In this section, we study the local stability and global attractivity of the equilibrium point of Eq. (E).
Lemma 3.1
If
\(\alpha \neq 1\), then Eq. (E) has a positive equilibrium point
$$ \overline{x}=f^{1/ ( 1\alpha ) } ( 1,1 ) . $$
(3.1)
Also, if
\(\alpha =1\)
and
\(f ( 1,1 ) \neq 1\), then Eq. (E) has only zero equilibrium point.
Proof
The equilibrium point of Eq. (E) is given by \(\overline{x}=f ( \overline{x},\overline{x} ) \). Since f is homogeneous with degree α, we obtain
$$ \overline{x}=\overline{x}^{\alpha }f ( 1,1 ) . $$
If \(\alpha \neq 1\), then \(\overline{x}=0\) or \(\overline{x}=f^{1/ ( 1\alpha ) } ( 1,1 ) \). Otherwise, if \(\alpha =1\) and \(f ( 1,1 ) \neq 1\), then we have \(\overline{x}=0\). Thus, the proof is completed. □
Theorem 3.1
The zero equilibrium point of Eq. (E) is locally asymptotically stable if
\(\alpha >1\), or
$$ \alpha =1\quad \textit{and}\quad \bigl\vert f_{u} ( 1,1 ) \bigr\vert + \bigl\vert f_{v} ( 1,1 ) \bigr\vert < 1. $$
(3.2)
Proof
Since f homogeneous with degree α, we have \(f_{u}\) and \(f_{v}\) are homogeneous with degree \(\alpha 1\). Now, if \(\alpha >1\), then we get \(f_{u} ( \overline{x},\overline{x} ) = \overline{x}^{\alpha 1}f_{u} ( 1,1 ) =0\) and \(f_{v} ( \overline{x},\overline{x} ) =0\), for \(\overline{x}=0\). Hence, \(\overline{x}=0\) is locally asymptotically stable. Next, if \(\alpha =1\), then \(f_{u} ( \overline{x},\overline{x} ) =f _{u} ( 1,1 ) \) and \(f_{v} ( \overline{x},\overline{x} ) =f_{v} ( 1,1 ) \). By using Theorem 1.3.7 in [24], we see that Eq. (E) is locally stable if
$$ \bigl\vert f_{u} ( 1,1 ) \bigr\vert + \bigl\vert f_{v} ( 1,1 ) \bigr\vert < 1, $$
which completes the proof. □
Theorem 3.2
The positive equilibrium point of Eq. (E) is locally asymptotically stable if
$$ \bigl\vert f_{u} ( 1,1 ) \bigr\vert + \bigl\vert f_{v} ( 1,1 ) \bigr\vert < f ( 1,1 ) $$
(3.3)
or
$$\begin{aligned} &0< \alpha < 1, \quad \textit{for } f_{u}>0,f_{v}>0; \\ &1< \alpha < 0, \quad \textit{for }f_{u}< 0,f_{v}< 0; \\ &2f_{u} ( 1,1 ) < ( 1+\alpha ) f ( 1,1 ), \quad \textit{for }f_{u}>0,f_{v}< 0; \\ &2f_{v} ( 1,1 ) < ( 1+\alpha ) f ( 1,1 ),\quad \textit{for }f_{u}< 0,f_{v}>0. \end{aligned}$$
Proof
The linearized equation of (E) about x̅ is the linear difference equation
$$ y_{n+1}= \frac{\partial f}{\partial u}\biggm _{ (\overline{x},\overline{x} ) }y_{nl}+ \frac{\partial f}{\partial v}\biggm _{ ( \overline{x},\overline{x} ) }y_{nk}. $$
(3.4)
From Theorem 1.3.7 in [24], Eq. (3.4) is locally stable if
$$ \bigl\vert f_{u} ( \overline{x},\overline{x} ) \bigr\vert + \bigl\vert f_{v} ( \overline{x},\overline{x} ) \bigr\vert < 1. $$
Using Corollary 2 in [7], we see that \(f_{u}\) and \(f_{v}\) are homogeneous with degree \(\alpha 1\). This implies
$$ \bigl\vert f_{u} ( 1,1 ) \bigr\vert + \bigl\vert f_{v} ( 1,1 ) \bigr\vert < f ( 1,1 ) . $$
If we admit that \(f_{u}>0\) and \(f_{v}>0\), then we obtain
$$ f_{u} ( \overline{x},\overline{x} ) +f_{v} ( \overline{x}, \overline{x} ) < 1. $$
(3.5)
From Euler’s homogeneous function theorem, we deduce that
$$ uf_{u}+vf_{v}=\alpha f, $$
(3.6)
which with (3.5) gives \(0<\alpha <1\). Similarly, if \(f_{u}<0\) and \(f_{v}<0\), then we get \(1<\alpha <0\).
In the case where \(f_{u}>0\) and \(f_{v}<0\), we have
$$ f_{u} ( \overline{x},\overline{x} ) f_{v} ( \overline{x}, \overline{x} ) < 1. $$
(3.7)
Combining (3.6) with (3.7), we get
$$ 2f_{u} ( 1,1 ) < ( 1+\alpha ) f ( 1,1 ) . $$
Finally, if \(f_{u}<0\) and \(f_{v}>0\), then we find
$$ 2f_{v} ( 1,1 ) < ( 1+\alpha ) f ( 1,1 ) . $$
Hence, the proof is completed. □
Example 3.1
Let the difference equation
$$ x_{n+1}=ax_{n}^{\alpha }+bx_{n1}^{\alpha }, $$
(3.8)
where α, a and b are real numbers \(,~a>0\), \(b>0\) and \(\alpha \neq 1\). Since \(f ( u,v ) =au^{\alpha }+bv^{\alpha }\), we get
$$ \alpha f_{u}>0 \quad \mbox{and} \quad \alpha f_{v}>0. $$
By using Theorem 3.2, the positive equilibrium point \(\overline{x}= ( a+b ) ^{1/ ( 1\alpha ) }\) of Eq. (3.8) is locally asymptotically stable if \(\vert \alpha \vert <1\). For example, for \(\alpha =0.6\), \(a=0.2\), \(b=0.7\), \(x_{1}=2.0\) and \(x_{0}=0.2\), the stable solution of (3.8) is shown in Fig. 2.
Remark 3.1
If \(\alpha =0\), then \(uf_{u}+vf_{v}=0\) and \(f_{u}f_{v}<0\). Thus, Eq. (3.8) is locally asymptotically stable if
$$ \bigl\vert f_{u} ( 1,1 ) \bigr\vert < \frac{1}{2}f ( 1,1 ) . $$
(3.9)
Theorem 3.3
Assume that
f
has nonpositive partial derivatives. Then Eq. (E) has a unique positive equilibrium
x̅
and every solution of Eq. (E) converges to
x̅.
Proof
Let \(( m,M ) \) is a solution of the system
$$\begin{aligned}& m = f ( M,M ) , \\& M = f ( m,m ) . \end{aligned}$$
This implies
$$\begin{aligned}& m = M^{\alpha }f ( 1,1 ) , \\& M = m^{\alpha }f ( 1,1 ) , \end{aligned}$$
and so
$$ \bigl( m^{\alpha +1}M^{\alpha +1} \bigr) f ( 1,1 ) =0. $$
Hence, we get \(m=M\). By Theorem 1.4.7 in [26], we see that every solution of Eq. (E) converges to x̅. Hence, the proof is completed. □
Remark 3.2
Assume that \(f_{u}>0\) and \(f_{v}<0\). By Theorem 1.4.5 in [26], if we were able to obtain a condition that ensures that
$$ \frac{f ( z,1 ) zf ( 1,z ) }{1z^{\alpha +1}} \neq 0, $$
(3.10)
for all \(z\in ( 0,\infty ) \), then the equilibrium point x̅ would be a global attractor of Eq. (E).
Remark 3.3
Assume that \(f\in C ( [ 0,\infty ) \times [ 0, \infty ) , [ 0,\infty ) ) \), \(f_{u}f_{v}>0\), \(\vert \alpha \vert <1,~l=0\) and \(k=1\). Then, by Euler’s homogeneous function theorem, we see that
$$\begin{aligned} u\vert f_{u}\vert +v\vert f_{v}\vert =&\vert uf _{u}+vf_{v}\vert \\ =&\vert \alpha \vert f \\ < &f, \end{aligned}$$
for all \(u,v\in ( 0,\infty ) \). Thus, by using Theorem 1.4.4 in [26], Eq. (E) has exactly one of the following three cases for all solutions (stability trichotomy):

(a)
\(\lim_{n\rightarrow \infty }x_{n} =\infty\) for \(x_{1}x_{0} \neq 0\).

(b)
\(\lim_{n\rightarrow \infty }x_{n} =0\) and Eq. (E) has only a zero equilibrium point.

(c)
\(\lim_{n\rightarrow \infty }x_{n} =\overline{x}\) for \(x_{1}x _{0}\neq 0\) and x̅ is the only positive equilibrium point.