fsolve a output variable is egual equal to the initial value I assigned

4 Ansichten (letzte 30 Tage)
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);

Antworten (1)

Alan Weiss
Alan Weiss am 4 Feb. 2013
Bearbeitet: Alan Weiss am 4 Feb. 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.
But you should also read the suggestions in this section of the documentation.
Good luck,
Alan Weiss
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