On a fourth order elliptic equation with supercritical exponent

R E S E A R C H

Open Access

On a fourth order elliptic equation with

supercritical exponent

Kamal Ould Bouh

*_{Correspondence:}

[email protected]; [email protected]

Department of Mathematics, Taibah University, P.O. Box 30002, Almadinah Almunawwarah, Kingdom of Saudi Arabia

Abstract

This paper is concerned with the semi-linear elliptic problem involving nearly critical exponent (Pε):

_∂

_ε

_ε

small, (Pε) has no sign-changing solutions with low energy which blow up at exactly three points. Moreover, we prove that (Pε) has no bubble-tower sign-changing solutions.

MSC: 35J20; 35J60

Keywords: nonlinear problem; critical exponent; sign-changing solutions; bubble-tower solution

1 Introduction and results

We consider the following semi-linear elliptic problem with supercritical nonlinearity:

(Pε)

⎧ ⎨ ⎩

_u=|u|p–+ε_{u in,}

u=u=  on∂,

whereis a smooth bounded domain inRn,n≥,εis a positive real parameter and

p+  = _n_–n is the critical Sobolev exponent for the embedding ofH_()∩H

() into

Lp+_().

When the biharmonic operator in (Pε) is replaced by the Laplacian operator, there are

many works devoted to the study of the counterpart of (Pε); see for example [–], and the

references therein.

Whenε< , many works have been devoted to the study of the solutions of (Pε) see

for example [–]. In the critical case, this problem is not compact, that is, whenε=  it corresponds exactly to the limiting case of the Sobolev embeddingH_()∩H

() into

Lp+(), and thus we lose the compact embedding. In fact, van Der Vorst showed in [] that (P) has no positive solutions ifis a starshaped domain. Whereas Ebobisse and Ould

Ahmedou proved in [] that (P) has a positive solution provided that some homology

group ofis non-trivial. This topological condition is sufficient, but not necessary, as examples of contractible domainson which a positive solution exists show [].

In the supercritical case,ε> , the problem (Pε) becomes more delicate since we lose

the Sobolev embedding which is an important point to overcome. The problem (Pε) was

studied in [] where the authors show that there is no one-bubble solution to the problem

and there is a one-bubble solution to the slightly subcritical case under some suitable con-ditions. However, we proved in [] that (Pε) has no sign-changing solutions which blow

up exactly at two points. In this work we will show the non-existence of sign-changing solutions of (Pε) having three concentration points.

We note that problem (Pε) has a variational structure. The related functional is

infJ(u), whereJ(u) :=

|u| 

(|u|p++ε₎/(p++ε),u∈H

_()∩H

(),u≡.

Jsatisfies the Palais-Smale condition in the subcritical case, while this condition fails in the critical case. Such a failure is due to the functions

δ(a,λ)(x) =c

λ(n–)/

( +λ|_x–a|₎(n–)/, c=

n(n– )n_{– }(n–)/_{,λ> ,a∈R}n_{. (.)}

cis chosen so thatδ(a,λ)is the family of solutions of the following problem:

u=up, u>  inRn. (.)

When we study problem (.) in a bounded smooth domain, we need to introduce the functionPδ(a,λ)which is the projection ofδ(a,λ)onH(). It satisfies

Pδ(a,λ)=δ(a,λ) in, Pδ(a,λ)=Pδ(a,λ)=  on∂.

These functions are almost positive solutions of (.). We denote byGthe Green’s function defined by,∀x∈,

G(x,·) =cnδx in, G(x,·) =G(x,·) =  on∂,

whereδxis the Dirac mass atxandcn= (n– )(n– )wn, withwnis the area of the unit sphere ofRn_{. We denote byHthe regular part ofG, that is,}

H(x,x) =|x–x|–n–G(x,x) for (x,x)∈.

Forx= (x,x)∈\, with={(y,y) :y∈}, we denote byM(x) the matrix defined by

M(x) = (mij)≤i,j≤, wheremii=H(xi,xi),m=m=G(x,x), (.)

and letρ(x) be its least eigenvalue.

The spaceH()∩H_() is equipped with the norm · and its corresponding inner product·,·defined by

u =

|u|

/

and u,v=

uv, u,v∈H()∩H_(). (.)

Theorem . Letbe any smooth bounded domain inRn_{,n≥.Ifis a regular value}

ofρ(x),then there existsε> ,such that,for eachε∈(,ε),problem(Pε)has no

sign-changing solutions uεwhich satisfy

uε=Pδ(aε,,λε,)–Pδ(aε,,λε,)+Pδ(aε,,λε,)+vε, (.)

with|uε|ε

∞is bounded and

⎧ ⎨ ⎩

aε,i∈, λε,id(aε,i,∂)→ ∞ for i= , , ,

Pδ(aε,i,λε,i),Pδ(aε,j,λε,j) → for i= j and vε →asε→.

The second result deals with the phenomenon of bubble-tower solutions for the bihar-monic problem (Pε) with supercritical exponent. We will give a generalization of the result

found in []. More precisely, we have the following.

Theorem . Letbe any smooth bounded domain inRn_{,n≥.There existsε}

> ,such

that,for eachε∈(,ε),problem(Pε)has no solutions uεof the form

uε=

k

i=

γiPδ(aε,i,λε,i)+vε, withλε,≤λε,≤ · · · ≤λε,kand|uε| ε

∞is bounded, (.)

where k≥,γi∈ {–, },aε,i∈,for each i≤j,λε,i|aε,i–aε,j|is bounded and asε→,

vε →,λε,id(aε,i,∂)→+∞, Pδ(aε,i,λε,i),Pδ(aε,j,λε,j) →for i=j,and if l∈ {/ k– ,k},

λε,l|aε,l–aε,l+| →,where l=min{q:γq=· · ·=γk}.

The proof of our results will be by contradiction. Thus, throughout this paper we will assume that there exist solutions (uε) of (Pε) which satisfy (.) or (.). In Section , we will

obtain some information as regards such (uε) which allows us to develop Section  which

deals with some useful estimates to the proof of our theorems. Finally, in Section , we combine these estimates to obtain a contradiction. Hence the proof of our results follows.

2 Preliminary results

In this section, we assume that there exist solutions (uε) of (Pε) which satisfy

uε=

k

i=

γiPδ(aε,i,λε,i)+vε, (.)

with |uε|ε

∞ is bounded, k≥,aε,i∈, and as ε→, vε →,λε,id(aε,i,∂)→+∞, Pδ(aε,i,λε,i),Pδ(aε,j,λε,j) → fori=j. Arguing as in [] and [], we see that foruεsatisfying

(.), there is a unique way to chooseαi,ai,λi, andvsuch that

uε=

k

i=

γiαiPδ(ai,λi)+v, (.)

with ⎧ ⎪ ⎨ ⎪ ⎩

αi∈R, αi→,

ai∈, λi∈R∗+, λid(ai,∂)→+∞,

v→ inH_()∩H

(), v∈E,

whereEdenotes the subspace ofH

() defined by

E:=w:w,ϕ= ,∀ϕ∈SpanPδi,∂Pδi/∂λi,∂Pδi/∂aj_i,i≤k;j≤n. (.)

Here,aj_idenotes thejth component ofaiand in the sequel, in order to simplify the nota-tions, we setδ(ai,λi)=δiandPδ(ai,λi)=Pδi. We always assume thatuε(which satisfies (.))

is written as in (.) and (.) holds. From (.), it is easy to see that the following remark holds.

Lemma .[] Let uε satisfying the assumption of the theorems. λi occurring in(.)

satisfies

λε_i → asε→for each i≤k. (.)

Remark .[, ] We recall the following estimate:

δε_i(x) –cε_λε_i(n–)/=Oεlog +λ_i|x–ai|

in. (.)

3 Some useful estimates

As usual in this type of problems, we first deal with thev-part ofuε, in order to show that

it is negligible with respect to the concentration phenomenon.

Lemma . The function v defined in(.),satisfies the following estimate:

v ≤cε+c

i(λidi)n–+

i=jεij(logε–ij )(n–)/n if n< ,

i(λidi)(n+)/– ε(n–)+

i=jε

(n+)/(n–)

ij (logεij–)(n+)/n if n≥,

where di:=d(ai,∂)for i≤k and for i=j,εijis defined by

εij=

λi λj

+λj λi

+λiλj|ai–aj| (–n)/

. (.)

Proof The proof is the same as that of Lemma . of [], so we omit it. Now, we state the crucial points in the proof of our theorems.

Proposition . Assume that n≥and letαi,aiandλibe the variables defined in(.)

with k= andγ= –γ=γ.We have

αic

n–  

H(ai,ai) λn_i– +

j=i

(–)i+jαjc

λi

∂εij ∂λi

+n–  

H(ai,aj) (λiλj)(n–)/

+αi

n–   cε

≤cε+c

⎧ ⎨ ⎩

k(λkdk)n–+

j=i(ε

n n–

ij logεij–+εij(logε–) (n–)

n _{) if n}≥_,

k(λkdk) +

j=iεij(logε–)/ if n= ,

(.)

Proof Let

Using [], we derive

Concerning the last integral, it can be written as

Observe that, forn≥, we havep–  = /(n– )≤, thus

It remains to estimate the second integral of (.). We have

Therefore, combining (.)-(.), and Lemma ., the proof of Proposition .

Proposition . Let n≥.We have the following estimate:

Proof The proof is similar to the proof of Proposition .. But there exist some integrals which have different estimates. We will focus in those integrals. In fact, (.), (.)-(.) are also true if we changeλ∂Pδ/∂λby (/λ)∂Pδ/∂a. It remains to deal with the other

equations. Following [], we get

The proof of Proposition . is thereby completed.

4 Proof of the theorems

Proof of Theorem 1.1

Arguing by contradiction, let us assume that problem (Pε) has solutions (uε) as stated in

Theorem .. Recall thatuεis written as

Furthermore, an easy computation shows that

On the other hand, following the proof of Proposition ., we have, for eachi= , , ,

(Fi) 

We distinguish many cases depending on the set

:=(i,j) :i=jandmin(λi,λj)|ai–aj|is bounded

and we will prove that all these cases cannot occur.

We remark that if (i,j)∈we deriveλi/λj→ or∞anddi/dj=  +o() asε→. Furthermore, the behavior ofεijdepends on the set. In fact we have, assuming that λi≤λj,

First we start by proving the following crucial lemmas.

Remark . Ordering theλi’s:λi≤λi≤λi, adding (Ei) + (Ei) + (Ei), and using (.),

Proof The proof will be by contradiction.

Proof of (i). Assume thatd/d→. In this case, we have

|a–a| ≥cd and ε=



(λλ|a–a|)(n–)/

which implies thatε=o((λd)–n+ (λd)–n). Using Remark ., we derive a

contradic-tion. In the same way, we prove thatd/d. Hence the proof of Claim (i) is completed.

Proof of (ii). Assume thatλ/λ→. By Claim (i), we have (λd)–=o((λd)–). Four

cases may occur.

Case .λ/λ or{(, ), (, )} ∩=φ. Using (.), (E) implies that

H(a,a)

λn_– +ε+ε+ε=o

 (λd)n–

+ε

.

By Claim (i) and (E), we obtainε=o((λd)–n). By Remark ., this case cannot occur.

Case .λ/λ→,{(, ), (, )} ∩=φ, andλ/λ →+∞. In this case, it is easy to

obtainε=o(ε+ε). Using Remark ., we derive a contradiction.

Case .λ/λ→, (, )∈, (, ) /∈, andλ/λ+∞. In this case, we see thatλ|a–

a|is bounded andλ|a–a| →+∞. Hence, we derive thatλ|a–a| →+∞, which

implies thatλk|a–a| →+∞fork= , . Thus

ε=

 +o() (λλ|a–a|)(n–)/

=

λ

λ

(n–)/

 +o() (λλ|a–a|)(n–)/

=o(ε).

Then by Remark ., we get a contradiction.

Case .λ/λ→, (, )∈, andλ/λ+∞. In this case, it is easy to getε=o(ε).

Using the formula [(E) + (E) – (E)], we deduce that ε=o(ε+ε), which implies

thatε=o(ε). Hence by Remark ., we derive a contradiction and Claim (ii) is thereby

completed.

Proof of (iii). Without loss of generality, we can assume thatd≤d. First, as in the proof

of Claim (i), we get|a–a| ≤cd. Now assume that|a–a|/d→, which implies

H(ai,ai)

λn_i– =o(ε) fori= , . Two cases may occur.

Case .λ≤λor{(, ), (, )} ∩=φ. Using (E), we obtain

H(a,a)

λn_– =o(ε), εi=o(ε) fori= ,  and ε=o(ε), and we derive a contradiction from (E).

Case . λ≤λ and{(, ), (, )} ∩= φ. Letk∈ {, } such that (,k)∈. Using

Claim (ii) and the fact thatλ≤λ, we derive thatεk≥c(λ/λk)(n–)/, which implies that

d∼dk,λ/λk→, andλ|a–ak|is bounded. Using (.) fori=k, we get

–λ|a–ak|ε

n n– k +

λλ

λk |a–a|ε

n n–  =o

 (λd)n–

+

r=j ε

n– n–

rj +ε

n– n–

. (.)

Sinceλ|a–ak|is bounded andε(λλ|a–ak|)(–n)/, we derive that

ε

n– n–  =o

 (λd)n–

+ε

n– n–  +ε

n– n–  +ε

n– n–

which implies that

ε=o

 (λd)n–

+ε+ε+ε

. (.)

By Remark ., we get a contradiction.

Lemma . There exists a positive constant c_such that

(i) c_λ≤λ;

(ii) di≥c for i= , .

Proof Without loss of generality, we can assume thatd≤d.

Proof of (i). Assume thatλ/λ→. First we claim thatd/d. In fact, arguing by

contradiction we assume thatd/d→, we getd→,|a–a| ≥cd, and|a–a| ≥cd.

Hence,{(, ), (, )} ∩=φ. From (E), we obtain

H(a,a)

λn_– +ε+ε+ε=o

 (λd)n–

+ 

(λd)n–

+ε

. (.)

Letνibe the outward normal vector atai. Sinced,d, and|a–a|are of the same order,

we have (see [] and [])

 λn_–

∂H(a,a)

∂ν ∼

c

(λd)n–

and ∂G(a,a)

∂ν ≤

. (.)

Using (F), we get /(λd)n–=o(ε(n–)/(n–)), which implies that /(λd)n–=o(ε). From

(E), we derive a contradiction. Hence our claim is proved.

Thus there exists a positive constantcso thatd≥cd. Now, since we have assumed

thatλ/λ→, Lemma . implies thatε=o((λd)–n). Finally, using Remark ., we

get a contradiction and the proof of Claim (i) follows.

Proof of (ii). Assume thatd→. Note that Claim (i) and (E) imply that (.) holds.

Now, following the proof of (i), we obtain a contradiction.

We turn now to the proof of Theorem .. By the previous lemmas, we know thatλand

λare of the same order,|a–a| ≥candλ≥cλi, fori= ,  wherecis a positive constant.

Hence, (E) implies that (.) holds. Furthermore, fori= ,  (Ei) implies that

H(ai,ai) λn_i– –

G(a,a)

(λλ)n–

=o

 (λd)n–

+ 

(λd)n–

+ε

. (.)

We denote byr(x) the eigenvector associated toρ(x) whose norm is . We point out that we can chooser(x) so that all their components are positive (see [] and []).

Leti=λ(–i n)/,= (,), andx= (a,a). From (.), we have

M(x)· t

The scalar product of (.) byr(x) gives

ρ(x)r(x)· t

=o(). (.)

Since the components ofr(x) are positive andλ,λare of the same order, there exists a

positive constantc, such thatr(x)· t ≥c> . Hence, we get

ρ(x) =o(). (.)

We deduce from (.) and (.) that

∂M

∂xi (x)·

t

=o(). (.)

Observe thatmay be written in the form

=βr(x) +r(x), withr(x)·r(x) = , r =o(β) andβ∼ . (.) Using (.), we get

∂M

∂xi

(x)·tr(x) +∂M ∂xi

(x)· r(x)

=o(). (.)

Sincedi≥cfori= ,  and|a–a| ≥c, the matrix ∂∂Mxi(x) is bounded.

Furthermore, we have r =o( ), which implies that ∂M

∂xi

(x)·tr(x) =o(). (.)

Let us consider the equality

M(x)·tr(x) =ρ(x)·tr(x)

and derivative it with respect toxi; we obtain ∂M

∂xi

(x)·tr(x) +M(x)∂ t_r ∂xi

(x) = ∂ρ ∂xi

(x)·tr(x) +ρ(x)∂ t_r ∂xi

(x).

The scalar product withr(x) gives

r(x)·∂M ∂xi

(x)·tr(x) = ∂ρ ∂xi

(x). (.)

Using (.), we obtain

∂ρ ∂xi

(x) =o(). (.)

Proof of Theorem 1.2

Arguing by contradiction, let us assume that problem (Pε) has solutions (uε) as stated in

Theorem .. From Section , these solutions have to satisfy (.) and (.). As in the proof of Proposition ., we have, for eachi= , . . . ,k,

wherecis a positive constant. Using (.), easy computations show that

ε(i–)j+εi(j+)=o(εij), ∀i<j,

The proof will depend on the value oflwhich is defined in the theorem.

Finally, using (.), (.), (.), and (E_) we obtain

H(a,a)

λn– 

=o

 (λd)n–

,

which gives a contradiction.

Case .l=k– . Using (.), an easy computation implies that

λk–

∂ε(k–)k ∂λk–

–λk ∂ε(k–)k

∂λk ≥

cε(k–)k. (.)

Then from (E_k–), (E_k), (.), (.), and the fact thatγk–γk=  andγk–γk–= – (since

l=k– ), we obtain

cε(k–)k+ε(k–)(k–)=o

ε+ 

(λd)n–

+

r=j εrj

. (.)

Now using (E_k) and (.) we get (.) and as before, (.) is satisfied. Hence we also derive a contradiction from (E_).

Case .l∈ {/ k,k– }. Recall that in this case we have assumed thatλl|al–al+| →. This

implies that

λl ∂εl(l+)

∂λl

=(n– )/εl(l+)

 +o(). (.)

Hence, using (E_l), the definition ofland (.) we obtain the first part of (.). The sec-ond part follows from (E_k) and the first one. Finally, as before we derive a contradiction from (E_).

Hence, our theorem is proved.

Competing interests

The author declares that they have no competing interests.

Acknowledgements

The author gratefully acknowledges the Deanship of Scientific Research at Taibah University on material and moral support, in particular by financing this research project.

Received: 26 August 2014 Accepted: 1 December 2014 Published:30 Dec 2014

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