Can this nearly-symmetric, dim-0, degree-2 polynomial system be linearly solved, or transformed to quadratic form?

Question

A common problem from chemical thermodynamics is to determine the vector $x_i$ with dimension $N$ that solves

$$n_i-x_i = s_i \frac{x_i}{\sum x_i}$$

for some constant vectors $n_i$ & $s_i$, subject to the constraints $\sum n_{i}=1$, $1>n_i>x_i>0$, and $s_i>0$, and (to exclude most degenerate cases) $s_i≠1$. When it exists, $x_i$ is unique.

The polynomial system is

$$ 0 = s_i x_i - n_i (\sum_j x_j) + x_i (\sum_j x_j)$$

With $u_i=1$ and diagonal matrix $S_{ii}=s_i$, the problem is

$$\mathbf n = \left(I\, + \, \left(\frac 1 {\mathbf u\cdot \mathbf x}\right)S\right) \cdot \mathbf x.$$

Is there an analytic expression for $\mathbf x$? Can the problem be transformed for quadratic programming, or even quadratically constrained programming? This system is very specifically constrained, and a related, more complicated problem has a partial solution.

Answer 1

Again a partial answer:

Claim. The constrained system has a solution if and only if the polynomial $$ P(t) = \sum_{i=1}^Nn_i\prod_{k\neq i}(s_k+t) - \prod_{k=1}^N(s_k+t) $$ has a zero in $(0,1)$. If $t$ is such a zero, then $x=(x_i)_{i=1}^N$ with $x_i = \frac{tn_i}{s_i+t}$ is a solution to the system.

Proof. As one easily can see, the polynomial equation $P(t) = 0$ transforms into the rational equation $$ \sum_{i=1}^N\frac{n_i}{s_i+t} = 1. $$ Assume that this equation has a solution $t\in (0,1)$ and set $x_i$ as in the claim. Then, surely, $0 < x_i < n_i$, and $\sum x_i = t$. Also $$ n_i-x_i = \left(1-\frac t{s_i+t}\right)n_i = \frac{s_in_i}{s_i+t} = s_i\frac{x_i}t, $$ which shows that $x$ is indeed a solution to the system. Conversely, let $x=(x_i)_{i=1}^N$ be a solution to the system. Set $t := \sum x_i$. Since $\sum x_i < \sum n_i=1$, we have $t\in (0,1)$. From $n_i-x_i = t^{-1}s_ix_i$ we obtain $tn_i-tx_i = s_ix_i$ and thus $x_i = \frac{tn_i}{s_i+t}$. So, $$ \sum_{i=1}^N\frac{n_i}{s_i+t} = \frac 1t\sum_{i=1}^N\frac{tn_i}{s_i+t} = 1, $$ which means that $t$ solves the rational equation and is thus a zero of $P$.

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And again, $x$ must be an eigenvector corresponding to a positive eigenvalue (namely, $\sum x_i$) of $\mathbf n\mathbf u^T-S$.