Biomedical Engineering Reference
In-Depth Information
In this way, the non-negativity of E(t) has been proven. From the second equation in
(32), it follows that
c H(r )
=σβ
I(0)S(0)
−
c H(r )
and then:
3
3
1
1
1
(
)
−
rt
−
rt
−
rt
c H(r ) e
−
e
+ β
I(0)S(0)e
1
3
3
S(t)
=
1
1
1
≥
0
(35)
σβ
I(t)
1
∀∈
t
by applying such a relation between
c and
c in (32) and by taking into
0
+
{
}
account that
cH(r)
>
0
since
r
< μ +
in
σ
,
γ
, S(0)
≥
0
and, I(t)
≥
0
and
1
1
−
rt
−
rt
. In this way, the non-negativity of S(t) has
been proven. Note that the function H(v) in (33) is an upper-open parabola zero-
valued for
e
−≥
e
0
∀∈
t
since
1
rr
<
1
3
0
+
3
{
}
.
(ii)
On one hand, if the control law (20) is used instead of that in (12) then the time
evolution of the infectious population is also given by (22) while the control action is
active. Thus, I(t)
v
= μ +σ
and
v
= μ + γ
so
H(r )
>
0
since
r
< μ +
in
σ
,
γ
2
1
in (22) implies directly the existence of a finite time
instant t at which the control (20) switches off. Obviously, the non-negativity of
I(t) , E(t) and S(t)
→
0
as t
→∞
[
]
is proved by following the same reasoning used in
the part (i) of the current theorem. The non-negativity of R(t)
∀∈
t
, t
f
[
]
is proven
by using continuity arguments. In this sense, if R(t) reaches negative values for
some
∀∈
t
, t
f
[
]
t
∈
, t
starting from an initial condition R(0)
≥
0
then R(t) passes through
f
[
)
zero, i.e., there exists at least a time instant
t
∈
, t
such that
R(t )
=
0
. Then, it
0
f
0
follows from (4) that:
2
3
μσβ+λ −λ μ+γ +λ
(
)
(
μ+γ
)
− μ+γ
(
)
R(t )
=γ
I(t )
+μ
NV(t )
=γ
I(t )
+
0
1
2
N
0
0
0
0
σβ
E(t )S(t )
β
+ λ +ω− μ−σ− γ
(
3
2 )S(t )
+σ
0
0
−
I(t )S(t )
(36)
2
0
0
0
I(t )
N
0
(
μ+γ
)
2
+
(2
μ+σ+γ μ+σ +λ −λ
)(
)
(2
μ+σ+γ
)
N
E(t )
+
1
2
0
β
I(t )
0
by introducing the control law (20), taking into account that V(t)
=
u(t)
and where
the fact that
I(t) E(t) S(t) N
+
+
=
, since
R(t )
=
0
, has been used. Moreover, the
0
0
0
0
[
]
non-negativity of I(t) , E(t) and S(t)
∀∈
t
, t
as it has been previously proven,
f
implies that
I(t )
≤
N
,
E(t )
≤
N
and
S(t )
≤
N
. Also,
I(t )
≥δ>
0
since
t
<
t
0
0
0
0
0
f
and from the definition of
t in (21). Then, one obtains:
2
μσβ+λ −λ μ+γ +λ
(
)
(
μ+γ
)
− μ+γ
(
)
3
R(t )
≥γ
I(t )
+
0
1
2
N
0
0
σβ
E(t )S(t )
0
0
(37)
+ λ +ω− μ−σ− γ−β
(
3
2
)S(t )
+σ
2
0
I(t )
0
2
(
μ+γ
)
+
(2
μ+σ+γ μ+σ +λ −λ
)(
)
(2
μ+σ+γ
)
E(t )
+
1
2
N
0
β
I(t )
0
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