Global Stability Analysis on the Dynamics of an SIQ Model with Nonlinear Incidence Rate

D. Jin and S. Lin (Eds.): Advances in FCCS, Vol. 2, AISC 160, pp. 561–565.

Model with Nonlinear Incidence Rate

Xiuxiang Yang1, Feng Li1, and Yuanji Cheng2 1

Department of Mathematics, Weinan Normal University, Weinan 714000 Shaanxi, China

2

School of Technology Malmo University, SE-205 06 Malmo, Sweden yangxiuxiang2000@yahoo.com.cn, lifeng5849@163.com,

yuanji.cheng@mah.se

Abstract. An SIQ epidemic model with isolation and nonlinear incidence rate is studied. We have obtained a threshold value

_R

and shown that there is only a disease free equilibrium point when

R

<

1

, and there is also an endemic equilibrium point if

_R

>

₁

. With the help of Liapunov function, we have shown that disease free- and endemic equilibrium point is globally stable.

Keywords: SIQ model, Isolation, nonlinear infectious rate, threshold, equilibrium point, global stability.

1 Introduction

In this paper, we shall as well exploit the Lyapunov direct method to study the global stability analysis on the dynamics of an SIQ model with nonlinear incidence rate. Assuming that the population has a very short immunity period, which is negligible, and whence there is a class of recovered and the models being studied here are of SIQ type. Assume that the total population is divided into: S = S(t) susceptible, I(t) infected, Q(t) isolated and R(t)recovered classes at time t. Let A be the constant migration rate of the population, and d be natural death rate of each kind,

α

₁ and

α

₂ be the infected death rate of infectious and isolated classes respectively, and

δ

_{be the transition rate} from infected to isolated classes,

r

,

ε

be immunity’s lose rate of infected respectively isolated classes. We assume that the constants

d

,

δ

,

A

are positive, and

r

,

ε

,

δ

,

α

1,

α

2 are nonnegative . Thus we consider the

' ' 1 ' 2

( )

1

( )

(

)

1

( )

(

)

IS

S t

A

dS

rI

Q

mI

IS

I t

r

d

I

mI

Q t

I

d

Q

β

_ε

β

_δ

_α

δ

ε

α

⎧⎪

_{= −}

₋

_{+ +}

⎪⎪

+

⎪⎪

⎪⎪

₌

_{− + + +}

⎨

₊

⎪⎪

⎪

₌

_{− + +}

⎪⎪

⎪⎪⎩

(1)

we see that N satisfies:

N t

'

( )

= −

A

dN

−

α

₁

I

−

α

₂

Q

≤ −

A

dN

，

thus

0

( )

dt

A

(1

dt

)

N t

N e

e

d

− −

≤

+

−

(2) It follows from the above estimates that the solutions of initial value problem for system (1) exist on

[0,

+∞

)

,

is invariant subset of (1).

2 Local Stability of Equilibrium

To find the equilibrium of system (1), we get the equilibrium

P

0

(

A

_d

, 0, 0)

and

P S I Q

₁

( , ,

_{1 1 1}

)

, let’s define the threshold number for (1)

1 1

(

)

d r

d

R

A

δ

α

β

+ + +

=

_{，Then we have that}

P

1 _{is in}

3

R

+ _{if and only if} R < . 1 1 Here 1 1 1

(

)

(1

)

r

d

S

δ

α

mI

β

+ + +

=

+

_,

1 2 2 I Q d δ ε α = + +

_,

1 1 2 1 1 2

(

)

(

)

(

)

(

)

A

d r

d

I

d

md r

d

d

d

β

δ α

δ

α

δ α

β

α

ε

α

−

+ + +

=

+

+ + +

+

+

+

+ +

Theorem 1. For the system (1), the infection-free equilibrium point

P

₀ is locally asymptotically stable if R > , If 1 1 R < , then the endemic equilibrium point 1 1

P

₁ is

locally asymptotically stable and

P

₀ becomes unstable.

Proof. First we consider the system (1) and the infectious free equilibrium

P

0, The

Jacobian matrix

J P

( )

0 ，Which has eigenvalues:

1

(

d

)

0

λ

= − + <

ε

_{， 2}

A

(

r

d

₁

)

d

β

λ

=

− + + +

δ

α

_，

λ

₃

= − + +

(

ε

d

α

₂

)

<

0

_， When R < , then 1 1

λ

2 is also negative, and hence

P

_{0 is locally stable.}

In the

P

₁,

β

S

₂

= + + +

(

r

d

δ α

₁

)(1

+

mI

₂

)

. Let

d

_i

= +

d

α

_i

,

i

=

1, 2

According to

J P

( )

₁ , the associated characteristic equation is

3 2

1 2 3

0

b

b

b

1 2

(1

2

)

2

(

1

)

(

2

)(1

2

)

0

b

=

β

I

+

d

+

mI

+

mI r

+ +

δ

d

+ +

ε

d

+

mI

>

2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2

(

)

(

)

(

)(

)(1

)

(

)(

)(1

)

(

)

0

b

I

d

dmI mI r

d

I

d

dmI

d

mI

mI r

d

d

mI

I

d

rmI

β

δ

β

ε

δ

ε

β

δ

=

+ +

+ +

+

+ +

+

+

+

+ +

+

+

+

+ −

>

3 2 2 2 1 2 2 2 2 2 2 1 2 2

(

)(1

)(

)(

)

((

)(1

)(

)

(1

) )

0

b

mI

d

mI

r

d

I

d

dmI

I

d

mI

d

rmI

mI

ε

δ

β

β

ε

δ

δε

=

+

+

+ +

+ +

+

+

+

+ −

−

+

>

To verify the Hurwitz condition b b1 2−b3> 0 , we first rewrite:

1 10 11 2

b

=

c

+

c mI

_, 2 2 20 21 2 22

(

2

)

b

=

c

+

c mI

+

c

mI

,

2 3 3 30 31 2 32

(

2

)

33

(

2

)

b

=

c

+

c mI

+

c

mI

+

c

mI

,

Where

c

10

= + +

d

ε

d

2

+

β

I

2 ，

c

11

= + + +

d

r

ε

d

2

+ +

δ

d

1 ,

c

20

=

d

(

ε

+

d

2

)

+ +

(

ε

d

2

+ +

δ

d

1

)

β

I

2_, 21

(

1

)(

2

)

(

2

)(

2 1

)

c

=

rd

+ + +

r

δ

d

ε

+

d

+

d

+

β

I

ε

+

d

+ +

δ

d

,

22

(

1

)(

2

)

(

2 1

)

c

= + +

r

δ

d

ε

+

d

+

d r

+ +

ε

d

+ +

δ

d

,

30

(

1 2 1 2

)

2

c

=

ε

d

+

δ

d

+

d d

β

I

,

31

(

1

)(

2

)

2(

1 2 1 2

)

2

c

=

d r

+ +

δ

d

ε

+

d

+

ε

d

+

δ

d

+

d d

β

I

,

32

2 (

1

)(

2

)

(

1 2 1 2

)

2

c

=

d r

+ +

δ

d

ε

+

d

+

ε

d

+

δ

d

+

d d

β

I

,

33

(

1

)(

2

)

c

=

d r

+ +

δ

d

ε

+

d

.

Then get 1 2 3 10 20 30

(

10 22 11 20 31

)

2

b b

− =

b

c c

−

c

+

c c

+

c c

−

c

mI

+

2 10 22 11 21 32 2

(

c c

+

c c

−

c

)(

mI

)

+

(

c c

_{11 22}

−

c

₃₃

)(

mI

₂

)

3 0

:

10 20 30

0,

1

:

10 22 11 20 31

0

C

=

c c

−

c

>

C

=

c c

+

c c

−

c

>

,

2

:

10 22 11 21 32

0,

C

=

c c

+

c c

−

c

>

C

₃

=

:

c c

_{11 22}

−

c

₃₃

> ,

0

_So 1

P

is locally asymptotically stable.

3 Global Stability of Equilibrium

Theorem 2. For the system (1), If

R >

₁

1

, then the disease free equilibrium

P

₀ of (2) is globally asymptotically stable; when

R <

₁

1

, then

P

₀ is unstable, and the endemic equilibrium point

P

is globally asymptotically stable.

Proof. we can easily proof the disease free equilibrium

P

0 of (1) is globally

asymptotically stable same as theorem. Following discuss asymptotic stability of

P

₁, we again change variable

N

= + +

S

I

Q

1 2 ' 1 ' 2

'( )

( )

[ (

)

(

)(1

)]

1

( )

(

)

N t

A

dN

I

Q

I

I t

N

I

Q

r

d

mI

mI

Q t

I

d

Q

α

α

β

δ

α

δ

ε

α

⎧

= −

−

−

⎪⎪

⎪⎪

⎪

₌

_{− −}

_{− + + +}

₊

⎨⎪

+

⎪⎪

⎪

₌

_{− + +}

⎪⎩

1

,

1

,

1

x

= −

N

N y

= −

I

I z

= −

Q Q

, then we deduce

1 2 2 2 2 2

'( )

'( )

(

(

(

))

)

1

(

)

'( )

(

)

x t

dx

y

z

y

I

y t

x

m r

d

y

z

m y

I

z t

y

d

z

α

α

β

β

δ α

β

δ

ε

α

⎧

= − −

−

⎪

₊

⎪

₌

₋

₊

_{+ + +}

₋

⎨

₊

₊

⎪

⎪

₌

_{− + +}

⎩

(3)

We have used the relation

(

r

+ + +

δ

d

α

1

)(1

+

mI

2

)

=

β

S

2

=

β

(

N

2

− −

I

2

Q

2

)

Choose the Liapunov function 2 2 2 2 3 1 2 2 2 ( ) ( ln(1 ) ) 2 2 2 2 z x y my x z V y I I ω ω ω − = + − + + + + 1 1 2 2 2 (3)

(

)

(

(

(

))

)

dV

x

dx

y

z

y

x

m r

d

y

z

dt

=

ω

− −

α

−

α

+

ω

β

−

β

+

+ + +

δ α

−

β

+

ω δ

3

z

(

y

− + +

(

ε

d

α

2

) )

z

+ −

(

x

z

)(

− − +

dx

(

δ α

1

)

y

+ +

(

ε

d z

) )

2 2 1 2 1 1 1 2 1 2 2 (1 ) ( ) ( 2 ) ( ( )) d ω x ω β ω α α δ xy ε d α ω xz ω β m r d δ α y = − + + − − − + + − − + + + +

+(α1+ +δ ω δ3 −ω β2 )y z−(ε+ d +(ε+ d +α ω2) 3)z2 If we choose

ω ω ω >

1

,

2

,

3

0

1 1 1 1 1 1 2 3 2 ( ) (ε 2 )d _, ω α α δ _, ω α ω = + _α ω = + + _β ω = _δ

_,

2 2 2 1 2 2 2 3 (3) (1 ) ( ( )) ( ( ) ) dV d x m r d y d d z dt = − +ω −ω β+ + + +δ α − + +ε ε+ +α ω

Which is clearly negative definite. Hence,

P S I Q

1

( , ,

1 1 1

)

is globally asymptotically

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