I have the following eigenvalues problem: $ xy'' + \lambda y = 0 $, $0 \leq x \leq 1$, $y(0)=y(1)=0$ that if we rewrite becomes $y''+ \lambda \frac{1}{x} y = 0$, a Sturm-Liouville problem with weight $w=1/x$.
The weight has a singularity at $x=0$, so the problem is singular. I don't know how to proceed with this kind of issue so I'll be thankful for any help. I need to find the eigenvalues and eigenfunctions.
If it helps, according to Wolfram Alpha the solution are the Bessel functions $y(x) = c_1\sqrt{λx} J_1(2 \sqrt{λx}) + 2 i c_2\sqrt{λx} Y_1(2 \sqrt{λx})$.
Best Answer
There is no difficulty here. Using the asymptotic form of the Bessel functions for small arguments, one can show that $\lim_{x\to 0}\sqrt{\lambda x}J_1(2\sqrt{\lambda x})=0$ and $\lim_{x\to 0}\sqrt{\lambda x}Y_1(2\sqrt{\lambda x})=-\frac{1}{\pi}.$ Therefore, the solution to the ODE that satisfies the boundary condition $y(0)=0$ is $y(x)=c_1\sqrt{\lambda x}J_1(2\sqrt{\lambda x})$. The boundary condition $y(1)=0$ yields the eigenvalue equation $J_1(2\sqrt{\lambda})=0$.
Addendum
Let me show how to obtain the solution given by WolframAlpha. First, we make the change of variables $y(x)=z(\xi)$, where $\xi=2\sqrt{\lambda x}$. Then, using the chain rule, we obtain $$ y'(x)=\frac{dz}{d\xi}\frac{d\xi}{dx}=z'(\xi)\sqrt{\frac{\lambda}{x}}= \frac{2\lambda}{\xi}z'(\xi), \tag{1} $$ $$ y''(x)=\frac{d}{d\xi}\left(\frac{2\lambda}{\xi}z'(\xi)\right)\frac{d\xi}{dx}=\frac{4\lambda^2}{\xi^2}\left(z''(\xi)-\frac{z'(\xi)}{\xi}\right). \tag{2} $$ Substituting $(2)$ in the ODE $xy''+\lambda y=0$ and simplifying, we obtain the ODE $$ \xi^2z''(\xi)-\xi z'(\xi)+\xi^2z(\xi)=0, \tag{3} $$ which looks like a Bessel differential equation, but has the wrong sign in the middle term. To change this sign, we rewrite $(3)$ in terms of the function $u(\xi)=\frac{z(\xi)}{\xi}$, so that $$ z=\xi u, \quad z'=\xi u'+u, \quad z''=\xi u''+2u'. \tag{4} $$ Substituting $(4)$ in $(3)$ and simplifying, we finally obtain a bona fide Bessel differential equation, $$ \xi^2u''+\xi u'+(\xi^2-1)u=0, \tag{5} $$ whose general solution is $u(\xi)=c_1J_1(\xi)+c_2Y_1(\xi)$. Therefore, the solution to the original ODE can be written as $$ y(x)=c_1\xi J_1(\xi)+c_2\xi Y_1(\xi)=c_1\sqrt{2\lambda x}J_1(\sqrt{2\lambda x})+c_2\sqrt{2\lambda x}Y_1(\sqrt{2\lambda x}). \tag{6} $$