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Solving a second order ode

  • Thread starter chwala
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  • #1
chwala
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Homework Statement:

##y"+y'\frac {1}{z}+y[\frac {z^2-n^2}{z^2}]=0##

Relevant Equations:

power series
let ##y= \sum_{k=-∞}^\infty a_kz^{k+c}##
##y'=\sum_{k=-∞}^\infty (k+c)a_kz^{k+c-1}##
##y"=\sum_{k=-∞}^\infty (k+c)(k+c-1)a_kz^{k+c-2}##
therefore,
##y"+y'\frac {1}{z}+y[\frac {z^2-n^2}{z^2}]=0## =##[\sum_{k=-∞}^\infty [(k+c)^2-n^2)]a_k + a_k-2]z^{k+c} ##
it follows that,
##(k+c)^2-n^2)]a_k + a_k-2=0##
if ##k=0, →c^2-n^2=0, a_0≠0##
##c^2=n^2, →c=±n ##(roots)
now considering case 1, if ##c=n##, and given k=1, then ##a_1=0##...this is the part that i am not getting.
i am ok with the rest of the steps...
 

Answers and Replies

  • #2
BvU
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therefore,
$$y"+y'\frac {1}{z}+y\Biggl [\frac {z^2-n^2}{z^2}\Biggr ]=0 $$ $$\Leftrightarrow \ (?) \\ \sum_{k=-∞}^\infty \Biggl [ (k+c)^2-n^2) a_k + a_k-2\Biggr ] z^{k+c} = 0 \\ \mathstrut \\ \mathstrut \\ $$

Doesn't seem right. I suppose you mean ##a_{k-2}##, not ##a_k - 2##. But: where is ##a_{k-1}## ?


[edit] can't get the scroll bar in the quote ! How come ?
 
  • #3
chwala
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yeah, it is supposed to be ##a_{k-2}##, i do not have ##a_{k-1} ## in my working, unless of course you want me to write all my steps...
 
  • #4
BvU
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do not have ##\ a_{k-1}\ ## in my working
I noticed that and I wondered what made the ##\ y'\over z\ ## series disappear :rolleyes: ?
 
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  • #5
BvU
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By the way, shouldn't it be ##\ a_{k+2}\ ## instead of ##\ a_{k-2}\ ## ?
 
  • #6
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And never mind my blabbing about ##\ y'\over z\ ##, it actually seems to fall out.
 
  • #7
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:smile: I suppose another o:) : Never mind my blabbing about ##a_{k+2}## either. I should retreat and only speak up after some sensible thinking :nb).
Sorry, Chwala !

By the way, I wonder why you sum starting from ##\ -\infty\ ##. Most analyses start at 0 !
 
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  • #8
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Took me a while to recognize your equation :cool: . Do you know what I mean ?
 
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  • #9
chwala
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Bvu sorry i was in a zoom meeting. I just used the summing formula from my university notes, ....i have ##a_{k-2}## and not ##a_{k+2}##
 
  • #10
chwala
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And never mind my blabbing about ##\ y'\over z\ ##, it actually seems to fall out.
no worries, we are a community i really appreciate your input...
 
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  • #11
Dr Transport
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The best way to go about this is to multiply the equation by [itex] z^2[/itex] to eliminate any division first. The way you went about it is convoluted and not posting the intermediate steps isn't helping.
 
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  • #12
chwala
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I can solve the problem to the end....the only area that i do not understand is where i indicated, case 1, ...##k=1, a_1=0## this is the only area that i need help...how is this valid? (the reference that i am using is my undergraduate university notes, that i am trying to refresh on...
 
  • #13
chwala
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can we say, from ##[(k+c)^2-n^2)]a_k+a_{k-2}##=0, then if ## c=n## & ##k=1## then,
##(1+2n)a_1+a_{-1}=0##
##a_1+2na_1+a_{-1}=0##
is this correct thinking...
 
  • #14
chwala
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The best way to go about this is to multiply the equation by [itex] z^2[/itex] to eliminate any division first. The way you went about it is convoluted and not posting the intermediate steps isn't helping.
and can we solve this problem using laplace?
 
  • #15
Dr Transport
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Series solution is still a valid way to go about it, but I as taught to not be dividing when using a series solution. It looks odd to me.
 
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  • #16
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Took me a while to recognize your equation :cool: . Do you know what I mean ?
the reference that i am using is my undergraduate university notes
I overlooked that one. Does it say there that you are usibg the Frobenius method to solve the Bessel equation ? (other link , link, solution with pictures, a whole chapter, ...).

And you should really start your power series from ##k=0##, not from ##k=-\infty##.

All this with just a little googling and without sitting down and seriously working out your core question ! Lazy me ... :wink:

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust :biggrin:
 
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  • #17
vela
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can we say, from ##[(k+c)^2-n^2)]a_k+a_{k-2}##=0, then if ## c=n## & ##k=1## then,
##(1+2n)a_1+a_{-1}=0##
##a_1+2na_1+a_{-1}=0##
is this correct thinking...
The problem is related to what the others have mentioned already a few times. The sum starts from ##k=0##, so the coefficient of the ##k=1## term is ##(1+2n)a_1##. There's no ##a_{-1}##.
 
  • #18
benorin
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If ##a_k## is only defined for ##k=0,1,2,\ldots## don't you think that for the equation ##\left[ (k+c)^2-n^2\right] a_k+a_{k-2}=0## the only admissible values of ##k## are ##k=2,3,4,\ldots## ? For ##k=2,\text{ and }c=n## this would lead to ##a_2=\tfrac{a_0}{4(n+1)}## and [omitting steps] in general

$$a_{2k}=\tfrac{(-1)^k a_0}{4^k(k!)^2 \binom{n+k}{k} }, \quad k\in\mathbb{Z}^+$$
and
$$a_{2k+1}=\tfrac{(-1)^k a_{1}}{\prod_{j=1}^{k}\left[ (2j+1) (2n+2j+1)\right]}, \quad k\in\mathbb{Z}^+$$

Note: ##a_{2k+1}## is "not pretty" in terms of binomial coefficients, but probably looks ok in terms of double factorials. Just need initial conditions now?

Edit: I messed up the formula for ##a_{2k}## first time around, fixed it!
 
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  • #19
chwala
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I overlooked that one. Does it say there that you are usibg the Frobenius method to solve the Bessel equation ? (other link , link, solution with pictures, a whole chapter, ...).

And you should really start your power series from ##k=0##, not from ##k=-\infty##.

All this with just a little googling and without sitting down and seriously working out your core question ! Lazy me ... :wink:

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust :biggrin:
i guess the lecturer must have mentioned the method but i probably did not indicate the method on my working...
 
  • #20
chwala
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If ##a_k## is only defined for ##k=0,1,2,\ldots## don't you think that for the equation ##\left[ (k+c)^2-n^2\right] a_k+a_{k-2}=0## the only admissible values of ##k## are ##k=2,3,4,\ldots## ? For ##k=2,\text{ and }c=n## this would lead to ##a_2=\tfrac{a_0}{4(n+1)}## and [omitting steps] in general

$$a_{2k}=\tfrac{(-1)^k a_0}{4^k(k!)^2 \binom{n+k}{k} }, \quad k\in\mathbb{Z}^+$$
and
$$a_{2k+1}=\tfrac{(-1)^k a_{1}}{\prod_{j=1}^{k}\left[ (2j+1) (2n+2j+1)\right]}, \quad k\in\mathbb{Z}^+$$

Note: ##a_{2k+1}## is "not pretty" in terms of binomial coefficients, but probably looks ok in terms of double factorials. Just need initial conditions now?

Edit: I messed up the formula for ##a_{2k}## first time around, fixed it!
i am quite fine with the values ##k=2,3,4##... and how to find the solution. At the end we shall have a series solution of form ##y=y_1 + y_2##, alternating for ##k=2,4,6##.....and ##k=3,5,7##...my interest on this question is on the value of ##k=1##, i do not understand how ##a_1=0## but of course i may be missing something from the question as indicated by the responses...
 
  • #21
chwala
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The problem is related to what the others have mentioned already a few times. The sum starts from ##k=0##, so the coefficient of the ##k=1## term is ##(1+2n)a_1##. There's no ##a_{-1}##.
I think i may be wrong with my limits, ##k## should start from ##0## and not ##-∞##, as indicated...
 
  • #22
chwala
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...now considering case 1, if ##c=n##, and given ##k=1##, then ##a_1=0##...this is the part that i am not getting.
i am ok with the rest of the steps...

This is the part that i need understanding ...lets assume as you have put it that ##k## starts from ##0→+∞## then how is it that when ##k=1##, that ##a_1=0##
or the other possibility is that my statement (my notes) is/are incorrect.
 
  • #23
haruspex
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...now considering case 1, if ##c=n##, and given ##k=1##, then ##a_1=0##...this is the part that i am not getting.
i am ok with the rest of the steps...

This is the part that i need understanding ...lets assume as you have put it that ##k## starts from ##0→+∞## then how is it that when ##k=1##, that ##a_1=0##
or the other possibility is that my statement (my notes) is/are incorrect.
Not sure that introducing c helps. Why not just write ##y=\Sigma_0a_kz^k## and allow that a0 and so could be zero?
Then you can write equations for the z0 term, the z1 term, and a general equation for zk, k>1.
The cases to be considered are then n=0, n=1, n=2...; a particular value of k is not a 'case'.
You should find that ak=0 for k<n, an is unconstrained, and for k>n ak is some (varying) multiple of ak-2
 
  • #24
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1589374007299.png
 
  • #25
chwala
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i think going with the above attached example, i may have an idea as to why ##a_1=0##, by using the indicial equations (14) and (15) as my comparison,
in my case, i have the indicial equation as,
##[(k+c)^2-n^2)]a_k + a_{k-2}=0## for some reason they chose to use the first part of the indicial equation...
when ##k=0## it follows that ##c=±n##, we may as well choose to ignore the ##-c## and go for ##c=n##
now for ##k=1##, we shall have;
##[(k+c)^2-n^2)]a_k=0## just like in equation(16) of attachment... and ##c=n## therefore,
##[(1+n)^2-n^2)]a_1=0##
##(1+2n)a_1=0## avoiding special case ##n=-0.5##, results into ##a_1=0##
 

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