Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 205, 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 FOURTH-ORDER p-LAPLACIAN
FUNCTIONAL DIFFERENTIAL EQUATIONS WITH
SIGN-VARIABLE COEFFICIENT
JIAYING LIU, WENBIN LIU, BINGZHUO LIU
Abstract. Using the theory of coincidence degree, we show the existence
of periodic solutions to the fourth-order p-Laplacian differential equations of
Liénard-type
φp (x00 ))00 + f (x(t))x0 (t) + α(t)g1 (x(t − τ1 (t, x(t))))
+ β(t)g2 (x(t − τ1 (t, x(t)))) = p(t).
The rate of growth of g1 (u) with respect to the variable u is allowed to be
greater than p − 1, and the coefficient β(t) is allowed to change sign.
1. Introduction
The study of the fourth-order differential equations is of great practical significance, whose classical application is to describe the equilibrium of elastic beams.
The study on periodic oscillations of the fourth-order differential equations has
gained more and more attention by many researchers, and some profound results
have been obtained (see [3, 7, 8, 10]). However, the results of periodic solutions to
a fourth order p-Laplacian delay differential equation are relatively rare.
In this article, we consider the existence of periodic solutions to the fourth-order
p-Laplacian differential equations with multiple deviating arguments:
φp (x00 ))00 + f (x(t))x0 (t) + α(t)g1 x(t − τ1 (t, x(t)))
(1.1)
+ β(t)g2 x(t − τ2 (t, x(t))) = p(t)
where p > 1, φp (s) = |s|p−2 s (s 6= 0), φp (0) = 0, α(t), β(t), p(t) ∈ C(R, R),
RT
RT
RT
p(t)dt = 0, 0 β(t)dt 6= 0, α(t) ≥ 0 (≤ 0) for t ∈ R, 0 α(t)dt > 0 (< 0),
0
α(t+T ) = α(t), β(t+T ) = β(t), p(t+T ) = p(t) τi ∈ C(R2 , R), τi (t+T, x) = τi (t, x),
gi ∈ C(R, R), i = 1, 2, T > 0.
In recent years, there have been a number of results on the existence of periodic
solutions of the second order p-Laplacian differential equations; see [1, 2, 5, 6, 9, 11]
2000 Mathematics Subject Classification. 34A12, 34C25.
Key words and phrases. p-Laplacian equation; periodic solution; multiple deviating argument;
Mawhin continuation theorem.
c
2013
Texas State University - San Marcos.
Submitted October 7, 2012. Published September 18, 2013.
Supported by grant 11271364 from the NNSF of China.
1
2
J. LIU, W. LIU, B. LIU
EJDE-2013/205
and the references therein. Cheung and Ren [9] studied the existence of periodic
solutions for the p-Laplacian delay equation
φp (x0 ))0 + f (x0 (t)) + βg(x(t − τ (t))) = e(t)
where β > 0 is a constant. Gao and Lu [2] studied the periodic solutions for the
p-Laplacian Rayleigh differential equation with a delay,
φp (x0 ))0 + f (x0 (t)) + β(t)g(x(t − τ (t))) = e(t) .
In 2007, Cheung and Ren [1] discussed the solvability of periodic problems for the
Lienard-type p-Laplacian delay differential equation
φp (x0 ))0 + f (x(t))x0 (t) + g(t, x(t − τ (t))) = e(t) ,
under the assumption
lim
|x|→∞
|g(x)|
= r ≥ 0.
|x|p−1
Motivated by the above works, we will present the existence of periodic solutions for
(1.1) by using Mawhin’s continuation theorem. Our main results are different from
those results in the literature. For instance, in our study we allow the growth rate
of g1 (u), with respect to u, to be greater than p − 1. Also we allow the coefficient
β(t) to change sign R.
2. Preliminaries
For simplicity, we use the following symbols in this article
CT = {x ∈ C(R, R) : x(t + T ) = x(t)},
CT1
|x|∞ = max |x(t)|,
t∈[0,T ]
= {x ∈ C (R, R) : x(t + T ) = x(t)}, kxk = max{|x|∞ , |x0 |∞ },
(
Z T
1/p
1,
0 < p ≤ 1,
|x|p =
|x(t)|p dt
, Dp =
p−1
2 , p > 1.
0
1
To state our main results, we introduce several technical lemmas.
Lemma 2.1 ([6]). Assume that Ω is an open bounded set in CT1 such that the
following three conditions hold:
(1) For each λ ∈ (0, 1), the equation
(φp (x00 ))00 = λf (t, x(t), x(t − µ(t)), x0 (t)),
(2.1)
has no T -periodic solution on ∂Ω, where f (t, x, y, z) ∈ C(R4 , R) and f (t +
T, ·, ·, ·)) = f (t, ·, ·, ·)).
(2) The equation
Z T
1
F (a) =
f (t, a, a, 0)dt = 0,
2π 0
has no solution on ∂Ω ∩ R.
(3) The Brouwer degree satisfies degB (F, Ω ∩ R, 0) 6= 0.
Then (2.1) has a T -periodic solution in Ω̄ when λ = 1.
EJDE-2013/205
PERIODIC SOLUTIONS
3
Lemma 2.2 ([11]). If ω(t) ∈ c1 (R, R) and ω(0) = ω(T ) = 0, then there holds
Z T
Z
T p T 0 p
|ω(t)|p dt ≤
|ω (t)| dt,
πp
0
0
where
(p−1)/p
Z
πp =
0
ds
2π(p − 1)1/p
=
.
p sin(π/p)
(1 − (p − 1)−1 sp )1/p
Lemma 2.3 ([4]). Let a, b, p > 0, then there holds
(a + b)p ≤ Dp (ap + bp ).
For the sake of convenience, we list the following assumptions which will be used
frequently in Section 3.
(H1) For i = 1, 2, there are positive constants ri , ri∗ , mi with m2 ≤ p − 1 and
m1 > p − 1 such that for |u| > 1 there hold
(i) r1 |u|m1 ≤ |g1 (u)| ≤ r2 |u|m1 and r1∗ |u|m2 ≤ |g2 (u)| ≤ r2∗ |u|m2 .
(ii) ugi (u) > 0.
r2∗ β̄ 1/m1
)
< 1.
(H2) A = D m1 ( ᾱr
1
1
(H3) There are constants γ, r3 > 0 and k0 ∈ Z such that m1 = r3 + p − 1 and
0 ≤ τ1 (t, x(t)) − k0 T ≤ max{
where q > 1 :
1
p
+
1
q
γq
, T },
1 + |x|r∞3 q
∀t ∈ [0, T ], x(t) ∈ C[0, T ].
=1
3. Main Results
Theorem 3.1. Suppose that (H1)–(H3). Then (1.1) has at least one T -periodic
solution if one of the following two conditions holds
(1) m2 = p − 1, ∆1 + ∆2 < 1,
(2) m2 < p − 1, ∆1 < 1,
where
m2 +1
p−1
r∗ βDm +1 T q
T m2 +1
∆2 = m2 2 +1 2
.
m
+1
2
2
(1 − A)
πp
RT
Proof. Without loss of generality, we assume α(t) ≥ 0, t ∈ R, 0 α(t)dt > 0, and
RT
β(t)dt > 0. Consider the homotopy equation
0
Dp−1 αr2 T q γ T p
,
∆1 = p−1
2 (1 − A)p−1 πp
φp (x00 ))00 + λf (x(t))x0 (t) + λα(t)g1 (x(t − τ1 (t, x(t))))
+ λβ(t)g2 (x(t − τ2 (t, x(t)))) = λp(t).
(3.1)
Suppose that x(t) is an arbitrary T -periodic solution of (3.1). Integrating both
sides of equation (3.1) on [0, T ] we obtain
Z T
Z T
α(t)g1 (x(t − τ1 (t, x(t))))dt = −
β(t)g2 (x(t − τ2 (t, x(t))))dt.
0
0
Applying the mean value theorem, then there exists a constant ξ ∈ [0, T ] such that
Z T
Z T
g1 (x(ξ − τ1 (ξ, x(ξ))))
α(t)dt = −
β(t)g2 (x(t − τ2 (t, x(t))))dt.
(3.2)
0
0
4
J. LIU, W. LIU, B. LIU
EJDE-2013/205
Now, we claim that the inequality
|x(ξ − τ1 (ξ, x(ξ)))| ≤ A|x|∞ + B
(3.3)
holds, where
r∗ β̄
A = D m1 ( 2 )1/m1 ,
1 ᾱr1
Z T
Z
ᾱ =
α(t)dt, β̄ =
0
B = D m1 (
Mg2 β̄ 1/m1
)
+ 1,
ᾱr1
|β(t)|dt,
Mg2 = max |g2 (u)|.
1
T
|u|≤1
0
In fact, if |x(ξ −τ1 (ξ, x(ξ)))| ≤ 1, then inequality (3.3) holds. If |x(ξ −τ1 (ξ, x(ξ)))| >
1, we define
E1 = {t ∈ [0, T ] : |x(t − τ1 (t, x(t)))| ≤ 1},
E2 = {t ∈ [0, T ] : |x(t − τ1 (t, x(t)))| > 1}.
It follows from (H1)(i) that
ᾱr1 |x(ξ − τ1 (ξ, x(ξ)))|m1 ≤
Z
0
Z
=
≤
T
β(t)g2 (x(t − τ2 (t, x(t))))dt
Z
+
β(t)g2 (x(t − τ2 (t, x(t))))dt
E1
E2
∗
m2
r2 β̄|x|∞ +
Mg2 β̄.
This implies that
1
1/m1
2
(r∗ β̄|x|m
∞ + Mg2 β̄)]
ᾱr1 2
m2
Mg β̄
r∗ β̄
m1
+ ( 2 )1/m1 ]
≤ D m1 [( 2 )1/m1 |x|∞
1
ᾱr1
ᾱr1
r2∗ β̄ 1/m1
Mg β̄
≤ D m1 (
)
|x|∞ + D m1 ( 2 )1/m1 .
1 ᾱr1
1
ᾱr1
|x(ξ − τ1 (ξ, x(ξ)))| ≤ [
Thus, it can be easily seen that (3.3) holds. Let
¯
ξ − τ1 (ξ, x(ξ)) = kT + ξ,
(3.4)
where k is an integer and ξ¯ ∈ [0, T ], thus we have
¯ = x(ξ).
¯
x(ξ − τ1 (ξ, x(ξ))) = x(kT + ξ)
Noting that
¯ +1
|x(t)| ≤ |x(ξ)|
2
Z
T
|x0 (s)|ds,
0
we have
|x|∞ = max |x(t)| ≤ A|x|∞ + B +
t∈[0,T ]
1
2
Z
T
|x0 (s)|ds,
0
which yields
RT
|x|∞ ≤
|x0 (s)|ds
B
+
.
2(1 − A)
1−A
0
(3.5)
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PERIODIC SOLUTIONS
5
On the other hand, multiplying both sides of (3.1) by x(t), and integrating on
[0, T ], we obtain
Z T
Z T
Z T
00
p
0
|x (t)| dt = −λ
f (x(t))x (t)x(t)dt − λ
α(t)g1 (x(t − τ1 (t, x(t))))x(t)dt
0
0
0
T
Z
−λ
Z
β(t)g2 (x(t − τ2 (t, x(t))))x(t)dt + λ
0
T
Z
≤λ
T
p(t)x(t)dt
0
α(t)|g1 (x(t − τ1 (t, x(t))))| |x(t) − x(t − τ1 (t, x(t)))|dt
0
T
Z
−λ
Z
α(t)g1 (x(t − τ1 (t, x(t))))x(t − τ1 (t, x(t)))dt
0
T
|β(t)g2 (x(t − τ2 (t, x(t))))x(t)|dt + p̄|x|∞ ,
+
0
(3.6)
RT
where p̄ = 0 |p(t)|dt.
By the condition (H1)(ii), we have
Z T
−λ
α(t)g1 (x(t − τ1 (t, x(t))))x(t − τ1 (t, x(t)))dt
0
Z
Z
= −λ
−λ
α(t)g1 (x(t − τ1 (t, x(t))))x(t − τ1 (t, x(t)))dt
E1
E2
Z
≤
α(t)|g1 (x(t − τ1 (t, x(t))))x(t − τ1 (t, x(t)))|dt
(3.7)
E1
≤ αMg1 ,
where Mg1 = max|u|≤1 |g1 (u)|. Using the condition (H1) again, we obtain
Z T
α(t)|g1 (x(t − τ1 (t, x(t))))||x(t) − x(t − τ1 (t, x(t)))|dt
0
Z
Z
=
+
α(t)|g1 (x(t − τ1 (t, x(t))))||x(t) − x(t − τ1 (t, x(t)))|dt
E1
E2
1
≤ αMg1 + αMg1 |x|∞ + αr2 max |x(t) − x(t − τ1 (t, x(t)))| × |x|m
∞ ,
t∈[0,T ]
and
Z
T
m2 +1
|β(t)||g2 (x(t − τ2 (t, x(t))))x(t)|dt ≤ βr2∗ |x|∞
+ βMg2 |x|∞ ,
0
RT
where Mg2 = max|u|≤1 |g2 (u)| and β = 0 |β(t)|dt. So (3.6) yields
Z T
∗
m2 +1
1
|x00 (t)|p dt ≤ αr2 max |x(t) − x(t − τ1 (t, x(t)))| × |x|m
∞ + βr2 |x|∞
0
t∈[0,T ]
+ (αMg1 + βMg2 + p)|x|∞ + 2αMg1
∗
m2 +1
1
= αr2 max |x(t) − x(t − τ1 (t, x(t)))| × |x|m
∞ + βr2 |x|∞
t∈[0,T ]
+ θ|x|∞ + K,
where θ = αMg1 + βMg2 + p and K = 2αMg1 .
(3.8)
6
J. LIU, W. LIU, B. LIU
EJDE-2013/205
Since x(0) = x(T ), there exists a constant ζ ∈ [0, T ] such that x0 (ζ) = 0. Let
ω(t) = x0 (t + ζ), then ω(0) = ω(T ) = 0. By Lemma 2.2, we have
Z T
Z
T p T 00 p |x0 (t)|p dt ≤
|x (t)| dt .
πp
0
0
From (H3) and Hölder’s inequality, we have
max |x(t) − x(t − τ1 (t, x(t)))|
t∈[0,T ]
= max |x(t) − x(t − τ1 (t, x(t)) + k0 T )|
t∈[0,T ]
Z
t
x0 (s)ds|
= max |
t∈[0,T ]
(3.9)
t−τ1 (t,x(t))+k0 T
≤ max |τ1 (t, x(t)) − k0 T |1/q
t∈[0,T ]
Z
t
1/p
|x0 (s)|p ds
t−τ1 (t,x(t))+k0 T
≤ max |τ1 (t, x(t)) − k0 T |1/q
∞
t∈[0,T ]
Z
T
1/p
|x0 (s)|p ds
.
0
Moreover, from (3.5) and by Hölder’s inequality, we have
RT 0
|x (s)|ds
B m2 +1
∗
0
2 +1
≤
r
β[
+
]
r2∗ β|x|m
∞
2
2(1 − A)
1−A
Z T
m2 +1 r∗ βD
m2 +1
r2∗ βDm2 +1
m2 +1 B
2
0
≤ m2 +1
+
|x
(s)|ds
2
(1 − A)m2 +1 0
(1 − A)m2 +1
m2 +1
Z
m2p+1
r2∗ βDm2 +1 T q
T m2 +1 T 00
p
≤ m2 +1
|x (s)| ds
2
(1 − A)m2 +1 πp
0
+
r2∗ βDm2 +1 B m2 +1
,
(1 − A)m2 +1
(3.10)
and
RT
|x0 (s)|ds
B +
2(1 − A)
1−A
(3.11)
Z T
1/p
1/q
θT
T θB
00
p
≤
.
|x (s)| ds
+
2(1 − A) πp
1−A
0
By of m1 = r3 + p − 1 and the condition (H3), and combining (3.9)-(3.11), we have
Z T
|x00 (t)|p dt
θ|x|∞ ≤ θ
0
0
p−1
m2 +1
≤ αr2 max |x(t) − x(t − τ1 (t, x(t)))||x|r∞3 |x|∞
+ βr2∗ |x|∞
+ θ|x|∞ + K
t∈[0,T ]
Z
RT
T
0
≤ αr2 γ(
p
1/p
|x (s)| ds)
0
Z
≤ αr2 γ(
T
|x0 (s)|p ds)1/p
0
+ αr2 γDp−1
Z
0
T
|x0 (s)|ds
B p−1
2 +1
+
]
+ βr2∗ |x|m
+ θ|x|∞ + K
∞
2(1 − A)
1−A
R
p−1
T
Dp−1 0 |x0 (s)|ds
[
0
2p−1 (1 − A)p−1
1/p B p−1
|x0 (s)|p ds
(1 − A)p−1
EJDE-2013/205
PERIODIC SOLUTIONS
T
+
θT 1/q
T 2(1 − A) πp
+
2
r2∗ βDm2 +1 T q
2m2 +1 (1 − A)m2 +1
Z
7
1/p
|x00 (s)|p ds
0
m +1
Z
m2 +1
T m2 +1 T 00
(
|x (s)|p ds) p
πp
0
r2∗ βDm2 +1 B m2 +1
Bθ
+
+K
(1 − A)m2 +1
1−A
p−1
Z
D αr B p−1 γ Z T
1/p
Dp−1 αr2 T q γ T 0
p−1
2
p
0
p
|x
|x
(s)|
ds
+
(s)|
ds
≤ p−1
2 (1 − A)p−1 0
(1 − A)p−1
0
m2 +1
Z
T
∗
q
m
+1
r βDm +1 T
T m2 +1
2
|x00 (s)|p ds) p
(
+ m2 2 +1 2
2
(1 − A)m2 +1 πp
0
Z
1/p
θT 1/q
T T 00
|x (s)|p ds
+
+C
2(1 − A) πp
0
p−1
Z
Dp−1 αr2 T q γ T p T 00
p
≤ p−1
|x
(s)|
ds
2 (1 − A)p−1 πp
0
Z
1/p
Dp−1 αr2 B p−1 γ T T 00
p
+
|x
(s)|
ds
(1 − A)p−1
πp
0
m2 +1
Z
m2p+1
∗
T m2 +1 T 00
r βDm +1 T q
p
|x
(s)|
ds
+ m2 2 +1 2
2
(1 − A)m2 +1 πp
0
Z T
1/q
1/p
T
θT
+
|x00 (s)|p ds
+C
2(1 − A) πp
0
Z T
Z T
m2p+1
00
p
= ∆1
|x (s)| ds + ∆2
|x00 (s)|p ds
+
0
0
Z
1/p
Dp−1 αr2 B p−1 γ T T 00
p
+
|x
(s)|
ds
(1 − A)p−1
πp
0
Z T
1/p
1/q
θT
T
+
|x00 (s)|p ds
+ C,
2(1 − A) πp
0
(3.12)
where
C=
r2∗ βDm2 +1 B m2 +1
Bθ
+
+ K.
(1 − A)m2 +1
1−A
RT
If m2 = p − 1 and ∆1 + ∆2 < 1, then from (3.12) it follows that 0 |x00 (t)|p dt is
bounded. If m2 < p − 1 and ∆1 < 1, then from m2p+1 < 1 and (3.12) we see that
R T 00 p
|x (t)| dt is also bounded. Thus, there exists a constant M > 0 such that
0
Z T
1/p
|x00 (t)|p dt
≤ M,
0
which shows that there exist positive numbers M0 and M1 such that
|x|∞ ≤ M0 ,
|x0 |∞ ≤ M1 .
Let
Ω = {x(t) ∈ CT1 : ||x|| < ρ},
8
J. LIU, W. LIU, B. LIU
EJDE-2013/205
where ρ > max{1, M0 , M1 }. Then the homotopy equation (3.1) has no T -periodic
solution on ∂Ω. In addition,
Z T
Z T
Z T
1
p(t)dt]
β(t)g2 (ρ)dt −
F (ρ) = − [
α(t)g1 (ρ)dt +
T 0
0
0
Z T
Z T
1
1
= − g1 (ρ)
α(t)dt − g2 (ρ)
β(t)dt.
T
T
0
0
It means that the second condition of Lemma 2.1 is satisfied, and F (ρ)F (−ρ) < 0
from (H1)(ii). Consequently, from Lemma 2.1 the equation (1.1) has at least one
T -periodic solution in Ω .
RT
Remark 3.2. If we replace the conditions α(t) > 0, 0 β(t)dt > 0 with α(t) < 0,
RT
RT
RT
β(t)dt < 0 or α(t) < 0, 0 β(t)dt > 0 or α(t) > 0, 0 β(t)dt < 0, we can obtain
0
the same conclusion as Theorem 3.1.
Remark 3.3. Condition (H1) can be replaced by
(H1’) For i = 1, 2, there are positive constants ri , ri∗ , mi , d with m2 ≤ p − 1 and
m1 > p − 1 such that
(i) r1 |u|m1 ≤ |g1 (u)| ≤ r2 |u|m1 and r1∗ |u|m2 ≤ |g2 (u)| ≤ r2∗ |u|m2 for all
|u| > d ≥ 1,
(ii) gi (u)(sgn u) > 0 for all |u| > d ≥ 1;
while the conclusion of Theorem 3.1 is still true.
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Jiaying Liu
Department of Mathematics, China University of Mining and Technology, Xuzhou,
Jiangsu 221116, China
E-mail address:
relinaliu@163.com
EJDE-2013/205
PERIODIC SOLUTIONS
9
Wenbin Liu (corresponding author)
Department of Mathematics, China University of Mining and Technology, Xuzhou,
Jiangsu 221116, China
E-mail address: wblium@163.com
Bingzhuo Liu
Department of Mathematics, China University of Mining and Technology, Xuzhou,
Jiangsu 221116, China
E-mail address: tuteng3839@163.com