Question

Let $\Phi: R^n \to R^n$ satisfy

$\Phi(x)=u+Ax+Q(x)$, with $x=(x_1, x_2,\ldots, x_n) \in R^n$. $u$ is a given positive vector, $A$ non negative matrix, and $Q(x)$ quadratic mapping with

$Q(x)_i=x_i(k_{i1}x_1+k_{i2}x_2+\ldots+k_{in}x_n)$, where all the $k_{ij}$ are nonnegative and at least one $k_{ij}, 1 \leq i, j \leq n $ is positive.

Suppose $\Phi(\mathbf{1})=\mathbf{1}$, where $\mathbf{1}$ is the vector each entry being 1.

How can I prove that there cannot be two distinct vectors u, v such that u, v are different from the vector $\mathbf{1}$ and $\Phi(v)=v, \Phi(u)=u$,

$v, u$ are vectors with each entry positive and no greater than 1.

Answer 1

The property does not hold as stated. Consider the case $u=(\frac{1}{2},\dots,\frac{1}{2}), A=\frac{1}{4}I,$ and $Q(x)_i = \frac{1}{4}x_i^2$. Then $\Phi$ maps $(x_1,\dots,x_n)$ to $\frac{1}{4}(x_1^2+x_1+2,\dots,x_n^2+x_n+2)$, and all $x$ with $x_i\in{1,2}$ are fixed points.

Answer 2

Are you sure that this holds in general? I have studied a very similar problem a few years ago (paper, arxiv), and there you can have up to $2^n$ solutions and you need a minimality property to characterize the physically meaningful solution. Maybe you can get something useful out of the paper.