How to solve 3 nonlinear system of equations?

I wrote a code in two files:

file: 1

function [fval,jac]= CSPW1(X)

x=X(1);
y=X(2);
z=X(3);

fval(1,1)=0.1315*x^2+0.4526*y^2+0.6793*x.*y+0.7549*z.*y-0.0378*z;
fval(2,1)=0.5983*x^2+1.5280*y^2+2.6025*x.*y+2.7582*y.*z-0.07785*z^2;
fval(3,1)=0.2335*x^3+1.0922*y^3+1.528*x.*y^2+1.6058*z.*y^2-0.07785*x.*z.^2-0.02595*z^3;

jac=[(263*x)/1000+(6793*y)/10000, (6793*x)/10000+(2263*y)/2500+(7549*z)/10000, (7549*y)/10000-189/5000;
     (263*x)/1000+(6793*y)/10000, (6793*x)/10000+(2263*y)/2500+(7549*z)/10000, (7549*y)/10000-189/5000;
     (263*x)/1000+(6793*y)/10000, (6793*x)/10000+(2263*y)/2500+(7549*z)/10000, (7549*y)/10000-189/5000];

File: 2

X0=[10; 10; 30];
maxIter=5;
tolX=1e-12;
X=X0;
Xold=X0;
for i=1:maxIter
    [f,j]=CSPW1(X);
    X=X-inv(j)*f;
    
    err(:,i)=abs(X-Xold);
    Xold=X;
    if (err(:,i)<tolX)
        break;
    end
end

But in answer it shows "Warning: Matrix is singular, close to singular or badly scaled. Results may be inaccurate. RCOND = NaN."

How to solve then?

Answer 1

[madhan ravi]

Try using fsolve()(link)

[Akshay Pratap Singh]

Thanks Madhan Ravi for quick reply. But it is also not being solved by fsolve(). If you can solve or modify the code for solution. It will be a great help to me.

Hoping to get the solution from you.

Thanks in advance

[madhan ravi]

I am not sure how robust the results are but :

options=optimoptions('fsolve','Display','iter');
options.MaxFunctionEvaluations = 3000;
options.OptimalityTolerance = 1e-12;
X=fsolve(@CSPW1,[10; 10; 30],options)
function [fval,jac]= CSPW1(X)
x=X(1);
y=X(2);
z=X(3);
fval=zeros(3,1);
fval(1,1)=0.1315*x^2+0.4526*y^2+0.6793*x.*y+0.7549*z.*y-0.0378*z;
fval(2,1)=0.5983*x^2+1.5280*y^2+2.6025*x.*y+2.7582*y.*z-0.07785*z^2;
fval(3,1)=0.2335*x^3+1.0922*y^3+1.528*x.*y^2+1.6058*z.*y^2-0.07785*x.*z.^2-0.02595*z^3;
jac=[(263*x)/1000+(6793*y)/10000, (6793*x)/10000+(2263*y)/2500+(7549*z)/10000, (7549*y)/10000-189/5000;
     (263*x)/1000+(6793*y)/10000, (6793*x)/10000+(2263*y)/2500+(7549*z)/10000, (7549*y)/10000-189/5000;
     (263*x)/1000+(6793*y)/10000, (6793*x)/10000+(2263*y)/2500+(7549*z)/10000, (7549*y)/10000-189/5000];
end

Gives:

Equation solved, inaccuracy possible.

The vector of function values is near zero, as measured by the default value
of the function tolerance. However, the last step was ineffective.

<stopping criteria details>


X =

    0.0036
   -0.0009
   -0.0000

[Walter Roberson]

When you return the jacobian you should have the option 'SpecifyObjectiveGradient','on'

[Walter Roberson]

Actual solution:

x = 3.4691311028950902695051613877565 y = -0.22793089975528930231491570459169 z = 5.0935669467576127190815147121607

This is one of the 7 solutions -- in particular the one that is smallest Euclidean distance to your starting point of [10 10 30] . The second closest solution is [0 0 0]

[madhan ravi]

Out of curiosity: How were the solutions found ? To learn ..