Research article Special Issues

An optimistic firefly algorithm-based deep learning approach for sentiment analysis of COVID-19 tweets

  • The unprecedented rise in the number of COVID-19 cases has drawn global attention, as it has caused an adverse impact on the lives of people all over the world. As of December 31, 2021, more than 2, 86, 901, 222 people have been infected with COVID-19. The rise in the number of COVID-19 cases and deaths across the world has caused fear, anxiety and depression among individuals. Social media is the most dominant tool that disturbed human life during this pandemic. Among the social media platforms, Twitter is one of the most prominent and trusted social media platforms. To control and monitor the COVID-19 infection, it is necessary to analyze the sentiments of people expressed on their social media platforms. In this study, we proposed a deep learning approach known as a long short-term memory (LSTM) model for the analysis of tweets related to COVID-19 as positive or negative sentiments. In addition, the proposed approach makes use of the firefly algorithm to enhance the overall performance of the model. Further, the performance of the proposed model, along with other state-of-the-art ensemble and machine learning models, has been evaluated by using performance metrics such as accuracy, precision, recall, the AUC-ROC and the F1-score. The experimental results reveal that the proposed LSTM + Firefly approach obtained a better accuracy of 99.59% when compared with the other state-of-the-art models.

    Citation: H. Swapnarekha, Janmenjoy Nayak, H. S. Behera, Pandit Byomakesha Dash, Danilo Pelusi. An optimistic firefly algorithm-based deep learning approach for sentiment analysis of COVID-19 tweets[J]. Mathematical Biosciences and Engineering, 2023, 20(2): 2382-2407. doi: 10.3934/mbe.2023112

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  • The unprecedented rise in the number of COVID-19 cases has drawn global attention, as it has caused an adverse impact on the lives of people all over the world. As of December 31, 2021, more than 2, 86, 901, 222 people have been infected with COVID-19. The rise in the number of COVID-19 cases and deaths across the world has caused fear, anxiety and depression among individuals. Social media is the most dominant tool that disturbed human life during this pandemic. Among the social media platforms, Twitter is one of the most prominent and trusted social media platforms. To control and monitor the COVID-19 infection, it is necessary to analyze the sentiments of people expressed on their social media platforms. In this study, we proposed a deep learning approach known as a long short-term memory (LSTM) model for the analysis of tweets related to COVID-19 as positive or negative sentiments. In addition, the proposed approach makes use of the firefly algorithm to enhance the overall performance of the model. Further, the performance of the proposed model, along with other state-of-the-art ensemble and machine learning models, has been evaluated by using performance metrics such as accuracy, precision, recall, the AUC-ROC and the F1-score. The experimental results reveal that the proposed LSTM + Firefly approach obtained a better accuracy of 99.59% when compared with the other state-of-the-art models.



    Nonlinear difference equations have long interested both mathematics and other sciences. Since these equations play a key role in many applications such as the natural model of a discrete process, they appear in many disciplines such as population biology, optics, economics, probability theory, genetics, psychology. See e.g., [1,2,3,4,5,6] and the references therein. For the last two decades, there has been interest in studying the global attractivity, the boundedness character and the periodic nature of nonlinear difference equations. For some recent results see, for example, [7,8,9,10,11,12,13,14,15,16]. However, for the last decade, some researchers have focused on the solvability of nonlinear difference equations and their systems. For some recent results see, for example, [17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33].

    In this paper, we consider the following system of second-order nonlinear difference equations

    xn+1=a1+ynxn1, yn+1=b1+xnyn1, nN0, (1.1)

    where the parameters a, b, and the initial values x1, x0, y1, y0 are arbitrary real numbers such that the solution {( xn,yn)}n1 exists. System (1.1) can be obtained systematically as follows. First, we consider the following difference equation

    xn+1=a+bxnc+dxn, adbc, d0, nN0, (1.2)

    where the parameters a, b, c, d, and the initial value x0 are arbitrary such that xn are defined. Equation (1.2) is called a first order linear fractional difference equation. Equation (1.2) is solvable by virtue of several changes of variables. The most common method is to transform Eq (1.2) into the second order linear equation by using the change of variables c+dxn=zn/zn1. For a detailed background on Eq (1.2), see e.g, [34,35]. Also, for other equations related to Eq (1.2), see [36,37,38,39]. A different case occurs when we get b=0. This case yields the following difference equation

    xn+1=ac+dxn, acd0, nN0. (1.3)

    Equation (1.3) can also be transformed into the second order linear equation by using the change of variables xn=zn1/zn and so is solvable. Some generalizations of Eq (1.3) can inherit its solvability property. For example, the following difference equation

    xn+1=ac+dxnxn1, acd0, nN0, (1.4)

    where the parameters a, c, d, and the initial values x1, x0 are arbitrary such that xn are defined, is also solvable by using the change of variables xn=zn1/zn. Hence, the general solutions of (1.2)–(1.4) follow from the general solutions of the associated linear equations and the corresponding changes of variables. Note that both Eq (1.2) and Eq (1.3) can be reduced equations with one parameter or two parameters. If we choose xn=cdun, adc2=α, and xn=cdvn, acdc=β, then they are reduced equations with one parameter in un and vn, respectively. Therefore, we can take c=d=1 under favorable conditions.

    Based on the above considerations, we investigate a two-dimensional generalization that maintains the solvability characteristic of Eq (1.4). So, we get a further generalization of (1.4), that is, the system given in (1.1). System (1.1) can also be transformed into a system of third-order linear equations by using the changes of variables xn=un1/vn, yn=vn1/un, and so can be solved. But, we will use a more practical method introduced firstly in [40] to solve the system.

    We need to the following two results in the sequel of our study.

    Lemma 1.1. [41] Consider the cubic equation

    P(z)=z3αz2βzγ=0. (1.5)

    Equation (1.5) has the discriminant

    Δ=α2β24β3+4α3γ+27γ2+18αβγ. (1.6)

    Then, the following statements are true:

    (i) If Δ<0, then the polynomial P has three distinct real zeros ρ1, ρ2, ρ3.

    (ii) If Δ=0, then there are two subcases:

    (a) if β=α23 and γ=α327 , then the polynomial P has the triple root ρ=α3,

    (b) if βα23 or γα327, then the polynomial P has the double root r and the simple root ρ.

    (iii) If Δ>0, then the polynomial P has one real root p and two complex roots re±iθ,θ(0,π).

    Theorem 1.1 (Kronecker's theorem). [42] If θ is irrational, the set of points un=nθ[nθ] is dense in the interval (0,1).

    In the above theorem, n is an integer and [nθ] is greatest integer function of nθ.

    This section, which contains our main results, is examined in three subsections.

    In this subsection, by using an interesting and practical method, we solve system (1.1). If we take a=0 in system (1.1), then we have xn=0 for every n1 and yn=b for every n2. If we take b=0 in system (1.1), then we have xn=a for every n2 and yn=0 for every n1. So, to enable the use of the method, we suppose ab0 in the sequel of our study.

    We start by writing system (1.1) in the following

    1x2n+1=1a+y2nx2n1a, (2.1)
    1x2n+2=1a+y2n+1x2na, (2.2)
    1y2n+1=1b+x2ny2n1b, (2.3)
    1y2n+2=1b+x2n+1y2nb (2.4)

    for every n0. By multiplying (2.1), (2.2), (2.3) and (2.4) by

    1nk=0x2k1nk=0y2k, (2.5)
     1nk=0x2kn+1k=0y2k1, (2.6)
    1nk=0x2knk=0y2k1, (2.7)
    1n+1k=0x2k1nk=0y2k, (2.8)

    we have the followings

    1n+1k=0x2k1nk=0y2k=1ank=0x2k1nk=0y2k+1an1k=0x2k1n1k=0y2k, (2.9)
    1n+1k=0x2kn+1k=0y2k1=1ank=0x2kn+1k=0y2k1+1an1k=0x2knk=0y2k1, (2.10)
    1nk=0x2kn+1k=0y2k1=1bnk=0x2knk=0y2k1+1bn1k=0x2kn1k=0y2k1 (2.11)
    1n+1k=0x2k1n+1k=0y2k=1bn+1k=0x2k1nk=0y2k+1bnk=0x2k1n1k=0y2k (2.12)

    for every n0, respectively. In fact, the equalities (2.9)–(2.12) constitute a linear system with respect to (2.5)–(2.8). Hence, we should solve (2.9)–(2.12). By using (2.9) in (2.12), (2.12) in (2.9) and similarly by using (2.10) in (2.11), (2.11) in (2.10), we have the following statements

    1n+1k=0x2k1n+1k=0y2k=1abnk=0x2k1nk=0y2k+2abn1k=0x2k1n1k=0y2k+1abn2k=0x2k1n2k=0y2k, (2.13)
    1n+1k=0x2k1nk=0y2k=1abnk=0x2k1n1k=0y2k+2abn1k=0x2k1n2k=0y2k+1abn2k=0x2k1n3k=0y2k, (2.14)
    1nk=0x2kn+1k=0y2k1=1abn1k=0x2knk=0y2k1+2abn2k=0x2kn1k=0y2k1+1abn3k=0x2kn2k=0y2k1, (2.15)

    and

    1n+1k=0x2kn+1k=0y2k1=1abnk=0x2knk=0y2k1+2abn1k=0x2kn1k=0y2k1+1abn2k=0x2kn2k=0y2k1 (2.16)

    for every n2. Note that the equations in (2.13)–(2.16) are linear with respect to (2.5), (2.8), (2.6) and (2.7), respectively, and they can be represented by the following third-order difference equation

    zn+11abzn2abzn11abzn2=0, n2, (2.17)

    whose characteristic equation is the following equation

    P(λ)=λ31abλ22abλ1ab=0. (2.18)

    By Lemma 1.1, we see that there are three cases to be considered.

    In this case P has three real distinct zeros denoted by ρ1, ρ2, ρ3, respectively. Hence, from (2.13) and (2.17), we can write

    zn=C1ρn1+C2ρn2+C3ρn3=1nk=0x2k1nk=0y2k (2.19)

    from which it follows that

    1x2n1y2n=C1ρn1+C2ρn2+C3ρn3C1ρn11+C2ρn12+C3ρn13, (2.20)

    where C1, C2, C3 are arbitrary real constants given by

    C1(x1,y0)=ρ2ρ3x1y2x3y4(ρ2+ρ3)x3y4+1(ρ1ρ2)(ρ1ρ3)x1y0x1y2x3y4,C2(x1,y0)=ρ1ρ3x1y2x3y4(ρ1+ρ3)x3y4+1(ρ2ρ1)(ρ2ρ3)x1y0x1y2x3y4,C3(x1,y0)=ρ1ρ2x1y2x3y4(ρ1+ρ2)x3y4+1(ρ3ρ1)(ρ3ρ2)x1y0x1y2x3y4,

    for every n0. By using (2.20) in the first equation of system (1.1), we have

    x2n+1=a(C1ρn1+C2ρn2+C3ρn3)C1(ρ1+1)ρn11+C2(ρ2+1)ρn12+C3(ρ3+1)ρn13 (2.21)

    for every n1. On the other hand, the first equation of system (1.1) can be written as follows

    y2n=ax2n+1x2n+1x2n1 (2.22)

    for every n0. By using (2.21) and its backward shifted one from n to n1 in (2.22), we have

    y2n=C1(ρ1+1)ρn21+C2(ρ2+1)ρn22+C3(ρ3+1)ρn23a(C1ρn1+C2ρn2+C3ρn3) (2.23)

    for every n0. Now, we consider Eq (2.16), which is linear with respect to (2.7) and (2.17). Hence, we have

    zn=C1ρn1+C2ρn2+C3ρn3=1nk=0x2knk=0y2k1 (2.24)

    from which it follows that

    1x2ny2n1=C1ρn1+C2ρn2+C3ρn3C1ρn11+C2ρn12+C3ρn13, (2.25)

    where C1, C2, C3 are arbitrary real constants given by

    C1=C1(y1,x0), C2=C2(y1,x0), C3=C3(y1,x0),

    for every n0. By using (2.25) in the second equation of system (1.1), we have

    y2n+1=b(C1ρn1+C2ρn2+C3ρn3)C1(ρ1+1)ρn11+C2(ρ2+1)ρn12+C3(ρ3+1)ρn13 (2.26)

    for every n1. On the other hand, the second equation of system (1.1) can be written as follows

    x2n=by2n+1y2n+1y2n1 (2.27)

    for every n0. By using (2.26) and its backward shifted one from n to n1 in (2.27), we have

    x2n=C1(ρ1+1)ρn21+C2(ρ2+1)ρn22+C3(ρ3+1)ρn23b(C1ρn1+C2ρn2+C3ρn3) (2.28)

    for every n0. Consequently, in the case 4ab<27, the representation forms of the general solution of system (1.1) are given by (2.21), (2.23), (2.26) and (2.28).

    In this case P(λ) has the simple root ρ and the double root r. Moreover, since ab=427, we have ρ=3/4 and r=3. Hence, from (2.13) and (2.17), we have

    zn=C1ρn+rn(C2+C3n)=1nk=0x2k1nk=0y2k (2.29)

    from which it follows that

    1x2n1y2n=C1ρn+rn(C2+C3n)C1ρn1+rn1(C2+C3(n1)), (2.30)

    where C1, C2, C3 are arbitrary real constants given by

    C1(x1,y0)=r2x1y2x3y42rx3y4+1(ρr)2x1y0x1y2x3y4,C2(x1,y0)=ρ(ρ2r)x1y2x3y4+2rx3y41(ρr)2x1y0x1y2x3y4,C3(x1,y0)=ρrx1y2x3y4(ρ+r)x3y4+1(rρ)rx1y0x1y2x3y4,

    for every n0. By substituting (2.30) in the first equation of system (1.1), we have

    x2n+1=a(C1ρn+rn(C2+C3n))C1(ρ+1)ρn1+rn1(C2(r+1)+C3(nr+n1)) (2.31)

    for every n1. On the other hand, by using (2.31) and its backward shifted one from n to n1 in (2.22), we have

    y2n=C1(ρ+1)ρn2+rn2(C2(r+1)+C3((n1)r+n2))a(C1ρn+rn(C2+C3n)) (2.32)

    for every n0. Now, by considering Eqs (2.16) and (2.17). we have

    zn=C1ρn+rn(C2+C3n)=1nk=0x2knk=0y2k1 (2.33)

    from which it follows that

    1x2ny2n1=C1ρn+rn(C2+C3n)C1ρn1+rn1(C2+C3(n1)), (2.34)

    where C1, C2, C3 are arbitrary real constants given by

    C1=C1(y1,x0), C2=C2(y1,x0), C3=C3(y1,x0),

    for every n0. By substituting (2.34) in the second equation of system (1.1), we have

    y2n+1=b(C1ρn+rn(C2+C3n))C1(ρ+1)ρn1+rn1(C2(r+1)+C3(nr+n1)) (2.35)

    for every n1. On the other hand, by using (2.35) and its backward shifted one from n to n1 in (2.27), it follows that

    x2n=C1(ρ+1)ρn2+rn2(C2(r+1)+C3((n1)r+n2))b(C1ρn+rn(C2+C3n)) (2.36)

    for every n0. Consequently, in the case 4ab=27, the representation forms of the general solution of system (1.1) are given by (2.31), (2.32), (2.35) and (2.36).

    In this case P(λ) has one real root and two complex roots denoted by ρ and re±iθ,θ(0,π), respectively. Hence, from (2.13) and (2.17), we have

    zn=C1ρn+rn(C2cosnθ+C3sinnθ)=1nk=0x2k1nk=0y2k (2.37)

    from which it follows that

    1x2n1y2n=C1ρn+rn(C2cosnθ+C3sinnθ)C1ρn1+rn1(C2cos(n1)θ+C3sin(n1)θ), (2.38)

    where C1, C2, C3 are arbitrary real constants given by

    C1(x1,y0)=r2x1y2x3y42rcosθx3y4+1(ρ22ρrcosθ+r2)x1y0x1y2x3y4,C2(x1,y0)=ρ(ρ2rcosθ)x1y2x3y4+2rcosθx3y41(ρ22ρrcosθ+r2)x1y0x1y2x3y4,C3(x1,y0)=ρr(rcos2θρcosθ)x1y2x3y4+(ρ2r2cos2θ)x3y4+rcosθρrsinθ(ρ22ρrcosθ+r2)x1y0x1y2x3y4,

    for every n0. By using (2.38) in the first equation of system (1.1), we have

    x2n+1=a(C1ρn+rn(C2cosnθ+C3sinnθ))C1(ρ+1)ρn1+rn1(C4cosnθ+C5sinnθ), (2.39)

    where C4=C2(r+cosθ)C3sinθ, C5=C3(r+cosθ)+C2sinθ, for every n1. On the other hand, by using (2.39) and its backward shifted one from n to n1 in (2.22), we have

    y2n=C1(ρ+1)ρn2+rn2(C4cos(n1)θ+C5sin(n1)θ)a(C1ρn+rn(C2cosnθ+C3sinnθ)) (2.40)

    for every n0. Now, by considering Eqs (2.16) and (2.17), we have

    zn=C1ρn+rn(C2cosnθ+C3sinnθ)=1nk=0x2knk=0y2k1 (2.41)

    from which it follows that

    1x2ny2n1=C1ρn+rn(C2cosnθ+C3sinnθ)C1ρn1+rn1(C2cos(n1)θ+C3sin(n1)θ), (2.42)

    where C1, C2, C3 are arbitrary real constants given by

    C1=C1(y1,x0), C2=C2(y1,x0), C3=C3(y1,x0),

    for every n0. By using (2.38) in the second equation of system (1.1), we have

    y2n+1=b(C1ρn+rn(C2cosnθ+C3sinnθ))C1(ρ+1)ρn1+rn1(C4cosnθ+C5sinnθ) (2.43)

    where C4=C2(r+cosθ)C3sinθ, C5=C3(r+cosθ)+C2sinθ, for every n1. On the other hand, by using (2.43) and its backward shifted one from n to n1 in (2.27), we have

    x2n=C1(ρ+1)ρn2+rn2(C4cos(n1)θ+C5sin(n1)θ)b(C1ρn+rn(C2cosnθ+C3sinnθ)) (2.44)

    for every n0. Consequently, in the case 4ab>27, the representation forms of the general solution of system (1.1) are given by (2.39), (2.40), (2.43) and (2.44).

    The representation forms given in the previous subsection are valid where the denominators are not zero. That is, we can obtain the set of initial values that make the solutions of the system undefined from the forms by equating their denominators to zero. This operation enables us to obtain a set of initial values that produce the well-defined solutions of system (1.1). In the following we give a theorem that helps us characterize such solutions.

    Theorem 2.1. Consider system (1.1). Then, the following statements are true:

    (a) If 4ab<27, then the forbidden set of system (1.1) is given by

    F={(x1,x0,y1,y0):αn=0 or βn=0 or αn=0 or βn=0},

    where

    αn=C1ρn1+C2ρn2+C3ρn3n0,βn=C1(ρ1+1)ρn11+C2(ρ2+1)ρn12+C3(ρ3+1)ρn13, n1,αn=C1ρn1+C2ρn2+C3ρn3,n0,βn=C1(ρ1+1)ρn11+C2(ρ2+1)ρn12+C3(ρ3+1)ρn13, n1.

    (b) If 4ab=27, then the forbidden set of system (1.1) is given by

    F={(x1,x0,y1,y0):αn=0 or βn=0 or αn=0 or βn=0},

    where

    αn=C1ρn+rn(C2+C3n)n0,βn=C1(ρ+1)ρn1+rn1(C2(r+1)+C3(nr+n1)), n1,αn=C1ρn+rn(C2+C3n),n0,βn=C1(ρ+1)ρn1+rn1(C2(r+1)+C3(nr+n1)), n1,

    and ρ=3/4, r=3.

    (c) If 4ab>27, then the forbidden set of system (1.1) is given by

    F={(x1,x0,y1,y0):αn=0 or βn=0 or αn=0 or βn=0},

    where

    αn=C1ρn+rn(C2cosnθ+C3sinnθ)n0βn=C1(ρ+1)ρn1+rn1(C4cosnθ+C5sinnθ), n1αn=C1ρn+rn(C2cosnθ+C3sinnθ),n0βn=C1(ρ+1)ρn1+rn1(C4cosnθ+C5sinnθ), n1.

    Proof. The proof is simple and follows by equalizing denominators of the representation forms obtained in the previous section to zero.

    By considering this theorem, we say that a well-defined solution of system (1.1) is a solution {(xn,yn)}n1 obtained using the initial values such that (x1,x0,y1,y0)R4F.

    In this subsection we study the long-term behavior of the solutions of system (1.1) by using the representation forms obtained in the first subsection. We analyze the solutions in the following cases of the parameter ab:

    i) Case 4ab<27: in this case we have ab(427,0).

    ii) Case 4ab=27: in this case we have ab=427.

    iii) Case 4ab>27: in this case we have ab(,427)(0,+).

    This case yields the following result.

    Theorem 2.2. Let {(xn,yn)}n1 be a well-defined solution of system (1.1). Suppose that 4ab<27. Then, the following statements are true:

    (a) If Ci0 for i{1,2,3} and |ρ|=max{|ρ1|,|ρ2|,|ρ3|}, then x2n+1aρρ+1 and y2nbρρ+1 as n.

    (b) If Ci0 for i{1,2,3} and |ρ|=max{|ρ1|,|ρ2|,|ρ3|} , then x2naρρ+1 and y2n+1bρρ+1 as n.

    (c) If Ci=0 and CjCk0 for i,j,k{1,2,3} with ijk and |ρ|=max{|ρj|,|ρk|}, then x2n+1aρρ+1 and y2nbρρ+1 as n.

    (d) If Ci=0 and CjCk0 for i,j,k{1,2,3} with ijk and |ρ|=max{|ρj|,|ρk|}, then x2naρρ+1 and y2n+1bρρ+1 as n.

    Proof. (a)–(b) Let us assume without losing generality that |ρ1|=max{|ρ1|,|ρ2|,|ρ3|}. Then, we have the following limits

    limnx2n=limnρn21ρn1C1(ρ1+1)+C2(ρ2+1)(ρ2ρ1)n2+C3(ρ3+1)(ρ3ρ1)n2b(C1+C2(ρ2ρ1)n+C3(ρ3ρ1)n)=ρ1+1bρ21.

    Since ρ1 is a zero of the polynomial P, we have the relation

    ρ1+1bρ21=aρ1ρ1+1

    from (2.18). Hence, we have

    limnx2n=aρ1ρ1+1

    and

    limnx2n+1=limnρn1ρn11a(C1+C2(ρ2ρ1)n+C3(ρ3ρ1)n)C1(ρ1+1)+C2(ρ2+1)(ρ2ρ1)n1+C3(ρ3+1)(ρ3ρ1)n1=aρ1ρ1+1,
    limny2n=limnρn21ρn1C1(ρ1+1)+C2(ρ2+1)(ρ2ρ1)n2+C3(ρ3+1)(ρ3ρ1)n2a(C1+C2(ρ2ρ1)n+C3(ρ3ρ1)n)=ρ1+1aρ21=bρ1ρ1+1,
    limny2n+1=limnρn1ρn11b(C1+C2(ρ2ρ1)n+C3(ρ3ρ1)n)C1(ρ1+1)+C2(ρ2+1)(ρ2ρ1)n1+C3(ρ3+1)(ρ3ρ1)n1=bρ1ρ1+1.

    The proofs of other cases are similar and so they will be omitted.

    Remark 2.1. Note that the cases |C1|+|C2|+|C3|=|Ci| and |C1|+|C2|+|C3|=|Cj|, i,j{1,2,3} are impossible. Because, for example, if C1=C2=0, then we need to the common solution of the system

    ρ2ρ3x1y2x3y4(ρ2+ρ3)x3y4+1=0, ρ1ρ3x1y2x3y4(ρ1+ρ3)x3y4+1=0.

    This case requires that ρ1=ρ2 which is a contradiction.

    Corollary 2.1. Suppose that 4ab<27. Then, every well-defined solution of system (1.1) has a finite limit point.

    This case yields the following result.

    Theorem 2.3. Let {(xn,yn)}n1 be a well-defined solution of system (1.1). Suppose that 4ab=27. Then, the following statements are true:

    (a) If |C2|+|C3|0, then x2n+13a2and y2n3b2 as n.

    (b) If |C2|+|C3|0, then x2n3a2 and y2n+13b2 as n.

    (c) If C2=C3=0 and C10, then x2n+13a and y2n3b as n.

    (d) If C2=C3=0 and C10, then x2n3a, and y2n+13b as n.

    Proof. (a)–(b) Since ρ=3/4 and r=3, we have |ρ|<|r|. So, from (2.31), (2.32), (2.35) and (2.36), we have the following limits

    limnx2n=limnrn2brnC1(ρ+1)(ρr)n2+C2(r+1)+C3((n1)r+n2)C1(ρr)n+(C2+C3n)=r+1br2=29b=3a2,
    limnx2n+1=limnarnrn1C1(ρr)n+C2+C3nC1(ρ+1)(ρr)n1+C2(r+1)+C3(nr+n1)=arr+1=3a2,
    limny2n=limnrn2arnC1(ρ+1)(ρr)n2+C2(r+1)+C3((n1)r+n2)C1(ρr)n+C2+C3n=r+1ar2=29a=3b2,
    limny2n+1=limnbrnrn1C1(ρr)n+C2+C3nC1(ρ+1)(ρr)n1+C2(r+1)+C3(nr+n1)=brr+1=3b2.

    The proofs of (c) and (d) are clear from the forms in (2.31), (2.32), (2.35) and (2.36).

    Corollary 2.2. Suppose that 4ab=27. Then, every well-defined solution of system (1.1) has a finite limit point.

    For this case we first prove the following lemma.

    Lemma 2.1. Suppose that 4ab>27 and the zeros of the polynomial P(λ) are ρ and re±iθ, r>0, θ(0,π). Then, the following statements are true:

    (a) If ab(0,+), then r<ρ

    (b) If ab(,427), then r>|ρ|

    Proof. Let 4ab>27. Then, since ρ and re±iθ are the zeros of the polynomial P(λ), the relations

    ρr2=1ab and ρ31ab(ρ+1)2=0

    are satisfied. We conclude from these relations that abρ>0. This implies that if ab<0, then ρ<0 and if ab>0, then ρ>0. Also, from (2.18), we have

    ρ21abρρ22abρρ1abρ=ρ2r2ρ22r2ρr2=0,

    which implies

    r=|ρρ+1|. (2.45)

    We must consider the following two cases:

    (a) If ab(0,+), then ρ>0 and so, from (2.45), we have

    r=ρρ+1<ρ.

    (b) If ab(,427), then we see from (2.45) that ρ<1. So, from (2.45), we have

    r=|ρρ+1|>|ρ|.

    So, the proof is completed.

    Remark 2.2. Note that the equality r=|ρ| is impossible. Because, in this case we have 1ab=ρ3 which yields ab= 18(,427).

    Theorem 2.4. Let {(xn,yn)}n1 be a well-defined solution of system (1.1). Suppose that 4ab>27. Then, the following statements are true:

    (a) If ab(0,+) and C10, then x2n+1aρρ+1 and y2nbρρ+1 as n.

    (b) If ab(0,+) and C10, then x2naρρ+1 and y2n+1bρρ+1 as n.

    (c) If ab(,427) and |C2|+|C3|0, then both x2n+1 and y2n are periodic or converge to a periodic solution or dense in R.

    (d) If ab(,427) and |C2|+|C3|0, then both x2n and y2n+1 are periodic or converge to a periodic solution or dense in R.

    Proof. (a)–(b) The proof follows from the formulas given in (2.39), (2.40), (2.43), (2.44) and Lemma 2.1 by taking the limit.

    (c)–(d) Since the proof is similar for the forms given in (2.39), (2.40), (2.43), (2.44), we only prove for (2.44). Suppose that C1=0. Then, from (2.44), we have

    x2n=1br2C4cos(n1)θ+C5sin(n1)θC2cosnθ+C3sinnθ, (2.46)

    where C4=C2(r+cosθ)C3sinθ, C5=C3(r+cosθ)+C2sinθ. Also, we see that the form given in (2.46) can be written as

    x2n=1br2(rcosθ+cos2θ+(rsinθ+sin2θ)tan(nθγ)),

    where γ is an arbitrary constant, which corresponds to the arbitrary constants C2, C3, and satisfies the equality

    cosγC2=sinγC3, π2γπ2.

    Now, we consider the following two cases:

    (i) If θ=pqπ such that p,q are co-prime integers, we have

    x2n=1br2(rcosθ+cos2θ+(rsinθ+sin2θ)tan(qθγ))=1br2(rcosθ+cos2θ+(rsinθ+sin2θ)tan(pπγ))=1br2(rcosθ+cos2θ+(rsinθ+sin2θ)tan(γ))

    which implies x2n+q=x2n. Suppose that C10. Then, since the inequality r>|ρ| holds, from (2.44), we have

    x2n=1br2C1(ρ+1)(ρr)n2+C4cos(n1)θ+C5sin(n1)θC1(ρr)n+C2cosnθ+C3sinnθ,

    which leads to

    x2n1br2C4cos(n1)θ+C5sin(n1)θC2cosnθ+C3sinnθ

    for large enough values of n. Hence, the sequence (x2n)n0 converges to a periodic solution obtained in the case C1=0.

    (ii) If θ=tπ such that t is irrational, then we have by virtue of the Kronecker's Theorem that the set {(nt[nt])π:nN0 and t is irrational} is dense in the interval (0,π). Hence, we have

    tan(nθγ)=tan(ntπγ)=tan(ntπ[nt]πγ)

    which implies the sequence (x2n)n0 is dense in (,+).

    In this paper we conducted a detailed analysis on all solutions of system (1.1). To do this analysis, we obtained the representation forms of general solution of the system by using a practical method. By using these forms, we characterized the well-defined solutions of system (1.1). Finally, we studied the long-term behavior of the well-defined solutions. We can summarize our results as follows:

    Consider system (1.1). Then,

    (a) If ab(427,0), then every well-defined solution of system (1.1) has a finite limit point.

    (b) If ab(0,+) and C1C10, then every well-defined solution of system (1.1) has a finite limit point.

    (c) If ab(,427) and (|C2|+|C3|)(|C2|+|C3|)0, then every well-defined solution of system (1.1) is periodic or converges to a periodic solution or dense in R2.

    This work was supported by the National Natural Science Foundation of China (No. 71601072), Key Scientific Research Project of Higher Education Institutions in Henan Province of China (No. 20B110006) and the Fundamental Research Funds for the Universities of Henan Province. The authors thank the participants of this study for their valuable contributions.

    The authors declare that there are no conflict of interest associated with this publication.



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