Dear Math enthusiasts,

I am struggeling with a problem for which a solution is already given to me, but I can just not see why it is true. Here is the setting: I am given a function $f(x,y,t)$. It's well behaved let's say. Smooth and things like that. Now, this function should be reexpressed in the following form $$f(x,y,t) = \int g(k_x,k_y,w) {\rm e}^{-\jmath (k_x x + k_y y - w t)} d k_x dk_y dw. \tag{1}\label{eq1}$$ If it were only this it would be very simple, $g$ is some sort of 3-D Fourier transform of $f$. However, the trouble is that the variables $k_x, k_y, w$ are not independent. They need to satisfy the relation $k_x^2 + k_y^2 = w^2$. Therefore, I would claim that the correct form of the above expression should be $$f(x,y,t) = \int \oint_{S(w)} g(k_x,k_y,w) {\rm e}^{-\jmath (k_x x + k_y y - w t)} d \begin{bmatrix} k_x\\ k_y\end{bmatrix} dw,\tag{2}\label{eq2}$$ where $S(w)$ is a circle of radius $w$, so that the inner integral goes over the perimeter of the circle and the outer over circle radii.

My question is essentially: given a target function $f(x,y,t)$, how do I find $g(k_x, k_y, w)$ such that \eqref{eq2} is true for every point $x,y,t$?

The reference I have for this simply redefines $g(k_y,k_y,w)$ into $h(k_x,w)$ since only two variables are independent (I'm assuming this means $h(k_x,w)=g(k_x,\pm \sqrt{w^2-k_x^2},w)$ though that's never written) and uses this in the first integral. This leads to $$f(x,y,t) = \int \int h(k_x,w) {\rm e}^{-\jmath (k_x x+k_y y - w t)} d k_x dw = \int \int \tilde{g}(k_x,y,w) {\rm e}^{-\jmath (k_x x - w t)} d k_x dw,\tag{3}\label{eq3}$$ where $\tilde{g}(k_x,y,w) = h(k_x,w) {\rm e}^{-\jmath k_y y} $ (again, omitting the argument $k_y$ for me can only mean the implicit relation $k_y = \pm \sqrt{w^2-k_x^2}$). From \eqref{eq3}, they claim that $f$ is the 2-D Fourier transform of $\tilde{g}$ along the first and third dimension so that all we need to do to find $\tilde{g}$ is $$ \tilde{g}(k_x,y,w) = \frac{1}{4\pi^2} \int \int f(x,y,t) {\rm e}^{\jmath (k_x x-wt)} dx dt \tag{4}\label{eq4}$$ which gives $h$ as $ h(k_x,w) = \tilde{g}(k_x,y,w) {\rm e}^{\jmath k_y y} $.

However, I have the feeling this is oversimplifying things a bit. I'm lacking rigor. My feeling is that the original problem \eqref{eq2} may not have a unique solution (due to the variable dependence) and a particular one was chosen here. Integration limits are always skipped which may be a delicate issue (after all, $k_x$ should never leave the interval $[-w,w]$, maybe this can be solved by defining $h$ zero outside this support). The fact that we cannot directly solve for $k_y$ (due to the $\pm$) troubles me. Overall I have a vague feeling that this may work but I cannot quite put my finger on it and really understand what's going on.

Would anyone be able to enlighten me how to treat such problems rigorously?

edit: A concrete example I am interested in is the function $f(x,y,t)={\rm e}^{-\jmath \left(\omega_0 t - \frac{\omega_0}{c}\sqrt{(x-x_0)^2+(y-y_0)^2}\right)}$. I'm awarding a bounty to anyone who can systematically explain me how to find the (set of) function(s) $g(k_x,k_y,\omega)$ that satisfy \eqref{eq2} for a given $f(x,y,t)$ everywhere. A concrete example may be helpful for the understanding, it can be the one I provided in this paragraph, but I'm also happy with any other non-trivial example, as long as it aids the understanding.

  • 1
    $\begingroup$ $f$ seems some kind of Radon transform of $g$: these problems are studied within the realm of integral geometry. $\endgroup$ Nov 7, 2018 at 16:04
  • $\begingroup$ Thanks for the suggestion. I added a tag. I know Radon transform only as integrals over lines, but you are right, this might be strongly related. $\endgroup$
    – Florian
    Nov 7, 2018 at 16:23

1 Answer 1


For this problem, it is much more convenient to introduce the polar coordinates ($\phi \in [0,2\pi)$) $$k_x = w \cos(\phi), \qquad k_y=w \sin(\phi)$$ which fulfill the constraint explicitly.

We then define $$f(x,y,t) = \int_0^\infty \int_{0}^{2\pi} g(w \cos\phi, w \sin\phi,w)e^{i w (t- x \cos\phi - y\sin\phi)} w d\phi dw . \tag{1} $$ Note that $w>0$ is required as otherwise the constraint $w^2 = k_x^2 + k_y^2$ does not have a solution.

We will apply, the 2D Fourier inversion theorem (in polar coordinates) $$ H(\rho, \theta) = \frac{1}{2\pi} \int_0^{\infty}\int_0^{2\pi} h(r,\phi) e^{-i \rho r \cos(\phi-\theta)} r\,d\phi \,dr \tag{2}$$ $$h(r,\phi) = \frac{1}{2\pi} \int_0^{\infty}\int_0^{2\pi} H(\rho,\theta) e^{i \rho r \cos(\phi-\theta)} \rho\,d\theta \,d\rho \tag{3}$$

We introduce the function (that implicitly depends on $t$) $$h(w,\phi;t) = 2\pi g(w \cos\phi, w \sin\phi, w) e^{i w t} .$$ Then, we see that (1) is equivalent to (2) with $$f(\rho \cos\theta,\rho \sin\theta ,t)= H(\rho, \theta;t).$$ The inverse expression is thus given by (3). In particular, we have that $$2\pi g(w \cos\phi, w \sin\phi, w) e^{i w t} =\frac{1}{2\pi} \int_0^{\infty}\int_0^{2\pi} f(\rho \cos\theta,\rho \sin\theta ,t) e^{i \rho w \cos(\phi-\theta)} \rho\,d\theta \,d\rho $$ or equivalently $$ g(w \cos\phi, w \sin\phi, w) =\frac{1}{(2\pi)^2} e^{-i w t}\int_0^{\infty}\int_0^{2\pi} f(\rho \cos\theta,\rho \sin\theta ,t) e^{i \rho w \cos(\phi-\theta)} \rho\,d\theta \,d\rho. \tag{4} $$ We note that $g$ is only specified on the points that fulfill the constraint!


Given the request of the OP, I will outline the result of a concrete example with $$f(x,y,t) = e^{i \omega_0(\sqrt{x^2+y^2} -t)}$$ which is inspired by the function given in the question; (In order to have a converging result, we assume that $\operatorname{Im}(\omega_0)>0$).

Using the inversion formula (4), one easily obtains $$ g(w \cos\phi, w \sin\phi, w) = \frac{i \omega_0 e^{-i (\omega_0+w) t}}{2\pi (w^2 -\omega_0^2)^{3/2}} .$$

  • $\begingroup$ @Florian: my response uses the 2D Fourier transform to obtain an inverse (I give explicit limits). Note that $g$ can only be obtained for the points that fulfill the constraints (parameterized by $w$ and $\phi$). This is of course to be expected. If you agree with the initial definition (1) (which I hope is the same as your equation (2)), then my post should give you the inverse. If you want, I can apply the transform to your example... $\endgroup$
    – Fabian
    Nov 17, 2018 at 18:28
  • $\begingroup$ Wow, thanks, this looks very helpful! Polar coordinates, why didn't I think of that. This solution disagrees with what I was given in (3) and (4), right? It would be nice to see how this compares with an example. Doesn't have to be the one I provided necessarily. $\endgroup$
    – Florian
    Nov 19, 2018 at 8:34
  • $\begingroup$ @Florian: I have added an example. You can easily check numerically that it works. I do not know how to compare it to your results, as I do not know how to understand $h(k_x,w) = \tilde{g}(k_x,y,w) {\rm e}^{i k_y y}$ (the right hand depends on $y$ whereas the left hand does not). $\endgroup$
    – Fabian
    Nov 19, 2018 at 14:52
  • $\begingroup$ Awesome, thanks! I think this helps. $\endgroup$
    – Florian
    Nov 19, 2018 at 15:24

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