About the exchangeability of variational and differential.

[Hongyi Zhao]

Hi all,

I'm a physics teacher China. Today, when I
state the principle of least action to my students, I meet a basic
problem which I cann't figure out:

Under what condition can we exchange the sequence of variational and
differential?

Say for the following example:

delta ( dq/dt ) = d ( delta q) / dt, where, the q is a funtion of t.

What's the condition for the above equality?

I thinks it should have some conditions, but I cann't figure it out.

Could you please give some hints? Thanks a lot in advance for your
time and patience :-)

Regards,
.: Hongyi Zhao [ hongyi.zhao AT gmail.com ] Free as in Freedom :.

[Rupert]

We want (delta/delta h) ( d(q+hq')/dt ) = (d/dt) ( (delta/delta h) (q
+hq') ).

So the problem is about when you are allowed to swap two partial
derivative operators with respect to two different variables when
dealing with a function of two variables. Do you know a sufficient
condition for this?

[Hongyi Zhao]

On Mon, 8 Oct 2012 07:56:28 -0700 (PDT), Rupert
<rupertm...@yahoo.com> wrote:

[snipped]

Thanks a lot for your quickly reply and warmly help;-)

>
>We want (delta/delta h) ( d(q+hq')/dt ) = (d/dt) ( (delta/delta h) (q
>+hq') ).

1- What's the meaning of q' used here? Why you use the combinational
function (q+hq')?

2- You mean the above equation is equivalent to my origal one, but I
cann't figure out from where or which point of view to duduce the
above eqution given by you, sorry for my poor mathematics. Could you
please give me some more hints?

>
>So the problem is about when you are allowed to swap two partial
>derivative operators with respect to two different variables when
>dealing with a function of two variables. Do you know a sufficient
>condition for this?

Essentially, the (partial) derivative operators are defined via
limits, so I think the sufficient condition for swap two (partial)
derivative operators is the corresponding two limits can be
exchangable. I think this should reqire the series based on the
function of two variables must be uniformly converged. Am I right?

Regards

[Rupert]

On Oct 9, 12:59 pm, Hongyi Zhao <hongyi.z...@gmail.com> wrote:
> On Mon, 8 Oct 2012 07:56:28 -0700 (PDT), Rupert
>

> <rupertmccal...@yahoo.com> wrote:
>
> [snipped]
>
> Thanks a lot for your quickly reply and warmly help;-)
>
>
>
> >We want (delta/delta h) ( d(q+hq')/dt ) = (d/dt) ( (delta/delta h) (q
> >+hq') ).
>
> 1- What's the meaning of q'  used here?  Why you use the combinational
> function (q+hq')?
>

q' is meant to represent a small change in the function q.

> 2- You mean the above equation is equivalent to my origal one,  but I
> cann't figure out from where or which point of view to duduce the
> above eqution given by you, sorry for my poor mathematics.  Could you
> please give me some more hints?
>

Look, to be honest I've never read a presentation of the variational
calculus, but I always assumed that if q is a function defined on a
compact interval [a,b], then saying delta q=0 means d/dh (q+hq')=0 at
h=0, for all C^1 functions q' which take the value zero at the
endpoints of the interval. Do you know any other definition of what it
means?

>
>
> >So the problem is about when you are allowed to swap two partial
> >derivative operators with respect to two different variables when
> >dealing with a function of two variables. Do you know a sufficient
> >condition for this?
>
> Essentially, the (partial) derivative operators are defined via
> limits,  so I think the sufficient condition for swap two  (partial)
> derivative operators is the corresponding two limits can be
> exchangable.   I think this should reqire the series based on the
> function of two variables must be uniformly converged.  Am I right?
>

It's a bit more complicated than that. Have a look at the early
chapters of Michael Spivak's "Calculus on Manifolds".