Solving a second order ode

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  • #26
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:
thats a nice one:biggrin::biggrin::biggrin:, you hit me under the belt lol...i need to upgrade my memory lol
 
  • #27
benorin
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Note: Sorry it took so long, been working on this post off and on all day. I just answered your earlier question about the solution to the recurrence relation, and I'm gonnna let you try to figure the rest out...

Given ##y"+\frac {1}{z} y' +\left(\tfrac{z^2-n^2}{z^2}\right) y=0##

Upon setting ##y = \sum_{k=0}^\infty a_{k}z^{k+c}##, we have:

$$\begin{gathered} \left(\tfrac{z^2-n^2}{z^2}\right) y =\left(\tfrac{z^2-n^2}{z^2}\right) \sum_{k=0}^\infty a_{k}z^{k+c} \\ = \sum_{k=0}^\infty a_{k}z^{k+c-2}(z^2 - n^2) \\ \end{gathered} $$

$$\frac {1}{z} y' = \tfrac{1}{z}\sum_{k=0}^\infty (k+c)a_{k}z^{k+c-1} = \sum_{k=0}^\infty (k+c)a_{k}z^{k+c-2}$$

$$y"=\sum_{k=0}^\infty (k+c)(k+c-1)a_kz^{k+c-2}$$

By the given DE, the sum of these three terms vanishes, to wit

$$\begin{gathered} y"+\tfrac{1}{z} y'+\left( \tfrac{z^2-n^2}{z^2}\right) y = 0 \\ \Rightarrow \sum_{k=0}^\infty [ (k+c)(k+c-1) + (k+c) - n^2 ]a_{k}z^{k+c-2}+ \sum_{k=0}^\infty a_{k}z^{k+c} =0 \\ \Rightarrow [c^2-n^2]a_{0}z^{c-2} +[(c+1)^2 -n^2]a_{1}z^{c-1} + \sum_{k=2}^\infty [ (k+c)^2 - n^2 ]a_{k}z^{k+c}+ \sum_{k=0}^\infty a_{k}z^{k+c} =0 \\ \Rightarrow [c^2-n^2]a_{0}z^{c-2} +[(c+1)^2 -n^2]a_{1}z^{c-1} + \sum_{k=0}^\infty \left\{ [ (k+c+2)^2-n^2)]a_{k+2} + a_{k}\right\} z^{k+c} = 0 \\ \end{gathered}$$

All the coefficients of powers of ##z## must vanish, and ##a_0\neq 0\Rightarrow c^2-n^2=0 \Rightarrow c = \pm n ##. Plugging these roots into the next coefficient gives, ##[(\pm n+1) ^2 -n^2]a_1=0\Rightarrow (1\pm 2n)a_1=0## but ##n\in\mathbb{Z}## so ##a_1=0##. Next plug the roots into the ##\text{k}^{th}## coefficient:

$$ [ (k\pm n+2)^2-n^2)]a_{k+2} + a_{k}=0\Rightarrow (k\pm n+2+n)(k\pm n+2-n)a_{k+2} = -a_{k}\Rightarrow a_{k+2}=\tfrac{-a_k}{(k+2)(k\pm 2n+2)}$$

From ##a_0\neq 0## and ##a_1=0## this recurrence simplifies to

$$a_{2k+2}=\tfrac{-a_{2k}}{(2k+2)(2k\pm 2n+2)}=\tfrac{-a_{2k}}{4(k+1)(k\pm n+1)}\wedge a_{2k+1}=0\forall k\in\mathbb{N}$$

Let's determine the pattern for ##a_{2k+2}## for small values of ##k## and then try to guess the solution to the recurrence:

$$\boxed{k=0}:\quad a_2=\tfrac{-a_{0}}{4\cdot 1(\pm n+1)}$$

$$\boxed{k=1}:\quad a_4=\tfrac{-a_{2}}{4\cdot 2(\pm n+2)}=\tfrac{(-1)^2 a_{0}}{4^2\cdot 2\cdot 1(\pm n+2)(\pm n+1)} $$

$$\boxed{k=2}:\quad a_6=\tfrac{-a_{4}}{4\cdot 3(\pm n+3)}=\tfrac{(-1)^3 a_{0}}{4^3\cdot 3\cdot 2\cdot 1(\pm n+3)(\pm n+2)(\pm n+1)} $$

My guess:

$$a_{2k+2}=\tfrac{(-1)^{k+1} a_{0}}{4^{k+1}(k+1)! \prod_{j=1}^{k+1}(\pm n+j)} $$

Whelp, my post has got you to this point

FrobeniusMethodPFCalcHW.png


in the Wikipedia article, I ad-libbed bit along the way to take the easier route, note we've already plugged in the roots, but they differ by an integer so you'll need to read on past that point to see how to derive the other linearly independent solution (after you determine which of the roots solves whichever equation (think it was a DE. Good luck!
 
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  • #28
chwala
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Note: Sorry it took so long, been working on this post off and on all day. I just answered your earlier question about the solution to the recurrence relation, and I'm gonnna let you try to figure the rest out...

Given ##y"+\frac {1}{z} y' +\left(\tfrac{z^2-n^2}{z^2}\right) y=0##

Upon setting ##y = \sum_{k=0}^\infty a_{k}z^{k+c}##, we have:

$$\begin{gathered} \left(\tfrac{z^2-n^2}{z^2}\right) y =\left(\tfrac{z^2-n^2}{z^2}\right) \sum_{k=0}^\infty a_{k}z^{k+c} \\ = \sum_{k=0}^\infty a_{k}z^{k+c-2}(z^2 - n^2) \\ \end{gathered} $$

$$\frac {1}{z} y' = \tfrac{1}{z}\sum_{k=0}^\infty (k+c)a_{k}z^{k+c-1} = \sum_{k=0}^\infty (k+c)a_{k}z^{k+c-2}$$

$$y"=\sum_{k=0}^\infty (k+c)(k+c-1)a_kz^{k+c-2}$$

By the given DE, the sum of these three terms vanishes, to wit

$$\begin{gathered} y"+\tfrac{1}{z} y'+\left( \tfrac{z^2-n^2}{z^2}\right) y = 0 \\ \Rightarrow \sum_{k=0}^\infty [ (k+c)(k+c-1) + (k+c) - n^2 ]a_{k}z^{k+c-2}+ \sum_{k=0}^\infty a_{k}z^{k+c} =0 \\ \Rightarrow [c^2-n^2]a_{0}z^{c-2} +[(c+1)^2 -n^2]a_{1}z^{c-1} + \sum_{k=2}^\infty [ (k+c)^2 - n^2 ]a_{k}z^{k+c}+ \sum_{k=0}^\infty a_{k}z^{k+c} =0 \\ \Rightarrow [c^2-n^2]a_{0}z^{c-2} +[(c+1)^2 -n^2]a_{1}z^{c-1} + \sum_{k=0}^\infty \left\{ [ (k+c+2)^2-n^2)]a_{k+2} + a_{k}\right\} z^{k+c} = 0 \\ \end{gathered}$$

All the coefficients of powers of ##z## must vanish, and ##a_0\neq 0\Rightarrow c^2-n^2=0 \Rightarrow c = \pm n ##. Plugging these roots into the next coefficient gives, ##[(\pm n+1) ^2 -n^2]a_1=0\Rightarrow (1\pm 2n)a_1=0## but ##n\in\mathbb{Z}## so ##a_1=0##. Next plug the roots into the ##\text{k}^{th}## coefficient:

$$ [ (k\pm n+2)^2-n^2)]a_{k+2} + a_{k}=0\Rightarrow (k\pm n+2+n)(k\pm n+2-n)a_{k+2} = -a_{k}\Rightarrow a_{k+2}=\tfrac{-a_k}{(k+2)(k\pm 2n+2)}$$

From ##a_0\neq 0## and ##a_1=0## this recurrence simplifies to

$$a_{2k+2}=\tfrac{-a_{2k}}{(2k+2)(2k\pm 2n+2)}=\tfrac{-a_{2k}}{4(k+1)(k\pm n+1)}\wedge a_{2k+1}=0\forall k\in\mathbb{N}$$

Let's determine the pattern for ##a_{2k+2}## for small values of ##k## and then try to guess the solution to the recurrence:

$$\boxed{k=0}:\quad a_2=\tfrac{-a_{0}}{4\cdot 1(\pm n+1)}$$

$$\boxed{k=1}:\quad a_4=\tfrac{-a_{2}}{4\cdot 2(\pm n+2)}=\tfrac{(-1)^2 a_{0}}{4^2\cdot 2\cdot 1(\pm n+2)(\pm n+1)} $$

$$\boxed{k=2}:\quad a_6=\tfrac{-a_{4}}{4\cdot 3(\pm n+3)}=\tfrac{(-1)^3 a_{0}}{4^3\cdot 3\cdot 2\cdot 1(\pm n+3)(\pm n+2)(\pm n+1)} $$

My guess:

$$a_{2k+2}=\tfrac{(-1)^{k+1} a_{0}}{4^{k+1}(k+1)! \prod_{j=1}^{k+1}(\pm n+j)} $$

Whelp, my post has got you to this point

View attachment 262799

in the Wikipedia article, I ad-libbed bit along the way to take the easier route, note we've already plugged in the roots, but they differ by an integer so you'll need to read on past that point to see how to derive the other linearly independent solution (after you determine which of the roots solves whichever equation (think it was a DE. Good luck!
Thanks for your insight, i know how to work on the rest of the steps leading to the solution....i keep on repeating this statement. My interest was only on how ##a_1=0##, ...only this part...my interest on this question is only and only on that part of the question...
 
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  • #29
benorin
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All the coefficients of powers of ##z## must vanish, and ##a_0\neq 0\Rightarrow c^2-n^2=0 \Rightarrow c = \pm n ##. Plugging these roots into the next coefficient gives, ##[(\pm n+1) ^2 -n^2]a_1=0\Rightarrow (1\pm 2n)a_1=0## but ##n\in\mathbb{Z}## so ##a_1=0##.
Like it says, because ##1\pm 2n\neq 0\forall n\in\mathbb{N}## hence we must have ##a_1=0## because there is no other way that the coefficient of ##z^{c-1}## can be zero.
 
  • #30
chwala
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yeah yeah and i was able to figure that out in post 25. Thank you sir.
 
  • #31
haruspex
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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.
I still do not understand your question here, and it looks to me that no one else has understood it either, which is why you are not getting a satisfactory answer.
What do you mean "when k=1"? As I wrote in post #23, that is not a 'case'.
I also suggested dropping c, so let's do that. You have ##z^2y''+zy'+y(z^2-n^2)=0## and ##y=\Sigma_{k=0}a_kz^k##.

##z^0: a_0n^2=0##
##z^1: a_1(1-n^2)=0##
##z^k, k>1: a_k(k^2-n^2)+a_{k-2}=0##

From this we get to deduce:
##a_n## can be chosen arbitrarily.
##a_k=0## if either k<n or k+n is odd.
##a_{n+2}=-\frac{a_n}{4n+4}##
Etc.

If I have not answered your question, please rephrase it in terms of that analysis.
 
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  • #32
chwala
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thanks for your asking haruspex, i asked in post 1 on how ##a_1=0##, that was my question, ...and i now know why as shown in my post 25 and also more insight has been given by benorin in post 28. My question has been answered fully sir.
 
  • #33
chwala
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I still do not understand your question here, and it looks to me that no one else has understood it either, which is why you are not getting a satisfactory answer.
What do you mean "when k=1"? As I wrote in post #23, that is not a 'case'.
I also suggested dropping c, so let's do that. You have ##z^2y''+zy'+y(z^2-n^2)=0## and ##y=\Sigma_{k=0}a_kz^k##.

##z^0: a_0n^2=0##
##z^1: a_1(1-n^2)=0##
##z^k, k>1: a_k(k^2-n^2)+a_{k-2}=0##

From this we get to deduce:
##a_n## can be chosen arbitrarily.
##a_k=0## if either k<n or k+n is odd.
##a_{n+2}=-\frac{a_n}{4n+4}##
Etc.

If I have not answered your question, please rephrase it in terms of that analysis.
Thanks haruspex, this is more clearer
 
  • #34
benorin
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@haruspex the problem stated to use the Frobenius method which is where the ##c## comes from.
 
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