Basin of Attraction through Invariant Curves and Dominant Functions

We study a second-order difference equation of the form z n+1 = z n F(z n−1 ) + h, where both F(z) and zF(z) are decreasing. We consider a set of invariant curves at h = 1 and use it to characterize the behaviour of solutions when h > 1 and when 0 < h < 1. The case h > 1 is related to the Y2K problem. For 0 < h < 1, we study the stability of the equilibrium solutions and find an invariant region where solutions are attracted to the stable equilibrium. In particular, for certain range of the parameters, a subset of the basin of attraction of the stable equilibrium is achieved by bounding positive solutions using the iteration of dominant functions with attracting equilibria.


Introduction
Second-order difference equations of the form where () is a continuous function, have been widely used in applications [1][2][3][4].Several models in mathematical biology take the form of (1) under the assumptions that () is decreasing and () is bounded and increasing [4,5].A well-known example is Pielou's difference equation which has been suggested to model the growth of a single species with delayed-density dependence [4,6].In Pielou's equation,  takes the form () = /(1 + ).Adding or subtracting a constant ℎ from (1) can be interpreted mathematically as a perturbation of the model, or biologically as constant stocking or constant yield harvesting [7][8][9].These meaningful applications motivate investigating the dynamics of the difference equation  +1 =    ( −1 ) ± ℎ,  ∈ N, ℎ > 0. (2) However, when the functions () and () are both decreasing, the corresponding difference equation becomes more abstract and so far little work has been done to investigate its dynamics.Although our ultimate goal is to reach a general theory for this type of difference equation, we find it interesting to consider () = /(−1+) as a prototype, and so we focus this work on the dynamics of the equation  +1 =    ( −1 ) + ℎ,  () =  −1 +  ,  ∈ N, ℎ,  > 0. (3) Among the important aspects of solutions of a difference equation are boundedness and global stability.On several occasions, the question of boundedness of all solutions of a particular difference equation was settled by finding invariant curves.Invariant curves of a second-order difference equation are plane curves on which forward orbits that start on a curve remain on the curve.Finding invariant curves, studying their properties and their relation with Liapunov functions is an active area of research [10][11][12][13].
In recent years, several results that give sufficient conditions for global stability or global attractivity of equilibrium solutions of difference equations have been established.Most of the results rely on the monotonicity of the function defined by the difference equation under investigation; see, for instance, [1,11,14,15] and the references therein.Another approach of establishing global attractivity is to relate solutions of the difference equation to the solutions of a onedimensional map where (under some conditions) the global attractivity of the equilibrium of the map implies the global attractivity of the equilibrium of the difference equation.This technique, which is sometimes called enveloping or dominance, is well-known in first-order difference equations [16], and it has been extended recently to tackle some higher order difference equations [17,18].
In this paper, we focus on (3), where , ℎ > 0 and the initial conditions are restricted to assure the existence of positive solutions (persistent solutions).Throughout our work, solutions are meant to be well-defined solutions.We consider a set of invariant curves and verify certain inequalities they satisfy.The inequalities describe the movement of solutions in the phase plane, which are ultimately used to prove that the larger positive equilibrium of (3) is a global attractor for a certain range of the parameter ℎ.When ℎ > 1, (3) is related (via a change of variables) to the Y2K difference equation where the global attractivity of its positive equilibrium was an open problem for several years till it was settled recently by Merino [19].Also, the case ℎ = 1 is related (via a change of variables) to Lyness equation in which solutions remain on invariant curves.The case 0 < ℎ < 1 of (3) exhibits very interesting dynamics and is discussed in Section 3.

The Case ℎ > 1
In this section, we consider (3) and assume ℎ > 1.Thus, throughout this section, we refer to the equation The substitutions Equation ( 6) is known in the literature as the Y2K problem [18].Observe that, for  > 0 and ℎ > 1, we have   > ℎ if and only if   > 0. Also, the restriction  > 0, ℎ > 1 on the (, ℎ)-parameters is equivalent to the restriction 0 <  <  on the (, )-parameters.In (6), it is obvious that positive initial conditions give rise to positive solutions, and there exists a unique positive equilibrium.Proving that the positive equilibrium is a global attractor with respect to the positive quadrant was an open question for over a decade till it was settled by Merino [19].Merino transformed (6) into where   =   .Then he used the function to prove the next crucial lemma, where  is the positive equilibrium of (7), which is given by Lemma 1 (see [19]).Consider  −1 ,   > 0, and then either Merino used this crucial lemma to show that the positive equilibrium of ( 7) is a global attractor.However, the given proof of Lemma 1 depends heavily on Mathematica code.So here, we present an alternative proof of Lemma 1, which is more trackable.But in order to keep things within the context of our work, we translate the curves   of (7) to the curves   of (4), which are given by where  = (1/)( 2 − ℎ) and  2 is the large equilibrium of (4).Observe that at ℎ = 1, ( −1 ,   ) = constant is in fact an invariant curve for Lyness equation after a transformation.
Proof.We consider  +1 >   and show that  +2 ≤   .Because then we need to show that We write Our aim here is to show that the R.H.S. is larger than or equal to 1.We proceed by taking two cases, namely, the following.( Case 1. Observe that the conditions we have on , , and  force all factors in the numerator and denominator to be positive.Furthermore, we have  >  and the fact that  < (1/)() gives  < .Next, write (15) as and observe that each one of the first three factors in the R.H.S. is larger than 1.Thus, we obtain Use Figure 1(a) to conclude that So, the R.H.S. of Inq. ( 19) is larger than one which completes the proof of Case 1.
Case 2. In this case, we have  > , and since  = /(−1) + ℎ, we obtain  > .To make all factors in both the numerator and denominator of ( 15) positives, we rewrite the equation as Again we show that the R.H.S. is larger than 1.To achieve this task, we rewrite (21) as Observe that the first and last factors are larger than 1, and therefore, we obtain Since () <  and () <  (see Figure 1(b)), we obtain Because the function is increasing on the interval [, ∞) as long as  ≥ ℎ, then  >  implies () > ().Therefore, as required, which completes the proof.

The Case 0 < ℎ < 1
In this section, we consider (3) and let 0 < ℎ < 1.Thus, throughout this section, we refer to the equation We investigate the stability of equilibrium solutions, the existence of periodic solutions, and the boundedness and persistence of solutions.Throughout our work, we use stability to denote local stability and persistence to denote positive solutions.We also find an invariant region and give a range of the parameters for which the region is part of the basin of attraction of the stable equilibrium.

Stability of Equilibria and Periodic Solutions.
The two equilibrium points  1 ,  2 of (27) satisfy 0 <  1 < ℎ and  2 > 1; furthermore, they are increasing in ℎ and satisfy  1  2 = ℎ.The linearized form of (27) at  2 is given by then (30) Thus, the eigenvalues of the linearized equation at  2 are complex and out of the unit circle which make  2 a repeller.Similarly, we find the eigenvalues of the linearized equation at  1 .We have Hence,  > 0 and 0 <  < 1, which give us that  1 is either locally stable or a saddle.Figure 2 illustrates this case in the parameter space.Before we address the issue of boundedness, let us have a look at the possibility of periodic solutions.We always use period to denote the prime period.We start with period-two solutions by considering the equations Observe that the two parabolas intersect in either two or four points and  1 ,  2 must be among them.We substitute and eliminate the factor that gives  1 and  2 to obtain This equation has two real solutions when (ℎ + 1 − ) 2 > 4( − 1)( − ℎ), which gives the region to the left of the curve ℎ 1 () in Figure 2. So, suppose (ℎ + 1 − ) 2 > 4( − 1)( − ℎ) and name the period-two solution {, }.If we go back to the case ℎ > 1, we find that min{, } < ℎ, which is obviously not in the region of global stability considered in [19].For 0 < ℎ < 1, it is possible that min{, } < 0 while max{, } > 0; however this is beyond our interest in this paper.On the other hand, it is possible that min{, } > 0. In fact, this takes place when 0 <  < 1 and ℎ ≥ , or when  ≥ 1 and ℎ 1 (b) ≤ ℎ ≤ .We formalize this discussion in the following result.
Next, by algebraic manipulations of the equations we find that period-three solutions exist and can be positive.However, we avoid high level of computations and just give an interesting example of period-three solution that we use in the sequel.Let ℎ := 1 − (1 − ) and consider  −1 = 0,  0 = ℎ, we obtain a period-three solution given by {0, ℎ, ℎ(1 − )}, which is nonnegative whenever  ≤ 1.
Finally, as a consequence of the boundedness and oscillation results that we establish later on, no periodic solutions of period higher than four exist.Thus, our discussion about the existence of periodic solutions ends by investigating the existence of period-four solutions.In fact, algebraic manipulations show that positive period-four solutions do not exist within the range of our parameters.

Boundedness of Nonnegative Solutions.
In this section, we prove that the only positive solution of (27) that satisfies   > 1 for all  ≥ −1 is the equilibrium solution  2 .Then we show that the remaining positive solutions are bounded.To achieve this task, we first establish a reversed form of the inequalities given in Lemma 1.We can follow the technique used to prove Lemma 1 (when ℎ > 1), but for the readers convenience, we give another elegant technique that serves as a handy tool for similar scenarios.We handle the expression  +2 −   using the points (, ).Thus, we need to consider where  is the map defined by ( −1 ,   ) = (  ,  +1 ).We illustrate these sets in Figure 3, and it is straightforward to check that while Now, write in which Ignore the common factors and keep in mind that (ℎ − 1) is negative and then define Proof.(i) Because  − (ℎ − 1)( − ℎ) > , then we obtain the first part of the statement.To show the second part of the statement, we use the assumption  ≥ ℎ and the fact that  ≥  ≥  2 to conclude that which completes the proof of part (i).
Proof.It is obvious that  1 and  3 are both positive.To show that  2 + 2√ 1  3 > 0, we use part (i) of Proposition 3 to obtain Then, we obtain Because 0 <  < ℎ < 1 <  2 and () = ( − ℎ + )/ is increasing in , we obtain Finally, observe that which implies Hence, the proof is complete.

Lemma 5. Consider (27) with
Proof.Let 0 < ℎ < 1 and assume  +1 <   .Because the common factor (ℎ − 1) that we ignored in (39) is negative, it is sufficient to show that Again, we take two cases as follows.

Attractivity of the Small Equilibrium.
After establishing an invariant region D ,ℎ that contains  1 when 0 < ℎ <  < 1, we aim to characterize the behavior of solutions within D ,ℎ .
From the fact that we observe that if  −1 ,   >  1 then  +1 <  When extrema of consecutive positive (and negative) semicycles form monotonic sequences, we obtain subsequences of the orbit that aids us in characterizing the orbit.This approach has been widely used to prove attractivity of equilibria [1,11,14,20].However, if we follow the extrema of positive semicycles of an orbit of (27), we find it is possible for an extreme value to overshoot (or undershoot) a previous one.For instance, let ℎ = 0.78,  = 0.99,  −1 = 0.06, and  0 = 0.23.In this case  1 ≈ 0.3181, and the orbit through which shows that the extrema of positive semicycles do not form a monotonic sequence.
Instead of bounding an orbit with the extrema of its semicycles, here we develop a technique that bounds the elements of semicycles by a monotonic sequence that does not form a subsequence of the orbit itself, and then we use the monotonic sequence to show the attractivity of  1 .This approach is closely related to the enveloping technique used by [17,18]; however, instead of finding one dominant function that goes through the equilibrium, we find a sequence of functions that evolve based on the semicycles of the solution.We write and we define the maps   as in which   is a fixed value that will be determined by an upper bound of |( −1 )|.In general, we need   to be increasing with a unique fixed point inside a certain region.Before we embark on our main result, we give some characteristics of   in the following proposition.
The condition for the existence of two fixed points is obvious, and to obtain    increasing in   , we must have   − ℎ − 1 negative.
The next result about the attractivity of the small fixed point of   will be used in the sequel.
then the equilibrium solution    of the difference equation  +1 =   (  ) is stable with a basin of attraction that contains the interval [0, V   ).Furthermore, the convergence to the equilibrium is monotonic.
Proof.Since   (0) > 0 and   () is increasing with two fixed points    < V   , a cobweb (stair-step) diagram shows the result.

Figure 1 :
Figure 1: (a) clarifies the proof of Case 1 while (b) clarifies the proof of Case 2. The scale on the axes is intentionally missing because the graphs represent the general situation.
Similarly, consecutive terms of a nonequilibrium solution that satisfy   <  1 form a negative semicycle.Thus, from (67), we conclude that the length of a semicycle is at most two.