# Derivatives of singularity functions

1. Jun 16, 2014

### iRaid

1. The problem statement, all variables and given/known data
Hello, I'm having trouble understanding this, seemingly simple, concept. Any help or input is appreciated.

Evaluate the following derivatives:
$$\frac{d}{dt} u(t-1)u(t+1)$$
$$\frac{d}{dt} r(t-6)u(t-2)$$
$$\frac{d}{dt} sin(4t)u(t-3)$$

2. Relevant equations

3. The attempt at a solution
1) $$\delta(t-1)u(t+1)+u(t-1)\delta(t+1)$$
2) $$r(t-6)\delta(t-2)+u(t-6)u(t-2)$$
3) $$sin(4t)\delta(t-3)+cos(4t)u(t-3)$$

Now, I don't really understand what to do from here...

2. Jun 16, 2014

### Ray Vickson

Before trying to differentiate the function $f(t) = u(t-1)u(t+1)$ you should first see if you can re-write $f(t)$ in a simpler form. For example, what does the graph of $f(t)$ look like?

Do the same type of preliminary analysis on all the other functions you are given.

3. Jun 16, 2014

### iRaid

Sorry, I'm honestly not sure what 2 singularity functions multiplied together would look like. I know that they're both step functions starting at t=-1 and stepping up at t=1.

4. Jun 17, 2014

### Ray Vickson

The unit step function is defined on the whole real line, so if it "starts" anywhere, it starts at -∞. For what values of $w$ is $u(w)$ equal to zero? Equal to 1? So, for what values of $t$ is $u(t-1)$ equal to zero, and to 1? Same question for $u(t+1)$. So, when is the product $u(t-1)u(t+1)$ equal to zero? When does it equal 1?

5. Jun 17, 2014

### vanhees71

One must be very careful multiplying distributions, and the unit-step function is in fact a distribution in this context. As Ray has already stated, you should first rewrite the product which has apparently a meaning when interpreted as usual functions rather than distributions in terms of a well-defined distribution. Note that there are no problem to use linear combinations of distributions. You only must not multiply two or more distributions!

6. Jun 17, 2014

### HallsofIvy

Surely you can multiply 1s and 0s! If t< -1 then t-1< -2 and t+ 1< 0 so both u(t-1) and u(t+ 1) are 0. u(t-1)u(t+1)= 0. If -1< t< 1 then t-1< 0 but t+1> 0 so u(t-1)= 0 and u(t+ 1)= 1. u(t-1)u(t+1)= 0. if t> 1 then t-1> 0 and t+ 1> 2> 0 so both u(t-1) and u(t+ 1) are 0. u(t-1)u(t+1)= 1.

u(t-1)u(t+1)= 0 for t< 1, 1 for $t\ge 1$. Equivalently, u(t-1)u(t+1)= u(t-1). With a little thought you should see that if a> b then u(t-a)u(t-b)= u(t- b).

7. Jun 17, 2014

### vanhees71

Again, that's a very subtle point! Of course, you can multiply 0's and 1's, but not unitstep distributions. You can multiply them as functions, but you cannot do operations of such products that only have meaning in the sense of distributions. This exercise is a very nice example. If you formally do so you get sometimes (not always) nonsensical results.

If in doubt, you have to check your distribution valued results with test functions!