# Existence of Multiple Solutions for a Class of Biharmonic Equations

Authors: Chunhan Liu; Jianguo Wang
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

By a symmetric Mountain Pass Theorem, a class of biharmonic equations with Navier type boundary value at the resonant and nonresonant case are discussed, and infinitely many solutions of the equations are obtained.


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## Body

## 1. Introduction and Main Results

In this paper, we study the following fourth-order elliptic equation:(1)Δ2u+cΔu=μh(x)|u|p-2u+f(x,u),x∈Ω,u=Δu=0,x∈∂Ω,
where Δ2 is the biharmonic operator, c is a constant, Ω⊂RN is a bounded smooth domain, 1<p<2, μ≥0 is a parameter, h∈L∞(Ω), h(x)≥0, h(x)≢0, f(x,s) is a continuous function on Ω-×R.This fourth-order semilinear elliptic problem can be considered as an analogue of a class of second-order problems which have been studied by many authors. A main tool of seeking solutions of the problem is the Mountain Pass Theorem (see [1–3]). In [4], Pei studied the following problem:
(2)Δ2u+cΔu=f(x,u),x∈Ω,u=Δu=0,x∈∂Ω,
and obtained at least three nontrivial solutions by using the minimax method and Morse theory.Problem (1) has been studied extensively in recent years; we refer the reader to [5–11] and the references therein. In [12], the author showed that the problem (1) admits at least three (or four or five) nontrivial solutions by using the minimax method and Mountain Pass Theory.However, to the best of author's knowledge, there have been very few results dealing with (1) using a symmetric Mountain Pass Theorem. This paper will make some contribution in the research field. In this paper, we study the problem (1) by a symmetric Mountain Pass Theorem at the resonant and nonresonant case and obtain infinitely many solutions of the equation.Consider eigenvalue problem(3)-Δu=λu,x∈Ω,u=0,x∈∂Ω.
Let us denote that λk(k∈N) are the eigenvalues and φk(k∈N) are the corresponding eigenfunctions of the eigenvalue problem (3). It is well known that 0<λ1<λ2≤λ3≤⋯≤λk→+∞, and the first eigenfunction φ1>0, x∈Ω.It is easy to see thatλk(λk-c), k=1,2,…, are eigenvalues of the problem
(4)Δ2u+cΔu=Λu,x∈Ω,u=Δu=0,x∈∂Ω,
and φk(k=1,2,…) are still the corresponding eigenfunctions.LetH=H2(Ω)∩H01(Ω) be the Hilbert space equipped with the inner product
(5)〈u,v〉H=∫Ω‍(ΔuΔv+∇u∇v)dx,
and the deduced norm
(6)∥u∥H=[∫Ω‍(|Δu|2+|∇u|2)dx]1/2.Suppose thatc<λ1. Let us define a norm of the space H as follows:
(7)∥u∥=[∫Ω‍(|Δu|2-c|∇u|2)dx]1/2.
It is easy to verify that the norm ∥·∥ is equivalent to the norm ∥·∥H on H, and for all u∈H, the following Poincaré inequality holds:
(8)∥u∥2≥λ1(λ1-c)|u|22,
where |u|22=∫Ω‍|u|2dx.Throughout this paper, the weak solutions of (1) are the critical points of the associated functional
(9)Φμ(u)=12(∫Ω‍(|Δu|2-c|∇u|2)dx)-μp∫Ω‍h(x)|u|pdx-∫Ω‍F(x,u)dx,∀u∈H,
where F(x,u)=∫0u‍f(x,t)dt.ObviouslyΦμ∈C1(H,R), and for all u, φ∈H,
(10)〈Φμ′(u),φ〉=∫Ω‍(ΔuΔφ-c∇u∇φdx)-μ∫Ω‍h(x)|u|p-2uφdx-∫Ω‍f(x,u)φdx.
In order to establish solutions for problem (1), we make the following assumptions:(F0)h∈L∞(Ω), h(x)≥0;(F1)f(x,0)=0, f(x,-t)=-f(x,t), for all x∈Ω, t∈R;(F2)  lim|t|→0f(x,t)/t=α, lim|t|→∞f(x,t)/t=β uniformly for a.e. x∈Ω, where 0≤α<λk(λk-c)<β, or β=λk(λk-c);(F3)  lim|t|→∞(f(x,t)t-2F(x,t))=-∞ uniformly in x∈Ω.Definition 1 (see [13]).
LetΦ∈C1(X,R), we say that Φ satisfies the Cerami condition at the level c∈R, if any sequence {un}⊂X with
(11)Φ(un)⟶c,(1+∥un∥)Φ′(un)⟶0,asn⟶∞
possesses a convergent subsequence; Φ satisfies the (C) condition if Φ satisfies (C)c for all c∈R.Lemma 2.
LetXk=span{φ1,φ2,…,φk}; H=Xk⊕Xk⊥; then
(12)∥u∥2≤λk(λk-c)|u|22,∀u∈Xk;∥u∥2≥λk+1(λk+1-c)|u|22,∀u∈Xk⊥,k≥2.Proof.
It is similar to the proof of Lemma 2.5 in [13].Theorem 3 (see [14] a symmetric Mountain Pass Theorem).
Suppose thatΦ∈C1(E,R) is even, Φ(θ)=0, and(i)
there existρ, α>0, and a finite dimensional linear subspace Z such that
(13)Φ|Z⊥∩∂Bρ(θ)≥α,(ii)
there exist a sequence of linear subspaceZ¯m, dim(Z¯m)=m, and Rm>0 such that
(14)Φ(x)≤0,∀x∈Z¯m∖BRm,m=1,2,….If Φ satisfies (PS) condition, then Φ possesses infinitely many distinct critical points corresponding to positive critical values.Remark 4.
IfΦ satisfies the (C) condition, Theorem 3 still holds.The main results of this paper are as follows.Theorem 5.
Assume that(F0)−(F2) hold, and c<λ1, λk(λk-c)<β, β is not an eigenvalue of (4); then there exist μ*>0 such that for μ∈(0,μ*), (1) has infinitely many solutions.Theorem 6.
Assume that(F0)−(F3) hold, and c<λ1, β=λk(λk-c); then there exist μ*>0 such that for μ∈(0,μ*), (1) has infinitely many solutions.


## 2. Proofs of Theorems

Proof of Theorem5.
(i) Assume that{un}⊂H is a (C) sequence, that is,
(15)(1+∥un∥)Φμ′(un)⟶0,Φμ(un)⟶c.
We claim that {un} is bounded. Assume as a contradiction that |un|2→∞. Setting vn=un/|un|2, then |vn|2=1. Without loss of generality, we assume
(16)vn⇀vinH,vn⟶vinL2(Ω),vn⟶va.e.x∈Ω.
From (15) we know that
(17)|f(x,un)|un|2∫Ω(ΔvnΔφ-c∇vn∇φ)dx‍-μ∫Ω‍h(x)|un|p-2un|un|2φdx-∫Ω‍f(x,un)|un|2φdx|=1|un|2|〈Φμ′(un),φ〉|≤1|un|2∥Φμ′(un)∥∥φ∥⟶0,∀φ∈H.
Next we consider the two possible cases: (a)v≠0, (b)v=0.
In case (a), from(F2) we derive
(18)|un|=|vn||un|2⟶∞,limn→∞f(x,un)un=β.
For vn→v in L2(Ω), we have
(19)limn→∞f(x,un)|un|2=limn→∞f(x,un)unvn=βv,a.e.x∈Ω.
In case (b), we have
(20)f(x,un)|un|2≤c|vn|⟶0.
Then
(21)limn→∞∫Ω‍f(x,un)φ|un|2dx=limn→∞∫Ω‍f(x,un)vnunφdx=∫Ω‍βvφdx,∀φ∈H.
By (17) and limn→∞∫Ω‍h(x)(|un|p-2un/|un|2)φdx=0, we have
(22)∫Ω‍(ΔvΔφ-c∇v∇φ)dx=∫Ω‍βvφdx,∀φ∈H.
We can easily see that v≢0. In fact, if v≡0, then |v|2=0, which contradicts limn→∞|vn|2=|v|2=1. Hence, β is an eigenvalue of the problem (4); this contradicts our assumption. Then {un} is bounded; there exist a subsequence of {un} (we can also denote by {un}) and u∈H, such that un⇀u in H. By Lemma 2 and (15), we have
(23)∥un-um∥=μ∫Ω‍h(x)|un-um|pdx+∫Ω‍f(x,un-um)(un-um)dx+o(1)∥un-um∥≤μ|h|∞∫Ω‍|un-um|pdx+∫Ω‍f(x,un-um)(un-um)dx+o(1)∥un-um∥,
where o(1)→0, and
(24)|∫Ω‍f(x,un-um)(un-um)dx|≤|f(x,un-um)|2|un-um|2,
as n,m→∞. Hence un→u in H. We verify Φμ(u) satisfies (C) condition.
(ii) There exists someρ,γ>0 such that Φμ|Xk⊥∩∂Bρ(θ)≥γ>0, where Bρ(θ)={u∈H:∥u∥≤ρ}.
By(F1) and (F2), taking σ∈(2,2*), for any given ε>0, there exists C1>0 such that
(25)F(x,u)≤12(α+ε)|u|2+C1|u|σ,∀x∈Ω,
where
(26)2*={2NN-2,N>2,+∞,N≤2.
Takingε>0 such that α+ε<λk(λk-c), combining Lemma 2, Poincaré inequality, and Sobolev embedding, we have
(27)Φμ(u)≥12(∫Ω‍(|Δu|2-c|∇u|2)dx)-μ|h|∞p∫Ω‍|u|pdx-∫Ω‍F(x,u)dx≥12∥u∥2-α+ε2|u|22-μ|h|∞p|u|pp-C1|u|σσ≥12(1-α+ελk+1(λk+1-c))∥u∥2-μC2∥u∥p-C3∥u∥σ=(12(1-α+ελk+1(λk+1-c))-μC2∥u∥p-2-C3∥u∥σ-212(1-α+ελk+1(λk+1-c)))∥u∥2,
where C2,C3 are constant.
Letg(t)=μC2tp-2+C3tσ-2; we claim that there exists t0 such that
(28)g(t0)<12(1-α+ελk+1(λk+1-c)).
It is easy to see thatg(t) has a minimum at t0=((μC2(2-p))/(C3(σ-2)))1/(σ-p); substituting t0 in g(t), we have
(29)g(t0)=μ(σ-2)/(σ-p)[(C2(C2(2-p)C3(σ-2)))(p-2)/(σ-p)+(C3(C2(2-p)C3(σ-2)))(σ-2)/(σ-p)]<12(1-α+ελk+1(λk+1-c)),
where 0<μ<μ*=[1/2(1-((α+ε)/(λk+1(λk+1-c))))/((C2(C2(2-p)/C3(σ-2)))(p-2)/(σ-p)+(C3(C2(2-p)/C3(σ-2)))(σ-2)/(σ-p))](2-σ)/(σ-p).
We take∥u∥=ρ=t0>0, then there exists γ>0 such that Φμ|Xk⊥∩∂Bρ(θ)≥γ>0, where Bρ(θ)={u∈H:∥u∥≤ρ}.
(iii) There existsRk>ρ, such that
(30)Φμ(u)≤0,∀u∈Xk∖BRk(θ),k=1,2,….
For β>λk(λk-c), (F2) implies that for any ε>0, there exists C4>0 such that
(31)F(x,u)≥12(β-ε)|u|2-C4.
Taking ε>0 such that β-ε>λk(λk-c). By Lemma 2 and (μ/p)∫Ω‍h(x)|u|pdx≥0, we have
(32)Φμ(u)=12(∫Ω‍(|Δu|2-c|∇u|2)dx)-μp∫Ω‍h(x)|u|pdx-∫Ω‍F(x,u)dx≤12∥u∥2-β-ε2|u|22+C4|Ω|≤12(1-β-ελk(λk-c))∥u∥2+C4|Ω|⟶-∞,∀u∈Xk,as∥u∥⟶+∞.
Hence there exists Rk>ρ such that
(33)Φμ(u)≤0,∀u∈Xk∖BRk(θ),k=1,2,….
Summing up the above proofs, Φμ satisfies all the conditions of Theorem 3 and Remark 4. Then the problem (1) has infinitely many solutions.Proof of Theorem6.
Similar to the proof of Theorem5(i), we have
(34)∫Ω‍(ΔvΔφ-c∇v∇φ)dx=∫Ω‍λkvφdx,∀φ∈H.
We can easily see that v≢0. In fact, if v≡0, then |v|2=0, which contradicts limn→∞|vn|2=|v|2=1. Then v is a corresponding eigenfunction of the eigenvalue λk; hence |un|→∞, a.e. x∈Ω.
By(F3), we have
(35)limn→∞(f(x,un)un-2F(x,un))=-∞uniformlyinx∈Ω.
It follows from Fatous Lemma that
(36)limn→∞∫Ω‍(f(x,un)un-2F(x,un))dx=-∞.
On the other hand, (15) implies that
(37)2c⟵2Φμ(un)-〈Φμ′(un),un〉=∫Ω‍(|Δun|2-c|∇un|2)dx-2μp∫Ω‍h(x)|un|pdx-∫Ω‍2F(x,un)dx-∫Ω‍(|Δun|2-c|∇un|2)dx+μ∫Ω‍h(x)|un|pdx+∫Ω‍f(x,un)undx=∫Ω‍(f(x,un)un-2F(x,un))dx-(2p-1)μ∫Ω‍h(x)|un|pdx,
where limn→∞(2/p-1)μ∫Ω‍h(x)|un|pdx=∞, which contradicts (36); hence {un} is bounded. Similar to the proof of Theorem 5(i), we have {un}→u in H. Hence we verify Φμ(u) satisfies (C) condition.
Similar to [15], let H(x,t)=F(x,t)-(1/2)λk(λk-c)t2, and f(x,t)=λk(λk-c)t+h(x,t); then
(38)lim|t|→∞2H(x,t)t2=0,lim|t|→∞(h(x,t)t-2H(x,t))=-∞.
It follows that for every N>0, there exists RN>0 such that
(39)h(x,t)t-2H(x,t)≤-N,∀t∈R,cccccccccccc|t|≥RN,a.e.x∈Ω.
For t>0, we have
(40)ddt[H(x,t)t2]=h(x,t)t-2H(x,t)t3;
over the interval [t,s]⊂[T,+∞), we have
(41)H(x,s)s2-H(x,t)t2≤N2(1s2-1t2).
Letting s→+∞, we see that H(x,t)≥(N/2), t∈R, t≥RN, a.e. x∈Ω. In a similar way, we have H(x,t)≥(N/2), t∈R, t≤-RN, a.e. x∈Ω. Hence
(42)lim|t|→∞H(x,t)=+∞.a.e.x∈Ω.
By Lemma 2 and h(x)≥0, we have
(43)Φμ(u)=12(∫Ω‍(|Δu|2-c|∇u|2)dx)-12λk(λk-c)∫Ω‍u2dx-μp∫Ω‍h(x)|u|pdx-∫Ω‍H(x,u)dx≤-∫Ω‍H(x,u)dx⟶-∞,∀u∈Xk,as∥u∥⟶+∞.
Hence there exists Rk>ρ such that
(44)Φμ(u)≤0,∀u∈Xk∖BRk(θ),k=1,2,….
Summing up the above proofs and Theorem 5(ii), Φμ satisfies all the conditions of Theorem 3 and Remark 4, then the problem (1) has infinitely many solutions.



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