MATLAB Answers

Javier
0

Why does fsolve seem not iterate towards the solution?

Asked by Javier
on 12 Sep 2019
Latest activity Commented on by Javier
on 13 Sep 2019
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

  7 Comments

Ok, with some constants I get
a1*x1*x2^2 + a2*x1 + a3*x2 = 0 (1st equation after division by x1)
b1*x2^2 + b2*x1*x2^2 + b3*x1 + b4 = 0 (2nd equation)
Now you can either solve the first or the second equation for x1 and insert the expression in the other one.
In the end, you'll get a polynomial of degree 4 in x2 that can be solved using "roots".
Thanks Torsten,
I will check this with roots. On the other hand, shouldn' the solver be able to reach those solutions as well?
If you have a good starting guess ...

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1 Answer

Answer by Torsten
on 12 Sep 2019
 Accepted Answer

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

  5 Comments

Now I understood, yes I followed your advice and I came up with a bi-quadratic in x2, I was just solving by roots, I see the same answer!
I have reviewing the original equations too, and I think I missed a term in my matlab expression of F2, it should suppose to be like this
F(:,2) = ((((x(:,2)./10).*Kgeometry).*(x(:,1)./100)).^2).*( imgZin - w.*Ls * (1-x(:,1)./200) ) + (((x(:,1)./100).*Cmax).^2).*( w.^2*(imgZin - w*Ls .* (1-x(:,1)./200)) ) + w.*(Cmax.*(x(:,1)./100) );
This is likely to break the bi-quadratic equation that you made, then should I go again for an interative process right?
Thanks again
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".
Hi Torsten, thanks again.
I have followed your advice, since F1 hasn't change, the relationship between x1 and x2 based on first equation is the same as you mentioned.
If I now replace the value of x1 into F2 as you suggests, I will get an polynom in order fith instead of fourth, which I will resolve with roots

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