Use fsolve with an elliptic integral

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carlos Hernando
carlos Hernando 2020-3-17
评论： carlos Hernando 2020-3-17
Hello,
I need to solve a non-linear equation with an eliptic integral in it. I tried to used fsolve but it don't want to work, any ideas. I copy the code below:
NI = 6.749914199660242e+05;
NI_ol = 1.749914199660242e+05;
Muo = 4*pi*10^-7; % (T*m/A)
a = 0.2073;
y = 0.15;
% Bos_c = Muo*(NI-NI_ol)*(a^2)/((z(1)^2 + a^2)^(1.5));
% m = (4*a*y)/((a+y)^2+z(1)^2);
%
% [K, E]= ellipke(m);
% Be = 2*Muo*NI*a*((2*m)^0.5)*(a*m*E/(2-2*m) + y*K - y*(2-m)*E/(2-2*m))/(2*pi*(2*a*y)^(1.5));
F=@(z) [Muo*(NI-NI_ol)*(a^2)/((z(1)^2 + a^2)^(1.5))...
- 2*Muo*NI*a*((2*m)^0.5)*(a*m*E/(2-2*m) + y*K - y*(2-m)*E/(2-2*m))/(2*pi*(2*a*y)^(1.5));...
m - (4*a*y)/((a+y)^2+z(1)^2);
[K, E]== ellipke(m)];
z_c = [0; 1000];
opts = optimoptions(@fsolve,'Algorithm', 'levenberg-marquardt');
neff = fsolve(F,z_c,opts);

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采纳的回答

Matt J
Matt J 2020-3-17
编辑：Matt J 2020-3-17
It does not make sense to have a relational expression like,
[K, E]== ellipke(m)
as one of your equations, expecially one that doesn't depend on any of your unknowns, z(i). Also, z(2) is not used anywhere in the system of equations. In other words, you have multiple equations in a single unknown z(1), so the system is not likely to have a solution.
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carlos Hernando
carlos Hernando 2020-3-17
Thank you Matt, unknowns z(3) and z(4) are inside equation 1, I cannot eliminated them. The code now looks like this:
NI = 6.749914199660242e+05;
NI_ol = 1.749914199660242e+05;
Muo = 4*pi*10^-7; % (T*m/A)
a = 0.2073;
y = 0.15;
F=@(z) [Muo*(NI-NI_ol)*(a^2)/((z(1)^2 + a^2)^(1.5))...
- 2*Muo*NI*a*((2*z(2))^0.5)*(a*z(2)*z(4)/(2-2*z(2)) - y*z(3) + ...
y*(2-z(2))*z(4)/(2-2*z(2)))/(2*pi*(2*a*y)^(1.5));
z(2) - (4*a*y)/((a+y)^2+z(1)^2);
z(3) - ellipticK(z(2));
z(4) - ellipticE(z(2))];
z_c = [2 0.5 1 1] ;
opts = optimoptions(@fsolve,'Algorithm', 'levenberg-marquardt');
fsolve(F,z_c,opts);
It seems to work, but is unable to find a solution, the following message appears:
Solver stopped prematurely.
fsolve stopped because it exceeded the function evaluation limit,
options.MaxFunctionEvaluations = 8.000000e+02.
Any idea how can I solve this?
carlos Hernando
carlos Hernando 2020-3-17
Ok I sort it out, just eliminating opts seems to work.
Thank you Matt

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