Consider a typical function written in standard $2\text{D}$ polar form:

\begin{equation} f(\underline{r})=f(r,\theta)=\sum_{n=-\infty}^{\infty} f_n(r) e^{in\theta} \end{equation}

executing the Laplacian of f, we have:

\begin{equation} \begin{split} \nabla^2 f(\underline{r})=& \Bigg(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}\Bigg) \sum_{n=-\infty}^{\infty} f_n(r) e^{in\theta}\\ =& \sum_{n=-\infty}^{\infty} \Bigg(\frac{d^2 f_n}{d r^2}+\frac{1}{r}\frac{df_n}{d r}-\frac{n^2 f_n}{r^2}\Bigg) e^{in\theta}\\ =&\sum_{n=-\infty}^{\infty} \nabla^2_n f_n\ e^{in\theta} \end{split} \end{equation}

where: \begin{equation} \nabla_n^2=\frac{d^2}{d r^2}+\frac{1}{r}\frac{d}{d r}-\frac{n^2}{r^2} \end{equation}

the 2D Fourier Transform of $\nabla^2 f(\underline{r})$ is given by the following series:

\begin{equation} \mathbb{F}_{2D}\Big\{ \nabla^2 f(\underline{r})\Big\}=\sum_{n=-\infty}^{\infty} 2\pi i^{-n} e^{in\psi} \int_{0}^{\infty} \Bigg(\frac{d^2 f_n}{d r^2}+\frac{1}{r}\frac{df_n}{d r}-\frac{n^2 f_n}{r^2}\Bigg)J_n(\rho r)\ rdr \end{equation}

A simple application of integration by parts along with the definition of a Bessel function gives:

\begin{equation} \int_{0}^{\infty} \nabla^2_n f_n J_n(\rho r)\ rdr = -\rho^2 \int_{0}^{\infty} f_n J_n(\rho r)\ rdr\ \ \ \ \ (*) \end{equation}

how could i proof the equation (*)?

  • $\begingroup$ Didn't you prove it with your calculations? $\endgroup$
    – md2perpe
    Commented Oct 1, 2020 at 16:13
  • $\begingroup$ No, i reported the derivation of the laplacian coefficient of the fourier series for the function f. But i don't have understand how i can dimostrate the equation (*) by only applying the integration by parts rule $\endgroup$
    – Marco Toni
    Commented Feb 17, 2021 at 12:23

1 Answer 1


The differential equation for the Bessel functions $J_n$, $$x^2 J_n''(x) + x J_n'(x) + (x^2-n^2) J_n(x) = 0$$ can be written as $$\left( x J_n'(x) \right)' + (x - \frac{n^2}{x}) J_n(x) = 0,$$ which we will use later.

The given left hand expression can be split into a couple of integrals: $$\begin{align} \int_{0}^{\infty} \nabla^2_n f_n \, J_n(\rho r) \, r \, dr &= \int_{0}^{\infty} \left( f_n''(r) + r^{-1} f_n'(r) - n^2 r^{-2} f_n(r) \right) \, J_n(\rho r) \, r \, dr \\ &= \int_{0}^{\infty} \left( r f_n''(r) + f_n'(r) - n^2 r^{-1} f_n(r) \right) \, J_n(\rho r) \, dr \\ &= \int_{0}^{\infty} \left( (r f_n'(r))' - n^2 r^{-1} f_n(r) \right) \, J_n(\rho r) \, dr \\ &= \int_{0}^{\infty} (r f_n'(r))' \, J_n(\rho r) \, dr - n^2 \int_{0}^{\infty} r^{-1} f_n(r) \, J_n(\rho r) \, dr \end{align}$$

For the first integral we use integration by parts. The boundary terms all vanish so $$\begin{align} \int_{0}^{\infty} (r f_n'(r))' \, J_n(\rho r) \, dr &= - \int_{0}^{\infty} r f_n'(r) \, \frac{\partial}{\partial r} \left(J_n(\rho r)\right) \, dr \\ &= - \int_{0}^{\infty} r f_n'(r) \, \rho J_n'(\rho r) \, dr \\ &= - \int_{0}^{\infty} f_n'(r) \, \rho r J_n'(\rho r) \, dr \\ &= \int_{0}^{\infty} f_n(r) \, \frac{\partial}{\partial r} \left(\rho r J_n'(\rho r)\right) \, dr \\ &= \int_{0}^{\infty} f_n(r) \, \rho \frac{\partial}{\partial (\rho r)} \left(\rho r J_n'(\rho r)\right) \, dr \\ &= \int_{0}^{\infty} f_n(r) \, \rho ( \frac{n^2}{\rho r} - \rho r) J_n(\rho r) \, dr \\ &= \int_{0}^{\infty} f_n(r) \, ( \frac{n^2}{r} - \rho^2 r) J_n(\rho r) \, dr \end{align}$$ where the rewritten differential equation for $J_n$ has been used.

Adding the second integral from the first calculation results in $$\int_{0}^{\infty} \nabla^2_n f_n \, J_n(\rho r) \, r \, dr = -\rho^2 \int_{0}^{\infty} f_n J_n(\rho r)\, r \, dr.$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .