Abstract
In this paper, we solve and study the global behavior of the admissible solutions of the difference equation
where \(a, b>0\) and the initial values \(x_{-2}\), \(x_{-1}\), \(x_{0}\) are real numbers.
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1 Introduction
In [11], the author obtained the solutions and determined the forbidden sets of the two difference equations
and
where the initial values \(x_{-2}\), \(x_{-1}\), \(x_{0}\) are real numbers.
In [3], the author determined the forbidden set and discussed the global behavior of the solutions of the difference equation
where a, b, c are positive real numbers and the initial conditions \(x_{-2}\),\(x_{-1}\),\(x_0\) are real numbers.
He also in [12] determined the forbidden sets and discussed the global behaviors of solutions of the two difference equations
where a, b, c are positive real numbers and the initial conditions \( x_{-2},x_{-1},x_0\) are real numbers.
Recently, there has been a great interest in studying properties of nonlinear and rational difference equations (see [1, 4,5,6,7,8,9,10, 12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37] and the references therein).
In this paper, we study the difference equation
where \(a, b>0\) and the initial values \(x_{-2}\), \(x_{-1}\), \(x_{0}\) are real numbers.
The transformation
reduces Eq. (1) into the difference equation
The solution of (3) depends on the relation between a and b. By solving Eq. (3) and after some calculations, the solution of Eq. (1) can be obtained.
2 Case \(b^2>4a\)
In this section, we assume that \(b^2>4a\). Suppose that
where \(\lambda _{-}=\frac{b}{2}-\frac{\sqrt{b^2-4a}}{2}\) and \(\lambda _{+}=\frac{b}{2}+\frac{\sqrt{b^2-4a}}{2}\), \( j=0,1,\ldots \).
We give the following Lemma without proof:
Lemma 2.1
The following identities are true:
-
(1)
$$\begin{aligned} -a\theta _{j}+b\theta _{j+1}=\theta _{j+2},\quad j=0,1,\ldots \end{aligned}$$
-
(2)
$$ \begin{gathered} - a( - ax_{0} \theta _{j} + x_{{ - 1}} \theta _{{j + 1}} ) + b( - ax_{0} \theta _{{j + 1}} + x_{{ - 1}} \theta _{{j + 2}} ) \hfill \\ \quad = - ax_{0} \theta _{{j + 2}} + x_{{ - 1}} \theta _{{j + 3}} ,\quad j = 0,1, \ldots \hfill \\ \end{gathered} $$
-
(3)
$$ \begin{gathered} - a( - ax_{{ - 1}} \theta _{j} + x_{{ - 2}} \theta _{{j + 1}} ) + b( - ax_{{ - 1}} \theta _{{j + 1}} + x_{{ - 2}} \theta _{{j + 2}} ) \hfill \\ \quad = - ax_{{ - 1}} \theta _{{j + 2}} + x_{{ - 2}} \theta _{{j + 3}} ,\quad j = 0,1, \ldots \hfill \\ \end{gathered} $$
Theorem 2.2
Let \(\{x_n\}_{n=-2}^\infty \) be an admissible solution of Equation (1). Then
where \(\nu =x_{0}x_{-1}x_{-2}\), \(\gamma _{-i}(n)=-ax_{-i}\theta _{n} +x_{-i-1}\theta _{n+1}\), \(i=0,1\) and \( j=0,1,\ldots \).
Proof
We can write the given solution (4) as
When \(m=0\),
Similarly
Suppose that the result is true for \(m>0\).
Then
Using Lemma (2.1), we get
Similarly we can show that
This completes the proof. \(\square \)
Consider the set
and
Theorem 2.3
The two sets \(D_1\) and \(D_2\) are invariant sets for Equation (1).
Proof
Let \((x_0,x_{-1},x_{-2})\in D_1\). We show that \((x_n,x_{n-1},x_{n-2})\in D_1\) for each \(n\in {\mathbb {N}}\). The proof is by induction on n. The point \((x_0,x_{-1},x_{-2})\in D_1\) implies
Now for \(n=1\), we have
Then we have
This implies that \((x_1,x_{0},x_{-1})\in D_1\).
Suppose now that \((x_n,x_{n-1},x_{n-2})\in D_1\). That is
Then
Then \((x_{n+1},x_{n},x_{n-1})\in D_1\). Therefore, \(D_1\) is an invariant set for Eq. (1).
By similar way, we can show that \(D_2\) is an invariant set for Eq. (1).
This completes the proof. \(\square \)
Theorem 2.4
Assume that \(\{x_n\}_{n=-2}^\infty \) is an admissible solution of Equation (1). The following statements are true:
-
(1)
If \(b>a+1\), then the solution \(\{x_n\}_{n=-2}^\infty \) converges to zero.
-
(2)
If \(b=a+1\), then we have the following:
-
(a)
If \(b<2\), then the solution \(\{x_n\}_{n=-2}^\infty \) converges to a finite limit.
-
(b)
If \(b>2\), then the solution \(\{x_n\}_{n=-2}^\infty \) converges to zero.
-
(a)
-
(3)
If \(b<a+1\), then we have the following:
-
(a)
If \(b<2\), then the solution \(\{x_n\}_{n=-2}^\infty \) is unbounded.
-
(b)
If \(b>2\), then the solution \(\{x_n\}_{n=-2}^\infty \) converges to zero.
-
(a)
Proof
We can write \(\theta _j=\lambda ^j_{+}\frac{(1- (\frac{\lambda _{-}}{\lambda _{+}})^j)}{\sqrt{b^2-4a}}\).
-
(1)
If \(b>a+1\), then \(\lambda _{+}>1\). That is \(\theta _m\rightarrow \infty \) as \(m\rightarrow \infty \). Also both of \(\gamma _{0}(m)\) and \(\gamma _{-1}(m)\) are unbounded. This implies that
$$\begin{aligned} x_{2m+1}=\frac{\nu }{\gamma _{0}(m)\gamma _{-1}(m+1)}\rightarrow 0\ \text {as}\ m\rightarrow \infty . \end{aligned}$$Similarly for \(x_{2m+2}\).
-
(2)
Suppose that \(b=a+1\), then either \(\lambda _{-}=1\) or \(\lambda _{+}=1\).
-
(a)
If \(b<2\), then \(\lambda _{+}=1\). It follows that \(\theta _m\rightarrow \frac{1}{\sqrt{b^2-4a}}\) as \(m\rightarrow \infty \). This implies that
$$\begin{aligned} x_{2m+1}\rightarrow \frac{\nu (b^2-4a)}{(-ax_0+x_{-1})(-ax_{-1}+x_{-2})}. \end{aligned}$$Similarly for \(x_{2m+2}\).
-
(b)
If \(b>2\), then \(\lambda _{-}=1\). It follows that \(\theta _m\rightarrow \infty \) as \(m\rightarrow \infty \). This implies that
$$\begin{aligned} x_{2m+1}=\frac{\nu }{\gamma _{0}(m)\gamma _{-1}(m+1)}\rightarrow 0\ \text {as}\ m\rightarrow \infty . \end{aligned}$$Similarly for \(x_{2m+2}\).
-
(a)
-
(3)
The proof is similar to that in (2) and is omitted.
This completes the proof. \(\square \)
Example (1)
Figure 1 shows that, if \(a=1.1,b=2.2\) (\(b>a+1\)), then a solution \(\{x_n\}_{n=-2}^\infty \) of Eq. (1) with \(x_{-2}=3\), \(x_{-1}=-1\) and \(x_{0}=2\) converges to zero.
Example (2)
Figure 2 shows that, if \(a=0.7,b=1.7\) (\(b=a+1\) and \(b<2\)), then a solution \(\{x_n\}_{n=-2}^\infty \) of Eq. (1) with \(x_{-2}=3\), \(x_{-1}=-1\) and \(x_{0}=2\) converges to
Example (3)
Figure 3 shows that, if \(a=0.2,b=1\) (\(b<a+1\) and \(b<2\)), then a solution \(\{x_n\}_{n=-2}^\infty \) of Eq. (1) with \(x_{-2}=3\), \(x_{-1}=-1\) and \(x_{0}=2\) is unbounded.
Example (4)
Figure 4 shows that, if \(a=3, b=3.7\) (\(b<a+1\) and \(b>2\)), then a solution \(\{x_n\}_{n=-2}^\infty \) of Eq. (1) with \(x_{-2}=3\), \(x_{-1}=-1\) and \(x_{0}=2\) converges to zero.
3 Case \(b^2=4a\)
During this section, we assume that \(b^2=4a\).
Theorem 3.1
Let \(\{x_n\}_{n=-2}^\infty \) be an admissible solution of Equation (1). Then
where \(\nu =x_{0}x_{-1}x_{-2}\), \(\mu _{-i}(n)=-bnx_{-i}+2(1+n)x_{-i-1}\), \(i=0,1\) and \( j=0,1,\ldots \).
Proof
It is sufficient to write the formula (6) as
as well as noting that
\(\square \)
Theorem 3.2
Assume that \(\{x_n\}_{n=-2}^\infty \) is an admissible solution of Equation (1). The following statements are true:
-
(1)
If \(b\ge 2\), then the solution \(\{x_n\}_{n=-2}^\infty \) converges to zero.
-
(2)
If \(b<2\), then the solution \(\{x_n\}_{n=-2}^\infty \) is unbounded.
Proof
Clear that both of \(\mu _{0}(m)\) and \(\mu _{-1}(m)\) are unbounded. Then (1) is satisfied directly. When \(b<2\), then \((\frac{2}{b})^m\rightarrow \infty \) as \(m\rightarrow \infty \).
Then using formulas (7) we can write
By applying L’Hospital’s rule, we get \(|x_{2m+1}|\rightarrow \infty \) as \(m\rightarrow \infty \).
Similarly for \(|x_{2m+2}|\).
This completes the proof. \(\square \)
Example (5)
Figure 5 shows that, if \(a=0.64,b=1.6\) (\(b^2=4a\) and \(b<2\)), then a solution \(\{x_n\}_{n=-2}^\infty \) of Eq. (1) with \(x_{-2}=3\), \(x_{-1}=-1\) and \(x_{0}=2\) is unbounded.
Example (6)
Figure 6 shows that, if \(a=2.25,b=3\) (\(b^2=4a\) and \(b>2\)), then a solution \(\{x_n\}_{n=-2}^\infty \) of Eq. (1) with \(x_{-2}=3\), \(x_{-1}=-1\) and \(x_{0}=2\) converges to zero.
4 Case \(b^2<4a\)
In this section, we study the final case when \(b^2<4a\) and provide the forbidden set of Eq. (1).
Theorem 4.1
Let \(\{x_n\}_{n=-2}^\infty \) be an admissible solution of Equation (1). Then
where \(\nu =x_{0}x_{-1}x_{-2}\), \(\beta _{-i}(n)= \sqrt{a}x_{-i}\sin n\alpha -x_{-i-1}\sin (n+1)\alpha \),
\(\alpha = \tan ^{-1}\frac{\sqrt{4a-b^2}}{b}\in ]0,\frac{\pi }{2}[\) and \(i=0,1\).
Proof
We can write the given solution (8) as
When \(m=0\),
Similarly
Suppose that the result is true for \(m>0\).
Then
Using the identity
we get
Similarly we can show that
This completes the proof. \(\square \)
In the following result, we show the existence of periodic solutions for Eq. (1).
Theorem 4.2
Assume that \(\{x_n\}_{n=-2}^\infty \) is an admissible solution of Equation (1) and let \(a=1\). If \(\alpha =\frac{l}{k}\pi \) is a rational multiple of \(\pi \) (with \(0<l<\frac{k}{2}\)), then \(\{x_n\}_{n=-2}^\infty \) is periodic with prime period 2k.
Proof
Suppose that \(a=1\). Using formula (9), we can write
But for each \(i=0,1\) we have
Then
Similarly, we can show that \(x_{2m+2k+2}=x_{2m+2}\).
This completes the proof. \(\square \)
Theorem 4.3
Assume that \(\{x_n\}_{n=-2}^\infty \) is an admissible solution of Equation (1). The following statements are true:
-
(1)
If \(a<1\), then the solution \(\{x_n\}_{n=-2}^\infty \) is unbounded.
-
(2)
If \(a>1\), then the solution \(\{x_n\}_{n=-2}^\infty \) converges to zero.
Proof
The proof is a direct consequence using formulas (9). \(\square \)
Example (7)
Figure 7 shows that, if \(a=1,b=\sqrt{\frac{3-\sqrt{5}}{2}}\) (\(b^2<4a\) and \(\alpha =\frac{2\pi }{5}\)), then a solution \(\{x_n\}_{n=-2}^\infty \) of Eq. (1) with \(x_{-2}=3\), \(x_{-1}=-1\) and \(x_{0}=2\) is periodic with prime period 10.
Example (8)
Figure 8 shows that, if \(a=1,b=\sqrt{3}\) (\(b^2<4a\) and \(\alpha =\frac{\pi }{6}\)), then a solution \(\{x_n\}_{n=-2}^\infty \) of Eq. (1) with \(x_{-2}=3\), \(x_{-1}=-1\) and \(x_{0}=2\) is periodic with prime period 12.
In the remainder of this section, we give the forbidden set of Eq. (1) (which depends on the relation between a and b).
Clear that, if \(x_0=0\) and \(x_{-1}x_{-2}\ne 0\), then \(x_{3}\) is undefined. If \(x_{-1}=0\) and \(x_{0}x_{-2}\ne 0\), then \(x_{5}\) is undefined. If \(x_{-2}=0\) and \(x_{0}x_{-1}\ne 0\), then \(x_{4}\) is undefined.
Therefore, the point \((x_{0},x_{-1},x_{-2})\) with \(x_{0}x_{-1}x_{-2}=0\) belongs to the forbidden set of Eq. (1).
The following result gives the forbidden sets of Eq. (1) for all values of \(a>0\) and \(b>0\).
Theorem 4.4
The following statements are true:
-
(1)
If \(b^2>4a\), then the forbidden set of equation (1) is
$$\begin{aligned} \begin{aligned} F_1=&\bigcup _{i=0}^2\{(u_0,u_{-1},u_{-2})\in {\mathbb {R}}^3:u_{-i}=0\}\cup \\ {}&\bigcup _{m=1}^\infty \left\{ (u_0,u_{-1},u_{-2})\in {\mathbb {R}}^3:u_{0}=\frac{u_{-1}}{a}\frac{\theta _{m+1}}{\theta _{m}}\right\} \cup \\ {}&\bigcup _{m=1}^\infty \left\{ (u_0,u_{-1},u_{-2})\in {\mathbb {R}}^3:u_{-1}=\frac{u_{-2}}{a}\frac{\theta _{m+1}}{\theta _{m}}\right\} . \end{aligned} \end{aligned}$$ -
(2)
If \(b^2=4a\), then the forbidden set of equation (1) is
$$\begin{aligned} \begin{aligned} F_2=&\bigcup _{i=0}^2\{(u_0,u_{-1},u_{-2})\in {\mathbb {R}}^3:u_{-i}=0\}\cup \\ {}&\bigcup _{m=1}^\infty \left\{ (u_0,u_{-1},u_{-2})\in {\mathbb {R}}^3:u_{0}=2\frac{1+m}{m}u_{-1}\right\} \cup \\ {}&\bigcup _{m=1}^\infty \left\{ (u_0,u_{-1},u_{-2})\in {\mathbb {R}}^3:u_{-1}=2\frac{1+m}{m}u_{-2}\right\} . \end{aligned} \end{aligned}$$ -
(3)
If \(b^2>4a\), then the forbidden set of equation (1) is
$$\begin{aligned} \begin{aligned} F_3=&\bigcup _{i=0}^2\{(u_0,u_{-1},u_{-2})\in {\mathbb {R}}^3:u_{-i}=0\}\cup \\ {}&\bigcup _{m=1}^\infty \left\{ (u_0,u_{-1},u_{-2})\in {\mathbb {R}}^3:u_{0}=\frac{\sin (m+1)\alpha }{\sqrt{a}\sin m \alpha }u_{-1}\right\} \cup \\&\bigcup _{m=1}^\infty \left\{ (u_0,u_{-1},u_{-2})\in {\mathbb {R}}^3:u_{-1}=\frac{\sin (m+1)\alpha }{\sqrt{a}\sin m \alpha }u_{-2}\right\} . \end{aligned}\end{aligned}$$
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Abo-Zeid, R., Kamal, H. Global behavior of a third order difference equation with quadratic term. Bol. Soc. Mat. Mex. 27, 23 (2021). https://doi.org/10.1007/s40590-021-00337-z
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DOI: https://doi.org/10.1007/s40590-021-00337-z