Fourier transform of Laplacian in polar coordinates

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.,

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

• Have a look at Hankel transform. Jun 29, 2021 at 9:07

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.