NetNews Usenet Archive 1992 #16/NN_1992_16.iso/spool/sci/physics/11291

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Path: sparky!uunet!dtix!darwin.sura.net!uvaarpa!murdoch!kelvin.seas.Virginia.EDU!crb7q
From: [e-mail] (Cameron Randale Bass)
Newsgroups: sci.physics
Subject: Re: Chaos
Summary: longer than it probably should be
Message-ID: <[e-mail]>
Date: 21 Jul 92 22:25:53 GMT
References: <[e-mail]> <[e-mail]> <[e-mail]>
Sender: [e-mail]
Organization: University of Virginia
Lines: 185

In article <[e-mail]> [e-mail] (John C. Baez) writes:
>In article <[e-mail]> [e-mail] (Cameron Randale Bass) writes:
>
>>     Work on existence and uniqueness and 'attractors' in 2-D with
>>     usually special forcings and/or boundary conditions is hardly
>>     of earth-shattering import.  Much more interesting from Foias
>>     is his work on the Hopf functional formulation of turbulence
>>     This is, of course, unrelated to 'chaos'.  Most people do not know
>>     that the Navier-Stokes equations have been 'solved' (formally
>>     by functional methods J. Rat. Mech. Anal. 1:87 (1952) and all
>>     that followed from that work).  This could be very interesting
>>     if we were better at functional integration.  I suggest that work in
>>     this area (of course, there are people working in this area)
>>     is probably more potentially earthshaking, and there are applications
>>     in other research backwaters (e.g. QM).
>
>Hmm... I suppose I should just read the reference, but I wonder in what
>sense the Navier-Stokes equation can be said to be "solved".  There is
>no good global existence theorem for solutions, for one thing.  It's
>true that the Navier-Stokes equation, like most decent PDE's, can be
>written as a ODE with values in an infinite-dimensional vector space

    Only in the sense that the problem has been recast formally
    in terms of a linear functional equation and 'solved' in terms
    of functional integrals.  Unfortunately, as far as that goes,
    it is nearly useless.  One hopes that if we were somehow on a
    better footing as far as the approximation of functional integrals
    then this would be a very useful path.

    And I believe that you are correct in that there are no widely applicable
    existence or uniqueness results except some fairly special
    ones by Temam and the boys (for example in Temam's 'Navier-Stokes
    Equations').

>df/dt = G(f)
>
>and then "solved" by an integral equation
>
>f(t) = f(0) + int_0^t G(f(s)) ds .
>
>While this technique is really useful for studying PDE it would be
>misleading to say to ordinary folks that the Navier-Stokes equation had
>been solved if this was all there was to it.  Of course, you say it was
>"solved" with quotes.   If it's possible to clarify without too much
>work that'd be nice.

     Yes, it would.  You are right about my use of the quotes.  It
     was for the reason that the 'solution' is completely analogous to
     what you have above.  However, it is even worse than this.  If we
     were given G in the above example, we'd stand a good chance of
     getting substantial information out of the 'solution'.  The G in
     the Hopf equation is cast in terms of functional integrals.

     For completness sake, a LaTeX copy of some of my notes for an old
     proposal in which I derive the Hopf functional formulation for
     Burgers' equation is included below for purposes of increasing network
     volume.

>By the way, it's very enjoyable to read you writing about this stuff - as
>opposed to the SSC - I guess I like physics more than politics.

     Thanks.  However, I find that boiling my blood occasionally keeps the
     arterial plaque down.

                                dale bass

     Quick and dirty derivation of the Hopf functional formulation of
     Burgers' equation (LaTeX format, I have no idea how well-checked
     it is either since I wrote it five years ago).

...

Burgers' equation gives us
\begin{equation}
{\partial u \over \partial t}
= Q(u) = - u {\partial u \over \partial x}
+ \mu {\partial^2 u \over \partial x^2}
\label{eho1.1}
\end{equation}
We use the characteristic functional
\begin{equation}
\Phi(y,t) = \int_\Omega e^{i (y,u)} P^t (du)
= \int_\Omega e^{i (y,u^t)} P (du)
\label{eho1.2}
\end{equation}
(The above follows from the
assertion that $u \in T^{-t} B \rightarrow u^t \in B$ so that
\begin{eqnarray}
P^t(B)& =& P(T^{-t} B) \\
P^t(e^{i(y,u)} du)
&=&
P(e^{i(y,u^t)} du)
\label{eho1.3}
\end{eqnarray}
and is very important in what follows.)
The characteristic functional $\Phi$ should be continuous and for all
values of $y(x)$
\begin{equation}
\Phi(0,t) = 1 \quad\quad |\Phi(y(x),t)| \leq 1.
\label{}
\end{equation}
(more on the function $y$
and the characteristic functional, compact support, zero at infinity etc...)

We can differentiate this functional with respect to time
(ordinary derivative)
\begin{eqnarray}
{\partial \Phi(y,t) \over \partial t}
&=& \int_\Omega i (y,u^t) \, e^{i (y,u^t)} P (du) \nonumber \\
&=&  i \int_\Omega (y,u^t) \, e^{i (y,u)} P^t (du) \nonumber \\
&=&  i \int_\Omega (y,Q) \, e^{i (y,u)} P^t (du) \nonumber \\
&=&  i \int_\Omega \left[\int_R y Q(u) \, dx\right] \,
e^{i (y,u)} P^t (du)
\label{eho1.4}
\end{eqnarray}
Change the order of integration to take $y$ out of the
$\Omega$ integral
\begin{equation}
{\partial \Phi(y,t) \over \partial t}
=  i \int_R y \left[\int_\Omega  Q(u) \,
e^{i (y,u)} P^t (du) \right] \, dx
\label{eho1.5}
\end{equation}
Substitute for $Q$ in this expression from Burgers' equation
\begin{equation}
\int_\Omega  Q(u) \, e^{i (y,u)} P^t (du)
=
\int_\Omega  \left( - u u_x + \mu u_{xx} \right) \, e^{i (y,u)} P^t (du)
\label{eho1.6}
\end{equation}
But $uu_x = (1/2) (u^2)_x$, so that term by term
for the first part of equation \ref{eho1.6}
\begin{equation}
{1 \over 2}
{\partial \over \partial x}
\int_\Omega  u^2 \, e^{i (y,u)} P^t (du)
=
{1 \over 2}
{\partial \over \partial x}
\int_\Omega
{\partial \over i \partial y(x) \, dx}
{\partial \over i \partial y(x) \, dx}
 \, e^{i (y,u)} P^t (du)
\label{eho1.7}
\end{equation}
For the second part, we have
\begin{equation}
\mu
{\partial^2 \over \partial x^2}
\int_\Omega  u \, e^{i (y,u)} P^t (du)
=
\mu
{\partial^2 \over \partial x^2}
\int_\Omega
{\partial \over i \partial y(x) \, dx}
 \, e^{i (y,u)} P^t (du)
\label{eho1.8}
\end{equation}
So finally, we can pull all of the functional derivatives out
of the integral and consolidate with the definition
\begin{equation}
\Phi(y,t) = \int_\Omega e^{i (y,u)} P^t (du)
\label{eho1.9}
\end{equation}
to get
\begin{equation}
{\partial \Phi \over \partial t} =
\int_R y(x)
\left[
{i \over 2}
{\partial \over \partial x}
{\partial^2 \Phi \over \partial y(x) \, dx \,\, \partial y(x) \, dx}
+
\mu
{\partial^2 \over \partial x^2}
{\partial \Phi \over i \partial y(x) \, dx} \right]
\, dx
\label{eho1.10}
\end{equation}
This is the Hopf functional form of Burger's equation,
--
C. R. Bass                                           [e-mail]
Department of Mechanical and Aerospace Engineering
University of Virginia
Charlottesville, Virginia                            [phone]