Why does fsolve seem not iterate towards the solution?
조회 수: 2 (최근 30일)
이전 댓글 표시
Hello all,
I am trying to sovle a two non-linear equation system using fsolve and dogleg method. My objective function along with its jacobian is like this
function [F jacF]= objective(x)
F(:,1) = ((((x(:,2)./10).*k).*(x(:,1)./100)).^2).*(rZ - Rs) +(( Cmax .* ( x(:,1)./100 ) ).^2).*( w.^2.*(rZ - Rs) ) - (((x(:,2)./10).*k).*(x(:,1)./100));
F(:,2) = (x(:,2).*k).^2.*(iZ - w.*Ls) + (x(:,2).*k).^2.*x(:,1).*((w.*Ls)./200) + x(:,1).*((w.*Ls)/200).*(w.*Cmax).^2 + (w.*Cmax).^2 .*(iZ -(w.*Ls));
if nargout > 1 % need Jacobian
jacF = [- k - (k.^2.*x(:,2).*x(:,1).*(Rs - rZ))./50, - (k.^2.*x(:,2).^2.*(Rs - rZ))./100 - (Cmax.^2.*w.^2.*(Rs - rZ))./100;
2.*k.^2.*x(:,2).*(iZ - Ls.*w) + (k.^2.*Ls.*x(:,2).*w.*x(:,1))./100,(Ls.*Cmax.^2.*w.^3)./200 + (Ls.*k.^2.*x(:,2).^2.*w)./200];
end
end
Then my configuration for fsolve looks like this
options = optimoptions('fsolve','Display','iter-detailed','PlotFcn',@optimplotfirstorderopt);
% options.StepTolerance = 1e-13;
options.OptimalityTolerance = 1e-12;
options.FunctionTolerance = 6e-11;
options.MaxIterations = 100000;
options.MaxFunctionEvaluations = 400;%*400;
options.Algorithm = 'trust-region-dogleg';%'trust-region'%'levenberg-marquardt';%
% options.FiniteDifferenceType= 'central';
options.SpecifyObjectiveGradient = true;
fun= @objective;
x0 = [x1',x2'];
% Solve the function fun
[gwc,fval,exitflag,output,jacobianEval] =fsolve(fun,x0,options);
Being the values of the equations
Rs =
0.1640
Ls =
1.1000e-07
Cmax =
7.0000e-11
w =
1.7040e+08
rZ =
12.6518
iZ =
14.5273
K =
0.1007
x0 =
70.56 0.0759
My problem comes because I don't understand why fsolve seems not to iteratate over x(:,1) as i was expecting. I do know that the solution for the above system and parameters should be x1=58.8 and x2=0.0775.
In order to test the convergence of the method I am setting as initial guess x0 = [x1*(1+20/100) 0.0759] = [70.56 0.0759] ( 20 percent error in x1 and a higer value on x2), but the solution given by fsolve is the initial point, why is this? Am I doing something incorrect in my settings?
Thanks in advance
댓글 수: 7
채택된 답변
Torsten
2019년 9월 12일
k = 0.1007;
rZ = 12.6518;
Rs = 0.164;
Cmax = 7.0e-11;
W = 1.704e8;
iZ = 14.5273;
Ls = 1.1e-7;
a1 = (k/1000)^2*(rZ-Rs);
a2 = (Cmax/100*W)^2*(rZ-Rs);
a3 = -k/1000;
b1 = k^2*(iZ-W*Ls);
b2 = k^2*W*Ls/200;
b3 = W*Ls/200*(Cmax*W)^2;
b4 = (Cmax*W)^2*(iZ-W*Ls);
A = (a2*b1-b2*a3+a1*b4)/(a1*b1);
B = (-a3*b3+a2*b4)/(a1*b1);
disk = A^2/4-B;
if disk >=0
x21squared = -A/2+sqrt(disk);
x22squared = -A/2-sqrt(disk);
end
solx1 = zeros(4,1);
solx2 = zeros(4,1);
iflag1 = 0;
if x21squared >= 0
iflag1 = 1;
solx2(1) = sqrt(x21squared);
solx2(2) = -sqrt(x21squared);
end
iflag2 = 0;
if x22squared >= 0
iflag2 = 1;
solx2(3) = sqrt(x22squared);
solx2(4) = -sqrt(x22squared);
end
solx1 = zeros(4,1);
if iflag1 == 1
solx1(1) = -a3/(a1*solx2(1)^2+a2);
solx1(2) = -a3/(a1*solx2(2)^2+a2);
end
if iflag2 == 1
solx1(3) = -a3/(a1*solx2(3)^2+a2);
solx1(4) = -a3/(a1*solx2(4)^2+a2);
end
if iflag1 == 1
solx1(1)
solx2(1)
solx1(2)
solx2(2)
a1*solx1(1)*solx2(1)^2+a2*solx1(1)+a3
b1*solx2(1)^2+b2*solx1(1)*solx2(1)^2+b3*solx1(1)+b4
a1*solx1(2)*solx2(2)^2+a2*solx1(2)+a3
b1*solx2(2)^2+b2*solx1(2)*solx2(2)^2+b3*solx1(2)+b4
end
if iflag2 == 1
solx1(3)
solx2(3)
solx1(4)
solx2(4)
a1*solx1(3)*solx2(3)^2+a2*solx1(3)+a3
b1*solx2(3)^2+b2*solx1(3)*solx2(3)^2+b3*solx1(3)+b4
a1*solx1(4)*solx2(4)^2+a2*solx1(4)+a3
b1*solx2(4)^2+b2*solx1(4)*solx2(4)^2+b3*solx1(4)+b4
end
댓글 수: 5
Torsten
2019년 9월 13일
This is likely to break the bi-quadratic equation that you made, then should I go again for an interative process right?
No. Insert the expression from F1 for x1 in F2, multiply by the denominator, order according to powers of x2 and use MATLAB's "roots" to solve for x2. This will come out much more stable than using "fsolve".
추가 답변 (0개)
참고 항목
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!