Convolution between distributions over the real number line, as it is mentioned in the optics course in my uni and also in wolfram is defined as:
$$f(t) * g(t)=\int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau$$
For convolutions, there's a nice theorem called Convolution Theorem which states for any distributions $f$ and $g$:
$$\mathcal{F}(f(t)g(t))=\mathcal{F}(f(t)) *\mathcal{F}(g(t))$$
In this link and also this, it was found that for a 2 dimensional convolution, it is expressed as follows:
$$f(x,y) * g(x,y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x',y')g(x-x',y-y')dx'dy'$$
So I suspect in general, an n dimensional convolution will be written as:
$$f(a_1,\dots,a_n) * g(a_1,\dots,a_n)=\int_{\mathbb{R}^n}f(a_1',\dots,a_n')g(a_1-a_1',\dots,a_n-a_n')d^na'$$
But what if I want to carry out some sort of "partial convolution", i.e.
$$?=\int_{-\infty}^{\infty}f(x',y')g(x-x')dx'\hspace{15mm}[1]$$
- Is there a notation in terms of $*$ for $[1]$, or generalisation of it? Or is it whenever we wrote the convolution symbol $*$, we always imply it is convolution for all the independent variables present?
For a concrete example of $[1]$, suppose I have a laser beam with Gaussian profile $\exp(-x^2)$ which passes through a rectangular slit modeled by a product of boxcar functions $\Pi_{-a,a}(x)\Pi_{-b,b}(y)$. The far field diffraction pattern produced then landed on another filter which is $\delta(y)$.
Therefore the final diffraction pattern $\phi$ after being Fourier transformed again by a lens placed behind the 2nd filter (In our optics course we use the convention such that the prefactor in the Fourier transform is $2\pi i$ thus the normalisation factor is 1), would been given as:
$$\phi(x,y)=\int_{-\infty}^{\infty}\mathcal{F}\left[\int_{-\infty}^{\infty}e^{-x^2}\Pi_{-a,a}(x-x')\Pi_{-b,b}(y)dx'\right]\delta(v')dv'$$
So question 1 was asking whether we can express the above expression in terms of $*$ thus we can apply Convolution Theorem and common Fourier transform pairs to simplify it.
For question 2, we know that if some 2 variable distributions $k$ and $l$ are separable, then there is a relation between convolution and multiplication
$$k(x)*l(y)=k(x)l(y)$$
But I am having some trouble try to convince myself by proving it for the following case:
$$e^{-x^2}*\delta(y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x'^2}\delta(y-x'-y')dx'dy'$$ $$=\int_{-\infty}^{\infty}e^{-x'^2}\delta(x')dx'=e^{-0^2}=1$$ $$\color{red}{\neq e^{-x^2}\delta(y)}$$
- What kind of rigorous treatment on distributions that was missing in the above physical usage (possibly abuse of notation) that can help rectify the problem highlighted in red?
