Solution of system of nonlinear equations

Question

Swami 2024-10-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2160085-solution-of-system-of-nonlinear-equations

编辑： Mr. Pavl M. 2024-10-17

I have following system of equations to find q1, q2, q3 and q4, which I am unable to solve using solve using matlab 'solve' or 'fsolve' commands. All the values should be positive and should not be exceeding 1.0. Are there any suggestions? Or should I implement newton method to solve it?

eqc1_pre = 91.907*q1 - 1.1978e-14*q3 + 3.534e-14*q1*q3 + 9.1309e+6*(0.007649*q1 + 0.03937*q3)*(0.0038245*q1^2 + 0.03937*q1*q3 + 0.18186*q3^2 - 996.13*q3 - 0.03937*q4 + 1.6325e+7) + 1.819e-13*q3^2 == 0
eqc2_pre = 2914.4*q2 - 875.08*q3 + 3645.7*q2*q3 + 1.3054e+7*(0.007649*q2 + 0.03937*q3)*(0.0038245*q2^2 + 0.03937*q2*q3 + 0.18186*q3^2 - 996.23*q3 - 0.03937*q4 + 1.6325e+7) + 18765.0*q3^2 == 0
eqc3_pre = 3.638e-13*q1*q3 - 875.08*q2 - 9.4963e+8*q3 - 18765.0*q4 - 1.1978e-14*q1 + 37530.0*q2*q3 + 9.1309e+6*(0.03937*q1 + 0.36373*q3 - 996.13)*(0.0038245*q1^2 + 0.03937*q1*q3 + 0.18186*q3^2 - 996.13*q3 - 0.03937*q4 + 1.6325e+7) + 1.3054e+7*(0.03937*q2 + 0.36373*q3 - 996.23)*(0.0038245*q2^2 + 0.03937*q2*q3 + 0.18186*q3^2 - 996.23*q3 - 0.03937*q4 + 1.6325e+7) + 1.767e-14*q1^2 + 1822.8*q2^2 + 260050.0*q3^2 + 7.7808e+12 == 0
eqc4_pre = - 1374.8*q1^2 - 14153.0*q1*q3 - 1965.6*q2^2 - 20235.0*q2*q3 - 158850.0*q3^2 + 8.701e+8*q3 + 34387.0*q4 - 1.4259e+13 == 0
 

4 个评论
显示 2更早的评论隐藏 2更早的评论

Mr. Pavl M. 2024-10-16

编辑：Walter Roberson 2024-10-17

在 MATLAB Online 中打开

clc
clear all;
global st=0; %start of constraint intverval
global lst=1; % end of constraint interval
% 79% Completed:
function F = equations(q)
q1 = q(1);
q2 = q(2);
q3 = q(3);
q4 = q(4);
pi = 3.141592653589793;
F(1) = 91.907945087554152399491554310275*q1 - 6.9530743154589888866367962498927e-42*q3 + 0.000000000000035339972092069079645555751006121*q1*q3 + 0.0000000000001818989403545856475830078125*q3^2 + 9130872.0607999999462988460436463*((25*q1*pi^2)/32258 + (5*q3)/127)*((25*pi^2*q1^2)/64516 + (5*q1*q3)/127 + (2579522850115110167785284050996441184169*q3^2)/14183735516765013986772738621315219456000 - (269437024701681245919274115996940028669*q3)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000);
F(2) = 3061.5545338275630699055886034165*q2 - 0.00000000000000000000000052628096743060987461514329634383*q3 + 3645.7069886924359708335725693911*q2*q3 + 18764.877243218995339070902448522*q3^2 + 13054497.238043855951388792124012*((25*q2*pi^2)/32258 + (5*q3)/127)*((25*pi^2*q2^2)/64516 + (5*q2*q3)/127 + (2579522850115110167785284050996441184169*q3^2)/14183735516765013986772738621315219456000 - (269464072971883821971214932126562044669*q3)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000);
F(3) = 0.000000000000363797880709171295166015625*q1*q3 - 0.00000000000000000000000052628096743060987461514329634383*q2 - 949606892.52341601879283837341561*q3 - 18764.877243218995520969842803107*q4 - 1960*q3*(127/(10*pi) + 127/20) - 6.9530743154589888866367962498927e-42*q1 + 37529.754486437990678141804897043*q2*q3 + 9130872.0607999999462988460436463*((5*q1)/127 + (2579522850115110167785284050996441184169*q3)/7091867758382506993386369310657609728000 - 269437024701681245919274115996940028669/270482702025760519408161296220160000)*((25*pi^2*q1^2)/64516 + (5*q1*q3)/127 + (2579522850115110167785284050996441184169*q3^2)/14183735516765013986772738621315219456000 - (269437024701681245919274115996940028669*q3)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000) + 13054497.238043855951388792124012*((5*q2)/127 + (2579522850115110167785284050996441184169*q3)/7091867758382506993386369310657609728000 - 269464072971883821971214932126562044669/270482702025760519408161296220160000)*((25*pi^2*q2^2)/64516 + (5*q2*q3)/127 + (2579522850115110167785284050996441184169*q3^2)/14183735516765013986772738621315219456000 - (269464072971883821971214932126562044669*q3)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000) + 0.00000000000001766998604603453982277787550306*q1^2 + 1822.8534943462179854167862846955*q2^2 + 260045.56651039855910887591026306*q3^2 + 7780857152948.6118200978386072826;
F(4) = - 139.29999999999999918073863084391*pi^2*q1^2 - 199.15857468494441639131927665645*pi^2*q2^2 - 14152.879999999999916763044893742*q1*q3 - 20234.511187990352705358038508296*q2*q3 - 158847.98154891714318971301560184*q3^2 + 870095677.35924348389019942898346*q3 + 34387.391187990352622121083402037*q4 - 14258923878318.356547643662453739;
end
function q = steffensen_method(F, q0, tol, max_iter)
global st;
global lst;
q = q0;
for iter = 1:max_iter
    Fq = F(q);
    q1 = q - Fq;
    Fq1 = F(q1);
    q2 = q1 - Fq1;
    Fq2 = F(q2);
    q_new = q - (Fq.^2) ./ (Fq2 - 2*Fq1 + Fq);
    if norm(q_new - q) < tol && all (q_new > st & q_new <= lst)
        q = q_new;
        return;
    end
    q = q_new;
end
warning('Steffensen method did not converge');
end
function secant_method_optimized()
global st;
global lst;
% Initial guesses
q0 = [1; 1; 1; 1];  % First initial guess
q1 = [1.1; 1.1; 1.1; 1.1];  % Second initial guess
% Tolerance and maximum iterations
tol = 1e-5;
max_iter = 1500;
% Secant method iteration
for iter = 1:max_iter
    % Evaluate function at q0 and q1
    F0 = equations(q0);
    F1 = equations(q1);
    % Secant step
    q_next = q1 - (F1 .* (q1 - q0)) / (F1 - F0+eps);
    % Update guesses
    q0 = q1;
    q1 = q_next;
    % Check for convergence
    if norm(q1 - q0,1) < tol && all (q1 > st & q1 <= lst)
        fprintf('Converged to solution in %d iterations\n', iter);
        disp(q1);
        return;
    end
end
warning('Secant Method Optimized, Failed to converge in %d iterations\n', max_iter);
end
function q = fixed_point_iteration(q0, tol, max_iter)
global st;
global lst;
q = q0;
for iter = 1:max_iter
    q_new = equations(q);
    if norm(q_new - q,1) < tol && all (q_new > st & q_new <= lst)
        q = q_new;
        return;
    end
    q = q_new;
end
warning('Fixed-point iteration did not converge');
end
%To be continued to fix it:
function q = secant_method(q0, q1, tol, max_iter)
global st;
global lst;
for iter = 1:max_iter
    Fq0 = equations(q0);
    Fq1 = equations(q1);
    q2 = q1 - Fq1 * ((q1 - q0)./(Fq1 - Fq0+eps));
    q2(isinf(q2)) = 5000000;
    q2 = q2(:,1);
    %if sum(sqrt(abs(q2.^2 - q1.^2))) < tol % or norm(q2 - q1) < tol
    if norm(q2-q1,1) < tol && all (q_new > st & q_new <= lst)
        q = q2;
        fprintf('Converged in %d iterations.\n', iter);
        return;
    end
    q0 = q1;
    q1 = q2;
end
q = q2;
warning('Secant method did not converge');
end
function q = newton_m(initial_guess, tol, max_iter)
global st;
global lst;
q = initial_guess;
for iter = 1:max_iter
    F = equations(q);
    J = jacobian(q);
    delta_q = sum(-J \ F');
    q = q + delta_q';
    if norm(delta_q) < tol && all (q > st & q <= lst)
        fprintf('Converged in %d iterations.\n', iter);
        return;
    end
end
fprintf('Did not converge within %d iterations.\n', max_iter);
end
function q = broyden_method(initial_guess, tol, max_iter)
global st;
global lst;
q = initial_guess;
B = eye(length(q));  % Initial Jacobian approximation (identity matrix)
for iter = 1:max_iter
    F = equations(q);
    F = F(:);  % Ensure F is a column vector
    delta_q = -B \ F;  % Solve for delta_q
    q_new = q + delta_q;  % Update q
    % Check for convergence
    if norm(delta_q) < tol && all(q_new > st & q_new <= lst)
        fprintf('Converged in %d iterations.\n', iter);
        q = q_new;
        return;
    end
    F_new = equations(q_new);
    F_new = F_new(:);  % Ensure F_new is a column vector
    y = F_new - F;
    delta_q = delta_q(:); % Ensure delta_q is a column vector
    B = B + ((y - B * delta_q)'* delta_q) / (delta_q' * delta_q);  % Update Jacobian approximation
    q = q_new;  % Update q
end
fprintf('Did not converge within %d iterations.\n', max_iter);
end
function q = broyden_methodq(initial_guess, tol, max_iter)
global st;
global lst;
q = initial_guess;
B = eye(length(q));  % Initial Jacobian approximation (identity matrix)
for iter = 1:max_iter
    F = equations(q);
    delta_q = -B\F';
    q_new = q + sum(delta_q');
    if norm(delta_q) < tol && all (q_new > st & q_new <= lst)
        fprintf('Converged in %d iterations.\n', iter);
        return;
    end
    F_new = equations(q_new);
    y = F_new - F;
    B = B + ((y - B*delta_q).*delta_q)/(delta_q*delta_q');
    q = q_new;
end
fprintf('Did not converge within %d iterations.\n', max_iter);
end
function q = latin_pgd_method(initial_guess, tol, max_iter)
global st;
global lst;
% for LATIN-PGD method implementation
% This is a simplified version and may need adjustments for your specific problem
q = initial_guess
for iter = 1:max_iter
    % Solve the local problem
    q_local = solve_local_problem(q);
    % Solve the global problem
    q_global = solve_global_problem(q_local);
    % Update the solution
    q_new = update_solution(q, q_local, q_global);
    % Check for convergence
    if norm(q_new - q) < tol && all (q_new > st & q_new <= lst)
        fprintf('Converged in %d iterations.\n', iter);
        q = q_new;
        return;
        fprintf('LATIN-PGD method is not fully implemented in this example.\n');
    end
end
end
function q_local = solve_local_problem(q)
% Placeholder for solving the local problem
% Implement the local problem solution here
q_local = q;  % Example placeholder
end
function q_global = solve_global_problem(q_local)
% Placeholder for solving the global problem
% Implement the global problem solution here
q_global = q_local;  % Example placeholder
end
function q_new = update_solution(q, q_local, q_global)
% Placeholder for updating the solution
% Implement the solution update here
q_new = (q_local + q_global) / 2;  % Example placeholder
end
function u_corr = latin_pgd_complete(initial_guess)
global st;
global lst;
% Problem parameters
maxIter = 400;        % Maximum number of iterations
tol = 1e-6;           % Tolerance for convergence
% Initial guess
u = initial_guess;
% LATIN-PGD iteration
for iter = 1:maxIter
    % Prediction step
    u_pred = prediction_step(u);
    % Correction step
    u_corr = correction_step(u_pred);
    % Check for convergence
    if norm(u_corr - u,3) < tol && all (q_new > st & q_new <= lst)
        fprintf('Converged to solution in %d iterations\n', iter);
        #disp(u_corr);
        return;
    end
    % Update solution
    u = u_corr;
end
warning('Failed to converge in %d iterations\n', maxIter);
end
function u = initial_guess()
% Define an initial guess for the solution
u = rand(4, 1);  % Example with random initial guess, adapted to 4 variables
end
function u_pred = prediction_step(u)
% Prediction step of the LATIN method
% Example: using a simplified linear model for prediction
% Here we assume a linear relationship for illustration purposes
J = jacobian(u);
F = equations(u);
u_pred = u - J./(eps+F); %F*inv(J);  % Update with actual prediction logic
end
function u_corr = correction_step(u_pred)
% Correction step of the LATIN method
% Example: applying correction using nonlinear residuals
F = equations(u_pred);
u_corr = u_pred - 0.01 * F;  % Update with actual correction logic
end
function root = bisection_method(a, b, tol, max_iter,q)
global st;
global lst;
if equations(q+a) * equations(q+b)' >= 0
    error('The function must have different signs at a and b');
end
for iter = 1:max_iter
    c = (a + b) / 2;
    q = q+c;
    root = equations(q);
    if root == 0 || (b - a) / 2 < tol && all (q > st & q <= lst)
        fprintf('Converged in %d iterations.\n', iter);
        return;
    end
    if equations(q)*equations(q)' < 0
        b = c;
    else
        a = c;
    end
end
warning('Did not converge within the maximum number of iterations');
end
%Whether next Jacobians are correct (Fill in with your computed Jacobians):
function J = jacobian(q)
%{
 ##q1 = q(1);
 ##q2 = q(2);
 ##q3 = q(3);
 ##q4 = q(4);
%}
pi = 3.141592653589793;
J = zeros(4, 4);
n = length(q);
J = zeros(n, n);
h = 1e-6;  % Step size for finite differences
for i = 1:n
    for j = 1:n
        q1 = q;
        q2 = q;
        q1(j) = q1(j) + h;
        q2(j) = q2(j) - h;
        J(i, j) = (equations(q1)(i) - equations(q2)(i)) / (2 * h);
    end
end
% Partial derivatives for the first equation
%{
##J(1, 1) = 91.907945087554152399491554310275 + 0.000000000000035339972092069079645555751006121*q3 + 9130872.0607999999462988460436463*((25*pi^2)/32258)*((25*pi^2*q1^2)/64516 + (5*q1*q3)/127 + (2579522850115110167785284050996441184169*q3^2)/14183735516765013986772738621315219456000 - (269437024701681245919274115996940028669*q3)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000) + 9130872.0607999999462988460436463*((25*q1*pi^2)/32258)*((50*pi^2*q1)/64516 + (5*q3)/127); ##J(1, 2) = 0;
##J(1, 3) = -6.9530743154589888866367962498927e-42 + 0.000000000000035339972092069079645555751006121*q1 + 0.000000000000363797880709171295166015625*q3 + 9130872.0607999999462988460436463*((5)/127)*((25*pi^2*q1^2)/64516 + (5*q1*q3)/127 + (2579522850115110167785284050996441184169*q3^2)/14183735516765013986772738621315219456000 - (269437024701681245919274115996940028669*q3)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000) + 9130872.0607999999462988460436463*((25*q1*pi^2)/32258)*((5*q1)/127 + (5159045700230220335570568101992882368338*q3)/14183735516765013986772738621315219456000 - (269437024701681245919274115996940028669)/270482702025760519408161296220160000);
##J(1, 4) = -9130872.060799999946298846043646
##% Partial derivatives for the second equation
##J(2, 1) = 0;
##J(2, 2) = 3061.5545338275630699055886034165 + 3645.7069886924359708335725693911*q3 + 13054497.238043855951388792124012*((25*pi^2)/32258)*((25*pi^2*q2^2)/64516 + (5*q2*q3)/127 + (2579522850115110167785284050996441184169*q3^2)/14183735516765013986772738621315219456000 - (269464072971883821971214932126562044669*q3)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000) + 13054497.238043855951388792124012*((25*q2*pi^2)/32258)*((50*pi^2*q2)/64516 + (5*q3)/127);
##J(2, 3) = -0.00000000000000000000000052628096743060987461514329634383 + 3645.7069886924359708335725693911*q2 + 37529.754486437990678141804897043*q2 + 13054497.238043855951388792124012*((5)/127)*((25*pi^2*q2^2)/64516 + (5*q2*q3)/127 + (2579522850115110167785284050996441184169*q3^2)/14183735516765013986772738621315219456000 - (269464072971883821971214932126562044669*q3)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000) + 13054497.238043855951388792124012*((25*q2*pi^2)/32258)*((5*q2)/127 + (5159045700230220335570568101992882368338*q3)/14183735516765013986772738621315219456000 - (269464072971883821971214932126562044669)/270482702025760519408161296220160000);
##J(2, 4) = -13054497.238043855951388792124012*((25*q2*pi^2)/32258)*((5)/127);
##% Partial derivatives for the third equation
##%J(3, 1) = 0.000000000000363797880709171295166015625*q3 - 6.9530743154589888866367962498927e-42 + 9130872.0607999999462988460436463*((5)/127)*((25*pi^2*q1^2)/64516 + (5*q1*q3)/127 + (2579522850115110167785284050996441184169*q3^2)/14183735516765013986772738621315219456000 - (269437024701681245919274115996940028669*q3)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000) + 9130872.0607999999462988460436463*((25*q1*pi^2)/32258)*((5*q1)/127 + (5159045700230220335570568101992882368338*q3)/14183735516765013986772738621315219456000 - (269437024701681245919274115996940028669)/270482702025760519408161296220160000);
##%J(3, 2) = -0.00000000000000000000000052628096743060987461514329634383 + 37529.754486437990678141804897043*q3 + 1822.8534943462179854167862846955*q2;
##%J(3, 3) = 0.000000000000363797880709171295166015625*q1 - 949606892.52341601879283837341561 - 1960*(127/(10*pi) + 127/20) + 37529.754486437990678141804897043*q2 + 9130872.0607999999462988460436463*((5*q1)/127 + (5159045700230220335570568101992882368338*q3)/14183735516765013986772738621315219456000 - (269437024701681245919274115996940028669)/270482702025760519408161296220160000)*((25*pi^2*q1^2)/64516 + (5*q1*q3)/127 + (5159045700230220335570568101992882368338*q3)/14183735516765013986772738621315219456000 - (269437024701681245919274115996940028669)/270482702025760519408161296220160000 - (5*q4)/127 + 1010469987441913066378883825611918454307/61897001964269013744956211200000) + 13054497.238043855951388792124012*((5*q2)/127 + (5159045700230220335570568101992882368338*q3)/14183735516765013986772738621315219456000 - (269464072971883821971214932126562044669)/270482702025760519408161296220160000)*((25*pi^2*q2^2)/64516 + (5*q2*q3)/127 + (5159045700230220335570568101992882368338*q3)/14183735516765013986772738621315219456000
%Fill-in with your:
%J(3,4) =
%J(4,1) =
%J(4,2) =
%J(4,3) =
%J(4,4) =
%}
end
% Initial guess for the variables q1, q2, q3, q4
initial_guess = [0; 6; 1; -1];
% Solve the equations
%[q, fval, info] = fsolve(@equations, initial_guess);
tol = 1e-7;  % Tolerance for convergence
max_iter = 500;  % Maximum number of iterations
% Run the Newton's method
solution1 = newton_m(initial_guess, tol, max_iter);
disp('Solution Newton-Rhapson method:');
disp(solution1);
solution2 = latin_pgd_complete(initial_guess);
disp('Solution LATIN-PGD method:');
disp(solution2);
solution3 = broyden_method(initial_guess, tol, max_iter);
disp('Solution Broyden method:');
disp(solution3);
q0 = [0; 1; 1; 1];  % Initial guess
q1 = [2; 0; -2; 2];  % Second guess
tol = 1e-4;         % Tolerance
max_iter = 1000;     % Maximum number of iterations
solution4 = secant_method(q0, q1, tol, max_iter);
disp('Solution Secant method:')
disp(solution4);
q0 = [0; 1; -1; 2];  % Initial guess
tol = 1e-4;         % Tolerance
max_iter = 10000;     % Maximum number of iterations
solution5 = fixed_point_iteration(q0, tol, max_iter);
disp('Solution fixed point iteration methode:')
disp(solution5);
%solution_latin_pgd = latin_pgd_method(initial_guess, tol, max_iter);
%disp('Solution using LATIN-PGD method:');
%disp(solution_latin_pgd);
disp('Secant method optimized')
secant_method_optimized();
disp('Bisections methode:')
a = 1e-14;
b = 1e12;
root = bisection_method(a, b, tol, max_iter,q0);
disp(root)
%Constructed from needing help code by
%https://independent.academia.edu/PMazniker
%+380990535261
%https://diag.net/u/u6r3ondjie0w0l8138bafm095b
%https://github.com/goodengineer
%https://orcid.org/0000-0001-8184-8166
%https://willwork781147312.wordpress.com/portfolio/cp/
%https://www.youtube.com/channel/UCC__7jMOAHak0MVkUFtmO-w
%https://nanohub.org/members/130066
%https://pangian.com/user/hiretoserve/
%https://substack.com/profile/191772642-paul-m

Please be attentive and careful with digital items, electronic artifacts and network logs and records.

If you correctly pay attention, review logs and records, you will see the truth.

I once completed full Newton Rhapson iterative procedure set for solving real world physics system of non-linear equations in TCE Matlab without any AI companionship and delivered it.

Who are they, others? What did they do? No, not that not. I do ok of course. Alternatives to Newton Rhapson:

There are several alternatives to the Newton-Raphson method for solving nonlinear equations.

I tested(checked) it in TCE.

A few specific functions to solution to your specific questions are fully completed by me. I need substantial support to complete it. It is ok and reasonable and even mission and normal life balancee supporting critical to hire me.

John D'Errico 2024-10-16

@Pavl M.

Answers is not a job seeking forum. Please do not put requests for someone to hire you in your posts. I've removed them for you.

Aquatris 2024-10-16

编辑：Aquatris 2024-10-16

@Pavl M. Did you even solve the problem that was asked? The question askes for solutions within [0 1] space where as, if I understood your confusing comment, your solution does not put any constraints to the parameters.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Alex Sha 2024-10-17

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2160085-solution-of-system-of-nonlinear-equations#answer_1533085

在 MATLAB Online 中打开

@Swami Between the range [0,1] you specified, most likely, there should be no real numbers solution，an approximate solution is:

q1: -31830.1002324947
q2: -31835.6758682725
q3: 6184.74933332136
q4: 336421377.881206

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

John D'Errico 2024-10-17

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2160085-solution-of-system-of-nonlinear-equations#answer_1533090

编辑：John D'Errico 2024-10-17

在 MATLAB Online 中打开

This will be literally impossible to do in double precision, given the huge dynamic range of those coefficients. But we can look to see if it is even likely a solution does exist.

syms q [4,1]
eqc_pre(1) = 91.907*q1 - 1.1978e-14*q3 + 3.534e-14*q1*q3 + 9.1309e+6*(0.007649*q1 + 0.03937*q3)*(0.0038245*q1^2 + 0.03937*q1*q3 + 0.18186*q3^2 - 996.13*q3 - 0.03937*q4 + 1.6325e+7) + 1.819e-13*q3^2;
eqc_pre(2) = 2914.4*q2 - 875.08*q3 + 3645.7*q2*q3 + 1.3054e+7*(0.007649*q2 + 0.03937*q3)*(0.0038245*q2^2 + 0.03937*q2*q3 + 0.18186*q3^2 - 996.23*q3 - 0.03937*q4 + 1.6325e+7) + 18765.0*q3^2;
eqc_pre(3) = 3.638e-13*q1*q3 - 875.08*q2 - 9.4963e+8*q3 - 18765.0*q4 - 1.1978e-14*q1 + 37530.0*q2*q3 + 9.1309e+6*(0.03937*q1 + 0.36373*q3 - 996.13)*(0.0038245*q1^2 + 0.03937*q1*q3 + 0.18186*q3^2 - 996.13*q3 - 0.03937*q4 + 1.6325e+7) + 1.3054e+7*(0.03937*q2 + 0.36373*q3 - 996.23)*(0.0038245*q2^2 + 0.03937*q2*q3 + 0.18186*q3^2 - 996.23*q3 - 0.03937*q4 + 1.6325e+7) + 1.767e-14*q1^2 + 1822.8*q2^2 + 260050.0*q3^2 + 7.7808e+12;
eqc_pre(4) = - 1374.8*q1^2 - 14153.0*q1*q3 - 1965.6*q2^2 - 20235.0*q2*q3 - 158850.0*q3^2 + 8.701e+8*q3 + 34387.0*q4 - 1.4259e+13;
vpa(eqc_pre(:),4)

UGH. But we can look at whether it is likely a solution even exists in the hyper-box [0,1]^4.

double(subs(eqc_pre,q,rand(4,1)))
ans = 1×4
1.0e+17 *

    0.0001    0.0001   -3.6066   -0.0001
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
double(subs(eqc_pre,q,rand(4,1)))
ans = 1×4
1.0e+17 *

    0.0000    0.0000   -3.6070   -0.0001
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
double(subs(eqc_pre,q,rand(4,1)))
ans = 1×4
1.0e+17 *

    0.0000    0.0000   -3.6077   -0.0001
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

However, I would point out that having substituted three sets of random numbers into those equations, almost always equation 3 tends to run about 3.6e17, and those computations were performed in high precision.

eq3 = matlabFunction(eqc_pre(3))
eq3 = function_handle with value:
    @(q1,q2,q3,q4)q1.*(-1.1978e-14)-q2.*8.7508e+2-q3.*9.4963e+8-q4.*1.8765e+4+q1.*q3.*3.638e-13+q2.*q3.*3.753e+4+(q1.*3.59483533e+5+q3.*3.321182257e+6-9.095563417e+9).*(q3.*(-9.9613e+2)-q4.*3.937e-2+q1.*q3.*3.937e-2+q1.^2.*3.8245e-3+q3.^2.*1.8186e-1+1.6325e+7)+q1.^2.*1.767e-14+q2.^2.*1.8228e+3+q3.^2.*2.6005e+5+(q2.*5.1393598e+5+q3.*4.74813142e+6-1.300478642e+10).*(q3.*(-9.9623e+2)-q4.*3.937e-2+q2.*q3.*3.937e-2+q2.^2.*3.8245e-3+q3.^2.*1.8186e-1+1.6325e+7)+7.7808e+12
eq3val = eq3(rand(1000000,1),rand(1000000,1),rand(1000000,1),rand(1000000,1));
min(eq3val)
ans = -3.6078e+17
max(eq3val)
ans = -3.6061e+17

This is enough to convince me that no solution exists in that region, since any of a set of 1e6 random points in the box NEVER see any deviation from a very narrow range, that is very far from zero.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 3

Mr. Pavl M. 2024-10-17

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2160085-solution-of-system-of-nonlinear-equations#answer_1533440

编辑：Mr. Pavl M. 2024-10-17

Here, I've solved it in more than 4 methods, for specific equations' system approximated solutions found with fixed interval constraints [a,b] added as per requirements plan:

A.1: Runs:

I've corrected, revamped functionality, amended the software program running codes presented in my comment above.

Working codes in Matlab compatible T.C.E. (fles SystemOfNonlinEqSolver1.m, newton_method.m, fronext.m, ...+):

Pictures are worth than thouthands words.I have the corresponding, according to specifications software program code maintaining running on file disk storage controller electronic unit processed machine. I developed them. They are for sale, if you need the software program codes running, contact me more + for more.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Solution of system of nonlinear equations

4 个评论
显示 2更早的评论隐藏 2更早的评论

回答（3 个）

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

Solution of system of nonlinear equations

4 个评论 显示 2更早的评论隐藏 2更早的评论

回答（3 个）

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

4 个评论
显示 2更早的评论隐藏 2更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论