[scilab-Users] Problem with division by zero

CRETE Denis denis.crete at thalesgroup.com
Tue Jul 19 19:29:04 CEST 2011


Hello,
it may have to do with the (non-physical?) expression of the ODE.
Besides the fact that the last term is mathematically ill-defined (starting with "+/"), in the case K=0, the second derivative of X is positive for x<%pi/2: it may then reach the value %pi/2, which makes denominators be 0.
Physically, these denominators (%pi/2-X) look very strange as they may govern divergence of the force in the last term.
However, it appears also in the first term, where divergence is prevented because of (1-sin X): it may than be necessary to rewrite the function (1-sinX)/(%pi/2-X), with the addition of the case X=%pi/2, where the function goes to 0 by continuity.
HTH
Denis

De : arnaud dujeu [mailto:fangorn813 at hotmail.com]
Envoyé : mardi 19 juillet 2011 11:46
À : users at lists.scilab.org
Objet : [scilab-Users] Problem with division by zero

Hi,

I am using scilab to solve a differential equation and I have an error ("error 27") I don't understand.

When K=1 pr K>1 it's okay but as soon as K<1 there is the error.

This is my code :


function [Xprime]=VanDerPol(t, X)



 Xprime(1) = X(2)

 Xprime(2) = g/R*(1-sin(X(1)))/(%pi/2-X(1))-(mu*(K)*R^2*((X(2))^2)/2)+/(R^2*mu*(%pi/2-X(1)))



endfunction



 t=linspace(0,4,3000); X0=[85*%pi/180;0];

 [u]=ode(X0,0,t,VanDerPol);



plot(t,u(1,:)*180/%pi,'r')

 plot(t,R*u(2,:))





And there is the Scilab message when K<1 :



-->exec('C:\Documents and Settings\ademollerat\My Documents\accelerometre\Ecoulement\VanDerPol-2.sci', -1)

Attention: redéfinition de la fonction: VanDerPol               . Utilisez funcprot(0) pour éviter ce message



 !--error 27

Division par zéro ...



at line       7 of function VanDerPol called by :

 [u]=ode(X0,0,t,VanDerPol);

at line      35 of exec file called by :

exec('C:\Documents and Settings\ademollerat\My Documents\accelerometre\Ecoulement\VanDerPol-2.sci', -1)





Maybe I have made a mistake or maybe it is a bug.



Thank you for helping me.



Best regards
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