# Homework Help: Sturm Liouville System

1. Jun 9, 2010

### sassie

1. The problem statement, all variables and given/known data

Evaluate and find the eigenvalues and eigenfunctions for:

$$-\frac{d}{dx}[(1+x^{2})\frac{dy}{dx}]=\lambda\frac{1}{1+x^{2}}y}$$

2. Relevant equations

3. The attempt at a solution

Okay, what I did was to expand the given equation to get:

$$((1+x^{2})^{2})y''+2x(1+x^{2})y' + \lambda y =0$$ <---- edit typo

But then I get on how to get the needed general solution. I have tried substitution by letting $$t=1+x^{2}$$ and then using the chain rule to get the first and second derivatives etc. I tried that to get a nice equation, but it ended up with a messier equation. Now I am really stuck and don't know what to do. Your help is very much appreciated.

Last edited: Jun 9, 2010
2. Jun 9, 2010

### Staff: Mentor

It looks like you made an error when you differentiated. I get
$$-(1 + x^2)y'' -2x \frac{dy}{dx} = \lambda \frac{1}{1 + x^2} y' - \frac{2x}{(1 + x^2)^2} y'$$

After multiplying both sides by (1 + x2)2 I don't get what you show.

3. Jun 9, 2010

### sassie

Okay, what I did was (sorry I did skip a few steps):

LHS = $$-(1+x^2)y''-2xy'-(\lambda \frac{1}{1 + x^2}) y=0$$

I then multiplied through by $$(1+x^2)$$

$$((1+x^{2})^{2})y''+2x(1+x^{2}) + \lambda y =0$$

Now, I understand how you got the left hand side of the equation you gave, I am a little confused about how you got the right hand side.

4. Jun 9, 2010

### sassie

But can anyone help me with this question? I've been working on it for 5 hours straight and still I can't a solution! This question is just so much harder than ones I have handled previously!

5. Jun 9, 2010

### lanedance

ok so the first step is to solve the general ode

then find the eigenvalues and functions from the general solution, using the boundary conditions, where are you up to?

EDIT - re-read, so you're trying to find the general solutions...

Last edited: Jun 9, 2010
6. Jun 9, 2010

### sassie

I haven't even got the general ODE and that's what I have been having problems with. I have absolutely no idea how to obtain it :'(

I know it should be easy but for the life of me, I can't do it

7. Jun 9, 2010

### sassie

Re: EDIT

Yes, so I am trying to get the general solutions to the given equation but I do not know how to.

8. Jun 9, 2010

### Staff: Mentor

I should not have differentiated the right side. I erred by thinking incorrectly that we were differentiating both sides. Sorry for the bad advice - it was late at night and I misread your equation.

Your equation still has an error, though. It is missing a y' factor in the second term.
$$((1+x^{2})^{2})y''+2x(1+x^{2})y' + \lambda y =0$$

Last edited: Jun 9, 2010
9. Jun 9, 2010

### sassie

Oh, okay! I was pondering why your RHS was the way it was. But can you help me with getting the equation into a form where I can the general solution? I REALLY need a hint to get me on the right track. I am sure I can solve for the eigenvalues and eigenfunctions myself, it's just been getting the general solution which has been causing me problems.

EDIT re: error sorry, that was a typo

Last edited: Jun 9, 2010
10. Jun 9, 2010

### lanedance

yeah thats tricky, mucked around with a few things, series was pretty nasty, even if you solve for the lambda = 0 case you get an inverse tangent...

one thing i did notice however is that if you use the chain rule with some substitution t(x), then using chain rule, with dashes for x derivative & dots for t derivative
$$y' = \frac{d}{dx}y(t(x)) = \frac{dy(t)}{dt}\frac{dt(x)}{dx} = \dot{y} t'$$

so then it would be nice to cancel the x term in the brackets, so i started with
$$t'(x) = \frac{dt(x)}{dx} = 1+x^2$$

and things evolved nicely form there....

Last edited: Jun 9, 2010
11. Jun 9, 2010

### lanedance

the nice part is that after cancelling the term, you're left with:
$$\frac{d}{dx}(\dot{y}) = \ddot{y} t'(x) = \ddot{y}(1+x^2)$$

and unless i missed something, everything simplifies a heap... ;)

12. Jun 9, 2010

### sassie

Okay, thanks. I'll try that out :)

13. Jun 9, 2010

### Staff: Mentor

sassie, see the end of my post #8. Your equation in the first post did have an error in it.

14. Jun 9, 2010

### TheFirstOrder

Thanks, I did notice that mistake :)

Hey lanedance, is it possible for you to put up more of your working? I seem to be getting some thing different

15. Jun 9, 2010

### lanedance

apologies - think i made a mistake writing down the ODE, so it doesn't cancel like i thought;(

16. Jun 9, 2010

### sassie

sorry, that was my fault. i wrote down the wrong ode, but mark44 pointed out the mistake. look's like i'm back at square 1.

17. Jun 9, 2010

### Staff: Mentor

I spent a little time reviewing S-L equations this morning. In most of the examples I saw there were boundary conditions. Are there boundary conditions in your problem?

18. Jun 9, 2010

### sassie

yes,

y(0)=0 and y(1)=0

19. Jun 9, 2010

### lanedance

so the original ODE is:
ok think i've may have made some headway on the ODE, the lambda x=0 case is easy to solve & gives an inverse tangent function, which may be a hint of what direction to go...

now assume we are going to simplify by making a substitution t(x), then
$$y' = \dot{y}t'$$
$$y'' = \ddot{y} (t')^2 + \dot{y} t''$$

so going back to the DE and substituting in gives
$$((1+x^{2})^{2})(\ddot{y}(t')^2 + \dot{y}t'') +2x(1+x^{2})\dot{y}t' + \lambda y =0$$

collecting terms
$$((t')^2(1+x^{2})^{2})\ddot{y} + (t''(1+x^2)^2+t'2x(1+x^2))\dot{y} + \lambda y =0$$

so then it looks worth trying
$$t' = \frac{1}{1+x^2}$$

then differentiating again gives
$$t'' = \frac{-2x}{(1+x^2)^2}$$

then see how that goes subbing in, hopefully will simplify down for a better DE for t...

if this works, you'll need to be careful with your boundary conditions, on determining the eigenvalues & eigenfunctions...

Last edited: Jun 9, 2010
20. Jun 10, 2010

### lanedance

so did you get to $\ddot{y}+\lambda y = 0$?