Existence of Multiple Solutions for a Class of Biharmonic Equations

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.

1. Introduction and Main Results

In this paper, we study the following fourth-order elliptic equation: Δ²𝑢+𝑐Δ𝑢=𝜇ℎ(𝑥)|𝑢|^𝑝−2𝑢+𝑓(𝑥,𝑢),𝑥∈Ω,𝑢=Δ𝑢=0,𝑥∈𝜕Ω, (1) where Δ² is the biharmonic operator, 𝑐 is a constant, Ω⊂𝑅^𝑁 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: Δ²𝑢+𝑐Δ𝑢=𝑓(𝑥,𝑢),𝑥∈Ω,𝑢=Δ𝑢=0,𝑥∈𝜕Ω, (2) 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 −Δ𝑢=𝜆𝑢,𝑥∈Ω,𝑢=0,𝑥∈𝜕Ω. (3) Let us denote that 𝜆_𝑘(𝑘∈𝑁) are the eigenvalues and 𝜑_𝑘(𝑘∈𝑁) are the corresponding eigenfunctions of the eigenvalue problem (3). It is well known that 0<𝜆₁<𝜆₂≤𝜆₃≤⋯≤𝜆_𝑘→+∞, and the first eigenfunction 𝜑₁>0, 𝑥∈Ω.

It is easy to see that 𝜆_𝑘(𝜆_𝑘−𝑐), 𝑘=1,2,…, are eigenvalues of the problem Δ²𝑢+𝑐Δ𝑢=Λ𝑢,𝑥∈Ω,𝑢=Δ𝑢=0,𝑥∈𝜕Ω, (4) and 𝜑_𝑘(𝑘=1,2,…) are still the corresponding eigenfunctions.

Let 𝐻=𝐻²(Ω)∩𝐻¹₀(Ω) be the Hilbert space equipped with the inner product ⟨𝑢,𝑣⟩_𝐻=_Ω(Δ𝑢Δ𝑣+∇𝑢∇𝑣)d𝑥, (5) and the deduced norm ‖𝑢‖_𝐻=_Ω|Δ𝑢|²+|∇𝑢|²d𝑥^1/2. (6)

Suppose that 𝑐<𝜆₁. Let us define a norm of the space 𝐻 as follows: ‖𝑢‖=_Ω|Δ𝑢|²−𝑐|∇𝑢|²d𝑥^1/2. (7) 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: ‖𝑢‖²≥𝜆₁𝜆₁−𝑐|𝑢|²₂, (8) where |𝑢|²₂=∫_Ω|𝑢|²𝑑𝑥.

Throughout this paper, the weak solutions of (1) are the critical points of the associated functional Φ_𝜇(𝑢)=12_Ω|Δ𝑢|²−𝑐|∇𝑢|²−𝜇d𝑥𝑝_Ωℎ(𝑥)|𝑢|^𝑝d𝑥−_Ω𝐹(𝑥,𝑢)d𝑥,∀𝑢∈𝐻, (9) where 𝐹(𝑥,𝑢)=∫^𝑢₀𝑓(𝑥,𝑡)𝑑𝑡.

Obviously Φ_𝜇∈𝐶¹(𝐻,𝑅), and for all 𝑢, 𝜑∈𝐻, Φ^′_𝜇(𝑢)=,𝜑_ΩΔ𝑢Δ𝜑−𝑐∇𝑢∇𝜑d𝑥−𝜇_Ωℎ(𝑥)|𝑢|^𝑝−2𝑢𝜑d𝑥−_Ω𝑓(𝑥,𝑢)𝜑d𝑥. (10) 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<𝜆_𝑘(𝜆_𝑘−𝑐)<𝛽, or 𝛽=𝜆_𝑘(𝜆_𝑘−𝑐); (𝐹₃)  lim_|𝑡|→∞(𝑓(𝑥,𝑡)𝑡−2𝐹(𝑥,𝑡))=−∞ uniformly in 𝑥∈Ω.

Definition 1 (see [13]). Let Φ∈𝐶¹(𝑋,𝑅), we say that Φ satisfies the Cerami condition at the level 𝑐∈𝑅, if any sequence {𝑢_𝑛}⊂𝑋 with Φ𝑢_𝑛⟶𝑐,1+𝑢_𝑛Φ^′𝑢_𝑛⟶0,as𝑛⟶∞ (11) possesses a convergent subsequence; Φ satisfies the (𝐶) condition if Φ satisfies (𝐶)_𝑐 for all 𝑐∈𝑅.

Lemma 2. Let 𝑋_𝑘=span{𝜑₁,𝜑₂,…,𝜑_𝑘}; 𝐻=𝑋_𝑘⊕𝑋^⟂_𝑘; then ‖𝑢‖²≤𝜆_𝑘𝜆_𝑘−𝑐|𝑢|²₂,∀𝑢∈𝑋_𝑘;‖𝑢‖²≥𝜆_𝑘+1𝜆_𝑘+1−𝑐|𝑢|²₂,∀𝑢∈𝑋^⟂_𝑘,𝑘≥2. (12)

Proof. It is similar to the proof of Lemma 2.5 in [13].

Theorem 3 (see [14] a symmetric Mountain Pass Theorem). Suppose that Φ∈𝐶¹(𝐸,𝑅) is even, Φ(𝜃)=0, and(i)there exist 𝜌, 𝛼>0, and a finite dimensional linear subspace 𝑍 such that Φ|_𝑍^⟂_{∩𝜕𝐵𝜌(𝜃)}≥𝛼, (13)(ii)there exist a sequence of linear subspace 𝑍_𝑚, dim(𝑍_𝑚)=𝑚, and 𝑅_𝑚>0 such that Φ(𝑥)≤0,∀𝑥∈𝑍_𝑚⧵𝐵_𝑅𝑚,𝑚=1,2,…. (14)If Φ satisfies (𝑃𝑆) condition, then Φ possesses infinitely many distinct critical points corresponding to positive critical values.

Remark 4. If Φ 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, 𝜆_𝑘(𝜆_𝑘−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=𝜆_𝑘(𝜆_𝑘−𝑐); then there exist 𝜇^∗>0 such that for 𝜇∈(0,𝜇^∗), (1) has infinitely many solutions.

2. Proofs of Theorems

Proof of Theorem 5. (i) Assume that {𝑢_𝑛}⊂𝐻 is a (𝐶) sequence, that is, 1+𝑢_𝑛Φ^′_𝜇𝑢_𝑛⟶0,Φ_𝜇𝑢_𝑛⟶𝑐. (15) We claim that {𝑢_𝑛} is bounded. Assume as a contradiction that |𝑢_𝑛|₂→∞. Setting 𝑣_𝑛=𝑢_𝑛/|𝑢_𝑛|₂, then |𝑣_𝑛|₂=1. Without loss of generality, we assume 𝑣_𝑛⇀𝑣in𝐻,𝑣_𝑛⟶𝑣in𝐿²(Ω),𝑣_𝑛⟶𝑣a.e.𝑥∈Ω. (16) From (15) we know that _ΩΔ𝑣_𝑛Δ𝜑−𝑐∇𝑣_𝑛∇𝜑d𝑥−𝜇_Ωℎ(𝑥)𝑢_𝑛^𝑝−2𝑢_𝑛𝑢_𝑛2𝜑d𝑥−_Ω𝑓𝑥,𝑢_𝑛𝑢_𝑛2=1𝜑d𝑥𝑢_𝑛2Φ^′_𝜇𝑢_𝑛≤1,𝜑𝑢_𝑛2Φ^′_𝜇𝑢_𝑛𝜑⟶0,∀𝜑∈𝐻. (17) Next we consider the two possible cases: (a)𝑣≠0, (b)𝑣=0.
In case (a), from (𝐹₂) we derive 𝑢_𝑛=𝑣_𝑛𝑢_𝑛2⟶∞,lim_𝑛→∞𝑓𝑥,𝑢_𝑛𝑢_𝑛=𝛽. (18) For 𝑣_𝑛→𝑣 in 𝐿²(Ω), we have lim_𝑛→∞𝑓𝑥,𝑢_𝑛𝑢_𝑛2=lim_𝑛→∞𝑓𝑥,𝑢_𝑛𝑢_𝑛𝑣_𝑛=𝛽𝑣,a.e.𝑥∈Ω. (19) In case (b), we have 𝑓𝑥,𝑢_𝑛𝑢_𝑛2≤𝑐𝑣_𝑛⟶0. (20) Then lim_𝑛→∞Ω𝑓𝑥,𝑢_𝑛𝜑𝑢_𝑛2d𝑥=lim_𝑛→∞Ω𝑓𝑥,𝑢_𝑛𝑣_𝑛𝑢_𝑛=𝜑d𝑥_Ω𝛽𝑣𝜑d𝑥,∀𝜑∈𝐻. (21) By (17) and lim_𝑛→∞∫_Ωℎ(𝑥)(|𝑢_𝑛|^𝑝−2𝑢_𝑛/|𝑢_𝑛|₂)𝜑𝑑𝑥=0, we have ^Ω=Δ𝑣Δ𝜑−𝑐∇𝑣∇𝜑d𝑥_Ω𝛽𝑣𝜑d𝑥,∀𝜑∈𝐻. (22) We can easily see that 𝑣≢0. In fact, if 𝑣≡0, then |𝑣|₂=0, which contradicts lim_𝑛→∞|𝑣_𝑛|₂=|𝑣|₂=1. Hence, 𝛽 is an eigenvalue of the problem (4); this contradicts our assumption. Then {𝑢_𝑛} is bounded; there exist a subsequence of {𝑢_𝑛} (we can also denote by {𝑢_𝑛}) and 𝑢∈𝐻, such that 𝑢_𝑛⇀𝑢 in 𝐻. By Lemma 2 and (15), we have 𝑢_𝑛−𝑢_𝑚=𝜇_Ωℎ(𝑥)𝑢_𝑛−𝑢_𝑚^𝑝+d𝑥_Ω𝑓𝑥,𝑢_𝑛−𝑢_𝑚𝑢_𝑛−𝑢_𝑚(1𝑢d𝑥+𝑜)_𝑛−𝑢_𝑚≤𝜇|ℎ|_∞Ω𝑢_𝑛−𝑢_𝑚^𝑝d𝑥+_Ω𝑓𝑥,𝑢_𝑛−𝑢_𝑚𝑢_𝑛−𝑢_𝑚(1𝑢d𝑥+𝑜)_𝑛−𝑢_𝑚, (23) where 𝑜(1)→0, and _Ω𝑓𝑥,𝑢_𝑛−𝑢_𝑚𝑢_𝑛−𝑢_𝑚≤𝑓d𝑥𝑥,𝑢_𝑛−𝑢_𝑚2𝑢_𝑛−𝑢_𝑚2, (24) as 𝑛,𝑚→∞. Hence 𝑢_𝑛→𝑢 in 𝐻. We verify Φ_𝜇(𝑢) satisfies (𝐶) condition.
(ii) There exists some 𝜌,𝛾>0 such that Φ_𝜇|_𝑋^⟂_{𝑘∩𝜕𝐵𝜌(𝜃)}≥𝛾>0, where 𝐵_𝜌(𝜃)={𝑢∈𝐻∶‖𝑢‖≤𝜌}.
By (𝐹₁) and (𝐹₂), taking 𝜎∈(2,2^∗), for any given 𝜀>0, there exists 𝐶₁>0 such that 𝐹()≤1𝑥,𝑢2(𝛼+𝜀)|𝑢|²+𝐶₁|𝑢|^𝜎,∀𝑥∈Ω, (25) where 2^∗=2𝑁𝑁−2,𝑁>2,+∞,𝑁≤2. (26)
Taking 𝜀>0 such that 𝛼+𝜀<𝜆_𝑘(𝜆_𝑘−𝑐), combining Lemma 2, Poincaré inequality, and Sobolev embedding, we have Φ_𝜇(𝑢)≥12_Ω|Δ𝑢|²−𝑐|∇𝑢|²−𝜇|ℎ|d𝑥_∞𝑝_Ω|𝑢|^𝑝d𝑥−_Ω𝐹()≥1𝑥,𝑢d𝑥2‖𝑢‖²−𝛼+𝜀2|𝑢|²₂−𝜇|ℎ|_∞𝑝|𝑢|^𝑝_𝑝−𝐶₁|𝑢|^𝜎_𝜎≥121−𝛼+𝜀𝜆_𝑘+1𝜆_𝑘+1𝑢‖−𝑐‖²−𝜇𝐶₂‖𝑢‖^𝑝−𝐶₃‖𝑢‖^𝜎=121−𝛼+𝜀𝜆_𝑘+1𝜆_𝑘+1−𝑐−𝜇𝐶₂‖𝑢‖^𝑝−2−𝐶₃‖𝑢‖^𝜎−2‖𝑢‖², (27) where 𝐶₂,𝐶₃ are constant.
Let 𝑔(𝑡)=𝜇𝐶₂𝑡^𝑝−2+𝐶₃𝑡^𝜎−2; we claim that there exists 𝑡₀ such that 𝑔𝑡₀<121−𝛼+𝜀𝜆_𝑘+1𝜆_𝑘+1−𝑐. (28)
It is easy to see that 𝑔(𝑡) has a minimum at 𝑡₀=((𝜇𝐶₂(2−𝑝))/(𝐶₃(𝜎−2)))^{1/(𝜎−𝑝)}; substituting 𝑡₀ in 𝑔(𝑡), we have 𝑔𝑡₀=𝜇^{(𝜎−2)/(𝜎−𝑝)}𝐶₂𝐶₂2−𝑝𝐶₃(𝜎−2)^{(𝑝−2)/(𝜎−𝑝)}+𝐶₃𝐶₂2−𝑝𝐶₃(𝜎−2)^{(𝜎−2)/(𝜎−𝑝)}<121−𝛼+𝜀𝜆_𝑘+1𝜆_𝑘+1,−𝑐 (29) where 0<𝜇<𝜇^∗=[1/2(1−((𝛼+𝜀)/(𝜆_𝑘+1(𝜆_𝑘+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 Φ_𝜇|_𝑋^⟂_{𝑘∩𝜕𝐵𝜌(𝜃)}≥𝛾>0, where 𝐵_𝜌(𝜃)={𝑢∈𝐻∶‖𝑢‖≤𝜌}.
(iii) There exists 𝑅_𝑘>𝜌, such that Φ_𝜇(𝑢)≤0,∀𝑢∈𝑋_𝑘⧵𝐵_𝑅𝑘(𝜃),𝑘=1,2,…. (30) For 𝛽>𝜆_𝑘(𝜆_𝑘−𝑐), (𝐹₂) implies that for any 𝜀>0, there exists 𝐶₄>0 such that 𝐹()≥1𝑥,𝑢2𝛽−𝜀|𝑢|²−𝐶₄. (31) Taking 𝜀>0 such that 𝛽−𝜀>𝜆_𝑘(𝜆_𝑘−𝑐). By Lemma 2 and (𝜇/𝑝)∫_Ωℎ(𝑥)|𝑢|^𝑝d𝑥≥0, we have Φ_𝜇(𝑢)=12_Ω|Δ𝑢|²−𝑐|∇𝑢|²−𝜇d𝑥𝑝_Ωℎ(𝑥)|𝑢|^𝑝d𝑥−_Ω𝐹()≤1𝑥,𝑢d𝑥2‖𝑢‖²−𝛽−𝜀2|𝑢|²₂+𝐶₄|Ω|≤121−𝛽−𝜀𝜆_𝑘𝜆_𝑘‖𝑢‖−𝑐²+𝐶₄|Ω|⟶−∞,∀𝑢∈𝑋_𝑘,as‖𝑢‖⟶+∞. (32) Hence there exists 𝑅_𝑘>𝜌 such that Φ_𝜇(𝑢)≤0,∀𝑢∈𝑋_𝑘⧵𝐵_𝑅𝑘(𝜃),𝑘=1,2,…. (33) Summing up the above proofs, Φ_𝜇 satisfies all the conditions of Theorem 3 and Remark 4. Then the problem (1) has infinitely many solutions.

Proof of Theorem 6. Similar to the proof of Theorem 5(i), we have ^Ω=Δ𝑣Δ𝜑−𝑐∇𝑣∇𝜑d𝑥_Ω𝜆_𝑘𝑣𝜑d𝑥,∀𝜑∈𝐻. (34) We can easily see that 𝑣≢0. In fact, if 𝑣≡0, then |𝑣|₂=0, which contradicts lim_𝑛→∞|𝑣_𝑛|₂=|𝑣|₂=1. Then 𝑣 is a corresponding eigenfunction of the eigenvalue 𝜆_𝑘; hence |𝑢_𝑛|→∞, a.e. 𝑥∈Ω.
By (𝐹₃), we have lim_𝑛→∞𝑓𝑥,𝑢_𝑛𝑢_𝑛−2𝐹𝑥,𝑢_𝑛=−∞uniformlyin𝑥∈Ω. (35) It follows from Fatous Lemma that lim_𝑛→∞Ω𝑓𝑥,𝑢_𝑛𝑢_𝑛−2𝐹𝑥,𝑢_𝑛d𝑥=−∞. (36)
On the other hand, (15) implies that 2𝑐⟵2Φ_𝜇𝑢_𝑛−Φ^′_𝜇𝑢_𝑛,𝑢_𝑛=_ΩΔ𝑢_𝑛²−𝑐∇𝑢_𝑛²−d𝑥2𝜇𝑝_Ωℎ(𝑥)𝑢_𝑛^𝑝d𝑥−_Ω2𝐹𝑥,𝑢_𝑛−d𝑥_ΩΔ𝑢_𝑛²−𝑐∇𝑢_𝑛²d𝑥+𝜇_Ωℎ(𝑥)𝑢_𝑛^𝑝𝑑𝑥+_Ω𝑓𝑥,𝑢_𝑛𝑢_𝑛=d𝑥_Ω𝑓𝑥,𝑢_𝑛𝑢_𝑛−2𝐹𝑥,𝑢_𝑛−2d𝑥𝑝−1𝜇_Ωℎ(𝑥)𝑢_𝑛^𝑝d𝑥, (37) where lim_𝑛→∞(2/𝑝−1)𝜇∫_Ωℎ(𝑥)|𝑢_𝑛|^𝑝𝑑𝑥=∞, which contradicts (36); hence {𝑢_𝑛} is bounded. Similar to the proof of Theorem 5(i), we have {𝑢_𝑛}→𝑢 in 𝐻. Hence we verify Φ_𝜇(𝑢) satisfies (𝐶) condition.
Similar to [15], let 𝐻(𝑥,𝑡)=𝐹(𝑥,𝑡)−(1/2)𝜆_𝑘(𝜆_𝑘−𝑐)𝑡², and 𝑓(𝑥,𝑡)=𝜆_𝑘(𝜆_𝑘−𝑐)𝑡+ℎ(𝑥,𝑡); then lim_|𝑡|→∞2𝐻(𝑥,𝑡)𝑡²=0,lim_|𝑡|→∞(ℎ(𝑥,𝑡)𝑡−2𝐻(𝑥,𝑡))=−∞. (38) It follows that for every 𝑁>0, there exists 𝑅_𝑁>0 such that ℎ()()|𝑡|𝑥,𝑡𝑡−2𝐻𝑥,𝑡≤−𝑁,∀𝑡∈𝑅,≥𝑅_𝑁,a.e.𝑥∈Ω. (39) For 𝑡>0, we have 𝑑𝐻()𝑑𝑡𝑥,𝑡𝑡²=ℎ(𝑥,𝑡)𝑡−2𝐻(𝑥,𝑡)𝑡³; (40) over the interval [𝑡,𝑠]⊂[𝑇,+∞), we have 𝐻(𝑥,𝑠)𝑠²−𝐻(𝑥,𝑡)𝑡²≤𝑁21𝑠²−1𝑡². (41) Letting 𝑠→+∞, we see that 𝐻(𝑥,𝑡)≥(𝑁/2), 𝑡∈𝑅, 𝑡≥𝑅_𝑁, a.e. 𝑥∈Ω. In a similar way, we have 𝐻(𝑥,𝑡)≥(𝑁/2), 𝑡∈𝑅, 𝑡≤−𝑅_𝑁, a.e. 𝑥∈Ω. Hence lim_|𝑡|→∞𝐻(𝑥,𝑡)=+∞.a.e.𝑥∈Ω. (42) By Lemma 2 and ℎ(𝑥)≥0, we have Φ_𝜇(𝑢)=12_Ω|Δ𝑢|²−𝑐|∇𝑢|²−1d𝑥2𝜆_𝑘𝜆_𝑘−𝑐_Ω𝑢²−𝜇d𝑥𝑝_Ωℎ(𝑥)|𝑢|^𝑝d𝑥−_Ω𝐻(𝑥,𝑢)d𝑥≤−_Ω𝐻(𝑥,𝑢)d𝑥⟶−∞,∀𝑢∈𝑋_𝑘,as‖𝑢‖⟶+∞. (43) Hence there exists 𝑅_𝑘>𝜌 such that Φ_𝜇(𝑢)≤0,∀𝑢∈𝑋_𝑘⧵𝐵_𝑅𝑘(𝜃),𝑘=1,2,…. (44) Summing up the above proofs and Theorem 5(ii), Φ_𝜇 satisfies all the conditions of Theorem 3 and Remark 4, then the problem (1) has infinitely many solutions.

Acknowledgments

This work was supported by the National Nature Science Foundation of China (10971179) and the Project of Shandong Province Higher Educational Science and Technology Program (J09LA55, J12LI53).