MICHAEL B. BAYLON MS CE
COE5310 ASSIGNMENT #2 1
COE5310: ADVANCED MATHEMATICAL METHODS
PROF. HOMER CO
ASSIGNMENT NUMBER 2:
TOPIC: POWER SERIES SOLUTION NEAR A REGULAR SINGULAR POINT
1. By power series near a regular singular point
0334 yyyx .............................................................................................................. eq. 1.0
SOLUTION:
Assume the solution of:
0n
rn
nxcy .............................................................................................. eq. 1.1
Getting the first and second derivative of the solution to substitute in the given O.D.E.:
0
1
n
rn
n xrncy ........................................................................................................ eq. 1.2a
0
21n
rn
n xrnrncy ...................................................................................... eq. 1.2b
Substitute equations 1.2a and 1.2b to the given O.D.E.:
0331400
1
0
1
n
rn
n
n
rn
n
n
rn
n xcxrncxrnrnc ............................... eq. 1.3
Shift index, i.e. from n to n-1, hence, the second term changes:
033141
1
1
0
1
n
rn
n
n
rn
n xcxrnrnrnc ........................................ eq. 1.4a
03141
1
1
0
1
n
rn
n
n
rn
n xcxrnrnc ....................................................... eq. 1.4b
Investigating eq. 1.4b for n=0: (notice that the second term starts at n=1)
arbitrary. is ;014 :0nfor 01
0 cxrrcr
The indicial equation is: 014 rr . ........................................................................................... eq. 1.4c
The indicial roots are: 0r and 4
1r .
If n1, thus the second term of eq. 1.4b will appear, thus the recurrence relation is derived:
0340314
1
2
1
1
1
nn
rn
n
rn
n
crnrnc
xcxrnrnc
114
31
nn c
rnrnc ................................................................................................. eq. 1.5
MICHAEL B. BAYLON MS CE
COE5310 ASSIGNMENT #2 2
Investigating eq. 1.5, for even and odd integers of n, since there are arbitrary numbers, respectively:
For n1, r=0 For n1,
1
23
12
01
14
31
113
31
72
31
31
31
kk ckk
c
cc
cc
cc
1
23
12
01
14
31
313
31
29
31
15
31
kk ckk
c
cc
cc
cc
For the r=0 indicial root values, multiply all the terms of the LHS and equate to that of RHS:
1210
terms terms
1321141173321
33331111
k
kk
kk cccckk
ccccc
......................... eq. 1.6
Eq. 1.6 simplies to, (thus, the cn for r=0):
0
1
14!
31c
mk
ck
m
kk
k
........................................................................................................... eq. 1.7
For the r=1/4 root values, multiply all the terms of the LHS and equate to that of RHS:
1210
terms terms
1321141395321
33331111
k
kk
kk cccckk
ccccc
......................... eq. 1.8
Eq. 1.8 simplies to, (thus, the cn of the odd integers):
0
1
14!
31c
mk
ck
m
kk
k
......................................................................................................... eq. 1.9a
Changing variables from n to k, thus eq. 1.1 becomes:
0n
rn
nxcy ...................................................................................................................... eq. 1.10
Since the given O.D.E. is linear, by superposition, the general solution is: (refer to eq. 1.10)
4/10 rr yyy
0
41
0 k
k
k
k
k
k xcxcy .................................................................................................. eq. 1.11a
MICHAEL B. BAYLON MS CE
COE5310 ASSIGNMENT #2 3
Substituting equations 1.7b and 1.9b to eq. 1.11a:
0
41
0
1
0
0
1
14!
31
14!
31
k
k
k
m
kk
k
k
k
m
kk
xc
mk
xc
mk
y .................................................. eq. 1.11b
Rearranging eq. 1.11b:
0
41
1
0
0
1
0
14!
31
14!
31
k
k
k
m
kk
k
k
k
m
kk
x
mk
cx
mk
cy ................................................. eq. 1.11c
MICHAEL B. BAYLON MS CE
COE5310 ASSIGNMENT #2 4
2. By power series near a regular singular point
052152 yyxyx ........................................................................................................ eq. 2.0
SOLUTION:
Assume the solution of:
0n
rn
nxcy .............................................................................................. eq. 2.1
Getting the first and second derivative of the solution to substitute in the given O.D.E.:
0
1
n
rn
n xrncy .............................................................................................................. eq. 2.2a
0
21n
rn
n xrnrncy ............................................................................................. eq. 2.2b
Substitute equations 2.2a and 2.2b to eq. 2.0:
052151200
1
0
2
n
rn
n
n
rn
n
n
rn
n xcxrncxxrnrncx .................... eq. 2.3a
Distribute x from eq. 2.3a:
0510512000
1
n
rn
n
n
rn
n
n
rn
n xcxrncxrnrnrnc ............ eq. 2.3b
Combining similar x terms in eq. 2.3b:
01253200
12
n
rn
n
n
rn
n xrncxrnrnc ......................................... eq. 2.3c
Shift index, i.e. from n to n-1, hence, the second term changes:
01253221
1
1
0
1
n
rn
n
n
rn
n xrncxrnrnc ..................................... eq. 2.4a
Investigating eq. 2.4a for n=0: (notice that the second term starts at n=1)
arbitrary. is ;032 :0nfor 01
0 cxrrcr
The indicial equation is: 032 rr . .......................................................................................... eq. 2.4b
The indicial roots are: 0r and 2
3r .
If n1, thus the second term of eq. 2.4 will appear, thus the recurrence relation is derived:
01225322
0125322
1
1
1
1
rncrnrnc
xrncxrnrnc
nn
rn
n
rn
n
1322
1225
nn c
rnrn
rnc ...................................................................................................... eq. 2.5
MICHAEL B. BAYLON MS CE
COE5310 ASSIGNMENT #2 5
Investigating eq. 2.5, for even and odd integers of n, since there are arbitrary numbers, respectively:
For n1, r=0 For n1,
1
23
12
01
32
125
93
55
72
35
51
15
kk ckk
kc
cc
cc
cc
033
110
021
010
11
110
23
12
01
cc
cc
cc
For the r=0 indicial root values, multiply all the terms of the LHS and equate to that of RHS:
1210
terms
1321321212753321
1253155553
k
k
kk cccckkkk
kccccc
eq. 2.6
Eq. 2.6 simplies to, (thus, the cn for r=0):
03212!53
ckkk
ck
k
.................................................................................................. eq. 2.7
For the r=-3/2 root values, cn simplifies to:
0100 10; cccc ................................................................................................................................ eq. 2.8
Changing variables from n to k, thus eq. 2.1 becomes:
0n
rn
nxcy ........................................................................................................................ eq. 2.9
Since the given O.D.E. is linear, by superposition, the general solution is: (refer to eq. 2.9)
2/30 rr yyy
0
23
0 k
k
k
k
k
k xcxcy .................................................................................................. eq. 2.10a
Substituting equations 2.7b and 2.8b to eq. 2.10a:
21
02
3
0
0
0 103212!
53xcxcxc
kkky
k
kk
.................................................. eq. 2.10b
Rearranging eq. 1.11b:
xxcx
kkkcy
k
kk
1013212!
532
3
0
0
0
......................................................... eq. 2.10c
MICHAEL B. BAYLON MS CE
COE5310 ASSIGNMENT #2 6
3. By power series near a regular singular point
0212 2 xyyxyx ......................................................................................................... eq. 3.0
SOLUTION:
Assume the solution of:
0n
rn
nxcy .............................................................................................. eq. 3.1
Getting the first and second derivative of the solution to substitute in the given O.D.E.:
0
1
n
rn
n xrncy ........................................................................................................ eq. 3.2a
0
21n
rn
n xrnrncy ...................................................................................... eq. 3.2b
Substitute equations 3.2a and 3.2b to the given O.D.E.:
0211200
12
0
2
n
rn
n
n
rn
n
n
rn
n xcxxrncxxrnrncx ................ eq. 3.3
Shift index, i.e. from n to n-2, hence, the second term changes:
012320
1
0
12
n
rn
n
n
rn
n xrncxrnrnc ....................................... eq. 3.4a
05223222
1
2
0
1
n
rn
n
n
rn
n xrncxrnrnc .................................... eq. 3.4b
0322 1 rnn xrnrnc Investigating eq. 3.4b for n=0 and n=1: (notice that the second term starts at n=1)
arbitrary. is ;032 :0nfor 01
0 cxrrcr
arbitrary. is ;0121 :1nfor 11 cxrrcr
The indicial equations are:
014 rr . ....................................................................................................................... eq. 3.5
0121 rr .................................................................................................................. eq. 3.6 The indicial roots are:
For c0: 0r and 4
1r .
For c1: 1r and 2
1r .
The recurrence relation is (for n2):
2322
522
nn c
rnrn
rnc ............................................................................................... eq. 3.7
MICHAEL B. BAYLON MS CE
COE5310 ASSIGNMENT #2 7
For even coefficients, n2:
For r=0 For r=1/4
222
46
24
02
342
54
96
7
54
3
12
1
kk ckk
kc
cc
cc
cc
222
46
24
02
5818
984
1925
154
1117
74
39
14
kk ckk
kc
cc
cc
cc
For r=0:
22420222642 349512642
54731
kkk cccc
kk
kccccc
02
2
2!34!2
54
ckk
m
ck
k
mk
01
234
54
!2
1c
m
m
kc
k
mkk
For r=1/4:
22420
terms
22264258191131825179
9815714444
k
k
kk cccckk
kccccc
01
25818
984 c
mm
mc
k
m
k
k
For odd coefficients, n2:
For r=-1 For r=1/2
1212
57
35
13
342
54
96
7
54
3
12
1
kk ckk
kc
cc
cc
cc
1212
57
35
13
234
122
615
52
411
32
27
12
kk ckk
kc
cc
cc
cc
MICHAEL B. BAYLON MS CE
COE5310 ASSIGNMENT #2 8
For r=-1:
1275311212753 349512642
54731
kkk ccccc
kk
kccccc
12
2
12!34!2
54
ckk
m
ck
k
mk
1
1
1234
54
!2
1c
m
m
kc
k
mkk
For r=1/2:
12531
terms
121275326423415117
125312222
k
k
kk cccckk
kccccc
k
m
k cm
m
kc
1
11234
12
!
1
The solution is:
0n
rn
nxcy
0
21
0
41
0
1
0 k
k
k
k
k
k
k
k
k
k
k
k xcxcxcxcy
0
21
1
1
0
41
1
0
0
1
1
1
0 1
0
34
12
!
1
5818
984
34
54
!2
1
34
54
!2
1
k
kk
mk
kk
m
k
k
kk
mk
k
kk
mk
xm
m
kcx
mm
mc
xm
m
kcx
m
m
kcy