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Existence of solutions for a class of degenerate quasilinear elliptic equation in with vanishing potentials
Boundary Value Problems volume 2013, Article number: 92 (2013)
We establish the existence of positive solution for the following class of degenerate quasilinear elliptic problem
where , , , , , and 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.
Consider the following degenerate quasilinear elliptic problem in :
where , , , , , and denote the Hardy-Sobolev’s critical exponent, is a bounded, nonnegative and vanishing potential and a continuous function with a subcritical growth at infinity. Here, is the completion of the with the norm . We impose the following hypotheses on f and V:
is a continuous function verifying
() , uniformly in x.
() There exists such that , uniformly in x.
() There exists such that , for all .
is a continuous function verifying
() , for all .
() There are and such that .
Remark 1.1 The conditions () and () imply the following:
Example 1.2 An example for the function f is given by
with and α given by ().
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
Consider first the case , that is is the p-Laplacian operator, and the potential is bounded from below by a positive constant .
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 .
Let us now consider the case and the potential bounded from below by a positive constant .
In this case, the equations arise in problems of existence of stationary waves for anisotropic Schrödinger equation (see ) and others problems (for example, see [6, 23]). We cite  for ; and [24, 25] for . For the case , we cite , for and ; and , for and .
The result presented here for and extends that one in  for and . In , 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 ). Now we state the main result of this work.
Theorem 1.3 Suppose that V and f satisfy, respectively, () and () and () to (). Then there is a constantsuch that the problem (P) has a positive solution, for all, beingthe maximum of the f in the ball ofcentered 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 the ball in centered in the origin with radius R. The weak (⇀) and strong (→) convergences are always taken as and means . The weighted spaces are denoted by . When , we denote the usual norm in , with . For , we use .
2 The auxiliary problem
As usual, since we are looking for positive solutions of problem (P), we set , for all . The hypothesis () allows us to consider the space
Associated to the problem (P), we define on E, the Euler-Lagrange functional
being . From the assumptions on f, it follows that I is with Gâteaux derivative
To obtain solutions of problem (P), we introduce some truncation of the function f. Consider , and define
Now we define the auxiliary problem:
Associated to the problem (AP), we define, on E, the Euler-Lagrange functional
being . From the assumptions on f, it follows that J is with Gâteaux derivative
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 , the norm and we define the functional given by . Here, is the completion of the with the norm .
Lemma 3.1 The functionalhas the mountain pass geometry, namely,
such that for .
such that and .
Proof By using the growth of f given in Remark 1.1 and the Caffarelli-Kohn-Nirenberg’s inequality (see ), we get , and hence . Since , there exists such that . Thus, we have for . By (), it follows that there exist and such that . Now implies . Since , there exists a large enough such that, taking , we have and . □
Lemma 3.2 The functional J has the mountain pass geometry, namely,
such that for .
such that and .
Proof From the definition of G, we have . Thus, like the previous lemma, we have for . Take the same of the proof of the previous lemma. Thus, and . With the same argument, we have . Since , there exists a large enough such that, taking , we have and . □
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 satisfies two previous lemmas. This allows to define the minimax levels c and by
respectively. Since in , we have , by their definitions. Now using the above lemma together with the mountain pass theorem [, Theorem 2.2], we conclude that there exists a Palais-Smale sequence ((PS) sequence for short) for J, i.e., satisfies and .
Lemma 3.3 Suppose () and () to () and letbe a (PS) sequence for the functional J. Thenis bounded in E.
Proof Define the set . In A, we have . By using (), we conclude that there is such that . So, we have
Now consider the set . In B, we have and . Then
Therefore, we get . In particular, the above equation holds for the (PS) sequence for J, and we have . On the other hand, we have , and , since . Thus, we get , for some constant . Then we have , which can be rewritten as
Assuming , equation (4) implies that . So , which is a contradiction. Therefore, is bounded in E. □
Lemma 3.4 Suppose () and () to (). 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 , by Lemma 3.3, there exists such that , because E is reflexive space.a Thus, it is enough to show that . We divide this task in the four claims below.
Claim 1 .
Claim 2 .
Claim 3 .
Claim 4 .
Assuming Claims 1 to 4 for now, we proceed with the proof of lemma.
Since , we have , and by Claim 1, we get
As , we get . Passing the limit in the above equation and using Claims 2, 3 and 4, we get
In order to prove the claims, for a given , we choose r satisfying, the following two conditions:
is such that in and , , for all .
Observe that condition 1 follows by integrability of and . Note that, when one of these conditions holds for some , it also verifies for every . 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 satisfying conditions 1 and 2 above. From the growth of g and the choice of r, we conclude:
, in ;
, in ;
Verification of claims
In all proofs, except for Claim 4, we consider separately integration in and .
Claim 1: In , by combining the dominated convergence theorem with the compact embedding of E in , and (see [, Theorem 2.1]), we obtain . On the other hand, when integrating in we do not have the compact embedding of E in . In this case, we first estimate , by using the cut-off function η defined before. As is also bounded we have , namely,
Since in , equation (7) holds in . Adding to this, the fact that in we have
From this, we obtain
Thus, we have
Using the boundedness of in E, i.e., , strong convergence of in and Hölder’s inequality we conclude that
Using this fact, we conclude that
Now, using Remark 3.5, we have .
Therefore, 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 the proof proceeds like in Claim 1. For integration in , we estimate the value of by using the Hölder’s inequality
From this fact, we get and the proof of this claim is completed.
Claim 4: Let be dual of p. Since ,b we see that and . As is bounded, we conclude that is also bounded. Moreover, satisfies the hypothesis of Lemma 3.6, below. So that we have a.e. , which give us a.e. . Using a theorem [, Theorem 13.44], we conclude that in . Thus, we have , and hence .
Lemma 3.6 Let E and J be respectively the space and functional defined in Section 2. Leta bounded sequence such thatin E and. Then, passing to a subsequence if necessary, we have, a.e. .
Using Lemmas 3.2, 3.3, 3.4 and the mountain pass theorem, in [, Theorem 2.4], we conclude that there exists 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 , for all .
Lemma 4.1 Any least energy solution u of (AP) satisfies the estimate.
Proof Since u is a critical point in the minimax level , by Lemma 3.3, we have . So that . □
Remark 4.2 The constant depends only on , θ and f.
Lemma 4.3 Let h be such thatis integrable, with. Consider, andnonnegative and continuous functions such that, for all. Letbe a weak solution of (AP 2):
Then there exists a constantsuch that.
Proof Given and , set , and
Then and using it as the test function in (AP 2). we have
By definition of , we get
Now, taking into account (14), we get
Using (17), (13) and (16), we have
Combining (18), (15) and (12), we obtain
Let S be the best constant for inequality , for all . (See .) By the inequality (19), the boundedness of H and the definitions of and S, we get . Since in , we have
Making and using the monotone convergence theorem, we get
Since , . Thus, we can consider for and, using the Hölder’s inequality, we have
being . Note that is independent of j. Thus, we get
For and , we have and . Applying this in (20), we have
By iterating, we have
Thus, we get
Then we can say that, for all , we have
Putting , we get
As σ depends on q, we have, by definition of M, that . □
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 constantsuch that, for all u positive solution of (AP).
Proof Take r from the definition of g and define and . Now define H and b by
We will show that, if u is solution of (AP), then u is weak solution of (AP 2).
In particular, we recall that , in A, and , in B. Using the fact that is the disjoint union of A and B and the definitions above, for a given , we write
Hence, u verifies , for all . From () and (), we obtain , for all , with . From the definition of H, it follows that , with . Taking , we have that is integrable. Then u satisfies the hypotheses of Lemma 4.3, that is,
Using the definition of the constant S and Lemma 4.1, we have
Lemma 4.6 Let r be as in the definition of g and consider. Let u a be any positive solution of (AP). Then we have
Proof Consider . By Lemma 4.5, we have . So , for . It follows that , in , and the function given by
is so that . Moreover, , because . Let us show now that , in . 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
Putting and , we have
By (23) and (24), we obtain, for all , that
In the case , in addition to the above arguments used, we also use the Hölder’s inequality. Then
Then for all . Thus, we have , in , which implies that , in . From this, we conclude that in , and lemma is proved. □
Proof of Theorem 1.3 We will show that in , for all solution u of (AP). By Remark 1 we have , which gives us . Now, note that the hypothesis () holds for all . Hence, for , we can use () and Lemma 4.6. Thus, for each x in , and , we have
Now, taking it follows that , for every x in , which give us , in . □
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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.
The authors declare that they have no competing interests.
The authors declare that all authors collaborated and dedicated the same amount of time in order to perform this article.
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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 with vanishing potentials. Bound Value Probl 2013, 92 (2013). https://doi.org/10.1186/1687-2770-2013-92
- Weak Solution
- Previous Lemma
- Energy Solution
- Mountain Pass
- Quasilinear Elliptic Equation