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MathGroup Archive 2005

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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[2][s][t] -
> >       a*s[t]^2 - b*s[t] - c == 0;
> > ddeqn =
> >    ((Integrate[#1, t] & ) /@
> >       Expand[Derivative[1][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[1][s][t]^
> >       2 == 0
> ...
> 
> As you can see, the conditions s[0]==0 and s'[0]==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] == 0, s'[0] == 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)         
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Crawley WA 6009                      mailto:paul at physics.uwa.edu.au 
AUSTRALIA                            http://physics.uwa.edu.au/~paul


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