Boundary value problems for a second-order difference equation involving the mean curvature operator

In this paper, we consider the existence of multiple solutions for discrete boundary value problems involving the mean curvature operator by means of Clark’s Theorem, where the nonlinear terms do not need any asymptotic and superlinear conditions at 0 or at infinity. Further, the existence of a positive solution has been considered by the strong comparison principle. As an application, some examples are given to illustrate the obtained results.

Because of the wide applications of difference equations in various research fields such as computer science, economics, biology, and other fields [6][7][8][9][10][11], many authors have obtained excellent results for difference equations, for example, positive solutions [12][13][14][15], homoclinic solutions [16][17][18][19][20][21], and ground-state solutions [22,23]. In particular, Guo and Yu [24] first used the critical-point theory to study the existence of a periodic solution for the following discrete problem where is the forward difference operator defined by u(t) = u(t + 1)u(t), 2 u(t) = ( u(t)), f (t, ·) ∈ C(R, R) for each t ∈ Z. Also, the critical-point theory is an important tool to deal with the existence of solutions for the discrete boundary value problems [25][26][27]. However, few works have been done concerning the discrete problems (1). In [15], Zhou and Ling proved the existence results for the boundary value problem where T is a given positive integer, f (t, ·) ∈ C(R, R) for each t ∈ Z(1, T). Under some suitable oscillating assumption on the nonlinearity f at infinity, they investigated the existence of infinitely many positive solutions.
The aim of this paper is to study the existence of multiple solutions for the following nonlinear difference equations with mean curvature operator where q(t) ∈ R + for each t ∈ Z(1, T) and λ > 0 is a positive parameter. Based on a version of Clark's Theorem [28,29], we investigate the existence of multiple solutions of (4).
Obviously, if u is a solution of (4), then -u is also a solution of (4) by (a 3 ). We say that ±u is a pair of solutions.

Preliminaries
Consider the T-dimensional real space From functional analysis theory, we know that E is a real Banach space. Moreover, we define the following two equivalent norms on E, Let J be a C 1 functional on E. A sequence {u n } ⊂ E is called a Palais-Smale sequence (P.S. sequence for short) for J if {J(u n )} is bounded and J (u n ) → 0 as n → ∞. We say J satisfies the Palais-Smale condition (P.S. condition for short) if any P.S. sequence for J possesses a convergent subsequence in E.
Let θ be the zero element of Banach space E. Let denote the family of sets A ⊂ E\{θ } such that A is closed in E and symmetric with respect to θ , i.e., u ∈ A implies -u ∈ A.
The following Clark's Theorem will be used to prove our main result. 28,29]) Let E be a real Banach space, J ∈ C 1 (E, R) with J even, bounded from below, and satisfying the P.S. condition. Suppose J(θ ) = 0, there is a set K ⊂ such that K is homeomorphic to S j-1 by an odd map, and sup K J < 0. Then, J possesses at least j distinct pairs of critical points.
First, the following comparison principle is necessary for the positive solutions.
The proof is completed.
Proof By the above assumptions, we have Combining with the monotonicity of φ c , we have u(t ± 1) = 0. The proof is completed.
In particular, let v = 0 in Lemma 2.2, the strong comparison principle is given by the two lemmas above.
Define the functionalĴ on E as follows: It is easy to verify thatĴ ∈ C 1 (E, R) and is even. By using u(0) = u(T + 1) = 0, we can compute the Frećhet derivative, for all u, v ∈ E. It is clear that the critical points ofĴ are the solutions of problem (12). In what follows, we will prove thatĴ has at least T distinct pairs of nonzero critical points by Lemma 2.1.
For any sequence {u n } ⊂ E, if {Ĵ(u n )} is bounded andĴ (u n ) → 0 as n → +∞, we claim that {u n } is bounded. In fact, there exists a positive constant C ∈ R such that |Ĵ(u n )| ≤ C. Since E is a finite-dimensional real Banach space, there is u 2 ≤ u ≤ 2 u 2 for all u ∈ E (see [30]). Assume that u n → +∞ as n → +∞, then ]. This is impossible, since C is a fixed constant. Thus, {u n } is bounded in E. This implies that {u n } has a convergent subsequence. Then, the functionalĴ satisfies the P.S. condition. Moreover, the coerciveness ofĴ, be a base of E and e i = 1 for each i ∈ Z(1, T). We define Obviously, θ / ∈ A(ρ), A(ρ) is closed in E and symmetric with respect to θ . We note that A(ρ) is homeomorphic to S T-1 for any ρ > 0. For u ∈ A(ρ), we see that For u ∈ A( α T+1 ), we note that u = θ andf (t, u(t)) = f (t, u(t)). By (a 2 ) and (a 3 ), then T t=1F (t, u(t)) andλ = (2+q * )α 2 2Tτ . By (a 2 ), we know τ > 0. If λ >λ, then where q * = max t∈Z(1,T) q(t). Since all conditions of Lemma 2.1 hold, problem (4) admits at least T distinct pairs of nontrivial solutions. The proof is completed.
Proof We consider the following problem Define the variational framework of problem (15) then we have Hence, J q * is coercive on E and has a global minimum point u 0 that is its critical point. Combining with Lemma 2.4, u 0 is a positive solution of problem (15). We take ε > 0 so small that u 1 (t) = εu 0 (t) < α. Define a continuous function as follows: By (14), there exist a μ ∈ [0, μ 1 + q * ) and M 1 > u 1 (t) such that Thus, and lim sup Next, we claim that the following problem admits a positive solution u and u > u 1 > 0. We define the following variational framework corresponding to problem (19) Using (18), there is one positive constant M such that Let η = μ 1 +q * μ . For η > λ > 0, we havê This shows thatĴ is also coercive. As the functional is coercive and continuous, it has a global minimum point u ∈ E, which is a critical point. By (17) and Lemma 2.5, u is a positive solution. Moreover, if we can show u > u 1 , then u must be one positive solution of problem (4). First, we assume that u ≤ u 1 for every t ∈ Z(1, T). Since by Lemma 2.2, we obtain u ≥ u 0 > u 1 , contradicting the assumption above. Secondly, we consider that u and u 1 are not ordered vectors. There exist some j 0 ∈ Z(1, T) such that u(j 0 ) < u 1 (j 0 ). Let j := max j 0 |j 0 ∈ Z(1, T) such that u(j 0 ) < u 1 (j 0 ) .
We note that u 1 (i -1) = 0 or u 1 (i) = 0 still satisfies the above cases. Obviously, when ε is taken sufficiently small, (23), (24), and (25) cannot hold. These are contradictions. Thus, u ≥ u 1 . u is one positive solution of problem (4). Moreover, we see that -u is a negative solution of problem (4) because of (a 3 ). This completes the proof.
In fact, in [30], such a problem can be found when κ = 0 and q(t) = 0 for each t ∈ Z(1, T) in Corollary 5.1. We see that the conditions of Theorem 3.1 are different from the conditions of Corollary 5.1 of [30], and we find more solutions of problem (26).