Analysis of an SIS epidemic model with treatment

Jinghai Wang; Qiaohong Jiang

doi:10.1186/1687-1847-2014-246

Research
Open Access
Published: 24 September 2014

Analysis of an SIS epidemic model with treatment

Jinghai Wang¹ &
Qiaohong Jiang¹

Advances in Difference Equations volume 2014, Article number: 246 (2014) Cite this article

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Abstract

An SIS epidemic model with saturated incidence rate and treatment is considered. According to different recovery rates, we use differential stability theory and qualitative theory to analyze the various kinds of endemic equilibria and disease-free equilibrium. Finally, we get complete configurations of different endemic equilibria and disease-free equilibrium.

1 Introduction and model

Infectious diseases have tremendous influence on human life and will bring huge panic and disaster to mankind once out of control. Every year millions of human beings suffer from or die of various infectious diseases. In order to predict the spreading of infectious diseases, many epidemic models have been proposed and analyzed in recent years (see [1–13]). Some new conditions should be considered into SIS model to extend the results.

Li et al. (see [13]) studied an SIS model with bilinear incidence rate $β S I$ and treatment. The model takes into account the medical conditions. The recovery of the infected rate is divided into natural and unnatural recovery rates. Because of the medical conditions, when the number of infected persons reaches a certain amount $I 0$ , the unnatural recovery rate will be a fixed value $δ I 0$ . The study of this model should be divided into two cases to discuss with $I ≤ I 0$ and $I > I 0$ . In this paper, we study an SIS model with saturated incidence rate $β S I 1 + α S$ and treatment, and we extend some recent results.

By a saturated incidence rate $β S I 1 + α S$ , we consider an SIS epidemic model which consists of the susceptible individuals $S (t)$ , the infectious individuals $I (t)$ and the total population $N (t)$ at time t:

{d S d t = A − d S − β S I 1 + α S + γ I + T (I), d I d t = β S I 1 + α S − (d + ε + γ) I − T (I), N (t) = S (t) + I (t),

(1.1)

where $T (I) = {δ I, if 0 ≤ I ≤ I 0, k, if I ≥ I 0$ ( $k = δ I 0$ ) is the rate at which infected individuals are treated; A is the recruitment rate of individuals (including newborns and immigrants) into the susceptible population; $β S I 1 + α S$ is the nonlinear incidence rate; d is the natural death rate; γ is the rate at which infected individuals are recovered; ε is the disease-related death rate, A, d, γ, δ, ε, α are all positive numbers.

Thus, if $0 ≤ I ≤ I 0$ , model (1.1) implies

{d S d t = A − d S − β S I 1 + α S + γ I + δ I, d I d t = β S I 1 + α S − (d + ε + γ) I − δ I .

(1.2)

If $I > I 0$ , model (1.1) implies

{d S d t = A − d S − β S I 1 + α S + γ I + k, d I d t = β S I 1 + α S − (d + ε + γ) I − k .

(1.3)

2 Existence of equilibria

Now, we study equilibria of model (1.1). Steady states of model (1.1) satisfy the following equations:

{A − d S − β S I 1 + α S + γ I + T (I) = 0, β S I 1 + α S − (d + ε + γ) I − T (I) = 0 .

(2.1)

We easily see that model (1.1) has a disease-free equilibrium $P 0 (A d, 0)$ .

If $0 < I ≤ I 0$ , it follows from equation (2.1) that

{A − d S − β S I 1 + α S + γ I + δ I = 0, β S I 1 + α S − (d + ε + γ) I − δ I = 0,

(2.2)

and if $I > I 0$ , we get

{A − d S − β S I 1 + α S + γ I + k = 0, β S I 1 + α S − (d + ε + γ) I − k = 0 .

(2.3)

From two equations of (2.2), we have

S = A − (d + ε) I d .

(2.4)

By substituting (2.4) into the second equation of (2.2), we obtain the following equations:

β A − β (d + ε) I d + A α − α (d + ε) I − (d + ε + γ + δ) = 0, (d + ε) [− β + α (d + ε + γ + δ)] I = (d + A α) (d + ε + γ + δ) − β A .

(2.5)

Let $R 0 = β A (d + A α) (d + ε + γ + δ)$ . We study equation (2.5) as follows.

If $α (d + ε + γ + δ) > β$ , $R 0 < 1$ holds if and only if

I = (d + A α) (d + ε + γ + δ) (1 − R 0) (d + ε) [− β + α (d + ε + γ + δ)] > 0

with

S = 1 d [A − (d + ε) (d + A α) (d + ε + γ + δ) (1 − R 0) (d + ε) [− β + α (d + ε + γ + δ)]] = − d + ε + γ + δ − β + α (d + ε + γ + δ) < 0 .

So this case need not be considered.

If $α (d + ε + γ + δ) < β$ , $R 0 > 1$ holds if and only if

I = (d + A α) (d + ε + γ + δ) (R 0 − 1) (d + ε) [β − α (d + ε + γ + δ)] > 0

with

S = d + ε + γ + δ β − α (d + ε + γ + δ) > 0 .

Then we get a positive equilibrium $P ∗ (S ∗, I ∗)$ of (1.2), where

I ∗ = (d + A α) (d + ε + γ + δ) (R 0 − 1) (d + ε) [β − α (d + ε + γ + δ)], S ∗ = d + ε + γ + δ β − α (d + ε + γ + δ) .

Furthermore, if $0 < I ∗ ≤ I 0$ , $P ∗ (S ∗, I ∗)$ is an endemic equilibrium of model (1.1) when

1 < R 0 ≤ 1 + (d + ε) [β − α (d + ε + γ + δ)] (d + A α) (d + ε + γ + δ) I 0 .

Define

n 0 = 1 + (d + ε) [β − α (d + ε + γ + δ)] (d + A α) (d + ε + γ + δ) I 0 .

Therefore model (1.1) has a disease-free equilibrium $P 0 (A d, 0)$ and has an endemic equilibrium $P ∗ (S ∗, I ∗)$ except the disease-free equilibrium $P 0 (A d, 0)$ when $1 < R 0 ≤ n 0$ .

By substituting (2.4) into the second equation of (2.3), we obtain the following equation:

(d + ε) [β − α (d + ε + γ)] I 2 + [(A α + d) (d + ε + γ) − β A − k α (d + ε)] I + k (A α + d) = 0 .

(2.6)

Let $b = (A α + d) (d + ε + γ) − β A − k α (d + ε)$ . We study equation (2.6) as follows.

If $β = α (d + ε + γ)$ , (2.6) has a positive root if $b < 0$ , then

I = k (A α + d) k α (d + ε) − d (d + ε + γ), S = − k (d + ε) + A (d + ε + γ) k α (d + ε) − d (d + ε + γ) < 0 .

So this case need not be considered.

If $β < α (d + ε + γ)$ , it follows from (2.6) that

(d + ε) [α (d + ε + γ) − β] I 2 + [β A + k α (d + ε) − (A α + d) (d + ε + γ)] I − k (A α + d) = 0 .

(2.7)

Then

Δ 1 = [β A + k α (d + ε) − (A α + d) (d + ε + γ)] 2 + 4 k (d + ε) [α (d + ε + γ) − β] (A α + d) > 0 .

Denoting two roots of (2.7) by $I 1$ and $I 2$ , we have

I 1 + I 2 = − β A + k α (d + ε) − (A α + d) (d + ε + γ) (d + ε) [α (d + ε + γ) − β], I 1 ⋅ I 2 = − k (A α + d) (d + ε) [α (d + ε + γ) − β] < 0 .

So (2.7) has only one positive root, denote it by $I 1$ ,

I 1 = b + Δ 1 2 (d + ε) [α (d + ε + γ) − β], S 1 = 1 d [A − (d + ε) I 1] .

Then $S 1 > 0$ holds only if

R 0 < (A α − d) (d + ε + γ) + k α (d + ε) − Δ 1 (A α + d) (d + ε + γ + δ) .

Define

n 1 = (A α − d) (d + ε + γ) + k α (d + ε) − Δ 1 (A α + d) (d + ε + γ + δ) .

The point $P 1 (S 1, I 1)$ satisfies (2.3), then $I 1 > I 0$ , i.e.,

b + Δ 1 2 (d + ε) [α (d + ε + γ) − β] > I 0,

we have

Δ 1 > − b + 2 (d + ε) [α (d + ε + γ) − β] I 0 .

(2.8)

Then $− b + 2 (d + ε) [α (d + ε + γ) − β] I 0 < 0$ .

And

R 0 < 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + γ + δ) − 2 (d + ε) [α (d + ε + γ) − β] I 0 (d + ε + γ + δ) (A α + d) .

Define

n 2 = 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + γ + δ) − 2 (d + ε) [α (d + ε + γ) − β] I 0 (d + ε + γ + δ) (A α + d) .

Then (2.8) holds only if

{− b + 2 (d + ε) [α (d + ε + γ) − β] I 0 ≥ 0, Δ 1 ≥ {− b + 2 (d + ε) [α (d + ε + γ) − β] I 0} 2 .

Then

n 2 < R 0 ≤ 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + γ + δ) + k (d + ε + γ + δ) I 0 + (d + ε) [β − α (d + ε + γ)] I 0 (A α + d) (d + ε + γ + δ),

define

n 3 = 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + γ + δ) + k (d + ε + γ + δ) I 0 + (d + ε) [β − α (d + ε + γ)] I 0 (A α + d) (d + ε + γ + δ) .

So, if $R 0 ≤ n 3$ and $R 0 ≤ n 1$ , $P 1 (S 1, I 1)$ is an endemic equilibrium, where

I 1 = b + Δ 1 2 (d + ε) [α (d + ε + γ) − β], S 1 = 1 d [A − (d + ε) I 1] .

If $β > α (d + ε + γ)$ , it is easy to see that (2.6) has no positive root if $b ≥ 0$ .

If $b < 0$ ,

Δ 2 = b 2 − 4 k (d + ε) (A α + d) [β − α (d + ε + γ)], b = − R 0 (d + ε + γ + δ) (A α + d) + (A α + d) (d + ε + δ + γ) − k α (d + ε) − δ (A α + d) .

Then $Δ 2 ≥ 0$ implies $b 2 ≥ 4 k (d + ε) (A α + d) [β − α (d + ε + γ)]$ , we get

R 0 ≤ 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) − 2 k (A α + d) (d + ε) [β − α (d + ε + γ)] (A α + d) (d + ε + δ + γ)

or

R 0 ≥ 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) + 2 k (A α + d) (d + ε) [β − α (d + ε + γ)] (A α + d) (d + ε + δ + γ) .

Define

n 4 = 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) − 2 k (A α + d) (d + ε) [β − α (d + ε + γ)] (A α + d) (d + ε + δ + γ), n 5 = 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) + 2 k (A α + d) (d + ε) [β − α (d + ε + γ)] (A α + d) (d + ε + δ + γ) .

At the same time, $b < 0$ holds if and only if $R 0 > 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ)$ .

Define

n 6 = 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) .

Therefore, if $R 0 ≥ n 5$ , we have $b < 0$ and $Δ 2 ≥ 0$ , then (2.6) has two positive roots $I 2$ , $I 3$ , where

I 2 = − b − Δ 2 2 (d + ε) [β − α (d + ε + γ)], I 3 = − b + Δ 2 2 (d + ε) [β − α (d + ε + γ)] .

Then $S i = 1 d [A − (d + ε) I i] > 0$ ( $i = 2, 3$ ) holds only if

R 0 < 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) + 2 A [β − α (d + ε + γ)] + Δ 2 (A α + d) (d + ε + δ + γ), R 0 < 1 − δ (A α + d) + k α (d + ε) + Δ 2 (A α + d) (d + ε + δ + γ) + 2 A [β − α (d + ε + γ)] (A α + d) (d + ε + δ + γ) .

Define

n 7 = 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) + 2 A [β − α (d + ε + γ)] + Δ 2 (A α + d) (d + ε + δ + γ), n 8 = 1 − δ (A α + d) + k α (d + ε) + Δ 2 (A α + d) (d + ε + δ + γ) + 2 A [β − α (d + ε + γ)] (A α + d) (d + ε + δ + γ) .

It is easy to see that $n 8 < n 7$ , which implies that (2.6) has two positive equilibrium points $P 2 (S 2, I 2)$ , $P 3 (S 3, I 3)$ if $R 0 < n 8$ , (2.6) has only one positive equilibrium point $P 2 (S 2, I 2)$ if $n 8 < R 0 < n 7$ , (2.6) has no positive equilibrium point if $R 0 ≥ n 7$ .

Now, we consider the conditions for $I i > I 0$ ( $i = 2, 3$ ).

I 2 > I 0 ⇒ − b − Δ 2 > 2 (d + ε) [β − α (d + ε + γ)] I 0 ⇒ b + 2 (d + ε) [β − α (d + ε + γ)] I 0 < 0 .

Then

R 0 > 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) + 2 (d + ε) [β − α (d + ε + γ)] I 0 (A α + d) (d + ε + δ + γ) .

Define

n 9 = 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) + 2 (d + ε) [β − α (d + ε + γ)] I 0 (A α + d) (d + ε + δ + γ) .

Furthermore,

− b − Δ 2 > 2 (d + ε) [β − α (d + ε + γ)] I 0 ⇒ {b + 2 (d + ε) [β − α (d + ε + γ)] I 0} 2 > Δ,

i.e.,

R 0 < 1 − δ (A α + d) + k α (d + ε) (A α + d) (d + ε + δ + γ) + k (d + ε + δ + γ) I 0 + (d + ε) [β − α (d + ε + γ)] I 0 (A α + d) (d + ε + δ + γ) .

Therefore, if $n 9 < R 0 < n 3$ , $I 2 > I 0$ holds.

Similarly, if $I 3 > I 0$ ,

b + 2 (d + ε) [β − α (d + ε + γ)] I 0 ≤ 0

or

{b + 2 (d + ε) [β − α (d + ε + γ)] I 0 > 0, Δ > {2 (d + ε) [β − α (d + ε + γ)] I 0 + b} 2,

we get $R 0 ≤ n 9$ or $R 0 > max (n 3, n 9)$ .

From the above discussion, we get the following conclusions.

Theorem 2.1 If $R 0 < 1$ , model (1.2) has only one disease-free equilibrium $P 0 (A d, 0)$ ; if $R 0 > 1$ , model (1.2) has a unique endemic equilibrium $P ∗ (S ∗, I ∗)$ except the disease-free equilibrium $P 0 (A d, 0)$ ; if $1 < R 0 < n 0$ , $P ∗ (S ∗, I ∗)$ is a unique endemic equilibrium of model (1.1).

Theorem 2.2 If $β < α (d + ε + γ)$ , then $P 1 (S 1, I 1)$ is a unique endemic equilibrium of model (1.3) if $R 0 ≤ n 1$ ; $P 1 (S 1, I 1)$ is a unique endemic equilibrium of model (1.1) if $R 0 ≤ n 1$ and $R 0 ≤ n 2$ .

If $β > α (d + ε + γ)$ , model (1.3) has two positive equilibrium points $P 2 (S 2, I 2)$ , $P 3 (S 3, I 3)$ if $R 0 < n 8$ ; model (1.3) has only one positive equilibrium point $P 2 (S 2, I 2)$ if $n 8 < R 0 < n 7$ ; model (1.3) has no positive point if $R 0 ≥ n 7$ ; $P 2 (S 2, I 2)$ is an endemic equilibrium of model (1.1) if $n 9 < R 0 < n 3$ ; $P 3 (S 3, I 3)$ is an endemic equilibrium of model (1.1) if $R 0 < n 9$ or $n 9 < R 0 < n 3$ .

If $β = α (d + ε + γ)$ , model (1.3) has no endemic equilibrium.

3 Stability of equilibria

Theorem 3.1 The disease-free equilibrium $P 0 (A d, 0)$ is stable if $R 0 < 1$ and is a saddle point if $R 0 > 1$ ; the endemic equilibrium $P ∗ (S ∗, I ∗)$ is a stable node if it exists; the endemic equilibrium $P 1 (S 1, I 1)$ is a stable node if it exists; if the endemic equilibrium points $P 2 (S 2, I 2)$ , $P 3 (S 3, I 3)$ exist, then $P 2 (S 2, I 2)$ is a stable node of model (1.1) if $− b − Δ 2 2 (d + α) [β − α (d + α + γ)] D 1 + D 2 > 0$ and $P 3 (S 3, I 3)$ is a stable node of model (1.1) if $− b + Δ 2 2 (d + α) [β − α (d + α + γ)] D 1 + D 2 > 0$ .

Proof The Jacobi matrix of model (1.2) is

J = [− d − β I (1 + α S) 2 − β S 1 + α S + δ + γ β I (1 + α S) 2 β S 1 + α S − (d + ε + δ + γ)],

then

J (P 0) = [− d − β A d + α A + γ + δ 0 β A d + α A − (d + ε + γ + δ)], det J (P 0) = d (d + ε + δ + γ) (1 − R 0), tr J (P 0) = − d − (d + ε + δ + γ) (1 − R 0) .

Thus, $P 0 (A d, 0)$ is a stable node if $R 0 < 1$ , and is a saddle point if $R 0 > 1$ .

For $P ∗ (S ∗, I ∗)$ ,

J (P ∗) = [− d − β I ∗ (1 + α S ∗) 2 − β S ∗ 1 + α S ∗ + δ + γ β I ∗ (1 + α S ∗) 2 β S ∗ 1 + α S ∗ − (d + ε + δ + γ)] = [− d − β I ∗ (1 + α S ∗) 2 − (d + ε) β I ∗ (1 + α S ∗) 2 0], tr J (P ∗) = − d − β I ∗ (1 + α S ∗) 2 < 0, det J (P ∗) = (d + ε) β I ∗ (1 + α S ∗) 2 > 0 .

So, $P ∗ (S ∗, I ∗)$ is a stable node if it exists.

The Jacobi matrix of model (1.3) is

J = [− d − β I (1 + α S) 2 − β S 1 + α S + γ β I (1 + α S) 2 β S 1 + α S − (d + ε + γ)] .

Then

det J (P 1) = d α [α (d + ε + γ) − β] S 12 + 2 d [α (d + ε + γ) − β] S 1 + d (d + ε + γ) + β A .

If $β < α (d + ε + γ)$ , $P 1 (S 1, I 1)$ does not exist, then $det J (P 1) > 0$ .

Because

tr J (P 1) = 1 (1 + α S 1) 2 [− β I 1 + β S 1 (1 + α S 1) − (1 + α S 1) 2 (2 d + ε + γ)] < 0,

then $P 1 (S 1, I 1)$ is a stable node if it exists.

Consider points $P 2 (S 2, I 2)$ , $P 3 (S 3, I 3)$ ,

J (P 2) = [− d − β I 2 (1 + α S 2) 2 − β S 2 1 + α S 2 + γ β I 2 (1 + α S 2) 2 β S 2 1 + α S 2 − (d + ε + γ)], tr (P 2) = − d − β I 2 (1 + α S 2) 2 − (d + ε + γ) + β S 2 1 + α S 2 tr (P 2) = 1 (1 + α S 2) 2 (d + ε) [− β A + β d S 2 + β S 2 (1 + α S 2) (d + ε) tr (P 2) = − (1 + α S 2) 2 (2 d + ε + γ) (d + ε)] tr (P 2) ≤ (β (d + ε) S 2 + β α (d + ε) S 22 − [(2 d + ε + γ) (d + ε) + 2 α (2 d + ε + γ) (d + ε) S 2 tr (P 2) = + (2 d + ε + γ) (d + ε) α 2 S 22]) / ((1 + α S 2) 2 (d + ε))

If $β > α (d + ε + γ)$ , then $tr (P 2) < 0$ ,

det J (P 2) = d α [α (d + ε + γ) − β] S 22 + 2 d [α (d + ε + γ) − β] S 2 + d (d + ε + γ) + β A = d α [α (d + ε + γ) − β] [A − (d + ε) I 2] 2 d 2 + 2 d [α (d + ε + γ) − β] A − (d + ε) I 2 d + d (d + ε + γ) + β A .

Because $I 2$ satisfies equation (2.6), we get

det J (P 2) = D 1 I 1 + D 2 = − b − Δ 2 2 (d + α) [β − α (d + α + γ)] D 1 + D 2,

where

D 1 = α (A α + d) (d + α + γ) − k α 2 (d + α) − α β A D 1 = + 2 α (d + α) A + 2 d (d + α) [β − α (d + α + γ)], D 2 = − α [β − α (d + α + γ)] A 2 − 2 d [β − α (d + α + γ)] A + d 2 (d + α + γ) + k (A α + d) .

If $− b − Δ 2 2 (d + α) [β − α (d + α + γ)] D 1 + D 2 > 0$ , $det J (P 2) > 0$ holds, then $P 2 (S 2, I 2)$ is a stable node if $β > α (d + ε + γ)$ and $− b − Δ 2 2 (d + α) [β − α (d + α + γ)] D 1 + D 2 > 0$ . Similarly, $P 3 (S 3, I 3)$ is a stable node if $β > α (d + ε + γ)$ and $− b + Δ 2 2 (d + α) [β − α (d + α + γ)] D 1 + D 2 > 0$ . This completes the proof. □

Theorem 3.2 If $δ < d$ , there is no limit cycle of model (1.1).

Proof Consider the Dulac function $D = 1 I$ . Note that

P = A − d S − β S I 1 + α S + γ I + T (I), Q = β S I 1 + α S − (d + ε + γ) I − T (I) .

If $0 < I ≤ I 0$ ,

∂ (D P) ∂ S + ∂ (D Q) ∂ I = − d I − β (1 + α S) 2 < 0 .

If $I > I 0$ , because $k = δ I 0$ ,

∂ (D P) ∂ S + ∂ (D Q) ∂ I = − d I − β (1 + α S) 2 + k I 2 = 1 I (δ I 0 I − d) − β (1 + α S) 2 < 1 I (δ − d) − β (1 + α S) 2 .

Thus $∂ (D P) ∂ S + ∂ (D Q) ∂ I < 0$ if $δ < d$ .

Then there is no limit cycle of model (1.1) if $δ < d$ . This completes the proof. □

Theorem 3.3 There is no limit cycle of model (1.1) if $β ≤ α (2 d + ε + γ)$ .

Proof If $I ≤ I 0$ , consider the Dulac function $D = 1 I$ . Note that

P = A − d S − β S I 1 + α S + γ I + T (I), Q = β S I 1 + α S − (d + ε + γ) I − T (I), ∂ (D P) ∂ S + ∂ (D Q) ∂ I = − d I − β (1 + α S) 2 < 0 .

If $I > I 0$ ,

∂ (P) ∂ S + ∂ (Q) ∂ I = − d − β I (1 + α S) 2 + β S (1 + α S) − (d + ε + γ) ∂ (P) ∂ S + ∂ (Q) ∂ I = 1 (1 + α S) 2 [− d (1 + α S) 2 − β I + β S (1 + α S) − (d + ε + γ) (1 + α S) 2] ∂ (P) ∂ S + ∂ (Q) ∂ I = 1 (1 + α S) 2 [− β I − (2 d + ε + γ) (1 + α S) 2 + β S (1 + α S)], (2 d + ε + γ) (1 + α S) 2 β S (1 + α S) = (2 d + ε + γ) (1 + α S) β S = (2 d + ε + γ) + α (2 d + ε + γ) S β S .

Thus $∂ (D P) ∂ S + ∂ (D Q) ∂ I < 0$ if $β ≤ α (2 d + ε + γ)$ .

Then there is no limit cycle of model (1.1) if $β ≤ α (2 d + ε + γ)$ . This completes the proof. □

4 Numerical simulation and conclusion

With different A, d, γ, δ, ε, α, it is easy to test and verify the above results, so numerical simulation is omitted. In this paper, we study an SIS model with saturated incidence rate $β S I 1 + α S$ and treatment. We get some relatively complex conclusions by stability theory and qualitative theory of differential equations. These conclusions will help policy makers to make decisions.

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Acknowledgements

The research was supported by the Fujian Nature Science Foundation under Grant No. 2014J01008.

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College of Mathematics and Computer Science, Fuzhou University, Fuzhou, Fujian, 350002, P.R. China
Jinghai Wang & Qiaohong Jiang

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Jinghai Wang
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Qiaohong Jiang
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Correspondence to Jinghai Wang.

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Wang, J., Jiang, Q. Analysis of an SIS epidemic model with treatment. Adv Differ Equ 2014, 246 (2014). https://doi.org/10.1186/1687-1847-2014-246

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Received: 11 August 2014
Accepted: 05 September 2014
Published: 24 September 2014
DOI: https://doi.org/10.1186/1687-1847-2014-246

Keywords

epidemic model
incidence rates
treatment
globally asymptotically stable