fsolve message "Equation solved, inaccuracy possible"
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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));
2 个评论
John D'Errico
2019-4-26
编辑：John D'Errico
2019-4-26
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
2019-4-26
采纳的回答
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
4 个评论
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
2018-9-18
更多回答（1 个）
Alex Sha
2019-4-22
0 个投票
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
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