I keep getting this error, " Failure in initial objective function evaluation. FSOLVE cannot continue."

1 Ansicht (letzte 30 Tage)
Hi, currently I trying to solve for a system of nonlinear equations (5 equatios in total).
Here is my function part
% Coeffecient
M1 = 0.5;
M2 = 0.75;
M3 = 0.95;
M4 = 1.102;
M5 = 1.102;
K = 1;
Kc = 1;
Kk = 3;
KR = 0.95;
G = 0.001;
GR = 0.1;
Gc = 0.005;
B = 0.15;
A= 1;
w= 1:0.1:2;
%Set up equation
function F = MyFunction(x)
F(1 )= x(1).*(-M1.*(w.^2) + K + Kc +KR + G.*1i.*w) - (A.*(exp(1i.*w)) + Kc.*x(4) + KR.*x(3));
F(2) = x(2).*(-M2.*(w.^2) + K + Kc +KR + G.*1i.*w) - (A.*(exp(1i.*w)) + Kc.*x(5) + KR.*x(3));
F(3) = x(3).*(-M3.*(w.^2) + 2.*kk + GR.*1i.*w) - (KR.*(x(1) + (x(2))));
F(4) = x(4).*(-M4.*(w.^2) + Kc + kk + Gc.*1i.*w) - B.*1i.*(w.^3).*((x(4)).^3) - A.*(exp(1i.*w)) - Kc.*x(1);
F(5) = x(5).*(-M5.*(w.^2) + Kc + kk + Gc.*1i.*w) - B.*1i.*(w.^3).*((x(4)).^3) - A.*(exp(1i.*w)) - Kc.*x(2);
end
Then I create a different file to call back to it
x0 = [1 1 1 1 1];
x = fsolve(@MyFunction,x0);
So the idea is to solve x1, x2, x3, x4, x5 as a function of w. But I have tried fixing w to 1 value, as well as move the coeffecient to the 2nd file. But I still could fix this message that MATLAB gave me " Failure in initial objective function evaluation. FSOLVE cannot continue"
Anybody have a way for me to fix this

Akzeptierte Antwort

Navya Seelam
Navya Seelam am 17 Feb. 2020
fsolve fails to continue because the function MyFunction doesnot not have access to the variables (M1, M2, M3...). You can use global variables for MyFunction to work without errors.
global param
param.M1 = 0.5;
param.M2 = 0.75;
param.M3 = 0.95;
param.M4 = 1.102;
param.M5 = 1.102;
param.K = 1;
param.Kc = 1;
param.Kk = 3;
param.KR = 0.95;
param.G = 0.001;
param.GR = 0.1;
param.Gc = 0.005;
param.B = 0.15;
param.A= 1;
param.w= 1;
x0 = [1 1 1 1 1];
fun=@MyFunction;
x = fsolve(fun,x0);
function F = MyFunction(x)
global param
F(1 )= x(1).*(-param.M1.*(param.w.^2) + param.K + param.Kc +param.KR + param.G.*1i.*param.w) - (param.A.*(exp(1i.*param.w)) + param.Kc.*x(4) + param.KR.*x(3));
F(2) = x(2).*(-param.M2.*(param.w.^2) + param.K + param.Kc +param.KR + param.G.*1i.*param.w) - (param.A.*(exp(1i.*param.w)) + param.Kc.*x(5) + param.KR.*x(3));
F(3) = x(3).*(-param.M3.*(param.w.^2) + 2.*param.Kk + param.GR.*1i.*param.w) - (param.KR.*(x(1) + (x(2))));
F(4) = x(4).*(-param.M4.*(param.w.^2) + param.Kc + param.Kk + param.Gc.*1i.*param.w) - param.B.*1i.*(param.w.^3).*((x(4)).^3) -param.A.*(exp(1i.*param.w)) - param.Kc.*x(1);
F(5) = x(5).*(-param.M5.*(param.w.^2) + param.Kc + param.Kk + param.Gc.*1i.*param.w) - param.B.*1i.*(param.w.^3).*((x(4)).^3) - param.A.*(exp(1i.*param.w)) - param.Kc.*x(2);
end
  1 Kommentar
Viet Tran
Viet Tran am 17 Feb. 2020
Bearbeitet: Viet Tran am 17 Feb. 2020
Thank you for answering, this method works for me but is there a way I can keep w varying instead of fixing it to 1 value?

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Weitere Antworten (1)

Dhananjay Kumar
Dhananjay Kumar am 17 Feb. 2020
Declare a 5x11 F in the beginning and use proper indexing on F like this:
F = ones(5,11);
F(1,:)= x(1).*(-M1.*(w.^2) + K + Kc +KR + G.*1i.*w) - (A.*(exp(1i.*w)) + Kc.*x(4) + KR.*x(3));
F(2,:) = x(2).*(-M2.*(w.^2) + K + Kc +KR + G.*1i.*w) - (A.*(exp(1i.*w)) + Kc.*x(5) + KR.*x(3));
F(3,:) = x(3).*(-M3.*(w.^2) + 2.*kk + GR.*1i.*w) - (KR.*(x(1) + (x(2))));
F(4,:) = x(4).*(-M4.*(w.^2) + Kc + kk + Gc.*1i.*w) - B.*1i.*(w.^3).*((x(4)).^3) - A.*(exp(1i.*w)) - Kc.*x(1);
F(5,:) = x(5).*(-M5.*(w.^2) + Kc + kk + Gc.*1i.*w) - B.*1i.*(w.^3).*((x(4)).^3) - A.*(exp(1i.*w)) - Kc.*x(2);

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