fsolve a output variable is egual equal to the initial value I assigned
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Hi guys I have a "little" problem with Matlab. I'm using fsolve for solve a system, 8 equations and 8 variables, but one of the output variables, b(1), is equal to the initial value I assigned. I tried to change the equations, but the result is always the same, what's happening? I post the code. Thanks for your help.
function F = bottom19(b)
global Fc0 ke k1 k2 k3 k4 k6 e1 e2 e3 e4 e6 T Fsupp Q0 Pco20 R Mc Mcu Mcu2o Mcuo alfa0 Ptot1 T0 tau
F = [1-b(6)-b(7)-b(8); Fc0-b(1)-k1*e1*tau*b(1)*(b(6))^0.54-k6*e6*tau*b(1)*b(8)-ke*tau*b(1); Fsupp*(alfa0-b(2))-k2*e2*b(2)*b(7)-k4*e4*b(2); -Fsupp*b(3)+0.5*Mcu2o/Mcuo*k2*e2*b(2)*b(7)-k3*e3*b(3)*b(7)+0.5*Mcu2o/Mcuo*k4*e4*b(2); -Fsupp*b(4)+2*Mcu/Mcu2o*k3*e3*b(3)*b(7); Q0*Pco20/(R*T0)-b(5)*b(6)*Ptot1/(R*T)-k1*e1*tau*b(1)*(b(6))^0.54/Mc+0.5/Mcuo*k2*e2*b(2)*b(7)+1/Mcu2o*k3*e3*b(3)*b(7)+1/Mc*k6*e6*tau*b(1)*b(8); -b(5)*b(7)*Ptot1/(R*T)+2*k1*e1*tau*b(1)*(b(6))^0.54/Mc-0.5/Mcuo*k2*e2*b(2)*b(7)-1/Mcu2o*k3*e3*b(3)*b(7); -b(5)*b(8)*Ptot1/(R*T)+0.25/Mcuo*k4*e4*b(2)-1/Mc*k6*e6*tau*b(1)*b(8); Fc0/1000+Q0*Pco20/(R*T0)*44-b(1)/1000-ke*tau*b(2)/1000-b(5)*Ptot1/(R*T)*(b(6)*44+b(7)*28+b(8)*32)+Fsupp/1000*(alfa0-b(2)-b(3)-b(4))]
options = optimset('Display','iter','MaxFunEvals',1e10,'TolFun',1e-4,'Maxiter', 1e15,'Algorithm',{'levenberg-marquardt',0.0005});
b0 = [0.1;0.01;0.2;0.005;0.02;0.8;0.03;0.002];
[b,fval,exitflag,output] = fsolve(@bottom19,b0,options);
Réponses (1)
Alan Weiss
le 4 Fév 2013
Modifié(e) : Alan Weiss
le 4 Fév 2013
I count 9 equations, not 8:
function F = bottom19(b)
global Fc0 ke k1 k2 k3 k4 k6 e1 e2 e3 e4 e6 T Fsupp Q0 Pco20 R Mc Mcu Mcu2o Mcuo alfa0 Ptot1 T0 tau
F = [1-b(6)-b(7)-b(8); % 1
Fc0-b(1)-k1*e1*tau*b(1)*(b(6))^0.54-k6*e6*tau*b(1)*b(8)-ke*tau*b(1); % 2
Fsupp*(alfa0-b(2))-k2*e2*b(2)*b(7)-k4*e4*b(2); % 3
-Fsupp*b(3)+0.5*Mcu2o/Mcuo*k2*e2*b(2)*b(7)-k3*e3*b(3)*b(7)+0.5*Mcu2o/Mcuo*k4*e4*b(2); % 4
-Fsupp*b(4)+2*Mcu/Mcu2o*k3*e3*b(3)*b(7); % 5
Q0*Pco20/(R*T0)-b(5)*b(6)*Ptot1/(R*T)-k1*e1*tau*b(1)*(b(6))^0.54/Mc+0.5/Mcuo*k2*e2*b(2)*b(7)+1/Mcu2o*k3*e3*b(3)*b(7)+1/Mc*k6*e6*tau*b(1)*b(8); % 6
-b(5)*b(7)*Ptot1/(R*T)+2*k1*e1*tau*b(1)*(b(6))^0.54/Mc-0.5/Mcuo*k2*e2*b(2)*b(7)-1/Mcu2o*k3*e3*b(3)*b(7); % 7
-b(5)*b(8)*Ptot1/(R*T)+0.25/Mcuo*k4*e4*b(2)-1/Mc*k6*e6*tau*b(1)*b(8); % 8
Fc0/1000+Q0*Pco20/(R*T0)*44-b(1)/1000-ke*tau*b(2)/1000-b(5)*Ptot1/(R*T)*(b(6)*44+b(7)*28+b(8)*32)+Fsupp/1000*(alfa0-b(2)-b(3)-b(4))] % 9
This might make a difference -- fsolve is designed to work on "square" systems, systems with exactly as many equations as variables.
Good luck,
Alan Weiss
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