Transcript Series Solutions of Linear DEs

CHAPTER 5
Series Solutions of Linear
Differential Equations
Contents
5.1 Solutions about Ordinary Points
5.2 Solution about Singular Points
5.3 Special Functions
Ch5_2
5.1 Solutions about Ordinary Point
Review of Power Series
Recall from that a power series in x – a has the form

n
2
c
(
x

a
)

c

c
(
x

a
)

c
(
x

a
)

n
0
1
2
n 0
Such a series is said to be a power series centered at
a.
Ch5_3
Convergence
lim N  S N ( x)  lim N  n0 Cn ( x  a)n
N
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.
Ch5_4
Absolute Convergence
Within its interval of convergence, a power series
converges absolutely. That is, the following
converges.

n
n0 | cn ( x  a)
|
Ratio Test
Suppose cn  0 for all n, and
Cn1 ( x  a)n1
Cn1
lim
 | x  a | lim
L
n
n C ( x  a )
n C
n
n
If L < 1, this series converges absolutely, if L > 1, this
series diverges, if L = 1, the test is inclusive.
Ch5_5
A Power Defines a Function

n
y

c
x
Suppose
 n 0 n
then
y '   n 0 n x

n1
, y"  n0 n(n  1) x n2

(1)
Identity Property
If all cn = 0, then the series = 0.
Ch5_6
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
3
5
x
x
x
x
e x  1    , sin x  x    
1! 2!
3! 5!
2
4
6
x
x
x
cos x  1     
(2)
2! 4! 6!
Ch5_7
Arithmetic of Power Series
Power series can be combined through the operations
of addition, multiplication and division.
e x sin x



x 2 x3 x 4
x3 x5
x7
 1  x      x  

 
2 6 24
6 120 5040



1 1 3  1 1 4  1
1 1 5

2
 (1) x  (1) x      x      x  
  x  
 6 2
 6 6
 120 12 24 
3
5
x
x
 x  x2    
3 30
Ch5_8
Example 1

n2
n
(
n

1
)
c
x
n
n2
Write 
series.
Solution
Since

 n(n  1)cn x
n2
n2

  cn x
n 0
n 1
 
c
x
as one power
n 0 n
n 1

 2．1c2 x   n(n  1)cn x
0
n 3
n2

  cn x n1
n 0
we let k = n – 2 for the first series and k = n + 1 for the
second series,
Ch5_9
Example 1 (2)
then we can get the right-hand side as


k 1
k 1
2c2   (k  2)( k  1)ck  2 x k   ck 1 x k
We now obtain

 n(n  1)cn x
n2
n2

  cn x
(3)
n 1
n 0
(4)

 2c2   [( k  2)( k  1)ck  2  ck 1 ]x k
k 1
Ch5_10
A Solution
Suppose the linear DE
a2 ( x) y  a1 ( x) y  a0 ( x) y  0
is put into
y  P( x) y  Q( x) y  0
(5)
(6)
DEFINITION 5.1
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.
Ch5_11
Polynomial Coefficients
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.
Ch5_12
THEOREM 5.1
Criterion for an Extra Differential
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,
y  n0 cn ( x  x0 )n

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.
Ch5_13
Example 2
Solve y"xy  0
Solution
We know there are no finite singular points.

Now, y  n0 cn x n and y"   n(n  1)cn x n2
n2
then the DE gives


n2
n 0
y  xy   cn n(n  1) x n2  x  cn x n


n2
n 0
  cn n(n  1) x n2   cn x n1
(7)
Ch5_14
Example 2 (2)
From the result given in (4),

y  xy  2c2  [(k  1)(k  2)ck 2  ck 1 ]x k  0 (8)
k 1
Since (8) is identically zero, it is necessary all the
coefficients are zero, 2c2 = 0, and
(k  1)(k  2)ck 2  ck 1  0 ,
k 1, 2 , 3 ,
Now (9) is a recurrence relation, since
(k + 1)(k + 2)  0, then from (9)
ck 1
ck 2  
, k 1, 2 , 3 ,
(k  1)(k  2)
(9)
(10)
Ch5_15
Example 2 (3)
Thus we obtain
c0
k  1, c3  
2．3
c1
k  2, c4  
3．4
c2
0
k  3, c5  
4．5
c3
1
k  4, c6  

c0
5．6 2．3．5．6
c4
1
k  5, c7  

c1
6．7 3．4．6．7
Ch5_16
Example 2 (4)
k  6,
k  7,
k  8,
k  9,
c5
c8  
0
7．8
c6
1
c9  

c0
8．9
2．3．5．6．8．9
c7
1
c10  

c1
9．10
3．4．6．7．9．10
c8
c11  
0
10．11
and so on.
Ch5_17
Example 2 (5)
Then the power series solutions are
y = c0y1 + c1y2
c0 3 c1 4
c0
y  c0  c1x  0 
x 
x 0
x6
2． 3
3． 4
2． 3． 5． 6
c1

x 7  0  ....
3． 4． 6． 7
Ch5_18
Example 2 (6)
y1 ( x)  1 
1 3
1
1
x 
x6 
x9  
2．3
2．3．5．6
2．3．5．6．8．9

(1) k
 1 
x 3k
k 1 2．3(3k  1)(3k )
1 4
1
1
7
y2 ( x )  1 
x 
x 
x10  
3．4
3．4．6．7
3．4．6．7．9．10

(1) k
 x
x3k 1
k 1 3．4(3k )(3k  1)
Ch5_19
Example 3
Solve ( x 2  1) y" xy' y  0
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



n2
n 1
n 0
( x 2  1)  n(n  1)cn x n2  x  ncn x n1   cn x n




n2
n2
n 1
n 0
  n(n  1)cn x n   n(n  1)cn x n2   ncn x n   cn x n
Ch5_20
Example 3 (2)

 2c2 x  c0 x  6c3 x  c1 x  c1 x   n(n  1)cn x
n
 2


0
0
n
k n



  n(n  1)cn x n2   ncn x n   cn x n
n
 4
n

2 
 2
 n
k n2
k n
k n

 2c2  c0  6c3 x  [k (k  1)ck  (k  2)( k  1)ck  2  kck  ck ]x k
k 2

 2c2  c0  6c3 x  [( k  1)( k  1)ck  (k  2)( k  1)ck  2 ]x k  0
k 2
Ch5_21
Example 3 (3)
From the above, we get 2c2－c0 = 0, 6c3 = 0 , and
(k  1)(k  1)ck  (k  2)(k  1)ck 2  0
Thus c2 = c0/2, ck+2 = (1 – k)ck/(k + 2)
Then
1
1
1
c4   c2  
c0   2 c0
4
2．4
2 2!
2
c5   c3  3
5
3
3
1．3
c6   c4  
c0  3 c0
6
2．4．6
2 3!
4
c7   c5  0
7
Ch5_22
Example 3 (4)
5
3．5
1．3．5
c8   c6  
c0   4 c0
8
2．4．6．8
2 4!
6
c9   c7  0
9
7
3．5．7
1．3．5．7
c10   c8 
c0 
c0
5
10
2．4．6．8．10
2 ．5!
and so on.
Ch5_23
Example 3 (5)
Therefore,
y  c0  c1x  c2 x 2  c3 x3  c4 x 4  c5 x5
 c6 x 6  c7 x 7  c8 x8  c9 x9  c10 x10  
1 2 1 4 1．3 6 1．3．5 8 1．3．5．7 10

c x
 c0 1  x  2 x  3 x  4 x 
x


1
5

2 2!
2 3!
2 4!
2 5!
 2

 c0 y1 ( x)  c1 y2 ( x)
1 2 
n 1 1．3．5( 2n  3) 2 n
y1 ( x)  1  x   (1)
x ,
n
2
2 n!
n2
| x | 1
y2 ( x )  x
Ch5_24
Example 4
If we seek a power series solution y(x) for
y  (1  x) y  0
we obtain c2 = c0/2 and the recurrence relation is
ck  ck 1
ck 2 
, k 1, 2 , 3 ,
(k  1)(k  2)
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
1
c2  c0
2
Ch5_25
Example 4 (2)
c1  c0
c0
1
c3 

 c0
2．3 2．3 6
c2  c1
c0
1
c4 

 c0
3．4
2．3．4 24
c3  c2
c0
1 1 1

c5 

     c0
4．5
4．5  6 2  30
and so on. Next, we choose c0 = 0, c1  0, then
1
c2  c0  0
2
Ch5_26
Example 4 (3)
c1  c0
c1
1
c3 

 c1
2．3 2．3 6
c2  c1
c1
1
c4 

 c1
3．4 3．4 12
c3  c2
c1
1
c5 


c1
4．5 4．5．6 120
and so on. Thus we have y = c0y1 + c1y2, where
1 2 1 3 1 4 1 5
y1 ( x)  1  x  x  x  x  
2
6
24
30
1 3 1 4
1 5
y2 ( x )  x  x  x 
x 
6
12
120
Ch5_27
Example 5
Solve y" (cos x) y  0
Solution
We see x = 0 is an ordinary point of the equation. Using
the Maclaurin series for cos x, and using

y  n0 cn x n , we find
y  (cos x) y
2
4
6



x
x
x
  n(n  1)cn x n2  1       cn x n
2! 4! 6!
n2

 n 0
1  2 
1  3

 2c2  c0  (6c3  c1 ) x  12c4  c2  c0  x   20c5  c3  c1  x  
2 
2 


0

Ch5_28
Example 5 (2)
It follows that
1
1
2c2  c0  0 , 6c3  c1  0 , 12c4  c2  c0  0 , 20c5  c3  c1  0
2
2
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
1 2 1 4
y1 ( x)  1  x  x  
2
12
1 3 1 5
y2 ( x )  1  x  x  
6
30
Ch5_29
5.2 Solutions about Singular Points
A Definition
A singular point x0 of a linear DE
a2 ( x) y  a1 ( x) y  a0 ( x) y  0
(1)
is further classified as either regular or irregular. This
classification depends on
(2)
y  P( x) y  Q( x) y  0
Ch5_30
DEFINITION 5.2
Regular/Irregular Singular Points
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.
Ch5_31
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,
( x  x0 ) y  ( x  x0 ) p( x) y  q ( x) y  0
2
(3)
where p, q are analytic at x = x0
Ch5_32
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
3
P( x) 
( x  2)( x  2)2
5
Q( x) 
( x  2)2 ( x  2)2
Ch5_33
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.
Ch5_34
THEOREM 5.2
Frobenius’ Theorem
If x = x0 is a regular singular point of (1), then there
exists one solution of the form


y  ( x  x0 ) r  Cn ( x  x0 ) n   Cn ( x  x0 ) n r
n 0
n 0
(4)
where the number r is a constant to be determined.
The series will converge at least on some interval
0 < x – x0 < R.
Ch5_35
Example 2: Frobenius’ Method
Because x = 0 is a regular singular point of
3xy  y  y  0
we try to find a solution
.
Now,

y  n0 cn x nr
(5)

y   (n  r )cn x nr 1
n 0

y   (n  r )( n  r  1)cn x n r 2
n 0
Ch5_36
Example 2 (2)
3 xy  y  y



n 0
n 0
n 0
 3 (n  r )( n  r  1)cn x n r 1   (n  r )cn x n r 1   cn x n r

  (n  r )(3n  3r  2)cn x
n 0
n  r 1

  cn x n r
n 0





r
1
n 1
n
 x r (3r  2)c0 x   (n  r )(3n  3r  2)cn x   cn x 
n 1
n 1 




k  n 1
k n 


r
1
k
 x r (3r  2)c0 x  [(k  r  1)(3k  3r  1)ck 1  ck ]x   0


k 0
Ch5_37
Example 2 (3)
which implies
r(3r – 2)c0 = 0
(k + r + 1)(3k + 3r + 1)ck+1 – ck = 0, k = 0, 1, 2, …
Since nothing is gained by taking c0 = 0, then
r(3r – 2) = 0
(6)
and
ck
ck 1 
,
(k  r  1)(3k  3r  1)
k  0, 1, 2, 
(7)
From (6), r = 0, 2/3, when substituted into (7),
Ch5_38
Example 2 (4)
ck
, k = 0,1,2,…
r1 = 2/3, ck 1 
(3k  5)(k  1)
ck
, k = 0,1,2,…
r2 = 0, ck 1 
(k  1)(3k  1)
(8)
(9)
Ch5_39
Example 2 (5)
From (8)
From(9)
c1
c0
c2 

8．2 2! 5．8
c2
c0
c3 

11．3 3! 5．8．11
c3
c0
c4 

14．4 4! 5．8．11．14

c0
cn 
n! 5．8．11  (3n  2)
c1
c0
c2 

2．4 2!1．4
c2
c0
c3 

3．7 3!1．4．7
c3
c0
c4 

4．10 4!1．4．7．10

(1)n c0
cn 
n!1．4．7  (3n  2)
Ch5_40
Example 2 (6)
These two series both contain the same multiple c0.
Omitting this term, we have

1

2/3
n
y1 ( x)  x 1  
x 
 n1 n! 5．8．11  (3n  2) 
0
(10)

1
n
y2 ( x)  x 1  
x 
 n1 n!1．4．7  (3n  2) 
(11)
Ch5_41
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 < 
Ch5_42
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.
Ch5_43
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)
2
2
x y  x[ xP ( x)] y   [ x Q( x)] y  0
After some substitutions, we find the indicial equation,
r(r – 1) + a0r + b0 = 0
(14)
Ch5_44
Example 3
Solve 2 xy"(1  x) y ' y  0
Solution
Let y  n0 cn x nr , then
2 xy  (1  x) y  y


n 0
n 0
 2 (n  r )( n  r  1)cn x n r 1   (n  r )cn x n r 1

  (n  r )cn x
n 0

n r

  cn x n r
n 0
  (n  r )( 2n  2r  1)cn x
n 0
n  r 1

  (n  r  1)cn x n r
n 0
Ch5_45
Example 3 (2)




r
1
n 1
n
 x r (2r  1)c0 x   (n  r )(2n  2r  1)cn x   (n  r  1)cn x 
n 1
n
0 




k  n 1
k n

r
1
k
 x r (2r  1)c0 x  [(k  r  1)(2k  2r  1)ck 1  (k  r  1)ck ]x 


k 0
which implies r(2r – 1) = 0
(k  r  1)(2k  2r  1)ck 1  (k  r  1)ck  0,
(15)
k  0, 1, 2,  (16)
Ch5_46
Example 3 (3)
From (15), we have r1 = ½ , r2 = 0.
Foe r1 = ½ , we divide by k + 3/2 in (16) to obtain
 ck
ck 1 
,
2(k  1)
k  0, 1, 2, 
(17)
k  0, 1, 2, 
(18)
Foe r2 = 0 , (16) becomes
 ck
ck 1 
,
2k  1
Ch5_47
Example 3 (4)
From (17)
 c0
c1 
2． 1
 c1
c0
c2 
 2
2．2 2 ．2!
 c2
 c0
c3 
 3
2．3 2 ．3!
 c3
c0
c4 
 4
2．4 2 ．4!

(1) n c0
cn  n
2 n!
From (18)
 c0
c1 
1
 c1 c0
c2 

3 1．3
 c2
 c0
c3 

5
1．3．5
 c3
c0
c4 

7
1．3．5．7

(1) n c0
cn 
1．3．5．7  (2n  1)
Ch5_48
Example 3 (5)
Thus for r1 = ½
n
n




(

1
)
(

1
)
y1 ( x)  x1/ 2 1   n x n    n x n1/ 2
 n1 2 n!  n0 2 n!
for r2 = 0

(1) n
y2 ( x )  1  
xn ,
n 1 1．3．5．7  ( 2n  1)
| x|
and on (0, ), the solution is y(x) = C1y1 + C2y2.
Ch5_49
Example 4
Solve xy" y  0
Solution
From xP = 0, x2Q = x, and the fact 0 and x are their own
power series centered at 0, we conclude a0 = 0, b0 = 0.
Then form (14) we have r(r – 1) = 0, r1 = 1, r2 = 0. In
other words, there is only a single series solution

(1)n n1
x 2 x3
y1 ( x)  
x  x    ...
2 12
n0 n!( n  1)!
Ch5_50
Three Cases
(1) If r1, r2 are distinct and do not differ by an integer,
there exists two linearly independent solutions of the
form:


n 0
n 0
y1 ( x)   cn x nr1 and y2 ( x)   bn x nr2
Ch5_51
(2) If r1 – r2 = N, where N is a positive integer, there
exists two linearly independent solutions of the form:

y1 ( x)   cn x nr1 , c0  0
(19)
n 0

y2 ( x)  Cy1 ( x) ln x   bn x nr2 , b0  0
(20)
n 0
Ch5_52
(3) If r1 = r2, there exists two linearly independent
solutions of the form:

y1 ( x)   cn x
n 0
n  r1
, c0  0
(21)

y2 ( x)  y1 ( x) ln x   bn x nr2
(22)
n 0
Ch5_53
Finding a Second Solution
If we already have a known solution y1, then the
second solution can be obtained by
y2 ( x)  y1 
e
  Pdx
2
y1
dx
(23)
Ch5_54
Example 5
Find the general solution of xy" y  0
Solution
From the known solution in Example 4,
1 2 1 3 1 4
y1 ( x)  x  x  x 
x 
2
12
144
we can use (23) to find y2(x). Here please use a CAS
for the complicated operations.
Ch5_55
Example 5 (2)
y2 ( x)  y1 ( x) 
  0 dx
e
dx
dx

y
(
x
)
1
  1 2 1 3 1 4 2
[ y1 ( x)]2
 x  2 x  12 x  144 x  
 y1 ( x) 
dx
 x 2  x3  5 x 4  7 x5  


12
12
1 1 7 19

 y1 ( x)   2    x   dx
x 12 72
x

1
7
19 2

 y1 ( x)    ln x  x 
x  
12
144
 x

1 7
19 2

y2 ( x)  y1 ( x) ln x  y1 ( x)   x 
x  
 x 12 144

Ch5_56
5.3 Special Functions
Bessel’s Equation of order v
x 2 y  xy  ( x 2  v 2 ) y  0
(1)
where v  0, and x = 0 is a regular singular point of
(1). The solutions of (1) are called Bessel functions.
Lengender’s Equation of order n
(2)
(1  x 2 ) y  2 xy  n(n  1) y  0
where n is a nonnegative integer, and x = 0 is an
ordinary point of (2). The solutions of (2) are called
Legender functions.
Ch5_57
The Solution of Bessel’s Equation
Because x = 0 is a regular singular point, we know
there exists at least one solution of the

form y  n0 cn x nr . Then from (1),
x 2 y  xy  ( x 2  v 2 ) y

  cn (n  r )( n  r  1) x
n r
n 0
 c0 (r  r  r  v ) x  x
2
2
 c0 (r  v ) x  x
2
2
r
r
r


  cn (n  r ) x
n r
n 0
r
  cn x
n r  2
v
 cn [(n  r )(n  r  1)  (n  r )  v
n 1
2
2
n
2
n 0

 cn [(n  r )  v ]x  x
n 1

r

 cn x nr
n 0
2
]x  x
n
r

 cn x n2
n 0

n 2
c
x
n
n 0
(3)
Ch5_58
From (3) we have the indicial equation r2 – v2 = 0, r1 =
v, r2 = −v. When r1 = v, we have
(1 + 2v)c1 = 0
(k + 2)(k + 2+ 2v)ck+2 + ck = 0
or
ck 2
 ck

,
(k  2)(k  2  2v)
k  0, 1, 2, 
(4)
The choice of c1 = 0 implies c3 = c5 = c7 = … = 0,
so for k = 0, 2, 4, …., letting k + 2 = 2n, n = 1, 2, 3, …,
we have
c2 n2
c2 n   2
(5)
2 n( n  v )
Ch5_59
Thus
c0
c2   2
2 ．1．(1  v)
c2
c0
c4   2
 4
2 ．2(2  v) 2 ．1．2(1  v)(2  v)
c4
c0
c6  2
 6
2 ．3(3  v)
2 ．1．2．3(1  v)( 2  v)(3  v)

c2 n
(1) n c0
 2n
,
2 n!(1  v)(2  v)  (n  v)
n  1, 2, 3, 
(6)
Ch5_60
We choose c0 to be a specific value
1
c0  v
2 (1  v)
where (1 + v) is the gamma function. See Appendix II.
There is an important relation:
(1 + ) = ()
so we can reduce the denominator of (6):
(1  v  1)  (1  v)(1  v)
(1  v  2)  (2  v)(2  v)  (2  v)(1  v)(1  v)
Ch5_61
Hence we can write (6) as
c2 n
(1)n
 2 nv
, n  0,1,2,...
2
n!(1  v  n)
Ch5_62
Bessel’s Functions of the First Kind
We define Jv(x) by

(1)
x

J v ( x)  
 
n 0 n! (1  v  n)  2 
n
2 nv
(7)
and

(1)
x

J v ( x )  
 
n 0 n! (1  v  n)  2 
n
2 n v
(8)
In other words, the general solution of (1) on (0, ) is
y = c1Jv(x) + c2J-v(x), v  integer
(9)
See Fig 5.3
Ch5_63
Fig 5.3
Ch5_64
Example 1
Consider the DE
x 2 y" xy'( x 2  1/4) y  0
We find v = ½, and the general solution on (0, ) is
y  c1J1/2 ( x)  c2 J 1/2 ( x)
Ch5_65
Bessel’s Functions of the Second Kind
If v  integer, then
cos v J v ( x)  J v ( x)
Yv ( x) 
sin v
(10)
and the function Jv(x) are linearly independent.
Another solution of (1) is y = c1Jv(x) + c2Yv(x).
As v  m, m an integer, (10) has the form 0/0. From
L’Hopital’s rule, the function
Ym ( x)  lim Yv ( x)
vm
and Jv(x) are linearly independent solutions of
x 2 y" xy'( x 2  m2 ) y  0
Ch5_66
Hence for any value of v, the general solution of (1) is
y  c1J v ( x)  c2Yv ( x)
(11)
Yv(x) is called the Bessel function of the second kind of
order v. Fig 5.4 shows y0(x) and y1(x).
Ch5_67
Fig 5.4
Ch5_68
Example 2
Consider the DE
x 2 y" xy'( x 2  9) y  0
We find v = 3, and from (11) the general solution on
(0, ) is
y  c1J 3 ( x)  c2Y3 ( x)
Ch5_69
DEs Solvable in Terms of Bessel Function
Let t = x,  > 0, in
x 2 y  xy  ( 2 x 2  v 2 ) y  0
then by the Chain Rule,
(12)
dy dy dt
dy


dx dt dx
dt
2
d 2 y d  dy  dt 
d
y
2
     
2
dt  dx  dx 
dx
dt 2
Ch5_70
Thus, (12) becomes
2
2
t
d
   2 y   t  dy  t 2  v 2 y  0
 
 
2
dt    dt
 




2
d
y dy 2 2
2
t
t  t v y  0
2
dt
dt
The solution of the above DE is y = c1Jv(t) + c2Yv(t)
Let t = x, we have
y = c1Jv(x) + c2Yv(x)
(13)
Ch5_71
Another equation is called the modified Bessel
equation order v,
x 2 y  xy  ( x 2  v 2 ) y  0
(14)
This time we let t = ix, then (14) becomes
2
d
y dy
2
2
2
t

t

(
t


)y  0
2
dt
dt
The solution will be Jv(ix) and Yv(ix). A real-valued
solution, called the modified Bessel function of the
first kind of order v is defined by

(15)
I ( x)  i J (ix)
Ch5_72
Analogous to (10), the modified Bessel function of
the second kind of order v  integer is defined by
 I  ( x)  I ( x)
(16)
K ( x) 
2
sin
and for any integer v = n,
K n ( x)  lim K ( x)
 n
Because Iv and Kv are linearly independent on (0, ),
the general solution of (14) is
(17)
y  c1I ( x)  c2 K ( x)
Ch5_73
We consider another important DE:
 2 2 2 c 2 a 2  p 2c 2 
1  2a
(18)

y 
y   b c x

y

0
,
p

0
2

x
x


The general solution of (18) is
y  x a [c1J p (bx c )  c2Yp (bx c )]
(19)
We shall not supply the details here.
Ch5_74
Example 3
Find the general solution of xy  3 y  9 y  0 on (0, )
Solution
Writing the DE as
3
9
y  y  y  0
x
x
according to (18)
1 – 2a = 3, b2c2 = 9, 2c – 2 = −1, a2 – p2c2 = 0
then a = −1, c = ½ . In addition we take b= 6, p = 2.
From (19) the solution is
y  x 1[c1J 2 (6 x1/ 2 )  c2Y2 (6 x1/ 2 )]
Ch5_75
Example 4
Recall the model in Sec. 3.8
mx  ket x  0,   0
You should verify that by letting
2 k t / 2
s
e
 n
we have
2
d
x
dx 2
2
s
s s x0
2
ds
ds
Ch5_76
Example 4 (2)
The solution of the new equation is
x = c1J0(s) + c2Y0(s),
If we resubstitute
2 k t / 2
s
e
 n
we get the solution.
 2 k t / 2 
 2 k t / 2 
x(t )  c1J 0 
e
e
  c2Y0 

 m

 m

Ch5_77
Properties
(1) J m ( x)  (1) m J m ( x)
(2) J m ( x)  (1) m J m ( x)
0 , m  0
(3) J m (0)  
1 , m  0
(4) lim x0 Ym ( x)  
Ch5_78
Example 5
Derive the formula
Solution
It follows from (7)

xJ 'v ( x)  vJ v ( x)  xJ v1 ( x)
(1) n (2n  v)  x 
xJ v ( x)  
 
n
!

(
1

v

n
)
 2
n 0

2 nv
(1)
 x
 
n 0 n! (1  v  n)  2 
n
 v
2 nv

(1) n  x 
 
n 0 n! (1  v  n)  2 
 2
n
2 nv
2 n  v 1

(1) n
x

 vJ v ( x)  x 
 
(
n

1
)!

(
1

v

n
)
2

n

1

k  n 1
Ch5_79
Example 5 (2)

(1)
x

 vJ v ( x)  x 
 
k 0 k!( 2  v  k )  2 
 vJ v ( x)  xJ v1 ( x)
k
2 k v 1
Ch5_80
The result in example 5 can be written as
v
J v ( x)  J v ( x)   J v1 ( x)
x
which is a linear DE in Jv(x). Multiplying both sides
the integrating factor x-v, then
d v
(20)
[ x J v ( x)]   x v J v1 ( x)
dx
It can be shown
d v
(21)
v
[ x J v ( x)]   x J v1 ( x)
dx
When y = 0, it follows from (14) that
(22)
J 0 ( x)   J1 ( x), Y0( x)  Y1 ( x)
Ch5_81
Spherical Bessel Functions
When the order v is half an odd number, that is,
1/2, 3/2, 5/2, …..
The Bessel function of the first kind Jv(x) can be
expressed as spherical Bessel function:

(1) n
 x
J1 / 2 ( x)  
 
n  0 n!(1  1 / 2  n)  2 
2 n 1 / 2
Since (1 + ) = () and (1/2) = ½, then
 1
 (2n  1)!
1   n   2n1

 2
 2
n!
Ch5_82
Hence

(1) n
J1/ 2 ( x)  
n 0 ( 2n  1)! 
n!
22 n1 n!
 x
 
 2
2 n 1 / 2
2  (1)n 2 n1

x

x n0 (2n  1)!
and
2
J1/ 2 ( x) 
sin x
x
2
J 1/ 2 ( x) 
cos x
x
(23)
(24)
Ch5_83
The Solution of Legender Equation
Since x = 0 is an ordinary point of (2), we use
n
y  
c
x
n 0 n
After substitutions and simplifications, we obtain
n(n  1)c0  2c2  0
(n  1)( n  2)c1  6c3  0
( j  2)( j  1)c j  2  (n  j )( n  j  1)c j  0
or in the following forms:
Ch5_84
n(n  1)
c2  
c0
2!
(n  1)( n  2)
c3  
c1
3!
(n  j )( n  j  1)
c j 2  
c j , j  2, 3, 4, 
(25)
( j  2)( j  1)
Using (25), at least |x| < 1, we obtain
n(n  1) 2 (n  2)n(n  1)( n  3) 4

y1 ( x)  c0 1 
x 
x
2!
4!

(n  4)( n  2)n(n  1)( n  3)( n  5) 6

x  
6!

Ch5_85
(n  1)( n  2) 3 (n  3)( n  1)( n  2)( n  4) 5
y2 ( x)  c1  x 
x 
x
3!
5!

(n  5)( n  3)( n  1)( n  2)( n  4)( n  6) 7

x   (26)
7!

Notices: If n is an even integer, the first series
terminates, whereas y2 is an infinite series.
If n is an odd integer, the series y2 terminates with xn.
Ch5_86
Legender Polynomials
The following are nth order Legender polynomials:
P0 ( x)  1,
P1 ( x)  x
1 2
1 3
(27)
P2 ( x)  (3 x  1),
P3 ( x)  (5 x )  3 x
2
2
1
1

2
P4 ( x)  (35 x  30 x  3), P5 ( x)  (63x5  70 x3  15 x)
8
8
Ch5_87
They are in turn the solutions of the DEs. See Fig 5.5
n  0 : (1  x 2 ) y  2 xy  0
n  1 : (1  x 2 ) y  2 xy  2 y  0
n  2 : (1  x 2 ) y  2 xy  6 y  0
n  3 : (1  x 2 ) y  2 xy  12 y  0

(28)

Ch5_88
Fig 5.5
Ch5_89
Properties
(1) Pn ( x)  (1) n Pn ( x)
(2) Pn (1)  1
(3) Pn (1)  (1) n
(4) Pn (0)  0, n odd
(5) P'n (0)  0, n even
Ch5_90
Recurrence Relation
Without proof, we have
(k  1) Pk 1 ( x)  (2k  1) xPk ( x)  kPk 1 ( x)  0
(29)
which is valid for k = 1, 2, 3, …
Another formula by differentiation to generate
Legender polynomials is called the Rodrigues’
formula:
1 dn 2
n
Pn ( x)  n
(
x

1
)
, n  0, 1, 2, ...
n
2 n! dx
(30)
Ch5_91