0
$\begingroup$

If we look at the Bessel ODE: $$x^2 y'' + xy' + (x^2 - \alpha^2)y = 0$$

Suppose I then put the solution to the above ODE as $J_{\alpha}(x)$ in the RHS, and try to solve the following ODE: $$x^2 y'' + xy' + (x^2 - \alpha^2)y = J_{\alpha}(x)$$ Obviously the solution to this equation is the solution to the homogeneous equation plus a specific solution.

Suppose I want to reiterate this procedure of inserting in the RHS of Bessel ODE the solution to the previous equation iteratively.

What would the solution at the $n$-th step will look like?

This is a pure math exercise I am thinking about, I don't think that someone thought of this before.

You can of course generalize it to any special function you might think of with a suitable ODE that defines it.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ How are you choosing which solution to take at each step? are you setting some initial data for $y,y'$? $\endgroup$
    – Sam Zbarsky
    Nov 21, 2019 at 21:24
  • $\begingroup$ No, I am not setting initial data. By choosing a solution, I assume in each iteration that the homogeneous ode is solved by $J_{\alpha}$. $\endgroup$
    – Alan
    Nov 21, 2019 at 21:33
  • 1
    $\begingroup$ That's fine, but how are you choosing what the inhomogenous term is? Any possible solution could be the "specific term" you mention; you need some way to decide which one you are taking. $\endgroup$
    – Sam Zbarsky
    Nov 22, 2019 at 14:55
  • $\begingroup$ @SamZbarsky let's assume something like: $y(-\alpha)=0$ and $y(\alpha)=1$ for all the iterations. $\endgroup$
    – Alan
    Apr 22, 2022 at 21:29

1 Answer 1

Reset to default
2
$\begingroup$

Using Variation of Parameters, your inhomogeneous equation has a particular solution

$$ \frac{\pi}{2} \left( {{Y}_{\alpha}\left(x\right)} \int \!{\frac { {{J}_{\alpha}\left(x\right)} ^{2}}{ x}}\,{\rm d}x- J_\alpha(x) \int \!{\frac {{{ Y}_{\alpha}\left(x\right)}{{ J} _{\alpha}\left(x\right)}}{x}}\,{\rm d}x \right) $$

Improve this answer
$\endgroup$
3
  • $\begingroup$ How would you find the solution to the $n$-th iteration of my problem? $\endgroup$
    – Alan
    Nov 22, 2019 at 5:50
  • $\begingroup$ If $F_n$ is a particular solution to the $n$'th iteration, you can take $$F_{n+1}(x) = \frac{\pi}{2} \left( Y_\alpha(x) \int \frac{J_\alpha(x) F_n(x)}{x}\; dx - J_\alpha(x) \int \frac{Y_\alpha(x) F_n(x)}{x}\; dx \right)$$ You may be able to use integration by parts to reduce iterated integrals to single integrals, but things are going to get ugly... $\endgroup$
    – Robert Israel
    Nov 22, 2019 at 13:26
  • $\begingroup$ by all means let get ugly... :-D $\endgroup$
    – Alan
    Apr 22, 2022 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.