On 10-07-2013, at 16:21, Raphaëlle Carraud <raphaelle.carr...@oc-metalchem.com> 
wrote:

> Bonjour,
> 
> Je souhaite  résoudre le couple d'équation différentielles suivant :
> 
>    0 = -dA + dB + 2*dC - 2*r1 - 2*r5
>    0 = dA + dD + r1 + r4
>    0 = K2 - C/B^2
>    0 = K3 - D/(A*B)
> 
>    0 = r5 + 2*r4 - dE
>    0 = r5 -dI
>    0 = -r5 - r4 - dG
>    0 = -r1/2 - dH
> 
> en ayant connaissance des valeurs initiales de dA, dB, dC, dE, dI, dG, dH, 
> r1, r2, r4, r5, K2, K3, A, B, C et D.
> 

If all initial values are known then plugging the values in the system will 
give 0 or not 0. There is nothing to "solve".

> J'ai essayé plusieurs fonctions mais comme je ne peux pas lui faire calculer 
> une des dérivée de laquelle découlerait les autre, il n'arrive pas à me 
> fournir la solution.
> Je n'ai pas vu d'exemple qui pourrai s'assimiler à celui-ci dans la 
> documentation.
> 

You will have to redo your query in English. Questions in French won't receive 
many replies.
My French is rudimentary but I'll try.

You have 8 equations and 17 variables.
So how do you propose to "solve" the system?

Assuming that the d? variables are differentials and that you want to solve for 
those:
you have 7 of these and 8 equations. So how to solve?

But the third and fourth equations have no d? variables, so the may even be 
inconsistent given the values of K2, K3, C, B, A, D.
So you have 6 equations for 7 d? variables. So how do you propose to solve for 
the d? variables?

Finally your system seems to be linear in the d? variables. You would be able 
to use R's solve()  if you can get your system to be a square system.

If your system is not square and underdetermined then you can use a Moore 
Penrose inverse to get a minimum norm solution 
(http://en.wikipedia.org/wiki/Moore–Penrose_pseudoinverse#Minimum-norm_solution_to_a_linear_system).
package MASS provides a function ginv().

Berend

> Est-il possible de résoudre ce problème sur R ?
> 
> Merci
> 
> Cordialement,
> Raphaëlle Carraud
> 
>       [[alternative HTML version deleted]]
> 
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