Re: solving this impasse could result in solving the Navier-Stokes

David Purvance <d.purvance@xxxxxxx> writes:


I have a right to be frustrated with both you and Robert. You both
have given little 2x2 counter examples that I show don't apply to the
NSE and then you back off and say your examples mean something more
general, which I fail to see. This is engaging in "Winnie the
Poohism" as far as I am concerned. And to prove it, rationally like
professional mathematicians, let Robert successfully reply to my
request that he specifically explain how his "now generalized" example
demonstrates that there is a gap in my reasoning. Otherwise you're
both copping out.

This is the way it works in mathematics: if you have a proof of
something, you must be able to justify each step of that proof. It's
not enough that you make true statements, those statements must follow
from the previous statements for the reasons given in the proof.

Now suppose you have a certain step in your proof:
<statement #1>
and therefore because <reason>
<statement #2>.

To claim that your proof is faulty I could say that <reason> is not
enough to justify going from <statement #1> to <statement #2>. It's not
really required, but to support my claim I might come up with an example
where <statement #1> and <reason> are true but <statement #2> is false.
It doesn't matter that this example is small, or not realistic, or
that it doesn't satisfy other conditions that you have about the
situation at hand. All it says is that <reason> does not justify
going from <statement #1> to <statement #2>. In fact, it's often an
advantage to have a small example, because it is easier to see what
is going on there than in something larger and more realistic. Now
it might be that you can use some other conditions to patch up the
proof, by showing how <statement #2> really does follow from
<statement #1> plus those extra conditions. I hope you can do it.
But until you do, your proof has a gap.

In the case at hand you have formula (7.18):
A(u) = sum_{n=0}^infty A_n t^n
(where A_n don't depend on t, but A(u) does).
You then diagonalize A(u) = V Lambda V^(-1) so that
Lambda = sum_{n=0}^infty V^(-1) A_n V t^n
= sum_{n=0}^infty W_n Lambda_n W_n^(-1) t^n
and you assert that each term on the right side must be a diagonal
matrix because
(1) the left side is diagonal, and
(2) each matrix element on the right side is multiplied by t^n.

Stephen pointed out that this step does not work because V might
depend on t. In support of Stephen's assertion, I gave an example
where I have a sum of constant matrices times powers of t; I
diagonalize the sum as A = V Lambda V^(-1), multiply on the left
by V^(-1) and on the right by V, and notice that the terms
V^(-1) A_n V are not all diagonal. The conclusion is that you
have not justified this step of your proof.

Now it's not up to me to provide an example that satisfies a whole
bunch of other conditions: it's up to you to show that those other
conditions do justify your step. Until you do, you don't have a
valid proof.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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