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MathGroup Archive 2006

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Re: FindRoot & NDSolve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64673] Re: [mg64650] FindRoot & NDSolve
  • From: Pratik Desai <pdesai1 at umbc.edu>
  • Date: Sun, 26 Feb 2006 05:07:55 -0500 (EST)
  • References: <200602250753.CAA13936@smc.vnet.net> <44008AE9.6070708@umbc.edu>
  • Sender: owner-wri-mathgroup at wolfram.com

Pratik Desai wrote:

> Takashi Inoue wrote:
>
>> Hi all,
>>
>> This is my first post to here.
>> I have a problem and want your help.
>>
>> Mathematica 4 did the following calculation with no pleblem,
>> while Mathematica 5(.2)  cannot do it.
>>
>> FindRoot[
>> (f /.NDSolve[ {D[f[x], {x,1}] - 2 x - a == 0, f[0] == 0}, f, {x, -3, 
>> 3} ][[1]])[2] == 4,
>> {a, -10, 10}
>>           ] 
>
> I think FindRoot requires a initial guess in 5.X and you need some 
> kind of looping for your different values of a
>
>
> Here are two possible alternatives (that I could think of) using 
> DSolve and NDSolve
>
> (*Using DSolve (more preferable perhaps because of the closed form 
> nature)*)
>
> TableForm[{#,x/.FindRoot[(f/.DSolve[{D[f[x], {x,1}]== 2 x + #, f[0] == 
> 0}, f, 
> x][[1,1]])[x]==4,{x,1}]//N}&/@Range[-10,10,1],TableHeadings->{Automatic,{"a","Root 
> for f[x,a]"}}]
>
> (*Using NDSolve  )
>
> Clear[f,x,a]
> Table[Reap[NDSolve[{D[f[x], {x,1}]== 2 x + a, f[0] == 0}, f, {x,-3,3}, 
> Method->{EventLocator, 
> "Event"->f[x]-4,"EventAction":>Sow[{a,x}]}],{a,-10,10,1}]

There is an error in the above code for NDSolve, here is the correct one
Table[NDSolve[{D[f[x], {x,1}]== 2 x + a, f[0] == 0}, f, {x,-3,3}, 
Method->{EventLocator, 
"Event"->f[x]-4,"EventAction":>Print[{a,x}]}],{a,-10,10,1}];



>
>
> Hope this helps
>
> Pratik
>
>> Takashi Inoue,Dept. Phys. Sophia University
>
>
>
>


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