Multiplicity Results for Weak Solutions of a Semilinear Dirichlet Elliptic Problem with a Parametric Nonlinearity

Tchalla, Ayékotan Messan Joseph; Tcharie, Kokou

doi:https://doi.org/10.1155/2022/6011860

International Journal of Mathematics and Mathematical Sciences

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Research Article | Open Access

Volume 2022 | Article ID 6011860 | https://doi.org/10.1155/2022/6011860

Multiplicity Results for Weak Solutions of a Semilinear Dirichlet Elliptic Problem with a Parametric Nonlinearity

Ayékotan Messan Joseph Tchalla¹and Kokou Tcharie¹

Academic Editor: Nawab Hussain

Received24 Aug 2022

Accepted22 Oct 2022

Published17 Nov 2022

Abstract

This paper deals with the existence of weak solutions to a Dirichlet problem for a semilinear elliptic equation involving the difference of two main nonlinearities functions that depends on a real parameter . According to the values of , we give both nonexistence and multiplicity results by using variational methods. In particular, we first exhibit a critical positive value such that the problem admits at least a nontrivial non-negative weak solution if and only if is greater than or equal to this critical value. Furthermore, for greater than a second critical positive value, we show the existence of two independent nontrivial non-negative weak solutions to the problem.

1. Introduction

In the last years, most works studied the existence, nonexistence, and multiplicity of nontrivial weak solutions of a semilinear Dirichlet problem of the form as follows:where is a bounded domain in , is a real parameter, and is a nonlinear function taking different forms. According to the values of , Ambrosetti et al. studied in [1], the existence and multiplicity of non-negative weak solutions of the problem (1) when with . For example, by using variational method, they show the existence of infinitely many solutions of the problem as follows:for and small. Later, Alama and Tarantello in [2] studied the semilinear Dirichlet problem (1) by searching non-negative solutions withwhere is a bounded domain with smooth boundary, and are suitable functions, and . In this case also, the authors show the influence of values of on the existence and multiplicity of weak solutions of the problem. These different studies on nonexistence, existence, and multiplicity results for nontrivial weak solutions depending on a parameter for a Dirichlet problem for a semilinear elliptic equation were extensively investigated in the literature (see, for e.g., [3–6] and the references therein). Similar results, depending on a real parameter, are obtained in the case of quasilinear elliptic equations in bounded domains or in entire space . For example, we can mention the papers [7–9], which are devoted to the unbounded case. In [7], the authors deal with the nonexistence and existence of nontrivial weak solutions of the quasilinear problem:where is a smooth exterior domain in , is the unit vector of the outward normal on , or , and , and are suitable functions. They showed in different cases that the existence of weak solutions of the problem depends on the values of relative to the value of some critical value.

In [8], Pucci and Radulescu studied the following problem in whole space:where satisfies is a parameter and with if and if . They obtained that the nonexistence and multiplicity of nontrivial weak solutions of this quasilinear elliptic equation are corresponding to the smallness and the largeness of , respectively. In [10], Autuori and Pucci extended the results in [8] by solving a more general quasilinear elliptic equation with the same variational method. Motivated by these previous results, we are concerned in this paper with the existence, nonexistence, and multiplicity of nontrivial weak solutions of the following Dirichlet problem for a semilinear elliptic equation:where , , is a bounded domain with smooth boundary, , are suitable non-negative functions, and is a real parameter. By taking inspiration on the method developed in [8, 10], we use variational arguments to study the existence and the multiplicity of nontrivial weak solutions of problem according to the values of the parameter . To obtain our results in this work, we require in the following assumptions:

is a function continuous on and of such that

Let us set

There exist such that and

is a function continuous on and of such that

Let us set

There exists such that and

Some examples of functions and in satisfying the previous assumptions , , and :(1)For , we can have and in () with .(2)Another example of functions and is the following: In this case, for all t > 0, we have(3)Let fixed in . For all ,. Hence, for all t > 0, we have

Thus, it is clear that the functions and of the Dirichlet problem of our present work generalize the functions and which appear in the main equation studied in [7, 8] or [10].

The main goal of this paper is the proof of the following two theorems:

Theorem 1. By the fulfillment of assumptions , , and , there exists a critical value such that the Dirichlet problem admits at least a nontrivial non-negative weak solution if and only if .

Theorem 2. Suppose that the assumptions , , and are fulfilled. Then, there exists a critical value satisfying such that for all , the Dirichlet problem admits at least two nontrivial non-negative weak solutions.

In Section 2, we talk about Orlicz spaces, which we will use in our work. In Section 3, we give different imbeddings between the working spaces of this paper and prove the nonexistence of a nontrivial weak solution when in is least than a positive number. The conditions for existence of weak solutions of are established in Section 4. Section 5 has devoted to prove Theorem 1, and Section 6 deals with the proof of Theorem 2.

2. Notions on Orlicz Spaces (See Chapter 8 in [11])

Definition 1. (definition of a N-function). Let be a real-valued function defined on and having the following properties:(a)(b) is nondecreasing, that is, implies (c) is right continuous, that is, if , then Then, the real-valued function defined on byis called an N-function.
Any such N-function has the following properties:(i) is continuous on (ii) is strictly increasing(iii) is convex(iv) and (iv)The function is strictly increasing on For any N-function and an open set , the Orlicz space is defined. When satisfies -condition, i.e.,for some constant , thenEndowed with the normwhich is called the Luxembourg norm, the Orlicz space is a Banach space. It is known that if , then with .
The complement of is given by the Legendre transformation as follows:We say that and are complementary N-functions of each other.
For all , we have the inequality .
From Young’s inequalitya generalized version of Hölder’s inequality is obtained as follows:

3. Preliminaries and Nonexistence of Nontrivial Solution for Small

By condition and the definition of and by condition and the definition of , the functions and are N-functions, with or , respectively.

Lemma 1. The -function satisfies the -condition.

Proof. In fact, by (12),Thus, by Theorem 4.1 in [12], it follows that satisfies the -condition.

Lemma 2. The -function satisfies the -condition.

Proof. In fact, by (17)and satisfies the -condition by Theorem 4.1 in [12].
Let us consider the Sobolev space , which is the completion of with the normLet denote the completion of with respect to the normwhereis the Luxembourg norm in the Orlicz space . The space is the space in which we will find our nontrivial weak solutions.

Lemma 3. The embeddings are continuous with and .

Moreover, and are compact for all such that .

Proof. The first imbedding that i.e., is followed from the definition of the norm in . The second imbedding is followed from Talenti’s work in [13]; is the Talenti constant. The imbedding compact for is obtained by Rellich’s Theorem. It follows that the mapping is compact for .

Lemma 4. Let , , , and . Then,

Proof. The proof is given in [14] (see Lemma 2.1 of [14]). In fact by integrating the inequalities (29) and (30) respectively, we get inequalities (34) and (35). From (34) and (35) and the definition of Luxembourg norm we get respectively (35) and (37).

Lemma 5. The space is imbedded continuously in with , wherewith and .

Proof. This proof is based on the proof of Theorem 8.12 in [11]. Fix , with , thenby (34) and (36). Thus, by taking , we haveBy the proof of (Theorem 8.12 [11]),Let such that . Let us set and It follows that and consequently is imbedded continuously in .

Lemma 6. The satisfies the following:

Proof. The result is followed from property (34). In fact, by (34), , we have

Lemma 7. The imbedding is continuous withwhere, .

Proof. The -function increases more slowly than near infinity and by (34) it follows that for all . By the proof of (Theorem 8.12 [11]),Let such that . Let us set and .It follows that and consequently is imbedded continuously in .

Lemma 8. The imbedding is continuous with

Moreover, the imbeddings and are compacts.

Proof. The continuity of the imbedding is followed from Lemmas 3 and 7. By Lemma 6 and Theorem 8.36 in [11], it follows that the imbedding is compact. It follows also that is compact because is continuous by Lemma 3.

Lemma 9. If is an element of and is a real such thatthen and there exists two positive constants and independent of such thatwhere and are two functions of .

Proof. Let us take and such that (49) hold. Thus, we have . As is a non-negative function in , it follows that . Let us show the second part of the Lemma. Firstly, by using respectively (29), (35), (48), and (49), we getThus, this last inequality yieldsSecondly, by (29), (30), (48), and (49) we haveLet us set and . Since , (34) and (36) implyLet or . By applying Young’s inequality for and and , we havewhere , , , and . Inequality (55) yieldsbeing . Hence, (54) yieldsAs , it follows, from (58), thatThus, by (53) and (59), we getFinally, the last inequality and (52) yield the result (50).

Definition 2. An element of is a weak solution of ifConsequently, the weak solutions of are exactly the critical points of the energy functional defined by

Lemma 10. If has a nontrivial weak solution , then , wherewith

Proof. If admits a nontrivial weak solution , then equality (49) is satisfied and by Lemma 9. Let us now show that . By inequalities (50) and (60) of the proof of Lemma 9, we get the result.
We claim that the set is not empty and bounded above. Indeed, by Lemma 9, for all , admit only trivial solution. Thus, . Now suppose that for all , there exists such that . Therefore, there exists a sequence of elements of such that For all , for all such that , admits only trivial solution. By hypothesis, the sequence tends to . Hence, for all , admits only trivial solution. This contradicts the Lemma 9.
Let us defineIt is clear that . In Section 5, we will prove that is the required critical value of the Theorem 1.

4. Basic Results for Existence of Nontrivial Solution

The results in the previous section require us to work from now on with .

Lemma 11. The energy functional in coercive on .

Proof. Let .By using Young’s inequality, as we use it in (55), we havewhere . Therefore, we haveIn conclusion, is coercive in .

Lemma 12. Let and respectively be the complements of -functions and . Then, we havewhere and are respectively the derivatives of and .

Proof. The results are obtained by using (29) and (30) and the proof of the point (2.7) of Lemma 2.5 in [14].

Proposition 1. The spaces and are reflexives and separables Banach spaces.

Proof. The N-functions and satisfy -condition respectively by Lemmas 1 and 2. Moreover, the inequalities (69) and (70) of Lemma 12 imply that the N-functions and satisfy the -condition. Thus, by (Theorem 8.20 and Remark 8.22 in [11]), the spaces and are reflexives and separables Banach spaces.

Proposition 2. is a separable reflexive Banach space.

Proof. Let , which we endow with the norm . Since and are reflexives Banach spaces, then is a separable and reflexive Banach space by Theorem 1.23 in [11]. Let us consider the operator defined by . is well defined, linear, and isometric. Therefore, is closed subspace of and so is separable and reflexive by Theorem 1.22 in [11]. Consequently, is a separable reflexive Banach space, being isomorphic to a separable, reflexive space. Finally, we conclude that is a separable reflexive Banach space because reflexivity and separability are preserved under equivalent norms.

Lemma 13. Let be a sequence in such that is bounded. Then, admits a weakly convergent subsequent in .

Proof. The proof comes from the coercivity of in and the reflexivity of the space .

Lemma 14. The functional is convex, of class and is particularly sequentially weakly lower semicontinuous in .

Proof. The convexity of functional is followed from the convexity of function in .
Let us show the continuity of on . Let be a sequence of which elements are in and let such that in . Let be an arbitrary subsequence of . The subsequence in and hence, by Lemma 3, in . By Theorem 4.9 in [15], there exists a subsequence of and a function such that a.e in as and a.e in for all . So a.e in as and a.e in for all . The dominated convergence theorem implies that in as . The subsequence being arbitrary, we deduce that in as . Then, we get the continuity of on and also, is sequentially weakly lower semicontinuous in by Corollary 3.9 in [15]. Moreover, is Gateaux-differentiable in and for all , we haveLet us finish the proof by showing that is continuous. Let be a sequence of functions in and such that in when . By a simple calculus we haveThis inequality yields the continuity of .

Lemma 15. The functional is convex, of class and sequentially weakly lower semicontinuous in . Furthermore, if is a sequence of elements of and belongs to such that converge weakly to in , then in .

Proof. The convexity of the functional is obvious. The continuity of in follows from Lemma 3. Thus, is sequentially weakly lower semicontinuous in by Corollary 3.9 in [15]. To complete the proof of the theorem, it suffices to show the last part of the theorem. Therefore, let us take and such that weakly in . Since is compact by lemma 3, we have and then in .

Proposition 3. (i) If converge strongly to in , then converge to in . (ii) If converge strongly to in , then converge to in .

Proof. (i) Suppose that converge strongly to in . Let be a fixed subsequence of the sequence . For all , we have , where is a subsequence of the sequence , which converge to in . Thus, , as . It follows that, there exists a subsequence of and a positive function such that a.e in and a.e in . As is continuous and strictly increasing in , we have a.e in . Hence, a.e in and a.e in . On other hand, there exists such thata.e in . Indeed, by the fact that a.e in and the function is continuous and increases strictly in , we haveBy (13) and since satisfy -condition, we havewhere and are constants. Thus, . In consequence, it yields that in , as . Due to arbitrariness of , we deduce that in , as . Hence, in .
(ii) The proof is analogous to (i).

Lemma 16. The functional is convex, of class and sequentially weakly lower semicontinuous in . Furthermore, if is a sequence of elements in and belongs to such that converge weakly to , then in .

Proof. The convexity of follows from the convexity of the positive function defined on . The function is also continuous and vanishes in zero. By Lemma 3 in [16], for all , for , such that, , we havewhere , , with .
Let be a sequence of and an element of such that in . belongs to by Lemma 8 and we have in implies in . So, a.e in . By (35), properties of function and Lemma 8, we haveAs in , there exists such that for all integer ,Since belongs to , for all and thus,Hence, by Theorem 2 in [16], it follows that , as . This assures the continuity of the functional . It then follows that is sequentially weakly lower semicontinuous in by Corollary 3.9 in [15]. Furthermore, is Gateaux-differentiable in and for all ,Let and such that in as . Let us show that in . By Hölder’s inequality, we haveSince in , by Lemma 8 and Proposition 3, it follows that and consequently, in . In particular, this shows that is class in and the proof of the Lemma is finished.

Lemma 17. The functional is convex, of class and sequentially weakly lower semicontinuous in . Moreover, if is a sequence of elements in and such that as , then in .

Proof. The convexity of follows from the convexity of the positive function defined on . The proof of continuity of in is analogous to the proof of continuity of in in the previous Lemma. In fact, by Lemma 3 in [16], for all , for , such that, , we havewhere , , with .
Let and such that in . By the definition of the space , we have , , and in . in implies in . So, a.e in . By (35) and properties of function , we haveAs in , there exists such that for all integer , . Since belongs to , for all and thus,Hence, by Theorem 2 in [16], it follows that as . This assures the continuity of the functional in . It then follows that is sequentially weakly lower semicontinuous in by Corollary 3.9 of [15]. Furthermore, is Gateaux-differentiable in and for all ,Let and such that in as . Let us show that in . By Hölder’s inequality, we haveAccording to Proposition 3 (ii), it follows that and consequently, in . Thus, is class in .
Let us show the last part of this Lemma. Let and such that in . We shall proof that in as . Let . For any integer , let us set and consider an arbitrary subsequence of the sequence . We have , where is a subsequence of . Then, weakly in when . By Lemma 3, it follows that in . Thus, there exists a subsequence of such that a.e in and a.e in , where . In consequence, for all , a.e in and there exists such that a.e in . In fact, by (30) and (36) we haveThus, . Hence,In conclusion, in , as , since is an arbitrary subsequence of the sequence .
For any , let us set

Lemma 18. For any fixed , the functional is in ( is the Banach dual space of . In particular, if , then .

Proof. Let us take . It is easy to show that is linear. Moreover, for all , by using Hölder inequality and imbedding properties of in and in , we haveThus, we conclude that is continuous on . Then, it follows that, yields .

Lemma 19. The functional is of class and sequentially weakly lower semicontinuous in , that is if in , then

Proof. Lemmas 14–17 imply . Let and such that in . The definition of and (89) allow us to writeSince in , Lemmas 14 and 15 yield respectively thatHence, by (92), we getBy (88) and (89), for all ,where , .
Multiplying (96) by and integrating the result over [0,1] with respect to , we obtainAccording to the assumptions and , we haveAfterwards, (29) and (30) implyBy (24), for all , we haveLet or . By applying Young’s inequality for and and , we havewhere we take , , , and . Inequality (101) yieldsbeing . Hence, (100) yieldsAs , it follows, from (104), thatwhereFrom (104) and (105), we obtainConsequently, (97) yieldsSince in , Lemmas 3 and 18 yield respectively and . Thus, and then, by (95), we get the claim (91).

5. Existence of Weak Solutions of for Large Values of

Let

We remark that . Indeed, by the fact that , (48) implies that

By the property (37), Lemma 5 and the monotony of function , we have

Thus,

Lemma 20. For all , the functional admits in a global nontrivial non-negative minimizer with negative energy, that is .

Proof. For each , Lemmas 11 and 19 and Corollary 3.23 in [15] assure to the existence of a global minimizer , that isSo, is a solution of . Let us prove that is nontrivial as soon as . For that we shall just prove that . Let . By the definition of , there exists a function , with , such thatHence,and thenTherefore,Hence, it follows that, for , equation has a nontrivial weak solution with . To complete the proof, we can assume a.e in because and .
Let us defineBy Lemma 20, it is clear that this definition is meaningful. Furthermore, it follows easily that .

Theorem 3. For any , the problem admits a nontrivial non-negative weak solution .

Proof. Fix . By definition of , there exists such that has a nontrivial critical point . Without loss of generality, we assume that a.e in , since is also a solution of . We can easily see that is a subsolution for . Let us consider the following minimization problem:We remark easily that is a closed and convex set. Thus, is weakly closed. Moreover, being is coercive in by Lemma 11, it follows that it is coercive in . Finally, is sequentially weakly lower semicontinuous in and so in . Hence, Corollary 3.23 of [15] implies that there exists such thatNow, we shall prove that is a solution of . Let and . Let us set and . It is clear that andHenceLet us setIt is obvious that . Since, is a subsolution of and , it turns out that . Consequently, asWe have(i)For , it is clear that . Hence by (122), we have .(ii)Also, when , we have in because in this case in . Then, (125) yieldswhere . We claim that . Indeed, and are in and and are in since . Moreover, by and , the functions and are continuous in . Thus, the functions , , , and are continuous in . Consequently, , , , and belong to . Finally, since . Therefore, the claim is obtained. Thus,since as . Then, (126) implies that as . So, by (122), it follows that as .
We deduce that, for all , that is for all . Thus, is a weak solution of in because is the completion of with respect to the norm . Finally, is nontrivial and non-negative, since .

Lemma 21. .

Proof. This proof is the same to that one did in the step 5 of the proof of (Theorem 1.1 in [7]). In fact, by Theorem 3, we have . Indeed, for all such that a weak nontrivial solution, we have and thus, . Let us show that . Suppose that . For , the problem cannot admit a nontrivial solution since this would contradict the minimality of . Thus, for all the unique solution of is . This assertion is still again impossible because it would contradict the maximality of . Hence, .

Theorem 4. The problem admits a nontrivial non-negative weak solution in .

Proof. Let be a strictly decreasing sequence converging to and be a nontrivial non-negative weak solution of . For all , we haveBy (29), (30), (49), and Lemma 4 we haveFurthermore, by inequality (60) and the monotony of there exists a constant such thatwhereThus, by monotony of sequence , we getTherefore, the sequences and are bounded, and it yields that the sequence is bounded. According to Theorem 3.18 in [15], Propositions 1 and 2 and Lemma 8 allow to extract, from the sequence , a subsequence still relabeled and satisfyingfor some . We claim that , which is clearly non-negative by (133), is the solution we are looking for. In fact, for all ,as , since by (133). Since in , Lemma 15 yields in particular that for all ,as . Moreover, Lemmas 16 and 17 imply that for all as . By passing to the limit in (128) as , we get by (134)–(136)for all . Hence, is a weak non-negative solution of . It remains to show that the solution is nontrivial. Since in by (133), Lemma 8 yields in particular that . Moreover, (52) applied to each , implies thatThus, by passing to the limit as , we getsince and . Consequently, is nontrivial and non-negative by (133).
Proof of Theorem 1. Section 3, Lemma 20, and Theorems 3 and 4 show the existence of such that for all , admits at least a nontrivial non-negative weak solution in .

6. Second Solution for Large Values of the Parameter

By variational methods, we prove, in this section, that the Dirichlet problem admits at least two nontrivial weak solutions if is sufficiently large.

Lemma 22. For any and there exist and such that for all , with .

Proof. Let and . Let be in . By (35) and (48) and monotony of the function ,So, to find the result of the Lemma, it is enough to take such that In Lemma 20, we have shown that for all , the problem admits a nontrivial non-negative weak solution which is above all a global minimizer of in , with . In this section, we are looking for a second nontrivial weak solution of when .

Lemma 23. There exists such that

Proof. (i)If , the inequality is verified for all .(ii)If and (or and ), the inequality is obtained by (30) and .(iii)Suppose now that with and . By hypothesis , the function is strictly increased in and consequently it is strictly increased in because it is an odd function. Thus, for with and , andLetIn conclusion, taking , we get the result of this Lemma.

Theorem 5 (see Theorem A.3 in [10]). Let and be two Banach spaces such that . Let be a functional with . Suppose that there exist and such that , , and for all with . Then, there exists a sequence such that for all where

Proof. One can find the proof of this theorem in the section (Appendix A in [10]).
Proof of Theorem 2. Let be a strictly positive fixed number such that . Then, let be the global minimizer of given by Lemma 20. Thanks to Lemma 22 and the fact that , the assumptions of Theorem 5 which is a variant of Ekeland’s variational principle are fulfilled for the energy functional of . Hence, there exists a sequence such thatwhereIn the sequel, we shall prove that the sequence strongly converges to some in and that is a second nontrivial non-negative weak solution of .

Step 1. By Lemma 11 (coercivity of on ), the sequence is bounded in and consequently, is bounded in . For the sequel, let us use the argument of the proof of Theorem 4. Hence, by Propositions 1 and 2 and Lemma 8 we can extract from the sequence , a subsequence still relabeled and satisfyingfor some . It remains to show that is a nontrivial solution of (), with . We haveBy the same argument used in the proof Theorem 4, we have, for all ,as . Hence, passing to the limit as in (151), using and the fact that as for all , we getso is a weak solution of ().

Step 2. We claim thatas . By hypothesis , the function is strictly increased in and consequently, it is strictly increased in because it is an odd function. Thus, we haveBy (150), in . Thus, in by Proposition 3. Therefore, by Hölder inequality, we getas . This gives the proof of the claim.

Step 3. In this step, we show that as . Let us setFor the same arguments, as for , we haveNote that as , since in and in as . Hence, by (154)as . Furthermore, we haveas . By (143) and (159), it follows thatas . Thus, by section 8.13 in [11],since the -function satisfied the -condition. Finally,by (160) and (162).

Step 4. Since in and , we get . So, is a second independent nontrivial weak solution of (), with . We can assume a.e. in , since is also a solution of () due to . This concludes the proof.

7. Conclusion

In this paper we studied the nonexistence, the existence, and multiplicity results for nontrivial weak solutions of the semilinear elliptic Dirichlet problem with () involving a positive parameter . Our main results are obtained in the Theorems 1 and 2 by using variational methods.

Data Availability

No data were used to support the findings of this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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Copyright © 2022 Ayékotan Messan Joseph Tchalla and Kokou Tcharie. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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