In 2D cartesian coordinate, we know the spatial Fourier transform of the Laplacian is

$F(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})=(ik_x)^2+(ik_y)^2$, where $k_x$ and $k_y$ are the spatial frequencies.

The Laplacian in polar coordinate is (assume no angular dependence):

$\nabla^2=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}$

My question is: Just as the Cartesian Fourier transform of the Laplacian, but what is the corresponding Fourier transform of the Laplacian operator in polar coordinate? i.e.,

$F(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r})=?$

  • $\begingroup$ Have a look at Hankel transform. $\endgroup$ Jun 29, 2021 at 9:07

1 Answer 1


Let assume that $f(x,y)$ has circular symmetry: $f(\rho \cos \theta,\rho \sin \theta)=g(\rho)$

Let compute Fourier transform of $f$: $$ \mathcal{F}(f)(\xi,\eta)=\frac{1}{2\pi}\int\int f(x,y)e^{-i(\xi x+\eta y)}dxdy $$ use cylindrical coordinates: $x=\rho \cos \theta$, $y=\rho \sin \theta$: $$ \mathcal{F}(f)(\xi,\eta)=\frac{1}{2\pi}\int_0^{\infty}\int_0^{2\pi} \rho g(\rho)e^{-i(\xi \rho \cos \theta +\eta \rho \sin \theta)}d\rho d\theta $$ In order to use trigonometric identity $\cos(\theta - \alpha)=\cos \theta \cos \alpha + \sin \theta \sin \alpha$ we also introduce cylindrical coordinates for $\xi,\eta$ : $\xi = s \cos \alpha$ and $\eta = s \sin \alpha$. You get: $$ \mathcal{F}(f)(\xi,\eta)=\frac{1}{2\pi}\int_0^{\infty}\int_0^{2\pi} \rho g(\rho)e^{-i\rho s \cos (\theta-\alpha) }d\rho d\theta = \frac{1}{2\pi}\int_0^{\infty}\int_0^{2\pi} \rho g(\rho)e^{-i\rho s \cos \beta }d\rho d\beta $$ We then use Bessel function integral formula ($s>0$): $$ J_0(s)=\frac{1}{2\pi} \int_0^{2\pi}e^{-i s \cos \beta}d\beta $$ to get the Hankel transform $\mathcal{H}(f)(s)$ of $f$ $$ \mathcal{F}(f)(\xi,\eta)=\mathcal{H}(f)(s)=\int_0^\infty \rho g(\rho) J_0(\rho s)d\rho $$

One can show that Hankel transform diagonalize Laplace operator in cylindrical coordinate. If $f$ has circular symmetry as before (and $g(\rho)\rightarrow 0$ when $\rho\rightarrow\infty$), let $F(s)=\mathcal{H}(f)(s)$, then: $$ \mathcal{H}(\frac{d^2 g}{d\rho^2}+\frac{1}{\rho}\frac{d g}{d\rho})(s) = -s^2 \mathcal{H}(g)(s) $$ This property is useful for solving Laplace equation in cylindrical coordinates.


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