Failure in initial user-supplied objective function evaluation. FSOLVE cannot continue

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Mojtaba Mohareri
Mojtaba Mohareri 2019-4-7
评论： Stephan 2019-4-7
My currenct script looks like this:
function m=F(x)
k11=x(1);
k12=x(2);
k13=x(3);
k14=x(4);
l=x(5);
f1(y1,y2,y3,y4,z1)=(-(z1^3)/(y3^2))*(3*(y2-y1+y3^(-1)-z1/10)^2+(1/5)*(y2-y1+y3^(-1)-(z1)/10))-y4;
f2(y1,y2,y3,y4,z1)=1/10*z1-y4;
f3(y1,y2,y3,y4,z1)=(z1^3)*(3*(y2-y1+y3^(-1)-z1/10)^2+(1/5)*(y2-y1+y3^(-1)-(z1)/10));
f4(y1,y2,y3,y4,z1)=y1-y3^(-1);
g(y1,y2,y3,y4,z1)=(y1-y3^(-1))^2+y4^2-1/10*z1;
J1(y1,y2,y3,y4,z1)=jacobian([f1,f2,f3,f4],[y1,y2,y3,y4]);
J1=J1(2,2,1,0,10);
J2(y1,y2,y3,y4,z1)=jacobian([f1,f2,f3,f4],[z1]);
J2=J2(2,2,1,0,10);
J3(y1,y2,y3,y4,z1)=jacobian([g],[y1,y2,y3,y4]);
J3=J3(2,2,1,0,10);
J4(y1,y2,y3,y4,z1)=jacobian([g],[z1]);
J4=J4(2,2,1,0,10);
k1=[k11;k12;k13;k14];
m(1)=f1(2,2,1,0,10)+0.44*(J1(1,:)*k1 +J2(1,:)*l)-k11;
m(2)=f2(2,2,1,0,10)+0.44*(J1(2,:)*k1 +J2(2,:)*l)-k12;
m(3)=f3(2,2,1,0,10)+0.44*(J1(3,:)*k1 +J2(3,:)*l)-k13;
m(4)=f4(2,2,1,0,10)+0.44*(J1(4,:)*k1 +J2(4,:)*l)-k14;
m(5)=g(2,2,1,0,10)+0.44*(J3*k1 +J4(1,:)*l);
end
x0=[1,1,1,1,1];
fsolve(@F,x0)
Still nothing seems to get this working, can anyone help me out?
Thanks in advance.
  4 个评论
显示 2更早的评论
Star Strider
Star Strider 2019-4-7
It’s difficult to understand what you‘re doing.
However, you need to avoid using symbolic variables in the function to use as your function argument to fsolve. Use the matlabFunction function to create a numeric function fsolve can use.
Mojtaba Mohareri
Mojtaba Mohareri 2019-4-7
编辑：Mojtaba Mohareri 2019-4-7
Actually I want to solve a system of equations with 5 unknowns k11,k12,k13,k14,l11 defined by equations m(1),...m(5). I don't know how to do that.

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采纳的回答

Stephan
Stephan 2019-4-7
Hi,
this runs - not sure if it is really efficient, but it works:
x0=[1,1,1,1,1];
fsolve(@F,x0)
function m=F(x)
syms y1 y2 y3 y4 z1 k11 k12 k13 k14 l11
h=1/1000;
f1(y1,y2,y3,y4,z1)=(-(z1^3)/(y3^2))*(3*(y2-y1+y3^(-1)-z1/10)^2+(1/5)*(y2-y1+y3^(-1)-(z1)/10))-y4;
f2(y1,y2,y3,y4,z1)=1/10*z1-y4;
f3(y1,y2,y3,y4,z1)=(z1^3)*(3*(y2-y1+y3^(-1)-z1/10)^2+(1/5)*(y2-y1+y3^(-1)-(z1)/10));
f4(y1,y2,y3,y4,z1)=y1-y3^(-1);
g(y1,y2,y3,y4,z1)=(y1-y3^(-1))^2+y4^2-1/10*z1;
J1(y1,y2,y3,y4,z1)=jacobian([f1,f2,f3,f4],[y1,y2,y3,y4]);
J1=double(J1(2,2,1,0,10));
J2(y1,y2,y3,y4,z1)=jacobian([f1,f2,f3,f4],z1);
J2=double(J2(2,2,1,0,10));
J3(y1,y2,y3,y4,z1)=jacobian(g,[y1,y2,y3,y4]);
J3=double(J3(2,2,1,0,10));
J4(y1,y2,y3,y4,z1)=jacobian(g,z1);
J4=double(J4(2,2,1,0,10));
k11=x(1);
k12=x(2);
k13=x(3);
k14=x(4);
l11=x(5);
k1=[k11;k12;k13;k14];
m(1)=double(h*(f1(2,2,1,0,10)+0.44*(J1(1,:)*k1 +J2(1,:)*l11))-k11);
m(2)=double(h*(f2(2,2,1,0,10)+0.44*(J1(2,:)*k1 +J2(2,:)*l11))-k12);
m(3)=double(h*(f3(2,2,1,0,10)+0.44*(J1(3,:)*k1 +J2(3,:)*l11))-k13);
m(4)=double(h*(f4(2,2,1,0,10)+0.44*(J1(4,:)*k1 +J2(4,:)*l11))-k14);
m(5)=double(g(2,2,1,0,10)+0.44*(J3*k1 +J4(1,:)*l11));
end
Best regards
Stephan

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