Power Series (2/3) Series Solution Near and Ordinary...
56
Class Notes 10: Power Series (2/3) Series Solution Near and Ordinary Point 182A – Engineering Mathematics
Transcript of Power Series (2/3) Series Solution Near and Ordinary...
Class Notes 10:
Power Series (2/3)
Series Solution Near and Ordinary Point
182A – Engineering Mathematics
Series Solution –
Ordinary Point and Singular Point – Introduction
0)()()(2
2
yxRdx
dyxQ
dx
ydxP
• Assume P, Q, R are polynomials with no common factors, and that we want to solve
the equation in a neighborhood of a point of interest x0
• If there is a common factor we divided it out before proceeding
Series Solution –Ordinary Point – Definition
• Ordinary Point (Definition) - The point x0 is called an ordinary point if P(x0) 0.
• Since P is continuous, P(x) 0 for all x in some interval about x0. For x in this
interval, divide the differential equation by P to get
• Since P(x0) is continues, there is an interval about x0 in which P(x0) is never zero
• Solve the equation in some neighborhood of a point of interest x0
• Since p and q are continuous, Theorem 3.2.1 indicates that there is a unique
solution, given initial conditions y(x0) = y0, y'(x0) = y0'
0)()()(2
2
yxRdx
dyxQ
dx
ydxP
)(
)()( ,
)(
)()( where
,0)()(2
2
xP
xRxq
xP
xQxp
yxqdx
dyxp
dx
yd
Ordinary Point – Examples
0))(sin(
023
0
yxyey
yyy
yy
x
][ • Every point in is an ordinary point
Series Solution –Singular Point – Definition
0)()()(2
2
yxRdx
dyxQ
dx
ydxP
)(
)()( ,
)(
)()( where
,0)()(2
2
xP
xRxq
xP
xQxp
yxqdx
dyxp
dx
yd
• Singular Point (Definition) - The point x0 is called an singular point if P(x0) = 0.
• Since P, Q, R are polynomials with no common factors, it follows that Q(x0) 0 or
R(x0) 0, or both.
• Then at least one of p or q becomes unbounded as x x0, and therefore Theorem
3.2.1 does not apply in this situation.
• Sections 5.4 through 5.8 deal with finding solutions in the neighborhood of a
singular point.
Singular Point – Example 1
0)ln(
)(
)()(
)(
yxyxy
xP
xRxP
xQ
0xatesdiscontinu
numberrealpositiveeveryatanalytic
numberrealeveryatanalytic
(x)P
R
xP
Q
ln
eqdiff.theofpointsingularaisat
inseriespowerabydrepresentebetcan'
00
ln
0
xx
x(x)P
R
Singular Point – Example 1
eqdiff.theofpointsingularaisat
inseriespowerabydrepresentebetcan'
00
ln
0
xx
x(x)P
R
Singular Point – Example 2
01
0
yyx
y
xyyyx
01
)(
)( x
xxP
xQatanalyticbetoFail
equationtheofpointsingularais0x
Singular Point – Example 3
01
6
1
2
062)1(
22
2
yx
yx
xy
yyxyx
pointsordinaryareresttheall;pointsingular1012 xx
Series Solution – Ordinary Point
• In order to solve our equation near an ordinary point x0,
we will assume a series representation of the unknown solution function y:
• As long as we are within the interval of convergence, this representation of y is
continuous and has derivatives of all orders.
0)()()(2
2
yxRdx
dyxQ
dx
ydxP
0
0 )()(n
n
n xxaxy
Theorem 5.3.1 Existence of Power Series Solution
• If x0 is an ordinary point of the differential equation
• Then the general solution for this equation is
where a0 and a1 are arbitrary, and y1, y2 are linearly independent series solutions
that are analytic at x0.
• The solutions y1, y2 form a fundamental set of solutions
• The power series solution converges at least on some interval defined by
• The radius of convergence for each of the series solutions y1 and y2 is at
least as large as the minimum of the radii of convergence of the series for p
and q.
0)()()(2
2
yxRdx
dyxQ
dx
ydxP
)()()()( 2110
0
0 xyaxyaxxaxyn
n
n
0xx
Minimum of the radii of convergence – Example
minFind
0)52( 2 yyxyxxFor
0)52(
1
)52( 22
y
xxy
xx
xy
1
0pointsordinaryanabout
0
0
x
xi
x
xxx 21052
2
12
• Because are two ordinary points of the diff. eq. the theorem
guarantees that we can find two power series centered around the two
1
0
0
0
x
x
0x
0
0 )()(n
n
n xxaxy
Minimum of the radii of convergence – Example
The radius of convergence
00 x 10 x
• Each power series for each will converge around at least with a
minimal radius of 0x
0
)(n
n
nxaxy
0
)1()(n
n
n xaxy
Power Solution Around 00 00 xx and
• For the sake of simplicity, find only power series solution about the
ordinary point
• If is also an ordinary point substitute the solution for
00 x
00 x 0xxx
0
)(n
n
nxaxy
0
0 )()(n
n
n xxaxy
Series Solution –
Ordinary Point – Example 1
xyy 0
1)(0)(;1)( xRxQxP
→ Every point is an ordinary point
Look for a solution in the form of a power series about 00 x
2
22
2
1
11
21
0
2
210
)1()1(2
2
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
xannxannay
xnaxnaxaay
xaxaxaxaay
substituting is the diff. eq.0)1(
02
2
y
n
n
n
y
n
n
n xaxann
Series Solution –
Ordinary Point – Example 1
2* nn
y
n
n
nxann
0
22
2
**
*
*
*)12)(2(
0)1)(2(00
2
y
n
n
n
y
n
n
n xaxann
0
2 0)1)(2(n
n
nn xaann
for this equation to be satisfactory for all x the coefficient of each power
of x must be zero
3,2,1,00)1)(2( 2 naann nn
2* nn
y
n
n
nxann
2
2)1(
Series Solution –
Ordinary Point – Example 1
3,2,1,00)1)(2( 2 naann nn
5
4
3
2
1
0
n
n
n
n
n
n
067
056
045
034
023
012
57
46
35
24
13
02
aa
aa
aa
aa
aa
aa
)67(
)56(
)45(
)34(
)23(
)12(
57
46
35
24
13
02
aa
aa
aa
aa
aa
aa
Series Solution –
Ordinary Point – Example 1
12345656
123434
12
046
024
02
aaa
aaa
aa
Even
123456767
1234545
23
157
135
13
aaa
aaa
aa
Odd
kn 2
12 kn
,4,3,2,1
)!2(
)1(02
k
ak
aak
kn
,4,3,2,1
)!12(
)1(112
k
ak
aak
kn
Series Solution –
Ordinary Point – Example 1
x
n
nn
x
n
nn
nn
nn
n
odd
nn
even
n
xn
axn
a
xn
xxxa
xn
xxa
xn
ax
n
a
xa
xa
xa
xa
xaay
sin
0
12
1
cos
0
2
0
1253
1
242
0
12120
5140312010
)!12(
)1(
)!2(
)1(
)!12(
)1(
!5!3
)!2(
)1(
!4!21
)!12(
)1(
)!2(
)1(
!5!4!3!2
Series Solution –
Ordinary Point – Example 1
0 yy
0)0(
1)0(For
y
y
1)0(
0)0(For
y
y
)cos()sin(
)sin()cos(
10
10
xaxay
xaxay
110)0(
001)0(
10
10
aay
aayxy sin
xy cos010)0(
101)0(
10
10
aay
aay
Series Solution –
Ordinary Point – Example 1
0
2
)!2(
)1()cos(
n
nn
xn
x
0
12
)!12(
)1()sin(
n
nn
xn
x
Series Solution –
Ordinary Point – Example 2 – Airy’s Equation
0 xyy
xxRxQxP )(0)(;1)(
Assume a solution
substituting the series in the diff eq.
0
1
00
2)1)(2(n
n
n
n
n
n
n
n
n xaxaxxann
)()1)(2(0
2
0
examplepreviousseen
n
n
n
n
n
xanny
xay
xyy
1~1~
nn
nn
Series Solution –
Ordinary Point – Example 2 – Airy’s Equation
start the sum from n=1
0 1~
~
1~2)1)(2(n n
n
n
n
n xaxann
1 1
12
0
2 )1)(2(12n n
n
n
n
n
n
xaxanna
For the equation to be satisfied for all x is some interval, the coefficients
of the same powers of x must be equal
12
12
2
00
2
)1)(2(
0012
nn
nn
aa
aann
axxa
oftermsingivenis
:1
:0
nxn
x
Series Solution –
Ordinary Point – Example 2 – Airy’s Equation
:1
:0
nxn
x
12
12
2
00
2
)1)(2(
0012
nn
nn
aa
aann
axxa
oftermsingivenis
3ofsteps
0
011852852
741
630
aaaaaaa
aaa
aaa
)1)(2(
12
nn
aa n
n
10,7,4,1n
11,8,5,2n
12,9,6,3n
Series Solution –
Ordinary Point – Example 2 – Airy’s Equation
The general formula
)3()13(9865324 0
3nn
aan n
10,7,4,1,,,For 9630 naaaa
9853298;
653265;
32
069
036
03
aaa
aaa
aa
Series Solution –
Ordinary Point – Example 2 – Airy’s Equation
11,8,5,2,,,For 10741 naaaa
1097643109;
764376;
43
1710
147
14
aaa
aaa
aa
The general formula
)13(310976434 1
13
nn
aan n
Series Solution –
Ordinary Point – Example 2 – Airy’s Equation
)13()3(1097643764343
)3()13(9865326532321
1374
1
363
0
nn
xxxxa
nn
xxxay
n
n
The General solution
Notes
- Both series converge for all x
- The general solution is xxyaxyay )()( 2110
Series Solution –
Ordinary Point – Example 2 – Airy’s Equation
oscillatory for x < 0
Decay in Amp; Increase in Freq
monotone for x > 0
xxyaxyay )()( 2110
)13()3(1097643764343
)3()13(9865326532321
1374
1
363
0
nn
xxxxa
nn
xxxay
n
n
Example – Airy’s Equation
01
00
)0(
)0(
yaxy
yaxy
0
0
)0(
)0(I.C.givenaFor
yxy
yxy
)13()3(1097643764343
)3()13(9865326532321
1374
1
363
0
nn
xxxxa
nn
xxxay
n
n
see also example 3, p 202 – Airy’s Equation for Xo=1
(ordinary point)
Series Solution –
Ordinary Point – Example 3
01
1
1
0)1(
22
2
yx
yx
xy
yyxyx
ixx 012
00 xFor
1
1xleastatconverge
2
2
1
1
0
)1(n
n
n
n
n
n
n
n
n
xcnny
xncy
xcy
Series Solution –
Ordinary Point – Example 3
01
1
1
0)1(
22
2
yx
yx
xy
yyxyx
ixx 012
00 xFor
1
1xleastatconverge
00 xFor
y
n
n
n
y
n
n
n
y
n
n
n xcxncxxcnnx
01
1
2
22 )1(1
Series Solution –
Ordinary Point – Example 3
0)1( 2 yyxyx 00 xFor
2
2
1
1
0
)1(;;n
n
n
n
n
n
n
n
n xannyxnayxay
012
2
2
)1)(2()1(k
k
k
k
k
k
k
k
k
n
k
k xcxkcxckkxckk
012
2
2
01
1
2
22
)1()1(
)1(1
n
n
n
n
n
n
n
n
n
n
n
n
y
n
n
n
y
n
n
n
y
n
n
n
xcxncxcnnxcnn
xcxncxxcnnx
Series Solution –
Ordinary Point – Example 3
2
1
1
0
0
21
1
1
21
1
3
0
0
2
2
0
1)1)(2()2)(3()1)(2()1(
k
k
k
k
k
k
kk
k
k
kkk
k
k
xcxcxc
xkcxcxckkxcxcxckk
nk
2
2
kn
nk nk nk
Start at 2k
Series Solution –
Ordinary Point – Example 3
0)1)(2()1()6()2(2
2302
k
k
kkkk xckcckkckkxccc
2
1
1
0
0
21
1
1
21
1
3
0
0
2
2
0
1)1)(2()2)(3()1)(2()1(
k
k
k
k
k
k
kk
k
k
kkk
k
k
xcxcxc
xkcxcxckkxcxcxckk
0)1)(2()1)(1()6()2(2
2302
k
k
kk xckkckkxccc
Series Solution –
Ordinary Point – Example 3
0)1)(2()1)(1(
06
02
2
3
02
kk ckkckk
c
cc
4,3,2 ,2
1
0
2
1
2
3
02
kck
kc
c
cc
kk
For the equation to be satisfied for all x is some interval, the coefficients
of the same powers of x must be equal
kx
x
x
1
0
0)1)(2()1)(1()6()2(
20
2
0
3
0
02
k
k
kk xckkckkxccc
Series Solution –
Ordinary Point – Example 3
4,3,2 ,2
12
kc
k
kc kk
07
4 5
!32
31
642
3
6
3 4
05
2 3
!22
1
42
1
4
1 2
57
03046
35
02024
cck
cccck
cck
cccck
050810
79
04068
!52
7531
108642
753
10
7 8
09
6 7
!42
531
8642
53
8
5 6
cccck
cck
cccck
Series Solution –
Ordinary Point – Example 3
xy
x
xx
n
nxy
n
n
n
n
2
2
212
11
11
!2
)32(531)1(
2
11
2110
1
10
5
8
4
6
3
4
2
2
0
5
5
4
4
3
3
2
210
2
1
!52
7531
!42
531
!32
31
!22
1
2
11
ycyc
xcxxxxxc
xcxcxcxcxccy
y
y
Series Solution –
Ordinary Point – Example 4 (Three Recurrence Relation)
0
0)1(
yxyy
yxy
2
2
0
)1(n
n
n
n
n
n
xcnny
xcy
0)1(002
2
y
n
n
n
xy
n
n
n
y
n
n
n xcxcxxcnn
0)1(00
1
2
2
n
n
n
n
n
n
n
n
n xcxcxcnn
00 xFor
Series Solution –
Ordinary Point – Example 4 (Three Recurrence Relation)
0
)1)(2()2)(3()1)(2(
21
1
1
0
0
0
2
1
1
1
0
2
2
1
1
3
0
0
2
k
k
k
kk
k
k
k
k
k
k
k
kk
xcxcxc
xcxc
xcnkxcxc
0)1(00
1
2
2
n
n
n
n
n
n
n
n
n xcxcxcnn
0:
2:
2
2
k
n
kn
nk
1:
0:
1
1
k
n
kn
nk nk
0)1)(2(01
1
0
2
k
k
k
k
k
k
k
k
k xcxcxckk
Start at 2k
Series Solution –
Ordinary Point – Example 4 (Three Recurrence Relation)
0)1)(2(622
0
12
0
013
0
20
k
k
kk
k
k xccxckkxccccc
0)1)(2( 12 kkk ccckk
02202
1 02 cccc
1030136
1 06 cccccc
)2)(1( 3, 2, 1
2
kk
ccck kk
k
Series Solution –
Ordinary Point – Example 4 (Three Recurrence Relation)
022
1cc
65 4
54 3
43 2
346
235
124
ccck
ccck
ccck
1036
1ccc
Series Solution –
Ordinary Point – Example 4 (Three Recurrence Relation)
0 ;0 10 cc 0 ;0 10 cc
022
1cc 0
2
102 cc
632
0013
cccc
632
1013
cccc
2443243
00124
ccccc
124343
11124
ccccc
302
1
6
1
5454
00235
ccccc
12065454
11235
ccccc
0
2110 n
n
nxcyycycy
5432
01030
1
24
1
6
1
2
11 xxxxcyc
543
121120
1
12
1
6
1xxxxcyc
Series Solution –
Ordinary Point – Example 5 (Non-polynomial Coefficients)
Ordinary Point 0cos yxy
00x
0n
n
nxcy
02
120
2
11262
!4!21201262
!6!4!21)1(cos
3
135
2
0241320
3
3
2
210
423
5
2
432
0
cos
642
2
2
xcccxcccxcccc
xcxcxccxx
xcxcxcc
xcxxx
xcnnyxy
y
n
n
n
xy
n
n
n
02
120
02
112
06
02
135
024
13
20
ccc
ccc
cc
cc
Series Solution –
Ordinary Point – Example 5 (Non-polynomial Coefficients)
Because the differential equation has no finite singular points, both
power series converge for
53
121
42
010
30
1
6
1 7 ,5 ,3 ,1
12
1
2
11 8 ,6 ,4 ,2 ,0
xxxcycn
xxcycn
2110 ycycy
15
04
13
02
30
1
12
1
6
1
2
1
cc
cc
cc
cc
xx
Series Solution –Ordinary Point – Example 6 (Legendre’s Equation)
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
2
2
1
1
0
)1(;;n
n
n
n
n
n
n
n
n xcnnyxncyxcy
1 min
1101
0)1~(~1
0)1~(~21
22
2
2
xxx
ynndx
dyx
dx
d
ynnyxyx
point Regular 00 x
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
2
)1~)(~(
~~~~
~~)1~(~2
2
1
1
0
0
1
1
1
3
0
2
0120
2
012
22
2
2
)1(
0
2
1
1
2
22
2
2
2
22
22
2
2
)1~(~2)1()1)(2(
)1~(~2232
)1~(~2)1()1)(2(
)1~(~2)1()1(
)1~(~2)1()1(
)1~(~2
k
kk
k
knknkkknnknn
kknn
nnkkk
k
kkkkk
k
k
k
k
k
k
k
k
k
k
k
k
nk
n
n
n
nk
n
n
n
nk
n
n
n
knnk
n
n
n
ynn
n
n
n
yx
n
n
n
yx
n
n
n
y
n
n
n
xxcnnkkkckk
xccnnxcxcc
xcnnxkcxckkxckk
xcnnxncxcnnxcnn
xcnnxncxxcnnxxcnn
ynnyxyxy
2 withStart2 withStart2 withStart
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
0)1~(~)1)(2(
)1~(~26)1~(~2
2
2
)2~)(1~()2~~(
11302
1
12
k
k
kk
cnncnn
xcknknckk
xcnncccnnc
0)1~(~)1)(2(
0)2~)(1~(6
0)1~(~2
2
13
02
kk cknknckk
cnnc
cnnc
For the equation to be satisfied for all x is some interval, the coefficients
of the same powers of x must be equal
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
,5 ,4 ,3 ,2
)1)(2(
1~~
2
kc
kk
knknc kk
157
046
135
024
!7
6~4~2~1~3~5~
67
6~5~ 5
!6
5~3~1~~2~4~
56
5~4~ 4
!5
4~2~1~3~
45
4~3~ 3
!4
3~1~~2~
34
3~2~ 2
cnnnnnn
cnn
ck
cnnnnnn
cnn
ck
cnnnn
cnn
ck
cnnnn
cnn
ck
1
!3
3
0
!2
2
6
)2~)(1~(
2
)1~(~
cnn
c
cnn
c
0)1~(~)1)(2(
0)2~)(1~(6
0)1~(~2
2
13
02
kk cknknckk
cnnc
cnnc
• Note for as even integer,
– y1 terminates with xn,
– y2 becomes infinite series
• Note for as odd integer (vice versa)
– y1 becomes infinite series
– Y2 terminates with xn,
• For example
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
753
12
642
01
!7
6~4~2~1~3~5~
!5
4~2~1~3~
!3
)2~)(1~()(
!6
5~3~1~~2~4~
!4
3~1~~2~
!2
1~~1)(
xnnnnnn
xnnnn
xnn
xcxy
xnnnnnn
xnnnn
xnn
cxy
n~
4~ n
42
0
42
013
35101
!4
7542
!2
541)( xxcxxcxy
753
12!7
1086311
!5
8631
!3
63)( xxxxcxy
n~
• Note For (positive), one of the two solutions will be n-th degree
polynomial solution to the Legendre’s equation
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
0~ n
753
12
642
01
!7
6~4~2~1~3~5~
!5
4~2~1~3~
!3
)2~)(1~()(
!6
5~3~1~~2~4~
!4
3~1~~2~
!2
1~~1)(
xnnnnnn
xnnnn
xnn
xcxy
xnnnnnn
xnnnn
xnn
cxy
• Because a constant multiple of a solution of the Legendre’s equation is also a
solution
• Then it is traditional to choose specific values for c0, c1 depending on whether n is an
even or odd positive integer respectively
• For
• For
• For example
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
10~0 cn
n
ncn
n
~42
1~3116,4,2~
2
~
0
01~1 cn
1~42
~3117,5,3~
2
1~
1
n
ncn
n
4~ n
330358
1
3
35101
42
311)( 2442
2
4
1
xxxxxy
• These specific n-th degree polynomial solutions are called Legendre
polynomials and are denoted by Pn(x)
• Note that for
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
n Legendre’s EquationLegendre Polynomial
(Particular Solution)
0
1
2
3
1)(0 xP 021 2 yxyx
0221 2 yyxyx
0621 2 yyxyx
01221 2 yyxyx
xxP )(1
132
1)( 2
2 xxP
xxxP 352
1)( 3
3
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
Series Solution –
Ordinary Point – Example 6 (Legendre’s Equation)
Legendre Polynomials
