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Transcript of - 9주차강의내용contents.kocw.net/KOCW/document/2015/chungang/kimjinhong/... · 2016-09-09 ·...
The power series method is the standard method for solving liner ODEs
with variable coefficients. It gives solutions in the form of power series.
0
2
020100 )()()()1(m
m
m xxaxxaaxxa
where, x ; variable ; coefficients (constants)210 ,, aaa
0x ; center (constant)
- Power series is an infinite series of the form
0
3
3
2
210)2(m
m
m xaxaxaaxa
we obtain a power series in powers of,00 xIf x
5.1 Power Series Method
Maclaurin series
0
3211
1
m
m xxxxx
!3!21
!
32
0
xxx
m
xe
m
mx
!4!21
)!2(
)1(cos
42
0
2 xx
m
xx
m
mm
!5!3)!12(
)1(sin
53
0
12 xxx
m
xx
m
mm
← m does not include negative or fractional values.
,1x Geometric series
Ideas of the Power Series Methods
For a given ODE
0)()( yxqyxpy
and are represented by power series of)(xp )(xq x (or of )0xx or polynomials.
We assume a solution in the form of a power series with unknown coefficients.
0
3
3
2
210)3(m
m
m xaxaxaaxay
0
2
321
1 32)4(m
m
m xaxaaxmaya
0
2
4
1
32
2 34232)1()4(m
m
m xaxaaxammyb
We collect like powers of and equate the coefficients of each powers of
to zero, starting with the constant terms.
xx
Ex. 1) Solve by power series.xyy 2
Let
3
3
2
2
0
10 xaxaxaaxaym
m
m
(3) and (4a) are inserted into the given eq. xyy 2
)(232 2
210
2
321 xaxaaxxaxaa
5
6
4
5
3
4
2
321 65432 xaxaxaxaxaa
5
4
4
3
3
2
2
10 22222 xaxaxaxaxa
,26,25,24,23,22,0 46352413021 aaaaaaaaaaa
Hence, ,0,0 53 aa and for the coefficients with even subscripts,
!33,
!22, 04
602
402
aaa
aaaaa
Thus,2
0
8642
0 )!4!3!2
1( xeaxxx
xay
More rapidly, (3) and (4) are used for ODE xyy 2
0
1
0
1
2
0
1 221m
m
m
m
m
m
m
m
m xaxaxxmaxa
To get the same general power on both sides, we make a shift of index
and into the above eq. ;2 sm sm
0
11
0
21 2)2(s
s
s
s
s
s xaxasa
⇒ 01 a and ,2)2( 2 ss aas
For ,)4/2()4/2(,0)3/2(,)2/2(,2,1,0 0241302 aaaaaaas
ss as
a2
22
or
2
0
8642
0 )!4!3!2
1( xeaxxx
xa
6
6
5
5
4
4
3
3
2
210 xaxaxaxaxaxaay
,)!4/1()8/2(,)!3/1()6/2(,0 0680465 aaaaaaa
Ex. 2) Solve 0 yy by power series.
Let
3
3
2
2
0
10 xaxaxaaxaym
m
m
0)1(0
2
2
m
m
m
m
m
m
xaxamm
To get the same general power on both sides, we make a shift of index
and into the above eq. ;2 sm sm
00
2)1)(2(s
s
s
s
s
s xaxass
⇒ ),1,0()2)(1(
2
sss
aa s
s
!545,
!434,
!323,
!212
135
024
113
002
aaa
aaa
aaa
aaa
and so on
xaxaxx
xaxx
ay sincos)!5!3
()!4!2
1( 10
53
1
42
0
5140312010
!5!4!3!2x
ax
ax
ax
axaay
Ex. 3) Solve 02 yyxy by power series.
Let
3
3
2
2
0
10 xaxaxaaxaym
m
m
02)1(01
2
2
m
m
m
m
m
m
m
m
m
xaxmaxamm
To get the same general power on both sides, we make a shift of index
and into the above eq. ;2 sm sm
0)12()1)(2(00
2
s
s
s
s
s
s xasxass
⇒ ),1,0()2)(1(
)12(2
sa
ss
sa ss
1350241
30
2!5
51
54
5,
!4
3
43
3,
32,
21aaaaaa
aa
aa
and so on
)!8
1173
!6
73
!4
3
!2
11( 8642
0
xxxxay
9753
1!9
13951
!7
951
!5
51
!3
1( xxxxxa
,!6
73
65!4
73
65
70046 aaaa
1157!7
951
76!5
951
76
9aaaa
5.2 Theory of Power Series Method
- Basic Concepts
0
2
020100 )()()()1(m
m
m xxaxxaaxxa
where, ; variablex ; coefficients (constants)210 ,, aaa
0x ; center (constant)
n
nn xxaxxaxxaaxs )()()()()2( 0
2
02010
2
02
1
01 )()()()3( n
n
n
nn xxaxxaxR
← remainder of (1) after term ofn
n xxa )( 0
If for some the sequence converges, say,1xx ),()(lim 11 xsxsnn
then the (1) is convergent at1xx , and
If that sequence diverges at
,)()( 0
0
11
m
m
m xxaxs
1xx , the series (1) is divergent at 1xx
series
)()()( 111 xRxsxs nn
- Convergence Interval. Radius of Convergence
With respect to the convergence of the power series (1) there are three cases,
the useless Case 1, the usual Case 2, the best Case 3, as follows.
Case 1 ; The series (1) converges always at 0xx because for 0xx all its
terms are zero, except for the first one, .0a Such a series is of no practical interest.
Case 2 ; The series (1) converges and these values form an interval, called the
convergence interval. If this interval is finite, it has the midpoint 0x
(6) Rxx 0
and the series (1) converges for all such thatx Rxx 0 and diverges for all x
such that .0 Rxx R is called the radius of convergence,
(7)m
m
m
mm
m a
aRbaRa 1lim/1)(lim/1)(
Case 3 ; The series converges for all .x It corresponds to R
Ex. 3) Case 3 ; Series
0
2
!21
!m
mx x
xm
xe
1
1lim/1
)!1(
!lim/1lim/1!/1 1
mm
m
a
aRma
mmm
m
mm
The series converges for all .x
Ex. 1) Case 1 ; Series
32
0
621! xxxxm m
m
0)1(lim/1!
)!1(lim/1lim/1! 1
m
m
m
a
aRma
mmm
m
mm
This series converges only at .0x Such a series is useless.
Ex. 2) Case 2 ; Geometric series )1(11
1
0
32
xxxxxx m
m
for all1ma m and .1R
Geometric series converges and represents )1/(1 x when 1x
Ex. 4) Case 3 ; Series
022
2
)!(2
)1(
mm
mm
m
x
22 )!(2/)1( ma mm
m
The series converges for all .x
2
22
2)1(2
1
1
)1(4
1lim/1
)1(
)!(2
])!1[(2
)1(lim/1lim/1
m
m
ma
aR
mm
m
m
m
mm
m
m
Ex. 5) Find the radius of convergence of the series
,1
)3(
ma
m
m
The series converges for 3/1x
0 1
)3(
m
mm
m
x
)/2(1
)/1(13
2
13
)3(
1
2
)3( 1
1
m
m
m
mm
ma
am
m
m
m
3/1)/2(1
)/1(1lim3/1lim/1 1
m
m
a
aR
mm
m
m
Ex. 6) Find the radius of convergence of the series
5126481
8
)1( 96
0
33 xxx
xm
m
m
m
8,8
1
8
8,
8
)1(1
1
Ra
aa
m
m
m
m
m
m
m
The series converges for ,83 x that is 2x
5.3 Legendre's Equation
Important equation in the field of physics is Legendre's equation,
(1)
Legendre's equation has power series solutions of the form
(2)
0)1(2)1( 2 ynnyxyx
0m
m
mxay
Substituting (2) and its derivatives into (1), and denoting ,)1( knn
To obtain the same general power
02 1
122 02)1()1(m
m
m
m m
m
m
m
m xakxmaxxammx
2 2 1 0
2 02)1()1(m m m m
m
m
m
m
m
m
m
m xakxmaxammxamm
sx in all four series, we set sm 2
in the first series and sm in the other three series.
02)1()1)(2(00 2 1
2
s
s
s
s s s
s
s
s
s
s
s xkaxsaxassxass
Term ;
Term ;
0x
1x
0)1(12 02 anna
0)]1(2[23 13 anna
Subsequently, 0)]1(2)1([)1)(2( 2 ss annsssass
),1,0(,)1)(2(
)1)((2
sa
ss
snsna ss
← recurrence relation, recursion formula
02!2
)1(a
nna
0
24
!4
)3)(1()2(
34
)3)(2(
annnn
ann
a
13!3
)2)(1(a
nna
1
35
!5
)4)(2)(1)(3(
45
)4)(3(
annnn
ann
a
By inserting these expressions for the coefficients into (2), we obtain
(5) )()()( 2110 xyaxyaxy
where
42
1!4
)3)(1()2(
!2
)1(1)( x
nnnnx
nnxy
53
2!5
)4)(2)(1)(3(
!3
)2)(1()( x
nnnnx
nnxxy
These series converges for .1x1y and 2y are not proportional and thus
linearly independent solutions.
.11 x
(5) is a general solutions of (1) on the interval