I have solved an optimization problem, end up with the following optimality equations $$\sum_{j=1}^J \frac{a_{j,n} x_n}{\sum_{i=1}^N a_{j,i} x_i}=\sum_{j=1}b_{j,n}, n=1\dots N $$ Here $\{x_n\}_{n=1}^N$ is the optimizer. $a$ and $b$ are values such that $ 0<a<1,0<b<1, \sum_n a_{j,n}=1$ and $\sum_n b_{j,n}=1$ for all j.
It looks like $\{x_n\}_{n=1}^N$ is not determined, because when I sum up both sides over $n$, I got $J=J$. However, the ratio $\frac{x_m}{x_n}$ should have a fixed value for any given $m,n$, but I could not get its closed form. Any help for finding this $\frac{x_m}{x_n}$ closed form will be appreciated!
Firstly, let $$c_{ji}=\dfrac{a_{ji}}{\sum\limits_{k=1}^Na_{jk}},\quad s_n=\sum\limits_{j=1}^J b_{jn},\quad y_n=x_n-x^{\text{prev}}_{n},\tag1$$ then $$\sum\limits_{i=1}^N c_{ji}=1,\quad \sum\limits_{j=1}^J\dfrac{c_{jn}x_n}{\sum\limits_{i=1}^{N}c_{ji}x_i} =s_n,\qquad (n=1\dots N),\tag2$$ wherein $$s_n=\sum\limits_{j=1}^J\dfrac{c_{jn}(x^{\text{prev}}_{n}+y_n)}{\sum\limits_{i=1}^{N}c_{ji}x^{\text{prev}}_{j}+\sum\limits_{i=1}^{N}c_{ji}y_i},$$ $$s_n=\sum\limits_{j=1}^J c_{jn}p_j x^{\text{prev}}_{n}+\sum\limits_{j=1}^J\,c_{jn}p_j y_n,\quad\text{where}\quad p_j=\dfrac1{\sum\limits_{i=1}^{N}c_{ji}x^{\text{prev}}_{n}+\sum\limits_{i=1}^{N}c_{ji}y_i}.\tag3$$
Suppose $\,|y_n|\ll|x_n|,\;$ then from $(3)$ should the iteration optimization scheme in the form of $$x_n^{(0)}=1,\quad p^{(0)}_j=1,\tag4$$ $$\sum\limits_{j=1}^{J}c_{ji}p^{(k)}_j y^{(k+1)}_n=s_n-\sum\limits_{j=1}^J c_{ji}p^{(k)}_j x_n^{(k)}, \quad x^{(k+1)}_n= x^{(k)}_n+y^{(k+1)}_n,\quad p_j^{(k+1)}=\dfrac1{\sum\limits_{j=1}^{N}c_{ji}x_i^{(k+1)}}.\tag5$$ Extraction of the greatest part of unknowns $\,x_n\,$ should improve convergency of the linear system and the iterations at all. So using the weights factor $\,p_j\,$ from the previous iteration looks correct, and this simplify optimization procedure. Also, the previous summation ($s_n\,$ instead $\,b_{jn}\,$) looks as evident step.