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I have been trying to figure this out for a while, and I was wondering if anyone had any ideas. I need to solve the following differential equation:

$m\frac{d^2 r}{dt^2}=\epsilon\delta'(r)$,

where $\delta(r)$ is the Dirac delta function, and $m,\epsilon$ are constants. What would be the best way to go about this?

Cheers

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  • $\begingroup$ $\delta'(r) = -r'(0)$, no? $\endgroup$
    – fgp
    Apr 11, 2014 at 17:32
  • $\begingroup$ Ok, I probably should have given some context. r here is referring to the relative position of two particles moving on the real line. $\delta(r)$ is the potential. This is an Euler-Lagrange equation. I can sort of see how your argument works. Are you calling $r$ here a test function? $\endgroup$
    – Josh Cork
    Apr 11, 2014 at 17:38
  • $\begingroup$ $r$ is a function of $t$, and $\delta(r)$ is dependent on $r$ $\endgroup$
    – Josh Cork
    Apr 11, 2014 at 17:39
  • $\begingroup$ What does $\delta(r)$ is the potential mean? For me, $\delta(r)$ is a real number - $\delta'$ is a linear functional from some space of (test) functions to $\mathbb{R}$, and since $r$ is a function (and we assume it's in whatever domain $\delta'$ has), the result of applying $\delta'$ to $r$ is a real number. Though maybe I'm missing something - I'm not a physicist... $\endgroup$
    – fgp
    Apr 11, 2014 at 17:43
  • $\begingroup$ The usual definition for the derivative of a distribution $\Lambda$ is that $\Lambda'(f) = - \Lambda(f')$, from which it would follows that $\delta'(r) = -\delta(r') = -r'(0)$ (since $\delta(f) = f(0)$). But again, maybe I'm missing something... $\endgroup$
    – fgp
    Apr 11, 2014 at 17:45

2 Answers 2

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My approach is based on two facts:

  1. $\delta $ is the distribitional derivative of the Heaviside function $H = \chi_{[0,\infty)}$.
  2. A distribution whose first derivative is identically $0$ is constant.

Thus, $mr'=\epsilon \delta +C = \epsilon H'+C$. By the same logic, $mr = \epsilon H +Ct+B$.

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I have tried a little can't sure correct or not, $$m\frac{d^2r}{dt^2}=\epsilon\delta'(r)$$ $$\frac{d}{dt}\left[ m\frac{dr}{dt} -\epsilon\delta \right]=0$$ $$\left[m\frac{dr}{dt} -\epsilon\delta \right]=C$$ $$\int dr-\frac{\epsilon}{m}\int\delta\,dr =\int C\, dt$$ $$r-\frac{\epsilon}{m}=Ct + C'$$

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  • $\begingroup$ You have integrated $\delta$ from $-\infty$ to $\infty$ instead of taking the primitive function of it. Also, the integral should be w.r.t. $t$, not $r$. $\endgroup$
    – md2perpe
    Jul 30, 2021 at 21:18

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