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To be more specific I want to show that impulse function in 2D can be represented as $β(r)=δ(r)/πr$.

Also I want to show that each projection of a two dimensional impulse function at the origin is a Delta function.

I could not find any useful information about these two problems, if anyone could help I will be appreciated.

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  • $\begingroup$ Hi! In 2D (or higher dimensions) the are point impulses like $\delta(x,y)$ or line impuleses like $\delta_2(x)$; the latter representing set of point impulses placed continuously on the $y$ axis. So in 2D polar coordinate system which kind of impulse representation are you up to? $\endgroup$ – Fat32 Oct 28 '18 at 17:14
  • $\begingroup$ @Fat32 There are also ring impulses: ${}^2\delta(r - r_0)$ :) $\endgroup$ – Andy Walls Oct 28 '18 at 17:17
  • $\begingroup$ @AndyWalls The circle impulse you mean $\delta_2(x^2+y^2-r_0^2)$ :-)). And what about $\delta_2(r +\cos(\theta) -1 )$ :-) $\endgroup$ – Fat32 Oct 28 '18 at 17:48
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So dealing with generalized functions like the Dirac delta requires some care, and when dealing with N-dimensional versions you need to be very explicit with your notation to keep things straight.

I'll denote the 2 dimensional delta function in polar coordinates at the origin as ${}^2\delta(r, \theta) = {}^2\delta(r)$, since for the special case of the origin, $\theta$ doesn't matter.

For this derivation, ${}^2\delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):

$${}^2\delta(r) = \lim_{\epsilon \rightarrow 0^+} \dfrac{1}{\pi\epsilon^2} \quad 0 < r < \epsilon$$

Using this limiting sequence of functions, the 2-D delta function at the origin in polar coordinates has the following property of "integrating to 1" for the 2-D integration:

$$\int_0^{2\pi} \int_0^{\infty} {}^2\delta(r) \space r \space \mathrm{d}r \space \mathrm{d}\theta = 1$$

To separate the $r$ and $\theta$ portions of ${}^2\delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:

$${}^1\delta_a(r) = \lim_{\epsilon \rightarrow 0^+} \dfrac{1}{\epsilon} \quad 0 < r < \epsilon$$

Using this limiting sequence of functions, the 1-D delta function at the origin has the following property of "integrating to 1" for the 1-D integration: $$\int_0^{\infty} {}^1\delta_a(r) \space \mathrm{d}r = 1$$

We can then equate the two above integrals to get the relationship between $ {}^2\delta(r)$ and ${}^1\delta_a(r)$

$$\begin{align}\\ \int_0^{\infty} {}^1\delta_a(r) \space \mathrm{d}r &= 1\\ &= \int_0^{2\pi} \int_0^{\infty} {}^2\delta(r) \space r \space \mathrm{d}r \space \mathrm{d}\theta\\ &= \int_0^{\infty} {}^2\delta(r) \space 2\pi r \space \mathrm{d}r\\ \end{align}$$

so we have

$${}^2\delta(r) = \dfrac{{}^1\delta_a(r)}{2\pi r}$$

This differs from your desired result by a factor of $2$, because I did not allow the 1 dimensional delta function to be symmertic around $0$, which wouldn't make sense for a polar origin where $r$ cannot be less than $0$ and $\theta$ ranges in $[0, 2\pi)$.

If one allows $r$ to go negative and $\theta$ to range in $[0, \pi)$, then one can show for a 1 dimensional delta function that is symmetric around $0$ $${}^2\delta(r) = \dfrac{{}^1\delta(r)}{\pi |r|}$$

(note the absolute value bars!)

This is a good example of why delta functions require careful definitions and considerations when making statements.

To show the projections of a 2-D delta function are themselves delta functions, start with a suitable limiting sequence of functions for ${}^2\delta(x,y)$, like an infinitely thin and tall rectangle function both in x and in y, such that

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} {}^2\delta(x,y) \space \mathrm{d}x \space \mathrm{d}y = 1$$

and the answer should fall out from the limiting sequence of functions you use.

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  • $\begingroup$ Hi thank you so much now I have better understanding of delta function I completely understand the first part, but not the second part I did not get what you mean by the answer should fall out from the limiting sequence of functions you use. Maybe I was not so clear, I want to show that each projection of a two dimensional impulse function at the origin is a δ(t). I could not figured out how to convert 2δ(x,y) dx dy to δ(t). I hope that this is not too much to ask I m kinda new at this subject one more thanks in advance. $\endgroup$ – truckdriver Oct 29 '18 at 17:55

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