Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 401740, 11 pages
http://dx.doi.org/10.1155/2013/401740
Research Article
Existence and Multiplicity of Solutions to a Boundary Value
Problem for Impulsive Differential Equations
Chunyan He,1 Yongzhi Liao,2 and Yongkun Li1
1
2
Department of Mathematics, Yunnan University, Kunming, Yunnan 650091, China
School of Mathematics and Computer Science, Panzhihua University, Panzhihua, Sichuan 617000, China
Correspondence should be addressed to Yongkun Li; yklie@ynu.edu.cn
Received 24 October 2012; Revised 20 December 2012; Accepted 21 December 2012
Academic Editor: Wan-Tong Li
Copyright © 2013 Chunyan He et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We investigate the existence and multiplicity of solutions to a boundary value problem for impulsive differential equations. By
using critical point theory, some criteria are obtained to guarantee that the impulsive problem has at least one solution, at least two
solutions, and infinitely many solutions. Some examples are given to illustrate the effectiveness of our results.
1. Introduction
In this paper, we will investigate the existence and multiplicity
of solutions to the boundary value problem for impulsive
differential equations:
󸀠
󵄨𝑝−2
󵄨
− (𝜌 (𝑡) 󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡)) + 𝑠 (𝑡) |𝑢 (𝑡)|𝑝−2 𝑢 (𝑡)
= 𝑓 (𝑡, 𝑢 (𝑡)) ,
𝑡 ≠ 𝑡𝑖 , a.e. 𝑡 ∈ [0, 𝑇] ,
󵄨
󵄨𝑝−2
Δ (𝜌 (𝑡𝑖 ) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡𝑖 )󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡𝑖 )) = 𝐼𝑖 (𝑢 (𝑡𝑖 )) ,
(1)
𝑖 = 1, 2, . . . , 𝑘,
Impulsive and periodic boundary value problems have
been studied extensively in the literature. There have been
many approaches to study periodic solutions of differential
equations, such as the method of lower and upper solutions,
fixed point theory, and coincidence degree theory (see [7–
10]). However, the study of solutions for impulsive differential
equations using variational method has received considerably
less attention (see, [11–18]). Variational method is, to the
best of our knowledge, novel and it may open a new
approach to deal with nonlinear problems with some type of
discontinuities such as impulses.
Teng and Zhang in [15] studied the existence of solutions
to the boundary value problem for impulsive differential
equations
𝑢 (0) = 𝑢 (𝑇) = 0,
where 𝑝 ≥ 2, 𝜌(𝑡), 𝑠(𝑡) ∈ 𝐿∞ [0, 𝑇], with ess inf 𝑡∈[0,𝑇] 𝜌(𝑡) > 0,
ess inf 𝑡∈[0,𝑇] 𝑠(𝑡) > 0, and 1 ≤ 𝜌(𝑡) < +∞; 0 < 𝑠(𝑡) < +∞;
0 = 𝑡0 < 𝑡1 ⋅ ⋅ ⋅ < 𝑡𝑘 < 𝑡𝑘+1 = 𝑇, Δ(𝜌(𝑡𝑖 )|𝑢󸀠 (𝑡𝑖 )|𝑝−2 𝑢󸀠 (𝑡𝑖 )) =
𝜌(𝑡𝑖+ )|𝑢󸀠 (𝑡𝑖+ )|𝑝−2 𝑢󸀠 (𝑡𝑖+ ) − 𝜌(𝑡𝑖− )|𝑢󸀠 (𝑡𝑖− )|𝑝−2 𝑢󸀠 (𝑡𝑖− ), 𝑓 : [0, 𝑇] ×
𝑅 → 𝑅 is continuous, 𝐼𝑖 : 𝑅 → 𝑅 are continuous.
Recently, there have been many papers concerned with
boundary value problems for impulsive differential equations. Impulsive effects exist widely in many evolution processes in which their states are changed abruptly at certain
moments of time. The theory of impulsive differential systems
has been developed by numerous mathematicians (see [1–6]).
󸀠
󵄨𝑝−2
󵄨
− (󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡)) = 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡)) ,
𝑡 ≠ 𝑡𝑖 , a.e. 𝑡 ∈ [0, 𝑇] ,
󸀠
Δ𝑢 (𝑡𝑖 ) = 𝐼𝑖 (𝑢 (𝑡𝑖 )) ,
(2)
𝑖 = 1, 2, . . . , 𝑙,
𝑢 (0) = 𝑢 (𝑇) = 0.
By using variational methods and iterative methods they
showed that there exists a solution for problem (2).
In this paper, we will need the following conditions.
2
Journal of Applied Mathematics
(A) 𝐹(𝑡, 𝑢) is measurable in 𝑡 for every 𝑢 ∈ R and
continuously differentiable in 𝑢 for a.e. 𝑡 ∈ [0, 𝑇] and
there exist 𝑎 ∈ 𝐶(R+ , R+ ), 𝑏 ∈ 𝐿1 (0, 𝑇; R+ ) such that
󵄨
󵄨󵄨
󵄨󵄨𝑓 (𝑡, 𝑢)󵄨󵄨󵄨 ≤ 𝑎 (|𝑢|) 𝑏 (𝑡) ,
|𝐹 (𝑡, 𝑢)| ≤ 𝑎 (|𝑢|) 𝑏 (𝑡) ,
(3)
for
all 𝑢 ∈ R and a.e. 𝑡 ∈ [0, 𝑇], where 𝐹(𝑡, 𝑢) =
𝑢
∫0 𝑓(𝑡, 𝑠)d𝑠.
(𝐻0 ) There exist constants 𝑎, 𝑏 > 0 and 𝑟 ∈ [0, 𝑝 − 1) such
that
󵄨
󵄨󵄨
𝑟
󵄨󵄨𝑓 (𝑡, 𝑢)󵄨󵄨󵄨 ≤ 𝑎 + 𝑏|𝑢| ,
for (𝑡, 𝑢) ∈ [0, 𝑇] × 𝑅.
for (𝑡, 𝑢) ∈ [0, 𝑇] × 𝑅.
(5)
(𝐻2 ) There exist constants 𝑎𝑖 , 𝑏𝑖 > 0 and 𝑟𝑖 ∈ [0, 𝑝 − 1) (𝑖 =
1, 2, . . . , 𝑘) such that
󵄨
󵄨󵄨
𝑟
󵄨󵄨𝐼𝑖 (𝑢)󵄨󵄨󵄨 ≤ 𝑎𝑖 + 𝑏𝑖 |𝑢| 𝑖 ,
for 𝑢 ∈ 𝑅.
(6)
(𝐻3 ) There exist constants 𝜇 > 𝑝 and 𝑀 > 0 such that
0 < 𝜇𝐹 (𝑡, 𝑢) ≤ 𝑢𝑓 (𝑡, 𝑢) ,
for 𝑡 ∈ [0, 𝑇] , |𝑢| ≥ 𝑀.
(7)
(𝐻4 ) lim𝑢 → 0 𝑓(𝑡, 𝑢)/|𝑢|𝑝−1 = 0 uniformly, for 𝑡 ∈ [0, 𝑇] and
𝑢 ∈ 𝑅.
(𝐻5 ) there exist 𝜆 > 𝑝 and 𝛽 > 𝜆 − 𝑝 such that
lim sup
|𝑥| → ∞
𝐹 (𝑡, 𝑥)
|𝑥|𝜆
< ∞ uniformly, for a.e. 𝑡 ∈ [0, 𝑇] ,
lim inf
𝑥𝑓 (𝑡, 𝑥) − 2𝐹 (𝑡, 𝑥)
|𝑥|𝛽
|𝑥| → ∞
> 0,
1
1
0
(1 − 𝑠𝑝 )1/𝑝
𝜋𝑝 = 2 ∫
(8)
Each eigenvalue 𝜆 𝑘 is simple with the associated eigenfunction
𝜑𝑘 (𝑥) = sin (
𝑘𝜋𝑝 𝑥
𝑝
𝑇
𝑇
󵄨𝑝
󵄨
‖𝑢‖𝑊1,𝑝 = (∫ 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 d𝑡)
0
0
,
1,𝑝
𝑢 ∈ 𝑊0 (0, 𝑇) .
𝑢∈
We recall some facts which will be used in the proof of
our main results. It has been shown, for instance, in [19] that
the set of all eigenvalues of the following problem:
󸀠
󵄨𝑝−2
󵄨
− (󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡)) + 𝜆|𝑢 (𝑡)|𝑝−2 𝑢 (𝑡) = 0,
(10)
𝑢 (0) = 𝑢 (𝑇) = 0
(14)
(15)
(16)
where 𝐶 = 𝜆1 is precisely the largest 𝐶 > 0 for which the above
𝑇
𝑇
inequality holds true. Then ∫0 |𝑢󸀠 (𝑡)|𝑝 d𝑡 − 𝜆 1 ∫0 |𝑢(𝑡)|𝑝 d𝑡 ≥
0 while it minimizes and equals to zero exactly on the ray
generated by the first eigenfunction sin𝑝 (𝜋𝑝 𝑡/𝑇).
Let us recall that
0
(11)
,
(0, 𝑇) .
𝑇
𝑇
󵄨𝑝
󵄨
∫ 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 d𝑡 ≥ 𝐶 ∫ |𝑢 (𝑡)|𝑝 d𝑡,
0
0
𝑇
𝑝
1,𝑝
𝑊0
1/𝑝
By Poincaré inequality:
|𝑢|𝑝 = (∫ |𝑢 (𝑡)|𝑝 d𝑡)
is given by the sequence of positive numbers
for 𝑘 = 1, 2, . . . ,
1/𝑝
0
0
) ,
(13)
In this section, we recall some basic facts which will be used
in the proofs of our main results. In order to apply the critical
point theory, we make a variational structure. From this
variational structure, we can reduce the problem of finding
solutions of (1) to the one of seeking the critical points of a
corresponding functional.
1,𝑝
In [20], the Sobolev space 𝑊0 (0, 𝑇) be the endowed with
the norm
𝑇
󵄨
󵄨𝑝
‖𝑢‖ = (∫ (𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 + 𝑠 (𝑡) |𝑢 (𝑡)|𝑝 ) d𝑡)
2 ∫ 𝐼𝑖 (𝑠) d𝑠 − 𝐼𝑖 (𝑡) 𝑡 ≥ 0 ∀𝑖 ∈ {1, 2, . . . , 𝑘} , |𝑡| ≥ 𝜁𝑖𝑗 . (9)
𝑇
for 0 ≤ 𝑥 ≤ 𝑇.
2. Preliminaries
𝑡
𝜆 𝑘 = (𝑝 − 1) (
),
𝑋𝑘 denotes the eigenspace associated to 𝜆 𝑘 , then 𝑊0 (0, 𝑇) =
⨁𝑖∈𝑁𝑋𝑖 .
An outline of this paper is given as follows. In the next
section, we present some preliminaries including some basic
knowledge and critical point theory. In Section 3, by using
the critical point theory, we will establish some sufficient
conditions for the existence of solutions of system (1). In
Section 4, some examples are given to verify and support the
theoretical findings.
(𝐻6 ) there exists 𝜁𝑖𝑗 > 0 such that
𝑘𝜋𝑝
(12)
Throughout the paper, we also consider the norm
uniformly for a.e. 𝑡 ∈ [0, 𝑇] .
𝑡 ≠ 𝑡𝑖 , 𝑡 ∈ [0, 𝑇] ,
d𝑠.
1,𝑝
(4)
(𝐻1 ) There exist constants 𝑐 > 0 and 𝜏 ∈ [𝑝, +∞) such that
󵄨
󵄨󵄨
𝜏−1
󵄨󵄨𝑓 (𝑡, 𝑢)󵄨󵄨󵄨 ≤ 𝑐 (|𝑢| + 1) ,
where
1/𝑝
,
‖𝑢‖∞ = max |𝑢 (𝑡)| . (17)
𝑡∈[0,𝑇]
We denote by | ⋅ |𝑝 the usual 𝐿𝑝 -norm. The 𝑛-dimensional
Lebesgue measure of a set 𝐸 ⊆ 𝑅𝑛 is denoted by |𝐸|. By
1,𝑝
the Sobolev embedding theorem, 𝑊0 (0, 𝑇) 󳨅→ 𝐿𝑟 [0, 𝑇]
Journal of Applied Mathematics
3
continuously for 𝑟 ∈ [1, +∞), and there exists 𝛾𝑟 > 0 such
that
|𝑢|𝑟 ≤ 𝛾𝑟 ‖𝑢‖ ,
1,𝑝
∀𝑢 ∈ 𝑊0 (0, 𝑇) .
1,𝑝
Proof. If 𝑢 ∈ 𝑊0 (0, 𝑇), it follows from the mean value
theorem that
(18)
1 𝑇
∫ 𝑢 (𝑠) d𝑠 = 𝑢 (𝜉) ,
𝑇 0
(19)
for some 𝜉 ∈ (0, 𝑇). Hence, for 𝑡 ∈ [0, 𝑇], by Hölder inequality
and Poincaré inequality
Lemma 1. There exist 𝐶1 , 𝐶2 > 0 such that
1,𝑝
𝐶1 ‖𝑢‖𝑊1,𝑝 ≤ ‖𝑢‖ ≤ 𝐶2 ‖𝑢‖𝑊1,𝑝 ,
0
∀𝑢 ∈ 𝑊0 (0, 𝑇) .
0
Proof. Since ess inf 𝑡∈[0,𝑇] 𝜌(𝑡) > 0, ess inf 𝑡∈[0,𝑇] 𝑠(𝑡) > 0,
1 ≤ 𝜌(𝑡) < +∞, and 0 < 𝑠(𝑡) < +∞, we have that ess
inf 𝑡∈[0,𝑇] 𝜌(𝑡) := 𝑚1 ≥ 1, ess inf 𝑡∈[0,𝑇] 𝑠(𝑡) := 𝑚2 > −𝜆 1 , where
𝜆 1 is a positive number and that there exists 𝑛1 ∈ (0, 1) such
that −𝑚2 ≤ 𝜆 1 (1 − 𝑛1 ). Thus, by Poincaré inequality, we have
𝑇
󵄨󵄨
󵄨󵄨
𝑡
󵄨
󵄨
|𝑢 (𝑡)| = 󵄨󵄨󵄨𝑢 (𝜉) + ∫ 𝑢󸀠 (𝑠) d𝑠󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝜉
𝑇
󵄨
󵄨
≤ 󵄨󵄨󵄨𝑢 (𝜉)󵄨󵄨󵄨 + ∫
0
𝑇
󵄨𝑝
󵄨
(1 − 𝑛1 ) ∫ 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 d𝑡 ≥ (1 − 𝑛1 ) 𝜆 1 ∫ |𝑢 (𝑡)|𝑝 d𝑡
0
0
𝑇
0
for all 𝑢 ∈
1/𝑝
𝑇
𝑇
󵄨
󵄨𝑝
‖𝑢‖𝑝 = ∫ 𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 d𝑡 + ∫ 𝑠 (𝑡) |𝑢 (𝑡)|𝑝 d𝑡
𝐶 1/𝑝
1
≤ (( ) + 𝑇1/𝑞 )
‖𝑢‖ .
𝑇
𝐶1
0
𝑇
󵄨𝑝
󵄨
≥ 𝑚1 ∫ 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 d𝑡
0
𝑝
= 𝑚1 ‖𝑢‖
(21)
1,𝑝
𝑊0
𝑝
𝑝
= 𝐶1 ‖𝑢‖
1,𝑝
𝑊0
1,𝑝
Take 𝑣 ∈ 𝑊0 (0, 𝑇) and multiply the two sides of the
equality
𝑇
𝑇
󵄨𝑝
󵄨
‖𝑢‖𝑝 = ∫ 𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 d𝑡 + ∫ 𝑠 (𝑡) |𝑢 (𝑡)|𝑝 d𝑡
0
󸀠
󵄨
󵄨𝑝−2
−(𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡)) + 𝑠 (𝑡) |𝑢 (𝑡)|𝑝−2 𝑢 (𝑡) = 𝑓 (𝑡, 𝑢 (𝑡))
(27)
0
𝑇
Hence, ‖𝑢‖∞ ≤ ((𝐶/𝑇)1/𝑝 + 𝑇1/𝑞 )(1/𝐶1 )‖𝑢‖ = 𝐶3 ‖𝑢‖. The
proof is complete.
.
On the other hand, by Poincaré inequality, one has
󵄩 󵄩
≤ 󵄩󵄩󵄩𝜌󵄩󵄩󵄩∞ ∫
0
𝑇
󵄨󵄨 󸀠 󵄨󵄨𝑝
󵄨󵄨𝑢 (𝑡)󵄨󵄨 d𝑡 + ‖𝑠‖∞ ∫ |𝑢 (𝑡)|𝑝 d𝑡
󵄨
󵄨
0
by 𝑣 and integrate it from 0 to 𝑇, we have
𝑇
‖𝑠‖∞
󵄨𝑝
󵄨
󵄩 󵄩
) ∫ 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 d𝑡
≤ (󵄩󵄩󵄩𝜌󵄩󵄩󵄩∞ +
𝜆1
0
(22)
𝑇
󸀠
󵄨
󵄨𝑝−2
− ∫ (𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡)) 𝑣 (𝑡) d𝑡
‖𝑠‖∞
𝑝
󵄩 󵄩
) ‖𝑢‖ 1,𝑝
= (󵄩󵄩󵄩𝜌󵄩󵄩󵄩∞ +
𝑊0
𝜆1
=
(26)
𝑇
𝐶 1/𝑝
󵄨𝑝
󵄨
≤ (( ) + 𝑇1/𝑞 ) (∫ 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 d𝑠)
𝑇
0
1,𝑝
Thereby, for every 𝑢 ∈ 𝑊0 (0, 𝑇),
0
󵄨󵄨 󸀠 󵄨󵄨
󵄨󵄨𝑢 (𝑠)󵄨󵄨 d𝑠
󵄨
󵄨
1/𝑝
󵄨󵄨
󵄨
𝑇
1 󵄨󵄨󵄨 𝑇
󵄨󵄨
󵄨󵄨 󸀠 󵄨󵄨𝑝
1/𝑞
󵄨
󵄨
󵄨
󵄨
≤ 󵄨󵄨∫ 𝑢 (𝑠) d𝑠󵄨󵄨 + 𝑇 (∫ 󵄨󵄨𝑢 (𝑡)󵄨󵄨 d𝑠)
𝑇 󵄨󵄨 0
󵄨󵄨
0
(20)
≥ − 𝑚2 ∫ |𝑢 (𝑡)|𝑝 d𝑡,
1,𝑝
𝑊0 (0, 𝑇).
(25)
0
𝑇
+ ∫ 𝑠 (𝑡) |𝑢 (𝑡)|𝑝−2 𝑢 (𝑡) 𝑣 (𝑡) d𝑡
𝑝
𝑝
𝐶2 ‖𝑢‖ 1,𝑝 .
𝑊
0
0
𝑇
Take 𝐶1 = (𝑚1 + 𝑛1 − 1)1/𝑝 , 𝐶2 = (‖𝜌‖∞ + ‖𝑠‖∞ /𝜆 1 )1/𝑝 ,
then
𝐶1 ‖𝑢‖𝑊1,𝑝 ≤ ‖𝑢‖ ≤ 𝐶2 ‖𝑢‖𝑊1,𝑝 .
0
(23)
0
The proof is complete.
1,𝑝
Lemma 2. There exists 𝐶3 > 0 such that if 𝑢 ∈ 𝑊0 (0, 𝑇),
then
‖𝑢‖∞ ≤ 𝐶3 ‖𝑢‖ .
(24)
= ∫ 𝑓 (𝑡, 𝑢 (𝑡)) 𝑣 (𝑡) d𝑡.
0
Moreover,
𝑇
󸀠
󵄨𝑝−2
󵄨
− ∫ (𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡)) 𝑣 (𝑡) d𝑡
0
𝑘
𝑡𝑖+1
= −∑ ∫
𝑖=0 𝑡𝑖
󸀠
󵄨𝑝−2
󵄨
(𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡)) 𝑣 (𝑡) d𝑡
(28)
4
Journal of Applied Mathematics
𝑘
󵄨
−
− 󵄨󵄨𝑝−2 󸀠 −
−
= ∑ (𝜌 (𝑡𝑖+1
) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡𝑖+1
)󵄨󵄨󵄨 𝑢 (𝑡𝑖+1 ) 𝑣 (𝑡𝑖+1
)
Using the continuity of 𝑓 and 𝐼𝑖 , 𝑖 = 1, 2, . . . , 𝑘, one has that
1,𝑝
1,𝑝
𝜑 ∈ 𝐶1 (𝑊0 (0, 𝑇), 𝑅). For any 𝑣 ∈ 𝑊0 (0, 𝑇), we have
𝑖=0
󵄨
󵄨𝑝−2
− 𝜌 (𝑡𝑖+ ) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡𝑖+ )󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡𝑖+ ) 𝑣 (𝑡𝑖+ )
−∫
𝑡𝑖+1
𝑡𝑖
𝑇
󵄨
󵄨𝑝−2
𝜑󸀠 (𝑢) 𝑣 = ∫ 𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡) 𝑣󸀠 (𝑡) d𝑡
0
󵄨𝑝−2
󵄨
𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡) 𝑣󸀠 (𝑡) d𝑡)
𝑇
+ ∫ 𝑠 (𝑡) |𝑢 (𝑡)|𝑝−2 𝑢 (𝑡) 𝑣 (𝑡) d𝑡
0
𝑘
󵄨
󵄨𝑝−2
= ∑ (𝜌 (𝑡𝑖+ ) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡𝑖+ )󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡𝑖+ ) − 𝜌 (𝑡𝑖− )
𝑘
󵄨
󵄨𝑝−2
󵄨𝑝−2
󵄨
×󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡𝑖− )󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡𝑖− )) − 𝜌 (𝑇) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑇)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑇) 𝑣 (𝑇)
󵄨
󵄨𝑝−2
+ 𝜌 (0) 󵄨󵄨󵄨󵄨𝑢󸀠 (0)󵄨󵄨󵄨󵄨 𝑢󸀠 (0) 𝑣 (0)
0
𝑘
𝑇
󵄨
󵄨𝑝−2
= ∑𝐼𝑖 (𝑢 (𝑡𝑖 )) 𝑣 (𝑡𝑖 ) + ∫ 𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡) 𝑣󸀠 (𝑡) d𝑡.
0
(29)
Combining (28), we have
𝑇
󵄨
󵄨𝑝−2
∫ 𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡) 𝑣󸀠 (𝑡) d𝑡
0
𝑘
+ ∫ 𝑠 (𝑡) |𝑢 (𝑡)|𝑝−2 𝑢 (𝑡) 𝑣 (𝑡) d𝑡 + ∑𝐼𝑖 (𝑢 (𝑡𝑖 )) 𝑣 (𝑡𝑖 )
0
𝑖=1
𝑇
= ∫ 𝑓 (𝑡, 𝑢 (𝑡)) 𝑣 (𝑡) d𝑡.
0
(30)
Considering the above, we introduce the following concept
solution for problem (1).
1,𝑝
Definition 3. We say that a function 𝑢 ∈ 𝑊0 (0, 𝑇) is a
solution of problem (1) if the identity
𝑇
󵄨𝑝−2
󵄨
∫ 𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡) 𝑣󸀠 (𝑡) d𝑡
0
𝑘
𝑇
+ ∫ 𝑠 (𝑡) |𝑢 (𝑡)|𝑝−2 𝑢 (𝑡) 𝑣 (𝑡) d𝑡 + ∑𝐼𝑖 (𝑢 (𝑡𝑖 )) 𝑣 (𝑡𝑖 )
0
𝑖=1
Thus, the solutions of problem (1) are the corresponding
critical points of 𝜑.
Definition 5. Let 𝑋 be a Banach space and let 𝜑 : 𝑋 →
(−∞, +∞). 𝜑 is said to be sequentially weakly lower semicontinuous if lim inf 𝑘 → +∞ 𝜑(𝑥𝑘 ) ≥ 𝜑(𝑥) as 𝑥𝑘 ⇀ 𝑥 in 𝑋.
Definition 6. Let 𝐸 be a Banach space and let 𝑐 ∈ 𝑅. For any
sequence {𝑢𝑘 } in 𝐸, if 𝜑(𝑢𝑘 ) is bounded and 𝜑󸀠 (𝑢𝑘 ) → 0
as 𝑘 → +∞ possesses a convergent subsequence, then we
say that 𝜑 satisfies the Palais-Smale condition (denoted by PS
condition for short). We say that 𝜑 satisfies the Palais-Smale
condition at level 𝑐 (denoted by (PS)𝑐 condition for short) if
there exists a sequence {𝑢𝑘 } in 𝐸 such that 𝜑(𝑢𝑘 ) → 𝑐 and
𝜑󸀠 (𝑢𝑘 ) → 0 as 𝑘 → +∞ implies that 𝑐 is a critical value of
𝜑.
Definition 7. Let 𝐸 be a Banach space and let 𝜑 : 𝐸 →
(−∞, +∞). 𝜑 is said to be coercive if 𝜑(𝑢) → +∞ as ‖𝑢‖ →
+∞.
Lemma 8 (see [12]). If 𝜑 is sequentially weakly lower semicontinuous on a reflexive Banach space 𝑋 and has a bounded
minimizing sequence, then 𝜑 has a minimum on 𝑋.
Definition 9. Let 𝑋 be a real Banach space with a direct sum
decomposition 𝑋 = 𝑋1 ⊕ 𝑋2 . The functional 𝜑 ∈ 𝐶1 (𝑋, 𝑅) is
said to have a local linking at 0, with respect to (𝑋1 , 𝑋2 ), if,
for some 𝑟 > 0,
(i) 𝜑(𝑢) ≥ 0, 𝑢 ∈ 𝑋1 , ‖𝑢‖ ≤ 𝑟,
(ii) 𝜑(𝑢) ≤ 0, 𝑢 ∈ 𝑋2 , ‖𝑢‖ ≤ 𝑟.
𝑇
= ∫ 𝑓 (𝑡, 𝑢 (𝑡)) 𝑣 (𝑡) d𝑡
0
0
𝑖=1
Definition 4. Let 𝑋 be a normed space. A minimizing
sequence for a function 𝜑 : 𝑋 → (−∞, +∞) is a sequence
𝑢𝑘 such that 𝜑(𝑢𝑘 ) → inf 𝜑 whenever 𝑘 → +∞.
𝑇
󵄨𝑝−2
󵄨
+ ∫ 𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡) 𝑣󸀠 (𝑡) d𝑡
𝑇
𝑇
+ ∑𝐼𝑖 (𝑢 (𝑡𝑖 )) 𝑣 (𝑡𝑖 ) − ∫ 𝑓 (𝑡, 𝑢 (𝑡)) 𝑣 (𝑡) d𝑡.
𝑖=1
𝑖=1
(33)
(31)
1,𝑝
holds for any 𝑣 ∈ 𝑊0 (0, 𝑇).
1,𝑝
Consider the functional 𝜑 : 𝑊0 (0, 𝑇) → 𝑅 defined by
𝑘
𝑢(𝑡𝑖 )
𝑇
1
𝐼𝑖 (𝑡) d𝑡 − ∫ 𝐹 (𝑡, 𝑢 (𝑡)) d𝑡. (32)
𝜑 (𝑢) = ‖𝑢‖𝑝 + ∑ ∫
𝑝
0
𝑖=1 0
If 𝜑 has a local linking at 0, then 0 is critical point
(the trivial one). Suppose, furthermore, that there are two
sequences of finite dimensional subspaces 𝑋11 ⊂ 𝑋21 ⊂ ⋅ ⋅ ⋅ ⊂
𝑋1 and 𝑋12 ⊂ 𝑋22 ⊂ ⋅ ⋅ ⋅ ⊂ 𝑋2 such that
𝑋1 = ⋃𝑋𝑛1 ,
𝑛
𝑋2 = ⋃𝑋𝑛2 .
𝑛
(34)
Journal of Applied Mathematics
5
Definition 10 (see [13, Definition 2.2]). Let 𝐼 ∈ 𝐶1 (𝑋, R).
The functional 𝐼 satisfies the (𝐶)∗ condition if every sequence
(𝑢𝛼𝑛 ) such that 𝛼𝑛 is admissible and
𝑢𝛼𝑛 ∈ 𝑋𝛼𝑛 ,
󵄨
󵄨
sup 󵄨󵄨󵄨󵄨𝐼 (𝑢𝛼𝑛 )󵄨󵄨󵄨󵄨 < ∞,
󵄩 󵄩
(1 + 󵄩󵄩󵄩󵄩𝑢𝛼𝑛 󵄩󵄩󵄩󵄩) 𝐼󸀠 (𝑢𝛼𝑛 ) 󳨀→ 0
𝑇
− ∫ (𝑎 |𝑢| + 𝑏|𝑢|𝑟+1 ) d𝑡
0
≥
𝑘
𝑟𝑖 +1
− 𝑀2 ∑‖𝑢‖∞
− 𝑎𝑇‖𝑢‖∞ − 𝑏𝑇‖𝑢‖𝑟+1
∞
(35)
contains a subsequence which converges to a critical point of
𝐼.
1
Lemma 11 (see [13]). Suppose that 𝜑 ∈ 𝐶 (𝑋, 𝑅) satisfies the
following assumptions:
(1) 𝜑 satisfies the (𝐶)∗ condition,
(2) 𝜑 has a local linking at 0,
(3) 𝜑 maps bounded sets into bounded sets,
(4) for every 𝑚 ∈ 𝑁, 𝜑(𝑢) → −∞ as ‖𝑢‖ → +∞, 𝑢 ∈
1
𝑋𝑚
⊕ 𝑋2 .
1
‖𝑢‖𝑝 − 𝑘𝑀1 ‖𝑢‖∞
𝑝
𝑖=1
≥
1
‖𝑢‖𝑝 − 𝑘𝑀1 𝐶3 ‖𝑢‖
𝑝
𝑘
𝑟 +1
− 𝑀2 ∑𝐶3𝑖 ‖𝑢‖𝑟𝑖 +1 − 𝑎𝑇𝐶3 ‖𝑢‖ − 𝑏𝑇𝐶3𝑟+1 ‖𝑢‖𝑟+1 ,
𝑖=1
(36)
1,𝑝
for all 𝑢 ∈ 𝑊0 (0, 𝑇). This implies that lim‖𝑢‖ → ∞ 𝜑(𝑢) = ∞,
and 𝜑 is coercive.
On the other hand, we show that 𝜑 is weakly lower
1,𝑝
semicontinuous. If {𝑢𝑘 }𝑘∈𝑁 ⊂ 𝑊0 (0, 𝑇), 𝑢𝑘 ⇀ 𝑢, then we
have that {𝑢𝑘 }𝑘∈𝑁 converges uniformly to 𝑢 on [0, 𝑇] and
lim inf 𝑘 → ∞ ‖𝑢𝑘 ‖ ≥ ‖𝑢‖. Thus
𝑘
𝑢 (𝑡𝑖 )
𝑘
1 󵄩 󵄩𝑝
lim inf 𝜑 (𝑢𝑘 ) = lim inf ( 󵄩󵄩󵄩𝑢𝑘 󵄩󵄩󵄩 + ∑ ∫
𝑘→∞
𝑘→∞
𝑝
𝑖=1 0
𝐼𝑖 (𝑡) d𝑡
𝑇
− ∫ 𝐹 (𝑡, 𝑢𝑘 (𝑡)) d𝑡)
Then 𝜑 has at least two critical points.
0
Lemma 12 (see [14]). Let E be a Banach space. Let 𝜑 ∈
𝐶1 (𝑋, 𝑅) be an even functional which satisfies the 𝑃𝑆 condition
and 𝜑(0) = 0. If 𝐸 = 𝑉 ⊕ 𝑌, where 𝑉 is finite dimensional, and
𝜑 satisfies the following conditions:
≥
𝑘
𝑢(𝑡𝑖 )
𝑇
1
𝐼𝑖 (𝑡) d𝑡 − ∫ 𝐹 (𝑡, 𝑢 (𝑡)) d𝑡
‖𝑢‖𝑝 + ∑ ∫
𝑝
0
𝑖=1 0
= 𝜑 (𝑢) .
(37)
(1) there exist constants 𝜌, 𝛼 such that 𝜑|𝜕𝐵𝜌 ∩𝑌 ≥ 𝛼, where
𝐵𝜌 = {𝑥 ∈ 𝐸 : ‖𝑥‖ < 𝜌},
(2) for each finite-dimensional subspace 𝑊 ⊂ 𝐸 there is
𝑅 = 𝑅(𝑊) such that 𝜑(𝑢) ≤ 0, for all 𝑢 ∈ 𝑊 with
‖𝑢‖ ≥ 𝑅.
1,𝑝
By Lemma 8, 𝜑 has a minimum point on 𝑊0 (0, 𝑇),
which is a critical point of 𝜑. Hence, problem (1) has at least
one solution. The proof is complete.
We readily have the following corollary.
Then 𝜑 has an unbounded sequence of critical values.
Corollary 14. Assume that (𝐴), (𝐻0 ), and (𝐻2 ) are satisfied
and 𝑓 and the impulsive functions 𝐼𝑖 (𝑖 = 1, 2, . . . , 𝑘) are
bounded. Then problem (1) has at least one solution.
3. Existence of Periodic Solutions
Lemma 15. Assume that (𝐻1 ), (𝐻2 ), and (𝐻3 ) are satisfied,
then 𝜑(𝑢) satisfies the 𝑃𝑆 condition.
Theorem 13. Assume that (𝐴), (𝐻0 ), and (𝐻2 ) are satisfied,
then problem (1) has at least one solution.
Proof. Let 𝑀1 = max{𝑎1 , 𝑎2 , . . . , 𝑎𝑘 }, 𝑀2 = max{𝑏1 , 𝑏2 , . . . ,
𝑏𝑘 }. By Lemma 2, we have
1,𝑝
Proof. Assume that {𝑢𝑛 } ⊂ 𝑊0 (0, 𝑇) satisfies that 𝜑(𝑢𝑛 ) is
bounded and 𝜑󸀠 (𝑢𝑛 ) → 0 as 𝑛 → +∞. We will prove that
the sequence {𝑢𝑛 } is bounded.
It follows from (𝐻1 ), (𝐻2 ), (𝐻3 ), and Lemma 2, we have
𝜇𝜑 (𝑢𝑛 ) − 𝜑󸀠 (𝑢𝑛 ) 𝑢𝑛
𝜑 (𝑢) =
≥
𝑘
𝑢(𝑡𝑖 )
𝑇
1
𝐼𝑖 (𝑡) d𝑡 − ∫ 𝐹 (𝑡, 𝑢 (𝑡)) d𝑡
‖𝑢‖𝑝 + ∑ ∫
𝑝
0
𝑖=1 0
𝑘
𝑢(𝑡𝑖 )
1
(𝑎𝑖 + 𝑏𝑖 |𝑢|𝑟𝑖 ) d𝑡
‖𝑢‖𝑝 − ∑ ∫
𝑝
𝑖=1 0
=(
𝜇
󵄩 󵄩𝑝
− 1) 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩
𝑝
𝑘
+ ∑ [𝜇 ∫
𝑖=1
𝑢𝑛 (𝑡𝑖 )
0
𝐼𝑖 (𝑡) 𝑑𝑡 − 𝐼𝑖 (𝑢𝑛 (𝑡𝑖 )) 𝑢𝑛 (𝑡𝑖 )]
6
Journal of Applied Mathematics
𝑇
for some constant 𝑐𝑝 (Lemma 4.2 in [21]) and using Schwarz
inequality, we have
− ∫ (𝜇𝐹 (𝑡, 𝑢𝑛 (𝑡)) − 𝑓 (𝑡, 𝑢𝑛 (𝑡)) 𝑢𝑛 (𝑡)) d𝑡
0
≥(
𝜇
󵄩 󵄩𝑝
− 1) 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩 − (𝜇 + 1)
𝑝
𝑇
𝑇
󵄨
󵄨𝑝
󵄨
󵄨𝑝
𝑐𝑝 ∫ 𝜌 (𝑡) 󵄨󵄨󵄨󵄨𝑢𝑛󸀠 (𝑡) − 𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 d𝑡 + 𝑐𝑝 ∫ 𝑠 (𝑡) 󵄨󵄨󵄨𝑢𝑛 (𝑡) − 𝑢 (𝑡)󵄨󵄨󵄨 d𝑡
0
0
󵄩󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩󵄩𝜑󸀠 (𝑢𝑛 ) − 𝜑󸀠 (𝑢)󵄩󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑢󵄩󵄩󵄩
𝑘
𝑟 +1 󵄩 󵄩𝑟 +1
󵄩 󵄩
× (𝑘𝑀1 𝐶3 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩 + 𝑀2 ∑𝐶3𝑖 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩 𝑖 )
𝑖=1
−∫
{|𝑢𝑛 |≤𝑢0 }
𝑘
− ∑ (𝐼𝑖 (𝑢𝑛 (𝑡𝑖 )) − 𝐼𝑖 (𝑢 (𝑡𝑖 ))) (𝑢𝑛 (𝑡𝑖 ) − 𝑢 (𝑡𝑖 ))
(𝜇𝐹 (𝑡, 𝑢𝑛 (𝑡)) − 𝑓 (𝑡, 𝑢𝑛 (𝑡)) 𝑢𝑛 (𝑡)) d𝑡
𝑖=1
𝑇
𝜇
󵄩 󵄩𝑝
≥ ( − 1) 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩 − (𝜇 + 1)
𝑝
+ ∫ (𝑓 (𝑡, 𝑢𝑛 (𝑡)) − 𝑓 (𝑡, 𝑢 (𝑡))) (𝑢𝑛 (𝑡) − 𝑢 (𝑡)) d𝑡.
0
𝑘
𝑟 +1 󵄩 󵄩𝑟 +1
󵄩 󵄩
× (𝑘𝑀1 𝐶3 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩 + 𝑀2 ∑𝐶3𝑖 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩 𝑖 ) − 𝐶4 .
(42)
Since
𝑖=1
(𝐼𝑖 (𝑢𝑛 (𝑡𝑖 )) − 𝐼𝑖 (𝑢 (𝑡𝑖 ))) (𝑢𝑛 (𝑡𝑖 ) − 𝑢 (𝑡𝑖 ))
(38)
1,𝑝
Hence, {𝑢𝑛 } is bounded in 𝑊0 (0, 𝑇).
1,𝑝
Since 𝑊0 (0, 𝑇) is a reflexive Banach space, passing to a
subsequence if necessary, we may assume that there is a 𝑢 ∈
1,𝑝
𝑊0 (0, 𝑇) such that
󵄨 󵄨
󵄨 󵄩
󵄩
󵄨
≤ (󵄨󵄨󵄨𝐼𝑖 (𝑢𝑛 (𝑡𝑖 ))󵄨󵄨󵄨 + 󵄨󵄨󵄨𝐼𝑖 (𝑢 (𝑡𝑖 ))󵄨󵄨󵄨) 󵄩󵄩󵄩𝑢𝑛 − 𝑢󵄩󵄩󵄩∞ .
(43)
By the assumption (𝐻1 ), we have
𝑇
∫ (𝑓 (𝑡, 𝑢𝑛 (𝑡)) − 𝑓 (𝑡, 𝑢 (𝑡))) (𝑢𝑛 (𝑡) − 𝑢 (𝑡)) d𝑡
0
𝑇
𝑢𝑛 ⇀ 𝑢 in
1,𝑝
𝑊0
󵄨 󵄨
󵄨
󵄨
󵄨 󵄨
≤ ∫ (󵄨󵄨󵄨𝑓 (𝑡, 𝑢𝑛 (𝑡))󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑡, 𝑢 (𝑡))󵄨󵄨󵄨) 󵄨󵄨󵄨𝑢𝑛 (𝑡) − 𝑢 (𝑡)󵄨󵄨󵄨 d𝑡
0
(0, 𝑇) ,
𝑢𝑛 󳨀→ 𝑢 in 𝐿𝑝 [0, 𝑇] ,
(39)
𝑇
󵄨
󵄨 󵄨𝜏−1
󵄨
≤ 𝑐 ∫ (2 + 󵄨󵄨󵄨𝑢𝑛 󵄨󵄨󵄨 + |𝑢|𝜏−1 ) 󵄨󵄨󵄨𝑢𝑛 (𝑡) − 𝑢 (𝑡)󵄨󵄨󵄨 d𝑡
0
{𝑢𝑛 } converges uniformly to 𝑢 on [0, 𝑇] .
󵄩
󵄩
󵄨 󵄨𝜏−1
≤ 𝐶5 󵄩󵄩󵄩𝑢𝑛 − 𝑢󵄩󵄩󵄩∞ (1 + 󵄨󵄨󵄨𝑢𝑛 󵄨󵄨󵄨𝜏−1 + |𝑢|𝜏−1
𝜏−1 ) .
Notice that
1,𝑝
From (39), it follows that 𝑢𝑛 → 𝑢 in 𝑊0 (0, 𝑇). Thus, 𝜑(𝑢)
satisfies the PS condition. The proof is complete.
(𝜑󸀠 (𝑢𝑛 ) − 𝜑󸀠 (𝑢) , 𝑢𝑛 − 𝑢)
𝑇
󵄨𝑝−2
󵄨
󵄨𝑝−2
󵄨
= ∫ 𝜌 (𝑡) (󵄨󵄨󵄨󵄨𝑢𝑛󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢𝑛󸀠 (𝑡) − 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡))
Lemma 16. Assume that (𝐴), (𝐻2 ), (𝐻5 ), and (𝐻6 ) are
satisfied, then 𝜑 satisfies the (𝐶)∗ condition.
0
× (𝑢𝑛󸀠 (𝑡) − 𝑢󸀠 (𝑡)) d𝑡
1,𝑝
Proof. Let {𝑢𝛼𝑛 } be a sequence in 𝑊0 (0, 𝑇) such that 𝛼𝑛 is
admissible and
󵄨
󵄨
𝑢𝛼𝑛 ∈ 𝑋𝛼𝑛 , sup 󵄨󵄨󵄨󵄨𝜑 (𝑢𝛼𝑛 )󵄨󵄨󵄨󵄨 < +∞,
(45)
󵄩 󵄩
(1 + 󵄩󵄩󵄩󵄩𝑢𝛼𝑛 󵄩󵄩󵄩󵄩) 𝜑󸀠 (𝑢𝛼𝑛 ) 󳨀→ 0,
𝑇
󵄨𝑝−2
󵄨
+ ∫ 𝑠 (𝑡) (󵄨󵄨󵄨𝑢𝑛 (𝑡)󵄨󵄨󵄨 𝑢𝑛 (𝑡) − |𝑢 (𝑡)|𝑝−2 𝑢 (𝑡))
0
× (𝑢𝑛 (𝑡) − 𝑢 (𝑡)) d𝑡
𝑘
then there exist a constant 𝐶4 > 0 such that
+ ∑ (𝐼𝑖 (𝑢𝑛 (𝑡𝑖 )) − 𝐼𝑖 (𝑢 (𝑡𝑖 ))) (𝑢𝑛 (𝑡𝑖 ) − 𝑢 (𝑡𝑖 ))
𝑖=1
𝑇
− ∫ (𝑓 (𝑡, 𝑢𝑛 (𝑡)) − 𝑓 (𝑡, 𝑢 (𝑡))) (𝑢𝑛 (𝑡) − 𝑢 (𝑡)) d𝑡.
0
(40)
Recalling the following well-known inequality: for any 𝑥, 𝑦 ∈
𝑅𝑁,
󵄨𝑝
󵄨
󵄨 󵄨𝑝−2
(|𝑥|𝑝−2 𝑥 − 󵄨󵄨󵄨𝑦󵄨󵄨󵄨 𝑦) (𝑥 − 𝑦) ≥ 𝑐𝑝 󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 ,
(44)
𝑝 ≥ 2,
(41)
󵄨󵄨
󵄨
󵄨󵄨𝜑 (𝑢𝛼𝑛 )󵄨󵄨󵄨 ≤ 𝐶4 ,
󵄨
󵄨
󵄩 󵄩
(1 + 󵄩󵄩󵄩󵄩𝑢𝛼𝑛 󵄩󵄩󵄩󵄩) 𝜑󸀠 (𝑢𝛼𝑛 ) ≤ 𝐶4 ,
(46)
for all large 𝑛. On the other hand, by (𝐻5 ), there are constants
𝐶5 > 0 and 𝜌1 > 0 such that
𝐹 (𝑡, 𝑥) ≤ 𝐶5 |𝑥|𝜆 ,
(47)
for all |𝑥| ≥ 𝜌1 and a.e. 𝑡 ∈ [0, 𝑇]. By (A) one has
|𝐹 (𝑡, 𝑥)| ≤ max 𝑎 (𝑠) 𝑏 (𝑡) ,
𝑠∈[0,𝜌1 ]
(48)
Journal of Applied Mathematics
7
for all |𝑥| ≤ 𝜌1 and a.e. 𝑡 ∈ [0, 𝑇]. It follows from (47) and
(48) that
|𝐹 (𝑡, 𝑥)| ≤ max 𝑎 (𝑠) 𝑏 (𝑡) + 𝐶5 |𝑥|𝜆 .
𝑠∈[0,𝜌1 ]
𝑡
(49)
𝑘
= 𝑝∑ ∫
𝑢𝛼𝑛 (𝑡𝑖 )
𝑖=1 0
𝑘
𝑟𝑖 +1
≤ 𝑀1 𝑘‖𝑢‖∞ + 𝑀2 ∑‖𝑢‖∞
(50)
𝑖=1
∀𝑖 = 1, 2, . . . , 𝑘, 𝑡 ∈ R. (55)
𝐼𝑖 (𝑡) d𝑡 − 𝑢𝛼𝑛 𝐼𝑖 (𝑢𝛼𝑛 )
𝑇
+ ∫ [𝑢𝛼𝑛 (𝑡) 𝑓 (𝑡, 𝑢𝛼𝑛 (𝑡)) − 𝑝𝐹 (𝑡, 𝑢𝛼𝑛 (𝑡))] d𝑡
0
𝑘
𝑘
≤ 𝑀1 𝑘𝐶3 ‖𝑢‖ + 𝑀2 𝐶3 ∑‖𝑢‖𝑟𝑖 +1
= ∑ (𝑝 ∫
𝑖=1
1,𝑝
𝑘
𝑢𝛼𝑛 (𝑡𝑖 )
1 󵄩󵄩 󵄩󵄩𝑝
󵄩󵄩𝑢𝛼𝑛 󵄩󵄩 = 𝜑 (𝑢𝛼𝑛 ) − ∑ ∫
𝐼𝑖 (𝑡) d𝑡
𝑝󵄩 󵄩
𝑖=1 0
𝑢𝛼𝑛 (𝑡𝑖 )
0
𝑖=1
for all 𝑢 ∈ 𝑊0 (0, 𝑇), where 𝑀1 = {𝑎1 , 𝑎2 , . . . , 𝑎𝑘 }, 𝑀2 =
{𝑏1 , 𝑏2 , . . . , 𝑏𝑘 }. Combining (49), (50), and Hölder’s inequality,
we have
𝐼𝑖 (𝑡) d𝑡 − 𝐼𝑖 (𝑢𝛼𝑛 (𝑡𝑖 )) 𝑢𝛼𝑛 (𝑡𝑖 ))
𝑇
+ ∫ [(∇𝐹 (𝑡, 𝑢𝛼𝑛 (𝑡)) , 𝑢𝛼𝑛 (𝑡))
0
−𝑝𝐹 (𝑡, 𝑢𝛼𝑛 (𝑡))] d𝑡
𝑇
󵄨 󵄨𝛽
𝛽
≥ −𝑘𝐶9 + 𝐶7 ∫ 󵄨󵄨󵄨󵄨𝑢𝛼𝑛 󵄨󵄨󵄨󵄨 d𝑡 − 𝐶7 𝜌2 𝑇 − 𝐶8
0
𝑇
𝑇
× ∫ 𝑏 (𝑡) d𝑡,
+ ∫ 𝐹 (𝑡, 𝑢𝛼𝑛 (𝑡)) d𝑡
0
0
𝑘
≤ 𝐶4 + 𝑀1 𝑘𝐶3 ‖𝑢‖ + 𝑀2 𝐶3 ∑‖𝑢‖
𝑟𝑖 +1
𝑇
𝑇
󵄨𝜆
󵄨
+ 𝐶5 ∫ 󵄨󵄨󵄨󵄨𝑢𝛼𝑛 (𝑡)󵄨󵄨󵄨󵄨 d𝑡 + max 𝑎 (𝑠) ∫ 𝑏 (𝑡) d𝑡
𝑠∈[0,𝜌 ]
0
0
1
𝑘
≤ 𝐶4 + 𝑀1 𝑘𝐶3 ‖𝑢‖ + 𝑀2 𝐶3 ∑‖𝑢‖𝑟𝑖 +1
for all large 𝑛. From (56), ∫0 |𝑢𝛼𝑛 |𝛽 d𝑡 is bounded. If 𝛽 > 𝜆, by
Hölder’s inequality, we have
𝑇
𝑇
󵄨 󵄨𝜆
󵄨 󵄨𝛽
∫ 󵄨󵄨󵄨󵄨𝑢𝛼𝑛 󵄨󵄨󵄨󵄨 d𝑡 ≤ 𝑇(𝛽−𝜆)/𝛽 (∫ 󵄨󵄨󵄨󵄨𝑢𝛼𝑛 󵄨󵄨󵄨󵄨 d𝑡)
0
0
󵄨𝜆
󵄨󵄨
󵄨󵄨𝑢𝛼𝑛 (𝑡)󵄨󵄨󵄨 d𝑡 + 𝐶6 ,
󵄨
󵄨
(51)
󵄨𝛽
󵄩 󵄩𝜆−𝛽 𝑇 󵄨
≤ 󵄩󵄩󵄩󵄩𝑢𝛼𝑛 󵄩󵄩󵄩󵄩∞ ∫ 󵄨󵄨󵄨󵄨𝑢𝛼𝑛 (𝑡)󵄨󵄨󵄨󵄨 d𝑡
0
for all large 𝑛, where 𝐶6 = max𝑠∈[0,𝜌1 ] 𝑎(𝑠) ∫0 𝑏(𝑡)d𝑡. On the
other hand, by (𝐻5 ), there exist 𝐶7 > 0 and 𝜌2 > 0 such that
(53)
(54)
(58)
Since 𝜉𝑖𝑗 ∈ [0, 1), 𝜆 − 𝛽 < 2, by (51) and (58), {𝑢𝛼𝑛 } is
1,𝑝
also bounded in 𝑊0 (0, 𝑇). Hence, {𝑢𝛼𝑛 } is also bounded
1,𝑝
in 𝑊0 (0, 𝑇). Going if necessary to a subsequence, we can
1,𝑝
assume that 𝑢𝛼𝛼 ⇀ 𝑢 in 𝑊0 (0, 𝑇). As the same the proof
𝑛
for all |𝑥| ≤ 𝜌2 and a.e. 𝑡 ∈ [0, 𝑇], where 𝐶8 = (2 +
𝜌2 )max𝑠∈[0,𝜌2 ] 𝑎(𝑠). Combining (52) and (53), one has
𝛽
(57)
󵄨𝛽
󵄩𝜆−𝛽 𝑇 󵄨
𝜆−𝛽 󵄩
≤ 𝐶1 󵄩󵄩󵄩󵄩𝑢𝛼𝑛 󵄩󵄩󵄩󵄩 ∫ 󵄨󵄨󵄨󵄨𝑢𝛼𝑛 (𝑡)󵄨󵄨󵄨󵄨 d𝑡.
0
(52)
for all |𝑥| ≥ 𝜌2 and a.e. 𝑡 ∈ [0, 𝑇]. By (A),
(𝑓 (𝑡, 𝑥) , 𝑥) − 2𝐹 (𝑡, 𝑥) ≥ 𝐶7 |𝑥|𝛽 − 𝐶7 𝜌2 − 𝐶8 𝑏 (𝑡) ,
.
𝑇
𝑇
󵄨𝜆
󵄨𝛽 󵄨
󵄨𝜆−𝛽
󵄨
󵄨
∫ 󵄨󵄨󵄨󵄨𝑢𝛼𝑛 (𝑡)󵄨󵄨󵄨󵄨 d𝑡 = ∫ 󵄨󵄨󵄨󵄨𝑢𝛼𝑛 (𝑡)󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨𝑢𝛼𝑛 (𝑡)󵄨󵄨󵄨󵄨 d𝑡
0
0
𝑇
󵄨
󵄨󵄨
󵄨󵄨𝑥𝑓 (𝑡, 𝑥) − 2𝐹 (𝑡, 𝑥)󵄨󵄨󵄨 ≤ 𝐶8 𝑏 (𝑡) ,
𝜆/𝛽
Since 𝜉𝑖𝑗 ∈ [0, 1) for all 𝑖 = 1, 2, . . . , 𝑘, by (51) and (57), {𝑢𝛼𝑛 }
1,𝑝
is bounded in 𝑊0 (0, 𝑇). If 𝛽 ≤ 𝜆, by Lemma 2, we obtain
𝑖=1
𝑥𝑓 (𝑡, 𝑥) − 2𝐹 (𝑡, 𝑥) ≥ 𝐶7 |𝑥|𝛽 ,
(56)
𝑇
𝑖=1
0
0
(𝑝 + 1) 𝐶4 ≥ 𝑝𝜑 (𝑢𝛼𝑛 ) − ⟨𝜑󸀠 (𝑢𝛼𝑛 ) , 𝑢𝛼𝑛 ⟩
󵄨󵄨 𝑘 |𝑢(𝑡𝑖 )|
󵄨󵄨 𝑘 |𝑢(𝑡𝑖 )|
󵄨󵄨
󵄨󵄨
󵄨󵄨∑ ∫
𝐼𝑖 (𝑡) d𝑡󵄨󵄨󵄨 ≤ ∑ ∫
(𝑎𝑖 + 𝑏𝑖 |𝑡|𝑟𝑖 ) d𝑡
󵄨󵄨
󵄨
󵄨󵄨𝑖=1 0
󵄨󵄨 𝑖=1 0
𝑇
2 ∫ 𝐼𝑖𝑗 (𝑠) d𝑠 − 𝐼𝑖𝑗 (𝑡) 𝑡 ≥ −𝐶9
Thus by (46), (54), and (55), we obtain
From (𝐻2 ) and Lemma 2, we have that
+ 𝐶5 ∫
for all 𝑥 ∈ R𝑁 and a.e. 𝑡 ∈ [0, 𝑇]. According to (𝐻6 ), there
exists 𝐶9 > 0 such that
1,𝑝
of Lemma 15, Therefore, 𝑢𝛼𝑛 → 𝑢 in 𝑊0 (0, 𝑇). Hence 𝜑
satisfies the (𝐶)∗ condition.
Theorem 17. Assume that (𝐴), (𝐻1 ), (𝐻2 ), (𝐻3 ), (𝐻5 ), and
(𝐻6 ) are satisfied and the following conditions hold.
8
Journal of Applied Mathematics
𝑇 𝛾
2𝑘𝑀1
− ∫ ( |𝑢|𝑝 +
|𝑢|) d𝑡
𝑝
𝑇
0
(𝐻7 ) There exist constants 𝛿 > 0, 𝐾 > 0, 𝛾 > 0 such that
𝛾 𝑝 2𝑘𝑀1
1
𝑝
𝑝+1
,
|𝑢| +
|𝑢| ≤ 𝐹 (𝑡, 𝑢) ≤
𝑝 |𝑢| + 𝐾|𝑢|
𝑝
𝑇
2𝑝𝛾𝑝
≤
𝛾 𝑝
|𝑢| + 𝑘𝑀1 ‖𝑢‖∞ + 𝑀2
𝑝 𝑝
(59)
𝑘
𝑟𝑖 +1
× ∑‖𝑢‖∞
−
for 𝑡 ∈ [0, 𝑇] , |𝑢| ≤ 𝛿.
𝑖=1
𝛾 𝑝
|𝑢| − 2𝑘𝑀1 ‖𝑢‖∞
𝑝 𝑝
𝑘
𝑟𝑖 +1
≤ −𝑘𝑀1 ‖𝑢‖∞ + 𝑀2 ∑‖𝑢‖∞
≤ 0.
(𝐻8 ) 𝐼𝑖 (𝑢) (𝑖 = 1, 2, . . . , 𝑘) is nondecreasing.
𝑖=1
(62)
Then the problem (1) has at least two critical points.
Proof. Let 𝑋𝑛1 = span{𝜆 𝑘+1 , 𝜆 𝑘+2 , . . . , 𝜆 𝑘+𝑛 }, 𝑋𝑛2 = 𝑋2 =
𝑗
(𝑋1 )⊥ for 𝑛 ∈ 𝑁. Then 𝑋𝑗 = ⋃𝑛∈𝑁 𝑋𝑛 , 𝑗 = 1, 2. If 𝑢 ∈ 𝑋1
𝑝
one has |𝑢|𝑝 ≤ 𝛾𝑝 ‖𝑢‖, and if 𝑢 ∈ 𝑋2 , we have ‖𝑢‖𝑝 ≤ 𝛾|𝑢|𝑝 .
Step 1. 𝜑 has a local linking at 0 with respect to (𝑋1 , 𝑋2 ).
For 𝑢 ∈ 𝑋1 , using (𝐻7 ) we have
𝑘
∑∫
𝑢(𝑡𝑖 )
𝑖=1 0
𝐼𝑖 (𝑡) d𝑡 ≥ 0.
Take 𝑟 = min{𝑟1 , 𝑟2 }. We know that 𝜑 has a local linking at 0
with respect to (𝑋1 , 𝑋2 ).
Step 2. 𝜑 maps bounded sets into bounded sets.
Assume ‖𝑢‖ ≤ 𝑅0 , where 𝑅0 is a constant. By (𝐻8 ), one
has
𝜑 (𝑢) ≤
𝑇
(60)
+ ∫ |𝐹 (𝑡, 𝑢 (𝑡))| d𝑡 ≤
0
𝑘
𝑟𝑖 +1
+ 𝑀2 ∑‖𝑢‖∞
+
It follows from (𝐻8 ) that
𝜑 (𝑢) =
𝑖=1
𝑘
𝑢(𝑡𝑖 )
𝑇
1
𝐼𝑖 (𝑡) d𝑡 − ∫ 𝐹 (𝑡, 𝑢 (𝑡)) d𝑡
‖𝑢‖𝑝 + ∑ ∫
𝑝
0
𝑖=1 0
1
1
𝑝+1
𝑝
≥ ‖𝑢‖𝑝 −
𝑝 |𝑢|𝑝 − 𝐾|𝑢|𝑝+1
𝑝
2𝑝𝛾𝑝
≥
1
1 𝑝 𝑝
𝑝+1
𝑝+1
‖𝑢‖𝑝 −
𝑝 𝛾𝑝 ‖𝑢‖ − 𝐾𝛾𝑝+1 ‖𝑢‖
𝑝
2𝑝𝛾𝑝
≤
0
𝑘
𝑢(𝑡𝑖 )
+ ∑∫
𝑖=1 0
𝑟 +1
(63)
𝑝+1
𝑖=1
(61)
≤
3 󵄩󵄩 󵄩󵄩𝑝
󵄩 󵄩
󵄩𝑅 󵄩 + 𝑘𝑀1 𝐶3 󵄩󵄩󵄩𝑅0 󵄩󵄩󵄩 + 𝑀2
2𝑝 󵄩 0 󵄩
𝑘
𝑝+1 󵄩 󵄩𝑝+1
𝑟 +1 󵄩 󵄩𝑟 +1
× ∑𝐶3𝑖 󵄩󵄩󵄩𝑅0 󵄩󵄩󵄩 𝑖 + 𝐾𝛾𝑝+1 󵄩󵄩󵄩𝑅0 󵄩󵄩󵄩
< ∞,
𝑖=1
𝑘
𝑢(𝑡𝑖 )
1
𝐼𝑖 (𝑡) d𝑡
‖𝑢‖𝑝 + ∑ ∫
𝑝
𝑖=1 0
𝑇
1
𝑝+1
‖𝑢‖𝑝 + 𝐾𝛾𝑝+1 ‖𝑢‖𝑝+1
2𝑝
3
‖𝑢‖𝑝 + 𝑘𝑀1 𝐶3 ‖𝑢‖ + 𝑀2
2𝑝
𝑘
Thus, 𝜑(𝑢) ≥ 0 for 𝑢 ∈ 𝑋1 with ‖𝑢‖ ≤ 𝑟1 , where 𝑟1 > 0 is small
enough.
For 𝑢 ∈ 𝑋2 , with ‖𝑢‖ ≤ 𝑟2 := 𝛿/𝐶3 , we have |𝑢| ≤ |𝑢‖∞ ≤
𝐶3 ‖𝑢‖ ≤ 𝛿 since dim 𝑋2 = 𝑘 < +∞. Thus, From (𝐻2 ) and
(𝐻8 ) it follows that
− ∫ 𝐹 (𝑡, 𝑢 (𝑡)) d𝑡 ≤
1
‖𝑢‖𝑝 + 𝑘𝑀1 ‖𝑢‖∞
𝑝
× ∑𝐶3𝑖 ‖𝑢‖𝑟𝑖 +1 + 𝐾𝛾𝑝+1 ‖𝑢‖𝑝+1
1
𝑝+1
=
‖𝑢‖𝑝 − 𝐾𝛾𝑝+1 ‖𝑢‖𝑝+1 .
2𝑝
𝜑 (𝑢) =
𝑘
𝑢(𝑡𝑖 )
1
𝐼𝑖 (𝑡) d𝑡
‖𝑢‖𝑝 + ∑ ∫
𝑝
𝑖=1 0
1
‖𝑢‖𝑝
𝑝
(𝑎𝑖 + 𝑏𝑖 |𝑢|𝑟𝑖 )
which implies that 𝜑 maps bounded sets into bounded sets.
Step 3. For every 𝑚 ∈ 𝑁, 𝜑(𝑢) → −∞ as ‖𝑢‖ → +∞, 𝑢 ∈
1
𝑋𝑚
⊕ 𝑋2 .
By (𝐻5 ), for any 𝑀0 > 0, there exists a constant ℎ(𝑀0 )
such that 𝐹(𝑡, 𝑢(𝑡)) ≥ 𝑀|𝑢|𝑝 − ℎ(𝑀0 ) for all (𝑡, 𝑢) ∈ [0, 𝑇] × 𝑅.
1
⊕ 𝑋2 ) is of finite dimension, there exists 𝛾 > 0
Since dim(𝑋𝑚
𝑝
𝑝
1
such that ‖𝑢‖ ≤ 𝛾|𝑢|𝑝 for all 𝑢 ∈ 𝑋𝑚
⊕ 𝑋2 , which implies
𝜑 (𝑢) ≤
𝑘
𝑢(𝑡𝑖 )
1
𝐼𝑖 (𝑡) d𝑡
‖𝑢‖𝑝 + ∑ ∫
𝑝
𝑖=1 0
− 𝑀|𝑢|𝑝𝑝 + ℎ (𝑀) 𝑇
≤
1
‖𝑢‖𝑝 + 𝑘𝑀1 ‖𝑢‖∞ + 𝑀2
𝑝
Journal of Applied Mathematics
9
𝑘
𝑖=1
≤ (
On the other hand, by the continuity of 𝐹(𝑡, 𝑢), 𝐹(𝑡, 𝑢) is
bounded on [0, 𝑇] × [−𝑀0 , 𝑀0 ], there exists 𝐾1 > 0 such that
𝑀 𝑝
‖𝑢‖ + ℎ (𝑀) 𝑇
𝛾
𝑟𝑖 +1
× ∑‖𝑢‖∞
−
󵄨 󵄨𝜇
𝐹 (𝑡, 𝑢) ≥ −𝐾1 ≥ 𝛼1 |𝑢|𝜇 − 𝛼1 󵄨󵄨󵄨𝑀0 󵄨󵄨󵄨 − 𝐾1 ,
1 𝑀
− ) ‖𝑢‖𝑝 + 𝑘𝑀1 𝐶3 ‖𝑢‖
𝑝
𝛾
𝑘
Combining (69) and (71), we have
𝐹 (𝑡, 𝑢) ≥ 𝛼1 |𝑢|𝜇 − 𝛼2 ,
𝑟 +1
+ 𝑀2 ∑𝐶3𝑖 ‖𝑢‖𝑟𝑖 +1 + ℎ (𝑀) 𝑇.
𝑖=1
(64)
Choosing 𝑀 > 𝛾/𝑝, we have 𝜑(𝑢) → −∞ as ‖𝑢‖ →
1
+∞, 𝑢 ∈ 𝑋𝑚
⊕ 𝑋2 .
Summing up the above, and by Lemma 16, 𝜑 satisfies all
the assumptions of Lemma 11. Hence by Lemma 11, problem (1) has at least one nontrivial solution. The proof of
Theorem 17 is completed.
Theorem 18. Assume that (𝐴), (𝐻1 ), (𝐻2 ), (𝐻3 ), and (𝐻4 ) are
satisfied and the following conditions hold.
(𝐻10 ) 𝐼𝑖 (𝑢) = −𝐼𝑖 (−𝑢) (𝑖 = 1, 2, . . . , 𝑘) and nondecreasing.
Then the problem (1) has an infinite number of nontrivial
solutions.
1,𝑝
Proof. 𝜑 ∈ 𝐶1 (𝑊0 (0, 𝑇), 𝑅), by (𝐻9 ) and (𝐻10 ), 𝜑 is an even
functional and 𝜑(0) = 0.
First, we verify the condition (2) of Lemma 12.
𝑢 ≥ 𝑀0 ,
𝜇 𝑓 (𝑡, 𝑢)
≥
,
𝑢 𝐹 (𝑡, 𝑢)
𝑢 ≤ −𝑀0 .
𝑢 ≥ 𝑀0 ,
𝑀
𝐹 (𝑡, 𝑢)
𝜇 ln 0 ≥ ln
,
−𝑢
𝐹 (𝑡, −𝑢0 )
(66)
𝑢 ≤ −𝑀0 .
That is,
𝐹 (𝑡, 𝑢) ≥ 𝐹 (𝑡, 𝑀0 ) (
𝑢 𝜇
) ,
𝑀0
𝑢 ≥ 𝑀0 ,
(67)
𝐹 (𝑡, 𝑢) ≥ 𝐹 (𝑡, −𝑀0 ) (
−𝑢 𝜇
) ,
𝑀0
𝑢 ≤ −𝑀0 .
(68)
𝜑 (𝑢) =
𝑘
𝑢(𝑡𝑖 )
1
𝐼𝑖 (𝑡) d𝑡
‖𝑢‖𝑝 + ∑ ∫
𝑝
𝑖=1 0
𝑇
0
𝑘
+ ∑∫
𝑢(𝑡𝑖 )
𝑖=1 0
−𝜇
(69)
𝛼1 = 𝑢0 min { min 𝐹 (𝑡, 𝑀0 ) , min 𝐹 (𝑡, −𝑀0 )} > 0. (70)
𝑡∈[0,𝑇]
1
‖𝑢‖𝑝
𝑝
(𝑎𝑖 + 𝑏𝑖 |𝑢|𝑟𝑖 ) d𝑡
𝑇
− ∫ (𝛼1 𝑢𝜇 − 𝛼2 ) d𝑡 ≤
0
1
‖𝑢‖𝑝
𝑝
(74)
𝑘
− 𝛼1 |𝑢|𝜇𝜇 + 𝛼2 𝑇 ≤
𝑘
1
‖𝑢‖𝑝 + 𝑘𝑀1 𝐶3 ‖𝑢‖
𝑝
𝜇
𝑟 +1
+ 𝑀2 ∑𝐶3𝑖 ‖𝑢‖𝑟𝑖 +1 − 𝛼1 𝐶10 ‖𝑢‖𝜇 + 𝛼2 𝑇,
𝑖=1
for every 𝑢 ∈ 𝑊. This implies that 𝜑(𝑢) → −∞ as 𝑢 ∈ 𝑊
and ‖𝑢‖ → ∞. So there exists 𝑅(𝑊) > 0 such that 𝜑 ≤ 0 on
for all 𝑢 ∈ 𝑊 with ‖𝑢‖ ≥ 𝑅.
In the following, we verify the condition (1) of Lemma 12.
1,𝑝
Let 𝑉 = 𝑋1 ⨁ 𝑋2 , 𝑌 = ⨁∞
𝑖=3 𝑋𝑖 , then 𝑊0 (0, 𝑇) =
𝑉 ⨁ 𝑌 and 𝑉 is finite dimensional. Using (𝐻10 ) we have
∑∫
𝑢(𝑡𝑖 )
𝑖=1 0
𝐼𝑖 (𝑡) d𝑡 ≥ 0.
(75)
By (𝐻3 ) and (𝐻4 ), we have
lim
|𝑢| ≥ 𝑀0 ,
(73)
By (𝐻2 ), (72), (73), and Lemma 2, we have
𝑢→0
where
𝑡∈[0,𝑇]
|𝑢|𝜇 ≥ 𝐶10 ‖𝑢‖ .
𝑘
Combining (67) and (68), we have
𝐹 (𝑡, 𝑢) ≥ 𝛼1 |𝑢|𝜇 ,
where 𝛼2 = 𝛼1 + 𝐾1 .
1,𝑝
For arbitrary finite-dimensional subspace 𝑊 ⊂ 𝑊0 (0,
𝑇), and any 𝑢 ∈ 𝑊, there exists 𝐶10 = 𝐶10 (𝑊) > 0 such that
𝑖=1
(65)
𝐹 (𝑡, 𝑢)
𝑢
≤ ln
,
𝑀0
𝐹 (𝑡, 𝑢0 )
(72)
𝑟𝑖 +1
+ 𝑘𝑀1 ‖𝑢‖∞ + 𝑀2 ∑‖𝑢‖∞
Integrating (65) for 𝑢 from [𝑀0 , 𝑢] and [𝑢, −𝑀0 ], respectively,
we have
𝜇 ln
∀ (𝑡, 𝑢) ∈ [0, 𝑇] × 𝑅,
− ∫ 𝐹 (𝑡, 𝑢 (𝑡)) d𝑡 ≤
(𝐻9 ) 𝑓(𝑡, 𝑢) = −𝑓(𝑡, −𝑢) for (𝑡, 𝑢) ∈ [0, 𝑇] × 𝑅.
𝜇 𝑓 (𝑡, 𝑢)
,
≤
𝑢 𝐹 (𝑡, 𝑢)
|𝑢| ≤ 𝑀0 . (71)
𝐹 (𝑡, 𝑢)
= 0.
𝑢𝑝
(76)
𝑝
Hence, for 𝜖 = 1/2𝑝𝛾𝑝 , there exists 𝛿 > 0 such that for every
𝑢 with |𝑢| ≤ 𝛿,
|𝐹 (𝑡, 𝑢)| ≤
1
𝑝
𝑝 |𝑢| .
2𝑝𝛾𝑝
(77)
10
Journal of Applied Mathematics
Hence, for any 𝑢 ∈ 𝑌 with ‖𝑢‖ ≤ 𝛿/𝐶3 , ‖𝑢‖∞ ≤ 𝛿, by (18),
(54) and (55), we have
𝑘
𝑢(𝑡𝑖 )
1
𝐼𝑖 (𝑡) d𝑡
𝜑 (𝑢) = ‖𝑢‖𝑝 + ∑ ∫
𝑝
𝑖=1 0
𝑇
− ∫ 𝐹 (𝑡, 𝑢 (𝑡)) d𝑡 ≥
0
𝑇
− ∫ 𝐹 (𝑡, 𝑢 (𝑡)) d𝑡 ≥
0
≥
1
‖𝑢‖𝑝
𝑝
1
1
𝑝
‖𝑢‖𝑝 −
𝑝 |𝑢|𝑝
𝑝
2𝑝𝛾𝑝
(78)
𝑝
Take 𝛼 = (1/2𝑝)(𝛿𝑝 /𝐶3 ), 𝜌 = 𝛿/𝐶3 , then
∀𝑢 ∈ 𝜕𝐵𝜌 ∩ 𝑌.
(79)
Hence, by Lemma 12 and Lemma 15, 𝜑 possesses infinite
critical points, that is, problem (1) has infinite nontrivial
solutions. The proof is complete.
4. Example
Example 19. Let 𝑝 = 2, 𝜌(𝑡) = 1, 𝑠(𝑡) = 𝑡, 𝑡1 = 1/2. Consider
the boundary value problem
−𝑢󸀠󸀠 (𝑡) + 𝑡𝑢 (𝑡) = 𝑡 + √𝑢 (𝑡),
a.e. 𝑡 ∈ [0, 1] ,
𝑢 (0) = 𝑢 (1) = 0,
(80)
Δ𝑢󸀠 (𝑡1 ) = 𝑢󸀠 (𝑡1+ ) − 𝑢󸀠 (𝑡1− ) = 1 + √𝑢 (𝑡).
It is easy to see that conditions (𝐻0 ) and (𝐻2 ) of Theorem 13
hold. According to Theorem 13, problem (80) has at least one
solution.
Example 20. Let 𝑝 = 2, 𝜌(𝑡) = 1, 𝑠(𝑡) = 2 − 𝑡, 𝑡1 = 1/3.
Consider the boundary value problem
−𝑢󸀠󸀠 (𝑡) + (2 − 𝑡) 𝑢 (𝑡) = (𝑢 (𝑡))2 ,
a.e. 𝑡 ∈ [0, 1] ,
𝑢 (0) = 𝑢 (1) = 0,
(81)
Δ𝑢󸀠 (𝑡1 ) = 𝑢󸀠 (𝑡1+ ) − 𝑢󸀠 (𝑡1− ) = 1 + √𝑢 (𝑡).
It is easy to check that all the conditions of Theorem 17 are
satisfied. Thus, according to Theorem 17, problem (81) has at
least two critical points.
Example 21. Let 𝑝 = 3, 𝜌(𝑡) = 1 + 2𝑡, 𝑠(𝑡) = 2 + 𝑡, 𝑡1 = 1/2.
Consider the boundary value problem
󸀠
󵄨
󵄨
− ((1 + 2𝑡) 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑢󸀠 (𝑡)) + (2 + 𝑡) 𝑢 (𝑡) = (𝑢 (𝑡))3 ,
a.e. 𝑡 ∈ [0, 1] , 𝑢 (0) = 𝑢 (1) = 0,
This work is supported by the National Natural Sciences
Foundation of China under Grant no. 10971183.
References
1
1 𝑝 𝑝
1
‖𝑢‖𝑝 −
‖𝑢‖𝑝 .
𝑝 𝛾𝑝 ‖𝑢‖ =
𝑝
2𝑝
2𝑝𝛾𝑝
𝜑 (𝑢) ≥ 𝛼,
Acknowledgment
(82)
Δ𝑢󸀠 (𝑡1 ) = 𝑢󸀠 (𝑡1+ ) − 𝑢󸀠 (𝑡1− ) = √3 𝑢 (𝑡).
It is easy to check that all the conditions of Theorem 18 are
satisfied. Therefore, according to Theorem 18, problem (82)
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