gradient of a delta function...

[Stephen Burks]

How on earth does one take a gradient of a delta function...

Say I have:

grad sub i operating on grad sub j operating on delta of (x - xo)

I have searched for a day for properties of delta functions and the only
thing that I found was that
x*d/dx of delta (x) =-delta'(x)

this is not worth anything though is it?

Any help would be greatly appreciated...

Thanks,
Stephen

[Robin Chapman]

In article <8q6u4v$deb$1...@lendl.cc.emory.edu>,

"Stephen Burks" <burks_...@hotmail.com> wrote:
> How on earth does one take a gradient of a delta function...
>
> Say I have:
>
> grad sub i operating on grad sub j operating on delta of (x - xo)
>

I'm not sure of the notation here, but say in the plane, we have
phi = delta(x - x_0, y - y_0). This is the distribution sending the
test function f to f(x_0,y_0). Differentiating phi with respect
to x will give the distribution taking each test function f to
- (d/dx)f(x_0,y_0) etc.

We can now take the exterior derivative (= "gradient") of phi
and get the "differential form" (d phi/dx) dx + (d phi/dy) dy.
These differential forms with distribution coefficients are called
currents. In the same fashion one can take exterior derivatives of these
1-currents to get 2-currents: d(phi_1 dx + d phi_2 dy)
= (d phi_2/dx - d phi_1/dy) dx /\ dy just as one does with forms.
One can do this on n-space or on manifolds and one gets a analogue
of the de Rham complex.

--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)

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[Martin Green]

Steven Burks wrote:

> How on earth does one take a gradient of a delta function...
>
> Say I have:
>
> grad sub i operating on grad sub j operating on delta of (x - xo)

I'm sure Robin Chapman has answered this question correctly.
Just as a point of clarification, I don't think it makes sense to
talk about the gradient of a 1-d delta function, as you have
here: delta of (x-x0). I'm sure you need at least a 2-d delta
function, like: delta of (x-x0, y-y0).

The gradient, like the gradient of any scalar "funtion", will
naturally be a vector "function".

Martin Green
http://www.onforeignsoil.com
Teach yourself Yiddish while reading the
exciting autobiography of Falk Zolf.

[Stephen Burks]

I am sorry for my bad notation, but x and x0 would be vectors in three
space... I really should have probably mentioned that earlier... Sorry...

Thanks,
Stephen