Re: [boost] Library proposal: odeint

Subject: Re: [boost] Library proposal: odeint
From: Karsten Ahnert (karsten.ahnert_at_[hidden])
Date: 2010-04-28 02:07:22

Hi,

the informations are ok, maybe it could be slightly changed to

Boost.Numeric.Odeint
Author(s): Karsten Ahnert, Mario Mulansky
Version:
State: Under development but usable
Last upload: 2010/04/27
Links:  Sansbox Documentation
Categories: Math&Numerics
Description: Odeint is a library for solving ordinary differential
equations. It provides explicit methods like Euler, various Runge-Kutta
solvers, as well as adaptive step-size integration and the
Burlisch-Stoer algorithm. Furthermore, solvers for Hamiltonian systems
are implemented. Further development will go in the direction of
implicit solvers, stiff problems and CUDA support.

Thank you very much,

Karsten

On 04/27/2010 11:08 PM, vicente.botet wrote:
> ----- Original Message ----- 
> From: "Karsten Ahnert" <karsten.ahnert_at_[hidden]>
> To: <boost_at_[hidden]>
> Sent: Tuesday, April 27, 2010 9:16 PM
> Subject: [boost] Library proposal: odeint
> 
> 
>>
>> Hi,
>>
>> some months ago, I proposed a library for solving ordinary differential
>> equations, which can be found at
>>
>> http://svn.boost.org/svn/boost/sandbox/odeint/
>>
>> Is it possible to add odeint to the Libraries Under Construction page?
>>
>> Best regards,
>>
> 
> Hi,
> 
> Sorry I miss your library. Do you agree with this
> 
> Boost.Numeric.Odeint
> Author(s): Karsten Ahnert, Mario Mulansky Version: 
> State: ???
> Last upload: 2010/04/27
> Links:  Sansbox Documentation
> Categories: Math&Numerics
> Description:Odeint is a library for solving initial value problems (IVP) of ordinary differential equations. Mathematically, these problems are formulated as follows: x'(t) = f(x,t), x(0) = x0. x and f can be vectors and the solution is some function x(t) fullfilling both equations above. Numerical approximations for the solution x(t) are calculated iteratively. The easiest algorithm is the Euler-Scheme, where starting at x(0) one finds x(dt) = x(0) + dt*f(x(0),0). Now one can use x(dt) and obtain x(2dt) in a similar way and so on. The Euler method is of order 1, that means the error at each step is ~ dt[superscript 2]. This is, of course, not very satisfying, which is why the Euler method is merely used for real life problems and serves just as illustrative example.
> 
> Let me know if you prefer other informations or modify the description?
> 
> Best,
> _____________________
> Vicente Juan Botet Escribá
> http://viboes.blogspot.com/
> 
> 
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