Any help how this equation ( Eq. B3 in paper ) :

$$ \phi’’(\eta) + \frac{6(1+\omega)}{1+3\omega)} \frac{1}{\eta} \phi’(\eta) + \omega k^2 \phi(\eta) =0 $$

Has been solved by Bessel’s function with a solution:

$ \phi(\eta) = y^{-\alpha} [ C_1 J_\alpha (y) + C_2 Y_\alpha (y) ], ~~~ y = \sqrt{\omega} k \eta,~~~~~ \alpha= \frac{1}{2} \left( \frac{5+3\omega}{1+3\omega} \right). $

I spent much time reading about Bessel’s function , but there are different cases. Here $\omega$ is a constant and $k$ is the wave number.

  • $\begingroup$ the ODE $\phi''(x)+(a/x)\phi'(x)+b\phi(x)=0$ has as the general solution the superposition of $x^{1/2-a/2}J_{a/2-1/2}(x\sqrt{b})$ and $x^{1/2-a/2}Y_{a/2-1/2}(x\sqrt{b})$, as you can check by noting that $x^{a/2-1/2}\phi(x)$ satisfies the Bessel ODE. $\endgroup$ Feb 12 at 11:28
  • $\begingroup$ @CarloBeenakker. I wonder an ODE like : $\phi’’(x) + (a/\eta) \phi’(x) + b \phi(x) =0 $ can be solved for instance by letting $\phi = e^{rt}$, then the solution will be for a simple quadratic equation. Why to use Bessel function? $\endgroup$
    – Dr. phy
    Feb 12 at 11:44
  • $\begingroup$ the exponential will not solve the equation, just try it. $\endgroup$ Feb 12 at 11:49
  • $\begingroup$ Yes, I see. Because the factor $ 1/\eta$ in the second term. But what about an ODE like: $ \phi’’(x) + a \phi’ (x) + k \phi( x) =0 $ , this can be solved by letting $\phi= e^{rx} $ which gives a real solution or $\phi= e^{i rx} $ which gives an imaginary solution, right? $\endgroup$
    – Dr. phy
    Feb 12 at 12:44
  • $\begingroup$ yes, that ODE has constant coefficients, your ODE does not $\endgroup$ Feb 12 at 13:49


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