Timeline for Integrating a product of a Dirac delta distribution and its derivative.
Current License: CC BY-SA 4.0
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17 events
| when toggle format | what | by | license | comment | |
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| Apr 21 at 8:16 | vote | accept | Fibonacci M | ||
| S Apr 20 at 12:34 | vote | accept | Fibonacci M | ||
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| S Apr 20 at 12:34 | vote | accept | Fibonacci M | ||
| S Apr 20 at 12:34 | |||||
| Apr 20 at 12:33 | vote | accept | Fibonacci M | ||
| S Apr 20 at 12:34 | |||||
| Apr 17 at 18:36 | answer | added | md2perpe | timeline score: 1 | |
| Apr 17 at 16:50 | answer | added | Mark Viola | timeline score: 3 | |
| Apr 17 at 13:08 | comment | added | Mark Viola | You are correct that for the distribution $T=\delta(x-y)\delta_y(|x-y|-\varepsilon)$ where $\varepsilon>0$, we have for any $\phi(x,y)\in C_C^\infty(\mathbb{R}^2)$ $$\langle T,\phi \rangle =0$$ The Dirac Delta $\delta(x-y)$ has support on $y=x$ and its derivative $\delta_y(|x-y|-\varepsilon)$ has support on $y=x\pm \varepsilon$. Inasmuch as these lines are disjoint for all $\varepsilon\ne0$, the distribution $T$ of their product is $0$. Hence, $\langle T,\phi\rangle=0$. | |
| Apr 17 at 11:47 | history | edited | Fibonacci M | CC BY-SA 4.0 |
added 38 characters in body
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| Apr 17 at 11:40 | history | edited | Fibonacci M | CC BY-SA 4.0 |
added 520 characters in body
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| Apr 17 at 11:30 | comment | added | Fibonacci M | @LL3.14, thanks, can you provide details? Why does the integral exist? Why is it zero? | |
| Apr 17 at 7:06 | answer | added | Cryo | timeline score: 2 | |
| Apr 16 at 22:09 | comment | added | LL 3.14 | Yes, that would be $0$ since indeed the support are disjoints ... | |
| Apr 16 at 15:39 | history | edited | Fibonacci M | CC BY-SA 4.0 |
Added clarification of f(x,y) and termed \delta(...) inside the integral as a generalized function.
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| Apr 16 at 15:01 | comment | added | Fibonacci M | Thanks Mark and Deane. I use only the integral forms and the expression of the Dirac delta generalized function as defined above; this is loosely referred to as a Dirac delta distribution by physicists. I would be more than happy to refer to other more rigorous definitions if it helps me resolve the above integral. @MarkViola, I have read that the product of two distributions can make sense if their "singular support" is disjoint, for example here: arxiv.org/pdf/1404.1778 in theorem 1. I was hoping to use a result along those lines. Thanks! | |
| Apr 16 at 14:46 | comment | added | Deane | What’s the definition of a distribution that you know and use? | |
| Apr 16 at 13:09 | history | edited | Fibonacci M | CC BY-SA 4.0 |
added 50 characters in body
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| Apr 16 at 12:54 | history | asked | Fibonacci M | CC BY-SA 4.0 |