Re: steady state finder! still works after 6 hours!

[Fausto Arinos de A. Barbuto]

On 19-07-2015 15:59, Thomas D. Dean wrote:
> After running for several minutes, python is using 100% of one CPU.
>
> Tom Dean

After 6.5 seconds Matlab returned this (plus a few intermediate lines):

--------------

eq1 =

Omegabar == (97*Omegabar)/100 - (9603*Omegabar^2*((200/(13*Qbar) + (100/Qbar - 100/99)^2 - 
2/1287)^(1/2) - 100/Qbar + 100/99))/(1000000*((3*(200/(13*Qbar) + (100/Qbar - 100/99)^2 - 
2/1287)^(1/2))/10 - (39*((200/(13*Qbar) + (100/Qbar - 100/99)^2 - 2/1287)^(1/2) - 100/Qbar + 
100/99)^2)/20 - 30/Qbar + (99*Omegabar*((200/(13*Qbar) + (100/Qbar - 100/99)^2 - 2/1287)^(1/2) - 
100/Qbar + 100/99))/10000 + (99*Omegabar*(1/Qbar - 1/99))/100 + 1/330)) - 
(9603*Omegabar^2*(1/Qbar - 1/99))/(10000*((3*(200/(13*Qbar) + (100/Qbar - 100/99)^2 - 
2/1287)^(1/2))/10 - (39*((200/(13*Qbar) + (100/Qbar - 100/99)^2 - 2/1287)^(1/2) - 100/Qbar + 
100/99)^2)/20 - 30/Qbar + (99*Omegabar*((200/(13*Qbar) + (100/Qbar - 100/99)^2 - 2/1287)^(1/2) - 
100/Qbar + 100/99))/10000 + (99*Omegabar*(1/Qbar - 1/99))/100 + 1/330)) + 3/100

eq2 =

-(Qbar*((3*(200/(13*Qbar) + (100/Qbar - 100/99)^2 - 2/1287)^(1/2))/10 - (39*((200/(13*Qbar) + 
(100/Qbar - 100/99)^2 - 2/1287)^(1/2) - 100/Qbar + 100/99)^2)/20 - 30/Qbar + 
(99*Omegabar*((200/(13*Qbar) + (100/Qbar - 100/99)^2 - 2/1287)^(1/2) - 100/Qbar + 100/99))/10000 
+ (99*Omegabar*(1/Qbar - 1/99))/100 + 1/330))/Omegabar == (301*Qbar)/19800 - 
(97*Qbar*((3*(200/(13*Qbar) + (100/Qbar - 100/99)^2 - 2/1287)^(1/2))/10 - (39*((200/(13*Qbar) + 
(100/Qbar - 100/99)^2 - 2/1287)^(1/2) - 100/Qbar + 100/99)^2)/20 - 30/Qbar + 
(99*Omegabar*((200/(13*Qbar) + (100/Qbar - 100/99)^2 - 2/1287)^(1/2) - 100/Qbar + 100/99))/10000 
+ (99*Omegabar*(1/Qbar - 1/99))/100 + 1/330))/(99*Omegabar) + 199/200

Warning: Possibly spurious solutions. [solvelib::checkVectorSolutions]

c1 =

   33.957166883749936637715845077171
   1.1555947357398960182102501664145
  0.90221275720710083684727996953277
  0.55013082193348723765560802959132
 -0.64295531959083241534459063178293


c2 =

 -192.03106809965518577905241671637
  146.18591007809787081269098818378
  85.835723712966844293022398493044
 -241.42196871797658022054707968869
 -42.107485608195023067359978886338

--------------

Not sure if those are the desired/expected answers, but the code ran almost "as it 
was"
-- I only had to comment the 2nd and the 3rd lines.

Please attempt to the warning message, though.

Fausto