Defining variables for fsolve

Question

Alex 2012-2-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/30184-defining-variables-for-fsolve

Hey all, quick question:

I've been trying to pass along data to an fsolve command all morning, but I keep running into errors every time I run the program. I've been looking up information on resolving these errors all morning, and I've had some success with fixing the issues, only to have new ones replace them. I can tell from the types of errors that I am getting that the error stems in the values that I'm using as variables in the equations in fsolve, but I don't know how to remedy the issue. I've read all the literature I could find, and I've modeled the program after several examples, but to no avail. Someone with more experience than me should be able to spot the problem easily enough. Can I get a recommendation? Here's the code:

% This program aims to back-solve for reflectance (R) using given values
% input by hand.
syms x y z
y_1  = 309;
n0_1 = 1;
%n1_1 = 1.580086;
n1_1 = x;
%k1_1 = 0;
k1_1 = y;
n2_1 = 5.07;
k2_1 = 3.62;
%d1_1 = 25;
d1_1 = z;
R_1  = .4335;
y_2  = 310;
n0_2 = 1;
%n1_2 = 1.579925;
n1_2 = x;
%k1_2 = 0;
k1_2 = y;
n2_2 = 5.07;
k2_2 = 3.56;
%d1_2 = 25;
d1_2 = z;
R_2  = .4294;
y_3  = 311;
n0_3 = 1;
%n1_3 = 1.579764;
n1_3 = x;
%k1_3 = 0;
k1_3 = y;
n2_3 = 5.08;
k2_3 = 3.53;
%d1_3 = 25;
d1_3 = z;
R_3  = .4277;
g1_1 = (n0_1.^2 - n1_1.^2 - k1_1.^2)./((n1_1 + n2_1).^2 + k1_1.^2);
g2_1 = (n1_1.^2 - n2_1.^2 + k1_1.^2 - k2_1.^2)./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
h1_1 = (2.*n0_1.*k1_1)./((n0_1 + n1_1).^2 + k1_1.^2);
h2_1 = (2.*(n1_1.*k2_1 - n2_1.*k1_1))./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
a_1  = (2.*pi().*k1_1.*d1_1)./y_1;
b_1  = (2.*pi().*n1_1.*d1_1)./y_1;
A_1  = 2.*(g1_1.*g2_1 + h1_1.*h2_1);
B_1  = 2.*(g1_1.*h2_1 - g2_1.*h1_1);
C_1  = 2.*(g1_1.*g2_1 - h1_1.*h2_1);
D_1  = 2.*(g1_1.*h2_1 + g2_1.*h1_1);
g1_2 = (n0_2.^2 - n1_2.^2 - k1_2.^2)./((n1_2 + n2_2).^2 + k1_2.^2);
g2_2 = (n1_2.^2 - n2_2.^2 + k1_2.^2 - k2_2.^2)./((n1_2 + n2_2).^2 + (k1_2 + k2_2).^2);
h1_2 = (2.*n0_2.*k1_2)./((n0_2 + n1_2).^2 + k1_2.^2);
h2_2 = (2.*(n1_2.*k2_2 - n2_2.*k1_2))./((n1_2 + n2_2).^2 + (k1_2 + k2_2).^2);
a_2  = (2.*pi().*k1_2.*d1_2)./y_2;
b_2  = (2.*pi().*n1_2.*d1_2)./y_2;
A_2  = 2.*(g1_2.*g2_2 + h1_2.*h2_2);
B_2  = 2.*(g1_2.*h2_2 - g2_2.*h1_2);
C_2  = 2.*(g1_2.*g2_2 - h1_2.*h2_2);
D_2  = 2.*(g1_2.*h2_2 + g2_2.*h1_2);
g1_3 = (n0_3.^2 - n1_3.^2 - k1_3.^2)./((n1_3 + n2_3).^2 + k1_3.^2);
g2_3 = (n1_3.^2 - n2_3.^2 + k1_3.^2 - k2_3.^2)./((n1_3 + n2_3).^2 + (k1_3 + k2_3).^2);
h1_3 = (2.*n0_3.*k1_3)./((n0_3 + n1_3).^2 + k1_3.^2);
h2_3 = (2.*(n1_3.*k2_3 - n2_3.*k1_3))./((n1_3 + n2_3).^2 + (k1_3 + k2_3).^2);
a_3  = (2.*pi().*k1_3.*d1_3)./y_3;
b_3  = (2.*pi().*n1_3.*d1_3)./y_3;
A_3  = 2.*(g1_3.*g2_3 + h1_3.*h2_3);
B_3  = 2.*(g1_3.*h2_3 - g2_3.*h1_3);
C_3  = 2.*(g1_3.*g2_3 - h1_3.*h2_3);
D_3  = 2.*(g1_3.*h2_3 + g2_3.*h1_3);
Guess = [1.58,0,25];
F  = @(x,y,z) [((g1_1.^2 + h1_1.^2).*(exp(2.*a_1)) + (g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + A_1.*(cos(2.*b_1)) + B_1.*(sin(2.*b_1)))./((exp(2.*a_1)) + (g1_1.^2 + h1_1.^2).*(g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + C_1.*(cos(2.*b_1)) + D_1.*(sin(2.*b_1))) - R_1; ...
           ((g1_2.^2 + h1_2.^2).*(exp(2.*a_2)) + (g2_2.^2 + h2_2.^2).*(exp(-2.*a_2)) + A_2.*(cos(2.*b_2)) + B_2.*(sin(2.*b_2)))./((exp(2.*a_2)) + (g1_2.^2 + h1_2.^2).*(g2_2.^2 + h2_2.^2).*(exp(-2.*a_2)) + C_2.*(cos(2.*b_2)) + D_2.*(sin(2.*b_2))) - R_2; ...
           ((g1_3.^2 + h1_3.^2).*(exp(2.*a_3)) + (g2_3.^2 + h2_3.^2).*(exp(-2.*a_3)) + A_3.*(cos(2.*b_3)) + B_3.*(sin(2.*b_3)))./((exp(2.*a_3)) + (g1_3.^2 + h1_3.^2).*(g2_3.^2 + h2_3.^2).*(exp(-2.*a_3)) + C_3.*(cos(2.*b_3)) + D_3.*(sin(2.*b_3))) - R_3];
nkd = fsolve(F, Guess);

Thanks everyone!

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Answer 1

Walter Roberson 2012-2-24

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https://ww2.mathworks.cn/matlabcentral/answers/30184-defining-variables-for-fsolve#answer_38639

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Your Guess has 2 elements, suggesting you are wanting to work with two variables, but your F is @(x,y,z) suggesting you want to work with three variables.

fsolve() is going to pass one variable to the function handle. That one variable will be a vector the same length as Guess.

It appears to me that your Guess should have three elements, and it appears to me that your function handle should be:

F  = @(X) double( subs( [((g1_1.^2 + h1_1.^2).*(exp(2.*a_1)) + (g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + A_1.*(cos(2.*b_1)) + B_1.*(sin(2.*b_1)))./((exp(2.*a_1)) + (g1_1.^2 + h1_1.^2).*(g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + C_1.*(cos(2.*b_1)) + D_1.*(sin(2.*b_1))) - R_1; ...
       ((g1_2.^2 + h1_2.^2).*(exp(2.*a_2)) + (g2_2.^2 + h2_2.^2).*(exp(-2.*a_2)) + A_2.*(cos(2.*b_2)) + B_2.*(sin(2.*b_2)))./((exp(2.*a_2)) + (g1_2.^2 + h1_2.^2).*(g2_2.^2 + h2_2.^2).*(exp(-2.*a_2)) + C_2.*(cos(2.*b_2)) + D_2.*(sin(2.*b_2))) - R_2; ...
       ((g1_3.^2 + h1_3.^2).*(exp(2.*a_3)) + (g2_3.^2 + h2_3.^2).*(exp(-2.*a_3)) + A_3.*(cos(2.*b_3)) + B_3.*(sin(2.*b_3)))./((exp(2.*a_3)) + (g1_3.^2 + h1_3.^2).*(g2_3.^2 + h2_3.^2).*(exp(-2.*a_3)) + C_3.*(cos(2.*b_3)) + D_3.*(sin(2.*b_3))) - R_3], {x y z}, {X(1) X(2) X(3)}) );

It would not be typical to use symbolic expressions with fsolve() !

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Alex 2012-2-24

My guess has three elements, 1.58, 0, and 25. Does zero not count? One of the solutions should be zero, so how do I define that? Also, what alternative would you suggest to symbolic variables? I've been using them because the actual equations are very long, so using this substitution method makes them easier to work with.