Series Solutions of Linear Differential Equations - Spada UNS

Series Solutions of Linear Differential Equations

CHAPTER 5

Contents

5.1 Solutions about Ordinary Points

5.2 Solution about Singular Points

5.3 Special Functions

5.1 Solutions about Ordinary Point

Review of Power Series

Recall from that a power series in x – a has the form

Such a series is said to be a power series centered at a.















^ 



2 2

1 0

0

) (

) (

)

(x a c c x a c x a

c

n

n n

Convergence

exists.

Interval of Convergence

The set of all real numbers for which the series converges.

Radius of Convergence

If R is the radius of convergence, the power series converges for |x – a| < R and

diverges for |x – a| > R.











  ^N 

n

n n

N N

N S x C x a

0 ( )

lim )

( lim

Absolute Convergence

Within its interval of convergence, a power series converges absolutely. That is, the following

converges.

Ratio Test

Suppose c_n  0 for all n, and

If L < 1, this series converges absolutely, if L > 1, this series diverges, if L = 1, the test is inclusive.



^_0| ( ^ ) |

n

n n x a c

C L a C

a x x

C

a x

C

n n n n

n

n n

n   



 _





 





1 1

1 | | lim

) (

) lim (

A Power Defines a Function Suppose

then

Identity Property

If all c_n = 0, then the series = 0.

(1)

) 1 (

"

,

' 0

2 0

1





^

 



  

 n

n n

n y n n x

x n y



^

 n 0

n nx c y

Analytic at a Point

A function f is analytic at a point a, if it can be

represented by a power series in x – a with a positive radius of convergence. For example:

 (2)































! 6

! 4

! 1 2

cos

! 5

! sin 3

! , 2

! 1 1

6 4

2

5 3

2

x x

x x

x x x

x x e^x x

Arithmetic of Power Series

Power series can be combined through the operations of addition, multiplication and division.



















 



 



  

 



 



 

 



 



 









 



    



 



     



24 1 12

1 120

1 6

1 6

1 2

1 6

) 1 1 ( )

1 (

5040 120

6 24

6 1 2

sin

5 3

2

5 4

3 2

7 5

3 4

3 2

x x x

x

x x

x x

x

x x

x x x

x x x

x e^x

Example 1

Write as one power series.

Solution Since

we let k = n – 2 for the first series and k = n + 1 for the second series,





^n_2 ^{( 1) 2  }_n0 n ^n1 n

nx c x

c n

n



^

  





















     



2 0 3 0

1 2

0 2 1

2 2 1 ( 1)

) 1 (

n n n n

n n n

n n

n n

nx c x c x n n c x c x

c n

n ．

then we can get the right-hand side as

(3) We now obtain

(4)



^







 

 







1 1

1 2

2 ( 2)( 1)

2

k k

k k

k

k x c x

c k

k c



 



  



























1

1 2

2

2 0

1 2

] )

1 )(

2 [(

2

) 1 (

k

k k

k

n n

n n n

n

x c

c k

k c

x c x

c n

n

Example 1 (2)

Suppose the linear DE

(5) is put into

(6)

A Solution

0 )

( )

( )

( _{1 0}

2 x y  a x y  a x y  a

0 )

( )

(   

  P x y Q x y y

A point x₀ is said to be an ordinary point of (5) if both P and Q in (6) are analytic at x₀. A point that is not an ordinary point is said to be a singular point.

DEFINITION 5.1

Since P and Q in (6) is a rational function, P = a₁(x)/a₂(x), Q = a₀(x)/a₂(x)

It follows that x = x₀ is an ordinary point of (5) if a₂(x₀)  0.

Polynomial Coefficients

A series solution converges at least of some interval defined by |x – x₀| < R, where R is the distance from x₀ to the nearest singular point.

If x = x₀ is an ordinary point of (5), we can always find two linearly independent solutions in the form of power series centered at x₀, that is,

THEOREM 5.1

Criterion for an Extra Differential



^ 

 0 ( 0)

n

n

n x x

c y

Example 2

Solve Solution

We know there are no finite singular points.

Now, and

then the DE gives

(7) 0

"xy  y



^

 n 0

n nx c

y



^

 

 2

) 2

1 (

"

n

n nx c n

n y





 

 

 























 

1 2

2 0

2

) 1 (

) 1 (

n n n

n

n n

n n n

n

x c x

n n c

x c x

x n

n c xy

y

Example 2 (2)

From the result given in (4),

(8) Since (8) is identically zero, it is necessary all the

coefficients are zero, 2c₂ = 0, and

(9) Now (9) is a recurrence relation, since

(k + 1)(k + 2)  0, then from (9)

(10)



^

      





 

1

1 2

2 [( 1)( 2) ] 0

2

k

k k

k c x

c k

k c

xy y

, 3 , 2 , 1 ,

0 )

2 )(

1

(k  k  c_k2  c_k1  k 

 , 3 , 2 , 1 ) ,

2 )(

1 (

1

2 



 

 ^

 k

k k

c_k c^k

Example 2 (3)

Thus we obtain ,

1

k 2 3

0

3 ．

c   c

,

 2

k 3 4

1

4 ．

c   c

,

 3

k 0

5 4

2

5   

． c c

,

 4

k ₆ ³ ₀

6 5 3 2

1 6

5c c

c   ．  ．．． ,

 5

k ⁴ 1

c c

c   

and so on.

Example 2 (4)

,

 6

k 0

8 7

5

8   

． c c

,

 7

k ₉ ⁶ ₀

9 8 6 5 3 2

1 9

8c c

c   ．   ．．．．．

,

 8

k ₁₀ ⁷ ₁

10 9

7 6 4 3

1 10

9c c

c   ．   ． ． ． ． ．

,

 9

k 0

11 10

8

11   

． c c

Example 2 (5)

Then the power series solutions are y = c₀y₁ + c₁y₂

....

7 0 6． 4． 3．

． 6

． 5

． 3 0 2

． 4 3

． 3 0 2

1 7

0 6 1 4

0 3 1

0





















c x

c x c x

c x x

c c

y



^







 













1

1 3

10 7

4 2

) 1 3

)(

3 ( 4

3

) 1 (

10 9

7 6 4 3

1 7

6 4 3

1 4

3 1 1

) (

k

k k

k x x k

x x

x x

y





．

．

．

．

．

．

．

．

．

．

Example 2 (6)



^

 

 













1

3

9 6

3 1

) 3 )(

1 3

( 3 2

) 1 1 (

9 8 6 5 3 2

1 6

5 3 2

1 3

2 1 1

) (

k

k k

k x k

x x

x x

y





．

．

．

．

．

．

．

．

．

．

Example 3

Solve

Solution

Since x² + 1 = 0, then x = i, −i are singular points. A

power series solution centered at 0 will converge at least for |x| < 1. Using the power series form of y, y’ and y”, then

0 '

"

) 1

(x²  y xy y 









  





 







































2

2 1 0

1 2

2

) 1 (

) 1 (

) 1 (

) 1 (

n n

n n

n n

n n n

n n n

n

x c x

nc x

c n

n x

c n

n

x c x

nc x

x c n

n x















 



 





 



 



n k n

n n n

k n

n n n

k n

n n

n k n

n n

x c x

nc x

c n

n

x c n

n x

c x

c x

c x

c x

c



























































2 2

2 4

2

2 1

1 3

0 0 0

2

) 1 (

) 1 (

6 2

Example 3 (2)







 



 











































2

2 3

0 2

2

2 3

0 2

0 ]

) 1 )(

2 (

) 1 )(

1 [(

6 2

] )

1 )(

2 (

) 1 (

[ 6

2

k

k k

k k

k k k

k k

x c

k k

c k

k x

c c

c

x c kc

c k

k c

k k x

c c

c

Example 3 (3)

From the above, we get 2c₂－c₀ = 0, 6c₃ = 0 , and

Thus c₂ = c₀/2, c_k+2 = (1 – k)c_k/(k + 2) Then

0 )

1 )(

2 (

) 1 )(

1

(k  k  c_k  k  k  c_k2 

2 0 0

2

4 2 2!

1 4

2 1 4

1 c c c

c      

． 5 3

2

3 5   c  c

3 0 0

4

6 2 3!

3 1 6

4 2

3 6

3c c c

c ．

．

． 









Example 3 (4)

and so on.

4 0 0

6

8 2 4!

5 3 1 8

6 4 2

5 3 8

5c c c

c ．．

．

．

．

．  









9 0 6

7 9   c  c

5 0 0

8

10 2 5!

7 5 3 1 10

8 6 4 2

7 5 3 10

7 c c c

c ．

．

．

．

．

．

．

．

．

． 







Example 3 (5)

Therefore,

) ( )

(

! 5 2

7 5 3 1

! 4 2

5 3 1

! 3 2

3 1

! 2 2

1 2

1 1

2 1 1

0

1 10

5 8

4 6

3 4

2 2

0

10 10 9

9 8

8 7

7 6

6

5 5 4

4 3

3 2

2 1

0

x y c x

y c

x c x

x x

x x

c

x c x

c x

c x

c x

c

x c x

c x

c x

c x c c

y





 

 

      































．

．

．

．

．

．

1

|

|

! , 2

) 3 2

( 5

3 ) 1

1 2 (

1 1 )

( ²

2

1 2

1   



^   



 x x

n x n

x

y ⁿ

n

n

n ．． 

x x

y ( ) 

Example 4

If we seek a power series solution y(x) for

we obtain c₂ = c₀/2 and the recurrence relation is

Examination of the formula shows c₃, c₄, c₅, … are

expresses in terms of both c₁ and c₂. However it is more complicated. To simplify it, we can first choose c₀  0, c₁ = 0. Then we have

,  3 , 2 , 1 ) ,

2 )(

1 (

1

2 





  ^

 k

k k

c c_k c^{k k}

0 )

1

(  

  x y y

0

2 2

1c c 

Example 4 (2)

and so on. Next, we choose c₀ = 0, c₁  0, then

0 0

1 2

4 24

1 4

3 2 4

3 c c c

c  c   

．

．

．

0 0

2 3

5 30

1 2

1 6

1 5

4 5

4 c c c

c c 



 



 

 ． ．

0 0

0 1

3 6

1 3

2 3

2 c c c

c  c   

．

．

2 0 1

0 2  c  c

Example 4 (3)

and so on. Thus we have y = c₀y₁ + c₁y₂, where

1 1

0 1

3 6

1 3

2 3

2 c c c

c  c   

．

．

1 1

1 2

4 12

1 4

3 4

3 c c c

c  c   

．

．

1 1

2 3

5 120

1 6

5 4 5

4 c c c

c  c   

．

．

．











 ^{2 3 4 5}

1 30

1 24

1 6

1 2

1 1 )

(x x x x x

y









 ^{3 4 5}

2

1 1

) 1

(x x x x x

y

Example 5

Solve

Solution

We see x = 0 is an ordinary point of the equation. Using the Maclaurin series for cos x, and using

, we find



^_

 n 0

n nx c y

0 )

(cos

" x y  y



^









 



 



    







 

2 0

6 4

2 2

! 6

! 4

! 1 2

) 1 (

) (cos

n n

n n n

n x x x c x

x c n

n

y x y

 20 1 12 1

) 6

(

2 ²  ³ 

  

 

 

  









 c c c c x c c c x c c c x 

Example 5 (2)

It follows that

and so on. This gives c₂ =－1/2c₀, c₃ =－1/6c₁, c₄ = 1/12c₀, c₅ = 1/30c₁,…. By grouping terms we get the general solution y = c₀y₁ + c₁y₂, where the convergence is |x| < , and

2 0 20 1

, 2 0

12 1 , 0 6

, 0

2c₂  c₀  c₃  c₁  c₄  c₂  c₀  c₅  c₃  c₁ 







 ^{2 4}

1 12

1 2

1 1 )

(x x x

y







 ^{3 5}

2 30

1 6

1 1 )

(x x x

y

5.2 Solutions about Singular Points

A Definition

A singular point x₀ of a linear DE

(1) is further classified as either regular or irregular. This classification depends on

(2) 0

) ( )

( )

( _{1 0}

2 x y  a x y  a x y  a

0 )

( )

(   

  P x y Q x y y

A singular point x₀ is said to be a regular singular

point of (1), if p(x) = (x – x₀)P(x), q(x) = (x – x₀)²Q(x) are both analytic at x₀ .

A singular point that is not regular is said to be irregular singular point.

DEFINITION 5.2

Regular/Irregular Singular Points

Polynomial Ciefficients

If x – x₀ appears at most to the first power in the

denominator of P(x) and at most to the second power in the denominator of Q(x), then x – x₀ is a regular singular point.

If (2) is multiplied by (x – x₀)²,

(3) where p, q are analytic at x = x₀

0 )

( )

( ) (

)

(x  x₀ ² y  x  x₀ p x y  q x y 

Example 1

It should be clear x = 2, x = – 2 are singular points of (x² – 4)²y” + 3(x – 2)y’ + 5y = 0

According to (2), we have

)2

2 )(

2 (

) 3

(   

x x x

P

2 2( 2) )

2 (

) 5

(   

x x x

Q

Example 1 (2)

For x = 2, the power of (x – 2) in the denominator of P is 1, and the power of (x – 2) in the denominator of Q is 2. Thus x = 2 is a regular singular point.

For x = −2, the power of (x + 2) in the denominator of P and Q are both 2.

Thus x = − 2 is a irregular singular point.

If x = x₀ is a regular singular point of (1), then there exists one solution of the form

(4) where the number r is a constant to be determined.

The series will converge at least on some interval 0 < x – x₀ < R.

THEOREM 5.2

Frobenius’ Theorem





^



 













0

0 0

0

0) ( ) ( )

(

n

r n n

n

n n

r C x x C x x

x x

y

Example 2: Frobenius’ Method

Because x = 0 is a regular singular point of

(5) we try to find a solution .

Now,

0 3xy  y  y 



^_ ^

 n 0

r n nx c y



^





 

 

0

) 1

(

n

r n nx c r n

y



^





 





 

0

) 2

1 )(

(

n

r n nx c r

n r

n y

Example 2 (2)















 











 





 































 

 

0 0

1

0 0

1 0

1

) 2 3

3 )(

(

) (

) 1 )(

( 3

3

n

r n n n

r n n

n

r n n n

r n n n

r n n

x c x

c r

n r

n

x c x

c r n x

c r

n r n

y y

y x

0 ]

) 1 3

3 )(

1 [(

) 2 3

(

) 2 3

3 )(

( )

2 3

(

1 1

0

1 1

1

1 1

0

 

 

       









































 



















k k k

r

n k n

n n n

k n

n n r

x c c

r k

r k

x c r

r x

x c x

c r

n r

n x

c r

r x













 







 



Example 2 (3)

which implies r(3r – 2)c₀ = 0

(k + r + 1)(3k + 3r + 1)c_k+1 – c_k = 0, k = 0, 1, 2, … Since nothing is gained by taking c₀ = 0, then

r(3r – 2) = 0 (6)

and

(7) From (6), r = 0, 2/3, when substituted into (7),

, 2 , 1 , 0 ),

1 3

3 )(

1

1 ( 







 

 k

r k

r k

c_k c^k

Example 2 (4)

r₁ = 2/3, k = 0,1,2,… (8)

r₂ = 0, k = 0,1,2,… (9)

), 1 )(

5 3

1  (  

 k k

c_k c^k

), 1 3

)(

1

1  (  

 k k

c_k c^k

Example 2 (5)

From (8) From(9)

) 2 3

( 7

4 1

!

) 1 ( )

2 3

( 11

8 5

!

10 7

4 1

! 4 10

4 14

11 8

5

! 4 4

14

7 4 1

! 3 7

3 11

8 5

! 3 3

11

4 1

! 2 4

2 8

5

! 2 2

8

0 0

0 3

4 0

3 4

0 2

3 0

2 3

0 1

2 0

1 2



 

 

























n n

c c n

n c c

c c c

c c c

c c c

c c c

c c c

c c c

n n

n  





．

．

．

．

．

．

．

．

．

．

．

．

．

．

．

．

．

．

．

．

．

．

Example 2 (6)

These two series both contain the same multiple c₀. Omitting this term, we have

(10)

(11)





 





^

1 3

/ 2

1 !5 8 11 (3 2)

1 1 )

(

n

xn

n x n

x

y ．． 





 





^

1 0

2 !1 4 7 (3 2)

1 1 )

(

n

xn

n x n

x

y ． ． 

Example 2 (7)

By the ratio test, both (10) and (11) converges for all finite value of x, that is, |x| < . Also, from the forms of (10) and (11), they are linearly independent. Thus the solution is

y(x) = C₁y₁(x) + C₂y₂(x), 0 < x < 

Indicial Equation

Equation (6) is called the indicial equation, where the values of r are called the indicial roots, or

exponents.

If x = 0 is a regular singular point of (1), then p = xP and q = x²Q are analytic at x = 0.

Thus the power series expansions

p(x) = xP(x) = a₀＋a₁x＋a₂x²＋…

q(x) = x²Q(x) = b₀＋b₁x＋b₂x²＋… (12) are valid on intervals that have a positive radius of

convergence.

By multiplying (2) by x², we have

(13) After some substitutions, we find the indicial equation,

r(r – 1) + a₀r + b₀ = 0 (14) 0

)]

( [

)]

(

[ ²

2 y   x xP x y  x Q x y  x

Example 3

Solve

Solution

Let , then



^

 

0 n

r n nx c y

















 











 











 







































 



 

0 0

1 0

0

0

1 0

1

) 1 (

) 1 2

2 )(

(

) (

) (

) 1 )(

( 2

) 1

( 2

n

r n n n

r n n n

r n n n

r n n

n

r n n n

r n n

x c r

n x

c r

n r

n

x c x

c r n

x c r n

x c r

n r n

y y

x y

x

0 '

) 1

(

"

2xy   x y y 

Example 3 (2)

which implies r(2r – 1) = 0 (15)

(16)



         













































 





















0

1 1

0

0 1

1

1 1

0

] ) 1 (

) 1 2

2 )(

1 [(

) 1 2

(

) 1 (

) 1 2

2 )(

( )

1 2

(

k

k k k

r

n k n

n n n

k n

n n r

x c r

k c

r k

r k

x c r

r x

x c r

n x

c r

n r

n x

c r

r x



 



 









 







 



, 2 , 1 , 0 ,

0 )

1 (

) 1 2

2 )(

1

(k  r  k  r  c_k1  k  r  c_k  k 

Example 3 (3)

From (15), we have r₁ = ½ , r₂ = 0.

Foe r₁ = ½ , we divide by k + 3/2 in (16) to obtain

(17) Foe r₂ = 0 , (16) becomes

(18) ,

2 , 1 , 0 ),

1 (

1 2 



 

 k

k

c_k c^k

, 2 , 1 , 0 1,

1 2 



 

 k

k c_k c^k