Find the solution of the coupled system of equations

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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[Dyuman Joshi]

What have you tried yet?

[Arad]

Dear joshi

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];
x = A \ B;

Warning: Matrix is singular to working precision.

But the answers are INF and NaN.

[Torsten]

Thus with identical expressions for F11 and F12, you want to approximate at the same time 0.86 and -2.3 (similar for F21,F22 and F31,F33) ? That doesn't make sense.

[Sam Chak]

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).

Answer 1

[Sam Chak]

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)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

c = 6×1

-0.4775 0.9998 0.1000 0.4791 -0.5947 0.7912

%% 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

Answer 2

[Sulaymon Eshkabilov]

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

lsqr converged at iteration 3 to a solution with relative residual 4.4e-16.

fprintf('Solutions: x = %f; y = %f; z = %f \n', SOL1')

Solutions: x = 0.157895; y = 1.097744; z = 0.556391 

%% 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')

Solutions: x = 0.157895; y = 1.097744; z = 0.556391 

SOL3 = linsolve(A,b);              % Using linsolve()
fprintf('Solutions: x = %f; y = %f; z = %f \n', SOL3')

Solutions: x = 0.157895; y = 1.097744; z = 0.556391 

Answer 3

[Sulaymon Eshkabilov]

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