Cómo resolver los seis elementos siguientes ecuaciones?

Question

Cómo obtener el exacto o soluciones numéricas de los seis elementos siguientes ecuaciones? $$\begin{cases} \frac{c_1}{1-x_1}+\frac{c_2}{1-x_2}+\frac{c_3}{1-x_3}=0\\ \frac{c_1}{1-x_4}+\frac{c_2}{1-x_5}+\frac{c_3}{1-x_6}=0\\ c_1\ln{\frac{x_1}{1-x_1}}+c_2\ln{\frac{x_2}{1-x_2}} +c_3\ln{\frac{x_3}{1-x_3}}=0\\ c_1\ln{\frac{x_4}{1-x_4}}+c_2\ln{\frac{x_5}{1-x_5}} +c_3\ln{\frac{x_6}{1-x_6}}=0\\ \frac{x_1(1-x_1)}{x_4(1-x_4)}=\frac{x_2(1-x_2)}{x_5(1-x_5)}=\frac{x_3(1-x_3)}{x_6(1-x_6)} \end{casos}$$ donde $ c_1,c_2,c_3 $ son constantes, y $x_i\in(0,1), i=1,2,3,4,5,6$. El uso de $y_i=1-x_i$ en lugar de $x_i$ hace que las ecuaciones un poco más conciso $$\begin{cases} \frac{c_1}{y_1}+\frac{c_2}{y_2}+\frac{c_3}{y_3}=0\\ \frac{c_1}{y_4}+\frac{c_2}{y_5}+\frac{c_3}{y_6}=0\\ c_1\ln{\frac{1-y_1}{y_1}}+c_2\ln{\frac{1-y_2}{y_2}} +c_3\ln{\frac{1-y_3}{y_3}}=0\\ c_1\ln{\frac{1-y_4}{y_4}}+c_2\ln{\frac{1-y_5}{y_5}} +c_3\ln{\frac{1-y_6}{y_6}}=0\\ \frac{y_1(1-y_1)}{y_4(1-y_4)}=\frac{y_2(1-y_2)}{y_5(1-y_5)}=\frac{y_3(1-y_3)}{y_6(1-y_6)} \end{casos}$$

Answer 1

Edit 3: Ya sabemos que $x_1 = x_4$ $x_2 = x_5$ y $x_3$ $x_6$ son superfluas, creo que la solución correcta para que el sistema anterior es el conjunto de $(x_1,x_2)$ pares relacionados por

$$ \beta = {{c_1} \over {1-x_1}} + {{c_2} \over {1-x_2}} \\ x_3 = 1 + { {c_3} \over {\beta} } \\ \Rightarrow c_1 \ln \left( { {x_1} \over {1-x_1} } \right) + c_2 \ln \left( { {x_2} \over {1-x_2} } \right) = -c_3 \ln \left( {{-\beta} \over {c_3} } - 1 \right). $$

Since $\ln$ is nonlinear, I don't think the above can be simplified meaningfully.

Edit 2: So I updated the cost function to only use $x_1$, $x_2$, $x_4$, and $x_5$ since both $x_3$ and $x_6$ can be expressed in terms of them. I still get a different solution that depends on the starting point (I choose the four starting points using a random number generator). However, the solution always has the relations that $x_1 = x_4$ and $x_2 = x_5$. This suggests to me that there are still two superfluous relations in the system and I'd guess that the the first two equations can be combined with the last two to remove another two variables so that problem reduces to finding $x_1$ and $x_2$. Until then, I'd say that the problem is currently poorly formulated.

Edit 1: So the below will only get you one particular solution. If I change the initial guess to [0.7; 0.8; rand(1); 0.7; 0.8; rand(1)], I get a different solution each time.

Original Answer:

Depending on the values of $c_1$, $c_2$, $c_3$, I can get solutions using the following Nelder-Mead simplex search in Matlab. Note that there is one difference with the equations given above: I assumed that $\ln ( x_6 / (1-x_3) )$ was a typo and that it should be $\ln ( x_6 / (1-x_6) )$, as that fits the pattern of the other equations.

Here is the cost function I used with some arbitrary values chosen for the constants:

function f = costfun(x)

c1 = 100;
c2 = -54;
c3 = 0.354;

if( any( x >= 1 ) || any( x <= 0 ) )
  f = inf;
  return;
end

f1 = ( c1 / ( 1 - x(1) ) + c2 / ( 1 - x(2) ) + c3 / ( 1 - x(3) ) ).^2;
f2 = ( c1 / ( 1 - x(4) ) + c2 / ( 1 - x(5) ) + c3 / ( 1 - x(6) ) ).^2;

f3 = ( c1 * log( x(1) / ( 1 - x(1) ) ) + ...
  c2 * log( x(2) / ( 1 - x(2) ) ) + c3 * log( x(3) / ( 1 - x(3) ) ) ).^2;
f4 = ( c1 * log( x(4) / ( 1 - x(4) ) ) + ...
  c2 * log( x(5) / ( 1 - x(5) ) ) + c3 * log( x(6) / ( 1 - x(6) ) ) ).^2;

f5 = ( x(1)*(1-x(1)) / x(4)/(1-x(4)) - x(2)*(1-x(2)) / x(5)/(1-x(5)) ).^2;
f6 = ( x(1)*(1-x(1)) / x(4)/(1-x(4)) - x(3)*(1-x(3)) / x(6)/(1-x(6)) ).^2;
f7 = ( x(2)*(1-x(2)) / x(5)/(1-x(5)) - x(3)*(1-x(3)) / x(6)/(1-x(6)) ).^2;

f = f1 + f2 + f3 + f4 + f5 + f6 + f7;

return;

I then used the following to estimate the solution:

[x, costval, exitflag] = fminsearch( @costfun, 0.5 * ones(6,1) );

For the above values of $c_1$, $c_2$, and $c_3$, I got the following estimate for $\mathbf{x}$ with a total squared error of $3.3112 \cdot 10^{-6}$:

x =

    0.7176
    0.8477
    0.1655
    0.7176
    0.8477
    0.1656

El siguiente gráfico muestra el error cuadrático para cada iteración de la Nelder-Mead búsqueda: