Solving non-linear equations (0)
1 次查看（过去 30 天）
显示 更早的评论
Dear all,
I solved non-linear equations using Matlab, but all values are close to 0.
I think this solution is not incorrect, and I would like to know my coding is correct or not.
In addition, I would like to get the value of Me1. Please let me know how to get this value.
Thank you very much in advance.
Sincerely yours,
J1
function F = costate2(x)
t = 0;
tlast = 30;
phi = 3.14;
ep = 0.1;
L1 = 157.5;
L2 = 799.5;
GHGs = 2954;
delta1 = 0.005;
gam1 = 0.10957;
eta11 = 0.40022;
eta21 = 0.99931;
ag1 = 20.49921;
alpha1 = 15.93413;
rhoq1 = 0.00710;
beta11 = 0.35;
beta21 = 0.63;
beta31 = 0.02;
g1 = 2.6005;
chi1 = 0.01;
xi1 = 0.02;
rhom1 = 0.0041773;
delta2 = 0.00581;
gam2 = 0.14990;
eta12 = 0.08;
eta22 = 0.59998;
ag2 = 21.49967;
alpha2 = 3.25702;
rhoq2 = 0.005;
beta12 = 0.60;
beta22 = 0.38;
beta32 = 0.02;
g2 = 2.8643;
chi2 = 0.01;
xi2 = 0.02;
rhom2 = -0.00800;
d = 0.05;
psi = 0.1;
K1 = 3486;
K2 = 1951;
Kg1 = 3.49;
Kg2 = 1.95;
Pw = 0.1;
Pg = 0.05;
GHG = 3162;
Q1 = 1496;
Q2 = 604;
E1 = 2216;
E2 = 2417;
a1 = 0.1;
a2 = 0.5;
b1 = 0.5;
b2 = 0.5;
Gbar = 12000;
P1 = 309;
P2 = 1338;
gam3 = 0.0492288;
gam4 = 0.0243931;
d1 = 1.2521405;
delta3 = 0.0250461;
lamda1 = x(1);
qg1 = x(2);
qp1 = x(3);
qe1 = x(4);
qc1 = x(5);
qk1 = x(6);
Gbar1 = (P1/(P1+P2))*Gbar;
Gbar2 = (P2/(P1+P2))*Gbar;
M1 = (g1*exp(-rhom1*t)-chi1*(Kg1/(K1+Kg1))^xi1)*E1;
W1 = (Pw-E1-Pg*(P1./(P1+P2))*Gbar)^2+4*beta31*Pw*Q1*M1*Pg;
J1 = (Pg*(P1./(P1+P2))*Gbar-Pw*E1)^2+4*beta31*Pw*Q1*M1*Pg;
Pe1 = ((Pw*E1-Pg*Gbar1) + sqrt(W1))/(2*E1);
Me1 = ((Pg*(P1./(P1+P2))*Gbar-Pw*E1)+sqrt(J1))/(2*Pg);
Me2 = (g2*exp(-rhom2*t)-chi2*(Kg2./(K2+Kg2))^xi2)*beta32*Q2;
Y1 = E1^2*Pe1^2+beta31*Pw*Q1*M1*Pg;
QQe1 = alpha1*exp(rhoq1*t)*(K1+Kg1)^beta11*L1^beta21*E1^beta31;
Op = Me1 - (P1./(P1+P2))*Gbar;
PP = P1*P2;
F(1) = x(1)*( phi/(2*tlast) )
-x(3)*( psi*(P1+P2)*Pg/(sqrt(PP)*Gbar) )*( (chi1*xi1*beta31*Pw*Q1*E1^2*Pe1*Kg1^xi1)/(Y1*(K1+Kg1)^(xi1+1)) )
-x(4)*chi1*xi1*E1*(Kg1^xi1)/(K1+Kg1)^(xi1+1)
-x(5)*a1*beta11*QQe1/(K1+Kg1)
-x(6)*b1*(beta31*Pw*Q1*Pg/(E1*Pe1^2))*( (chi1*xi1*beta31*Pw*Q1*E1^2*Pe1*Kg1^xi1)/(Y1*(K1+Kg1)^(xi1+1)) );
F(2) = x(2)*( phi/(2*tlast) + 1/ag1 )
-x(2).^2/(x(1)*2*ag1)
-x(1)*( 1/(2*ag1) - Pg*( (chi1*xi1*beta31*Pw*Q1*E1^2*Pe1*Kg1^xi1)*(1/(K1+Kg1)-1/Kg1)/(Y1*(K1+Kg1)^xi1) ) )
-x(3)*psi*(P1+P2)*Pg*(1/(sqrt(PP)*Gbar))* ( (chi1*xi1*beta31*Pw*Q1*E1^2*Pe1*Kg1^xi1)*(1/(K1+Kg1)-1/Kg1)/(Y1*(K1+Kg1)^xi1) )
-x(4)*chi1*xi1*E1*((Kg1/(K1+Kg1))^xi1)*(1/(K1+Kg1)-1/Kg1)
-x(5)*a1*beta11*QQe1/(K1+Kg1)
+x(6)*b1*beta31*Pw*Q1*Pg*(1/(E1*Pe1^2))*( (chi1*xi1*beta31*Pw*Q1*E1^2*Pe1*Kg1^xi1)*(1/(K1+Kg1)-1/Kg1)/(Y1*(K1+Kg1)^xi1) )
;
F(3) = x(3)*( phi/(2*tlast) )
+x(1)*( (Me1-Gbar1) - Pg*beta31*Pw*Q1*M1*(Me1-Gbar1)/Y1 )
+x(6)*b1*beta31*Pw*Q1*M1*(Me1-Gbar1)/Y1;
F(4) = x(4)*( phi/(2*tlast) )
+x(1).^((eta11*gam1+gam1)/(eta11*gam1+gam1-1)) * eta21*(ep/eta11)^(eta11*gam1/(eta11+gam1-1)) * GHGs^(-eta21*gam1/(eta11*gam1+gam1-1)) * GHG^(((eta21-eta11)*gam1-gam1+1)/(eta11*gam1+gam1-1));
F(5) = x(5)*( phi/(2*tlast) )
-x(1)*(1-Pg*beta31*Pw+M1*E1*Pe1/Y1)
-x(3)*psi*(P1+P2)*Pg*(1/(sqrt(PP)*Gbar))*beta31*Pw+M1*E1*Pe1/Y1
-x(6)*b1*beta31*Pw+M1*E1*Pe1/Y1;
F(6) = x(6)*( phi/(2*tlast) )
+x(1)*Pg*M1*(1/E1)*beta31*Pw*Q1*Pg*(Me1-Gbar1)/Y1
-x(3)*psi*(P1+P2)*Pg*(1/(sqrt(PP)*Gbar))*M1*(1/E1)*beta31*Pw*Q1*Pg*(Me1-Gbar1)/Y1
-x(4)*M1/E1
-x(5)*a1*beta31*QQe1/E1;
fun = @costate2;
x0 = [1,1,1,1,1,1];
x = fsolve(fun,x0)
3 个评论
Walter Roberson
2018-8-18
You did not present the equations in symbolic form such as an image of the equations, so as far as we as onlookers are concerned, the equations are defined by your current code, and so must be correct. We would need some other representation of the equations in order to be able to check the code against the equations.
Star Strider
2018-8-18
Since ‘Me1’ does not appear to be a function of ‘x’, to display its value, put this line after the ‘Me1’ assignment:
fprintf(1, '\n\t\tMe1 = %23.15E\n', Me1)
in context:
Me1 = ((Pg*(P1./(P1+P2))*Gbar-Pw*E1)+sqrt(J1))/(2*Pg);
fprintf(1, '\n\t\tMe1 = %23.15E\n', Me1)
It will display the value of ‘Me1’ (several times) in the Command Window.
The result will be:
Me1 = 1.476154199966165E+02
回答（5 个）
Redwood
2018-8-19
编辑：Walter Roberson
2018-8-19
1 个评论
Walter Roberson
2018-8-19
You are missing continuation marks for your eq* variables.
eq1 = lamda1*( phi/(2*tlast) )
-qp1*( psi*(P1+P2)*Pg/(sqrt(PP)*Gbar) )*( (chi1*xi1*beta31*Pw*Q1*E1^2*Pe1*Kg1^xi1)/(Y1*(K1+Kg1)^(xi1+1)) )
is not considered all one long expression: it is considered an assignment, followed by the calculation and display (but no assignment) of the expression on the second line.
You would need
eq1 = lamda1*( phi/(2*tlast) ) ...
-qp1*( psi*(P1+P2)*Pg/(sqrt(PP)*Gbar) )*( (chi1*xi1*beta31*Pw*Q1*E1^2*Pe1*Kg1^xi1)/(Y1*(K1+Kg1)^(xi1+1)) )
Redwood
2018-8-19
编辑：Walter Roberson
2018-8-19
2 个评论
Walter Roberson
2018-8-19
You accidentally wrote
sol = solve( eq1,eq2, eq3. eq4, eq5, eq6 );
instead of
sol = solve( eq1,eq2, eq3, eq4, eq5, eq6 );
Alex Sha
2019-7-31
try numerical solution:
lamda1: 1.48312225679967E-24
qp1: 54185.6537859035
qn1: -133.888674663691
qc1: 0.188502621518643
qk1: -6630.74381075639
qg1: 282.666798246712
Me1=76.2725002858213
0 个评论
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
您也可以从以下列表中选择网站：
美洲
- América Latina (Español)
- Canada (English)
- United States (English)
欧洲
- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)
- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- United Kingdom (English)
亚太
- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)