August 30, 2013 18:18 WSPC/S0218-1274 1350144
Sliding Bifurcation and Global Dynamics of a Filippov Epidemic Model with Vaccination
b
22
.
= β[(γ − θ)+α(µ + θ)]
+ α(µ + θ + p + pf )[α(µ + θ) − (γ − θ)],
then ∆
2
> 0isequivalenttoS
c
>S
+
c2
or S
c
<S
−
c2
and ∆
2
< 0 if and only if S
−
c2
<S
c
<S
+
c2
, where
S
−
c2
=
b
22
−
b
2
22
− b
21
b
13
b
21
,
S
+
c2
=
b
22
+
b
2
22
− b
21
b
13
b
21
.
Moreover, the two roots defined in (12)arepositive
if and only if S
−
c3
<S
c
<S
−
c4
,where
S
−
c3
=
(γ − θ)+α(µ + θ)
β + α(µ + θ + p + pf)
,
S
−
c4
=
µ + θ
µ + θ + p + pf
.
Note that
R
02
> 1 ⇒ R
01
> 1 ⇒ S
−
c1
<S
+
c3
<S
+
c1
(13)
R
02
> 1 ⇒ S
−
c2
<S
−
c3
<S
+
c2
. (14)
Denote g
1
(S
c
)
.
=∆
1
, then g
1
(S
c
) is quadratic and
increasing for S
c
>S
+
c3
. Simple calculation yields
g
1
(S
+
c3
) < 0,g
1
(S
+
c1
)=0,
g
1
(S
+
c4
) > 0,S
+
c4
>S
+
c3
,
which result in S
+
c4
>S
+
c1
.Setg
2
(S
c
)
.
=∆
2
, then the
similar discussion gives S
−
c4
>S
+
c2
. Thus we obtain
S
−
c1
<S
+
c3
<S
+
c1
<S
+
c4
,
S
−
c2
<S
−
c3
<S
+
c2
<S
−
c4
.
(15)
It follows that if S
+
c1
<S
c
<S
+
c4
and S
+
c2
<S
c
<S
−
c4
are satisfied, Eqs. (9)and(10) possess two distinct
positive real roots, respectively, and 0 <I
2
min
<
I
1
min
<I
1
max
<I
2
max
.
Therefore, the sliding mode domain of Filippov
system (6) may consist of some of the following pos-
sible five sliding segments
Σ
1
s
= {(S
c
,I) | I
2
min
<I<I
1
min
},
Σ
2
s
= {(S
c
,I) | I
1
max
<I<I
2
max
},
Σ
3
s
= {(S
c
,I) | 0 <I<I
1
min
},
Σ
4
s
= {(S
c
,I) | 0 <I<I
2
max
},
Σ
5
s
= {(S
c
,I) | I
2
min
<I<I
2
max
}.
Now, we shall present various sliding domains
under different combinations of parameters S
+
c1
,
S
+
c2
,S
+
c4
,S
−
c4
. Note that on one hand, S
+
c1
>S
+
c3
and g
2
(S
c
) >g
1
(S
c
)forS
c
>S
+
c3
imply that
g
2
(S
c
) >g
1
(S
c
)forS
c
>S
+
c1
; on the other hand,
we have g
1
(S
c
) > 0forS
c
>S
+
c1
and g
1
(S
+
c1
)=0.
These result in g
2
(S
c
) > 0forS
c
≥ S
+
c1
. We also
note that g
2
(S
+
c2
)=0, so S
+
c1
>S
+
c2
holds true.
Thus,accordingto(15), we only need to discuss
the relationship of the parameters S
+
c1
and S
−
c4
, and
investigate the following six cases.
(A
1
) S
−
c4
>S
+
c1
,S
+
c1
<S
c
<S
−
c4
.
It follows from the above discussion that I
i
min
,I
i
max
(i =1, 2) are well defined and satisfy 0 <I
2
min
<
I
1
min
<I
1
max
<I
2
max
in such scenarios, as indicated
in Fig. 1(A
1
). Therefore, there are two pieces of slid-
ing segments Σ
1
s
and Σ
2
s
. Then the sliding mode
domain can be defined as Σ
s
=Σ
1
s
∪ Σ
2
s
.
(A
2
) S
−
c4
>S
+
c1
,S
−
c4
<S
c
<S
+
c4
.
It follows from S
c
>S
−
c4
that ∆
2
> 0, which means
I
2
max
,I
2
min
are well defined and I
2
max
> 0,I
2
min
< 0.
Further, since S
+
c1
<S
c
<S
+
c4
, one gets that I
1
min
and I
1
max
are well defined and I
1
max
> 0,I
1
min
> 0,
as indicated in Fig. 1(A
2
). Furthermore,
S
c
>S
+
c1
⇒ S
c
>S
+
c3
⇒ ∆
2
> ∆
1
⇒ I
1
max
<I
2
max
.
Then, in such case, there are also two pieces of
sliding segments Σ
2
s
and Σ
3
s
. Therefore, the sliding
mode domain can be defined as Σ
s
=Σ
2
s
∪ Σ
3
s
.
(A
3
) S
−
c4
>S
+
c1
,S
c
>S
+
c4
.
In such scenarios, we have
S
c
>S
+
c4
⇒ ∆
1
> 0,I
1
max
> 0,I
1
min
< 0
S
c
>S
+
c4
⇒ S
c
>S
−
c4
⇒ ∆
2
> 0,I
2
max
> 0,I
2
min
< 0.
Moreover, it follows from the similar discussion in
Case (A
2
)thatS
c
>S
+
c4
implies I
1
max
<I
2
max
, so
the sliding segment domain appears as Σ
s
=Σ
2
s
,as
indicated in Fig. 1(A
3
).
(A
4
) S
−
c4
<S
+
c1
,S
−
c4
<S
c
<S
+
c1
.
Note that S
c
>S
−
c4
implies ∆
2
> 0andI
2
max
>
0,I
2
min
< 0. Further, it follows from the above dis-
cussion and S
c
<S
+
c1
that there is no positive real
roots for Eq. (9). Therefore, the sliding segment
domain is Σ
s
=Σ
4
s
in this case, as indicated in
Fig. 1(A
4
).
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