A non-linear discrete-time dynamical system related to epidemic SISI model
AA NON-LINEAR DISCRETE-TIME DYNAMICAL SYSTEMRELATED TO EPIDEMIC SISI MODEL
S. K. SHOYIMARDONOV
Abstract.
We consider SISI epidemic model with discrete-time. Thecrucial point of this model is that an individual can be infected twice.This non-linear evolution operator depends on seven parameters and weassume that the population size under consideration is constant, so deathrate is the same with birth rate per unit time. Reducing to quadraticstochastic operator (QSO) we study the dynamical system of the SISImodel. Introduction
In [5] SISI model is considered in continuous time as a spread of bovinerespiratory syncytial virus (BRSV) amongst cattle. They performed an equi-librium and stability analysis and considered an applications to Aujesky’sdisease (pseudorabies virus) in pigs. In [1] SISI model was considered as anexample and characterised the conditions for fixed point equation. In theboth these works it was assumed that the population size under considera-tion is a constant, so the per capita death rate is equal to per capita birthrate.Let us consider SISI model [1]:(1.1) dSdt = b ( S + I + S + I ) − µS − β A ( I, I ) S dIdt = − µI + β A ( I, I ) S − αI dS dt = − µS + αI − β A ( I, I ) S dI dt = − µI + β A ( I, I ) S where S − density of susceptibles who did not have the disease before, I − density of first time infected persons, S − density of recovereds, I − densityof second time infected persons, b − birth rate, µ − death rate, α − recoveryrate, β − susceptibility of persons in S , β − susceptibility of persons in S ,k − infectivity of persons in I , k − infectivity of persons in I . Moreover, A ( I, I ) denotes the so-called force of infection, A ( I, I ) = k I + k I P Mathematics Subject Classification.
Key words and phrases.
Quadratic stochastic operator, fixed point, discrete-time, SISImodel, epidemic. a r X i v : . [ m a t h . D S ] A ug YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 2 and P = S + I + S + I denotes the total population size. Here we do somereplacements: x = SP , u = IP , y = S P , v = I P In (1.1) we assume that b = µ and by substituting x, u, y, v we have(1.2) dxdt = b − bx − β A ( u, v ) x dudt = − bu + β A ( u, v ) x − αu dydt = − by + αu − β A ( u, v ) y dvdt = − bv + β A ( u, v ) y where all parameters are non-negative. We notice that ddt ( x + u + y + v ) =0 , from this we deduce that the total population size is constant over timeand therefore we assume x + u + y + v = 1 .2. Quadratic Stochastic Operators
The quadratic stochastic operator (QSO) [3], [4] is a mapping of the stand-ard simplex.(2.1) S m − = { x = ( x , ..., x m ) ∈ R m : x i ≥ , m (cid:88) i =1 x i = 1 } into itself, of the form(2.2) V : x (cid:48) k = m (cid:88) i =1 m (cid:88) j =1 P ij,k x i x j , k = 1 , ..., m, where the coefficients P ij,k satisfy the following conditions(2.3) P ij,k ≥ , P ij,k = P ji,k , m (cid:88) k =1 P ij,k = 1 , ( i, j, k = 1 , ..., m ) . Thus, each quadratic stochastic operator V can be uniquely defined by acubic matrix P = ( P ij,k ) mi,j,k =1 with conditions (2.3).Note that each element x ∈ S m − is a probability distribution on (cid:74) , m (cid:75) = { , ..., m } . Each such distribution can be interpreted as a state of the corres-ponding biological system.For a given λ (0) ∈ S m − the trajectory (orbit) { λ ( n ) ; n ≥ } of λ (0) underthe action of QSO (2.2) is defined by λ ( n +1) = V ( λ ( n ) ) , n = 0 , , , ... The main problem in mathematical biology consists in the study of theasymptotical behaviour of the trajectories. The difficulty of the problemdepends on given matrix P . Definition 1.
A QSO V is called regular if for any initial point λ (0) ∈ S m − ,the limit lim n →∞ V n ( λ (0) ) exists, where V n denotes n -fold composition of V with itself (i.e. n timeiterations of V ). YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 3 Reduction to QSO
In this paper we study the discrete time dynamical system associated tothe system (1.2).Define the evolution operator V : S → R , ( x, u, y, v ) (cid:55)→ (cid:0) x (1) , u (1) , y (1) , v (1) (cid:1) (3.1) V : x (1) = x + b − bx − β A ( u, v ) xu (1) = u − bu + β A ( u, v ) x − αuy (1) = y − by + αu − β A ( u, v ) yv (1) = v − bv + β A ( u, v ) y where A ( u, v ) = k u + k v. Note that if k = k = 0 then A ( u, v ) = 0 andoperator (3.1) becomes linear operator which is well studied.By definition the operator V has a form of QSO, but the parameters ofthis operator are not related to P ij,k . Here to make some relations with P ij,k we find conditions on parameters of (3.1) rewriting it in the form (2.2) (asin [6],[7]). Using x + u + y + v = 1 we change the form of the operator (3.1)as following: V : x (1) = x (1 − b )( x + u + y + v ) + b ( x + u + y + v ) − β ( k u + k v ) xu (1) = u (1 − b − α )( x + u + y + v ) + β ( k u + k v ) xy (1) = y (1 − b )( x + u + y + v ) + αu ( x + u + y + v ) − β ( k u + k v ) yv (1) = v (1 − b )( x + u + y + v ) + β ( k u + k v ) y From this system and QSO (2.2) for the case m = 4 we obtain the followingrelations:(3.2) P , = , P , = b − β k , P , = b , P , = b , P , = − b − α , P , = α , P , = − b + α − β k , P , = − b − β k , P , = − b , P , = b − β k , P , = b , P , = b , P , = − b − α + β k , P , = − b − α , P , = − b , P , = α , P , = − b , P , = − b + β k , P , = b , P , = b , P , = b , P , = β k , P , = − b − α , P , = α , P , = − b , P , = β k , P , = − b , other P ij,k =0 . Proposition 2.
We have V (cid:0) S (cid:1) ⊂ S if and only if the non-negative para-meters b, α, β , β , k , k verify the following conditions (3.3) α + b ≤ , β k ≤ , β k ≤ ,b + β k ≤ , | b − β k | ≤ , | b − β k | ≤ , | b − β k | ≤ , | α + b − β k | ≤ , | α − b − β k | ≤ . Moreover, under conditions (3.3) the operator V is a QSO. YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 4
Proof.
The proof can be obtained by using equalities (3.2) and solving in-equalities ≤ P ij,k ≤ for each P ij,k . (cid:3) Remark . In the sequel of the paper we consider operator (3.1) with para-meters b, α, β , β , k , k which satisfy conditions (3.3). This operator maps S to itself and we are interested to study the behaviour of the trajectory ofany initial point λ ∈ S under iterations of the operator V. Fixed points of the operator (3.1)
To find fixed points of operator V given by (3.1) we have to solve V ( λ ) = λ. Finding fixed points of the operator (3.1).
Denote λ = (1 , , , , λ = (0 , , , , λ = (0 , , , , λ = (0 , , , , Λ = { λ = ( x, u, y, v ) ∈ S : u = v = 0 } , Λ = { λ = ( x, u, y, v ) ∈ S : u = 0 } , Λ = { λ = ( x, u, y, v ) ∈ S : x = 0 } , Λ = { λ = ( x, u, y, v ) ∈ S : x = u = 0 } ,λ = (cid:16) bβ k , β k − bβ k , , (cid:17) , λ = (cid:16) b + αβ k , b ( β k − b − α ) β k ( b + α ) , α ( β k − b − α ) β k ( b + α ) , (cid:17) ,λ = (cid:16) bb + β A , bβ A ( b + β A )( b + α ) , αbβ A ( b + β A )( b + β A )( b + α ) , αβ β A ( b + β A )( b + β A )( b + α ) (cid:17) , where A is a positive solution of the equation(4.1) bβ k ( b + β A )( b + α ) + αβ β k A ( b + β A )( b + β A )( b + α ) By the following proposition we give all possible fixed points of the oper-ator V. Proposition 4.
Let
F ix ( V ) be set of fixed points of the operator (3.1). Then F ix ( V ) = { λ }{ λ , λ , λ } , if b = 0 { λ , λ } (cid:83) Λ , if b = α = 0Λ , if b = β = β = 0 { λ } (cid:83) Λ , if b = α = β = 0 , β > { λ , λ } (cid:83) Λ , if b = β = 0 , β > , α > , k k > S , if b = α = k = k = 0 or b = α = β = β = 0 { λ , λ } , if b > , α = 0 , β k > b { λ , λ } , if b > , α > , β = 0 , β k > b + α { λ , λ } , if αbβ β k k > Proof.
Recall that a fixed point of the operator V is a solution of V ( λ ) = λ. From this straightforward λ i , i = 1 , . For the other cases we assume that b > . Now we find λ . If y = v = 0 , α = 0 , then by (3.1) we have y (1) = 0 , v (1) = 0 , and A ( u, v ) = k u. Using this and u (1) = u we obtain u = β k − bβ k . Bysubstituting them to x (1) = x we get x = bβ k . Of course, for positiveness of
YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 5 u it requests that β k > b > . Similarly, by the conditions to parameterswe can find easily the next fixed point λ . For the interior fixed point λ we request that all parameters are positive.First, using x (1) = x we have x = bb + α , from this and by u (1) = u we get u = β Axb + α = bβ A ( b + β A )( b + α ) . Similarly, by y (1) = y we have y = αub + β A = αbβ A ( b + β A )( b + β A )( b + α ) , and from v (1) = v we get v = β Ayb = αβ β A ( b + β A )( b + β A )( b + α ) . In this case from A ( u, v ) = k u + k v we obtain the quadratical equation(4.1). Thus, Proposition is proved. (cid:3) Note that the set of positive solutions of (4.1) is non-empty when β k ≥ b + α (see the statements after Conjecture 2). For example, α = 0 . , b =0 . , β = 0 . , β = 0 . , k = k = 1 . Then the equation (4.1) has the form A − A − and the positive solution is A = √ ≈ . . Type of the fixed point λ .Definition 5. [2]. A fixed point p for F : R m → R m is called hyperbolic ifthe Jacobian matrix J = J F of the map F at the point p has no eigenvalueson the unit circle.There are three types of hyperbolic fixed points:(1) p is an attracting fixed point if all of the eigenvalues of J ( p ) are lessthan one in absolute value.(2) p is an repelling fixed point if all of the eigenvalues of J ( p ) are greaterthan one in absolute value.(3) p is a saddle point otherwise. Proposition 6.
Let λ be the fixed point of the operator V. Then λ = nonhyperbolic , if b = 0 or β k = b + α attractive , if b > β k < b + α saddle , if b > β k > b + α Proof.
The Jacobian of the operator (3.1) is: J = − b − β A − β k x − β k xβ A − b − α + β k x β k x α − β k y − b − β A − β k y β k y β A − b + β k y Then at the fixed point λ the Jacobian is J ( λ ) = − b − β k − β k − b − α + β k β k α − b
00 0 0 1 − b and the eigenvalues of this matrix are µ = 1 − b, µ = 1 − b − α + β k . Bythe conditions (3.3) we have µ ≥ , µ ≥ . It is easy to se that if b = 0 or β k = b + α then µ = 1 or µ = 1 respectively, if b > , β k < b + α thenthe fixed point λ is an attracting, otherwise saddle point. (cid:3) YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 6
Remark . The type of other fixed points is not studied and the type of fixedpoint λ will be useful for some results in below.5. The limit points of trajectories
In this section we study the limit behavior of trajectories of initial point λ (0) ∈ S under operator (3.1), i.e the sequence V n ( λ (0) ) , n ≥ . Note thatsince V is a continuous operator, its trajectories have as a limit some fixedpoints obtained in Proposition 4.5.1. Case no susceptibility of persons ( β = β = 0 ). We study herethe case where in the model there is no susceptibility of persons.
Proposition 8.
For an initial point λ = (cid:0) x , u , y , v (cid:1) ∈ S (except fixedpoints)the trajectory (under action of operator (3.1)) has the following limit lim n →∞ V ( n ) ( λ ) = λ if α = b = 0( x , , − x − v , v ) if b = 0 , α > λ if b > Proof. If β = β = 0 then the operator (3.1) has the following form:(5.1) V : x (1) = x + b − bxu (1) = u (1 − b − α ) y (1) = y − by + αuv (1) = v (1 − b ) If b = α = 0 then every point is fixed point, so this case is clear. If b =0 , α > then by (5.1) we get x ( n ) = x , v ( n ) = v and u ( n ) = u (1 − α ) n → . Moreover, y (1) = y + αu ≥ y, so the sequence y ( n ) has a limit. From x ( n ) + u ( n ) + y ( n ) + v ( n ) = 1 it follows the proof of this case. If b > then the sequences u ( n ) , v ( n ) have zero limits. In addition, from x ( n +1) = x ( n ) + b (1 − x ( n ) ) one obtains that limit of the sequence x ( n ) is 1 (since b > ). Thus, the Proposition is proved. (cid:3) Case no susceptibility of persons in S ( β = 0 , β > ).Proposition 9. For an initial point λ = (cid:0) x , u , y , v (cid:1) ∈ S (except fixedpoints) the trajectory (under action of the operator (3.1)) has the followinglimit lim n →∞ V ( n ) ( λ ) = ( x , u , , − x − u ) if α = b = 0 λ if b > , α = 0( x , , ¯ y, − x − ¯ y ) if b = 0 , α > , k = 0( x , , , − x ) if b = 0 , α > , k > λ if b > , α > where ¯ y = ¯ y ( λ ) YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 7
Proof. If β = 0 then the operator (3.1) is(5.2) V : x (1) = x + b (1 − x ) u (1) = u (1 − b − α ) y (1) = y − by + αu − β ( k u + k v ) yv (1) = v − bv + β ( k u + k v ) y Case : b = α = 0 . We assume that A ( u , v ) (cid:54) = 0 , otherwise, A ( u ( n ) , v ( n ) ) =0 , ∀ n ∈ N, and limit point of the operator (5.2) is initial point λ . The proofof this case is coincides with second case of Proposition 10.
Case: b > , α = 0 . In this case u ( n ) = u (1 − b ) n has zero limit, and wehave x (1) = x + b (1 − x ) ≥ x, y (1) = y − by − β ( k u + k v ) y ≤ y, so the sequences x ( n ) , y ( n ) have limits and consequently, v ( n ) also has limit.Let ¯ x be a limit of x ( n ) . Then from x ( n +1) = x ( n ) + b (1 − x ( n ) ) we get limitand it follows that ¯ x = 1 since b > . Thus, limit of the considering operatoris λ = (1 , , , . Case: b = 0 , α > , k = 0 . Then x ( n ) = x , u ( n ) = u (1 − α ) n → and v (1) = v + β k uy ≥ v, i.e., the sequence v ( n ) has limit, so y ( n ) also has limit.But limits of y ( n ) and v ( n ) depend on initial point λ . Case: b = 0 , α > , k > . Here also, as previous case, x ( n ) = x , u ( n ) = u (1 − α ) n → and the sequences y ( n ) , v ( n ) have limits. Let ¯ y, ¯ v be limitsof y ( n ) and v ( n ) respectively. If we take limit from both side of the followingequality v ( n +1) = v ( n ) + β ( k u ( n ) + k v ( n ) ) y ( n ) then we have β ¯ v ¯ y = 0 , i.e., ¯ y = 0 , because, v ( n ) increasing sequence, so ¯ v (cid:54) = 0 (Note that A ( u , v ) (cid:54) = 0 ). Thus, ¯ v = 1 − x . Case: b > , α > . Then u ( n ) = u (1 − b − α ) n → , and x (1) = x + b (1 − x ) ≥ x ≥ x, i.e., the sequence x ( n ) has limit ¯ x . From x ( n +1) = x ( n ) + b (1 − x ( n ) ) weget limit and it obtains that ¯ x = 1 . Moreover, from the x ( n ) + u ( n ) + y ( n ) + v ( n ) = 1 we have that y ( n ) + v ( n ) → . In addition, every terms of the bothsequences are non-negative, so the limits of the sequences y ( n ) and v ( n ) existand zero. Thus, the proof of the Proposition is completed. (cid:3) Case no birth (death) rate and recovery rate ( b = α = 0 ).Proposition 10. For an initial point λ = (cid:0) x , u , y , v (cid:1) ∈ S (except fixedpoints) the trajectory (under action of the operator (3.1)) has the followinglimit lim n →∞ V ( n ) ( λ ) = λ if k = k = 0( x , u , , − x − u ) if β = 0 , β > , k + k > , − y − v , y , v ) if β > , β = 0 , k + k > , u , , − u ) if β > , β > , k k > Proof. If b = 0 , α = 0 then the operator (3.1) is YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 8 (5.3) V : x (1) = x − β ( k u + k v ) xu (1) = u + β ( k u + k v ) xy (1) = y − β ( k u + k v ) yv (1) = v + β ( k u + k v ) y From the equations of the operator (5.3) we have that the sequences x ( n ) , u ( n ) ,y ( n ) , v ( n ) are monotone, so they have limits. The case k = k = 0 is clear.If β = 0 , β > , k + k > then x ( n ) = x , u ( n ) = u , y ( n +1) = y ( n ) − β ( k u ( n ) + k v ( n ) ) y ( n ) . If A ( u , v ) = k u + k v (cid:54) = 0 , then A ( u ( n ) , v ( n ) ) (cid:54) = 0 , otherwise, A ( u ( n ) , v ( n ) ) = k u ( n ) + k v ( n ) = k u + k v = 0 , so A ( u ( n ) , v ( n ) ) = k u ( n ) + k v ( n ) has non-zero limit. We assume that lim n →∞ y ( n ) = y (cid:54) = 0 . Then from y ( n +1) = y ( n ) − β ( k u ( n ) + k v ( n ) ) y ( n ) we get limit and it is contradiction to lim n →∞ A ( u ( n ) , v ( n ) ) (cid:54) = 0 , so we have aproof of the second case. Similarly, for cases β > , β = 0 , k + k > and β > , β > , k k > one can complete the prove of this Proposition. (cid:3) Case no recovery rate and infectivity of persons in I ( α =0 , k = 0 ).Proposition 11. For an initial point λ = (cid:0) x , u , y , v (cid:1) ∈ S (except fixedpoints) the trajectory (under action of the operator (3.1)) has the followinglimit lim n →∞ V ( n ) ( λ ) = (cid:40) λ if β k ≤ b or u = 0 λ if β k > b and u > Proof.
Here we consider the case b > , otherwise it coincides with previousProposition cases. If α = k = 0 then the operator (3.1) is(5.4) V : x (1) = x + b − bx − β k uxu (1) = u − bu + β k uxy (1) = y − by − β k uyv (1) = v − bv + β k uy From the system (5.4) we have y (1) = y − by − β k uy ≤ y (1 − b ) , i.e., y ( n ) ≤ y (1 − b ) n , so y ( n ) → as n → ∞ . Moreover, from the sum of lasttwo equations of (5.4) we get lim n →∞ ( y ( n +1) + v ( n +1) ) = (1 − b ) lim n →∞ ( y ( n ) + v ( n ) ) = ( y + v ) lim n →∞ (1 − b ) n = 0 , thus, the sequence v ( n ) converges to zero.–If β k = 0 then u ( n ) = u (1 − b ) n has zero limit and from this we havethat the sequence x ( n ) has limit one. Thus, the limit of the operator V is λ . YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 9 –If < β k ≤ b then from u (1) = u − ( b − β k x ) u ≤ u we get that thesequence u ( n ) has limit. We assume that lim n →∞ u ( n ) = ¯ u (cid:54) = 0 , then fromthis and u ( n +1) = u ( n ) − ( b − β k x ( n ) ) u ( n ) we have lim n →∞ x ( n ) = bβ k , andfrom this ¯ u = 1 − bβ k = β k − bβ k . But the condition β k ≤ b is contradictionof positiveness of ¯ u, so u ( n ) has zero limit. Hence, limit point of the operatoris λ . –If β k > b and u > . From first two equations of the operator (5.4)we formulate a new operator:(5.5) W : (cid:40) x (1) = x + b − bx − β k uxu (1) = u − bu + β k ux Here we normalize the operator (5.5) as following:(5.6) W : x (1) = x + b − bx − β k uxx + u + b − b ( x + u ) u (1) = u − bu + β k uxx + u + b − b ( x + u ) For this operator x ( n ) + u ( n ) = 1 , n ≥ , so from x (1) + u (1) = 1 we have (cid:40) x (2) = x (1) + b − bx (1) − β k u (1) x (1) u (2) = u (1) − bu (1) + β k u (1) x (1) Thus, operators W and W have same dynamics. From first equation of thelast system we obtain x (2) = (1 − b − β k ) x (1) + β k ( x (1) ) + b. If we denote x (1) = x, x (2) = f b,β k ( x ) then we get f b,β k ( x ) = b + (1 − b − β k ) x + β k x . Definition 12. (see [2], p. 47) Let f : A → A and g : B → B be two maps. f and g are said to be topologically conjugate if there exists a homeomorphism h : A → B such that, h ◦ f = g ◦ h . The homeomorphism h is called atopological conjugacy.Let F µ ( x ) = µx (1 − x ) be quadratic family (discussed in [2]) and f b,β k ( x ) = b + (1 − b − β k ) x + β k x . Lemma 13.
Two maps F µ ( x ) and f b,β k ( x ) are topologically conjugate for µ = β k − b + 1 .Proof. We take the linear map h ( x ) = px + q and by Definition 12 we shouldhave h ( F µ ( x )) = f b,β k ( h ( x )) , i.e., pµx (1 − x ) + q = b + ( px + q )(1 − b − β k ) + β k ( px + q ) from this identity we get − pµ = β k p pµ = p (1 − b − β k ) + 2 pqβ k q = q (1 − b − β k ) + β k q + b ⇒ p = − µβ k q = µ − b + β k β k β k q − ( b + β k ) q + b = 0 The roots of the equation β k q − ( b + β k ) q + b = 0 are q = 1 , q = bβ k . If we choose q = 1 then by q = µ − b + β k β k we have µ = β k − b + 1 . Then
YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 10 the homeomorphism is h ( x ) = b − β k − β k x + 1 . Moreover, since b < β k ≤ we have < µ < . (cid:3) The importance of this Lemma is that if two maps are topologically con-jugate then they have essentially the same dynamics (see [2], p. 53). Theoperator f b,β k ( x ) = b + (1 − b − β k ) x + β k x has two fixed points p = 1 and p = bβ k . In addition, f (cid:48) b,β k ( x ) = 1 − b − β k + 2 β k x, from thisand β k > b it obtains that the fixed point p = 1 is repelling, p = bβ k isattractive. Moreover, for any initial point x ∈ (0 , and for µ ∈ (1 , thetrajectory of the operator F µ ( x ) converges to the attractive fixed point (see[2], p. 32). Thus, for the case β k > b the limit point of the operator (5.4)is λ . (cid:3) Case no only susceptibility of persons in S ( β = 0 , β > ).Proposition 14. For an initial point λ = (cid:0) x , u , y , v (cid:1) ∈ S (except fixedpoints) the trajectory (under action of the operator (3.1)) has the followinglimit lim n →∞ V ( n ) ( λ ) = ( x , , − x − v , v ) if b = 0 , α > and k u + k v = 0(0 , , − v , v ) if b = 0 , α > , k v > x, , − ¯ x − v , v ) if b = k v = 0 , k u > , α > λ if bα > , k u + k v = 0 ,λ if bα > , k v = 0 , β k ≤ b + α, where ¯ x = ¯ x ( λ ) Proof.
Here we consider the case β > , otherwise, this proposition is samewith Proposition 8. We note that the case b = α = 0 is considered inProposition 10. If β = 0 then the operator (3.1) is(5.7) V : x (1) = x + b − bx − β ( k u + k v ) xu (1) = u − bu − αu + β ( k u + k v ) xy (1) = y − by + αuv (1) = v (1 − b ) Case: b = 0 , α > , k u + k v = 0 . From A ( u , v ) = k u + k v = 0 wehave the following simple cases:–If k = k = 0 then x ( n ) = x , v ( n ) = v and u ( n ) = u (1 − α ) n → . –If k = v = 0 then v ( n ) = 0 , x ( n ) = x , so u ( n ) = u (1 − α ) n → and y ( n ) → − x . –If u = k = 0 , then u ( n ) = 0 , v ( n ) = v , so x ( n ) = x , y ( n ) = y = 1 − x . –If u = v = 0 , then u ( n ) = 0 , v ( n ) = 0 , so x ( n ) = x , y ( n ) = y = 1 − x . Case: b = 0 , α > , k v > then v ( n ) = v and x (1) = x − β ( k u + k v ) x ≤ x, y (1) = y + αu ≥ y, i.e., the sequences x ( n ) , y ( n ) , v ( n ) have limits, so u ( n ) also has limit. From the equation y ( n +1) = y ( n ) + αu ( n ) we get limit andby α > it obtains that u ( n ) converges to zero. Moreover, by the secondequation of the system (5.7), u ( n +1) = (1 − α ) u ( n ) + β ( k u ( n ) + k v ( n ) ) x ( n ) , if we take a limit from two sides then we have β k v ¯ x, from this and YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 11 k v > we have ¯ x = 0 , where ¯ x is a limit of the sequence x ( n ) . Case: b = k v = 0 , k u > , α > . Here also as previous case, allsequences have limits and v ( n ) = v . From y ( n +1) = y ( n ) + αu ( n ) we getlimit and by α > it obtains that u ( n ) converges to zero. But, the limit lim n →∞ x ( n ) = ¯ x depends on initial conditions x , u . Case: b > , α > , k u + k v = 0 . –If k = k = 0 then x (1) = x + (1 − x ) b ≥ x, u ( n ) = u (1 − b − α ) n → and from v ( n ) = v (1 − b ) n → we get that all sequences have limits. Since b > we have that x ( n +1) = x ( n ) (1 − b ) + b has limit 1.–If k = v = 0 then v ( n ) = 0 , u ( n ) = u (1 − b − α ) n → and as previouscase x ( n ) → . –If u = k = 0 then u ( n ) = 0 , y ( n ) = y (1 − b ) n → and from v ( n ) → implies x ( n ) → . –If u = v = 0 then u ( n ) = 0 , v ( n ) = 0 , y ( n ) = y (1 − b ) n → so x ( n ) → . Case: b > , α > , k v = 0 , β k ≤ b + α. – If k = 0 then u (1) = u − ( b + α − β k x ) u ≤ u, i.e., the sequence u ( n ) has limit. Let us to show existence the limit of y ( n ) . We assume that y (1) ≤ y, i.e., by − αu ≥ . If we show that by (1) − αu (1) ≥ then it obtains that y ( n ) has limit. We check this condition: by (1) − αu (1) = b ( y − by + αu ) − α ( u − bu − αu + β k ux ) == by − αu + bαu − b y + bαu + α u − αβ k ux == by − αu − b ( by − αu ) + αu ( b + α − β k x ) == ( by − αu )(1 − b ) + αu ( b + α − β k x ) ≥ . Thus, the sequences u ( n ) and y ( n ) have limits and from v ( n ) → we get thatall sequences have limits. For the case β k ≤ b + α fixed point λ is unique,so it must be limit point.–If v = 0 then the proof is same with case k = 0 . Thus, the Propositionis proved.For the case b > , α > , k v > if we assume that the sequences x ( n ) , u ( n ) , y ( n ) have limits then for β k ≤ b + α fixed point λ is uniqueglobally attracting, for β k > b + α fixed point λ is saddle and fixed point λ must be limit point of the operator V, because there is no other fixedpoints. In addition, we consider some numerical simulations for these twocases ( β k ≤ b + α and β k > b + α ). In Fig. 5.1 the trajectory convergesto λ , and in Fig 5.2 the trajectory converges to λ . Therefore we formulatethe following conjecture:
Conjecture 1. If β = 0 then for an initial point λ = (cid:0) x , u , y , v (cid:1) ∈ S (except fixed points) the trajectory has the following limit lim n →∞ V ( n ) ( λ ) = (cid:40) λ if β k ≤ b + α, bα > and k v > λ if u + v > and β k > b + α, bα > (cid:3) YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 12
Figure 5.1. α =0 . , b = 0 . , β =0 . , β = 0 , k =1 , k = 0 . , x =0 . , u =0 . , y = 0 . , lim n →∞ V ( n ) ( λ ) = λ . Figure 5.2. α =0 . , b = 0 . , β =0 . , β = 0 , k =1 , k = 0 . , x =0 . , u =0 . , y = 0 . , lim n →∞ V ( n ) ( λ ) = λ . The following Conjecture also formulated for the limit point of the oper-ator (3.1) with nonzero parameters.
Conjecture 2. If αbβ β k k > then for an initial point λ = (cid:0) x , u , y , v (cid:1) ∈ S (except fixed points) the trajectory has the following limit lim n →∞ V ( n ) ( λ ) = (cid:40) λ if u = v = 0 or β k ≤ b + α, b ( b + α ) ≥ αβ k λ if u + v > and β k > b + α The case u = v = 0 is clear, i.e., the limit of the operator V is λ . For the case u + v > let us assume that all sequences have limits andconsider limit behaviour in below. First, we denote by f ( x ) , g ( x ) : (5.8) f ( x ) = b + β x, g ( x ) = bβ k b + α + αβ β k x ( b + β x )( b + α ) . Then the roots of the equation (4.1) are roots of the equation f ( x ) = g ( x ) . We consider graphical solutions of this equation.
Case : β k > b + α . In this case bβ k b + α > b and the graphic of thefunction g ( x ) has horizontal asymptote y = β ( bk + αk ) b + α = const, so theequation f ( x ) = g ( x ) has unique positive solution (Fig. 5.3). Moreover, for β k > b + α fixed point λ is saddle fixed point, so λ must be limit pointof the operator (3.1). Case : β k < b + α. In this case, slope of the f ( x ) is T anϕ = β and slopeof a tangent at the point x = 0 of g ( x ) is T anψ = αβ β k b ( b + α ) . It is clear that,if
T anϕ ≥ T anψ, i.e., β ≥ αβ β k b ( b + α ) or b ( b + α ) ≥ αβ k then f ( x ) = g ( x ) does not have positive solution (Fig.5.4). Case : β k = b + α. In this case the equation f ( x ) = g ( x ) has solution x = 0 , i.e., operator (3.1) has unique fixed point λ . YNAMICAL SYSTEM RELATED TO EPIDEMIC SISI MODEL 13
Figure 5.3. β k >b + α Figure 5.4. β k αβ k Figure 5.5. α = 0 . , b =0 . , β = 0 . , β =0 . , k = 1 . , k =1 . , x = 0 . , u =0 . , y = 0 . , lim n →∞ V ( n ) ( λ ) = λ . Figure 5.6. α = 0 . , b =0 . , β = 0 . , β =0 . , k = 0 . , k =1 . , x = 0 . , u =0 . , y = 0 . , lim n →∞ V ( n ) ( λ ) = λ . References [1] Johannes M ¨ u ller, Christina Kuttler. Methods and models in mathematical biology .Springer, 2015, 721 p. 1[2] R.L. Devaney,
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Subcritical endemic steady states in math-ematical models for animal infections with incomplete immunity . Math.Biosc.165, 1-25pp, (2000). 1[6] U.A. Rozikov, S.K. Shoyimardonov,
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Sobirjon Shoyimardonov. V.I.Romanovskiy institute of mathematics, 81,Mirzo Ulug’bek str., 100125, Tashkent, Uzbekistan.
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