Re: Nonlinear Differential Equation

*To*: mathgroup at smc.vnet.net*Subject*: [mg14157] Re: Nonlinear Differential Equation*From*: Alan Lewis <alan at enfs.com>*Date*: Mon, 28 Sep 1998 18:57:22 -0400*Organization*: @Home Network*References*: <6unfnj$o89@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi David, Interesting problem; I evaluated the DSolve below and got an expression with two constants c[1] and c[2]. Applying your boundary conditions (manually) I think the constants are c[1]=1/4-I Pi/2 and c[2]=0. Then by playing around with the integral, I'll guess that the exact amplitude and period are given by the expressions A and P below. (Technically, you have to take eps->0 in the expression for P). You'll see these numbers are pretty close to the ones you quoted. Hope this helps. DSolve[w''[x]+2 w[x]/(1-w[x]^2)==0,w[x],x] A=Sqrt[1-E^(-1/2)]; P[eps_]:= N[2 Sqrt[2]NIntegrate[(1/2+ Log[1-t^2])^(-1/2),{t,0,A-eps}]]; N[A] 0.627271 P[10^(-7)] 3.70814 Alan Lewis David Djajaputra wrote: > > Hello, > > I'm just wondering if anyone knows how to solve this DE: > > y''(x) = - 2 y(x) / [1 - y^2(x)], y(0) = 0, y'(0) = 1. > > The function is periodic. Mathematica says that it's transcendental but > it cannot solve it. I'm curious if the amplitude and/or the period can > be obtained in closed forms. By the way, if you plot it the amplitude > is > > about 0.6 and the period is about 3.7. > > Much thanks in advance, > > David