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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


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