I wonder whether the system of equations and inequations below have a solution. If there are solutions, what are they? A numerical solution is also desired. $$\begin{cases} \frac{c_1}{1-x_1}+\frac{c_2}{1-x_2}+\frac{c_3}{1-x_3}=0\\ \frac{c_1}{1-x_4}+\frac{c_2}{1-x_5}+\frac{c_3}{1-x_6}=0\\ c_1\ln{\frac{x_1}{1-x_1}}+c_2\ln{\frac{x_2}{1-x_2}} +c_3\ln{\frac{x_3}{1-x_3}}=0\\ c_1\ln{\frac{x_4}{1-x_4}}+c_2\ln{\frac{x_5}{1-x_5}} +c_3\ln{\frac{x_6}{1-x_6}}=0\\ \frac{x_1(1-x_1)}{x_4(1-x_4)}=\frac{x_2(1-x_2)}{x_5(1-x_5)}=\frac{x_3(1-x_3)}{x_6(1-x_6)}>1 \end{cases}$$ where $ c_1,c_2,c_3 $ are constants, and $x_i\in(0,1), i=1,2,3,4,5,6$.


A simple set of solutions are any $(c_{1},c_{2},c_{3})$ such that $c_{1}+c_{2}+c_{3}=0$, $x_{1}=x_{2}=x_{3}=\frac{1}{2}$, $x_{4}=x_{5}=x_{6}<\frac{1}{2}$.

  • $\begingroup$ Thanks for your answer. Now I know when $c_1+c_2+c_3=0$, there are solutions. What if $c_1+c_2+c_3!=0$? $\endgroup$ Feb 8 '14 at 2:22
  • $\begingroup$ @YangzheLau: That is the Mathematica command. The Latex is \neq to get, $c_1+c_2+c_3 \neq 0$. $\endgroup$ Feb 15 '18 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.