# Homework Help: Second Derivative of Heaviside Function

1. Dec 3, 2012

### Choragos

Hello all. In short, I am wondering what the second derivative of the Heaviside function (let's say H[(0)]) would be. I'm presuming that it's undefined (or more accurately, zero everywhere but at x=0), but I would like to know if that is correct.

Essentially, I am attempting to extend a proof which uses f(x) where f''(x) is defined. I would like to extend this proof to a piecewise-constant f(x). However, one of the requirements is that f'(x) >> f''(x) >> f(n)(x). If H''[(0)] is undefined, than this approach will not work.

My attempt at a solution is as follows. Given a general, differentiable function f(x), then

$\int$f(x)δ(n)dx $\equiv$ -$\int$$\frac{∂f}{∂x}$δ(n-1)(x)dx

I am, however, unclear as to the meaning of δ(n-1), especially when n=1.

2. Dec 3, 2012

### Staff: Mentor

The Heaviside function is defined everywhere, but it's not continuous at x = 0. What are you referring to when you say "it's undefined", the Heaviside function or its 2nd derivative? What you wrote isn't clear.

If x < 0, H'(x) = 0. If x > 0, H'(x) = 0, so on these intervals, H''(x) would also be zero.
Both H' and H'' are not defined at x = 0.

3. Dec 4, 2012

### Choragos

Thanks for your reply. I meant that the second derivative did not exist at x=0.

My notation is sloppy (my apologies, I'm a geophysicist and a poor mathematician), but shouldn't H'=δ, where δ is the Dirac delta function? Proceeding from that, then the question becomes δ'=? That is why I was investigating the identity above, but wasn't really understanding what it was telling me.

I agree that H'' = 0 everywhere but at x=0. I suppose H''(0) is undefined.

4. Dec 4, 2012

### Staff: Mentor

One definition is
$$H(x) = \int_{-\infty}^x δ(t)dt$$
"although this expansion may not hold (or even make much sense) for x = 0." http://en.wikipedia.org/wiki/Heaviside_step_function
If you differentiate both sides above, H'(x) = δ(x), using the FTC.

I was look at H(x) from the perspective of being a step function instead of as an integra.