我有6个未知数的6个方程组，我试图用fsolve解决它。但是出现了一件奇怪的事情。当我将6个unq中的6个保持为6并且调用fsolve时，他们的求解器不会收敛并且非常慢。但是当我用5个未知量减少5中的方程式时（我可以通过用其余的方程代替其余的方程式）并使用＆＃39; levenberg-marquardt＆＃39; ，求解器真正收敛快速并达成解决方案。虽然我得到的解决方案很好，但我不确定它们是否正确，我无法理解为什么会发生这种情况。我还尝试在6x6情况下将初始值设置为我从5x6情况得到的解决方案，但fsolve再次没有收敛。我不知道这是一个数学问题还是解决问题
有什么想法？ 谢谢 p.s我没有发布代码，它真的很大，我的系统首先用符号编写，然后我用数字替换符号变量，以便用fsolve解决它。
function f=examplegeneral(x);
beta11=0.5; beta12=0.5; beta21=0.5; beta22=0.5; b11=0.5; b12=0.5; b21=0.5; b22=0.5; v1=0.5; v2=0.5;
f(1)=x(2)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^beta12 - v1*(b12*x(1)^2 + b11)^(1/2) - x(2)*(x(4) + x(6))*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 1);
f(2)=x(3)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^beta22 - v2*(b22*x(1)^2 + b21)^(1/2) - x(3)*(x(5) + x(6))*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 1);
f(3)=(x(2)*(x(4) + x(6))*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2))/beta12 - (x(2)*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2)*((b12*x(1)*v1)/(b12*x(1)^2 + b11)^(1/2) - x(2)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 1) + (x(2)*(x(4) + x(6))*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2))/beta12))/(beta12*((b12^2*x(1)^2*v1)/(b12*x(1)^2 + b11)^(3/2) - (b22*v2)/(b22*x(1)^2 + b21)^(1/2) - (b12*v1)/(b12*x(1)^2 + b11)^(1/2) + (b22^2*x(1)^2*v2)/(b22*x(1)^2 + b21)^(3/2) + (x(2)*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2))/beta12 + (x(3)*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2))/beta22));
f(4)=(x(3)*(x(5) + x(6))*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2))/beta22 - (x(3)*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2)*((b22*x(1)*v2)/(b22*x(1)^2 + b21)^(1/2) - x(3)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 1) + (x(3)*(x(5) + x(6))*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2))/beta22))/(beta22*((b12^2*x(1)^2*v1)/(b12*x(1)^2 + b11)^(3/2) - (b22*v2)/(b22*x(1)^2 + b21)^(1/2) - (b12*v1)/(b12*x(1)^2 + b11)^(1/2) + (b22^2*x(1)^2*v2)/(b22*x(1)^2 + b21)^(3/2) + (x(2)*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2))/beta12 + (x(3)*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2))/beta22));
f(5)=(x(2)*(x(4) + x(6))*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2))/beta12 - (((x(2)*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2))/beta12 + (x(3)*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2))/beta22)*((x(2)*(x(4) + x(6))*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2))/beta12 + (x(3)*(x(5) + x(6))*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2))/beta22))/((b12^2*x(1)^2*v1)/(b12*x(1)^2 + b11)^(3/2) - (b22*v2)/(b22*x(1)^2 + b21)^(1/2) - (b12*v1)/(b12*x(1)^2 + b11)^(1/2) + (b22^2*x(1)^2*v2)/(b22*x(1)^2 + b21)^(3/2) + (x(2)*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2))/beta12 + (x(3)*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2))/beta22) + (x(3)*(x(5) + x(6))*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2))/beta22;
f(6)=1/((b12^2*x(1)^2*v1)/(b12*x(1)^2 + b11)^(3/2) - (b22*v2)/(b22*x(1)^2 + b21)^(1/2) - (b12*v1)/(b12*x(1)^2 + b11)^(1/2) + (b22^2*x(1)^2*v2)/(b22*x(1)^2 + b21)^(3/2) + (x(2)*(beta12 - 1)*(1/beta11)^beta11*((x(1) + x(4) + x(6))/beta12)^(beta12 - 2))/beta12 + (x(3)*(beta22 - 1)*(1/beta21)^beta21*((x(1) + x(5) + x(6))/beta22)^(beta22 - 2))/beta22);
然后我打电话
x0=[0.8,0.05,0.05,0.01,0.01,-0.01]; % This is the vector of initial values.
options = optimset('Display','iter','MaxFunEvals',1e+15,'MaxIter',1000000000);
[sol,exitflag,fval]=fsolve('examplegeneral',x0,options)