       Re: NDSolve solution does not fulfill boundary condition ??

• To: mathgroup at smc.vnet.net
• Subject: [mg121121] Re: NDSolve solution does not fulfill boundary condition ??
• From: Peter Pein <petsie at dordos.net>
• Date: Mon, 29 Aug 2011 07:43:39 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <j3fg96\$fb4\$1@smc.vnet.net>

```Am 29.08.2011 09:48, schrieb Alex:
> Hello everybody,
>
> I use Mathematica 7 and try to solve and plot a 2nd order ODE like this:
>
> sol = NDSolve[{0 == 4800*O2''[x] + 1500*Exp[-(100*x)^2] - 2000*O2[x],
> O2'[-0.5] == 0, O2'[0.5] == 0}, O2, {x, -1, 1}]
> Plot[O2[x] /. sol, {x, -1, 1}, PlotRange -> All]
>
> The solution looks ok, i.e. the slope of the solution is 0 at 0.5 and
> -0.5, as required by the boundary conditions. But if I slightly change
> the parameter values from 1500 to 1000 like this:
>
> sol = NDSolve[{0 == 4800*O2''[x] + 1000*Exp[-(100*x)^2] - 2000*O2[x],
> O2'[-0.5] == 0, O2'[0.5] == 0}, O2, {x, -1, 1}]
>
> I get a solution with slope=0 at -0.5, but with a positive slope at
> 0.5 !!??
>
> How can this be ?  What am I doing wrong ?
>
> Many thanks,
>
> axel
>

Hi Axel,

your solution is an even function. Try solving from 0 to 1 and define

o2[x_]=O2[Abs[x]] /. First[NDSolve[...,Method->"ImplicitRungeKutta"]]

and keep in mind you're using inexact numbers, or use DSolve[{...}, x]
in place of NDSolve[{...}, {x, -1, 1}].

O2'[{-1,1}/2] /. First[
DSolve[{0 == 4800*O2''[x] + 1500*Exp[-(100*x)^2] - 2000*O2[x],
O2'[-(1/2)] == 0, O2'[1/2] == 0}, O2, x]]

will give you exactly {0, 0}

Peter

```

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