I'm having a fairly large conceptual block on my understanding of eigenfunctions - I've tried out two different methods that yield embarrassingly different answers, and I would appreciate if someone could glance at them quickly and tell me which method is completely wrong and maybe where the flaw lies.
Short background: we are considering Poisson's equation in a disk, with homogenous boundary conditions $ u(R,\theta) = 0$:
$$ \nabla^2 \phi = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2} u_{\theta \theta} = - f(r, \theta), \space r < R$$
Method 1 (a bit suspect)
I expanded $ f(r,\theta)$ in terms of eigenfunctions:
$$ f(r,\theta) = \sum_{\lambda}\space [f_{\lambda}(r) ] \space[\psi_{\lambda}(\theta)] $$
As f is periodic I found the coefficients $f_{\lambda}(r)$ through a real Fourier series expansion
So we want an eigenfunction of the form
$$ \phi_{\lambda}(r, \theta) = R_{\lambda}(r) * \psi_{\lambda}(\theta) $$
Using the property of the eigenfunction, $ \nabla^2 \phi_{\lambda} = -\lambda \phi_{\lambda}$, I substituted this into Poisson's equation:
$$ -\lambda \phi_{\lambda} = -f(r,\theta) $$
After expanding this out + using orthogonality, I found:
$$ \sum_{\lambda} \space \lambda [R_{\lambda}(r) ] \space[\psi_{\lambda}(\theta)] = \sum_{\lambda} \space [f_{\lambda}(r) ] \space[\psi_{\lambda}(\theta)]$$
$$ \phi_{\lambda}(r, \theta) = \sum_{\lambda} [\frac{f_{\lambda}(r)}{\lambda}] [\psi_{\lambda}(\theta)]$$
Where I can evaluate the $ f_{\lambda}(r)$ as Fourier Coefficients.
Method 2 (I am leaning toward this)
Exactly the same buildup, only this time I substitute the separated solution into Poisson's equation directly.
$$ \phi_{\lambda}(r, \theta) = R_{\lambda}(r) * \psi_{\lambda}(\theta) $$
The eigenfunctions of the $-\frac{\partial^2}{\partial \theta^2}$ operator are $ \psi_{\lambda}(\theta) = [A_n \space \mathrm{cos}(n\theta) + B_n \space \mathrm{sin}(n\theta)]$, so:
$$ [R_{\lambda}''(r) + \frac{1}{r} R_{\lambda}'(r) - \frac{n^2}{r^2} R_{\lambda}(r)] * [A_n \space \mathrm{cos}(n\theta) + B_n \space \mathrm{sin}(n\theta)] = - f(r, \theta) $$
I think $f$ can be expanded as a Fourier series, like so:
$$ f(r,\theta) = \frac{\alpha_0 (r)}{2} + \sum_{n=1}^{\infty}[\alpha_n (r) cos(n\theta) + \beta(r) sin(n\theta)]$$
So I get two (ugly) ODEs for $R(r)$ by orthogonality, which I can then use to fully determine $\phi(r,\theta)$
$$ [A_n][R_{\lambda}''(r) + \frac{1}{r} R_{\lambda}'(r) - \frac{n^2}{r^2} R_{\lambda}(r)] = -\alpha_n(r), n \geq 1$$
$$ [A_0][R_{\lambda}''(r) + \frac{1}{r} R_{\lambda}'(r)] = \frac{-\alpha_0(r)}{2}, n = 0$$
I know that one of these methods is flat out wrong, but I'm struggling to see where I went wrong, can anyone help me shed some light on the problem?