Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning...

Advanced Engineering Mathematics, 7th Edition Peter V. O’Neil

© 2012 Cengage Learning Engineering. All Rights Reserved.

CHAPTER 4

Series Solutions

© 2012 Cengage Learning Engineering. All Rights Reserved. 2

• Sometimes we can solve an initial value problem explicitly. For example, the problem

has the unique solution

CHAPTER 4 : Page 129

2 1; (0) 3y y y

21( ) (1 5 )2

xy x e


• This solution is in closed form, which means that it is a finite algebraic combination of elementary functions (such as polynomials, exponentials, sines and cosines, and the like).



• We may, however, encounter problems for which there is no closed form solution. For example,

has the unique solution

• This solution (while explicit) has no elementary, closed form expression.


2; (0) 4xy e y x y

2

0( ) 4 .

x xxe e ey x e e d e


• In such a case, we might try a numerical approximation. However, we may also be able to write a series solution that contains useful information. In this chapter, we will deal with two kinds of series solutions: power series (Section 4.1) and Frobenius series (Section 4.2).



4.1 Power Series Solutions


A function f is called analytic at x0 if f (x) has a power series representation in some interval (x0 − h, x0 + h) about x0. In this interval,

where the an’s are the Taylor coefficients of f (x) at x0:

00

( ) ( ) ,nn

n

f x a x x

( )0

1 ( ).!

nna f x

n



• Here n! (n factorial) is the product of the integers from 1 through n if n is a positive integer, and 0! = 1 by definition. The symbol f (n)(x0) denotes the nth derivative of f evaluated at x0.




• As examples of power series representations, sin(x) expanded about 0 is

for all x, and the geometric series is

for −1 < x < 1.


2 1

0

1sin( )2 1 !

n

n

x xn

0

11

n

n

xx



• An initial value problem having analytic coefficients has analytic solutions. We will state this for the first- and second-order cases when the differential equation is linear.



THEOREM 4.1

1. 1. If p and q are analytic at x0, then the problem

has a unique solution that is analytic at x0.

2. If p, q, and f are analytic at x0, then the problem

has a unique solution that is analytic at x0.


0 0( ) ( ); ( )y p x y q x y x y

0 0( ) ( ) ( ); ( ) , ( )y p x y q x y f x y x A y x B



• We are therefore justified in seeking power series solutions of linear equations having analytic coefficients. This strategy may be carried out by substituting into the differential equation and attempting to solve for the ans.


00

nnn

y a x x


EXAMPLE 4.1

• We will solve

• We can solve this using an integrating factor, obtaining

This is correct, but it involves an integral we cannot evaluate in closed form.


121

y xyx

2 2 2

0

1( ) .1

xx xy x e e d ce


EXAMPLE 4.1

• For a series solution, let

• Then

with the summation starting at 1, because the derivative of the first term a0 of the power series for y is zero.

CHAPTER 4 : 0

0

nn

n

y a x

1

1

nn

n

y na x


EXAMPLE 4.1

• Substitute the series into the differential equation to obtain

(4.1)

CHAPTER 4 : Page 130-131

1 1

1 0

12 .1

n nn n

n n

na x a xx


EXAMPLE 4.1

• We would like to combine these series and factor out a common power of x to solve for the an ’s. To do this, write 1/(1−x) as a power series about 0 as

for −1 < x < 1. Substitute this into equation (4.1) to obtain

(4.2)


0

11

n

n

xx

1 1

1 0 0

2 .n n nn n

n n n

na x a x x


EXAMPLE 4.1

• Now rewrite the series so that they all contain powers xn . This is like a change of variables in the summation index. First,

and next,


1 21 2 3 1

1 0

2 3 1 ,n nn n

n n

na x a a x a x n a x

1 2 30 1 2

0

11

2 2 2 2

2 .

nn

n

nn

n

a x a x a x a x

a x


EXAMPLE 4.1

• Now equation (4.2) can be written as

(4.3)

• These rearrangements allow us to combine these summations for n = 1, 2, ··· and to write the n = 0 terms separately to obtain

(4.4)


1 10 1 0

1 2 0.n n nn n

n n n

n a x a x x

1 1 11

(( 1) 2 1) 1 0.nn n

n

n a a x a


EXAMPLE 4.1

• Because the right side of equation (4.4) is zero for all x in (−1, 1), the coefficient of each power of x on the left, as well as the constant term a1 − 1, must equal zero. This gives us

and


1 1( 1) 2 1 0 for 1,2,3,n nn a a n

1 1 0.a


EXAMPLE 4.1

• Then a1 = 1, and

• This is a recurrence relation for the coefficients, giving an+1 in terms of a preceding coefficient an−1.


1 11 (1 2 ) for 1,2,3, .

1n na a nn


EXAMPLE 4.1

• Now solve for some of the coefficients using this recurrence relation:


2 0

3 1

4 2

0 0

1( 1) (1 2 ),21 1( 2) (1 2 ) ,3 31( 3) (1 2 )41 1 (1 1 2 ) ,4 2

n a a

n a a

n a a

a a


EXAMPLE 4.1

and so on.


5 3

06 4

7 5

1 1 1( 4) (1 2 ) (1 2/3) ,5 5 3

11( 5) 1 2 ,6 61 1( 6) (1 2 ) ,7 21

n a a

an a a

n a a


EXAMPLE 4.1

• With the coefficients computed thus far, the solution has the form

• This has one arbitrary constant, a0, as expected. By continuing to use the recurrence relation, we can compute as many terms of the series as we like.


2 30 0

4 50

6 70

1 1( ) (1 2 )2 3

1 12 31 11 .6 21

y x a x a x x

a x x

a x x


EXAMPLE 4.2

• We will find a power series solution of

expanded about x0 = 0.• Substitute into the differential equation.

This will require that we compute


2 0y x y

0 n

nny a x

1 2

1 2

and 1 .n nn n

n n

y na x y n na x


EXAMPLE 4.2

• Substitute these power series into the differential equation to obtain

or

(4.5)


2 2

2 0

1 0n nn n

n n

n na x x a x

2 2

2 0

1 0n nn n

n n

n n a x a x


EXAMPLE 4.2

• We will shift indices so that the power of x in both summations is the same, allowing us to combine terms from both summations. One way to do this is to write

and


22

2 0

1 2 1n nn n

n n

n na x n n a x

22

0 2

n nn n

n n

a x a x


EXAMPLE 4.2

• Now equation (4.5) is

• We can combine the terms for n ≥ 2 in one summation. This requires that we write the n = 0 and n = 1 terms in the last equation separately, or else we lose terms:


2 20 2

2 1 0.n nn n

n n

n n a x a x

02 3 2 2

2

2 2(3) [( 2)( 1) + ] 0.nn n

n

a x a x n n a a x


EXAMPLE 4.2

• The left side can be zero for all x in some interval (−h, h) only if the coefficient of each power of x is zero:

and

• The last equation gives us

(4.6)


2 3 0a a

2 2( 2)( 1) 0 for 2.n nn n a a n

2 21 for 2,3, .

( 2)( 1) n na a nn n


EXAMPLE 4.2

• This is a recurrence relation for the coefficients of the series solution, giving us a4 in terms of a0, a5 in terms of a1, and so on. Recurrence relations always give a coefficient in terms of one or more previous coefficients, allowing us to generate as many terms of the series solution as we want. To illustrate, use n =2 in equation (4.6) to obtain


4 0 01 1

(4)(3) 12a a a


EXAMPLE 4.2

• With n = 3,

• In turn, we obtain


5 1 11 1 .

(5)(4) 20a a a

6 21 0

(6)(5)a a


EXAMPLE 4.2

because a2 = 0,

because a3 = 0,

and so on. CHAPTER 4 : Page 133

7 31 0

(7)(6)a a

8 4 0 0

9 5 1 1

1 1 1 ,(8)(7) (56)(12) 672

1 1 1 ,9 8 72 20 1440

a a a a

a a a a


EXAMPLE 4.2

• Thus far, we have the first few terms of the series solution about 0:


42 30 1 0

5 6 7 8 91

4 80

5 91

1( ) 0 012

1 1 1 +0 +020 672 1440

1 1 112 6721 1 .20 1440

y x a a x x x a x

a x x x x x

a x x

a x x x


EXAMPLE 4.2

• This is the general solution, since a0 and a1 are arbitrary constants. Because a0 = y(0) and a1 = y(0), a unique solution is determined by specifying these two constants.



Homework for 4.1

• 1, 3, 6, 8



4.2 Frobenius Solutions

• We will focus on the differential equation(4.7)

• If P(x) 0 on some interval, then we can divide by P(x) to obtain the standard form

(4.8)


( ) ( ) ( ) ( ).P x y Q x y R x y F x

( ) ( ) ( ).y p x y q x y f x




If P(x0) = 0, we call x0 a singular point of equation (4.7). This singular point regular if

are analytic at x0. A singular point that is not regular is an irregular singular point.

20 0

( )( ) and ( ) ( ) ( )

Q x R xx x x xP x P x


EXAMPLE 4.3

has singular points at 0 and 2. Now

is not analytic (or even defined) at 0, so 0 is an irregular singular point.


3 2 2( 2) 5( 2)( 2) 3 0x x y x x y x y

3 2 2

( ) 5 ( 2)( 2) 5 2( 0)( ) ( 2) 2

Q x x x x xxP x x x x x


EXAMPLE 4.3

• But

and

are both analytic at 2, so 2 is a regular singular point of this differential equation.


3

( ) 5( 2) ( 2)( )

Q x xxP x x

2 ( ) 3( 2)( )

R xxP x x



• We will not treat the case of an irregular singular point. If equation (4.7) has a regular singular point at x0, there may be no power series solution about x0, but there will be a Frobenius series solution, which has the form

with c0 0. We must solve for the coefficients cn and a number r to make this series a solution. We will look at an example to get some feeling for how this works and then examine the method more critically.


00

( ) ( )n rn

n

y x c x x


EXAMPLE 4.4

• Zero is a regular singular point of

• Substitute to obtain


2 5 ( 4) 0.x y xy x y

0n r

nny c x

0 0

1

0 0

( )( 1) 5( )

4 0.

n r n rn n

n n

n r n rn n

n n

n r n r c x n r c x

c x c x


EXAMPLE 4.4

• Notice that the n = 0 term in the proposed series solution is c0xr , which is not constant if c0 0, so the series for the derivatives begins with n = 0 (unlike what we saw with power series). Shift indices in the third summation to write this equation as


0 0

11 0

1 5( )

4 0.

n r n rn n

n n

n r n rn n

n n

n r n r c x n r c x

c x c x


EXAMPLE 4.4

• Combine terms to write

• Since we require that c0 0, the coefficient of xr is zero only if

This is called the indicial equation and is used to solve for r , obtaining the repeated root r = −2.


0

11

[ 1 5 4]

( )( 1) 5( r) 4 ] 0.

r

n rn n n n

n

r r r c x

n r n r c n c c c x

( 1) 5 4 0.r r r


EXAMPLE 4.4

• Set the coefficient of xn+r in the series equal to zero to obtain

or, with r = −2,


1( )( 1) 5( ) 4 0n n n nn r n r c n r c c c

1( 2)( 3) 5( 2) 4 0.n n n nn n c n c c c


EXAMPLE 4.4

• From this we obtain the recurrence relation

• This simplifies to


11 for 1, 2, .

( 2)( 3) 5( 2) 4n nc c nn n n

12

1 for 1,2, . n nc c nn


EXAMPLE 4.4

• Solve for some coefficients:

and so on.CHAPTER 4 : Page 136

1 0

2 1 0 02

3 2 02

4 3 02

1 1 14 4 21 19 2 3

1 116 (2 3 4)

c c

c c c c

c c c

c c c


EXAMPLE 4.4

• In general,

for n =1, 2, 3, ···. We have found the Frobenius solution

for x 0. This series converges for all nonzero x.CHAPTER 4 : Page 136

02

1( 1)!

nnc c

n

2 1 20

20 2

0

1 1 14 36 576

11!

n n

n

y x c x x x x

c xn



• Usually, we cannot expect the recurrence equation for cn to have such a simple form.

• Example 4.4 shows that an equation with a regular singular point may have only one Frobenius series solution about that point. A second, linearly independent solution is needed. The following theorem tells us how to produce two linearly independent solutions. For convenience, the statement is posed in terms of x0 = 0.



THEOREM 4.2

• Suppose 0 is a regular singular point of

• Then(1) The differential equation has a Frobenius

solution

with c0 0. This series converges in some interval (0, h) or (−h, 0).


( ) ( ) ( ) 0.P x y Q x y R x y

0

( ) n rn

n

y x c x


THEOREM 4.2

• Suppose that the indicial equation has real roots r1 and r2 with r1 ≥ r2. Then the following conclusions hold.

(2) If r1 − r2 is not a positive integer, then there are two linearly independent Frobenius solutions

with c0 0 and c∗0 0. These solutions are valid at least in an interval (0, h) or (−h, 0).


1 2*1 2

0 0

( ) and ( )n r n rn n

n n

y x c x y x c x


THEOREM 4.2

(3) If r1 − r2 = 0, then there is a Frobenius solution

with c0 0 , and there is a second solution

These solutions are linearly independent on some interval (0, h).


11

0

( ) n rn

n

y x c x

1*2 1

1

ln .n rn

n

y x y x c x


THEOREM 4.2

(4) If r1 − r2 is a positive integer, then there is a Frobenius solution

with c0 0 , and there is a second solution

with c∗0 0. y1 and y2 are linearly independent solutions on some interval (0, h).


11

0

( ) .n rn

n

y x c x

2*2 1

0

( ) ( ) ln( ) n rn

n

y x ky x x c x



• The method of Frobenius consists of using Frobenius series and Theorem 4.2 to solve equation (4.7) in some interval (−h, h), (0, h), or (−h, 0), assuming that 0 is a regular singular point.




• Proceed as follows:• Step 1.

– Substitute into the differential equation, and solve for the roots r1 and r2 of the indicial equation for r . This yields a Frobenius solution (which may or may not be a power series).


0n r

nny c x



• Step 2. – Depending on which of Cases (2), (3), or (4) of

Theorem 4.2 applies, the theorem provides a template for a second solution which is linearly independent from the first. Once we know what this second solution looks like, we can substitute its general form into the differential equation and solve for the coefficients and, in Case (4), the constant k.




• We will illustrate the Cases (2), (3), and (4) of the Frobenius theorem. For case (2), Example 4.5, we will provide all of the details. In Cases (3) and (4) (Examples 4.6, 4.7, and 4.8), we will omit some of the calculations and include just those that relate to the main point of that case.



EXAMPLE 4.5 Case 2 of the Frobenius Theorem

• We will solve


2 1 12 0.2 2

x y x x y x y



• It is routine to check that 0 is a regular singular point. Substitute the Frobenius series to obtain


0n r

nny c x

2 1

0 0 0

1

0 0

1( )( 1) ( ) 2( )2

1 0.2

n r n r n rn n

n n n

n r n rn n

n n

n r n r c c n r c x n r x

c x c x



• In order to be able to factor xn+r from most terms, shift indices in the third and fourth summations to write this equation as


1 11

0 0 0

1 1( )( 1) 2 12 2

1 1( 1) 0.2 2

n rn n n n n

n

r

n r n r c n r c n r c c c x

r r c c r c x



• This equation will hold if the coefficient of each power of x is zero:

(4.9)

and for n =1, 2, 3, ···,

(4.10)


01 1( 1) 02 2

r r r c

1 11 1( )( 1) ( ) 2 1 0.2 2n n n n nn r n r c n r c n r c c c



• Assuming that c0 0, an essential requirement of the method, equation (4.9) implies that

(4.11)


1 1( 1) 0.2 2

r r r



• This is the indicial equation for this differential equation. It has the roots r1 = 1 and r2 = −1/2. This puts us in case 2 of the Frobenius theorem. From equation (4.10), we obtain the recurrence relation

for n =1, 2, 3, ···.


11 12 2

1 2( 1)=( )( 1) ( )n n

n rc cn r n r n r



• First put r1 = 1 into the recurrence relation to obtain

for n = 1, 2, 3, ···.


132

2 1n n

nc cn n



• Some of these coefficients are

and so on.


1 0 0

2 1 0 0

3 2 0 0

3 6 ,5 / 2 55 5 6 6 ,7 7 5 7

7 14 6 4 ,27 / 2 27 7 9

c c c

c c c c

c c c c



• One Frobenius solution is

Because r1 is a nonnegative integer, this first Frobenius series is actually a power series about 0.


2 3 41 0

6 6 4( ) .5 7 9

y x c x x x x



• For a second Frobenius solution, substitute r = r2 = −1/2 into the recurrence relation. To avoid confusion with the first solution, we will denote the coefficients c∗n instead of cn .We obtain

for n = 1, 2, 3, ···.


32* *

131 1 1 12 2 2 2 2

1 2n n

nc c

n n n



• This simplifies to

for n = 1, 2, 3, ···. It happens in this example that c∗1 = 0, so each c∗n =0 for n = 1, 2, 3, ···, and the second Frobenius solution is

for x > 0.CHAPTER 4 : Page 139

* *

132

2 2n n

nc n cn n

* 1/ 2 * 1/ 22 0

0

nn

n

y x c x c x



• We will solve

• In Example 4.4, we found the indicial equation

with repeated root r1 = r2 = −2 and the recurrence relation

for n =1, 2, ···.CHAPTER 4 : Page 139

2 5 ( 4) 0.x y xy x y

( 1) 5 4 0r r r

12

1n nc c

n



• This yielded the first Frobenius solution


21 0 2

0

2 1 20

11!

1 1 1 .4 36 576

n n

n

y x c xn

c x x x x



• Conclusion (3) of Theorem 4.2 tells us the general form of a second solution that is linearly independent from y1(x). Set


* 22 1

1

( ) ( ) ln( ) .nn

n

y x y x x c x



• Note that on the right, the series starts at n = 1, not n = 0. Substitute this series into the differential equation and find after some rearranging of terms that


* 2 * 21 1

1 1

* 1 * 2

1 1

21 1 1

4 2 ( 2)( 3) 5( 2)

4

ln 5 4 0.

n nn n

n n

n nn n

n n

y xy n n c x n c x

c x c x

x x y xy x y



• The bracketed coefficient of ln(x) is zero because y1 is a solution. Choose c∗0 = 1 (we need only one second solution), shift the indices to write , and substitute the series for y1(x) to obtain


* 11

nnn

c x

1 * 1 * * * *1 12 2

2

2

4 1 2 12 2 2 3 5 2 4

! !

0.

n n

n n n nn

n

x c x n n n c n c c cn n

x

* 212

nnn

c x



• Set the coefficient of each power of x equal to 0. From the coefficient of x−1, we have c∗1 = 2. From the coefficient of xn−2, we obtain (after some routine algebra)

or

for n =2, 3, 4, ···.CHAPTER 4 : Page 140

2 * *12

2 10

!

n

n nn n c cn

* *12 2

1 2( 1) ( !)

n

n nc cn n n



• With this, we can calculate as many coefficients as we want, yielding


2 1

2 3

2 3 11( ) ( ) ln( )4 108

25 137 .3456 432,000

y x y x x xx

x x



• The next two examples illustrate Case (4) of the theorem, first with k = 0 and then k 0.



EXAMPLE 4.7 Case 4 of Theorem 4.2 with k = 0

• We will solve

• There is a regular singular point at 0. Substitute to obtain


22 2 0.x y x y y

0n r

nny c x

0

11

( ( 1) 2)

( )( 1) ( 1) 2 ] 0.

r

n rn n n

n

r r c x

n r n r c n r c c x



• The indicial equation is r 2 − r − 2 = 0 with roots r1 = 2, r2 = −1. Now r1 − r2 = 3, putting us in Case (4) of the theorem. From the coefficient of xn+r , we obtain the general recurrence relation

for n =1, 2, 3, ···.


1( )( 1) ( 1) 2 0n n nn r n r c n r c c



• For a first solution, use r = 2 to obtain the recurrence relation

for n = 1, 2, ···. Using this, we obtain a first solution


11

( 3)n nnc c

n n

2 2 3 4 51 0

1 3 1 1 1( ) 1 .2 20 30 168 1120

y x c x x x x x x



• Now we need a second, linearly independent solution. Put r = −1 into the general recurrence relation to obtain

for n = 1, 2, ···.


* * *1( 1)( 2) ( 2) 2 0n n nn n c n c c



• When n = 3, this gives c∗2 =0, which forces c∗n = 0 for n = 2, 3, ···. But then

• Substitute this into the differential equation to obtain


* *2 0 1

1( ) .y x c cx

2 * 3 2 * 2 * * * *0 0 1 0 0 1

1(2 ) ( ) 2 2 0.x c x x c x c c c cx



• Then c∗1 = −c∗0/2, and a second solution is

with c∗0 arbitrary but nonzero. The functions y1 and y2 form a fundamental set of solutions. In these solutions, there is no y1(x) ln(x) term.


*2 0

1 1( )2

y x cx


EXAMPLE 4.8 Case 4 of Theorem 4.2 with k 0

• We will solve

which has a regular singular point at 0. Substitute

and rearrange terms to obtain


0,xy y

0n r

nny c x

2 1 10 1

1

( ) [( )( 1) ] 0.r n rn n

n

r r c x n r n r c c x



• The indicial equation is r2 − r = 0, with roots r1 = 1, r2 = 0. Here r1 − r2 = 1, a positive integer, putting us in Case (4) of the theorem. The general recurrence relation is

for n = 1, 2, ···.


1( )( 1) 0n nn r n r c c



• With r =1, this is

and some of the coefficients are

and so on.CHAPTER 4 : Page 141-142

11 .

( 1)n nc cn n

1 0

2 1 0

3 2 0

1 ,2

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

1 1 ,3(4) 2(3)(2)(3)(4)

c c

c c c

c c c



• In general,

for n = 1, 2, 3, ···, and one Frobenius solution is


01

!( 1)!nc cn n

11 0

0

2 3 40

1( ) !( 1)!1 1 1 .

12 144

n

n

y x c xn n

c x x x xx



• For a second solution, put r = 0 into the general recurrence relation to obtain

for n=1, 2, ···.


1( 1) 0n nn n c c



• If we put n = 1 into this, we obtain c0 = 0, violating one of the conditions for the method of Frobenius. Here we cannot obtain a second solution as a Frobenius series. Theorem 4.2, Case (4), tells us to look for a second solution of the form


*2 1

0

( ) ln( ) .nn

n

y x ky x c x



• Substitute this into the differential equation to obtain


* 21 1 1 2

2

*1

0

1 1 ln( ) 2 ( 1)

ln( ) 0.

nn

n

nn

n

x ky x ky ky n n c xx x

ky x c x



• Now

because y1 is a solution. For the remaining terms, let c0 = 1 in y1(x) for convenience (we need only one more solution) to obtain


1 1ln( )[ ] 0,k x xy y

* 1 *

20 =0 =2 =0

1 12 ( 1) 0.!( 1)!!

n n n nn n

n n n n

k x k x c n n x c xn nn



• Shift indices in the third summation to write


20 =0

* *1

1 0

1 12!( 1)!!

( 1) 0.

n n

n n

n nn n

n n

k x k xn nn

c n n x c x



• Then

• This implies that k − c∗0 = 0, so


* 0 * *0 12

1

2(2 ) ( 1) 0.( !) !( 1)!

nn n

n

k kk k c x n n c c xn n n

*0.k c



• Furthermore, the recurrence relation is

for n =1, 2, ···.Since c∗0 can be any nonzero number, we will for convenience let c∗0 = 1. For a particular solution, we may also choose c∗1 = 0. These give us


2 3 42 1

3 7 35( ) ln( ) 1 .4 36 1728

y x y x x x x

* *1

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

n kc cn n n n


Homework for 4.2

• 1, 4, 5, 9, 10


Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning...

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Transcript of Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning...