Why i call this function of equation? help with fsolve. thanks

2 vues (au cours des 30 derniers jours)
sensei
sensei le 23 Nov 2016
Commenté : sensei le 23 Nov 2016
S_IT = 51.2e-4;
S_ID = 8.2e-4;
S_IE = 520e-4;
BW = 53;
tau_G = 40;
tau_I = 55;
A_G = 0.8;
k_12 = 0.066;
k_a1 = 0.006;
k_b1 = S_IT*k_a1;
k_a2 = 0.06;
k_b2 = S_ID*k_a2;
k_a3 = 0.03;
k_b3 = S_IE*k_a3;
k_e = 0.138;
V_I = 0.12*BW;
V_G = 0.16*BW;
F_01 = 0.0097*BW;
EGP_0 = 0.0161*BW;
%u = 0.0954119*BW;
r = 5; %[mmol/L]
%U_G=y(2)/tau_G;
%U_I=y(4)/tau_I;
G= y(11)/V_G;
D = 0;
if(G>=4.5)
F_01c=F_01;
else
F_01c=F_01*G/4.5;
end
if(G>=9)
F_R=0.003*(G-9)*V_G;
else
F_R=0;
end
f1=(A_G*D)-(y(1)/tau_G); %D1
f2=(y(1)/tau_G)-(y(2)/tau_G); %D2
f3=y(3)-(y(4)/tau_I); %u & S1
f4=(y(4)/tau_I)-(y(5)/tau_I); %S2
f5=((y(5)/tau_I)/V_I)-(k_e*y(6)); %I
f6=(k_b1*y(6))-(k_a1*y(7)); %x1
f7=(k_b2*y(6))-(k_a2*y(8)); %x2
f8=(k_b3*y(6))-(k_a3*y(9)); %x3
f9=(y(7)*y(11))-((k_12+y(8))*y(10)); %Q2
f10=-(F_01c+F_R)-(y(7)*y(11))+(k_12*y(10))+(y(2)/tau_G)+(EGP_0*(1-y(9))); %Q1
f11=(y(11)/V_G)-r;
%f6=y(9)*y(5)-(k_12+y(10))*y(6); %Q2
%f7=y(4)/tau_I/V_I-k_e*y(7); %I
%f8=k_b1*y(7)-k_a1*y(8); %x1
%f9=k_b2*y(7)-k_a2*y(9); %x2
%f10=k_b3*y(7)-k_a3*y(10); %x3
F = [f1;f2;f3;f4;f5;f6;f7;f8;f9;f10;f11];
2nd script
y0 = [0 0 0 0 0 0 0 0 0 0 0];
options=optimset('Display','off');
y = fsolve(@function_hovorka_,y0,options);

Réponse acceptée

Walter Roberson
Walter Roberson le 23 Nov 2016
That first bit of code needs to be stored in a file named function_hovorka_.m and you need to add at the very beginning
function F = function_hovorka_(y)

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