Science > Physics > Independent/Dependent Phases 2: Classical Logistic Equation
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Science > Physics |
| User: |
"OsherD" |
| Date: |
04 Oct 2005 01:29:51 PM |
| Object: |
Independent/Dependent Phases 2: Classical Logistic Equation |
From Osher Doctorow
The revised Theorem now reads:
Theorem For the generalized standardized Logistic Differential
equation:
1) dy/dt = B(t)y(1 - y), B(t) > 0
the second derivative Dtt(y) > 0 iff B'(t)/B^2(t) > 2y - 1 or
equivalently:
2) y < (1/2) + B'(t)/(2B(t)^2)
The proof has already been given.
Let's apply this to the classical standardized Logistic Differential
equation:
3) dy/dt = ky(1 - y), 0 < = y < = 1, k > 0
Here B(t) = k > 0, B'(t) = 0 for all t, so (2) becomes:
4) y < 1/2
which is precisely the result from Dyt(y) = Dy(dy/dt) = 0.
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Independent/Dependent Phases 2: Classical Logistic Equation |
04 Oct 2005 01:45:44 PM |
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From Osher Doctorow
Are there any examples for which the conditions of the Theorem are not
met? Yes.
Let's consider the generalized standardized Logistic Differential
equation with B(t) = exp(-t):
1) dy/dt = exp(-t)y(1 - y)
Then B' (t) = -exp(-t) and B' (t)/B^2(t) = -exp(-t)/[exp(-t)]^2 =
-1/exp(-t) = -1/(1/exp(t)) = -exp(t). Then B' /B^2 = -exp(t) > 2y - 1
iff 2y < 1 - exp(t) or y < (1/2) - (1/2)exp(t). For t = 0, this says y
< 0. For t > 0, this says y < negative number, both of which violate
the definition of y.
When B(t) is replaced above by exp(t), the condition always holds, but
the acceleration Dtt(y) doesn't change from positive for any t > = 0.
However, asymptotically as t --> infinity, we get y < 1/2
approximately, as expected from Dyt.
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Independent/Dependent Phases 2: Classical Logistic Equation |
04 Oct 2005 01:52:29 PM |
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From Osher Doctorow
I meant to say "which is precisely the result from Dyt(y) = Dy(dy/dt),"
not the same thing "= 0", in two postings back in the line below
equation (4).
Osher Doctorow
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