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Modeling and analysis of the dynamics of an excessive gambling problem with modified fractional operator
            it is easy to prove that the remaining state vari-  Theorem 6. The gambling problem-free equilib-
            ables are non-negative for all t > 0. Therefore,  rium point E 0 is locally asymptotically stable when
            the solutions of system (7) are non-negative for  R 0 < 1 and unstable when R 0 > 1.
            all t ≥ 0.

                                                              Proof. To prove the local stability, we consider
            3.2. Stability analysis of the model
                                                              the Jacobian matrix of the system (7) at E 0 as
            In this section, the basic reproduction number    follows:
            and stability analyses of endemic and no problem                     J (E 0 ) =
                                                                                               
            gambling equilibrium points are discussed.                   a 11 a 12 a 13 a 14  0
                                                                         a 21      a 23 a 24
                                                                              0              0  
                                                                                               
                                                                          0   0   a 33 a 34  0    ,    (25)
            3.2.1. No problem gambling equilibrium point                                       
                                                                          0    0    0         0
                                                                                       a 44    
            To find the gambling problem-free equilibrium                 0   a 52 a 53 a 54 a 55
            point of the model, we set the derivatives of all
            compartments to zero and assume that there are
            no individuals with gambling problem ( M =
            0, P = 0 and R = 0). The system reduces to:
                                                                  where
                     
                                 γ 1 A
                      Λ − α 1 N     − ψN = 0,
                     
                     
                                 T
                                                       (24)      a 11 = −(α 1 γ 1 − ℘), a 12 = −(℘ + ψ)
                             γ 1 A
                     
                     
                      α 1 N     − (℘ + ψ) A = 0.
                     
                              T                                            ℘ + ψ             ℘ + ψ
                                                                 a 13 = −γ 2     , a 14 = −γ 3
            Solving these equations yield:                                   γ 1               γ 1

                            α 1 γ 1 − ℘ − ψ
                  E 0 =  s,              s, 0, 0, 0 .            a 21 = α 1 γ 1 − ℘ − ψ,
                                ℘ + ψ
                          Λ                                              ℘ + ψ             α 1 γ 1 − ℘ − ψ
            where s =           and α 1 γ 1 > ℘ to ensure posi-  a 23 = γ 2     + λ − α 2 γ 2
                       α 1 γ 1 − ℘
                                                                           γ 1                 α 1 γ 1
            tivity of N 0 and α 1 γ 1 > ℘+ψ for positivity of A 0 .
            This equilibrium represents a state where there              ℘ + ψ             α 1 γ 1 − ℘ − ψ
            are no individuals with gambling problems and        a 24 = γ 3     + κ − α 2 γ 3          ,
                                                                           γ 1                 α 1 γ 1
            the population consists only of non-gamblers (N)
            and non-problem gamblers (A).                        a 25 = 0, a 52 = ℘, a 53 = ς, a 54 = ζ

                                                                             α 1 γ 1 − ℘ − ψ
            3.2.2. The basic reproduction number                 a 33 =  α 2 γ 2            − (λ + ς + ψ)
                                                                                  α 1 γ 1
            The basic reproduction number R 0 is a critical
            threshold in epidemiological models, representing               α 1 γ 1 − ℘ − ψ
            the average number of new ”infections” (in this      a 34 =  α 2 γ 3            + δ,
                                                                                  α 1 γ 1
            case, new problem gamblers) generated by a sin-
            gle addicted gambler in a fully susceptible popu-    a 44 = −(κ + δ + ζ + ψ), a 55 = −ψ
            lation.
                                                              In this Jacobian matrix, three of the eigenvalues
                                                              are negative, that is m 1 = −ψ, m 2 = −(κ+δ+ζ+

                        α 2 γ 2 q α 2 γ 3 q   u −δ
                  F =                 , V =                   ψ) and m 3 = − (λ + ς + ψ) (1 − R 0 ) for R 0 < 1.
                          0      0            0   v
                                                              The remaining eigenvalues can be obtained from
            where                                             the characteristic equation:
                        α 1 γ 1 − ℘ − ψ                                       2
                   q =              , u = λ + ς + ψ,                        m + d 1 m + d 2 = 0.         (26)
                            α 1 γ 1
                                                              Since the coefficients d 1 = α 1 γ 1 − ℘ > 0 for
                           v = κ + δ + ζ + ψ.
                                                              α 1 γ 1 > ℘ and d 2 = (℘ + ψ)(α 1 γ 1 − ℘ − ψ) > 0 for
            Using the next-generation matrix method, the ba-  α 1 γ 1 > ℘+ψ. The solutions of Equation (26) have
            sic reproduction number of the proposed model is  negative real parts. Since the Im(m i = 0, i =
            the dominant eigenvalue of FV  −1 . Therefore,                                           υπ
                                                              1, 2, 3, 4, then clearly | arg(m i )| = π >  , υ ∈
                                                                                                      2
                                  α 1 γ 1 − ℘ − ψ             (0, 1). According to the Matignon criterion, 38  the
                       R 0 = α 2 γ 2            .
                                 α 1 γ 1 (λ + ς + ψ)          equilibrium point E 0 is stable when R 0 < 1.
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