Existence of solutions for fourth order elliptic equations of Kirchhoff type on R N
Fanglei Wang
B1, Tianqing An
1and Yukun An
21College of Science, Hohai University, Nanjing, 210098, P. R. China
2Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing, 210016, P. R. China
Received 13 February 2014, appeared 4 August 2014 Communicated by Patrizia Pucci
Abstract. In this paper, we study the positive solutions to a class of fourth order elliptic equations of Kirchhoff type on RN by using variational methods and the truncation method.
Keywords: Kirchhoff type problems, variational methods, truncation method.
2010 Mathematics Subject Classification: 35J65, 35J50.
1 Introduction
The purpose of this work is to study the existence of positive solutions for the fourth order elliptic equations:
∆2u−
a+b Z
RN
|∇u|2dx
∆u+cu = f(u) (1.1)
where N > 4, ∆2 is the biharmonic operator, and ∇u denotes the spatial gradient of u, and a, b, c are positive constants. Usually, the proof is based on either variational approach or topological methods. For example, in [7, 8, 13], T. F. Ma, F. Wang et al. applied the varia- tional methods to study the existence and multiplicity of solutions for a nonlocal fourth order equation of Kirchhoff type:
u0000−M R1
0 |u0|2dx
u00 =h(x)f(x,u), u(0) =u(1) =0, u00(0) =u00(1) =0,
and (
∆2u−M R
Ω|∇u|2dx
∆u= f(x,u), inΩ, u=∆u=0, on∂Ω,
whereΩ⊂ RN is a bounded smooth domain,M: R→ Ris continuous, and satisfies
BE-mail: wang-fanglei@hotmail.com.
This research is supported by Natural Science Foundation of JiangSu Province (Grant No. 1014-51314411)
(H) For somem0 >0, M(t)>m0,∀ t >0. In addition, there existm0 >m0 andt0 >0, such that M(t) =m0, ∀ t>t0.
In [10], by using the fixed point theorems in cones of ordered Banach spaces, T. F. Ma studied the existence of positive solutions for
u0000−M Z 1
0
|u0|2dx
u00= h(x)f(x,u,u0).
In [9,11] also fourth order problems with nonlinear boundary conditions are studied. In this case the Kirchhoff function is possibly degenerate and multiplies lower order terms rather than the leading fourth order term. More recently, in [14], F. Wang et al. studied the existence of nontrivial solutions for the fourth order elliptic equations:
(∆2u−λ a+bR
Ω|∇u|2dx
∆u= f(x,u), inΩ,
u=0, ∆u=0, on∂Ω,
where λ is a positive parameter, a, b are positive constants, Ω ⊂ RN is a bounded smooth domain, f: Ω×R → R is locally Lipschitz continuous. The authors show that there exists a λ∗ such that the fourth order elliptic equation has nontrival solutions for 0< λ<λ∗ by using the mountain pass techniques and the truncation method.
In addition, the problem treated here presents the Kirchhoff function multiplying a lower order term, while in general it multiplies the leading (fourth order) operator. Some results related to problems involving Kirchhoff functions in front of lower order terms are obtained in [2,3]. Some other Kirchhoff problems are also been studied. For example, in [1,5,6,15], the authors studied the existence of positive solutions of second order non-degenerate Kirchhoff- type problems; in [4], F. Colasuonno and P. Pucci studied a higher order elliptic Kirchhoff equation, under Dirichlet boundary conditions. The novelty there is to take the Kirchhoff function possibly zero at zero, that is to cover also the degenerate case.
The object of this paper is to study the existence of a positive solution to the fourth order elliptic equation (1.1) of Kirchhoff type onRN by using variational methods. In particular, we use a cut-off functional to obtain bounded (PS)-sequences. The main result can be described as follows.
Theorem 1.1. Assume that the following conditions hold:
(H1) f: R+ →R+is continuous, f(t)≡0, if t≤0and satisfies f(t)≤C(1+tp) ∀ t∈ R+, where1< p< NN+−44 if N>4;
(H1) limt→0 f(t) t =0;
(H1) limt→+∞ f(t)
t = +∞.
Then there exists b∗ >0such that problem(1.1)has at least one positive solution for0≤ b<b∗.
2 Preliminaries
In this section, we show examples how theorems, definitions, lists and formulae should be formatted.
Let H = {u ∈ H2(RN) : u(x)is radial}, where H2(RN) is the usual Sobolev space. We equip Hwith the inner product
(u,v) =
Z
RN
(∆u∆v+a∇u∇v+cuv)dx, and the deduced norm
kuk2=
Z
RN
|∆u|2dx+a Z
RN
|∇u|2+cu2dx.
For the Kirchhoff problem (1.1), the associated function is defined onHas follows J(u) = 1
2kuk2+b 4
Z
RN
|∇u|2dx 2
−
Z
RNF(u)dx where F(t) = Rt
0 f(s)ds. By (H1), we know that J is well defined, and is C1. To overcome the difficulty of finding bounded Palais–Smale sequences for the associate functional J, we modify the functional J as follows
JλT(u) = 1
2kuk2+ b 4ψ
kuk2 T2
Z
RN
|∇u|2dx 2
−λ Z
RNF(u)dx (2.1) where T>0, the cut-off functionalψ(t)is defined by
ψ(t) =1, t ∈[0, 1], 0≤ψ(t)≤1, t ∈(1, 2), ψ(t) =0, t ∈[2,+∞), kψ0(t)k∞≤2.
The following lemma is important to our arguments.
Lemma 2.1 ([12]). Let (X,k · k)be a Banach space and I ⊂ R+ an interval. Consider the family of C1functionals on X
Jλ = A(u)−λB(u), λ∈ I,
with B nonnegative and either A(u)→∞or B(u)→∞askuk →∞and such that Jλ(0) =0.
For anyλ∈ I, we set
Γλ= {γ∈C([0, 1],X):γ(0) =0, Jλ(γ(1))<0}. If for everyλ∈ I, the setΓλis nonempty and
cλ = inf
γ∈Γλ max
t∈[0,1]Jλ(γ(t))>0, then for almost everyλ∈ I there is a sequence{un} ⊂X such that
(i) {un}is bounded;
(ii) Jλ(un)→cλ;
(iii) Jλ0(un)→0in the dual X−1 of X.
Throughout this paper, let A(u) = 1
2kuk2+ b 4ψ
kuk2 T2
Z
RN
|∇u|2dx 2
, B(u) =
Z
RN F(u)dx.
Now, we show that JλT satisfies the conditions of Lemma2.1.
Lemma 2.2. Γλ6=∅for allλ∈ I = [δ, 1], whereδ ∈(0, 1)is a positive constant.
Proof. We chooseφ∈ C0∞(RN)satisfying the following conditions:
φ≥0, kφk=1,
suppφ⊂ B(0,R)for someR>0.
By (H3), we have that for anyC1>0 with δC1R
B(0,R)φ2dx> 12, there existsC2>0 such that F(t)≥C1t2−C2, t ∈R+.
Then fort2>2T2, we have JλT(tφ) = 1
2ktφk2+ b 4ψ
ktφk2 T2
Z
RN
|∇(tφ)|2dx 2
−λ Z
RNF(tφ)dx
= 1
2ktφk2−λ Z
RNF(tφ)dx
≤ 1 2t2−λ
Z
RNC1t2φ2dx+C3
≤ 1
2t2−λC1t2 Z
B(0,R)φ2dx+C3
≤ 1
2t2−δC1t2 Z
B(0,R)φ2dx+C3.
Iftis sufficiently large, we have JλT(tφ)<0. The proof is completed.
Lemma 2.3. There exists a constant c>0such that cλ ≥c>0for anyλ∈ I.
Proof. By (H1) and (H2), we see that for anyε>0, there exists a constantCe>0 such that for allt ∈R+, one has
F(t)≤ 1
2εt2+Cetp+1. Furthermore, combining with the Sobolev inequality, we have
JλT(u) = 1
2kuk2+ b 4ψ
kuk2 T2
Z
RN
|∇u)|2dx 2
−λ Z
RNF(u)dx
≥ 1
2kuk2− e 2
Z
RN
|u|2dx−Ce Z
RN
|u|p+1dx
≥ 1
2 −e 2
kuk2−Ce
Z
RN
|u|p+1dx.
Then foresufficiently small, there exists ρ > 0 such that JλT(u)> 0 for any λ ∈ I, u ∈ H with 0<kuk ≤ρ. In particular, forkuk=ρ, we have JλT(u)≥c>0. Fixλ∈ I andγ∈Γλ. By the definition ofΓλ,kγ(1)k> ρ. By continuity, there existstγ ∈(0, 1)such thatkγ(tγ)k= ρ.
Therefore, for any λ∈ I, we have
cλ≥ inf
γ∈ΓλJλT(γ(tγ))≥ c>0.
Lemma 2.4. For anyλ∈ I and8bT2 ≤1, each bounded (PS)-sequence of the functional JλT admits a convergent subsequence.
Proof. Letλ∈ I and{un}be a bounded (PS)-sequence of JλT, namely
{un}is bounded, {JλT(un)}is bounded, (JλT)0(un)→0 inH0. Up to a subsequence, there existsu ∈Hsuch that
un*u inH,
un→u in Lp+1(RN), un→u a.e. in RN. By the definition of JλT, we get
(JλT)0(un),un−u
=
"
1+ b 2T2ψ0
kunk2 T2
Z
RN
|∇un|2dx 2#
(un,un−u) +bψ
kunk2 T2
Z
RN
|∇un|2dx Z
RN
∇un· ∇(un−u)dx
−λ Z
RN f(un)(un−u)dx.
Furthermore, from ψ0
k
uk2 T2
R
RN|∇u|2 dx2
≤8T4, 8bT2≤1, we easily obtain
"
1+ b 2T2ψ0
kunk2 T2
Z
RN
|∇un|2dx 2#
(un−u,un−u)
=(JλT)0(un),un−u
−
"
1+ b 2T2ψ0
kunk2 T2
Z
RN
|∇un|2dx 2#
(u,un−u)
−bψ
kunk2 T2
Z
RN
|∇un|2dx Z
RN
∇(un−u)· ∇(un−u)dx
−bψ
kunk2 T2
Z
RN
|∇un|2dx Z
RN
∇u· ∇(un−u)dx+λ Z
RN f(un)(un−u)dx
≤(JλT)0(un),un−u
−
"
1+ b 2T2ψ0
kunk2 T2
Z
RN
|∇un|2dx 2#
(u,un−u)
−bψ
kunk2 T2
Z
RN
|∇un|2dx Z
RN
∇u· ∇(un−u)dx+λ Z
RN f(un)(un−u)dx.
Firstly, it is clear to know that ((JλT)0(un),un−u)→ 0; secondly, since un * u in H, then we have
"
1+ b 2T2ψ0
kunk2 T2
Z
RN
|∇un|2dx 2#
(u,un−u)→0;
thirdly, by the fact that the imbedding H ,→ He is continuous (see [6, p. 2287]), where He = u∈ L2(RN):∇u∈[L2(RN)]N is endowed with the normkuk
He = (R
RN|∇u|2dx)12, we have un *uinHe and this implies
bψ
kunk2 T2
Z
RN
|∇un|2dx Z
RN
∇u· ∇(un−u)dx→0.
Finally, from (H1) and (H2), it follows that for any ε > 0, there exists a constant Ce > 0 such that for allt ∈R+, one has
f(t)≤εt+Ce|t|p. Then, we obtain
Z
RN f(un)(un−u)dx
≤
Z
RN
|f(un)||(un−u)|dx
≤ε|un|L2|un−u|L2 +Ce Z
RN
|un|p|un−u|dx
≤ekunkkun−uk+Ce|un|p
Lp+1|un−u|Lp+1
→0.
Therefore, we can get
"
1+ b 2T2ψ0
kunk2 T2
Z
RN
|∇un|2dx 2#
(un−u,un−u)→0, which implies thatun→uinH.
From Lemmas2.1–2.4, we obtain the following result.
Lemma 2.5. Let 8bT2 ≤ 1, then for almost every λ ∈ I, there exists uλ ∈ H\{0} such that (JλT)0(uλ) =0and JλT(uλ) =cλ.
Lemma 2.6. Let N≥5. If u∈His a critical point of JλT(u), namely, u a weak solution of 1+ b
2T2ψ0
kuk2 T2
Z
RN
|∇u|2dx 2!
Lu−bψ kuk2
T2 Z
RN
|∇u|2dx∆u= λf(u), where Lu=∆2u−a∆u+cu, then the following Pohozaev identity holds:
λN Z
RN F(u)dx = b(N−2)
2 ψ
kuk2 T2
Z
RN
|∇u|2dx 2
+ 1
2 + b 4T2ψ0
kuk2 T2
×
×
(N−4)
Z
RN
|∆u|2dx+a(N−2)
Z
RN
|∇u|2dx+cN Z
RN
|u|2dx
.
Proof. LetT(t):H→Hbe a family of transformations such that T(t)u(x) =ux
t
, t>0, and consequently
T(1) =id.
Ifu∈His a critical point of JλT, then we have JλT(T(t)u) = t
N−4
2 Z
RN
|∆u|2dx+ at
N−2
2 Z
RN
|∇u|2dx+ct
N
2 Z
RN
|u|2dx + bt
2N−4
4 ψ
R
RNtN−4|∆u|2+atN−2|∇u|2+ctN|u|2dx T2
!Z
RN
|∇u|2dx 2
−λtN Z
RNF(u)dx and
0= ∂
∂tJλT(T(t)u)|t=1
= N−4 2
Z
RN
|∆u|2dx+ a(N−2) 2
Z
RN
|∇u|2dx+ cN 2
Z
RN
|u|2dx +b(2N−4)
4 ψ
kuk2 T2
Z
RN
|∇u|2dx 2
+ b 4T2ψ0
kuk2 T2
Z
RN
|∇u|2dx 2
×
×
(N−4)
Z
RN
|∆u|2dx+a(N−2)
Z
RN
|∇u|2dx+cN Z
RN
|u|2dx
−λN Z
RNF(u)dx.
Then, ifN≥5, the Pohozaev identity of the fourth order elliptic equation:
1+ b 2T2ψ0
kuk2 T2
Z
RN
|∇u|2dx 2!
∆2u−a∆u+cu
−bψ kuk2
T2 Z
RN
|∇u|2dx∆u= λf(u), takes the form
λN Z
RN F(u)dx= 1
2+ b 4T2ψ0
kuk2 T2
×
×
(N−4)
Z
RN
|∆u|2dx+a(N−2)
Z
RN
|∇u|2dx+cN Z
RN
|u|2dx
+ b(2N−4)
4 ψ
kuk2 T2
Z
RN
|∇u|2dx 2
.
Lemma 2.7. Let unbe a critical point of JλTn at level cλn. Then for T>0sufficiently large, there exists b∗ = b∗(T)with8b∗T2 ≤1such that for any b ∈[0,b∗), up to a subsequence,kunk< T.
Proof. According to Lemma2.5, there exists a sequence{λn} ⊂Iwithλn→1−, and{un} ⊂H such that
JλTn(un) =cλn,
JλTn0
(un) =0.
Firstly, since(JλTn)0(un) =0, from Lemma2.6, it follows that λnN
Z
RNF(un)dx
= b(2N−4)
4 ψ
kunk2 T2
Z
RN
|∇un|2dx 2
+ 1
2+ b 4T2ψ0
kunk2 T2
×
×
(N−4)
Z
RN
|∆un|2dx+a(N−2)
Z
RN
|∇un|2dx+cN Z
RN
|un|2dx
.
(2.2)
Secondly, usingJλTn(un) =cλn, we have that N
2 kunk2+ Nb 4 ψ
kunk2 T2
Z
RN
|∇un|2dx 2
−λnN Z
RN F(un)dx= cλnN (2.3) Finally, from (2.2) and (2.3), we have
2
Z
RN
|∆un|2dx+a Z
RN
|∇un|2dx
≤
2+ b T2ψ0
kunk2 T2
Z
RN
|∆u|2dx+a
1+ b 2T2ψ0
kunk2 T2
Z
RN
|∇u|2dx
= N
2kunk2+ Nb 4T2ψ0
kunk2 T2
kunk2+b(2N−4)
4 ψ
kunk2 T2
Z
RN
|∇un|2dx 2
−λnN Z
RN F(un)dx
= N
2kunk2+ Nb 4T2ψ0
kunk2 T2
kunk2+b(2N−4)
4 ψ
kunk2 T2
Z
RN
|∇un|2dx 2
+cλnN− N
2 kunk2− Nb 4 ψ
kunk2 T2
Z
RN
|∇un|2dx 2
=cλnN+ Nb 4T2ψ0
kunk2 T2
kunk2+ b(N−4)
4 ψ
kunk2 T2
Z
RN
|∇un|2dx 2
. Now we show some estimates on the right-hand side.
cλn ≤max
t JλTn(tφ)
≤max
t
1
2ktφk2+b 4ψ
ktφk2 T2
Z
RN
|∇tφ)|2dx 2
−λ Z
RNF(tφ)dx
≤ 1 2t2−λ
Z
RNC1t2φ2dx+C3
≤max
t
1
2t2−λC1t2 Z
B(0,R)φ2dx+C3+max
t
b 4ψ
t2 T2
Z
RN
|∇φ|dx 2
t4
≤max
t
1
2t2−δC1t2 Z
B(0,R)φ2dx+C3+max
t
b 4ψ
t2 T2
t4.
Sinceψ t2
T2
=0 fort2 ≥2T2, we can obtain
cλn ≤C3+bT4, Nb
4T2ψ0
kuk2 T2
kuk2≤ Nb, b(N−4)
4 ψ
kunk2 T2
Z
RN
|∇un|2dx 2
≤b(N−4)T4. Then we have
2 Z
RN
|∆un|2dx+a Z
RN
|∇un|2dx≤ C3+bT4+b(N−4)T4+Nb.
Finally, we show that there existsT >0 such that kunk ≤ T. On the contrary, there exists no subsequence of {un} which is uniformly bounded by T, namely, kunk > T. By (H1) and (H2), we have
kunk2+ b 2T2ψ0
kunk2 T2
Z
RN
|∇un|2dx 2
kunk2+bψ
kunk2 T2
Z
RN
|∇un|2dx 2
=λn Z
RN
f(un)undx
≤ε|un|2L2 +Ce|un|2L∗2∗. Furthermore, we have
(1−e)kunk2≤Ce|un|2L∗2∗ − b 2T2ψ0
kunk2 T2
Z
RN
|∇un|2dx 2
kunk2
≤C4|∇un|2L∗2 +8bT2
≤C4
C3+bT4+b(N−4)T4+Nb a
2
∗ 2
+8bT2. Then we get the following inequality
T<kunk ≤C5
C3+bT4+b(N−4)T4+Nb2
∗
2 +8bT2.
However, this inequality in not true for sufficiently large T with 8bT2 < 1, and this implies the conclusion.
Proof of Theorem1.1. LetTandb∗ = 1
8T2 be defined as in Lemma2.7, andun be a critical point for JλT
n at levelcλn. Then from Lemma2.7, we know thatkunk ≤T. So JλTn(un) = 1
2kunk2+ b 4ψ
kunk2 T2
Z
RN
|∇un|2dx 2
−λn Z
RN F(un)dx
= 1
2kunk2+ b 4
Z
RN
|∇un|2dx 2
−λn Z
RNF(un)dx.
Sinceλn →1, we have J0(un),v
=(JλTn)0(un),v
+ (λn−1)
Z
RN f(un)v dx, v∈H,
which implies that J0(un)→0. Then combining with the boundedness of {un}, we show that {un}also is a (PS)-sequence of J. By Lemma 2.4, {un}has a convergent subsequence {unk} withunk →u. Consequently, J0(u) =0. According to Lemma2.3, we have
J(u) = lim
k→∞J(unk) = lim
k→∞JλT
nk(unk)≥c>0, anduis a positive solution of (1.1) by (H1). The proof is completed.
Acknowledgements
The authors thank an anonymous referee for a careful reading and some helpful comments, which greatly improve the manuscript.
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