Vol. 20 (2019), No. 1, pp. 153–168 DOI: 10.18514/MMN.2019.2583
MULTIPLE SOLUTIONS FOR ASYMPTOTICALLY LINEAR 2p-ORDER HAMILTONIAN SYSTEMS WITH IMPULSIVE
EFFECTS
YUCHENG BU Received 04 April, 2018
Abstract. In this paper, we are concerned with2p-order Hamiltonian systems with impulsive effects. We investigate the variational structure associated to this system. In addition, we obtain some results of multiple solutions for asymptotically linear2p-order Hamiltonian systems via variational methods and critical point theorems. Meanwhile, some examples are presented to illustrate our main results.
2010Mathematics Subject Classification: 34B15; 34B37; 58E30
Keywords: Hamiltonian systems, impulse, critical points, periodic boundary conditions
1. INTRODUCTION AND MAIN RESULTS
In recent years, variational methods have been introduced to investigate various impulsive differential equations since papers [10,13] appeared. As one kind of widely applicable differential equations, Hamiltonian systems with impulsive effects have been also concerned on widely and many new corresponding results have been obtained, see for instance [3,7–9,11,12,15]. However, in aforementioned papers Hamiltonian systems are second order. To the best of our knowledge, few authors have considered asymptotically linear2p-order Hamiltonian systems with impulsive effects. One difficulty is that the suitable impulsive effects associated to this system have been not found. Another is that the suitable critical point theorems have been not applied. In this paper, we present such impulsive conditions. More precisely, we investigate multiple solutions for
8 ˆˆ
<
ˆˆ :
. 1/pC1u.2p/C rV .t; u/D0; a:e: t2Œ0; T ;
u.j /.0/Du.j /.T /; j D0; 1; ; 2p 1;
.ui.2p j /.tk//DIij k.ui.j 1/.tk//;
iD1; 2; ; N; j D1; 2; ; p; kD1; 2; ; q;
(1.1)
The author was supported in part by the National Natural Science Foundation of China, Grant No.
11171157.
c 2019 Miskolc University Press
wherepis a positive integer,u.t /D.u1.t /; u2.t /; ; uN.t //,V WŒ0; T RN !R is measurable with respect tot, for everyu2RN, continuously differentiable inu, for almost everyt2Œ0; T ,0Dt0< t1< < tq< tqC1DT,tk.kD1; 2; ; q/are the instants where the impulses occur,.ui.2p j /.tk//Dui.2p j /.tkC/ ui.2p j /.tk/ andIij kWR!R.iD1; 2; ; N; j D1; 2; ; p; kD1; 2; ; q/are continuous.
From now on, we writeA,BandCasf1; 2; ; Ng;f1; 2; ; pgandf1; 2; ; qg respectively. In addition,Ls.RN/stands for the space of symmetric matrices of order N andIN is the unit matrix inLs.RN/. For anyA1; A22Ls.RN/, we denote by A1A2 ifA2 A1is positively semi-definite, and denote byA1< A2 ifA2 A1 is positively definite. For anyA1; A22L1.Œ0; T ILs.RN//, we denote byA1A2
ifA1.t /A2.t /fora:e: t 20; T Œ, and denote byA1< A2ifA1A2andA1.t / <
A2.t /on a subset of0; T Œwith nonzero measure.
Let us have the space H1pWDn
u2Hp.0; T ŒIRN/ju.j /.0/Du.j /.T /; j D0; 1; ; p 1o with the inner product
hu; vi D Z T
0
.u.p/.t /; v.p/.t //C.u.t /; v.t // dt; 8u; v2H1p;
where.;/denotes the inner product inRN. The corresponding norm is defined by kuk D
Z T
0 ju.p/.t /j2C ju.t /j2dt
!12
; 8u2H1p: Suppose thatIij kandV satisfy that the following some conditions:
.I1/ EveryIij k.i 2A; j 2B; k2C/is bounded andIij k.0/D0.
.I2/ EveryIij k.i 2A; j 2B; k2C/is odd.
.V1/ V .t; u/is twice continuously differentiable inufor a.e.t2Œ0; T .
.V2/ rV .t; 0/0and setA0.t /Du2V .t; 0/.
.V3/ There existA1; A22L1.Œ0; T ILs.RN//andr > 0such that A1.t /Du2V .t; u/A2.t /
for everyu2RN withjuj r, and a.e.t2Œ0; T .
.V4/ V .t; u/DV .t; u/for everyu2RN and a.e. t2Œ0; T .
.V5/ V .t; 0/0.
Our main results are the following two theorems.
Theorem 1. Suppose that.I1/; .V1/ .V3/withip.A1/Dip.A2/ > 0,
p.A2/D0, andip.A1/2 Œip.A0/; ip.A0/Cp.A0/hold. Then,.1:1/ has at least one nontrivial weak solution. Further, ifp.A0/D0andˇ
ˇip.A1/ ip.A0/ˇ ˇpN, then problem.1:1/has two nontrivial weak solutions.
Theorem 2. Suppose that.I1/; .I2/; .V1/ .V5/withip.A1/Dip.A2/ > 0, p.A2/D0, andp.A0/D0hold. Then, problem.1:1/has at leastjip.A1/ ip.A0/j distinct pairs of nontrivial weak solutions.
Remark1. Hereip.A/andp.A/are called the index and nullity ofArespectively.
Indeed, for anyA2L1.Œ0; T ILs.RN//, we define A.u; v/D
Z T 0
.u.p/.t /; v.p/.t // dt Z T
0
.A.t /u.t /; v.t // dt; 8u; v2H1p: For anyx; y2H1pifA.x; y/D0, we say thatxandy A-orthogonal. IfH1; H2are the two subsets ofH1pand for anyx2H1andy2H2,A.x; y/D0, we say thatH1
andH2 A-orthogonal. In addition,H1p has aA-orthogonal decompositionH1pD H1pC.A/˚H1p0.A/˚H1p .A/ such thatA is positive definite, zero and negat- ive definite on H1pC.A/; H1p0.A/ andH1p .A/ respectively. Moreover, H1p0.A/
andH1p .A/are finitely dimensional. Hence, for anyA2L1.Œ0; T ILs.RN//, we define ip.A/DH1p .A/, p.A/DdimH1p0.A/. These results are the immediate conclusions of Proposition2:1:1, Definition2:1:2in [5].
This paper is organized as follows. In Section 2, we first recall several critical point theorems. Then, we investigate the variational structure associated to problem .1:1/inH1p. Finally, we quote the two lemmas which are crucial in our argument.
In Section 3, we verify our main results by applying variational methods and critical point theorems when V satisfies the generalized asymptotically linear conditions.
Our results extends some conclusions directly in [4]. Analogously, by new definition of weak solution, one can be dealt with problem.1:1/whenV satisfies some other conditions, such as the convex potential condition, the even type potential condition, the Ahmad-Lazer-Paul type coercive condition and its several generalizations, the sublinear potential condition, the superquadratic potential condition, the subquadratic potential condition and the asymptotically quadratic potential condition. In Section 4, we present some examples in order to illustrate our results.
2. PRELIMINARIES
LetX be a Hilbert space. We first recall some critical point theorems in critical point theory. These theorems are due to K. C. Chang.
Theorem 3( [1, Theorem 4.3.4]). Letf 2C1.X;R/be even andf .0/D0. As- sumef satisfiesPS condition and
(i) there is an m-dimensional subspaceX1and a constantr > 0such that sup
x2X1\@Ur
f .x/ < 0;
(ii) there is a j-codimensional subspaceX2such that
xinf2X2
f .x/ > 1:
Thenf has at leastm j distinct pairs of critical points providedm j > 0.
Theorem 4([1, Theorem 4.3.6]). Letf 2C1.X;R/be even and f .0/D0. As- sumef satisfiesPS condition and
(i) there is a j-codimensional subspaceX1and two constantsr; ˛ > 0such that f .x/˛ f or any x2X1\@Ur;
(ii) there is a m-dimensional subspaceX2and a constantR > 0such that f .x/0 f or any x2X2nUR:
Thenf has at leastm j distinct pairs of critical points providedm j > 0.
The last one is called three solutions theorem and can be verified by Theorem5:1, Theorem5:2and Corollary5:2in [2]. One can find its proof in [6].
Theorem 5 ([6, Proposition 5.5.2]). Assume f 2C2.X;R/ and satisfies PS- condition,f00.x/is Fredholm with finite Morse index for each critical pointx2X andf0.0/D0. Suppose there is a positive integer such that
2Œm .f00.0//; m0.f00.0//Cm .f00.0//
andHq.X; faIR/ŠıqRfor some regular valuea < I.0/, whereıq D
1; ıD; 0; ı¤: Then,f have a critical pointx0¤0. Moreover, if0is a non-degenerate critical point, andm0.f00.x0// j m .f00.0//j, thenf have another critical pointx1¤x0; 0.
Remark 2. Herefa D fx2Xjf .x/a; a2Rg. The Morse nullity and Morse index off atx2X are defined as dim.kerf00.x//and the supremum of the dimen- sions of the vector subsequence ofXin whichf00.x/is negative definite respectively.
Both are denoted bym0.f00.x//andm .f00.x//respectively.
Next, we investigate the variational structure of (1.1). This idea comes from papers [10,14].
Ifu2H1p, thenu.2p j /i .tkC/andu.2p j /i .tk/.i2A; j 2B; k2C/may not exist.
This leads to impulsive effects.
Let us takey2H1p, multiply both sides of the equation in (1.1) byy, and integrate between0andT, then
Z T 0
.. 1/pC1u.2p/; v/ dtC Z T
0
.rV .t; u/; v/ dt D0: (2.1) Moreover, combingu.2p 1/.0/Du.2p 1/.T /, one has
Z T 0
.. 1/pC1u.2p/; v/ dt
D. 1/pC1
q
X
kD0
Z tkC1
tk
.u.2p/; v/ dt
D. 1/pC1
q
X
kD0
.u.2p 1/.tkC1/; v.tkC1// .u.2p 1/.tkC/; v.tkC//
. 1/pC1
q
X
kD0
Z tkC1 tk
.u.2p 1/;v/ dtP
D. 1/pC1
N
X
iD1 q
X
kD0
.u.2p 1/i .tkC1/; vi.tkC1// .u.2p 1/i .tkC/; vi.tkC//
C. 1/pC2
q
X
kD0
Z tkC1
tk
.u.2p 1/;v/ dtP
D. 1/pC1
.u.2p 1/.T /; v.T // .u.2p 1/.0C/; v.0C//
C. 1/pC2
N
X
iD1 q
X
kD1
.u.2p 1/i .tk//vi.tk/C. 1/pC2 Z T
0
.u.2p 1/;v/ dtP
D Z T
0
. 1/pC2.u.2p 1/;v/ dtP C. 1/pC2
N
X
iD1 q
X
kD1
Ii1k.ui.tk//vi.tk/:
Similarly, combiningu.2p j /.0/Du.2p j /.T /,j 2Bnf1g, one has Z T
0
.. 1/pC2u.2p 1/;v/ dtP
D Z T
0
.. 1/pC3u.2p 2/;v/ dtR C. 1/pC3
N
X
iD1 q
X
kD1
Ii 2k.uPi.tk//vPi.tk/;
; Z T
0
.. 1/2pu.pC1/; v.p 1// dt
D Z T
0
.. 1/2pC1u.p/; v.p// dtC. 1/2pC1
N
X
iD1 q
X
kD1
Iipk.u.p 1/i .tk//vi.p 1/.tk/:
Hence,
Z T 0
.u.p/; v.p// dt Z T
0
.rV .t; u/; v/ dt
C
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCpIij k.u.j 1/i .tk//vi.j 1/.tk/D0: (2.2) Definition 1. Say thatu2H1p is a weak solution for (1.1) if (2.2) holds for any v2H1p.
Consider the functional'WH1p!Rdefined by putting '.u/WD1
2 Z T
0 ju.p/.t /j2dt Z T
0
V .t; u.t // dtC
N
X
iD1 p
X
jD1 q
X
kD1
Z u.ji 1/.tk/ 0
Iij k.s/ ds:
(2.3) Then,'is Gateaux differential at anyO u2H1p and
h'0.u/; vi D Z T
0
.u.p/; v.p// dt Z T
0
.rV .t; u/; v/ dt
C
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCpIij k.u.j 1/i .tk//v.j 1/i .tk/ (2.4) for anyrV .t; u/2H1p. Hence, the following lemma holds by (2.4) and Definition1.
Lemma 1. Ifu2H1pis a critical point of', thenuis a weak solution for (1.1).
Finally, we quote the two important lemmas. The first lemma is identical with Proposition in [6] whenT D1 and it’s proof is absolutely similar to the proof of Proposition 5.3.1.
Lemma 2. If.V1/-.V3/hold, then for any" > 0, there existsAWŒ0; T RN ! Ls.RN/andgWŒ0; T RN !RN such that
rv.t; u/DA.t; u/uCg.t; u/; (2.5) where
A1 "IN A.t; u/A2C"IN;for anyu2RN a.e.t2Œ0; T ; (2.6) A.; u.//2L1.Œ0; T ;Ls.RN//for allu2L2.Œ0; T ;RN/; (2.7) and
g.; u.//2L1.Œ0; T ;RN/is bounded for allu2L2.Œ0; T ;RN/: (2.8) Remark3. Say thenV satisfies asymptotically linear conditions if (2.5)-(2.8) hold.
Lemma2shows that if.V1/-.V3/hold, thenF satisfies these conditions.
The second lemma is a immediate corollary of Proposition2:1:3in [5].
Lemma 3. For anyA2L1.Œ0; T ILs.RN//, we have (i)p.A/is the dimension of the solution subspace of
. 1/pC1u.2p/CA.t /uD0; t2Œ0; T ;
u.j /.0/Du.j /.T /; j D0; 1; ; 2p 1; (2.9) andp.A/2 f0; 1; ; pNg.
(ii)ip.A/DP
<0p.ACIN/.
(iii) Ifip.A/D0, then Z T
0
ˇ ˇ ˇu.p/.t /
ˇ ˇ ˇ
2
dt Z T
0
.A.t /u.t /; u.t // dt; 8u2H1p: And the equality holds if and only ifu2H1p0.A/.
(iv)p.A/Dm0.A/,ip.A/Dm .A/.
For anyA1; A22L1.Œ0; T ILs.RN//, we have
(v) If A1A2, thenip.A1/ip.A2/ andip.A1/Cp.A1/ip.A2/Cp.A2/; if A1< A2, thenip.A1/Cp.A1/ip.A2/.
(vi) Ifip.A1/Dip.A2/ > 0,p.A2/D0, thenH1pDH1p .A1/˚H1pC.A2/.
3. PROOFS OF MAIN RESULTS
Proof of Theorem1. By Theorem 5 and Lemma 3.iv/, we complete the whole proof by three steps.
Step 1:'2C2.H1p;R/and0is a non-degenerate critical point of'.
If.I1/and.V1/hold, then by the continuity ofIij k.i2A; j 2B; k2C/, for every u2H1p,'00.u/is determined by
h'00.u/v; wi D Z T
0
.w.p/.t /; v.p/.t // dt Z T
0
.Du2V .t; u.t //w.t /; v.t // dt
C
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCpIij k.u.j 1/i .tk//w.j 1/i .tk/ (3.1) for all v; w2H1p. Moreover, we have'2C2.H1p;R/. Meanwhile, (2.4) together with.I1/and.V2/implies'0.0/D0. Namely,0is a critical point of'. In addition, by (3.1), we have
h'00.0/v; wi D Z T
0
.w.p/.t /; v.p/.t // dt Z T
0
.A0.t /w.t /; v.t // dt:
If'00.0/vD0, then for allw2H1p, Z T
0
.w.p/.t /; v.p/.t // dt Z T
0
.A0.t /w.t /; v.t // dtD0:
It implies thatvis a solution of (2.9) whereADA0. Because ofp.A0/D0, we get vD0. The only thing left to do is to prove that for allw2HTp,'00.0/vDwhas one solution inH1p. Indeed,'00.0/DId KwhereKWH1p!H1p defined by
hKv; wi D Z T
0
.A0.t /w.t /; v.t // dt 8v; w2H1p:
Considering thatKis compact, we obtain thatR.'00.0//DH1pfrom ker'00.0/D f0g. Hence,'00.0/has a bounded inverse. Namely,0is a non-degenerate critical point of '.
Step 2:'satisfiesPS-condition.
If.I1/and.V1/ .V3/withi.A1/Di.A2/ > 0,.A2/D0hold, then' satisfies PS-condition.
Letf'.un/gbe a bounded sequence such that'0.un/!0. We first prove thatfung is bounded inHTp. Indeed, by.2:4/, we have
Z T 0
.u.p/n ; v.p// dtD h'0.un/; vi C Z T
0
.rV .t; un/; v/ dt
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCpIij k.u.j 1/ni .tk//v.j 1/i .tk/: (3.2) LetF DC.Œ0; T IRN/with the normkuk1D max
t2Œ0;T ju.t /j. We only need to verify that fung will be bounded in F. If not, we can assume kunk1 ! C1 and set vnDun=kunk1.
By (3.2) and (2.5), we get Z T
0
.u.p/n ; v.p// dt
D kunk11h'0.un/; vi C Z T
0
.A.t; un/vn.t /; v/ dtC kunk11 Z T
0
.g.t; un/; v/ dt
kunk11
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCpIij k.u.j 1/i .tk//vi.j 1/.tk/: (3.3) It implies thatfvngis bounded inH1p. Without loss of generality, we assumevn* v0
inH1p, thenvn!v0inF. By mean ofkvnk1D1, we havekv0k1D1. In addition, (3.3) can become
Z T 0
.u.p/n ; v.p// .A.t; un/vn; v/
dt
D kunk11h'0.un/; vi C kunk11 Z T
0
.g.t; un/; v/ dt
kunk11
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCpIij k.u.j 1/i .tk//vi.j 1/.tk/: (3.4) However, by.2:6/and.2:7/, there existsAQ2L1.Œ0; T ILs.RN//such that
Z T 0
.A.t; un/v.t /; w.t // dt ! Z T
0
.A.t /v.t /; w.t // dtQ 8v; w2L2.Œ0; T ;RN/;
(3.5) by going to subsequences if necessary, and
A1 "IN QAA2C"IN: (3.6) Since'0.un/!0andkunk1! C1, the right side of.3:4/tends to zero..3:4/and .3:5/imply that
Z T 0
.v0.p/.t /; v.p/.t // .A.t /vQ 0.t /; v.t //
dt D0; 8v2H1p: (3.7) Hence,v0is the solution of the following problem
(. 1/pC1v.2p/.t /C QA.t /v.t /D0;
v.j /.0/Dv.j /.T /; j D0; 1; ; 2p 1: (3.8) By (3.6) and Lemma 3.v/, we have ip.A/Q ip.A2C"IN/ and ip.A/Q Cp.A/Q ip.A2C"IN/Cp.A2C"IN/. Butp.A2C"IN/Dp.A2/D0. Hence,p.A/Q D0 and.3:8/has only trivial solution. This contradictskv0k1D1.
Next, we prove thatfungcontains a convergent subsequence. Sincefungis bounded, there exists a subsequencefunmgsuch thatunm * u0inH1p. Thenunm !u0uni- formly inŒ0; 1and
unm!u0 (3.9)
inL2.Œ0; T /;RN/. By.3:2/, we have Z T
0
.u.p/n
m; v.p// dtD h'0.unm/; vi C Z T
0
.rV .t; unm/; v/ dt
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCpIij k.unm
.j 1/
i .tk//vi.j 1/.tk/:
Moreover,'0.unm/!0implies Z T
0
.u.p/0 ; v.p// dtD Z T
0
.rV .t; u0/; v/ dt
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCpIij k.u.j 1/0
i .tk//vi.j 1/.tk/; 8v2H1p:
Hence, Z T 0
u.p/nm.t / u.p/0 .t /
; v.p/.t / dt
D h'0.unm/; vi C Z T
0
..rV .t; unm.t // rV .t; u0.t /// ; v.t // dt
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCp
Iij k.unm
.j 1/
i .tk// Iij k.u.j 1/0
i .tk/
vi.j 1/.tk/:
It follows that kunm u0k
D ku.p/nm u.p/0 kL2C kunm u0kL2
D sup
kvk1
Z 1 0
u.p/nm.t / u.p/0 .t /
; v.p/
dtC kunm u0kL2
D sup
kvk1
'0.unm/; vC Z 1
0
V0.t; unm/ V0.t; u0/ v dt
N
X
iD1 p
X
jD1 q
X
kD1
. 1/jCp
Iij k.unm
.j 1/
i .tk// Iij k.u.j 1/0
i .tk/
v.j 1/i .tk/ 1 A
By'0.unm/!0,.V1/, the continuity ofIij k.i2A; j 2B; k2C/and.3:9/, we have kunm u0k !0. Namely,unm!u0inH1p.
Step 3: For some regular valuea < '.0/,
Hq.H1p; 'aIR/ŠıqR; where Dip.A1/: (3.10) By Lemma2, we know that
rV .t; u/DA.t; u/uCg.t; u/; (3.11) where
A1 "IN A.t; u/A2C"IN
andg.t; u/is bounded since.V1/ .V3/withip.A1/Dip.A2/ > 0hold. In addition, if letA3.t /DA1.t / "IN; A4.t /DA2.t /C"IN, then
A3.t /A.t; u/A4.t /: (3.12)
Moreover, letH1DH1p .A3/; H2DH1pC.A4/, then H1p DH1˚H2by Lemma 3.vi /. Denote
kuk1WD. A3.u; u//12;8u2H1; kuk2WDA4.u; u//12;8u2H2:
From Lemma3.i i i /, we can verify thatk k1andk k2are equivalent tok konH1
andH2respectively. Hence, by.2:4/, for anyuDu1Cu2withu12H1; u22H2, we have
h'0.u/; .u2 u1/i D
Z T 0
.ju.p/2 .t /j2 ju.p/1 .t /j2/ dtC Z T
0
A.t; u.t //u.t /.u2.t / u1.t // dt Z T
0
.g.t; u.t //; .u2.t / u1.t /// dtC
M
X
lD1 N
X
kD1
Ikl.uk.tl//.u2 u1/k.tl/
D Z T
0 ju.p/2 .t /j2dt Z T
0
.A.t; u.t //u2.t /; u2.t // dt Z T
0
.g.t; u.t //; u2.t // dt Z T
0 ju.p/1 .t /j2dtC Z T
0
.A.t; u.t //u1.t /; u1.t // dtC Z T
0
.g.t; u.t //; u1.t // dt
C
M
X
lD1 N
X
kD1
Ikl.uk.tl//uk2.tl/
M
X
lD1 N
X
kD1
Ikl.uk.tl//uk1.tl/
Z T
0 ju.p/2 .t /j2dt Z T
0
.A4u2.t /; u2.t // dt Z T
0
.g.t; u.t //; u2.t // dt Z T
0 ju.p/1 .t /j2dtC Z 1
0
.A3u1.t /; u1.t // dtC Z T
0
.g.t; u.t //; u1.t // dt
C
M
X
lD1 N
X
kD1
Ikl.uk.tl//uk2.tl/
M
X
lD1 N
X
kD1
Ikl.uk.tl//uk1.tl/
D ku2k22
Z T 0
.g.t; u.t //; u2.t // dtC ku1k21C Z T
0
.g.t; u.t //; u1.t // dt
C
M
X
lD1 N
X
kD1
Ikl.uk.tl//uk2.tl/
M
X
lD1 N
X
kD1
Ikl.uk.tl//uk1.tl/
C1ku2k2CC2ku1k2 C3ku2k C4ku1k CC5 (3.13) whereCi 2RC.i D1; 2/, RCis a set of positive constants, Ci 2R.i D3; 4; 5/. It follows that there existsR0> 0such that
h'0.u/; u2 u1i> 1; 8u2H1p
withku2k> R0 orku1k> R0. Denote by MD.H2\ NUR0/˚H1, whereUNR0 is a closed ball with center 0and radiusR0. Since' is decreasing along vector field V .u/D u2Cu1 for every uDu2Cu162M, we can define the flow .t; u/D
e tu2Cetu1, and the timeTuarriving atMsatisfiese Tuku2k DR0. Set .t; u2Cu1/D
(u2Cu1; forkuk R0; .Tut; u/; forkuk> R0:
We can verify that for any a > '.0/ large enough, .t; u/ is a deformation retract from.H1p; 'a/to.M;M\'a/. Hence,
Hq.H1p; 'aIR/ŠHq.M;M\'aIR/: (3.14) On the other hand, by (3.11), we have
V .t; u/D Z T
0
.rV .t; su/; u/ dsCV .t; 0/
D Z T
0
.A.t; su/su; u/ dsC Z T
0
.g.t; su/; u/ dsCV .t; 0/:
Thus, Z T
0
V .t; u/ dt
D Z T
0
Z T 0
.A.t; su/su; u/ dsC Z T
0
.g.t; su/; u/ dsCV .t; 0/
! dt
D Z T
0
Z T 0
.A.t; su/su; u/ ds dtC Z T
0
Z T 0
.g.t; su/; u/ ds dtC Z T
0
V .t; 0/ dt
Z T
0
Z T 0
.A3su; u/ ds dtC Z T
0
Z T 0
.g.t; su/; u/ ds dtC Z T
0
V .t; 0/ dt
D1 2
Z T 0
.A3u; u/ dtC Z T
0
Z T 0
.g.t; su/; u/ ds dtC Z T
0
V .t; 0/ dt because of.3:12/. Hence, for anyuDu1Cu22M, we have
'.u/ 1 2
Z T
0 ju.p/j2dt 1 2
Z T 0
.A3u; u/ dt Z T
0
Z T 0
.g.t; su/; u/ ds dt Z T
0
V .t; 0/ dtC
N
X
iD1 p
X
jD1 q
X
kD1
Z u.ji 1/.tk/ 0
Iij k.s/ ds
1 2
Z T 0
.ju.p/1 j2C ju.p/2 j2C2.u.p/1 ; u.p/2 / dt 1
2 Z T
0
.A3u1; u1/ dt 1
2 Z T
0
.A3u2; u2/ dt Z T
0
.A3u1; u2/ dt Z T
0
Z T 0
.g.t; su/; u1/ ds dt
Z T 0
Z T 0
.g.t; su/; u2/ ds dtCC6
D 1 2
Z T 0
ju.p/1 j2 .A3u1; u1/ dtC1
2 Z T
0
ju.p/2 j2 .A3u2; u2/ dt
C Z T
0
.u.p/1 ; u.p/2 / .A3u1; u2/ dt
Z T 0
Z T 0
.g.t; su/; u1/ ds dt Z T
0
Z T 0
.g.t; su/; u2/ ds dtCC6
DA3.u1; u1/CA3.u2; u2/ Z T
0
Z T 0
.g.t; su/; u1/ ds dt Z T
0
Z T 0
.g.t; su/; u2/ ds dtCC6
C7ku1k2CC8ku1k CC9; (3.15)
whereC72RC; Ci2R.iD6; 8; 9/.
By (3.12), we get0 < A3.u; u/ A4.u; u/. Sincek k1 is equivalent tok k onH1andH1is finitely dimensional,. A4.;//12 is also the norm ofH1. Similar to (3.15), we have
'.u/ C10ku1k2CC11ku1k CC12; (3.16) whereC102RC; Ci2R.iD11; 12/.
(3.15) and (3.16) show that for anyuDu1Cu22M,
'.u/! 1 if and only if ku1k ! C1
uniformly inu22H2\ NUR0. Thus, there existT > 0; a1< a2< T; R1> R2> R0
such that
.H2\ NUR0/˚.H1nUR1/'a1\M.H2\ NUR0/˚.H1nUR2/'a2\M:
Let .t; u/De tu2Cetu1, for everyu2M\.'a2n'a1/,'. .t; u//is continuous with respect tot,'. .0; u//D'.u/ > a1and'. .t; u//! 1.t! C1/, so there exists uniquelytDT1.u/such that .t; u/2M\'a1and'. .t; u//Da1. Define
1.t; u/D
(u; foru2M\'a1; .T1.u/t; u/; foru2M\.'a2n'a1/ and
2.t; u/D
(u; forku1k R1; u2Ct u1C.1 t /kRu1
1ku1; forku1k< R1:
By the map.t; u/D2.t; 1.t; u//we can verify that.H2\ NUR0˚.H1nUR1/IR/
is a strong deformation retract of.M\'a2IR/. Hence,
Hq.M;M\'a2IR/ŠHq..H2\ NUR0/˚H1; .H2\ NUR0/˚.H1nUR1/IR/
ŠHq.H1\ NUR1; @.H1\UR1/IR/
ŠıqR: (3.17)
(3.14) and (3.17) imply that (3.10) holds.
Now all the conditions of Theorem5are satisfied, and the corresponding conclu-
sions hold. The proof is complete.
Proof of Theorem2. By the proof of Theorem 1, we know that'2C1.H1p;R/, and ' satisfies PS-condition when .I1/; .V1/ .V3/ with ip.A1/Dip.A2/ > 0, p.A2/D0hold. Moreover, we find that'is even and'.0/D0when.I1/; .I2/; .V4/ and.V5/hold. In fact,
'. u/
D1 2
Z T
0 j u.p/j2dtC
N
X
iD1 p
X
jD1 q
X
kD1
Z u.ji 1/.tk/ 0
Iij k.s/ ds Z T
0
V .t; u/ dt:
D1 2
Z T
0 ju.p/j2dtC
N
X
iD1 p
X
jD1 q
X
kD1
Z u.ji 1/.tk/ 0
Iij k. s1/ ds1
Z T 0
V .t; u/ dt:
D1 2
Z T
0 ju.p/j2dtC
N
X
iD1 p
X
jD1 q
X
kD1
Z u.ji 1/.tk/ 0
Iij k.s1/ ds1
Z T 0
V .t; u/ dt D'.u/:
Now it suffices to consider the two possibilities: ip.A0/ > ip.A1/andip.A0/ <
ip.A1/. In fact, we only consider the second case because the first case can be in- vestigated as before by Theorem 3. For small" > 0satisfyingp.A0C"IN/D0D p.A1 "IN/,ip.A0C"IN/Dip.A0/andip.A1 "IN/Dip.A1/, similar to.3:13/
and.3:15/, we can verify that for anyx2H1pC.A0C"IN/, '.u/C13kuk2CC14kuk CC15; and for anyx2H1p .A1 "IN/,
'.u/ C16kuk2CC17kuk CC18;
whereCi2RC.iD13; 16/; Ci 2R.iD14; 15; 17; 18/. LetjDip.A0/; mDip.A1/.
Hence, for these two subspacesH1pC.A0C"IN/andH1p .A1 "IN/ofH1p, The-
orem 4.i /.i i /hold. The proof is complete.
4. EXAMPLES
For the sake of simplicity, we only consider (1.1) for the case of N DpDqD T D1. In addition, lett1D12 andI111.s/DjsjCs 1,s2R.
Example1. Consider the following problem 8
ˆ<
ˆ: R
u.t /C rV .t; u/D0;
u.0/Du.1/;u.0/P D Pu.1/;
.u.tP 1//D Pu.t1C/ u.tP 1/DI111.u.t1//;
(4.1)
whereV .t; u/D42e 10t u2C2e 10t Œ.u 6/ln.u2C1/C12uarctanuC2arctanu 2u.
Clearly,jI111.s/j 1andI111.0/D0.V .t; u/isC2inufor everyt2Œ0; T and Du2V .t; u/D2e t4.8C2uu2CC121/. Then,72e 10t < Du2V .t; u/ < 92e 10t asjuj>
5. In addition, A0.t /D202e 10t . By Lemma3, we obtain thati1.7:52e 4t/D i1.8:52e 4t/D3,i1.A0/D5; 1.A0/D0by simple calculations. Hence, (4.1) has two nontrivial weak solutions by Theorem1.
Example2. Consider.4:1/, whereV .t; u/D722e 10t u2 82e 10t Œ 12ln.u2C 1/Cuarctanu. By Example 1, I111 satisfies .I1/. Meanwhile, I111 is odd. V satisfies .V1/-.V2/ and is even in u for every t 2Œ0; T . In addition, V .t; 0/0 and Du2V .t; u/D2e 10t .7 u28C1/. Then 6:22e 10t < Du2V .t; u/ < 72e 10t as juj> 3. Meanwhile, A0.t /D 2e 10t . i1.6:22e 10t /Di1.72e 10t /D3, 1.72e 10t /Di1.A0/D1.A0/D0. Hence, (4.1) has at least three distinct pairs of weak solutions by Theorem2.
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Author’s address
Yucheng Bu
Nanjing Normal University, School of Mathematical Science, Qixia district, Wenyuan road no.1, Nanjing, China, and Zhenjiang College, Danyang Normal School, Danyang district, Qiliang road no.1, Zhenjiang, China
E-mail address:ychbu@126.com
