I have been given this recently in PDE class involving the solutions to the Bessel fucntion in Sturm-Liouville form, asking for Eigenvalues and Eigenfunctions:
$ (xy')'+\lambda x y = 0 \space \space \space \space 0<a<x<b $
$ y(a)=y(b)=0 $
Here is where my problems start: If I expand the equation I get:
$ xy''+y'+\lambda x y = 0 \to x^2y''+xy'+\lambda x^2 y = 0 $ now depending on whether or not the eigenvalues are positive or negative (checking zero is not one is fairly simple) we obtain the general solution to the Bessel equation or modified Bessel equation of order zero:
$ y(x) = AJ_0(\sqrt{\lambda}x) + BY_0(\sqrt{\lambda}x) $ or
$ y(x) = AI_0(\sqrt{\lambda}x) + BK_0(\sqrt{\lambda}x) $
So there is not the matter of boundedness at zero resulting in eliminating the singular solution from the linear combination so my problem here is basically applying the boundary conditions like this:
$ AJ_0(\sqrt{\lambda}a) + BY_0(\sqrt{\lambda}a) = AJ_0(\sqrt{\lambda}b) + BY_0(\sqrt{\lambda}b) = 0 $
so this equation and deriving $ \lambda $ and the actual eigenfunctions from this is beyond my ability, and that's where I am stuck and need the help. Thanks all.
Best Answer
You can show that the eigenvalue must be positive even without solving the differential equation. The Bessel equation in Sturm-Lioville form is
$${\left( {xy'} \right)^\prime } = - \lambda xy$$
Multiply by $y$ and integrate from $a$ to $b$ to obtain
$$\int_a^b {y{{\left( {xy'} \right)}^\prime }dx} = - \lambda \int_a^b {x{y^2}dx} $$
now use integration by parts for the left side
$$\left. {xyy'} \right|_a^b - \int_a^b {x{{\left( {y'} \right)}^2}dx} = - \lambda \int_a^b {x{y^2}dx} $$
using the boundary conditions and solving for $\lambda $ leads to
$$\lambda = {{\int_a^b {x{{\left( {y'} \right)}^2}dx} } \over {\int_a^b {x{y^2}dx} }}$$
considering $b>a>0$ (this must happen to satisfy the Sturm-Liouville assumption $x>0$) you can easily conclude that
$$\lambda \ge 0$$
but $\lambda $ cannot be zero too! The reason is that by the above formula $\lambda $ is zero if and only if ${y'}$ is zero and this implies $y = Const$ but by the boundary conditions you conclude that $y=0$. It means that $\lambda $ is zero if and only if $y$ is zero. Hence, not leading to a nonzero solution. Finally, we proved that
$$\lambda > 0$$