Frobenius method examples, page 1 of 3

Examples on Frobenius methodExample 1.

B C #B #B C B #B # C œ !# ww # w #a b a bSolution.At , the equation has a regular singular point:B œ !

C # C " C œ !ww w# # #B B B

ˆ ‰ ˆ ‰#

By the recipe,

C œ + B!8œ!

_

88<

C œ 8 < + Bw 8<"

8œ!

_

8!a b C œ 8 < 8 < " + Bww 8<#

8œ!

_

8!a ba bInto the equation, aligning the powers:

B C œ 8 < 8 < " + B# ww 8<

8œ!

_

8!a ba b #BC œ # 8 < + Bw 8<

8œ!

_

8! a b #B C œ # 8 " < + B# w 8<

8œ"

_

8"! a b B C œ + B# 8<

8œ#

_

8#! #BC œ #+ B!

8œ"

_

8"8<

#C œ #+ B!8œ!

_

88<

Sum up (infinite sum from , separate terms for )8 œ # 8 œ !ß 8 œ "

!c da b a ba ba b a b a b8œ#

_

8 8" 88<8 < 8 < " # 8 < # + # 8 " < # + + B

+c d c da b a b a b< < " + #<+ #+ B " < <+ # " < + #<+ #+ #+ B! ! ! " " ! ! "< <"

œ !

Collect the terms:!c da ba b a ba b8œ#

_

8 8" 8#8<8 < " 8 < # + % # 8 < + + B

< " < # + B < " <+ # #< + B œ !a ba b a ba b a b! " !< <"

Taking , guaranteed to give a series solution, we get< œ #+ œ +" !ß and the recursion is + œ8

#8+ +8 8"8" 8#a b

It is a homogeneous problem, so we are free to choose .+ œ "!

Anticipating division by factorials, we might change the variable to + œ8

,8" x

8a b

Frobenius method examples, page 2 of 3

so that the recursion becomes, after simplification, , œ "ß , œ #ß! "

, œ #, ,8 8" 8

The solution of the recursion is easy to find: 1, so, œ 8 8

+ œ œ88" "8" x 8xa b

and the solution of the ODE is

C B œ œ B / Þ"8œ!

_B8x

# Ba b ! 8#

Second solution:

where C œ C .B [ œ # .B œ B /# "[ #C B

# #B' '"# expŠ ‹ˆ ‰

C œ B / .B œ B / .B œ B/## B # B BB / "

B / B' # #B

% #B #'

The general solution is then C œ - B / - B/" #

# B B

______________________________________________________________Example 2. This time a formula for the solution was not found by computer algebra software,so I will settle for a more modest task:Find the first three terms of the Frobenius solution around forB œ !

BC / C œ !ww B

Solution. For easier handling of the series, write

B C B/ C œ !# ww B

Again,

C œ + B !8œ!

_

88<

C œw !a b8œ!

_

88<"8 < + B

C œww !a ba b8œ!

_

88<#8 < 8 < " + B

B C œ 8 < 8 < " + B# ww 8<

8œ!

_

8!a ba band we take only the first three terms:B C œ < < " + B < " <+ B < # < " + B á# ww < <" <#

! " # a b a b a ba bB/ C œ B á + B + B + B áB < <" <#B B

"x #x ! " # Š ‹a b# $

œ + B + + B + + + B á! " ! # " !<" <# <$" " "

"x "x #xˆ ‰ ˆ ‰

total:! œ < < " + B < " <+ + B < # < " + + + B áa b a b a ba b a ba b! " ! # " !

< <" <#

Since actually we can take the larger of the integer values of is + Á !ß + œ "ß < < œ "Þ! !

Frobenius method examples, page 3 of 3

The equations simplify to + œ "ß!

#+ + œ !" !

'+ + + œ !# " !

with solution + œ "ß + œ ß + œ ! " #

" "# "#

and the series starts with C œ B B B á" "

# "## $

Remark. The computer returned some more coefficients: + œ ß + œ ß + œ$ % &

" #$ "*("%% #))! )'%!!

and a pattern is not clear.

Second solution?

By the formula, C œ C .Bß# "[C

'"#

and since the coefficient at in the differential equation is 0, we can takeCw

[ œ !.B œ "exp ' C œ C .B# "

"

B " B B á'

# #" "# "#

#ˆ ‰ œ B " B B á B á .Bˆ ‰ ˆ ‰" " " " "" $

# "# B B "# %# '

#

œ B " B B á B B B ሠ‰ˆ ‰" " " "" $# "# B "# )

# #ln œ C B " B B á"

"#

#ln ˆ ‰The other solution does have a logarithmic term.