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.

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Sumera Yamin
Sumera Yamin 2018-5-25
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);

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Star Strider
Star Strider 2018-5-25
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.
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Walter Roberson
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.

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