The Asymptotic Results of a SARS Epidemic Model without Quarantine

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THE ASYMPTOTIC RESULTS OF A SARS EPIDEMIC MODEL WITHOUT QUARANTINE

SZE-BI HSU AND LIH-ING W. ROEGER

Abstract. In this short article, we present the continuing work on a SARS model without quarantine by Hsu and Hsieh [SIAM J. Appl. Math., 66 (2006), 627–647]. We find the relation between the initial susceptible population S0 and the asymptotic susceptible population S∞.

1. Introduction

The following is a differential equation model for severe acute respiratory syn-drome (SARS) without quarantine proposed by Hsu and Hsieh [1].

S′ = −βIS E+ I + S c 1 + a(P + R + D), E′ = βIS E+ I + S c 1 + a(P + R + D)−µE, I′ = µE −(σ1+ ρ1+ γ3)I, (1) P′ = γ3I −(σ2+ ρ2)P, R′ = σ1I+ σ2P, D′ = ρ1I+ ρ2P.

The initial conditions are S(0) = S0 >0, I(0) = I0 >0, E(0) = P (0) = R(0) =

D(0) = 0, and S(t) + E(t) + I(t) + P (t) + R(t) + D(t) ≡ N = I0+ S0. The

vari-ables S(t), E(t), I(t), and P (t) are the number of susceptible individuals, infected asymptomatic individuals, infected individuals with onset of symptoms, and iso-lated probable SARS cases at time t. R(t) is the cumulative number of discharged SARS patients and D(t), the cumulative number of SARS deaths, at time t. Sys-tem (1) is an SARS epidemic model without quarantine. To see the details of the modeling process and the related model with quarantine, please refer to [1].

Hsu and Hsieh [1] had done some analysis for this model. We will state their results. The disease-free equilibrium (DFE) for the system in (S, E, I, P, R, D) is

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(S∗ ,0, 0, 0, R∗ , D∗ ) with S∗ +R∗ +D∗

= N ; the endemic equilibrium is (0, 0, 0, 0, R#, D#)

with R#+ D#= N . The basic reproduction number R

0 for this model is

R0= 1

σ1+ ρ1+ γ3

βc 1 + a(N − S∗).

If R0<1, the disease-free equilibrium is locally asymptotically stable and if R0>1,

unstable. Hsu and Hsieh [1] had obtained the following results.

THEOREM 1. For the SARS model with quarantine (1), the solutions have the following asymptotic properties: S(t) → S∞ ≥0, R(t) → R∞>0, D(t) → D∞>

0, I(t) → 0, E(t) → 0, and P (t) → 0 as t → ∞.

THEOREM 2. Consider system (1). Let ˜β =1+aNβc and q = σ1+ ρ1+ γ3.

(i) If q > ˜β, then S(t) → S∞>0 as t → ∞.

(ii) If q < ˜β, then S(t) → 0 as t → ∞.

The above theorems state that the asymptotic dynamics are actually global so that we can write the disease-free equilibrium or the endemic equilibrium to be (S∞,0, 0, 0, R∞, D∞) and (0, 0, 0, 0, R∞, D∞) depending on R0. In the following

section, we will give details of finding the relation between S0and S∞.

We note that in the classical Kermack-Mckendric SIR model, the asymptotic state S∞ satisfies a transcendental equation [2, 3], so does S∞ obtained in this

model (1).

2. Main Results

We can integrate the equations of P , R, and D in (1) from t = 0 to t = ∞. Since I(∞) = 0 = P (∞), we have (2) γ3 Z ∞ 0 I(t)dt = (σ2+ ρ2) Z ∞ 0 P(t)dt, (3) R∞= σ1 Z ∞ 0 I(t)dt + σ2 Z ∞ 0 P(t)dt, and (4) D∞= ρ1 Z ∞ 0 I(t)dt + ρ2 Z ∞ 0 P(t)dt. By substituting (2) into (3) and (4), we obtain

R∞= (σ1+ σ2 γ3 (σ2+ ρ2)) Z ∞ 0 I(t)dt,

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and D∞= (ρ1+ ρ2 γ3 (σ2+ ρ2)) Z ∞ 0 I(t)dt. Let (5) r= σ1+ σ2 γ3 (σ2+ρ2) ρ1+ ρ22γ3 2) . Then we have (6) R∞ D∞ = r.

Let V = S + E + I. Then P + R + D = N − V , and system (1) can be simplified so that the two equations for S and V are

S′ = −βIS V c 1 + a(N − V ), V′ = −qI. Then we have dS dV = βc q S V(1 + a(N − V )).

Applying the method of separation of variables and integrating both sides of the equation leads to ln S S0  = βc q(1 + aN )ln  V V0(1 + a(N − V ))  . Since V0= S0+ E0+ I0= N , and V∞= S∞, S∞ satisfies the following

(7)  S∞ S0 q(1+aN )βc = S∞ N(1 + a(N − S∞)) . We can show that the two equations

(8) y= x

S0

q(1+aN )βc

and y = x

N(1 + a(N − x))

only intersect at x = 0 and 0 < x = S∞ < S0 as seen in Figure 1. We obtain the

results in Figure 1 by choosing the parameters β = 1, c = 0.4, a = 0.0013, µ = 0.14, σ1= σ2 = 0.2, ρ1= ρ2= 0.1, γ3= 0.4, and the initial conditions S0= 10, I0= 1,

and E0= P0= R0= D0= 0. We found that S∞= 8.8105 which also satisfies (7).

In case (i) of Theorem 2, since S∞+ R∞+ D∞= N , and R∞ and D∞ satisfied

(6), we can find all three values, S∞, R∞, and D∞. In case (ii) of Theorem 2, all

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0 50 100 150 200 8.5 9 9.5 10 Time S 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1

Figure 1. The top figure shows S∞= 8.8105. The number is also

the x value of the intersection of the equations (8) in the bottom figure.

eventually everyone gets infected and recovered or died. Since R∞+ D∞= N , it

is also easy to find R∞ and D∞. Therefore, we have the following results.

THEOREM 3. Let ˜β and q are defined as in Theorem 2 and r as in (5).

(i) If q > ˜β, then S∞>0 and satisfies (7), R∞=

r(N −S∞)

1+r , and D∞=N −S

a+r .

(ii) If q < ˜β, then S∞= 0, R∞= 1+rrN , and D∞= a+rN .

References

[1] Sze-Bi Hsu and Ying-Hen Hsieh. Modeling Intervention Measures and Severity-dependent Public Response During Severe Acute Respiratory Syndrome Outbreak, SIAM J. Appl. Math., 66 (2006), 627–647. Add references:

[2] J. D. Murray. Mathematical Biology. Springer-Verlag, New York, 1989.

[3] P. E. Waltman. Deterministic Threshold Models in the Theory of Epidemics. Lecture Notes in Biomathematics, Vol 1, Springer-Verlag, New York, 1974.

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Sze-Bi Hsu, Department of Mathematics, National Tsing Hua University, Hsinchu, Taiwan

E-mail address: sbhsu@math.nthu.edu.tw

Lih-Ing W. Roeger, Department of Mathematics and Statistics, Box 41042, Texas Tech University, Lubbock, TX 79409, Tel: 806-742-2580, Fax: 806-742-1112

數據

Figure 1. The top figure shows S ∞ = 8.8105. The number is also

Figure 1.

The top figure shows S ∞ = 8.8105. The number is also p.4

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