How to solve 3 simultaneous algebraic equations with a equality constraint.

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If someone could help me to plot x1 vs t from the information. Please if someone could give any idea.
%Initial conditions
x1=140; x2=140; x3=140;
%Equations
x1 =t*x1+x2+t*x3;
x2 = 2*t*x1+t*x2+x3;
x3 = t*x1+x2+x3;
% Equality constraint
x1+x2+x3=420;
  3 Commenti
MSM Farhan
MSM Farhan il 27 Lug 2021
N1=976:1:10000;
Q1=976:1:10000;
N=reshape (N1, 95, 95);
Q=reshape (q1, 95, 95);
Z0=4
Z1=1
f0=9.0336
f1=-0.3275
D1=10
D2=10
d1=10
d2=10
H1=10
H2=10
n1=10
K1=10
c11=N*Z1-f1/4-(3/4)*cos(2*q)*(2*f0+ f1)-2*(sin(q).^2-cos(q).^2)*D1+4*(cos(q).^2)*
(cos (q).^2-3*(sin(q).^2))*d1+H1*sin(q)+H2*cos (q)-2*n1+4*K1*sin(2*q)
c22=N*Z1-f1/4-(3/4 )*cos(2*q)*(2*f0+f1)-2*(sin(q).^2-cos(q).^2)*D2+4*(cos(q).^2)*(cos
(q).^2-3*(sin(q).^2))*d2+H1*sin(q)+H2*cos (q)-2*n1+4*K1*sin(2*q)
c21=-N*Z1+f1/4-(3/4 )*cos(2*q)*f1+2*n1
a1=-(3/4 )*sin(2*q)*(f0+f1)+D1*sin(2*q)+2*d1*(cos(q).^2)*sin(2*q)-H1*cos(q)+H2*sin(q)- 2*K1*cos(2*q)
a2=-(3/4 )*sin(2*q)*(f0+ f1)+D2*sin(2*q)+2*d2*(cos(q).^2 )*sin(2*q)-H1*cos(q)+H2*sin(q)- 2*K1*cos(2*q)
C11=(c21+c22)/(2*(c11*c22-c21.^2))
C21=(c11+c21)/(2*(c21.^2-c11*c22))
e1=(a2-a1)*C11
e2=(a2-a1)*C21
E0=-N*Z0+f0/4-N*Z1-f1/4+(3/4)*cos(2*q)*(f0+f1)-(cos(q).^2)*(D1+D2)-(cos(q).^4)*(d1+d2)-2*(H1 *sin(q)+H2*cos(q)-n1+K1*sin(2*q)
E11=-(3/4)*sin(2*q)*(f0+f1)*(e1+e2)+sin(2*q)*(D1*e1+D2*e2)+2*(cos(q).^2)*sin(2*q)* (d1*e1+d2*e2)-H1*cos (q)*(e1 + e2)+H2 *sin(q)*(e1+e2)-2*K1*(e1+e2)
E2=(1/2)*(N*Z1-f1/4)*((e1-e2).^2)-(3/8)*cos(2*q)*(2*f0*(e1.^2+e2.^2)+f1*((e1+e2).^2))-(sin(q).^2-cos(q).^2)*(D1 *e1.^2+D2*e2.^2)+2*(cos(q).^2*(cos(q).^2-3*(sin(q).^2))* (d1*e1.^2+d2*e2.^2)+(1/2)*H1*sin(q)*(e1.^2+e2.^2)+(1/2)*H2*cos(q)*(e1.^2+e2.^2)-n1* ((e1-e2).^2)+2*K1*sin(2*q)*(e1.^2+e2.^2)
E3=(1/8)*sin(2*q)*(4*f0*(e1.^3+e2.^3)+f1*((e1+e2).^3))-(4/3)*cos(q)*(sin(q)*(D1*e1.^3+ D2*e2.^3)-4*cos(q)*sin(q)*((5/3)cos(q).^2-(sin(q).^2)*(d1*e1.^3+d2*e2.^3)+(1/6)* H1*cos(q)*(e1.^3+e2.^3)-(1/6)*H2*sin (q)*(e1.^3+e2.^3)+(4/3)*K1*cos(2*q)*(e1.^3+ e2.^3)
E4=-(1/24)*(N*Z1-f1/4)*((e1-e2).^4)+(1/32)*cos(2*q)*(8* f0*(e1.^4 +e2.^4)+f1*((e1+e2).^4))-(1/3)*cos(2*q)*(D1*e1.^4+D2*e2.^4)-(1/3)*cos(2*q)*(5*cos(q).^2-3*sin(q).^2)*(d1*e1.^4+d2*e2.^4)- (1/24)*H1*sin(q)*(e1.^4+e2.^4)-(1/24)*H2*cos(q)*(e1.^4 +e2.^4)+(1/12)*n1*((e1-e2).^4)-(2/3)*K1*sin (2*q)*(e1.^4+e2.^4)
R=E0+E1+E2+E3+E4;
Figure (1)
h=surf (R);
h=xlabel (‘J/\omega‘);
h=ylabel (‘angle \theta (radians)‘);
h=Zlabel (‘E(\theta) /\omega‘);
Walter Roberson
Walter Roberson il 27 Lug 2021
That does not appear to be related? please open a new question, and when you do please be more clear what you are asking for.

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Risposta accettata

Walter Roberson
Walter Roberson il 23 Giu 2021
You cannot usefully plot x1 vs t. Your system defines three specific sets of points, two of which are complex-valued
syms x1 x2 x3 t
eqn = [x1 == t*x1+x2+t*x3;
x2 == 2*t*x1+t*x2+x3;
x3 == t*x1+x2+x3;
x1+x2+x3==420]
eqn = 
sol = solve(eqn,[x1, x2, x3, t], 'maxdegree', 3)
sol = struct with fields:
x1: [3×1 sym] x2: [3×1 sym] x3: [3×1 sym] t: [3×1 sym]
[sol.x1, sol.x2, sol.x3, sol.t]
ans = 
vpa(ans,10)
ans = 
so the only real-valued solution is x1 about -235, x2 about 732, x3 about -76, and t about 3.1 .
You might perhaps be expecting all-positive results, but look at your equations:
x3 = t*x1+x2+x3;
x3 appears with coefficient 1 on both sides, so you can subtract it from both sides, leading to
0 == t*x1 + x2
and if t and x1 and x2 are all positive, then that equation cannot be satisfied. It can potentially be satisfied if t and x2 are both 0
If you substitute t = 0 into your first three equations, you can come out with a consistent solution only if x1 = x2 = x3 = 0. However, that does not satisfied the constraint. This establishes that there is no consistent solution for arbitrary times.
  20 Commenti
Walter Roberson
Walter Roberson il 17 Lug 2021
syms x1 x2 x3 x_L t
eqns = [
x1 == t*(10*abs((18*x2)/125 - (24*x1)/125 + 111/50)^2*sign((18*x2)/125 - (24*x1)/125 + 111/50) + 10*abs((18*x2)/125 - (24*x1)/125 + 111/50)^(1/2)*sign((18*x2)/125 - (24*x1)/125 + 111/50)) - t*(10*abs((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100)^2*sign((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100) + 10*abs((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100)^(1/2)*sign((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100)) + 140
x2 == 2*t*(10*abs((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100)^2*sign((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100) + 10*abs((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100)^(1/2)*sign((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100)) - t*(10*abs((18*x2)/125 - (21*x3)/100 + 91/100)^2*sign((18*x2)/125 - (21*x3)/100 + 91/100) + 10*abs((18*x2)/125 - (21*x3)/100 + 91/100)^(1/2)*sign((18*x2)/125 - (21*x3)/100 + 91/100)) - t*(10*abs((18*x2)/125 - (24*x1)/125 + 111/50)^2*sign((18*x2)/125 - (24*x1)/125 + 111/50) + 10*abs((18*x2)/125 - (24*x1)/125 + 111/50)^(1/2)*sign((18*x2)/125 - (24*x1)/125 + 111/50)) + 140
x3 == t*(10*abs((18*x2)/125 - (21*x3)/100 + 91/100)^2*sign((18*x2)/125 - (21*x3)/100 + 91/100) + 10*abs((18*x2)/125 - (21*x3)/100 + 91/100)^(1/2)*sign((18*x2)/125 - (21*x3)/100 + 91/100)) - t*(10*abs((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100)^2*sign((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100) + 10*abs((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100)^(1/2)*sign((24*x1)/125 - (36*x2)/125 + (21*x3)/100 - 313/100)) + 140
x_L==0.002*x1*x1
x1+x2+x3==420+x_L]
eqns = 
E2 = lhs(eqns)-rhs(eqns)
E2 = 
F = matlabFunction(E2, 'vars', {[x1,x2,x3,x_L], t});
T = linspace(0,3,500);
nT = length(T);
sols = zeros(nT, 4);
x0 = [140, 140, 140, 140^2*0.002];
options = optimoptions(@fsolve, 'Algorithm', 'levenberg-marquardt', 'display', 'none');
have_warned = false;
have_warned2 = false;
for tidx = 1 : nT
[thissol, ~, exitflag, output] = fsolve(@(x) F(x,T(tidx)), x0, options);
sols(tidx, :) = thissol;
x0 = thissol;
if exitflag <= 0
if ~have_warned
warning('solution failure code %d starting at time = %g', exitflag, T(tidx));
have_warned = true;
disp(output)
end
elseif ~have_warned2
warning('Solutions resumed at time = %g', T(tidx));
have_warned2 = true;
end
end
Warning: solution failure code -2 starting at time = 0
iterations: 3 funcCount: 20 stepsize: 0.0071 cgiterations: [] firstorderopt: 2.1854e-04 algorithm: 'levenberg-marquardt' message: '↵No solution found.↵↵fsolve stopped because the last step was ineffective. However, the vector of function↵values is not near zero, as measured by the value of the function tolerance. ↵↵<stopping criteria details>↵↵fsolve stopped because the sum of squared function values, r, is changing by less ↵ than options.FunctionTolerance = 1.000000e-06 relative to its initial value.↵However, r = 3.670370e+02, exceeds sqrt(options.FunctionTolerance) = 1.000000e-03.↵↵'
plot(T, sols);
legend({'x1', 'x2', 'x3', 'x_L'});
Despite the graph, the messages tell you there was no solution.
Walter Roberson
Walter Roberson il 18 Lug 2021
There does not appear to be any solution for that system of equations.

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Più risposte (1)

Ildeberto de los Santos Ruiz
You only need to express in terms of t and plot that relationship:
syms x1 x2 x3 t
x3 = solve(x1+x2+x3 == 420,x3)
EQ1 = x1 == t*x1+x2+t*x3;
EQ2 = x2 == 2*t*x1+t*x2+x3;
[x1,x2] = solve(EQ1,EQ2)
ezplot(x1,[0,2])
  5 Commenti
Ildeberto de los Santos Ruiz
Modificato: Ildeberto de los Santos Ruiz il 21 Lug 2021
Original post from @Shiv:
If someone could help me to plot x1 vs t from the information.
Walter Roberson
Walter Roberson il 21 Lug 2021
And the information given includes three equations plus one constraint equation.
syms x1 x2 x3 t
x3 = solve(x1+x2+x3 == 420,x3)
x3 = 
EQ1 = x1 == t*x1+x2+t*x3;
EQ2 = x2 == 2*t*x1+t*x2+x3;
EQ3 = x3 == t*x1+x2+x3;
[x1_12,x2_12] = solve(EQ1,EQ2)
x1_12 = 
x2_12 = 
[x1_13,x2_13] = solve(EQ1,EQ3)
x1_13 = 
x2_13 = 
[x1_23,x2_23] = solve(EQ2,EQ3)
x1_23 = 
x2_23 = 
ezplot(x1_12,[-2,5])
hold on
ezplot(x1_13,[-2 5])
ezplot(x1_23,[-2 5])
hold off
legend({'EQ1,EQ2', 'EQ1,EQ3', 'EQ2,EQ3'}, 'location', 'southwest');
Three very different pairwise answers. It looks like there might be a common answer near t = 3; let us see:
ezplot(x1_12,[2.5,3.5])
hold on
ezplot(x1_13,[2.5,3.5])
ezplot(x1_23,[2.5,3.5])
hold off
legend({'EQ1,EQ2', 'EQ1,EQ3', 'EQ2,EQ3'}, 'location', 'southwest');
tsol = vpasolve(x1_12 == x1_13, 3)
tsol = 
X1 = subs([x1_12, x1_13, x1_23], t, tsol(2))
X1 = 
X2 = subs([x2_12, x2_13, x2_23], t, tsol(2))
X2 = 
So far, so good, the pairs of equation seem to check out.
X3 = subs(x3, [x1, x2], [X1(1), X2(1)])
X3 = 
That is, there is only one real-valued solution to all of the equations simultaneously.

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