Re: NDSolve problem - solution does not satisfy boundary conditions
- To: mathgroup at smc.vnet.net
- Subject: [mg41082] Re: NDSolve problem - solution does not satisfy boundary conditions
- From: "Bill Bertram" <wkb at ansto.gov.au>
- Date: Thu, 1 May 2003 05:00:13 -0400 (EDT)
- Organization: Australian Nuclear Science and Technology Organisation
- References: <b8o28d$p31$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
"Martin Manscher" <please.respond at in.newsgroup> wrote in message news:b8o28d$p31$1 at smc.vnet.net... > I am trying to solve a nonlinear initial value problem numerically using > NDSolve from Mathematica 3.0. The differential equation contains > fourth-order space derivatives and first-order time derivatives. The > function evolving is a function of one space coordinate and time; b(z,t). > The I specify an initial value b(z,0) and boundary conditions b(0,t) and > b(L/2,t), which are consistent. I know that there is a solution, since this > is merely a first step to test the method used in a paper I have read. In > the paper, the solution evolved essentially to the effect of increasing the > initial disturbance 0.5*Cos[2*Pi*z/L] smoothly in about 6 time units. The > boundary values are supposed to kee the solution periodic. > > However, the solution produced by NDSolve does not satisfy the boundary > condition Derivative[1, 0][b][0, t] == 0, but develops a finite slope almost > immediately. I have tried all sorts of remedies, such as forcing Gear's > method, increasing WorkingPrecision, DifferenceOrder, MaxSteps, > StartingStepSize etc. etc. > > Does anyone have any ideas? Am I doing something wrong? Hi Martin, Seems to work with Mathematica 4.2. Well almost, it does give a warning that the boundary and initial conditions are inconsistent. The solution looks OK and seems to satisfy the boundary and initial conditions. However there is not much difference between the solution at t = 0 and the solution at t = 10! I don't know why mathematica issues the warning, as far as I can tell, the inital condition does satisfy all of the boundary conditions. Cheers, Bill