Why are the eigenvalues of this Sturm Liouville problem all positive? I have this problem in my book$$(\rho R')'+\lambda \rho R=0$$ with the solution that is finite for $\rho \to 0$ and $R(\rho=1)=0$. The book says "it can be shown that the eigenvalues are all positive, of the for $\lambda=\mu^2$.It doesn't show how. So here's my try
MY APPROACH
$\lambda = 0$: $(\rho R')'=0$ which gives $\rho R' = A$ which then gives $R'=\frac{A}{\rho}$ and therefore $R=A\ln{(\rho)}+B$ Now the natural logarithm goes to minus infinity when $\rho$ approaches $0$ and therefore we know $A=0$. Hence $R=B$. However, we also know that $R(\rho=1)=0$ and this tells us that $B=0$. Hence for $\lambda=0$ we only have trivial solutions.
$\lambda <0$: Suppose $\lambda=-\mu^2$, then we have $(\rho R')'-\mu^2\rho R=0$

However I don't know how to continue. Any idea on how to show that the eigenvalues are all positive?

 A: Multiply by $R$ and integrate from $0$ to $1$:
$$ 0=\int_0^1 ((\rho R')'R + \lambda \rho R^2) \, d\rho. $$
Now integrate the first term by parts:
$$ \int_0^1 ((\rho R')'R \, d\rho = \left[ \rho R'R \right]_0^1 - \int_0^1 \rho R'^2 \, d\rho. $$
The first term vanishes if we assume that $R'$ is bounded as well, so
$$ 0 = -\int_0^1 \rho R'^2 \, d\rho + \lambda  \int_0^1 \rho R^2 \, d\rho. $$
Both integrals are strictly positive, so $\lambda$ must also be. This argument will work for any Sturm–Liouville problem, provided it is properly formulated.
A: One solution of Bessel's equation (without the right endpoint condition) is
$$
    R_{\lambda}(r)=\sum_{k=0}^{\infty}\lambda^k\frac{(-1)^k\rho^{2k}}{2^{2k}(k!)^2}.         
$$
A second linearly independent solution is obtained by variation of parameters, from which it can be shown that the second solution $S_{\lambda}$ is singular at $\rho=0$ because $R_{\lambda}(\rho)=1$ at $\rho=0$:
$$
        S_{\lambda}(\rho) = R_{\lambda}(\rho)\int\frac{1}{\rho R_{\lambda}(\rho)}d\rho.
$$
So any eigenvalue of your problem satisfying the right endpoint equation is a solution of the power series equation $R_{\lambda}(1)=0$. Of course $\lambda$ must be real, and it is obvious from the form of the power series in $\lambda$ that $\lambda \ge 0$ must hold because all terms of the power series $R_{\lambda}(1)$ are positive for $\lambda$ real and negative. What also comes out of this argument is the fact that the eigenvalues have no finite point of accumulation because they are the zeros of an entire function.
