Sequences of positive homoclinic solutions to difference equations with variable exponent
Robert Stegli\'nski, Magdalena Nockowska-Rosiak

TL;DR
This paper investigates the existence of infinitely many positive homoclinic solutions for a second-order difference equation involving a variable exponent p_k-Laplacian, using critical point theory and Ricceri's variational principle.
Contribution
It introduces a novel approach to find multiple positive homoclinic solutions in difference equations with variable exponents using advanced variational methods.
Findings
Proved existence of infinitely many positive homoclinic solutions
Applied critical point theory to variable exponent difference equations
Utilized Ricceri's variational principle effectively
Abstract
We study the existence of infinitely many positive homoclinic solutions to a second-order difference equation on integers with -Laplacian. To achieve our goal we use the critical point theory and the general variational principle of Ricceri.
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Sequences of positive homoclinic solutions to difference equations with variable exponent
Robert Stegliński and Magdalena Nockowska-Rosiak111Corresponding author
Institute of Mathematics, Lodz University of Technology,
ul. W\a’olcza\a’nska 215, 90-924, Ł\a’od\a’z, Poland
emails: [email protected], [email protected]
Abstract. We study the existence of infinitely many positive homoclinic solutions to a second-order difference equation on integers with -Laplacian. To achieve our goal we use the critical point theory and the general variational principle of Ricceri.
Keywords: Difference equations; Laplacian; variational methods; infinitely many solutions.
1 Introduction
In this paper we consider the nonlinear second-order difference problem of the form
[TABLE]
In the whole paper we assume with , , , , and are continuous functions for . Moreover, we use the notation , , . We say that a solution to equation given in (1.1) is homoclinic if
Difference equations represent the discrete counterpart of ordinary differential equations and are usually studied in connection with numerical analysis. Equation given in (1.1) is the discrete counterpart of the following
[TABLE]
The above equation is interesting because of its wide applications for example in nonlinear elasticity or fluid dynamics. Moreover, equation (1.2) becomes the stationary nonlinear Schrödinger equation for and . As its continuous counterpart, equation given in (1.1) for has many applications in various areas of physics.
The variational method is very powerful tool in study of boundary value problems to difference equations. Many authors have applied different results of the critical point theory to prove existence and multiplicity results for solutions to discrete nonlinear problems. Studying such problems on bounded discrete intervals allows to search for solutions in a finite-dimensional Banach space (see [5, 6, 17, 19]). To study such problems on unbounded intervals directly by variational methods, [10] and [16] introduced coercive weight functions which allow to preserve of certain compactness properties on -type spaces. That method was used in the following papers [9, 12, 21, 22, 23, 24, 25].
The goal of the present paper is to establish sufficient conditions for the existence of a sequence of positive solutions to problem (1.1) which norm tend to infinity or zero. Infinitely many solutions for a constant were obtained in [25] by employing Nehari manifold methods, in [12, 24] by applying a variant of the fountain theorem, in [21, 22, 23] by use of the Ricceri’s theorem (see [3, 20]) and in [22, 23] by applying a direct argumentation. The existence of a nontrivial or infinitely many solutions to problem (1.1) with variable exponent were proved in [1, 7, 11] by the mountain pass theorem. In [1, 7] authors assumed some symmetry conditions on nonlinear part of the problem to get infinitely many solutions to it. Moreover, in [1] there were considered different sequences of variable exponent on the left side of equation given in (1.1). In this paper, similarly to [21, 22, 23], the nonlinearity has a suitable behavior at infinity or zero, without any symmetry conditions.
We will study problem (1.1) under the following assumptions:
, for all , as ;
for all ;
for any ;
there exist sequences such that and for every and ;
there exist sequences such that and for every and ;
;
;
;
;
;
,
where is the primitive function of , that is for , , and
[TABLE]
Notice that , because and is bounded.
Equation given in (1.1) is considered in certain reflexive Banach space defined in the next Section. This equation is the Euler-Lagrange equation of certain action functional which is well defined on space . Elements of this space automatically satisfy boundary condition in problem (1.1). In this paper we will prove the following theorems:
Theorem 1**.**
Assume that , , and are satisfied. Moreover, assume that one of conditions or is satisfied. Then, problem (1.1) possesses a sequence of positive solutions in whose norms tend to infinity.
Theorem 2**.**
Assume that , , and are satisfied. Moreover, assume that one of conditions or is satisfied. Then, problem (1.1) has a sequence of positive solutions in whose norms tend to 0.
Theorem 3**.**
Assume that , , and are satisfied and assume that one of the conditions or holds. Then, problem (1.1) possesses a sequence of positive solutions in whose norms tend to zero.
Theorem 4**.**
Assume that , , and are satisfied and assume that one of the conditions or holds. Then, problem (1.1) has a sequence of positive solutions in whose norms tend to infinity.
To get our goal in Theorems 1, 2 a direct variational approach is used, introduced in [8] and then used in such papers as [13, 14, 15, 18, 23]. In the proof for Theorems 3, 4 the general variational principle of Ricceri is used, which was applied in [2, 4, 21, 23].
The plan of the paper is as follows: Section 2 is devoted to the abstract framework, in Section 3 proofs of Theorems 1-4 and examples are presented.
2 Abstract framework
Recall that the sequence is bounded with . We introduce
[TABLE]
with norm
[TABLE]
and
[TABLE]
with norm
[TABLE]
Proposition 1**.**
For we have
[TABLE]
where and is defined in (1.3).
Proof.
Let and . Choose such that . Then , and so , for any . This implies that . Passing to we get . ∎
Theorem 5**.**
[11]** and are a reflexive Banach space and embeddings and are compact.
Let
[TABLE]
It is easy to see that
Proposition 2**.**
[11]** For we have the following relations:
- (i)
* ;*
- (ii)
* ;*
- (iii)
* ;*
- (iv)
, as , as .
Now, we define a functional on by formula
[TABLE]
Note that implies that is well defined. By the definition of and Proposition 2 we get
Proposition 3**.**
For we have the following inequalities:
- (i)
;
- (ii)
;
- (ii)
.
Lemma 6**.**
Assume that is satisfied. Then with
[TABLE]
for all . Moreover, is a sequentially weakly lower semicontinuous functional on .
Proof.
The standard proof is omitted, see [7], [10], [23].
∎
The following notation will be used in the rest of the paper:
[TABLE]
for , .
Remark 1**.**
If is satisfied, then is a continuous function on for . Moreover, notice that for , . This implies that if one of conditions , , , , , is satisfied for , then this conditions is satisfied for , too.
We define a functional on by formula
[TABLE]
Lemma 7**.**
Assume that , hold. Then with
[TABLE]
for all . Moreover, is a sequentially weakly continuous functional on .
Proof.
In [23] there was shown that if is satisfied, then for any functional given by formula , , is continuously differentiable on . This and continuous emending gives that .
Now we prove that is a sequentially weakly continuous functional on . Let in . Set . By mean value theorem, there exists for any , which belongs to interval with ends , such that . Hence
[TABLE]
From the above estimation and the fact that emending is compact, we have that for any subsequence of there exists such that and , as . Hence as . ∎
Let be a functional associated to problem (1.1) defined by
[TABLE]
where is given by (2.2) and is given by (2.4).
Proposition 4**.**
Assume that , and are satisfied, then and is a sequentially weakly lower semicontinuous functional on . Moreover, every critical point of is a positive homoclinic solution to problem (1.1).
Proof.
By Lemma 6 and Lemma 7 we get that and is a sequentially weakly lower semicontinuous functional on . It is easy, to see that, for a critical point of , we have
[TABLE]
Hence , for and is nonnegative. It means that is a homoclinic solution to (1.1). If there exists such that , then by we get that
[TABLE]
By non-negativity of each term of the above equality and we get that . By induction we get that , . It means, that every nontirivial critical point of is a positive solution to problem (1.1). ∎
3 Proofs of main Theorems
3.1 Proofs of Theorems 1, 2
Proof of Theorem 1.
Let and as in (2.6). By Proposition 4 we need to find a sequence of critical points of whose norms tend to infinity.
Let be sequences which satisfy assumption .
We define the set
[TABLE]
for every .
We have divided the rest of the proof into few steps.
Step 1. Firstly, we show that, for every , the functional is bounded from below on and its infimum on is attained.
Let . Note that
[TABLE]
for any . By the definition of and (3.1) we have
[TABLE]
for . Thus, is bounded from below on . Let and be a sequence in such that for all .
By Propositions 2 and 3, if , then
[TABLE]
It means that is bounded in . Taking into account that is a reflexive Banach space and is a weakly closed subset of , we get that weakly converges in to some . By the sequentially weakly lower semicontinuity of we conclude that .
Step 2. Now, we show that if , , then, for all .
Fix . Let and suppose that . Put
[TABLE]
Thus, .
We define the truncation function by formula . Now, put . Clearly, and for every . Thus .
[TABLE]
Since is a Lipschitz function with Lipschitz-constant equal to 1, we get
[TABLE]
Moreover, as for all , for all , and for all , we have
[TABLE]
Next, we estimate . Firstly, by the definition we have for , , and consequently . As we have for , and by , we have , for , hence
[TABLE]
Consequently,
[TABLE]
Combining above inequalities we get that
[TABLE]
But , since . So, every term in is equal to zero. In particular, , hence
[TABLE]
By , we get that for every and for every . By the definition of sets and , we must have , this contradicts to ; therefore .
Step 3. Now, we show that if , then is a critical point of for every .
It is sufficient to show that is a local minimum point of in . Assuming the contrary, consider a sequence which converges to and for all . From this inequality it follows that for any . Since in , due to (2.1), in as well. Choose a positive such that . Then, there exists such that for every . By using previous Step and taking into account the choice of the number , we conclude that for all and , which contradicts to the fact .
Step 4. Now we show that , where , .
Firstly, we assume that holds. Without lost of generality we can assume that
[TABLE]
There exists such that
[TABLE]
Hence we get the existence of sequences , such that , , and
[TABLE]
for all . Without lost of generality, we may assume that for all . We define a sequence such that, for every , , if and if . It is clear that Hence
[TABLE]
which gives . Now, assume that holds, i.e.
[TABLE]
There exist and such that
[TABLE]
Then, there exists a sequence of real numbers such that and
[TABLE]
for all . Without lost of generality, we may assume that for all . Thus, take in a sequence such that, for every , and for every . Then, and
[TABLE]
which gives .
Now, we are ready to end the proof of Theorem 1. From Proposition 4 and the above Steps we have infinitely many pairwise distinct positive homoclinic solutions to (1.1) with . To finish the proof, we will prove that as . On the contrary suppose, that there exists a subsequence of which is bounded in . From reflexivity of there exists such that , as , by choosing a subsequence, if necessary. From sequentially weakly lower semicontinuity of on we get
[TABLE]
a contradiction. ∎
Proof of Theorem 2.
In analogously way to the proof of Theorem 1 we can prove the existence of a sequence of positive solutions to (1.1) such that , where and satisfies . Moreover, for any , . As in Step 4 from Theorem 1 we get that , . Moreover, we get that
[TABLE]
Above and , imply that . To finish the proof, notice that
[TABLE]
for . Hence , as . ∎
3.2 Proofs of Theorems 3, 4
We start this section with presentation of main tool in our proofs.
Theorem 8**.**
[3]**, [20] Let be a reflexive real Banach space, let be two continuously differentiable functionals with coercive, i.e. and a sequentially weakly lower semicontinuous functional and a sequentially weakly upper semicontinuous functional. For every , let us put
[TABLE]
and
[TABLE]
Let for all . Then
- i)
If then, for each , the following alternative holds: either
there is a global minimum of which is a local minimum of , or
there is a sequence of pairwise distinct critical points of , with , which weakly converges to a global minimum of .
- ii)
If then, for each , the following alternative holds: either
* has a global minimum, or*
there is a sequence of critical points of , with .
Proof of Theorem 3.
Let be as in (2.2) and be as in (2.4). By Lemma 6, is a continuously differentiable, sequentially weakly lower semicontinuous functional on and by Proposition 3 it is easy to see that is coercive. is a continuously differentiable and sequentially weakly upper semicontinuous functional on . Let be as in (2.6). To apply Theorem 8 to the function , we show that . By the definition of we get that . Let
[TABLE]
By the definition of and assumption we get that , where is given by (1.3). Let be a sequence such that and
[TABLE]
Set
[TABLE]
for every . Then, and , as . By Proposition 3, if and for , then we have and hence
[TABLE]
and
[TABLE]
From above and we have
[TABLE]
for . Hence
[TABLE]
Now, we show that the point in Theorem 8 does not hold, i.e. that the unique global minimum 0 of is not a local minimum of .
Firstly, we assume that is satisfied. Without lost of generality, there exist sequence and sequence such that , and
[TABLE]
for any . Now we define a sequence , where for , and otherwise. It is easy to see that for any , ,
[TABLE]
and . We prove that . Notice that form (3.4) we have
[TABLE]
Without lost of generality, we may assume that for all . Since and the limit in (3.8) is finite, we get
[TABLE]
Therefore, we get uniformly for all . Hence and so
Now, we assume that holds. There exist and sequence such that and
[TABLE]
for any . Moreover, we define a sequence , where for , and otherwise. As in the previous part, we can show that , and , as .
From above it follows that 0 is not a local minimum of and, by from Theorem 8, there is a sequence of pairwise distinct critical points of with . The last means that , and by Proposition 3, strongly converges to zero. ∎
Proof of Theorem 4.
In analogously way to the proof of Theorem 3 we can prove that and we exclude condition in Theorem 8. It means that does not have a global minimum and, by from Theorem 8, there is a sequence of critical points of with . By Propositions 2 and 3 we have . The proof is complete. ∎
3.3 Examples
Example 1**.**
Consider a problem
[TABLE]
where is any bounded sequence such that . Fix such that . For , is defined by
[TABLE]
with sequences defined by
[TABLE]
Here is the indicator of . It is easily seen that is satisfied, are continuous for and is satisfied. Notice that is a decreasing sequence of positive numbers which is convergent to 0 and for , hence it is enough to check for . For we get that
[TABLE]
for such that for any . Notice that and
[TABLE]
and
[TABLE]
It means that conditions are satisfied and by Theorem 3 problem (3.5) with (3.6) and (3.7) admits a sequence of positive solutions in whose norms tend to zero. Notice that is not satisfied in this case.
Remark 2**.**
Fix . If we define for and
[TABLE]
with sequences defined in (3.7), then , , and are satisfied, but is not satisfy.
Remark 3**.**
Let us note that Theorem 2 and Theorem 3 are independent. Let . Fix such that . Let us replace in (3.7) by
[TABLE]
Then functions given by (3.6) are continuous for and , are satisfied. An easy computation shows that
[TABLE]
and
[TABLE]
This means that we can not apply Theorem 3, but Theorem 2 works. On the other hand, it is easy to see that we can modify in the way, that for some (or even infinitely many) we have for all and the limits (3.8), (3.9) do not change. Therefore, such do not satisfy and can not be used in Theorem 2.
Example 2**.**
Consider problem (3.5) with where is any bounded sequence such that . Fix such that . For is defined by
[TABLE]
with sequences defined by
[TABLE]
It is easily seen that are continuous for and is satisfied. For any the set is finite, so is satisfied. Notice that and
[TABLE]
and
[TABLE]
It means that conditions are satisfied and by Theorem 4 problem (3.5) with (3.10) and (3.11) admits a sequence of positive solutions in whose norms tend to infinity. Notice that is not satisfied in this case.
Remark 4**.**
In analogously way to Remarks 2 and 3 we can construct an example which show that Theorems 1, 4 are independent and an example of a function which satisfies , , , and does not satisfy .
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