adams’typeinequalityandapplicationtoaquasilinear ... · 2019. 7. 30. · the adams’ inequality...

Annali di Matematica Pura ed Applicata (1923 -) (2019) 198:1331–1349https://doi.org/10.1007/s10231-018-00821-w

Adams’ type inequality and application to a quasilinearnonhomogeneous equation with singular and vanishingradial potentials inR4

Sami Aouaoui1 · Francisco S. B. Albuquerque2

Received: 3 April 2018 / Accepted: 22 December 2018 / Published online: 4 January 2019© Fondazione Annali di Matematica Pura ed Applicata and Springer-Verlag GmbH Germany, part of SpringerNature 2019

AbstractIn this paper, we establish some Adams’ type inequality for weighted second-order Sobolevspaces in four dimensions. The weights are radial and can have a singular or decayingbehavior. This inequality is used to study somenonhomogeneous quasilinear elliptic equation.

Keywords Adams’ inequality · Singular or decaying weights · Radial functions ·Nonhomogeneous quasilinear elliptic equation · Exponential critical growth

Mathematics Subject Classification 35A23 · 35B33 · 35J30 · 35J35 · 35J91

1 Introduction andmain result

Let� be a bounded domain inRN , N ≥ 2. The famous classical Trudinger–Moser inequalityfor the limiting case of Sobolev inequalities (see [25,36]) states that, for all u ∈ W 1,N

0 (�)

and α > 0, we have

sup|∇u|N≤1

∫�

eα|u| NN−1 dx

{< +∞, when α ≤ αN

= +∞, when α > αN ,(1.1)

where αN = Nω1

N−1N−1, |∇u|N is the Dirichlet norm, that is, |∇u|N := (∫

�|∇u|N dx

) 1N and

ωN−1 is the measure of the unit sphere in RN . In addition, for any bounded domain �, the

supremum in (1.1) can be attained. This result has been generalized and extended in many

B Sami [email protected]

Francisco S. B. [email protected]

1 Institut Supérieur des Mathématiques Appliquées et de l’Informatique de Kairouan, Avenue AssadIben Fourat, 3100 Kairouan, Tunisia

2 Departamento de Matemática, Universidade Estadual da Paraíba, CEP 58429-500 Campina Grande,PB, Brazil

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http://crossmark.crossref.org/dialog/?doi=10.1007/s10231-018-00821-w&domain=pdf

http://orcid.org/0000-0003-3693-5820

1332 S. Aouaoui, F. S. B. Albuquerque

directions in recent years. For example, some extended versions of the Trudinger–Moserinequality to unbounded domains in Euclidean spaceswere established in [1,10,12,13,22,28].

The Adams’ inequality is a natural extension of the Trudinger–Moser inequality for thehigher-order derivatives. In 1988, Adams [2] established the following result: Form ∈ N and� an open bounded set of RN such that m < N , there exists a positive constant Cm,N suchthat

sup

u∈Wm, Nm0 (�), |∇mu| N

m≤1

∫�

eβ0|u| NN−m dx ≤ Cm,N |�| , (1.2)

where

β0 = β0(m, N ) := N

ωN−1

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

[π

N2 2m�(m

2 )

�(N−m2

)] N

N−m

, if m is even

[π

N2 2m�

(m+12

)

�(N−m+1

2

)] N

N−m

, if m is odd

and

∇mu :={

�m2 u, m even

∇�m−12 u, m odd

denotes the mth-order gradient of u. Furthermore, inequality (1.2) is sharp, that is, for anyβ > β0, the integral in (1.2) can be made as large as possible.

Plainly, the Adams’ inequality in its form (1.2) cannot be extended to unbounded domains.For t ≥ 0, set

φ(t) = et −j Nm

−2∑j=0

t j

j ! , m ∈ N, N > m, j Nm

= min

{j ∈ N, j ≥ N

m

}.

A first attempt to generalize Adams’ inequality to unbounded domains is due to Ozawa who,in [26], proved the existence of two positive constants α and C such that

∫RN

φ(α |u| N

N−m

)dx ≤ C |u|

NmNm

, ∀ u ∈ Wm, Nm (RN ),∣∣∇mu

∣∣Nm

≤ 1. (1.3)

In the particular case (which is the most interesting in our present work) where N = 4 andm = 2, inequality (1.3) takes the form

∫R4

(eαu2 − 1) dx ≤ C |u|22 , ∀ u ∈ H2(R4), |�u|2 ≤ 1. (1.4)

In 2013, Ruf and Sani [29] established the following very useful result: Let m = 2k aneven integer less than N . Then, there exists a constant Cm,N > 0 such that, for any domain� ⊂ R

N , we have

sup

u∈Wm, Nm0 (�), |(−�+I )ku| N

m≤1

∫�

φ(β0 |u| N

N−m

)dx ≤ Cm,N , (1.5)

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Adams’ type inequality and application to a quasilinear… 1333

and this inequality is sharp. In the particular case where N = 4, m = 2 and � = R4,

inequality (1.5) becomes

supu∈H2(R4), |(−�+I )u|2≤1

∫R4

(e32π

2u2 − 1)dx < +∞. (1.6)

Many other improvements of theAdams’ inequality in bounded and unbounded domains havebeen proved. We can refer to [17,19,23,24,35,37]. Among them, we can cite the followinginequality, proved by Masmoudi and Sani [24]:

∫R4

e32π2u2 − 1

(1 + |u|)2 dx ≤ C |u|22 , ∀ u ∈ H2(R4), |�u|2 ≤ 1. (1.7)

Moreover, (1.7) is no longer true if the exponent 2 is replaced by a real number p < 2.Furthermore, as itwasmentioned in [24], inequalities (1.6) and (1.3) canbederived from (1.7).Another improvement of Adams’ inequality (1.2) in bounded domains has been providedby Tarsi [35]. In fact, in [35], the author obtained some interesting embeddings in Zygmundspaces. One result of these embeddings is the following stronger version ofAdams’ inequality(1.2) in the particular case where m = 2 and N = 4: Let � a bounded domain of R4, thenthere exists a constant C > 0 such that

supu∈H2(�)∩H1

0 (�), |�u|2≤1

∫�

e32π2u2 dx ≤ C |�| , (1.8)

and this inequality is also sharp. Observing that H20 (�) is strictly contained in H2(�) ∩

H10 (�), we can see that (1.8) shows that the optimal exponent 32π2 depends only on the

first-order trace. Now, concerning the singular case, by introducing singular weights of theform 1

|x |α , we have the following inequality proved by Lam and Lu [17] which will be of

big interest in our work: Let 0 ≤ α < 4 and � be a bounded domain in R4. Then, for all

0 ≤ β ≤ 32π2(1 − α

4

), we have

supu∈H2(�)∩H1

0 (�), |�u|2≤1

∫�

eβu2

|x |α dx < +∞. (1.9)

When β > 32π2(1 − α

4

), the supremum is infinite.

Our present work can be considered as a contribution in this direction. In fact, the mainpurpose of this paper is to establish a new inequality of Adams’ type where some radialweights that can be singular at the origin, unbounded or vanishing at infinity are involved andwe will apply it to study some elliptic quasilinear equation by using a variational frameworkbased on a nonstandard Orlicz space. Here, we have to highlight the fact that our resultsconcern radial functions.

In a more precise way and throughout this work, we consider some radial weight functionsV (|x |) and K (|x |) satisfying the following assumptions:

(V ) V : (0,∞) → R is continuous, V (r) > 0 for all r > 0 and there exist a0 > −4, a > −6such that

lim supr→0+

V (r)

ra0< +∞ and lim inf

r→+∞V (r)

ra> 0;

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(K ) K : (0,∞) → R is continuous, K (r) > 0 for all r > 0 and there exist b0 > −4, b < asuch that

lim supr→0+

K (r)

rb0< ∞ and lim sup

r→+∞K (r)

rb< ∞.

This kind of weight functions was firstly introduced by Su et al. [33] (see also [34]) inthe study of a nonlinear Schrödinger equation. In [33], the authors studied the existence ofsolution for the problem

⎧⎨⎩

−�u + V (|x |)u = K (|x |) f (u),

x ∈ RN ,

|u(x)| → 0 as |x | → ∞,

where the nonlinearity considered was f (s) = |s|q−2s, with 2 < q < 2∗ = 2NN−2 for

N ≥ 3 and 2 < q < ∞ if N = 2. Succeeding this study, Albuquerque et al. [4] studied theabove problem in the critical case suggested by Trudinger–Moser inequality (1.1). The aboverestrictions on V and K describing the behavior of the weights near the origin and infinityare crucial to obtain our embedding result (see Lemma 2.3) as well as Adams’ inequality(see Theorem 1.2).

Example 1.1 1. Typical models of weights V and K satisfying (V )− (K ), respectively, canbe found, for instance, in [3,4,6], and they are of the form

•

V (r) ={rγ1 , if r ≤ 1,rγ2 , if r ≥ 1,

and K (r) = rγ3 ,

with γ1 > −4, γ2 > −4 and −4 < γ3 < γ2.•

A1

1 + rγ1≤ V (r) ≤ A2 and 0 < K (r) ≤ A3

1 + rγ2,

for positive constants A1, A2, A3, with γ1 ∈ (0, 6) and γ2 > γ1.

2. One can also take

V (r) = 3 + cos r

rand K (r) = 1 + e−r2 sin r

3r52

.

Clearly, if we take a = a0 = −1 and b = b0 = − 52 , then V and K are singular at the

origin and vanishing at infinity.

In order to state our results, we need to introduce some notations. If 1 ≤ q < ∞, wedefine the weighted Lebesgue spaces

Lq(R4; K ) :={u : R4 → R, u is measurable and

∫R4

K (|x |)|u|q dx < ∞}

and

Lq(R4; V ) :={u : R4 → R, u is measurable and

∫R4

V (|x |)|u|q dx < ∞}

,

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endowed, respectively, with the norms

|u|Lq (R4;K ) =(∫

R4K (|x |)|u|q dx

) 1q

and |u|Lq (R4;V ) =(∫

R4V (|x |)|u|q dx

) 1q

.

Let C∞0 (R4) be the set of smooth functions with compact support and D1,2(R4) the closure

of C∞0 (R4) under the norm

|∇u|2 =(∫

R4|∇u|2 dx

) 12

.

Observe that since 4 > 2, then one can define D1,2(R4) as follows

D1,2(R4) = {u ∈ L1loc(R

4), |∇u| ∈ L2(R4)}.

Denote by D1,2rad (R4) the subspace of the radial functions in D1,2(R4). Set

E :={u ∈ D1,2

rad (R4) ∩ L2(R4; V ), �u ∈ L2(R4)}

,

which is a Hilbert space endowed with the inner product

〈u, v〉 :=∫R4

[�u�v + ∇u∇v + V (|x |)uv] dx, u, v ∈ E,

and the corresponding norm given by

‖u‖ :=(∫

R4

[|�u|2 + |∇u|2 + V (|x |)u2] dx) 1

2

, u ∈ E .

Our Adams’ type inequality is stated in the following theorem.

Theorem 1.2 Assume that (V ) − (K ) hold. Then, for any u ∈ E and α > 0, we have that

K (|x |)(eαu2 − 1

)∈ L1(R4). Furthermore, if α ≤ α0 := min{32π2, 8π2(4 + b0)}, there

holds

supu∈E, ‖u‖≤1

∫R4

K (|x |)(eαu2 − 1

)dx < ∞. (1.10)

Moreover, in the case where −4 < b0 ≤ 0, if we also assume that lim infr→0+

K (r)

rb0> 0, then

the value α0 is optimal, that is,


∫R4

K (|x |)(eαu2 − 1

)dx = +∞, ∀ α > α0.

The outline of the paper is as follows In the forthcoming section, we establish someembedding results involving appropriate weighted Lebesgue spaces. In Sect. 3, we proveour main result, Theorem 1.2. Section 4 is devoted to study a nonhomogeneous quasilinearelliptic problem using our Adams’ type inequality.

Throughout this paper, we use | · |p to denote the norm of the Lebesgue space L p(R4), 1 ≤p ≤ ∞, and the symbols C,C0,C1,C2, . . . to denote positive generic constants (possiblydifferent).

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2 Embedding results: compactness

In the following, we denote by B(x, R) ⊂ R4 the open ball centered at x ∈ R

4 with radiusR > 0 and, to simplify notations, we set BR := B(0, R), Bc

R := R4 \ BR and BR \ Br

denotes the annulus with interior radius r and exterior radius R. Here is some new variant ofthe well-known radial Lemma of Strauss [32].

Lemma 2.1 Assume that (V ) holds. Then, there exist R0 > 0 and C > 0 such that for allu ∈ E,

|u(x)| ≤ C‖u‖|x |−(a+6)/4, |x | > R0. (2.1)

Proof The proof follows some arguments developed in [4,33,34]. For the convenience of thereader, we give it in detail. By a standard density argument, it suffices to prove (2.1) for radialfunctions u ∈ C∞

0 (R4). Let ρ = |x | and ϕ : [0,+∞) → R be such that ϕ(ρ) = u(|x |).Since a > −6, one has

d

dρ

[ρ(a+6)/2ϕ2(ρ)

]= a + 6

2ρ(a+4)/2ϕ2(ρ) + 2ρ(a+6)/2ϕ(ρ)ϕ′(ρ) ≥ 2ρ(a+6)/2ϕ(ρ)ϕ′(ρ).

(2.2)

It follows from (V ) that there exist R0 > 0 and C0 > 0 such that

V (|x |) ≥ C0|x |a, for all |x | ≥ R0. (2.3)

Then for ρ > R0, by (2.2), (2.3) and the Hölder’s inequality, it follows

ρ(a+6)/2ϕ2(ρ) ≤ 2∫ ∞

ρ

s(a+6)/2|ϕ(s)||ϕ′(s)| ds = 2∫ ∞

ρ

(|ϕ′(s)|s3/2) (sa/2|ϕ(s)|s3/2) ds

≤ 2

(∫ ∞

ρ

|ϕ′(s)|2s3 ds)1/2 (∫ ∞

ρ

sa |ϕ(s)|2s3 ds)1/2

≤ C

(∫Bc

ρ

|∇u|2 dx)1/2 (∫

Bcρ

V (|x |)u2 dx)1/2

≤ C‖u‖2.

Thus, we conclude that |u(x)| ≤ C‖u‖|x |−(a+6)/4, for all |x | > R0, which completes theproof. ��Remark 2.2 One can easily observe that in the last proof, we have proved a more preciseresult than the one stated in Lemma 2.1. In reality, we prove the existence of two positiveconstants C and R0 such that

|u(x)| ≤ C |x |− a+64 V(u), ∀ |x | > R0,

where

V(u) =(∫

R4|∇u|2 dx +

∫R4

V (|x |)u2 dx) 1

2

.

Now, using Lemma 2.1 and taking profit of some results established in [33] (see also [4]),we have the following embedding result which will be used in the sequel of the paper.

Lemma 2.3 Suppose that (V ) and (K ) hold. Then, the space E is continuously and compactlyimmersed in Lq(R4; K ) for each 2 ≤ q < ∞.

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Proof For q ≥ 2, consider the quantity Sq := infu∈E\{0}

‖u‖2|u|2

Lq (R4;K )

. To prove the continuity

of the embedding E ↪→ Lq(R4; K ), it suffices to show that Sq > 0. Suppose the contrary,that is, Sq = 0. Then, there exists a sequence (un) ⊂ E satisfying

|un |Lq (R4;K ) = 1 and limn→∞ ‖un‖2 = 0.

By the hypothesis (K ), there exists a positive constant C1 such that

K (|x |) ≤ C1|x |b, for all |x | ≥ R0. (2.4)

Now, for R > R0, by Lemma 2.1, (2.3) and (2.4), one has∫BcR

K (|x |)|un |q dx ≤ C1

∫BcR

|x |b|un |q dx = C1

∫BcR

|x |b−a |un |q−2|x |au2n dx

≤ C2‖un‖q−2∫BcR

|x |b−a−(q−2) a+64 V (|x |)u2n dx .

Since a > −6, b < a and q ≥ 2, then b − a − (q−2)(a+6)4 < 0. Thus, we obtain∫

BcR

K (|x |)|un |q dx ≤ C2Rb−a−(q−2) a+6

4 ‖un‖q . (2.5)

Again due to (K ), there exist 0 < r0 < 1/2 and C3 > 0 such that

K (|x |) ≤ C3|x |b0 , for all 0 < |x | < r0. (2.6)

Now, we shall estimate the integral∫Br0

K (|x |)|un |q dx . For that aim, consider σ > 1 be

such that b0σ > −4 and define a radial cutoff function ψ ∈ C∞0 (B1) verifying

0 ≤ ψ ≤ 1 in B1, ψ ≡ 1 in B1/2, ψ ≡ 0 in B1 \ B3/4 and max {|∇ψ |, |�ψ |} ≤ C in B1.

By (2.6), Hölder’s inequality, and the continuous embedding H20 (B1) ↪→ L

qσσ−1 (B1), we

have ∫Br0

K (|x |)|un |q dx ≤ C3

∫Br0

|x |b0 |un |q dx = C3

∫Br0

|x |b0 |ψun |q dx

≤ C3

(∫Br0

|x |b0σ dx

)1/σ (∫Br0

|ψun |qσ

σ−1 dx

) σ−1σ

≤ C4rb0+ 4

σ

0

(∫B1

|ψun |qσ

σ−1 dx

) σ−1σ ≤ C5r

b0+ 4σ

0

(∫B1

|�(ψun)|2 dx) q

2

≤ C6rb0+ 4

σ

0

(∫B1

|un�ψ |2 dx +∫B1

|∇ψ · ∇un |2 dx +∫B1

|ψ�un |2 dx

) q2

≤ C7rb0+ 4

σ

0

(∫B3/4\B1/2

|un |2 dx +∫B1

|∇un |2 dx +∫B1

|�un |2 dx) q

2

≤ C8rb0+ 4

σ

0

(∫B3/4\B1/2

V (|x |)|un |2 dx +∫B1

|∇un |2 dx +∫B1

|�un |2 dx) q

2

≤ C9rb0+ 4

σ

0 ‖un‖q ,

(2.7)

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where the fact that infB3/4\B1/2

V > 0 has been used. Next, we shall estimate the integral on

BR \Br0 . By the continuity of the functions V and K and the continuous embedding H2(BR \Br0) ↪→ Lq(BR \ Br0), we infer∫

BR\Br0K (|x |)|un|q dx ≤ C10

∫BR\Br0

|un |q dx ≤ C11‖un‖q . (2.8)

Combining (2.5), (2.7) and (2.8), we deduce that∫R4

K (|x |)|un |q dx ≤ C12

(Rb−a−(q−2) a+6

4 + rb0+ 4

σ

0 + 1

)‖un‖q = on(1).

Hence,

limn→∞

∫R4

K (|x |)|un|q dx = 0,

which contradicts the fact that |un |Lq (R4;K ) = 1. Consequently, Sq > 0 and this proves thecontinuity of the embedding. To prove the compacity, let (un) be a bounded sequence in E .Thus, there exists u ∈ E such that, up to subsequence, un⇀u weakly in E . Without loss ofgenerality, we may suppose that u = 0. As in (2.5), we have∫

BcR

K (|x |)|un |q dx ≤ C13Rb−a−(q−2) a+6

4 ‖un‖q ≤ C14Rb−a−(q−2) a+6

4 .

Since b − a − (q−2)(a+6)4 < 0, given ε > 0, we can take R > 0 sufficiently large such that

∫BcR

K (|x |)|un |q dx <ε

3. (2.9)

On the other hand, by (2.7), we have∫Br0

K (|x |)|un |q dx ≤ C15rb0+ 4

σ

0 ‖un‖q ≤ C16rb0+ 4

σ

0 .

Thus, for r0 chosen sufficiently small, it follows∫Br0


3. (2.10)

Furthermore, by the compactness of the embedding H2(BR\Br0) into Lq(BR\Br0) togetherwith the fact that inf

BR\Br0K > 0, there exists n0 ∈ N such that if n > n0, then

∫BR\Br0


3. (2.11)

From (2.9), (2.10) and (2.11), we deduce that∫R4

K (|x |)|un |q dx < ε, ∀n > n0.

Therefore, un → 0 in Lq(R4; K ) and this finalizes the proof of Lemma 2.3. ��Remark 2.4 Besides their important role in order to establish our Adams’ inequality, in thisremark we give some additional comments concerning the ranges of the parameters a, a0, band b0.

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1. The condition b0 > −4 is necessary to get the continuity of the embeddings E ↪→Lq(R4; K ), q ≥ 2. In fact, suppose that b0 ≤ −4 and K (r) ≡ rb0 . Take a radialfunction ϕ ∈ C∞

0 (R4) such that inf0<r<ε0

ϕ(r) > 0 for some ε0 > 0 sufficiently small. We

have∫ +∞

0K (r)ϕ2(r)r3 dr ≥

∫ ε0

0rb0+3ϕ2(r) dr = +∞.

Consequently, E is not continuously immersed in L2(R4; K ).

2. The condition a0 > −4 is necessary to obtain that C∞0,rad(R

4) :={u ∈ C∞

0 (R4) :u is radial

}⊂ E which plays a capital role in our functional setting.

3. The condition b < a is necessary to obtain the compactness of the embeddings E ↪→Lq(R4; K ), q ≥ 2. In fact, suppose that V (r) = K (r) = 1, ∀ r ∈ (0,+∞). Thus, one

can choose a = 0. Clearly, lim supr→+∞

K (r)

rb< +∞, ∀ b ≥ 0. Therefore, it suffices to note

that, in this case, E = H2rad(R

4) and to use the fact that the embedding H2rad(R

4) ↪→L2(R4) is not compact.

4. Finally, the condition a > −6 is purely technical and it is highly needed to get our radialLemma 2.1 which is widely used in the proof of our main results. This is the best range ifwe want to control the radial function by some (negative) exponent of the variable’s normin R

4. The authors believe that this range is quite good and almost the best accessibleone.

3 Adams’ inequality: the Proof of Theorem 1.2

Let α ≤ α0. Recall that by the hypothesis (K ), we have

K (|x |) ≤ C |x |b, for |x | ≥ R0, and K (|x |) ≤ C |x |b0 , for 0 < |x | ≤ r0. (3.1)

Let R > 1 to be chosen later and set

I1(α, u) =∫BR

K (|x |)(eαu2 − 1

)dx, I2(α, u) =

∫BcR

K (|x |)(eαu2 − 1

)dx .

Thus,∫R4

K (|x |)(eαu2 − 1

)dx = I1(α, u) + I2(α, u).

Now, we are going to estimate I1(α, u) and I2(α, u). First, using (3.1) and the continuity ofthe embedding E ↪→ L2(R4; K ), for u ∈ E, one has

I2(α, u) =∫BcR

K (|x |)∞∑j=1

α j u2 j

j ! dx =∫BcR

K (|x |)∞∑j=2

α j u2 j

j ! dx + α

∫BcR

K (|x |)u2 dx

≤ C0

∫BcR

|x |b∞∑j=2

α j u2 j

j ! dx + C1‖u‖2 = C0

∞∑j=2

α j

j !∫BcR

|x |bu2 j dx + C1‖u‖2.

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By the virtue of Lemma 2.1 and using the fact that a > −6 and b − j a+62 + 4 < b − a < 0,

for all j ≥ 2, we get∫BcR

|x |bu2 j ≤ (C‖u‖)2 j∫BcR

|x |b− j a+62 dx = 2π2 (C‖u‖)2 j

∫ ∞

Rsb− j a+6

2 +3 ds

= 2π2 (C‖u‖)2 j Rb− j a+62 +4

j a+62 − b − 4

≤ 2π2

(a − b)Ra−b (C‖u‖)2 j ,

and then

I2(α, u) ≤ 2π2C0

(a − b)Ra−b

∞∑j=2

(αC2‖u‖2) j

j ! + C1‖u‖2

= 2π2C0

(a − b)Ra−b

(eαC2‖u‖2 − 1 − αC2‖u‖2

)+ C1‖u‖2.

Hence, I2(α, u) < +∞, ∀ u ∈ E . Moreover,


I2(α, u) < +∞, ∀ α > 0. (3.2)

Next, we shall estimate the integral I1(α, u). Here, two cases have to be analyzed:Case 1 b0 ≥ 0. In view of (3.1), there exists a positive constant C depending on R such

that

I1(α, u) ≤ C∫BR

eαu2 dx . (3.3)

Let v(x) = u(x) − u(R), x ∈ BR . For x ∈ BR, we have

u2(x) = (v(x) + u(R))2 = v2(x) + u2(R) + 2v(x)u(R) ≤ (1 + u2(R))v2(x) + u2(R) + 1.

Using Remark 2.2, then for R > R0, we get

u2(x) ≤(1 + CR− a+6

2 V2(u))

v2(x) + CR− a+62 ‖u‖2 + 1, ∀ x ∈ BR .

Choosing R large enough such that CR− a+62 ≤ 1, we infer

u2(x) ≤(1 + CR− a+6

2 V2(u))

v2(x) + ‖u‖2 + 1, ∀ x ∈ BR .

Putting that last inequality in (3.3), we obtain

I1(α, u) ≤ C1eα‖u‖2

∫BR

eα

(1+CR− a+6

2 V2(u)

)v2(x)

dx . (3.4)

On the other hand, it is easy to see that v ∈ H2(BR) ∩ H10 (BR). Hence, we can apply the

classical Adams’ inequality (see (1.8)) in (3.4) to deduce that

I1(α, u) < +∞, ∀ u ∈ E .

Now, let u ∈ E be such that ‖u‖2 = |�u|22 + V2(u) ≤ 1. Set w(x) =√1 + CR− a+6

2 V2(u)

v(x), x ∈ BR . By construction, we have∫BR

|�v|2 dx=∫BR

|�u|2 dx ≤ 1 − V2(u).

123


Hence,∫BR

|�w|2 dx =(1 + CR− a+6

2 V2(u)) ∫

BR

|�v|2 dx ≤(1 + CR− a+6

2 V2(u)) (

1 − V2(u)).

Choosing R sufficiently large such that CR− a+62 ≤ 1, it follows∫

BR

|�w|2 dx ≤ (1 + V2(u)) (1 − V2(u)

) ≤ 1.

Having in mind that w ∈ H2(BR) ∩ H10 (BR), we infer from (1.8) that∫

BR

eαw2dx ≤ C |BR | , ∀ α ≤ 32π2,

where |BR | = ∫BRdx . That inequality together with (3.4) implies that


I1(α, u) < +∞, ∀ α ≤ 32π2.

Case 2 −4 < b0 < 0. Since 0 < r0 < R0 < R, by (3.1) we obtain

I1(α, u) =∫Br0

K (|x |)(eαu2 − 1) dx +∫BR\Br0

K (|x |)(eαu2 − 1) dx

≤ C0

∫Br0

|x |b0eαu2 dx + C∫BR\Br0

eαu2 dx

≤ C0

∫BR

|x |b0eαu2 dx + C∫BR

eαu2 dx .

(3.5)

From the first case, we have


∫BR

eαu2 dx < +∞, ∀ α ≤ 8π2(4 + b0) ≤ 32π2. (3.6)

On the other hand, a similar inequality to (3.4) can be immediately established, that is∫BR

|x |b0 eαu2 dx ≤ C2eα‖u‖2

∫BR

eαw2(x)

|x |−b0dx . (3.7)

Now, using again the fact that w ∈ H2(BR) ∩ H10 (BR) and that |�w|L2(BR) ≤ 1, one can

apply the singular Adams’ inequality for bounded domains (1.9) to deduce that∫BR

eαw2(x)

|x |−b0dx ≤ sup

u∈W 2,2N (BR), |�u|L2(BR )

≤1

∫BR

eαu2(x)

|x |−b0dx < +∞, ∀ α ≤ 8π2(4 + b0),

(3.8)

where W 2,2N (BR) = H2(BR) ∩ H1

0 (BR). Combining (3.8), (3.7), (3.6) and (3.5), we obtain


I1(α, u) < +∞, ∀ α ≤ 8π2(4 + b0).

This ends the proof of the first part of Theorem 1.2. Now, we arrive to the sharpness of

inequality (1.10). Assume that −4 < b0 ≤ 0, lim infr→0+

K (r)

rb0> 0. We claim that


∫R4

K (|x |)(eαu2 − 1

)dx = +∞, ∀ α > α0 = 8π2(4 + b0). (3.9)

123


We will consider a family of test functions that has been introduced and used in [23]. For0 < ε < 1, we define

uε(x) =

⎧⎪⎪⎨⎪⎪⎩

√− log ε

32π2 − |x |2√−8π2ε log ε

+ 1√−8π2 log ε

, if |x | < ε1/4,

− log|x |√−2π2 log ε

, if ε1/4 ≤ |x | ≤ 1,

ηε, if |x | > 1,

where ηε ∈ C∞0 (B2) is such that

ηε(x) = 0,∂ηε

∂ν(x) = 1√−2π2 log ε

, ∀ x ∈ R4, |x | = 1,

and |ηε |∞ , |∇ηε |∞ and |�ηε |∞ are all O(

1√− log ε

)as ε → 0+. Clearly, uε ∈ D1,2

rad (R4)

and∫R4 |�uε |2 dx < +∞. An easy computation gives

|∇uε |22 = O

( −1

log ε

), |�uε |22 = 1 + O

( −1

log ε

). (3.10)

On the other hand since lim supr→0+

V (r)

ra0< +∞, then, without loss of generality, one has

V (|x |) ≤ C0 |x |a0 , ∀ 0 < |x | < r0.

Recall that r0 has been chosen such that r0 < 12 . Thus,

∫R4

V (|x |)u2ε (x) dx ≤ C1(− log ε)

∫ ε1/4

0sa0+3 ds +

(−C1

log ε

)∫ ε1/4

0sa0+3

(1 − s2√

ε

)2ds

+(−C1

log ε

)∫ 1

ε1/4sa0+3 |log s| ds + C1 |ηε |2∞

∫ 2

1sa0+3 ds.

(3.11)

Since a0 > −4, (3.11) implies that uε ∈ E . Taking again into account that a0 > −4, we can

easily see that (log ε)εa0+44 → 0, as ε → 0+. On the other hand, one has

(−C1

log ε

)∫ ε1/4

0sa0+3

(1 − s2√

ε

)2ds ≤

(−C1

log ε

)∫ ε1/4

0sa0+3 ds → 0, as ε → 0+.

Also, the fact that a0 > −4 implies that the function s �−→ sa0+3 |log s| is integrable on]0, 1[. Consequently,

(−C1

log ε

)∫ 1

ε1/4sa0+3 |log s| ds → 0, as ε → 0+.

Finally, noting that |ηε |∞ → 0, as ε → 0+, we deduce from (3.11) that∫R4

V (|x |)u2ε(x) dx → 0, as ε → 0+. (3.12)

Combining (3.10) and (3.12), it follows

‖uε‖2 → 1, as ε → 0+.

123


Next, by the condition added concerning the behavior of K near 0, there exists a positiveconstant C2 such that

K (|x |) ≥ C2 |x |b0 , ∀ 0 < |x | < r0.

Set uε = uε‖uε‖ . Observing that uε(x) ≥√− log ε

32π2

‖uε‖ , ∀ x ∈ R4, |x | < ε1/4, we infer

∫R4

K (|x |)(eαuε

2 − 1)dx ≥ C2

∫|x |<ε1/4

|x |b0(e

−α log ε

32π2‖uε‖2 − 1

)dx

= C2

(e

−α log ε

32π2‖uε‖2 − 1

)ε

b0+44

= C2εb0+44 − α

32π2‖uε‖2 − C2εb0+44 .

(3.13)

Since b0 + 4 > 0 and ‖uε‖2 → 1, as ε → 0+, from (3.13) we deduce that

limε→0+

∫R4

K (|x |)(eαuε

2 − 1)dx = +∞, ∀ α > 8π2(b0 + 4). (3.14)

Taking into account that


∫R4

K (|x |)(eαu2 − 1

)dx ≥

∫R4

K (|x |)(eαuε

2 − 1)dx, ∀ 0 < ε < 1,

then our sharpness result can immediately be derived from (3.14). A consequence of Theorem1.2, which will be very useful in our application, is given by the following corollary:

Corollary 3.1 Under the assumptions of Theorem 1.2, if u ∈ E is such that ‖u‖ ≤ (α0α

) 12 ,

then there exists a constant C = C(α) > 0 independent of u such that∫R4

K (|x |)(eαu2 − 1

)dx ≤ C .

4 An application to a quasilinear elliptic problem

In this section, we will apply Theorem 1.2 and Lemma 2.3 to study the following ellipticsemilinear nonhomogeneous equation

�2u − �u + V (|x |)u = K (|x |) f (u) + V (|x |)h(|x |), x ∈ R4, (4.1)

where (F1) lims→0

f (s)

s= 0.

(F2) There exist C0 > 0, β > 0 and s0 > 0 such that

| f (s)| ≤ C0eβs2 , ∀ |s| ≥ s0.

(H1) h ∈ L2rad(R

4; V ).

Nowadays, the importance of the study of higher-order elliptic equations and especiallythose involving the biharmonic operator (i.e., of fourth order) is well confirmed. In reality,such equations arise in applications from conformal geometry and mathematical physicsbecause they can be used to describe themechanical vibrations of an elastic plate, the travelingwaves in a suspension bridge or the static deflection of an elastic plate in a fluid. For thisaspect of the polyharmonic equations, the reader is invited to see [11,15,16,21] and references

123


therein. Many works have dealt with elliptic equations similar to (4.1) but with some strongcoercivity conditions on the potential V in order to recover the compactness lost by theunboundedness of the domain. We refer, for example, to [27,38,39]. Very recently, the casewhen the nonlinearity enjoys an exponential growth at infinity has known a big interest. Wecan cite [7–9,18,20,30,31,37]. But, to the very best of our knowledge, it seems that there isno work dealing with the existence of solutions for nonhomogeneous biharmonic equationsinvolving simultaneously an exponential nonlinearity and vanishing potentials. In fact, thiscase needs a newAdams’ inequality appropriate to this new situation. In this section, profitingof the Adams’ inequality established in Theorem 1.2, we prove two existence results for theequation (4.1).

A weak solution of Eq. (4.1) is a function u ∈ E such that∫R4

(�u�v + ∇u · ∇v + V (|x |)uv) dx =∫R4

(K (|x |) f (u) + V (|x |)h(|x |)) v dx,

for all v ∈ E . Clearly, the weak solutions of (4.1) correspond to the critical points of the C1

energy functional I : E → R defined by

I (u) = ‖u‖22

−∫R4

(K (|x |)F(u) + V (|x |)h(|x |)u) dx,

where F stands for the primitive of f vanishing at 0, that is, F(s) =∫ s

0f (t) dt , s ∈ R.

Here, we state our existence results.

Proposition 4.1 Assume that (V ) − (K ), (F1) − (F2) and (H1) hold. Then, there existsm > 0 such that Eq. (4.1) has at least one nontrivial solution u such that I (u) < 0 providedthat 0 < |h|L2(R4;V ) < m.

In the second existence result, we prove that if we add two hypotheses, then Eq. (4.1) has aweak solution of positive energy. The first one:

(F3) There exist μ > 0 and p > 2 such that

F(s) ≥ μ |s|p , ∀ s ∈ R,

has been introduced by Cao [10] and it prescribes the growth of f near the origin. The secondone is the well-known Ambrosetti–Rabinowitz (AR) condition:

(F4) There exists θ > 2 such that

θF(s) ≤ f (s)s, ∀ s ∈ R.

In this case, we have the following existence result.

Proposition 4.2 In addition to the hypotheses in Proposition 4.1, assume that (F3) and (F4)hold. Then, there exists μ0 > 0 such that, if 0 ≤ |h|L2(R4;V ) < m and μ > μ0, then Eq.(4.1) has a weak solution of positive energy.

4.1 Weak solution of negative energy

In this part, we prove Proposition 4.1. As a first step in the proof, we claim that there existγ > 0, ρ > 0 and m > 0 such that, if |h|L2(R4;V ) < m, then

I (u) ≥ γ, ∀ u ∈ E, ‖u‖ = ρ. (4.2)

123


By (F1) and (F2), for all ε > 0, one can find cε > 0 such that

|F(s)| ≤ εs2 + cεs3(eβs2 − 1

), ∀ s ∈ R. (4.3)

Let r1 > 1 and r2 > 1 be such that 1r1

+ 1r2

= 1. By Hölder’s inequality and (4.3), for allu ∈ E, we infer∫

R4K (|x |)F(u) dx ≤ ε

∫R4

K (|x |)u2 dx

+ c′ε

(∫R4

K (|x |) |u|3r1 dx) 1

r1(∫

R4K (|x |)

(eβr2u2 − 1

)dx

) 1r2

.

The continuity of the embeddings E ↪→ L2(R4; K ) and E ↪→ L3r1(R4; K ) together withCorollary 3.1 implies

∫R4

K (|x |)F(u) dx ≤ C1ε ‖u‖2 + C2(ε) ‖u‖3 , ∀ u ∈ E, ‖u‖ ≤(

α0

βr2

) 12

. (4.4)

By (4.4), one has

I (u) ≥(1

2− C1ε

)‖u‖2 − C2(ε) ‖u‖3 − |h|L2(R4;V ) ‖u‖ , ∀ u ∈ E, ‖u‖ ≤

(α0

βr2

) 12

.

Choosing ε = 14C1

, we obtain

I (u) ≥ 1

4‖u‖2 − C3 ‖u‖3 − |h|L2(R4;V ) ‖u‖ , ∀ u ∈ E, ‖u‖ ≤

(α0

βr2

) 12

.

Taking 0 < ρ < min

(1

8C3,(

α0βr2

) 12)

, it follows

I (u) ≥ ρ(ρ

8− |h|L2(R4;V )

), ∀ u ∈ E, ‖u‖ = ρ.

Consequently, one can take 0 < m <ρ8 and (4.2) follows.Now, by (4.2), for |h|L2(R4,V ) < m,

we have

infu∈E, ‖u‖=ρ

I (u) ≥ γ > 0 = I (0) ≥ infu∈E, ‖u‖<ρ

I (u).

Thus, one can apply the Ekeland’s variational principle (see [14]) to assert the existence ofa (PS) sequence (un) ⊂ {v ∈ E, ‖v‖ ≤ ρ} of I at the level

d = infu∈E, ‖u‖≤ρ

I (u),

that is,

I (un) → d, I ′(un) → 0, as n → +∞.

By (F1) and (F2), one has∫R4

K (|x |)( f (un))2 dx ≤ C4

∫R4

K (|x |)(u2n + u2n

(e2βu

2n − 1

))dx

≤ C4

∫R4

K (|x |)u2n dx + C5

(∫R4

K (|x |)u4n dx) 1

2(∫

R4K (|x |)

(e4βu

2n − 1

)) 12

dx .

(4.5)

123


Plainly, one could choose ρ small enough such that ρ ≤(

α04β

) 12.Hence, in view of Corollary

3.1 and taking the continuity of the embeddings E ↪→ L2(R4; K ) and E ↪→ L4(R4; K ) intoaccount, we deduce from (4.5) that

supn∈N

∫R4

K (|x |)( f (un))2 dx < +∞. (4.6)

Since (un) is bounded in E, then, up to a subsequence, there exists u ∈ E such that un⇀uweakly in E . Using (4.6) together with Hölder’s inequality, we infer∣∣∣∣∫R4

K (|x |) f (un)(un − u) dx

∣∣∣∣ ≤(supn∈N

∫R4

K (|x |)( f (un))2 dx) 1

2 |un − u|L2(R4;K ) .

Now, the compactness of the embedding E ↪→ L2(R4; K ) established in Lemma 2.3 implies

limn→+∞

∫R4

K (|x |) f (un)(un − u) dx = 0. (4.7)

Since⟨I ′(un), un − u

⟩→ 0, as n → +∞, then in view of (4.7) it follows∫R4

(�un�(un − u) + ∇un · ∇(un − u) + V (|x |)un(un − u)) dx → 0, n → +∞.

On the other hand, the weak convergence of (un) to u in E leads to∫R4

(�u�(un − u) + ∇u · ∇(un − u) + V (|x |)u(un − u)) dx → 0, n → +∞.

Therefore,

‖un − u‖2 → 0, n → +∞.

By the continuity of I and I ′, we deduce that I (u) = d and I ′(u) = 0. It remains to show

that d < 0. Consider a function ϕ0 ∈ E such that ϕ0 �= 0 and∫R4

V (|x |)h(|x |)ϕ0 dx > 0.

Observe that by (4.4), one has

1

t

∫R4

K (|x |)F(tϕ0) dx → 0, t → 0+.

Hence,

I (tϕ0)

t→ −

∫R4

V (|x |)h(|x |)ϕ0 dx < 0, t → 0+.

Therefore, I (tϕ0) < 0 for t sufficiently small. This ends the proof of Proposition 4.1.

4.2 Weak solution of positive energy

In this subsection, we prove Proposition 4.2. For that aim, let ϕ0 be the nontrivial functionin E used in the proof of Proposition 4.1. By (F3), we have

I (tϕ0) ≤ t2‖ϕ0‖22

− t∫R4

V (|x |)h (|x |) ϕ0 dx − μt p∫R4

K (|x |) |ϕ0|p dx, ∀ t ≥ 0.

Since p > 2, then one can find t1 > 0 sufficiently large such that t1 >ρ

‖ϕ0‖ and I (t1ϕ0) < 0,where ρ is given by (4.2). Hence, by the mountain pass theorem without the Palais–Smale

123


condition (see [5]), there exists a sequence (un) ⊂ E such that I ′(un) → 0 and I (un) → c,as n → +∞, with

c = infg∈�

max0≤t≤1

I (g(t)), � = {g ∈ C([0, 1], E), g(0) = 0, g(1) = t1ϕ0} .

Clearly, c ≥ γ. We claim that c = c(μ) → 0 as μ → +∞. By the definition of c, we have

c ≤ maxt≥0

(‖ϕ0‖22

t2 −(

μ

∫R4

K (|x |) |ϕ0|p dx)t p)

=(2A1

pA2

) 2p−2

A1

(1 − 2

p

),

where A1 = ‖ϕ0‖22 , A2 = μ

∫R4 K (|x |) |ϕ0|p dx . Our claim follows immediately. In the

presence of the well-knownAmbrosetti–Rabinowitz (AR) (condition (F4)), the boundednessof the (PS) sequence (un) is evident. In fact, it suffices, as usual, to estimate the sequence

I (un) − 1

θ

⟨I ′(un), un

⟩.

By Young’s inequality, there exists a positive constant C1 such that∫R4

V (|x |) |h(|x |)| |un | dx ≤ 1

2

(1

2− 1

θ

)‖un‖2 + C1 |h|2L2(R4;V )

, ∀ n ∈ N. (4.8)

By the boundedness of the sequence (un) in E, it follows

I (un) = I (un) − 1

θ

⟨I ′(un), un

⟩+ on(1)

=(1

2− 1

θ

)‖un‖2 −

(1 − 1

θ

)∫R4

V (|x |)h (|x |) un dx + on(1).

Hence, (1

2− 1

θ

)‖un‖2 ≤ I (un) +

∫R4

V (|x |) |h(|x |)| |un | dx + on(1). (4.9)

Putting (4.8) in (4.9), we infer

1

2

(1

2− 1

θ

)‖un‖2 ≤ I (un) + C1 |h|2L2(R4;V )

+ on(1).

Having in mind that I (un) → c, then after passing to the limit in the last inequality, weobtain

lim supn→+∞

‖un‖ ≤(

2c12 − 1

θ

) 12

+(

2C112 − 1

θ

) 12

|h|L2(R4;V ) . (4.10)

Since c → 0 as μ → +∞, then one can find μ0 > 0 large enough such that

(2c

12 − 1

θ

) 12

<1

2

(α0

4β

) 12

, ∀ μ > μ0. (4.11)

On the other hand, it is clear that one could choose m > 0 small enough such that

(2C112 − 1

θ

) 12

m <1

2

(α0

4β

) 12

, ∀ μ > μ0. (4.12)

123


Consequently, for μ > μ0 and |h|L2(R4;V ) < m, by (4.12), (4.11) and (4.10), we get

lim supn→+∞

‖un‖ <

(α0

4β

) 12

. (4.13)

In view of (4.5), we see that (4.13) implies that (4.6) and (4.7) hold true. Proceeding exactlyas in the proof of Proposition 4.1, we immediately deduce that, up to subsequence, (un) isstrongly convergent to some point u ∈ E . Therefore, I ′(u) = 0 and I (u) = c ≥ γ > 0. Thisends the proof of Proposition 4.2.

Acknowledgements The authors are very grateful to the anonymous referee for his(her) careful reading of themanuscript and his(her) insightful and constructive remarks and comments that helped to clarify the contentand improve the presentation of the results in this paper. The second author is supported by Programa deIncentivo à Pós-Graduacão e Pesquisa(PROPESQ) Edital 2015, UEPB.

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