Power Series Methods† National Chiao Tung University Chun-Jen Tsai 11/25/2019 † Chapter 6 in the textbook.

Power Series Methods

^†

National Chiao Tung University Chun-Jen Tsai 11/25/2019

Power Series

 A power series in (x – a) is a series of the form

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

 A power series is convergent at a value x  I if the limit of partial sums exists, i.e.

The interval of convergence, I, of a power series is the set of all numbers where the series converges.

. )

( )

( )

( _{0 1 2} ²

0       



^_n c_n x a ⁿ c c x a c x a







 

 _{N n}^N c_n x a ⁿ x

f ( ) lim ₀ ( )

Ratio Test

 Convergence of a power series can often be checked by the ratio test: suppose c_n  0 for all n in

and that

If L < 1, the series converges absolutely; if L > 1 the series diverges; and if L = 1 the test is inconclusive.

, )

0 (



^_n c_n x  a ⁿ

. ) lim

(

)

lim ( ¹

1

1 L

c a c

a x x c

a x c

n n n n

n

n n

n   



 _





 





Radius of Convergence (1/2)

 A power series f(x) = (x – a)ⁿ has a radius of convergence



, such that f(x) converges for |x – a| <



and diverges for |x – a| >



^.

 If



> 0, f(x) converges for |x – a| <



, and diverges for

|x – a| >



. If f(x) converges only at its center a, then



= 0. If it converges for all x,



= .

 The ratio test is inconclusive at an end point a 



^.

Radius of Convergence (2/2)

 Given the power series g(x) = xⁿ, if the limit

exists, then

 If  = 0, then g(x) diverges for all x  0.

 If 0 <  < , g(x) converges if |x| < , and diverges if |x| > .

 If  = , then g(x) converges for all x.

1

lim

 

 

n n

n c

 c

Power Series of a Function f(x)

 The Taylor series of a function f(x) is defined as

If y converges to f(x) for all x in some open interval containing a, then we say that the function f(x) is analytic at x = a.

 Polynomials are analytic; a rational function is analytic wherever the denominator is not zero.

 Arithmetic of power series

 The operations of addition, multiplication, and division can be applied to power series as in polynomials.

 If f and g are analytic at a, so are f+g, f·g, and f/g (if g(a)  0).

. )

! ( ) (

0 )



^ ( 

 _n ⁿ x a ⁿ

n a y f

Examples of Power Series

 By Taylor ( a) or Maclaurin (a = 0) series expansions, common functions can be written in power series forms:



















 







 











 

 









































3 2

0

3 2

1

1

5 3

0

1 2

3 2

0

1 1 1

3 2

! ) 1 ) (

1 ln(

! 5

! 3 )!

1 2

( ) 1 sin (

! 3

! 1 2

!

x x

x x x

x x x

n x x

x x x

n x x

x x x

n e x

n

n n

n n n

n n n

x n

Example: Adding Two Power Series

 Write as one power

series.

Solution:

. ] )

1 )(

2 [(

2

) 1 )(

2 (

2

) 1 (

1 2 )

1 (

1

1 2

2

1 1 1

2 2

0

1 3

2 0

2 0

1 2

2

















  



 



 





 



 



 









































k

k k

k

k

k k k

k k

n

n n n

n n n

n n n

n n

x c

c k

k c

x c x

c k

k c

x c x

c n

n x

c x

c x

c n

n

Power Series Method

 The power series method for solving a DE consists of substituting the power series

in the DE and determining the coefficients c₀, c₁, … so that the equation satisfies.

 If , then  .

 If , for all x in the interval of convergence, then a_n = b_n for all n  0.



^





0 n

n nx c y

Example: y' + 2y = 0

 Since , and  ,

we have

Perform change of index n to align xⁿ,

We have a recurrence relation c_n+1 = –2c_n/(n+1), n  0.

 c_n = (–2)ⁿc₀/n!, n  1 

. 0 2

0 1

1 







^









n

n n n

n

nx c x

nc



^{( 1) 2}



^0.

0 2

) 1 (

0

1 0

0

1      



 



^

 







 

n

n n n

n

n n n

n

n x c x n c c x

c n

! . ) 2 ) (

(

0



^ 0



 

n

n n

n x x c

y

Power Series Solutions

 Suppose the linear 2^nd-order DE

a₂(x)y" + a₁(x)y' + a₀(x)y = 0 is put into the standard form

y" + P(x)y' + Q(x)y = 0, then:

A point x = x₀ is said to be an ordinary point of the DE if both P(x) and Q(x) are analytic at x₀. A point that is not an ordinary point is said to be a singular point of the equation.

Example: Ordinary, Singular Points

 Every finite value of x is an ordinary point of y" + (e^x) y' + (sin x) y = 0.

 x = 0 is a singular point of the DE y" + (e^x) y' + (ln x) y = 0.

Polynomial-Coefficient DEs

 Recall that a polynomial is analytic at any value x, and a rational function is analytic except at points where its denominator is zero. Thus, a 2^nd-order polynomial-

coefficient DE has singular points when a₂(x) = 0, since

 Examples

 The Euler equation ax²y + bxy' + cy = 0 has a singular point at x = 0.

 The equation (x² + 1)y + xy – y = 0 has singular points at x = i.

). (

) ) (

Q(

) , (

) ) (

(

2 0 2

1

x a

x x a

x a

x x a

P  

Solutions Near an Ordinary Point

 Theorem: If x = x₀ is an ordinary point of a₂(x)y + a₁(x)y + a₀(x)y = 0,

we can always find two linearly independent solutions in the form of a power series centered at x₀, that is,

A series solution converges at least on some interval defined by | x – x₀ | <



^{, where}



is the distance from x₀ to the nearest singular point.

 Note that for | x – x₀ | 



, y(x) may or may not converge.

Further investigations are required.

. )

0 ( 0



^ 

 _n c_n x x ⁿ y

Ex: (x

²

– 4)y + 3xy+ y = 0, y(0) = 4, y(0) =1

 Note that the singular points are 2, there should be a solution with radius of convergence at least 2.

Since

Therefore,

Combining the terms, we have .

add dummy terms for n = 0, 1 change of index n add dummy term n = 0







^

 



 

     

 _{n 0}c_nxⁿ, y _{n 1}nc_nx^{n 1}, y _{n 2}n(n 1)c_nx^{n 2}

y and

. 0 3

) 1 (

4 )

1 (

0 1

2

2 2













  



^











 

 n

n n n

n n n

n n n

n

nx n n c x nc x c x

c n

n

. 0 3

) 1 )(

2 (

4 )

1 (

0 0

0

2 0















  



^









 



 n

n n n

n n n

n n

n

n

nx n n c x nc x c x

c n

n

Ex: (x

²

– 4)y + 3xy+ y = 0 (2/2)

 When n = 0, 2, 4, …, we have

When n = 1, 3, 5, …, we have

Therefore, the solution is

and y(0) = c₀ = 4, y(0) = c₁ = 1.

) . 2 ( 4 2 4

) 1 2

( 5 3 , 1

6 4 2 4

5 3

4 , 2 4 3

2,

4 ^{3 2 0}

0 2 6

0 4

0

2 c

n c n

c c c c

c c _{n n}



 









 

 





 



 

 

) . 1 2

( 3 1 4

) 2 ( 6 4 , 2

7 5 3 4

6 4 2

5, 3 4

4 2

3, 4

2

1 1

3 2 7 1

2 5 1

3 1 c

n c n

c c c c

c c _{n n}









 

 







 





 

  _



 

) . 1 2

( 3 1 2

!

! 2

) 1 2

( 5 3 1 1

) (

1

1 2 1

1

2

0 3 



 









 





 



   





 

^



 

 n

n n n

n

n x

n x n

c n x

c n x

y 



Translated Series Solutions

 If the initial condition is given at x₀ other than zero, we have to assume the general solution form

This way, we can obtain the IVP solution with y(x₀) = c₀ and y(x₀) = c₁ easily.

 As an alternative, we can translate the equation by letting t = x – x₀ and assume the solution form:

. ) (

) (

0



^ 0







n

n

n x x

c x

y



^





0 n

n nx c y

Ex: ^(t

²

–2t–3)y+3(t–1)y+y = 0, y(1) = 4, y(1) = –1

 Perform a change of variable to the DE by x = t – 1:

t² – 2t – 3 = (x + 1)² – 2(x + 1) – 3 = x² – 4, and

Hence, the DE becomes (x²–4)d²y/dx² + 3x(dy/dx) + y = 0.

 Same DE as the previous one.

Substituting x = t–1 into the previous solution, we get

which converges if –1 < t < 3.

.

, ₂

2 2

2

dx y d dt

dx dx

dy dx

d dt

y d dx dy dt

dx dx dy dt

dy  



 



 



 



 























 ^{2 3} ( 1)⁴

32 ) 3

1 6 (

) 1 1 2 (

) 1 1 (

4 )

(t t t t t

y

Example: y + (cos x) y = 0 (1/2)

 Since



^









 

 



    









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





^





    



 



   









 _{2 3 4} ^{2 2 4} _{0 1 2} ²

! 4

! 1 2

12 6

2 x x c c x c x

x c x

c c

2 0 12 1

) 6

(

2 _{2 0 3 1 4 2 0}  ²  



 



  









 c c c c x c c c x 

Example: y + (cos x) y = 0 (2/2)

We have:

Therefore:

with region of convergence | x | < .

x y1

3

2 1 0

x 1

0 -1 -2 -3

y2

,  2 0

12 1 , 0 6

, 0

2c₂  c₀  c₃  c₁  c₄  c₂  c₀ 



    







 ^{2 4} ₂ ^{3 5}

1 30

1 6

) 1 ( 12 ,

1 2

1 1 )

(x x x y x x x x

y

Solutions about Singular Points

 Let y(x) = Sc_nxⁿ, if P(x) is not analytic at 0, its power series form Sb_kx^k will not converge to P(0) at 0 given any b_k. However, it is possible that

(Sb_kx^k) (Snc_nx^n–1) may still converge to P(x)y(x).

In short, even if x = 0 is a singular point, the power series expression of

y + P(x)y + Q(x)y may still converge to zero.

Regular Singular Points

 Assume that a DE in the standard form

y + P(x)y + Q(x)y = 0 has a singular point at x₀. If there are two functions p(x) and q(x), both are analytic at x₀, such that p(x) = (x – x₀)P(x) and

q(x) = (x – x₀)²Q(x), the original DE can be rewritten as:

then, we call x = x₀ a regular singular point of the DE.

Otherwise, x = x₀ is a irregular singular point.

, ) 0

(

) ( )

(

) (

2 0 0

 



 

 y

x x

x y q

x x

x y p

Remarks on Singularity of P and Q

 If x = 0 is a singular point, the power series expansion of P(x) at 0 approaches .

 However, if P(x) grows slower than 1/x when x  0,

then xP(x) is convergent. That is, p(x) = xP(x) is analytic at 0. Similarly, q(x) is analytic at 0 if Q(x) grows slower than 1/x².

 Note that, for the DE

x = 0 is a regular singular point if p(x) and q(x) are polynomials.

, ) 0

( )

(

2 



 y

x x y q

x x y p

Example: Singular Points

 For (x² – 4)²y + 3(x – 2)y + 5y = 0, x = 2 and x = –2 are singular points. We have

and

 Obviously x = 2 is a regular singular point, and x = –2 is an irregular singular point.

)2

2 )(

2 (

) 3

(   

x x x

P .

) 2 (

) 2 (

) 5

( _{2 2}



 

x x x

Q

Example: Non-polynomial p(x), q(x)

 The DE x⁴y + (x² sin x)y + (1 – cos x)y = 0 can be expressed as

Since x = 0 is not an ordinary point and

are both analytic (convergent) at 0, thus x = 0 is a regular singular point.

. / 0

) cos 1

( /

sin

2

2 

 

 

 y

x

x y x

x x y x

! , 5

! 1 3

! 5

! 3 1

) sin (

4 2

5 3



    



 



   



 x x x x

x x x

x x p

! , 6

! 4

! 2

1

! 4

! 1 2

1 1 cos

) 1 (

4 2

4 2

2

2 x x  x x 

x x

x x

q    



 



 



 



   



 



Solution near Singular Points

 For a constant-coefficient Cauchy-Euler equation x²y + p₀xy + q₀y = 0,

where p₀ and q₀ are constants, we can assume that y(x) = x^r is a solution  r is a root of the equation:

r(r – 1) + p₀r + q₀ = 0.

If we have coefficient functions p(x) and q(x) instead, is it possible that

is a solution?





^



 







0 0

) (

n

r n n n

n n

r c x c x

x x

y

Method of Frobenius

 If x = x₀ is a regular singular point of the differential equation a₂(x)y + a₁(x)y + a₀(x)y = 0, then there exists at least one solution of the form

where the number r is a constant to be determined.

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

, ) (

) (

) (

0

0 0

0

0

 

^



 













n

r n n

n

n n

r c x x c x x

x x y

Example: 3xy + y – y = 0 (1/3)

 Solution: let , we have

Therefore,

We have:

. )

1 )(

( )

(

0

2 0

1





^





 





     



 

n

r n n n

r n

nx y n r n r c x

c r n

y and

0 ]

) 1 3

3 )(

1 [(

) 2 3

( 3

0

1 1

0  



 



       









^

 



k

k k k

r r r c x k r k r c c x

x

y y

y x























 0, 0,1, 2,

) 1 3

3 )(

1 (

0 )

2 3

(

1 0

k c

c r

k r

k

c r

r

k k

Example: 3xy + y – y = 0 (2/3)

 Hence,

Substituting r = 0 and r = 2/3 into the recurrence eq.,







 





 



  , 0,1, 2,

) 1 3

3 )(

1 (

3 / 2 , 0

1 k

r k

r k c c

r

k k

 

 













 

 



 







 

 



 









) 2 3

( 7 4 1

! 2

) 1 (

3 ) 1 , (

0

) 2 3

( 11 8 5

! )

1 (

2 ) 1 (

, 3 3 / 2

0 1

0 1

n n

c c k

k c c

r

n n

c c k

k c c

r

n k

k

n k

k





Example: 3xy + y – y = 0 (3/3)

 Let c₀ = 1, we have two series solutions

Since y₁(x) and y₂(x) are linearly independent on the entire axis, y(x) = k₁y₁(x) + k₂y₂(x) is the general solution of the DE on any interval not containing the origin (note that 0⁰ is undefined).

. )

2 3

( 7

4 1

! 1 1

) (

) 2 3

( 11 8 5

! 1 1

) (

1 0

2

1 3

/ 2 1













 









 





 









 















n

n n

n

n x x n

x y

n x x n

x y





Indicial Equation

 The equation derived from the coefficient of the

smallest degree of x in the Frobenius method is the indicial equation.

 The solutions of the indicial equation with respect to r are called the indicial roots.

Frobenius Series Solutions (1/2)

 Theorem: If x = 0 is a regular singular point of x²y + xp(x)y + q(x)y = 0.

Let



> 0 denote the minimum of the radii of convergence of the power series

Let r₁ and r₂ be the (real) roots, with r₁  r₂, of the indicial equation

r(r – 1) + p₀r + q₀ = 0.

Then, we have the following properties:

. )

( )

(



₀



^₀



 

 _n p_nxⁿ q x _n q_nxⁿ x

p and

Frobenius Series Solutions (2/2)

1. For x > 0, there exists a solution of the form

corresponding to the larger root r₁.

2. If r₁  r₂ and r₁ – r₂ Z⁺, then there exists a 2^nd linearly independent solution for x > 0 of the form

corresponding to the smaller root r₂.

3. The radii of convergence of the solutions are at least



^,

the nearest distance to the nearby singular point.

) 0 (

)

( ₀

0

1  ¹



^ 



a x

a x

x y

n

n n r

) 0 (

)

( ₀

0

2  ²



^ 



b x

b x

x y

n

n n r

Example: 2xy + (1 + x) y + y = 0

 Since x²y + ½ x(1+x) y + ½ xy = 0, p(x) = (1+x)/2 and q(x) = x/2  p₀ = ½ and q₀ = 0  r² – r/2 = 0,  r = 0, ½.

For r₁ = ½, let , then .

For r₂ = 0, let , we have

The general solution is y = c₁y₁(x) + c₂y₂(x).

). 1 2

( 7 5 3 1

) 1

( ₀









 

n b b

n

n 

Example: xy + 2y + xy = 0 (1/3)

 When r₁ – r₂ is a positive integer, the Frobenius solution is only guaranteed for r₁. However, in this example, we still have two solutions even if r₁ – r₂ = 1.

The DE can be written as

The indicial equation r(r – 1) + 2r = 0 has roots 0, –1.

Start with r₂ = –1, we have Hence,

 Check the coefficients of x^–2 and x^–1!

. 2 0

2

2 



 y

x y x

y x

. )

( ¹



₀



^₀ ¹

 



 

 x _n c_nxⁿ _n c_nxⁿ x

y

. 0 )

1

(  ² 







^{ n n c xn  c xn}

Example: xy + 2y + xy = 0 (2/3)

The first two terms gives us 0·c₀= 0 and 0·c₁= 0, which means c₀ and c₁ can be arbitrary constants.

Thus, the recurrence relation c_n = –c_n–2/n(n–1), n  2 can be divided into two groups of coefficients:

Therefore, a general solution is

. 1 )!,

1 2

( ) 1 ( )!

2 (

) 1

( ₁

1 0 2

2 



 

  _ n

n c c

n c c

n n

n

n and for

)! . 1 2

( ) 1 ( )!

2 (

) 1 ( )

(

0

1 2 1

0

2 0

0 1











 











 

 



n

n n

n

n n n

n n

n x x

c n

x x

c

x c x

x y

Example: xy + 2y + xy = 0 (3/3)

 Now, if you pay attention, you will recognize that the solution is simply

y(x) = x^–1(c₀ cos x + c₁ sin x).

The graph of the solution is:

2

^nd

Solution by Reduction of Order

 If there is only one solution in Frobenius form for y + P(x)y + Q(x)y = 0,

we can find the 2^nd solution by reduction of order.

Recall that the reduction of order formula tells us ) .

) ( ( )

( ₂

1 ) ( 1

2 ^{x  y x}



^e_y^ _x ^dx

y

dx x P

Summary of Indicial Roots (1/2)

 Case I: r₁ and r₂ are distinct, r₁ – r₂  N, for some

integer N  exists two linearly independent solutions of the form

 Case II: r₁ – r₂ = N, for some integer N  exist two linearly independent solutions of the form

Note that C could be zero.

. )

( )

( _{0 2 0}

1



¹



^ ²

 



 

 _n c_nx^{n r} y x _n c_nx^{n r} x

y and

0 . ,

ln ) ( )

(

0 ,

) (

0 0

1 2

0 0

1

2 1

















 













n

r n n n

r n n

b x

b x

x Cy x

y

c x

c x

y

Summary of Indicial Roots (2/2)

 Case III: If r₁ = r₂, there exists two linearly independent solutions of the form

ln . ) ( )

(

0 ,

) (

1 1 2

0 0

1

1 1















 













n

r n n n

r n n

x b x

x y x

y

c x

c x

y

Bessel’s Equations

 Bessel’s equation of order v  0 is defined as x²y + xy + (x² – v²)y = 0.

The solutions are called Bessel functions of order v.

 Bessel’s functions first appear in 1764 when Euler was studying the vibration of drum membrane. Later, the

functions appears in many physics problems, from fluid equations to planet motions.

Gamma Function (x)

 The gamma function (or generalized factorial function) is defined as

For x > 0, we have (x+1) = x (x).

. )

( 0



^{ }1 ^



 x t^x e ^tdt

Solution of Bessel’s Equation (1/2)

 Let the solution be , we have

The indicial equation is r² – v² = 0, pick r = v



^{( )}



^.

) (

) (

0

2 1

2 2

2 2

0

2 2

2





^



 





















n

n n r

n n

n r

r x c n r v x x c x

x v r

c

y v x

y x y

x



^{( 2)( 2 2 )}



^0.

) 2 1 (

) 2 (

0

2 2

1

1 0

2





 



      









 





 











k

k k k

v

n n

n n v

n n

v

x c c

v k

k x

c v x

x c x

x v n

n c x

Solution of Bessel’s Equation (2/2)

Therefore, we have







 





 





 , 0,1, 2,...

) 2 2 )(

2 (

0 )

2 1 (

2

1

v k k

k c c

c v

k k

















 







 











 

 





 













 2 ( 1)

if 1 ) , 1 (

! 2

) 1 ( )

1 (

! 2

) 1 (

) 1 (

,...

3 , 2 , 1 ) ,

( ) 2

)(

1 (

! 2

) 1 (

0

2 0 2

0 2

0 2

7 5

3 1

c v v

n n

v n n

v c

v n n v

v n

c c

c c

c c

v v

n

n n

n n

n

n 



Bessel Functions of the 1

^st

Kind (1/2)

 The solutions of Bessel’s Equation can be written as

Similarly, starting from r = –v, we have

J_v(x) and J_–v(x) are called Bessel’s functions of the first kind of order v and –v.





^



 



 



 











 



0

2

0

2

2 ! (1 ) 2

) 1 ) (

(

n

v n n

n

v n n v

x n

v x n

c x

J

2 . ) 1

(

!

) 1 ) (

(

0

2

0

2

2





^



 





 



 











 

 

n

v n n

n

v n n v

x n

v x n

c x

J

Bessel Functions of the 1

^st

Kind (2/2)

 Now, we want to find the general solution of the Bessel’s DE. Notice that r₁ – r₂ = 2v:

1. If 2v ≠ integer, then J_v(x) and J_{– v}(x) are linearly independent.

2. If 2v = 2m + 1, m is an integer, then J_m+1/2(x) and J_–m–1/2(x) are still linearly independent.

3. If 2v = 2m, m is an integer, then J_m(x) and J_{– m}(x) are linear dependent solutions of Bessel’s DE.

 must find another solution!