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MathGroup Archive 2005

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Re: function of a function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg62668] Re: function of a function
  • From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
  • Date: Wed, 30 Nov 2005 05:40:40 -0500 (EST)
  • Organization: Uni Leipzig
  • References: <dmha20$932$1@smc.vnet.net><dmhfhd$bit$1@smc.vnet.net> <dmjrv9$638$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

"Narasimham" <mathma18 at hotmail.com> schrieb im 
Newsbeitrag news:dmjrv9$638$1 at smc.vnet.net...
| Jens-Peer Kuska wrote:
|
| > it can't work because f [0] ==1 given in your 
differential equation
| > f ' [0]==f [1] and NDSolve[] can't find the 
value for
| > f[1] until it has integrated the equation.
|
| ???
|

You don't understand, that in

EQ= { f'[x] == f[f[x]], f[0]== 1} ; 
NDSolve[EQ,f,{x,0,2}];

the right hand side must be evaluated by NDSolve[] 
at x==0 and this
gives
f'[x]==f[f[x]] /. x->0
f'[0]==f[f[0]]

and with your initial condition f[0]==1
one ends up with
f'[0]==f[1]
and this can only evaluated when NDSolve[] has 
escaped from x==0
but it can't, because it is not able to get a 
numerical value for the
derivative at x==0 ???


| > The nested dependence is equivalent to an 
infinite
| > system of ordinary differential equations and 
it seems to be
| > hard to do this by a finte computer.
|
| I cannot understand this. In the following two 
examples the first one
| works, not the second.
|
| Clear[x,f,EQ];
| EQ={f'[x] == f[Cos[x]],f[0]== 1};
| NDSolve[EQ,f,{x,0,4}];
| f[x_]=f[x]/.First[%];
| Plot[f[x],{x,0,4}];
|
| Clear[x,f,EQ];
| EQ={f'[x] == Cos[f[x]],f[0]== 1};
| NDSolve[EQ,f,{x,0,4}];
| f[x_]=f[x]/.First[%];
| Plot[f[x],{x,0,4}];

the first example can't work (and it does not) 
because

EQ={f'[x] == f[Cos[x]],f[0]== 1};

at x==0 gives
f'[0]==f[Cos[0]]

and

f'[0]==f[1]

but nobody can find f[1]

That has *nothing* to do with Mathematica because 
it is a
functional equation that can't be solved. Even the 
simpler case

f'[x]==g[f[x-tau]]

has an infinite number of degrees of freedom 
because you have to
supply a function for f[] in the interval 
[x0-tau,x0] to integrate
the equation numerical.

Regards
  Jens





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