Why does solving my equation result in the message "No complex subexpressions allowed in real mode."?

4 次查看（过去 30 天）
Goncalo Costa
Goncalo Costa 2023-9-12
评论： Torsten 2023-9-13
I have an equation with a bunch of complex values in it, and with a variable in it that experimentally must be a real number. I am trying to find this real variable, p_w.
The code is as follows:
%loading the constants
eps_w = load('eps_w.mat');
eps_w = eps_w.eps_w; %complex number
eps_l = load('eps_lat.mat');
eps_l = eps_l.eps_lat; %complex number
eps_exp = load('eps_exp.mat');
eps_exp = eps_exp.refind_eps; %row of 225 complex numbers
%I am trying to solve the equation below and I know that p_w is real and
%equal or greater than 0.
for x = 1:length(eps_exp)
clear p_w
syms p_w
equation = eps_exp(x) - (p_w.*eps_w.^(1.3) + (1-p_w).*eps_l.^(1/3)).^3 == 0 ;
assume(p_w > 0)
p_water(x,:)= solve(equation, p_w, 'Real' , true)
end
Whenever I run this I get the following message:
"Error using mupadengine/feval (line 195): No complex subexpressions allowed in real mode.
Error in solve (line 293) : sol = eng.feval('solve', eqns, vars, solveOptions);
Error in (...) (line 20): p_water(x,:)= solve(equation, p_w, 'Real', true)"
Does this simply mean that there are no possible real solutions to the equation? From my pov there should be, as can be seen from the easier-to-read version of the equation, below. All epsilon values are complex, so there should be a value for p_w using real numbers.
If I am correct, then why am I getting the error message?
Previous attempts:
I have tried to simply manipulate the equation, as shown in one of the answers below, but I only get complex values when I do so.

请先登录，再回答此问题。

回答（2 个）

Torsten
Torsten 2023-9-12
编辑：Torsten 2023-9-12
Why don't you simply solve for pw ?
pw = (eps_exp^(1/3) - eps_l^(1/3))/(eps_w^(1/3)-eps_l^(1/3))
Note that you use eps_w^(1.3) in your code, not eps_w^(1/3).
  3 个评论
显示 1更早的评论
Goncalo Costa
Goncalo Costa 2023-9-13
I have data from a measurement, , and I have a model that I am trying to fit in it, the model equation is .
I have quantities (which are attached above) for and I am trying to find which values of could make .
The way I thought of doing this was by basically equalling one to the other, as shown below, and try to get a the value for that will get me the subtraction closest to 0 (or exactly 0 if possible).
I believe that could be a real number, because it is solely a factor and all other terms on the left and right side of the equation are complex numbers.

请先登录，再进行评论。

Torsten
Torsten 2023-9-13
编辑：Torsten 2023-9-13
Ran in:
Next time, please report the real problem right at the beginning.
%loading the constants
eps_w = load('eps_w.mat');
eps_w = eps_w.eps_w; %complex number
eps_l = load('eps_lat.mat');
eps_l = eps_l.eps_lat; %complex number
eps_exp = load('eps_exp.mat');
eps_exp = eps_exp.refind_eps; %row of 225 complex numbers
eps_exp = eps_exp.';
A = [real(eps_w.^(1/3)-eps_l.^(1/3))*ones(size(eps_exp));imag(eps_w.^(1/3)-eps_l.^(1/3))*ones(size(eps_exp))];
b = [real(eps_exp.^(1/3)-eps_l.^(1/3));imag(eps_exp.^(1/3)-eps_l.^(1/3))];
pw = A\b
pw = 0.0771
  4 个评论
显示 2更早的评论
Torsten
Torsten 2023-9-13
pw - determined as A\b - minimizes the expression
sum_{i=1}^{i=225} (pw*real(eps_w.^(1/3)-eps_l.^(1/3))-real(eps_exp(i).^(1/3)-eps_l.^(1/3)))^2 +
sum_{i=1}^{i=225} (pw*imag(eps_w.^(1/3)-eps_l.^(1/3))-imag(eps_exp(i).^(1/3)-eps_l.^(1/3)))^2

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Formula Manipulation and Simplification 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by