Quantum Gravity Via Expansion-Contraction 12.0: Riccati Coefficients Yield Expansion-Contraction Transition at Golden Ratio (Phi)

**OsherD** · #1 August 22nd 06 posted to sci.physics

From Osher Doctorow

Let's look at the Riccati Differential equation:

1) dy/dt = A(t) + B(t)y + C(t)y^2

and let's choose B(t) = 1, A(t) = exp(kt), C(t) = -exp(-kt), k 0.
Then (1) becomesL

2) dy/dt = exp(kt) + y - exp(-kt)y^2

I remind readers that the Golden Ratio (phi) is the positive solution
of:

3) x^2 - x - 1 = 0

which is in fact:

4) x = (1 + sqrt(5))/2

Looking at the left hand side dy/dt of (1) or (2), we get increase
versus decrease in time, or expansion versus contraction, when dy/dt
changes from + to -, with dy/dt = 0 in between. Likewise, when dy/dt
changes from - to +, we get decrease followed by increase.

From (2), dy/dt 0 is equivalent to:

5) exp(kt) + y - exp(-kt)y^2 0

which is to say:

6) exp(kt) + y - y^2/exp(kt) 0

which is to say:

7) exp(2kt) + y exp(kt) y^2

Let's rewrite (7) as:

8) [exp(kt)]^2 + y exp(kt) y^2

It is natural, as for example in the Logistic Differential equation
subcase of the Riccati Differential equation, to restrict y to [0, 1]
or some such interval, and let's take a look now at y in (-1, 0).
Surely y^2 1 in (-1, 0), and also y = -/y/ in (-1, 0), so let's write
(8) as:

9) [exp(kt)]^2 - /y/ exp(kt) y^2

which is equivalent to:

10) [exp(kt)]^2 /y/exp(kt) + y^2

To guarantee (10), a simple condition (although perhaps not the only
one) is:

11) [exp(kt)]^2 exp(kt) + 1 exp(kt)/y/ + y^2

The second inequality holds always when y is in (-1, 0), and the first
inequality is on one side of the boundary:

12) [exp(kt)]^2 = exp(kt) + 1

which readers will recognize more easily as (letting x = exp(kt)):

13) x^2 - x - 1 = 0

which yields the positive solution x = phi (the Golden Ratio). When x
phi, it is easy to prove that the first inequality of (11) holds because (d/dx)(x^2 - x - 1) = 2x - 1 0 iff x 1/2 and for x = exp(kt) with k 0, we have x 1/2 always. So the question is, when is x = exp(kt) phi? We have:

14) exp(kt) = phi iff kt = log(phi)

and with exp(kt) increasing for k 0 with t, we have:

15) exp(kt) phi iff t log(phi)/k

So late in time (t log(phi)/k ), we get expansion of y via dy/dt 0.

Osher Doctorow

**OsherD** · #2 August 22nd 06 posted to sci.physics

From Osher Doctorow

Assuming that t is nonnegative and begins at 0, notice what happens at
t = 0:

1) dy/dt = exp(kt) + y - exp(-kt)y^2 = 1 + y - y^2 = 1 -/y/ - y^2

To have dy/dt negative at t = 0, we requi

2) /y/ + y^2 1

This holds on (-1, 0) for y if /y/^2 1/2, since on (-1, 0) as with
(0, 1), /y/ /y/^2 = y^2. But /y/^2 1/2 iff /y/ sqrt(1/2) =
sqrt(2)/2 which is approximately 0.71. So for -1 y 0.71
approximately, dy/dt is negative at t = 0.

Osher Doctorow

**Tom** · #3 August 22nd 06 posted to sci.physics

"OsherD" wrote in message
ps.com...
From Osher Doctorow

Assuming that t is nonnegative and begins at 0, notice what happens at
t = 0:

1) dy/dt = exp(kt) + y - exp(-kt)y^2 = 1 + y - y^2 = 1 -/y/ - y^2

the solution is wrong. and why did you replace y with -/y/ ?

To have dy/dt negative at t = 0, we requi

what is the reason to have dy/dt negative at t=0 ??

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