Getting Green's Function for Laplace's Equation in Cylindrical Coordinates I am trying to understand a derivation for finding the Green's function of Laplace's eq in cylindrical coordinates.
Let the Green's function be written as: $G(r,\theta,z,r',\theta',z') = G(\mathbf{r},\mathbf{r}')$
We know that in cylindrical coordinates
$$\nabla^{2}G(\mathbf{r},\mathbf{r}')=\delta(\mathbf{r}-\mathbf{r}')=\frac{1}{r}\delta(r-r')\delta(\theta-\theta')\delta(z-z')$$
Using the cylindrical Laplacian we can then write:
$$\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial G}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 G}{\partial \theta^2} + \frac{\partial^2 G}{\partial z^2} = \frac{1}{r}\delta(r-r')\delta(\theta-\theta')\delta(z-z')$$
Using the identities:
$$\delta(\theta-\theta') = \frac{1}{2\pi}\sum^{\infty}_{m=-\infty}e^{im(\theta-\theta')}$$
$$\delta(z-z') = \frac{1}{2\pi}\int^{\infty}_{-\infty}e^{ik(z-z')}dk$$
We find that
$$G(\mathbf{r},\mathbf{r}') = \frac{1}{4\pi^2}\sum^{\infty}_{m=-\infty}\int^{\infty}_{-\infty}g_{m}(r,r')e^{ik(z-z')}e^{im(\theta-\theta')}dk$$
$$\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial g_{m}}{\partial r}\right) - \left(k^2+\frac{m^2}{r^2} \right)g_{m} = \frac{1}{r}\delta(r-r')$$
I'm getting confused on the last step. It seems that the Green's function is assumed to be $G(r,\theta,z,r',\theta',z') = R(r)Q(\theta)Z(z)$ and this is plugged into the cylindrical Laplacian but I'm not seeing how to get $G(r,\theta,z,r',\theta',z')$ in the form shown or how to get that ODE in terms of $g_{m}$. Are there any ideas on how to do this intermediate step?
 A: The expression
$$G(\mathbf{r},\mathbf{r}') = \frac{1}{4\pi^2}\sum^{\infty}_{m=-\infty}\int^{\infty}_{-\infty}g_{m}(r,r')e^{ik(z-z')}e^{im(\theta-\theta')}dk\tag{∗}$$
is nothing but a Fourier expansion of some function of $z$ and $\theta$, with $g_m(r,r')$ being the expansion coefficients. It's not something you should get: it's an ansatz.
Now, to get your ODE you can go the following way. Use your identities to express the whole RHS of your PDE (the product of deltas) in the form
$$\sum_{m=-\infty}^\infty\int_{-\infty}^\infty...\mathrm{d}k.\tag1$$
Then insert $G(\mathbf r, \mathbf r')$ in the form $(*)$ into the PDE. Apply each differential operator termwise, and then move all the summands-integrands under the single sum-integral sign.
You now get the equation
$$\sum_{m=-\infty}^\infty\int_{-\infty}^\infty\exp\big(ik(z-z')+im(\theta-\theta')\big)f_\mathrm{ODE}(k,m,r)\,\mathrm{d}k=0,\tag2$$
where $f_\mathrm{ODE}(k,m,r)$ is the $\mathrm{LHS}-\mathrm{RHS}$ of your ODE.
The LHS of $(2)$ is a Fourier expansion of some function of $z-z'$ and $\theta-\theta'$, where $f_\mathrm{ODE}(k,m,r)$ denotes the Fourier coefficients. By uniqueness of Fourier expansion, the only way for the whole LHS of $(2)$ to be zero is for $f_\mathrm{ODE}$ to be zero for all $k$ and $m$. Since we also want this equation to hold for all $r$, we finally get
$$f_\mathrm{ODE}=0\;\forall\, k,m,r,\tag3$$
which is equivalen to your ODE.
A: To make the typing easier I will set $r'$, $\theta'$, and $z'$ to zero (wlog). So, now G is considered to be a function of just $r$, $\theta$, and $z$ (wlog).
Any[1] function of $r$, $\theta$, and $z$ can be expanded in a double fourier series like:
$$
G(r,\theta, z) = \frac{1}{2\pi}\sum_m \frac{1}{2\pi}\int dk g_{mk}(r)e^{ikz}e^{im\theta}\;,
$$
where $k$ is a continuous parameter and $m$ is a discrete parameter.
Plug this expansion into your equation for $G$ and find:
$$
\frac{1}{2\pi}\sum_m \frac{1}{2\pi}\int dk e^{ikz}e^{im\theta} \left[\frac{1}{r}\frac{d}{dr}r\frac{dg_{mk}(r)}{dr}
-\frac{n^2}{r^2}g_{mk}(r) - k^2 g_{mk}(r)\right]
=\frac{1}{r}\delta(r)\frac{1}{2\pi}\sum_{m'} \frac{1}{2\pi}\int d{k'} e^{ik'z}e^{im'\theta}
$$
Because the $e^{im\theta}$ and $e^{ikz}$ are a complete orthonormal set, each side of the equation holds term by term for all $m$ and $k$. So we can "drop the sums" over the dummy variable on each side and we find:
$$
\left[\frac{1}{r}\frac{d}{dr}r\frac{dg_{mk}(r)}{dr}
-\frac{m^2}{r^2}g_{mk}(r) - k^2 g_{mk}(r)\right]
=\frac{1}{r}\delta(r)
$$

[1] Some caveats may apply, but we don't care.
