Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 140, pp. 1–8.
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 FUNCTIONAL
DIFFERENTIAL EQUATIONS WITH TWO DEVIATING
ARGUMENTS
CHANGXIU SONG, XUEJUN GAO
Abstract. Using the theory of coincidence degree, we prove the existence
of periodic solutions for the p-Laplacian functional differential equations with
deviating arguments.
1. Introduction
In recent years, the existence of periodic solutions for the Duffing equation,
Rayleigh equation and Liénard equation has received a lot of attention; see [1, 2, 4, 5,
6, 7, 8]. For example, Liu [5] studied periodic solutions for the p-Laplacian Liénard
equation with a deviating argument. Using Mawhin’s continuation theorem, some
results on the existence of periodic solution are obtained. But the p-Laplacian
Liénard equation with two deviating arguments has been studied far less often.
In this article, we study the existence of periodic solutions for the following
Liénard equation with two deviating arguments:
(φp (x0 (t)))0 + f (x(t))x0 (t) + g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) = e(t),
(1.1)
where f, τ1 , τ2 , e ∈ C(R, R); g1 , g2 ∈ C(R2 , R); τ1 (t), τ2 (t), g1 (t, x), g2 (t, x), e(t) are
periodic functions with period T ; φp (·) is the p-Laplacian operator, 1 < p < ∞.
By using the theory of coincidence degree, we obtain some results to guarantee the
existence of periodic solutions. Even for p = 2, the results in this paper are also
new.
In what follows, the Lp −norm in Lp ([0, T ], R) is defined by
Z T
|x(t)|p dt)1/p ,
kxkp = (
0
∞
∞
and the L −norm in L ([0, T ], R) is kxk∞ = maxt∈[0,T ] |x(t)|. Let the Sobolev
space W 1,p ([0, T ], R] be denoted by W .
2000 Mathematics Subject Classification. 34B15.
Key words and phrases. p-Laplacian operator; periodic solutions; coincidence degree;
deviating arguments.
c
2011
Texas State University - San Marcos.
Submitted March 5, 2011. Published October 27, 2011.
Supported by grants 10871052 and 109010600 NNSF of China, and by grant
10151009001000032 from NSF of Guangdong.
1
2
C. SONG, X. GAO
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Lemma 1.1 ([9]). Suppose u ∈ W and u(0) = u(T ) = 0. Then
kuk∞ ≤ (T /2)1/q ku0 kp .
The following Mawhin’s continuous theorem is useful in obtaining the existence
of T -periodic solutions of (1.1).
Lemma 1.2 ([3]). Let X and Y be two Banach spaces. Suppose that L : D(L) ∈
X → Y is a Fredholm operator with index zero and N : X → Y is L-compact on Ω,
where Ω is an open bounded subset of X. Moreover, assume that all the following
conditions are satisfied:
(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{JQN, Ω ∩ ker L, 0} =
6 0, where J : Im Q → ker L is an isomorphism,
then equation Lx = N x has a solution on Ω ∩ D(L).
2. Main results
To use coincidence degree theory in the study of T -periodic solutions for (1.1),
we rewrite (1.1) in the form
x0 (t) = φq (y(t))
y 0 (t) = −f (x(t))x0 (t) − g1 (t, x(t − τ1 (t))) − g2 (t, x(t − τ2 (t))) + e(t).
(2.1)
If z(t) = (x(t), y(t))T is a T -periodic solution of (2.1), then x(t) must be a T periodic solution of (1.1). Thus, the problem of finding a T -periodic solution for
(1.1) reduces to finding one for (2.1).
We set the following notation: T > 0 is a constant, CT = {x ∈ C(R, R) :
x(t + T ) ≡ x(t)} with the norm kxk∞ = maxt∈[0,T ] |x(t)|, X = Z = {z = (x, y) ∈
C(R, R2 ) : z(t) ≡ z(x + T )} with the norm kzk = max{kxk∞ , kyk∞ }. Clearly, X
and Z are Banach spaces. Also let L : Dom L ⊂ X → Z be defined by
0 x (t)
0
(Lz)(t) = z (t) =
,
y 0 (t)
and N : X → Z defined by
φq (y(t))
(N z)(t) =
−f (x(t))x0 (t) − g1 (t, x(t − τ1 (t))) − g2 (t, x(t − τ2 (t))) + e(t)
RT
It is easy to see that ker L = R2 , Im L = {z ∈ Z : 0 z(s)ds = 0}. So L is a
Fredholm operator with index zero. Let P : X → ker L and Q : Z → Im Q be
defined by
Z
Z
1 T
1 T
Pu =
u(s)ds, u ∈ X; Qv =
v(s)ds, v ∈ Z,
T 0
T 0
and let Kp denote the inverse of L|ker P ∩Dom L . Obviously, ker L = Im Q = R2 and
Z t
Z Z
1 T t
(Kp z)(t) =
z(s)ds −
z(s) ds dt.
(2.2)
T 0 0
0
From this equality, one can easily see that N is L-compact on Ω, where Ω is an
open bounded subset of X.
Theorem 2.1. Suppose that there exist constants d > 0 r1 ≥ 0 and r2 ≥ 0 such
that
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PERIODIC SOLUTIONS
3
(H1) g1 (t, u) + g2 (t, v) − e(t) > 0 for all t ∈ R, | max{u, v}| > d;
2 (t,x))|
1 (t,x)|
≤ r1 ; limx→−∞ supt∈[0,T ] |g|x|
≤ r2 .
(H2) limx→−∞ supt∈[0,T ] |g|x|
p−1
p−1
Then (1.1) has at least one T -periodic solution, if 4(r1 + r2 )T (T /2)p/q < 1.
Proof. Consider the parametric equation
(Lz)(t) = λ(N z)(t), λ ∈ (0, 1).
(2.3)
x(t)
Let z(t) =
be a possible T -periodic solution of (2.3) for some λ ∈ (0, 1).
y(t)
One can see x = x(t) is a T -periodic solution of the equation
(φp (x0 (t)))0 + λp−1 f (x(t))x0 (t) + λp g1 (t, x(t − τ1 (t))) + λp g2 (t, x(t − τ2 (t))) = λp e(t).
(2.4)
Integrating both sides of (2.4) over [0, T ], we have
Z T
[g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)]dt = 0,
(2.5)
0
which implies that there exists η ∈ [0, T ] such that
g1 (η, x(η − τ1 (η))) + g2 (η, x(η − τ2 (η))) − e(η) = 0.
From assumption (H1), we know that there exists ξ ∈ R such that |x(ξ)| ≤ d.
Let ξ = kT +t0 , where t0 ∈ [0, T ] and k is an integer. Let χ(t) = x(t+t0 )−x(t0 ).
Then χ(0) = χ(T ) = 0 and χ ∈ W 1,p ([0, T ], R). By Lemma 1.1, we have
T
T
kxk∞ ≤ kχk∞ + d ≤ ( )1/q kχ0 kp + d = ( )1/q kx0 kp + d.
2
2
On the other hand, multiplying the two sides of (2.4) by x(t) and integrating them
over [0, T ], we obtain
Z T
−kx0 kpp = −λp
x(t)[g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)]dt;
o
i.e.,
kx0 kpp
Z
≤ kxk∞
T
|g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)|dt.
(2.6)
0
From assumption (H2), there exists a constant ρ > 0 such that
|g1 (t, x)| ≤ r1 |x|p−1 ,
|g2 (t, x)| ≤ r2 |x|p−1 ,
∀t ∈ R, x < −ρ.
Let
E1 = {t ∈ [0, T ] : max{x(t − τ1 (t)), x(t − τ2 (t))} < −ρ},
E2 = {t ∈ [0, T ] : max{x(t − τ1 (t)), x(t − τ2 (t))} > ρ},
E3 = {t ∈ [0, T ] : | max{x(t − τ1 (t)), x(t − τ2 (t))}| ≤ ρ},
E4 = {t ∈ [0, T ] : −ρ ≤ x(t − τ1 (t)) ≤ ρ, −ρ ≤ x(t − τ2 (t)) ≤ ρ},
E5 = {t ∈ [0, T ] : x(t − τ1 (t)) < −ρ, −ρ ≤ x(t − τ2 (t)) ≤ ρ},
E6 = {t ∈ [0, T ] : −ρ ≤ x(t − τ1 (t)) ≤ ρ, x(t − τ2 (t)) < −ρ}.
By (2.4) it is easy to see that
Z T
[g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)]dt = 0.
0
(2.7)
4
C. SONG, X. GAO
EJDE-2011/140
Hence
Z
|g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)|dt
E2
Z
[g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)]dt
Z Z
=−
+
[g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)]dt
E1
E3
Z Z
≤
+
|g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)|dt.
=
E2
E1
E3
From the above inequality and (2.7) we obtain
Z T
|g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)|dt
0
Z Z
≤2
+
|g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)|dt
E1
E3
Z Z Z
Z
≤2
+
|g1 (t, x(t − τ1 (t)))|dt + 2
+
|g2 (t, x(t − τ2 (t)))|dt
E1
T
E3
E1
E3
Z
|e(t)|dt
+2
0
Z
|x(t − τ1 (t)|p−1 dt + 2r2
≤ 2r1
Z
E1
|x(t − τ2 (t)|p−1 dt
E1
Z
T
Z
(|g1 (t, x(t − τ1 (t)))| + |g2 (t, x(t − τ2 (t)))|)dt + 2
|e(t)|dt
E3
0
Z
Z
Z
≤ 2(r1 + r2 )T kxkp−1
+
+
|g1 (t, x(t − τ1 (t)))|
∞ +2
+2
E4
E5
Z
+ |g2 (t, x(t − τ2 (t)))| dt + 2
E6
T
|e(t)|dt
0
≤ 2(r1 +
r2 )T kxkp−1
∞
Z
|x(t − τ1 (t)|p−1 dt
+ 2T (g1ρ + g2ρ ) + 2r1
E5
Z
|x(t − τ2 (t)|p−1 dt + 2T g1ρ + 2
+ 2T g2ρ + 2r2
Z
E6
≤ 4(r1 +
r2 )T kxkp−1
∞
T
|e(t)|dt
0
Z
T
|e(t)|dt,
+ 4T (g1ρ + g2ρ ) + 2
0
where
g1ρ =
max
t∈[0,T ],|x|≤ρ
|g1 (t, x(t − τ1 (t)))|,
g2ρ =
max
t∈[0,T ],|x|≤ρ
|g2 (t, x(t − τ2 (t)))|.
From (2.6) and the above inequality, we have
Z T
kx0 kpp ≤ kxk∞
|g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)|dt
0
≤ kxk∞ [4(r1 +
r2 )T kxkp−1
∞
Z
T
|e(t)|dt]
+ 4T (g1ρ + g2ρ ) + 2
0
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PERIODIC SOLUTIONS
5
p
T
= 4(r1 + r2 )T ( )1/q kx0 kp + d + [4T (g1ρ + g2ρ )
2
Z T
T
+2
|e(t)|dt] ( )1/q kx0 kp + d .
2
0
Case (1): kx0 (t)k = 0, from (2.6) we see kxk∞ ≤ d.
Case (2): kx0 (t)k > 0, then we know that
d
T
T
T
[( )1/q kx0 kp + d]p = ( )p/q kx0 kpp [1 + ( )−1/q 0
]p .
2
2
2
kx (t)kp
From mathematical analysis, there is a constant δ > 0 such that
(1 + x)p < 1 + (1 + p)x,
If
d
( T2 )−1/q kx0 (t)k
>
p
T −1/q
d
If ( 2 )
kx0 (t)kp
δ, then we have kx0 kp <
∀x ∈ [0, δ].
(2.8)
( T2 )−1/q dδ .
≤ δ, by (2.8) we know that
T
T
T
[( )1/q kx0 kp + d]p ≤ ( )p/q kx0 (t)kpp + (p + 1)( )p−1)/q dkx0 (t)kpp−1 .
2
2
2
By (2.9), we obtain
(2.9)
T
T
kx0 kpp ≤ 4(r1 + r2 )T ( )p/q kx0 (t)kpp + (p + 1)( )p−1)/q dkx0 (t)kpp−1
2
2
Z T
T
+ (4T (g1ρ + g2ρ ) + 2
|e(t)|dt) ( )1/q kx0 kp + d .
2
0
As p > 1, 4(r1 + r2 )T ( T2 )p/q < 1, there exists a constant R2 > 0 such that kx0 kp ≤
R2 .
Let R1 = max{( T2 )−1/q dδ , R2 }. Then we have kxk∞ ≤ ( T2 )1/q R1 := R0 . By the
second equation of (2.1) we obtain
y 0 (t) = −f (x(t))x0 (t) − λg1 (t, x(t − τ1 (t))) − λg2 (t, x(t − τ2 (t))) + λe(t).
Hence
Z
T
0
Z
|y (t)|dt ≤ fR0
0
T
T
Z
0
|x (t)|dt + T g1R0 + T g2R0 +
|e(t)|dt
0
0
≤ fR0 T 1/q kx0 kp + T g1R0 + T g2R0 +
T
Z
|e(t)|dt
0
≤ fR0 T 1/q R1 + T g1R0 + T g2R0 +
Z
T
|e(t)|dt := R3 ,
0
where
fR0 = max |f (s)|,
|s|≤R0
g1R0 =
max
t∈[0,T ],s≤R0
|g1 (t, s)|,
g2R0 =
max
t∈[0,T ],s≤R0
|g2 (t, s)|.
RT
By the first equation of (2.1) we have 0 φq (y(t))dt = 0, which implies there exists
a constant t1 ∈ [0, T ] such that y(t1 ) = 0. So
Z t
Z T
0
|y(t)| = |
y (s)ds| ≤
|y 0 (s)|ds ≤ R3 ,
t1
and kyk∞ ≤ R3 .
0
6
C. SONG, X. GAO
EJDE-2011/140
Let R4 > max{R0 , R3 }, Ω = {z ∈ Z : kzk < R4 }, then Lz 6= λN Z, for all
z ∈ Dom L ∩ ∂Ω, λ ∈ (0, 1). Since
Z 1 T
φq (y(t))
QN z =
,
−f (x(t))x0 (t) − g1 (t, x(t − τ1 (t))) − g2 (t, x(t − τ2 (t))) + e(t)
T 0
for any z ∈ ker L ∩ ∂Ω, if QN z = 0, we obtain y = 0, |x| = R4 > d. But when
|x| = R4 , we know that −g1 (t, x) − g2 (t, x) + e(t) < 0, which yields a contradiction.
So conditions (1) and (2) of Lemma 1.2 is satisfied.
Define the isomorphism J : Im Q → ker L as follows:
J(x, y)T = (−y, x)T .
Let H(µ, z) = µx + (1 − µ)JQN z, (µ, z) ∈ [0, 1] × Ω, then we have
RT
µx + (1 − µ) T1 0 [g1 (t, x) + g2 (t, x) − e(t)]dt
H(µ, z) =
,
µy + (1 − µ)φq (y)
where (µ, z) ∈ [0, 1]×(ker L∩∂Ω). It is obvious that H(µ, z) = µx+(1−µ)JQN z 6=
0 for (µ, z) ∈ [0, 1] × (ker L ∩ ∂Ω). Hence
deg{JQN, Ω ∩ ker L, 0} = deg{I, Ω ∩ ker L, 0} = 1 6= 0.
So the condition (3) of Lemma 1.2 is satisfied. By applying Lemma 1.2, we conclude
that equation Lz = N z has a solution z(t) = (x(t), y(t))T ; i.e., (1.1) has a T periodic solution x(t).
Theorem 2.2. Suppose that there exist constants d > 0, r1 ≥ 0, and r2 ≥ 0 such
that (H1) holds and
(H2*) limx→+∞ supt∈[0,T ]
|g1 (t,x)|
|x|p−1
|g2 (t,x))|
|x|p−1
r2 )T ( T2 )p/q
≤ r1 ; limx→+∞ supt∈[0,T ]
Then (1.1) has at least one T -periodic solution, if 4(r1 +
≤ r2 .
< 1.
Theorem 2.3. Suppose that p > 2 and there exist constants d > 0, b1 ≥ 0, b2 ≥ 0
such that (H1) holds and
(H3) |gi (t, u) − gi (t, v)| ≤ bi |u − v| for all t, u, v ∈ R, i = 1, 2.
Then (1.1) has at least one T -periodic solution.
Proof. By the proof of Theorem 2.1, we have
T
T
kxk∞ ≤ kχk∞ + d ≤ ( )1/q kχ0 kp + d = ( )1/q kx0 kp + d,
2
2
and
Z
T
kx0 kpp ≤ kxk∞
|g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)|dt.
0
From assumption (H3), we have
Z T
|g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)|dt
0
Z
T
T
Z
|g1 (t, x(t − τ1 (t))) − g1 (t, 0)|dt +
≤
0
Z
+ |g2 (t, x(t − τ2 (t))) − g1 (t, 0)|dt +
|g1 (t, 0)|dt
0
T
Z
|g2 (t, 0)|dt +
0
Z
≤ b1
T
Z
|x(t − τ1 (t))|dt +
0
T
|e(t)|dt
0
Z
|g1 (t, 0)|dt + b2
0
T
T
|x(t − τ2 (t))|dt
0
EJDE-2011/140
PERIODIC SOLUTIONS
Z
T
Z
|g2 (t, 0)|dt +
+
0
7
T
|e(t)|dt
0
≤ T (b1 + b2 )kxk∞ + T b
where b = max{|g1 (t, 0)| + |g2 (t, 0)| + |e(t)|}. Thus,
kx0 kpp ≤ kxk∞ [T (b1 + b2 )kxk∞ + T b]
T
T
≤ T (b1 + b2 )[( )1/q kx0 kp + d]2 + T b( )1/q kx0 kp + T bd.
2
2
As p > 2, there exists a constant R2 > 0 such that kx0 kp ≤ R2 . The rest of the
proof is same to Theorem 2.1 and is omitted.
Corollary 2.4. Suppose that p = 2 and conditions (H1), (H3) hold. Then (1.1)
2
has at least one T -periodic solution, if T (b1 + b2 )( T2 ) q < 1.
Remark 2.5. If condition (H1) is replaced by
(H1*) g1 (t, u) + g2 (t, v) − e(t) < 0 for all t ∈ R, | max{u, v}| > d,
then the results in this article still hold.
Example 2.6. Consider the equation
(φ3 (x0 (t)))0 + ex(t) x0 (t) + g1 (t, x(t − sin t)) + g2 (t, x(t − cos t)) =
1
sin t,
π
(2.10)
where p = 3, T = 2π, τ1 (t) = sin t, τ2 (t) = cos t,
( 2
esin t x3 + π1 sin t,
x ≥ 0,
g1 (t, x) =
2
2
x
sin t
+ π1 sin t, x < 0,
18eπ 3 e
(
2
ecos t x3 ,
x ≥ 0,
g2 (t, x) =
2
2
x
cos t
, x < 0.
18eπ 3 e
By (2.10), we can get d = 1/10 (Actually, d can be an arbitrarily small positive),
r1 = r2 = 1/(18π 3 ), 4(r1 +r2 )T (T /2)p/q < 1 and check that (H1)–(H2) hold. Thus,
according to Theorem 2.1, equation (2.10) has at least one 2π-periodic solution.
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Changxiu Song
School of Applied Mathematics, Guangdong University of Technology, Guangzhou
510006, China
E-mail address: scx168@sohu.com
Xuejun Gao
School of Applied Mathematics, Guangdong University of Technology, Guangzhou
510006, China
E-mail address: gaoxxj@163.com