Re: Relatively simple, but problematic, non-linear ODE

*To*: mathgroup at smc.vnet.net*Subject*: [mg72002] Re: Relatively simple, but problematic, non-linear ODE*From*: dh <dh at metrohm.ch>*Date*: Thu, 7 Dec 2006 06:26:22 -0500 (EST)*Organization*: hispeed.ch*References*: <el685m$2p9$1@smc.vnet.net>

Hi Alan, your diff. equ. has a singular solution y==2, that is a solution that is tangential to a family of solutions. At a point with y==2 we have 2 solutions runing through this point and every numerical procedure will have problems. You can convince yourself by calculating and analytical solution: rho = 0; z = 3.2; res = y[x] /. NDSolve[{y'[x]^2 - rho y'[x] y[x] + (1/4) y[x]^2 == 1,y[0] == 0}, y[x], {x, 0, z}]; Plot[Evaluate[res], {x, -3, z}] Daniel Alan wrote: > I am trying to find positive, non-decreasing solutions y(z) > on the interval (0,z), z > 0 to the ODE: > > (y')^2 - rho y y' + (1/4) y^2 = 1, with y[0] = 0. > > where y' = dy/dz, and the constant parameter |rho| < 1. > There are generally two solutions. The one I want has the > same sign as z. > > Here is my code: > > FSolver[z_, rho_] := > Module[{solnpair, vals, ans}, > solnpair = y[x] /. NDSolve[{y'[x]^2 - rho y'[x] y[x] + (1/4) y[x]^2 > == 1, > y[0] == 0}, y[x], {x, 0, z}]; > vals = Re[solnpair /. x -> z]; > ans = Select[vals, (Sign[#] == Sign[z]) &][[1]]; > Return[ans]; > > You can see that y[x] = 2 is also a solution to the ODE. > The solution I want apparently switches to y = 2 > at some critical value of z. The problem is that this causes Mathematica to > choke. > For example, for rho = 0, the exact solution is y(z) = 2 Sin(z/2) > for |z| < Pi and y(z) = 2 for larger z. I can fix things for rho = 0, since > I > am very confident about the answer. But I am less confident about > the solution for rho not equal to zero. > > For example, run my fragment with z = 3, rho = 1/2 to see the problems. > Any suggestions on how to get some reliable output > from NDSolve for this problem will be much appreciated. > > Thanks! > alan > >

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