Re: Solving Weissinger's ODE
- To: mathgroup at smc.vnet.net
- Subject: [mg104885] Re: [mg104842] Solving Weissinger's ODE
- From: Mark McClure <mcmcclur at unca.edu>
- Date: Thu, 12 Nov 2009 06:06:45 -0500 (EST)
- References: <200911110929.EAA29439@smc.vnet.net>
On Wed, Nov 11, 2009 at 4:29 AM, Virgil Stokes <vs at it.uu.se> wrote: > I can not see why the following does not work as expected, > > s = NDSolve[{t * (y[t])^2 * (y'[t])^3 - (y[t])^3 * (y'[t])^2 + t * > (t^2 + 1) * y'[t] - t^2 *y[t] == 0, y[1] == Sqrt[3/2]}, y[t], {t, 1, 10}] > > Note, the solution to this nonlinear, non-autonomous, implicit ODE for > initial condition y[1] = Sqrt[3/2] is just y[t] = Sqrt[t^2 + 1]. First, I think the correct solution is y(t) = sqrt(t^2 + 1/2). In[1]:= y[t_] = Sqrt[t^2 + 1/2]; t*(y[t])^2*(y'[t])^3 - (y[t])^3*(y'[t])^2 + t*(t^2 + 1)*y'[t] - t^2*y[t] // Simplify // InputForm Out[2]//InputForm= 0 Second, I see no problem with Mathematica's numerical solution. In[3]:= Clear[y]; s = NDSolve[{ t*(y[t])^2*(y'[t])^3 - (y[t])^3*(y'[t])^2 + t*(t^2 + 1)*y'[t] - t^2*y[t] == 0, y[1] == Sqrt[3/2]}, y[t], {t, 1, 10}]; y[t_] = y[t] /. s[[1]]; NIntegrate[(y[t] - Sqrt[t^2 + 1/2])^2, {t, 1, 10}] Out[6]= 1.43523*10^-13 One bit of strangeness: NDSolve returns a list of three solutions. Does this violate any particular existence and uniqueness theorem? Perhaps not, as two solutions are complex. Mark McClure
- References:
- Solving Weissinger's ODE
- From: Virgil Stokes <vs@it.uu.se>
- Solving Weissinger's ODE