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The function $e^{-a|x|}$ has a discontinuous first derivative at $x=0$, but seemingly a continuous second derivative there. Formally applying the product rule, we find that its second derivative is $a^2e^{-a|x|}$$-$but it turns out that the correct derivative is actually $a^2e^{-a|x|}-2a\delta(x)$.

  • How can we obtain the Dirac delta term in the second derivative?
  • What property of this function is causing the Dirac delta term to appear?
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    $\begingroup$ $e^{-a|x|} = e^{-ax}1_{x > 0} + e^{ax}1_{x < 0}$. The (distributional) derivative of $1_{x > 0}$ is $\delta(x)$, and for the multiplication of a distribution ($1_{x > 0} $) with a $C^\infty$ function ($e^{-ax}$) the product rule for the derivative works fine. $\endgroup$ – reuns Oct 23 '16 at 5:40
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    $\begingroup$ Thus $(e^{-ax}1_{x > 0})' = -a e^{-ax}1_{x > 0}+e^{-ax} \delta(x) = -a e^{-ax}1_{x > 0}+ \delta(x) $ $\endgroup$ – reuns Oct 23 '16 at 5:41
  • $\begingroup$ "but seemingly a continuous second derivative there' ?? So we have a discontinuous function with a derivative? Please explain. $\endgroup$ – Did Oct 23 '16 at 5:54
  • $\begingroup$ @Did: I simply mean that the graph of the second derivative seems continuous at $x=0$, nothing more. $\endgroup$ – a-cyclohexane-molecule Oct 23 '16 at 14:16
  • $\begingroup$ ?? How do you "graph" the second derivative? $\endgroup$ – Did Oct 23 '16 at 15:51
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The delta function comes due to the non-differentiability of the absolute value function at the point $0$. In that case, a delta function (centered at zero) gets added. Furthermore, the coefficient of the delta function is the "jump" of the function at the point i.e. the right limit minus the left limit at the point. In this case, it is $-2a$, hence we see that the factor $-2a\delta(x)$ gets added to the derivative.

More formally:

Given that the function is continuously differentiable except at a point $p$, then: $$ Df(\phi) = f'(\phi) + J_f(p)* \delta_{p}(\phi) $$

Where $\phi$ is a $C^\infty$ function with compact support, and $Df$ is the distributional derivative, and $f'$ is the classical derivative treated as a distribution, and $J_f(p)$ is the jump at $p$ which I described in the case earlier.

The proof of this is a little involved (rigorously, that is), but not difficult.

EDIT:

As Did pointed out, your logic prior to this is wrong, because if your function is not continuous, then it can't be differentiable. However, there is a concept of weak derivative, and I think you should read this up before you actually understand the terms of this answer (distributions, the topic is called). Furthermore, the delta function, well what is it? It's a a distribution, not a function, and when I answered this question I thought you had this background. Besides, do read up, because the subject is fascinating.

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