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

$$\frac{n_i-x_i}{1-\sum x_i} = k_i \frac{x_i}{\sum x_i}$$

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

With $s_i=1$ and diagonal matrix $K_{ii}=k_i$, the problem is

$$\mathbf n = \left(I\, + \, \left(\frac 1 {\mathbf s\cdot \mathbf x} - 1\right)K\right) \cdot \mathbf x.$$

Is there a closed-form expression for $x_i$? Can the problem be transformed into standard form for quadratic programming?


This corresponds to solving with "volatilities" in chemical language, but I suspect the general case of "saturation pressures" actually has a simpler solution.


This is only a partial answer. Set $A := ns^T$. Multiplying your equation with $s^Tx = \sum x_i$ gives $$ (ns^T)x = Kx + s^Tx(I-K)x, $$ hence $$ (I-K)^{-1}(A-K)x = (\sum x_i)x. $$ Thus, we have $Bx=(\sum x_i)x$, where $B := (I-K)^{-1}(A-K)$. Hence,

Claim: The original system (with constraints) has a solution if and only if $B$ has a positive eigenvalue with a corresponding positive eigenvector. In this case, if $Bu=\lambda u$ with $\lambda > 0$ and $u>0$, then $x = \lambda(\sum u_i)^{-1}u$ is a solution of the original system.

  • $\begingroup$ Thank you. I suspect the more general problem is actually more easily solved, and have asked it here. $\endgroup$ – alexchandel Oct 19 '19 at 22:32
  • $\begingroup$ Can the related problem (solving with "saturation pressures and finite liquid volumes" in chemical language) be solved similarly, $(n\,s^T - K)x = (s^Tx)x - (v^Tx) K x$ with $v_i>0$ and $0 < v^Tx < 1$? $\endgroup$ – alexchandel Jan 6 at 22:52

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