Solving a second order ode

[chwala]

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...

 

[BvU]

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 ?

 

[chwala]

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...

 

[BvU]

do not have $\ a_{k-1}\$ in my working

I noticed that and I wondered what made the $\ y'\over z\$ series disappear ?

 

[BvU]

By the way, shouldn't it be $\ a_{k+2}\$ instead of $\ a_{k-2}\$ ?

 

[BvU]

And never mind my blabbing about $\ y'\over z\$, it actually seems to fall out.

 

[BvU]

I suppose another : Never mind my blabbing about $a_{k+2}$ either. I should retreat and only speak up after some sensible thinking .
Sorry, Chwala !

By the way, I wonder why you sum starting from $\ -\infty\$. Most analyses start at 0 !

 

[BvU]

Took me a while to recognize your equation . Do you know what I mean ?

 

[chwala]

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 $a_{k+2}$

 

[chwala]

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...

 

[Dr Transport]

The best way to go about this is to multiply the equation by $z^2$ to eliminate any division first. The way you went about it is convoluted and not posting the intermediate steps isn't helping.

 

[chwala]

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...

 

[chwala]

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...

 

[chwala]

The best way to go about this is to multiply the equation by $z^2$ 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?

 

[Dr Transport]

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.

 

[BvU]

Took me a while to recognize your equation . 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 ...

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust

 

[vela]

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 $a_{-1}$.

 

[benorin]

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!

 

[chwala]

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 ...

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust

i guess the lecturer must have mentioned the method but i probably did not indicate the method on my working...

 

[chwala]

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...

 

[chwala]

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...

 

[chwala]

...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.

 

[haruspex]

...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 z^k, 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 a_k=0 for k<n, a_n is unconstrained, and for k>n a_k is some (varying) multiple of a_k-2

 

[chwala]

 

[chwala]

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 $a_1=0$

 

[chwala]

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 ...

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust

thats a nice one, you hit me under the belt lol...i need to upgrade my memory lol

 

[benorin]

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

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!

 

[chwala]

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...

 

[benorin]

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.

 

[chwala]

yeah yeah and i was able to figure that out in post 25. Thank you sir.

 

[haruspex]

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.

 

[chwala]

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.

 

[chwala]

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

 

[benorin]

@haruspex the problem stated to use the Frobenius method which is where the $c$ comes from.