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
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