In a stable steady state at most one $x_i$ takes the "lower" solution

For convenience, let $\beta_i=1/r_i$. The Jacobian matrix of the system defined by equation 1, at a stationary point $x$ in which none of the species has $x_i=0$, is given by

$$J_{i,j}(x)=-P_i + \delta_{i,j} Q_i, i,j=1..n$$

where $P_i=2 d_i \alpha D \beta_i$, and $Q_i=d_i \left( 1 -2 \beta_i x_i \right)$. We assume that $\alpha>0$ and note that $P_i>0$. Eigenvalues $\lambda$ of $J$ are solutions to the equation $\det\left( J - \lambda I_n \right) =0$.

$$\det\left( J - \lambda I_n \right) = \begin{vmatrix}Q_1 - P_1 - \... ...&&\\ \vdots &&&&\\ -P_n &\cdots&\cdots&-P_n&Q_n - P_n - \lambda \end{vmatrix}$$

$$\det\left( J - \lambda I_n \right) = \left( \Pi_{i=1}^n P_i \righ... ...1\\ -1&\cdots&\cdots&-1&\frac{Q_n}{P_n} -1 -\frac{ \lambda}{P_n} \end{vmatrix}$$

$$\det\left( J - \lambda I_n \right) =\left( \Pi_{i=1}^n P_i \right... ...ac{\lambda}{P_1}&0&\cdots&0&\frac{Q_n}{P_n} -\frac{ \lambda}{P_n} \end{vmatrix}$$

With $A=1-\frac{1}{\frac{Q_1}{P_1} - \frac{\lambda}{P_1} }$,

$$\det\left( J - \lambda I_n \right) = \left( \Pi_{i=1}^n P_i \righ... ...s&\ddots&0\\ -1&0&\cdots&0&\frac{Q_n}{P_n} -\frac{ \lambda}{P_n} \end{vmatrix}$$

With $B_i=\frac{Q_i}{P_i} - \frac{\lambda}{P_i}$,

$$\det\left( J - \lambda I_n \right) = \left( \Pi_{i=1}^n P_i \righ... ...ots&\vdots\\ \vdots &\vdots&\ddots&\ddots&0\\ -1&0&\cdots&0&B_n \end{vmatrix}$$

For $n\ge 2$, let

$$L_n= \begin{vmatrix} A & -1 & \cdots & \cdots&-1\\ -1 & B_2 & 0&... ...1&0&B_3&\cdots&0\\ \vdots&0&\cdots&\ddots&0\\ -1&0&\cdots&0&B_n \end{vmatrix}$$

By developing with respect to the last column, $L_n=B_n L_{n-1} - \left(-1\right) ^ {n-1}C_{n-1}$, where

$$C_{n-1}= \begin{vmatrix} -1&\cdots&\cdots&\cdots&-1\\ B_2&0&\cdo... ...\cdots&0\\ \vdots&\ddots&\ddots&\ddots&0\\ 0&\cdots&0&B_{n-1}&0 \end{vmatrix}$$

By developing with respect to the last row, $C_{n-1}=-B_{n-1}C_{n-2}$ for $n\ge 4$. Since $C_2=\begin{vmatrix}-1 & -1 B_2&0 \end{vmatrix}=B_2$, by induction $C_n=\left(-1\right)^n\Pi_{i=2}^n B_i$.

Therefore, $L_n=B_n L_{n-1}-\Pi_{i=2}^{n-1} B_i$ for $n \ge 3$. Since $L_2=\begin{vmatrix}A & -1 -1 & B_2 \end{vmatrix}=A B_2 -1$, it can be shown by induction that

$$L_n=A \Pi_{i=2}^n B_i - \Sigma_{i=2}^n \Pi_{j=2,j\ne i}^n B_j, \mathrm{for} n \ge 2$$

Since $A B_1 = B_1 -1$,

$$B_1 L_n=\left(B_1-1\right)\Pi_{i=2}^n B_i - B_1 \Sigma_{i=2}^n \Pi_{j=2,j\ne i}^n B_j = \Pi_{j=1}^n B_j - \Sigma_{i=1}^n \Pi_{j=1,j\ne i}^n B_j,$$

for $n\ge 2$.

Thus, $\det\left( J - \lambda I_n \right) = \left( \Pi_{i=1}^n P_i \right) \left(\Pi_{j=1}^n B_j - \Sigma_{i=1}^n \Pi_{j=1,j\ne i}^n B_j \right)$, and any eigenvalue $\lambda$ of $J$ satisfies

$$P(\lambda)= \Sigma_{i=1}^n \Pi_{j=1,j\ne i}^n B_j - \Pi_{j=1}^n B_j =0,$$ (4)

with $B_i=\frac{Q_i}{P_i} - \frac{\lambda}{P_i}$. Suppose without loss of generality that $Q_n$ and $Q_{n-1}$ are respectively the largest and second-largest $Q_i$. Suppose in addition that these largest values are unique (the case where they are not will be dealt with below). Then $P(Q_n)$ has the same sign as $(-1)^{n-1}$, and $P(Q_{n-1})$ has the same sign as $(-1)^n$. Thus, $\exists t \in ] Q_{n-1} , Q_n[ \mathrm{s.t.} P(t)=0$. Therefore, at any stable steady-state, $Q_{n-1}<0$, and therefore $\forall i \ne n, Q_i<0$, meaning

$$d_i \left( 1 -2 \beta_i x_i \right)<0$$

$$\forall i<n, x_i>\frac{1}{2\beta_i}=\frac{r_i}{2}$$ (5)

Therefore, at any stable steady-state, any $x_i$ with $i<n$ is at the higher solution of equation 2, and

$$\forall i<n, x_{i}=\frac{r_i + \sqrt{r_{i}^2-4\alpha D^2}}{2}$$

If $Q_n$ or $Q_{n-1}$ are not unique in the re-numbering scheme discussed above, then the nonunique value is a root of P and hence cannot be positive. Therefore at most one $Q_i$ can be positive and it is possible to renumber for a non-strict version of inequality 5 to hold. Strictness follows follows since $\alpha\ne0$.