Singularity of the Extremal Solution for Supercritical Biharmonic Equations with Power-Type Nonlinearity?

Baishun LAIZhengxiang YANYinghui ZHANG

1 Introduction

In the previous two decades,positive solutions to the second order semilinear elliptic problem

have attracted a lot of interest(see[1–5]and references therein).Here,we only mention the work by Joseph and Lundgren[2].In their well known work,Joseph and Lundgren gave a complete characterization of all positive solutions of(1.1)in the case g(u)=euor g(u)=(1+au)p,ap>0, λ >0 and ? being a unit ball in Rn.In particular,they found a remarkable phenomenon for g(u)=euand n>2:Either(1.1)has at most one solution for each λ or there is a value of λ for which in finitely many solutions exist.In the case of a power nonlinearity the same alternative is valid if n≥3 and p>As a subsequent step,Lions(see[3,Section 4.2(c)])suggested to study positive solutions to systems of semilinear elliptic equations.So it is an important task to gain a deeper understanding for related higher order problems.

In this paper,we study a semilinear equation involving the bilaplacian operator and a power type nonlinearity

where B?Rnis the unit ball,λ>0 is an eigenvalue parameter,n≥5 and p≥The subcritical case phas been done and many beautiful important results have been proved.In what follows,we will summarize some of the results obtained by[7–8].For convenience,we introduce the following notions.

De finition 1.1We say that u∈Lp(B)is a solution of(1.2)if u≥0 and if for all φ∈C4with φ|?B=|?φ||?B=0,one has

We call u singular if u ? L∞(B),and regular if u∈ L∞(B).A radial singular solution u=u(r)of(1.2)is called weakly singular ifexists.

Note that by standard regularity theory for the biharmonic operator,any regular solution u of(1.2)satis fies u∈Note also that by the positivity preserving property of?2in the ball(see[10])any solution of(1.2)is positive,see also[11]for a generalized statement.This property is known to fail in general domains.For this reason,we restrict ourselves to the ball.Hence,the sub-and super-solution method applies as well as monotone iterative procedures.

De finition 1.2We call a solution u of(1.2)minimal if u≤v a.e.in B for any other solution v of(1.2).

We also denote by λ1>0 the first eigenvalue for the biharmonic operator with Dirichlet boundary conditions

It is known from the positivity preserving property and Jentzsch’s(or Krein-Rutman’s)theorem that λ1is isolated and simple and the corresponding eigenfunction φ1does not change sign.

De finition 1.3We say a weak solution of(1.2)is stable(resp.semi-stable)if

is positive(resp.non-negative).

To illuminate the motivations,we need the following notations which will be used throughout the paper.Set

and

withand the number pcsuch that when p=pc

Now we summarize some of the well-known results as follows.

Theorem 1.1(see[7–8])There existssuch that the following terms hold:

(i)For λ ∈ (0,λ?),(1.2)admits a minimal stable regular solution,denoted by uλ.This solution is radially symmetric and strictly decreasing in r=|x|.

(ii)For λ = λ?,(1.2)admits at least one not necessarily bounded solution,which is called extremal solution u?.

(iii)For λ > λ?,(1.2)admits no(not even singular)solutions.

Theorem 1.2(see[9])Assume that

Then u?is regular.

From Theorem 1.2,we know that the extremal solution of(1.2)is regular for a certain range of p and n.At the same time,they left an open problem:If

is u?singular?

In this paper,by constructing a semi-stable singular(B)-weak sub-solution of(1.2),we prove that,if p is large enough,the extremal solution is singular for dimensions n≥13 and complete part of the above open problem.Our result is stated as follows.

Theorem 1.3There exists p0(n)>1 large enough such that for p≥p0(n),the unique extremal solution of(1.2)is singular for dimensions n≥13,in which case on the unit ball B.

From the technical point of view,one of the obstacle is the well-known difficulty of extracting energy estimates for solutions of fourth order problems from their stability properties.Besides,for the corresponding second order problem(1.1),the starting point was an explicit singular solution for a suitable eigenvalue parameter λ which turned out to play a fundamental role for the shape of the corresponding bifurcation diagram(see[12]).When turning to the biharmonic problem(1.2)the second boundary condition=0 prevents to find an explicit singular solution.This means that the method used to analyze the regularity of the extremal solution for second order problem could not carry to the corresponding problem for(1.2).In this paper,in order to overcome the second obstacle,we use improved and non-standard Hardy-Rellich inequalities recently established by Ghoussoub-Moradifam in[13]to construct a semi-stable singular(B)-weak sub-solution of(1.2).

This paper is organized as follows.In the next section,some preliminaries are reviewed.In Section 3,we will show that the extremal solution u?in dimensions n≥ 13 is singular by constructing a semi-stable singular(B)-weak sub-solution of(1.2).

2 Preliminaries

First we give some comparison principles which will be used throughout this paper.

Lemma2.1(see[10])If u∈satis fies

then u≥0 in BR.Here and in what follows,BRis denoted by the ball of radius R centered at 0.

Lemma 2.2Let u∈L1(BR)and suppose that

for all φ ∈such that φ ≥ 0 in= =0.Then u ≥ 0 in BR.Moreover,u≡0 or u>0 a.e.in BR.

For a proof,see Lemma 17 in[11].

Lemma 2.3If u∈H2(BR)is radial,?2u≥0 in BRin the weak sense,that is

and≥ 0,≤ 0,then u ≥ 0 in BR.

ProofFor the sake of completeness,we include a brief proof here.We only deal with the case R=1 for simplicity.Solve

in the sense u1∈(B)and?u1?φdx=?u?φdx for all φ∈(B).Then u1≥0 in B by Lemma 2.2.

Let u2=u?u1so that?2u2=0 in B.De fine f= ?u2.Then?f=0 in B and since f is radial,we find that f is a constant.It follows that u2=ar2+b.Using the boundary conditions,we deduce a+b≥0 and a≤0,which imply u2≥0.

Now we give a notion of(B)-weak solutions,which is an intermediate class between classical and weak solutions.

De finition 2.1We say that u is an(B)-weak solution of(1.2)if(1+u)p∈L1(B)and if

We say that u is an(B)-weak super-solution(resp.(B)-weak sub-solution)of(1.2)if for ?≥0 the equality is replaced with≥ (resp.≤)and u≥0(resp.≤),≤0(resp.≥)on?B.

We also need the following comparison principle.

Lemma 2.4Let u1,u2∈H2(BR)with(1+u1)p,(1+u2)p∈L1(BR).Assume that u1is stable and

in the H2(BR)-weak sense,i.e.,

and?2u2≥λ(1+u2)pin BRin the similar weak sense.Suppose also

Then

ProofDe fine ω :=u1?u2.Then by the Moreau decomposition(see[14])for the biharmonic operator,there exist ω1,ω2∈(BR),with ω = ω1+ω2,ω1≥ 0 a.e.,?2ω2≤ 0 in the(BR)-weak sense and

By Lemma 2.3,we have that ω2≤ 0 a.e.in BR.

Given now 0≤φ∈(BR),we have that

where f(u)=(1+u)p.Since u1is semi-stable and by density,one has

Since ω1≥ ω,one also has

which once re-arrange gives

where=f(u1)?f(u2)?f′(u1)(u1?u2).The strict convexity of f gives≤ 0 and<0 whenever u1≠u2.Since ω1≥ 0 a.e.in BR,one sees that ω ≤ 0 a.e.in BR.The inequality u1≤u2a.e.in BRis then established.

The following variant of Lemma 2.4 also holds.

Lemma 2.5Let u1,u2∈H2(BR)be radial with(1+u1)p,(1+u2)p∈L1(BR).Assume?2u1≤λ(1+u1)pin BRin the sense of(2.1)and?2u2≥λ(1+u2)pin BRin the same weak sense.Suppose u1|?B≤ u2|?Band |?B≥ |?Band suppose also that u1is semi-stable.Then u1≤u2in BR.

ProofWe solve for∈(B)such that

ByLemma2.3 it follows that≥u1?u2.Next we apply the Moreau decomposition to bu,that is bu=w+v with w,v∈(BR),w≥0,?2v≤0 in BRand∫BR?w?vdx=0.Then the argument follows that of Lemma 2.4.

Lemma2.6Let u be a semi-stable(B)-weak solution of(1.2)and U be an (B)-supersolution of(1.2).Then if u is a classical solution andμ1(u)=0,we have u=U.

ProofSince u is a classical solution,it is easy to see that the in fimum inμ1(u)is attained at some φ.The function φ is then the first eigenfunction of ?2? λf′(u)in(B),where f(u)=(1+u)p.Now we show that ? is of fixed sign.Using the Moreau decomposition,one has

where ?i∈(B)for i=1,2,and

in the(B)-weak sense.If ? changes sign,then ?1? 0 and ?2<0 in B.We can write now

in view of ?1?2< ??1?2in a set of positive measure,leading to a contradiction.

So we can assume ? ≥ 0,and by the Boggio’s principle,we have ? >0 in B.For 0 ≤ t≤ 1,de fine

where ? is the above first eigenfunction.Since f is convex,one sees that

for every t≥0.Since g(0)=0 and

we get that

Since f′′(u)? >0 in B,we finally get that U=u a.e.in B.

From this lemma,we immediately obtain the following corollary.

Corollary 2.1(i)When u?is a classical solution,then μ1(u?)=0 and u?is the unique(B)-weak solution of(1.2);

(ii)If v is a singular semi-stable(B)-weak solution of(1.2),then v=u?and λ = λ?.

Proof(i)Since the function u?is a classical solution,and by the implicit function theorem,we have that μ1(u?)=0 to prevent the continuation of the minimal branch beyond λ?.ByLemma2.4,u?is then the unique(B)-weak solution of(1.2).

(ii)Assume now that v is a singular semi-stable(B)-weak solution of(1.2).If λ < λ?,then by the uniqueness of the semi-stable solution,we have v=uλ.So v is not singular and a contradiction arises.By Theorem A(iii)we have that λ = λ?.Since v is a semi-stable(B)-weak solution of(1.2)and u?is an(B)-weak super-solution of(1.2),we can apply Lemma 2.4 to get v≤ u?a.e.in B.Since u?is also a semi-stable solution,we can reverse the roles of v and u?in Lemma 2.4 to see that v≥ u?a.e.in B.So equality v=u?holds and the proof is complete.

3 Proof of Theorem 1.3

Inspired by the work of[16],we will first show the following upper bound on u?.

Lemma 3.1If n≥13 and p>pc,then u?≤for x ∈ B.

ProofRecall from Theorem 1.1 that K0≤ λ?.We now claim that uλ≤for all λ ∈ (K0,λ?).Indeed, fix such a λ and assume by contradiction that

From the boundary conditions,one has

Hence,

Now consider the following problem:

Then uλis a super-solution to above problem while eu is a sub-solution to the same problem.Moreover for n≥13,we have

and

Sois semi-stable and we deduce that uλ>by Lemma 2.4,and a contradiction arises in view of the fact

The proof is done.

In order to prove Theorem 1.3,we need a suitable Hardy-Rellich type inequality which was established by Ghoussoub-Moradifam in[13].It is stated as follows.

Lemma 3.2Let n≥5 and B be the unit ball in Rn.Then there exists C>0,such that the following improved Hardy-Rellich inequality holds for all φ∈(B):

Lemma 3.3Let n≥5 and B be the unit ball in Rn.Then the following improved Hardy-Rellich inequality holds for all φ∈(B):

As a consequence,the following improvement of the classical Hardy-Rellich inequality holds:

We now give the following lemma which is crucial for the proof of the Theorem 1.3.

Lemma 3.4Suppose there exist λ′>0 and a radial function u ∈ H2(B)∩ (B{0})such that u ? L∞(B)and

and

for either β > λ′or β = Then u?is singular and

ProofFirst,we prove λ?≤ λ′.Noting that the stability inequality(set φ =u)and u∈(B{0})yield(1+u)p∈L1(B),we easily see that u is a weak sub-solution of(1.2).If now λ′< λ?,by Lemma 2.5,u would necessarily be below the minimal solution uλ′,which is a contraction since u is singular while uλ′is regular.

Suppose first that β = λ′=and that n ≥ 13.From the above we have λ?≤We get from Lemma 3.1 and the improved Hardy-Rellich inequality that there exists C>0 so that for all ?∈(B),

It follows thatμ1(u?)>0 and u?must therefore be singular since otherwise,one could use the implicit function theorem to continue the minimal branch beyond λ?.

Suppose now that

We claim that

To prove this,we shall show that for every λ < λ?,

Indeed,we have

Now by the choice of α,we have αp+1λ′< λ?.To prove(3.4),it suffices to prove it for αp+1λ′< λ < λ?.Fix such λ and assume that(3.4)is not true.Then

is non-empty.Since= α ?1>0=uλ(1),we have

and≤.Now consider the following problem:

Then uλis a solution to above problem whileis a sub-solution to the same problem.Moreover,is stable since λ < λ?and

we deduce≤ uλin BR1,which is impossible,sinceis singular while uλis smooth.This establishes(3.3).From(3.3)and the above inequalities,we have

Thus

This is not possible if u?is a smooth function by the implicit function theorem.

Proof Theorem 1.3 Uniqueness and the upper bound estimate of the extremal solution u?have been proven by Corollary 2.1 and Lemma 3.1.Now we only prove that u?is a singular solution of(1.1)for n≥13.In order to achieve this,we shall find a singular H-weak sub-solution of(1.1),denoted by ωm(r),which is stable,according to the Lemma 3.4.

Choosing

since ωm(1)=(1)=0,we have

For any m fixed,when p→+∞,we have

and

Note that

with

(1)Let m=2 and n≥32,then we can prove that

So(3.5)≥0 is valid as long as

At the same time,we have(since a1+≤ a1+a2≤ 1 in[0,1])

Let β =(λ′+ ε)K0,where ε is arbitrary sufficienty small.We need finally here

For that,it is sufficient to have for p?→+∞,

So(3.7)≥0 holds only for n≥32 when p?→+∞.Moreover,for p large enough

Thus it follows from Lemma 3.4 that u?is singular with λ′=e2K0, β =(e2K0+ ε(n,p))and λ?≤ e2K0.

(2)Assume 13≤n≤31.We shall show that u=ω3.5satis fies the assumptions of Lemma 3.4 for each dimension 13≤n≤31.Using Maple,for each dimension 13≤n≤31,one can verify that inequality(3.5)≥ 0 holds for λ′given by Table 1.Then,by using Maple again,we show that there exists β > λ′such that

The above inequality and the improved Hardy-Rellich inequality(3.1)guarantee that the stability condition holds for β > λ′.Hence by Lemma 3.4 the extremal solution is singular for 13 ≤ n ≤ 31,where the value of λ′and β are shown in Table 1.

Table 1

Remark 3.1The improved Hardy-Rellich inequality(3.1)is crucial to prove that u?is singular in dimensions n≥13.Indeed by the classical Hardy-Rellich inequality and u:=w2,Lemma 3.4 only implies that u?is singular n dimensions(n≥32).

AcknowledgementThe first author would like to thank his advisor Prof.Yi Li for his constant support and encouragement.

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