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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>
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