1D Fourier transform of 2D function with shift

Question

Say you have a 2D function $f\left(x,t\right)$ whose 1D Fourier transform in time is \[F\left(x,\omega\right)=\int_{-\infty}^{\infty}f\left(x,t\right)e^{-i \omega t}dt.\] And you now introduce a time-dependent translation $x \rightarrow x + vt$ where $v$ is a constant. How is the 1D Fourier transform of the translated function, \[\int_{-\infty}^{\infty}f\left(x+vt,t\right)e^{-i \omega t}dt,\] related to the original 1D Fourier transform, $F\left(x,\omega\right)$?

I know that for 2D Fourier transforms the effect of the translation in the time domain is a translation in the frequency domain: \[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x+vt,t\right)e^{-i \omega t}e^{-i k x} \,dt \, dx = F\left(k,\omega + vk\right).\] But I'm not sure how to do the 1D case.

Answer 1

This is just a hint and not full solution.

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I think easiest way to approach this is by linear algebra

In this light, what we are doing is in practice a linear coordinate transformation:

$$\begin{bmatrix}x_1\\t_1\end{bmatrix} = {\bf M}\begin{bmatrix}x\\t\end{bmatrix}$$

What is our mystery matrix $\bf M$?

$${\bf M}=\begin{bmatrix}1&v\\0&1\end{bmatrix}$$

and you will see that

$$\begin{bmatrix}x_1\\t_1\end{bmatrix}={\bf M}\begin{bmatrix}x\\t\end{bmatrix} = \begin{bmatrix}1&v\\0&1\end{bmatrix}\begin{bmatrix}x\\t\end{bmatrix}=\begin{bmatrix}x+vt\\t\\\end{bmatrix}$$

We know that multidimensional F-T is linear, orthogonal and separable in the variables.

How can this help us now?

Answer 2

I will expand my comment a bit. I am used to work with analytic functions, so I will assume $f$ is analytic. My reasoning here is that if there is a simple formula, it will work for analytic functions, so if we find it here, we may be able to prove that it holds in the general setting. So we can get some insight using this, but definitely cannot be a proper proof. (Also I will keep passing infinite sums and derivatives in and out of integrals so some care is needed to make sense of what follows, but assuming that the functions are "nice enough" everything works.)

Ok, first as an example let's look at a function of one variable. Let $f:\mathbb{C}\to\mathbb{C}$ be our function and $F$ its Fourier transform. Then we have: $$f(t+c) = \sum_{n\ge0}\frac{1}{n!}f^{(n)}(t)c^n$$ and $$ \mathcal{F}[f(t+c)](\omega) = \int_{-\infty}^{\infty}f(t+c)e^{i \omega t}dt \\ = \int_{-\infty}^{\infty}\sum_{n\ge0}\frac{1}{n!}f^{(n)}(t)c^ne^{i \omega t}dt \\ = \sum_{n\ge0}\frac{c^n}{n!}\int_{-\infty}^{\infty}f^{(n)}(t)e^{i \omega t}dt \\ = \sum_{n\ge0}\frac{(i\, c\, \omega)^n}{n!} F(\omega) \\ = F(\omega) \sum_{n\ge0}\frac{(i\, c\, \omega)^n}{n!} \\ = e^{i\,c\,\omega} F(\omega) $$ as expected. Admittedly this is not the best way to prove this property of the Fourier transform, but I just wanted to convince you that it works.

Let's go to the actual question now. Let's define $$F\left(x,\omega\right)=\int_{-\infty}^{\infty}f\left(x,t\right)e^{i \omega t}dt.$$ Then we get $$\partial_x F\left(x,\omega\right)=\partial_x\int_{-\infty}^{\infty}f(x,t)e^{i \omega t}dt=\int_{-\infty}^{\infty}f^{(1,0)}(x,t)e^{i \omega t}dt.$$ As in my comment we can write $$f(x+ut,t)= \sum_{n\ge0} \frac{u^n t^n}{n!}f^{(n,0)}(x,t).$$ So we get: $$\int_{-\infty}^{\infty}f(x+vt,t)e^{-i \omega t}dt = \int_{-\infty}^{\infty} \sum_{n\ge0} \frac{u^n t^n}{n!}f^{(n,0)}(x,t) e^{-i \omega t}dt \\ = \sum_{n\ge0} \frac{u^n}{n!}\partial_x^n \int_{-\infty}^{\infty} f(x,t) t^ne^{i \omega t}dt \\ = \sum_{n\ge0} \frac{i^{-n}u^n}{n!}\partial_x^n \partial_\omega^n \int_{-\infty}^{\infty} f(x,t) e^{i \omega t}dt \\ = \sum_{n\ge0} \frac{i^{-n}u^n}{n!}\partial_x^n \partial_\omega^n F(x,\omega). $$

I cannot really think a way to simplify this last sum, though it seems that there should be one. If I figure something I will add it later.

Answer 3

Look at this simple trick: denoting $G(x,v,\omega) \equiv \int_{-\infty}^{\infty}f(x+vt,t)e^{-i \omega t} \,dt$, your formula for two dinensional Fourier transformation may be rewritten as $$\int_{-\infty}^{\infty} G(x,v,\omega) e^{-i k x} \, dx = F\left(k,\omega + vk\right).$$

Considering $v$ and $\omega$ as parameters, $F_{v,\omega}(k)$ is just a 1D Fourier transformation of $G_{v,\omega}(x)$. Thus you may recover $G_{v,\omega}$ as inverse Fourier transformation of $F_{v,\omega}$ for any parameters $v$ and $\omega$.