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Submitted by:
Rakesh Kumar,
(Deptt. Of Mathematics),
P.G.G.C.G, Sec-11, Chandigarh.
Chapter 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
210
0
)()()( axcaxccaxcn
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.
0lim ( ) lim ( )
N n
N N N nnS x c x a
Absolute Convergence
Within its interval of convergence, a power series
converges absolutely. That is, the following
converges.
Ratio Test
Suppose cn 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
nn axc
1
1 1( )lim | | lim
( )
n
n n
nn nn n
c x a cx a L
c x a c
A Power Defines a Function
Suppose
then
Identity Property
If all cn = 0, then the series = 0.
1 2
0 0' and " ( 1) (1)n n
n nn ny c nx y c n n x
0n
nnxcy
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!21cos
!5!3sin ,
!2!11
642
532
xxxx
xxxx
xxex
Arithmetic of Power Series
Power series can be combined through the operations
of addition, multiplication and division.
303
24
1
12
1
120
1
6
1
6
1
2
1
6
1)1()1(
5040120624621
sin
532
5432
753432
xxxx
xxxxx
xxxx
xxxx
xex
Example 1 Adding Two Power Series
Write as one power series.
Solution:
Since
we let k = n – 2 for the first series and k = n + 1 for the
second series,
0
1
2
2)1(n
n
nn
n
n xcxcnn
2 0 3 0
120
2
12 )1(12)1(n n n n
n
n
n
n
n
n
n
n xcxcnnxcxcxcnn ．
series starts with
x for n = 3
↓
series starts with
x for n = 0
↓
Example 1 (2)
then we can get the right-hand side as
(3)
We now obtain
(4)
1 1
122 )1)(2(2k k
k
k
k
k xcxckkc
1
122
2 0
12
])1)(2[(2
)1(
k
k
kk
n n
n
n
n
n
xcckkc
xcxcnn
same
same
Suppose the linear DE(5)
is put into(6)
A Solution
0)()()( 012 yxayxayxa
0)()( yxQyxPy
A point x0 is said to be an ordinary point of (5) if both
P(x) and Q(x) in (6) are analytic at x0. A point that is
not an ordinary point is said to be a singular point.
Definition 5.1.1 Ordinary and Singular Points
Since P(x) and Q(x) in (6) is a rational function,
P(x) = a1(x)/a2(x), Q(x) = a0(x)/a2(x)
It follows that x = x0 is an ordinary point of (5) if
a2(x0) 0.
Polynomial Coefficients
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,
A series solution converges at least of some interval
defined by |x – x0| < R, where R is the distance from x0
to the closest singular point.
Theorem 5.1.2 Existence of Power Series Solutions
0 0 )(n
n
n xxcy
Example 2 Power Series Solutions
Solve
Solution:
We know there are no finite singular points.
Now, and
then the DE gives
(7)
0" xyy
0n
n
nxcy
2
2)1("n
n
nxcnny
0
1
2
2
2 0
2
)1(
)1(
n
n
n
n
n
n
n n
n
n
n
n
xcxnnc
xcxxnncxyy
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
122 0])2)(1[(2k
k
kk xcckkcxyy
,3,2,1,0)2)(1( 12 kcckk kk
,3,2,1,)2)(1(
12
kkk
cc k
k
Example 2 (3)
Thus we obtain
,1k32
03
．
cc
,2k43
14
．
cc
,3k 054
25
．
cc
,4k 03
66532
1
65c
cc
．．．．
,5k 14
77643
1
76c
cc
．．．．
← c2 is zero
and so on.
Example 2 (4)
,6k 087
58
．
cc
,7k 06
9986532
1
98c
cc
．．．．．．
,8k 17
101097643
1
109c
cc
．．．．．．
,9k 01110
811
．
cc
← c5 is zero
← c8 is zero
Example 2 (5)
Then the power series solutions are
y = c0y1 + c1y2
....07．6．4．3
6．5．3．20
4．33．20
71
60413010
xc
xc
xc
xc
xccy
1
13
1074
2
)13)(3(43
)1(
1097643
1
7643
1
43
11)(
k
kk
xnn
x
xxxxy
．
．．．．．．．．．
Example 2 (6)
1
3
963
1
)3)(13(32
)1(1
986532
1
6532
1
32
11)(
k
kk
xnn
xxxxy
．
．．．．．．．．．
Example 3 Power Series Solution
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( 2 yxyyx
012
2
2
2 01
122
)1()1(
)1()1(
n
n
n
n
n
n
n
n
n
n
n
n
n n
n
n
n
n
n
n
n
xcxncxcnnxcnn
xcxncxxcnnx
nk
n
n
n
nk
n
n
n
nk
n
n
n
nk
n
n
n
xcxncxcnn
xcnnxcxcxcxcxc
22
2
4
2
2
113
0
0
0
2
)1(
)1(62
Example 3 (2)
2
2302
2
2302
0])1)(2()1)(1[(62
])1)(2()1([62
k
k
kk
k
k
kkkk
xckkckkxccc
xckcckkckkxccc
Example 3 (3)
From the above, we get 2c2－c0 = 0, 6c3 = 0 , and
Thus c2 = c0/2, c3 = 0, ck+2 = (1 – k)ck/(k + 2)
Then
0)1)(2()1)(1( 2 kk ckkckk
02024!22
1
42
1
4
1cccc
．
05
235 cc
03046!32
31
642
3
6
3cccc
．
．．
07
457 cc
← c3 is zero
← c5 is zero
Example 3 (4)
and so on.
04068!42
531
8642
53
8
5cccc
．．
．．．
．
09
679 cc
050810!52
7531
108642
753
10
7cccc
．
．．．
．．．．
．．
← c7 is zero
Example 3 (5)
Therefore,
)()(
!52
7531
!42
531
!32
31
!22
1
2
11
2110
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
210
xycxyc
xcxxxxxc
xcxcxcxcxc
xcxcxcxcxccy
．．．．．．
1||,!2
)32(531)1(
2
11)( 2
2
12
1
xxn
nxxy n
nn
n ．．
xxy )(2
Example 4 Three-Term Recurrence Relation
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.
,3,2,1,)2)(1(
12
kkk
ccc kk
k
0)1( yxy
Example 4 (2)
Then we have
and so on. Next, we choose c0 = 0, c1 0, then
0012
424
1
43243c
cccc
．．．
0023
530
1
2
1
6
1
5454c
cccc
．．
0001
36
1
3232c
cccc
．．
02
102 cc
022
1cc
Example 4 (3)
and so on. Thus we have y = c0y1 + c1y2, where
1101
36
1
3232c
cccc
．．
1112
412
1
4343c
cccc
．．
1123
5120
1
65454c
cccc
．．．
5432
130
1
24
1
6
1
2
11)( xxxxxy
543
2120
1
12
1
6
1)( xxxxxy
Example 5 ODE with Nonpolynomial
Coefficients
Solve
Solution:
We see x = 0 is an ordinary point of the equation. Using
the Maclaurin series for cos x, and using
we find
,0
n
n
nxcy
0)(cos" yxy
2 0
6422
!6!4!21)1(
)(cos
n n
n
n
n
n xcxxx
xcnn
yxy
0
2
120
2
112)6(2 3
135
2
0241302
xcccxcccxcccc
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
02
120,0
2
112,06,02 1350241302 cccccccccc
42
112
1
2
11)( xxxy
53
230
1
6
11)( xxxy
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)()()( 012 yxayxayxa
0)()( yxQyxPy
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 5.2.1 Regular/Irregular Singular Points
Polynomial Coefficients
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)()()()( 0
2
0 yxqyxpxxyxx
Example 1 Classification of Singular Points
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)(
xxxP
22 )2()2(
5)(
xxxQ
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 5.2.1 Frobenius’ Theorem
0
0
0
00 )()()(n
rn
n
n
n
n
r xxcxxcxxy
Example 2 Two Series Solutions
Because x = 0 is a regular singular point of
(5)
we try to find a solution . Now,
03 yyyx
0
1)(n
rn
n xcrny
0
2)1)((n
rn
n xcrnrny
0n
rn
n xcy
Example 2 (2)
00
1
00
1
0
1
)233)((
)()1)((3
3
n
rn
n
n
rn
n
n
rn
n
n
rn
n
n
rn
n
xcxcrnrn
xcxcrnxcrnrn
yyyx
0])133)(1[()23(
)233)(()23(
0
1
1
0
1
1
1
11
0
k
k
kk
r
nk
n
n
n
nk
n
n
n
r
xccrkrkxcrrx
xcxcrnrnxcrrx
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
(7)
From (6), r = 0, 2/3, when substituted into (7),
,2,1,0,)133)(1(
1
krkrk
cc k
k
Example 2 (4)
r1 = 2/3, k = 0, 1, 2, … (8)
r2 = 0, k = 0, 1, 2, … (9)
,)1)(53(
1
kk
cc k
k
,)13)(1(
1
kk
cc k
k
Example 2 (5)
From (8) From(9)
)23(741!
)1(
)23(1185!
10741!4104141185!4414
741!3731185!3311
41!24285!228
1115
00
034
034
023
023
012
012
01
01
nn
cc
nn
cc
ccc
ccc
ccc
ccc
ccc
ccc
cc
cc
n
nn
．．．．
．．．．．．．．
．．．．．．
．．．．
．．
Example 2 (6)
These two series both contain the same multiple c0.
Omitting this term, we have
(10)
(11)
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 <
1
3/2
1)23(1185!
11)(
n
nxnn
xxy．．
1
0
2)23(741!
11)(
n
nxnn
xxy．．
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(x) and q = x2Q(x) 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)]([)]([ 22 yxQxyxxPxyx
Example 3 Two Series Solutions
Solve
Solution:
Let , then
0n
rn
nxcy
00
1
00
0
1
0
1
)1()122)((
)(
)()1)((2
)1(2
n
rn
n
n
rn
n
n
rn
n
n
rn
n
n
rn
n
n
rn
n
xcrnxcrnrn
xcxcrn
xcrnxcrnrn
yyxyx
0')1("2 yyxxy
Example 3 (2)
which implies r(2r – 1) = 0 (15)
(16)
0
1
1
0
0
1
1
11
0
])1()122)(1[()12(
)1()122)(()12(
k
k
kk
r
nk
n
n
n
nk
n
n
n
r
xcrkcrkrkxcrrx
xcrnxcrnrnxcrrx
,2,1,0,0)1()122)(1( 1 kcrkcrkrk kk
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(2
1
k
k
cc k
k
,2,1,0,12
1
k
k
cc k
k
Example 3 (4)
From (17) From (18)
1
1．20
10
1
cc
cc
)12(7531
)1(
!2
)1(
75317!4242
5315!3232
313
!2222
00
0344
034
0233
023
0122
012
n
cc
n
cc
ccc
ccc
ccc
ccc
ccc
ccc
n
nn
n
n
．．．
．．．．．
．．．．
．．．
Example 3 (5)
Thus for r1 = ½
for r2 = 0
and on (0, ), the solution is y(x) = C1y1(x) + C2y2(x).
0
2/1
1
2/1
1!2
)1(
!2
)1(1)(
n
n
n
n
n
n
n
n
xn
xn
xxy
||,)12(7531
)1(1)(
1
2 xxn
xyn
nn
．．．
Example 4 Only One Series Solutions
Solve
Solution:
From xP(x) = 0, x2Q(x) = 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
0" yxy
...144
1
122)!1(!
)1()( 4
321
0
1
xxx
xxnn
xy n
n
n
Three Cases
(1) If r1, r2 are distinct and do not differ by an integer,
there exists two linearly independent solutions of the
form:
0
2
0
121 )( and )(
n
rn
n
n
rn
n xbxyxcxy
(2) If r1 – r2 = N, where N is a positive integer, there
exists two linearly independent solutions of the form:
(19)
(20) 0 ,ln)()(
0 ,)(
0
0
12
0
0
1
2
1
bxbxxCyxy
cxcxy
n
rn
n
n
rn
n
(3) If r1 = r2, there exists two linearly independent
solutions of the form:
(21)
(22) ln)()(
0 ,)(
0
12
0
0
1
2
1
n
rn
n
n
rn
n
xbxxyxy
cxcxy
Finding a Second Solution
If we already have a known solution y1, then the
second solution can be obtained by
(23)
)(
)()(2
1
12 dxxy
exyxy
Pdx
Example 5 Example 4 Revised—Using a
CAS
Find the general solution of
Solution:
From the known solution in Example 4,
we can use (23) to find y2(x). Here please use a CAS
for the complicated operations.
0" yxy
432
1144
1
12
1
2
1)( xxxxxy
Example 5 (2)
2
1
21
54321
2
432
12
1
0
12
144
19
12
7ln
1)(
72
19
12
711)(
12
7
12
5)(
144
1
12
1
2
1)(
)]([)()(
xxxx
xy
dxxxx
xy
xxxx
dxxy
xxxx
dxxydx
xy
exyxy
dx
2
112144
19
12
71)(ln)()( xx
xxyxxyxy
5.3 Special Functions
Bessel’s Equation of order v
(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)
where n is a nonnegative integer, and x = 0 is an
ordinary point of (2). The solutions of (2) are called
Legender functions.
0)( 222 yvxyxyx
0)1(2)1( 2 ynnyxyx
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 . Then from (1),
(3)
0n
rn
nxcy
0
2
1
2222
0
0
22
1
22
0
00
22
00
222
])[()(
])()1)([(
)(
)()1)((
)(
n
n
n
r
n
n
n
rr
n
n
n
rn
n
n
r
r
n
rn
n
n
rn
n
n
rn
n
n
rn
n
xcxxvrncxxvrc
xcxxvrnrnrncx
xvrrrc
xcvxcxrncxrnrnc
yvxyxyx
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 (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
(5)
,2,1,0,)22)(2(
2
k
vkk
cc k
k
)(22
222
vnn
cc n
n
Thus
(6)
,3,2,1,)()2)(1(!2
)1(
)3)(2)(1(3212)3(32
)2)(1(212)2(22
)1(12
2
02
6
0
2
46
4
0
2
24
2
02
nvnvvn
cc
vvv
c
v
cc
vv
c
v
cc
v
cc
n
n
n
．．．．
．．．
．．
We choose c0 to be a specific value
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(2
10
vc
v
)1()1)(2()2()2()21(
)1()1()11(
vvvvvv
vvv
Hence we can write (6) as
... ,2 ,1 ,0 ,)1(!2
)1(22
n
nvnc
vn
n
n
Bessel’s Functions of the First Kind
We define Jv(x) by
(7)
and
(8)
In other words, the general solution of (1) on (0, ) is
y = c1Jv(x) + c2J-v(x), v integer (9)
See Fig 5.3.1.
0
2
2)1(!
)1()(
n
vnn
v
x
nvnxJ
0
2
2)1(!
)1()(
n
vnn
v
x
nvnxJ
Example 1 General Solution: v Not an
Integer
Consider the DE
We find v = ½, and the general solution on (0, ) is
0)1/4('" 22 yxxyyx
)()( 1/221/21 xJcxJcy
Bessel’s Functions of the Second Kind
If v integer, then
(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
and Jv(x) are linearly independent solutions of
v
xJxJvxY vv
vsin
)()(cos)(
)(lim)( xYxY vmv
m
0)('" 222 ymxxyyx
Hence for any value of v, the general solution of (1)
is
(11)
Yv(x) is called the Bessel function of the second
kind of order v. Fig 5.3.2 shows y0(x) and y1(x).
)()( 21 xYcxJcy vv
Example 2 General Solution: v an Integer
Consider the DE
We find v = 3, and from (11) the general solution on
(0, ) is
0)9('" 22 yxxyyx
)()( 3231 xYcxJcy
DEs Solvable in Terms of Bessel Functions
Let t = x, > 0, in
(12)
then by the Chain Rule,
0)( 2222 yvxyxyx
dt
dy
dx
dt
dt
dy
dx
dy
2
22
2
2
dt
yd
dx
dt
dx
dy
dt
d
dx
yd
Thus, (12) becomes
The solution of the above DE is y = c1Jv(t) + c2Yv(t)
Let t = x, we have
y = c1Jv(x) + c2Yv(x) (13)
0
0
22
2
22
22
2
22
2
yvtdt
dyt
dt
ydt
yvtdt
dyt
dt
ydt
Another equation is called the modified Bessel
equation order v,
(14)
This time we let t = ix, then (14) becomes
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)
0)( 222 yvxyxyx
0)( 22
2
22 yt
dt
dyt
dt
ydt
)()( ixJixI
Analogous to (10), the modified Bessel function of
the second kind of order v integer is defined by
(16)
and for any integer v = n,
Because Iv and Kv are linearly independent on (0, ),
the general solution of (14) is
(17)
sin
)()(
2)(
xIxIxK
)(lim)( xKxKn
n
)()( 21 xKcxIcy
We consider another important DE:
(18)
The general solution of (18) is
(19)
We shall not supply the details here.
0 ,021
2
2222222
py
x
cpaxcby
x
ay c
)]()([ 21
c
p
c
p
a bxYcbxJcxy
Example 3 Using (18)
Find the general solution of on (0, ).
Solution:
Writing the DE as
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
093 yyyx
093
yx
yx
y
)]6()6([ 2/1
22
2/1
21
1 xYcxJcxy
Example 4 The Aging Spring Revised
Recall the model in Sec. 3.8
You should verify that by letting
we have
0 ,0 xkexm t
2/ 2 te
m
ks
02
2
22 xs
ds
dxs
ds
xds
Example 4 (2)
The solution of the new equation is
x = c1J0(s) + c2Y0(s),
If we resubstitute
we get the solution.
2/ 2 te
m
ks
2/
02
2/
01
22)( tt e
m
kYce
m
kJctx
Properties
(i)
(ii)
(iii)
(iv)
)()1()( xJxJ m
m
m
)()1()( xJxJ m
m
m
0,1
0,0)0(
m
mJm
)(lim
0xYmx
Example 5 Derivation Using Series
Definition
Derive the formula
Solution:
It follows from (7)
1
1
12
0
2
0
2
0
2
2)1()!1(
)1()(
2)1(!
)1(2
2)1(!
)1(
2)1(!
)2()1()(
nk
n
vnn
v
n
vnn
n
vnn
n
vnn
v
x
nvnxxvJ
x
nvn
nx
nvnv
x
nvn
vnxJx
)()()( 1 xxJxvJxJx vvv
)()(2)2(!
)1()( 1
0
12
xxJxvJx
kvkxxvJ vv
k
vkk
v
The result in example 5 can be written as
which is a linear DE in Jv(x). Multiplying both sides the integrating factor x-v, then
(20)
It can be shown(21)
When y = 0, it follows from (14) that(22)
)()()( 1 xJxJx
vxJ vvv
)()]([ 1 xJxxJxdx
dv
v
v
v
)()]([ 1 xJxxJxdx
dv
v
v
v
,)()( 10 xJxJ )()( 10 xYxY
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:
Since (1 + ) = () and (1/2) = ½, then
0
2/12
2/12)2/11(!
)1()(
n
nn x
nnxJ
!2
)!12(
2
11
12 n
nn
n
Hence
and
(23)
(24)
0
12
0
2/12
12
2/1)!12(
)1(2
2
!2
)!12(!
)1()(
n
nn
n
n
n
n
xnx
x
n
nn
xJ
xx
xJ
xx
xJ
cos2
)(
sin2
)(
2/1
2/1
The Solution of Legendre Equation
Since x = 0 is an ordinary point of (2), we use
After substitutions and simplifications, we obtain
or in the following forms:
0n
n
nxcy
0)1)(()1)(2(
06)2)(1(
02)1(
2
31
20
jj cjnjncjj
ccnn
ccnn
(25)
Using (25), at least |x| < 1, we obtain
6
42
01
!6
)5)(3)(1()2)(4(
!4
)3)(1()2(
!2
)1(1)(
xnnnnnn
xnnnn
xnn
cxy
,4,3,2,)1)(2(
)1)((
!3
)2)(1(
!2
)1(
2
13
02
jcjj
jnjnc
cnn
c
cnn
c
jj
(26)
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.
!7
)6)(4)(2)(1)(3)(5(
!5
)4)(2)(1)(3(
!3
)2)(1()(
7
53
12
xnnnnnn
xnnnn
xnn
xcxy
Legendre Polynomials
The following are nth order Legendre polynomials:
(27)
)157063(8
1)(),33035(
8
1)(
)35(2
1)(),13(
2
1)(
)(,1)(
35
5
2
4
3
3
2
2
10
xxxxPxxxP
xxxPxxP
xxPxP
They are in turn the solutions of the DEs. See Fig
5.3.5
(28)
0122)1(:3
062)1(:2
022)1(:1
02)1(:0
2
2
2
2
yyxyxn
yyxyxn
yyxyxn
yxyxn
Properties
(i)
(ii)
(iii)
(iv)
(v)
)()1()( xPxP n
n
n
1)1( nP
n
nP )1()1(
odd ,0)0( nPn
even ,0)0(' nP n
Recurrence Relation
Without proof, we have
(29)
which is valid for k = 1, 2, 3, …
Another formula by differentiation to generate
Legendre polynomials is called the Rodrigues’
formula:
(30)
0)()()12()()1( 11 xkPxxPkxPk kkk
... ,2 ,1 ,0 ,)1(!2
1)( 2 nx
dx
d
nxP n
n
n
nn
THANK YOU
Submitted by:
Rakesh Kumar,
(Deptt. Of Mathematics),
P.G.G.C.G, Sec-11, Chandigarh.
