Is solving Poisson's equation in Polars different from Cartesian?

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I'm having trouble figuring out how to separate variables in polar coordinates in 2D. In cartesian coordinates it is fairly simple to use eigenfunction ideas because I can group together the x, y variables as a single $\mathbf{r}$, but I am unsure about polar coordinates as I have to deal with an $r, \theta$ terms. I would very much appreciate it if someone could aid my understanding of this problem.

My thinking

For a normal plane region using cartesian coordinates:

$$ \nabla^2 \phi = - f(\mathbf{r}), \space r < R$$

Expanding $\space f$ in terms of eigenfunctions, we seek a solution on terms of eigenfunctions of the Laplacian, and use the property of the eigenfunction, $\nabla^2 \phi_{\lambda}(\mathbf{r}) = - \lambda \phi_{\lambda}(\mathbf{r})$

$$ \nabla^2\sum_{\lambda} \space U_{\lambda} \space \phi_{\lambda}(\mathbf{r}) = \sum_{\lambda} \space f_{\lambda} \space \phi_{\lambda}(\mathbf{r}) $$

$$ \sum_{\lambda} - \lambda\space U_{\lambda} \space \phi_{\lambda}(\mathbf{r}) = \sum_{\lambda} \space f_{\lambda} \space \phi_{\lambda}(\mathbf{r}) $$

So I can write the series expansion of the solution as

$$ u(\mathbf{r}) = \sum_{\lambda} \frac{-f}{\lambda} \phi_{\lambda}(\mathbf{r}) $$

But for polar coordinates I can't group $x,y$ as one $\mathbf{r}$, I need to separate the $r, \theta $ terms:

$$ f(r,\theta) = \sum_{\lambda}\space f_{\lambda}(r) \space \phi_{\lambda}(\theta) $$ $$ u(r, \theta) = \sum_{\lambda}\space U_{\lambda}(r) \space \phi_{\lambda}(\theta) $$

Poisson's equation becomes $$ u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2} u_{\theta \theta} = -f(r, \theta), \space r < R$$

I think I have to consider the eigenfunctions $\phi_{\lambda}(\theta)$ instead of the eigenfunctions $\phi_{\lambda}(\mathbf{r})$ this time - so I don't think I can use the previous approach and have to expand $f(r,\theta) $ as a Fourier Series, then substitute $u(r, \theta)$ into Poisson's equation and match the radial coefficients $ F(r) $ and $U(r) $ where $F(r)$ are Fourier coefficients and depend on $r$

Then I solve the resulting PDEs to get an expression for $u(r, \theta)$ , does my thought process seem ok?