Existence of Weak Solutions for Elliptic Nonlinear System in RN
1Department of Mathematics, University Tebessa, Tebessa, Algeria
We study the nonuniformly elliptic, nonlinear system Under growth and regularity conditions on the nonlinearities f and g, we obtain weak solutions in a subspace of the Sobolev space by applying a variant of the Mountain Pass Theorem.
Keywords: nonuniformly elliptic, nonlinear systems, mountain pass theorem, weakly continuously differentiable functional
International Journal of Partial Differential Equations and Applications, 2014 2 (2),
Received May 06, 2014; Revised May 28, 2014; Accepted May 28, 2014Copyright © 2013 Science and Education Publishing. All Rights Reserved.
Cite this article:
- Bouali, Tahar, and Rafik Guefaifia. "Existence of Weak Solutions for Elliptic Nonlinear System in RN." International Journal of Partial Differential Equations and Applications 2.2 (2014): 32-37.
- Bouali, T. , & Guefaifia, R. (2014). Existence of Weak Solutions for Elliptic Nonlinear System in RN. International Journal of Partial Differential Equations and Applications, 2(2), 32-37.
- Bouali, Tahar, and Rafik Guefaifia. "Existence of Weak Solutions for Elliptic Nonlinear System in RN." International Journal of Partial Differential Equations and Applications 2, no. 2 (2014): 32-37.
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We study the nonuniformly elliptic, nonlinear system
System , There, under appropriate growth and regularity conditions on the functions and , the weak solutions are exactly the critical points of a functional defined on a Hilbert space of functions in . In the scalar case, the problem
with and , has been studied by Mihailescu and Radulescu . In this situation, the authors overcome the lack of compactness of the problem by using the the Caffarelli-Kohn-Nirenberg inequality
In this paper, we consider which may be a nonuniformly elliptic system. We shall reduce to a uniformly elliptic system by using appropriate weighted Sobolev spaces. Then applying a variant of the Mountain pass theorem in , we prove the existence of weak solutions of system in a subspace of (,).
To prove our main results, we introduce the following some hypotheses:
There exists a function such that , for all ,
for all there exists a positive constant such that
for all ,
There exists a constant such that
for all ,
Let be the usual Sobolev space under the norm
Consider the subspace
Then is a Hilbert space with the norm
It is clear that
and the embeddings are continuous. moreover, the embedding is compact see ). we now introduce the space
endowed with the norm
Since for all we have with and1.2. Proposition
The set is a Hilbert space with the inner product
Proof. It suffices to check that any Cauchy sequences in converges to . Indeed, let be a Cauchy sequence in Then
and is bounded. Moreover, by Remark , is also a Cauchy sequence in . Hence the sequence converges to ; i.e,
It follows that converges to and converges to in Therefore converges to and converges to for almost everywhere . Applying Fatou's lemma we get
Hence Applying again Fatou's lemma
We conclude that converges to in1.3. Definition
We say that is a weak solution of system if
Our main result is stated as follows.1.4. Theorem
Let and are satisfied, the system has at least one non-trivial weak solution in .
This theorem will be proved by using variational techniques based on a variant of the Mountain pass theorem in . Let us define the functional given by
2. Existence of weak solutions
In general, due to the functional may be not belong to (in this work, we do not completely care whether the functional belongs to or not).This means that we cannot apply directly the Mountain pass theorem by Ambrosetti-Rabinowitz (see ), we recall the following concept of weakly continuous differentiability. Our approach is based on a weak version of the Mountain pass theorem by Duc (see ).2.1. Definition
Let be a functional from a Banach space in to . We say that is weakly continuously differentiable on if and only if the following conditions are satisfied is continuous on . For any , there exists a linear map from into such that For any , the map , vi is continuous on .
We denote by the set of weakly continuously differentiable functionals on . It is clear that , where is the set of all continuously Frechet differentiable functionals on . The following proposition concerns the smoothness of the functional2.2. Proposition
Under the assumptions of Theorem 1.4, the functional given by is weakly continuously differentiable on and
Proof. By conditions -- and the embedding is continuous, it can be shown (cf. [, Theorem A.VI]) that the functional is well-defined and of class . Moreover, we have
Next, we prove that is continuous on . Let be a sequence converging to in , where , ., Then
and is bounded. Observe further that
Similarly, we obtain
From the above inequalities, we obtain
Thus is continuous on . Next we prove that for all
Indeed, for any, any and we have
Applying Lebesgue's Dominated convergence theorem we get
Similarly, we have
Combining - , we deduce that
Thus is weakly differentiable on .
Let be fixed. We now prove that the map is continuous on . Let be a sequence converging to in . We have
It follows by applying Cauchy's inequality that
Thus the map , i is continuous on and we conclude that functional is weakly continuously differentiable on . Finally, is weakly continuously differentiable on2.3. Remark
From Proposition we observe that the weak solutions of system correspond to the critical points of the functional given by Thus our idea is to apply a variant of the Mountain pass theorem in  for obtaining non-trivial critical points of and thus they are also the non-trivial weak solutions of system2.4. Proposition
The functional given by satisfies the Palais-Smale condition.
Proof Let be a sequence in H such that
First, we prove that is bounded in . We assume by contradiction that is not bounded in . Then there exists a subsequence of such that as . By assumption
it follows that
Letting since we deduce that which is a contradiction. Hence is bounded in .
Since is a Hilbert space and is bounded in , there exists a subsequence of weakly converging to in . Moreover, since the embedding is continuous, is weakly convergent to in . We shall prove that
since converges weakly to in ; i.e,
for all this implies that converges weakly to in
Applying [, Theorem 1.6], we obtain
Thus is proved. We now prove that
Indeed, by , we have
where are positive constants.
Set We have and
On the other hand, using the continuous embeddings together with the interpolation inequality , it follows that
Since the embedding is compact we have as Hence as and is proved.
On the other hand, by and it follows
Hence, by the convex property of the functional we deduce that
Relations and imply
Finally, we prove that converges strongly to in . Indeed, we assume by contradiction that is not strongly convergent to in . Then there exist a constant and a subsequence of such that for any Hence
With the same arguments as in the proof of , and remark that the sequence converges weakly to in , we have
Hence letting , from ) and we infer that
Relations and imply which is a contradiction. Therefore, we conclude that converges strongly to in and satisfies the Palais -Smale condition on
To apply the Mountain pass theorem we shall prove the following proposition which shows that the functional has the Mountain pass geometry.2.5. Proposition
There exist and such that for all with There exists such that and
Proof. From , it is easy to see that
where in view of . It follows from that
By using the embeddings , with simple calculations we infer from that for small enough This implies .
By , for each compact set there exists such that
Let having compact support, for large enough from we have
where , Then and imply
Proof of Theorem 1.4. It is clear that . Furthermore, the acceptable set
where is given in Proposition is not empty (it is easy to see that the function by Proposition. and Propositions - , all assumptions of the Mountain pass theorem introduced in  are satisfied. Therefore there exists such that
and for all is a weak solution of system . The solution is a non-trivial solution by The proof is complete.
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