I was asked the following question:
Find the general solution in terms of Bessel functions:
$$t^2x'' + x' + x = 0, \quad t < 0, \text{ Hint: } s = 2\sqrt{t}$$
My approuch
I think that what I have to do is transform the given equation in one that has a form of a Bessel equation, and for that it must be used the hint.
- $\frac{ds}{dt} = t^{-1/2}$
- $\frac{dx}{dt} = \frac{dx}{ds}\frac{ds}{dt} = t^{-1/2} \frac{dx}{ds}$
- $\frac{d^2x}{dt^2} = \frac{d}{dt} \Big( \frac{dx}{dt} \Big) = \frac{d}{dt} \Big( t^{-1/2} \frac{dx}{ds} \Big) = \frac{d}{dt}t^{-1/2} \frac{dx}{ds} + t^{-1/2}\frac{d}{dt} \Big( \frac{dx}{ds} \Big) = \frac{-1}{2} t^{-3/2} \frac{dx}{ds} + t^{-1/2} \frac{d}{ds} \Big( \frac{dx}{ds} \Big) \frac{ds}{dt} = \frac{-1}{2} t^{-3/2} \frac{dx}{ds} + t^{-1} \frac{d^2x}{ds^2}$
So, in the given equation, now we have:
$$ \begin{align} t^2\Big( \frac{-1}{2} t^{-3/2} \frac{dx}{ds} + t^{-1} \frac{d^2x}{ds^2} \Big) + t^{-1/2}\frac{dx}{ds} + x &= 0 \\ \frac{-1}{2}t^{1/2}\frac{dx}{ds} + t\frac{d^2x}{ds^2} + t^{-1/2}\frac{dx}{ds} + x &= 0 \\ \frac{-1}{2} \frac{s}{2} \frac{dx}{ds} + \frac{s^2}{4} \frac{d^2x}{ds^2} + \frac{2}{s} \frac{dx}{ds} + x &= 0 \\ s^2 \frac{d^2x}{ds^2} + \Big( \frac{8}{s} - s \Big)\frac{dx}{ds} + 4x &= 0 \end{align} $$
Which doesn't have a Bessel-equation form. So please, can anybody enlighten me? Thanks in advanced!
$\endgroup$ 51 Answer
$\begingroup$We see that $t=0$ is not a regular singularity of the equation$$x''+\dfrac{1}{t^2}x'+\dfrac{1}{t^2}x=0$$then this is not a Bessel differential equation.
$\endgroup$ 2
