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Re: NDSolve and vector functions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg87881] Re: NDSolve and vector functions
  • From: Szabolcs Horvát <szhorvat at gmail.com>
  • Date: Sat, 19 Apr 2008 03:35:02 -0400 (EDT)
  • Organization: University of Bergen
  • References: <fu9vl0$ie2$1@smc.vnet.net>

dh wrote:
> Hello all,
> 
> does anybody know a way to compute a vector valued function using 
> 
> NDSolve without explicitely specifying all vector components. Here is a 
> 
> simple example: Although NDSolve[{p'[t]==p[t],p[0]=={1,0}},p,{t,0,1}] works,
> 
> NDSolve[{p'[t]==p[t]+{1,1},p[0]=={1,0}},p,{t,0,1}]
> 
> does not work because p[t] in "p[t]+{1,1}"  is treated as a scalar and 
> 
> the expression is evaluated to {1+p[t],1+p[t]} what is clearly not 
> 
> intended. Even in "IdentityMatrix[2].p[t]+{1,1}" 
> 
> "IdentityMatrix[2].p[t]" is treated like a scalar and added to the 
> 
> components of {1,1}.
> 
> do I miss something???

Hi Daniel,

Obviously, there are workarounds, but they might not be pretty ... I 
think that this is a fundamental limitation of Mathematica ... Most 
functions (incuding Plus) assume that symbols represent scalars ...

A workaround is to define a new `plus' function that is only evaluated 
when its arguments are vectors:

vecPlus[arg__?VectorQ] := Plus[arg]

Now this works:

f = p /. First@
    NDSolve[{p'[t] == vecPlus[p[t], {1, 1}], p[0] == {1, 0}},
     p, {t, 0, 1}]

Plot[f[x], {x, 0, 1}]

One disadvantage of this approach is that it prevents NDSolve from 
transforming the ODE symbolically.  But this is really a limitation 
stemming from using vector functions, and not a problem introduced by 
vecPlus.  A more serious problem is that most probably vecPlus cannot be 
compiled, so this might slow NDSolve down.

Szabolcs


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