Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 148, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
PERIODIC SOLUTIONS FOR P-LAPLACIAN NEUTRAL
FUNCTIONAL DIFFERENTIAL EQUATIONS WITH MULTIPLE
DEVIATING ARGUMENTS
AOMAR ANANE, OMAR CHAKRONE, LOUBNA MOUTAOUEKKIL
Abstract. By means of Mawhin’s continuation theorem, we prove the existence of periodic solutions for a p-Laplacian neutral functional differential
equation with multiple deviating arguments
(ϕp (x0 (t) − c(t)x0 (t − r)))0
= f (x(t))x0 (t) + g(t, x(t), x(t − τ1 (t)), . . . , x(t − τm (t))) + e(t).
1. Introduction
In recent years, periodic solutions involving the scalar p-Laplacian have been
studied extensively by many researchers. Lu and Ge [4] discussed sufficient conditions for the existence of periodic solutions to second order differential equation,
with a deviating argument,
x00 (t) = f (t, x(t), x(t − τ (t)), x0 (t)) + e(t).
Recently, Pan [5] studied the equation
x(n) (t) =
n−1
X
bi x(i) (t) + f (t, x(t), x(t − τ1 (t)), . . . , x(t − τm (t))) + p(t).
i=1
Feng, Lixiang and Shiping [2] investigated the existence of periodic solutions for a
p-Laplacian neutral functional differential equation
(ϕp (x0 (t) − c(t)x0 (t − r)))0 = f (x(t))x0 (t) + β(t)g(x(t − τ (t))) + e(t),
where c(t) and β(t) are allowed to change signs.
The purpose of this article is to study the existence of periodic solution for
p-Laplacian neutral functional differential equation
(ϕp (x0 (t) − c(t)x0 (t − r)))0
= f (x(t))x0 (t) + g(t, x(t), x(t − τ1 (t)), . . . , x(t − τm (t))) + e(t).
(1.1)
Where p > 1 is a fixed real number. The conjugate exponent of p is denoted by
q; i.e., p1 + 1q = 1. Let ϕp : R → R be the mapping defined by ϕp (s) = |s|p−2 s
2000 Mathematics Subject Classification. 34K15, 34C25.
Key words and phrases. Periodic solution; neutral differential equation;
deviating argument; p-Laplacian; Mawhin’s continuation.
c
2012
Texas State University - San Marcos.
Submitted May 17, 2012. Published August 29, 2012.
1
2
A. ANANE, O. CHAKRONE, L. MOUTAOUEKKIL
EJDE-2012/148
for s 6= 0, and ϕp (0) = 0, f, e, c ∈ C(R, R) are continuous T -periodic functions
defined on R and T > 0 , r ∈ R is a constant with r > 0, g ∈ C(Rm+2 , R) and
g(t + T, u0 , u1 , . . . , um ) = g(t, u0 , u1 , . . . , um ), for all (t, u0 , u1 , . . . , um ) ∈ Rm+2 ,
τi ∈ C(R, R)(i = 1, 2, . . . , m) with τi (t + T ) = τi (t).
2. Preliminaries
For convenience, define CT = {x ∈ C(R, R) : x(t + T ) = x(t)} with the norm
|x|∞ = max |x(t)|t∈[0,T ] . Clearly CT is a Banach space. We also define a linear
operator
A : CT → CT , (Ax)(t) = x(t) − c(t)x(t − r),
(2.1)
(
1
if 1 < p ≤ 2,
and constant Cp =
2p−2 if p > 2.
To simplify the studying of the existence of periodic solution for (1.1) we cite
the following lemmas.
Lemma 2.1 ([2]). Let p ∈]1, +∞[ be a constant, s ∈ C(R, R) such that s(t + T ) ≡
s(t), for all t ∈ [0, T ]. Then for for all u ∈ C 1 (R, R) with u(t + T ) ≡ u(t), we have
Z T
Z T
|u(t) − u(t − s(t))|p dt ≤ 2( max |s(t)|)p
|u0 (t)|p dt.
0
t∈[0,T ]
0
Lemma 2.2 ([2]). Let B : CT → CT , (Bx)(t) = c(t)x(t − r). Then B satisfies the
following conditions
(1) kBk ≤ |c|∞ .
RT
RT
(2) ( 0 |[B j x](t)|p dt)1/p ≤ |c|j∞ ( 0 |x(t)|p dt)1/p , ∀x ∈ CT , p > 1, j ≥ 1.
Lemma 2.3 ([2]). If |c|∞ < 1, then A, defined by (2.1), has continuous bounded
inverse A−1 with the following properties:
(1) kA−1 k ≤ 1/(1 − |c|∞ ),
P∞ Qj
(2) (A−1 f )(t) = f (t) + j=1 i=1 c(t − (i − 1)r)f (t − jr), for all f ∈ CT ,
RT
RT
1
(3) 0 |(A−1 f )(t)|p dt ≤ ( 1−|c|
)p 0 |f (t)|p dt for all f ∈ CT .
∞
Now, we recall Mawhin’s continuation theorem which will be used in our study.
Let X and Y be real Banach spaces and L : D(L) ⊂ X → Y be a Fredholm operator
with index zero. Here D(L) denotes the domain of L. This means that Im L is closed
in Y and dim ker L = dim(Y / Im L) < +∞. Consider the supplementary subspaces
X1 and Y1 and such that X = ker L ⊕ X1 and Y = Im L ⊕ Y1 and let P : X → ker L
and Q : Y → Y1 be natural projections. Clearly, ker L ∩ (D(L) ∩ X1 ) = {0}, thus
the restriction Lp := L|D(L)∩X1 is invertible. Denote the inverse of Lp by K.
Now, let Ω be an open bounded subset of X with D(L) ∩ Ω 6= ∅, a map N :
Ω → Y is said to be L-compact on Ω. If QN (Ω) is bounded and the operator
K(I − Q)N : Ω → Y is compact.
Lemma 2.4 ([3]). . Suppose that X and Y are two 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 : Ω → Y is L-compact on Ω. If all of the following conditions
hold:
(1) Lx 6= λN x, ∀x ∈ ∂Ω ∩ D(L), λ ∈]0, 1];
(2) N x 6∈ Im L for all x ∈ ∂Ω ∩ ker L; and
EJDE-2012/148 PERIODIC SOLUTIONS
3
(3) deg{JQN, Ω ∩ ker L, 0} =
6 0, where J : Im Q → ker L is an isomorphism.
Then the equation Lx = N x has at least one solution on Ω ∩ D(L).
To use Mawhin’s continuation theorem to study the existence of T -periodic solution for (1.1), we rewrite (1.1) in the system
x01 (t) = [A−1 ϕq (x2 )](t),
x02 (t) = f (x1 (t))[A−1 ϕq (x2 )](t)
(2.2)
+ g(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t))) + e(t).
Where q > 1 is constant with p1 + 1q = 1. Clearly, if x(t) = (x1 (t), x2 (t))T is a
T -periodic solution to equation set (2.2), then x1 (t) must be a T -periodic solution
to equation (1.1). Thus, to prove that (1.1) has a T -periodic solution, it suffices to
show that equation set (2.2) has a T -periodic solution.
Now, we set X = Y = {x = (x1 (t), x2 (t))T ∈ C(R, R2 ) : x1 ∈ CT , x2 ∈ CT } with
the norm kxk = max{|x1 |∞ , |x2 |∞ }. Obviously, X and Y are two Banach spaces.
Meanwhile, let
0
x1
0
L : D(L) ⊂ X → Y, Lx = x =
.
(2.3)
x02
and N : X → Y be defined by
[N x](t)
[A−1 ϕq (x2 )](t)
.
=
f (x1 (t))[A−1 ϕq (x2 )](t) + g(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t))) + e(t)
(2.4)
It is easy to see that (2.2) can be converted to the abstract equation Lx = N x.
Moreover, from the definition of L, we see that ker L = R2 , Im L = {y : y ∈
RT
Y, 0 y(s)ds = 0}. So L is a Fredholm operator with index zero.
Let projections P : X → ker L and Q : Y → Im Q be defined by
Z
Z
1 T
1 T
Px =
x(s)ds, Qy =
y(s)ds,
T 0
T 0
and let K represent the inverse of L|ker P ∩D(L) . Clearly, ker L = Im Q = R2 and
Z T
[Ky](t) =
G(t, s)y(s)ds,
(2.5)
0
where
(
G(t, s) =
s
T,
s−T
T
if 0 ≤ s < t ≤ T
, if 0 ≤ t ≤ s ≤ T.
From (2.4) and (2.5), it is not hard to find that N is L-compact on Ω, where Ω is
an arbitrary open bounded subset of X.
Lemma 2.5 ([6]). If ω ∈ C 1 (R, R) and ω(0) = ω(T ) = 0 then
Z T
Z
T p T 0 p
p
|ω(t)| dt ≤
|ω (t)| dt,
πp
0
0
where
Z
πp = 2
0
(p−1)1/p
ds
(1 −
sp
1/p
p−1 )
=
2π(p − 1)1/p
.
p sin( πp )
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A. ANANE, O. CHAKRONE, L. MOUTAOUEKKIL
EJDE-2012/148
Lemma 2.6 ([7]). Ω ⊂ Rn is an open bounded set, and symmetric with respect to
0 ∈ Ω. If f ∈ C(Ω, Rn ) and f (x) 6= µf (−x) for all x ∈ ∂Ω and all µ ∈ [0, 1], then
deg(f, Ω, 0) is an odd number.
Lemma 2.7 ([2]). If c(t) ∈ CT is not a constant function, |c|∞ < 1/2,
(Ax)(t) = x(t) − c(t)x(t − r) ≡ d1 ,
(2.6)
where d1 is a nonzero constant, and x(t) ∈ CT , then
(1) x(t) = A−1 d1 is not a constant function,
RT
(2) 0 (A−1 d1 )(t)dt 6= 0.
3. Main results
For the next theorem, we assume that the following conditions:
(H1) There is a constant d > 0 such that:
(1) g(t, u0 , u1 , . . . , um ) > |e|∞ , for all (t, u0 , u1 , . . . , um ) ∈ [0, T ] × Rm+1
with ui > d (i = 0, 1, . . . , m).
(2) g(t, u0 , u1 , . . . , um ) < −|e|∞ , for all (t, u0 , u1 , . . . , um ) ∈ [0, T ] × Rm+1
with ui < −d (i = 0, 1, . . . , m).
(H2) The function g has the decomposition
g(t, u0 , u1 , . . . , um ) = h1 (t, u0 ) + h2 (t, u0 , . . . , um ),
Pm
p−1
such that u0 h1 (t, u0 ) ≥ l|u0 |n , |h2 (t, u0 , . . . , um )| ≤
+ β,
i=0 αi |ui |
where n, l, αi (i = 0, . . . , m), β are non-negative constants with n ≥ p.
Theorem 3.1. Assume (H1)–(H2). Then, (1.1) has at least one T -periodic solution, if |c|∞ < 1/2 and if
m
p T p
T X
1
|c|∞ (1 + |c|∞ )p−1 + 2p−1 δ
+ Cp 21/q
αi |τi |p−1
< 1,
∞
1 − |c|∞
πp
πp i=1
Pm
where δ = max(Cp i=1 αi − α0 − l, 0).
Proof. Let Ω1 = {x ∈ X : Lx = λN x, λ ∈]0, 1]} if x(·) = (x1 (·), x2 (·))T ∈ Ω1 , then
from (2.3) and (2.4), we have
x01 (t) = λ[A−1 ϕq (x2 )](t),
x02 (t) = λf (x1 (t))[A−1 ϕq (x2 )](t)
(3.1)
+ λg(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t))) + λe(t).
From the first equation in (3.1), we have x2 (t) = ϕp ( λ1 (Ax01 )(t), together with the
second formula of (3.1), which yields
[ϕp ((Ax01 )(t))]0 = λp−1 f (x1 (t))x01 (t)
+ λp g(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t))) + λp e(t).
(3.2)
Integrating both sides of (3.2) on the interval [0, T ] and applying integral mean
value theorem, then there exists a constant t0 ∈ [0, T ] such that
Z
1 T
g(t, x1 (t0 ), x1 (t0 − τ1 (t0 )), . . . , x1 (t0 − τm (t0 ))) = −
e(t)dt.
(3.3)
T 0
We can prove that there is t1 ∈ [0, T ] such that |x1 (t1 )| ≤ d.
If |x1 (t0 )| ≤ d, then taking t1 = t0 such that |x1 (t1 )| ≤ d.
EJDE-2012/148 PERIODIC SOLUTIONS
5
If |x1 (t0 )| > d. It follows from (H1) that there is some i ∈ {1, 2, . . . , m} such
that |x1 (t0 − τi (t0 ))| ≤ d. Since x1 (t) is continuous for t ∈ R and x1 (t + T ) = x1 (t),
so there must be an integer k and a point t1 ∈ [0, T ] such that t0 − τi (t0 ) = kT + t1 .
So |x1 (t1 )| = |x1 (t0 − τi (t0 ))| ≤ d. Then, we have
Z t
Z t
|x1 (t)| = |x1 (t1 ) +
x01 (s)ds| ≤ d +
|x01 (s)|ds, t ∈ [t1 , t1 + T ],
t1
t1
and
Z
t1
|x1 (t)| = |x1 (t − T )| = |x(t1 ) −
x01 (s)ds| ≤ d +
Z
t1
|x01 (s)|ds,
t ∈ [t1 , t1 + T ].
t1 −T
t−T
Combining the above two inequalities, we obtain
|x1 |∞ = max |x1 (t)| =
t∈[0,T ]
max
t∈[t1 ,t1 +T ]
|x1 (t)|
Z
Z t1
o
n
1 t 0
|x1 (s)|ds +
|x01 (s)|ds
d+
2 t1
t∈[t1 ,t1 +T ]
t−T
Z T
1
≤d+
|x01 (s)|ds.
2 0
≤
max
(3.4)
On the hand, multiplying both sides of (3.2) by x1 (t) and integrating it from 0 to
T , we obtain
Z T
[ϕp ((Ax01 )(t))]0 x1 (t)dt
0
=λ
p−1
+λ
p
Z
T
f (x1 (t))x01 (t)x1 (t)dt
0
Z T
(3.5)
g(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t)))x1 (t)dt
0
+ λp
Z
T
e(t)x1 (t)dt.
0
On the other hand we have
Z T
[ϕp ((Ax01 )(t))]0 x1 (t)dt
0
Z
T
ϕp ((Ax01 )(t))x01 (t)dt
=−
0
Z
(3.6)
T
ϕp ((Ax01 )(t))[x01 (t)
=−
−
c(t)x01 (t
− r) +
c(t)x01 (t
− r)]dt
0
Z
=−
T
|(Ax01 )(t)|p dt −
0
Z
T
c(t)x01 (t − r)ϕp ((Ax01 )(t))dt.
0
Meanwhile,
Z
0
T
f (x1 (t))x01 (t)x1 (t)dt = 0.
(3.7)
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A. ANANE, O. CHAKRONE, L. MOUTAOUEKKIL
EJDE-2012/148
Substituting (3.6)-(3.7) into (3.5) we obtain
T
Z
|(Ax01 )(t)|p dt
0
Z
T
c(t)x01 (t − r)ϕp ((Ax01 )(t))dt
=−
0
−λ
p
Z
(3.8)
T
g(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t)))x1 (t)dt
0
− λp
Z
T
e(t)x1 (t)dt.
0
In view of (H2), we obtain
Z
T
|(Ax01 )(t)|p dt
0
Z
T
c(t)x01 (t − r)ϕp ((Ax01 )(t))dt
=−
0
−λ
p
= −λp
T
Z
g(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t)))x1 (t)dt − λ
Z
0
T
p
Z
T
e(t)x1 (t)dt
0
h1 (t, x1 )x1 (t)dt
0
− λp
T
Z
h2 (t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t)))x1 (t)dt
0
− λp
T
Z
e(t)x1 (t)dt.
0
(3.9)
Define ∆1 = {t ∈ [0, T ] : |x1 (t)| ≤ 1}, ∆2 = {t ∈ [0, T ] : |x1 (t)| > 1}, in view of
(H2) again we have
−λp
Z
0
T
Z
T
|x1 (t)|n dt
0
Z Z
p
= −λ l
+
|x1 (t)|n dt
∆1
∆2
Z
p
≤ −λ l
|x1 (t)|n dt
∆2
Z
p
≤ −λ l
|x1 (t)|p dt
h1 (t, x1 )x1 (t)dt ≤ −λp l
∆2
T
= −λp l
Z
|x1 (t)|p dt + λp l
0
p
Z
≤ −λ l
0
Z
∆1
T
|x1 (t)|p dt + lT.
(3.10)
|x1 (t)|p dt
EJDE-2012/148 PERIODIC SOLUTIONS
7
Substituting (3.10) into (3.9),
Z T
|(Ax01 )(t)|p dt
0
T
Z
|ϕp ((Ax01 )(t))||x01 (t
≤ |c|∞
+λ
|x1 (t)|p dt
− r)|dt − λ l
0
p
T
Z
p
0
(3.11)
T
Z
|h2 (t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t)))||x1 (t)|dt
0
+ λp |e|∞
T
Z
|x1 (t)|dt + lT.
0
Moreover, by using Hölder’s inequality and Minkowski inequality, we obtain
Z T
|ϕp ((Ax01 )(t))||x01 (t − r)|dt
0
≤
T
Z
|ϕp ((Ax01 )(t))|q dt
=
T
|(Ax01 )(t)|p dt
T
1/q Z
0
=
|x01 (t − r)|p dt
|x01 (t)|p dt
1/p
0
T
h Z
|x01 (t)
c(t)x01 (t
−
p
− r)| dt
1/p ip/q Z
T
h Z
|x01 (t)|p dt
1/p
|x01 (t)|p dt
1/p
+
Z
0
≤
|x01 (t)|p dt
1/p
T
|c(t)x01 (t − r)|p dt
+ |c|∞
T
|x01 (t)|p dt
1/p ip/q Z
1/p
T
|x01 (t)|p dt
1/p
0
0
Z
|x01 (t)|p dt
0
Z
0
p−1
T
1/p ip/q Z
0
T
h Z
T
0
0
≤
1/p
0
0
Z
T
1/q Z
T
= (1 + |c|∞ )
|x01 (t)|p dt.
0
(3.12)
By (3.11) and (3.12) and combining with (H2) and Lemma 2.1, we obtain
Z T
|(Ax01 )(t)|p dt
0
≤ |c|∞ (1 + |c|∞ )p−1
Z
T
|x01 (t)|p dt − λp l
0
+ λp
Z
Z
T
|x1 (t)|p dt
0
T
|h2 (t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t)))||x1 (t)|dt
0
+ λp |e|∞
Z
T
|x1 (t)|dt + lT
0
≤ |c|∞ (1 + |c|∞ )p−1
Z
T
|x01 (t)|p dt + λp (α0 − l)
Z
0
+λ
p
Z
0
m
T X
T
|x1 (t)|p dt
0
p−1
αi |x1 (t − τi (t))|
Z
p
T
|x1 (t)|dt + λ (β + |e|∞ )
|x1 (t)|dt + lT
0
i=1
p−1
Z
≤ |c|∞ (1 + |c|∞ )
0
T
|x01 (t)|p dt
p
Z
+ λ (α0 − l)
0
T
|x1 (t)|p dt
8
A. ANANE, O. CHAKRONE, L. MOUTAOUEKKIL
p
+ λ Cp
m
X
Z
+ λp Cp
T
|x1 (t − τi (t)) − x1 (t)|p−1 |x1 (t)|dt
αi
0
i=1
m
X
Z
T
|x1 (t)|p dt + λp (β + |e|∞ )
αi
T
Z
|x01 (t)|p dt + λp (Cp
0
+ λ Cp
|x1 (t)|dt + lT
0
≤ |c|∞ (1 + |c|∞ )p−1
p
αi
T
Z
0
i=1
m
X
EJDE-2012/148
m
X
p
|x1 (t − τi (t)) − x1 (t)| dt
1/q Z
0
i=1
|x1 (t)|p dt
0
i=1
T
Z
T
Z
αi + α0 − l)
T
|x1 (t)|p dt
1/p
0
T
Z
+ λp (β + |e|∞ )T 1/q
|x1 (t)|p dt
1/p
+ lT
0
≤ |c|∞ (1 + |c|∞ )p−1
T
Z
|x01 (t)|p dt + δ
0
+ Cp
m
X
αi 21/q |τi |p−1
∞
T
Z
|x1 (t)|p dt
0
T
Z
|x01 (t)|p dt
+ (β + |e|∞ )T 1/q
T
Z
|x1 (t)|p dt
1/p
0
0
i=1
T
1/q Z
|x1 (t)|p dt
1/p
+ lT.
(3.13)
0
Let ω(t) = x1 (t + t1 ) − x1 (t1 ), then ω(T ) = ω(0) = 0 and from Lemma 2.1 we see
that
Z T
T p Z T
T p Z T
0
p
p
|ω (t)| dt =
|x01 (t)|p dt.
(3.14)
|ω(t)| dt ≤
πp
πp
0
0
0
By (3.14) and the Minkowski inequality, we obtain
Z T
1/p
1/p Z T
p
|ω(t) + x1 (t1 )|p dt
|x1 (t)| dt
=
0
0
≤
T
Z
|ω(t)|p dt
1/p
+
Z
T πp
|x1 (t1 )|p dt
1/p
(3.15)
0
0
≤
T
T
Z
|x01 (t)|p dt
1/p
+ dT 1/p .
0
Substituting (3.15) into (3.13) yields
Z T
|(Ax01 )(t)|p dt
0
p−1
Z
T
≤ |c|∞ (1 + |c|∞ )
|x01 (t)|p dt
+δ
h T Z
0
+ Cp 21/q
m
X
αi |τi |p−1
∞
i=1
+ (β + |e|∞ )T 1/q
h
πp
h T Z T
πp
p−1
≤ |c|∞ (1 + |c|∞ )
h T Z
+2
πp
T
|x01 (t)|p dt
T
|x01 (t)|p dt
1/p
ip
0
1/p
+ dT 1/p
i Z
0
|x01 (t)|p dt
+ dT 1/p
T
|x01 (t)|p dt
1/q
0
1/p
i
+ dT 1/p + lT
0
p−1
m
iZ T
X
T p
1/q T
p−1
δ
+ Cp 2
αi |τi |∞
|x01 (t)|p dt
πp
πp i=1
0
EJDE-2012/148 PERIODIC SOLUTIONS
1/q
+ Cp 2
m
X
1/p
αi |τi |p−1
∞ dT
9
T
Z
|x01 (t)|p dt
1/q
0
i=1
Z T
1/p
1/q T
|x01 (t)|p dt
+ (β + |e|∞ )T
πp 0
+ 2p−1 δdp T + (β + |e|∞ )dT + lT.
(3.16)
By applying the third part of Lemma 2.2, we obtain
p Z T
Z T
Z T
1
|x01 (t)|p dt =
|(Ax01 )(t)|p dt. (3.17)
|(A−1 Ax01 )(t)|p dt ≤
1 − |c|∞
0
0
0
Then, substituting (3.17) into (3.16), we have
Z T
p h
T p
1
|c|∞ (1 + |c|∞ )p−1 + 2p−1 δ
|x01 (t)|p dt ≤
1
−
|c|
π
∞
p
0
m
iZ T
X
T
+ Cp 21/q
αi |τi |p−1
|x01 (t)|p dt
∞
πp i=1
0
Pm
1/q
1/p Z T
Cp 21/q i=1 αi |τi |p−1
∞ dT
0
p
|x
(t)|
dt
+
1
(1 − |c|∞ )p
0
1/p
(β + |e|∞ )T 1/q πTp Z T
2p−1 δdp T
0
p
|x
+
+
(t)|
dt
1
p
(1 − |c|∞ )
(1 − |c|∞ )p
0
lT
(β + |e|∞ )dT
+
.
+
p
(1 − |c|∞ )
(1 − |c|∞ )p
(3.18)
As
m
i
p h
1
T X
T p
+ Cp 21/q
|c|∞ (1 + |c|∞ )p−1 + 2p−1 δ
αi |τi |p−1
< 1,
∞
1 − |c|∞
πp
πp i=1
1/p < 1, 1/q < 1, then from (3.18), there exists a constant M > 0 such that
Z T
|x01 (t)|p dt ≤ M.
(3.19)
0
Which together with (3.4) gives
1
|x1 |∞ ≤ d + T 1/q M 1/p =: M1 .
2
(3.20)
Again, from the first equation in (3.1), we have
Z T
(A−1 ϕq (x2 ))(t)dt = 0.
0
Then there is a constant η ∈ [0, T ], such that (A−1 ϕq (x2 ))(η) = 0, which together
with the second part of lemma 2.3 gives
(A−1 ϕq (x2 ))(η) = ϕq (x2 (η)) +
j
∞ Y
X
j=1 i=1
c(η − (i − 1)r)ϕq (x2 (η − jr)) = 0,
10
A. ANANE, O. CHAKRONE, L. MOUTAOUEKKIL
EJDE-2012/148
|x2 (η)|q−1 = |ϕq (x2 (η))|
j
∞ Y
X
=
c(η − (i − 1)r)ϕq (x2 (η − jr))
≤
j=1 i=1
∞
X
|c|j∞ |x2 |q−1
∞
j=1
=
|c|∞
|x2 |q−1
∞ .
1 − |c|∞
It follows that
|x2 (η)| ≤
|c|
1/(q−1)
∞
|x2 |∞ .
1 − |c|∞
(3.21)
Let Mf = max|u|≤M1 |f (u)|, Mg = maxt∈[0,T ],|u0 |≤M1 ,...,|um |≤M1 |g(t, u0 , . . . , um )|
and from (3.1), we have
x02 (t) = λf (x1 (t))x01 (t) + λg(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t))) + λe(t),
and
Z
T
|x02 (t)|dt
0
T
Z
|f (x1 (t))x01 (t)|dt
≤
T
Z
|g(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t)))|dt
+
0
0
T
Z
|e(t)|
+
0
Z
T
|x01 (t)|dt + T (Mg + |e|∞ )
≤ Mf
≤ Mf T
0
1/q
M 1/p + T (Mg + |e|∞ ) =: M2 .
(3.22)
By (3.21) and (3.22)
Z
|x2 (t)| = |x2 (η) +
t
x02 (s)ds| ≤
η
Z T
1/(q−1)
|c|
∞
|x2 |∞ +
|x02 (s)|ds
1 − |c|∞
0
|c|
1/(q−1)
∞
≤
|x2 |∞ + M2 ,
1 − |c|∞
(3.23)
t ∈ [0, T ] .
|c|∞
Since |c|∞ < 12 , ( 1−|c|
)1/(q−1) < 1, and (3.23), it follows that there exists a
∞
positive constant M3 such that
|x2 |∞ ≤ M3 .
(3.24)
Let Ω2 = {x ∈ ker L, QN x = 0}. If x ∈ Ω2 then x ∈ R2 is a constant vector, and
1
T
Z
T
[A−1 ϕq (x2 )](t)dt = 0,
0
Z
1 T
[f (x1 (t))[A−1 ϕq (x2 )](t)
T 0
+ g(t, x1 (t), x1 (t − τ1 (t)), . . . , x1 (t − τm (t))) + e(t)]dt = 0.
(3.25)
EJDE-2012/148 PERIODIC SOLUTIONS
11
By the first formula in (3.25) and the second part of Lemma 2.7, we have x2 = 0.
Which together with the second equation in (3.25) yields
Z
1 T
[g(t, x1 , x1 , . . . , x1 ) + e(t)]dt = 0.
T 0
In view of (H1), we see that |x1 | ≤ d. Now, we let Ω = {x|x = (x1 , x2 )T ∈ X, |x1 | <
M1 + d, |x2 | < M3 + d}, then Ω1 ∪ Ω2 ⊂ Ω. So from (3.20) and (3.24), it is easy to
see that conditions (1) and (2) in Lemma 2.4 are satisfied.
Next, we verify the condition (3) in Lemma 2.4. To do this, we define the
isomorphism J : Im Q → ker L, J(x1 , x2 )T = (x1 , x2 )T . Then
!
R
1 T
[A−1 ϕq (x2 )](t)dt
T
0
JQN (x) = 1 R T
,
−1
ϕq (x2 )](t) + g(t, x1 , x1 , . . . , x1 ) + e(t)]dt
T 0 [f (x1 (t))[A
x ∈ ker L ∩ Ω. By Lemma 2.6, we need to prove that
JQN (x) 6= µ(JQN (−x)),
∀x ∈ ∂(Ω ∩ ker L),
T
µ ∈ [0, 1]
T
Case1. If x = (x1 , x2 ) ∈ ∂(Ω∩ker L)\{(M1 +d, 0) , (−M1 −d, 0)T }, then x2 6=
RT
0 which, together with the second part of Lemma 2.7, gives us 0 [A−1 ϕq (x2 )](t)dt 6=
0,
1 Z T
1 Z T
−1
[A ϕq (x2 )](t)dt
[A−1 ϕq (−x2 )](t)dt < 0,
T 0
T 0
obviously, for all µ ∈ [0, 1], JQN (x) 6= µ(JQN (−x)).
Case2. If x = (M1 + d, 0)T or x = (−M1 − d, 0)T then
0
R
JQN (x) = 1 T
T 0 [g(t, x1 , x1 , . . . , x1 ) + e(t)]dt
which, together with (H1), yields JQN (x) 6= µ(JQN (−x)) for all µ ∈ [0, 1]. Thus,
condition (3) of Lemma 2.4 is also satisfied. Therefore, by applying Lemma 2.4, we
conclude that the equation Lx = N x has at least one T -periodic solution on Ω, so
(1.1) has at least one T -periodic solution. This completes the proof.
4. Example
In this section, we provide an example to illustrate Theorem 3.1. Let us consider
the equation
(ϕp (x0 (t) − 0.1 sin(20πt)x0 (t − r)))0
(4.1)
cos 20πt
sin 20πt
= f (x(t))x0 (t) + g(t, x(t), x(t −
), x(t −
)) + cos(20πt),
90
100
where p = 3, T = 1/10, c(t) = 0.1 sin(20πt), τ1 (t) = cos(20πt)/90, τ2 (t) =
sin(20πt)/100,
3
g(t, u, v, w) = u3 2 + sin(20πt) +
sgn(v)v 2 + sgn(w)w2 | cos(20πt)|,
225
e(t) = cos(20πt). Therefore we can choose l = 1, d = 1, α0 = 0, α1 = α2 = 0, 014.
We can easily check condition (H1), (H2) in Theorem 3.1 hold. We can check that
m
i
X
p h
1
T p
T
|c|∞ (1 + |c|∞ )p−1 + 2p−1 δ
+ Cp 21/q
αi ( )|τi |p−1
< 1.
∞
1 − |c|∞
πp
πp
i=1
By Theorem 3.1, equation (4.1) has at least one
1
10 -periodic
solution.
12
A. ANANE, O. CHAKRONE, L. MOUTAOUEKKIL
EJDE-2012/148
References
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Anal. 65 (2006), 2003-2012.
[2] Liang Feng, Guo Lixiang, Lu Shiping; Existence of periodic solutions for a p-Laplacian
neutral functional differential equation, Nonlinear Analysis 71 (2009), 427-436.
[3] R. E. Gaines, J. L. Mawhin; Coincidence Degree and Nonlinear Differential Equations,
Springer Verlag, Berlin, 1977.
[4] S. Lu, W. Ge; Sufficient conditions for the existence of Periodic solutions to some second
order differential equation with a deviating argument, J. Math. Anal. Appl 308 (2005), 393419.
[5] Lijun. Pan; periodic solutions for higher order differential equation with a deviating argument,
J. Math. Anal. Appl 343 (2008), 904-918.
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Aomar Anane
Université Mohamed I, Faculté des Sciences, Département de Mathématiques et Informatique, Oujda, Maroc
E-mail address: anane@sciences.univ-oujda.ac.ma
Omar Chakrone
Université Mohamed I, Faculté des Sciences, Département de Mathématiques et Informatique, Oujda, Maroc
E-mail address: chakrone@yahoo.fr
Loubna Moutaouekkil
Université Mohamed I, Faculté des Sciences, Département de Mathématiques et Informatique, Oujda, Maroc
E-mail address: loubna anits@yahoo.fr