Electronic Journal of Differential Equations, Vol. 2009(2009), No. 17, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
PERIODIC SOLUTIONS OF NON-AUTONOMOUS SECOND
ORDER SYSTEMS WITH p-LAPLACIAN
ZHIYONG WANG, JIHUI ZHANG
Abstract. We prove the existence of periodic solutions for non-autonomous
second order systems with p-Laplacian. Our main tools are the minimax methods in critical point theory. Our results are new, even when p = 2.
1. Introduction
In this article concerns the existence of periodic solutions for the problem
d
(|u̇(t)|p−2 u̇(t)) = ∇F (t, u(t)) a.e. t ∈ [0, T ],
dt
(1.1)
u(0) − u(T ) = u̇(0) − u̇(T ) = 0,
where p > 1, T > 0, F : [0, T ] × RN → R satisfies the following assumption:
(A) F (t, x) is measurable in t for every x ∈ RN and continuously differentiable
in x for a.e. t ∈ [0, T ], and there exist a ∈ C(R+ , R+ ), b ∈ L1 (0, T ; R+ ),
such that
|F (t, x)| ≤ a(|x|)b(t),
|∇F (t, x)| ≤ a(|x|)b(t)
for all x ∈ R and a.e. t ∈ [0, T ].
Considerable attention has been paid to the periodic solutions of problem (1.1)
for p = 2, in recent years. The firsts to consider this problem when p = 2 were
Berger and Schechter [12] in 1977, they proved the existence of solutions to problem
(1.1) for p = 2 under the condition that F (t, x) → ±∞ as |x| → ∞ uniformly for
a.e. t ∈ [0, T ]. Subsequently, using the variational methods, many existence results
are obtained, we refer the readers to [1, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15] and the
references therein. However, there are few papers discussing periodic solutions for
second order systems with p-Laplacian. In [2], Tian has established the existence
results for problem (1.1) by the dual least action principle. Moreover, if problem
(1.1) with nonlinear boundary conditions, the existence of periodic solutions has
also been proved in [3] by means of the least action principle and the mountain
pass lemma.
For p = 2, under the assumptions that there exists h(t) ∈ L1 (0, T ; R+ ) such that
N
|∇F (t, x)| ≤ h(t)
2000 Mathematics Subject Classification. 34B15, 47J30, 58E05.
Key words and phrases. Periodic solutions; Hamiltonian system; p-Laplacian.
c
2009
Texas State University - San Marcos.
Submitted August 26, 2008. Published January 20, 2009.
1
2
Z. WANG, J. ZHANG
EJDE-2009/17
for all x ∈ RN and a.e. t ∈ [0, T ], and that
Z T
F (t, x)dt → ±∞
as |x| → ∞,
0
Mawhin and Willem in [1] have shown that problem (1.1) admitted a periodic
solution. After that, Tang in [4] generalized these results to the sublinear case.
Concertely speaking, it is assumed that the nonlinearity satisfied the following
restrictions
|∇F (t, x)| ≤ k(t)|x|α + p(t)
1
|x|2α
Z
for all x ∈ RN and a.e. t ∈ [0, T ],
(1.2)
T
F (t, x)dt → ±∞ as |x| → ∞,
(1.3)
0
here, k(t), p(t) ∈ L1 (0, T ; R+ ) and α ∈ [0, 1). Under these conditions, periodic
solutions of problem (1.1) with p = 2 have been obtained. In addition, Tang
in [5] first introduced the local α-coercive conditions, that is, there exists q(t) ∈
L1 (0, T ; R+ ) and a subset E of [0, T ] with meas(E) > 0 such that
F (t, x)
≤ q(t)
|x|2α
(1.4)
for all x ∈ RN and a.e. t ∈ [0, T ], and that
F (t, x)
→ −∞
|x|2α
as |x| → ∞
(1.5)
for a.e. t ∈ E, to deal with the generalized Josephson-type systems.
An interesting question naturally arises: In all the results [1, 4, 5] discussed
above, the nonlinearity is required to grow at infinity at most like |x|α , is it possible
to handle nonlinearity with faster increase at infinity and get the similar results of
[1, 4, 5]. In the present paper, we will focus on this problem and give a positive
answer. Here, we emphasize that our results are new even for p = 2.
We now state our main theorems.
Theorem 1.1. Suppose that F satisfies assumption (A) and the following conditions:
(H1) There exists a bounded nonincreasing positive function ω ∈ C((0, ∞); R+ )
with the properties:
ω(t)
(i) lim inf t→∞ ω(t
for some γ ∈ (0, 1),
γ) > 0
(ii) ω(t) → 0, ω(t)t → ∞ as t → ∞.
Moreover, there exist f ∈ L2 (0, T ; R+ ) and g ∈ L1 (0, T ; R+ ) such that
|∇F (t, x)| ≤ f (t)[ω(|x|)|x|]p−1 + g(t)
for all x ∈ RN and a.e. t ∈ [0, T ];
(H2) There exists a bounded non-increasing positive function ω ∈ C((0, ∞); R+ )
which satisfies the conditions (i), (ii) and
Z T
1
F (t, x)dx → −∞ as |x| → ∞.
[ω(|x|)|x|]p 0
EJDE-2009/17
PERIODIC SOLUTIONS
3
Then problem (1.1) has at least one solution in the set
WT1,p := u : [0, T ] → RN , u is absolutely continuous,
u(0) = u(T ), u̇ ∈ Lp (0, T ; RN ) .
This set is a Banach space with the norm
Z T
Z T
1/p
kuk :=
|u(t)|p dt +
|u̇(t)|p dt
0
0
for u ∈ WT1,p .
Remark 1.2. Obviously, Theorem 1.1 does not satisfy the corresponding conditions in [6, 7, 8, 9, 10, 11, 12, 13, 14, 15]. Furthermore, there are functions F (t, x)
that satisfy our Theorem 1.1 and do not satisfy the corresponding results in [1, 4, 5]
even for p = 2. For example, let
∇F (t, x) = −
|x|p−2 x
+ d(t),
[ln(2 + |x|2 )]p−1
∀t ∈ [0, T ], x ∈ RN ,
where d(t) ∈ L1 (0, T ; R). Let ω(|x|) = 1/ ln(2 + |x|2 ), γ = 1/2, a straightforward
computation shows that F (t, x) satisfies all the conditions of our Theorem 1.1.
However, it is clear that F (t, x) neither satisfies (1.2), (1.3) nor (1.4), (1.5) for any
α ∈ [0, 1) even for p = 2.
Theorem 1.3. Suppose that F satisfies assumption (A) and (H1). Moreover assume F satisfies the following conditions:
(H2*) (1) There exists a bounded nonincreasing and positive function ω in
C((0, ∞); R+ ) and constant C > 0 such that
ω(t)
(iii) lim supt→∞ ω(t
γ ) < +∞ for some γ ∈ (0, 1);
2
2
(iv) t ≤ Cω (t)t as t → ∞;
(2) There exists r(t) ∈ L1 (0, T ; R+ ) such that
F (t, x)
≤ r(t)
[ω(|x|)|x|]p
for all x ∈ RN and a.e. t ∈ [0, T ];
(3) There exists a subset E of [0, T ] with meas(E) > 0 such that for a.e.
t∈E
F (t, x)
→ −∞ as |x| → ∞.
[ω(|x|)|x|]p
Then problem (1.1) has at least one solution in WT1,p .
Remark 1.4. There are functions F (t, x) that satisfy our Theorem 1.3 and not do
not satisfy the corresponding results in [1, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15].
For example, let
F (t, x) = −θ(t)
|x|p
,
[ln(2 + |x|2 )]p−1
∀t ∈ [0, T ], x ∈ RN ,
where
(
sin 2πt
T ,
θ(t) =
0,
t ∈ [0, T /2],
t ∈ [T /2, T ].
Take ω(|x|) = 1/ ln(2 + |x|2 ), γ = 1/2, E = [T /6, T /4], by simple computation,
F (t, x) satisfies our Theorem 1.3. However, here F (t, x) also neither satisfies (1.2),
(1.3) nor (1.4), (1.5) for any α ∈ [0, 1) even for p = 2.
4
Z. WANG, J. ZHANG
EJDE-2009/17
2. Preliminaries
For convenience, we will denote various positive constants as Ci , i = 0, 1, 2, . . . .
Define functional ϕ on WT1,p by
Z
Z T
1 T
ϕ(u) =
|u̇(t)|p dt +
F (t, u(t))dt
(2.1)
p 0
0
for u ∈ WT1,p . It follows from assumption (A) that functional ϕ is continuously
differentiable on WT1,p , moreover, one has
Z T
0
[(|u̇(t)|p−2 u̇(t), v̇(t)) + (∇F (t, u(t)), v(t))]dt
(ϕ (u), v) =
0
WT1,p .
for all u, v ∈
It is well known that the solutions to problem (1.1) correspond
to the critical points of the functional ϕ.
RT
For u ∈ WT1,p , let ū := T1 0 u(t)dt and ũ(t) := u(t) − ū, then we have
kũk∞ ≤ C0 ku̇kLp
(Sobolev’s inequality),
kũkLp ≤ C0 ku̇kLp
(Wirtinger’s inequality),
where kũk∞ := max0≤t≤T |ũ(t)|.
To proof of our main theorems, we need the following auxiliary results.
Lemma 2.1 ([6]). Suppose that G satisfies assumption (A) and E is a measurable
subset of [0, T ]. Assume that
G(t, x) → +∞
as |x| → ∞
for a.e. t ∈ E. Then for every δ > 0, there exists subset Eδ of E with meas(E\Eδ ) <
δ such that
G(t, x) → +∞ as |x| → ∞
uniformly for all t ∈ Eδ .
Lemma 2.2 ([5]). For every constant β > 0, there exists a constant mβ > 0 such
that
meas{t ∈ [0, T ]||ū| < mβ kūk} < β for all ū 6= 0.
The proof of the above lemma follows from the first part in [5, Lemma 3], we
omit the details.
Lemma 2.3. Suppose that (H1) holds, then there exists a non-increasing positive
function ω̃(t) ∈ C((0, ∞); R+ ) which satisfies the following conditions:
(a) ω̃(t) → 0, ω̃(t)t → ∞ as t → ∞;
(b) k∇F (t, u(t))kL1 ≤ kf kL2 [ω̃(kuk)kuk]p−1 + kgkL1 ;
(c) If (H2) holds, then
Z T
1
F (t, ū)dt → −∞ as kūk → ∞;
[ω̃(kūk)kūk]p 0
kũk
< +∞, then
(d) If (H2*) holds, moreover assume that lim supkuk→∞ ω̃(kuk)kuk
Z T
1
F (t, u(t))dt → −∞ as kuk → ∞.
[ω̃(kuk)kuk]p 0
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PERIODIC SOLUTIONS
5
Proof. Let A := {t ∈ [0, T ]||u(t)| ≥ (T −1/p kuk)1/2 }, for u(t) 6= 0, by (H1 ) and the
Hölder’s inequality, we have
k∇F (t, u(t))kL1
Z T
[f (t)(ω(|u(t)|)|u(t)|)p−1 + g(t)]dt
≤
0
≤ kf kL2
≤ kf kL2
T
hZ
[ω(|u(t)|)|u(t)|]2(p−1) dt
0
hZ
[ω(|u(t)|)|u(t)|]2(p−1) dt +
A
≤ kf kL2
T
hZ
i1/2
+ kgkL1
Z
[ω(|u(t)|)|u(t)|]2(p−1) dt
i1/2
+ kgkL1
[0,T ]\A
2(p−1)
ω((T −1/p kuk)1/2 )|u(t)|
dt
0
+ (sup ω(s))2(p−1)
s>0
Z
T
(T −1/p kuk)p−1 dt
i1/2
+ kgkL1
0
1/2
2(p−1)
≤ kf kL2 C1 ω((T −1/p kuk)1/2 )kuk
+ C2 kukp−1
+ kgkL1
h i p−1
2
+ kgkL1 .
≤ kf kL2 C3 ω 2 ((T −1/p kuk)1/2 )kuk2 + kuk
1/2
Take ω̃(t) := C3 ω 2 ((T −1/p t)1/2 ) + 1t
, t > 0, then
k∇F (t, u(t))kL1 ≤ kf kL2 [ω̃(kuk)kuk]p−1 + kgkL1 .
Obviously, ω̃(t) satisfies (a) due to the properties of ω(t).
Next, we come to check condition (c). Note that lim inf t→∞ ω(t)/ω(tγ ) > 0 for
γ ∈ (0, 1), we define
ξ := lim inf
t→∞
ω 2 (T −1/p t)
ω 2 ((T −1/p t)1/2 )
> 0,
if (H2) holds, for any M > 0, we get
Z
T
F (t, u(t))dt ≤ −M [ω(|u(t)|)|u(t)|]p + C4 ,
(2.2)
0
which implies that for ū 6= 0
p
F (t, ū)dt
−M ω(|ū|)|ū| + C4
p ≤ p/2
[ω̃(kūk)kūk]
C3 ω 2 (T −1/p kūk)1/2 kūk2 + kūk
p
−M ω(T −1/p kūk)T −1/p kūk + C4
=
p/2 .
C3 ω 2 (T −1/p kūk)1/2 kūk2 + kūk
RT
0
(2.3)
By the definition of ξ, there exists R > 0 such that
ω2
ω 2 (T −1/p t)t2
ξ
≥
2
(T −1/p t)1/2 t2 + t
as t ≥ R.
(2.4)
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Z. WANG, J. ZHANG
EJDE-2009/17
Therefore,
Z T
1
F (t, ū)dt
[ω̃(kūk)kūk]p 0
p/2
−M T −1 2ξ ω 2 ((T −1/p kūk)1/2 )kūk2 + kūk
+ C4
≤
p/2
2
−1/p
1/2
2
C3 ω ((T
kūk) )kūk + kūk
≤ −C5 M as kūk → ∞,
(2.5)
condition (c) holds.
Finally, we show that (d) is true. From Lemma 2.2, for all ū 6= 0, we have
meas t ∈ [0, T ]|ω(|ū|)|ū| < mβ ω(|ū|)kūk = mβ ω(T −1/p kūk)kūk < β.
Consequently, for all ū 6= 0, let B := {t ∈ [0, T ]|ω(|ū|)|ū| ≥ mβ ω(T −1/p kūk)kūk},
then we have meas([0, T ]\B) < β.
By (H2*)(3) and Lemma 2.1, there exists subset Eδ of E with meas(E\Eδ ) < δ
such that
F (t, x)
→ −∞ as |x| → ∞
(2.6)
[ω(|x|)|x|]p
uniformly for all t ∈ Eδ . Then, we find
meas(B ∩ Eδ ) ≥ meas(Eδ ) − meas([0, T ]\B) ≥ meas(E) − δ − β > 0
(2.7)
for δ and β small enough. By (2.6), for every η > 0, there exists L > 0 such that
F (t, x)
≤ −η
[ω(|x|)|x|]p
for all |x| ≥ L and a.e. t ∈ Eδ . Furthermore, applying assumption (H2∗ )(2) yields
F (t, x) ≤ −η[ω(|x|)|x|]p + r∗ (t)
for all x ∈ RN and a.e. t ∈ Eδ , where r∗ (t) := (sups>0 ω(s))p Lp r(t). Noting the
ω(t)
ω(t)
definition of ω̃(t), 0 < lim inf t→∞ ω(t
γ ) ≤ lim supt→∞ ω(tγ ) < +∞ and assumption
(H2∗ )(1)(iv), we deduce that there exist C6 , C7 > 0 such that
C6 ω(T −1/p kūk)kūk ≤ ω̃(kūk)kūk
= [C3 (ω 2 ((T −1/p kūk)1/2 )kūk2 + kūk)]1/2
≤ C7 ω(T −1/p kūk)kūk as kūk → ∞,
which implies
Z
Z
Z
p
F (t, ū)dt ≤ −η
[ω(|ū|)|ū|] dt +
r∗ (t)dt
B∩Eδ
B∩Eδ
B∩Eδ
Z
Z
−1/p
p
≤ −η
[ω(T
kūk)mβ kūk] dt +
r∗ (t)dt
B∩Eδ
B∩Eδ
Z
Z
mβ p
p
(ω̃(kūk)kūk) dt +
r∗ (t)dt
≤ −η
C
7
B∩Eδ
B∩Eδ
Z
p
∗
≤ −ηC8 [ω̃(kūk)kūk] +
r (t)dt as kūk → ∞,
B∩Eδ
(2.8)
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PERIODIC SOLUTIONS
7
and
Z
Z
r(t)[ω(|ū|)|ū|]p dt
F (t, ū)dt ≤
[0,T ]\(B∩Eδ )
[0,T ]\(B∩Eδ )
T
Z
r(t)[ω(T −1/p kūk)T −1/p kūk]p dt
≤
0
T
T −1
≤
(ω̃(kūk)kūk)p r(t)dt
C6
0
Z T
r(t)dt as kūk → ∞.
≤ C9 [ω̃(kūk)kūk]p
Z
(2.9)
0
So, it follows from (2.8) and (2.9) that
Z T
Z T
1
lim sup
F (t, ū)dt ≤ −ηC8 + C9
r(t)dt,
p
kūk→∞ [ω̃(kūk)kūk]
0
0
which implies that
1
[ω̃(kūk)kūk]p
Z
T
F (t, ū)dt → −∞ as kūk → ∞.
(2.10)
0
On the other hand, from (b) and the assumptions of (d), we get
Z
Z T
Z TZ 1
T
F (t, u(t))dt −
F (t, ū)dt ≤
|∇F (t, ū + sũ)||ũ| ds dt
0
0
0
0
≤ kũk∞ [kf kL2 (ω̃(kuk)kuk)p−1 + kgkL1 ]
(2.11)
≤ C10 [ω̃(kuk)kuk]p−1 kũk + C11 kũk.
As a consequence, note lim supkuk→∞
C12 := lim sup kuk→∞
h
1
p
[ω̃(kuk)kuk]
Z
kũk
ω̃(kuk)kuk
< +∞, we then have
T
Z
F (t, u(t))dt −
0
0
T
i
F (t, ū)dt < +∞.
(2.12)
In addition, for kuk → ∞, one knows
1=
kūk + kũk
kūk
kũk
kūk
kuk
=
=
+
· ω̃(kuk) =
.
kuk
kuk
kuk ω̃(kuk)kuk
kuk
This, in conjunction with (2.10)-(2.12), gives
Z T
1
lim sup
F (t, u(t))dt
p
kuk→∞ [ω̃(kuk)kuk]
0
Z T
1
≤ lim sup
F (t, ū)dt + C12
p
kuk→∞ [ω̃(kuk)kuk]
0
Z T
1
F (t, ū)dt + C12 → −∞,
= lim sup
p
kūk→∞ [ω̃(kūk)kūk]
0
which completes the proof.
(2.13)
(2.14)
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Z. WANG, J. ZHANG
EJDE-2009/17
3. Proofs of theorems
Now, we give the proofs of the main results.
Proof of Theorem 1.1. First, we prove that ϕ satisfies the (PS) condition. Suppose
that {un } is a (PS) sequence for ϕ, that is, ϕ0 (un ) → 0 as n → ∞ and {ϕ(un )} is
bounded. It follows from Wirtinger’s inequality that
ku̇n kLp ≤ kũn k ≤ (C0 + 1)1/p ku̇n kLp
(3.1)
for all n. By virtue of the properties of ω̃(t), one has
ω̃(kū + ũk) ≤ min{ω̃(kūk), ω̃(kũk)},
(3.2)
which implies
Z
T
(∇F (t, un (t)), ũn (t))dt
0
≤ kũn k∞ kf kL2 (ω̃(kun k)kun k)p−1 + kgkL1
≤ kũn k∞ kf kL2 (ω̃(kūn + ũn k)kūn + ũn k)p−1 + kgkL1
≤ kũn k∞ kf kL2 (ω̃(kūn k)kūn k + ω̃(kũn k)kũn k)p−1 + kgkL1
p−1
≤ C13 kũn k [ω̃(kūn k)kūn k]
(3.3)
p−1
+ C13 kũn k [ω̃(kũn k)kũn k]
+ C14 kũn k
for all n. Thus, by (3.1) and (3.3), we get
kũn k ≥ (ϕ0 (un ), ũn )
Z T
Z
=
|u̇n (t)|p dt +
0
T
(∇F (t, un (t)), ũn (t))dt
0
(3.4)
p−1
≥ C15 kũn kp − C13 kũn k [ω̃(kūn k)kūn k]
p−1
− C13 kũn k [ω̃(kũn k)kũn k]
− C14 kũn k.
Assume that {kũn k} is unbounded; that is, kũn k → ∞ as n → ∞. Since ω̃(t) → 0
as t → ∞, it follows from (3.4) that we can find a constant C16 > 0 such that
C16 ω̃(kūn k)kūn k ≥ kũn k,
(3.5)
which implies
[C3 (ω 2 ((T −1/p kūn k)1/2 )kūn k2 + kūn k)]1/2 = ω̃(kūn k)kūn k → ∞ as n → ∞.
Since ω is bounded, this leads to
kūn k → ∞
as n → ∞.
(3.6)
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PERIODIC SOLUTIONS
9
On the other hand, kūn + sũn k ≥ kūn k, s ∈ [0, 1], by Lemma 2.3 (b) and (3.5), we
see that
Z T
[F (t, un (t)) − F (t, ūn )]dt
0
≤
Z
0
T
1
Z
(∇F (t, ūn + sũn (t)), ũn (t)) ds dt
0
Z
T
Z
1
|∇F (t, ūn + sũn (t))| ds dt
≤ kũn k∞
0
0
p−1
≤ kf kL2 [ω̃(kūn + sũn k)kūn + sũn k]
kũn k∞ + kgkL1 kũn k∞
p−1
≤ kf kL2 [ω̃(kūn k)kūn k + ω̃(kūn k)kũn k]
p−1
≤ C17 kũn k[ω̃(kūn k)kūn k]
kũn k∞ + kgkL1 kũn k∞
+ C17 kũn k[ω̃(kūn k)kũn k]p−1 + C18 kũn k
≤ C19 [ω̃(kūn k)kūn k]p + C20 [ω̃(kūn k)kūn k]p [ω̃(kūn k]p−1 + C21 ω̃(kūn k)kūn k,
(3.7)
which implies that
Z T
Z T
1
p
[F (t, un (t)) − F (t, ūn )]dt +
F (t, ūn )dt
ϕ(un ) = ku̇n kLp +
p
0
0
≤ C22 [ω̃(kūn k)kūn k]p + C20 [ω̃(kūn k)kūn k]p [ω̃(kūn k]p−1
Z T
+ C21 ω̃(kūn k)kūn k +
F (t, ūn )dt
(3.8)
0
h
≤ [ω̃(kūn k)kūn k]p C22 + C20 [ω̃(kūn k]p−1 +
1
+
[ω̃(kūn k)kūn k]p
Z
T
C21
[ω̃(kūn k)kūn k]p−1
i
F (t, ūn )dt → −∞ as kūn k → ∞.
0
This contradicts the boundedness of ϕ(un ). So
{kũn k} is bounded.
(3.9)
Suppose that {kūn k} is unbounded and {kũn k} is bounded. With the similar
manner above, we deduce that
Z T
Z T
1
p
ϕ(un ) = ku̇n kLp +
[F (t, un (t)) − F (t, ūn )]dt +
F (t, ūn )dt
p
0
0
Z T
≤ C23 kũn kp + C17 kũn k[ω̃(kūn k)kũn k]p−1 + C18 kũn k +
F (t, ūn )dt
0
≤ C24 + C25 [ω̃(kūn k)kũn k]p−1 + C26 [ω̃(kūn k)]p−1 +
Z
T
F (t, ūn )dt
0
C26
C24
C25
+
+
p
[ω̃(kūn k)kūn k]
ω̃(kūn k)kūn k ω̃(kūn k)kūn kp
Z T
i
1
+
F (t, ūn )dt → −∞ as kūn k → ∞,
[ω̃(kūn k)kūn k]p 0
(3.10)
which also contradicts the boundedness of ϕ(un ). Then
≤ [ω̃(kūn k)kūn k]p
h
{kūn k} is also bounded.
(3.11)
10
Z. WANG, J. ZHANG
EJDE-2009/17
From (3.9) and (3.11), we have {kun k} is bounded, thus ϕ satisfies the (PS) condition.
RT
Since WT1,p = RN ⊕ W̃T1,p , where W̃T1,p := {u ∈ WT1,p | 0 u(t)dt = 0}. Next, we
shall prove that
ϕ(u) → +∞ as kuk → ∞ in W̃T1,p .
(3.12)
1
.
In fact, since ω̃(t) → 0 as t → ∞, then there exists A > 0 such that ω̃(A) ≤ 2pC
29
In a similar way to (3.7), we have
Z
T
[F (t, u(t)) − F (t, A)]dt
0
≤ C17 kũk[ω̃(A)A]p−1 + C17 kũk[ω̃(A)kũk]p−1 + C18 kũk
≤ C27 kũk + C17 ω̃(A)kũkp
≤ C28 ku̇kLp + C29 ω̃(A)ku̇kpLp
1
≤ C28 ku̇kLp + ku̇kpLp ,
2p
which implies
1
ϕ(u) = ku̇kpLp +
p
Z
T
Z
T
[F (t, u(t)) − F (t, A)]dt −
0
F (t, A)dt
0
1
1
≥ ku̇kpLp − ku̇kpLp − C28 ku̇kLp −
p
2p
T
Z
(3.13)
F (t, A)dt
0
for all u ∈ W̃T1,p . By Wirtinger’s inequality, one has
kuk → ∞ ⇔ ku̇kLp → ∞ on W̃T1,p .
Hence, (3.12) follows from (3.13). On the other hand, by (H2),
ϕ(u) → −∞
as |u| → ∞
in RN .
(3.14)
Combine (3.12) and (3.14), applying saddle point theorem, then problem (1.1) has
at least one solution in WT1,p .
Proof of Theorem 1.3. We commence by showing that ϕ satisfies (PS) condition.
Let {un } be a sequence in WT1,p such that {ϕ(un )} is bounded and ϕ0 (un ) → 0
as n → ∞. If {un } is unbounded, without loss of generality, we may assume that
kun k → ∞ as n → ∞. It follows from Lemma 2.3 (b) that
Z
T
(∇F (t, un (t)), ũn (t))dt ≤ kũn k∞ [kf kL2 [ω̃(kun k)kun k]p−1 + kGkL1 ]
(3.15)
0
p−1
≤ C30 [ω̃(kun k)kun k] kũn k + C31 kũn k.
Hence, we have
kũn k ≥ (ϕ0 (un ), ũn )
Z T
Z
=
|u̇n (t)|p dt +
0
T
(∇F (t, un (t)), ũn (t))dt
0
≥ C32 kũn kp − C30 [ω̃(kun k)kun k]p−1 kũn k − C31 kũn k,
which implies
kũn k
< +∞.
kun k→∞ ω̃(kun k)kun k
lim sup
(3.16)
EJDE-2009/17
PERIODIC SOLUTIONS
Therefore, by Lemma 2.3 (d), one has
Z T
1
F (t, un (t))dt → −∞ as kun k → ∞.
[ω̃(kun k)kun k]p 0
11
(3.17)
However, by the boundedness of ϕ(un ) and (3.16), we get
R
Z T
1 T
p
ϕ(un )
1
p 0 |u̇n (t)| dt F
(t,
u
(t))dt
=
−
n
[ω̃(kun k)kun k]p 0
[ω̃(kun k)kun k]p
[ω̃(kun k)kun k]p
R
1 T
p
p 0 |u̇n (t)| dt
(3.18)
≤
[ω̃(kun k)kun k]p
C33 kũn kp
≤
< +∞,
[ω̃(kun k)kun k]p
which contradicts (3.17). So, ϕ satisfies the (PS) condition.
As in the proof of Theorem 1.1, we can obtain
ϕ(u) → +∞ as kuk → ∞ in W̃T1,p .
Furthermore, from (2.10), it is easy to see that
ϕ(u) → −∞
as |u| → ∞
in RN .
Thus, using the saddle point theorem, problem (1.1) has at least one solution in
WT1,p .
Acknowledgments. This Project was Supported by Foundation of Major Project
of Science and Technology of Chinese Education Ministry, SRFDP of Higher Education and NSF of Education Committee of Jiangsu Province and Foundation of
Nanjing University of Information Science & Technology. The authors would like
to thank the anonymous referee for his/her valuable suggestions.
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Zhiyong Wang
Department of Mathematics, Nanjing University of Information Science and Technology, Nanjing 210044, Jiangsu, China
E-mail address: mathswzhy1979@gmail.com
Jihui Zhang
Institute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal University, Nanjing 210097, Jiangsu, China
E-mail address: jihuiz@jlonline.com