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Is it alright to approximate a Delta distribution with an exponential like this:

$$\delta(x-1) = \omega\,e^{-\omega (x-1)}, \hspace{1cm} x \geq 1,$$

where, $\omega \gg 1 $, and, $$\int_1^{\infty} f(x)\,\delta(x-1) = f(1).$$ Also, what are the caveats of differentiating this distribution, e.g., can I write, $$\frac{\partial}{\partial x}\delta(x-1) = -\omega\,\delta(x-1).$$

Thanks.

devanshu shekhar
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2 Answers2

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The approximation you use for $\delta(x)$ is $D_\omega(x) = \omega e^{-\omega x} H(x),$ where $H$ is the Heaviside step function. The derivative of this is $$ D_\omega'(x) = -\omega^2 e^{-\omega x} H(x) + \omega e^{-\omega x} \delta(x) = -\omega D_\omega(x) + \omega \delta(x) . $$

md2perpe
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  • Hi, thanks for your answer. However, can we say that $D'{\omega}(x)$ is approximately zero, since $D{\omega}(x) \approx \delta (x)$? – devanshu shekhar Jul 08 '23 at 10:07
  • @devanshushekhar. No, the difference $D_\omega - \delta$ is small when $\omega$ is large, but then you multiply this with the large value $\omega$ and get a not so small distribution. Actually, $\delta - D_\omega \approx \frac{1}{\omega} \delta'.$ – md2perpe Jul 08 '23 at 13:29
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The Dirac Delta has a standard regularization by the family of normed Gaussians $$\delta_\varepsilon (x) = \frac{1} {\sqrt{2\pi \varepsilon^2}} \ e^{- \frac{x^2}{2 \varepsilon^2}}$$ that has unit 1-norm and falls off to zero sudenly for $|x| > 3 \varepsilon$. The central points of a regularization of the $\delta$-distribution:

  1. Approximate point support at $x=0$,
  2. approximate unit step function of its indefinite integral,
  3. indefinite differentiability (smoothness) to implement the unlimited algebra of partial differentiation of products with smooth functions of compact support.

Check your approach against the thre points.

Roland F
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  • Hi, I have considered Gaussian distribution earlier, bu the problem is its first derivative would yield an extra x, which when evaluated at 0 makes everything 0. How do I deal with this issue? – devanshu shekhar Jun 11 '23 at 06:26
  • Thats correct, it reprents $\delta'$ as a dipole distribution of two opposite charges going to infinitiy with distance. 0. Any derivative $\delta^{(n)}$ maps to the derivative of the smooth factor of compact support under the integral, yielding $f^{(n)}(0)$. – Roland F Jun 11 '23 at 07:33
  • What I mean is, let $f^{(1)}(x)$ at $x=0$ is not zero, but since $\delta^{(1)}(x) = x \delta{(x)}$, $f(x) x $ at $x = 0$ becomes 0. So is it correct to approximate $\delta(x)$ with a Gaussian? – devanshu shekhar Jun 11 '23 at 08:40
  • $ f \delta' != x f \delta$. Try $f=(x-1)^2\delta' (x) \to - 2(x-1) \delta(x)$ – Roland F Jun 11 '23 at 09:03
  • Yes, but what if I want to differentiate the $\delta(x)$ itself, rather than using integrating by parts and differentiating $f(x)$, since I know the functional form of $\delta(x)$, which is a Gaussian. In that case, we'll get $-x(x-1)^2 \delta(x)$, which is quite different than what we get the other way. Am I missing something? – devanshu shekhar Jun 11 '23 at 09:12
  • @devanshushekhar. Why do you think that $\delta'(x) = x\delta(x)$? That's not true. – md2perpe Jun 15 '23 at 08:39
  • Well, actually, I meant $\delta'(x)=\frac{x}{\epsilon^2} \delta(x).$ Because, when you differentiate the Gaussian function, we get the extra factor of $\frac{x}{\epsilon^2}$, and the remaining is the delta according to the definition. Does this make sense? – devanshu shekhar Nov 02 '23 at 10:48