       Re: nonlinear differential equation

• To: mathgroup at smc.vnet.net
• Subject: [mg54756] Re: nonlinear differential equation
• From: Paul Abbott <paul at physics.uwa.edu.au>
• Date: Tue, 1 Mar 2005 01:58:12 -0500 (EST)
• Organization: The University of Western Australia
• References: <cvhequ\$qft\$1@smc.vnet.net> <200502250618.BAA02402@smc.vnet.net> <cvrqae\$p3s\$1@smc.vnet.net> <cvulfi\$bit\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```In article <cvulfi\$bit\$1 at smc.vnet.net>, Peter Pein <petsie at arcor.de>
wrote:

> DrBob wrote:
>
> > Yikes!!! Good luck inverting the functions involved.
> >
> > Off[Solve::verif, Solve::tdep]
> > deqn = Derivative[s][t] -
> >       a*s[t]^2 - b*s[t] - c == 0;
> > ddeqn =
> >    ((Integrate[#1, t] & ) /@
> >       Expand[Derivative[s][t]*
> >         #1] & ) /@ deqn
> > s /. DSolve[{%}, s, t]
> > (-c)*s[t] - (1/2)*b*s[t]^2 -
> >     (1/3)*a*s[t]^3 +
> >     (1/2)*Derivative[s][t]^
> >       2 == 0
> ...
>
> As you can see, the conditions s==0 and s'==v0 hold only if v0==0.

Actually, this restriction is artificial -- I think it is introduced via
the step where the differential equation is multiplied by s'[t] and
integrated. Clearly, the differential equation with given initial
conditions for v0 nonzero has a solution:

nsol = NDSolve[{-8 s[t]^2 + 1 + 9 s[t] + s''[t] == 0,
s == 0, s' == 1}, s, {t, 0, 5}]

Plot[Evaluate[{s[t], s'[t], s''[t]} /. nsol], {t, 0, 5}]

The solutions look to me like some combination of Jacobi elliptic
functions.

Cheers,
Paul

--
Paul Abbott                                   Phone: +61 8 6488 2734
School of Physics, M013                         Fax: +61 8 6488 1014
The University of Western Australia      (CRICOS Provider No 00126G)
35 Stirling Highway
Crawley WA 6009                      mailto:paul at physics.uwa.edu.au
AUSTRALIA                            http://physics.uwa.edu.au/~paul

```

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