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On the solutions of a maxtype system of difference equations of higher order
Advances in Difference Equations volume 2020, Article number: 213 (2020)
Abstract
In this paper, we study the following maxtype system of difference equations of higher order:
where \(A,B\in (0, +\infty )\), \(t,s\in \{1,2,\ldots \}\) with \(\gcd (s,t)=1\), the initial values \(x_{d},y_{d},x_{d+1},y_{d+1}, \ldots , x_{1}, y_{1}\in (0,+ \infty )\) and \(d=\max \{t,s\}\).
1 Introduction
Concrete nonlinear difference equations and systems have attracted some recent attention (see, e.g., [1–39]). One of the classes of such equations/systems are maxtype difference equations/systems. For some results of solutions of many maxtype difference equations and systems, such as eventual periodicity, the boundedness character and attractivity, see, e.g. [1–5, 7–9, 11–16, 18–25, 28–30, 32–36, 38, 39] and the references therein. Our purpose in this paper is to study the eventual periodicity of the following maxtype system of difference equation of higher order:
where \(A,B\in {\mathbf{{R}}}_{+}\equiv (0,+\infty )\), \(t,s\in {\mathbf{{N}\equiv }\{\mathbf{1},\mathbf{2},\ldots \}}\) with \(\gcd (s,t)=1\), the initial values \(x_{d},y_{d}, x_{d+1},y_{d+1}, \ldots , x_{1}, y_{1}\in \mathbf{{R}}_{+}\) and \(d=\max \{t,s\}\).
When \(t=1\) and \(s=2\), (1.1) reduces to the maxtype system of difference equations
Fotiades and Papaschinopoulos in [5] showed that every positive solution of (1.2) is eventually periodic.
In 2012, Stević [23] obtained in an elegant way the general solution to the following maxtype system of difference equations:
for the case \(x_{0},y_{0}\geq A>0\) and \(y_{0}/x_{0}\geq \max \{A,1/A\}\).
In [35], Sun and Xi studied the following maxtype system of difference equations:
where \(A,B\in { \mathbf{{R}}}_{+}\), \(m,r,t\in {\mathbf{{N}}}\) and the initial values \(x_{d},y_{d},x_{d+1},y_{d+1}, \ldots , x_{1}, y_{1}\in {\mathbf{{R}}}_{+}\) with \(d=\max \{m,r,t\}\) and showed that every positive solution of (1.4) is eventually periodic with period 2m.
When \(m=r=t=1\) and \(A=B\), (1.4) reduces to the maxtype system of difference equations
Yazlik et al. [39] in 2015 obtained in an elegant way the general solution of (1.5).
In 2012, Stević [24] studied the following maxtype system of difference equations:
where \(s,l,m_{j},k^{(j)}_{i,t}\in {\mathbf{{N}}}\) (\(j,t\in \{1,2,\ldots ,l \}\)) and \(f_{ji}:\mathbf{{R}}_{+}^{l}\times {\mathbf{{ N}}}_{0}\longrightarrow \mathbf{{R}}_{+}\) (\(j\in \{1,\ldots ,l\}\) and \(i\in \{1,\ldots ,m_{j}\}\)), and showed that every positive solution of (1.6) is eventually periodic with (not necessarily prime) period s if \(f_{ji}\) satisfy some conditions.
Moreover, Stević et al. [29] in 2014 investigated the following maxtype system of difference equations:
where \(s,l,m_{j},t_{j},k^{(j)}_{i_{j},h}\in { \mathbf{{N}}}\) (\(j,h\in \{1,2,\ldots ,l\}\)), \((\sigma (1),\ldots ,\sigma (l))\) is a permutation of \((1,\ldots ,l)\) and \(f_{ji_{j}}:\mathbf{{R}}_{+}^{l}\times {\mathbf{{ N}}}_{0}\longrightarrow \mathbf{{R}}_{+}\) (\(j\in \{1,\ldots ,l\}\) and \(i_{j}\in \{1,\ldots ,m_{j}\}\)). They showed that every positive solution of (1.7) is eventually periodic with period sT for some \(T\in \mathbf{{N}}\) if \(f_{ji_{j}}\) satisfy some conditions.
2 Main results and proofs
In this section, we study the eventual periodicity of positive solutions of system (1.1). Let \(\{(x_{n},y_{n})\}_{n\geq d}\) be a solution of (1.1) with the initial values \(x_{d},y_{d},x_{d+1},y_{d+1},\ldots , x_{1}, y_{1}\in { \mathbf{{R}}}_{+}\).
Lemma 2.1
If\(x_{n}=A\)eventually, then\(y_{n}\)is a periodic sequence with period 2seventually. If\(y_{n}=B\)eventually, then\(x_{n}\)is a periodic sequence with period 2seventually.
Proof
Assume that \(x_{n}=A\) eventually. By (1.1) we see
which implies \(y_{n}y_{ns}\geq A\) eventually and
Then, for any \(0\leq i\leq 2s1\), \(y_{2ns+i}\) is eventually nonincreasing.
We claim that, for every \(0\leq i\leq 2s1\), \(y_{2ns+i}\) is a constant sequence eventually. Assume on the contrary that, for some \(0\leq i\leq 2s1\), \(y_{2ns+i}\) is not a constant sequence eventually. Then there exists a sequence of positive integers \(k_{1}< k_{2}<\cdots \) such that, for any \(n\in { \mathbf{{N}}}\), we have
which implies \(y_{2sk_{n+1}+is}>y_{2sk_{n}+is}\) for any \(n\in { \mathbf{{N}}}\). This is a contradiction. Thus \(y_{n}\) is a periodic sequence with period 2s eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □
Lemma 2.2
If\(A\geq B\geq 1/A\), then\(x_{2(n+1)t+i}\leq x_{2nt+i}\)for any\(n\geq t+s\)and\(i\in { {\mathbf{{N}}}}_{0}\). If\(B\geq A\geq 1/B\), then\(y_{2(n+1)t+i}\leq y_{2nt+i}\)for any\(n\geq t+s\)and\(i\in { {\mathbf{{N}}}}_{0}\).
Proof
Assume that \(A\geq B\geq 1/A\). By (1.1) we see that \(x_{n}\geq A\) and \(y_{n}\geq B\) for any \(n\in { \mathbf{{N}}}_{0}\), and
Since \(B/x_{2(n+1)t+is}\leq B/A\leq 1\leq A\) and \(x_{2(n+1)t+is}y_{2(n+1)t+its}\geq AB\geq 1\) for \(2(n+1)t+i\geq t+s\), we obtain
The second case follows from the previously proved one by interchanging letters. The proof is complete. □
Theorem 2.1
Let\(AB>1\). If\(A\geq B\), then\(x_{n}=A\)eventually and\(y_{n}\)is a periodic sequence with period 2seventually. If\(B> A\), then\(y_{n}=B\)eventually and\(x_{n}\)is a periodic sequence with period 2seventually.
Proof
Assume that \(A\geq B\). For any \(0\leq i\leq 2t1\) and \(n\in { \mathbf{{N}}}_{0}\), we have
By Lemma 2.2 we may let \(\lim_{n\longrightarrow \infty }x_{2nt+i}=A_{i}\). Note that
and
Thus we have \(x_{n}=A\) eventually. By Lemma 2.1, we see that \(y_{n}\) is a periodic sequence with period 2s eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □
In the following, we assume \(AB=1\). For any \(i\in { \mathbf{{N}}}_{0}\), let
and
Then \(A_{i}\geq A\) and \(B_{i}\geq B\).
Lemma 2.3
If\(A\geq B=1/A\)and\(A_{i}>A\)for some\(i\in { {\mathbf{{N}}}}_{0}\), then, for any\(k\in { {\mathbf{{N}}}}\), \(x_{2nt+ks+i}\)and\(y_{2ntt+ks+i}\)are constant sequences eventually. If\(B\geq A=1/B\)and\(B_{i}>B\)for some\(i\in { {\mathbf{{N}}}}_{0}\), then, for any\(k\in { {\mathbf{{N}}}}\), \(y_{2nt+ks+i}\)and\(x_{2ntt+ks+i}\)are constant sequences eventually.
Proof
Assume that \(A\geq B\) and \(A_{i}>A\) for some \(i\in { \mathbf{{N}}}_{0}\). Since \(A_{i}>A\), it follows from Lemma 2.2 and (1.1) that
From this we have
This implies
and
and
Since
and
we see that \(x_{2nt+s+i} =A\) eventually. Note that
from which it follows that
and
and
If \(x_{2nt+2s+i}>A\) eventually, then, in a similar fashion, we obtain:
 (1)
\(x_{2nt+3s+i} =A\) eventually and \(y_{2ntt+3s+i} =B \) eventually.
 (2)
\(x_{2nt+4s+i}\) and \(y_{2ntt+4s+i}\) are constant sequences eventually.
If \(x_{2nt+2s+i}=A\) eventually, then \(y_{2ntt+2s+i} =A/B\) eventually, and
and
From this we see that if \(A=B\), then
and if \(A>B\), then
since
Using induction and arguments similar to the ones developed in the above given proof, we can show that, for any \(k\in { \mathbf{{N}}}\), \(x_{2nt+ks+i}\) and \(y_{2ntt+ks+i}\) are constant sequences eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □
Lemma 2.4
If\(A=1/B>B\)and for some\(i\in { {\mathbf{{N}}}}_{0}\), \(x_{2nt+i}>A\)eventually and\(A_{i}=A\), then, for any\(k\in { {\mathbf{{N}}}}\), \(x_{2nt+ks+i}\)and\(y_{2ntt+ks+i}\)are constant sequences eventually. If\(B=1/A>A\)and for some\(i\in { {\mathbf{{N}}}}_{0}\), \(y_{2nt+i}>B\)eventually and\(B_{i}=B\), then, for any\(k\in { {\mathbf{{N}}}}\), \(y_{2nt+ks+i}\)and\(x_{2ntt+ks+i}\)are constant sequences eventually.
Proof
Assume that \(A=1/B>B\) and for some \(i\in { \mathbf{{N}}}_{0}\), \(x_{2nt+i}>A\) eventually and \(A_{i}=A\). By (1.1) we have
and
and
Then we see that \(x_{2nt+s+i} =A\) eventually. From this and \(y_{2ntt+i}\geq A^{2}\) eventually it follows that
and
and
Thus \(x_{2nt+2s+i}\) and \(y_{2ntt+2s+i}\) are constant sequences eventually. Using arguments similar to the ones developed in the proof of Lemma 2.3, we can show that, for any \(k\in { \mathbf{{N}}}\), \(x_{2nt+ks+i}\) and \(y_{2ntt+ks+i}\) are constant sequences eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □
Theorem 2.2

(1)
Assume\(A=1/B>B\). Then one of the following statements holds.
 (i)
\(x_{n}=A\)eventually and\(y_{n}\)is a periodic sequence with period 2seventually.
 (ii)
Ifsis odd, then\(x_{n}\), \(y_{n}\)are periodic sequences with period 2teventually.
 (iii)
Ifsis even, then\(x_{n}\)is a periodic sequence with period 2teventually and\(y_{n}\)is a periodic sequence with period\(2st\)eventually.
 (i)

(2)
Assume\(B=1/A>A\). Then one of the following statements holds.
 (i)
\(y_{n}=B\)eventually and\(x_{n}\)is a periodic sequence with period 2seventually.
 (ii)
Ifsis odd, then\(x_{n}\), \(y_{n}\)are periodic sequences with period 2teventually.
 (iii)
Ifsis even, then\(y_{n}\)is a periodic sequence with period 2teventually and\(x_{n}\)is a periodic sequence with period\(2st\)eventually.
 (i)
Proof
Assume that \(A=1/B>B\). If \(x_{n}=A\) eventually, then by Lemma 2.1 we see that \(y_{n}\) is a periodic sequence with period 2s eventually. Now we assume that \(x_{n}\neq A\) eventually. Then we have \(A_{i}>A\) (or \(x_{2nt+i}>A\) eventually and \(A_{i}=A\)) for some \(0\leq i\leq 2t1\).
If s is odd, then \(\gcd (2t,s)=1\). Thus, for every \(j\in \{0,1,2,\ldots ,2t1\} \), there exist some \(1\leq i_{j}\leq 2t\) and integer \(\lambda _{j}\) such that \(i_{j}s=\lambda _{j} 2t+j\) since \(\{rs:0\leq r\leq 2t1\}=\{0,1,2,\ldots ,2t1\}\) (mod 2t). By Lemma 2.3 and Lemma 2.4 we see that, for any \(k\in { \mathbf{{N}}}\), \(x_{2nt+ks+i}\) and \(y_{2ntt+ks+i}\) are constant sequences eventually. Thus, for any \(0\leq r\leq 2t1\), \(x_{2nt+r}\) and \(y_{2nt+r}\) are constant sequences eventually, which implies that \(x_{n}\), \(y_{n}\) are periodic sequences with period 2t eventually.
In the following, we assume that s is even with \(s=2s'\). Then \(\gcd (t,s')=1\) and t is odd. Thus, for every \(j\in \{0,1,2,\ldots ,t1\} \), there exist some \(1\leq i_{j}\leq t\) and integer \(\lambda _{j}\) such that \(i_{j}s'=\lambda _{j} t+j\) and \(i_{j}s=\lambda _{j} 2t+2j\).
If \(x_{2nt+i}\neq A\) eventually for some \(i\in \{0,2,\ldots \}\) and \(x_{2nt+l}\neq A\) eventually for some \(l\in \{1,3,\ldots \}\), then by Lemma 2.3 and Lemma 2.4 we see that, for any \(k\in { \mathbf{{N}}}\), \(x_{2nt+ks+i}\), \(y_{2ntt+ks+i}\), \(x_{2nt+ks+l}\) and \(y_{2ntt+ks+l}\) are constant sequences eventually. Thus, for any \(0\leq r\leq 2t1\), \(x_{2nt+r}\) and \(y_{2nt+r}\) are constant sequences eventually, which implies that \(x_{n}\), \(y_{n}\) are periodic sequences with period 2t eventually.
If \(x_{2nt+i}\neq A\) eventually for some \(i\in \{0,2,\ldots \}\) and \(x_{2nt+l}=A\) eventually for any \(l\in \{1,3,\ldots \}\), then by Lemma 2.3 and Lemma 2.4 we see that, for any \(k\in { \mathbf{{N}}}\), \(x_{2nt+ks+i}\) and \(y_{2ntt+ks+i}\) are constant sequences eventually. This implies that, for every \(r\in \{0,1,2,\ldots ,2t1\}\), \(x_{2nt+r}\) is constant sequence eventually and for every \(l\in \{1,3,\ldots \}\), \(y_{2nst+l}\) is constant sequence eventually. By (1.1) we see that there exists \(N\in { \mathbf{{N}}}\) such that, for any \(n\geq N\) and \(r\in \{0,2,\ldots \}\),
Then we have \(y_{2nt+r}y_{2nt+rs}\geq A\). Thus, for any \(n\geq N\) and \(l\in \{1,3,\ldots \}\) and \(k\in { \mathbf{{N}}}\),
Then, for every \(n\geq N\) and \(l\in \{1,3,\ldots \}\), we have
We claim that, for every \(n\geq N\) and \(l\in \{1,3,\ldots \}\), \(\{y_{2nt+t+2ks+l}\}_{k\in { \mathbf{{N}}}}\) is a constant sequence eventually. Assume on the contrary that, for some \(n\geq N\) and some \(l\in \{1,3,\ldots \}\), \(\{y_{2nt+t+2ks+l}\}_{k\in { \mathbf{{N}}}}\) is not a constant sequence eventually. Then there exists a sequence of positive integers \(k_{1}< k_{2}<\cdots \) such that, for any \(r\in { \mathbf{{N}}}\), we have
which implies \(y_{2nt+t+2k_{r}s+ls}>y_{2nt+t+2k_{r1}s+ls}\) for any \(r\in { \mathbf{{N}}}\). This is a contradiction. Take \(ps>N\). Then \(y_{2nst+t+l}=y_{2pst+t+2(ntpt)s+l}\) is a constant sequence eventually for any \(l\in \{1,3,\ldots \}\). From the above we see that \(y_{n}\) is a periodic sequence with period \(2st\) eventually.
In a similar fashion, we can show that if \(x_{2nt+i}=A\) eventually for any \(i\in \{0,2,\ldots \}\) and \(x_{2nt+l}\neq A\) eventually for some \(l\in \{1,3,\ldots \}\), then also statement (1(iii)) holds.
The second case follows from the previously proved one by interchanging letters. The proof is complete. □
Now we assume that \(A=B=1\). Then, for any \(0\leq i\leq 2t1\) and \(n\in { \mathbf{{N}}}_{0}\), we have \(1\leq x_{2(n+1)t+i}\leq x_{2nt+i}\) eventually and \(1\leq y_{2(n+1)t+i}\leq y_{2nt+i}\) eventually.
Lemma 2.5
Let\(A=B=1\)and\(s\geq t\). Then the following statements hold.
 (1)
If\(A_{i}=1\), then\(B_{t+i}=1\). If\(B_{i}=1\), then\(A_{t+i}=1\).
 (2)
If\(x_{N}=1\)for some\(N\in { {\mathbf{{N}}}}\)and\(A_{2nt+N+ks}=1\)for any\(k,n\in { {\mathbf{{N}}}}\), then\(x_{2nt+N+ks}=y_{2nt+t+ks+N}=1\)for any\(k,n\in { {\mathbf{{N}}}}\). If\(y_{N}=1\)for some\(N\in { {\mathbf{{N}}}}\)and\(B_{2nt+N+ks}=1\)for any\(k,n\in { {\mathbf{{N}}}}\), then\(y_{2nt+N+ks}=x_{2nt+t+ks+N}=1\)for any\(k,n\in { {\mathbf{{N}}}}\).
 (3)
Ifsis even and\(\gcd (s,t)=1\), then\(1\in \{x_{n}:n\in \{0,2,\ldots \}\}\cup \{y_{t+n}:n\in \{0,2,\ldots \}\}\)and\(1\in \{x_{n}:n\in \{1,3,\ldots \}\}\cup \{y_{t+n}:n\in \{1,3,\ldots \}\}\).
 (4)
\(1\in \{x_{n}:n\in { {\mathbf{{N}}}}\}\cup \{y_{n}:n\in { {\mathbf{{N}}}}\}\).
Proof
(1) Assume that \(A_{i}=1\). Assume on the contrary that \(B_{t+i}>1\). It follows from (1.1) that
This implies
This is a contradiction. The second case follows from the previously proved one by interchanging letters.
(2) If \(x_{N}=1\) for some \(N\in { \mathbf{{N}}}\), then \(x_{2nt+N}=1\) for any \(n\in { \mathbf{{N}}}\). It follows from (1.1) that
and
and
Thus \(x_{2nt+N+s}=y_{2nt+t+s+N}=1\) for any \(n\in { \mathbf{{N}}}\). In a similar fashion, we can show that \(x_{2nt+N+ks}=y_{2nt+t+ks+N}=1\) for any \(k,n\in { \mathbf{{N}}}\). The second case follows from the previously proved one by interchanging letters.
(3) If s is even and \(\gcd (s,t)=1\), then t is odd. Assume on the contrary that \(1\notin \{x_{n}:n\in \{0,2,\ldots \}\}\cup \{y_{t+n}:n\in \{0,2, \ldots \}\}\). Then it follows from (1.1) that, for any \(n\in { \mathbf{{N}}}\),
and
Thus
This is a contradiction.
(4) Case (4) is treated similarly to case (3). The proof is complete. □
Theorem 2.3
Let\(A=B=1\)and\(s\geq t\). Then one of the following statements holds.
 (1)
\(x_{n}=1\)eventually and\(y_{n}\)is a periodic sequence with period 2seventually.
 (2)
\(y_{n}=1\)eventually and\(x_{n}\)is a periodic sequence with period 2seventually.
 (3)
\(x_{n}\), \(y_{n}\)are periodic sequences with period 2teventually.
Proof
If \(x_{n}=1\) (or \(y_{n}=1\)) eventually, then by Lemma 2.1 we see that \(y_{n}\) (or \(x_{n}\)) is a periodic sequence with period 2s eventually. Now we assume that \(x_{n}\neq1\) eventually. Then we have \(A_{i}>1\) for some \(0\leq i\leq 2t1\) or \(\lim_{n\longrightarrow \infty }x_{n}=1\).
If s is odd, then \(\gcd (2t,s)=1\). Thus, for every \(j\in \{0,1,2,\ldots ,2t1\} \), there exist some \(1\leq i_{j}\leq 2t\) and integer \(\lambda _{j}\) such that \(i_{j}s=\lambda _{j} 2t+j\). By Lemma 2.3 and Lemma 2.5 we see that, for any \(k\in { \mathbf{{N}}}\), \(x_{2nt+ks+i}\) and \(y_{2ntt+ks+i}\) are constant sequences eventually, or for some \(N\in { \mathbf{{N}}}\), \(x_{2nt+N+ks}=y_{2nt+t+ks+N}=1\) for any \(k,n\in { \mathbf{{N}}}\), or for some \(N\in { \mathbf{{N}}}\), \(y_{2nt+N+ks}=x_{2nt+t+ks+N}=1\) for any \(k,n\in { \mathbf{{N}}}\). Thus, for any \(0\leq r\leq 2t1\), \(x_{2nt+r}\) and \(y_{2nt+r}\) are constant sequences eventually, which implies that \(x_{n}\), \(y_{n}\) are periodic sequences with period 2t eventually.
In the following, we assume that s is even with \(s=2s'\), then \(\gcd (t,s')=1\) and t is odd. Thus, for every \(j\in \{0,1,2,\ldots ,t1\} \), there exist some \(1\leq i_{j}\leq t\) and integer \(\lambda _{j}\) such that \(i_{j}s=\lambda _{j} 2t+2j\).
If \(A_{i}>1\) for some \(i\in \{0,2,\ldots \}\), then by Lemma 2.3 we see that, for any \(k\in { \mathbf{{N}}}\), \(x_{2nt+ks+i}\) and \(y_{2ntt+ks+i}\) are constant sequences eventually. If \(A_{i}=1\) for any \(i\in \{0,2,\ldots \}\), then by Lemma 2.5 we have \(B_{t+i}=1\) for any \(i\in \{0,2,\ldots \}\) and \(x_{2nt+i+ks}=y_{2ntt+ks+i}=1\) for any \(k\in { \mathbf{{N}}}\) eventually. In a similar fashion, also we can show that, for any \(i\in \{1,3,\ldots \}\), \(x_{2nt+ks+i}\) and \(y_{2ntt+ks+i}\) are constant sequences eventually for any \(k\in { \mathbf{{N}}}\), or \(x_{2nt+i+ks}=y_{2ntt+ks+i}=1\) for any \(k\in { \mathbf{{N}}}\) eventually for any \(i\in \{1,3,\ldots \}\) and \(k\in { \mathbf{{N}}}\). Thus, for any \(0\leq r\leq 2t1\), \(x_{2nt+r}\) and \(y_{2nt+r}\) are constant sequences eventually. This implies that \(x_{n}\), \(y_{n}\) are periodic sequences with period 2t eventually.
Using the previously proved one by interchanging letters, also we can show that if \(y_{n}\neq1\) eventually, then \(x_{n}\), \(y_{n}\) are periodic sequences with period 2t eventually. The proof is complete. □
In Example 3.1 of [37], we showed that the equation
has a positive solution \(z_{n}\) (\(n\geq t\)) with \(1< z_{n+1}< z_{n}\) for any \(n\geq t\) and \(\lim_{n\longrightarrow \infty }z_{n}=1\).
From Example 3.1 of [37], we obtain the following theorem.
Theorem 2.4
Let\(A\leq 1\)and\(B\leq 1\)and\(s< t\). Assume\(z_{n}\) (\(n\geq t\)) is a positive solution of (2.45) with\(1< z_{n+1}< z_{n}\)for any\(n\geq t\)and\(\lim_{n\longrightarrow \infty }z_{n}=1\). Then equation (1.1) have a solution\((x_{n},y_{n})\)with\(1< x_{n+1}=y_{n+1}=z_{n+1}< x_{n}=y_{n}=z_{n}\)for any\(n\geq t\)and\(\lim_{n\longrightarrow \infty }x_{n}=\lim_{n\longrightarrow \infty }y_{n}=1\).
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The authors would like to thank the referees for their valuable comments and suggestions.
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The research was supported by NNSF of China (11761011, 71862003) and NSF of Guangxi (2018GXNSFAA294010) and SF of Guangxi University of Finance and Economics (2019QNB10).
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Su, G., Han, C., Sun, T. et al. On the solutions of a maxtype system of difference equations of higher order. Adv Differ Equ 2020, 213 (2020). https://doi.org/10.1186/s13662020026732
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DOI: https://doi.org/10.1186/s13662020026732