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Math Help - eigenvalue problem

  1. #1
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    eigenvalue problem

    Please, help me with this one:

    \psi ''(z)+(\frac{20}{cosh^{2}z}+\lambda )\psi(z)=0
    \psi '(0)=0
    I try it, and got the following:

    y=coshz
    \frac{d\psi }{dz}=\frac{dx}{dy}sinhz
    (y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{dx}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0

    and i stuck here....
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  2. #2
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    I get the following DE:

     \frac{dy}{dz} = sinh z
     \frac{d\psi}{dy}=\frac{d\psi}{dz} \frac{ dz}{dy}= \frac{\psi '(z)}{sinh z}= \frac{\psi '(z)}{\sqrt{y^2-1}}
     \frac{d^2\psi}{dy^2}= \frac{\psi ''(z)}{y^2-1} -\frac{\psi '(z)}{(y^2-1)^2} y
    Plugging back I get:
     (y^2-1)\psi ''(y) + \frac{y \psi '(y)}{\sqrt{y^2-1}} + (\frac{20}{y^2}+\lambda)\psi (y)

    Though I might have mistaken somewhere.
    Last edited by InvisibleMan; January 1st 2011 at 10:14 AM. Reason: correction
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  3. #3
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    Quote Originally Posted by sinichko View Post
    Please, help me with this one:

    \psi ''(z)+(\frac{20}{cosh^{2}z}+\lambda )\psi(z)=0
    \psi '(0)=0
    I try it, and got the following:

    y=coshz
    \frac{d\psi }{dz}=\frac{dx}{dy}sinhz
    (y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{dx}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0

    and i stuck here....
    Not sure if this is what is wanted, but you could try for a solution of the form \psi(y) = y^\alpha. That will give you two eigenvalues.
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  4. #4
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    Sorry, there is a mistake in post #2, line 3, last term.

    <br />
\displaystyle {<br />
\frac{d}{dy}\frac{1}{\sqrt{y^2-1}}=- \: \frac{y}{\sqrt{(y^2-1)^3}}<br />
}<br />
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  5. #5
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    Your ODE can be solved using a 4th order Darboux transformation. Do you have a second IC for \psi? Also is \lambda > 0 or < 0?
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  6. #6
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    Quote Originally Posted by Danny View Post
    Your ODE can be solved using a 4th order Darboux transformation. Do you have a second IC for \psi? Also is \lambda > 0 or < 0?

    \lambda < 0 and real
    also i know the answer:

    \lambda_{1} =-16, <br /> <br />
  \lambda_{2} =-16, i don't have a second IC for \psi...
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  7. #7
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    Thanks, but how to continue?
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  8. #8
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    Quote Originally Posted by InvisibleMan View Post
    I get the following DE:

     \frac{dy}{dz} = sinh z
     \frac{d\psi}{dy}=\frac{d\psi}{dz} \frac{ dz}{dy}= \frac{\psi '(z)}{sinh z}= \frac{\psi '(z)}{\sqrt{y^2-1}}
     \frac{d^2\psi}{dy^2}= \frac{\psi ''(z)}{y^2-1} -\frac{\psi '(z)}{(y^2-1)^2} y
    Plugging back I get:
     (y^2-1)\psi ''(y) + \frac{y \psi '(y)}{\sqrt{y^2-1}} + (\frac{20}{y^2}+\lambda)\psi (y)

    Though I might have mistaken somewhere.
    Thanks, but how to continue?
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  9. #9
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    Quote Originally Posted by sinichko View Post
    Thanks, but how to continue?
    Frist, where did this problem come from and how did you know that \lambda = -16. Are there other values?

    Second, are you familar with Wronskians?
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  10. #10
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    The equation (y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0 has eigenfunctions \psi(y) = y^5 and \psi(y) = y^{-4}, with corresponding eigenvalues \lambda = -25 and \lambda=-16 (see my comment #3 above).
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  11. #11
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    Quote Originally Posted by Danny View Post
    Frist, where did this problem come from and how did you know that \lambda = -16. Are there other values?

    Second, are you familar with Wronskians?

    I took it from an article: problem (4.8)

    eigenvalue problem-article.jpg


    Wronskian sounds familiar...
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  12. #12
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    Quote Originally Posted by Opalg View Post
    The equation (y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0 has eigenfunctions \psi(y) = y^5 and \psi(y) = y^{-4}, with corresponding eigenvalues \lambda = -25 and \lambda=-16 (see my comment #3 above).
    thanks, can you write please first steps of your solution, in the article he got a little different answer...

    \lambda_{1} =-16, <br /> <br />
  \lambda_{2} =-4
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  13. #13
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    Quote Originally Posted by sinichko View Post
    Quote Originally Posted by Opalg View Post
    The equation (y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0 has eigenfunctions \psi(y) = y^5 and \psi(y) = y^{-4}, with corresponding eigenvalues \lambda = -25 and \lambda=-16 (see my comment #3 above).
    thanks, can you write please first steps of your solution, in the article he got a little different answer...

    \lambda_{1} =-16, <br /> <br />
  \lambda_{2} =-4
    It seemed obvious to me that there should be solutions of the form \psi(y) = y^\alpha, for some values of \alpha. The reason is that when you differentiate a power of y, it reduces the power by 1 (and when you differentiate twice, the power goes down by 2). So when you put \psi(y) = y^\alpha in the expression (y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi, each term will be a multiple of either y^\alpha or y^{\alpha-2}.

    In fact, if \psi(y) = y^\alpha then \frac{d\psi}{dy} = \alpha y^{\alpha-1} and \frac{d^2\psi}{dy^2} = \alpha(\alpha-1) y^{\alpha-2}. Thus

    \begin{aligned}(y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi &= \alpha(\alpha-1)(y^{2}-1)y^{\alpha-2}+ \alpha y^{\alpha}+\left ( \frac{20}{y^{2 }} + \lambda\right )y^{\alpha} \\ &= \bigl(\alpha(\alpha-1) + \alpha + \lambda\bigr)y^{\alpha} + \bigl(-\alpha(\alpha-1) + 20\bigr)y^{\alpha-2}.\end{aligned}

    If that is 0 for all y then (putting the coefficients of y^\alpha and y^{\alpha-2} equal to 0), you get \alpha(\alpha-1) = 20, from which \alpha = 5 or -4; and \lambda = -\alpha^2, from which \lambda = -25 or -16.
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  14. #14
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    Thank you very much
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