# 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.