The Study of SIR Model

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(1)國立臺灣師範大學數學系碩士班碩士論文. 指導教授：張幼賢. 博士. The Study of SIR Model. 研究生：莊曜豪. 中華民國一百零二年六. 月.

(2) The Study of SIR Model Yao Hao Chuang 莊曜豪 June 11, 2013. Abstract. In this thesis, we use a pair of coupled upper and lower solution to the SIR(susceptible, infective and removed) model. We show that there exist two equilibrium, one is diseasefree equilibrium and another one is stationary positive equilibrium, which means that the disease becomes stable. Keywords: SIR model, upper and lower solutions. 1.

(3) Contents 1. Introduction. 3. 2. SIR system for the case n=1. 4. 3. SIR system with n group. 12. 4. References. 16. Page 2.

(4) 1 Introduction Multigroup models have beeb proposed in the literate to describe the transmission of infective disease in heterogeneous populations, and much research has been done on various forms of multigroup models, see e.g.[1,2,3,6,7]. In this thesis, we study a multigroup SIR model which was proposed in [4], Guo et al. Let Sk (t), Ik (t) and Rk (t) denote the susceptible, infective and recovered population at time t in the k-th group, k = 1, 2, · · · , n, respectively. If we don't consider the fractions who are immune or vaccinated, and suppose the death rates of Sk , Ik and Rk in the k-th group are different, then the model can be written as :  n   Sk′ (t) = Λk − ∑ βk j Sk (t)I j (t) − dk Sk (t)    j=1  n Ik′ (t) = ∑ βk j Sk (t)I j (t) − (εk + γk )Ik (t) k = 1, 2, · · · , n   j=1     R′ (t) = γ I (t) − δ R (t) k k k k k. (1.1). The parameters in the model are summarized in the following list: Λk : influx of individuals in to the k-th group. βk j : transmission coefficient between compartments Sk and I j dk : death rate of S compartment in the k-th group. εk : death rate of I compartment in the k-th group δk : death rate of R compartment in the k-th group γk : recovery rate of infectious individuals in the k-th group The value of all the parameters are assumed to be nonnegative and dk , Λk > 0, dk 6 {εk , δk } for all k. The rest of thesis is organized as follows. In section 2, we find the equilibrium points of the model for the case n = 1, and construct two pairs of upper and lower solutions as initial iteration functions. By using monotone iterative method, we construct two sequences which converge to the equilibrium points, respectively. In section 3, we extend the result to the general n multigroup case.. Page 3.

(5) 2 SIR system for the case n=1 First we consider the case n = 1 in (1.1), that is, the equation becomes    S′ (t) = Λ − β S(t)I(t) − dS(t)   I ′ (t) = β S(t)I(t) − (ε + γ )I(t)     R′ (t) = γ I(t) − δ R(t). (2.1). To simplify the analysis, we change the variation. Let U(s) =. d s S( ), Λ d. V (s) =. β s I( ), d d. W (s) =. β s R( ), γ d. then system (2.1) becomes Λ d dU d dS dt d 1 1 Λ = = (Λ − β SI − dS) = (Λ − β U V − d U) = 1 −UV −U, ds Λ dt ds Λ d Λ d β d dV β dI dt β 1 β Λ d d βΛ ε +γ = = (β SI − (ε + γ )I) = 2 (β U V − (ε + γ ) V ) = 2 UV − V, ds d dt ds d d d d β β d d dW β dR dt β 1 β d γ δ = = (γ I − δ R) = (γ V − δ R) = V − W. ds γ dt ds γ d dγ β β d. βΛ ε +γ δ , ρ = , and ρ = then system (2.1) is equivalent to 2 3 d2 d d dU(s) = 1 −U(s)V (s) −U(s) ds dV (s) = ρ1U(s)V (s) − ρ2V (s) ds dW (s) = V (s) − ρ3W (s) ds. Set ρ1 =           . (2.2). Replacing variable s by t, we can write (2.2) with initial data u0 = (U(0),V (0),W (0)) = (U0 ,V0 ,W0 ) to the following semilinear initial value problem :   du(t) + Au(t) = f (t, u(t)) dt  u(t0 ) = u0      1 0 0 U(t) 1 −U(t)V (t)           where u(t) =  V (t) , A =  0 ρ2 0 , f (t, u(t)) =  ρ1U(t)V (t)      W (t) 0 −1 ρ3 0. Page 4. (2.3)     .

(6) Definition 2.1. We say that a function f : [0, ∞) × X → X satisfies local Lipschitz condition in u, uniformly in t on bounded intervals if for every t ′ > 0 and constant c > 0 there is a constant L(c,t ′ ) such that | f (t, u) − f (t, v)| 6 L(c,t ′ )|u − v|. (2.4). holds for all u, v ∈ X with ∥u∥ 6 c, ∥v∥ 6 c and t ∈ [0,t ′ ].. Since A is a bounded linear operator and f satisfies Lipschitz condition, if the initial value problem has a classical or strong solution which also is a solution of the integral equation u(t) = T (t − t0 )u0 +. ∫ t t0. T (t − s) f (s, u(s))ds,. (2.5). where T (t) = et(−A) . Definition 2.2. A continuous solution u of the integral equation will be called a mild solution of the initial value problem. Theorem 2.3. ([9],p.189) Let −A be the infinitesimal generator of a C0 semigroup T (t) on X. If f : [t0 , T ] × X → X is continuously differentiable, then the mild solution is a classical solution of the initial value problem.. Clearly, f (t, u(t)) = (1 −U(t)V (t), ρ1U(t)V (t), 0) is continuously differentiable, so the solution of (2.3) exists, and it can be represent as. u(t) = e−(t−t0 )A u. ∫ t. 0+. e(s−t)A f (s, u(s))ds. t0. It is not hard to see that the equilibrium points of (2.2) are u = (u1 , u2 , u3 ) = (1, 0, 0) and. (. ρ2 ρ1 − ρ2 ρ1 − ρ2 , , ). ρ1 ρ2 ρ2 ρ3. We only focus on the nonnegative solution, we assume ρ1 − ρ2 > 0, i.e., we suppose that. βΛ > 1. (ε + γ )d βΛ < 1, then there has only one equilibrium point (1,0,0), which means the infection (ε + γ )d (ε + γ )d will die out in the long run. We called the value R0 ≡ to be the basic reproduction βΛ number. If. Page 5.

(7) In this section, we want to find two pairs of initial functions and using them to construct two pair sequence of functions. Each pair sequence of functions converge to a unique solution, respectively. For this purpose, we will use following notations and preliminaries. Definition 2.4. We write the vector u in the split form u = (ui , [u]ai , [u]bi ) and rewrite the function fi as fi (t, x, u) = fi (t, x, ui , [u]ai , [u]bi ), a vector function f = ( f1 , · · · fN ) is said to possess a quasimonotone property if for each i there exist nonnegative integers ai , bi with ai + bi = N − 1 such that fi (t, ui , [u]ai , [u]bi ) is monotone nondecreasing in [u]ai and it is monotone nonincreasing in [u]bi .. Let u = (u1 , u2 , u3 ) = (S(t), I(t), R(t)) and f (t, u(t)) = ( f1 , f2 , f3 ) = (1 −U(t)V (t) −U(t), ρ1U(t)V (t) − ρ2V (t),V (t) − ρ3W (t)) Now we can rewrite the system (2.3) as (ui )t = fi (t, x, ui , [u]ai , [u]bi ), (ui )0 = u0. (i = 1, 2, 3),. where fi satisfies Lipschitz condition. For applying the monotone iterative method, we need to show that f1 , f2 , f3 are quasimonotone: For fixed U(t) and V1 (t) > V2 (t), (1 −UV1 −U) − (1 −UV2 −U) = U(V2 −V1 ) 6 0. Similarly, for fixed V (t) and U1 (t) > U2 (t), we have (ρ1U1V − ρ2V ) − (ρ1U2V − ρ2V ) = ρ1 (U1 −U2 )V > 0. For fixed W (t) and V1 (t) > V2 (t), we have (V1 − ρ3W ) − (V2 − ρ3W ) = V1 −V2 > 0.. Page 6.

(8) This shows that f1 , f2 , f3 possess quasimonotone property.. ˜ V˜ , W˜ ) and ub = Now we want to find two differentiable solutions u˜ = (u˜1 , u˜2 , u˜3 ) = (U, b Vb , W b ), which satisfy: (b u1 , ub2 , ub3 ) = (U,    ˜ ai , [b u]bi )  (u˜i )t > fi (t, u˜i , [u]  (b ui )t 6 fi (t, ubi , [b u]ai , [u] ˜ bi )     u˜ (0) > 0 > ub (0) i i. (2.6). Definition 2.5. A pair of function u˜ , ub are called coupled upper and lower solutions of (2.2) if u˜ > ub and satisfies (2.6) . We usually called u˜ to be upper solution and ub to be lower solution. We denote the sector ⟨b u, u⟩ ˜ ≡ {u ∈ C; ub 6 u 6 u}. ˜ b Vb , W b ), ˜ V˜ , W˜ ) and ub = (U, In this thesis, we hope to find two differentiable solutions u˜ = (U, which satisfy:  ′  b   U˜ > 1 − U˜ V − U˜     b′ 6 1 − U bV˜ − U b  U       V˜ ′ > ρ1U˜ V˜ − ρ2V˜      bVb − ρ2Vb  Vb ′ 6 ρ1U   W˜ ′ > V˜ − ρ3W˜     b ′ 6 Vb − ρ3W b  W      b ˜  U(0) > U0 > U(0)       V˜ (0) > V0 > Vb (0)      W˜ (0) > W > W b (0). (2.7). 0. For proving next lemma, we need to modify our forcing term function f1 , f2 and f3 . Precisely, we are going to find a continuous function ci (t) such that fi (t, ui , [u]ai , [u]bi ) − fi (t, vi , [u]ai , [u]bi ) > −ci (t)(ui − vi )for. ubi 6 vi 6 ui 6 u˜i. i = 1, 2, 3. (2.8). Let Fi (t, ui , [u]ai , [u]bi ) = ci ui + fi (t, ui , [u]ai , [u]bi ).. Page 7.

(9) Under these notations, the system (2.2) can be written as    U ′ + c1U = c1U + 1 −UV −U = F1 (t, ui , [u]ai , [u]bi )   V ′ + c2V = c2V + ρ1UV − ρ2V = F2 (t, ui , [u]ai , [u]bi )     W ′ + c W = c W +V − ρ W = F (t, u , [u] , [u] ) 3. 3. 3. 3. i. ai. (2.9). bi. Lemma 2.6. Let fi (t, u) be Holder ¨ continuous in t and Lipschitz continuous in u ∈ ⟨b u, u⟩, ˜ then the function Fi (u) (i = 1, 2, 3) is Holder ¨ continuous for any u ∈ ⟨b u, u⟩. ˜ Moreover, if f = ( f1 , · · · , fN ) is quasimonotone in ⟨b u, u⟩ ˜ and satisfies (2.7), then for any u, v ∈ ⟨b u, u⟩ ˜ with u > v, we have Fi (t, ui , [u]ai , [v]bi ) − Fi (t, vi , [v]ai , [u]bi ) > 0 Proof. |( fi (u)(t)) − ( fi (u)(t ′ ))| = | fi (t, x, u(t, x)) − fi (t ′ , x′ , u(t ′ , x′ ))| 6 Hi |t − t ′ |α /2 + Ki |u(t) − u(t ′ , )| 6 (Hi + Ki∗ )|t − t ′ |α /2 where Hi , Ki are Holder ¨ coefficient and Lipschitz constant of fi , and Ki∗ is a positive constant. This shows that Fi (u) is Holder ¨ continuous. For any u, v ∈ ⟨b u, u⟩, ˜ Fi (ui , [u]ai , [v]bi ) − Fi (vi , [v]ai , [u]bi ) = [ci (ui − vi ) + fi (ui , [u]ai , [v]bi ) − fi (vi , [u]ai , [v]bi )] +[ fi (vi , [u]ai , [v]bi ) − fi (vi , [v]ai , [v]bi )] +[ fi (vi , [v]ai , [v]bi ) − fi (vi , [v]ai , [u]bi )] By (2.7), the first bracket is nonnegative. By quasimonotone property of f , the second and third bracket are also nonnegative. The conclusion of this lemma is established.. Now, we are looking for two pairs of coupled upper and lower solution. If R0 < 1, then there has only one equilibrium (1, 0, 0) . In this case, we can choose u˜ = (Ω,V0 e(ρ1 Ω−ρ2 )t ,W0 e(ρ1 Ω−ρ2 )t ), 1 V0 + ρ2 − ρ3 )}, and ub = (0, 0, 0) as a pair of coupled upper and where Ω = max{1,U0 , ( ρ1 W0 lower solution. If R0 > 1, then ρ1 > ρ2 , In this case there has additional positive equilibrium ρ2 ρ1 − ρ2 ρ1 − ρ2 ( , , ). We see that (1, 0, 0) is unstable and the positive equilibrium is globally ρ1 ρ2 ρ2 ρ3 asymptotically stable([3],proposition 3.1 and theorem. 3.3). For the positive equilibrium, we find a pair of upper and lower solution u˜ = (Ω,V0 eat ,W0 eat ) and ub = (0, ε eη t + ε , ε eη t + ε ), reV0 1 ρ2 spectively, where ε 6 min{{U0 ,V0 ,W0 } , Ω = max{U0 , , }, a = max{ρ1 Ω − ρ2 , − ρ3 } ε ρ1 W0 Page 8.

(10) and η = min{−2ρ2 , 2(1 − ρ3 )} < 0. We will use this coupled upper and lower solution as initial iterations to construct two monotone convergent sequence {u(k) } and {u(k) }, which converges to the unique solution of (2.2) from above and below, respectively. It is obviously that ε eη t + ε dose not approach to zero as t → ∞. Thus, the solution will converge to the positive equilibrium ρ2 ρ1 − ρ2 ρ1 − ρ2 ( , , ) as t → ∞ . This means that the infection will be able to spread in the ρ1 ρ2 ρ2 ρ3 population. To see this , let u(0) = u˜ and u(0) = ub as two initial functions to construct two sequences (k). (k). (k). (k). {u(k) } = {u1 , · · · , uN } and u(k) = {u1 , · · · , uN } from following iteration process. ′. Li ui (k) ≡ ui (k) + ci u(k) = Fi (t, x, ui (k−1) , [u(k−1) ]ai , [u(k−1) ]bi ) ′. Li ui (k) ≡ ui (k) + ci u(k) = Fi (t, x, ui (k−1) , [u(k−1) ]ai , [u(k−1) ]bi ). (2.10) (2.11). where u(0) = u˜ and u(0) = ub can be chosen either one pair of coupled upper and lower solution in the preceding paragraph. Lemma 2.7. (positive lemma)([8],p.19) If w is differentiable and satisfies the relation wt + cw > 0 w(0, x) > 0 where c = c(t) is any bounded function, then w > 0. Lemma 2.8. The upper and lower sequenses {u(k) } and {u(k) } possess the monotone property ub 6 u(k) 6 u(k+1) 6 u(k+1) 6 u(k) 6 u˜. (2.12). Moreover, u(k) and u(k) are coupled upper and lower solution for each k.. Proof. Let w = u(0) − u(1) = u˜ − u(1) . By (2.9) and (2.10) and the property of upper solution, w = (w1 , · · · , wN ) satisfies the differential relation (0). Li wi = Li u˜i − Fi (ui , [u(0) ]ai , [u(0) ]bi ) = (u˜i )t − fi (u˜i , [u] ˜ ai , [b u]bi ) > 0 This imply that wi > 0, that is u(1) 6 u(0) . An analogous argument for lower solution gives u(1) > u(0) . Let w = u(1) − u(1) . By (2.9), (2.10) and Lemma 2.6, (1). Li w i. (0). (0). = Fi (ui , [u(0) ]ai , [u(0) ]bi ) − Fi (ui , [u(0) ]ai , [u(0) ]bi ) > 0. Page 9.

(11) The positive lemma implies that u(1) > u(1) . The above results show that u(0) 6 u(1) 6 u(1) 6 u(0) . By induction, assume u(k−1) 6 u(k) 6 u(k) 6 u(k−1). (2.13). In view of Lemma 2.6, the function w(k) = u(k) − u(k+1) satisfies the relation (k). (k−1). Li wi = Fi (ui. (k). , [u(k−1) ]ai , [u(k−1) ]bi ) − Fi (ui , [u(k) ]ai , [u(k) ]bi ) > 0. The positive lemma implies that u(k) > u(k+1) . Similar, by (2.9) and (2.10), the functions w(k) = u(k+1) − u(k) and w(k) = u(k+1) − u(k+1) satisfies following respective equations (k). (k−1). (k). (k). Li w(k) = Fi (ui , [u(k) ]ai , [u(k) ]bi ) − Fi (ui. , [u(k−1) ]ai , [u(k−1) ]bi ). Li w(k) = Fi (ui , [u(k) ]ai , [u(k) ]bi ) − Fi (ui , [u(k) ]ai , [u(k) ]bi ). (2.14) (2.15). By (2.11) and Lemma 2.6 the right hand side of (2.13) and (2.14) are both nonnegative. The positive lemma ensures that wi (k) > 0 and wi (k) > 0. This proves relation (2.12) when k is replaced by k + 1. The monotone property (2.11) follows by the principle of induction. Now for each fixed k, the relation Fi (t, ui , [u]ai , [v]bi ) − Fi (t, vi , [v]ai , [u]bi ) > 0 hold. Since ′. Li ui (k) ≡ ui (k) + ci u(k) = Fi (t, x, ui (k−1) , [u(k−1) ]ai , [u(k−1) ]bi ) and ′. Li ui (k) ≡ ui (k) + ci u(k) = Fi (t, x, ui (k−1) , [u(k−1) ]ai , [u(k−1) ]bi ) the monotone property (2.11) imply that Li ui (k) = Fi (ui (k−1) , [u(k−1) ]ai , [u(k−1) ]bi ) > Fi (ui (k) , [u(k) ]ai , [u(k) ]bi ) and Li ui (k) = Fi (ui (k−1) , [u(k−1) ]ai , [u(k−1) ]bi ) 6 Fi (ui (k) , [u(k) ]ai , [u(k) ]bi ). This leads to the relations (k). (k). (ui )t > fi (ui , [u(k) ]ai , [u(k) ]bi ). Page 10.

(12) and (k). (k). (ui )t > fi (ui , [u(k) ]ai , [u(k) ]bi ) This completes the proof of the lemma.. From this lemma, the pointwise limits lim u(k) (t) = u(t) and. k→∞. lim u(k) (t) = u(t). k→∞. (2.16). exist and satisfy the relation ub 6 u 6 u 6 u˜. Theorem 2.9. ([6],p.435)Let fi (t, u) satisfy the global Lipschitz condition | fi (t, u) − fi (t, v)| 6 Ki |u − v| for u, v ∈ RN. (2.17). Then the problem has a unique solution u. Moreover, u is the limit of the sequence {u(k) } given by the equation (k). (ui )t = fi (t, uk−1 ).. To demonstrate that system (2.2) has a unique solution, we need only to show u = u. The following theorem gives uniqueness result for the system. Theorem 2.10. Let u, ˜ ub be coupled upper and lower solution of (2.2), and let f ≡ ( f1 , · · · , fN ) be quasimonotone in ⟨b u, u⟩ ˜ and satisfies the condition (2.7) and (2.16). Then there exists a unique solution u∗ ∈ ⟨b u, u⟩. ˜ Moreover, the sequence {u(k) } , {u(k) } with u(0) = u˜ and u(0) = ub both converge monotonenically to u∗ from above and below, respectively.. Proof. By (2.15) and the continuity of Fi , the pointwise limits (k). lim Fi (t, ui , [u(k) ]ai , [u(k) ]bi ) = Fi (t, ui , [u]ai , [u]bi ). k→∞. (k). lim Fi (t, ui , [u(k) ]ai , [u(k) ]bi ) = Fi (t, ui , [u]ai , [u]bi ). k→∞. exist. The limits u and u satisfy the coupled equations (ui )t = fi (t, ui , [u]ai , [u]bi ). (2.18). (ui )t = fi (t, ui , [u]ai , [u]bi ). (2.19). Page 11.

(13) and the initial conditions in (2.2). To show that u and u are solutions of (2.2), it suffices to prove that u = u. For this purpose, we consider the 2N coupled system (vi )t = fi (t, vi , [v]ai , [w]bi ). (2.20). (wi )t = fi (t, wi , [w]ai , [v]bi ). (2.21). By identifying the 2N vector u = (u1 , · · · , u2N ) with (u1 , · · · , uN ) = (v1 , · · · , vN ), (uN+1 , · · · , u2N ) = (w1 , · · · , wN ) the above system may be expressed in the form (ui )t = fi (t, u1 , · · · , u2N ) Since fi satisfies a global Lipschitz condition in u, the theorem 2.10 guarantees the existence of the unique solution (v∗ , w∗ ) to the system (2.19) and (2.20). It is easily seen by symmetry that (w∗ , v∗ ) is also a solution of (2.19) and (2.20). By the uniqueness property, (v∗ , w∗ ) = (w∗ , v∗ ) which leads to w∗ = v∗ . However by (2.17) and (2.18), the function (u, u) is clearly a solution of (2.19) and (2.20). It follows again by the uniqueness property that (u, u) = (v∗ , w∗ ) = (v∗ , v∗ ). This shows that u = u. Now we can conclude that u = u ≡ u∗ is the unique solution of (2.2) in the sector ⟨b u, u⟩. ˜. 3 SIR system with n group In this section we consider the n group model,  n  ′ (t) = Λ −  βk j Sk (t)I j (t) − dk Sk (t) S ∑ k  k   j=1  n Ik′ (t) = ∑ βk j Sk (t)I j (t) − (εk + γk )Ik (t) k = 1, 2, · · · , n   j=1     R′ (t) = γ I (t) − δ R (t) k k k k k accompany with initial data u0 = (S1 (0), I1 (0), R1 (0), · · · , Sn (0), In (0), Rn (0)) = (S10 , I10 , R01 , · · · , Sn0 , In0 , R0n ).. Page 12. (3.1).

(14) Similarly, we can write (3.1) to the semilinear initial value problem :   du(t) + Au(t) = f (t, u(t)) dt  u(t0 ) = u0 . where . . . S1 (t). d 0 0  1     0 ε1 + γ1 0 I1 (t)       0 R1 (t)  −γ1 δ1   ..   .. . , A =  .      Sn (t)  dn 0 0      In (t)  0 εn + γ n 0   Rn (t) 0 −γn δn   n Λ − ∑ β1 j S1 (t)I j (t)  1 j=1    n    β1 j S1 (t)I j (t)  ∑   j=1       0     ... f (t, u(t)) =     n    Λn − ∑ βn j Sn (t)I j (t)    j=1   n     β S (t)I (t) ∑ n j 1 j   j=1   0.         u(t) =        . (3.2).          and       . It is obviously that A is a bounded linear operator and f satisfies Lipschiz condition. If the initial value problem has a classical or strong solution , it must satisfied of the integral equation u(t) = T (t − t0 )u0 +. ∫ t t0. T (t − s) f (s, u(s))ds,. where T (t) = et(−A) . by thm 2.3, since f : [t0 , T ] × X → X is continuously differentiable, the mild solution u(t) is a classical solution of the initial value problem. Clearly, f (t, u(t)) = (1 −U(t)V (t), ρ1U(t)V (t), 0) is continuously differentiable, so the solution of (3.2) exists, and it can be represent as. u(t) = e−(t−t0 )A u. We can see that there is a disease-free equilibrium (S1 , I1 , R1 , · · · , Sn , In , Rn ) = (. Λn Λ1 , 0, 0, · · · , , 0, 0). d1 dn. Page 13. ∫ t. 0+. t0. e(s−t)A f (s, u(s))ds.

(15) and another positive equilibrium.([4]) Denote ub = (Sb1 , Ib1 , Rb1 , · · · , Sbn , Ibn , Rbn ). u˜ = (S˜1 , I˜1 , R˜ 1 , · · · , Sñ , Iñ , R˜ n ) and. be a pair of upper and lower solutions of (3.2). Similarly as the previous section, we want to find two differentiable solutions u˜ = (u˜1 , · · · , u˜3n ) = (S˜1 , I˜1 , R˜ 1 , · · · , Sñ , Iñ , R˜ n ) and ub = (b u1 , · · · , ub3n ) = (Sb1 , Ib1 , Rb1 , · · · , Sbn , Ibn , Rbn ), which satisfy:    (u˜ ) > fi (t, u˜i , [u] ˜ ai , [b u]bi )   i t (b ui )t 6 fi (t, ubi , [b u]ai , [u] ˜ bi )     u˜ (0) > 0 > ub (0) i i. (3.3). That is S˜k ,I˜k ,R˜ k ,Sbk ,Ibk ,Rbk need satisfy following inequalities:  n ′ >Λ −  ˜  S βk j S˜k Ibj − dk S˜k ∑ k  k   j=1   n   ′ 6Λ − b  S βk j Sbk I˜j − dk Sbk ∑ k  k   j=1   n   ′  I˜k > ∑ βk j S˜k I˜j − (εk + γk )I˜k    j=1   n   ′  Ib 6 ∑ βk j Sbk Ibj − (εk + γk )Ibk (k = 1, 2, · · · , n) k.                          . j=1. R˜ ′k. (3.4). > γk I˜k − δk R˜ k. Rb′k 6 γk Ibk − δk Rbk S˜k > Sk0 > Sbk I˜k > Ik0 > Ibk R˜ k > R0k > Rbk. For the disease-free equilibrium, we choose ub = (0, 0, 0, · · · , 0, 0, 0), and find u˜. n Λk from S˜k′ > Λk − ∑ βk j S˜k Ibj − dk S˜k = Λk − dk S˜k . In fact, we can choose any constant S˜k > . dk j=1 Λk 0 Let Ω = max { , Sk } and choose 16k6n dk. S˜k = Ω for. k = 1, · · · , n.. Now we are looking for an upper solution I˜k for Ik , i.e., it have to satisfies I˜k′ >. n. n. j=1. j=1. ∑ βk j S˜k I˜j − (εk + γk )I˜k = Ω ∑ βk j I˜j − (εk + γk )I˜k Page 14.

(16) For simple, let β = max {βk j }, ε = min {εk } and γ = min {γk }. 16k6n. 16 j,k6n. 16k6n. It sufficient to solve n. I˜k′ > Ω ∑ β I˜j − (ε + γ )I˜k . j=1. Set I˜k = V0 eat (k = 1, · · · , n) where a > nΩβ − (ε + γ ) n. then it satisfies the inequality I˜k′ > Ω ∑ β I˜j − (ε + γ )I˜k j=1. Similarly, we are going to find an upper solution R˜ k for Rk , i.e., it satisfies R˜ ′k > γk I˜k − δk R˜ k = γkV0 eat − δk R˜ k For simple, let δ = min {δk }, and set 16k6n. R˜ k = W0 eat , where a satisfies a >. 1 (γ V0 − δ W0 ) W0. Then we can choose u˜ = (Ω,V0 eat ,W0 eat , · · · , Ω,V0 eat ,W0 eat ) and. ub = (0, 0, 0, · · · , 0, 0, 0). Λk 0 1 , Sk } and a = max {nΩβ − (ε + γ ), (γ V0 − δ W0 )}. It is easy to see W0 16k6n dk 16k6n that u˜ and ub be a pair of coupled upper and lower solution.. where Ω = max {. ˜ I, ˜ I, ˜ R, ˜ · · · , S, ˜ R) ˜ and Using same argument, for the positive equilibrium, we choose u˜ = (S, b I, b I, b R, b · · · , S, b R) b (such that I(t), b R(t) b not approach to zero as t → ∞) as a pair of coupled ub = (S, upper and lower solution. Let Λ∗ = max{Λk }, Λ∗ = min{Λk },. β ∗ = max{β jk }, β∗ = min{βk }, d ∗ = max{dk }, d∗ = min{dk }, (ε + γ )∗ = max{εk + γk }, (ε + γ )∗ = min{εk + γk }, S∗ = max{Sk0 }, S∗ = min{Sk0 } I ∗ = max{Ik0 }, I∗ = min{Ik0 } R∗ = max{R0k }, R∗ = min{R0k }. Page 15.

(17) For searching a pair of coupled upper and lower solution for system (3.1), we need to solve following inequality system.   S˜′ > Λ∗ − nβ∗ S˜Ib− d∗ S˜        Sb′ 6 Λ∗ − nβ ∗ SbI˜ − d ∗ Sb       I˜′ > nβ ∗ S˜I˜ − (ε∗ + γ∗ )I˜     ′ ∗ ∗  b   Ib 6 nβ∗ SIb− (ε + γ )Ib      R˜ ′ > γ ∗ I˜ − δ∗ R˜      Rb′ 6 γ Ib− δ ∗ Rb ∗   S˜ > S∗       Sb 6 S∗       I˜ > I ∗       Ib 6 I∗       R˜ > R∗      Rb 6 R ∗ For example, we may choose u˜ = (Ω, I ∗ eat , R∗ eat , · · · , Ω, I ∗ eat , R∗ eat ) and. ub = (0, ε eη t + ε , ε eη t + ε , · · · , 0, ε eη t + ε , ε eη t + ε ). γ ∗I∗ Λ∗ } , a = max{nβ ∗ Ω−(ε∗ + γ∗ ), ∗ − δ∗ }, ε 6 min{S∗ , I∗ , R∗ } nβ ∗ ε + d∗ R and η = min{−2(ε ∗ + γ ∗ ), 2(γ∗ − δ ∗ )}. where Ω = max{S∗ , I ∗ , R∗ ,. With analogous argument discussed in section 2, it is easy to show that the system (3.1) has a unique solution u∗ in the sector ⟨b u, u⟩ ˜ which converges to the positive equilibrium as t → ∞.. 4 References [1] E. Beretta, V. Capasso (1986).Global stability result for a multigroup SIR epidemic model. Mathematical Ecology,,world scientific, Singapore, 1986, 317-342 [2] Ji Chunyan, Jiang Daqing, Yang Qingshan, Shi Ningzhong . Dynamics of Multigroup SIR Epidemic Model with Stochastic Perturbation, Automatica, Volume 48, Issue 1, January 2012, Page 16.

(18) 121–131 [3] ZL Feng, WZ. Huang, C. Castillo-Chavez (2005). Global behaviour of a multi-group SIS epidemic model with age structure. Journal of Differential Equation, 218, 292-324. [4] HB Guo, MY. Li, ZS. Shuai (2006). Global stability of the endemic equilibrium of multigroup SIR epidemic models. Canadian Applied Math Quarterly, 14, 259-284. [5] HW. Hethcote (1978). An immunization model for a heterogeneous population. Yheoretical population biology, 14, 338-349. [6] W. Huang, KL. Cooke, C. Castillo-Chavez(1992). Stability and bifurcation for a multiplegroup model for the dynamics of HIV/AIDS transmission. SIAM Journal on Applied Mathematics, 52, 835-854. [7] C. Koide , H. Seno(1996). Sex ratio features of two-group SIR model for asymmetric transmission of heterosexual disease. Mathematical and computer modelling, 23, 67-91. [8] C. V. Pao. Nonlinear Parabolic and Elliptic Equations, Plenum Pub Corp. [9] A. Pazy. Semigroups of Linear Operators and Applications to Partial Differential Equations, Springer, Applied Mathematical Sciences Volume 44.. Page 17.

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