Solving non linear equations

Hi all,

The folloiwng code solves non linear equations for T1, T2, T3 and T4 as well as for J1, J2 and J3. I am only interested on the tempreture:

It returns an array solution that includes several answers for each T. How Can I obtain the exact solution (one single soution) for each T?

syms J1 J2 J3 T1 T2 T3 T4
Jm = 5077.12;
Js = 301.32;
    Je = 330.136;
    A2 = 449200;
    A4 = 519000;
    Fms = 0.305;
    Fm1 = 0.45;
    Fme = 0.245;
    F1s = 0.610;
    F1m = 0.389;
    eps = 0.85;
    K2 = 15;
    L2 = 0.03;
eq1 = (Jm - Js)*(A2*Fms) + (Jm - J1)*(A2*Fm1) + (Jm - Je)*(A2*Fme) == 0;
eq2 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (J1 - Js)*(A4*F1s) + (J1 - Jm)*(A4*F1m) == 0;
eq3 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (T1-T2)*K2*A4/L2 == 0;
eq4 = -(T1 - T2)*K2/L2 + (5.67e-8*T2^4 - J2)*eps/(1-eps) == 0;
eq5 = -(5.67e-8*T2^4 - J2)*eps/(1-eps) + (J2 - J3) == 0;
eq6 = -(J2-J3) + (5.67e-8*T3^4 - J3)*eps/(1-eps) + 185.95 == 0;
eq7 = -(5.67e-8*T3^4 - J3)*eps/(1-eps) + (T3 - T4)*K2/L2 == 0;

eqs = [eq1, eq2, eq3, eq4, eq5, eq6, eq7];
vars = [J1, J2, J3, T1, T2, T3, T4];

sol = solve(eqs, vars);
T1_val = real(double(sol.T1))

T1_val = 16×1

684.3838 684.3838 684.3838 684.3838 0 0 0 0 0 0

T2_val = real(double(sol.T2))

T2_val = 16×1

666.9262 666.9262 666.9262 666.9262 -17.4576 -17.4576 -17.4576 -17.4576 -17.4576 -17.4576

T3_val = real(double(sol.T3))

T3_val = 16×1

456.1979 -0.0000 -0.0000 -456.1979 508.4679 508.4679 42.8456 42.8456 -42.8456 -42.8456

T4_val = real(double(sol.T4))

T4_val = 16×1

439.1121 -17.0857 -17.0857 -473.2836 491.3822 491.3822 25.7598 25.7598 -59.9313 -59.9313

Answer 1

[Torsten]

Each of the 16 quadruples (T1(i),T2(i),T3(i),T4(i)) (i = 1,...,16) constitutes a solution for your system of equations.

You must check which of the 16 quadruples are physical. E.g. the first quadruple (with the corresponding values for J1, J2 and J3 printed) would be

syms J1 J2 J3 T1 T2 T3 T4
Jm = 5077.12;
Js = 301.32;
    Je = 330.136;
    A2 = 449200;
    A4 = 519000;
    Fms = 0.305;
    Fm1 = 0.45;
    Fme = 0.245;
    F1s = 0.610;
    F1m = 0.389;
    eps = 0.85;
    K2 = 15;
    L2 = 0.03;
eq1 = (Jm - Js)*(A2*Fms) + (Jm - J1)*(A2*Fm1) + (Jm - Je)*(A2*Fme) == 0;
eq2 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (J1 - Js)*(A4*F1s) + (J1 - Jm)*(A4*F1m) == 0;
eq3 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (T1-T2)*K2*A4/L2 == 0;
eq4 = -(T1 - T2)*K2/L2 + (5.67e-8*T2^4 - J2)*eps/(1-eps) == 0;
eq5 = -(5.67e-8*T2^4 - J2)*eps/(1-eps) + (J2 - J3) == 0;
eq6 = -(J2-J3) + (5.67e-8*T3^4 - J3)*eps/(1-eps) + 185.95 == 0;
eq7 = -(5.67e-8*T3^4 - J3)*eps/(1-eps) + (T3 - T4)*K2/L2 == 0;
eqs = [eq1, eq2, eq3, eq4, eq5, eq6, eq7];
vars = [J1, J2, J3, T1, T2, T3, T4];
sol = solve(eqs, vars);
J1_val = real(double(sol.J1(1)))

J1_val = 1.0899e+04

J2_val = real(double(sol.J2(1)))

J2_val = 9.6771e+03

J3_val = real(double(sol.J3(1)))

J3_val = 948.2529

T1_val = real(double(sol.T1(1)))

T1_val = 684.3838

T2_val = real(double(sol.T2(1)))

T2_val = 666.9262

T3_val = real(double(sol.T3(1)))

T3_val = 456.1979

T4_val = real(double(sol.T4(1)))

T4_val = 439.1121

[Ghanim]

Thank you very much!

Answer 2

[Walter Roberson]

syms J1 J2 J3 T1 T2 T3 T4
Jm = 5077.12;
Js = 301.32;
    Je = 330.136;
    A2 = 449200;
    A4 = 519000;
    Fms = 0.305;
    Fm1 = 0.45;
    Fme = 0.245;
    F1s = 0.610;
    F1m = 0.389;
    eps = 0.85;
    K2 = 15;
    L2 = 0.03;
eq1 = (Jm - Js)*(A2*Fms) + (Jm - J1)*(A2*Fm1) + (Jm - Je)*(A2*Fme) == 0;
eq2 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (J1 - Js)*(A4*F1s) + (J1 - Jm)*(A4*F1m) == 0;
eq3 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (T1-T2)*K2*A4/L2 == 0;
eq4 = -(T1 - T2)*K2/L2 + (5.67e-8*T2^4 - J2)*eps/(1-eps) == 0;
eq5 = -(5.67e-8*T2^4 - J2)*eps/(1-eps) + (J2 - J3) == 0;
eq6 = -(J2-J3) + (5.67e-8*T3^4 - J3)*eps/(1-eps) + 185.95 == 0;
eq7 = -(5.67e-8*T3^4 - J3)*eps/(1-eps) + (T3 - T4)*K2/L2 == 0;

eqs = [eq1, eq2, eq3, eq4, eq5, eq6, eq7];
vars = [J1, J2, J3, T1, T2, T3, T4];

sol = solve(eqs, vars);

vals = double(subs([T1, T2, T3, T4], sol));
valid_vals = vals(all(vals > 0, 2),:)

valid_vals = 1×4

684.3838 666.9262 456.1979 439.1121

[Ghanim]

Thank you very much!