Mathematical methods of Physics, ODE

1. Dec 8, 2011

fluidistic

1. The problem statement, all variables and given/known data
I must find the constant K such that $y''-\left ( \frac{1}{4}+\frac{K}{x} \right )y=0$ for x>0 has a non trivial solution that is worth 0 when x tends to 0 and when x tends to infinity.

2. Relevant equations
Frobenius method.

3. The attempt at a solution
I proposed a solution of the form $y(x)=\sum_{n=0}^{\infty}a_nx^{n+s}$.
This gave me $y''(x)=\sum _{n=2}^\infty (n+s)(n+s-1)a_nx^{n+s-2}$.
Plugging y and y'' into the DE I get that $\sum _{n=2}^\infty (n+s)(n+s-1)a_nx^{n+s-2} - \frac{1}{4}\sum_{n=0}^{\infty}a_nx^{n+s}+K\sum_{n=0}^{\infty}a_nx^{n+s-1}=0$.
Now I must equate the powers of x inside each series in order to put all the arguments of the series under a single series.
So let's say I want to have as common power of x, n+s.
This gives me $\sum_{n=0}^{\infty}a_{n+2}x^{n+s} (n+s+2)(n+s+1)-\frac{1}{4}\sum_{n=0}^{\infty}a_nx^{n+s}+K\sum_{n=1}^{\infty}a_{n+1}x^{n+s}=0$.
I separate the first term of the series in order to get the indicial equation.
Hence $Ka_0 x^{s-1}+\sum_{n=0}^{\infty}x^{n+s} [a_{n+2}(n+s+2)(n+s+1)-\frac{1}{4}a_n+Ka_{n+1}]=0$.
So that the indicial equation is $Ka_0=0$. I think this implies that $a_0=0$.
Setting the argument of the infinite series equal to 0 gives me the recurrence relation $a_{n+2}(n+s+2)(n+s+1)-\frac{a_n}{4}+Ka_{n+1}=0$.
Now I know I should get all terms in function of $a_1$, but I don't know how to mathematically show this.
For example if I calculate $a_2$, I get that it's worth $-\frac{Ka_1}{(s+1)(s+2)}$. I'd get $a_3$ in terms of $a_2$ and $a_1$ and thus of $a_1$ alone, etc.
So how can I proceed further?

EDIT: Nevermind, I've made lots of errors, I'm going to redo all.

Last edited: Dec 8, 2011
2. Dec 8, 2011

fluidistic

The indicial equation is $x^{s-2}a_0 s(s-1)=0$. Now I know that $a_0$ can't be 0 so that s(s+1)=0 (indicial equation) which gives $s=0$ or $s=1$.
I also get $a_1=-\frac{Ka_0}{s(s+1)}$.
And again a non beautiful recurrence relation: $a_{n+2}= \left ( \frac{a_n}{4}-Ka_{n+1} \right ) \frac{1}{(n+s+1)(n+s+2)}$.
I'm still confronted to the same problem as before, how to express $a_{n+2}$ in terms of $a_{n}$ only.
Hmm now that I think about it, s can't be 0 otherwise $a_1$ isn't definied. So that s=1.

I guess I have to find a patern by calculating a few first coefficient $a_n$. I'm trying this, but this doesn't come easily to me.

Last edited: Dec 8, 2011
3. Dec 8, 2011

vela

Staff Emeritus
When you calculate y'', the index should still start at n=0. Keep in mind that s isn't necessarily an integer, so you can't assume differentiation will cause certain terms to vanish.

Try using the approach used on

http://quantummechanics.ucsd.edu/ph130a/130_notes/node236.html

where you factor out the asymptotic behavior first, and then look for a series solution.

4. Dec 8, 2011

fluidistic

Hi vela, yes I'm aware of my mistake for n=0, n=2. I've redone everything and fixed it for post 2.
Also I'm not really understanding Frobenius's method. Why a solution of the form $\sum _{n=0}^{\infty}a_n x^{n+s}$ rather than just $\sum _{n=0}^{\infty}a_n x^{n}$?
Also I wonder why in all my problems I never get $a_{n+1}$ in function of $a_n$ but instead I get $a_{n+2}$ in function of both $a_{n+1}$ and $a_n$ which doesn't simplificate things out. In all the examples I've read on the internet, they always get $a_{n+1}$ in function of $a_n$.

5. Dec 9, 2011

vela

Staff Emeritus
It's because the coefficient of y in the differential equation is singular at x=0, and you're expanding your series about this point. If the coefficients weren't singular at the point you're expanding about, you could use the plain old Taylor series-type expansion.

It's because you're trying to solve hard equations.

One trick you can try is to factor out certain behaviors of the solution. In this problem, when x gets large, the differential equation becomes approximately
$$y''-\frac{1}{4}y = 0,$$ which has solutions $e^{\pm x/2}$. Because you want the solution to vanish at infinity, you eliminate the positive-exponent solution and posit that the solution can be written in the form $y=f(x)e^{-x/2}$. Substitute this back into the original differential equation to find the differential equation for f, and then try solving for f using the Frobenius method. This time you will find a recursion relation that's easier to deal with.

6. Dec 9, 2011

fluidistic

Ah ok thank you very much, very clear explanation.
The DE then becomes $ce^{-x^2/2} \left [ \frac{f(x)}{4}-f'(x)+f''(x) \right ]- \left [ \frac{1}{4}+\frac{K}{x} \right ] f(x)ce^{-x^2/2}=0$.
After simplificating things out, this is equivalent to the DE $f''(x)-f'(x)-\frac{Kf(x)}{x}=0$. I'm going to try Frobenius's method on this DE. I hope I didn't make any mistake.

7. Dec 9, 2011

fluidistic

After a lot of algebra I reach that the indicial equation gives $s=0$ or $s=1$ but I can discard s=0 because otherwise $a_1$ cannot be definied.
I get that $a_{n+1}= \frac{a_n (n+K)}{n(n+1)}$.
I've calculated the first 3 terms, setting $a_0=1$ for convenience.
I noticed a patern and I think that in general, $a_n=\frac{(K+n)!}{n!(n+1)!(K+1)}$.
So that $f(x)=\sum_{n=0}^\infty \frac{x^n}{(K+1)} \frac{(K+n)!}{n!(n+1!)}$.
But instead of considering $y(x)=f(x)e^{-x/2}$, I considered $y(x)=f(x)ce^{-x/2}$.
Let me know if you think I made an error, so that I post all my work.

8. Dec 9, 2011

vela

Staff Emeritus
This should be
$$a_{n+1} = \frac{n+K+1}{(n+1)(n+2)}a_n$$but otherwise your results look fine. Now in the large-n limit, you have
$$\frac{a_{n+1}}{a_n} \sim \frac{1}{n}$$which is the same behavior as the series for ex. This means if your solution for f(x) is an infinite series, f(x) will behave like ex when x is large, which is exactly what you don't want because that means y(x) ~ ex/2, which diverges as x→∞. So you don't want f(x) to be an infinite series. How can you achieve this?

9. Dec 9, 2011

fluidistic

Whoops for what I wrote. I copied from my draft the wrong relation (I get yours for s=1), which would be valid in case s=0 was allowed.
Abour your question, maybe truncate the series... but I don't know up to which n. I could keep I think in principles any finite power of x. But what's the difference if I truncate after n=3 and n=120, I don't really know. Do I get a better "approximation" of the solution? If so, I don't really understand why taking an infinite series doesn't lead to the "real" solution. I know it will diverge and that if I truncate it it won't diverge but no more than this.

10. Dec 9, 2011

vela

Staff Emeritus
You're not finding the solution. You're finding a whole set of solutions. If you truncate at n=3, that's one solution, for a certain value of K=K3. If you truncate at n=120, that's a different solution, for a different value of K=K120. The point is that you can find these truncated solutions for only certain values of K. These are exactly the Ks you're looking for. It's only for these values of K that you can find solutions which satisfy the boundary conditions.

11. Dec 10, 2011

fluidistic

Ok I see.
Let's say I truncate at n=0 in order to look for $K_0$. This means $f(x)=\frac{K_0!}{K_0+1}$. So $y_0 (x)=ce^{-x/2}\frac{K_0!}{K_0+1}$.
So $y''_0(x)=\frac{c}{4}e^{-x/2}\frac{K_0!}{K_0+1}$.
Plugging this into the original DE, I reach $ce^{-x/2}\frac{K_0!K_0}{x(K_0+1)}=0$. This means $K_0=0$. Is this possible?

I asked a friend about this problem yesterday, he said I should get 2 functions as solution because it's a DE of second order. He said I should truncate the infinite series when n is such that $a_{n+1}=0$ but $a_n \neq 0$. So in this case it would be $n=-s-K$. He also said we use Frobenius method when we want to know how the solution function behaves in the singular points. Something you can't get with a simple Taylor series (this comments makes sense to me).
I'd appreciate if you could comment on what he told me... he's just a student like me.

12. Dec 10, 2011

vela

Staff Emeritus
If the series truncates at the first term, you have f(x) = a0x1. (Remember your solution is for s=1.) Therefore, $y_0 = xe^{-x/2}$. If you plug this into the original DE, you get (1+K)e-x/2=0, which is satisfied when K=-1. This is consistent with what your friend said below, namely that K=-(s+n), which is equal to -1 when n=0.

You don't get a second solution because you eliminated it by requiring the wave function remain finite at the origin and at infinity.

Yup, that's right.

13. Dec 10, 2011

fluidistic

Ok thank you vela for these explanations.
I've some troubles however to calculate $K_0$.
$a_{n+1}=a_n \frac{(n+s+K)}{(n+s)(n+s+1)}$. I take s=1 so $a_{n+1}=a_n \frac{(n+1+K)}{(n+1)(n+2)}$ in agreement with you.
Now if $a_0=1$, $a_1=\frac{K+1}{2}$, $a_2=a_1 \frac{K+2}{2\cdot 3}$, $a_3=\frac{(K+1)(K+2)(K+3)}{2\cdot 1 \cdot 2 \cdot 3 \cdot 3 \cdot 4}$.
In general $a_n = \frac{(K+n)!}{n!(n+1)!K}$ which sligtly differ from my previous post (this current one looks more correct to me).
Thus $f(x)= \sum_{n=0}^{\infty } \frac{x^n (K+n)!}{n!(n+1)!K}$. If I truncate this series at the first term ($n=0$), I get $f(x)=\frac{K_0!}{K_0}=(K_0-1)!$.
So $y_0(x)=ce^{-x/2}(K_0-1)!$. I don't understanding why I don't get your value "x" for f(x).

14. Dec 10, 2011

vela

Staff Emeritus
The numerator would be (K+n)!/K! if K were a non-negative integer, but K actually turns out to be negative.

It's easier if you simply solve
$$a_{n+1} = \frac{K+n+1}{(n+1)(n+2)} a_n = 0$$for k, assuming an≠0.

15. Dec 11, 2011

fluidistic

Ah ok thank you, I think I finally understand now.