# Existence of Positive Solutions of Semilinear Biharmonic Equations

Authors: Yajing Zhang; Yinmei Lü; Ningning Wang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

This paper is concerned with the existence of positive solutions of semilinear biharmonic problem whose associated functionals do not satisfy the Palais-Smale condition.


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## Body

## 1. Introduction

We consider the semilinear biharmonic problem(1)Δ2u=f(x,u)inΩ,u=Δu=0on⁡∂Ω,
where Ω is a bounded domain with smooth boundary in ℝN,N≥5.Problems of this type have been studied in [1–7]. In [6] Liu and Wang studied (1) when f(x,t) is asymptotically linear with respect to t at infinity. In order to find critical points of the functional Φ associated with (1), one usually applies the Mountain Pass Theorem proposed by Ambrosetti and Rabinowitz [8]. For applying the theorem, one often requires the following condition, that is, for some θ>2 and M>0:
(2)0<θF(x,t)≤tf(x,t)for|t|≥M,
where F(x,t)=∫0tf(x,s)ds. Condition (2) is important for ensuring that each Palais-Smale sequence is bounded in H2(Ω)∩H01(Ω). We say Φ satisfies the Palais-Smale condition (henceforth denoted by (PS)) if any sequence {un} for which Φ(un) is bounded and Φ′(un)→0 as n→∞ possesses a convergent subsequence. Note that the nonlinear term f(x,t) is asymptotically linear, not superlinear, with respect to t at infinity, which means that condition (2) cannot be assumed in their case. Lin et al. use some tricks and techniques to prove that the (PS) sequence is bounded. Then they use Mountain Pass Theorem to get a positive solution to (1).In [2] Ramos and Rodrigues considered (1) with the nonlinearity f(x,t)=μt+a(x)g(t), where μ is a real parameter, a∈C1(Ω¯) changes sign in Ω, and g∈C1(ℝ) is subcritical and has a superlinear behavior both at zero and at infinity. They extended for the biharmonic operator results that were obtained for the corresponding second order problem in [9]. Their assumptions on f do not seem to imply suitable compactness properties (namely, the so-called Palais-Smale condition) for the corresponding functional, if one uses a variational argument. Moreover, due to the absence of sign in the nonlinear term, it is not clear whether the geometric structure of the functional associated with (1) falls into one of the usual schemes used in critical point theory.In this paper, we suppose thatf satisfies the following:(H1)
f∈C1(Ω¯×ℝ), ∂ft′/∂xi∈C(Ω¯×ℝ)(1≤i≤n), and f(x,t)≥0 if t≥0;(H2)
f(x,0)=ft′(x,0)=0 for all x∈Ω¯;(H3)
there existT>0 and 1<p<(N+4)/(N-4) such that
(3)|ft′(x,t)|≤C|t|p-1,|∇xf(x,t)|≤C|t|p,|∇xft′(x,t)|≤C|t|p-1for |t|≥T and x∈Ω¯;(H4)
there existμ>0 and T>0 such that f(x,t)≥μtp for all x∈Ω¯ and t≥T.This type of hypotheses assumed here does not imply the (PS) condition and does not fit in the condition that implies a priori bounds. Recently, de Figueiredo and Yang [10], Liu et al. [11], and Ramos et al. [9] have considered semilinear second order elliptic problems without the (PS) condition. Our assumptions (H1)–(H4) exactly come from [10]. In [9, 10] the link between the Morse index and the L∞ bounds of solutions is shown. In [12] Bahri and Lions mentioned that bounds on Morse indices are useful in some problems to prove the Palais-Smale compactness condition.For the reader's convenience, we give an example:f(x,t)=tp[2+sin(ln⁡t)cos⁡|x|2] for x∈Ω¯,t≥0, where p>1. Due to advances of our method and our interest in positive solutions, without loss of generality, we may assume that f(x,t)=0 for t≤0. It is easy to see that f(x,t) satisfies the conditions (H1)–(H4). Moreover, it is obvious that f(x,t) does not satisfy the hypothetical conditions on nonlinearity in [2].Our main result is the following.Theorem 1.
Supposef satisfies (H1)–(H4). Problem (1) has at least a positive solution.The organization of the paper is as follows. In Section2 we prove some new nonlinear Liouville type theorems which may be useful in other situations. In Section 3 we prove Theorem 1. Firstly, we apply the Mountain Pass Theorem to a suitable sequence of truncated problems. In particular, it follows that the Morse index of the solutions of the truncated problems is finite. We use this fact and the blow-up argument to show that the sequence of the truncated problems is bounded. A version of the well-known Pohozaev identity is in turn essential. Throughout this paper, the constant C will denote various generic constants.


## 2. Liouville Type Theorems

ForR>2r>0, let ψr,R∈C0∞(ℝN) be a cut-off function satisfying
(4)ψr,R(x)=1,x∈BR∖B2r,ψr,R(x)=0,x∈Br∪B2Rc,|∇ψr,R|≤CR,x∈BRc,|D2ψr,R(x)|≤CR2,x∈BRc.
Define
(5)J′(u)v=∫ℝNΔuΔvdx-∫ℝNQ(x)upvdx,hhhhhhhhhhhhhhhhhhh∀v∈C0∞(ℝN),
and then
(6)J′′(u)φ2=∫ℝN|Δφ|2dx-p∫ℝNQ(x)up-1φ2dx,hhhhhhhhhhhhhhhhhhhhhhh∀φ∈C0∞(ℝN).Lemma 2.
Suppose thatQ is a function satisfying 0<μ≤Q≤C, where μ and C are constants. Let u be a nonnegative solution of the following problem:
(7)Δ2u=Q(x)upinℝN,
with finite Morse index, where 1<p<(N+4)/(N-4). Then there exists r0>0 such that J′′(u)(ψr0,R2u)2≥0, ∀R>2r0.The Morse index of solutions of (7) is defined as the dimension of the negative space corresponding to the spectral decomposition of the operator Δ2-pQup-1.Proof of Lemma2.
Suppose the assertion is false. Then forr1>0, there exists R1>2r1 such that J′′(u)(ψr1,R12u)2<0 and for r2>2R1, we may find R2>2r2 such that J′′(u)(ψr2,R22u)2<0. Then the supports of ψr1,R12u and ψr2,R22u are disjoint, so the Morse index of u is larger than or equal to 2. Iterating the argument, we may get a contradiction since the Morse index of u is supposed to be finite.Proposition 3.
Letu be a bounded nonnegative solution with finite Morse index of (7). Then both ∥Δu∥L2(ℝN) and ∥u∥Lp+1(ℝN) are finite.Proof.
By Lemma2, there exists a r0>0 such that
(8)J′′(u)(ψr0,R2u)2≥0,∀R>2r0.
That is,
(9)∫ℝN[4(ψr0,RΔψr0,R+|∇ψr0,R|2)2u2+16ψr0,R2(∇ψr0,R∇u)2+ψr0,R4|Δu|2+16(ψr0,RΔψr0,R+|∇ψr0,R|2)×uψr0,R∇ψr0,R∇u+8ψr0,R3Δu∇ψr0,R∇u+4(ψr0,RΔψr0,R+|∇ψr0,R|2)uψr0,R2Δu4(ψr0,RΔψr0,R+|∇ψr0,R|2)2]dx≥p∫ℝNQ(x)up+1ψr0,R4dx.
Multiplying (7) by uψr0,R4, we find
(10)∫ℝN[(3ψr0,R2|∇ψr0,R|2+ψr0,R3Δψr0,R)ψr0,R4+8ψr0,R3Δu∇u∇ψr0,R+4(3ψr0,R2|∇ψr0,R|2+ψr0,R3Δψr0,R)uΔu]dx=∫ℝNQ(x)up+1ψr0,R4dx.
From (9) and (10), it follows that
(11)∫ℝN‍[4(ψr0,RΔψr0,R+|∇ψr0,R|2)2u2+16ψr0,R2(∇ψr0,R∇u)2+16(ψr0,RΔψr0,R+|∇ψr0,R|2)uψr0,R∇ψr0,R∇u(ψr0,RΔψr0,R+|∇ψr0,R|2)2]dx≥(p-1)∫ℝN(ψr0,R4|Δu|2+8ψr0,R3Δu∇u∇ψr0,R+4uψr0,R3Δψr0,RΔu)dx+(12p-4)∫ℝNψr0,R2|∇ψr0,R|2uΔudx≥(p-1)∫ℝN[(εψr0,R4|Δu|2+1εψr0,R2u2|Δψr0,R|2)ψr0,R4|Δu|2-4(εψr0,R4|Δu|2+1εψr0,R2|∇u|2|∇ψr0,R|2)-2(εψr0,R4|Δu|2+1εψr0,R2u2|Δψr0,R|2)]dx-(6p-2)∫ℝN(εψr0,R4|Δu|2+1ε|∇ψr0,R|4u2)dx,
where ε>0 is small enough. Consequently,
(12)∫ℝN‍ψr0,R4|Δu|2dx≤C∫ℝN‍[(ψr0,RΔψr0,R+|∇ψr0,R|2)2u2hhhhhhhhh+ψr0,R2|∇ψr0,R|2|∇u|2hhhhhhhhh+(ψr0,RΔψr0,R+|∇ψr0,R|2)uψr0,R∇ψr0,R∇uhhhhhhhhh+ψr0,R2u2|Δψr0,R|2+|∇ψr0,R|4u2(ψr0,RΔψr0,R+|∇ψr0,R|2)2(ψr0,RΔψr0,R+|∇ψr0,R|2)2]dx.
Using the value of ψr0,R, we get
(13)∫B2R‍ψr0,R4|Δu|2dx≤C∫B2R[|∇ψr0,R|2ψr0,R2|Δψr0,R|2u2+|∇ψr0,R|4u2hhhhhhh+ψr0,R2|∇u|2|∇ψr0,R|2]dx=C[∫B2r0∖Br0‍(ψr0,R2|Δψr0,R|2+C|∇ψr0,R|4)u2dxhhhh+∫B2R∖BR(ψr0,R2|Δψr0,R|2+|∇ψr0,R|4)u2dxhhhh+∫B2Rψr0,R2|∇u|2|∇ψr0,R|2dx∫B2r0∖Br0‍(ψr0,R2|Δψr0,R|2+C|∇ψr0,R|4)]≤C(1+1R4∫B2Ru2dx+1R2∫B2Rψr0,R2|∇u|2dx).
Using the interpolation inequality (see [13]), we obtain
(14)∫B2Rψr0,R4|Δu|2dx≤C(1+1R4∫B2Ru2dx).
From (9) and (10), it follows that
(15)(p-1)∫ℝNQ(x)up+1ψr0,R4dx≤∫ℝN[4(ψr0,RΔψr0,R+|∇ψr0,R|2)2u2hhhhhh+16ψr0,R2(∇ψr0,R∇u)2hhhhhh+16(ψr0,RΔψr0,R+|∇ψr0,R|2)uψr0,R∇ψr0,R∇uhhhhhh-8ψr0,R2|∇ψr0,R|2uΔu(ψr0,RΔψr0,R+|∇ψr0,R|2)2]dx.
Using the value of ψr0,R again, we have
(16)∫BRQ(x)up+1ψr0,R4dx≤C∫B2R[(ψr0,RΔψr0,R+|∇ψr0,R|2)2u2hhhhhhhhh+ψr0,R2(∇ψr0,R∇u)2hhhhhhhhh+(ψr0,RΔψr0,R+|∇ψr0,R|2)uψr0,R∇ψr0,R∇uhhhhhhhhh-ψr0,R2|∇ψr0,R|2uΔu(ψr0,RΔψr0,R+|∇ψr0,R|2)2]dx.
Estimating the right side of (16) by the argument exactly as above and using (14), we have
(17)∫BRQ(x)up+1ψr0,R4dx≤C(1+1R4∫B2Ru2dx),
where C does not depend on R.
Sinceψr0,R=1 over BR∖B2r0,
(18)∫BRQ(x)up+1ψr0,R4dx=∫B2r0Q(x)up+1ψr0,R4dx+∫BR∖B2r0Q(x)up+1ψr0,R4dx=∫BRQ(x)up+1dx-∫B2r0Q(x)up+1(1-ψr0,R4)dx.
By (17), we get
(19)∫BRQ(x)up+1dx≤∫B2r0Q(x)up+1(1-ψr0,R4)dx+C(1+1R4∫B2Ru2dx).
Hence
(20)∫BRup+1dx≤C(1+1R4∫B2Ru2dx),
since μ≤Q(x)≤C and u is bounded.
We will prove∫ℝN‍|u|p+1dx<+∞. Assume to the contrary that ∫ℝN‍|u|p+1dx is not finite; by (20), we obtain
(21)∫BR|u|p+1dx≤CR4∫B2Ru2dx,
for large R.
Using Hölder’s inequality, we get(22)∫B2Ru2dx≤C(∫B2R|u|p+1dx)2/(p+1)RN(p-1)/(p+1),
Substituting (22) for the right-hand side in (21) gives. 							(23)∫BR|u|p+1dx≤CR-4+(N(p-1)/(p+1))(∫B2R|u|p+1dx)2/(p+1).
Let α=-4+(N(p-1)/(p+1)), β=2/(p+1), and I(R)=∫BR‍|u|p+1dx. Then iterating (23), we get
(24)I(R)≤Cγ2αδRαγI(2k+1R)βk+1,
where γ=1+β+⋯+βk, δ=β+2β2+⋯+kβk.
Sinceu is bounded, the left side of (24) is of the order RN, while the right side is of the order RM, where
(25)M=α1-βk+11-β+Nβk+1⟶α1-β,ask⟶+∞,
which yields a contradiction since α<0.
As above using (14), we obtain
(26)∫BR|Δu|2dx≤C(1+1R4∫B2Ru2dx).
Combining (26) and (22), we have
(27)∫BR|Δu|2dx≤C[1+R-4+(N(p-1)/(p+1))(∫B2R|u|p+1dx)2/(p+1)].
Using the already proved fact that ∫ℝN‍|u|p+1dx is finite, we obtain ∫ℝN‍|Δu|2dx<+∞.Using an approach similar to the method used in the proof of Proposition3, we prove the following proposition.Proposition 4.
Suppose thatQ is a function satisfying 0<μ≤Q≤C, where μ and C are constants. Let u be a bounded nonnegative solution with finite Morse index of
(28)Δ2u=Q(x)upinℝ+N,u=Δu=0on∂ℝ+N,
where 1<p<(N+4)/(N-4) and ℝ+N={x=(x1,…,xN)∈ℝ+N:xN>0}. Then both ∥Δu∥L2(ℝ+N) and ∥u∥Lp+1(ℝ+N) are finite.Proposition 5.
Letu be as in Proposition 3. Suppose that
(29)∫ℝN|Δu|2dx≥ν∫ℝNQ(x)up+1dx,
where ν>1. Then u≡0.Proof.
By Proposition3 and (7), we obtain
(30)∫ℝN|Δu|2dx=∫ℝNQ(x)up+1dx.
By (29), it yields
(31)∫ℝNQ(x)up+1dx≥ν∫ℝNQ(x)up+1dx.
Then u≡0.Similar arguments are used to prove the following proposition.Proposition 6.
Letu be as in Proposition 4. Suppose that
(32)∫ℝ+ℕ|Δu|2dx≥ν∫ℝ+NQ(x)up+1dx,
where ν>1. Then u≡0.For future reference, we have the following result at the end of this section.Lemma 7.
Letu be a nonnegative solution of problem (7) with finite Morse index. Then there exists r0>0 such that for R>2r0 one has
(33)R∫∂BR(|∇u|2+|Δu|2+|∇(Δu)|2+Q(x)up+1)dS≤CR(N(p-1)/(p+1))-4(∫ℝNup+1)2/(p+1).Proof.
By Proposition3, ∫ℝN‍up+1dx<∞. We proceed as in [10, 12]. Denote by i(Q,u) the Morse index of u with respect to the operator Δ2-pQup-1. Suppose that 0≤i(Q,u)<m. Let ψi,i=1,2,…, be smooth functions such that
(34)ψ1=1on⁡B1,supp⁡ψ1⊂B3/2,ψ1=1on⁡Di∶={x∈ℝN:2(i-1)≤|x|≤2i-1},hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhfori≥2,supp⁡ψi⊂Ai∶={x∈ℝN:2i-52<|x|<2i-12},hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhfori≥2,|∇ψi|≤C,|Δψi|≤C,fori≥1.
Denote ψi,R=ψi(x/R) for R>0. For each R>0, there exists i=i(R) such that
(35)∫ℝN[4(ψi,RΔψi,R+|∇ψi,R|2)2u2hhhh+16ψi,R2(∇ψi,R∇u)2+ψi,R4|Δu|2hhhh+16(ψi,RΔψi,R+|∇ψi,R|2)uψi,R∇ψi,R∇uhhhh+8ψi,R3Δu∇ψi,R∇uhhhh+4(ψi,RΔψi,R+|∇ψi,R|2)uψi,R2Δu(ψi,RΔψi,R+|∇ψi,R|2)2]dx≥p∫ℝNQ(x)up+1ψi,R4dx.
So we deduce as (12) and (15) that
(36)∫ℝNψi,R4|Δu|2dx≤C∫ℝN‍[(ψi,RΔψi,R+|∇ψi,R|2)2u2hhhhhhhhh+ψi,R2(∇ψi,R∇u)2+(ψi,RΔψi,R+|∇ψi,R|2)hhhhhhhhh×uψi,R|∇ψi,R|·|∇u|+ψi,R2|∇u|2|∇ψi,R|2hhhhhhhhh+ψi,R2u2|Δψi,R|2+|∇ψi,R|4u2(ψi,RΔψi,R+|∇ψi,R|2)2]dx,∫ℝNQ(x)up+1ψi,R4dx≤C∫ℝN‍[(ψi,RΔψi,R+|∇ψi,R|2)2u2+ψi,R2(∇ψi,R∇u)2hhhhhhhhh+(ψi,R|Δψi,R|+|∇ψi,R|2)uψi,R|∇ψi,R|·|∇u|hhhhhhhhh+8ψi,R2|∇ψi,R|2u|Δu|(ψi,RΔψi,R+|∇ψi,R|2)2]dx.
By (36), we have
(37)∫ℝN‍(ψi,R4|Δu|2+Q(x)up+1ψi,R4)dx≤C∫ℝN‍[(ψi,R|Δψi,R|+|∇ψi,R|2)2u2+ψi,R2|∇ψi,R|2|∇u|2hhhhhhhhh+(ψi,R|Δψi,R|+|∇ψi,R|2)uψi,r|∇ψi,R|·|∇u|hhhhhhhhh+ψi,R2|∇ψi,R|2u|Δu|+ψi,R2|Δψi,R|2u2hhhhhhhhh+|∇ψi,R|4u2(ψi,R|Δψi,R|+|∇ψi,R|2)2]dx≤C∫ℝN‍[(ψi,R|Δψi,R|+|∇ψi,R|2)2u2hhhhhhhhh+ψi,R2|∇ψi,R|2|∇u|2hhhhhhhhh+(ψi,R|Δψi,R|+|∇ψi,R|2)uψi,R|∇ψi,R|hhhhhhhhh·|∇u|+ψi,R2|Δψi,R|2u2+|∇ψi,R|4u2hhhhhhhhh+12εψi,R4|Δu|2+12ε|∇ψi,R|4u2]dx,
where ε>0 is small enough.
Thus(38)∫ℝN‍(ψi,R4|Δu|2+Q(x)up+1ψi,R4)dx≤C∫ℝN‍[(ψi,R|Δψi,R|+|∇ψi,R|2)2u2+ψi,R2|∇ψi,R|2|∇u|2hhhhhhhhh+(ψi,R|Δψi,R|+|∇ψi,R|2)uψi,R|∇ψi,R|hhhhhhhhh·|∇u|+ψi,R2|Δψi,R|2u2+|∇ψi,R|4u2(ψi,RΔψi,R+|∇ψi,R|2)2]dx≤C∫ℝN‍[(ψi,R|Δψi,R|+|∇ψi,R|2)2u2+ψi,R2|∇ψi,R|2|∇u|2hhhhhhhhh+ψi,R2|Δψi,R|2u2+|∇ψi,R|4u2(ψi,R|Δψi,R|+|∇ψi,R|2)2]dx.
Integrating by parts, we obtain
(39)∫ℝNuψi,R4Δudx=-∫ℝN(|∇u|2ψi,R4+4ψi,R3u∇u∇ψi,R)dx.
Consequently,
(40)∫ℝN‍|∇u|2ψi,R4dx≤C(∫ℝN‍|Δu|uψi,R4dx+∫ℝNψi,R3u|∇u|·|∇ψi,R|dx)≤C(∫ℝN(ψi,R4|Δu|2+u2ψi,R4)dx+ε∫ℝN‍|∇u|2ψi,R4dx+ε-1∫ℝNψi,R2u2|∇ψi,R|2dx).
Hence
(41)∫ℝN|∇u|2ψi,R4≤C(∫ℝNψi,R4|Δu|2dx+∫ℝNu2ψi,R4dx+∫ℝNu2|∇ψi,R|4dx)≤C(∫ℝN‍(ψi,R4|Δu|2+u2|∇ψi,R|4)dx+∫ℝN‍|Δ(uψi,R2)|2dx)≤C(∫ℝN‍(ψi,R4|Δu|2+u2|∇ψi,R|4)dxhhhhhh+∫ℝN‍(ψi,R2Δu+2uψi,RΔψi,R+2uψi,RΔψi,Rhhhhhhhhhhhh+2u|∇ψi,R|2+4ψi,R∇u∇ψi,R)2dx)≤C∫ℝN‍(ψi,R4|Δu|2+u2|∇ψi,R|4+u2ψi,R2|Δψi,R|2hhhhhhhhh+ψi,R2|∇u|2|∇ψi,R|2)dx.
Multiplying (7) by ψi,R4Δu and integrating by parts, we obtain
(42)∫ℝN‍ψi,R4|∇(Δu)|2dx=-4∫ℝNψi,R3Δu∇ψi,R·∇(Δu)dx-∫ℝNQupψi,R4Δudx.
Consequently,
(43)∫ℝN‍ψi,R4|∇(Δu)|2dx≤C∫ℝN‍[ψi,R3|Δu|·|∇ψi,R|·|∇(Δu)|+Qupψi,R4|Δu|]dx≤C∫ℝN‍[εψi,R4|∇(Δu)|2+ε-1ψi,R2|Δu|2|∇ψi,R|2hhhhhhhhh+Q2u2pψi,R4+ψi,R4|Δu|2]dx.
Hence
(44)∫ℝNψi,R4|∇(Δu)|2dx≤C[1R2∫ℝNψi,R2|Δu|2dx+∫ℝN(Qup+1ψi,R4+ψi,R4|Δu|2)dx].
By (38)–(44), we have
(45)∫ℝN‍(ψi,R4+ψi,R4+ψi,R4|∇(Δu)|2+Q(x)up+1ψi,R4)dx≤[1R2∫ℝN‍ψi,R2|Δu|2dx+∫ℝN‍(ψi,R4|Δψi,R|2+|∇ψi,R|4)u2dx+∫ℝNψi,R2|∇ψi,R|2|∇u|2dx].
By the definition of ψi,R, it follows that
(46)∫RDi‍(|∇u|2+|Δu|2+|∇(Δu)|2+Q(x)up+1)dx≤CR4∫RAi∖RDiu2dx.
This implies by Hölder’s inequality that
(47)∫RDi‍(|∇u|2+|Δu|2+|∇(Δu)|2+Q(x)up+1)dx≤1R4(∫RAi∖RDiup+1dx)2/(p+1)×[mes(RAi∖RDi)](p-1)/(p+1)≤CR(N(p-1)/(p+1))-4(∫ℝN‍up+1dx)2/(p+1).
Hence, there exists R¯∈(R,(2m-1)R) such that
(48)R∫∂BR¯(|∇u|2+|Δu|2+|∇(Δu)|2+Q(x)up+1)dS≤CR(N(p-1)/(p+1))-4(∫ℝN‍up+1dx)2/(p+1).
The assertion follows.


## 3. Proof of Theorem1

Let us first note that hypotheses (H1)–(H4) imply that there exist a sequence{tn},tn→+∞ as n→∞ and a continuous function c such that
(49)ft′(x,tn)>0,lim⁡n→∞ft′(x,tn)tnp-1=c(x)hhhhhhhhhhhhhuniformlyinx∈Ω.
Using L'Hospital's rule, we obtain
(50)lim⁡n→∞f(x,tn)tnp=lim⁡n→∞ft′(x,tn)ptnp-1=1pc(x).
Without loss of generality, we may assume that f(x,t)=0, if t≤0. Define a truncation of f by
(51)fn(x,t)={f(x,tn)-1ptnft′(x,tn)+1pft′(x,tn)tnp-1tpift>tn,f(x,t)if0≤t≤tn,0ift<0.
Note that fn∈C1 in the variable t.Let us consider the truncated problem(52)Δ2u=fn(x,u)inΩ,u=Δu=0on⁡∂Ω.Lemma 8.
Problem (52) possesses at least one positive solution with finite Morse index.Proof.
Consider the functional associated with problem (52)
(53)Jn(u)=12∫Ω‍|Δu|2dx-∫Ω‍Fn(x,u)dx,
where Fn(x,t)=∫0t‍fn(x,s)ds. We will use the Mountain Pass Theorem by Ambrosetti and Rabinowitz [8] to obtain existence result for problem (52). One can easily check that there exist θ>2 and M>0 such that for |t|≥M,
(54)0<θFn(x,t)≤tfn(x,t).
We note that condition (54) is important for ensuring that Jn has a Mountain Pass geometry and satisfies the Palais-Smale condition. So, using the Mountain Pass Theorem, we obtain a nontrivial weak solution un of (52). By Lemma B3 in [14], un is a classical solution of (52). By the maximum principle for Δ2 with Navier boundary conditions we get that un is positive. The geometry of the Mountain Pass, described in [15, 16], implies that the Morse indices of un are less than or equal to 1. Thus Lemma 8 is proven.Letg be a function satisfying (H1)–(H3), and consider the problem
(55)Δ2u=g(x0+ax,bu)inΩ,u=Δu=0,on⁡∂Ω,
where a,b are positive constants and x0∈ℝN.Lemma 9.
Letu be a solution of (55). Then for any ball BR(0)⊂Ω one has
(56)b-1N∫BRG(x0+ax,bu)dx+R∫∂BR[2∂u∂n∂(Δu)∂n-〈∇(Δu),∇u〉]dS+(N-2)∫∂BRΔu∂u∂ndS+12R∫∂BR‍|Δu|2dS+ab-1∫BR〈x,∇xG(x0+ax,bu)〉dx=N-42∫BR|Δu|2dx+b-1R∫∂BR‍G(x0+ax,bu)dS,
where G(x,u)=∫0u‍g(x,t)dt and n denotes the unit outward normal to ∂BR.Proof.
By standard procedures, one can prove the Pohozaev type identity. We give the proof for completeness and for the reader's convenience.
By Proposition 2.2 in [17], we have
(57)∫BR‍Δ2u〈x,∇u〉dx=-12∫∂BR‍|Δu|2〈x,n〉dS+∫∂BR‍{∂(Δu)∂n〈x,∇u〉+∂u∂n〈x,∇(Δu)〉hhhhhhhhhh-〈∇(Δu),∇u〉〈x,n〉∂(Δu)∂n}dS+N2∫BR‍|Δu|2dx+(N-2)∫BR‍〈∇(Δu),∇u〉dx.
It is clear that
(58)g(x0+ax,bu)〈x,∇u〉=b-1div⁡(xG(x0+ax,bu))-b-1NG(x0+ax,bu)-ab-1〈x,∇xG(x0+ax,bu)〉.
By (55), we have
(59)∫BR‍Δ2u〈x,∇u〉dx=∫BR‍g(x0+ax,bu)〈x,∇u〉dx.
Substituting (57) and (58) into (59) and using the divergence theorem, we find
(60)-12∫∂BR‍|Δu|2〈x,n〉dS+∫∂BR‍{∂(Δu)∂n〈x,∇u〉+∂u∂n〈x,∇(Δu)〉hhhhhhhhhh-〈∇(Δu),∇u〉〈x,n〉∂(Δu)∂n}dS+N2∫BR‍|Δu|2dx+(N-2)∫BR‍〈∇(Δu),∇u〉dx=b-1∫∂BR‍G(x0+ax,bu)〈x,n〉dS-b-1N∫BR‍G(x0+ax,bu)dx-ab-1∫BR‍〈x,∇xG(x0+ax,bu)〉dx.
Using the Green's formula, we obtain
(61)∫BR‍〈∇(Δu),∇u〉dx=-∫BR‍Δu·Δudx+∫∂BR‍Δu∂u∂ndS.
Substituting (61) into (60) and using n=x/R on ∂BR, we obtain (56).If it happens that∥un∥≤tn, for some n, then un is also a solution of (1), and the proof of Theorem 1 will be completed. Thus it suffices to prove the following proposition. We prove the proposition by the blow-up technique of Gidas and Spruck [18].Proposition 10.
Suppose thatun is a solution of (52) with finite Morse index. Then there exists a tn>0 such that ∥un∥∞≤tn.Proof.
Assume by contradiction that there does not exist such atn. So we should have ∥un∥>tn, for all n. Then ∥un∥→∞ as n→∞.
LetMn=max⁡Ω¯un(x)=un(xn),xn∈Ω. Define
(62)vn(y)=Mn-1un(xn+Mn-((p-1)/4)y),y∈Ωn∶=Mn(p-1)/4(Ω-xn),
which satisfies
(63)Δ2vn=Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn)inΩn,vn=Δvn=0on⁡∂Ωn,
and vn≤1 in Ω¯n, vn(0)=1.
Due to compactness ofΩ¯ we may also assume that xn→x0∈Ω¯. So there are two cases to be considered, x0∈Ω and x0∈∂Ω.
Case
1. x0∈Ω.
GivenR>0 there is an n0∈ℕ such that B¯2R(0)⊂Ωn for all n≥n0. By the Lp-estimates due to Agmon et al. [19], we have that for all γ>1(64)∥vn∥W4,γ(BR)≤C{∥Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn)∥Lγ(B2R)+∥vn∥Lγ(B2R)∥Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn)∥∥Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn)∥}.
By assumptions (H1)–(H3) and the definition of fn, it follows that
(65)|fn(x,t)|≤C(1+|t|p),
and then for large n,
(66)|Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn)|≤Mn-pC(1+|Mnvn|p)≤C(Mn-p+|vn|p)≤2C.
So we obtain that
(67)∥vn∥W4,γ(BR)≤Cuniformlyinn.
Choosing γ>n, it follows from standard embedding theorems that {vn} is uniformly bounded in C3,α(B¯R),0<α<1. By the Schauder estimate of Agmon et al. [19] one has
(68)∥vn∥4,α,B(1/2)R≤C{∥Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn)∥α,BR+∥vn∥0,BR∥Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn)∥α,BR∥Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn)∥}.
Next we claim that
(69)∥Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn)∥α,BR≤C.
In order to do that we write
(70)fn(xn+mn-((p-1)/4)y,Mnvn(y))-fn(xn+mn-((p-1)/4)z,Mnvn(z))=[fn(xn+mn-((p-1)/4)y,Mnvn(y))-fn(xn+mn-((p-1)/4)y,Mnvn(z))]+[fn(xn+mn-((p-1)/4)y,Mnvn(z))-fn(xn+mn-((p-1)/4)z,Mnvn(z))]∶=I1+I2.
Then we have
(71)|I1|≤|∂fn∂t(xn+Mn-((p-1)/4)y,Mnvn)hh×Mn(vn(y)-vn(z))∂fn∂t|≤C(1+|Mnwn|p-1)Mn|vn(y)-vn(z)|≤C(1+Mnp)|y-z|α.
According to the definition of fn, we divide the estimate of I1 into three cases.(i)
IfMnvn(z)≤T, then since fn is C1 we get
(72)|I2|≤CMn-((p-1)/4)|y-z|.(ii)
IfT≤Mnvn(z)≤tn, we use condition (H3) to get
(73)|I2|≤|∇xf(xn+Mn-((p-1)/4)wn,Mnvn(z))|·Mn-((p-1)/4)|y-z|≤CMn-((p-1)/4)|Mnvn(z)|p·|y-z|.(iii)
Iftn≤Mnvn(z), by the definition of fn, we have
(74)|I2|≤|f(xn+Mn-((p-1)/4)y,tn)-f(xn+Mn-((p-1)/4)z,tn)|+1ptn|ft′(xn+M-((p-1)/4)y,tn)hhhhhhh-ft′(xn+M-((p-1)/4)z,tn)|+1p(Mnvn(z))p|ft′(xn+Mn-((p-1)/4)y,tn)tnp-1hhhhhhhhhhhhhhh-1pft′(xn+Mn-((p-1)/4)z,tn)tnp-1|≤|∇xf(xn+Mn-((p-1)/4)wn,tn)|·Mn-((p-1)/4)|y-z|+1ptn|∇xft′(xn+Mn-((p-1)/4)w~n,tn)|·Mn-((p-1)/4)|y-z|+1p·1tnp-1Mnp|∇xft′(xn+Mn-((p-1)/4)w~n,tn)|·Mn-((p-1)/4)|y-z|≤CMn(3p+1)/4|y-z|.By (71)–(74), we obtain
(75)|I1+I2|≤Mnp|y-z|,
which proves (69), and therefore
(76)∥vn∥4,α,B(1/2)R≤Cuniformlyinn.
Using Arzela-Ascoli Theorem and (69) and (76), we obtain a subsequence of vn still denoted by vn, such that
(77)vn⟶vinC4,α′(B(1/2)R),(78)Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn(y))⟶A(y)hhhhhhhhhhhhhhhhhhhhhhinC0,α′(B(1/2)R),
where 0<α′<α<1, as n→∞.
Assume thatβ=lim⁡n→∞tn/Mn. By (65), we have
(79)Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn(y))≤CMn-p(Mnvn)p=Cvnp.
By (77), we have
(80)vn(y)<β,y∈{y:v(y)<β},
for large n. Consequently,
(81)vn≤tnMn.
That is,
(82)Mnvn≤tn.
Then, by the definition of fn and the assumption (H4), we get
(83)μvnp≤Mn-pfn(xn+Mn-((p-1)/4)y,Mnvn(y)),
for large n. Combining (79) and (83) and letting n→∞, we obtain
(84)μvp(y)≤A(y)≤Cvp(y),y∈{y:v(y)<β}.
Define
(85)Q(y)={liminf⁡z→yz∈ωA(z)vp(z)ify∈ω:={y:v(y)=0},A(y)v-p(y)ify∈{y:0<v(y)<β},1pc(x0)ify∈{y:v(y)=β}.
Then there exist positive constants σ and γ such that σ≤Q(y)≤γ, ∀∈B(1/2)R. Passing to the limit in (63) and using (77) and (78), we conclude that v satisfies
(86)Δ2v=Q(y)vpinB(1/2)R.
By a diagonal process, it follows that
(87)Δ2v=Q(y)vpinℝN.
Next we claim that the Morse index of v is finite. If y∈B(1/2)R∖Ω, by the L'Hospital's rule we have
(88)Q(y)vp(y)=lim⁡n→∞fn(xn+Mn-((p-1)/4)y,Mnvn)Mnp=lim⁡n→∞1pMnp-1[vn∂∂tfn(xn+Mn-((p-1)/4)y,Mnvn)hhhhhhhhhhhh-p-14∇xfn(xn+Mn-((p-1)/4)y,Mnvn)hhhhhhhhhhhh×Mn-((p+3)/4)yvn∂∂t].
By (H3) for y∈B(1/2)R,
(89)|Mn-p+1∇xfn(xn+Mn-((p-1)/4)y,Mnvn)Mn-((p+3)/4)y|≤CMn-((p-1)/4).
Then, by (88), we get
(90)Mn-p+1∂∂tfn(xn+Mn-((p-1)/4)y,Mnvn)⟶pQ(y)vp-1(y).
If y∈ω∩B(1/2)R, then vn(y)→0 as n→∞. By assumption (H3), we have
(91)|Mn-p+1∂∂tfn(xn+Mn-((p-1)/4))y,Mnvn|≤Mn-p+1C(Mnvn)p-1=Cvnp-1⟶0
as n→0. Therefore, (90) holds for all x∈B(1/2)R. By the diagonal process, one knows that (90) holds also in ℝN and it converges uniformly on compact sets of ℝN as n→∞.
The uniform convergence ofvn to v on compact sets implies that the Morse index of v is finite. To handle this, we set
(92)Jn(u)φ2=∫ℝN‍|Δφ|2dx-∫ℝN‍Mn-p+1∂∂tfn(xn+Mn-((p-1)/4)x,Mnu)φ2dx,hhhhhhhhhhhhhhhhhhhhhhhhhh∀φ∈C0∞(ℝN).
Let ψ∈C0∞(ℝN) be such that J′′(v)ψ2<0. The uniform convergence of vn to v on compact sets and the fact that (90) holds on compact sets imply
(93)Jn′′(vn)ψ2<0
for large n. Since the Morse index of vn is finite, it follows easily that the Morse index of v is finite. Proposition 3 implies that ∥v∥Lp+1(ℝN) is finite. We claim that there exists a λ>1 such that
(94)∫ℝN|Δv|2dx≥λ∫ℝNQvp+1dy.
Then Proposition 5 yields v≡0, which contradicts v(0)=1.
Now, we prove (94). Applying Lemma 9 to (63) in the ball BR(0) for fixed R>0, we obtain
(95)Mn-p-1N∫BRFn(xn+Mn-((p-1)/4)y,Mnvn)dy+R∫∂BR[2∂vn∂n∂(Δvn)∂n-〈∇(Δvn),∇vn〉]dS+(N-2)∫∂BRΔvn∂vn∂n+12R∫∂BR|Δvn|2dS+Mn-((5p+3)/4)×∫BR〈y,∇xFn(xn+Mn-((p-1)/4)y,Mnvn)〉dy=N-42∫BR|Δvn|2dy+Mn-p-1R×∫∂BRFn(xn+Mn-((p-1)/4)y,Mnvn)dS.
By (H3), we estimate
(96)|〈y,∇xFn(xn+Mn-((p-1)/4)y,Mnvn)〉|≤CRMnp+1.
Therefore
(97)Mn-((5p+3)/4)|∫BR‍〈y,∇xFn(xn+Mn-((p-1)/4)y,Mnvn)〉dy|≤CRN+1Mn-((p-1)/4),
which tends to zero as n→∞.
Using a similar argument that leads to (69) we can prove
(98)∥Mn-p-1Fn(xn+Mn-((p-1)/4)y,Mnvn)∥α,BR≤Cuniformlyinn.
Then its limit exists as n→∞. Using L’Hospital’s rule as (88) we get
(99)Mn-p-1Fn(xn+Mn-((p-1)/4)y,Mnvn)⟶1p+1Q(y)vp+1(y)hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhuniformlyinBR,
as n→∞. Therefore, by the Lebesgue dominated convergence theorem, we obtain
(100)lim⁡n→∞Mn-p-1∫BRFn(xn+Mn-((p-1)/4)y,Mnvn)dy=1p+1∫BRQ(y)vp+1dy,lim⁡n→∞Mn-p-1∫∂BRFn(xn+Mn-((p-1)/4)y,Mnvn)dS=1p+1∫∂BRQ(y)vp+1(y)dS.
Letting n→∞ in (95), we get
(101)Np+1∫BRQ(y)vp+1(y)dy+R∫∂BR[2∂v∂n∂(Δv)∂n-〈∇(Δv),∇v〉]dS+(N-2)∫∂BRΔv∂v∂ndS+12R∫∂BR|Δv|2dS=N-42∫BR|Δv|2dy+Rp+1∫∂BRQ(y)vp+1(y)dS.
By Lemma 7, there exists R≥2r0 such that
(102)R∫∂BR(|∇v|2+|Δv|2+|∇(Δv)|2+Q(y)vp+1)dS≤CR(N(p-1)/(p+1))-4(∫ℝNvp+1dy)2/(p+1).
Since (N(p-1)/(p+1))-4<0, this implies that
(103)lim⁡R→∞R∫∂BR(|∇v|2+|Δv|2+|∇(Δv)|2+Q(y)vp+1)dS=0.
Taking the limit R→∞ in (101), one has
(104)2NN-4·1p+1∫ℝNQ(y)vp+1(y)dy=∫ℝN‍|Δv|2dy.
Assertion (94) follows.
Case
2. x0∈∂Ω.
Two cases may occur: eitherd(xn,∂Ω)Mn-((p-1)/4)→+∞ or d(xn,∂Ω)Mn-((p-1)/4)→L≥0 as n→∞.
Ifd(xn,∂Ω)Mn-((p-1)/4)→+∞ as n→∞, then for all R>0 the ball BR is contained in Ωn for n large enough. We also obtain a contradiction as in Case 1.
Ifd(xn,∂Ω)Mn-((p-1)/4)→L≥0 as n→∞, by the blow-up argument we get a solution v of
(105)Δ2v=Q(y)vpinΠ,v=Δv=0,on⁡∂Π,
with v≤1 in Π, v(0)=1, and the Morse index being finite, where Π={x∈ℝN:xN>-L}. We may deduce as Case 1 that v≡0. This is a contradiction since v(0)=1. Thus the proof of Proposition 10 is completed.



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*Source: 624328-2014-04-09.xml*