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Existence of solutions for a class of degenerate quasilinear elliptic equation in R N with vanishing potentials

Abstract

We establish the existence of positive solution for the following class of degenerate quasilinear elliptic problem

( P ) { L u a p + V ( x ) | x | a p | u | p 2 u = f ( u ) in  R N , u > 0 in  R N ; u D a 1 , p ( R N ) ,

where L u a p =div( | x | a p | u | p 2 u), 1<p<N, <a< N p p , aea+1, d=1+ae, and p := p (a,e)= N p N d p denote the Hardy-Sobolev’s critical exponent, V is a bounded nonnegative vanishing potential and f has a subcritical growth at infinity. The technique used here is a truncation argument together with the variational approach.

MSC:35B09, 35J10, 35J20, 35J70.

1 Introduction

Consider the following degenerate quasilinear elliptic problem in R N :

( P ) { L u a p + V ( x ) | x | a p | u | p 2 u = f ( u ) in  R N , u > 0 in  R N ; u D a 1 , p ( R N ) ,

where L u a p =div( | x | a p | u | p 2 u), 1<p<N, <a< N p p , aea+1, d=1+ae, and p := p (a,e)= N p N d p denote the Hardy-Sobolev’s critical exponent, V: R N R is a bounded, nonnegative and vanishing potential and f:RR a continuous function with a subcritical growth at infinity. Here, D a 1 , p ( R N ) is the completion of the C 0 ( R N ) with the norm |u|= ( R N | x | a p | u | p d x ) 1 p . We impose the following hypotheses on f and V:

f:RR is a continuous function verifying

( f 1 ) lim sup s 0 + s f ( s ) | x | a p s p <, uniformly in x.

( f 2 ) There exists α(p, p ) such that lim sup s s f ( s ) | x | a p s α <, uniformly in x.

( f 3 ) There exists θ>p such that θF(s)sf(s), for all s>0.

V: R N R is a continuous function verifying

( V 1 ) V(x)0, for all x R N .

( V 2 ) There are Λ>0 and r ¯ >1 such that inf | x | > r ¯ V(x) | x | p 2 [ N p ( a + 1 ) ] ( p 1 ) ( N p ) Λ.

Remark 1.1 The conditions ( f 1 ) and ( f 2 ) imply the following:

| s f ( s ) | C 0 | x | a p | s | α with α ( p , p ) , for all sR.
(1)

Example 1.2 An example for the function f is given by

f(t)={ | x | a p t λ 1 , if  0 t 1 , | x | a p t α 1 , if  t 1 ,

with λ> p and α given by ( f 2 ).

By vanishing potential we mean a potential that vanish on some bounded domain or become very close to zero at infinity. An important example for a such potential is given by

V(x)={ 0 , if  | x | r ¯ 1 , Λ r ¯ p 2 [ N p ( a + 1 ) ] ( p 1 ) ( N p ) ( | x | r ¯ + 1 ) , if  r ¯ 1 < | x | r ¯ , Λ | x | p 2 [ N p ( a + 1 ) ] ( p 1 ) ( N p ) , if  | x | r ¯ ,

with Λ>0.

Consider first the case a=0, that is L u a p is the p-Laplacian operator, and the potential is bounded from below by a positive constant V 0 >0.

Equations involving the p-Laplacian operator appear in many problems of nonlinear diffusion. Just to mention, in nonlinear optics, plasma physics, condensed matter physics and in modeling problems in non-Newtonian fluids. For more information on the physical background, we refer to [1].

For the case p=2, we cite [212], and references therein. In [13], in addition to the above assumptions, the authors consider a local condition, namely,

min x Ω ¯ V< min x Ω V,

where Ω R N is a open bounded set, instead of the global condition imposed by Rabinowitz in [12]. For p2, see [1417].

Now, consider that V is the zero mass case, that is lim | x | V(x)=0. When p=2, we cite [1820] and the recent paper [21] by Alves and Souto.

Let us now consider the case a0 and the potential bounded from below by a positive constant V 0 >0.

In this case, the equations arise in problems of existence of stationary waves for anisotropic Schrödinger equation (see [22]) and others problems (for example, see [6, 23]). We cite [22] for p=2; and [24, 25] for p2. For the case V0, we cite [26], for p=2 and a0; and [27], for p2 and a=0.

The result presented here for 1<p<N and a0 extends that one in [21] for p=2 and a=0. In [21], the presence of Hilbertian structure and some compact embeddings provide the convergence of the gradient. In the case studied here, with the absence of this structure, we do not obtain the convergence so directly. To overcome this problem, we use a result found in [28, 29], whose ideas come from [30, 31], when the domain is a smooth and bounded. In addition to this difficulty, there are others. For instance, in the present situation, our space is no longer Hilbert, which forces us to obtain new estimates. Since the problem involves singular terms, the estimates are more refined and for which the principal ingredient is the Caffarelli-Kohn-Nirenberg’s inequality (see [32]). Now we state the main result of this work.

Theorem 1.3 Suppose that V and f satisfy, respectively, ( V 1 ) and ( V 2 ) and ( f 1 ) to ( f 3 ). Then there is a constant Λ = Λ ( V ,θ,p, c 0 )>0such that the problem (P) has a positive solution, for allΛ Λ , being V the maximum of the f in the ball of R N centered in the origin with radius 1.

In order to prove this theorem, we first build an auxiliary problem (AP), and then we solve the problem (AP) using variational methods. To finish, we show that the solution of (AP) is also a solution of (P). These steps are the content of the next three sections.

Hereafter, C is a positive constant which can change value in a sequence of inequalities. We denote B R = B R (0) the ball in R N centered in the origin with radius R. The weak () and strong (→) convergences are always taken as n and A f means A f(x)dx. The weighted L p spaces are denoted by L α p (A)={u: R N R: A | x | α | u | p <}. When α=0, we denote L p ( A ) the usual norm in L p (A), with 1p. For A= R N , we use p .

2 The auxiliary problem

As usual, since we are looking for positive solutions of problem (P), we set f(t)=0, for all t0. The hypothesis ( V 1 ) allows us to consider the space

E= { u D a 1 , p ( R N ) : R N V | x | a p | u | p < } ,

with norm

u= ( R N | x | a p | u | p + V | x | a p | u | p ) 1 p .

Associated to the problem (P), we define on E, the Euler-Lagrange functional

I(u)= R N | x | a p | u | p +V | x | a p | u | p R N F(u),

being F(s)= 0 s f(t)dt. From the assumptions on f, it follows that I is C 1 with Gâteaux derivative

I (u)v= R N | x | a p | u | p 2 uv+V | x | a p | u | p 2 uv R N f(u)v,vE.

To obtain solutions of problem (P), we introduce some truncation of the function f. Consider k= p θ θ p >p, r ¯ >1 and define

g(x,t)={ f ( t ) , | x | r ¯ , f ( t ) , | x | > r ¯  and  f ( t ) V k | x | a p | t | p 2 t , V k | x | a p | t | p 2 t , | x | > r ¯  and  f ( t ) > V k | x | a p | t | p 2 t .
(2)

Now we define the auxiliary problem:

( AP ) { L u a p + V ( x ) | x | a p | u | p 2 u = g ( x , u ) in  R N , u > 0 in  R N ; u D a 1 , p ( R N ) .
(3)

Associated to the problem (AP), we define, on E, the Euler-Lagrange functional

J(u)= 1 p R N | x | a p | u | p +V | x | a p | u | p R N G(x,u)= 1 p u p R N G(x,u),

being G(x,s)= 0 s g(x,t)dt. From the assumptions on f, it follows that J is C 1 with Gâteaux derivative

J (u)v= R N | x | a p | u | p 2 uv+V | x | a p | u | p 2 uv R N g(x,u)v,vE.

3 Solving the problem (AP)

In this section, we show that the problem (AP) has a least energy solution, but first we define some minimax levels. To begin with we set in the space D a 1 , p ( B 1 ), the norm u= ( B 1 | x | a p | u | p + V | x | a p | u | p ) 1 p and we define the functional I 0 given by I 0 (u)= 1 p B 1 | x | a p | u | p + V | x | a p | u | p B 1 F(u). Here, D a 1 , p ( B 1 ) is the completion of the C 0 ( B 1 ) with the norm |u|= ( B 1 | x | a p | u | p ) 1 p .

Lemma 3.1 The functional I 0 has the mountain pass geometry, namely,

  1. 1.

    r 0 , ρ 0 >0 such that I 0 (u) ρ 0 for u= r 0 .

  2. 2.

    e 0 D a 1 , p ( B 1 ) such that e 0 r 0 and J( e 0 )0.

Proof By using the growth of f given in Remark 1.1 and the Caffarelli-Kohn-Nirenberg’s inequality (see [32]), we get B 1 F(u) B 1 c 0 | x | a p | u | p c 0 u p , and hence I 0 (u) 1 p u p c 0 u p . Since p >p, there exists r 0 such that ρ 0 := 1 p r 0 p c r 0 p >0. Thus, we have I 0 (u) ρ 0 for u= r 0 . By ( f 3 ), it follows that there exist θ>p and C>0 such that F(s)C | s | θ . Now u 0 D a 1 , p ( B 1 ) implies I 0 (t u 0 ) t p p u 0 p t θ C B 1 | u 0 | θ . Since θ>p, there exists a t 0 large enough such that, taking e 0 = t 0 u 0 , we have I 0 ( e 0 )<0 and e 0 r 0 . □

Lemma 3.2 The functional J has the mountain pass geometry, namely,

  1. 1.

    r 1 , ρ 1 >0 such that J(u) ρ 1 for u= r 1 .

  2. 2.

    e 1 E such that e 1 r 1 and J( e 1 )0.

Proof From the definition of G, we have R N G(x,u) R N F(u). Thus, like the previous lemma, we have J(u) ρ 1 := 1 p r 1 p c r 1 p >0 for u= r 1 . Take the same u 0 D a 1 , p ( B 1 ) of the proof of the previous lemma. Thus, u 0 E and G(x,u)=F(u). With the same argument, we have J(t u 0 ) t p p u 0 p t θ C R N | u 0 | θ . Since θ>p, there exists a t 1 large enough such that, taking e 1 = t 1 u 0 , we have J( e 1 )<0 and e 1 r 1 . □

Next, we are going to define two minimax levels, which will play an important role in our arguments. Note that is possible to take t such that e=t u 0 satisfies two previous lemmas. This allows to define the minimax levels c and c ¯ by

c= inf γ Γ max t [ 0 , 1 ] J ( γ ( t ) ) with Γ= { γ C ( [ 0 , 1 ] , E ) : γ ( 0 ) = 0  and  γ ( 1 ) = e }

and

c ¯ = inf γ Γ max t [ 0 , 1 ] I 0 ( γ ( t ) ) with Γ= { γ C ( [ 0 , 1 ] , D a 1 , p ( B 1 ) ) : γ ( 0 ) = 0  and  γ ( 1 ) = e }

respectively. Since J( u 0 ) I 0 ( u 0 ) in D a 1 , p ( B 1 ), we have c c ¯ , by their definitions. Now using the above lemma together with the mountain pass theorem [[33], Theorem 2.2], we conclude that there exists a Palais-Smale sequence ((PS) sequence for short) ( u n )E for J, i.e., ( u n ) satisfies J( u n )c and J ( u n )0.

Lemma 3.3 Suppose ( V 1 ) and ( f 1 ) to ( f 3 ) and let( u n )Ebe a (PS) sequence for the functional J. Then( u n )is bounded in E.

Proof Define the set A={x R N :|x|R or f(u(x)) V ( x ) k | x | a p | u ( x ) | p 2 u(x)}. In A, we have G(x,u)=F(u). By using ( f 3 ), we conclude that there is θ>p such that G(x,u)+ 1 θ ug(x,u)0. So, we have

Now consider the set B= A c ={x R N :|x|>R and f(u(x))> V ( x ) k | x | a p | u ( x ) | p 2 u(x)}. In B, we have B ug(x,u)>0 and G(x,u)= V p k | x | a p | u | p . Then

Therefore, we get J(u) 1 θ J (u)u ( p 1 ) p k u p . In particular, the above equation holds for the (PS) sequence ( u n ) for J, and we have J( u n ) 1 θ J ( u n ) u n ( p 1 ) p k u n p . On the other hand, we have J( u n )c, θ>p>1 and J ( u n ) u n u n 0, since J ( u n )0. Thus, we get J( u n ) 1 θ J ( u n ) u n M+ 1 θ u n M+ u n , for some constant M>0. Then we have ( p 1 ) p k u n p M+ u n , which can be rewritten as

u n ( ( p 1 ) u n p 1 p k ) pkM.
(4)

Assuming u n , equation (4) implies that (p1) u n p 1 pk0. So u n ( p k p 1 ) 1 p 1 , which is a contradiction. Therefore, u n is bounded in E. □

Lemma 3.4 Suppose ( V 1 ) and ( f 1 ) to ( f 3 ). Then the functional J satisfies the Palais-Smale condition, i.e., every (PS) sequence has a convergent subsequence.

Proof Observe that, given the (PS) sequence ( u n ), by Lemma 3.3, there exists uE such that u n u, because E is reflexive space.a Thus, it is enough to show that u n u. We divide this task in the four claims below.

  1. 1.

    Claim 1 R N u n g(x, u n ) R N ug(x,u).

  2. 2.

    Claim 2 R N ug(x, u n ) R N ug(x,u).

  3. 3.

    Claim 3 R N V | x | a p | u n | p 2 u n u R N V | x | a p | u | p .

  4. 4.

    Claim 4 R N | x | a p | u n | p 2 u n u R N | x | a p | u | p .

Assuming Claims 1 to 4 for now, we proceed with the proof of lemma.

Since J ( u n ) u n 0, we have R N | x | a p | u n | p +V | x | a p | u n | p R N u n g(x, u n )= o n (1), and by Claim 1, we get

lim sup n u n p = R N ug(x,u).
(5)

As J ( u n )u0, we get R N | x | a p | u n | p 2 u n u+V | x | a p | u n | p 2 u n u R N ug(x, u n )= o n (1). Passing the limit in the above equation and using Claims 2, 3 and 4, we get

u p = R N | x | a p | u | p +V | x | a p | u | p = R N ug(x,u).
(6)

Using equations (5) and (6), we have u n u. □

In order to prove the claims, for a given ϵ>0, we choose r satisfying, the following two conditions:

  1. 1.

    max{ B 2 r B r | x | a p | u | p , B 2 r c V | x | a p | u | p }ϵ.

  2. 2.

    η= η r C 0 ( B r c ) is such that η1 in B 2 r c and 0η1, |η| 2 r d , for all x R N .

Observe that condition 1 follows by integrability of | x | a p | u | p and V | x | a p | u | p . Note that, when one of these conditions holds for some r 0 , it also verifies for every r r 0 . Thus, we can choose an r that satisfies both conditions.

Remark 3.5 From now on, we consider the function g defined in (2) with r ¯ =r satisfying conditions 1 and 2 above. From the growth of g and the choice of r, we conclude:

  1. 1.

    g(x,t)=f(t), G(x,t)=F(t) in B r ;

  2. 2.

    g(x,t) V k | x | a p | t | p 2 t, G(x,t) V p k | x | a p | t | p in B r c ;

  3. 3.

    B 2 r c |ug(x,u)| B 2 r c |u| V k | x | a p | u | p 1 1 k B 2 r c V | x | a p | u | p <ϵ.

Verification of claims

In all proofs, except for Claim 4, we consider separately integration in B 2 r and B 2 r c .

Claim 1: In B 2 r , by combining the dominated convergence theorem with the compact embedding of E in L α r ( B 2 r ), 1r< p and α<(a+1)r+N(1 r p ) (see [[34], Theorem 2.1]), we obtain B 2 r u n g(x, u n ) B 2 r ug(x,u). On the other hand, when integrating in B 2 r c we do not have the compact embedding of E in L α r ( B 2 r c ). In this case, we first estimate B 2 r c | x | a p | u n | p +V | x | a p | u n | p , by using the cut-off function η defined before. As (η u n ) is also bounded we have J ( u n )(η u n )0, namely,

R N | x | a p | u n | p 2 u n (η u n )+V | x | a p | u n | p 2 u n η u n = o n (1)+ R N ηg(x, u n ) u n .
(7)

Since η0 in B r , equation (7) holds in B r c . Adding to this, the fact that g(x,t) V k | x | a p | t | p 2 t in B r c we have

From this, we obtain

Thus, we have

(8)

Using the boundedness of ( u n ) in E, i.e., u n C, strong convergence of ( u n ) in L a p p ( B 2 r B r ) and Hölder’s inequality we conclude that

lim sup n B 2 r B r | x | a p | u n | | u n | p 1 C ( B 2 r B r | x | a p | u | p ) 1 p C ω N d N ( 2 r ) d ( B 2 r B r | x | a p | u | p ) 1 p ,
(9)

where ω N is the volume of the unit sphere in R N . Therefore, given ϵ>0, by the choice of r and by equations (9) and (8), we infer that

Hence,

lim sup n B 2 r c | x | a p | u n | p +V | x | a p | u n | p =0.
(10)

Using this fact, we conclude that

lim sup n B 2 r c u n g ( x , u n ) lim sup n 1 k B 2 r c V | x | a p | u n | p lim sup n 1 k B 2 r c | x | a p | u n | p + V | x | a p | u n | p = 0 .

Now, using Remark 3.5, we have lim sup n B 2 r c | u n g(x, u n )ug(x,u)|=0.

Therefore, B 2 r c u n g(x, u n ) B 2 r c ug(x,u) and the proof of Claim 1 is completed.

Claim 2: The proof of this fact is made as in the proof of Claim 1.

Claim 3: In B 2 r the proof proceeds like in Claim 1. For integration in B 2 r c , we estimate the value of B 2 r V | x | a p | u n | p 2 u n u by using the Hölder’s inequality

B 2 r c V | x | a p | u n | p 2 u n u ( B 2 r c V | x | a p | u | p ) 1 p ( B 2 r c V | x | a p | u n | p ) 1 p C ( B 2 r c V | x | a p | u n | p ) 1 p C ( B 2 r c | x | a p | u n | p + V | x | a p | u n | p ) 1 p .
(11)

Using equations (10) and (11), we have

lim sup n B 2 r c V | x | a p | u n | p 2 u n u lim sup n C ( B 2 r c | x | a p | u n | p + V | x | a p | u n | p ) 1 p =0.

From this fact, we get B 2 r c V | x | a p | u n | p 2 u n u B 2 r c V | x | a p | u | p and the proof of this claim is completed.

Claim 4: Let p be dual of p. Since u D a 1 , p ( R N )={u: R N R N : | x | a u L p ( R N )and | x | a u L p ( R N )},b we see that h= | x | a |u| L p ( R N ) and w n = | x | a ( p 1 ) | u n | p 2 u n L p ( R N ). As u n is bounded, we conclude that w n L p ( R N ) is also bounded. Moreover, ( u n ) satisfies the hypothesis of Lemma 3.6, below. So that we have u n u a.e. x R N , which give us w n w= | x | a ( p 1 ) | u | p 2 u a.e. x R N . Using a theorem [[35], Theorem 13.44], we conclude that w n w in L p ( R N ). Thus, we have R N w n h R N wh, and hence R N | x | a p | u n | p 2 u n u R N | x | a p | u | p .

Lemma 3.6 Let E and J be respectively the space and functional defined in Section  2. Let( u n )Ea bounded sequence such that u n uin E and J ( u n )0. Then, passing to a subsequence if necessary, we have u n u, a.e. x R N .

The proof of this lemma follows using the same ideas made in [30] and [31], for a bounded domain. It can be found in [[28], Claim 1] and [[29], Lemma 1].

Using Lemmas 3.2, 3.3, 3.4 and the mountain pass theorem, in [[36], Theorem 2.4], we conclude that there exists uE which is a critical point for the functional J, in the minimax level c. Moreover, u is the least energy solution to the problem (AP).

4 The solution of (AP) is solution of (P)

Now, our aim is to show that the solution found in the previous section is also a solution of the problem (P). It is sufficient to verify f(u) V k | x | a p | u | p 2 u, for all x B r c .

Lemma 4.1 Any least energy solution u of (AP) satisfies the estimate u p p k c ¯ p 1 .

Proof Since u is a critical point in the minimax level c c ¯ , by Lemma 3.3, we have ( p 1 ) p k u p J(u) 1 θ J (u)u=c c ¯ . So that u p p k c ¯ p 1 . □

Remark 4.2 The constant p k c ¯ p 1 depends only on V , θ and f.

Lemma 4.3 Let h be such that | x | a p | h | q is integrable, withpq>N. ConsiderH: R N ×RR, andb: R N Rnonnegative and continuous functions such that|H(x,s)|h(x) | x | a p | s | p 1 , for alls>0. LetvEbe a weak solution of (AP 2):

L v a p +b | x | a p | v | p 2 v=H(x,v) in R N .

Then there exists a constantM=M(q, h L a p q ( R N ) )>0such that v M | x | a v p .

Proof Given mN and β>1, set A m ={x R N : | v | β 1 m}, B m = R N A m and

v m ={ v | v | p ( β 1 ) in  A m , m p v in  B m .

Thus

v m ={ ( p β p + 1 ) | v | p ( β 1 ) v in  A m , m p v in  B m .

Then v m E and using it as the test function in (AP 2). we have

R N | x | a p | v | p 2 v v m +b | x | a p | v | p 2 v v m = R N H(x,v) v m .
(12)

By definition of v m , we get

(13)

and

R N b | x | a p | v | p 2 v v m = A m b | x | a p | v | p β + m p B m b | x | a p | v | p >0.
(14)

Using equations (13) and (14), we have

(15)

Putting

w m ={ v | v | β 1 in  A m , m v in  B m ,

we have

w m ={ β | v | β 1 v in  A m , m v in  B m .

Thus,

R N | x | a p | w m | p = β p A m | x | a p | v | p ( β 1 ) | v | p + m p B m | x | a p | v | p .
(16)

Now, taking into account (14), we get

R N b | x | a p | w m | p = A m b | x | a p | v | p β + m p B m b | x | a p | v | p = R N b | x | a p | v | p 2 v v m .
(17)

Using (17), (13) and (16), we have

(18)

Combining (18), (15) and (12), we obtain

R N | x | a p | w m | p +b | x | a p | w m | p β p R N H(x,v) v m .
(19)

Let S be the best constant for inequality ( R N | x | a p | u | p ) p p S R N | x | a p | u | p , for all u D a 1 , p ( R N ). (See [32].) By the inequality (19), the boundedness of H and the definitions of v m and S, we get ( A m | x | a p | w m | p ) p p S β p R N h(x) | x | a p | v | p β . Since | w m |= | v | β in A m , we have

( A m | x | a p | v | p β ) p p S β p R N h(x) | x | a p | v | p β .

Making m and using the monotone convergence theorem, we get

( R N | x | a p | v | p β ) p p S β p R N h(x) | x | a p | v | p β .

Since pq>N, σ= N q 1 ( N p ) >1. Thus, we can consider β= σ j for j=1,2,3, and, using the Hölder’s inequality, we have

| x | a σ j v p σ j p σ j S σ j p R N h ( x ) | x | a p | v | p σ j S σ j p [ R N | x | a p | h ( x ) | q ] 1 q [ R N | | x | a σ j 1 | v | | p σ j q 1 ] 1 q 1 M 0 σ j p [ R N | | x | a σ j 1 | v | | p σ j q 1 ] 1 q 1 M 0 σ j p | x | a σ j 1 v p σ j q 1 p σ j ,

being M 0 =S [ R N | x | a p | h ( x ) | q ] 1 q . Note that M 0 is independent of j. Thus, we get

| x | a σ j v p σ j M 0 1 p σ j σ j σ j | x | a σ j 1 v p σ j q 1 for all j=1,2,3,.
(20)

For j=1 and j=2, we have pσ q 1 = p and p σ 2 q 1 = p σ. Applying this in (20), we have

| x | a σ v p σ M 0 1 p σ σ 1 σ | x | a v p and | x | a σ 2 v p σ 2 M 0 1 p σ 2 σ 2 σ 2 | x | a σ v p σ .

By iterating, we have

| x | a σ 2 v p σ 2 σ 1 σ + 2 σ 2 M 0 1 p ( 1 σ + 1 σ 2 ) | x | a v p .

Thus, we get

| x | a σ j v p σ j σ 1 σ + 2 σ 2 + + j σ j M 0 1 p ( 1 σ + 1 σ 2 + + 1 σ j ) | x | a v p for all j=1,2,3,.

Then we can say that, for all t p , we have

v t σ σ ( σ 1 ) 2 M 0 1 p ( σ 1 ) | x | a v p .

Putting M:= σ σ ( σ 1 ) 2 M 0 1 p ( σ 1 ) , we get

v = lim t v t lim t M | x | a v p =M | x | a v p .

As σ depends on q, we have, by definition of M, that M=M(q, h L a p q ( R N ) )>0. □

Remark 4.4 In the previous lemma, the constant M does not depend on the potential b of the problem (AP 2).

Lemma 4.5 There exists a constant M 1 >0such that u M 1 , for all u positive solution of (AP).

Proof Take r from the definition of g and define A={x R N :|x|r or f(u(x)) V ( x ) k | x | a p | u ( x ) | p 2 u(x)} and B= R N A={x R N :|x|>r and f(u(x))> V ( x ) k | x | a p | u ( x ) | p 2 u(x)}. Now define H and b by

H(x,t)={ f ( t ) in  A , 0 in  B andb(x)={ V ( x ) in  A , ( 1 1 k ) V ( x ) in  B .

We will show that, if u is solution of (AP), then u is weak solution of (AP 2).

In particular, we recall that H(x,u)=f(u)=g(x,u), in A, and g(x,u)= V k | x | a p | u | p 2 u, in B. Using the fact that R N is the disjoint union of A and B and the definitions above, for a given ϕE, we write

Hence, u verifies R N | x | a p | u | p 2 uϕ+b | x | a p | u | p 2 uϕ= R N H(x,u)ϕ, for all ϕE. From ( f 1 ) and ( f 2 ), we obtain |f(s)| c 0 | x | a p | s | α 1 , for all s>0, with α(p, p ). From the definition of H, it follows that |H(x,u)||f(u)|h(x) | x | a p | u | p 1 , with h(x)= c 0 | u | α p . Taking q= p α p , we have | x | a p | h | q = | x | a p | u | p that is integrable. Then u satisfies the hypotheses of Lemma 4.3, that is,

u M | x | a u p .
(21)

Using the definition of the constant S and Lemma 4.1, we have

| x | a u p = ( R N | x | a p | u | p ) 1 p ( S R N | x | a p | u | p ) 1 p ( S u p ) 1 p ( S p k c ¯ p 1 ) 1 p .
(22)

Combining equations (21) and (22), we obtain u M ( S p k c ¯ p 1 ) 1 p := M 1 . □

Lemma 4.6 Let r be as in the definition of g and consider R 0 r. Let u a be any positive solution of (AP). Then we have

u(x) u R 0 N p p 1 | x | N p ( a + 1 ) p 1 M 1 R 0 N p p 1 | x | N p ( a + 1 ) p 1 for all x B R 0 c .

Proof Consider v(x)= M 1 R 0 N p p 1 | x | N p ( a + 1 ) p 1 . By Lemma 4.5, we have u M 1 . So uv, for |x|= R 0 . It follows that ( u v ) + =0, in |x|= R 0 , and the function given by

w={ 0 , | x | < R 0 , ( u v ) + , | x | R 0

is so that w D a 1 , p ( R N ). Moreover, wE, because u,vE. Let us show now that ( u v ) + =0, in |x| R 0 . Taking w as the test function, using the hypotheses on g and V and the fact that u is positive solution of (AP), we have

R N | x | a p | u | p 2 u w = R N g ( x , u ) w R N V | x | a p | u | p 2 u w = B R 0 c g ( x , u ) w B R 0 c V | x | a p | u | p 2 u w = ( 1 k 1 ) B R 0 c V | x | a p | u | p 2 u w 0 .
(23)

Here, we considered that k>1. Using the radially symmetric form of the operator L v a p (see [37, 38]), we have that div( | x | a p | v | p 2 v)=0 in B R 0 c . In the weak form, it is

B R 0 c | x | a p | v | p 2 vϕ=0for all ϕE.

Hence,

R N | x | a p | v | p 2 vw= B R 0 c | x | a p | v | p 2 vw=0.
(24)

Putting A={x R N :|x| R 0  and u(x)>v(x)} and B= R N A, we have

w={ u v in  A , 0 in  B .

By (23) and (24), we obtain, for all 1<p<N, that

A | x | a p [ | u | p 2 u | v | p 2 v ] [ u v ] = A [ | x | a p | u | p 2 u | x | a p | v | p 2 v ] w = R N | x | a p | u | p 2 u w R N | x | a p | v | p 2 v w 0 .
(25)

Consider 2p<N. Using the Tolksdorf’s inequality (see [[39], Lemma 2.1] or [[31], Lemma 4.1]) and equation (25), we obtain

R N | x | a p | w | p = A | x | a p | u v | p c A | x | a p [ | u | p 2 u | v | p 2 v ] [ u v ] 0 .

In the case 1<p<2, in addition to the above arguments used, we also use the Hölder’s inequality. Then

Then R N | x | a p | w | p 0 for all 1<p<N. Thus, we have w=0, in R N , which implies that ( u v ) + =0, in |x| R 0 . From this, we conclude that uv in R N , and lemma is proved. □

Proof of Theorem 1.3 We will show that f(u) V k | x | a p | u | p 2 u in B r c , for all solution u of (AP). By Remark 1 we have |sf(s)| C 0 | x | a p | s | p , which gives us f ( u ) | x | a p | u | p 2 u C 0 | u | P 2 N p . Now, note that the hypothesis ( V 3 ) holds for all R 0 >R. Hence, for R 0 =r>R, we can use ( V 3 ) and Lemma 4.6. Thus, for each x in B r c , and Λ =k C 0 M 1 P 2 N p R 0 p 2 p 1 , we have

f ( u ) | x | a p | u | p 2 u C 0 | u | P 2 N p C 0 | M 1 R 0 N p p 1 | x | N p ( a + 1 ) p 1 | P 2 N p = C 0 M 1 P 2 N p R 0 p 2 p 1 V V | x | p 2 [ N p ( a + 1 ) ] ( p 1 ) ( N p ) = Λ k V V | x | p 2 [ N p ( a + 1 ) ] ( p 1 ) ( N p ) Λ Λ V k .

Now, taking Λ Λ it follows that f ( u ) | x | a p | u | p 2 u V k , for every x in B r c , which give us f(u) V k | x | a p | u | p 2 u, in B r c . □

Endnotes

a The proof of this fact follows by using the same ideas made in the usual Sobolev spaces; see [40].

b The proof of this fact follows by using the same ideas made in the case D 1 , p ( R N ); see [41].

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Acknowledgements

The authors would like to thank the referees for valuable comments and suggestions in improving this article. OHM was partially supported by CNPq/Brazil and Fapemig/Brazil (CEX-APQ 00025-11). RSV was partially supported by Capes/Brazil.

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Bastos, W.D., Miyagaki, O.H. & Vieira, R.S. Existence of solutions for a class of degenerate quasilinear elliptic equation in R N with vanishing potentials. Bound Value Probl 2013, 92 (2013). https://doi.org/10.1186/1687-2770-2013-92

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Keywords

  • Weak Solution
  • Previous Lemma
  • Energy Solution
  • Mountain Pass
  • Quasilinear Elliptic Equation