Reverse-Order Lower and Upper Functions for Periodic Problems of Second-Order Singular Difference Equations

Lemma 1. Let there exist positive functions α, β ∈ E, such that

and α(t) ≤ β(t) for t ∈ [1, T] _ℤ. Then there exists at least one positive T-periodic solution to (1).

A function α ∈ E (resp., β ∈ E) verifying (6) (resp, (7)) is called lower (resp, upper) function (solution) of (1). When the order between the lower and the upper functions is the inverse, an additional hypothesis is needed.

Definition 2. A function φ ∈ E and φ ≥ 0 is said to verify the property (P) if the implication

holds.

Lemma 3. Let there exist positive functions α, β ∈ E satisfying (6), (7), and β(t) ≤ α(t), t ∈ [1, T] _ℤ. Moreover, there exists φ ∈ E with the property (P), such that

where β(t) ≤ u(t) ≤ v(t) ≤ α(t) for t ∈ [0, T + 1] _ℤ. Then there exists at least one positive T-periodic solution to (1).

Proof. From the condition (9), it follows that

That is, the nonlinearity is increasing.

Define the operator T : E → E as the unique solution of problem (1) as follows:

where G(t, s) is the Green’s function of As φ satisfies the property (P), it follows that G(t, s) > 0. Now we divide the proof into three steps.

Step  1. We show

where K = {u ∈ E∣β ≤ u ≤ α} is a nonempty bounded closed subset in E.

In fact, for u ∈ K, set w = Tu(t). From the definitions of α, β, and K, combining with (9), we have

Using property (P), we get β ≤ w.

Analogously, we can prove that w ≤ α. Thus, (13) holds.

Step  2. Let u₁ = Tη₁, u₂ = Tη₂, where η₁, η₂ ∈ K satisfy β ≤ η₁ ≤ η₂ ≤ α. Then we claim that

In fact, let ω = u₂ − u₁, it follows from (10) and (11) that

Step  3. The sequences {α_n} and {β_n} are obtained by recurrence:

From the results of Steps  1 and  2, it follows that Moreover, from the definition of T, we get

This together with (18), we can easily get that there exists C_α depending only on α but not on t and n, such that |α_n | ≤ C_α, so {α_n} is bounded in E. Similarly, {β_n} is bounded in E. Therefore, we can conclude that {α_n} and {β_n} converge uniformly to the extremal solutions u of the problem (1). Subsequently, there exists at least one positive T-periodic solutions of (1).

Lemma 4 (see [7].)Let us assume φ ∈ E, φ≢0 and the following conditions holds:

Then φ verifies the property (P).

Lemma 5 ([13, Lemma  2.2]). Given v ∈ E, then

where Moreover, (21) is fulfilled as an equality if and only if v is a constant function.

Theorem 6. Let G > 0, F < 0, let functions w, σ ∈ E be such that the equalities

are fulfilled and let there exist x₀ ∈ (0, +∞) such that where Moreover, define and assume that φ(t) = μg(t)/β^1+μ(t) verifies the property (P). Then problem (1) has at least one positive T-periodic solution.

Proof. Let β be defined by (28). Then β ∈ E and in view of (24) and (25), we have

Moreover, according to (26) and (27) Now (29) and (30) imply Consequently, β is an upper function to (1).

Further, we can choose x₁ ∈ (0, x₀) such that

and put Then α ∈ E, and in view of (24) and (25) we have Moreover, according to (27) and (32) Now (34) and (35) imply Consequently, α is a lower function to (1) and according to (30) and (35) we have

Furthermore, note that the function

is nondecreasing for y ≥ β. Therefore we have whenever β(t) ≤ u(t) ≤ v(t) for t ∈ [1, T] _ℤ, hence, we get Thus, the assertion follows from Lemma 3.

Remark 7. Note that for every q ∈ E such that $$, the periodic solution v of the equation

is given by the Green formula: where c ∈ ℝ. Therefore, the periodic functions w and σ with properties (24) and (25) exist and, moreover, are unique up to a constant term, the value of which has no influence on the validity of the condition (26). A similar observation can be made in relation to the formulations of the theorems given below.

Theorem 8. Let μ > λ, H > 0, G > 0, F = 0, let functions w, σ ∈ E be such that (24) and (25) are fulfilled and let there exist x₀ ∈ (0, +∞) such that

where m_w and m_σ are defined by (27). Moreover, assume that φ(t) = μg(t)/β^1+μ(t) verifies the property (P), where β is given by (28). Then problem (1) has at least one positive T-periodic solution.

Proof. Note that the inequality μ > λ implies

Therefore, analogously to the proof of Theorem 6, one can show that there exist lower and upper functions α, β satisfying (37). Consequently, the assertion follows from Lemma 3 with φ(t) = μg(t)/β^1+μ(t).

Corollary 9. Let μ > λ, H > 0, G > 0, and let w ∈ E be such that (24) is fulfilled. Let

where m_w is given by (27) and Moreover, let us define and assume that φ(t) = μg(t)/β^1+μ(t) verifies the property (P). Then problem (2) has at least one positive solution.

Proof. Put f ≡ 0 and

Then the assertion follows from Theorem 8.

Corollary 10. Let G > 0, F < 0, let σ ∈ E be such that (25) is fulfilled, and let

where m_σ is defined by (27) and Moreover, if holds. Then problem (4) has at least one positive T-periodic solution.

Proof. Put H = 0,

and define a function β by (28). Let β(t) = (G/|F | ) ^1/μ − M_σ + σ(t) for t ∈ [1, T] _ℤ. Then (51) guaranties that φ(t) = μg(t)/β^1+μ(t) satisfies the property (P). Moreover, (51) yields (25). Therefore, the assertion follows from Theorem 6.

Corollary 11. Let G > 0, F < 0 and

Then problem (4) has at least one positive T-periodic solution.

Proof. According to Lemma 5,

Then (53) implies (51). Consequently, the assertion follows from Corollary 10.

Example 12. Let us consider the boundary value problem:

where T > 3 is an integer.

Obviously, (1/u)−(1/4)→+∞ as u → 0. Let g ≡ 1, f ≡ −1/4T, μ = 1. Then G = T, F = −1/4 and

Thus, from Corollary 11, problem (55) has at least one positive T-periodic solution.