What technique is to use for differentiation regarding $\delta$-functions?

Question

Interesting differentiation regarding $\delta$-functions. Let we define a function $h(x)=e^{-ax}H(x)+e^{ax}H(-x)$ and we want to find the $n$-th derivative. \begin{align} h(x)&=e^{-ax}H(x)+e^{ax}H(-x)\\ h'(x)&=e^{-ax}\delta(x)-ae^{-ax}H(x)-e^{ax}\delta(x)+ae^{ax}H(-x)=ae^{ax}H(-x)-ae^{-ax}H(x)\\ \end{align} I believe this step has no problem, but the following difference arises. If we start from the resulting expression of the first derivative, then \begin{align} h''(x)&=(h'(x))'=-e^{ax}a\delta(x)+a^2e^{ax}H(-x)-e^{-ax}a\delta(x)+a^2e^{-ax}H(x)=a^2e^{ax}H(-x)+a^2e^{-ax}H(x)-(e^{-ax}+e^{ax})a\delta(x) \end{align} but if we start before we discard $(e^{-ax}-e^{ax})\delta(x)=0$ which has the derivative $-a(e^{-ax}+e^{ax})\delta(x)+(e^{-ax}-e^{ax})\delta'(x)$, then \begin{align} h''(x)&=-a(e^{-ax}+e^{ax})\delta(x)-e^{ax}a\delta(x)+a^2e^{ax}H(-x)-e^{-ax}a\delta(x)+a^2e^{-ax}H(x)=a^2e^{ax}H(-x)+a^2e^{-ax}H(x)-(e^{-ax}+e^{ax})a\delta(x)\\ &=a^2e^{ax}H(-x)+a^2e^{-ax}H(x)-2ae^{-ax}\delta(x). \end{align}

Though the results are equal due to $\delta$'s property, but what are the strictly correct way to proceed further.

Could be any easier way to arrive at the $n$-derivatives?

Answer 1

If $h$ is a distribution given by

$$h(x)=e^{-ax}H(x)+e^{ax}H(-x)$$

then the distributional derivative, $h'$, of $h$ is given by

$$h'(x)=\underbrace{e^{-ax}\delta(x)-e^{ax}\delta(x)}_{=0}+ae^{ax}H(-x)-ae^{-ax}H(x)\tag1$$

Now, if we differentiate the first two terms on the right-hand side of $(1)$ we find that

$$\begin{align} 0&=(e^{-ax}\delta(x)-e^{ax}\delta(x))'\\\\ &=(-2\sinh(ax)\delta(x))'\\\\ &=-2\sinh(ax)\delta'(x)-2a\cosh(ax)\delta(x)\\\\ &=2a\cosh(ax)\delta(x)-2a\cosh(ax)\delta(x)\\\\ &=0 \end{align}$$

as expected!

Answer 2

For an efficient way to calculate the $n$^th derivative of $e^{-|x|}$ the identity given in the post you found can be used.

You first have $h^{(0)}(x)=h(x)=e^{-|x|}$ and $h^{(1)}(x)=h'(x)=\operatorname{sign}(x)e^{-|x|}.$ Then $h''(x)-h(x)=-2\delta(x)$ gives $h^{(2)}(x) = h''(x) = h(x)-2\delta(x)=e^{-|x|}-2\delta(x)$ and also implies $$ h^{(3)}(x)-h^{(1)}(x)=-2\delta'(x) \\ h^{(4)}(x)-h^{(2)}(x)=-2\delta''(x) \\ h^{(5)}(x)-h^{(3)}(x)=-2\delta'''(x) \\ \vdots $$ From this we can conclude for example that $$ h^{(5)}(x) =h^{(3)}(x)-2\delta'''(x) =\left( h^{(1)}(x) - 2\delta'(x) \right) - 2\delta'''(x) = \operatorname{sign}(x)e^{-|x|} - 2\delta'(x) - 2\delta'''(x). $$ Generally we find that $$ h^{(n)}(x) = \begin{cases} e^{-|x|} - 2\delta^{(n-2)}(x) - \cdots - 2\delta^{(0)}(x) & \text{if $n$ even} \\ \operatorname{sign}(x)e^{-|x|} - 2\delta^{(n-2)}(x) - \cdots - 2\delta^{(1)}(x) & \text{if $n$ odd} \end{cases} $$