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I am studying torsional Alfven waves in spicules. In this concern I have encountered the following equation:

$ \left(1-m^2 e^{-αz}\right)y''(z)+\left(4π i m e^{-αz}-1/h\right)y'(z)+\left(\frac{1}{4h^2} +4π^2 ω^2 e^{-αz}\right)y(z)=0 $

where $i=\sqrt{-1}$. The boundary conditions are: $y(0)=0$, $y(1)=0$, and the constants are: $h=0.23, α=1.84, m=0.25$. We need to find $ω_1$ and $ω_2$ and $y_1$ and $y_2$ i.e. fundamental and first harmonic eigenvalue and eigenfunction. If you want to guess the initial value of $ω$, you can set it $ω=1$.

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  • $\begingroup$ Since you have numeric values for your constants $h$, $m$, and $\alpha$, would you be satisfied with numerical results? If so, You could write $y(z)=\sum_{k=1}^\infty a_k \sin(k \pi z)$, and convert your eigenvalue problem to a discrete one in terms of the $a_k$'s. To solve it numerically, just truncate the sum at some sufficiently high value $1\le k \le M$. $\endgroup$
    – Yoav Kallus
    Commented Nov 22, 2014 at 18:50
  • $\begingroup$ Dear Yoav I need only the walues of w1 and w2. Maybe I plot w1/w2 with respect to some values of m and alpha. The i symbol is confusing me. without it, it is possible to solve it by shooting method! I will check in the same way as you described. $\endgroup$
    – Hossein Ebadi
    Commented Nov 22, 2014 at 19:32
  • $\begingroup$ I think you can still use the shooting method, but instead of a bracketing root finding method (like bisection method), you will have to use a method that doesn't rely on bracketing. I think you could even use Newton's method, since you can write down $dy(1)/d\omega$ as an integral that you can evaluate numerically. $\endgroup$
    – Yoav Kallus
    Commented Nov 22, 2014 at 19:39
  • $\begingroup$ PS, I just noticed on Wikipedia a link to the Lehmer-Schur algorithm which is described as an extension of the bisection method to the complex plane. I don't know anything about it, but you should check if it does what you want. $\endgroup$
    – Yoav Kallus
    Commented Nov 22, 2014 at 19:41
  • $\begingroup$ Your equation becomes hypergeometric under the substitution $u=\exp(-\alpha z)$. $\endgroup$
    – Michael Renardy
    Commented Nov 23, 2014 at 4:09

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