The following theorem proposes conditions for persistence and boundedness for the positive solution \((x_n,y_n)\) of (3).
Theorem 1
Every positive solution \((x_n,y_n)\) of (3) is bounded and persists whenever \(bde^{-\alpha _1-\alpha _2}<1\).
Proof
\(x_n \ge \alpha _1, y_n \ge \alpha _2\), \(n=3,4,\ldots .\)
Hence, \((x_n,y_n)\) of system (3) persists.
Also, (3) becomes
$$\begin{aligned} x_{n+1}&\le \alpha _1 + a e ^{-\alpha _1} + b e^{-\alpha _2} [\alpha _2 +{\mathrm{d}}x_{n-1}e^{-x_{n-2}}+ce^{-y_{n-2}}].\nonumber \\&\le A + bdx_{n-1}e^{-\alpha _1-\alpha _2} \end{aligned}$$
(4)
where \(A= \alpha _1 + a e ^{-\alpha _1} + b \alpha _2e^{-\alpha _2}+ bce^{-\alpha _2-\alpha _2}\).
Similarly,
$$\begin{aligned} y_{n+1} \le C + bdy_{n-1}e^{-\alpha _1-\alpha _2} \end{aligned}$$
(5)
where \(C= \alpha _2 + c e ^{-\alpha _2} + d \alpha _1e^{-\alpha _1}+ ade^{-\alpha _1-\alpha _1}\).
Now, consider the difference equations
$$\begin{aligned} z_{n+1}&= A + Bz_{n-1}.\nonumber \\ v_{n+1}&= C + Dv_{n-1}, \end{aligned}$$
(6)
where \(B=D=bde^{-\alpha _1-\alpha _2}<1\). Therefore, an arbitrary solution \((z_n, v_n)\) of (6) can be written as
$$\begin{aligned} z_n&= r_1B^{n/2} + r_2(-1)^nB^{n/2}+\frac{A}{1-B} , \quad n=0,1,2,\ldots \end{aligned}$$
(7)
$$\begin{aligned} v_n&= s_1B^{n/2} + s_2(-1)^nB^{n/2}+\frac{C}{1-B} , \quad n=0,1,2,\ldots \end{aligned}$$
(8)
where \(r_1\), \(r_2\) rely on the initial conditions \(z_{-1}\), \(z_0\) and \(s_1\), \(s_2\) rely on the initial conditions \(v_{-1}\), \(v_0\). Hence, \((z_n, v_n)\) is bounded.
Let us examine the solution \((z_n, v_n)\) such that \(z_{-1}=x_{-1}, z_0=x_0,v_{-1}=y_{-1}, v_0=y_0.\)
Hence by induction, \(x_n \le z_n\) and \(y_n \le v_n, n=0,1,2,\ldots\).
Therefore, we get \((x_n, y_n)\) is bounded. \(\square\)
The following two theorems confirm the existence of invariant boxes of (3).
Theorem 2
Let \(bde^{-\alpha _1-\alpha _2}<1\). Let \((x_n, y_n)\) denote a positive solution of (3). Then \(\displaystyle [\alpha _1,\frac{\alpha _1 + a e ^{-\alpha _1} + b \alpha _2e^{-\alpha _2}+ bce^{-\alpha _2-\alpha _2}}{(1-bde^{-\alpha _1-\alpha _2})}]\) \(\displaystyle \times [\alpha _2,\frac{\alpha _2 + c e ^{-\alpha _2} + d \alpha _1e^{-\alpha _1}+ ade^{-\alpha _1-\alpha _1}}{(1-bde^{-\alpha _1-\alpha _2})}]\) is an invariant set for (3).
Proof
Let \(I_1=\displaystyle [\alpha _1,\frac{\alpha _1 + a e ^{-\alpha _1} + b \alpha _2e^{-\alpha _2}+ bce^{-\alpha _2-\alpha _2}}{(1-bde^{-\alpha _1-\alpha _2})}]\) and \(\displaystyle I_2=[\alpha _2,\frac{\alpha _2 + c e ^{-\alpha _2} + d \alpha _1e^{-\alpha _1}+ ade^{-\alpha _1-\alpha _1}}{(1-bde^{-\alpha _1-\alpha _2})}]\).
Let \(x_{-1}, x_{0} \in I_1\) and \(y_{-1}, y_{0} \in I_2.\)
Then
$$\begin{aligned} x_{1}&\le \alpha _1+ ae ^{-\alpha _1} + be^{-\alpha _2}y_0\\&\le \alpha _1+ ae ^{-\alpha _1} + be^{-\alpha _2}\displaystyle \left[ \frac{\alpha _2 + c e ^{-\alpha _2} + d \alpha _1e^{-\alpha _1}+ ade^{-\alpha _1-\alpha _1}}{1-bde^{-\alpha _1-\alpha _2}}\right] . \end{aligned}$$
Hence, we get \(x_1 \le \displaystyle \frac{\alpha _1 + a e ^{-\alpha _1} + b \alpha _2e^{-\alpha _2}+ bce^{-\alpha _2-\alpha _2}}{1-bde^{-\alpha _1-\alpha _2}}.\), i.e., \(x_{1} \in I_1\). Similarly, we get \(y_1 \in I_2.\)
Hence, the proof follows by applying the method of induction. \(\square\)
Theorem 3
Let \(bde^{-\alpha _1-\alpha _2}<1\). Consider the intervals
$$\begin{aligned} I_3= \displaystyle \left[ \alpha _1,\frac{\alpha _1 + a e ^{-\alpha _1} + b \alpha _2e^{-\alpha _2}+ bce^{-\alpha _2-\alpha _2}+\epsilon }{1-bde^{-\alpha _1-\alpha _2}}\right] \end{aligned}$$
and
$$\begin{aligned} I_4=\displaystyle \left[ \alpha _2,\frac{\alpha _2 + c e ^{-\alpha _2} + d \alpha _1e^{-\alpha _1}+ ade^{-\alpha _1-\alpha _1}+\epsilon }{1-bde^{-\alpha _1-\alpha _2}}\right] \end{aligned}$$
where \(\epsilon\) is an arbitrary positive number. If \((x_n,y_n)\) is any arbitrary solution of (3), then there exists an \(N \in {\mathbb {N}}\) such that \(x_n \in I_3\) and \(y_n \in I_4, n \ge N\).
Proof
Let \((x_n,y_n)\) denote an arbitrary solution of (3).
Then by Theorem 1, \(\limsup _{n \rightarrow \infty }{x_n}=M< \infty\) and \(\limsup _{n \rightarrow \infty }{y_n}=L< \infty\).
Hence from Theorem 1, \(x_{n+1} \le A + bdx_{n-1}e^{-\alpha _1-\alpha _2}\) and \(y_{n+1}\le C + bdy_{n-1}e^{-\alpha _1-\alpha _2}\)
Hence \(\displaystyle M \le \frac{A}{1-bde^{-\alpha _1-\alpha _2}}\), and \(\displaystyle L \le \frac{C}{1-bde^{-\alpha _1-\alpha _2}}\).
Hence, there exists an \(N \in {\mathbb {N}}\) such that the theorem holds. \(\square\)
Now we prove a lemma which is an alteration of Theorem 1.16 of [14].
Lemma 4
Let [a, b] and [c, d] denote intervals of real numbers. Let \(f:[a,b]\times [c,d]\times [c,d] \rightarrow [a,b]\) and \(g:[a,b]\times [a,b]\times [c,d] \rightarrow [c,d]\) be continuous functions. Consider the difference system
$$\begin{aligned} x_{n+1}&= f(x_{n-1},y_n,y_{n-1}),\nonumber \\ y_{n+1}&= g(x_n,x_{n-1},y_{n-1}), \quad n=0,1,2,\ldots \end{aligned}$$
(9)
such that the initial values \(x_{-1},x_0 \in [a,b]\) and \(y_{-1}, y_0 \in [c,d]\). (or \(x_{n_0},x_{n_0+1} \in [a,b],\) \(y_{n_0},y_{n_0+1} \in [c,d], n_0 \in {\mathbb {N}}\)). Suppose the following are true.
-
1.
If f(x, y, z) is nonincreasing in x, f(x, y, z) is nondecreasing in y and f(x, y, z) is nonincreasing in z.
-
2.
If g(x, y, z) is nondecreasing in x, g(x, y, z) is nonincreasing in y and g(x, y, z) is nonincreasing in z.
-
3.
If \((m_1,M_1,m_2,M_2) \in [a,b]^2\times [c,d]^2\) satisfies the systems \(m_1=f(M_1,m_2,M_2),\) \(M_1=f(m_1,M_2,m_2)\) and \(m_2=g(m_1,M_1,M_2), M_2=g(M_1,m_1,m_2)\) then \(M_1=m_1\) and \(M_2=m_2\),
then there exists a unique equilibrium solution \(({\bar{x}},{\bar{y}})\) of (9) with \({\bar{x}} \in [a,b]\), \({\bar{y}} \in [c,d]\). Also every solution of (9) converges to \(({\bar{x}},{\bar{y}})\).
Proof
Set \(m_1^{-1}=a, m_1^{0}=a,m_2^{-1}=c, m_2^{0}=c.\)
$$\begin{aligned} M_1^{-1}=b, M_1^{0}=b,M_2^{-1}=d, M_2^{0}=d. \end{aligned}$$
For each \(i \ge 0\), let \(m_1^{i+1}=f(M_1^{i-1},m_2^i,M_2^{i-1}), M_1^{i+1}=f(m_1^{i-1},M_2^i,m_2^{i-1})\) and
$$\begin{aligned} m_2^{i+1}=g(M_1^i,m_1^{i-1},m_2^{i-1}), M_2^{i+1}=g(m_1^i,M_1^{i-1},M_2^{i-1}). \end{aligned}$$
Hence \(m_1^1= f(M_1^{-1},m_2^0,M_2^{-1}) \le f(m_1^{-1},M_2^0,m_2^{-1})= M_1^{1} ,\) and
$$\begin{aligned} m_2^{1}=g(m_1^0,M_1^{-1}, M_2^{-1}) \le g(M_1^0,m_1^{-1},m_2^{-1})= M_2^1. \end{aligned}$$
Therefore,
$$\begin{aligned} M_1^{-1}\ge & {} M_1^0 \ge M_1^1 \ge m_1^1 \ge m_1^0 \ge m_1^{-1} \quad {\hbox {and}}\\ M_2^{-1}\ge & {} M_2^0 \ge M_2^1 \ge m_2^1 \ge m_2^0 \ge m_2^{-1}. \end{aligned}$$
Also \(m_1^{0}=a \le x_n \le b =M_1^0, n\ge 0\) and \(m_2^{0}=c \le y_n \le d =M_2^0, n\ge 0\).
For all \(n\ge 0\), we have
$$\begin{aligned} m_1^{1}&= f(M_1^{-1},m_2^0,M_2^{-1}) \le f(x_{n-1},y_n,y_{n-1}) \le f(m_1^{-1},M_2^0,m_1^{-1}) = M_1^1.\\ m_2^{1}&= g(m_1^0,M_1^{-1},M_2^{-1}) \le g(x_n,x_{n-1},y_{n-1}) \le g(M_1^0,m_1^{-1},M_2^0) =M_2^{1}. \end{aligned}$$
Hence \(m_1^{1} \le x_n \le M_1^1, n\ge 1\) and \(m_2^{1} \le y_n \le M_2^1, n\ge 1\).
We then obtain by induction that for \(i \ge 0\), the following are true.
-
1.
\(a=m_1^{-1}\le m_1^0 \le m_1^1 \ldots \le m_1^{i-1}\le m_1^{i}\le M_1^{i}\ldots \le M_1^1 \le M_1^0 \le M_1^{-1}=b\).
-
2.
\(c=m_2^{-1}\le m_2^0 \le m_2^1 \ldots \le m_2^{i-1}\le m_2^{i}\le M_2^{i}\ldots \le M_2^1 \le M_2^0 \le M_2^{-1}=d\).
-
3.
\(m_1^{i} \le x_n \le M_1^i, n\ge 1\) and \(m_2^{i} \le y_n \le M_2^i, n\ge 1\).
Set \(m_1= \lim _{i \rightarrow \infty } m_1^{i} , m_2= \lim _{i \rightarrow \infty } m_2^{i}\) and \(M_1= \lim _{i \rightarrow \infty } M_1^{i} , M_2= \lim _{i \rightarrow \infty } M_2^{i}\).
Since f and g are continuous, we get \(m_1=f(M_1,m_2,M_2), M_1=f(m_1,M_2,m_2)\) and \(m_2=g(m_1,M_1,M_2), M_2=g(M_1,m_1,m_2).\)
Hence \(M_1=m_1={\bar{x}}\) and \(M_2=m_2={\bar{y}}\), from which we get the proof. \(\square\)
The following theorem proposes conditions for the convergence of the equilibrium solution of (3).
Theorem 5
Suppose
$$\begin{aligned} bde^{-\alpha _1-\alpha _2}<1, ce^{-\alpha _2}<1, ae^{-\alpha _1}<1 \end{aligned}$$
(10)
and
$$\begin{aligned}& \frac{bde^{-\alpha _1-\alpha _2}}{[1-bde^{-\alpha _1-\alpha _2}]^2} \frac{[1-bde^{-\alpha _1-\alpha _2}+ \alpha _2 +ce^{-\alpha _2}+{\mathrm{d}}\alpha _1e^{-\alpha _1} +ade^{-\alpha _1-\alpha _1}]}{[1-ae^{-\alpha _1}]} \\&\quad \times \frac{[1-bde^{-\alpha _1-\alpha _2}+ \alpha _1 +ae^{-\alpha _1}+b\alpha _2e^{-\alpha _2} +bce^{-\alpha _2-\alpha _2}]}{ [1-ce^{-\alpha _2}]}<1. \end{aligned}$$
(11)
Then (3) has a unique positive equilibrium \(E({\bar{x}},{\bar{y}})\). Also, every solution of (3) converges to \(E({\bar{x}},{\bar{y}})\).
Proof
Let \(f: {\mathbb {R}}^+ \times {\mathbb {R}}^+ \times {\mathbb {R}}^+ \rightarrow {\mathbb {R}}^+,g: {\mathbb {R}}^+ \times {\mathbb {R}}^+ \times {\mathbb {R}}^+ \rightarrow {\mathbb {R}}^+\) be continuous functions such that \(f(x,y,z)= \alpha _1 + ae^{-x} +bye^{-z}\), \(g(x,y,z)=\alpha _2 + ce^{-z} +dxe^{-y}\).
Let \(M_1,m_1,M_2,m_2\) be positive real numbers satisfying
$$\begin{aligned} m_1&= \alpha _1 +ae^{-M_1} + bm_2e^{-M_2}, M_1=\alpha _1 +ae^{-m_1} + bM_2e^{-m_2} \end{aligned}$$
and
$$\begin{aligned} m_2=\alpha _2 +ce^{-M_2}+ {\mathrm{d}}m_1e^{-M_1} , M_2=\alpha _2 +ce^{-m_2}+{\mathrm{d}}M_1e^{-m_1}. \end{aligned}$$
(12)
Therefore, \(M_1-m_1=a[e^{-m_1}-e^{-M_1}] + b[M_2e^{-m_2}-m_2e^{-M_2}].\)
$$\begin{aligned} M_1-m_1=a[e^{-m_1}-e^{-M_1}] + be^{-m_2-M_2}[M_2e^{M_2}-m_2e^{m_2}]. \end{aligned}$$
(13)
Also, there exists a \(\zeta\) , \(m_2 \le \zeta \le M_2\) satisfying
$$\begin{aligned} M_2e^{M_2}-m_2e^{m_2}= (1+\zeta ) e^\zeta (M_2-m_2). \end{aligned}$$
(14)
From (13) and (14), we get
$$\begin{aligned} M_1-m_1=a[e^{-m_1}-e^{-M_1}] + be^{-m_2-M_2+\zeta }(1+\zeta )[M_2-m_2]. \end{aligned}$$
(15)
Now, \(a[e^{-m_1}-e^{-M_1}] = ae^{-m_1-M_1}[e^{M_1}-e^{m_1}].\)
Also there exists a \(\lambda\), \(m_1 \le \lambda \le M_1\) satisfying
$$\begin{aligned} a[e^{-m_1}-e^{-M_1}] = a e^{{-m_1-M_1+\lambda }}[M_1-m_1]. \end{aligned}$$
(16)
Since \(M_1,m_1 \ge \alpha _1\) and \(\lambda \le M_1,\)
$$\begin{aligned} a[e^{-m_1}-e^{-M_1}] \le ae^{-\alpha _1}[M_1-m_1]. \end{aligned}$$
(17)
Thus, from (15) and (17) we get,
$$\begin{aligned} M_1-m_1 \le ae^{-\alpha _1}[M_1-m_1] + be^{-m_2-M_2+\zeta }(1+\zeta )[M_2-m_2]. \end{aligned}$$
(18)
Since \(M_2,m_2 \ge \alpha _2\) and \(\zeta \le M_2\), (18) becomes
$$\begin{aligned} M_1-m_1 \le ae^{-\alpha _1}[M_1-m_1] + be^{-\alpha _2}(1+\zeta )[M_2-m_2]. \end{aligned}$$
(19)
, i.e.,
$$\begin{aligned}{}[1-ae^{-\alpha _1}][M_1-m_1] \le be^{-\alpha _2}(1+\zeta )[M_2-m_2]. \end{aligned}$$
(20)
Also, (12) can be written as
$$\begin{aligned} M_2&= \alpha _2 +ce^{-m_2}+{\mathrm{d}}[\alpha _1 +ae^{-m_1}+ bM_2e^{-m_2}]e^{-m_1}. \end{aligned}$$
(21)
$$\begin{aligned} M_2&\le \frac{\alpha _2 +ce^{-\alpha _2}+ {\mathrm{d}}\alpha _1e^{-\alpha _1}+ ade^{-\alpha _1-\alpha _1}}{1-bde^{-\alpha _1-\alpha _2}}. \end{aligned}$$
(22)
Since \(\zeta \le M_2\) we get,
$$\begin{aligned} \zeta \le \frac{\alpha _2 +ce^{-\alpha _2}+ {\mathrm{d}}\alpha _1e^{-\alpha _1}+ade^{-\alpha _1-\alpha _1}}{1-bde^{-\alpha _1-\alpha _2}}. \end{aligned}$$
(23)
Therefore, (20) becomes
$$\begin{aligned}&[1-ae^{-\alpha _1}][M_1-m_1]\nonumber \\&\quad \le be^{-\alpha _2}\left[ \frac{1-bde^{-\alpha _1-\alpha _2}+ \alpha _2 +ce^{-\alpha _2}+{\mathrm{d}}\alpha _1e^{-\alpha _1} +ade^{-\alpha _1-\alpha _1}] }{1-bde^{-\alpha _1-\alpha _2}}\right] [M_2-m_2]. \end{aligned}$$
(24)
Similarly, we get
$$\begin{aligned}&[1-ce^{-\alpha _2}][M_2-m_2]\nonumber \\&\quad \le {\mathrm{d}}e^{-\alpha _1}\left[ \frac{1-bde^{-\alpha _1-\alpha _2}+ \alpha _1 +ae^{-\alpha _1}+b\alpha _2e^{-\alpha _2} +bce^{-\alpha _2-\alpha _2}] }{1-bde^{-\alpha _1-\alpha _2}}\right] [M_1-m_1]. \end{aligned}$$
(25)
From (24) and (25), we get
$$\begin{aligned}&\displaystyle [M_1-m_1]\nonumber \\&\quad \displaystyle \le \frac{bde^{-\alpha _1-\alpha _2}}{[1-(bde^{-\alpha _1-\alpha _2)}]^2} \frac{[1-bde^{-\alpha _1-\alpha _2}+ \alpha _2 +ce^{-\alpha _2}+{\mathrm{d}}\alpha _1e^{-\alpha _1} +ade^{-\alpha _1-\alpha _1}]}{[1-ae^{-\alpha _1}]}\nonumber \\&\qquad \times \frac{[1-bde^{-\alpha _1-\alpha _2}+ \alpha _1 +ae^{-\alpha _1}+b\alpha _2e^{-\alpha _2} +bce^{-\alpha _2-\alpha _2}]}{ [1-ce^{-\alpha _2}]}[M_1-m_1]. \end{aligned}$$
(26)
Therefore from (11) and (26), we get \(M_1=m_1\) and \(M_2=m_2\).
Therefore by applying Lemma 4, the result is obtained. \(\square\)
In the next theorem, we derive conditions for the global asymptotic stability of the equilibrium solution of (3).
Theorem 6
Assume (10) and (11) holds.
-
1.
Let \((a+ac+c)<1\). If \((1+{\bar{x}})(1+{\bar{y}})< \displaystyle \frac{1-(a+ac+c)}{bd}\), then the unique equilibrium \(E({\bar{x}},{\bar{y}})\) is globally asymptotically stable.
-
2.
If \((a+c+ac+bd)+ bd[\frac{A}{1-B}+\frac{C}{1-B}+\frac{AC}{(1-B)^2}]<1\), where A, B and C are defined as in (4) and (5), then the unique equilibrium \(E({\bar{x}},{\bar{y}})\) is globally asymptotically stable.
Proof
First we show that \(E({\bar{x}},{\bar{y}})\) is locally asymptotically stable in both the cases. The Jacobian \(JF({\bar{x}},{\bar{y}})\) about the equilibrium point \(E({\bar{x}},{\bar{y}})\) is given by
$$\begin{aligned} \begin{bmatrix} 0 &{}\quad -ae^{-{\bar{x}}} &{}\quad be^{-{\bar{y}}} &{}\quad -b{\bar{y}}e^{-{\bar{y}}}\\ 1 &{}\quad 0 &{}\quad 0&{}\quad 0\\ {\mathrm{d}}e^{-{\bar{x}}} &{}\quad -d{\bar{x}}e^{-{\bar{x}}} &{}\quad 0 &{}\quad -ce^{-{\bar{y}}}\\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 \end{bmatrix}. \end{aligned}$$
Hence the characteristic equation of the Jacobian \(JF({\bar{x}},{\bar{y}})\) about the equilibrium point \(E({\bar{x}},{\bar{y}})\) is given by
$$\begin{aligned}&\displaystyle -\lambda ^4 + \lambda ^2(-ce^{-{\bar{y}}} + bde^{-{\bar{x}}}e^{-{\bar{y}}} -ae^{-{\bar{x}}})\\&\quad + \lambda (-bd{\bar{y}}e^{-{\bar{x}}}e^{-{\bar{y}}} -bd{\bar{x}}e^{-{\bar{x}}}e^{-{\bar{y}}}) + bd{\bar{x}}{\bar{y}}e^{-{\bar{x}}}e^{-{\bar{y}}}- ace^{-{\bar{x}}}e^{-{\bar{y}}} =0. \end{aligned}$$
Then
$$\begin{aligned}&|-ce^{-{\bar{y}}}|+ |bde^{-{\bar{x}}}e^{-{\bar{y}}}| + |ae^{-{\bar{x}}}|\\&\quad + |bd{\bar{y}}e^{-{\bar{x}}}e^{-{\bar{y}}}| + |bd{\bar{x}}e^{-{\bar{x}}}e^{-{\bar{y}}}| + |bd{\bar{x}}{\bar{y}}e^{-{\bar{x}}}e^{-{\bar{y}}}| + |ace^{-{\bar{x}}}e^{-{\bar{y}}}| <1 \end{aligned}$$
is satisfied whenever
$$\begin{aligned} |c|+ |bd| + |a| + |bd{\bar{y}}| + |bd{\bar{x}}| + |bd{\bar{x}}{\bar{y}}| + |ac| <1. \end{aligned}$$
(27)
-
1.
From (27), we get
$$\begin{aligned} (1+{\bar{x}})(1+{\bar{y}})< \displaystyle \frac{1-(a+ac +c)}{bd}. \end{aligned}$$
(28)
Hence, by (28) and Remark 1.3.1 of [15], we get the result.
-
2.
Since \(E({\bar{x}},{\bar{y}})\) is the equilibrium point of (3), we get
$$\begin{aligned} {\bar{x}} \le \alpha _1 + a e ^{-\alpha _1} + b e^{-\alpha _2} [\alpha _2 +d{\bar{x}}e^{-\alpha _1}+ce^{-\alpha _2}]. \end{aligned}$$
, i.e.,
$$\begin{aligned} {\bar{x}} \le \frac{A}{(1-bde^{-\alpha _1-\alpha _2})}. \end{aligned}$$
(29)
Similarly
$$\begin{aligned} {\bar{y}} \le \frac{C}{(1-bde^{-\alpha _1-\alpha _2})}. \end{aligned}$$
(30)
Substituting (29), (30) in (27), we get
$$\begin{aligned} (a+c+ac+bd)+ bd\left[ \frac{A}{1-B}+\frac{C}{1-B}+\frac{AC}{(1-B)^2}\right] <1. \end{aligned}$$
Hence by Remark 1.3.1 of [15], we get the result.
Therefore by using Theorem 5, we obtain the conditions for global asymptotic stability. \(\square\)