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Hello So I am currently trying to prove that $$\frac{d^2}{dx^2}(e^{-a|x|}) = a^2 e^{-a|x|} - 2a\delta(x)$$ I have made a proof and was wondering if my proof would be considered valid

Proof $$\frac{d}{dx}(e^{-a|x|}) = -a e^{-a|x|} - a\delta(x)$$ because using the product rule, the derivative of $-a|x|$ is $-a\delta(x)$ then $$\frac{d^2}{dx^2}(e^{-a|x|}) = a^2 e^{-a|x|} - a\delta(x) - a\delta (x) = a^2 e^{-a|x|} - 2a\delta(x)$$ Would this proof be correct or am I making wrong assumptions?

Continuum
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Really can't follow what you're saying here and it doesn't seem right. In terms of distributions, we have $$ \frac{d}{dx}|x| = \operatorname{sgn}(x) $$ and $$ \frac{d}{dx}\operatorname{sgn}(x) = 2\delta(x),$$ so differentiating gives $$ \frac{d}{dx}\left(e^{-a|x|}\right) = -a\operatorname{sgn}(x) e^{-a|x|}$$ and then by the product rule, $$ \frac{d^2}{dx^2}\left(e^{-a|x|}\right) = a^2\operatorname{sgn}(x)^2 e^{-a|x|} -2a\delta(x) e^{-a|x|} =a^2e^{-a|x|} -2a\delta(x)$$ where in the last equality we used the fact that $\operatorname{sgn}(x)^2 = 1$ and that $\delta(x) f(x) = \delta(x)f(0).$

spaceisdarkgreen
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  • This makes more sense! I was under the assumption that d/dx |x| = delta(x) which is probably not true! – Continuum Oct 23 '17 at 01:58
  • Nope, it is $\operatorname{sgn}(x)$ which is $+1$ for positive $x$ and $-1$ for negative $x$ (and $0$ for $x=0$ though the value here doesn't particularly matter.) – spaceisdarkgreen Oct 23 '17 at 02:05