We know that $$\frac{d^2}{dx^2}\left(\frac{1}{|x|}\right) =-4\pi\delta(x)$$ where $\delta(x)$ is Dirac delta distribution. $$\Rightarrow \lim_{x\to 0}\frac{d^2}{dx^2}\left(\frac{1}{|x|}\right) =-\infty$$ $$\frac{d^2}{dx^2}\left(\frac{1}{\sqrt{x^2 + a^2}}\right) =\frac{3 x^2}{(a^2 + x^2)^{5/2}}-\frac{1}{(a^2 + x^2)^{3/2}}$$ $$\lim_{a\to 0}\frac{d^2}{dx^2}\left(\frac{1}{\sqrt{x^2 + a^2}}\right) \approx\frac{3x^2}{|x|^5}-\frac{1}{|x|^3}=\frac{2}{|x|^3}$$ $$\lim_{x\to 0}\lim_{a\to 0}\frac{d^2}{dx^2}\left(\frac{1}{\sqrt{x^2 + a^2}}\right) \approx\color{red}{+\infty}$$ Shouldn't the red part be $-\infty$? Where did something go wrong?
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$\frac{d^2}{dx^2}\left(\frac{1}{|x|}\right) =-4\pi\delta(x)$ doesn't make much sense – reuns Feb 14 '22 at 20:03
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@reuns I took the 3D formula and did $r\to |x|$ is that wrong? – Kasi Reddy Sreeman Reddy Feb 14 '22 at 20:04
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If $1/|x|$ is the distribution on $\Bbb{R}^3$ then why do you say "in 1D" and what is the meaning of $d/dx^2$ – reuns Feb 14 '22 at 20:05
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@reuns I was thinking that by replacing $r\to |x|$ in the 3D solution I will get a 1D soltuion. But now I am not sure if it was correct. – Kasi Reddy Sreeman Reddy Feb 14 '22 at 20:15
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The map sending $f\in D(\Bbb{R})$ to $F(x_1,x_2,x_3)=f(x_1)$ doesn't have its image in $D(\Bbb{R}^3)$ so the distributions on $\Bbb{R}^3$ don't give canonically some distributions on $\Bbb{R}$. If the distribution is compactly supported or if you restrict to test functions supported on some disk then you can send $f$ to $F(x_1,x_2,x_3)=f(x_1) \phi(x_1,x_2,x_3)$ where $\phi \in D(\Bbb{R}^3)$ and $\phi=1$ on the support. – reuns Feb 14 '22 at 20:18
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https://physics.stackexchange.com/questions/44515/electric-field-and-electric-potential-of-a-point-charge-in-2d-and-1d similar question I just found. – Kasi Reddy Sreeman Reddy Feb 14 '22 at 20:20
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If you want to take the Laplacian of $1/r$ in 3D, you need to take the 3D Laplacian correctly. This is not the right way to do it. The fundamental solution in 1D is different. – Ian Feb 14 '22 at 20:21
2 Answers
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In 1D, the fundamental solution is given by the absolute value.
$$ {d^2\over dx^2} |x| = {d\over dx} ( -{\bf 1}_{]-\infty,0[}+{\bf 1}_{]0,\infty[}) = 2\delta(x). $$
The expression you give at the start of your question is related to the laplacian of the radial function $(x,y,z) \mapsto {1\over \sqrt{x^2+y^2+z^2}}$ in 3D. You can use the following formula to compute it outside the origin. Setting $r = \sqrt{x^2+y^2+z^2}$ and for $f : [0,\infty[ \rightarrow {\bf R}$,
$$\Delta f(r) = {1\over r^2}{\partial\over \partial r} \Bigl(r^2 {\partial f \over \partial r}\Bigr).$$
coudy
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Can you share a link to the proof of the 1st equation? – Kasi Reddy Sreeman Reddy Feb 14 '22 at 20:23
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Intuitively I understood how the middle equation gives the RHS. Integrating over 2 times delta gives the difference between -1 and +1. – Kasi Reddy Sreeman Reddy Feb 14 '22 at 20:25
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1This follows from the fact that the derivative of the Heaviside function is the Dirac distribution. A proper justification makes use of the theory of Schwartz distributions. This is discussed for example at https://math.stackexchange.com/questions/13898/how-to-prove-that-the-derivative-of-heavisides-unit-step-function-is-the-dirac – coudy Feb 14 '22 at 20:40
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Writing @coudy's answer more clearly. Not rigorous but intuitively correct.
Let $f=\frac{d}{dx}|x|$. Then $f(x>0)=1$ and $f(x<0)=-1$ then \begin{align} f(\epsilon)-f(-\epsilon)&=2=\int_{-\epsilon}^{\epsilon}2\delta(x)dx\\ \implies \frac{df}{dx}&=\frac{d^2|x|}{dx^2}=2\delta(x)\\ \implies \frac{d^2}{dx^2}\left(\frac{|x|}{2}\right)&=\delta(x) \end{align}