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) 35 Stirling Highway Crawley WA 6009 mailto:paul at physics.uwa.edu.au AUSTRALIA http://physics.uwa.edu.au/~paul