Why does fsolve seem not iterate towards the solution?

Question

Javier 2019-9-12

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此问题的直接链接

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评论： Javier 2019-9-13

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

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Javier 2019-9-12

编辑：Javier 2019-9-12

在 MATLAB Online 中打开

Hi Torsten, thanks for replying, I will get this

Value of res

res =
   1.4989e-04   2.5817e-04

Output for fsolve after being called

                                         Norm of      First-order   Trust-region
 Iteration  Func-count     f(x)          step         optimality    radius
        1     3.40966e-07                      2.77e-05               1
        2     3.40966e-07              1       2.77e-05               1
        3     3.40966e-07           0.25       2.77e-05            0.25
        4     3.40966e-07         0.0625       2.77e-05          0.0625
        5     3.40966e-07       0.015625       2.77e-05          0.0156
        6     3.40966e-07     0.00390625       2.77e-05         0.00391
        7     3.40966e-07    0.000976562       2.77e-05        0.000977
        8     3.40966e-07    0.000244141       2.77e-05        0.000244
        9     3.40966e-07    6.10352e-05       2.77e-05         6.1e-05
       10     3.40966e-07    1.52588e-05       2.77e-05        1.53e-05
       11     3.40966e-07     3.8147e-06       2.77e-05        3.81e-06
       12     3.40966e-07    9.53674e-07       2.77e-05        9.54e-07
       13     3.40966e-07    2.38419e-07       2.77e-05        2.38e-07
       14     3.40966e-07    5.96046e-08       2.77e-05        5.96e-08
       15     3.40966e-07    1.49012e-08       2.77e-05        1.49e-08
       16     3.40966e-07    3.72529e-09       2.77e-05        3.73e-09
       17     3.40966e-07    9.31323e-10       2.77e-05        9.31e-10
       18     3.40966e-07    2.32831e-10       2.77e-05        2.33e-10
       19     3.40966e-07    5.82077e-11       2.77e-05        5.82e-11
fsolve stopped because the relative norm of the current step, 8.131781e-13, is less than 
max(options.StepTolerance^2,eps) = 1.000000e-12. The sum of squared function values,
r = 3.409656e-07, is less than sqrt(options.FunctionTolerance) = 1.000000e-06.
Optimization Metric                                             Options
relative norm(step) =   8.13e-13               max(StepTolerance^2,eps) =   1e-12 (default)
r =   3.41e-07                                  sqrt(FunctionTolerance) =  1.0e-06 (selected)

Javier 2019-9-12

Thanks Torsten,

I will check this with roots. On the other hand, shouldn' the solver be able to reach those solutions as well?

Torsten 2019-9-12

If you have a good starting guess ...

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Answer 1

Torsten 2019-9-12

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/480027-why-does-fsolve-seem-not-iterate-towards-the-solution#answer_391563

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

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Torsten 2019-9-13

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".

Javier 2019-9-13

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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Why does fsolve seem not iterate towards the solution?

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Why does fsolve seem not iterate towards the solution?

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