# Existence of Multiple Solutions for Fourth-Order Elliptic Problem

Authors: Hua Gu; Tianqing An
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

By using the variant fountain theorem, we study the existence of multiple solutions for a class of superquadratic fourth-order elliptic problem with Navier boundary value condition.


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## Body

## 1. Introduction

Consider the following fourth-order boundary value problem:(1)Δ2u+cΔu=g(x,u)inΩ,u=Δu=0on⁡∂Ω,
where Δ2 denotes the biharmonic operator, Ω⊂RN(N>4) is a bounded domain with smooth boundary, and g∈C(Ω¯×R,R).The fourth-order elliptic equations which contain a biharmonic operator can describe the static form change of beam or the motion of rigid body. Thus the fourth-order elliptic equations are widely applied in physics, oceanics, aerospace engineering and other engineering. In [1], Lazer and Mckenna considered the biharmonic problem:
(2)Δ2u+cΔu=d[(u+1)+-1]inΩ,u=Δu=0on⁡∂Ω,
where u+=max⁡{u,0} and d∈R. They pointed out that this type of nonlinearity furnishes a model to study traveling waves in suspension bridges. Afterwards, in [2], they have proved the existence of 2k-1 solutions when N=1 and d>λi(λi-c) ({λi}i≥1 is the sequence of the eigenvalues of -Δ in H01(Ω)) by the global bifurcation method. In [3] the existence of a negative solution of (2) was proved when d>λ1(λ1-c) by using the Leray-Schauder degree. In particular, in [1, 4] the authors observed that problem (2) was interesting also when the nonlinearity (u+1)+-1 was replaced by a somewhat more general function g(·,u). In [5], Micheletti and Pistoia used a variational linking theorem to investigate the existence of two solutions for a more general nonlinearity g(·,u). Moreover, by using a variational result, they and Saccon also showed the existence of three solutions for some special g(·,u) (see [6]). Next year, in [7], Micheletti and Saccon obtained two results about the existence of two nontrivial solutions and four nontrivial solutions by the similar variational approach, depending on the position of a suitable parameter with respect to the eigenvalues of the linear part. In recent years, more researchers have used variational approach to investigate the fourth-order elliptic equations. In [8], Xu and Zhang studied the existence of positive solutions of problem (1) when g satisfied the local superlinearity and sublinearity condition and c<λ1 by the classical mountain pass theorem. Recently, in [9], Pu et al. used the least action principle, the Ekeland variational principle, and the mountain pass theorem to prove the existence and multiplicity of solutions of (1) when g(x,u)=a(x)|u|s-2u+f(x,u)(a∈L∞(Ω),s∈(1,2)). For other related results, see [8–14] and the references therein. Here, we emphasize that most authors considered the case c<λ1.The variant fountain theorems established in [15] have been used in the study of a class of semilinear elliptic equations (see [16, 17]) and the investigation of the Hamiltonian system (see [18, 19]). Inspired by [9, 17], we will use the variant fountain theorem to investigate the problem (1). More precisely, we make the following assumptions.(S1)
There exist constantsd1>0 and 1<ν<(N+4)/(N-4) such that
(3)|g(x,u)|≤d1(1+|u|ν),∀(x,u)∈Ω×R.(S2)
G(x,u)≥0 for all (x,u)∈Ω×R and
(4)liminf⁡|u|→∞G(x,u)|u|2=∞,uniformlyforx∈Ω.Here,G(x,u):=∫0u‍g(x,s)ds is the primitive of the nonlinearity g.(S3)
There exist constantsϱ>(2N/(N+4))ν, L>0 and d3>0 such that
(5)ug(x,u)-2G(x,u)≥d3|u|ϱ∀|u|≥L,x∈Ω.Our main result is the following theorem.Theorem 1.
Assume that(S1)–(S3) hold and G(x,u) is even in u. Then problem (1) possesses infinitely many solutions.Remark 2.
In Theorem1, we do not assume c<λ1, which is widely used in the investigation of the fourth-order equations. As is known, the so-called global Ambrosetti-Rabinowitz condition (AR-condition for short) is introduced by Ambrosetti and Rabinowitz in [20] and wildly used to the existence of infinitely many solutions for superquadratic situation: there is a constant α>2 such that, for all u≠0 and x∈Ω, the nonlinearity is assumed to satisfy
(6)0<αG(x,u)≤ug(x,u).
In fact, if we choose
(7)G(x,u)=H(x)(|u|μ+(μ-2)|u|μ-ɛsin2(|u|ɛɛ)),
where ɛ∈(0,μ-2), H∈C(Ω¯), and H(x)>0 for all x∈Ω¯. Then it is easy to see that G satisfies the conditions (S1)–(S3) in Theorem 1 with μ=3, ν=2, ɛ=0.1, ϱ=2.9, and N=5, but G does not satisfy the AR-condition (6).Remark 3.
By(S1), we can obtain that there exists a constant d2>0 such that
(8)|G(x,u)|≤d1(|u|+|u|ν+1)+d2,∀(x,u)∈Ω×R.
And by (S3), there exists a constant d4>0 such that
(9)ug(x,u)-2G(x,u)≥d3|u|ϱ-d4,∀(x,u)∈Ω×R.


## 2. Preliminaries

In this section, we will establish the variational setting for our problem and state a variant fountain theorem.LetE=H2(Ω)∩H01(Ω) be the Hilbert space equipped with the inner product
(10)(u,v)E=∫Ω‍ΔuΔvdx
and the norm
(11)∥u∥E=(u,v)E1/2.A weak solution of problem (1) is a u∈E such that
(12)∫Ω‍(ΔuΔv-c〈∇u,∇v〉)dx-∫Ω‍g(x,u)vdx=0
for any v∈E. Here and in the sequel, 〈·,·〉 always denotes the standard inner product in RN. Let Φ:E→R be the functional defined by
(13)Φ(u)=12∫Ω‍(|Δu|2-c|∇u|2)dx-∫Ω‍G(x,u)dx.
It is well known that a critical point of the functional Φ in E corresponds to a weak solution of problem (1).Letλi(i=1,2,…) be the eigenvalues of -Δ in H01(Ω). Then the eigenvalue problem
(14)Δ2u+cΔu=μuinΩ,u=Δu=0on⁡∂Ω,
has infinitely many eigenvalues μi=λi(λi-c), i=1,2,….Define a selfadjoint linear operatorA:L2(Ω)→L2(Ω) by
(15)(Au,v)2=∫Ω‍(ΔuΔv-c〈∇u,∇v〉)dx
with domain D(A)=E. Here, (·,·)2 denotes the inner product in L2(Ω) and in the sequel L2(Ω) is simply denoted by L2. Then the sequence of eigenvalues of A is just {μi}(i=1,2,…). Denote the corresponding system of eigenfunctions by {en}; it forms an orthogonal basis in L2.Denote(16)n-=#{i∣μi<0},n0=#{i∣μi=0},n¯=n-+n0.
Here, #{·} denotes the cardinal of a set. Let
(17)L-=span⁡{e1,…,en-},L0=span⁡{en-+1,…,en¯},L+=(L-⊕L0)⊥=span⁡{en¯+1,…,}.¯
Decompose L2 as
(18)L2=L-⊕L0⊕L+.
Then E also possesses the orthogonal decomposition
(19)E=E-⊕E0⊕E+
with
(20)E-=L-,E0=L0,E+=E∩L+=span⁡{en¯+1,…,}¯.We define onE a new inner product and the associated norm by
(21)(u,v)=(Au+,v+)2-(Au-,v-)2+(u0,v0)2,∥u∥=(u,u)1/2.
Therefore, Φ can be written as
(22)Φ(u)=12(∥u+∥2-∥u-∥2)-Ψ(u),
where Ψ(u)=∫Ω‍G(x,u)dx for all u=u-+u0+u+∈E=E-⊕E0⊕E+. Then Φ and Ψ are continuously differentiable.Direct computation shows that(23)Ψ′(u)v=∫Ω‍g(x,u)vdxΦ′(u)v=(u+,v+)-(u-,v-)-Ψ′(u)v
for all u,v∈E with u=u-+u0+u+ and v=v-+v0+v+, respectively. It is known that Ψ′:E→E is compact.Denote by∥·∥p the usual norm of Lp≡Lp(Ω) for all 1≤p≤2N/(N-4); then by the Sobolev embedding theorem, there exists a τp>0 such that
(24)∥u∥p≤τp∥u∥,∀u∈E.
Noting that the constants ν and ϱ appeared in (S1) and (S3) satisfies
(25)1+ν<2NN-4,ϱϱ-ν<2NN-4.To prove our main result Theorem1, we need an abstract critical point theorem found in [15].LetE be a Banach space with the norm ∥·∥ and E=⊕j∈NXj¯ with dim⁡Xj<∞ for any j∈N. Set Yk=⊕j=1kXj and Zk=⊕j=k∞Xj¯. Consider the following C1-functional Φλ:E→R defined by
(26)Φλ(u):=A(u)-λB(u),λ∈[1,2].Theorem 4 (see [15, Theorem 2.1]).
Assume that the functionalΦλ defined above satisfies the following:(F1)
Φλ maps bounded sets to bounded sets for λ∈[1,2], and Φλ(-u)=Φλ(u) for all (λ,u)∈[1,2]×E;(F2)
B(u)≥0 for all u∈E; moreover, A(u)→∞ or B(u)→∞ as ∥u∥→∞;(F3)
there existrk>ρk>0 such that
(27)αk(λ):=inf⁡u∈Zk,∥u∥=ρkΦλ(u)>βk(λ):=max⁡u∈Yk,∥u∥=rkΦλ(u),vvvvvvvvvvvvvvvvvvvvivvvvv∀λ∈[1,2].Then
(28)αk(λ)≤ζk(λ):=inf⁡γ∈Γkmax⁡u∈BkΦλ(γ(u)),∀λ∈[1,2],
where Bk={u∈Yk:∥u∥≤rk} and Γk:={γ∈C(Bk,E):γisodd,γ|∂Bk=id}. Moreover, for a.e. λ∈[1,2], there exists a sequence {umk(λ)}m=1∞ such that
(29)sup⁡m∥umk(λ)∥<∞,Φλ′(umk(λ))⟶0,vvvvvvvivΦλ(umk(λ))⟶ζk(λ)vvvvvvvvvvvivvvvasm⟶∞.In order to apply this theorem to prove our main result, we define the functionalsA, B, and Φλ on our working space E=H2(Ω)∩H01(Ω) as follows:
(30)A(u)=12∥u+∥2,B(u)=12∥u-∥2+∫Ω‍G(x,u)dx,(31)Φλ(u)=A(u)-λB(u)=12∥u+∥2-λ(12∥u-∥2+∫0T‍G(x,u)dx)
for all u=u-+u0+u+∈E=E-+E0+E+ and λ∈[1,2]. Then Φλ∈C1(E,R) for all λ∈[1,2] and
(32)Φλ′(u)v=(u+,v+)-λ((u-,v-)+∫Ω‍g(x,u)vdx).
Let Xj=span⁡{ej}, j=1,2,…. Note that Φ1 is just equal to the functional Φ defined in (22).


## 3. Proof of Theorem1

In this section we firstly establish the following two lemmas and then give the proof of Theorem1.Lemma 5.
Assume that(S1) and (S2) hold. Then B(u)≥0 for all u∈E. Furthermore, A(u)→∞ or B(u)→∞ as ∥u∥→∞.Proof.
SinceG(x,u)≥0, by (30), it is obvious that B(u)≥0 for all u∈E.
By the similar method used in the proof of Lemma 2.6 of [17], for any finite-dimensional subspace F⊂E, there exists a constant ϵ>0 such that
(33)m({x∈Ω:|u|≥ϵ∥u∥})≥ϵ,∀u∈F∖{0},
where m(·) is the Lebesgue measure in RN.
Now for the finite-dimensional subspaceE-⊕E0⊂E, there exists a constant ϵ corresponding to the one in (33). Let
(34)Λu={x∈Ω:|u|≥ϵ∥u∥},∀u∈E-⊕E0∖{0}.
Then m(Λu)≥ϵ. By (S2), there exist positive constants d5 and R1 such that
(35)G(x,u)≥d5|u|2,∀x∈Ω,|u|≥R1.
Note that
(36)|u(x)|≥R1,∀x∈Λu
for any u∈E-⊕E0 with ∥u∥≥R1/ϵ. Combining (35) and (36), for any u∈E-⊕E0 with ∥u∥≥R1/ϵ, we have
(37)B(u)=12∥u-∥2+∫Ω‍G(x,u)dt≥∫Λu‍G(x,u)dt≥∫Λu‍d5|u|2dt≥d5ϵ2∥u∥2·m(Λu)≥d5ϵ3∥u∥2,
which implies that
(38)B(u)⟶∞as∥u∥⟶∞on⁡E-⊕E0.
Combining this with E=E-⊕E0⊕E+ and (30), we have
(39)A(u)⟶∞orB(u)⟶∞as∥u∥⟶∞.
The proof is completed.Lemma 6.
Let(S1), (S2) be satisfied. Then there exist a positive integer k1 and two sequences rk>ρk→∞ as k→∞ such that
(40)αk(λ):=inf⁡u∈Zk,∥u∥=ρkΦλ(u)>0,∀k≥k1,(41)βk(λ):=max⁡u∈Yk,∥u∥=rkΦλ(u)<0,∀k∈N,
where Yk=⊕j=1kXj=span⁡{e1,e2,…,ek} and Zk=⊕j=k∞Xj¯=span⁡{ek,ek+1,…}¯ for all k∈N.Proof.
Step
1. We first prove (40).
By virtue of (8) and (31), for any u∈E+(42)Φλ(u)≥12∥u∥2-2∫Ω‍G(x,u)dx≥12∥u∥2-2d1(∥u∥1+∥u∥ν+1ν+1)-2d2·m(Ω),vvvvvvvvvvvvvvvivvvvvvvvvvv∀λ∈[1,2],
where d1, d2 are the constants in (8). Let
(43)ιν+1(k)=sup⁡u∈Zk,∥u∥=1∥u∥ν+1,∀k∈N.
Then
(44)ιν+1(k)⟶0ask⟶∞
since E is compactly embedded into Lν+1. Note that
(45)Zk⊂E+,∀k≥n¯+1,
where n¯ is the integer given in (16). Combining (24), (42), (43), and (45), for k≥n¯+1, we have
(46)Φλ(u)≥12∥u∥2-2d1τ1∥u∥-2d2·m(Ω)-2d1ιν+1ν+1(k)∥u∥ν+1,∀(λ,u)∈[1,2]×Zk,
where τ1 is the constant given in (24). By (44), there exists a positive integer k1≥n¯+1 such that
(47)ρk≔(16d1ιν+1ν+1(k))1/(1-ν)>max⁡{16d1τ1+1,16d2·m(Ω)},∀k≥k1
since ν>1. Clearly,
(48)ρk⟶∞ask⟶∞.
Combining (46) and (47), direct computation shows
(49)αk(λ):=inf⁡u∈Zk,∥u∥=ρkΦλ(u)≥ρk24>0,∀k≥k1.
Step
2. We then prove (41).
Note that for anyk∈N, Yk is of finite dimension, so we can choose M1>0 sufficiently large such that
(50)∥u∥≤M1(∫Ω|u|2)1/2,∀u∈Yk.
By (S2) and (8), for the former M1, there exists a M2>0 such that
(51)G(x,u)≥M12|u|2-M2,∀(t,u)∈[0,T]×RN.
Consequently, by (50) and (51), we have
(52)Φλ(u)≤12∥u+∥2-12∥u-∥2-∫Ω‍G(x,u)dt≤12∥u+∥2-12∥u-∥2-M12∫Ω‍|u|2dt+M2·m(Ω)≤12∥u+∥2-12∥u-∥2-M12(1M12∥u+∥2+1M12∥u0∥2)+M2·m(Ω)≤-12∥u+∥2-12∥u-∥2-∥u0∥2+M2·m(Ω)≤-12∥u∥2+M2·m(Ω)
for all u=u-+u0+u+∈Yk. Now for any k∈N, if we choose
(53)rk>max⁡{ρk,2M2·m(Ω)},
then (52) implies
(54)βk(λ):=max⁡u∈Yk,∥u∥=rkΦλ(u)<0,∀k∈N.
The proof is completed.Now we prove our main result Theorem1.Proof of Theorem1.
In view of (8), (24), and (31), Φλ maps bounded sets to bounded sets uniformly for λ∈[1,2]. By virtue of the evenness of G(x,u) in u, it holds that Φλ(-u)=Φλ(u) for all (λ,u)∈[1,2]×E. Therefore the condition (F1) of Theorem 4 holds. Lemma 5 shows that the condition (F2) holds, whereas Lemma 6 implies that condition (F3) holds for all k≥k1, where k1 is given in Lemma 6. Thus, by Theorem 4, for each k≥k1 and a.e. λ∈[1,2], there exists a sequence {umk(λ)}m=1∞⊂E such tha
(55)sup⁡m∥umk(λ)∥<∞,Φλ′(umk(λ))⟶0,vvvvvvvvΦλ(umk(λ))⟶ζk(λ)vvvvvvivvvvvvvvvasm⟶∞,
where
(56)ζk(λ):=inf⁡γ∈Γkmax⁡u∈BkΦλ(γ(u)),∀λ∈[1,2]
with Bk={u∈Yk:∥u∥≤rk} and Γk:={γ∈C(Bk,E):γisodd,γ|∂Bk=id}.
Moreover, by the proof of Lemma6, we have
(57)ζk(λ)∈[α¯k,ζ¯k],∀k≥k1,
where ζ¯k:=max⁡u∈BkΦ1(u) and α¯k:=ρk2/4→∞ as k→∞ by (48).
Since the sequence{umk(λ)}m=1∞ obtained by (55) is bounded, it is clear that, for each k≥k1, we can choose λn→1 such that the sequence {umk(λn)}m=1∞ has a strong convergent subsequence.
In fact, without loss of generality, assume that(58)umk(λn)-⟶u0k(λn)-,umk(λn)0⟶u0k(λn)0,vvvvvvvvvvvivumk(λn)+⇀u0k(λn)+vvvvvvvvvvvvvvivvvasm⟶∞,(59)umk(λn)⇀u0k(λn)asm⟶∞
for some u0k(λn)=u0k(λn)-+u0k(λn)0+u0k(λn)+∈E=E-⊕E0⊕E+ since dim⁡(E-⊕E0)<∞.
Note that(60)Φλn′(umk(λn))=umk(λn)+-λn(umk(λn)-+Ψ′(umk(λn))),vvvvvvvvvvvvvvvvvvvvvvvvvvvvv∀n∈N.
That is,
(61)umk(λn)+=Φλn′(umk(λn))+λn(umk(λn)-+Ψ′(umk(λn))),vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv∀m∈N.
In view of (55), (58), (59), and the compactness of Ψ′, the right-hand side of (61) converges strongly in E and hence umk(λn)+→u0k(λn)+ in E. Together with (58), {umk(λn)}m=1∞ has a strong convergent subsequence in E.
Without loss of generality, we assume(62)lim⁡m→∞umk(λn)=unk,∀n∈N,k≥k1.
This together with (55) and (57) yields
(63)Φλn′(unk)=0,Φλn(unk)∈[α¯k,ζ¯k],vvvviivvv∀n∈N,k≥k1.
Now we claim that the sequence {unk}n=1∞ in (63) is bounded in E and possesses a strong convergent subsequence with the limit uk∈E for each k≥k1. For the sake of notational simplicity, throughout the remaining proof of Theorem 1 we always denote un=unk.
Now we claim that{un} is bounded in E. Otherwise, going to a subsequence if necessary, we can assume that ∥un∥→∞ as n→∞. By (9), we have
(64)2Φλn(un)-Φλn′(un)un=λn∫Ω‍[g(x,un)un-2G(x,un)]dx≥d3∫Ω‍|un|ϱdx-d4·m(Ω),
which yields that
(65)∫Ω‍|un|ϱdx∥un∥⟶0asn⟶∞.
Write un=un-+un0+un+∈E-⊕E0⊕E+. It follows from (S1), (24), (25), (32), and the Hölder inequality that
(66)Φλn′(un)un+=∥un+∥2-λn∫Ω‍g(x,un)un+dx≥∥un+∥2-2∫Ω‍|g(x,un)|·|un+|dx≥∥un+∥2-d1∫Ω‍|un+|dx-d1∫Ω‍|un|ν|un+|dx≥∥un+∥2-d1∥un+∥1-d1(∫Ω‍(|un|ν)ϱ/νdx)ν/ϱ·(∫Ω‍|un+|ϱ/(ϱ-ν)dx)(ϱ-ν)/ϱ≥∥un+∥2-c1∥un+∥-c2∥un∥ϱν·∥un+∥
for any n∈N. Here and in the sequel, we denote ci>0(i=1,2,…) for different positive constants. Since ϱ>(2N/(N+4))ν and N≥5, we have ν<ϱ. So, by (65) we get
(67)∥un+∥∥un∥⟶0asn⟶∞.
Similarly, we have
(68)∥un-∥∥un∥⟶0asn⟶∞.
By (S3), there also exist constants d6>0 and d7>0 such that
(69)ug(x,u)-2G(x,u)≥d6|u|-d7,∀(x,u)∈Ω×R.
So we get
(70)2Φλn(un)-Φλn′(un)un=λn∫Ω‍[g(x,un)un-2G(x,un)]dx≥d6∫Ω‍|un|dx-d7·m(Ω)≥d6∫Ω‍(|un0|-|un+|-|un-|)dx-d7·m(Ω)≥c3∥un0∥-c4(∥un-∥+∥un+∥)-c5
keeping in mind that dim⁡E0<∞ and (24). Hence, by (67) and (68), we get
(71)∥un0∥∥un∥⟶0asn⟶∞.
Then we arrive at
(72)1=∥un∥∥un∥≤∥un-∥+∥un0∥+∥un+∥∥un∥⟶0asn⟶∞
which is a contradiction. Thus, {un} is bounded in E. Then the proof that {un} has a strong convergent subsequence is the same as the preceding proof of {umk(λn)}m=1∞.
Now for eachk≥k1, by (63), the limit uk is just a critical point of Φ=Φ1 with Φ(uk)∈[α¯k,ζ¯k]. Since α¯k→∞ as k→∞ in (57), we get infinitely many nontrivial critical points of Φ. Therefore, system (1) possesses infinitely many nontrivial solutions.



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