Applied mathematics, when to leave out delta function?

Question

I cant figure out when I´m supposed to ignore the $\delta(t)$ function in the answer.

$\theta(t)$ is The Heaviside function

I have three examples:

1. Let $f(t) = e^t\theta(t)$ and find $f'$

Answer: $(e^t\theta(t))' = (e^t)'\theta(t) + e^t(\theta(t))' = e^t\theta(t) + e^t\delta(t) = e^t\theta(t) + \delta(t)$

Hence $\delta$(t) stay.

2. Let $f(t) = e^{2t}\theta(t)$ and find $f'$

Answer: $(e^{2t}\theta(t))' = (e^{2t})'\theta(t) + e^{2t}(\theta(t))' = 2e^{2t}\theta(t) + e^{2t}\delta(t) = 2e^{2t}\theta(t) + \delta(t)$

Again $\delta(t)$ stays.

3. Let $f(t) = t\theta(t)$ and find $f'$

Answer: $(t\theta(t))' = t'\theta(t) + t(\theta(t))' = \theta(t) + t\delta(t) = \theta(t)$

Here $\delta(t)$ is just removed in the last step? Why is it okay to remove it?

(These writings are from an previous exam and I've just entered the answers from the key)

Yes, exaclty. I should add that. I just know it as the "stepfunction". — netigger, Mar 14 '13 at 10:55
the delta function is $0$ everywhere but at $0$. The function $t$ is $0$ at $0$, hence $t\delta(t)$ is $0$ everywhere — user27182, Mar 14 '13 at 11:38
@DavidEverlöf Actually, "step function" is a better name than "Heaviside function", because it names the function after its properties and not after some obscure guy who used it (https://en.wikipedia.org/wiki/Heaviside_step_function). I'm just more used to the other name. — Elmar Zander, Mar 14 '13 at 12:14

score 5 · Accepted Answer · answered Mar 14 '13 at 11:03

5

$f(t)\delta(t) = f(0)\delta(t)$

answered Mar 14 '13 at 11:03

Rhys

1

Hm, guess that makes sense. – netigger Mar 14 '13 at 11:10
1

It's easy to prove, remembering that the delta 'function' is only defined through integration (it's a distribution, not a function). – Rhys Mar 14 '13 at 11:17
Fine for me, as long as I understand how to answer the questions.. Thanks – netigger Mar 14 '13 at 11:20

1 Answers1