Re: Riccati Type 1 Solutions

From: Osher Doctorow (mdoctorow_at_comcast.net)
Date: 08/27/04 

Date: Fri, 27 Aug 2004 14:09:35 +0000 (UTC)

On 26 Aug 04 13:08:47 -0400 (EDT), Osher Doctorow wrote:
>According to my present calculations, (1) can be converted to:
>
>2) ab I[1/(u^2 + mu + h)]du = cdt
>
>for m (not necessarily integer), h, c constants and I...du the
>indefinite integral, and if u^2 + mu + h factors into say (u+c1)
>times (u + c2) for constants c1, c2, then u is a rational express-
>ion in exp(rt) for r constant.

Let's consider two simple examples.

1) u = texp(t)
2) u = (1/t)exp(t)

Differentiating u in (1) yields:

3) du/dt = texp(t) + exp(t) = u + u/t = u(1 + 1/t) = A(t)u

where A(t) = (1 + 1/t). Here du/dt > 0 everywhere.

Next, differentiating u in (2) yields:

4) du/dt = [texp(t) - exp(t)]/t^2 = exp(t)/t - exp(t)/t^2
= u - u/t = u(1 - 1/t) = A(t)u

where A(t) = 1 - 1/t. Here du/dt > 0 iff t > 1 and du/dt < 0 iff
0 < t < 1 where t is time. This is certainly not oscillating, but
neither is there pure increase or pure decrease - there is rather
decrease followed by increase as t increases, with a minimum at t
= 1. If t is taken as a [0, 1] time variable with 1 representing
"infinity", we do have a pure decrease regime.

The Riccati differential equation does represent pure expansion
and pure contraction, but also the first followed by the second
or vice versa on a finite (and typically small) number of
occasions, for example one or two occasions. Again, the value of
limiting A(t) to t^2 or t or t^(-1) or t^(-2) in polynomial
extremes is similar to that of a line or simple closed curve in
discriminating between two sets.

Osher Doctorow