Variation of Dirac delta function

[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?

 

[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

 

[friend]

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

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

 

[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.

 

[friend]

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.

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

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

 

[friend]

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

[itex]x\frac{{d\delta (x)}}{{dx\,\,\,\,\,\,\,\,\,\,\,}} = - \delta (x).[/itex] OR [itex]\frac{{d\delta (x)}}{{dx\,\,\,\,\,\,\,\,\,\,\,}} = - \frac{{\delta (x)}}{x}[/itex]

So,
[tex]\frac{{\partial \delta (S(x))}}{{\partial S}} = - \frac{{\delta (S(x))}}{{S(x)}}[/tex]
And since
[tex]\delta S(x) = \frac{{dS(x)}}{{dx}}dx[/tex]
We have,
[tex]\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[/tex]
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.