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Mathematical Study of a Class of Epidemiological Models with Multiple Infectious Stages

  • S. Bowong EMAIL logo , A. Temgoua , Y. Malong and J. Mbang

Abstract

This paper deals with the mathematical analysis of a general class of epidemiological models with multiple infectious stages for the transmission dynamics of a communicable disease. We provide a theoretical study of the model. We derive the basic reproduction number R0 that determines the extinction and the persistence of the infection. We show that the disease-free equilibrium is globally asymptotically stable whenever R01, while when R0>1, the disease-free equilibrium is unstable and there exists a unique endemic equilibrium point which is globally asymptotically stable. A case study for tuberculosis (TB) is considered to numerically support the analytical results.

JEL Classification: 34A34; 34D23; 34D40; 92D30

Appendix A: Proof of Theorem 3

Herein, we give the proof of Theorem 3 on the GAS of the endemic equilibrium Q=(S,E,Ii,,L,R) where S*, E, Ii (1 ≤ i ≤ n), L* and R* are defined as in eq. (16).

Consider the following Lyapunov function:

(19)U(S,E,Ik,L,R)=(SSlnS)+a1(EElnE)+a2(LLlnL)+a3(RRlnR)+k=1nbk(IkIklnIk),

where ai (i = 1,2,3) and bk (k = 1,…,n) are defined as

(20)a1=k(1r1)A1p+(1p)k(1r1),b1=A1A1p+(1p)k(1r1),a3=A1γA3[A1p+(1p)k(1r1)],a2=a3δ+b1λ+βlSA2,bn=a2εn+a3αn+βnSBn,bk=a2εk+a3αk+bk+1ηk+βkSBk,k=2,3,,n1.

The time derivative of U(S,E,Ik,L,R) satisfies

(21)U˙=1SSS˙+a11EEE˙+a21LLL˙+a31RRR˙+k=1nbk1IkIkI˙k,=1SSΛSk=1nβkIk+βlLμS+a11EE(1p)Sk=1nβkIk+βlLa1A1E+a1A1E+a21LLk=1nεkIka2A2L+a2A2L+a31RRk=1nαkIk+a3δLa3δLRRa3A3R+a3A3R+b11I1I1pSk=1nβkIk+βlL+k(1r1)E+γRb1(1r1)EI1I1b1γREEb1λLEEb1B1I1+b1B1I1+k=2nbk1IkIkηk1Ik1k=2nbkBkIk+k=2nbkBkIk.

By considering model system (2) at the endemic equilibrium, one has

(22)Λ=Sk=1nβkIk+βlL+μS,a1A1E=a1(1p)Sk=1nβkIk+βlL,a2A2L=a2k=1nεkIk,a3A3R=a3k=1nαkIk+a3δL,b1B1I1=b1pSk=1nβkIk+βlL+b1k(1r1)E+b1γR+b1λL,bkBkIk=bkηk1Ik1,k=1,,n1.

Introducing eq. (22) into eq. (21) yields

(23)U˙=μ(SS)2S+Sk=1nβkIk+βlL1SS+Sk=1nβkIk+βlL[1+a1(1p)+b1p]a1(1p)k=1nβkIk+βlLSEEb1pk=1nβkIk+βlLSI1I1+[a1A1+b1k(1r1)]E+[a2A2+a3δ+b1λ+βlS]L+[a3A3+b1γ]R+k=1n1[a2εk+a3αk+bk+1ηkbkBk+βkS]Ik+[a2εn+a3αnbnBn+βnS]In+Sk=1nβkIk+βlL[a1(1p)+b1p]+b1k(1r1)E1EEI1I1+b1λL1LLI1I1+a3δL1LLRR+a3k=1nαkIk1RRIkIk+a2k=1nεkIk1LLIkIk+b1γR1I1I1RR+k=2nbkηk1Ik11IkIkIk1Ik1.

Now, let (u1,u2,u3,u4,vk)=EE,RR,LL,SS,IkIk, k = 1,…,n. Then, eq. (23) becomes

(24)U˙=μ(SS)2S+Sk=1nβkIk+βlL1u4+Sk=1nβkIk+βlL[1+a1(1p)+b1p]a1(1p)k=1nβkIkvk+βlLu3Su1u4b1pk=1nβkIkvk+βlLu3Sv1u4+[a1A1+b1k(1r1)]E+[a2A2+a3δ+b1λ+βlS]L+[a3A3+b1γ]R+k=1n1[a2εk+a3αk+bk+1ηkbkBk+βkS]Ik+[a2εn+a3αnbnBn+βnS]In+Sk=1nβkIk+βlL[a1(1p)+b1p]+b1k(1r1)E1v1u1+b1λL1v1u3+a3δL1u2u3+a3k=1nαkIk1u2vk+a2k=1nεkIk1u3vk+b1γR1v1u2+k=2nbkηk1Ik11vkvk1.

The coefficients ai et bk are chosen such that the coefficients of Sk=1nβkIk+βlL, E, L , I1, Ik (k = 2,…,n - 1), In, and R are equal to zero, that is,

(25)1+a1(1p)+b1p=0,a1A1+b1k(1r1)=0,a2A2+a3δ+b1λ+βlS=0,a3A3+b1γ=0,a2ε1+a3α1+b2η1b1B1+β1S=0,a2εk+a3αk+bk+1ηkbkBk+βkS=0,k=2,,n1,a2εn+a3αnbnBn+βnS=0.

Now, multiplying the fifth equation of (25) by I1 and using the expression of b1B1I1 defined as in eq. (22) yields

(26)a2ε1I1+a3α1I1+b2η1I1b1B1I1+β1SI1=a2ε1I1+a3α1I1+b2η1I1b1pSk=1nβkIk+βlLb1γRb1λL+β1SI1.

On the other hand, from eq. (22), one has

(27)a1A1Ea1(1p)Sk=1nβkIk+βlL+a3A3Ra3k=1nαkIka3δL+a2A2La2k=1nεkIk+k=1n1bkBkIkk=1n1bkηk1Ik1+bnBnInbnηn1In1=0.

Adding the above equation in the right-hand side of eq. (26) gives

(28)a2ε1I1+a3α1I1+b2η1I1b1B1I1+β1SI1=[a1A1b1k(1r1)]E+[b1γ+a3A3]R+[b1pa1(1p)+1]k=1nβkIk+[b1pa1(1p)+1]βlSL+k=2n1[βkSa2εka3αkbk+1ηk+bkBk]Ik+[βlSb1λa3δ+a2A2]L+[βnS+bnBna2εna3αn]In=0.

Note that when the fifth equation of (25) is satisfied, then all equations of (25) are also satisfied. This is why in the sequel, we only consider the following system of equations:

(29)1+a1(1p)+b1p=0,a1A1+b1k(1r1)=0,a2A2+a3δ+b1λ+βlS=0,a3A3+b1γ=0,a2εk+a3αk+bk+1ηkbkBk+βkS=0,k=2,,n1,a2εn+a3αnbnBn+βnS=0.

Solving the above equations give the expressions of ai and bk defined as in eq. (20). Thus, after plugging the expressions of ai and bk given as in eq. (20) into eq. (24), one obtains

(30)U˙=μ(SS)2S+Sk=1nβkIk+βlL2u4a1(1p)k=1nβkIkvk+βlLu3Su1u4b1pk=1nβkIkvk+βlLu3Sv1u4+b1k(1r1)E1v1u1+b1λL1v1u3+a3δL1u2u3+a3k=1nαkIk1u2vk+a2k=1nεkIk1u3vk+b1γR1v1u2+k=2nbkηk1Ik11vkvk1.

Using the fact that a1(1p)+b1p=1, eq. (30) becomes

(31)U˙=μ(SS)2S+a1(1p)k=1nβkSIk2u4u1vku4+b1pk=1nβkIkS2u4v1vku4+a1(1p)βlLS2u4u1u3u4+b1pβlLS2u4v1u3u4+b1k(1r1)E1v1u3+b1λL1v1u3+a3δL1u2u3+a3k=1nαkIk1u2vk+a2k=1nεkIk1u3vk+b1γR1v1u2+k=2nbkηk1Ik11vkvk1.

Now, multiplying the second, third, fourth, and fifth equations of (29) by E*, L*, R*, Ik (k = 2,…,n - 1), and In, respectively, gives

(32)a1A1E+b1k(1r1)E=0,a2A2L+a3δL+b1λL+βlSL=0,a3A3R+b1γR=0,a2εkIk+a3αkIk+bk+1ηkIkbkBkIk+βkSIk=0,k=2,,n1,a2εnIn+a3αnInbnBnIn+βnSIn=0.

Using the expressions of a1A1E, a2A2L, a3A3R, bkBkIk (k = 2,…,n - 1) and bnBnIn given as in eq. (22), then eq. (32) becomes

(33)a1(1p)Sk=1nβkIk+βlL+b1k(1r1)E=0,a2k=1nεkIk+a3δL+b1λL+βlSL=0,a3k=1nαkIka3δL+b1γR=0,a2εkIk+a3αkIk+bk+1ηkIkbkηk1Ik1+βkSIk=0,k=2,,n1,a2εnIn+a3αnInbnηn1In1+βnSIn=0.

Now, let F1(u), F2(u), F3(u), and Gk(u) (k = 2,…,n) where u=(u1,u2,u3,vk)T be n + 2 functions to be determined later. Then, multiplying the first, second, third, fourth, and fifth equations in (33) by F1(u), F2(u), F3(u), and Gk(u) (k = 2,…,n - 1) and Gn(u), respectively, yields

(34)a1(1p)k=1nβkSIkF1(u)a1(1p)SβlSLF1(u)+b1k(1r1)EF1(u)=0,a2k=1nεkIkF2(u)+a3δLF2(u)+b1λLF2(u)+[a1(1p)+b1p]βlSLF2(u)=0,a3k=1nαkIkF3(u)a3δLF3(u)+b1γRF3(u)=0,a2εkIkGk(u)+a3αkIkGk(u)+bk+1ηkIkGk(u)bkηk1Ik1Gk(u)+[a1(1p)+b1p]βkSIkGk(u)=0,k=2,,n1,a2εnInGn(u)+a3αnInGn(u)bnηn1In1Gn(u)+[a1(1p)+b1p]βnSInGn(u)=0.

Adding eq. (34) to the right-hand side of (31) gives

(35)U˙=μ(SS)2S+a1(1p)β1SI12u4u1v1u4F1+a1(1p)k=2nβkIk2u4u1vku4F1+Gk+b1pβ1SI12u41u4+b1pk=2nβkIkS2u4v1vku4+Gk+a1(1p)βlLS2u4u1u3u4+F2F1+b1pβlLS2u4v1u3u4+F2+b1k(1r1)E1v1u1+F1+b1λL1v1u3+F2+a3δL1u2u3+F2F3+a3α1I11u2v1F3+a3k=2nαkIk1u2vkF3+Gk+a2ε1I11u3v1F2+a2k=2nεkIk1u3vk+GkF2+b1γR1v1u2+F3+b2η1I11v2v1G2+k=3nbkηk1Ik11vkvk1+Gk1Gk.

Now, we shall choose the functions F1(u), F2(u), F3(u), F4(u), and Gk(u), k = 2,…,n which make U˙ nonpositive. To do so, the functions F1(u), F2(u), F3(u), and F4(u) are chosen such that the coefficients of b1k(1r1)E, a2ε1I1, b1γR, b2η1I1 are equal to zero, and the function Gk(u) is chosen such that the coefficient of b2η1I1 is equal to zero and the coefficient of αkIk is nonpositive, that is,

(36)F1(u)=1+v1u1,F2(u)=1u3v1,F3(u)=1+v1u2andGk(u)=k1v2v1v3v2vkvk1,k=2,,n.

After plugging these expressions given as in eq. (36) into eq. (35), one finally obtains

(37)U˙=μ(SS)2S+a1(1p)β1SI13u4u1v1u4v1u1+a1(1p)k=2nβkIkk+2u4u1vku4v1u1v2v1v3v2vkvk1+b1pβ1SI12u41u4+b1pk=2nβkIkSk+2u4v1vku4v2v1v3v2vkvk1+a1(1p)βlLS4u4u1u3u4u3v1v1u1+b1pβlLS3u4v1u3u4u3v1+b1λL2v1u3u3v1+a3δL3u2u3u3v1v1u2+a3α1I12u2v1v1u2+a3k=2nαkIkk+2u2vkv1u2v2v1v3v2vkvk1+a2k=2nεkIkkv2v1v3v2vkvk1u3vk+u3v11.

Since the arithmetic mean exceeds the geometric mean, the following inequalities hold:

(38)3u4u1v1u4v1u10,k+2u4u1vku4v1u1v2v1v3v2vkvk10,2u41u40,k+2u4v1vku4v2v1v3v2vkvk10,4u4u1u3u4u3v1v1u10,3u4v1u3u4u3v10,2v1u3u3v10,3u2u3u3v1v1u20,2u2v1v1u20,k+2u2vkv1u2v2v1v3v2vkvk10.

Now, let

(39)Dk=u3v1+(k1)v2v1+v3v2++vkvk1+u3vk,k=2,,n.

The next step is to show that the functions Dk is nonpositive for all v1,u1,u3,vkR+ for each value of k. Others expressions are negative by applying Corollary 1. By using Lemma 5 in Appendix B with w = k,  Y = v1,  yi=vi+1,i=1,...,k1, and  yk=u3, when viu3,i=1,...,k, one has Dk ≤ 0. Thus, U˙0 for all S,E,Ik,L,R ≥ 0, provided that S,E,Ik,L,R are positive, where the equality U˙=0 holds only on the straight line S = S*, E/E=Ik/Ik=L/L=R/R, k = 1,…,n. It is easy to see that for model system (2), Q* is the only equilibrium state on this line. Therefore, by Lyapunov–LaSalle asymptotic stability theorem [38, 39, 40, 41], the positive equilibrium state Q* is globally asymptotically stable in the positive region ΩR+n+4, except on the S-axis which is the stable manifold for the fixed point Q0. This achieves the proof.

   □

Appendix B: Useful inequalities

In this appendix, we give inequalities which are necessary to demonstrate that the time derivative of the Lyapunov function (19) is nonpositive. A key tool is the arithmetic geometric means inequality, which we state here.

Lemma 4

(Arithmetic geometric means inequality): Letz1,,zwbe positive real numbers. Then,

(40)z1zwwz1++zww.

Furthermore, exact equality only occurs ifz1==zw.   □

An immediate consequence of the arithmetic geometric means inequality follows.

Corollary 1

Lety1,,ywbe positive real numbers such asy1yw=1 . Then

(41)w(y1+y2++yw)0,

Furthermore, exact equality only occurs ify1==yw

We also have the following result

Lemma 5

LetYy1ywbe positive real numbers. Then

(42)ykY+(w1)y1Y+y2y1++ywyw10.

   □

Proof

We have

ywY+(w1)y1Y+y2y1++ywyw1=ywYYy1YY+y2y1y1++ywyw1yw1,ywYYy1YY+y2y1Y++ywyw1Y,0.

   □

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Received: 2017-11-07
Accepted: 2019-10-21
Published Online: 2019-11-22
Published in Print: 2020-05-26

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