With two different initial values I am getting two different answers in fsolve. How to decide which one is more reliable?

function F = FeMnC300newuipf3variable(x)
F(1) =10.3014287+log(x(3))+3.37E+04*x(3)+(-3.37E+04*x(3)^2)/2-19.1333*0.03592*x(3)- log(x(1)) -21.4105*x(1)+10.41166*x(2)+(21.4105*x(1)^2)/2-10.41166*x(1)*x(2)+(1.99159*x(2)^2)/2;
F(2) =-0.269548173-log(x(2))+19.1333*x(3) +(-3.37E+04*x(3)^2)/2-19.1333*0.03592*x(3)+10.41166*x(1)-1.99159*x(2)+(21.4105*x(1)^2)/2-10.41166*x(1)*x(2)+(1.99159*x(2)^2)/2;
F(3) =-0.643539436+log(1-0.03592-x(3))-log(1-x(2)-x(1))+(-3.37E+04*x(3)^2)/2-19.1333*0.03592*x(3)+(21.4105*x(1)^2)/2-10.41166*x(1)*x(2)+(1.99159*x(2)^2)/2;
end

First initial value- x0 = [0.002 0.4 0.000000001]

Second initial value- x0 = [0.01 0.4 0.000000001]

[Alex Sha]

There is one solution only?

x1	0.00215170099218674
x2	0.406209428191469
x3	9.4346283948294E-10

[Vikash Sahu]

Kindly explain how did you come to conclusion that there is only one solution and what initial guess have you put.

Answer 1

[John D'Errico]

When you have different variables with such a large discrepancy in magnitude, the problem is not which solution is more correct, but in how we can improve the solver's efficacy on such a problem.

The trick is to transform the problem so that all variables will be of a similar magnitude. That is, if you think x(3) will be 8 orders of magnitude smaller then the other variables, then scale x(3). I might change do the effective substitution:

u(1) = x(1)

u(2) = x(2)

u(3) = x(3)/1e-8

Now u(3) will be of the same rough order of magnitude as the other variables. So a starting value for u(3) might now be 0.1.

Now use fsolve to look for a solution.

u = fsolve(@FeMnC300newuipf3variable,[0.002 0.4 0.1])

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.

u = 1×3

0.0022 0.4062 0.0943

u = fsolve(@FeMnC300newuipf3variable,[0.01 0.4 0.1])

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.

u = 1×3

0.0022 0.4062 0.0943

Now you can see that fsolve seems happy, regardless of where you started, giving the same result from either start point. To recover x, just multiply the third element of u by 1e-8.

function F = FeMnC300newuipf3variable(u)
F(1) =10.3014287+log(u(3)*1e-8)+3.37E+04*u(3)*1e-8+(-3.37E+04*u(3)^2*1e-16)/2-19.1333*0.03592*u(3)*1e-8- log(u(1)) -21.4105*u(1)+10.41166*u(2)+(21.4105*u(1)^2)/2-10.41166*u(1)*u(2)+(1.99159*u(2)^2)/2;
F(2) =-0.269548173-log(u(2))+19.1333*u(3)*1e-8 +(-3.37E+04*u(3)^2*1e-16)/2-19.1333*0.03592*u(3)*1e-8+10.41166*u(1)-1.99159*u(2)+(21.4105*u(1)^2)/2-10.41166*u(1)*u(2)+(1.99159*u(2)^2)/2;
F(3) =-0.643539436+log(1-0.03592-u(3)*1e-8)-log(1-u(2)-u(1))+(-3.37E+04*u(3)^2*1e-16)/2-19.1333*0.03592*u(3)*1e-8+(21.4105*u(1)^2)/2-10.41166*u(1)*u(2)+(1.99159*u(2)^2)/2;
end

Always beware large differences in magnitude between variables when you are working in floating point arithmetic.

Finally, could I have used vpasolve to solve this more accurately? Wel, yes. But solving a problem to 40 decimal digits is a bit silly, when the coefficients are known to only 4 significant digits.

[Vikash Sahu]

I would really appreciate if you can explain how to solve above equation using vpasolve.

Answer 2

[AndresVar]

The results are almost the same, but you can take a look at the display for each iteration see fsolve documentation, here is an example:

problem.options = optimoptions('fsolve','Display','iter','PlotFcn',@optimplotfirstorderopt);
problem.objective = @FeMnC300newuipf3variable;
problem.solver = 'fsolve';

problem.x0 = [0.002 0.4 0.000000001];
fsolve(problem)

problem.x0 = [0.01 0.4 0.000000001];
fsolve(problem)

The first guess has f(x) closer to 0.

You can also see why the solver converged and increase the tolerances or maximum iterations to see wether the solutions converge regardless of initial guess.