4

The first proof that comes to my mind is that the units of left side and right side does not match. Delta functions take an input and spit out an output of units = 1/[units of input].

What would be the rigorous proof of this?

Statement to be proven: (also given below):

$$\delta \left ( r \right )\space \space \neq \space \space \delta \left ( x \right )\delta \left ( y \right )$$

NerdyNerdie
  • 43
  • Remember that delta function is not actually a function. Try to construct a proper radial delta in a distributional sense (how should it act on an arbitrary function $f(r)$?) – Yuriy S Feb 13 '21 at 17:36
  • I edited the question. So the proper definition is integral(f(r)delta(r)dr) = f(0) if the integration line crosses the origin right? – NerdyNerdie Feb 13 '21 at 17:45
  • Actually the obvious definition of the radial 2D delta function would be $\delta(\sqrt{x^2+y^2})$ which is ill defined due to the bad regularity of the function $\sqrt{x^2+y^2}$ at the origin. – Ian Feb 13 '21 at 17:48
  • https://math.stackexchange.com/q/56939/8157 – Giuseppe Negro Feb 13 '21 at 17:49
  • @GiuseppeNegro Can you please elaborate a bit more? The non equality is just due to the factor of of gradient of "r"? Or I think the direction of approach to the origin really matters in this integration? – NerdyNerdie Feb 13 '21 at 18:01
  • @ian Actually, $\delta(x-x')\delta(y-y')=\frac{\delta(r-r')\delta(\theta-\theta')}{r}$ where $x=r\cos(\theta)$, $y=r\sin(\theta)$, $x'=r'\cos(\theta')$, and $y'=r'\sin(\theta')$. – Mark Viola Feb 13 '21 at 18:05
  • 1
    @NerdyNerdie See THIS – Mark Viola Feb 13 '21 at 18:08
  • @MarkViola. That is exactly what I was looking for. Thanks. – NerdyNerdie Feb 13 '21 at 18:10
  • 1
    @NerdyNerdie You're welcome. My pleasure. – Mark Viola Feb 13 '21 at 18:12
  • 1
    @MarkViola: I disagree with the answer given to that question. The dirac delta at the end of the interval of integration is a problem. See my answer here. – robjohn Feb 13 '21 at 18:32
  • @NerdyNerdie: I was clumsily trying to redirect you to robjohn’s excellent answer there. Meanwhile, robjohn himself showed up here and solved the problem. TL;DR: just forget about that – Giuseppe Negro Feb 13 '21 at 18:50
  • @robjohn: so you mean that, in that answer, the one you disagree with, the result is correct but the proof is not, for $r’\ne 0$? – Giuseppe Negro Feb 13 '21 at 19:05
  • @robjohn What part of that answer was wrong?? – Mark Viola Feb 13 '21 at 20:04
  • @MarkViola: It says there that $\delta(x),\delta(y) = \frac1{2\pi r}\delta(r)$, but in my answer, and as mentioned in another MSE answer, with MathWorld and support in what I think is a respected book, it should be $\delta(x),\delta(y) = \frac1{\pi r}\delta(r)$. – robjohn Feb 13 '21 at 21:35
  • @robjohn Would you say that the Unilateral Laplace Transform of the Dirac Delta is $1/2$? Or would you say that it is $1$? The issue is interpreting the notation $\int_0^\infty \delta(t) e^{-st},dt$. Does this mean, $\langle \delta_0 , \phi\rangle$, where $\phi(t)=e^{-st}H(t)$. If so, then inasmuch as $\phi$ is not a suitable test function, and is not even continuous at $0$, this is a meaningless object as a distribution. We can adopt a convention that $\int_0^\infty \delta(x) \phi(x),dx=C\phi(0)$ with any value of $C$ chosen. And there's a non-symmetric regularization for such a $C$ – Mark Viola Feb 13 '21 at 22:21

2 Answers2

3

Delta Function in Two Dimensions

The problem with this question is that the radial delta function is not $\delta(r)$, where $\delta$ is the usual $1$-dimensional delta function.

An approximation of $\delta(x)$ is $ne^{-\pi n^2x^2}$, so $$ \begin{align} \delta(x)\,\delta(y) &=ne^{-\pi n^2x^2}ne^{-\pi n^2y^2}\tag{1a}\\ &=n^2e^{-\pi n^2r^2}\tag{1b} \end{align} $$ Now, let's see what the integral of $n^2e^{-\pi n^2r^2}$ is (in polar coodinates) $$ \begin{align} \int_0^{2\pi}\int_0^\infty n^2e^{-\pi n^2r^2}\,r\,\mathrm{d}r\,\mathrm{d}\theta &=\int_0^{2\pi}\int_0^\infty\frac1{2\pi}\,e^{-u}\,\mathrm{d}u\,\mathrm{d}\theta\tag{2a}\\ &=1\tag{2b} \end{align} $$ Explanation:
$\text{(2a)}$: $u=\pi n^2r^2$
$\text{(2b)}$: evaluate the integrals

So, it seems that $n^2e^{-\pi n^2r^2}$ is an approximation of the radial delta function. Taking the limit says that $\delta(x)\,\delta(y)$ is the radial delta function.

The Radial Delta Function is not $\boldsymbol{\delta(r)}$

However, if we use $\delta(r)$ in place of $\delta(x)\,\delta(y)$, we get $$ \begin{align} \int_0^{2\pi}\int_0^\infty ne^{-\pi n^2r^2}\,r\,\mathrm{d}r\,\mathrm{d}\theta &=\int_0^{2\pi}\int_0^\infty\frac1{2\pi n}\,e^{-u}\,\mathrm{d}u\,\mathrm{d}\theta\tag{2a}\\ &=\frac1n\tag{2b} \end{align} $$ which tends to $0$.

The Radial Dirac Delta Function $$ \begin{align} \int_0^{2\pi}\int_0^\infty\overbrace{\frac{n}{\pi r}e^{-\pi n^2r^2}}^{\large\frac1{\pi r}\delta(r)}\,r\,\mathrm{d}r\,\mathrm{d}\theta &=\int_0^{2\pi}\int_0^\infty \frac1{\pi}e^{-\pi r^2}\,\mathrm{d}r\,\mathrm{d}\theta\tag{3a}\\ &=1\tag{3b} \end{align} $$ Thus, the radial delta function is $\frac1{\pi r}\delta(r)$.

robjohn
  • 345,667
  • It's that $\delta(x) \delta(y)$ is the right thing that yields $f(0,0)$ when integrated over $\mathbb{R}^2$ against $f(x,y)$. $\delta(r)$ is not the same thing: indeed as the OP says it does not even have the same units. – Ian Feb 13 '21 at 18:00
  • 1
    @Ian: yes, I see the problem is that $\delta(r)$ is not the radial dirac delta. I have modified my answer to reflect this. – robjohn Feb 13 '21 at 18:13
  • Whatever the missing factor of $\frac12$ is, I like a lot this approach to the Dirac delta. If we interpret it as a limit of Gaussians, we can perform all sort of manipulation on these without “magic tricks” with Jacobians. It feels much more robust. – Giuseppe Negro Feb 13 '21 at 19:07
  • $\int_0^{2\pi}\int_0^\infty\frac1\pi\delta(r),\mathrm{d}r,\mathrm{d}\theta=\frac12\int_0^{2\pi}\int_{-\infty}^\infty\frac1\pi\delta(r),\mathrm{d}r,\mathrm{d}\theta=1$ – robjohn Feb 13 '21 at 19:32
  • this answer cites MathWorld (formula 46), which cites "The Fourier Transform And Its Applications" by Ronald Bracewell, whose 3rd edition gives this identity on page 89: $$^2\delta(x,y)=\frac{\delta(r)}{\pi|r|}$$ To preserve symmetry and additivity, I usually see $\int_{-\infty}^0\delta(x),\mathrm{d}x=\int_0^\infty\delta(x),\mathrm{d}x=\frac12$. – robjohn Feb 13 '21 at 20:35
  • @robjohn The result $1/2$ is a consequence of the chosen symmetric regularization. That is not a requirement. In fact, we may choose a regularization such that $\int_0^\infty \delta(x)\phi(x),dx=\phi(0)$. This is the cased used in practice for the Laplace Transform and for problems in engineering and physics. – Mark Viola Feb 13 '21 at 22:24
  • As I said in my earlier comment, "The dirac delta at the end of the interval of integration is a problem." I usually use the even distribution, but we can use $\lim\limits_{u\to0^+}\delta(x-u)$ here as well. – robjohn Feb 14 '21 at 03:13
1

Since$$\iint f(r,\,\theta)\delta(r)\delta(\theta)drd\theta=f(O)=\iint f(x,\,y)\delta(x)\delta(y)\underbrace{dxdy}_{rdrd\theta},$$we have the dimensionally homogeneous result $\delta(r)\delta(\theta)=r\delta(x)\delta(y)$. @MarkViola's comment has noted a more general finding we ca get by writing $f(p)$ for some point $p$ as two different double integrals.

J.G.
  • 115,835
  • Should there be an $r$ in the LHS integral? – Ian Feb 13 '21 at 18:29
  • @Ian No, that's the whole point: in $\delta(r)\delta(\theta)drd\theta=\delta(x)\delta(y)dxdy$, the Jacobian doesn't cancel (that would be dimensionally heterogeneous). – J.G. Feb 13 '21 at 18:41
  • I see now, $\delta(r) dr$ are all the dimensional factors in the LHS and combine to be dimensionless already, which is what you have on the other side. – Ian Feb 13 '21 at 18:51
  • @J.G. I see that now. So delta(theta) is not actually a delta function in our common sense (it does not blow up at theta = 0), it is just a constant. – NerdyNerdie Feb 13 '21 at 18:55