Find the solution of the coupled system of equations
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Arad
am 14 Jan. 2024
Bearbeitet: Sulaymon Eshkabilov
am 15 Jan. 2024
I have several equations with coefficients such as c_i, d_i, etc. I need to obtain the coefficients using the least squares method, subject to the following conditions:
- I would appreciate any help in determining these coefficients using the least squares method given the above constraints.
- Additionally, what would be a better or more efficient way to obtain these coefficients?
Thank you.
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F11 = c00 + d00 - (2*c2)/3 - (2*c5)/9 + ((4*c4)/7 + 4*(c1) + 3*(c3) - 6*c1*c3 + (16*c1*c4)/7 - (12*c3*c4)/7 + (c00 - d00)^2)^(1);
F12 = c00 + d00 - (2*c2)/3 - (2*c5)/9 - ((4*c4)/7 + 4*(c1) + 3*(c3) - 6*c1*c3 + (16*c1*c4)/7 - (12*c3*c4)/7 + (c00 - d00)^2)^(1);
F21 = c00 + d00 + 2*c2 + (2*c5)/3 + ((24*c4)/7 + 12*(c1) + 3*(c3) + 18*c1*c3 + (144*c1*c4)/7 + (36*c3*c4)/7 + (c00 - d00)^2)^(1);
F22 = c00 + d00 + 2*c2 + (2*c5)/3 - ((24*c4)/7 + 12*(c1) + 3*(c3) + 18*c1*c3 + (144*c1*c4)/7 + (36*c3*c4)/7 + (c00 - d00)^2)^(1);
F31 = c00 + d00 - c2 + (2*c5)/3 + (4*(c1^20282)^(1) - (c1*c3)/50706 + 4*(c3^2/8112)^(1) + (c00 - d00)^2)^(1);
F33 = c00 + d00 - c2 + (2*c5)/3 - (4*(c1^20282)^(1) - (c1*c3)/50706 + 4*(c3^2/8112)^(1) + (c00 - d00)^2)^(1);
F11 = 0.86;
F12 = -2.3;
F21 = 6.8;
F22 = -6.3;
F31 = 0.3;
F32 = -0.4;
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4 Kommentare
Sam Chak
am 15 Jan. 2024
Hi @Torsten, they do look like 3 pairs of identical equations to me at first glance. However, when I displayed them in code format (horizontal single line), you will see the plus-minus signs in each pair.
Another issue is that the system is considered underdetermined because there are fewer equations (6: F11, F12, F21, F22, F31, F32) than unknown variables (7: c00, d00, c1, c2, c3, c4, c5).
Akzeptierte Antwort
Sam Chak
am 15 Jan. 2024
Hi @Arad
If you set a specific value for the variable d00, the system of 6 nonlinear equations can be solved using the fsolve() command.
fun = @system;
c0 = ones(6, 1);
c = fsolve(fun, c0)
%% A system of nonlinear equations
function F = system(c)
% 6 unknown variables
c00 = c(1);
c1 = c(2);
c2 = c(3);
c3 = c(4);
c4 = c(5);
c5 = c(6);
% Constants
d00 = 0; % <-- assumed, find out this value
F1 = 0.86;
F2 = -2.3;
F3 = 6.8;
F4 = -6.3;
F5 = 0.3;
F6 = -0.4;
% 6 Nonlinear equations
F(1) = c00 + d00 - (2*c2)/3 - (2*c5)/9 + ((4*c4)/7 + 4*(c1) + 3*(c3) - 6*c1*c3 + (16*c1*c4)/7 - (12*c3*c4)/7 + (c00 - d00)^2)^(1) - F1;
F(2) = c00 + d00 - (2*c2)/3 - (2*c5)/9 - ((4*c4)/7 + 4*(c1) + 3*(c3) - 6*c1*c3 + (16*c1*c4)/7 - (12*c3*c4)/7 + (c00 - d00)^2)^(1) - F2;
F(3) = c00 + d00 + 2*c2 + (2*c5)/3 + ((24*c4)/7 + 12*(c1) + 3*(c3) + 18*c1*c3 + (144*c1*c4)/7 + (36*c3*c4)/7 + (c00 - d00)^2)^(1) - F3;
F(4) = c00 + d00 + 2*c2 + (2*c5)/3 - ((24*c4)/7 + 12*(c1) + 3*(c3) + 18*c1*c3 + (144*c1*c4)/7 + (36*c3*c4)/7 + (c00 - d00)^2)^(1) - F4;
F(5) = c00 + d00 - c2 + (2*c5)/3 + (4*(c1^20282)^(1) - (c1*c3)/50706 + 4*(c3^2/8112)^(1) + (c00 - d00)^2)^(1) - F5;
F(6) = c00 + d00 - c2 + (2*c5)/3 - (4*(c1^20282)^(1) - (c1*c3)/50706 + 4*(c3^2/8112)^(1) + (c00 - d00)^2)^(1) - F6;
end
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Weitere Antworten (2)
Sulaymon Eshkabilov
am 14 Jan. 2024
Applying the LS method for solving such systems is relatively straight forward, e.g.:
b1 = 3x + 2y-3z
b2 = 2x-y+5z
b3 = -3x+6y-2z
b1 = 1; b2 = 2; b3 = 5;
% Step 1. Define A matrix:
A = [3 2 -3; 2 -1 5; -3 6 -2];
b1 = 1; b2 = 2; b3 = 5;
% Step 2. Define b matrix:
b = [b1;b2;b3];
% Step 3. Determine solution tolerance
tol = 1e-15;
% Step 4. Solve the system of [A]*{X} = [b] where [A] is coefficients, [b]
% is the column matrix (also called a system response), {X} is unknowns
% How to apply LS method:
SOL1 = lsqr(A,b, tol); % Using the least squares method
fprintf('Solutions: x = %f; y = %f; z = %f \n', SOL1')
%% NB: \ operator or linsolve() can be also used for such systems:
SOL2 = A\b; % Using backslash (\)
fprintf('Solutions: x = %f; y = %f; z = %f \n', SOL2')
SOL3 = linsolve(A,b); % Using linsolve()
fprintf('Solutions: x = %f; y = %f; z = %f \n', SOL3')
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Sulaymon Eshkabilov
am 15 Jan. 2024
Bearbeitet: Sulaymon Eshkabilov
am 15 Jan. 2024
The accepted answer is NOT the least squares method as mentioned in the question itself.
F11 = 0.86;
F12 = -2.3;
F21 = 6.8;
F22 = -6.3;
F31 = 0.3;
F32 = -0.4;
A = [1, 1, 0, 0, -2/3, -2/9;
1, 1, 0, 0, -2/3, -2/9;
1, 1, 2, 0, 0, 2/3;
1, 1, 2, 0, 0, 2/3;
1, 1, -1, 0, -1, 2/3;
1, 0, 0, 0, 0, 0;];
B = [F11; F12; F21; F22; F31; F32];
tol = 1e-7;
SOLUTION = lsqr(A,B, tol) % Using the least squares method
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