# Find eigenvalues of $x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi = 0$ with boundary conditions $\phi(1) = \phi(b) = 0$

Since this is an equidimensional equation, determine all positive eigenvalues. \begin{align*} x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi &= 0 \\ \phi(1) &= 0 \\ \phi(b) &= 0 \\ \end{align*} The given answer that I need to find the solution for is for $$n = 1,2,\ldots$$: \begin{align*} \lambda_n &= \left( \frac{n \pi}{\ln b} \right)^2 \\ \end{align*}

If we multiply by $$1/x$$ and rearrange we can get this in Sturm-Liouville form with $$p(x) = x, q(x) = 0, \sigma(x) = 1/x$$:

\begin{align*} x \frac{d^2\phi}{dx^2} + \frac{d\phi}{dx} + \frac{1}{x} \lambda \phi &= 0 \\ \frac{d}{dx} \left( x \frac{d\phi}{dx} \right) + \lambda \frac{1}{x} \phi &= 0 \\ \end{align*}

If we look at the Rayleigh quotient we can see there are no negative eigenvalues and we can manually verify that zero is not an eigenvalue.

Assume a trial solution of $$\phi(x) = x^m, \phi'(x) = m x^{m-1}, \phi''(x) = m(m-1)x^{m-2}$$ such that:

\begin{align*} x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi &= 0 \\ m (m-1) x^m + m x^m + \lambda x^m &= 0 \\ \left( m^2 + \lambda \right) x^m &= 0 \\ \lambda &= -m^2 \\ \end{align*}

Since we know that $$\lambda > 0$$, then $$m^2 < 0$$, which means that $$m = i r$$ where $$r \in \mathbb{R}$$. However, this can't satisfy the boundary conditions:

\begin{align*} \phi(x) &= e^{ir} \\ \phi(1) = 0 &\neq 1^{ir} = 1 \\ \end{align*}

I'm stuck on what else to try.

You have to take the change of variable $$y=\ln x$$ and the equation becomes $$\frac{d^2\phi}{dy^2}+\lambda\phi=0$$ that is a standard form having as a solution $$\phi(y)=A\cos\sqrt{\lambda} y+B\sin\sqrt{\lambda} y.$$ I think now you can go on from here.