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MathGroup Archive 2000

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Re: Q: NDSolve

  • To: mathgroup at
  • Subject: [mg21576] Re: Q: NDSolve
  • From: Bojan Bistrovic <bojanb at>
  • Date: Sat, 15 Jan 2000 02:04:19 -0500 (EST)
  • Organization: Old Dominion Universityaruba
  • References: <85mmgb$>
  • Sender: owner-wri-mathgroup at

Christoph Handel wrote:
>    howdy,
> is there a nice way to solve an equation like this:
> x'[t] == aMatrix[t] . x[t]
> where x ist a Vector.
> I tried something like this:
> NDSolve[{x'[t]== aMatrix[t] . x[t], x[0]=={1,2,3...}},x,{t,0,42}]
> Greetings
>         Christoph
> --
> the adress is valid
> for faster reply use handel at the same host

This might not be "a nice" way, but it works:


In[2]:= MyEqual[arg1_==arg2_]:=

In[3]:= SetAttributes[MyApplyFunction,Listable]
In[4]:= MyApplyFunction[f_,t_]:=f[t]/;MyfunctionTest[f]
In[5]:= MyApplyFunction[f_,t_]:=f/;Not[MyfunctionTest[f]]
In[6]:= MyThread[xx_/;Head[Head[xx]]===List]:=

In[7]:= MyThread[xx_[t_]/;Head[Head[xx]]===Derivative]:=

In[8]:= MyThread[xx_==yy_]:=MyThread[xx]==MyThread[yy]
In[9]:= MyThread[xx_ .yy_]:=MyThread[xx].MyThread[yy]

In[10]:= x={{f1},{f2}};
In[11]:= A={{2, t},{t+3,5}};


Out[12]:={{f1 -> InterpolatingFunction[{{0.,10.}},"<>"],
    f2 -> InterpolatingFunction[{{0.,10.}},"<>"]}}

You'll probably want to modify MyfunctionTest to fit your needs.

Bye, Bojan

Bojan Bistrovic,                       bojanb at  
Old Dominion University, Physics Department,      Norfolk, VA

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