Re: NDSolve[]
- To: mathgroup at smc.vnet.net
- Subject: [mg123534] Re: NDSolve[]
- From: Frederick Bartram <bartramf at acm.org>
- Date: Sat, 10 Dec 2011 07:29:59 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201112091058.FAA03928@smc.vnet.net>
- Reply-to: bartramf at acm.org
Well, if your stumped then I don't know if I can help
Sorry, if this is a trivial answer but from the documentation the two
forms are not strictly equivalent.
From the documentation for "Derivative": f' is equivalent to
Derivative[1][f].
So
> NDSolve[{Derivative[x[t], t] == -y[t] - x[t]^2,
> Derivative[y[t], t] == 2*x[t] - y[t], x[0] == y[0] == 1},
{x,
> y}, {t, 10}]
should be
NDSolve[{Derivative[1][x][t] == -y[t] - x[t]^2,
Derivative[1][y][t] == 2*x[t] - y[t], x[0] == y[0] == 1},
{x,
y}, {t, 10}]
Another example of behavior that is not quite what you might expect...
*-------------------------------------------------
* Frederick Bartram
* PGP key id: 0x63fa758 keyserver: http://keyserver.pgp.com
*/
> Z
> Subject: [mg123513] NDSolve[]
> To: mathgroup at smc.vnet.net
>
> For the same equations, why does the first method as following give
> the error but the other one give the desired result?
>
> NDSolve[{Derivative[x[t], t] == -y[t] - x[t]^2,
> Derivative[y[t], t] == 2*x[t] - y[t], x[0] == y[0] == 1},
{x,
> y}, {t, 10}]
>
> NDSolve[{x'[t] == -y[t] - x[t]^2, y'[t] == 2*x[t] - y[t],
> x[0] == y[0] == 1}, {x, y}, {t, 10}]
>
- References:
- [no subject]
- From: "Steven M. Christensen" <steve@smc.vnet.net>
- [no subject]