Hello, I want to solve system of non linear equations using fsolve. There are thre equations and two unknowns K and L. looking at my code, please advice how to best do this. Also please advice if this is the best way or there is an alternate option.

1 次查看（过去 30 天）

Sumera Yamin 2018-5-25

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/402574-hello-i-want-to-solve-system-of-non-linear-equations-using-fsolve-there-are-thre-equations-and-two

评论： Walter Roberson 2018-6-1

Ld= 0.8194  %constant
% cos^2(K*L)+ Ld*K*sin(K*L)cos(K*L)=2.2 two non linear equations
% cos^2(K*L)=0.299
% Ld*cos^2(K*L)+sin(K*L)cos(K*L)/K=0.262
cos^2(K*L)+ Ld*K*sin(K*L)cos(K*L)-2.2=0
cos^2(K*L)-0.299=0
Ld*cos^2(K*L)+(sin(K*L)cos(K*L))/K-0.262=0
function F=calib(K,L)
F(1)=cos(K*L)^2+ Ld*K*sin(K*L)*cos(K*L)-2.2;
F(2)=cos(K*L)^2-0.299;
F(3)=Ld*cos^2(K*L)+(sin(K*L)cos(K*L))/K-0.262;
fun = @calib (K,L);
initial_val=[2.73,0.6]  % initial value of K and L respectively
x = fsolve(fun,initial_val);

2 个评论
显示无隐藏无

Star Strider 2018-5-25

In the future, please highlight the code in your questions or comments, then click on the {}Code button to format it.

Sumera Yamin 2018-5-28

Thank you for the guidance. in future i will take care about coding format

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Star Strider 2018-5-25

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/402574-hello-i-want-to-solve-system-of-non-linear-equations-using-fsolve-there-are-thre-equations-and-two#answer_321958

在 MATLAB Online 中打开

You have the correct approach and are close to the correct solution. There are some errors in your code for ‘calib’ that I corrected. (Check to be certain that they do what you want.) I also created it as an anonymous function, for convenience.

The optimization functions all expect a vector of parameters, not separate parameters. I changed ‘fun’ to provide the correct calling convention, without having to change ‘calib’.

This runs:

Ld= 0.8194  %constant
calib = @(K,L) [cos(K*L).^2 + Ld*K*sin(K*L).*cos(K*L)-2.2;  cos(K*L).^2-0.299;  Ld*cos(K*L).^2 + (sin(K*L).*cos(K*L))/K-0.262];
fun = @(b) calib(b(1),b(2));
initial_val=[2.73, 0.6]  % initial value of K and L respectively
[x,fval] = fsolve(fun,initial_val);

You may need to experiment with it to actually solve your equations, since the end values are not near zero.

14 个评论
显示 12更早的评论隐藏 12更早的评论

Sumera Yamin 2018-5-31

在 MATLAB Online 中打开

xi_xf_R11.png

ok, but i do not understand what is mathematically wrong with the system that it doesnot have a solution. So in theory, "K" and "L" represents real physical quantities for which we know the estimated initial values. The whole problem goes like this. I have my data x and y, i plotted the data does the curve fitting and the equations above are transfer functions of my data. So, i want to calibrate the "K" and "L" values which fits my data according to above equation. This is explained in the following code. For simplicity, i have again reduced the no of equations to 3. Can you please comment what is mathematically wrong with my system?

if true
  % code
data=[0  0;
  0.05 1.108646244630E-01;
  0.10 2.217423074817E-01;
  0.15 3.325947375398E-01;
  0.20 4.434863433851E-01;
  0.25 5.543595496420E-01;
  0.30 6.652338361973E-01;
  0.35 7.761094191116E-01;
  0.40 8.869865144820E-01;
  0.45 9.978653384221E-01;
  0.50 1.108746107036E+00];
k=10.9251;                   %fixed
xdata=data(:,1);
ydata=data(:,2);
% where data is related by the following euation.
xdata=(cos(K*L).^2- ld*K*sin(K*L)*cos(K*L))*ydata;
%when i plot x and y and do linear curve fit as shown in attached fig, the slope is 2.2. So,
cos(K*L).^2- ld*K*sin(K*L)*cos(K*L) = 2.2;  % 2.2 is the slope from the linear curve fitting to the above data
%Similarly for othersets of data yields following equations
%cos(K*L).^2 = 0.299; 
%Ld*cos(K*L).^2 + (sin(K*L).*cos(K*L))/K = 0.262;
% so i intend to solve these 3 equations using fval to get the Value of K and L from
% data. My sloution should be closed to values given in initial_val. This
% time use initial value as given below.
 initial_val=[5.46,0.14]
end

Star Strider 2018-5-31

在 MATLAB Online 中打开

There may not be anything ‘mathematically wrong’ with your system. There may not be a solution, especially since ‘K’ and ‘L’ are arguments to trigonometric functions.

If you create a matrix version of ‘calib’:

calibm = @(K,L) [cos(K.*L).^2 + Ld*K.*sin(K.*L).*cos(K.*L)-2.2*ones(size(K));  cos(K.*L).^2-0.299*ones(size(K));  Ld*cos(K.*L).^2 + (sin(K.*L).*cos(K.*L))./K-0.262*ones(size(K))];

and then plot contours of each component matrix:

Nv = 50;
v = linspace(0,1,Nv);
[Km,Lm] = ndgrid(v*10, v);
Z = calibm(Km,Lm);
figure
contour(Km, Lm, Z(1:Nv,:), [0 0], '-r')
hold on
contour(Km, Lm, Z((1:Nv)+Nv,:), [0 0], '-g')
contour(Km, Lm, Z((1:Nv)+2*Nv,:), [0 0], '-b')
hold off
grid on
xlabel('K')
ylabel('L')
legend('F(1)', 'F(2)', 'F(3)', 'Location','SW')

you will see that at no one point do all of the zero-contours have the same ‘K’ and ‘L’ values. Your system simply has no solution, at least in this region. If you do a global search, you may be able to find a solution. I leave that to you to experiment with.

(The code for the plots is not efficient. However it is a ‘one-off’ designed to demonstrate the problem, and is not necessary for the optimization.)

Star Strider 2018-6-1

Not stupid at all. The ga (link) function is the genetic algorithm implementation in the Global Optimization Toolbox.

Walter Roberson 2018-6-1

You appear to be trying to match experimental data to theoretical equations as if each entry were exact and the model was perfect. You should instead be looking for parameters that minimize the overall error. For that task, I recommend the Curve Fitting Toolbox.

请先登录，再进行评论。