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

[Doug Stewart]

On Mon, Jul 20, 2015 at 7:01 AM, Doug Stewart <address@hidden> wrote:

On Mon, Jul 20, 2015 at 12:14 AM, Fausto Arinos de A. Barbuto <address@hidden> wrote:

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

This shows that the code is ok, since the second values are the same as Tom got.

I also tried:

xx6=solve(eq1,Qbar)

and 

xx7=solve(eq2,Qbar)

they also ran  for a while and then I broke out of it.

This example should be reported to the python symbolic group.

Doug

--

DAS

I also tried one that is impossible to solve:

syms x y

eq1= y== x*log(x)+sin(x)

w=solve(eq1,x)

and it reported quickly that it could not solve it.

So sympi is trying to solve it and it is solvable.

Sympi just seems to get lost!!!

--

DAS