[Date Index]
[Thread Index]
[Author Index]
Re: function of a function
 To: mathgroup at smc.vnet.net
 Subject: [mg62668] Re: function of a function
 From: "JensPeer Kuska" <kuska at informatik.unileipzig.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: ownerwrimathgroup at wolfram.com
Hi,
"Narasimham" <mathma18 at hotmail.com> schrieb im
Newsbeitrag news:dmjrv9$638$1 at smc.vnet.net...
 JensPeer 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[xtau]]
has an infinite number of degrees of freedom
because you have to
supply a function for f[] in the interval
[x0tau,x0] to integrate
the equation numerical.
Regards
Jens
Prev by Date:
Re: Bypassing builtin functions in differentiation
Next by Date:
Re: Re: function of a function
Previous by thread:
Re: function of a function
Next by thread:
Re: Re: function of a function
 