Sturm-Liouville Problem $((x^2+1)y’)’+\frac{\lambda }{x^2+1}y=0, y(0)=y(1)=0. $

Question

How to find eigenvalues and eigenfunctions of Sturm-Liouville Problem $$((x^2+1)y’)’+\frac{\lambda }{x^2+1}y=0, y(0)=y(1)=0 $$ In the question hint is given as $$\text{Let}~ x=\tan(t).$$

Now, as we put $x=tan(t)$, it’s converted to $$z’’+\lambda cos^2(t)z=0, z(0)=z(\frac{\pi}{4})=0$$ Which is not so easy to solve . Please help. Thank you.

Answer 1

You can get directly a simplified form $$ [(1+x^2)D]^2y+\lambda y=0. $$ The derivative operator can be made into a pure derivative $$ (1+x^2)D=(1+x^2)\frac{d}{dx}=\frac{d}{d(\arctan x)} $$ Thus it makes sense to solve the equation first in the variable $t=\arctan x$ or $x=\tan(t)$, where the equation then reduces to $$ z''(t)+\lambda z(t)=0,~~~z(t)=y(x)=y(\tan t) $$

Answer 2

It is true that

$$ \frac{d}{dx} = \frac{1}{1 + x^2}\frac{d}{dt} $$

But your mistake was

$$ \frac{d}{dx}\left(\left(1+x^2\right)\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{dy}{dt}\right) \neq \frac{d^2y}{dt^2} $$

Instead it is equal to

$$ \frac{1}{1+x^2}\frac{d}{dt}\left(\frac{dy}{dt}\right) = \frac{1}{1 + x^2}\frac{d^2y}{dt^2} $$

So you end up with

$$ y''(t) + \lambda y(t) = 0 $$