe right is . This proves (21b).

(c), (d) We perform the differentiation in (21a). Then we do the same in (21b) andmultiply the result on both sides by . This gives

(a*)

(b*) .

Substracting (b*) from (a*) and dividing the result by gives (21c). Adding (a*) and(b*) and dividing the result by gives (21d).

E X A M P L E 2 Application of Theorem 1 in Evaluation and Integration

Formula (21c) can be used recursively in the form

for calculating Bessel functions of higher order from those of lower order. For instance, so that can be obtained from tables of and (in App. 5 or, more accurately, in Ref. [GenRef1] in App. 1).

To illustrate how Theorem 1 helps in integration, we use (21b) with integrated on both sides. Thisevaluates, for instance, the integral

.

A table of (on p. 398 of Ref. [GenRef1]) or your CAS will give you

.

Your CAS (or a human computer in precomputer times) obtains from (21), first using (21c) with ,that is, then (21c) with , that is, . Together,J2 � 2x�1J1 � J0� � 1J3 � 4x�1J2 � J1,

� � 2J3

�18

# 0.128943 � 0.019563 � 0.003445

J3

I � �2

1

x�3J4(x) dx � �x�3J3(x) 2 21

� �

1

8 J3(2) � J3(1)

� � 3J1J0J2

J2(x) � 2J1(x)>x � J0(x),

J��1(x) �2�

x J�(x) � J��1(x)

�x�x�

��x��1J� � x�Jr� � �x�J��1

�x��1J� � x�Jr� � x�J��1

x2�

�x��J��1(x)�� 1

(x��J�)r � a�

m�1

(�1)mx2m�1

22m��1(m � 1)! (� � m � 1)� a

�

s�0

(�1)s�1x2s�1

22s��1s! (� � s � 2) .

m � s � 1m! � m(m � 1)!x�x��

(x�J�)r � a�

m�0

(�1)m2(m � �)x2m�2��1

22m��m! (� � m � 1)� x�x��1

a�

m�0

(�1)mx2m

22m��1m! (� � m) .

�� 1 (� � m � 1) � (� � m) (� � m)

x2��1

x�J�(x) � a�

m�0

(�1)mx2m�2�

22m��m! (� � m � 1) .

x2�x�


c05.qxd 10/28/10 1:33 PM Page 192

This is what you get, for instance, with Maple if you type int . And if you type evalf(int ), you obtain0.003445448, in agreement with the result near the beginning of the example.

Bessel Functions with Half-Integer Are ElementaryWe discover this remarkable fact as another property obtained from the series (20) andconfirm it in the problem set by using Bessel’s ODE.

E X A M P L E 3 Elementary Bessel Functions with . The Value

We first prove (Fig. 111)

(22)

The series (20) with is

The denominator can be written as a product AB, where (use (16) in B)

here we used (proof below)

(23) .

The product of the right sides of A and B can be written

.

Hence

J1>2(x) �B

2px

a�

m�0

(�1)mx2m�1

(2m � 1)!�B

2px

sin x.

AB � (2m � 1)2m (2m � 1) Á 3 # 2 # 11p � (2m � 1)!1p

�(12) � 1p

� (2m � 1)(2m � 1) Á 3 # 1 # 1p ;

B � 2m�1�(m � 32) � 2m�1(m � 1

2)(m � 12) Á

32

# 12�(1

2)

A � 2mm! � 2m(2m � 2)(2m � 4) Á 4 # 2,

J1>2(x) � 1x a�

m�0

(�1) mx2m

22m�1>2m! �(m � 32)

�B

2

x a

�

m�0

(�1) mx2m�1

22m�1m! �(m � 32)

.

� � 12

(a) J1>2(x) �B

2px

sin x, (b) J�1>2(x) �B

2px

cos x.

�( 12

)� � 12

, � 32

, �52

, Á�J�

�J�

�( Á )( Á )

� �18

J1(2) � 14

J0(2) � 7J1(1) � 4J0(1).

� �18 32J1(2) � 2J0(2) � J1(2)4 � 38J1(1) � 4J0(1) � J1(1)4

I � x�3(4x�1(2x�1J1 � J0) � J1) 2 21


x2π 4π 6π0

1

Fig. 111. Bessel functions and J�1>2J1>2

c05.qxd 10/29/10 10:56 PM Page 193

This proves (22a). Differentiation and the use of (21a) with now gives

This proves (22b). From (22) follow further formulas successively by (21c), used as in Example 2.We finally prove by a standard trick worth remembering. In (15) we set . Then

and

We square on both sides, write v instead of u in the second integral, and then write the product of the integralsas a double integral:

We now use polar coordinates r, by setting Then the element of area is and we have to integrate over r from 0 to and over from 0 to (that is, over the first quadrant of theuv-plane):

By taking the square root on both sides we obtain (23).

General Solution. Linear DependenceFor a general solution of Bessel’s equation (1) in addition to we need a second linearlyindependent solution. For not an integer this is easy. Replacing by in (20), wehave

(24) .

Since Bessel’s equation involves , the functions and are solutions of the equationfor the same . If is not an integer, they are linearly independent, because the first termsin (20) and in (24) are finite nonzero multiples of and . Thus, if is not an integer,a general solution of Bessel’s equation for all is

This cannot be the general solution for an integer because, in that case, we havelinear dependence. It can be seen that the first terms in (20) and (24) are finite nonzeromultiples of and , respectively. This means that, for any integer , we havelinear dependence because

(25) .(n � 1, 2, Á )J�n(x) � (�1)n Jn(x)

� � nx��x�

� � n

y(x) � c1J�(x) � c2J��(x)

x � 0�x��x�

��J��J��2

J��(x) � x��a�

m�0

(�1)mx2m

22m��m! (m � � � 1)

��J�

�

a 1

2b

2

� 4�p>2

0�

�

0

e�r 2

r dr du � 4 #p

2 ��

0

e�r 2

r dr � 2a�1

2b e�r

2 `�

0� p.

p>2u�

du dv � r dr duu � r cos u, v � r sin u.u

a 1

2b

2

� 4��

0

e�u2

du ��

0

e�v2

dv � 4��

0�

�

0

e�(u2�v2) du dv.

a1

2b � �

�

0

e�tt �1>2 dt � 2��

0

e�u2

du.

dt � 2u dut � u2 (1

2) � 1p

[1x J1>2(x)]r �B

2p

cos x � x1>2 J�1>2(x).

� � 12


c05.qxd 10/28/10 1:33 PM Page 194

P R O O F To prove (25), we use (24) and let approach a positive integer n. Then the gammafunction in the coefficients of the first n terms becomes infinite (see Fig. 553 in App.A3.1), the coefficients become zero, and the summation starts with . Since inthis case by (18), we obtain

(26)

The last series represents , as you can see from (11) with m replaced by s. Thiscompletes the proof.

The difficulty caused by (25) will be overcome in the next section by introducing furtherBessel functions, called of the second kind and denoted by .Y�

�

(�1)nJn(x)

(m � n � s).J�n(x) � a�

m�n

(�1) mx2m� n

22m�nm! (m � n)!� a

�

s�0

(�1)n�sx2s�n

22s�n (n � s)! s!

(m � n � 1) � (m � n)!m � n

�


1. Convergence. Show that the series (11) converges forall x. Why is the convergence very rapid?

2–10 ODES REDUCIBLE TO BESSEL’S ODEThis is just a sample of such ODEs; some more follow inthe next problem set. Find a general solution in terms of and or indicate when this is not possible. Use theindicated substitutions. Show the details of your work.

2.

3.

4.

5. Two-parameter ODE

6.

7.

8.

9.

10.

11. CAS EXPERIMENT. Change of Coefficient. Findand graph (on common axes) the solutions of

for (or as far as you get usefulgraphs). For what k do you get elementary functions?Why? Try for noninteger k, particularly between 0 and 2,to see the continuous change of the curve. Describe thechange of the location of the zeros and of the extrema ask increases from 0. Can you interpret the ODE as a modelin mechanics, thereby explaining your observations?

12. CAS EXPERIMENT. Bessel Functions for Large x.

(a) Graph for on common axes.n � 0, Á , 5Jn(x)

k � 0, 1, 2, Á , 10

ys � kx�1 yr � y � 0, y(0) � 1, yr(0) � 0,

(y � x�u, x� � z)x2 ys � (1 � 2�)xyr � �2(x2� � 1 � �2)y � 0

xys � (2� � 1)yr � xy � 0 (y � x��u)

(2x � 1 � z)(2x � 1) 2ys � 2(2x � 1)yr � 16x(x � 1)y � 0

x2 ys � xyr � 14 (x2 � 1)y � 0 (x � 2z)

x2ys � 14 (x � 3

4) y � 0 (y � u1x, 1x � z)

(lx � z)x2 ys � xyr � (l2x2 � �2)y � 0

ys � (e�2x � 19)y � 0 (e�x � z)

xys � yr � 14 y � 0 (1x � z)

x2 ys � xyr � (x2 � 449)y � 0

J��

J�


(b) Experiment with (14) for integer n. Using graphs,find out from which on the curves of (11)and (14) practically coincide. How does changewith n?

(c) What happens in (b) if (Our usual notationin this case would be .)

(d) How does the error of (14) behave as a func-tion of x for fixed n? [Error exact value minusapproximation (14).]

(e) Show from the graphs that has extrema where. Which formula proves this? Find further

relations between zeros and extrema.

13–15 ZEROS of Bessel functions play a key role inmodeling (e.g. of vibrations; see Sec. 12.9).

13. Interlacing of zeros. Using (21) and Rolle’s theorem,show that between any two consecutive positive zerosof there is precisely one zero of .

14. Zeros. Compute the first four positive zeros of and from (14). Determine the error and comment.

15. Interlacing of zeros. Using (21) and Rolle’s theorem,show that between any two consecutive zeros of there is precisely one zero of .

16–18 HALF-INTEGER PARAMETER: APPROACHBY THE ODE

16. Elimination of first derivative. Show that with gives from the ODE

the ODE

not containing the first derivative of u.

us � 3q(x) � 14 p(x)2 � 1

2 pr(x)4 u � 0,

p(x)yr � q(x)y � 0ys �

v(x) � exp (�12 � p(x) dx)

y � uv

J1(x)J0(x)

J1(x)J0(x)

Jn�1(x)Jn(x)

J1(x) � 0J0(x)

�

�n � 1

2?

xn

x � xn

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5.5 Bessel Functions Y (x). General SolutionTo obtain a general solution of Bessel’s equation (1), Sec. 5.4, for any , we now introduceBessel functions of the second kind , beginning with the case .

When , Bessel’s equation can be written (divide by x)

(1) .

Then the indicial equation (4) in Sec. 5.4 has a double root . This is Case 2 in Sec.5.3. In this case we first have only one solution, . From (8) in Sec. 5.3 we see thatthe desired second solution must be of the form

(2)

We substitute and its derivatives

into (1). Then the sum of the three logarithmic terms , and is zerobecause is a solution of (1). The terms and (from ) cancel. Hencewe are left with

2 Jr0 � a�

m�1

m(m � 1) Am xm�1 � a�

m�1

m Am xm�1 � a�

m�1

Am xm�1 � 0.

xys and yrJ0>x�J0>xJ0

x J0 ln xx Js0 ln x, Jr0 ln x

ys2 � Js0 ln x �2Jr0x

�

J0

x2 � a�

m�1

m (m � 1) Am xm�2

yr2 � Jr0 ln x �J0

x � a�

m�1

mAm xm�1

y2

y2(x) � J0(x) ln x � a�

m�1

Am xm.

J0(x)r � 0

xys � yr � xy � 0

n � 0� � n � 0Y�(x)�

n

17. Bessel’s equation. Show that for (1) the substitutionin Prob. 16 is and gives

(27) x2u� � (x2 � 1_4 � �2)u � 0.

18. Elementary Bessel functions. Derive (22) in Example 3from (27).

19–25 APPLICATION OF (21): DERIVATIVES,INTEGRALS

Use the powerful formulas (21) to do Probs. 19–25. Showthe details of your work.

19. Derivatives. Show that

20. Bessel’s equation. Derive (1) from (21).

J0(x) � J1(x)>x, Jr2(x) � 12[J1(x) � J3(x)].

Jr1(x) �Jr0(x) � �J1(x),

y � ux�1>221. Basic integral formula. Show that

22. Basic integral formulas. Show that

23. Integration. Show that (The last integral is nonelemen-

tary; tables exist, e.g., in Ref. [A13] in App. 1.)

24. Integration. Evaluate .

25. Integration. Evaluate .�J5(x) dx

�x�1J4(x) dx

xJ0(x) ��J0(x) dx.�x2J0(x) dx � x2J1(x) �

�J��1(x) dx � �J��1(x) dx � 2J�(x).

�x��J��1(x) dx � �x��J�(x) � c,

�x�J��1(x) dx � x�J�(x) � c.


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SEC. 5.5 Bessel Functions Y� (x). General Solution 197

Addition of the first and second series gives The power series of isobtained from (12) in Sec. 5.4 and the use of in the form

Together with and this gives

(3*)

First, we show that the with odd subscripts are all zero. The power occurs only inthe second series, with coefficient . Hence . Next, we consider the even powers

. The first series contains none. In the second series, gives the termIn the third series, . Hence by equating the sum of the

coefficients of to zero we have

.

Since , we thus obtain successively.

We now equate the sum of the coefficients of to zero. For this gives

thus .

For the other values of s we have in the first series in , hence, in the second , and in the third We thus obtain

For this yields

thus

and in general

(3) .

Using the short notations

(4)

and inserting (4) and into (2), we obtain the result

(5) � J0(x) ln x �14

x2 �3

128 x4 �

1113,824

x6 � � Á .

y2(x) � J0(x) ln x � a�

m�1

(�1)m�1hm

22m(m!)2 x2m

A1 � A3 � Á � 0

m � 2, 3, Áh1 � 1 hm � 1 �12

� Á �1m

m � 1, 2, ÁA2m �(�1)m�1

22m(m!)2 a1 �12

�13

� Á �1mb ,

A4 � � 3

12818 � 16A4 � A2 � 0,

s � 1

(�1)s�1

22s(s � 1)! s!� (2s � 2)2A2s�2 � A2s � 0.

m � 1 � 2s � 1.m � 1 � 2s � 1m � s � 1(3*) 2m � 1 � 2s � 1

A2 � 14�1 � 4A2 � 0,

s � 0x2s�1

A3 � 0, A5 � 0, Á ,A1 � 0

s � 1, 2, Á(2s � 1)2A2s�1 � A2s�1 � 0,

x2sm � 1 � 2s(2s � 1)2A2s�1x2s.

m � 1 � 2sx2sA1 � 0A1

x0Am

a�

m�1

(�1)mx2m�1

22m�2m! (m � 1)!� a

�

m�1

m2Am xm�1 � a�

m�1

Am xm�1 � 0.

�Am xm�1�m2Am xm �1

Jr0(x) � a�

m�1

(�1)m2mx2m�1

22m (m!)2� a

�

m�1

(�1)mx2m�1

22m�1m! (m � 1)!.

m!>m � (m � 1)!Jr0(x)�m2Amxm�1.

c05.qxd 10/28/10 1:33 PM Page 197

Since and are linearly independent functions, they form a basis of (1) for .Of course, another basis is obtained if we replace by an independent particular solutionof the form , where and b are constants. It is customary to choose

and , where the number is the so-calledEuler constant, which is defined as the limit of

as s approaches infinity. The standard particular solution thus obtained is called the Besselfunction of the second kind of order zero (Fig. 112) or Neumann’s function of orderzero and is denoted by . Thus [see (4)]

(6)

For small the function behaves about like ln x (see Fig. 112, why?), and

Bessel Functions of the Second Kind For a second solution can be obtained by manipulations similar to thosefor , starting from (10), Sec. 5.4. It turns out that in these cases the solution alsocontains a logarithmic term.

The situation is not yet completely satisfactory, because the second solution is defineddifferently, depending on whether the order is an integer or not. To provide uniformityof formalism, it is desirable to adopt a form of the second solution that is valid for allvalues of the order. For this reason we introduce a standard second solution definedfor all by the formula

(7)(a)

(b)

This function is called the Bessel function of the second kind of order or Neumann’sfunction7 of order . Figure 112 shows and .

Let us show that and are indeed linearly independent for all (and ).For noninteger order , the function is evidently a solution of Bessel’s equation

because and are solutions of that equation. Since for those the solutionsand are linearly independent and involves , the functions and areY�J�J��Y�J��J�

�J�� (x)J�(x)Y�(x)�

x � 0�Y�J�

Y1(x)Y0(x)��

Yn(x) � lim�:n

Y�(x).

Y�(x) �1

sin �p [J�(x) cos �p � J��(x)]

�Y�(x)

�

n � 0� � n � 1, 2, Á

Yn(x)

Y0(x) : �� as x : 0.Y0(x)x � 0

Y0(x) �2p

c J0(x) aln x2

� gb � a�

m�1

(�1)m�1hm

22m(m!)2 x2m d .Y0(x)

1 �12 � Á �

1s � ln s

g � 0.57721566490 Áb � g � ln 2a � 2>pa (� 0)a( y2 � bJ0)

y2

x � 0y2J0


7 CARL NEUMANN (1832–1925), German mathematician and physicist. His work on potential theory usinginteger equation methods inspired VITO VOLTERRA (1800–1940) of Rome, ERIK IVAR FREDHOLM (1866–1927)of Stockholm, and DAVID HILBERT (1962–1943) of Göttingen (see the footnote in Sec. 7.9) to develop the fieldof integral equations. For details see Birkhoff, G. and E. Kreyszig, The Establishment of Functional Analysis, HistoriaMathematica 11 (1984), pp. 258–321.

The solutions are sometimes denoted by ; in Ref. [A13] they are called Weber’s functions; Euler’sconstant in (6) is often denoted by C or ln .g

N�(x)Y�(x)

c05.qxd 11/4/10 12:19 PM Page 198

SEC. 5.5 Bessel Functions Y� (x). General Solution 199

linearly independent. Furthermore, it can be shown that the limit in (7b) exists and is a solution of Bessel’s equation for integer order; see Ref. [A13] in App. 1. We shallsee that the series development of contains a logarithmic term. Hence and

are linearly independent solutions of Bessel’s equation. The series developmentof can be obtained if we insert the series (20) in Sec. 5.4 and (2) in this sectionfor and into (7a) and then let approach n; for details see Ref. [A13]. Theresult is

(8)

where , and [as in (4)] ,

hm � 1 �12

� Á �1m

, hm�n � 1 �12

� Á �1

m � n.

h0 � 0, h1 � 1x � 0, n � 0, 1, Á

� x�n

p a

n�1

m�0

(n � m � 1)!22m�nm!

x2m

a�

m�0

(�1)m�1(hm � hm�n)

22m�nm! (m � n)! x2mYn(x) �

2p

Jn(x) aln x

2� gb �

xn

p

�J�� (x)J�(x)Yn(x)

Yn(x)Jn(x)Yn(x)

Yn

–0.5

0.5

0 5 x

Y0

Y1

10

Fig. 112. Bessel functions of the second kind and (For a small table, see App. 5.)

Y1.Y0

For the last sum in (8) is to be replaced by 0 [giving agreement with (6)].Furthermore, it can be shown that

.

Our main result may now be formulated as follows.

T H E O R E M 1 General Solution of Bessel’s Equation

A general solution of Bessel’s equation for all values of (and ) is

(9)

We finally mention that there is a practical need for solutions of Bessel’s equation thatare complex for real values of x. For this purpose the solutions

(10)H�

(2)(x) � J�(x) � iY�(x)

H�(1)(x) � J�(x) � iY�(x)

y(x) � C1J�(x) � C2Y�(x).

x � 0�

Y�n(x) � (�1)nYn(x)

n � 0

c05.qxd 10/28/10 1:33 PM Page 199

1. Why are we looking for power series solutions of ODEs?

2. What is the difference between the two methods in thischapter? Why do we need two methods?

3. What is the indicial equation? Why is it needed?

4. List the three cases of the Frobenius method, and giveexamples of your own.

5. Write down the most important ODEs in this chapterfrom memory.

1–9 FURTHER ODE’s REDUCIBLE TO BESSEL’S ODE

Find a general solution in terms of and . Indicatewhether you could also use instead of . Use theindicated substitution. Show the details of your work.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. CAS EXPERIMENT. Bessel Functions for Large x.It can be shown that for large x,

(11)

with defined as in (14) of Sec. 5.4.

(a) Graph for on common axes. Arethere relations between zeros of one function andextrema of another? For what functions?

(b) Find out from graphs from which on thecurves of (8) and (11) (both obtained from your CAS)practically coincide. How does change with n?xn

x � xn

n � 0, Á , 5Yn(x)

�

Yn(x) � 22>(px) sin (x � 12

np� 14 p)

xys � 5yr � xy � 0 (y � x3u)

ys � k2x4y � 0 (y � u1x, 13 kx3 � z)

ys � k2x2y � 0 (y � u1x, 12 kx2 � z)

xys � yr � 36y � 0 (121x � z)

4xys � 4yr � y � 0 (1x � z)

ys � xy � 0 ( y � u1x, 23x3>2 � z)

9x2 ys � 9xyr � (36x4 � 16)y � 0 (x2 � z)

xys � 5yr � xy � 0 ( y � u>x2)

x2 ys � xyr � (x2 � 16) y � 0

Y�J��

Y�J�

(c) Calculate the first ten zeros , offrom your CAS and from (11). How does the error

behave as m increases?

(d) Do (c) for and . How do the errorscompare to those in (c)?

11–15 HANKEL AND MODIFIED BESSEL FUNCTIONS

11. Hankel functions. Show that the Hankel functions (10)form a basis of solutions of Bessel’s equation for any .

12. Modified Bessel functions of the first kind of order

are defined by . Showthat satisfies the ODE

(12)

13. Modified Bessel functions. Show that has therepresentation

(13) .

14. Reality of . Show that is real for all real x (andreal ), for all real , and where n is any integer.

15. Modified Bessel functions of the third kind (sometimescalled of the second kind) are defined by the formula (14)below. Show that they satisfy the ODE (12).

(14) .K�(x) �p

2 sin �p 3I��(x) � I�(x)4

I�n(x) � In(x),x � 0I�(x) � 0�I�(x)I�

I�(x) � a�

m�0

x2m��

22m��m! (m � � � 1)

I�(x)

x2 ys � xyr � (x2 � �2) y � 0.

I�

I� (x) � i��J� (ix), i � 1�1�

�

Y2(x)Y1(x)

Y0(x)xm, m � 1, Á , 10


C H A P T E R 5 R E V I E W Q U E S T I O N S A N D P R O B L E M S

6. Can a power series solution reduce to a polynomial?When? Why is this important?

7. What is the hypergeometric equation? Where does thename come from?

8. List some properties of the Legendre polynomials.

9. Why did we introduce two kinds of Bessel functions?

10. Can a Bessel function reduce to an elementary func-tion? When?

8HERMANN HANKEL (1839–1873), German mathematician.


are frequently used. These linearly independent functions are called Bessel functions ofthe third kind of order or first and second Hankel functions8 of order .

This finishes our discussion on Bessel functions, except for their “orthogonality,” whichwe explain in Sec. 11.6. Applications to vibrations follow in Sec. 12.10.

��

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11–20 POWER SERIES METHODOR FROBENIUS METHOD

Find a basis of solutions. Try to identify the series asexpansions of known functions. Show the details of yourwork.

11.

12.

13. (x � 1)2 ys � (x � 1) yr � 35y � 0

xys � (1 � 2x) yr � (x � 1) y � 0

ys � 4y � 0

14.

15.

16.

17.

18.

19.

20. xys � yr � xy � 0

ys �14x

y � 0

xys � 3yr � 4x3 y � 0

xys � (x � 1) yr � y � 0

x2 ys � 2x3 yr � (x2 � 2) y � 0

x2 ys � xyr � (x2 � 5) y � 0

16(x � 1)2 ys � 3y � 0

Summary of Chapter 5 201

SUMMARY OF CHAPTER 5Series Solution of ODEs. Special Functions

The power series method gives solutions of linear ODEs

(1)

with variable coefficients p and q in the form of a power series (with any center ,e.g., )

(2) .

Such a solution is obtained by substituting (2) and its derivatives into (1). This givesa recurrence formula for the coefficients. You may program this formula (or evenobtain and graph the whole solution) on your CAS.

If p and q are analytic at (that is, representable by a power series in powersof with positive radius of convergence; Sec. 5.1), then (1) has solutions ofthis form (2). The same holds if

are analytic at and so that we can divide by and obtain the standardform (1). Legendre’s equation is solved by the power series method in Sec. 5.2.

The Frobenius method (Sec. 5.3) extends the power series method to ODEs

(3)

whose coefficients are singular (i.e., not analytic) at , but are “not too bad,”namely, such that a and b are analytic at . Then (3) has at least one solution ofthe form

(4) y(x) � (x � x0)ra�

m�0

am(x � x0)m � a0(x � x0)r � a1(x � x0)r�1 � Á

x0

x0

ys �a(x)

x � x0 yr �

b(x)

(x � x0)2 y � 0

h�h�(x0) � 0,x0

h�(x)ys � p�(x)yr � q�(x)y � 0

h, � p, � q� inx – x0

x0

y(x) � a�

m�0

am(x � x0)m � a0 � a1(x � x0) � a2(x � x0)2 � Á

x0 � 0x0

ys � p(x) yr � q(x)y � 0

c05.qxd 10/28/10 1:33 PM Page 201

where r can be any real (or even complex) number and is determined by substituting(4) into (3) from the indicial equation (Sec. 5.3), along with the coefficients of (4).A second linearly independent solution of (3) may be of a similar form (with differentr and ’s) or may involve a logarithmic term. Bessel’s equation is solved by theFrobenius method in Secs. 5.4 and 5.5.

“Special functions” is a common name for higher functions, as opposed to theusual functions of calculus. Most of them arise either as nonelementary integrals [see(24)–(44) in App. 3.1] or as solutions of (1) or (3). They get a name and notationand are included in the usual CASs if they are important in application or in theory.Of this kind, and particularly useful to the engineer and physicist, are Legendre’sequation and polynomials (Sec. 5.2), Gauss’s hypergeometric equationand functions F(a, b, c; x) (Sec. 5.3), and Bessel’s equation and functions and

(Secs. 5.4, 5.5).Y�

J�

P0 , P1, Á

am


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series solutions of odes. special functions · series solutions of odes. special functions ......

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