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MathGroup Archive 1998

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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


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