Abstract
In the present study, we show that the second-order nonlinear difference equation
where \(a\in \left[ 0,\infty \right) \) and the initial values \(x_{-2}\), \(x_{-1}\) are real numbers, is solvable in closed form and that solutions can be analyzed in detail by the Fibonacci numbers.
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1 Introduction
An important problem in the theory of difference equations is to determine the behavior of the solutions of equations. There are some methods of doing this. The most basic one of these methods is undoubtedly to find a closed formula for the solutions of equations. By doing so, one can reach more concrete results.
While solvability is a natural property for linear difference equations with constant coefficients and their systems, it is an amazing characteristic for nonlinear equations and systems. However, the number of solvable nonlinear difference equations among others is quite small. Discovering the mysteries of solvable nonlinear equations and systems motivates experts to work on them. Recently, there has been intense interest in the solvability of difference equations and their systems. See, for example, [1, 4,5,6,7,8,9,10,11,12,13,14,15,16, 19, 20, 23, 24, 26, 33,34,35,36, 41, 42] and the references cited within them.
Some classes of solvable difference equations and their systems have been investigated for twenty years. Among these, the difference equations defined using trigonometric identities are quite interesting. It has long been known that they are generally solvable. Solvable difference equation classes can also be defined using hyperbolic identities, which are known for their similarity to trigonometric identities.
In [21], Rhouma worked on the solvability of many equations including the hyperbolic cotangent-type difference equations
Although not in detail, he also touched upon the equation, which is the main object of this paper and is in the abstract. In [22], Rhouma gave the closed-form solutions to the cotangent-type difference equation
and the linear difference equation
and the relationship between them, and also the closed-form solutions to the hyperbolic cotangent-type difference equations
and
In [2], Abu-Saris et al. studied globally asymptotically stability of the positive equilibrium point of the hyperbolic cotangent-type difference equation
where \(a\in \left[ 0,\infty \right) \), k is a nonnegative integer and \( x_{-k},x_{-k+1},...,x_{0}\in \left( 0,\infty \right) \). See, also, [37,38,39] for some system of the hyperbolic cotangent-type difference equations. In [29], Stević et al. presented a natural method for solving the hyperbolic cotangent-type difference equation
where \(k, l \in \mathbb {N}\), parameter a, and initial values \(x_{-j}\), \(j=\overline{1, t}\), \(t=\max \{k, l\}\), are real or complex numbers. They presented the case k = 1, l = 2 as a concrete example, and studied it in detail, and gave several interesting formulas for solutions and objects used in the analysis of the equation. In [30], the authors considered eight systems of difference equations of the hyperbolic cotangent-type
where \(a \in [0,+\infty )\), the sequences \(p_{n}\), \(q_{n}\), \(r_{n} \), \(s_{n}\) are some of the sequences \(x_{n}\) and \(y_{n}\), with real initial values \(x_{-i},y_{-i}\), \(\left( i=0,1\right) \). They mainly showed that systems are solvable and described two methods for showing the solvability. Also, they conducted a detailed semi-cycle analysis. In [31], the same authors examined the other eight systems of difference equations, which are the rest parts of the whole, in the same way. There are also solvable difference equations and systems associated with other classes of functions in the literature. For example, for systems defined by power functions, see [25, 27, 28]. For hyperbolic cosine-type difference equations, see [32]. For sine-type difference equations, see [3]. Also, one can find some classic methods on the solvability of some difference equations and systems in [18].
When \(a=1\), the right hand side of Eq. (3) resembles the right hand side of the equality
which is the formula of the hyperbolic cotangent of the sum of x and y. For this reason, we say that Eq. (3) is an equation in hyperbolic cotangent-type. It is worth noting that if the change of variables
is applied to Eq. (1), Eq. (2), Eq. (3) and Eq. (4), all of them are reduced to the form of the identity (6). For further hyperbolic cotangent-type equations and their systems, see [21, 22, 40].
Consider the following equality
which is the formula of the hyperbolic cotangent of \(y-x\). In this case, it is quite natural that the idea that the above equations resemble the identity in (6) makes us think of the following second-order hyperbolic cotangent-type difference equation
where \(a\in \left[ 0,\infty \right) \) and the initial values \(x_{-2}\), \( x_{-1}\) are real numbers. In the present study, we show that Eq. (9) is solvable in closed form and that solutions can be analyzed in detail by the Fibonacci numbers.
In this study, with the notation \(\mathbb {N}_{0}\), we denote the set \(\mathbb {N} \cup \{0\}\). Also, if a sequence \(\left( a_{n}\right) \) converges to a by decreasing with greater values than a, then we denote this case by
while if it converges to a by increasing with less values than a, then we denote this case by
1.1 Auxiliary results
To catch the aim of the paper, we give the following auxiliary information. The following definition helps to characterize the existence of solutions to a difference equation.
Definition 1
Consider the following difference equation
where \(k \in \mathbb {N}\), the initial values \(x_{-1},x_{-2},\ldots ,x_{-k}\) are real numbers and D is the domain of the function g. The forbidden set of Eq. (10) is given by
When studies on the solvability of difference equations are examined, it is seen that such studies generally focus on linear difference equations with constant coefficient. We now consider the special case when the function g is linear. In this case, Eq. (10) takes the form of
where the coefficients \(a_{j}\in \mathbb {R}\) \(\left( j\in \{1,2,\dots ,k\}\right) \) are constants. It has long been known that the higher-order linear homogeneous difference equation with constant coefficients is solvable. Some special cases of Eq. (11) are frequently encountered in the literature. A typical special case of Eq. (11) is the equation
which corresponds to the case \(k=2\) and \(a_{1}=a_{2}=1\). The characteristic polynomial of Eq. (12) is given by
whose roots are
If we choose \(x_{-2}=0\), \(x_{-1}=1\), then the formula of the general solution of (12) becomes
We know that \(x_{n}\) is a term of the well-known Fibonacci sequence. That is, if the nth Fibonacci number is \(F_{n}\), then \( x_{n}=F_{n+2}\). Another special case of Eq. (11) is the equation
which corresponds to the case \(k=2\) and \(a_{1}=-1\), \(a_{2}=1\). The characteristic polynomial of Eq. (14) is given by
whose roots are
If we again choose\(y_{-2}=0\), \(y_{-1}=1\), then the formula of the general solution of (14) becomes
In this case, we see that \(y_{n}=\left( -1\right) ^{n+3}x_{n}\). That is, if the nth Fibonacci number is \(F_{n}\), then \(y_{n}=\left( -1\right) ^{n+3}F_{n+2}\). The numbers in the type of \(\left( -1\right) ^{n+1}F_{n}\) are called Fibonacci numbers with negative index and denoted by \(F_{-n}\).
Now we give a background for the naturally occurring Fibonacci numbers in the formulas in the sequel of our work. Because these numbers, thanks to their interesting properties, make it very easy for us to explain the behavior of the solutions.
Lemma 2
[17]Let \(F_{n}\) be nth Fibonacci number. Then the following statements hold:
-
(a)
\(\left( \frac{1+\sqrt{5}}{2}\right) ^{-\left( 2n+1\right) }=\frac{1+ \sqrt{5}}{2}F_{2n+1}-F_{2n+2}\).
-
(b)
\(\left( \frac{1+\sqrt{5}}{2}\right) ^{-\left( 2n+2\right) }=F_{2n+3}- \frac{1+\sqrt{5}}{2}F_{2n+2}\).
-
(c)
For every \(n\in \mathbb {Z}\), \(F_{3n}\) is even integer, and \(F_{3n+1}\) and \(F_{3n+2}\) are odd integer.
-
(d)
\(\lim _{n\rightarrow \infty }\frac{F_{n+1}}{F_{n}}=\frac{1+\sqrt{5}}{2}\).
2 Solvability and general solution of Eq. (9)
In this section, we give the general solution in closed form of Eq. (9 ) in both the cases \(a=0\) and \(a>0\).
2.1 Case \(a=0\)
Suppose that \(a=0\). Then Eq. (9) becomes
If \(x_{-2}=0\), then \(x_{0}=x_{1}=0\), and therefore \(x_{2}\) is undefined. If \( x_{-1}=0\), then \(x_{0}=0\), and therefore \(x_{1}\) is undefined. Hence, from now on, we suppose that \(x_{-2}x_{-1}\ne 0\). The fact that \( x_{-2}x_{-1}\ne 0\) does not guarantee that the solution is well-defined. So, for now, we can only accept this. Let \(x_{n}\ne 0\) for every \(n\ge -2\). In this case, we can apply the change of variables
to Eq. (15), and so the equation reduces to Eq. (14). If the initial values \(y_{-2}\), \(y_{-1}\) are nonzero real numbers, then Eq. (14) has the solution
where \(n\ge -2\). By using the Fibonacci numbers with negative index, we can rewrite (17) as follows:
Hence, by using (18) and the change of variables (16), the general solution to Eq. (15) is obtained as
2.2 Case \(a>0\)
Suppose that \(a>0\). Then, Eq. (9) is in its original form. Let us first present some of our observations regarding the solutions of the equation. Note that if \(x_{-i}=\sqrt{a}\) for any \(i\in \{1,2\}\), then from (9) we can see that
Repeating this procedure, from Eq. (9), one can easily see that Eq. (9) is eventually periodic with period 2. That is, the solution of Eq. (9) is
Similarly, one can easily see that if \(x_{-i}=-\sqrt{a}\) for any \(i\in \{1,2\}\), then Eq. (9) is eventually periodic with period 2, that is, the solution of Eq. (9) is
From Eq. (9), we have
Suppose that \(\left( x_{-i}+\sqrt{a}\right) \left( x_{-i}-\sqrt{a}\right) \ne 0\) for any \(i\in \{1,2\}\). Then, with the Eqs. in (22)-(23), it can be easily shown that \(\left( x_{n}+\sqrt{a}\right) \left( x_{n}-\sqrt{a}\right) \ne 0\). By dividing (22) by (23), we reach the equation of product type
By applying the change of variables
where \(\left( x_{n}+\sqrt{a}\right) \left( x_{n}-\sqrt{a}\right) \ne 0\), to Eq. (24), we obtain the following difference equation
or equivalently
Let
Then, by (27), we rewrite Eq. (26) as
from which along with \(n\rightarrow n-1\) into itself, we can write
where
Suppose that the following equality
where
have been proved for \(n\ge m\ge 2\). Then, employing Eq. (26) with \( n\rightarrow n-m\) into Eq. (31), we obtain
for \(n\ge m+1\), where
from which along with (29), (30), (33), (34) and induction method, we can say that (31) and (32) hold for all natural numbers n and m such that \(2\le m\le n\). It is clear that (31) also provides for \(1\le m\le n\). From (34) and for \( m=n+1\), Eq. (31) becomes
On the other hand, from (32), the recurrence relation of the sequence \(\left( a_{1}^{\left( m\right) }\right) _{m\in \mathbb {N}_{0}}\) can be written as follows
It is clear that by using (27) and the recurrence relation in (36), one can easily obtain the initial values of Eq. (36) as
Note that (36) with the initial values in (37) is the recurrence relation of Fibonacci sequence with negative index. That is,
where \(F_{n}\) is nth Fibonacci number. Hence, by using (38) in (35) and also (35) in (25) and performing some simple calculations, we get the general solution of Eq. (9) as follows
where \(n\in \mathbb {N}_{0}\). Formula (39) is also provided for \(n=-2\) and \(n=-1\). So we can assume \(n\ge -2\).
The following theorem serves to characterize the well-defined solutions of Eq. (9).
Theorem 3
The forbidden set of Eq. (9) is given by
if \(a=0\), and it is given by
if \(a>0\).
Proof
The formulas in (19) and (39) can be rewritten as
and
respectively. The proof immediately follows from (40) and (41 ) by equalizing to 0 their denominators. \(\square \)
3 Asymptotic behavior of the solutions
In this section, we study the asymptotic behavior of the solutions of Eq. ( 9). Firstly, we suppose that \(a=0\) and the sequence \(\left( x_{n}\right) _{n\ge -2}\) is a well-defined solution of Eq. (9), that is, \(\left( x_{-2},x_{-1}\right) \notin S_{1}\). Then, from (40), we have
for every \(n\ge -1\). Let \(x_{-1}=\frac{1+\sqrt{5}}{2}x_{-2}\). Then, from (a) and (b) of Lemma 2, ( 42) and (43), we get
and
from which it follows that
Let \(x_{-1}\ne \frac{1+\sqrt{5}}{2}x_{-2}\). Then, from (42) and ( 43), we have
Hence, by the fact that \(\lim _{n\rightarrow \infty }F_{n}=\infty \) and (d) of Lemma 2 and by taking limit both sides of above equations as \(n\rightarrow \infty \), it follows that
Suppose that \(a>0\) and the sequence \(\left( x_{n}\right) _{n\ge -2}\) is a well-defined solution of Eq. (9), that is, \(\left( x_{-2},x_{-1}\right) \notin S_{2}\). From (41), we obtain
and
Let
Then the character of \(\left( x_{2n}\right) _{n\ge -1}\) depends on that of the sequence \(\left( r_{n}\right) _{n\ge -1}\), and the character of \(\left( x_{2n+1}\right) _{n\ge -1}\) depends on that of the sequence \(\left( s_{n}\right) _{n\ge -1}\). Moreover, we rewrite \(r_{n}\) and \(s_{n}\) as
and
and
Then the characters of the sequences \(\left( x_{6n}\right) _{n\ge 0}\), \(\left( x_{6n+2}\right) _{n\ge 0}\), \(\left( x_{6n+4}\right) _{n\ge -1}\) depend on that of the sequence \(\left( r_{3n}\right) _{n\ge 0}\), \(\left( r_{3n+1}\right) _{n\ge 0}\), \(\left( r_{3n+2}\right) _{n\ge -1}\), and the characters of the sequences \(\left( x_{6n+1}\right) _{n\ge 0}\), \(\left( x_{6n+3}\right) _{n\ge 0}\), \(\left( x_{6n+5}\right) _{n\ge -1}\) depend on that of the sequence \(\left( s_{3n}\right) _{n\ge 0}\), \(\left( s_{3n+1}\right) _{n\ge 0}\), \(\left( s_{3n+2}\right) _{n\ge -1}\), respectively.
It is obvious from Eq. (9) that if \(x_{-2}=x_{-1}=0\), then \(x_{0}\) and \(x_{1}\) is undefined and so \(x_{n}\) is undefined for \(n\ge 0\).This can also be seen from (54) and (58). Therefore, we suppose that \(\left( x_{-2}\right) ^{2}+\left( x_{-1}\right) ^{2}\ne 0\). Consider the function \(h\left( x\right) =\frac{x+\sqrt{a}}{x-\sqrt{a}}\). It is clear that \(h\left( 0\right) =-1\). We will consider the cases where the initial values are equal to zero separately. Let \(x_{-2}=0\) and \(x_{-1}\ne 0\). Then, by using (c) of Lemma 2, ( 54)-(59) become
and
respectively. Let \(x_{-2}\ne 0\) and \(x_{-1}=0\). Then, by using (c) of Lemma 2, (54)-(59) become
and
respectively. Also, the following cases are provided for the function h.
(i) If \(x\in \left( \sqrt{a},\infty \right) \), then \(h\left( x\right) \in \left( 1,\infty \right) \).
(ii) If \(x\in \left( 0,\sqrt{a}\right) \), then \(h\left( x\right) \in \left( -\infty ,-1\right) \).
(iii) If \(x\in \left( -\sqrt{a},0\right) \), then \(h\left( x\right) \in \left( -1,0\right) \).
(iv) If \(x\in \left( -\infty ,-\sqrt{a}\right) \), then \(h\left( x\right) \in \left( 0,1\right) \).
According to (i)-(iv), we consider the eight cases.
If \(x_{-2}=0\) and \(x_{-1}\in \left( \sqrt{a},\infty \right) \), then \(\frac{ x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}>1\) and so, from (60)-(65), we have
If \(x_{-2}=0\) and \(x_{-1}\in \left( 0,\sqrt{a}\right) \), then \(\frac{x_{-1}+ \sqrt{a}}{x_{-1}-\sqrt{a}}<-1\) and so, from (60)-(65), we have the limits in (72).
If \(x_{-2}=0\) and \(x_{-1}\in \left( -\sqrt{a},0\right) \), then \(-1<\frac{ x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<0\) and so, from ((60)-(65), we have
If \(x_{-2}=0\) and \(x_{-1}\in \left( -\infty ,-\sqrt{a}\right) \), then \(0< \frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<1\) and so, from (60)-(65), we have the limits in (73).
If \(x_{-1}=0\) and \(x_{-2}\in \left( \sqrt{a},\infty \right) \), then \(\frac{ x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}>1\) and so, from (66)-(71), we have the limits in (73).
If \(x_{-1}=0\) and \(x_{-2}\in \left( 0,\sqrt{a}\right) \), then \(\frac{x_{-2}+ \sqrt{a}}{x_{-2}-\sqrt{a}}<-1\) and so, from (66)-(71), we have the limits in (73).
If \(x_{-1}=0\) and \(x_{-2}\in \left( -\sqrt{a},0\right) \), then \(-1<\frac{ x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}<0\) and so, from(66)-(71), we have the limits in (72).
If \(x_{-1}=0\) and \(x_{-2}\in \left( -\infty ,-\sqrt{a}\right) \), then \(0< \frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}<1\) and so, from (66)-(71), we have the limits in (72).
Let \(x_{-2}x_{-1}\ne 0\). Then, since \(x_{-2}x_{-1}\ne 0\), according to (i)-(iv), we consider the sixteen cases and some sub-cases.
3.1 Case 1: \(x_{-2},x_{-1}\in \left( \sqrt{a},\infty \right) \)
Suppose that \(x_{-2},x_{-1}\in \left( \sqrt{a},\infty \right) \). Then \(\frac{ x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}\) \(>1\) and \(\frac{x_{-1}+\sqrt{a}}{x_{-1}- \sqrt{a}}>1\). There are three sub-cases to consider.
Sub-case 1.1: Let
Then, from (a) and (b) of Lemma 2, (46) and (74), we have
Similarly, from (a) and (b) of Lemma 2, (47) and (74), we have
Therefore, for n large enough,
This means that the sequence \(\left( r_{n}\right) _{n\ge -1}\) is monotonically decreasing and less than 1, and so it converges to 0, while the sequence \(\left( s_{n}\right) _{n\ge -1}\) is monotonically increasing and greater than 1, and so it diverges to \(+\infty \). In this case, from (44) and (45), we have that
Sub-case 1.2: Let
Then, from (a) and (b) of Lemma 2, (46)-(47) and (75), we have
and
That is, both the sequences \(\left( r_{n}\right) _{n\ge -1}\) and \(\left( s_{n}\right) _{n\ge -1}\) converge to 1 by decreasing with values greater than 1. In this case, from (44) and (45), we have that
Sub-case 1.3: Let
Then, from (a) and (b) of Lemma 2, (46) and (78), we have
and
Similarly, from (a) and (b) of Lemma 2, (47) and (78), we have
and
Therefore, we have
This means that the sequence \(\left( r_{n}\right) _{n\ge -1}\) is monotonically increasing and greater than 1, and so it diverges to \(\infty \), while the sequence \(\left( s_{n}\right) _{n\ge -1}\) is monotonically decreasing and less than 1, and so it converges to 0. In this case, from (44) and (45), we have that
3.2 Case 2: \(x_{-2},x_{-1}\in \left( 0,\sqrt{a}\right) \)
Suppose that \(x_{-2},x_{-1}\in \left( 0,\sqrt{a}\right) \). Then \(\frac{ x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}\) \(<-1\) and \(\frac{x_{-1}+\sqrt{a}}{x_{-1}- \sqrt{a}}<-1\). Therefore, by using (c) of Lemma 2, \(r_{3n}\), \(r_{3n+1}\), \(r_{3n+2}\) and \(s_{3n}\), \(s_{3n+1}\), \(s_{3n+2}\) become
and
respectively. There are three sub-cases to consider.
Sub-case 2.1. Let
By using (a) and (b) of Lemma 2, (79)-(81) and (85), we obtain
By using (a) and (b) of Lemma 2, (82)-(84) and (85), we obtain
Hence we have that
In this case, from (54)-(59), we have the limits in (72).
Sub-case 2.2. Let
Then, by using (a) and (b) of Lemma 2, (79)-(81) and ( 86), we obtain
By using (a) and (b) of Lemma 2, (82)-(84) and (86), we obtain
Hence we have that
In this case, from (54)-(59), we have that
Sub-case 2.3. Let
By using (a) and (b) of Lemma 2, (79)-(81) and (93), we obtain
By using (a) and (b) of Lemma 2, (82)-(84) and (93), we obtain
Hence we have that
In this case, from (54)-(59), we have the limits in (73).
3.3 Case 3: \(x_{-2},x_{-1}\in \left( -\sqrt{a},0\right) \)
Suppose that \(x_{-2},x_{-1}\in \left( -\sqrt{a},0\right) \). Then, \(-1<\frac{ x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}\) \(<0\) and \(-1<\frac{x_{-1}+\sqrt{a}}{ x_{-1}-\sqrt{a}}<0\). Since the formulas are the same, we here use (79)-(81) and (82)-(84). In this case, by repeating the operations in Sub-case 2.1 and Sub-case 2.3, we again obtain the same results for Sub-case 3.1 and Sub-case 3.3, respectively. But, because of the case of the initial values, we obtain a different result for Sub-case 3.2. In fact, only the sign changes.
Sub-case 3.1. Let
Then, we have the limits in (72).
Sub-case 3.2. Let
Then, we have the limits
Sub-case 3.3. Let
Then, we have the limits in (73).
3.4 Case 4: \(x_{-2},x_{-1}\in \left( -\infty ,-\sqrt{a} \right) \)
Suppose that \(x_{-2},x_{-1}\in \left( -\infty ,-\sqrt{a}\right) \). Then, \(0< \frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}\) \(<1\) and \(0<\frac{x_{-1}+\sqrt{a}}{ x_{-1}-\sqrt{a}}<1\). Since the formulas are the same, we here use (46) and (47). By repeating the operations in Sub-case 1.1 and Sub-case 1.3, we again obtain the same results for Sub-case 4.1 and Sub-case 4.3, respectively. But, because of the case of the initial values, we obtain a different result for Sub-case 4.2. Only the sign changes.
Sub-case 4.1. Let
Then, we have the limits
Sub-case 4.2. Let
Then, we have that both the sequences \(\left( r_{n}\right) _{n\ge -1}\) and \( \left( s_{n}\right) _{n\ge -1}\) converge to \(-1\) by increasing with values less than \(-1\) and so
Sub-case 4.3. Let
Then, we have the limits
3.5 Case 5: \(x_{-2}\in \left( \sqrt{a},\infty \right) \) and \( x_{-1}\in \left( 0,\sqrt{a}\right) \)
Suppose that \(x_{-2}\in \left( \sqrt{a},\infty \right) \) and \(x_{-1}\in \left( 0,\sqrt{a}\right) \). Then, \(\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}\) \( >1\) and \(\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<-1\). In this case, using (c) of Lemma 2, from (48)-(50), \(r_{3n}\), \(r_{3n+1}\), \(r_{3n+2}\) become
Similarly, by using (c) of Lemma 2, from (51)-(53), \(s_{3n}\), \(s_{3n+1}\), \(s_{3n+2}\) become
There are three sub-cases to consider.
Sub-case 5.1. Let
Then, by using (a) and (b) of Lemma 2, (100)-(102), and (106), we obtain
By using (a) and (b) of Lemma 2, (103)-(105), and ( 106), we obtain
Hence we have that
In this case, from (54)-(59), we have the limits in (72).
Sub-case 5.2. Let
Then, by using (a) and (b) of Lemma 2, (100)-(105), and (107), we obtain
and
Hence we have that
In this case, from (54)-(59), we have that
Sub-case 5.3. Let
Then, by using (a) and (b) of Lemma 2, (100)-(102), and (108), we obtain
By using (a) and (b) of Lemma 2, (103)-(105), and ( 108), we obtain
Hence we have that
In this case, from (54)-(59), we have the limits in (73).
3.6 Case 6: \(x_{-2}\in \left( 0,\sqrt{a}\right) \) and \( x_{-1}\in \left( \sqrt{a},\infty \right) \)
Suppose that \(x_{-2}\in \left( 0,\sqrt{a}\right) \) and \(x_{-1}\in \left( \sqrt{a},\infty \right) \). Then \(\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}<-1\) and \(\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}>1\). In this case, by using (c) of Lemma 2, from (48)-(50), \(r_{3n}\), \(r_{3n+1}\), \( r_{3n+2}\) become
Similarly, by using (c) of Lemma 2, from (51)-(53), \(s_{3n}\), \(s_{3n+1}\), \(s_{3n+2}\) become
There are three sub-cases to consider.
Sub-case 6.1. Let
Then, by using (a) and (b) of Lemma 2, (109)-(111), and (115), we obtain
By using (a) and (b) of Lemma 2, (112)-(114), and ( 115), we obtain
Hence we have that
In this case, from (54)-(59), we have the limits in (72).
Sub-case 6.2. Let
Then, by using (a) and (b) of Lemma 2, (109)-(114), and (116), we obtain
and
Hence we have that
In this case, from (54)-(59), we have that
Sub-case 6.3. Let
Then, by using (a) and (b) of Lemma 2, (109)-(111), and (117), we obtain
By using (a) and (b) of Lemma 2, (112)-(114), and ( 117), we obtain
Hence we have that
In this case, from (54)-(59), we have the limits in (73).
3.7 Case 7: \(x_{-2}\in \left( \sqrt{a},\infty \right) \) and \( x_{-1}\in \left( -\sqrt{a},0\right) \)
Suppose that \(x_{-2}\in \left( \sqrt{a},\infty \right) \) and \(x_{-1}\in \left( -\sqrt{a},0\right) \). Then, \(\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}\) \( >1\) and \(-1<\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<0\). In this case, by using (c) of Lemma 2, from (48)-(50), we again obtain (100)-(105). Therefore, with a direct calculation, we obtain
By using these limits in (54)-(59), we have the limits in (73).
3.8 Case 8: \(x_{-2}\in \left( -\sqrt{a},0\right) \) and \( x_{-1}\in \left( \sqrt{a},\infty \right) \)
Suppose that \(x_{-2}\in \left( -\sqrt{a},0\right) \) and \(x_{-1}\in \left( \sqrt{a},\infty \right) \). Then, \(-1<\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}} <0 \) and \(\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}>1\). In this case, using (c) of Lemma 2, from (48)-(50), \(r_{3n}\), \(r_{3n+1}\), \(r_{3n+2}\) become (109)-(111) and \(s_{3n}\), \(s_{3n+1}\), \( s_{3n+2}\) become (112)-(114). Therefore, with a direct calculation, we obtain
By using these limits in (54)-(59), we have the limits in (72).
3.9 Case 9: \(x_{-2}\in \left( \sqrt{a},\infty \right) \) and \( x_{-1}\in \left( -\infty -\sqrt{a}\right) \)
Suppose that \(x_{-2}\in \left( \sqrt{a},\infty \right) \) and \(x_{-1}\in \left( -\infty -\sqrt{a}\right) \). Then, \(\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{ a}}\) \(>1\) and \(0<\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<1\). In this case, from (46)-(47), with a direct calculation, we obtain
By using these limits in (44)-(45), we have
3.10 Case 10: \(x_{-2}\in \left( -\infty -\sqrt{a}\right) \) and \(x_{-1}\in \left( \sqrt{a},\infty \right) \)
Suppose that \(x_{-2}\in \left( -\infty -\sqrt{a}\right) \) and \(x_{-1}\in \left( \sqrt{a},\infty \right) \). Then, \(0<\frac{x_{-2}+\sqrt{a}}{x_{-2}- \sqrt{a}}<1\) and \(\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}\) \(>1\). In this case, from (46)-(47), with a direct calculation, we obtain
By using these limits in (44)-(45), we have
3.11 Case 11: \(x_{-2}\in \left( 0,\sqrt{a}\right) \) and \( x_{-1}\in \left( -\sqrt{a},0\right) \)
Suppose that \(x_{-2}\in \left( 0,\sqrt{a}\right) \) and \(x_{-1}\in \left( - \sqrt{a},0\right) \). Then, \(\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}\) \(<-1\) and \(-1<\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<0\). In this case, using (c) of Lemma 2, from (48)-(50), we again obtain ( 79)-(84). Therefore, with a direct calculation we obtain
By using these limits in (54)-(59), we have the limits in (73).
3.12 Case 12: \(x_{-2}\in \left( -\sqrt{a},0\right) \) and \( x_{-1}\in \left( 0,\sqrt{a}\right) \)
Suppose that \(x_{-2}\in \left( -\sqrt{a},0\right) \) and \(x_{-1}\in \left( 0, \sqrt{a}\right) \). Then, \(-1<\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}<0\) and \( \frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<-1\). In this case, by using (c) of Lemma 2, from (48)-(50), we again obtain (79 )-(84). Therefore, with a direct calculation, we obtain
By using these limits in (54)-(59), we have the limits in (72).
3.13 Case 13: \(x_{-2}\in \left( 0,\sqrt{a}\right) \) and \( x_{-1}\in \left( -\infty -\sqrt{a}\right) \)
Suppose that \(x_{-2}\in \left( 0,\sqrt{a}\right) \) and \(x_{-1}\in \left( -\infty -\sqrt{a}\right) \). Then, \(\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}\) \( <-1\) and \(0<\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<1\). In this case, by using (c) of Lemma 2, from (48)-(50), we again obtain (109)-(114). Therefore, with a direct calculation we obtain
By using these limits in (54)-(59), we have the limits in (73).
3.14 Case 14: \(x_{-2}\in \left( -\infty -\sqrt{a}\right) \) and \(x_{-1}\in \left( 0,\sqrt{a}\right) \)
Suppose that \(x_{-2}\in \left( -\infty -\sqrt{a}\right) \) and \(x_{-1}\in \left( 0,\sqrt{a}\right) \). Then, \(0<\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}} <1 \) and \(\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<-1\). In this case, by using (c) of Lemma 2, from (48)-(50), we again obtain (100)-(105). Therefore, with a direct calculation we obtain
By using these limits in (54)-(59), we have the limits in (72).
3.15 Case 15: \(x_{-2}\in \left( -\sqrt{a},0\right) \) and \( x_{-1}\in \left( -\infty ,-\sqrt{a}\right) \)
Suppose that \(x_{-2}\in \left( -\sqrt{a},0\right) \) and \(x_{-1}\in \left( -\infty ,-\sqrt{a}\right) \). Then, \(0<\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}} <1\) and \(-1<\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}<0\). In this case, by using (c) of Lemma 2, from (48)-(50), we again obtain (109)-(114). There are three sub-cases to consider.
Sub-case 15.1. Let
Then, by using (a) and (b) of Lemma 2, (109)-(114), and the assumption (118), we obtain
In this case, from (54)-(59), we have the limits in (72).
Sub-case 15.2. Let
Then, by using (a) and (b) of Lemma 2, (109)-(114), and the assumption (119), we obtain
In this case, from (54)-(59), we have that
Sub-case 15.3. Let
Then, by using (a) and (b) of Lemma 2, (109)-(114), and the assumption (120), we obtain
In this case, from (54)-(59), we have the limits in (73).
3.16 Case 16: \(x_{-2}\in \left( -\infty ,-\sqrt{a}\right) \) and \(x_{-1}\in \left( -\sqrt{a},0\right) \)
Suppose that \(x_{-2}\in \left( -\infty ,-\sqrt{a}\right) \) and \(x_{-1}\in \left( -\sqrt{a},0\right) \). Then, \(0<\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}} <1\) and \(-1<\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}<0\). In this case, by using (c) of Lemma 2, from (48)-(50), \(r_{3n}\), \( r_{3n+1}\), \(r_{3n+2}\) become (100)-(102) and \(s_{3n}\), \( s_{3n+1}\), \(s_{3n+2}\) become (103)-(105). There are three sub-cases to consider.
Sub-case 16.1. Let
Then, by using (a) and (b) of Lemma 2, (100)-(105), and (121), we obtain
In this case, from (54)-(59), we have the limits in (72).
Sub-case 16.2. Let
Then, by using (a) and (b) of Lemma 2, (100)-(105), and (122), we obtain
In this case, from (54)-(59), we have that
Sub-case 16.3. Let
Then, by using (a) and (b) of Lemma 2, (100)-(105), and (123), we obtain
In this case, from (54)-(59), we have the limits in (73).
As a summary of the above considerations, we give the following result.
Theorem 4
Suppose that \(a\in \left[ 0,\infty \right) \) and the sequence \( \left( x_{n}\right) _{n\ge -2}\) is a well-defined solution of Eq. (9 ), that is, if \(a=0\), then \(\left( x_{-2},x_{-1}\right) \notin S_{1}\), and if \(a>0\), then \(\left( x_{-2},x_{-1}\right) \notin S_{2}\). Then, the following statements are true:
-
(a)
If \(a=0\), then the general solution to Eq. (9) is given by formula (19).
-
(b)
If \(a=0\) and \(x_{-1}=\frac{1+\sqrt{5}}{2}x_{-2}\), then \(x_{n}\rightarrow \)sgn\((x_{-2})\infty \), as \(n\rightarrow \infty \).
-
(c)
If \(a=0\) and \(x_{-1}\ne \frac{1+\sqrt{5}}{2}x_{-2}\), then \( x_{n}\rightarrow 0\), as \(n\rightarrow \infty \).
-
(d)
If \(a>0\), then the general solution to Eq. (9) is given by formula (39).
-
(e)
If \(a>0\) and \(\left( x_{-2}\pm \sqrt{a}\right) \left( x_{-1}\mp \sqrt{a} \right) =0\), then every solution of Eq. (9) is eventually periodic with prime period two.
-
(f)
If \(a>0\) and \(\left| \frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}} \right| \ne \left| \frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}} \right| ^{\frac{1+\sqrt{5}}{2}}\), then every solution of Eq. (9) converges to its periodic points \(\sqrt{a}\) and \(-\sqrt{a}\).
-
(g)
If \(a>0\) and \(\left| \frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}} \right| =\left| \frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}\right| ^{ \frac{1+\sqrt{5}}{2}}\), then Eq. (9) has unbounded solutions.
4 Numerical examples
In this section, to support and validate our theoretical results, we provide some numerical examples that include its simulations.
Example 5
Consider the following difference equation
with the parameter \(a=0\) and the initial values \(x_{-2}=0.9,\) \(x_{-1}=1.7.\) Then the following figure shows that every solution of Eq. (124) tends to 0.
The plot of Eq. (124) with \(a=0\)
Example 6
Consider the following difference equation
with the parameter \(a=0\) and the initial values \(x_{-2}=1.9,\) \(x_{-1}=1.9\left( \frac{1+\sqrt{5}}{2}\right) .\) Then the following figure shows that every solution of Eq. (125) are divergent.
The plot of Eq. (125) with \(a=0\)
Example 7
Consider the following difference equation
with the parameter \(a=6.4\) and the initial values \(x_{-2}=1.61,\) \(x_{-1}=0.49.\) Since \(x_{-2}, x_{-1} \in \left( 0,\sqrt{a}\right) ,\) we have \(\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}} <-1\), \(\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}} <-1\) and \(-\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}=4.50068 > \left( -\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}\right) ^{\frac{1+\sqrt{5}}{2}}=1.88667\). From (73), the following figure shows that \(\left( x_{n}\right) \) converges to a two-periodic solution. Moreover, for \(i\in \{0,1,2\},\) the sequences \(x_{6n+2i}\rightarrow \sqrt{a}\) as \(n\rightarrow \infty \) and the sequences \(x_{6n+2i+1}\rightarrow -\sqrt{a}\) as \(n\rightarrow \infty .\)
The plot of Eq. (126) with \(a=6.4\)
Example 8
Consider the following difference equation
with the parameter \(a=0.5\) and the initial values \(x_{-2}=-0.23,\) \(x_{-1}=-5.44.\) Since \(x_{-2}\in \left( -\sqrt{a},0\right) \) and \(x_{-1} \in \left( -\infty ,-\sqrt{a}\right) ,\) we have \(-1<\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}} <0\), \(0<\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}} <1\) and \(-\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}=0.509127 < \left( \frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}\right) ^{\frac{1+\sqrt{5}}{2}}=0.655061\). From (72), the following figure shows that \(\left( x_{n}\right) \) converges to a two-periodic solution. Moreover, for \(i\in \{0,1,2\},\) the sequences \(x_{6n+2i}\rightarrow -\sqrt{a}\) as \(n\rightarrow \infty \) and the sequences \(x_{6n+2i+1}\rightarrow \sqrt{a}\) as \(n\rightarrow \infty .\)
The plot of Eq. (127) with \(a=0.5\)
Example 9
Consider the following difference equation
with the parameter \(a=6.4\) and the initial values \(x_{-2}=\sqrt{a}\frac{-1+\left( -\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}\right) ^{\frac{1+\sqrt{5}}{2}}}{1+\left( -\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}\right) ^{\frac{1+\sqrt{5}}{2}}},\) \(x_{-1}=1.9.\) Since \(x_{-2}, x_{-1} \in \left( 0,\sqrt{a}\right) ,\) we have \(\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}} <-1\), \(\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}} <-1\) and \(-\frac{x_{-2}+\sqrt{a}}{x_{-2}-\sqrt{a}}= \left( -\frac{x_{-1}+\sqrt{a}}{x_{-1}-\sqrt{a}}\right) ^{\frac{1+\sqrt{5}}{2}}\). From Sub-case 2.2, the following figure shows that some sub-sequences of the sequence \(\left( x_{n}\right) \) are divergent while others converge to zero.
The plot of Eq. (128) with \(a=6.4\)
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Tollu, D.T., Yazlık, Y. Solvability and solution character of a hyperbolic cotangent-type difference equation of second-order. J. Appl. Math. Comput. (2024). https://doi.org/10.1007/s12190-024-02085-5
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DOI: https://doi.org/10.1007/s12190-024-02085-5