Mathematical methods of Physics, ODE

[fluidistic]

Homework Statement

For what values of K does the DE $xy''-2xy'+(K-3x)y=0$ (1) has a bounded solution in $(0, \infty)$?

Homework Equations

Not so sure, Frobenius method maybe.

The Attempt at a Solution

First, I check what happens when x tends to infinity. I see that the DE behaves like $\phi ''-2 \phi '-3 \phi =0$. Where $\phi (x)=c_1e^{3x}+c_2e^{-x}$. I checked out and this indeed satisfies the DE (2).
I propose a solution to (1) of the form $y(x)=f(x) \phi (x)$. So I calculated y' and y'' and replaced all this into the DE (1) and after a lot of algebra I reached that the DE (1) is equivalent to the DE $f''(x)+4f'(x)+f(x) \left ( \frac{Kc_1e^{3x}}{x} \right ) \cdot \left ( \frac{1}{c_1e^{3x}+c_2e^{-x}} \right )=0$ (3) where these $c_1$ and $c_2$ differs from the ones given in the expression for $\phi (x)$ but this doesn't matter, they are constants. Now I must say I'm not really eager to use Frobenius's method on the DE (3). Mainly due to the denominator. I don't really know how to proceed... any help will be appreciated as usual.

 

[LCKurtz]

Not sure if it will help, but surely you wouldn't want to include the e^3x in your calculations because it isn't bounded. I would try just the e^-x part, but no guarantees; I haven't tried it myself.

 

[fluidistic]

Not sure if it will help, but surely you wouldn't want to include the e^3x in your calculations because it isn't bounded. I would try just the e^-x part, but no guarantees; I haven't tried it myself.

Whoops, you're absolutely right! I forgot about this.

 

[fluidistic]

I progressed on the problem.
I skip the algebra, and I must solve $f''(x)-4f'(x)+\frac{K}{x}f(x)=0$. I use Frobenius's method on it and this gave me the indicial equation s=0 (that I discard later) and s=1.
I reach the recurrence relation $a_{n+1}=\frac{-a_n [4(n+s)+K]}{(n+2)(n+1)}$.
The values of K they ask for is when $a_{n+1}=0$ and $a_n \neq 0$. So $K=-4n-4$ with n going from 0 to infinity. (This is, I hope, the answer to the problem).
However say I want to find the general terms $a_n$ in $\sum _{n=0}^{\infty} a_nx^{n+1}$.
I'm currently trying to find a pattern, by setting $a_0=1$ between the $a_n$'s but I'm not getting sucessful in it.
$a_0=1$, $a_1=-\frac{4+K}{2}$, $a_2=\left ( \frac{4+K}{2} \right )\left ( \frac{8+K}{3\cdot 2} \right )$, $a_3=-\frac{(4+K)}{2} \left ( \frac{8+K}{3\cdot 2} \right ) \left ( \frac{12+K}{4\cdot 3} \right )$.

So far I realize that $a_n=\frac{(-1)^n}{n!(n+1)!}$ multiplied by something that I'm not sure (and don't find yet). Maybe something similar to $\frac{(4n+K)!}{(4n+K-1)(4n+K-2)(4n+K-3)}$. Can someone help me on this task?

 

[vela]

Did you perhaps make a sign error? I find the recursion relation to be
$$a_{n+1} = -\frac{K-4(n+1)}{(n+1)(n+2)} a_n.$$

 

[fluidistic]

Did you perhaps make a sign error? I find the recursion relation to be
$$a_{n+1} = -\frac{K-4(n+1)}{(n+1)(n+2)} a_n.$$

Yes I think I did, though now I get a different answer from ours.
From $f''(x)-4f'(x)+\frac{K}{x}f(x)=0$ and $f(x)=\sum_{n=0}^{\infty}a_n x^{n+s}$, I reach that it's equivalent to $\sum_{n=-1}^{\infty}a_{n+1}(n+s)(n+s+1) x^{n+s-2}-4\sum_{n=0}^{\infty}a_n(n+s) x^{n+s-1}-K \sum_{n=0}^{\infty}a_n x^{n+s-1}=0$
$\Rightarrow a_0s(s-1)x^{s-2}+\sum_{n=0}^{\infty}a_n x^{n+s-1} [-4(n+s)-K]+a_{n+1}(n+s+1)(n+s)x^{n+s-1}=0$.
So $-a_n [4(n+s)+K]+a_{n+1}(n+s+1)(n+s)=0\Rightarrow a_{n+1}=\frac{a_n[4(n+s)+K]}{(n+s+1)(n+s)}$.

Mod note: fixed LaTeX

 

[vela]

Yes I think I did, though now I get a different answer from ours.
From $f''(x)-4f'(x)+\frac{K}{x}f(x)=0$ and $f(x)=\sum_{n=0}^{\infty}a_n x^{n+s}$, I reach that it's equivalent to $$\sum_{n=-1}^{\infty}a_{n+1}(n+s)(n+s+1) x^{n+s-2}-4\sum_{n=0}^{\infty}a_n(n+s) x^{n+s-1}-K \sum_{n=0}^{\infty}a_n x^{n+s-1}=0$$

You changed the sign on the K term, so you should get
$$a_0s(s-1)x^{s-2}+\sum_{n=0}^{\infty}a_n x^{n+s-1} [K-4(n+s)]+a_{n+1}(n+s+1)(n+s)x^{n+s-1}=0.$$Then you made one more sign error which got rid of the alternating sign.

 

[fluidistic]

You are right, I made a typo when I copied the DE when I turned the page of my draft for the sign of K. I now reach your result.
I'm trying to find a pattern by calculating the first terms, assuming $a_0=1$.
$a_1=-\frac{(K-4)}{2}, a_2=\frac{(K-4)(K-8)}{2\cdot 3\cdot 2}$, etc.
My attempt is $a_n=\frac{(-1)^n(K-4)(K-8)...(K-4n)}{n!(n+1)!}$ but this doesn't work for $a_0$ since I get K instead of 1.

 

[vela]

If you require that the series terminates, you know K has the form K=4m where m=1, 2, 3, ...

 

[fluidistic]

If you require that the series terminates, you know K has the form K=4m where m=1, 2, 3, ...

Yes indeed. $K-4(n+1)=0 \Rightarrow k=4(n+1)$ with n=0, 1, etc. which is equivalent to K=4m for m=1, 2, etc.
I was just trying to do more than what they asked me. I wanted to express $a_n$ in terms of n only rather than $a_{n+1}$ in terms of $a_n$. I don't think it really matters anyway, but since I've a weakness in doing it I wanted to eradicate it.

 

[vela]

I didn't do this calculation out carefully yesterday so I may have made a mistake, but I think you can use the fact that K=4m to express a_n in terms of n and m.