(ProQuest: ... denotes non-US-ASCII text omitted.)
Chunyan He 1 and Yongzhi Liao 2 and Yongkun Li 1
Recommended by Wan-Tong Li
1, Department of Mathematics, Yunnan University, Kunming, Yunnan 650091, China
2, School of Mathematics and Computer Science, Panzhihua University, Panzhihua, Sichuan 617000, China
Received 24 October 2012; Revised 20 December 2012; Accepted 21 December 2012
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.
1. Introduction
In this paper, we will investigate the existence and multiplicity of solutions to the boundary value problem for impulsive differential equations: [figure omitted; refer to PDF] where p ...5;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 ...4; ρ (t ) < + ∞ ; 0 <s (t ) < + ∞ ; 0 = t 0 < t 1 ... < t k < t k +1 =T , Δ ( ρ ( t i ) | u [variant prime] ( t i ) | p -2 u [variant prime] ( t i ) ) = ρ ( t i + ) | u [variant prime] ( t i + ) | p -2 u [variant prime] ( t i + ) - ρ ( t i - ) | u [variant prime] ( t i - ) | p -2 u [variant prime] ( t i - ) , f : [0 ,T ] ×R [arrow right]R is continuous, I i :R [arrow right]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]).
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 [figure omitted; refer to PDF] 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.
(A) F (t ,u ) is measurable in t for every u ∈ ... and continuously differentiable in u for a.e. t ∈ [0 ,T ] and there exist a ∈C ( ... + , ... + ) , b ∈ L 1 (0 ,T ; ... + ) such that [figure omitted; refer to PDF] for all u ∈ ... and a.e. t ∈ [0 ,T ] , where F ( t ,u ) = ∫ 0 u ...f ( t ,s ) ds .
( H 0 ): There exist constants a ,b >0 and r ∈ [0 ,p -1 ) such that [figure omitted; refer to PDF]
( H 1 ): There exist constants c >0 and τ ∈ [p , + ∞ ) such that [figure omitted; refer to PDF]
( H 2 ): There exist constants a i , b i >0 and r i ∈ [ 0 ,p -1 ) (i =1,2 , ... ,k ) such that [figure omitted; refer to PDF]
( H 3 ): There exist constants μ >p and M >0 such that [figure omitted; refer to PDF]
( H 4 ): lim u [arrow right]0 f (t ,u ) / |u | p -1 =0 uniformly, for t ∈ [0 ,T ] and u ∈R .
( H 5 ): there exist λ >p and β > λ -p such that [figure omitted; refer to PDF]
( H 6 ): there exists ζ ij >0 such that [figure omitted; refer to PDF]
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: [figure omitted; refer to PDF] is given by the sequence of positive numbers [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Each eigenvalue λ k is simple with the associated eigenfunction [figure omitted; refer to PDF] X k denotes the eigenspace associated to λ k , then W 0 1 ,p (0 ,T ) = ... i ∈N X i ¯ .
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.
2. Preliminaries
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.
In [ 20], the Sobolev space W 0 1 ,p (0 ,T ) be the endowed with the norm [figure omitted; refer to PDF]
Throughout the paper, we also consider the norm [figure omitted; refer to PDF] By Poincaré inequality: [figure omitted; refer to PDF] where C ... = ... λ 1 is precisely the largest C ... > ...0 for which the above inequality holds true. Then ∫ 0 T ... | u [variant prime] (t ) | p dt - λ 1 ∫ 0 T ... |u (t ) | p dt ... ...5; ...0 while it minimizes and equals to zero exactly on the ray generated by the first eigenfunction sin p ( π p t /T ) .
Let us recall that [figure omitted; refer to PDF] We denote by | · | p the usual L p -norm. The n -dimensional Lebesgue measure of a set E ⊆ R n is denoted by |E | . By the Sobolev embedding theorem, W 0 1 ,p (0 ,T ) ... L r [0 ,T ] continuously for r ∈ [1 , + ∞ ) , and there exists γ r >0 such that [figure omitted; refer to PDF]
Lemma 1.
There exist C 1 , C 2 >0 such that [figure omitted; refer to PDF]
Proof.
Since ess inf t ∈ [0 ,T ] ρ (t ) >0 , ess inf t ∈ [0 ,T ] s (t ) >0 , 1 ...4; ρ ( t ) < + ∞ , and 0 <s (t ) < + ∞ , we have that ess inf t ∈ [0 ,T ] ρ (t ) : = m 1 ...5;1 , ess inf t ∈ [0 ,T ] s (t ) : = m 2 > - λ 1 , where λ 1 is a positive number and that there exists n 1 ∈ (0,1 ) such that - m 2 ...4; λ 1 (1 - n 1 ) . Thus, by Poincaré inequality, we have [figure omitted; refer to PDF] for all u ∈ W 0 1 ,p (0 ,T ) . Thereby, for every u ∈ W 0 1 ,p (0 ,T ) , [figure omitted; refer to PDF]
On the other hand, by Poincaré inequality, one has [figure omitted; refer to PDF]
Take C 1 = ( m 1 + n 1 -1 ) 1 / p , C 2 = ( || ρ || ∞ + ||s || ∞ / λ 1 ) 1 / p , then [figure omitted; refer to PDF]
The proof is complete.
Lemma 2.
There exists C 3 >0 such that if u ∈ W 0 1 ,p (0 ,T ) , then [figure omitted; refer to PDF]
Proof.
If u ∈ W 0 1 ,p (0 ,T ) , it follows from the mean value theorem that [figure omitted; refer to PDF] for some ξ ∈ (0 ,T ) . Hence, for t ∈ [0 ,T ] , by Hölder inequality and Poincaré inequality [figure omitted; refer to PDF] Hence, ||u || ∞ ...4; ( (C /T ) 1 /p + T 1 /q ) ( 1 / C 1 ) ||u || = C 3 ||u || . The proof is complete.
Take v ∈ W 0 1 ,p (0 ,T ) and multiply the two sides of the equality [figure omitted; refer to PDF] by v and integrate it from 0 to T , we have [figure omitted; refer to PDF] Moreover, [figure omitted; refer to PDF] Combining ( 28), we have [figure omitted; refer to PDF] Considering the above, we introduce the following concept solution for problem ( 1).
Definition 3.
We say that a function u ∈ W 0 1 ,p (0 ,T ) is a solution of problem ( 1) if the identity [figure omitted; refer to PDF] holds for any v ∈ W 0 1 ,p (0 ,T ) .
Consider the functional [straight phi] : W 0 1 ,p (0 ,T ) [arrow right]R defined by [figure omitted; refer to PDF] Using the continuity of f and I i , i =1,2 , ... ,k , one has that [straight phi] ∈ C 1 ( W 0 1 ,p (0 ,T ) ,R ) . For any v ∈ W 0 1 ,p (0 ,T ) , we have [figure omitted; refer to PDF] Thus, the solutions of problem ( 1) are the corresponding critical points of [straight phi] .
Definition 4.
Let X be a normed space. A minimizing sequence for a function [straight phi] :X [arrow right] ( - ∞ , + ∞ ) is a sequence u k such that [straight phi] ( u k ) [arrow right]inf [straight phi] whenever k [arrow right] + ∞ .
Definition 5.
Let X be a Banach space and let [straight phi] :X [arrow right] ( - ∞ , + ∞ ) . [straight phi] is said to be sequentially weakly lower semicontinuous if lim inf k [arrow right] + ∞ [straight phi] ( x k ) ...5; [straight phi] (x ) as x k ...x in X .
Definition 6.
Let E be a Banach space and let c ∈R . For any sequence { u k } in E , if [straight phi] ( u k ) is bounded and [straight phi] [variant prime] ( u k ) [arrow right]0 as k [arrow right] + ∞ possesses a convergent subsequence, then we say that [straight phi] satisfies the Palais-Smale condition (denoted by PS condition for short). We say that [straight phi] satisfies the Palais-Smale condition at level c (denoted by (PS ) c condition for short) if there exists a sequence { u k } in E such that [straight phi] ( u k ) [arrow right]c and [straight phi] [variant prime] ( u k ) [arrow right]0 as k [arrow right] + ∞ implies that c is a critical value of [straight phi] .
Definition 7.
Let E be a Banach space and let [straight phi] :E [arrow right] ( - ∞ , + ∞ ) . [straight phi] is said to be coercive if [straight phi] (u ) [arrow right] + ∞ as ||u || [arrow right] + ∞ .
Lemma 8 (see [ 12]).
If [straight phi] is sequentially weakly lower semicontinuous on a reflexive Banach space X and has a bounded minimizing sequence, then [straight phi] has a minimum on X .
Definition 9.
Let X be a real Banach space with a direct sum decomposition X = X 1 [ecedil]5; X 2 . The functional [straight phi] ∈ C 1 (X ,R ) is said to have a local linking at 0 , with respect to ( X 1 , X 2 ) , if, for some r >0 ,
(i) [straight phi] (u ) ...5;0 , u ∈ X 1 , ||u || ...4;r ,
(ii) [straight phi] (u ) ...4;0 , u ∈ X 2 , ||u || ...4;r .
If [straight phi] 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 X 1 1 ⊂ X 2 1 ⊂ ... ⊂ X 1 and X 1 2 ⊂ X 2 2 ⊂ ... ⊂ X 2 such that [figure omitted; refer to PDF]
Definition 10 (see [ 13, Definition 2.2]).
Let I ∈ C 1 (X , ... ) . The functional I satisfies the (C ) * condition if every sequence ( u α n ) such that α n is admissible and [figure omitted; refer to PDF] contains a subsequence which converges to a critical point of I .
Lemma 11 (see [ 13]).
Suppose that [straight phi] ∈ C 1 (X ,R ) satisfies the following assumptions:
(1) [straight phi] satisfies the (C ) * condition,
(2) [straight phi] has a local linking at 0,
(3) [straight phi] maps bounded sets into bounded sets,
(4) for every m ∈N , [straight phi] (u ) [arrow right] - ∞ as ||u || [arrow right] + ∞ , u ∈ X m 1 [ecedil]5; X 2 .
Then [straight phi] has at least two critical points.
Lemma 12 (see [ 14]).
Let E be a Banach space. Let [straight phi] ∈ C 1 (X ,R ) be an even functional which satisfies the PS condition and [straight phi] (0 ) =0 . If E =V [ecedil]5;Y , where V is finite dimensional, and [straight phi] satisfies the following conditions:
(1) there exist constants ρ , α such that [straight phi] | ∂ B ρ ∩Y ...5; α , where B ρ = {x ∈E : ||x || < ρ } ,
(2) for each finite-dimensional subspace W ⊂E there is R =R (W ) such that [straight phi] (u ) ...4;0 , for all u ∈W with ||u || ...5;R .
Then [straight phi] has an unbounded sequence of critical values.
3. Existence of Periodic Solutions
Theorem 13.
Assume that (A ) , ( H 0 ) , and ( H 2 ) are satisfied, then problem ( 1) has at least one solution.
Proof.
Let M 1 =max { a 1 , a 2 , ... , a k } , M 2 =max { b 1 , b 2 , ... , b k } . By Lemma 2, we have [figure omitted; refer to PDF] for all u ∈ W 0 1 ,p (0 ,T ) . This implies that lim ||u || [arrow right] ∞ [straight phi] (u ) = ∞ , and [straight phi] is coercive.
On the other hand, we show that [straight phi] is weakly lower semicontinuous. If { u k } k ∈N ⊂ W 0 1 ,p (0 ,T ) , u k ...u , then we have that { u k } k ∈N converges uniformly to u on [0 ,T ] and liminf k [arrow right] ∞ || u k || ...5; ||u || . Thus [figure omitted; refer to PDF]
By Lemma 8, [straight phi] has a minimum point on W 0 1 ,p (0 ,T ) , which is a critical point of [straight phi] . Hence, problem ( 1) has at least one solution. The proof is complete.
We readily have the following corollary.
Corollary 14.
Assume that (A ) , ( H 0 ) , and ( H 2 ) are satisfied and f and the impulsive functions I i (i =1,2 , ... ,k ) are bounded. Then problem ( 1) has at least one solution.
Lemma 15.
Assume that ( H 1 ) , ( H 2 ) , and ( H 3 ) are satisfied, then [straight phi] (u ) satisfies the PS condition.
Proof.
Assume that { u n } ⊂ W 0 1 ,p (0 ,T ) satisfies that [straight phi] ( u n ) is bounded and [straight phi] [variant prime] ( u n ) [arrow right]0 as n [arrow right] + ∞ . We will prove that the sequence { u n } is bounded.
It follows from ( H 1 ) , ( H 2 ) , ( H 3 ) , and Lemma 2, we have [figure omitted; refer to PDF] Hence, { u n } is bounded in W 0 1 ,p (0 ,T ) .
Since W 0 1 ,p (0 ,T ) is a reflexive Banach space, passing to a subsequence if necessary, we may assume that there is a u ∈ W 0 1 ,p (0 ,T ) such that [figure omitted; refer to PDF] Notice that [figure omitted; refer to PDF] Recalling the following well-known inequality: for any x ,y ∈ R N , [figure omitted; refer to PDF] for some constant c p (Lemma 4.2 in [ 21]) and using Schwarz inequality, we have [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] By the assumption ( H 1 ) , we have [figure omitted; refer to PDF] From ( 39), it follows that u n [arrow right]u in W 0 1 ,p (0 ,T ) . Thus, [straight phi] (u ) satisfies the PS condition. The proof is complete.
Lemma 16.
Assume that ( A ) , ( H 2 ) , ( H 5 ) , and ( H 6 ) are satisfied, then [straight phi] satisfies the (C ) * condition.
Proof.
Let { u α n } be a sequence in W 0 1 ,p (0 ,T ) such that α n is admissible and [figure omitted; refer to PDF] then there exist a constant C 4 >0 such that [figure omitted; refer to PDF] for all large n . On the other hand, by ( H 5 ) , there are constants C 5 >0 and ρ 1 >0 such that [figure omitted; refer to PDF] for all |x | ...5; ρ 1 and a.e. t ∈ [0 ,T ] . By (A ) one has [figure omitted; refer to PDF] for all |x | ...4; ρ 1 and a.e. t ∈ [0 ,T ] . It follows from ( 47) and ( 48) that [figure omitted; refer to PDF]
From ( H 2 ) and Lemma 2, we have that [figure omitted; refer to PDF] for all u ∈ W 0 1 ,p (0 ,T ) , where M 1 = { a 1 , a 2 , ... , a k } , M 2 = { b 1 , b 2 , ... , b k } . Combining ( 49), ( 50), and Hölder's inequality, we have [figure omitted; refer to PDF] for all large n , where C 6 = max s ∈ [0 , ρ 1 ] a (s ) ∫ 0 T ...b (t )dt . On the other hand, by ( H 5 ) , there exist C 7 >0 and ρ 2 >0 such that [figure omitted; refer to PDF] for all |x | ...5; ρ 2 and a.e. t ∈ [0 ,T ] . By (A ) , [figure omitted; refer to PDF] for all |x | ...4; ρ 2 and a.e. t ∈ [0 ,T ] , where C 8 = (2 + ρ 2 ) max s ∈ [0 , ρ 2 ] a (s ) . Combining ( 52) and ( 53), one has [figure omitted; refer to PDF] for all x ∈ ... N and a.e. t ∈ [0 ,T ] . According to ( H 6 ) , there exists C 9 >0 such that [figure omitted; refer to PDF] Thus by ( 46), ( 54), and ( 55), we obtain [figure omitted; refer to PDF] for all large n . From ( 56), ∫ 0 T ... | u α n | β dt is bounded. If β > λ , by Hölder's inequality, we have [figure omitted; refer to PDF] Since ξ ij ∈ [0,1 ) for all i =1,2 , ... ,k , by ( 51) and ( 57), { u α n } is bounded in W 0 1 ,p (0 ,T ) . If β ...4; λ , by Lemma 2, we obtain [figure omitted; refer to PDF] Since ξ ij ∈ [0,1 ) , λ - β <2 , by ( 51) and ( 58), { u α n } is also bounded in W 0 1 ,p (0 ,T ) . Hence, { u α n } is also bounded in W 0 1 ,p (0 ,T ) . Going if necessary to a subsequence, we can assume that u α α n ...u in W 0 1 ,p (0 ,T ) . As the same the proof of Lemma 15, Therefore, u α n [arrow right]u in W 0 1 ,p (0 ,T ) . Hence [straight phi] satisfies the (C ) * condition.
Theorem 17.
Assume that (A ) , ( H 1 ) , ( H 2 ) , ( H 3 ) , ( H 5 ) , and ( H 6 ) are satisfied and the following conditions hold.
( H 7 ): There exist constants δ >0 , K >0 , γ >0 such that [figure omitted; refer to PDF]
( H 8 ): I i ( u ) (i =1,2 , ... ,k ) is nondecreasing.
Then the problem ( 1) has at least two critical points.
Proof.
Let X n 1 =span { λ k +1 , λ k +2 , ... , λ k +n } , X n 2 = X 2 = ( X 1 ) [perpendicular] for n ∈N . Then X j = ... n ∈N ... X n j ¯ ,j =1,2 . If u ∈ X 1 one has |u | p ...4; γ p ||u || , and if u ∈ X 2 , we have ||u || p ...4; γ |u | p p .
Step 1. [straight phi] has a local linking at 0 with respect to ( X 1 , X 2 ) .
For u ∈ X 1 , using ( H 7 ) we have [figure omitted; refer to PDF] It follows from ( H 8 ) that [figure omitted; refer to PDF] Thus, [straight phi] (u ) ...5;0 for u ∈ X 1 with ||u || ...4; r 1 , where r 1 >0 is small enough.
For u ∈ X 2 , with ||u || ...4; r 2 : = δ / C 3 , we have |u | ...4; |u || ∞ ...4; C 3 ||u || ...4; δ since dim X 2 =k < + ∞ . Thus, From ( H 2 ) and ( H 8 ) it follows that [figure omitted; refer to PDF] Take r =min { r 1 , r 2 } . We know that [straight phi] has a local linking at 0 with respect to ( X 1 , X 2 ) .
Step 2. [straight phi] maps bounded sets into bounded sets.
Assume ||u || ...4; R 0 , where R 0 is a constant. By ( H 8 ) , one has [figure omitted; refer to PDF] which implies that [straight phi] maps bounded sets into bounded sets.
Step 3. For every m ∈N , [straight phi] (u ) [arrow right] - ∞ as ||u || [arrow right] + ∞ , u ∈ X m 1 [ecedil]5; X 2 .
By ( H 5 ) , for any M 0 >0 , there exists a constant h ( M 0 ) such that F (t ,u (t ) ) ...5;M |u | p -h ( M 0 ) for all (t ,u ) ∈ [0 ,T ] ×R . Since dim ( X m 1 [ecedil]5; X 2 ) is of finite dimension, there exists γ >0 such that ||u || p ...4; γ |u | p p for all u ∈ X m 1 [ecedil]5; X 2 , which implies [figure omitted; refer to PDF] Choosing M > γ / p , we have [straight phi] (u ) [arrow right] - ∞ as ||u || [arrow right] + ∞ , u ∈ X m 1 [ecedil]5; X 2 .
Summing up the above, and by Lemma 16, [straight phi] satisfies all the assumptions of Lemma 11. Hence by Lemma 11, problem ( 1) has at least one nontrivial solution. The proof of Theorem 17is completed.
Theorem 18.
Assume that (A ) , ( H 1 ) , ( H 2 ) , ( H 3 ) , and ( H 4 ) are satisfied and the following conditions hold.
( H 9 ) : f (t ,u ) = -f (t , -u ) for (t ,u ) ∈ [0 ,T ] ×R .
( H 10 ) : I i (u ) = - I i ( -u ) (i =1,2 , ... ,k ) and nondecreasing.
Then the problem ( 1) has an infinite number of nontrivial solutions.
Proof.
[straight phi] ∈ C 1 ( W 0 1 ,p (0 ,T ) ,R ) , by ( H 9 ) and ( H 10 ), [straight phi] is an even functional and [straight phi] (0 ) =0 .
First, we verify the condition (2) of Lemma 12. [figure omitted; refer to PDF] Integrating ( 65) for u from [ M 0 ,u ] and [u , - M 0 ] , respectively, we have [figure omitted; refer to PDF] That is, [figure omitted; refer to PDF] Combining ( 67) and ( 68), we have [figure omitted; refer to PDF] where [figure omitted; refer to PDF] On the other hand, by the continuity of F (t ,u ) , F (t ,u ) is bounded on [0 ,T ] × [ - M 0 , M 0 ] , there exists K 1 >0 such that [figure omitted; refer to PDF] Combining ( 69) and ( 71), we have [figure omitted; refer to PDF] where α 2 = α 1 + K 1 .
For arbitrary finite-dimensional subspace W ⊂ W 0 1 ,p (0 ,T ) , and any u ∈W , there exists C 10 = C 10 (W ) >0 such that [figure omitted; refer to PDF] By ( H 2 ) , ( 72), ( 73), and Lemma 2, we have [figure omitted; refer to PDF] for every u ∈W . This implies that [straight phi] (u ) [arrow right] - ∞ as u ∈W and ||u || [arrow right] ∞ . So there exists R (W ) >0 such that [straight phi] ...4;0 on for all u ∈W with ||u || ...5;R .
In the following, we verify the condition (1) of Lemma 12.
Let V = X 1 ... X 2 , Y = ... i =3 ∞ X i ¯ , then W 0 1 ,p (0 ,T ) =V ...Y and V is finite dimensional. Using ( H 10 ) we have [figure omitted; refer to PDF] By ( H 3 ) and ( H 4 ) , we have [figure omitted; refer to PDF] Hence, for ... = 1 / 2p γ p p , there exists δ >0 such that for every u with |u | ...4; δ , [figure omitted; refer to PDF] Hence, for any u ∈Y with ||u || ...4; δ / C 3 , ||u || ∞ ...4; δ , by ( 18), ( 54) and ( 55), we have [figure omitted; refer to PDF] Take α = ( 1 / 2p ) ( δ p / C 3 p ) , ρ = δ / C 3 , then [figure omitted; refer to PDF] Hence, by Lemma 12and Lemma 15, [straight phi] 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 , t 1 =1 /2 . Consider the boundary value problem [figure omitted; refer to PDF] It is easy to see that conditions ( H 0 ) and ( H 2 ) of Theorem 13hold. According to Theorem 13, problem ( 80) has at least one solution.
Example 20.
Let p =2 , ρ (t ) =1 , s (t ) =2 -t , t 1 =1 /3 . Consider the boundary value problem [figure omitted; refer to PDF] It is easy to check that all the conditions of Theorem 17are 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 , t 1 =1 /2 . Consider the boundary value problem [figure omitted; refer to PDF] It is easy to check that all the conditions of Theorem 18are satisfied. Therefore, according to Theorem 18, problem ( 82) has infinite nontrivial solutions.
Acknowledgment
This work is supported by the National Natural Sciences Foundation of China under Grant no. 10971183.
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Copyright © 2013 Chunyan He et al. 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.
Abstract
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.
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Neither ProQuest nor its licensors make any representations or warranties with respect to the translations. The translations are automatically generated "AS IS" and "AS AVAILABLE" and are not retained in our systems. PROQUEST AND ITS LICENSORS SPECIFICALLY DISCLAIM ANY AND ALL EXPRESS OR IMPLIED WARRANTIES, INCLUDING WITHOUT LIMITATION, ANY WARRANTIES FOR AVAILABILITY, ACCURACY, TIMELINESS, COMPLETENESS, NON-INFRINGMENT, MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Your use of the translations is subject to all use restrictions contained in your Electronic Products License Agreement and by using the translation functionality you agree to forgo any and all claims against ProQuest or its licensors for your use of the translation functionality and any output derived there from. Hide full disclaimer