Re: NDSolve Problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg91835] Re: NDSolve Problem*From*: dh <dh at metrohm.ch>*Date*: Tue, 9 Sep 2008 06:59:15 -0400 (EDT)*References*: <g9r4d5$crm$1@smc.vnet.net>

Hi, we all know that a numerical solution is not 100% correct. Know, we must distinguish between absolute and relative error. E.g. An error of 1 may be negligable if the function value is 10^6, but hugh if the function is 1. Therefore, if you plot relative errors, you will see that NDSolve is not too bad: Res={(-x1'[t]+x2[t])/x2[t],(-x2'[t]+c^2 x1[t]-(c^2+p^2) Sin[p t])/x2'[t]}; Plot[Evaluate[Res/.sol/.dp],{t,0.0,2}] Concerning your second question, where you get something like 10^-16 instead of zero, this is the old problem of cancelation in sums. As soon as you calulate with machine precision your numbers are only approximative. If you then subtract 2 large numbers that are only approximately the same, you will not get zero. hope this will help, Daniel pratip wrote: > Lets take this system of ODE and solve with DSolve > > exsol = DSolve[{x1'[t] == x2[t], > x2'[t] == c^2 x1[t] - (c^2 + p^2) Sin[p t], x1[0] == 0, > x2[0] == Pi}, {x1[t], x2[t]}, t] // FullSimplify > > solution is as follows > {{x1[t] -> Sin[p t] + ((-p + \[Pi]) Sinh[c t])/c, > x2[t] -> p Cos[p t] + (-p + \[Pi]) Cosh[c t]}} > > Here is the values of the parameters > > dp = {c -> 100, p -> 2} > > Now solve with NDSolve > > sol = NDSolve[{x1'[t] == x2[t], > x2'[t] == c^2 x1[t] - (c^2 + p^2) Sin[p t], x1[0] == 0, > x2[0] == Pi} /. dp, {x1, x2}, {t, 0, 2}] > > Lets Plot the residual error > Res = {-x1'[t] + x2[t], -x2'[t] + c^2 x1[t] - (c^2 + p^2) Sin[p t]}; > Plot[Evaluate[Res /. sol /. dp], {t, 0, 2}] > > One can see the exponential increase in error. Please tell me if it is possible to manage this error for any parameter values. I want to know how I can use NDSolve to simulate such ODE with very little residual error just like any other normal system of ODE. Should I choose some special method options? > > Another very funny thing > > der = D[exsol, t]; > Res /. exsol /. der /. dp // Flatten // FullSimplify > > gives {0,0}. But lets take the residual with out simplifying. > res = Res /. exsol /. der /. dp // Flatten; > Plot[Evaluate[res], {t, 0, 1.2}, PlotStyle -> Thick] > > The Plot shows how badly exponential error shows up in the boundary. Is it a problem with Plot command? why it can not understand without simplifying that "res={0,0}". I hope you guys have an answer. > -- Daniel Huber Metrohm Ltd. Oberdorfstr. 68 CH-9100 Herisau Tel. +41 71 353 8585, Fax +41 71 353 8907 E-Mail:<mailto:dh at metrohm.com> Internet:<http://www.metrohm.com>