Electronic Journal of Differential Equations, Vol. 2010(2010), No. 100, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
PERIODIC SOLUTIONS FOR A SECOND-ORDER NEUTRAL
DIFFERENTIAL EQUATION WITH VARIABLE PARAMETER
AND MULTIPLE DEVIATING ARGUMENTS
BO DU, XIAOJING WANG
Abstract. By employing the continuation theorem of coincidence degree theory developed by Mawhin, we obtain periodic solution for a class of neutral differential equation with variable parameter and multiple deviating arguments.
1. Introduction
Neutral functional differential equations (in short NFDEs) are an important
research subject of functional differential equations and provide good models in
many fields including physics, mechanics, biology and economics (see [1, 2, 3, 4, 5]).
With such clear indications of the importance of NFDEs in the applications, it is
not surprising that the subject has undergone a rapid development in the previous
twenty years. Particularly, in recent years the problems of periodic solution for
second-order NFDEs have been studied by many authors. In [6], by employing the
continuation theorem of coincidence degree theory, Lu and Ge studied the following
second-order NFDE:
(x(t) + cx(t − r))00 + f (x0 (t)) + g(x(t − τ (t))) = p(t).
After that, Lu and Gui [7] went still one step further to study the above equation
in the critical case and obtained more profound results. Furthermore, in [8] Lu
and Ren investigated the second-order NFDE with multiple deviating arguments
as follows:
d2
(u(t) − ku(t − τ )) = f (u(t))u0 (t) + α(t)g(u(t)) + Σnj=1 βj (t)g(u(t − γj (t))) + p(t).
dt2
The authors used new techniques and methods for multiple deviating arguments and
obtained some new results. In very recent years, p-Laplacian NFDEs were studied
by some researchers. In [9]-[10], Zhu and Lu studied the following p-Laplacian
NFDEs:
0
(ϕp [(x(t) − cx(t − σ))0 ]) + g(t, x(t − τ (t))) = e(t)
and
0
(ϕp [(x(t) − cx(t − σ))0 ]) = f (x(t))x0 (t) + Σnj=1 βj (t)g(x(t − γj (t))) + p(t).
2000 Mathematics Subject Classification. 34B15, 34B13.
Key words and phrases. Mawhin’s continuation theorem; periodic solution; neutral;
variable parameter.
c
2010
Texas State University - San Marcos.
Submitted April 12, 2010. Published July 21, 2010.
1
2
B. DU, X. WANG
EJDE-2010/100
However, for all the above papers they obtained the existence of periodic solution to
NFDEs based on the properties of neutral operator A. In 1995, Zhang [11] obtained
the following results. Define A on CT
A : CT → CT , [Ax](t) = x(t) − cx(t − τ ), ∀t ∈ R,
where CT = {x : x ∈ C(R, R), x(t + T ) ≡ x(t)}, c is constant. When |c| 6= 1, then
A has a unique continuous bounded inverse A−1 satisfying
(P
j
if |c| < 1, ∀f ∈ CT ,
−1
j≥0 c f (t − jτ ),
[A f ](t) =
P
− j≥1 c−j f (t + jτ ), if |c| > 1, ∀f ∈ CT .
Obviously, we have
1
;
(1) kA−1 k ≤ |1−|ck
RT
RT
1
−1
(2) 0 |[A f ](t)|dt ≤ |1−|ck
|f (t)|dt, ∀f ∈ CT ;
R0 T
RT
1
−1
2
(3) 0 |[A f ](t)| dt ≤ |1−|ck 0 |f (t)|2 dt, ∀f ∈ CT .
When c is a variable c(t), we have obtained the properties of the neutral operator
A : CT → CT , [Ax](t) = x(t) − c(t)x(t − τ ) in [12]. We note that there are
few results on the existence of periodic solutions to second-order neutral equations
for the cases of a variable c(t). The purpose of this article is to investigate the
existence of periodic solution for the second-order NFDE with variable parameter
and multiple deviating arguments by using the properties of the operator A in
[12] and Mawhin’s continuation theorem. Here we use the same technique, but
our results extend and complement the existing ones. We will study the following
NFDE:
n
X
00
(x(t) − c(t)x(t − τ )) +
βj (t)g(x(t − γj (t))) = e(t),
(1.1)
j=1
2
where g ∈ C(R, R); c ∈ C (R, R) with c(t) = c(t+T ) and |c(t)| =
6 1; e(t), βj (t), γj (t)
are T −periodic functions on R (j = 1, 2, . . . , n); τ, T > 0 are given constants.
In this article, we assume that e(t) is not a constant function on R. Furthermore,
we suppose that γj ∈ C 1 (R, R) with γj0 (t) < 1, ∀t ∈ R, (j = 1, 2, . . . , n). It is obvious
that the function t − γj (t) has a unique inverse denoted by µj (t), (j = 1, 2, . . . , n).
Let
Z
1 T
βj (µj (t)
Γ(t) = Σnj=1
,
h̄
=
h(s)ds.
1 − γj0 (µj (t))
T 0
2. Preliminary
In this section, we give some lemmas which will be used in this paper.
Lemma 2.1 ([12]). If |c(t)| =
6 1, then operator A has continuous inverse A−1 on
CT , satisfying: (1)
(
P∞ Qj
f (t) + j=1 i=1 c(t − (i − 1)τ )f (t − jτ ), c0 < 1, ∀f ∈ CT ,
−1
P∞ Qj+1 1
[A f ](t) =
(t+τ )
− fc(t+τ
i=1 c(t+iτ ) f (t + jτ + τ ), σ > 1, ∀f ∈ CT ,
j=1
) −
(2)
Z
(
T
−1
|[A
0
f ](t)|dt ≤
RT
1
1−c0 R 0 |f (t)|dt,
T
1
σ−1 0 |f (t)|dt,
c0 < 1, ∀f ∈ CT ,
σ > 1, ∀f ∈ CT ,
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PERIODIC SOLUTIONS
3
where
c0 = max |c(t)|,
σ = min |c(t)|,
t∈[0,T ]
t∈[0,T ]
c1 = max |c0 (t)|.
t∈[0,T ]
Let X and Y be two Banach spaces and let L : D(L) ⊂ X → Y be a a linear
operator, Fredholm operator with index zero (meaning that Im L is closed in Y
and dim ker L = codim Im L < +∞. If L is a Fredholm operator with index
zero, then there exist continuous projectors P : X → X, Q : Y → Y such that
Im P = ker L, Im L = ker Q = Im(I − Q) and LD(L)∩ker P : (I − P )X → Im L is
invertible. Denote by Kp the inverse of LP .
Let Ω be an open bounded subset of X, a map N : Ω̄ → Y is said to be Lcompact in Ω̄ if QN (Ω̄) is bounded and the operator Kp (I − Q)N (Ω̄) is relatively
compact. We first give the famous Mawhin’s continuation theorem.
Lemma 2.2 ([13]). Suppose that X and Y are Banach spaces, and L : D(L) ⊂
X → Y , is a Fredholm operator with index zero. Furthermore, Ω ⊂ X is an open
bounded set and N : Ω̄ → Y is L-compact on Ω̄. if all the following conditions hold:
(1) Lx 6= λN x, for all x ∈ ∂Ω ∩ D(L), and all λ ∈ (0, 1),
(2) N x ∈
/ Im L, for all x ∈ ∂Ω ∩ ker L,
(3) deg{QN, Ω ∩ ker L, 0} =
6 0,
Then the equation Lx = N x has a solution on Ω̄ ∩ D(L).
Define the linear operator L : D(L) ⊂ CT → CT as Lx = (Ax)00 , and a nonlinear
operator N : CT → CT ,
n
X
Nx = −
βj (t)g(x(t − γj (t))) + e(t),
j=1
where D(L) = {x|x ∈
CT1 }.
For x ∈ ker L, we have (x(t) − c(t)x(t − τ ))00 = 0. Then
x(t) − c(t)x(t − τ ) = c˜1 t + c˜2 ,
where c˜1 , c˜2 ∈ R. Since x(t) − c(t)x(t − τ ) ∈ CT , then c˜1 = 0. Let ϕ(t) be a solution
RT
of x(t) − c(t)x(t − τ ) = 1 and 0 ϕ2 (t)dt 6= 0. We get
Z T
ker L = {a0 ϕ(t), a0 ∈ R}, Im L = {y|y ∈ CT ,
y(s)ds = 0}.
0
Obviously, Im L is a closed in CT and dim ker L = codim Im L = 1, So L is a
Fredholm operator with index zero. Define continuous projectors P, Q
RT
x(t)ϕ(t)dt
ϕ(t),
P : CT → ker L, (P x)(t) = 0R T
ϕ2 (t)dt
0
Z
1 T
Q : CT → CT / Im L, Qy =
y(s)ds.
T 0
Let
LP = L|D(L)∩ker P : D(L) ∩ ker P → Im L,
then
L−1
P = Kp : Im L → D(L) ∩ ker P.
Since Im L ⊂ CT and D(L) ∩ ker P ⊂ CT1 , so Kp is an embedding operator. Hence
Kp is a completely operator in Im L. By the definitions of Q and N , it follows that
QN (Ω̄) is bounded on Ω̄. Hence nonlinear operator N is L-compact on Ω.
4
B. DU, X. WANG
EJDE-2010/100
3. Existence of periodic solution for (1.1)
For convenience when applying Lemma 2.1 and Lemma 2.2, we introduce some
notation and sate some assumptions:
CT = {x ∈ C(R, R) : x(t + T ) = x(t), ∀t ∈ R},
|ϕ|0 = max |ϕ(t)|,
t∈[0,T ]
∀ϕ ∈ CT ,
CT1 = {x ∈ C 1 (R, R) : x(t + T ) = x(t), ∀t ∈ R},
kϕk = max {|ϕ|0 , |ϕ0 |0 },
t∈[0,T ]
∀ϕ ∈ CT1 ,
where | · |0 and k · k are the norms of CT and CT1 respectively. Obviously, CT , CT1
are both Banach space.
(H1) Γ(t) > 0, for all t ∈ R;
(H2) lim|x|→+∞ |g(x)|
|x| ≤ r ∈ [0, ∞);
(H3) There exists a positive constant d such that xg(x) > 0, whenever |x| > d.
RT
RT
Theorem 3.1. Suppose that 0 e(s)ds = 0, 0 ϕ2 (s)ds 6= 0, |c(t)| =
6 1 for all
t ∈ R, and assumptions (H1)–(H3) hold, where ϕ(t) is a solution of x(t) − c(t)x(t −
τ ) = 1. Then (1.1) has at least one T -periodic solution, if
v
u X
n
1
c1 T
T 1/2 u
tT
< 1 for c0 < ,
|βj |0 (1 + c0 )r +
1 − c0
1 − c0
2
j=1
or if
v
u X
n
T 1/2 u
c1 T
tT
<1
|βj |0 (1 + c0 )r +
σ−1
σ−1
j=1
for σ > 1.
Proof. Take Ω1 = {x ∈ D(L) : Lx = λN x, λ ∈ (0, 1)}. For x ∈ Ω1 , we have
(x(t) − c(t)x(t − τ ))00 + λ
n
X
βj (t)g(x(t − γj (t))) = λe(t).
(3.1)
j=1
We claim that there exists a point ξ ∈ R such that
|x(ξ)| ≤ d,
(3.2)
where d is a constant which is independent with λ. Integrating two sides of (3.1)
over the interval [0, T ],
n Z T
X
βj (t)g(x(t − γj (t)))dt = 0;
j=1
0
i.e.,
Z
T
Γ(t)g(x(t))dt = 0.
0
By mean value theorem for integrals, there exists a point ξ1 ∈ [0, T ] such that
g(x(ξ1 ))Γ̄T = 0.
EJDE-2010/100
PERIODIC SOLUTIONS
5
By Γ̄ 6= 0, then g(x(ξ1 )) = 0. From the assumption (H3), the inequality (3.2) holds.
Furthermore we have
Z
T
|x0 (t)|dt.
|x(t)| ≤ d +
(3.3)
0
From the conditions
v
u X
n
T 1/2 u
c1 T
tT
|βj |0 (1 + c0 )r +
< 1,
1 − c0
1
− c0
j=1
and
v
u X
n
T 1/2 u
c1 T
tT
|βj |0 (1 + c0 )r +
< 1,
σ−1
σ−1
j=1
there exists a constant ε1 > 0 such that
v
u n
1/2 u X
c1 T
T
tT
|βj |0 (1 + c0 )(r + ε1 ) +
< 1,
1 − c0
1
− c0
j=1
or
v
u X
u n
T
c1 T
tT
|βj |0 (1 + c0 )(r + ε1 ) +
< 1.
σ−1
σ
−1
j=1
(3.4)
1/2
(3.5)
For such a constant ε1 , by (H2), there exists a constant ρ > 0 such that
|g(u)| ≤ (r + ε1 )|u|,
|u| > ρ > d.
(3.6)
Let
E1j = {t|t ∈ [0, T ], |x(t − γj (t))| ≤ ρ},
E2j = {t|t ∈ [0, T ], |x(t − γj (t))| > ρ},
for j = 1, 2 . . . , n. Multiplying both sides of (3.1) by (Ax)(t) and integrating over
[0, T ], from (3.3) and (3.6), we obtain
Z T
|(Ax)0 (t)|2 dt
0
T
Z
=λ
0
n
X
Z
βj (t)g(x(t − γj (t)))(Ax)(t)dt − λ
n
X
Z
e(t)(Ax)(t)dt
0
j=1
≤ |Ax|0
T
|βj (t)kg(x(t − γj (t)))|dt
E1j j=1
Z
n
X
+ |Ax|0
|βj (t)kg(x(t − γj (t)))|dt + T |Ax|0 |e|0
E2j j=1
≤T
n
X
|βj |0 gρ |Ax|0 + T
j=1
n
X
≤ T
n
X
|βj |0 |Ax|0 (r + ε1 )|x|0 + T |Ax|0 |e|0
j=1
n
X
|βj |0 gρ (1 + c0 ) + T |e|0 (1 + c0 ) |x|0 + T
|βj |0 (1 + c0 )(r + ε1 )|x|20
j=1
Z
≤ k1
0
j=1
T
|x0 (t)|dt + k2
Z
T
2
|x0 (t)|dt + k3 ,
0
(3.7)
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B. DU, X. WANG
EJDE-2010/100
where
gρ =
k1 = T
n
X
max
|x(t−γj (t))|≤ρ
|g(x(t − γj (t)))|,
|βj |0 gρ (1 + c0 ) + T |e|0 (1 + c0 ) + 2T
j=1
n
X
|βj |0 (1 + c0 )(r + ε1 )d,
j=1
k2 = T
n
X
|βj |0 (1 + c0 )(r + ε1 ),
j=1
k3 = T
n
X
|βj |0 gρ (1 + c0 )d + T |e|0 (1 + c0 )d + T
j=1
n
X
|βj |0 (1 + c0 )(r + ε1 )d2 .
j=1
From (Ax0 )(t) = (Ax)0 (t) + c0 (t)x(t − τ ), (3.7) and Lemma 2.1, if c0 < 12 , we have
Z T
Z T
0
|x (t)|dt =
|(A−1 Ax0 )(t)|dt
0
0
1
≤
1 − c0
Z
1
1 − c0
Z
T
|(Ax0 )(t)|dt
0
T
c1 T
|x|0
1 − c0
0
Z
Z T
1/2
T 1/2 T
c1 T
c1 T d
≤
|(Ax)0 (t)|2 dt
+
|x0 (t)|dt +
1 − c0 0
1 − c0 0
1 − c0
Z T
Z T
2
i1/2
T 1/2 h
≤
k1
|x0 (t)|dt + k2
|x0 (t)|dt + k3
1 − c0
0
0
Z T
c
T
d
c1 T
1
|x0 (t)|dt +
.
+
1 − c0 0
1 − c0
≤
|(Ax)0 (t)|dt +
By (3.4), there exists a constant M1 > 0 which is independent with λ such that
Z T
|x0 (t)|dt ≤ M1 .
0
Similarly, for σ > 1, by (3.5), there exists a constantM10 > 0 which is independent
with λ such that
Z T
|x0 (t)|dt ≤ M10 .
0
Combining (3.3) with the above two inequalities, we obtain
|x|0 ≤ d + max{M1 , M10 } := M2 .
From
(Ax00 )(t) = (Ax)00 (t) + 2c0 (t)x0 (t − τ ) + c00 (t)x(t − τ ),
if c0 < 1/2, we have
Z T
|x00 (t)|dt
0
Z
=
0
T
|[A−1 Ax00 ](t)|dt
EJDE-2010/100
Z
T
≤
0
PERIODIC SOLUTIONS
7
|(Ax00 )(t)|
dt
1 − c0
T
|(Ax)00 (t) + 2c0 (t)x0 (t − τ ) + c00 (t)x(t − τ )|
dt
1 − c0
0
Z
Z T
n
1 TX
≤
|βj (t)kg(x(t − γj (t)))|dt +
|e(t)|dt + 2c1 M1 + c2 M2 T
1 − c0 0 j=1
0
Z
=
≤
n
X
1
(
|βj |0 T gM2 + T |e|0 + 2c1 M1 + c2 M2 T ) := M3 ;
1 − c0 j=1
if σ > 1, we have
Z T
|x00 (t)|dt ≤
0
n
1 X
(
|βj |0 T gM2 + T |e|0 + 2c1 M1 + c2 M2 T ) := M30 ,
σ − 1 j=1
where gM2 = max|x|≤M2 |g(x)|, c2 = maxt∈[0,T ] |c00 (t)|. Since x ∈ Ω1 , so x(0) = x(T )
and there exists a point η ∈ [0, T ] such that x0 (η) = 0. Then
Z t
x0 (t) = x0 (η) +
x00 (s)ds,
η
|x0 |0 ≤
Z
T
|x00 (t)|dt ≤ max{M3 , M30 } := M4 .
0
Then
kxk = max {|x|0 , |x0 |0 } ≤ max{M2 , M4 }.
t∈[0,T ]
Hence Ω1 is bounded.
Take Ω2 = {x ∈ ker L ∩ CT1 : N x ∈ Im L}, for all x ∈ Ω2 , then x(t) = a0 ϕ(t),
a0 ∈ R satisfying
Z T
Γ(t)g(a0 ϕ(t))dt = 0.
(3.8)
0
When c0 < 1/2, we have
ϕ(t) = A−1 (1) = 1 +
j
∞ Y
X
c(t − (i − 1)τ )
j=1 i=1
≥1−
j
∞ Y
X
c0
j=1 i=1
c0
1 − c0
1 − 2c0
=
:= δ1 > 0.
1 − c0
=1−
Then we have a0 ≤ d/δ1 . Otherwise, for all t ∈ [0, T ], a0 ϕ(t) > d, from assumption
(H3), we have
Z T
Γ(t)g(a0 ϕ(t))dt > 0
0
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B. DU, X. WANG
EJDE-2010/100
which is contradiction to (3.8). When σ > 1, we have
∞ j+1
ϕ(t) = A−1 (1) = −
XY
1
1
−
c(t + τ ) j=1 i=1 c(t + iτ )
∞ j+1
≤−
1 XY 1
−
σ j=1 i=1 σ
1
:= δ2 < 0.
σ−1
Then we have a0 ≤ −d/δ2 . Otherwise, for all t ∈ [0, T ], a0 ϕ(t) < −d, from
assumption (H3), we have
Z T
Γ(t)g(a0 ϕ(t))dt < 0
=−
0
which is contradiction to (3.8). Then we have
|x| = |a0 ϕ(t)| ≤ max{
d
d
, − }|ϕ|0 .
δ1
δ2
Hence Ω2 is a bounded set.
Let Ω ⊃ Ω1 ∪ Ω2 be a bounded set. For x ∈ ∂Ω ∪ D(L), ∀λ ∈ (0, 1), we have
Lx 6= λN x. For all x ∈ ∂Ω ∩ ker L, we have N x ∈
/ Im L. Hence the conditions (1)
and (2) of Lemma 2.2 hold. It remains to verify conditions (3) of Lemma 2.2. Now,
for x ∈ ∂Ω ∩ ker L, take the homotopy
(
R T Pn
Pn ¯
−µx − T1 (1 − µ) 0
βj )xg(x) > 0;
j=1 βj (t)g(x)dt, if (
R T Pn
H(x, µ) =
Pj=1
n
1
µx − T (1 − µ) 0
if ( j=1 β¯j )xg(x) < 0.
j=1 βj (t)g(x)dt,
Clearly,
(
Pn
Pn
−µx − (1 − µ)g(x) j=1 β¯j , if ( j=1 β¯j )xg(x) > 0;
H(x, µ) =
Pn ¯
Pn
µx − (1 − µ)g(x) j=1 βj ,
if ( j=1 β¯j )xg(x) < 0.
For x ∈ ∂Ω ∩ ker L and µ ∈ [0, 1], xH(x, µ) 6= 0. So we have
Z
n
1 TX
βj (t)g(x)dt, Ω ∩ ker L, 0
deg{QN, Ω ∩ ker L, 0} = deg −
T 0 j=1
= deg{−x, Ω ∩ ker L, 0} =
6 0.
Applying Lemma 2.2, we reach the conclusion.
As an application, we consider the following example.
Example 3.1. Consider the equation
u(t − 12 cos t)
1
1
(2 − sin t)x(t − τ ))00 + (1 + sin t)
10
2
80000
1
u(t
−
sin
t)
1
2
+ (1 − sin t)
= sin t,
2
80000
(x(t) −
where
c(t) =
1
(2 − sin t),
10
β1 (t) = 1 +
1
sin t,
2
β2 (t) = 1 −
1
sin t,
2
(3.9)
EJDE-2010/100
PERIODIC SOLUTIONS
9
1
1
cos t, γ2 (t) = sin t, e(t) = sin t, T = 2π.
2
2
From simple calculations, we have
1
3
3
1
3
, c1 =
, |β1 |0 = , |β2 |0 = , r =
.
c0 =
10
10
2
2
80000
Let µ1 (t) and µ2 (t) be the inverses of t − 21 cos t and t − 21 sin t respectively. We
have
β1 (µ1 (t))
β2 (µ2 (t))
Γ(t) =
+
1 − γ10 (µ1 (t)) 1 − γ20 (µ2 (t))
γ1 (t) =
=
1+
1+
=1+
and
1
2
1
2
sin µ1 (t)
1 − 21 sin µ2 (t)
+
sin µ1 (t) 1 − 21 cos µ2 (t)
1 − 21 sin µ2 (t)
>0
1 − 21 cos µ2 (t)
v
u X
n
c1 T
T 1/2 u
tT
|βj |0 (1 + c0 )r +
≈ 0.96 < 1.
1 − c0
1
− c0
j=1
Applying Theorem 3.1, Equation (3.9) has at least one 2π-periodic solution.
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Department of Mathematics, Huaiyin Normal University, Huaian Jiangsu, 223300,
China
E-mail address, Bo Du: dubo7307@163.com
E-mail address, Xiaojing Wang: wwxxjj@126.com