On heteroclinic cycles of competitive maps via carrying simplices

Abstract

We concentrate on the effects of heteroclinic cycles and the interplay of heteroclinic attractors or repellers on the boundary of the carrying simplices for low-dimensional discrete-time competitive systems. Based on the existence of the carrying simplex for the competitive mapping, we provide the criteria on stability of the heteroclinic cycle. This result can be seen as a discrete counterpart of that for the continuous-time systems. Several concrete discrete-time competition models are further analyzed, which do admit heteroclinic cycles. The criteria on the stability of the heteroclinic cycle for each model are also given, which are comparable with the corresponding continuous-time models.

Appendix: Existence of the carrying simplex

Let $W\subseteq \mathbf {R}^n$ be an open neighborhood containing $\mathbf {R}^n_+$, and let $T:W(\subseteq \mathbf {R}^n)\rightarrow \mathbf {R}^n$ be a $C^1$-map given by the form (1) so that $T\mathbf {R}^n_+\subseteq \mathbf {R}^n_+$ and $T^{-1}(\mathbf {R}^n_+)\subseteq \mathbf {R}^n_+$. For any subset $A\subseteq \mathbf {R}^n_+$, we denote by $\mathrm {Int}A$ (resp. $\partial A$) the interior (resp. boundary) of A relative to $\mathbf {R}^n_+$ for short.

We further assume the following properties of T:

$({{\varUpsilon }}\mathbf{0})$ :

$G_i(0)>1$ for any $i\in N$; and moreover, $G_i(x)>0$ whenever $x=(x_1,\ldots ,x_n)\in \mathbf {R}^n_+$ with $x_i>0$.

$({{\varUpsilon }}\mathbf{1})$ :

For each i, $T|_{\dot{H}_{\{i\}}^+}$ has a globally attracting fixed point $q_{\{i\}}=q_ie_{\{i\}}$ (called an axial fixed point), and set $q=\sum q_{\{i\}}=(q_1,\ldots ,q_n)$.

$({{\varUpsilon }}\mathbf{2})$ :

There exists a lower order convex subset S with $[0,q]\subseteq S \subseteq \mathbf {R}_+^n,$ which satisfies that $T(\mathbf {R}_+^n)\subseteq S$ and

$$\begin{aligned} Tx<_I Ty \Longrightarrow x\ll _I y,\quad \text {whenever }x\in S^+_I, y\in \dot{S}^+_I\quad \text {and}\quad \emptyset \ne I\subseteq N. \end{aligned}$$

$({{\varUpsilon }}\mathbf{3})$ :

For any $x,y\in [0,q]$ with $y=Tx$, $[0,y]\subseteq T[0,x]$.

$({{\varUpsilon }}\mathbf{4})$ :

For any $x\in [0,q]\cap H_I^+$, the matrix $[\partial G_i/\partial x_j(x)]_{i,j\in I}$ has strictly negative entries.

Our main result in the Appendix is the following:

Theorem 6

(Carrying Simplex) Assume that $({\varUpsilon }0$)–$({\varUpsilon }4)$ hold. Then $T|_{\mathbf {R}^n_+}$ admits a compact global attractor ${\varGamma }$, which satisfies the following property:

• (i) ${\varGamma }$ is lower order convex, i.e., $[0,x]\subseteq {\varGamma }$ whenever $x\in {\varGamma };$

• (ii) $\partial {\varGamma }$ is a positively invariant compact set and is homeomorphic to the $(n-1)$-dimensional standard probability simplex ${\varDelta }^{n-1}$ via radial projection; 

• (iii) $\mathrm{Int}{\varGamma }\subseteq \{y\in {\varGamma }:\alpha (y)=\{0\}$, for any negative orbit in ${\varGamma }$ through $y\};$

• (iv) For any $x\in \mathbf {R}^n_+{\setminus }\{0\}$, $\omega (x)\subseteq \partial {\varGamma }$.

Assume further that T is locally injective in an open neighborhood V of [0, q] relative to W (hence relative to $\mathbf {R}^n$). Then the above statements (i)–(iv) remain true even if $({\varUpsilon }0)$–$({\varUpsilon }3)$ and

$({{\varUpsilon }}\mathbf{4'})$ :

For any $x\in [0,q]\cap H_I^+$, the diagonal entries of $[\partial G_i/\partial x_j(x)]_{i,j\in I}$ are strictly negative, while the off-diagonal ones are non-positive

hold. Moreover, $\partial {\varGamma }$ is invariant,

$$\begin{aligned} \mathrm{Int}{\varGamma }=\{y\in {\varGamma }:\alpha (y)=\{0\},\quad \text {for any negative orbit in }{\varGamma }\text { through }y \} \end{aligned}$$

and for any $x\in \mathbf {R}^n_+{\setminus } \{0\}$, there exists some $y\in \partial {\varGamma }$ such that

$$\begin{aligned} \Vert T^kx-T^ky\Vert \rightarrow 0\quad \text { as }\,k\rightarrow +\infty . \end{aligned}$$

(17)

We will break the proof of the above theorem into several Lemmas.

Lemma 2

If conditions $({\varUpsilon }1)$ and $({\varUpsilon }2)$ hold, then $T[0,q]\subseteq [0,q]$. Moreover, for any bounded set $U\subseteq \mathbf {R}^n_+$, the positive orbit $O^+(U)$ is bounded and $\omega (U)\subseteq [0,q]$.

Proof

The proof is very similar to the arguments in Smith (1986, Lemma 3.2 and Proposition 3.4). For the sake of completeness, we provide the details here. Given any positive number $a>0$ and $i\in N$ with $ae_{\{i\}}\in S$, we claim that, for any $y=(y_1,\ldots ,y_n)\in S$, if $y_i\leqslant a$, then $T_i(y) \le T_i(ae_{\{i\}})$. Indeed, if such $y\in H_{\{i\}}^+$, then clearly $T_i(y)\le T_i(ae_{\{i\}})$ (Otherwise, one has $T_i(y)>T_i(ae_{\{i\}})$. By (${\varUpsilon }2$), we obtain that $a<y_i$, a contradiction). If such y satisfies that its ith coordinate $y_i=0$, then obviously $T_i(y)=0<T_i(ae_{\{i\}})$. As a consequence, one can now assume that such y satisfies $y_i>0$ and $y_j>0$, for some $j\ne i$. Let $I\subseteq N$ be the nonempty index set such that $y\in \dot{S}^+_I$. Then $ae_{\{i\}}\in S^+_I$. Suppose that $T_i(y)> T_i(ae_{\{i\}})$. Then it is easy to see that $Ty>_I T(ae_{\{i\}})$, so by (${\varUpsilon }$2) we have $y\gg _I ae_{\{i\}}$, and hence $a<y_i$, which contradicts $y_i\leqslant a$. Thus we have proved the claim.

Now, for any $y\in [0,q]$, one can use the claim above to obtain that $T_i(y)\leqslant q_i$ because $y_i\leqslant q_i$. So $T[0,q]\subseteq [0,q]$.

Moreover, for any bounded set $U\subseteq \mathbf {R}_+^n$, let $c_i=\sup \{(Ty)_i:y\in U\}<+\infty$ for each i. Without loss of generality, we assume that $c_i>0$ for all i. Recall that $T(\mathbf {R}_+^n)\subseteq S$ and S is lower order convex, so $c_i e_{\{i\}}\in S$. Then one can repeatedly utilize the claim above to obtain that

$$\begin{aligned} T^{k+1}_i(y)\leqslant T^{k}_i(c_i e_{\{i\}})\quad \text { for any }y\in U\quad \text {and}\quad k=0,1,\ldots . \end{aligned}$$

(18)

Recall that $T^k_i(c_i e_{\{i\}})\rightarrow q_i$ as $k\rightarrow \infty$. Then there is an integer $k_0>0$ such that $T^{k+1}_i(y)\le q_i+1$ for any $k\ge k_0$ and $y\in U$. This implies that the positive orbit $O^+(U)$ is bounded. Furthermore, by (18) and $T^k_i(c_i e_{\{i\}})\rightarrow q_i$ as $k\rightarrow \infty$ again, we obtain that $\omega (U)\subseteq [0,q]$.

Proposition 1

Let $({\varUpsilon }1)$–$({\varUpsilon }2)$ hold. Let also ${\varGamma }:=\bigcap ^{\infty }_{k=1}T^k[0,q]\subseteq [0,q]\subseteq S$. Then we have

• (i) ${\varGamma }$ is a non-empty compact invariant set which contains all the axial fixed points $q_{\{i\}}, 1\leqslant i\leqslant n$. Moreover, $\omega (U)\subseteq {\varGamma }$ for any bounded set $U\subseteq \mathbf {R}^n_+$, and ${\varGamma }$ is a global attractor for $T|_{\mathbf {R}^n_+};$

• (ii) If additionally $({\varUpsilon }3)$ holds, then ${\varGamma }$ is lower order convex, i.e., $[0,x]\subseteq {\varGamma }$ whenever $x\in {\varGamma };$

• (iii) If $({\varUpsilon }1)$–$({\varUpsilon }3)$ hold, then $\partial {\varGamma }$ is a compact positively invariant set. Furthermore, $\partial {\varGamma }$ is weakly unordered in the sense that, for any $\emptyset \ne I\subseteq N$, there do not exist two points $x_1,x_2\in \partial {\varGamma }\cap H_I^+$ such that $x_1\ll _I x_2$.

Proof

(i) By Lemma 2, $T[0,q]\subseteq [0,q]$. Since $T^k[0,q]$ is compact and $0\in {\varGamma }\ne \emptyset$, ${\varGamma }$ is a nonempty compact set. Note also that

$$\begin{aligned} T{\varGamma }=T(\cap _kT^k[0,q])\subseteq \cap _k T(T^k[0,q])={\varGamma }. \end{aligned}$$

Then ${\varGamma }$ is positively invariant under T. On the other hand, given any $x\in {\varGamma }$ and each integer $l\ge 1$, there exists $x_l\in [0,q]$, such that $x=T^lx_l=T(T^{l-1}x_l)$. Because $T^{l-1}x_l\in [0,q]$ for any l, we may assume without loss of generality that $T^{l-1}x_l\rightarrow y\in [0,q]$. Moreover, for a fixed k, $T^{l-1}x_l\in T^k[0,q]$ for any $l\ge k+2$, which implies that $y\in T^k[0,q]$. Consequently, $y\in {\varGamma }$ and $x=Ty$, that is, $T{\varGamma }={\varGamma }$. Furthermore, for any bounded $U\subseteq \mathbf {R}_+^n$, it follows from $\omega (U)\subseteq [0,q]$ (by Lemma 2) and the total invariance of $\omega (U)$ that $\omega (U)\subseteq \cap _kT^k[0,q]$, i.e., $\omega (U)\subseteq {\varGamma }$.

Since $T:\mathbf {R}_+^n\rightarrow \mathbf {R}_+^n$ is continuous and point-dissipative, it follows from Hale (1988, Theorem 2.4.6) that there is a global attractor ${\varGamma }_1$ of T in $\mathbf {R}_+^n$. We will prove that ${\varGamma }$ is the global attractor by showing that ${\varGamma }={\varGamma }_1$. In fact, for the bounded set ${\varGamma }$ obtained above, one has $\omega ({\varGamma })\subseteq {\varGamma }_1$. Recall that $T{\varGamma }={\varGamma }$. Then ${\varGamma }=\omega ({\varGamma })\subseteq {\varGamma }_1$. On the other hand, one also has $\omega ({\varGamma }_1)\subseteq {\varGamma }$ by the above result, which implies that ${\varGamma }_1=\omega ({\varGamma }_1)\subseteq {\varGamma }$. Thus, ${\varGamma }={\varGamma }_1$, which implies that ${\varGamma }$ is the global attractor for $T|_{\mathbf {R}^n_+}$.

(ii) For any $x\in {\varGamma }$ and integer $k\ge 0$, the definition of ${\varGamma }$ implies that there exists $y_k\in [0,q]$ such that $T^k y_k=x$. If additionally (${\varUpsilon }$3) holds, then

$$\begin{aligned} {[0,x]}=[0,T^k y_k]\subseteq T^k[0,y_k]\subseteq T^k[0,q], \end{aligned}$$

which implies that $[0,x]\subseteq {\varGamma }$, i.e., ${\varGamma }$ is lower order convex.

(iii) Clearly, ${\varGamma }$ is compact. We now show that $\partial {\varGamma }$ is positively invariant. Without loss of generality, we assume that the interior $\mathrm{Int}{\varGamma }\ne \emptyset$ (otherwise, $\partial {\varGamma }={\varGamma }$ is invariant). Suppose that there is an $x\in \partial {\varGamma }$ such that $Tx\in \mathrm{Int}{\varGamma }$. Then one can find some $y\in {\varGamma }\cap \dot{\mathbf {R}}^n_+$, nearby Tx, such that $Tx\ll y$. Since ${\varGamma }$ is invariant, we choose some $w\in {\varGamma }$ with $Tw=y$. By (${\varUpsilon }$2), it follows that $x\ll w$. Then the lower order convexity of ${\varGamma }$ implies that $x\in \mathrm{Int}{\varGamma }$, a contradiction. Thus, $\partial {\varGamma }$ is positively invariant.

Finally, by the lower order convexity of ${\varGamma }$, it is clear that there do not exist two points $x_1,x_2\in \partial {\varGamma }$ such that $x_1\ll x_2$. Moreover, for each $\emptyset \ne I\subseteq N$, note that the hypotheses (${\varUpsilon }$2)–(${\varUpsilon }$3) still hold with $\mathbf {R}_+^n$ and q replaced by $H_I^+$ and $q_I$, respectively. Here the i-th coordinate $(q_I)_i$ of $q_I$ satisfies that $(q_I)_i=q_i$ if $i\in I$ and $(q_I)_i=0$ otherwise. Then one can repeat the arguments above to obtain the compact invariant set ${\varGamma }_I$ satisfying (i)–(iii) in $H_I^+$. Furthermore, it is easy to see that ${\varGamma }_I={\varGamma }\cap H_I^+$ and $\partial _{H_I^+}{\varGamma }_I=\partial {\varGamma }\cap H_I^+$. In particular, there do not exist two points $x_1,x_2\in \partial {\varGamma }\cap H_I^+$ such that $x_1\ll _I x_2$.

Remark 6

By the definition of ${\varGamma }$, it is clear that $p=(p_1,\ldots ,p_n)\in {\varGamma }$ implies that $p_i\le q_i$ for any $i\in N$. Furthermore, we claim that if the coordinate $p_i=q_i$ for some i, then $p=q_{\{i\}}$. (Otherwise, let $I\subseteq N$ be the index set such that $p\in \dot{H}_I^+$. Then $p>_Iq_{\{i\}}$. Since ${\varGamma }$ is invariant, there is a $p^*\in {\varGamma }$ such that $Tp^*=p>_Iq_{\{i\}}=Tq_{\{i\}}$. By virtue of (${\varUpsilon }$2), $p^*\gg _I q_{\{i\}}$, which entails that $p^*\notin [0,q]$, a contradiction to ${\varGamma }\subseteq [0,q]$.)

By virtue of this claim, one can obtain that $\partial {\varGamma }$ is equal to the boundary $\partial _{[0,q]}{\varGamma }$ of ${\varGamma }$ relative to [0, q], that is, $\partial {\varGamma }=\partial _{[0,q]}{\varGamma }$. (Indeed, it is clear that $\partial _{[0,q]}{\varGamma }\subseteq \partial {\varGamma }$. Suppose that there is some $z\in \partial {\varGamma }{\setminus } \partial _{[0,q]}{\varGamma }$. Then $z\ne q_{\{i\}}$ for any i, because $q_{\{i\}}\in \partial _{[0,q]}{\varGamma }$. Noticing also that $z\in \partial {\varGamma }\subseteq {\varGamma }$, the claim above then yields that $z\ll q$. So any small neighborhood of z relative to $\mathbf {R}^n_+$ is also a neighborhood of z relative to [0, q]. Thus, $z\in \partial {\varGamma }$ implies that $z\in \partial _{[0,q]}{\varGamma }$, contradicting to $z\notin \partial _{[0,q]}{\varGamma }$.) Therefore, one also has $\mathrm {Int}{\varGamma }=\mathrm {Int}_{[0,q]}{\varGamma }$. As a consequence, we hereafter always write $\mathrm {Int}{\varGamma }$ and $\partial {\varGamma }$ relative to [0, q] without any confusion.

The following remark is about the assumption (${\varUpsilon }$0).

Remark 7

The first statement of (${\varUpsilon }$0) guarantees that, for any index set $\emptyset \ne I\subseteq N$, $x\ll _I Tx$ whenever $x\in \dot{H}_I^+$ is sufficiently close to 0. As a consequence, we hereafter always fix some $0\ll x_0\in [0,q]$ nearby 0, such that, $w\ll _I Tw$ whenever $w\in [0,x_0]\cap \dot{H}_I^+$ for any $I\ne \emptyset$. In particular, the origin 0 is the unique fixed point of T in $[0,x_0]$.

Meanwhile, the second statement of (${\varUpsilon }$0) implies that $T^{-1}(H_I^+)\subseteq H_I^+$ for any $I\subseteq N$. In particular, $T^{-1}(\{0\})=\{0\}$.

Lemma 3

Let $({\varUpsilon }0)$–$({\varUpsilon }3)$ hold. Then $\mathrm{Int}{\varGamma }\ne \emptyset$. Moreover, for any $x\in \dot{H}^+_I$ with $I\ne \emptyset$, $x\in \mathrm{Int}{\varGamma }$ if and only if $x\ll _Iw_x$, where $w_x\in \partial {\varGamma }$ is the (unique) point of intersection of the ray $L_{x}:=\{tx:t\ge 0\}$ with $\partial {\varGamma }$.

Proof

Let $x_0\gg 0$ be in Remark 7. Clearly, $x_0\ll Tx_0$ and $x_0\le q$. Combining with (${\varUpsilon }$3), we obtain that $[0,x_0]\subseteq [0,Tx_0]\subseteq T[0,x_0]\subseteq T[0,q]$. So,

$$\begin{aligned}{}[0,x_0]\subseteq T[0,x_0]\subseteq \cdots \subseteq T^m[0,x_0]\subseteq T^m[0,q] \end{aligned}$$

(19)

for any $m\ge 0$. Hence, $[0,x_0]\subseteq {\varGamma }$, which implies that

$$\begin{aligned} \left[ 0,\frac{x_0}{2}\right] \subseteq \mathrm{Int}{\varGamma }. \end{aligned}$$

(20)

Thus, $\mathrm{Int}{\varGamma }\ne \emptyset$.

Let $x\in \dot{H}^+_I$ with $I\ne \emptyset$. Note that ${\varGamma }$ is bounded and $tx\in [0,x_0]\subseteq {\varGamma }$ for small $t\ge 0$. Then there exists at least one point $w_x\in \partial {\varGamma }\cap L_x$. By the non-ordering property of $\partial {\varGamma }$ in Proposition 1(iii), $w_x$ is unique. If $x\in \mathrm{Int}{\varGamma }$, then it is easy to see that $x\ll _I w_x$ (otherwise, $w_x\ll _I x$. The lower order convexity of ${\varGamma }$ in Proposition 1(ii) then implies that $w_x\in \mathrm{Int}{\varGamma }$, a contradiction). On the other hand, if $x\ll _I w_x$, then we will show $x\in \mathrm{Int}{\varGamma }$. Suppose $x\notin \mathrm{Int}{\varGamma }$. Then the weakly unordered property of $\partial {\varGamma }$ implies that $x\notin {\varGamma }$, and hence $w_x\notin {\varGamma }$ by the lower order convexity of ${\varGamma }$, contradicting that $w_x\in \partial {\varGamma }$. Thus, we have proved that $x\in \mathrm{Int}{\varGamma }$ if and only if $x\ll _Iw_x$.

Lemma 4

Let $({\varUpsilon }0)$–$({\varUpsilon }3)$ hold. Let also $B_0:=\bigcup _{k\ge 0}T^k[0,\frac{x_0}{2}]$. Then

• (i) $B_0$ is an invariant and lower order convex subset of ${\varGamma };$

• (ii) $B_0= \{y\in {\varGamma }$: there exists a negative orbit in ${\varGamma }$ of y such that $\alpha (y)=\{0\} \};$

• (iii) $\partial B_0$ is compact, positively invariant and weakly unordered, i.e., for any $\emptyset \ne I\subseteq N$, there do not exist two points $x_1,x_2\in \partial B_0\cap H_I^+$ such that $x_1\ll _I x_2$.

Proof

(i) Clearly, $B_0$ is positively invariant. By Remark 7 and (${\varUpsilon }$3), one has that $[0,x_0/2]\subseteq T[0,x_0/2]\subseteq \cdots \subseteq T^m[0,x_0/2]$ for any $m\ge 1$, which implies that $B_0$ is negatively invariant, hence $B_0$ is invariant. Moreover, $B_0\subseteq {\varGamma }$ is immediate from (20) and the invariance of ${\varGamma }$. For any $y\in B_0$, there exist some $k\ge 1$, $v\in [0,\frac{x_0}{2}]$ such that $y=T^kv$. So, by (${\varUpsilon }$3), $[0,y]\subseteq [0,T^kv]\subseteq T^k[0,v]\subseteq T^k[0,\frac{x_0}{2}]\subseteq B_0,$ which implies that $B_0$ is lower order convex.

(ii) Set $M=\{y\in {\varGamma }$: there exists a negative orbit in ${\varGamma }$ of y such that $\alpha (y)=\{0\} \}$. Given any $y\in M$, let $\{y(-m),m\ge 0\}$ be a negative orbit in ${\varGamma }$ through y such that $\alpha (y)=\{0\}$. Then $y(-m)\in [0,\frac{x_0}{2}]$ for some m sufficiently large. So, $y=T^my(-m)\in T^m[0,\frac{x_0}{2}]$, which implies that $M\subseteq B_0$. On the other hand, for any $v \in [0,x_0]$ and any negative orbit $\{v(-m),m\ge 0\}$ in ${\varGamma }$ through v, it follows from (${\varUpsilon }$2) that $\{v(-m):m\ge 1\}$ is a strictly-decreasing sequence in $[0,x_0]$. By Remark 7, it entails that $v(-m)\rightarrow 0$ as $m\rightarrow \infty$ and hence, $[0,x_0]\subseteq M$. Now for any $y\in B_0$, there exist some $k\ge 1$ and $v\in [0,\frac{x_0}{2}]$ such that $T^kv=y$. So, $\{v(-m):m\ge 0\}\cup \{Tv,T^2v,\ldots ,T^{k-1}v,y\}$ forms a negative orbit in ${\varGamma }$ through y, where $\{v(-m)\}$ is a negative orbit in ${\varGamma }$ through v. Since $v(-m)\rightarrow 0$, such a negative orbit through y possesses a trivial $\alpha (y)=\{0\}$. Thus, $B_0\subseteq M$, and we have proved that $M=B_0$.

(iii) Clearly, $\overline{B_0}$ is also invariant by continuity of T and $\overline{B_0}\subseteq {\varGamma }$. We now show that $\partial B_0$ is positively invariant. Note that $\mathrm{Int}B_0\ne \emptyset$. Suppose that there is an $x\in \partial B_0$ such that $Tx\in \mathrm{Int}B_0$. Then one can find some $y\in B_0\cap \dot{\mathbf {R}}^n_+$ so close to Tx that $Tx\ll y$. Since $B_0$ is invariant, we choose some $w\in B_0$ with $Tw=y$. Then ${\varUpsilon }$2) implies that $x\ll w$. Hence $x\in \mathrm{Int}B_0$ by the lower order convexity of $B_0$, a contradiction. Thus, $\partial B_0$ is positively invariant. The proof of the remaining statement is similar as in Proposition 1(iii). We omit it here.

Lemma 5

Assume that $({\varUpsilon }0)$–$({\varUpsilon }3)$ hold. Then for any $y\in {\varGamma }{\setminus }\{0\}$ and any negative-orbit $\{y(-k),k\ge 0\}$ through y, the $\alpha$-limit set $\alpha (y)$ associated to $\{y(-k)\}$ satisfies: either (i) $\alpha (y)\cap [0,\frac{x_0}{2}]=\emptyset$, or (ii) $\alpha (y)=\{0\}$.

Proof

Choose some $\emptyset \ne I\subseteq N$ such that $y\in \dot{H}_I^+$. By the second statement of (${\varUpsilon }$0) and form (1), we can write such a negative-orbit as $\{y(-k):k\ge 0\}\subseteq \dot{H}_I^+$ such that $y(0)=y$ and $Ty(-k-1)=y(-k)$. Suppose that $z\in \alpha (y)\cap [0,\frac{x_0}{2}]\ne \emptyset$. Then there exists a sequence $k_m\rightarrow \infty$ such that $y(-k_m)\in [0,x_0]\subseteq [0,q]$ for all $k_m$ sufficiently large. So, it follows from the first statement of Lemma 2 that $\{y(-k):k\ge 0\}\subseteq [0,q]$, and hence, $\{y(-k):k\ge 0\}\subseteq \dot{S}_I^+$.

Choose some $k_*\in \{k_m\}$ such that $y(-k_*)\in [0,x_0]$ (and hence, by Remark 7, $y(-k_*)\ll _I Ty(-k_*)=y(-k_*+1)$. So for any $k\ge k_*$, one has $T^{k-k_*}y(-k)=y(-k_*)\ll _I Ty(-k_*)=T^{k-k_*}y(-k+1)$. Since $\{y(-k):k\ge 0\}\subseteq \dot{S}_I^+$, (${\varUpsilon }2$) yields that $y(-k)\ll _I y(-k+1)$. Thus $\{y(-k):k\ge 0\}$ is an eventually decreasing sequence, which implies that $\alpha (y)$ is a fixed point of T in $[0,x_0]$. So $\alpha (y)=\{0\}$.

Lemma 6

Assume that $({\varUpsilon }0)$–$({\varUpsilon }3)$ hold. Then for any $y\in \mathbf {R}^n_+{\setminus }\{0\}$, $\omega (y)\cap [0,\frac{x_0}{2}]=\emptyset$.

Proof

By Proposition 1, we already know that $\omega (y)\subseteq {\varGamma }$ for any $y\in \mathbf {R}^n_+{\setminus }\{0\}$. Suppose that there are some $y_0\in \dot{H}_I^+(I\ne \emptyset )$ and a sequence $k_m\rightarrow \infty$ such that $T^{k_m}y_0\rightarrow z\in \omega (y_0)\cap [0,\frac{x_0}{2}]\ne \emptyset$. By Remark 7 again, $T^{k_m}y_0\ll _I T^{k_m+1}y_0$ for all $k_m$ sufficiently large. So, for each $k\ge 1$, one can find a $k_m\ge k$ such that $T^{k_m-k}(T^ky_0)=T^{k_m}y_0\ll _I T^{k_m+1}y_0=T^{k_m-k}(T^{k+1}y_0)$. It then follows from (${\varUpsilon }$2) that $T^ky_0\ll _I T^{k+1}y_0$ (Since $T^k y_0\in S$ as $k \ge 1$). Thus, $\{T^ky_0:k\ge 0\}$ is an eventually increasing sequence in ${\varGamma }$, which implies that $T^ky_0\rightarrow z\gg _I0$ as $k\rightarrow \infty$. So z is a non-zero fixed point in $[0,\frac{x_0}{2}]$, contradicting Remark 7. We have proved the lemma.

Remark 8

Up to now, we would like to point out here that Proposition 1 (i.e., the existence of ${\varGamma }$, with $\partial {\varGamma }$ being positively invariant) and Lemmas 3–6 were obtained only under the assumptions (${\varUpsilon }0$)–(${\varUpsilon }3$). In order to show that $\partial {\varGamma }$ attracts all non-trivial orbits, we need the further assumption (${\varUpsilon }$4) or (${\varUpsilon }$4’), by which one can obtain the following two technical lemmas, motivated by Shen and Wang (2008), which are crucial for the proof of the attractivity of $\partial {\varGamma }$.

Lemma 7

Assume that $({\varUpsilon }0)$–$({\varUpsilon }3)$ and $({\varUpsilon }4')$ hold. Given $\emptyset \ne I\subseteq N$, let $x,y\in \dot{H}_I^+\cap [0,q]$ satisfying $x \ll _I y$. Let also $\{x(-m):m\ge 0\}$ and $\{y(-m):m\ge 0\}$ be negative orbits of x and y in [0, q], respectively. If $\liminf _{m\rightarrow \infty }||y(-m)||>0$, then

$$\begin{aligned} \liminf _{m\rightarrow +\infty }\Vert x(-m)-y(-m)\Vert >0. \end{aligned}$$

(21)

Proof

Since $0\in H_I^+$, $x,y\in \dot{H}_I^+$ and $0\ll _I x\ll _I y$, by (${\varUpsilon }$0) and (${\varUpsilon }$2) we get that $x(-m),y(-m)\in \dot{H}_I^+$ and $0\ll _I x(-m)\ll _I y(-m)$ for any $m\ge 1$.

For each $i\in I$, let $K_i(-m)=\dfrac{x_i(-m)}{y_i(-m)}$ for $m=0,1,\ldots$. Then

$$\begin{aligned} K_i(-m)-K_i(-m+1)=K_i(-m)\cdot \frac{G_i(y(-m))-G_i(x(-m))}{G_i(y(-m))}, \end{aligned}$$

for $m\ge 1$. Here the denominator is nonzero due to $0\ll _I y(-m)$ and the second statement of (${\varUpsilon }$0). Recall that $\{x(-m):m\ge 0\},\{y(-m):m\ge 0\}\subseteq [0,q]$ (where q is from (${\varUpsilon }$1)) and $[\partial G_i/\partial x_j]_{i,j\in I}\le 0$ on $[0,q]\cap H_I^+$ (see (${\varUpsilon }$4’)). Then we obtain that $K_i(-m)\le K_i(-m+1)$. Consequently,

$$\begin{aligned} K_i(-m)\le \cdots \le K_i(-1)\le K_i(0)=\dfrac{x_i}{y_i}<1 \end{aligned}$$

for any $m\ge 1$. Set $K_*=\min _{i\in I}|K_i(0)-1|>0$. Then

$$\begin{aligned} \Vert x(-m)-y(-m)\Vert&=\sum _{i\in I}|x_i(-m)-y_i(-m)| \\&=\sum _{i\in I}|K_i(-m)-1|\cdot |y_i(-m)|\\&\ge K_*\sum _{i\in I}|y_i(-m)|=K_*\Vert y(-m)\Vert . \end{aligned}$$

So if $\liminf _{m\rightarrow +\infty }\Vert y(-m)\Vert >0$, then we obtain that $\liminf _{m\rightarrow +\infty }\Vert x(-m)-y(-m)\Vert >0$.

Lemma 8

Assume that $({\varUpsilon }0)$–$({\varUpsilon }4)$ hold. Given $\emptyset \ne I\subseteq N$, let $x,y\in \dot{H}_I^+\cap [0,q]$ satisfying $x \ll _I y$. Let also $\{x(-m):m\ge 0\}$ and $\{y(-m):m\ge 0\}$ be negative orbits of x and y in [0, q], respectively. Assume further that there exists a sequence $m_k\rightarrow \infty$ such that $x(-m_k)\rightarrow x_*$ and $y(-m_k)\rightarrow y_*$ as $k\rightarrow \infty$. Then one has $x_*=0$.

Proof

As the proof of Lemma 7, we have $x(-m),y(-m)\in \dot{H}_I^+$ and $0\ll _I x(-m)\ll _I y(-m)\in [0,q]$ for any $m\ge 0$. Suppose that $x_*\ne 0$. Then one has $y_*\ge x_* \ne 0$. By Lemma 5, $\alpha (y)\cap [0,\frac{x_0}{2}]=\emptyset$, hence $\liminf _{m\rightarrow \infty }||y(-m)||>0$. Now, Lemma 7 entails that (21) holds, i.e., there is a real number $d>0$ such that

$$\begin{aligned} \liminf _{m\rightarrow +\infty }\Vert x(-m)-y(-m)\Vert =d>0. \end{aligned}$$

(22)

As a consequence, $x_*,y_*\in H_I^+$ and $x_*<y_*$ with $||x_*-y_*||\ge d>0$.

Thus, one can find nonempty index subsets $I_*\subseteq I_1\subseteq I$ such that $x_*\in \dot{H}_{I_*}^+$ and $y_*\in \dot{H}_{I_1}^+$. Now, for each $l\ge 1$, it follows from (${\varUpsilon }$0) that $T^l x_*\in \dot{H}_{I_*}^+$, $T^l y_*\in \dot{H}_{I_1}^+$. Moreover, we have $T^l x_*=\lim _{k\rightarrow +\infty } T^l x(-m_k)=\lim _{k\rightarrow +\infty } x(l-m_k)$. Similarly, one has $T^l y_*=\lim _{k\rightarrow +\infty } y(l-m_k).$ Note that $x(l-m_k)\ll _I y(l-m_k)\in [0,q]$ for all $m_k$ sufficiently large. Then it holds that $T^l x_*\le T^l y_*\le q$. Again, by virtue of (22), $T^l x_*< T^l y_*$ and $\Vert T^l x_* - T^l y_*\Vert \ge d > 0$. Therefore, together with (${\varUpsilon }$2), we have obtained that $0\ll _{I_*} T^l x_*\ll _{I_1} T^l y_*\le q$, for any $l\ge 0$.

For each $i\in I_1$ and $l\ge 0$, define $K_i(l):=\dfrac{(T^l x_*)_i}{(T^l y_*)_i}$. Obviously, $0<K_i(l)<1$ for $i\in I_*$, and $K_i(l)=0$ for $i\in I_1{\setminus } I_*$. Moreover,

$$\begin{aligned} K_i(l+1)-K_i(l)=K_i(l)\cdot \frac{G_i(T^{l}x_*)-G_i(T^{l}y_*)}{G_i(T^{l}y_*)}. \end{aligned}$$

(23)

Since $[\frac{\partial G_i}{\partial x_j}]_{i,j\in I_1}\le 0$ on $[0,q]\cap H_{I_1}^+$, one has $K_i(l+1)\ge K_i(l)$. In particular, for any $i\in I_*\ne \emptyset$, we have

$$\begin{aligned} K_i(l+1)-K_i(l)&=\frac{K_i(l)}{G_i(T^{l}y_*)}\sum ^n_{j=1}\int ^1_0\frac{\partial G_i}{\partial x_j}(sT^{l}x_*+(1-s)T^{l}y_*)ds\cdot (T^{l}x_*-T^{l}y_*)_j\\&\ge \frac{\delta _i\cdot K_i(0)}{G_i(T^{l}y_*)}\cdot \sum _{j\in I_1}(T^{l}y_*-T^{l}x_*)_j\\&= \frac{\delta _i\cdot K_i(0)}{G_i(T^{l}y_*)}\cdot ||T^ly_*-T^lx_*||\ge \frac{\delta _i\cdot K_i(0)\cdot d}{G_i(T^{l}y_*)}, \end{aligned}$$

where $K_i(0)>0$ and $\delta _i=\min \nolimits _{\begin{array}{c} j\in I_1\\ z\in [0,q] \end{array}}{|\frac{\partial G_i}{\partial x_j}(z)|}.$ By virtue of (${\varUpsilon }$4), it is clear that $\delta _i>0$. Hence,

$$\begin{aligned} K_i(l)-K_i(0)=\sum ^{l-1}_{j=0}[K_i(j+1)-K_i(j)]\ge \dfrac{\delta _i\cdot K_i(0)\cdot d}{G_*}\cdot l, \end{aligned}$$

where $G_*=\max _{x\in [0,q]}(\sum _{i\in I^*}G_i(x))>0$. Then

$$\begin{aligned} K_i(l)\ge K_i(0)+\dfrac{\delta _i\cdot K_i(0)\cdot d}{G_*}\cdot l\rightarrow +\infty \end{aligned}$$

as $l\rightarrow \infty$, a contradiction to the fact that $K_i(l)\le 1$ for any $l\ge 0$. Thus, we have shown $x_*=0$, which completes our proof.

Proposition 2

Assume that $({\varUpsilon }0)$–$({\varUpsilon }4)$ hold. Then we have the following two statements:

• (i) $\mathrm{Int}{\varGamma }\subseteq \{y\in {\varGamma }:\alpha (y)=\{0\},$ for any negative orbit of y in ${\varGamma }\}\subseteq B_0$;

• (ii) For any $x\in \mathbf {R}^n_+{\setminus }\{0\}$, $\omega (x)\subseteq \partial {\varGamma }$.

Proof

(i) It suffices to show that any non-zero element of $\mathrm{Int}{\varGamma }$ is contained in $\{y\in {\varGamma }:\alpha (y)=\{0\},\text { for any negative orbit of }y \text { in }{\varGamma }\}$. For any such non-zero $y\in \mathrm{Int}{\varGamma }$, let I be the nonempty index subset such that $y\in \dot{H}_I^+$. Then it follows from Lemma 3 that $y\ll _I w_y$, where $w_y=\partial {\varGamma }\cap L_y$. Suppose that there is a negative orbit $\{y(-k):k\ge 0\}$ in ${\varGamma }$ through y such that $\alpha (y)\ne \{0\}$. Then Lemma 5 ensures that $\alpha (y)\cap [0,\frac{x_0}{2}]=\emptyset$.

Note also that $w_y\in \partial {\varGamma }\subseteq {\varGamma }$, the invariance of ${\varGamma }$ implies that there is a negative orbit $\{w_y(-k):k\ge 0\}\subseteq {\varGamma }$ through $w_y$. By (${\varUpsilon }$2), the second statement of (${\varUpsilon }$0) and form (1), we obtain that $0\ll _Iy(-k)\ll _I w_y(-k)$ for any $k\ge 0$, where $\{w_y(-k),k\ge 0\}$ is the negative orbit contained in ${\varGamma }$ through $w_y$. By virtue of Lemma 8, we obtain that $0\in \alpha (y)$, a contradiction to $\alpha (y)\cap [0,\frac{x_0}{2}]=\emptyset$. Thus, we have proved $\mathrm{Int}{\varGamma }\subseteq \{y\in {\varGamma }:\alpha (y)=\{0\},\text { for any negative orbit of }y \text { in }{\varGamma }\}$. The last inclusion in (i) follows from Lemma 4(ii).

(ii) Again by Proposition 1, one has $\omega (x)\subseteq {\varGamma }$. Suppose that there is an $x\in \mathbf {R}^n_+{\setminus }\{0\}$ such that $\omega (x)\nsubseteq \partial {\varGamma }$. Then we choose a $z\in \omega (x)\cap \mathrm{Int}{\varGamma }$. Since $0\notin \omega (x)$ by Lemma 6, one can find a nonempty index set I such that $z\in \dot{H}_I^+$. By Lemma 3, we have $z\ll _Iw_z$, where $\{w_z\}=\partial {\varGamma }\cap L_z$. By the invariance of $\omega (x)$ and ${\varGamma }$, there is a negative orbit $\{z(-k):k\ge 0\}\subseteq \omega (x)$ (resp. a negative orbit $\{w(-k):k\ge 0\}\subseteq {\varGamma }$) through z (resp. $w_z$). Moreover, (${\varUpsilon }$2) also implies that $z(-k)\ll _I w(-k)$ for any $k\ge 0$. Without loss of generality, we assume that $z(-k)\rightarrow z_*\in \omega (x)$. Again by Lemma 6, $z_*\notin [0,\frac{x_0}{2}]$. On the other hand, it follows from Lemma 8 that $z_*=0$, a contradiction. We have completed the proof.

Remark 9

We here emphasize that all the previous Propositions and Lemmas are obtained under (${\varUpsilon }0$)–(${\varUpsilon }4$), and without the local injectivity assumption. In the following lemma, we need to impose the additional condition that T is locally injective in an open subset V of $\mathbf {R}^n$ with $[0,q]\subseteq V\subseteq W$. As we will see in the following Lemma 9, the local injectivity of T in V will guarantee the invariance of $\partial {\varGamma }$ and $\partial B_0$. Without such assumption, we can only obtain the positive invariance of them.

Lemma 9

Let $({\varUpsilon }0)$–$({\varUpsilon }3)$ hold. Assume further that T is locally injective in an open subset V of $\mathbf {R}^n$ with $[0,q]\subseteq V\subseteq W$. Then we have

• (i) $\partial {\varGamma }$ is invariant and unordered, i.e., any two points in $\partial {\varGamma }$ cannot be related by “$\le$”.

• (ii) $B_0$ is an open subset in ${\varGamma }$, and hence, $B_0\subseteq \mathrm{Int}{\varGamma }$. Moreover,

  $$\begin{aligned} B_0=\{y\in {\varGamma }:\alpha (y)=\{0\},\text { for any negative orbit in } {\varGamma }\text { of }y \}, \end{aligned}$$

  i.e., $B_0$ is the basin of repulsion.

• (iii) $\partial B_0$ is invariant and unordered, i.e., any two points in it cannot be related by “$\le$”.

Proof

(i) In order to show the invariance of $\partial {\varGamma }$, by Proposition 1(iii), one only needs to prove the negative invariance of $\partial {\varGamma }$. Suppose that there is some $z\in \partial {\varGamma }$ such that its pre-image $T^{-1}(\{z\})\cap \mathrm{Int}{\varGamma }\ne \emptyset$. Then choose a $z_1\in T^{-1}(\{z\})\cap \mathrm{Int}{\varGamma }$. Since T is locally injective in the open subset V of $\mathbf {R}^n$, the domain invariance theorem (Dold 1972) implies that T is an open mapping in V (It deserves to point out that the domain invariance Theorem is not valid if one only assumes that T is locally injective in [0, q], instead of V). This means that, for any $x_0\in V$ and any neighborhood $B(x_0)$ of $x_0$ in V, $TB(x_0)$ will contain an open neighborhood $B(Tx_0)$ of $Tx_0$ in $\mathbf {R}^n$.

Since $z_1\in \mathrm{Int}{\varGamma }\subseteq [0,q]\subseteq V$, we choose a neighborhood $B(z_1)$ of $z_1$ in V so small that $B(z_1)\cap \mathbf {R}^n_+\subseteq {\varGamma }$. Moreover, $TB(z_1)$ contains an open neighborhood B(z) of z in $\mathbf {R}^n$. Because B(z) is open in $\mathbf {R}^n$, one can find some $y\in B(z)$ such that $z\ll y$. Note also that $B(z)\subseteq TB(z_1)$, there is a $y_1\in B(z_1)$ such that $Ty_1=y$. Since $y \in \mathbf {R}^n_+$ and $T^{-1}(\mathbf {R}^n_+)\subseteq \mathbf {R}^n_+$, one has $y_1\in B(z_1)\cap \mathbf {R}^n_+\subseteq {\varGamma }$. Therefore, $y=Ty_1\in {\varGamma }$. Recall that $z\ll y$, it then implies that $z\in \mathrm{Int}{\varGamma }$, a contradiction to $z\in \partial {\varGamma }$. Thus, we have proved that $\partial {\varGamma }$ is invariant. Moreover, together with (${\varUpsilon }$2), the invariance of $\partial {\varGamma }$ clearly implies that it is unordered.

(ii) For any $z\in B_0$, there exist some integer $m\ge 0$ and some $0\le y\ll x_0/2$ such that $T^my=z$. Since T is an open mapping in V (hence $T^m$), now, one can choose a small neighborhood $V_z$ of z in $\mathbf {R}^n$ such that $({V_z\cap \mathbf {R}^n_+})\subseteq T^m(U_y\cap \mathbf {R}^n_+)\subseteq T^m[0,x_0/2]\subseteq B_0$, where $U_y$ is a small neighborhood of y in $\mathbf {R}^n$ with $(U_y\cap \mathbf {R}^n_+)\subseteq [0,x_0/2]$. Thus, $B_0$ is an open subset in ${\varGamma }$, and hence, $B_0\subseteq \mathrm{Int}{\varGamma }$.

Next, we will prove $B_0= \{y\in {\varGamma }:\alpha (y)=\{0\}$, for any negative orbit in ${\varGamma }$ of $y \}$. To this end, by Lemma 4(ii), it suffices to prove that for any $y\in \mathrm{Int}{\varGamma }$, there is a unique $y_1\in {\varGamma }$ such that $Ty_1=y$. Suppose that one can find two distinct $y_1,y_2\in {\varGamma }$ s.t. $Ty_1=y=Ty_2$. By the invariance of $\partial {\varGamma }$, we have $y_1,y_2\in \mathrm{Int}{\varGamma }$. Since $y_1,y_2,y\in \mathrm{Int}{\varGamma }\subseteq V$ and T is a local homeomorphism in V, there exist open neighborhoods $V_i\subseteq V$ of $y_i$, $V_*\subseteq V$ of y s.t. $V_i\cap \mathbf {R}^n_+\subseteq [0,q]$ and $T:V_i\rightarrow V_*,i=1,2,$ are homeomorphisms. Now, consider the homeomorphism $T:V_1\rightarrow V_*$, choose $z_n\gg y$ with $z_n\rightarrow y$ (hence $z_n\in \mathbf {R}^n_+$). Recall that $T^{-1}(\mathbf {R}^n_+)\subseteq \mathbf {R}^n_+$ and $T:V_1\rightarrow V_*$ is homeomorphic. Then there is a sequence $x_n\in \mathbf {R}_+^n\cap V_1\subseteq [0,q]$ such that $Tx_n=z_n$ and $x_n\rightarrow y_1$. Consequently, $Tx_n\gg y=Ty_2$. By (${\varUpsilon }$2), one has $x_n\ge y_2$, which implies that $y_1\ge y_2.$ Similarly, one can obtain $y_2\ge y_1$. So $y_1=y_2$, a contradiction. Thus, we have proved the uniqueness of $y_1$. This imples that for any $y\in B_0\subseteq \mathrm{Int}{\varGamma }$, there is a unique negative orbit $\{y(-m):m\ge 0\}\subseteq {\varGamma }$ through y. So, Lemma 4 directly entails that

$$\begin{aligned} B_0= \{y\in {\varGamma }:\alpha (y)=\{0\},\text { for any negative orbit in } {\varGamma }\text { of }y \}. \end{aligned}$$

(iii) By Lemma 4(iii), it suffices to prove the negative invariance of $\partial B_0$. Note that $\overline{B_0}$ and $B_0$ are invariant and $B_0$ is open. Then this can be done by the similar arguments in (i).

Now we are ready to prove Theorem 6.

Proof of Theorem 6

Under the assumptions (${\varUpsilon }$0)–(${\varUpsilon }$4), the statements of (i)-(iv) are obtained directly from Propositions 1 and 2. Moreover, by Proposition 1(iii), the map $x\mapsto w_x$ from $\{x\in \mathbf {R}_+^n:||x||=1\}$ to $\partial {\varGamma }$ is well defined, where $w_x$ is the point of intersection of the ray $L_{x}=\{tx:t\ge 0\}$ with $\partial {\varGamma }$, and it is also injective and onto $\partial {\varGamma }$ (by (20)). The inverse mapping from $\partial {\varGamma }$ onto $\{x\in \mathbf {R}_+^n:||x||=1\}$ is clearly continuous, hence it is a homeomorphism by the compactness of the two sets. Thus, $\partial {\varGamma }$ is homeomorphic to the unit sphere in $\mathbf {R}_+^n$, which in turn, is homeomorphic to ${\varDelta }^{n-1}$.

Hereafter, we always assume that T is locally injective in an open subset V of $\mathbf {R}^n$ with $[0,q]\subseteq V\subseteq W$. By Proposition 2(i) and Lemma 9, (${\varUpsilon }$0)–(${\varUpsilon }$4) immediately imply that $\partial {\varGamma }$ is invariant and

$$\begin{aligned} \mathrm{Int}{\varGamma }=\{y\in {\varGamma }:\alpha (y)=\{0\},\text { for any negative orbit in }{\varGamma }\text { through }y \}=B_0. \end{aligned}$$

(24)

Moreover, we claim that (24), as well as (i)–(iv), can still hold even under the weaker assumption (${\varUpsilon }4$’), instead of (${\varUpsilon }4$), when T is locally injective in V. Indeed, under (${\varUpsilon }$0)–(${\varUpsilon }$3), Lemma 9 implies that $\partial {\varGamma }$ and $\partial B_0$ are invariant and unordered. In the following paragraph, we will prove that $\partial B_0=\partial {\varGamma }$. But, before doing it, we first show how it implies the claim. In fact, for any $y\in \mathrm{Int}{\varGamma }$, let I be the nonempty index subset such that $y\in \dot{H}_I^+$. Then it follows from Lemma 3 that $y\ll _I w_y\in \partial {\varGamma }=\partial B_0$. Hence, one can find some $z\in B_0$ so close to $w_y$ that $y\ll z$. Lemma 4(i) then entails that $y\in B_0$. So, $\mathrm {Int}{\varGamma }\subseteq B_0$. Thus, by Lemma 9(ii), one can also obtain (24). Now by Proposition 1, Lemma 9 and (24), (i)–(iii) still hold under (${\varUpsilon }$4’). As for (iv), suppose that there is an $x\in \mathbf {R}^n_+{\setminus }\{0\}$ such that $\omega (x)\nsubseteq \partial {\varGamma }$. Then we choose a $y\in \omega (x)\cap \mathrm{Int}{\varGamma }=\omega (x)\cap B_0$. By the invariance of $\omega (x)$, there exists a negative orbit $\{y(-k):k\ge 0\}$ through y such that $\{y(-k)\}\subseteq \omega (x)\subseteq {\varGamma }$ for all $k\ge 0$. Since $y\in B_0$, (24) entails that $0\in \omega (x)$, which contradicts Lemma 6. Thus, we have proved (iv), that is, $\omega (x)\subseteq \partial {\varGamma }$ for all $x\in \mathbf {R}^n_+{\setminus }\{0\}$. This confirms our claim above.

So, in order to prove the claim, it suffices to show $\partial B_0=\partial {\varGamma }$ under (${\varUpsilon }$4’), which will be done by induction on n. If $n=1$, then it is clear that $\partial B_0=\partial {\varGamma }=\{q_{1}\}$. From now on, we assume $n > 1$. Then the induction hypothesis is that $\partial B_0=\partial {\varGamma }$ holds for systems in $\mathbf {R}^m_+$ for any $m < n$. Therefore, fix $n>1$, the induction hypothesis implies that for any $\phi \ne I\subseteq N$, we have $\partial B_0\cap H^+_I=\partial {\varGamma }\cap H^+_I$. Hence, $\partial B_0\cap \partial \mathbf {R}^n_+ =\partial {\varGamma }\cap \partial \mathbf {R}^n_+$. So, one only needs to prove that $\partial B_0\cap \dot{\mathbf {R}}^n_+=\partial {\varGamma }\cap \dot{\mathbf {R}}^n_+$. Note that $\forall u\in \dot{\mathbf {R}}^n_+, |u|=1$, the ray $L_{u}=\{tu:t\ge 0\}$ meets $\partial B_0$ and $\partial {\varGamma }$ in points x and y. If there exist $x\ne y$ such that $0\ll x\ll y$, then by the invariance of $\partial B_0$ and $\partial {\varGamma }$ there exist negative orbits $\{x(-k):k\ge 0\}\subseteq \partial B_0, \{y(-k):k\ge 0\}\subseteq \partial {\varGamma }$ such that $x(-k)\ll y(-k)$, $\forall k\ge 1$, and $\liminf _{k\rightarrow +\infty }\Vert x(-k)-y(-k)\Vert =d>0$ by Lemma 7. Choose a sequence of integers $m_k\rightarrow +\infty$ such that $x_*=\lim _{k\rightarrow +\infty }x(-m_k)$, and $y_*=\lim _{k\rightarrow +\infty }y(-m_k)$, so $x_*<y_*$. Note that $x_*\in \partial B_0$ and $y_*\in \partial {\varGamma }$. Suppose $x_*\in \partial \mathbf {R}^n_+$, then $x_*\in \partial B_0\cap \partial \mathbf {R}^n_+=\partial {\varGamma }\cap \partial \mathbf {R}^n_+$, hence $x_*\in \partial {\varGamma }$, which contradicts that $\partial {\varGamma }$ is unordered. Consequently, one may assume that $0\ll x_*<y_*$. Furthermore, one can obtain that $T^l x_*< T^l y_*$ and $\Vert T^l x_* - T^l y_*\Vert \ge d > 0$ for any $l\ge 0$. For $i\in N, l\ge 0$, set $K_i(l):=\frac{(T^{l} x_*)_i}{(T^l y_*)_i}$. Obviously, $0<K_i(l)<1$ for any $i\in N$. Then it follows from (23) that $K_i(l)\ge K_i(l-1)$ for any $i\in N$; and moreover, by (${\varUpsilon }$4’), we have

$$\begin{aligned} K_i(l+1)-K_i(l)&=\frac{K_i(l)}{G_i(T^{l}y_*)}\sum ^n_{j=1}\int ^1_0\frac{\partial G_i}{\partial x_j}(sT^{l}x_*+(1-s)T^{l}y_*)ds\cdot (T^{l}x_*-T^{l}y_*)_j\\&\ge \frac{\delta \cdot K_i(0)}{G_i(T^{l}y_*)}\cdot (T^{l}y_*-T^{l}x_*)_i \ge \frac{\delta K_*}{G_*}\cdot (T^{l} y_*-T^{l} x_*)_i, \end{aligned}$$

where $\delta =\min _{\begin{array}{c} i\in N\\ x\in [0,q] \end{array}}{|\frac{\partial G_i}{\partial x_i}(x)|}>0$ (see (${\varUpsilon }4$’)), $K_*=\min _{i\in N}K_i(0)>0$, and $G_*=\max _{x\in [0,q]}(\sum _{i=1}^n G_i(x))>0$. Consequently,

$$\begin{aligned} \sum ^n_{i=1}K_i(l) \ge \sum ^n_{i=1}K_i(0)+\frac{\delta K_*}{G_*}\sum ^{l-1}_{j=0}\Vert T^j y_*-T^j x_*\Vert \ge \sum ^n_{i=1}K_i(0)+\frac{\delta K_*}{G_*}dl\rightarrow +\infty \end{aligned}$$

as $l\rightarrow +\infty$, which contradicts $\sum ^n_{i=1}K_i(l)\le n$. Therefore, it yields that $x=y$. Thus, we have proved $\partial B_0=\partial {\varGamma }$.

Finally, we will show (17). Given any $x\in {\mathbf {R}^n_+}{\setminus }\{0\}$, one can find some index set $\emptyset \ne I\subseteq N$ such that $x\in \dot{H}^+_I$. By Proposition 1(iii), the ray from the origin through $T^kx$ will meet $\partial {\varGamma }$ at the unique point $w_k\in \dot{H}^+_I$, $\forall k\ge 0$. So one of the following three alternatives will occur:

• (a) There exists some $k_0\ge 0$, such that $T^{k_0}x=w_{k_0}$;

• (b) There exists some $k_0\ge 0$, such that $T^{k_0}x\ll _I w_{k_0}$;

• (c) $T^{k}x\gg _I w_k$ for any $k\ge 0$.

For case (a), since $\partial {\varGamma }$ is invariant, one can choose $y\in T^{-k_0}(w_{k_0})\cap \partial {\varGamma }$, and hence, (17) holds immediately. For case (b), Lemma 3 implies that $T^{k_0}x\in \mathrm{Int}{\varGamma }$. Recall that $\mathrm{Int}{\varGamma }$ is invariant. Then it entails that $T^kx\in \mathrm{Int}{\varGamma }$ for all $k\ge k_0$. Without loss of generality, we hereafter assume that $k_0=0$.

Now, for each $k\ge 0$, define $M_+(T^kx)=\{y\in \partial {\varGamma }:y\ge T^kx\}$. Clearly, $M_+(T^kx)\subseteq \partial {\varGamma }$ is a compact set, and moreover, $M_+(T^kx)\ne \emptyset$ by Lemma 3. Note that $T^kx\notin \partial {\varGamma }$. Then $M_+(T^kx)=\{y\in \partial {\varGamma }:y> T^kx\}$. Together with the invariance of $\partial {\varGamma }$, (${\varUpsilon }$2) then implies that

$$\begin{aligned} \emptyset \ne T^{-1}(M_+(T^{k+1}x))\cap \partial {\varGamma }\subseteq M_+(T^{k}x),\quad \text {for all } k\ge 0. \end{aligned}$$

(25)

Now we define

$$\begin{aligned} D_1(T^kx):=T^{-1}(M_+(T^{k}x))\cap \partial {\varGamma },\quad \text {for } k\ge 0, \end{aligned}$$

and

$$\begin{aligned} D_{j+1}(T^{k}x):=T^{-1}(D_{j}(T^{k+1}x))\cap \partial {\varGamma }, \quad \text {for } j\ge 1,k\ge 0. \end{aligned}$$

By the invariance of $\partial {\varGamma }$, $D_j(T^kx)$ is a nonempty compact subset in $\partial {\varGamma }$ for any $j\ge 1$ and $k\ge 0$. Then one has the following properties for $D_j(T^kx)$:

• (P1) $D_{j+1}(T^kx)\subseteq D_j(T^kx)\subseteq \cdots \subseteq D_1(T^kx)\subseteq M_+(T^{k-1}x)$, for any $j,k\ge 1$;

• (P2) If $z\in D_{j}(Tx)$, then $T^kz\in D_{j-k}(T^{k+1}x)$ for any $j\ge 1$ and $0\le k<j$.

As a matter of fact, the last inclusion in (P1) is directly from the definition of $D_1(T^kx)$ and (25). In order to prove $D_{j+1}(T^kx)\subseteq D_j(T^kx)$ in (P1), by the iterated definition of $D_{j}(T^{k}x)$, it suffices to show that $D_2(T^kx)\subseteq D_1(T^kx)$ for any $k\ge 0$, and this can be easily verified by (25). Moreover, (P2) can be checked directly by the definition of $D_{j}(T^kx)$.

Now we can prove (17) for some $y\in \partial {\varGamma }$. Indeed, by virtue of (P1), we define $D_x=\bigcap _{j\ge 1}^\infty D_j(Tx)$. Clearly, $D_x$ is a nonempty compact subset in $\partial {\varGamma }$ satisfying $D_x\subseteq M_+(x)$. Choose a $y\in D_x$. Then $y\in D_j(Tx)$ for any $j\ge 1$. By virtue of (P1)–(P2), it follows that $T^ky\in D_{j-k}(T^{k+1}x)\subseteq M_+(T^{k}x)$ for any $k<j$. As a consequence, we obtain that $T^kx<T^ky$ for any $k\ge 0$. Suppose that $\Vert T^ky-T^kx\Vert \nrightarrow 0$, then one can choose a subsequence $\{k_i\}$ such that $T^{k_i}y\rightarrow y_0\in \omega (y)\subseteq \partial {\varGamma }$, $T^{k_i}x \rightarrow x_0\in \omega (x)\subseteq \partial {\varGamma }$ and $y_0>x_0$, contradicting that $\partial {\varGamma }$ is unordered. Thus, we have proved (17).

For case (c), a similar argument as in the last part of the previous paragraph shows the result. We have completed the proof of Theorem 6. $\square$

Remark 10

It deserves to mention that the assumption $({\varUpsilon }4)$ is needed (see in Lemma 8), if we do not have the local injectivity of T in V (see also in Remark 9). However, a careful examination of the arguments after (24) yields that, if the local injectivity of T is assumed, then $\partial {\varGamma }$ and $\partial B_0$ are invariant and unordered with respect to “$\le$”. Based on this, one can replace $({\varUpsilon }4)$ by the weaker assumption (${\varUpsilon }4$’) to obtain (i)–(iv) in Theorem 6.

In the following, we will give a simple example, in which $[\partial G_i/\partial x_j(x)]_{i,j\in I}$ has non-positive off-diagonal entries [i.e., (${\varUpsilon }4$’) holds], which admits a carrying simplex.

Example

Consider the following two-dimensional map:

$$\begin{aligned} T(x)=\left( \frac{2x_1}{1+x_1+b_{12}x_2^2},\frac{2x_2}{1+b_{21}x_1+x_2}\right) ,\quad b_{12}>0,\;0<b_{21}\le \frac{1}{2},\;x\in \mathbf {R}^2_+. \end{aligned}$$

(26)

Let $G_1(x)=\frac{2}{1+x_1+b_{12}x_2^2}$, $G_2(x)=\frac{2}{1+b_{21}x_1+x_2}$ and $G(x)=(G_1(x),G_2(x))$. It is clear that $G(x)\gg 0$ in a small open neighborhood $W\subseteq \mathbf {R}^2$ of $\mathbf {R}^2_+$. Now $T_i(x)=x_iG_i(x)$ can be defined on W with $T(\mathbf {R}^2_+)\subseteq \mathbf {R}^2_+$ and $T^{-1}(\mathbf {R}^2_+)\subseteq \mathbf {R}^2_+$. Note that $\partial G_1(x)/\partial x_2=-\frac{4b_{12}x_2}{(1+x_1+b_{12}x_2^2)^2}\le 0$ for any $x\in \mathbf {R}^2_+$ and $\partial G_1(x)/\partial x_2=0$ for $x\in H_{\{1\}}^+$. On the other hand, $\forall x\in \mathbf {R}^2_+$, $\partial G_i(x)/\partial x_j<0$ for $i\ne 1,j\ne 2$. So, (${\varUpsilon }$4’) holds [(${\varUpsilon }$4) does not]. Clearly, $G_i(0)>1$ and $T|_{\dot{H}_{\{i\}}^+}$ has a globally attracting fixed point $q_{\{i\}}=e_{\{i\}}$. Hence, (${\varUpsilon }$0) and (${\varUpsilon }$1) hold for T. Set $q=(1,1)$ and $S=[0, 2)\times [0,2)$. Then $T(\mathbf {R}^2_+)\subseteq S$ and $[0,q]\subseteq S$.

A routine computation yields that for any $x\in S$,

$$\begin{aligned} \det DT(x)=\frac{4(1+b_{21}x_1+b_{12}x_2^2(1-b_{21}x_1))}{(1+x_1+b_{12}x^2_2)^2(1+b_{21}x_1+x_2)^2} >0. \end{aligned}$$

Hence, $DT(x)^{-1}$ exists for every $x \in S$ and T is locally homeomorphic on S. Together with $T^{-1}(\{0\})=\{0\}$, Lemma 3.4 in Chow and Hale (1982, p. 27) implies that T is also homeomorphic on S. On the other hand, it is not difficult to calculate that

$$\begin{aligned} D(T|_{H^+_I})(x)^{-1}=(DT(x)^{-1})_I\gg 0, \quad \forall \emptyset \ne I\subseteq N, x\in \dot{S}^+_I. \end{aligned}$$

Hence (${\varUpsilon }$5) holds on S. Now Corollary 2 implies that T admits a carrying simplex ${\varSigma }$.

Postscript After the first version of this manuscript had been submitted, we noticed the detailed work in Ruiz-Herrera (2013, Appendix) containing result similar to the one we present in the Appendix here. We thank one of the referees to remind us for this. Comparing with (Ruiz-Herrera 2013), we first show that T can still admit a global attractor ${\varGamma }$ only under assumptions (${\varUpsilon }$1)–(${\varUpsilon }$2). We further study the geometrical structure of ${\varGamma }$ and $\partial {\varGamma }$, as well as the global attractivity of $\partial {\varGamma }$, under assumptions (${\varUpsilon }$0)–(${\varUpsilon }$4) without the local injectivity assumption. The local injectivity assumption is only needed for the proof of the negative invariance of $\partial {\varGamma }$ and the property (17). Moreover, we have replaced the positivity of G(x) in the whole domain $\mathbf {R}^n_+$ by (${\varUpsilon }$0), and negativity of the entries of DG(x) in the whole domain $\mathbf {R}^n_+$ by the weaker condition (${\varUpsilon }$4) (or (${\varUpsilon }$4’)).