Clay Six
- Van Warren |
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Problem
6 in the Clay
million dollar prize series concerns the Navier
Stokes equations. They
are simultaneously the most simple to state (one line of math),
I would like to see Navier stokes expanded
using interval arithmetic. Highly uncertain flow
is subject to chaos. NS
rewritten in interval arithmetic looks like:
where
(p,p’)
= p +/- p’. read in the ordinary sense, a quantity p plus or
minus its associated uncertainty. Where
the quantity is a vector, so is the uncertainty. Interval
Arithmetic The
rules for adding, subtracting, multiplying and dividing in interval
arithmetic are: (a, a’) + (b, b’) = (a + b, a’ + b’) (a, a’) – (b, b’) = (a – b, a’ +
b’) (a, a’) * (b, b’) = (a * b, 2(a’ * |b| + a *
|b’|)) (a, a’) / (b, b’) = (a / b, 2(a’ / |b| + a *
|b’|)) This
is very beautiful math, since it propagates what is “knowable” as
part of the algebra. A related
equation, the second law of thermodynamics is involved.
Second
Law Considerations Currently
the second law of thermodynamics is an inequality. It
is was originally my conjecture that to solve Clay problem 6, one
must first An
increase in entropy is a measure of the degree to which the system
has gone from a more ordered, a +
bi, where a is the value, b is how much
it could vary and I is a unit that identifies the uncertainty. Interval
numbers are not complex numbers, the rules
are different for their arithmetic as we saw above from the four
basic ops. Vectors
of the form ai + bj are
to forms like a + bi like namespaces of the form index0, index1 are
to index, index1 Sidebars: 1) if someone ever asks you for an answer that has a numerical
value, you should report your certainty with it. 2) If a policeman ever writes down
your speed on a ticket without also reporting an uncertainty, you
should ask the judge to throw out the ticket. 3) You might want the upper uncertainty to be different than the lower uncertainty, rather than trying to center uncertainty. One
could represent a system’s degree of in-for-med-ness not as
a scalar but rather as a vector field of forcing functions with potentially
infinite complexity. If
this bothers you as it does me, you might
want to use the fact that increasing a systems informed-ness decreases
the value b of uncertainty at every point in its vector fields and
just run with that. History The
mathematics of finite differences came after those of continuous
fields mainly because sand pounders started making computers chips
lately. If finite differences had come first (as in Newton and Leibniz
not bothering to take the limit as Dx goes to zero), then we might have gotten to figuring this
out quicker and never have bothered to do continuum assumption mathematics.
The reason I make this unclear remark is
that if we are carrying finite differences around, the uncertainty
math and interval arithmetic fits right in. If you feel lost, do
not despair. Consider
interval addition. It makes sense, the new
number is just the sum of the originals and the sum of the uncertainties.
Uncertainties are always positive or zero, if you are perfectly certain.
Negative uncertainty is something for smarter people than I to chase after.
If your uncertainty is larger than the number itself then that is
interesting too. The derivation of the multiplication version is
so beautiful you must try it yourself lest I spoil
the fun. Just draw the 2D plane. The absolute value signs are necessary
to guarantee that uncertainties are not negative. Toy
Example Solutions
to the common quadratic equation ax2 + bx +
c = 0, are
not closed with respect to the real numbers, Let’s
solve the quadratic equation using interval arithmetic with real
numbers. Neglecting uncertainty in x itself which
complicates the solution the answer is: (-(b,b’)
+/- sqrt( (b,b’)^2 – 4
(a,a’) (c,c’)))/2(a,a’) The
interesting thing is what happens when the discriminant: (b,b’)^2 – 4
(a,a’) (c,c’) is negative. It can be the case that the center values produce a positive or
zero result, but including the uncertainty creates the possibility
of an undefined result with respect to the real numbers. This is
very interesting. Conclusions
and a Conjecture Solutions
to the Navier stokes equations depend on the boundary conditions
that define the flow conditions coming into and out of some characteristic
control volume. An
informal Statement of the Clay problem is: Show
that a box named Navier Stokes has smooth outputs for all time for
smooth inputs. My response is: The
shape of the outputs depend on the shape of the box in which Navier
Stokes is enforced. My conjecture is that
there exists a box in which non-smooth outputs exist for smooth inputs
and these outputs include shock waves, turbulence and other chaotic
phenomena that we observe in much simpler equations such as non-linear
pendulums. The scale of these phenomena is just the scale associated
with the uncertainty. Reynold’s number is the dimensionless parameter associated
with producting the expected behavior of
the flow. It would be an interesting exercise to relate Reynold’s number
to uncertainty in the rewritten Navier stokes. |
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