Rouché's theorem [21], we conclude that all roots of Eq. (4) have
negative real parts when νA ½0; ν
0
Þ. This proves the lemma. □
Lemma 3. Let
λðνÞ¼αðνÞþiωðνÞ be a root of Eq. (4) satisfying
αðν
n
Þ¼0 and ωðν
n
Þ¼ω
n
. For each fixed τ 4 1=2, Then α′ðν
n
Þ4 0.
Proof. Taking the derivation with respect to
ν in Eq. (4),wehave
d
λ
d
ν
1
¼
1
p
n
d
λ
dk
1
¼
3
λ
2
2ða
1
þa
2
Þ
λ
þa
1
a
2
kðb
1
þb
2
Þe
λτ
þk
τ
½b
1
ð
λ
a
2
Þþb
2
ð
λ
a
1
Þe
λτ
p
n
½b
1
ð
λ
a
2
Þþb
2
ð
λ
a
1
Þe
λτ
¼
½3
λ
2
2ða
1
þa
2
Þ
λ
þa
1
a
2
e
λτ
kðb
1
þb
2
Þ
p
n
½b
1
ð
λ
a
2
Þþb
2
ð
λ
a
1
Þ
þ
ντ
:
It follows from Eq. (6) that
Re
dλ
dν
1
ν
¼
ν
n
¼
1
M
½3
ω
2
n
cos ðω
n
τÞ2ða
1
þa
2
Þω
n
sin ð ω
n
τÞa
1
a
2
cos ðω
n
τÞ
þkðb
1
þb
2
Þða
2
b
1
þa
1
b
2
Þ
½3
ω
2
n
sin ð ω
n
τÞþ2 ða
1
þa
2
Þω
n
cos ðω
n
τÞa
1
a
2
sin ðω
n
τÞðb
1
þb
2
Þω
n
þ
ν
n
τ
¼
1
M
3
ω
2
n
½ða
2
b
1
þa
1
b
2
Þcos ðω
n
τÞðb
1
þb
2
Þω
n
sin ð ω
n
τÞ
2ða
1
þa
2
Þω
n
½ða
2
b
1
þa
1
b
2
Þsin ðω
n
τÞ
þðb
1
þb
2
Þω
n
cos ðω
n
τÞa
1
a
2
½ða
2
b
1
þa
1
b
2
Þcos ð ω
n
τÞ
ðb
1
þb
2
Þω sin ð ω
n
τÞþkðb
1
þb
2
Þða
2
b
1
þa
1
b
2
Þ
þν
n
τ
¼
1
kM
f
ω
4
n
ða
1
þa
2
Þa
1
a
2
ða
1
þa
2
Þω
2
n
þk
2
ðb
1
þb
2
Þða
2
b
1
þa
1
b
2
Þ
þk
2
τ½ða
2
b
1
þa
1
b
2
Þ
2
þðb
1
þb
2
Þ
2
g4 0;
where M ¼ða
2
b
1
þa
1
b
2
Þ
2
þω
2
n
ðb
1
þb
2
Þ
2
4 0. This completes the
proof. □
Based on Lemmas 1–3 and applying the Hopf bifurcation
theorem for delayed differential equations in [22], we can obtain
the following theorem.
Theorem 1. For each fixed
τ4 1=2,
(1) When
νA ð0; ν
0
Þ, the non-zero equilibrium x
n
is asymptotically
stable.
(2) When
ν4 ν
0
, the non-zero equilibrium x
n
is unstable.
(3) When
ν ¼ ν
n
; n ¼ 0; 1; 2; …, the system (1) undergoes a Hopf
bifurcation.
3. Hopf bifurcation control via state feedback controller
In the previous section, we analyze the Hopf bifurcation of the
uncontrolled system (1) by choosing the parameter
ν as the
bifurcation parameter. The aim of this section is to introduce a
state feedback controller to control the creation of the Hopf
bifurcation for the system (1) at any desired parameter location
ν. The system (1) added a controller becomes
_
x
1
ðtÞ¼k
1
x
1
ðt τÞ
1pðtÞ
α
2
x
1
ðtÞ
κ
1
x
1
ðtÞpðtÞ
;
_
x
2
ðtÞ¼k
2
x
2
ðt τÞ
1pðtÞ
α
2
x
2
ðtÞ
κ
2
x
2
ðtÞpðtÞ
;
_
pðtÞ¼
νpðtÞ½x
1
ðt τÞþx
2
ðt τÞcm
1
ðpðtÞp
n
Þ
m
2
ðpðtÞp
n
Þ
2
m
3
ðpðtÞp
n
Þ
3
;
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
ð10Þ
where m
i
; i ¼ 1; 2; 3 are positive feedback gain parameters.
Let u
1
ðtÞ¼x
1
ðtÞx
n
1
, u
2
ðtÞ¼x
2
x
n
2
, u
3
ðtÞ¼pðtÞp
n
, then linear-
izing system Eq. (10) at the equilibrium x
n
¼ðx
n
1
; x
n
2
; p
n
Þ produces
_
u
1
ðtÞ¼a
1
u
1
ðtÞþb
1
u
3
ðtÞ;
_
u
2
ðtÞ¼a
2
u
2
ðtÞþb
2
u
3
ðtÞ;
_
u
3
ðtÞ¼ku
1
ðt τÞþku
2
ðt τÞm
1
u
3
ðtÞ;
8
>
<
>
:
ð11Þ
whose corresponding characteristic equation is
λ
3
þðm
1
a
1
a
2
Þλ
2
þða
1
a
2
m
1
a
1
m
1
a
2
Þλþm
1
a
1
a
2
½ðλa
1
Þb
2
þðλa
2
Þb
1
ke
λτ
¼0: ð12Þ
Here, i
ω
c
ðω
c
4 0Þ is a root of (12) iff
kðb
1
þb
2
Þ
ω
c
sin ð
ω
c
τ
Þþkða
2
b
1
þa
1
b
2
Þ cos ð
ω
c
τ
Þ¼ða
1
þa
2
m
1
Þð
ω
c
Þ
2
m
1
a
1
a
2
;
kðb
1
þb
2
Þ
ω
c
cos ð
ω
c
τ
Þþkða
1
b
2
þa
2
b
1
Þ sin ð
ω
c
τ
Þ¼ð
ω
c
Þ
3
þða
1
a
2
m
1
a
1
m
1
a
2
Þ
ω
c
:
(
ð13Þ
The above equations can be rewritten as
½ða
2
1
b
2
þa
2
2
b
1
Þðω
c
Þ
2
þðb
1
þb
2
Þðω
c
Þ
4
þm
1
ða
1
b
1
þa
2
b
2
Þðω
c
Þ
2
þm
1
ða
2
1
a
2
b
2
þa
1
a
2
2
b
1
Þ sin ðωτÞ;
¼½ða
1
b
1
þa
2
b
2
Þðω
c
Þ
3
ða
2
b
1
þa
1
b
2
Þa
1
a
2
ω
c
þm
1
ðb
1
þb
2
Þðω
c
Þ
3
þm
1
ða
2
1
b
2
þa
2
2
b
1
Þω
c
cos ðωτÞ:
Let
ℏ
c
ðωÞ¼h
c
1
ðωÞh
c
2
ðωÞ;
where h
c
1
¼½ða
2
1
b
2
þa
2
2
b
1
Þω
2
þðb
1
þb
2
Þω
4
þm
1
ða
1
b
1
þa
2
b
2
Þω
2
þ
m
1
ða
2
1
a
2
b
2
þa
1
a
2
2
b
1
Þ sin ðωτÞ and h
c
2
¼½ða
1
b
1
þa
2
b
2
Þω
3
ða
2
b
1
þ
a
1
b
2
Þa
1
a
2
ωþm
1
ðb
1
þb
2
Þω
3
þm
1
ða
2
1
b
2
þa
2
2
b
1
Þω cos ðωτÞ.
Similar to the analysis in the uncontrolled system, we obtain
the following lemmas.
Lemma 4. When m
1
4 0, then
(1) Eq. (12) has a pair of simple purely imaginary roots 7i
ω
c
n
, ω
c
n
A
ð2n
π=τ ; ð 2nπ þπÞ=τÞ, when ν ¼ν
c
n
, n ¼ 0; 1; 2; …;
(2) all the roots of Eq. (12) have negative real parts when
νA ð0; ν
c
0
Þ,
where
ν
c
n
¼
1
p
n
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð
ω
c
n
Þ
6
þða
2
1
þa
2
2
þm
2
1
Þð
ω
c
n
Þ
4
þða
2
1
a
2
2
þm
2
1
a
2
1
þm
2
1
a
2
2
Þð
ω
c
n
Þ
2
ðb
1
þb
2
Þ
2
ð
ω
c
n
Þ
2
þða
2
b
1
þa
1
b
2
Þ
2
s
;
n ¼0; 1; 2; …:
Lemma 5. If m
1
4 0, then ν
c
0
4 ν
0
.
Proof. Obviously,
ω
0
, ω
c
0
A ð0; π=τÞ. Since ða
2
1
b
2
þa
2
2
b
1
Þω
2
þðb
1
þ
b
2
Þω
4
o 0 and m
1
ða
1
b
1
þa
2
b
2
Þω
2
þm
1
ða
2
1
a
2
b
2
þa
1
a
2
2
b
1
Þ4 0, then
h
1
ðωÞo h
c
1
ðωÞ and ðh
1
ðωÞÞo 0asωA ðo; π=τÞ. Besides, the ampli-
tude of h
2
is smaller than that of h
c
2
for ða
1
b
1
þa
2
b
2
Þω
3
ða
2
b
1
þa
1
b
2
Þa
1
a
2
ωo 0 and m
1
ðb
1
þb
2
Þω
3
þm
1
ða
2
1
b
2
þa
2
2
b
1
Þo 0. By
taking the derivative of h
1
, h
c
1
, h
2
and h
c
2
, it is obvious that h
2
and
h
c
2
reach the first trough more quickly within a short time, and
then monotonically increase until
ω ¼π=τ (See Fig. 2).
When
ωA ð0; π=ð2τÞÞ, h
c
2
o h
2
o 0, so the curve h
2
firstly inter-
sects the curve h
1
, then the curve h
c
2
intersects the curve h
1
and
finally the curve h
c
2
intersects the curve h
c
1
. Therefore, ω
0
o ω
c
0
.
Since
ω
4
0
þða
2
1
þa
2
2
Þω
2
0
þa
2
1
a
2
2
ðb
1
þb
2
Þ
2
þ
ða
2
b
1
þa
1
b
2
Þ
2
ω
2
0
o
ðω
c
0
Þ
4
þða
2
1
þa
2
2
Þðω
c
0
Þ
2
þa
2
1
a
2
2
ðb
1
þb
2
Þ
2
þ
ða
2
b
1
þa
1
b
2
Þ
2
ð
ω
c
0
Þ
2
o
ðω
c
0
Þ
4
þða
2
1
þa
2
2
þm
2
1
Þðω
c
0
Þ
2
þða
2
1
a
2
2
þm
2
1
a
2
1
þm
2
1
a
2
2
Þ
ðb
1
þb
2
Þ
2
þ
ða
2
b
1
þa
1
b
2
Þ
2
ð
ω
c
0
Þ
2
;
Thus,
ν
c
0
4 ν
0
. □
W. Xu et al. / Neurocomputing 129 (2014) 232–245234