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Using WKB(J) method to obtain asymptotics for an elastic string?

  1. Jun 3, 2009 #1

    The differential equation is question that models an elastic string is:
    [tex]\frac {\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x}\left( p(x) \frac{\partial u}{\partial x}\right)= 0[/tex]

    Taking [tex]u(x,t) = e^{-i\lambda t} y(x) [/tex]
    I simplify above diff.eqn. to: [tex]\lambda^2 y(x) + \frac{\partial}{\partial x}\left( p(x) y'(x)\right)= 0[/tex]

    Then taking a WKB ansatz for [tex]y(x,\lambda)[/tex] as follows: [tex]y(x,\lambda) = e^{-i\lambda \xi(x)} \sum_{n=0}^{\infty} \frac{Y_{n}(x) }{\lambda^n}[/tex], I end up with;

    [tex]\lambda^2 e^{-i\lambda \xi(x)} \sum_{n=0}^{\infty}\frac{Y_{n}(x) }{\lambda^n}
    + \frac{\partial}{\partial x}\left[p(x)e^{-i\lambda \xi(x)}\left(\sum_{n=0}^{\infty}\frac{Y_n'(x)}{\lambda^n} - i\lambda\xi(x)\sum_{n=0}^{\infty}\frac{Y_n(x)}{\lambda^n}\right)\right][/tex]

    The Problem: I wonder if this last equation can be simplified? - I'm trying to obtain an recurrence relation for the eigenfunctions [tex]Y_n(x)[/tex] and their derivatives.

    If it can't be simplified, what adjustments / changes are needed for this above method in order to obtain a recurrence relation?

    Thanks, dev00790
    Last edited: Jun 3, 2009
  2. jcsd
  3. Jun 3, 2009 #2
    The sign of the second equation is wrong, it should be read [itex]-\lambda^2 y(x)[/itex]. Appart from that, what is your question?
  4. Jun 3, 2009 #3
    Above post is edited, including further working and my question.

    Replying to your question. I've checked my working, and i believe

    [tex] \frac{\partial^2 u}{\partial t^2} = -\lambda^2 e^{-i\lambda t}y(x)[/tex]

    [tex] \frac{\partial u}{\partial x} = e^{-i\lambda t}y'(x)[/tex]

    Substituting into [tex] \frac {\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x}\left( p(x) \frac{\partial u}{\partial x}\right)= 0 [/tex] then taking [tex]-e^{-i\lambda t}[/tex] outside bracket, and canceling by this factor this simplifies to
    [tex] \lambda^2 y(x) + \frac{\partial}{\partial x}\left( p(x) y\'(x)\right)= 0 [/tex] ?
  5. Jun 3, 2009 #4
    Just curious. If the above linear DE can be solve for y(x) why do we need to consider that WKB ansatz (whatever it is) ?
  6. Jun 3, 2009 #5
    I've been using a WKB ansatz to deduce a recurrence relation in terms of [tex]Y_n[/tex] and [tex]Y_n{'}[/tex] and hence formulae for [tex]Y_n[/tex], as its the method I have been using to derive an asymptotic expansion of [tex]y(x,\lambda)[/tex] as n tends to infinity for the following equation:

    [tex]y^{''}+ \lambda^2 p(x)y= 0[/tex].
    This gave the following reccurence relation (with simplification): [tex]i\xi^{''}Y_{n+1}+2i\xi{'}Y'_{n+1}+Y_n{''}=0[/tex]

    So I tried this method with [tex] \frac {\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x}\left( p(x) \frac{\partial u}{\partial x}\right)= 0 [/tex] to see if method can still be used, if not how to change it / and or have to use a different method? - If so why?

    Then hit the mentioned problem stated in my first post. Can anyone help with this please?
    Last edited: Jun 3, 2009
  7. Jun 3, 2009 #6
    The equation d/dx[p(x)dy/dx]+λ2y(x)=0 can be transformed, by the variable changes dz=p(x)dx, w(z)=y(x(z)), to the form:


    Now you can apply the WKB approximation which is rather accurate if p is not step-wise.

    The latter equation can be also transformed in a Schroedinger-like form

    u'' + Vu +λ2u = 0

    and solved by the perturbation theory (see Mors, Feshbah book Methods of Theoretical Physics, V. 1). When n>>1, |λn2u|>>|Vu|, so the asymptotic solutions for un and λn are the solutions of the "non perturbed" equation:

    u'' + λ2u = 0

    Last edited: Jun 3, 2009
  8. Jun 3, 2009 #7
    Bob_for_short, the following point you made;

    but this is not the same equation as the one above: [tex]\frac{\partial}{\partial x}\left(p(x)\frac{dy}{dx}\right) + \lambda^2 y(x)= 0 [/tex]

    Can one apply a transformation of variables to this equation as it involves partial differentiation? If so what would this transformation be? Thanks.
    Last edited: Jun 3, 2009
  9. Jun 3, 2009 #8
    After the time and space coordinate separation you obtain the ordinary rather than partial derivatives. Your y(x) depends only on x. So my transformation applies well.

  10. Jun 3, 2009 #9
    Ok, so I try and do this transformation:

    Using w(z)=y(x(z)), I get w'(z) = y'(x(z)).x'(z) = [y'(x(z))]/p(x) using [tex]\frac{dx(z)}{dx}=\frac{1}{p(x)}[/tex]

    Thus w''(z) = y'(x(z)).x''(z) + x'(z).y''(x(z))

    I cannot see how u arrive at [tex]\frac{d^2w}{dz^2}+\lambda^2 p^{-1}w=0[/tex] though?
    Please explain this in detail. Thanks.
  11. Jun 4, 2009 #10
    I made a mistake, sorry. In fact dx=pdz or dz=p-1dx. Then

    d/dx=z'd/dz=p-1d/dz; pd/dx=d/dz, OK?

    y=w, OK?

    So you directly obtain : p-1d2w/dz22w=0

    or d2w/dz22pw=0

  12. Jun 18, 2009 #11
    On Perturbation Theory for the Sturm-Liouville Problem with Variable Coefficients by Vladimir Kalitvianski: http://arxiv.org/abs/0906.3504. Lots of interesting results.
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