fsolve message "Equation solved, inaccuracy possible"

Hi

I want to solve a system of nonlinear equations using fsolve. But I get "Equation solved, inaccuracy possible." and all x are incorrect. By changing initial guess, nothing improves!

I appreciate your help

Regards,

Fun = @xisolver1;
x0 = [1e-4 1e-4 1e-4 1e-4 1e-4 1e-2 1e-2];
% options = optimoptions('fsolve','OptimalityTolerance',1e-20, 'FunctionTolerance', 1e-20);
options = optimoptions('fsolve','Display','iter','TolFun',1e-40,'TolX',1e-40);
xr = fsolve(Fun, x0, options);

function y = xisolver1(x)
T = 31.65 + 275.15;

lnK1 = 132.899 + (-13445.9/T) + -22.4773*(log10(T)/log10(exp(1)));
lnK2 = 216.05 + (-12431.7/T) + -35.4819*(log10(T)/log10(exp(1)));
lnK3 = 231.465 + (-12092.1/T) + -36.7816*(log10(T)/log10(exp(1)));

K1 = power(exp(1),lnK1);
K2 = power(exp(1),lnK2);
K3 = power(exp(1),lnK3);

KS1 = 1/K1;
KS2 = 1/K2;
KS3 = 1/K3;

% mole
n_CO2 = 6.3746;
n_K2CO3 = 9.7119;
n_H2O = 358.8905;
n_T = n_CO2 + n_K2CO3 + n_H2O;

% mass - gr/s
m_T = 8088.2946;

% molecular weight
MW_CO2 = 44.0095;
MW_CO3 = 60.0089;
MW_HCO3 = 61.0168;
MW_OH = 17.0073;
MW_H3O = 19.0232;
MW_H2O = 18.01528;
MW_K = 39.098;

MW_K2CO3 = 138.2055;
MW_KHCO3 = 100.1151;

% mole fraction
X_CO2T = (n_CO2 + n_K2CO3)/n_T;
X_K__ = 2*(n_K2CO3/n_T);

% mole number
N_CO2T = (n_CO2 + n_K2CO3);

y(1) = K1*(x(6)^2) - 1*x(4)*x(5);
y(2) = K2*x(3)*x(6) - 1*x(5)*x(2);
y(3) = K3*x(1)*(x(6)^2) - 1*x(5)*x(3);
y(4) = x(7) + x(5) - (2*x(2) + x(3) + x(4));

y(5) = x(1) + x(2) + x(3) - ...
    N_CO2T/(m_T/(x(1)*MW_CO2 + x(2)*MW_CO3 + x(3)*MW_HCO3 + x(4)*MW_OH + ...
    x(5)*MW_H3O + x(6)*MW_H2O + x(7)*MW_K));

y(6) = x(1) + x(2) + x(3) + x(4) + x(5) + x(6) + x(7) - 1;

y(7) = x(7) - (2*n_K2CO3)/(m_T/(x(1)*MW_CO2 + x(2)*MW_CO3 + x(3)*MW_HCO3 + x(4)*MW_OH + ...
    x(5)*MW_H3O + x(6)*MW_H2O + x(7)*MW_K));

[John D'Errico]

You should understand that

log10(T)/log10(exp(1))

is equivalent to the simple

log(T)

That is, log(T) is the NATURAL LOG of T?\

As well, why would you do this?

K1 = power(exp(1),lnK1);

You seem to understand that exp(1) yields the number e. So just use

K1 = exp(lnK1);

[sina]

You're absolutly right, I modified that.

Do you have any idea about the question?

Answer 1

[Matt J]

But the solution you got solves the equations quite well and also zeros the optimality measure quite well. If the wrong solution solves the equations, then the equations are to blame.

                                         Norm of      First-order   Trust-region
 Iteration  Func-count     f(x)          step         optimality    radius
     0          8        0.959589                             1               1
     1         16     2.73535e-08       0.901598       2.81e-06               1
     2         24     5.66403e-09      0.0230512       1.72e-06            2.25
     3         32     1.82444e-11     0.00549156       9.15e-08            2.25
     4         40     3.05915e-16    0.000351564       3.73e-10            2.25
     5         48     8.60818e-21    0.000134172       3.73e-13            2.25

[sina]

Hi Matt,

I got the same solution with the following message

Equation solved, inaccuracy possible.

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

<stopping criteria details>

In order to check the accuracy of the x, I solve y(1) to get K1, but K1 is not the same as used in the previous calculation. This can be seen for K2 and K3. Because of that, I need to improve the x(1..7). But How, it's my question, even I tried to eleminate x(1,3,6) but still was not accurate!

[Matt J]

Try increasing the weight on the first 3 equations

y(1) = K1*(x(6)^2) - 1*x(4)*x(5);
y(2) = K2*x(3)*x(6) - 1*x(5)*x(2);
y(3) = K3*x(1)*(x(6)^2) - 1*x(5)*x(3);

y(1:3)=1e12*y(1:3);

[sina]

This worked better, However, because of K value I don't know maybe this is normal, and we can accept x

[sina]

It worked however I got "Equation solved, inaccuracy possible." but I think for my case I can ignore the error!

Answer 2

[Alex Sha]

Multi-solutions:

1:

x1: 1.00788811123075E-6

x2: 0.0083586258586415

x3: 0.0330975342164431

x4: 0.000243072462467804

x5: 4.25824266670525E-13

x6: 0.908241901178143

x7: 0.0500578583957681

2:

x1: 0.0398069369652828

x2: 4.09226387610729E-13

x3: -6.59445321197498E-9

x4: -7.80181990889573E-15

x5: -0.0480652696493024

x6: 0.960193076222411

x7: 0.0480652630556597

3:

x1: -0.00867531212733977

x2: 3.28114366134408E-10

x3: 0.0504921742289807

x4: -1.14968645806141E-9

x5: -1.22255237713138E-9

x6: 0.907690964984408

x7: 0.0504921749580755

[sina]

Hi Alex Sha,

x should be positive. How did you solve it?