I have this second-order differential equation:

$$x''(t) + \frac{1}{(\tau + t)}x'(t) + k^2x(t) = 0$$

I want to make the solution to this ODE amenable to a closed form Bessel function, and so a suggested way is to make a change of variables so that we can compare the differential equation above to the transformation equation below: (where this $x$ is analogous to my $t$, and this $y$ is analogous to my $x(t)$)

Transformation equation:

enter image description here

The goal (or atleast the way I did it for a simple function) was to compare and identify what values the parameters $\alpha, \beta, C, m$ must have so that the form of differential equation is captured by a Bessel function that makes use of these parameters (such as a linear combination of $x^{\alpha}J_m(Cx^{\beta})$ and $x^{\alpha}Y_m(Cx^{\beta})$). This method allowed me to solve a simple equation like the Airy equation. But if I try to do that in this case, the moment I divide the boxed equation on both sides by $x^2$, you get a $\frac{1}{x}$ as the co-efficient for the first-derivative term, which doesn't represent the form of my differential equation's 2nd term (which has $\frac{1}{\tau + t}$ as its coefficient).

I am wondering if I am missing something here, or perhaps there's an intermediary step that's required before I can use this method. Ultimately, I just need a solution to that differential equation that is represented as a Bessel function.


1 Answer 1



Changing variable $u=k(\tau+t)$ would lead to a very simple Bessel equation in $x(u)$

  • $\begingroup$ Sorry if I seem imperceptive, but what does the variable $r$ represent? Do we have to define two variables? Or were you referring to $\tau$? I'm quite new to Bessel transformations, so I'm not sure how I would implement the variable change to the original differential equation for x(u) $\endgroup$
    – Ferreroire
    Dec 6, 2016 at 7:06
  • $\begingroup$ @Ferreroire. Sorry for the typo ! I did misred the equation (I am almost blind). Cheers. $\endgroup$ Dec 6, 2016 at 7:08
  • $\begingroup$ Oh, no worries, thanks! Am I on the right track if I do something like $x''(u) + kux'(u) + k^2x(u) = 0$? $\endgroup$
    – Ferreroire
    Dec 6, 2016 at 7:12
  • 1
    $\begingroup$ @Ferreroire : sorry for the typo. The change of variable is $u=\tau+t$ . Hense $du=dt \quad\to\quad x(t)=X(u) \quad\to\quad \frac{dx}{dt}=\frac{dX}{du} \quad\to\quad \frac{d^2x}{dt^2}=\frac{d^2X}{du^2}$ . The first form of the ODE $\frac{d^2x}{dt^2}+\frac{1}{\tau+t}\frac{dx}{dt}+k^2x(t)=0$ is transformed to $\frac{d^2X}{du^2}+\frac{1}{u}\frac{dX}{du}+k^2X(u)=0$ , which is a standard form of Bessel's ODE. $\endgroup$
    – JJacquelin
    Dec 6, 2016 at 13:42
  • 1
    $\begingroup$ @Ferreroire : You can also do the change of variable is $v=k(\tau+t)$ . Hense $dv=kdt \quad\to\quad x(t)=X(v) \quad\to\quad \frac{dx}{dt}=k\frac{dX}{dv} \quad\to\quad \frac{d^2x}{dt^2}=k^2\frac{d^2X}{dv^2}$ . The first form of the ODE $\frac{d^2x}{dt^2}+\frac{1}{\tau+t}\frac{dx}{dt}+k^2x(t)=0$ is transformed to $\frac{d^2X}{dv^2}+\frac{1}{v}\frac{dX}{dv}+X(v)=0$ , which solution is $c_1J_0(v)+c_2Y_0(v)=c_1J_0(k(\tau+t))+c_2Y_0(k(\tau+t))$... $\endgroup$
    – JJacquelin
    Dec 6, 2016 at 13:53

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