Re: multistep iterative methods
- To: mathgroup at smc.vnet.net
- Subject: [mg40335] Re: multistep iterative methods
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Wed, 2 Apr 2003 04:34:14 -0500 (EST)
- Organization: Universitaet Leipzig
- References: <b6bnnc$3t$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
the most efficent way ist to add a step size and order
control .. but that was not your challenge.
It is an oxymoron to ask for a *most efficient* ode solver
without a step size control. Since the variation of the order
is so easy with multistep methods it is a little work to
add an order control ..
Regards
Jens
Selwyn Hollis wrote:
>
> I'd like to throw this out as a challenge to the group: What's the most
> efficient way to implement in Mathematica an explicit multistep
> iterative method such as, say, the 4-step Adams-Bashforth method for
> solving y' = f(t,y):
>
> y[k+1]:= y[k] + (h/24)*(55*f[k] - 59*f[k-1] + 37*f[k-2] - 9*f[k-3])
>
> where y[0], y[1], y[2], y[3] are "given," and f[i] denotes f[t0 +i*h,
> y[i]]. The desired output would be the list
>
> {y[0], y[1], y[2], ... , y[n]}.
>
> A suitable toy problem is
>
> y' = -2t*y^2, y(0) = 1,
>
> with h = 0.01, n = 1000 (?), and the starting values taken from the
> exact solution y = 1/(1+t^2):
>
> y[0]=1, y[1] = 0.9999, y[2] = 0.9996, y[3] = .999101.
>
> Thanks in advance.
>
> -------
> Selwyn Hollis