Research Article | Open Access
Sonia Ben Othman, Rym Chemmam, Habib Mâagli, "Asymptotic Behavior of Ground State Radial Solutions for -Laplacian Problems", Journal of Mathematics, vol. 2013, Article ID 409329, 7 pages, 2013. https://doi.org/10.1155/2013/409329
Asymptotic Behavior of Ground State Radial Solutions for -Laplacian Problems
Let , we take up the existence, the uniqueness and the asymptotic behavior of a positive continuous solution to the following nonlinear problem in , , , , where , is a positive differentiable function in and is a positive continuous function in such that there exists satisfying for each in , , and such that .
Let and . The following differential equation: has been studied with various boundary conditions, where is a continuous function in , differentiable and positive in , is a nonnegative continuous function in , and , (see [1–15]).
In this paper, our main purpose is to obtain the existence of a unique positive solution to the following boundary value problem: and to establish estimates on such solution under an appropriate condition on .
The study of this type of (1) is motivated by [5, 15]. Namely, in the special case , and , , the authors in  studied (1) and gave some uniqueness results. In this work, we consider a wider class of weights and we aim to extend the study of (1) in , to .
The pure elliptic problem of type has been investigated by several authors with zero Dirichlet boundary value; we refer the reader to [16–24] and the references therein. More recently, applying Karamata regular variation theory, Chemmam et al. gave in  the asymptotic behavior of solutions of problem . In this work, we aim to extend the result established in  to the radial case associated to problem .
To simplify our statements, we need to fix some notations and make some assumptions. Throughout this paper, we shall use , to denote the set of Karamata functions defined on by where is a positive constant and such that .
It is clear that if and only if is a positive function in such that For two nonnegative functions and on a set , we write , , if there exists a constant such that , for each .
The letter will denote a generic positive constant which may vary from line to line.
Furthermore, we point out that if is a nonnegative continuous function in , then the function defined on by is the solution of the problem
Here, the function is continuous in , differentiable and positive in such that with .
The function is required to satisfy the following hypothesis.
() is a positive measurable function on such that with and the function such that .
Remark 2. We need to verify condition in hypothesis , only if , (see Lemma 6 below).
As a typical example of function satisfying , we quote the following.
Example 3. Put . Then for and or and , the function satisfies .
Now, we are ready to state our main result.
The main body of the paper is organized as follows. In Section 2, we establish some estimates and we recall some known results on functions belonging to . Theorem 4 is proved in Section 3. The last Section is reserved to some applications.
2. Key Estimates
In what follows, we are going to give estimates on the functions and , where is a function satisfying and is the function given by (12). First, we recall some fundamental properties of functions belonging to the class , taken from [17, 22].
Lemma 5. Let and . Then one has and .
Lemma 6 (Karamata’s theorem). Let and be a function in . Then one has the following properties:(i)If , then converges and (ii)If , then diverges and
Lemma 7. Let . Then there exists such that for and
Lemma 8. Let , then one has If further converges, one has
Now, we are able to prove the following propositions which play a crucial role in this paper.
Proposition 9. Let be a function satisfying . Then one has for where is the function defined by
Proof. For , we have
To prove the result, it is sufficient to show that for .
Since the function is continuous and positive in , we have Now, assume that , then we have It follows from Lemma 7 that To reach our estimates, we consider the following cases.
(i) If , then it follows from Lemma 6 that Now, using Lemma 5 and again Lemma 6, we deduce that
(ii) If , then it follows from Lemma 6, that . So, since , we have
(iii) If , then for each , we have Since the function is in , then using the fact that and Lemma 6, we obtain that
(iv) If , we have by Lemma 6 that this yields to Hence, we reach the result by combining (22) with the estimates stated in each case above. This completes the proof.
Proposition 10. Let be a function satisfying and let be the function given by (12). Then for , one has
Proof. Let and , we obtain by simple calculus that for , where So, one can see that where . Then, using Lemmas 5, 7, and 8, we obtain that and . Hence, it follows from Proposition 9 that where is the function defined in (19) by replacing by and by . This ends the proof.
3. Proof of Theorem 4
3.1. Existence and Asymptotic Behavior
Let be a function satisfying and let be the function given by (12). By Proposition 10, there exists a constant such that for each We look now at the existence of positive solution of problem satisfying (11).
For the case , the existence of a positive continuous solution to problem is due to . Now, we look to the existence result of problem when and we give precise asymptotic behavior of such solution for . For that, we split the proof into two cases.
Case 1 (). Let be a positive continuous solution of problem . So, in order to obtain estimates (11) on the function , we need the following comparison result.
Lemma 11. Let and such that Then .
Proof. Suppose that for some . Then there exists , such that and for with or .
By an elementary argument, we have On the other hand, since , then , for each . This yields to
Using further (40), we deduce that the function is nondecreasing on with . Hence, from the monotonicity of , we obtain that the function is nondecreasing on with and . This yields to a contradiction, which completes the proof.
Now, we are ready to prove (11). Put and . It follows from (7) that the function satisfies According to (37), we obtain by simple calculation that and satisfy, respectively, (38) and (39). Thus, we deduce by Lemma 11 that This implies (11) by using (37).
Case 2 (). Put and let Obviously, the function belongs to and so is not empty. We consider the integral operator on defined by We shall prove that has a fixed point in , to construct a solution of problem . For this aim, we look at first that . Let , then we have for each This together with (37) implies that Since and , then leaves invariant the convex . Moreover, since , then the operator is nondecreasing on . Now, let be the sequence of functions in defined by Since , we deduce from the monotonicity of that for , we have Thanks to the monotone convergence theorem, we deduce that the sequence converges to a function which satisfies We conclude that is a positive continuous solution of problem satisfying (11).
Assume that satisfies . For , the uniqueness of solution to problem follows from Lemma 11. Thus in the following, we look at the case . Let Let and be two positives solutions of problem in . Then there exists a constant such that This implies that the set is not empty. Now, put , then we aim to show that .
Suppose that , then we have So, we have in , which implies that the function is nondecreasing on with . Hence from the monotonicity of , we obtain that the function is nondecreasing on with . This implies that . On the other hand, we deduce by symmetry that . Hence . Now, since and , we have . This yields to a contradiction with the fact that . Hence, and then .
4.1. First Application
Let be a positive measurable function in satisfying for where the real numbers and satisfy one of the following two conditions:(i) and ,(ii) and . Using Theorem 4, we deduce that problem has a positive continuous solution in satisfying the following.(i)If , then for (ii)If and , then for (iii)If and , then for (iv)If and , then for (v)If , then for (vi) If and , then for
4.2. Second Application
Let be a function satisfying and let . We are interested in the following nonlinear problem: Put , then by a simple calculus, we obtain that satisfies Using Theorem 4, we deduce that problem (64) has a unique solution such that Consequently, we deduce that (63) has a unique solution satisfying
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