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Homework Help: Sturm Liouville System

  1. Jun 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Evaluate and find the eigenvalues and eigenfunctions for:

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

    2. Relevant equations



    3. The attempt at a solution

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

    [tex]((1+x^{2})^{2})y''+2x(1+x^{2})y' + \lambda y =0[/tex] <---- edit typo

    But then I get on how to get the needed general solution. I have tried substitution by letting [tex]t=1+x^{2}[/tex] 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. jcsd
  3. Jun 9, 2010 #2

    Mark44

    Staff: Mentor

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

    After multiplying both sides by (1 + x2)2 I don't get what you show.
     
  4. Jun 9, 2010 #3
    Okay, what I did was (sorry I did skip a few steps):

    LHS = [tex]-(1+x^2)y''-2xy'-(\lambda \frac{1}{1 + x^2}) y=0[/tex]

    I then multiplied through by [tex](1+x^2)[/tex]

    [tex]((1+x^{2})^{2})y''+2x(1+x^{2}) + \lambda y =0[/tex]

    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.
     
  5. Jun 9, 2010 #4
    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!
     
  6. Jun 9, 2010 #5

    lanedance

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    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
  7. Jun 9, 2010 #6
    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
     
  8. Jun 9, 2010 #7
    Re: EDIT

    Yes, so I am trying to get the general solutions to the given equation but I do not know how to.
     
  9. Jun 9, 2010 #8

    Mark44

    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.
    [tex]((1+x^{2})^{2})y''+2x(1+x^{2})y' + \lambda y =0[/tex]
     
    Last edited: Jun 9, 2010
  10. Jun 9, 2010 #9
    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
  11. Jun 9, 2010 #10

    lanedance

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    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
    [tex] y' = \frac{d}{dx}y(t(x)) = \frac{dy(t)}{dt}\frac{dt(x)}{dx} = \dot{y} t' [/tex]

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

    and things evolved nicely form there....
     
    Last edited: Jun 9, 2010
  12. Jun 9, 2010 #11

    lanedance

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

    and unless i missed something, everything simplifies a heap... ;)
     
  13. Jun 9, 2010 #12
    Okay, thanks. I'll try that out :)
     
  14. Jun 9, 2010 #13

    Mark44

    Staff: Mentor

    sassie, see the end of my post #8. Your equation in the first post did have an error in it.
     
  15. Jun 9, 2010 #14
    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
     
  16. Jun 9, 2010 #15

    lanedance

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    apologies - think i made a mistake writing down the ODE, so it doesn't cancel like i thought;(
     
  17. Jun 9, 2010 #16
    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.
     
  18. Jun 9, 2010 #17

    Mark44

    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?
     
  19. Jun 9, 2010 #18
    yes,

    y(0)=0 and y(1)=0
     
  20. Jun 9, 2010 #19

    lanedance

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    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
    [tex] y' = \dot{y}t' [/tex]
    [tex]y'' = \ddot{y} (t')^2 + \dot{y} t'' [/tex]

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

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

    so then it looks worth trying
    [tex] t' = \frac{1}{1+x^2} [/tex]

    then differentiating again gives
    [tex] t'' = \frac{-2x}{(1+x^2)^2} [/tex]

    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
  21. Jun 10, 2010 #20

    lanedance

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    so did you get to [itex] \ddot{y}+\lambda y = 0 [/itex]?
     
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