I am trying to solve $\nabla^2 u = 0$ in cylindrical polar coordinates (and radial symmetry) $$ \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial z^2}=0$$
with boundary conditions $\partial u / \partial z = 1 $ for $ 0 \geq r \geq a , \ \ 0$ otherwise (for $z=0$ and $z=1$).
I am trying to use the Fourier transform to get an ODE and solve it, however I don't know any tricks to find the Fourier transform of the term
$$\frac{1}{r} \frac{\partial u}{\partial r}$$
Is it even possible to solve this using the Fourier transform? I am choosing it because the transform of the boundary condition is simple: $sin(a \omega)/\pi \omega$ and this can be applied at $z=0$ and $z=1$ once I get the ODE.
You can think of the Fourier transform in 2D, transformed to polar coordinates, assuming that the function does not depend on $\theta$. If $\xi=\hat{x}\xi_1+\hat{y}\xi_2$, $$ \hat{f}(\xi) = \frac{1}{2\pi}\int\int f(r)e^{-ix\xi_1-iy\xi_2}dxdy \\ = \frac{1}{2\pi}\int_{0}^{\infty}\int_{0}^{2\pi}f(r)e^{-i\xi_1 r\cos\theta-i\xi_2 r\sin\theta}d\theta rdr. $$ Letting $\xi = \hat{x}\rho\cos\phi+\hat{y}\rho\sin\phi$, \begin{align} \hat{f}(\xi)&=\frac{1}{2\pi}\int_{0}^{\infty}f(r)\int_{0}^{2\pi}e^{-i\rho r\cos\theta\cos\phi-i\rho r\sin\theta\sin\phi}d\theta rdr \\ &= \frac{1}{2\pi}\int_{0}^{\infty}f(r)\int_{0}^{2\pi}e^{-i\rho r\cos(\theta-\phi)}d\theta rdr\\ &= \int_{0}^{\infty}f(r)\frac{1}{2\pi}\int_{0}^{2\pi}\sin(\rho r\sin\theta)d\theta rdr \\ &= \int_{0}^{\infty}f(r)\frac{1}{\pi}\int_{0}^{\pi}\sin(\rho r\sin\theta)d\theta rdr \\ &=\int_{0}^{\infty}f(r)J_0(\rho r)rdr, \end{align} where $J_0$ is the zero order Bessel function $$ J_0(r) = \frac{1}{\pi}\int_{0}^{\pi}\sin(r\sin\theta)d\theta. $$ This the Hankel transform of order $0$, and it is the same as the ordinary Fourier transform of a function that depends on the radial coordinate only, and not on the polar angular coordinate. The inverse Fourier transform is the same as the forward tranform. So the inversion formula is $$ f(r)=\int_0^{\infty}J_0(\rho r)\left(\int_{0}^{\infty}f(r)J_0(\rho r)r dr\right)\rho d\rho $$ Just as the Fourier transform converts differentiation in multiplication, the Hankel transform converts the Bessel operator $\frac{d^{2}}{dr^2}+\frac{1}{r}\frac{d}{dr}$ into multiplication by $\rho^2$: \begin{align} \mathscr{H}_{0}f & =\int_{0}^{\infty}J_{0}(\rho r)f(r)rdr \\ \mathscr{H}_{0}\left\{\frac{df^2}{dr^2}+\frac{1}{r}\frac{df}{dr}\right\} & = \int_{0}^{\infty}J_{0}(\rho r)\frac{d}{dr}\left(r\frac{df}{dr}\right)dr \\ & = \left.rJ_0(\rho r)\frac{df}{dr}\right|_{r=0}^{\infty}-\int_{0}^{\infty}r\frac{d J_0(\rho r)}{dr}\frac{df}{dr}dr \\ & = \int_{0}^{\infty}\frac{d}{dr}\left(r\frac{dJ_0(\rho r)}{dr} \right)f(r)dr \\ & = \left. \left[r J_0(\rho r)\frac{df}{dr}-rf(r)\frac{dJ_0(\rho r)}{dr}\right] \right|_{r=0}^{\infty} \\ & +\int_{0}^{\infty}\left[\frac{d^2 J_{0}(\rho r)}{dr^2}+\frac{1}{r}\frac{d J_0(\rho r)}{dr}\right]f(r)rdr \\ & = \left. \left[r J_0(\rho r)\frac{df}{dr}-rf(r)\frac{d}{dr}J_0(\rho r) \right] \right|_{r=0}^{\infty}-\rho^2 \mathscr{H}_0 f. \end{align} The above requires showing that $$ \left(\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr}\right)J_0(\rho r) = -\rho^2 J_0(\rho r). $$ Using the above formula for $J_0$ allows you to do that. In other words, $J_0(\rho r)$ is an eigenfunction of the Bessel operator $\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr}$. Assuming the evaluation terms vanish at $r=0$ and at $r=\infty$, your PDE is transformed into $$ -\rho^2\mathscr{H}_0 u+\frac{\partial^2}{\partial z^2}\mathscr{H}_{0} u = 0. $$