Existence of positive solutions for a class of p&q elliptic problems with critical growth on RN

J. Math. Anal. Appl. 378 (2011) 507–518

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Existence of positive solutions for a class of p&q elliptic problems with critical growth on R N Giovany M. Figueiredo 1 Universidade Federal do Pará, Faculdade de Matemática, CEP: 66075-110, Belém, PA, Brazil

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 29 May 2010 Available online 15 February 2011 Submitted by Goong Chen

This paper is concerned with the existence of positive solutions to the class of p&q elliptic problems with critical growth type

      ∗ − div a |∇ u | p |∇ u | p−2 ∇ u + b |u | p |u | p−2 u = λ f (u ) + |u |γ −2 u , u ( z ) > 0,

∀x ∈ R N ,

where λ is a positive parameter, a : R → R is a function of C 1 class and b, f : R → R are continuous functions. © 2011 Elsevier Inc. All rights reserved.

1. Introduction The purpose of this article is to investigate the existence of positive solutions for the following class of quasilinear problem

⎧       ∗ ⎨ − div a |∇ u | p |∇ u | p −2 ∇ u + b |u | p |u | p −2 u = λ f (u ) + |u |γ −2 u in R N , ⎩ u ∈ X , 1 < p < NN, u ( z ) > 0, ∀ z ∈ R ,

( P λ)

where γ ∗ and X will be stated later. Let us introduce the set W as being the collection of all functions k : R+ → R+ satisfying the following properties: There exist constants a0 , b0 > 0, a1 , b1  0, q > p such that

a0 + H (b1 )a1 t

q− p p

 k(t )  b0 + b1 t

q− p p

for all t  0,

(k1 )

where H (s) = 1 if s > 0 and H (s) = 0 if s = 0. There exist constants α and θ such that γ < θ < γ ∗ and

K (t ) 

1

α

k(t )t

with 1 <

for all t  0, where K (t ) = The function

t 0

q p

α<

θ p

(k2 )

,

k(s) ds and where

γ = (1 − H (b1 )) p + H (b1 )q and γ ∗ = (1 − H (b1 )) p ∗ + H (b1 )q∗ .

 

t → k t p t p −2 is increasing.

1

E-mail address: [email protected]. Supported by CNPq/PQ 300705/2008-5.

0022-247X/$ – see front matter doi:10.1016/j.jmaa.2011.02.017

©

2011 Elsevier Inc. All rights reserved.

(k3 )

508

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

The hypotheses on functions a, b and f are the following: a, b ∈ W and the nonlinearity f : R → R is a continuous function and since we are looking for positive solutions, we suppose that

f (t ) = 0 for all t < 0.

( f1)

Moreover, we assume the following growth conditions at the origin and at infinity:

lim

| f (t )|

|t |→0 |t | p −1

=0

( f2)

and there exists s ∈ (γ , γ ∗ ) verifying

| f (t )| = 0. |t |→∞ |t |s−1 lim

( f3)

In this article, we use the classical Palais–Smale condition. Related with this condition, we suppose that f verifies the well-known Ambrosetti–Rabinowitz superlinear condition,

t 0 < θ F (t ) =

f (ξ ) dξ  t f (t ) for all t > 0,

( f4)

0

where θ appeared in (a2 ). Our main result is Theorem 1.1. Assume that a ∈ C 1 (R+ , R+ ) ∩ W , b ∈ C (R+ , R+ ) ∩ W and that the conditions ( f 1 )–( f 4 ) hold. Then, there exists λ∗ > 0, such that problem ( P λ ) has a ground-state positive solution in C 1,α (R N ), with 0 < α < 1, for all λ  λ∗ . Now, we will give some examples of functions a and b in order to illustrate the degree of generality of the kind of operators studied here. Example 1.1. Considering a(t ) = b(t ) = 1, we have that a, b ∈ W with a0 = b0 = 1 and b1 = 0 and a1 > 0. Hence, Theorem 1.1 is valid for the problem

− p u + |u | p −2 u = λ f (u ) + |u | p

∗ −2

in R N .

u

Note that, in this case, Theorem 1.1 is the main result in [1] with λ = 1. q− p p

Example 1.2. Considering a(t ) = b(t ) = 1 + t valid for the problem

, we have that a, b ∈ W with a0 = b0 = a1 = b1 = 1. Hence, Theorem 1.1 is

− p u − q u + |u | p −2 u + |u |q−2 u = λ f (u ) + |u |q Example 1.3. Considering a(t ) = 1 +

1

(1+t )

∗ −2

in R N .

u

and b(t ) = 1, we have that a ∈ W with a0 = 1, b0 = 2 and b1 = 0, a1 > 0 and

p −2 p

b ∈ W with a0 = b0 = 1 and b1 = 0 and a1 > 0. Hence, Theorem 1.1 is valid for the problem

− div |∇ u | p −2 ∇ u +



|∇ u | p −2 ∇ u (1 + |∇ u | p )

Example 1.4. Considering a(t ) = 1 + t

p −2 p

q− p p

+

+ | u | p −2 u = λ f ( u ) + | u | p

1

(1+t )

and b(t ) = 1 + t

p −2 p

∗ −2

q− p p

u

in R N .

, we have that a ∈ W with a0 = 1, b0 = 2 and

b1 = a1 = 1 and b ∈ W with a0 = a1 = b0 = b1 = 1. Hence, Theorem 1.1 is valid for the problem



− p u − q u − div

|∇ u | p −2 ∇ u (1 + |∇ u | p )



p −2 p

+ |u | p −2 u + |u |q−2 u = λ f (u ) + |u |q

∗ −2

u

in R N ,

or still more complex problems, for example: Example 1.5. Considering a(t ) = b(t ) = 1 + t

− p u − q u − div

|∇ u | p −2 ∇ u (1 + |∇ u | p )

p −2 p

q− p p



+

1

(1+t )



+ |u |

p −2 p

p −2

, we have that a, b ∈ W with a0 = 1, b0 = 2 and b1 = a1 = 1 q −2

u + |u |

u+

| u | p −2 u (1 + |u | p )

p −2 p

= λ f (u ) + |u |q

∗ −2

u

in R N .

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

509

Other combinations can be made with the functions presented in the examples above, generating very interesting elliptic problems from the mathematical point of view. The quasilinear equation of the type p&q-Laplacian has received special attention in the last years, see for example the articles [6–8,10,11,13,14,17,20] and the references therein. The existence and multiplicity of solutions of quasilinear problem



    − div a |∇ u | p |∇ u | p −2 ∇ u = f (x, u ) in Ω, u = 0 on ∂Ω,

where Ω is a bounded domain of R N , were studied by J.M. B do O (see [7,8]). In [7] the author showed a result of multiplicity using a Z2 version of the Mountain Pass Theorem [18], f being a function with subcritical exponential growth. In [8] the author used an argument of minimization, f being a function with subcritical growth. The subcritical problem



− p u − q u + v (x)|u | p −2 u + w (x)|u |q−2 u = λ f (x, u ) in Ω, u = 0 on ∂Ω,

was studied by L. Cherfils and V. Il’yasov in [6]. In this article, the authors showed a result of existence and nonexistence using a variational principle. In [17], Medeiros and Perera showed the existence of two solutions for the problem



− p u − q u = λ|u | p −2 u − μ|u |q−2 u + f (x, u ) in Ω, u = 0 on ∂Ω.

The first solution was obtained via Mountain Pass Theorem and the second solution was obtained via cohomological linking theorem. In [14], the problem with critical growth on bounded domain of R N was treated by Li and Guo. The authors showed a result of multiplicity of solutions for the problem



− p u − q u = |u | p u = 0 on ∂Ω.

∗ −2

u + μ|u |r −2 u

in Ω,

The case on R N was studied in [10] by He and Li. More precisely, the authors studied the subcritical problem



− p u − q u + m|u | p −2 u + n|u |q−2 u = f (x, u ) in R N ,     u ∈ W 1, p R N ∩ W 1,q R N ,

using the Mountain Pass Theorem and the concentration–compactness principle of Lions [15]. Other solutions’ existence’s results of p&q-problems can be seen in [13,20]. For a result of regularity, see [11]. Moreover, that class of equations comes, for example, from a general reaction–diffusion system:



ut = div D (u )∇ u + c (x, u ), p −2

(1.1)

q−2

where D (u ) = (|∇ u | + |∇ u | ). This system has a wide range of applications in physics and related sciences, such as biophysics, plasma physics and chemical reaction design. In such applications, the function u describes a concentration, the first term on the right-hand side of (1.1) corresponds to the diffusion with a diffusion coefficient D (u ); whereas the second one is the reaction and relates to source and loss processes. Typically, in chemical and biological applications, the reaction term c (x, u ) is a polynomial of u with variable coefficients (see [11,13,20]). Our theorem extends or complements the articles above, because we consider a more general class of operators, f has a critical growth and problem ( P λ ) is on R N . 2. Variational framework We say that u ∈ X with u > 0 on R N is a weak solution of the problem ( P λ ) if it verifies







a |∇ u | p |∇ u | p −2 ∇ u ∇φ dx +

RN



RN





b |u | p |u | p −2 u φ dx − λ



RN



f (u )φ dx −

| u |γ

∗ −2

u φ dx = 0

RN

for all φ ∈ X , where X denotes the Sobolev space W 1, p (R N ) ∩ W 1,γ (R N ) endowed with the norm

u = u 1, p + H (b1 ) u 1,q , where

 u m 1,m = RN

 |∇ u |m dx + RN

|u |m dx.

510

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

We will look for solutions of ( P λ ) by finding critical points of the C 1 -functional I : X → R given by

I (u ) =



1





p







B |u | p dx − λ

p

RN

RN

Note that



1

I (u )φ =



1

A |∇ u | p dx +



RN



a |∇ u | p |∇ u | p −2 ∇ u ∇φ dx +

p

F (u ) dx −

1





γ∗

γ∗

u + dx. RN



b |u | p |u | p −2 u φ dx − λ

p

RN



1

RN





γ ∗ −1

f (u )φ dx −

RN

u+

φ dx,

RN

for all φ ∈ X . In order to use critical point theory we firstly derive the results related to the Palais–Smale compactness condition. We say that a sequence (un ) is a Palais–Smale sequence for the functional I if

I (u n ) → c ∗ and



  I (un ) → 0 in ( X ) , where c ∗ = infη∈Γ maxt ∈[0,1] I (η(t )) > 0 and Γ := {η ∈ C ([0, 1], X ): η(0) = 0, I (η(1)) < 0}. If every Palais–Smale sequence of I has a strong convergent subsequence, then one says that I satisfies the Palais–Smale condition ((PS) for short). Firstly one proves that functional I has the geometry of Mountain Pass Theorem. Lemma 2.1. For each λ > 0, the functional I satisfies the following conditions: (i) There exist r, ρ > 0 such that:

I (u )  ρ (ii) There exists e ∈

with u = r .

B rc (0)

with I (e ) < 0.

Proof. (i) By ( f 2 ) and ( f 3 ), we get

I (u ) 



1





p



1

A |∇ u | p dx +



p

RN

RN

a0



p RN

−λ

C

 |u |s dx −

s RN

1

γ∗



a1

|∇ u | p dx + H (b1 )

|∇ u |q dx +

q

p

|u | p dx − λ

p

p



C

 |u |s dx −

s

q



C

 |u |s dx −

s RN

 |u |s dx −

s RN

By the Sobolev embedding we get

1

γ∗



∗

|u |γ dx. RN

< 1 and, hence,

I (u )  C 1 u 1, p + H (b1 ) u 1,q − λ

1

γ∗



∗

|u |γ dx. RN

a1

1

γ∗



∗

|u |γ dx. RN

1

γ∗

 RN

|u |q dx − λ

q

∗

|u |γ dx.

 RN

RN

(q− p )

C

RN

∗

Choosing 0 < r = u < 1, we get u 1, p

I (u )  C 2 u q − λ

s

|u | p dx + H (b1 )

RN

Hence,

|u |s dx −

|u |γ dx.

q

q



RN

I (u )  C 1 u 1, p + H (b1 ) u 1,q − λ



C



a0

RN



So







RN

Now, by (k1 ) we derive

I (u ) 



B |u | p dx − λ



 |u | p dx

p RN

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

I (u )  C 2 u q − λC 3 u s − C 4 u γ

511

∗

and, since that q < s < γ ∗ , the lemma is proved. (ii) From ( f 4 ), there exist C 4 , C 5 > 0 such that

F (t )  C 4 t θ − C 5 ,

∀t > 1.

∈ C 0∞ (R N ) with

Thus, fixing φ

1

I (t φ) 





p

A t |∇φ|

p

φ > 0 in R N , we get p





1

dx +

b0t p





 dx − C 4 t

p

RN



− C 4t θ

θ

φ θ dx + C 5 | supp φ| − RN

|φ| p dx +

p

b1t q q

tγ



∗

γ∗

∗

|u |γ dx. RN

 |φ|q dx

H (b1 )

RN

∗  tγ

γ∗



b0t p

|∇φ|q dx +

q

φ dx + C 5 | supp φ| −

θ RN



b1t q

|∇φ| p dx + RN

Since γ ∗

p

RN

From (k1 ) we derive

I (t φ) 

p

B t |φ|

p

RN



RN

∗

|u |γ dx.

RN

> θ > γ , there exists t¯ > 1 such that e = t¯φ satisfies I (e ) < 0 and e > ρ . 2

We devote the rest of this section to show that c ∗ is attained by a positive function. We start by defining the best ∗ constant of the Sobolev embedding W 1,γ (R N ) → L γ (R N ) as



 ∗ |u |γ dx = 1 .



|∇ u |γ dx: u ∈ X ,

S := inf RN

RN

As in [5] and arguing as in [2], we are able to compare the minimax level c ∗ with a suitable number which involves the constant S. Lemma 2.2. If the conditions (k1 )–(k3 ) and ( f 1 )–( f 4 ) hold, then there exists λ∗ > 0 such that c ∗ ∈ (0, ( θ1 − γ1∗ )(a0 S ) N /γ ) for all λ  λ∗ . Proof. Considering φ ∈ C 0∞ (R N ) with φ > 0, there exists t λ > 0 verifying I (t λ φ) = maxt 0 I (t φ) and t λ φ ∈ N , that is,





 a |t λ ∇φ| |t λ ∇φ| p dx +



p

RN

  b |t λ φ| p |t λ φ| p dx = λ

RN

Using (k1 ), we have



RN

RN



p

q



RN

p



q



γ∗

 RN

which implies that (t λ ) is bounded. Thus, there exists a sequence (λn ) ⊂ R such that

t λn → t 0  0 when λn → +∞. Note that if t 0 > 0 then there exists K > 0 such that



K  b0



p

|t λn ∇φ| dx + b1

RN

and

 λn RN

|t λn ∇φ| dx + b0

RN

γ∗



q



RN

∗

φ γ dx → +∞

f (t λn φ)t λn φ dx + t λn

RN

 p

|t λn φ| dx + b1 RN

∗

φ γ dx

|t λn φ|q dx

∗

φ γ dx. RN

|φ|q dx  t λ

RN



f (t λ φ)t λ φ dx + t λ

RN

|φ| p dx + b1t λ

RN

γ∗

|t λ φ|q dx  λ

RN

|∇φ|q dx + b0t λ

RN





RN

|∇φ| p dx + b1t λ

t λ b0

RN

|t λ φ| p dx + b1

Since b1  0, then using (2.1) we get

∗

φ γ dx.

f (t λ φ)t λ φ dx + t λ

 |t λ ∇φ|q dx + b0



γ∗

RN



|t λ ∇φ| p dx + b1

b0



(2.1)

512

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

which is an absurd. Hence we conclude that t 0 = 0. Thus, if we define thus

0 < c ∗  max I



t ∈[0,1]



η∗ (t ) = I (t λ φ)  b0t λp





q

|∇φ| p dx + t λ b1

RN

η∗ (t ) = te for t ∈ [0, 1], it follows that η∗ ∈ Γ and p



q

|∇φ|q dx + b0t λ

RN



|φ| p dx + b1t λ

RN

|φ|q dx.

RN

This way, if λ is large enough we derive p



b0t λ



q

p

|∇φ| p dx + t λ b1

RN

which leads to

RN



0 < c∗ <

1

θ

−



|∇φ|q dx + b0 t λ

RN





q

|φ| p dx + b1t λ

|φ|q dx <

1

θ

−

RN

1



γ∗

(a0 S ) N /γ ,



1

γ∗

(a0 S ) N /γ .

2

Remark 2.3. Note that, from the lemma above, if λ → ∞, then c ∗ → 0. Lemma 2.4. Let (un ) be a sequence in X such that I (un ) → c ∗ and I (un ) → 0 as n → ∞. Then i) un  u in X . ii) There exists λ∗ > 0 such that I (u ) = 0 for all λ  λ∗ . iii) un  0 for n ∈ N. Proof. i) We shall prove that (un ) is bounded in X . Indeed, from ( f 3 ) we get





C 1 + u n  I (u n ) − So





C 1 + u n 

1



1

I (u n )u n .

θ





A |∇ un | p dx −

p

1





θ

RN



a |∇ un | p |∇ un | p dx +



1 p

RN





B |un | p dx − RN

1







b |un | p |un | p dx.

θ RN

By (k2 ) we derive







C 1 + u n 

1 pα

−

1

 







a |∇ un | p |∇ un | p dx +

θ RN







b |un | p |un | p dx.

RN

From (k1 ) we have





C 1 + u n 



1 pα

−

1

θ





a0



|∇ un | p + |un | p dx +

RN



1 pα

−

1



θ

 H (b1 )a1



|∇ un |q + |un |q dx.

RN

Hence,





p

q

C 1 + un  C 1 un 1, p + C 2 H (b1 ) un 1,q .

(2.2)

Thus, if b1 = 0, then (un ) is bounded in X . If b1 > 0, suppose, for contradiction, that, up to a subsequence, un → +∞. We consider several cases: a) un 1, p → +∞ and un 1,q → +∞; b) un 1, p → +∞ and un 1,q is bounded; c) un 1, p is bounded and un 1,q → +∞. q− p

q

p

In the first case, for n sufficiently large, un 1,q  1 and un 1,q  un 1,q . Thus, recalling (2.2),





p

p



C 1 + un  C 1 un 1, p + C 2 un 1,q  C 3 un 1, p + un 1,q which is an absurd.

p

= C 3 u n p ,

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

In case b), by ( f 4 ), we have











C 1 + un 1, p + un 1,q = C 1 + un  Thus, we derive



C

1 p

un 1, p

+

1

+

p −1

un 1, p

un 1,q



p

un 1, p



1 p

−

1 p

1

θ

−

1

θ

513

p un 1, p .

.

Since p − 1 > 0, passing to the limit as n → ∞, we obtain 0 < ( 1p − θ1 )  0, which is an absurd. The last case is similar to the case b). Thus un  u in X . (R N ) for 1  m < γ ∗ , ii) Since, up to a subsequence, un → u in L m loc





|un |γ φ dx → RN

and

|u |γ φ dx

RN



 s

|u |s u φ dx.

|un | φ dx → RN

RN

Hence, from generalized Lebesgue’s Theorem









b |un | p |un | p φ dx → RN

and





b |u | p |u | p φ dx

(2.3)

RN



 f (un )un φ dx → RN

f (u )u φ dx,

(2.4)

RN

for all φ ∈ X . Moreover, un (x) → u (x) a.e. in R N and recalling a result due to Brezis and Lieb [4] (see also [9, Lemma 4.6])



|un |γ −2 u φ dx →

RN

|u |γ −2 u φ dx,

RN



|u n |

s −2



|u |s−2 u φ dx

u φ dx →

RN

and



RN

 | u n |γ

∗ −2

 | u |γ

u φ dx →

RN

∗ −2

u φ dx.

(2.5)

RN

Hence, from generalized Lebesgue’s Theorem



  b |un | p |un | p −2 u φ dx →

RN

and



(2.6)

 f (un )u φ dx →

RN

f (u )u φ dx,

(2.7)

RN

for all φ ∈ X . We claim that ∗



|un |γ φ dx → RN



b |u | p |u | p −2 u φ dx

RN







RN

∗

|u |γ φ dx.

(2.8)

514

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

In order to prove the claim we note that, taking a subsequence, we may suppose that ∗



∗

|∇ un |γ  |∇ u |γ + μ and |un |γ  |u |γ + ν



weak∗ -sense of measures .

Using the concentration–compactness principle due to Lions (cf. [16, Lemma 1.2]), we obtain an at most countable index set Λ, sequences (xi ) ⊂ R N , (μi ), (νi ) ⊂ (0, ∞), such that

ν=





μ

ν i δ xi ,

i ∈Λ

∗

μi δxi and S νiγ /γ  μi ,

(2.9)

i ∈Λ

for all i ∈ Λ, where δxi is the Dirac mass at xi ∈ R N . Now, for every  > 0, we set ψ (x) := ψ((x − xi )/) where ψ ∈ C 0∞ (R N , [0, 1]) is such that ψ ≡ 1 on B 1 (0), ψ ≡ 0 on R N \ B 2 (0) and |∇ψ|∞  2. Since (ψ un ) is bounded, I (un )(ψ un ) → 0, that is,





a |∇ un | RN

p



|∇ un |



p −2





∇ un · ∇ψ dx + 

 p

ψ a |∇ un | |∇ un | p dx −

=− RN





b |un | p |un | p −2 un ψ dx

RN

  ψ b |un | p |un | p dx + λ

RN



 f (x, un )ψ un dx +

RN

∗

ψ |un |γ dx + on (1).

RN

Since by (k1 ) a(t ), b(t )  a0 > and arguing as in [3], we can prove that



lim





a |∇ un |

lim

→0 n→∞

p



|∇ un |

p −2



∇ un · ∇ψ dx = 0

RN

and

 lim

   p p −2 b |u n | |u n | un · ψ dx = 0.

 lim

→0 n→∞

RN

Moreover, since un → u in L m (R N ) for all 1  m < loc expression to obtain

γ ∗ and ψ has compact support, we can let n → ∞ in the above





ψ d ν  RN

a0 ψ dμ.

RN

 → 0 we conclude that νi  a0 μi . It follows from (2.9) that νi  (a0 S )N /γ . Thus, we derive

1 1 − ∗ (a0 S ) N /γ . νi  θ γ

Letting

(2.10)

Now we shall prove that the above expression cannot occur, and therefore the set Λ is empty. Indeed, arguing by contradiction, let us suppose that νi  ( θ1 − γ1∗ )(a0 S ) N /γ for some i ∈ Λ. Thus,

c ∗ = I (u n ) −

1

I (un )un + on (1).

θ

From ( f 4 ), (k1 ) and (k2 ) we have



c∗ 

1

θ

−



1

γ∗



1

θ

−



1

γ∗

ψ (xi )νi =

i ∈Λ

∗



|un |γ dx → RN

RN

−

1



∗

ψ |un |γ dx + on (1).

γ∗ B  ( xi )

1

θ

−

1

γ∗

which does not make sense for all λ > λ∗ . Hence



1

θ

RN

Letting n → ∞, we get

c∗ 



∗

|un |γ dx + on (1) 

∗

|u |γ dx.

 i ∈Λ



νi 

1

θ

−

1

γ∗

(a0 S ) N /γ ,

Λ is empty and it follows that

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

The next step is to prove that





a |∇ un |

p



|∇ un |

p −2



∇ un ∇φ dx =

RN



515



a |∇ u | p |∇ u | p −2 ∇ u ∇φ dx + on (1),

(2.11)

RN

for all φ ∈ X . To this end, we will prove the inequality below:

 









C |x − y | p  a |x| p |x| p −2 x − a | y | p | y | p −2 y , x − y , for all x, y ∈ R N . Indeed, firstly note that N   p  p −2       p  p −2    a |x| |x| x − a | y | p | y | p −2 y , x − y = a |x| |x| x j − a | y | p | y | p −2 y j ( x j − y j ) j =1

and for all z, ξ ∈ R

N

we get

N N     ∂   p  p −2  a | z| | z| z j ξi ξ j = ( p − 2)| z| p −4 a | z | p z i z j ξi ξ j ∂ zi

i , j =1

i , j =1

N 

+





a | z | p | z | p −2 δ i , j ξ i ξ j + p

i , j =1

N 





a | z| p | z|2p −4 zi z j ξi ξ j .

i , j =1

Hence N N    ∂   p  p −2  a | z| | z| z j ξi ξ j = ( p − 2)| z| p −4 a | z| p z i z j ξi ξ j ∂ zi

i , j =1

i , j =1

N 

+









a | z| p | z| p −2 |ξ |2 + pa | z| p | z|2p −4

i , j =1

N 

z i z j ξi ξ j .

i , j =1

Since



N 

z i z j ξi ξ j =

i , j =1

N 

2 z jξ j

,

j =1

we have N  ∂   p  p −2  a | z| | z| z j ξi ξ j = ∂ zi



i , j =1

N 

2 z jξ j

   

  | z| p −4 ( p − 2)a | z| p + pa | z| p | z| p + a | z| p | z| p −2 |ξ |2 .

j =1

By (k3 ), we derive N    ∂   p  p −2  a | z| | z| z j ξi ξ j  a | z| p | z| p −2 |ξ |2 . ∂ zi

(2.12)

i , j =1

Moreover, if | y |  |x|, we have

1 |x − 2

y |  | y | and for t ∈ [0, 14 ] we get

   y + t (x − y )  | y | − t |x − y |  1 |x − y |. 4

Making z = x − y and ξ = x − y, from direct calculations we get N   









a |x| p |x| p −2 x j − a | y | p | y | p −2 y j (x j − y j ) =

j =1

1  N ∂   p  p −2  a | z| | z| z j ξi ξ j . ∂ zi 0 i , j =1

Using (2.12) we derive

 p   p −2   p  p −2     a |x| |x| x − a | y | p | y | p −2 y , x − y  a  y + t (x − y )  y + t (x − y ) |x − y |2 .

516

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

By (k1 ) we conclude

 









a |x| p |x| p −2 x − a | y | p | y | p −2 y , x − y 

a0 4

|x − y | p −2 |x − y |2 =

a0 4

|x − y | p .

Now, considering

 







P N = a |∇ un | p |∇ un | p −2 ∇ un − a |∇ u | p |∇ u | p −2 ∇ u , ∇ un − ∇ u



and ψ ∈ C 0∞ (R N ) such that ψ ≡ 1 in B 1 (0) and ψ ≡ 0 in R N \ B 2 (0), we have

0

a0





4 B 1 ( 0)

Hence

0

a0



|∇ un − ∇ u | p dx 

P N dx  B 1 ( 0)







|∇ un − ∇ u | p dx 

4

P N ψ dx.

RN





a |∇ un | p |∇ un | p ψ dx −

RN

B 1 ( 0)





a |∇ un | p |∇ un | p −2 ∇ un ∇ u ψ dx + on (1).

RN

Using (2.3), (2.4), (2.5), (2.6), (2.7) and (2.8) we get

0

a0



|∇ un − ∇ u | p dx  I (un )(un ψ) − I (un )(u ψ) = on (1).

4 B 1 ( 0)

Thus





∂u ∂ un → ∂ xi ∂ xi

in L p B 1 (0)

and, up to a subsequence,

∂u ∂ un (x) → (x) a.e. in R N . ∂ xi ∂ xi Using also a result due to Brezis and Lieb [4] (see also [9, Lemma 4.6]), we conclude that (2.11) holds. Hence, I (u )ψ = 0 for all ψ ∈ X and for λ  λ∗ . iii) In view of ( f 1 ) and (a1 ), we have

on (1) = I (un )un− = −



RN



 −p ∇ u  dx − b1 n

 −a0 RN

Hence, un− 1, p

= u − n



+

 





p p a ∇ un−  ∇ un−  dx −







RN

 − q ∇ u  dx − a0 n

RN

 



p p b un−  un−  dx



 −p u  dx − H (b1 )



n

RN

 − q     u  dx = −u −  p − H (b1 )u − q . n n 1, p n 1,q

RN

− 1,q = on (1) which implies u n = on (1). Thus, we can easily compute



I (un ) = I un + on (1) and





I (un ) = I un+ + on (1). Thus, we will assume hereafter that (un ) is nonnegative.

2

The next proposition is a version of Lions’ results [15] to problem with p&q-Laplacian. Proposition 2.5. Let (un ) ⊂ X be a (PS)c∗ sequence for I with un  0 in X . Then we have either: a) un → 0 in X or b) there exist a sequence ( yn ) ∈ R N and constants R , β > 0 such that



lim inf

n→+∞ B R ( yn )

|un |γ dx  β > 0.

G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

Proof. Suppose that b) does not occur. Thus, lim infn→+∞ sup y ∈RN un → 0 in L (R m

), for all m ∈ (γ , γ ∗ ). This fact implies that

N



 B R ( y)

517

|un |γ dx = 0. Using Lemma 8.4 in [15], we get

f (un )un dx → 0. RN

It follows that





a |∇ un |

p



 p

|∇ un | dx +

RN

RN

Since, up to a subsequence,





RN



∗

|un |γ dx + on (1).

RN





a |∇ un | p |∇ un | p dx + and

  b |un | p |un | p dx =





b |un | p |un | p dx → L λ

RN



∗

|un |γ dx → L λ . RN

If L λ → 0 as λ → ∞, then

C



p un 1, p

q  H (b1 ) un 1,q

+







a |∇ un |

p



 p

RN





b |un | p |un | p dx = on (1)

∇ un | dx + RN

and therefore

u n → 0. If there exists M > 0, independent of λ, such that L λ  M, then

on (1) + c ∗ =

1







A |∇ un | p dx +

p



on (1) + c ∗ 

1 pα





B |un | p dx −

p

RN

By (k2 ) we get



1

RN

−

1



Lλ  γ∗



1 pα

−

1

γ∗

1

γ∗



∗

|un |γ dx. RN

M > 0,

which is an absurd from Remark 2.3. Hence

u n → 0.

2

2.1. Proof of Theorem 1.1 By Lemma 2.4, there exists u ∈ X such that I (u ) = 0 and u  0. Suppose that u ≡ 0. Adapting arguments from [12], we conclude that u ∈ L ∞ (R N ) ∩ C 1,α (R N ) for some 0 < α < 1, and therefore it follows from Harnack’s inequality [19] that u (x) > 0 for all x ∈ R N . If u ≡ 0, then un no converges strongly to zero, because for the contrary case, we get c ∗ = 0. Thus, from Proposition 2.5, there is a sequence ( yn ) ∈ R N and R , α > 0 such that



lim inf

n→+∞ B R ( yn )

|un |γ dx > β.

(2.13)

Now, letting  un (x) = un (x + yn ), using the invariance of R N for translation, by a routine calculus we obtain  u n = u n , I ( un ) = I (un ) and I ( un ) = on (1). Then, there exists  u such that  un   u weakly in X and as before it follows that I ( u ) = 0. Now, by (2.13), taking a subsequence and R bigger we conclude that  u is nontrivial and the proposition is proved. Acknowledgments The author would like to thank UAME/UFCG – Campina Grande, and especially Professor Claudianor Oliveira Alves for his attention and friendship. This article was done while he was visiting that institution.

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G.M. Figueiredo / J. Math. Anal. Appl. 378 (2011) 507–518

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