How to solve the six elements equations below?

Question

How to get the exact or numerical solutions of the six elements equations below? $$\begin{cases} \frac{c_1}{1-x_1}+\frac{c_2}{1-x_2}+\frac{c_3}{1-x_3}=0\\ \frac{c_1}{1-x_4}+\frac{c_2}{1-x_5}+\frac{c_3}{1-x_6}=0\\ c_1\ln{\frac{x_1}{1-x_1}}+c_2\ln{\frac{x_2}{1-x_2}} +c_3\ln{\frac{x_3}{1-x_3}}=0\\ c_1\ln{\frac{x_4}{1-x_4}}+c_2\ln{\frac{x_5}{1-x_5}} +c_3\ln{\frac{x_6}{1-x_6}}=0\\ \frac{x_1(1-x_1)}{x_4(1-x_4)}=\frac{x_2(1-x_2)}{x_5(1-x_5)}=\frac{x_3(1-x_3)}{x_6(1-x_6)} \end{cases}$$ where $ c_1,c_2,c_3 $ are constants, and $x_i\in(0,1), i=1,2,3,4,5,6$. Using $y_i=1-x_i$ instead of $x_i$ makes the equations a little more succinct $$\begin{cases} \frac{c_1}{y_1}+\frac{c_2}{y_2}+\frac{c_3}{y_3}=0\\ \frac{c_1}{y_4}+\frac{c_2}{y_5}+\frac{c_3}{y_6}=0\\ c_1\ln{\frac{1-y_1}{y_1}}+c_2\ln{\frac{1-y_2}{y_2}} +c_3\ln{\frac{1-y_3}{y_3}}=0\\ c_1\ln{\frac{1-y_4}{y_4}}+c_2\ln{\frac{1-y_5}{y_5}} +c_3\ln{\frac{1-y_6}{y_6}}=0\\ \frac{y_1(1-y_1)}{y_4(1-y_4)}=\frac{y_2(1-y_2)}{y_5(1-y_5)}=\frac{y_3(1-y_3)}{y_6(1-y_6)} \end{cases}$$

Answer 1

Edit 3: Since we now know that $x_1 = x_4$ and $x_2 = x_5$ and that $x_3$ and $x_6$ are superfluous, I believe that the correct solution to the above system is the set of $(x_1,x_2)$ pairs related by

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

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

The below plot shows the squared error for each iteration of the Nelder-Mead search: