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Defining variables for fsolve

2 vues (au cours des 30 derniers jours)
Alex
Alex le 24 Fév 2012
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!

Réponse acceptée

Walter Roberson
Walter Roberson le 24 Fév 2012
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() !
  2 commentaires
Alex
Alex le 24 Fév 2012
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.
Walter Roberson
Walter Roberson le 24 Fév 2012
Ah, with your spacing I read the 0,25 as 0.25 . 0 is fine as a guess, but it would be easier to read as 0, 25

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