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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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