# Variation of Dirac delta function

1. Jun 4, 2012

### friend

Is it possible to take the variation of the Dirac delta function, by that I mean take the functional derivative of the Dirac delta function?

2. Jun 4, 2012

### charbel

yes you can (its laplace transform is s) and you can even take the derivative of this one although in practice i m not really sure how you can use it

3. Oct 2, 2012

### friend

Could someone describe this procedure in math symbols? Or maybe point me to a link? Thanks.

4. Oct 2, 2012

### Mute

I suppose in principle you could do it. I doubt it would be useful.

$$\frac{\delta~\delta(t)}{\delta~\delta(t')} = \delta(t-t').$$

$$\frac{\delta~\delta(t)}{\delta \eta(t')} = 0,$$
where $\eta(t')$ is an arbitrary function not related to the dirac delta function.

5. Oct 3, 2012

### friend

You ask how it would be useful. Let me give some context. We know
$$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1$$
Then suppose that the dirac delta is composed with some other function, f(x). We get
$$\int_{ - \infty }^{ + \infty } {\delta (f(x))f'(x)dx} = 1$$
Now, we also know that the variation of a constant is zero,
$${\rm{\delta }}[{\rm{constant]}} = 0$$
So we should have,
$${\rm{\delta }}[\int_{ - \infty }^{ + \infty } {\delta (f(x))f'(x)dx} {\rm{]}} = {\rm{\delta }}[1{\rm{]}} = 0$$
But the variation of a definite integral is the definite integral of the variation. A variation is like differentiation which commutes with integration. So we have,
$${\rm{\delta }}[\int_{ - \infty }^{ + \infty } {\delta (f(x))f'(x)dx} {\rm{]}} = \int_{ - \infty }^{ + \infty } {{\rm{\delta }}[\delta (f(x))f'(x)]dx}$$
And in order for this to be identically zero, then the integrand must be zero, or
$${\rm{\delta }}[\delta (f(x))f'(x)] = 0$$
Thus the question about variations of dirac delta functions.

I suppose you could use the chain rule to break this down to
$${\rm{\delta }}[\delta (f(x))]f'(x) + \delta (f(x)){\rm{\delta }}[f'(x)] = 0$$
So what's the next step? Is it true that
$${\rm{\delta }}[\delta (f(x))] = \frac{d}{{dx}}\{ \delta (f(x))\} \,{\rm{\delta }}x$$

6. Oct 9, 2012

### friend

As I understand it, if
$$F\left[ {y\left( x \right),z\left( x \right)} \right]$$
then
$$\delta F = \frac{{\partial F}}{{\partial y}}\delta y + \frac{{\partial F}}{{\partial z}}\delta z$$
So the variation of the Dirac delta of a function, S(x), would be
$$\delta \delta (S(x)) = \frac{{\partial \delta (S(x))}}{{\partial S}}\delta S(x)$$
And from the wikipedia.com site

$x\frac{{d\delta (x)}}{{dx\,\,\,\,\,\,\,\,\,\,\,}} = - \delta (x).$ OR $\frac{{d\delta (x)}}{{dx\,\,\,\,\,\,\,\,\,\,\,}} = - \frac{{\delta (x)}}{x}$

So,
$$\frac{{\partial \delta (S(x))}}{{\partial S}} = - \frac{{\delta (S(x))}}{{S(x)}}$$
And since
$$\delta S(x) = \frac{{dS(x)}}{{dx}}dx$$
We have,
$$\delta \delta (S(x)) = \frac{{\partial \delta (S(x))}}{{\partial S}}\delta S(x) = - \frac{{\delta (S(x))}}{{S(x)}}\delta S(x) = - \frac{{\delta (S(x))}}{{S(x)}}\frac{{dS(x)}}{{dx\,\,\,\,\,\,\,\,\,}}dx$$
Does this all seem right so far? Or is there something special about the Dirac delta that these techniques don't apply? I don't work with the math of variation very often. So I'd appreciate some help. Thank you.

Last edited: Oct 9, 2012