Periodic Boundary Value Problems for First-Order Impulsive Functional Integrodifferential Equations with Integral-Jump Conditions

Definition 1. We say that the functions α, β   ∈ E are lower and upper solutions of PBVP (4) if there exist M > 0, W ≥ 0, N ≥ 0, L ≥ 0, L_k ≥ 0, and 0 ≤ σ_k ≤ τ_k ≤ t_k − t_k−1 such that

where where

Denote l = max⁡{k : t ≥ t_k, k = 1,2, …}. We prove the comparison principle by using the following lemma (see [20]).

Lemma 2. Let r ∈ {t₀, t₁, …, t_m}, c_k ≥ 0,and0 ≤ σ_k ≤ τ_k ≤ t_k − t_k−1, γ_k, k = 1, …, m, be constants and let q ∈ PC(J, R), x ∈ PC¹(J, R). If

then, for t ∈ (r, T],

Lemma 3. Assume that x ∈ E satisfies

where constants M > 0, W ≥ 0, N ≥ 0, L ≥ 0, L_k ≥ 0, and 0 ≤ σ_k ≤ τ_k ≤ t_k − t_k−1, k = 1,2, …, m. If where $$, t ∈ J₀, and $$ are given by Definition 1 with β = x and then x(t) ≤ 0, t ∈ J.

Proof. Case 1. One has x(0) ≥ x(T). Suppose that there exists t^* ∈ J such that x(t^*) > 0 and distinguish two cases.

Case (a). x(t) ≥ 0 for all t ∈ J, x≢0; then

so that x is nondecreasing in J, and then x(T) ≥ x(0). Since x(0) ≥ x(T), then x is a constant function x(t) = C > 0, which implies that getting a contradiction.

Case (b). x(t) < 0 for some t ∈ J. Let inf⁡_t∈Jx(t) = −λ < 0; then there exists $$, for some i such that $$ or $$. Without loss of generality, we only consider $$, and for the case $$ the proof is similar.

From (11), it is easy to see that

We consider the inequalities From Lemma 2, we have

Let $$ in (18); then

so that

If x(0) > 0, then (20) with A_k ≥ 1, B_k ≥ 0 for all k implies

This contradicts the condition (12).

Suppose that x(0) ≤ 0. If $$ for t^* ∈ (t_ν, t_ν+1], then Lemma 2 provides that

Since and integrating (11) from t^* into t_ν+1, we obtain Hence, We note that if t ∈ (t_v, t_v+1], then t_l = t^*.

Let $$ in (25); then

From (26), we have which gives contradicting condition (12). For the case $$, the proof is similar.

Case 2. One has x(0) < x(T). Set v(t) = x(t) + ((T − t)/T)(x(T) − x(0)). It follows that v(0) = x(T) = v(T), and for t ∈ J₀,

and for t = t_k, k = 1,2, …, m, In view of Case 1, we see that v(t) ≤ 0 on J, and therefore x(t) ≤ 0 on J. This completes the proof.

Corollary 4. Assume that x ∈ E satisfies

where constants M > 0, W ≥ 0, N ≥ 0, L ≥ 0, L_k ≥ 0, and 0 ≤ σ_k ≤ τ_k ≤ t_k − t_k−1, k = 1,2, …, m. If where and $$, A_k are given by Lemma 3, then x(t) ≤ 0, for t ∈ J.

Corollary 5. Assume that x ∈ E satisfies

where constants M > 0, W ≥ 0, N ≥ 0, L ≥ 0, L_k ≥ 0, and 0 ≤ σ_k ≤ τ_k ≤ t_k − t_k−1, k = 1,2, …, m, and condition (32) holds. Then x(t) ≤ 0, for t ∈ J.

Lemma 6. A function x ∈ E is a solution of (35) if and only if x ∈ PC(J, R) is a solution of the following impulsive integral equation:

where $$ and

Proof. If x(t) is a solution of (35), by directly integrating, we obtain

If x(t) is a solution of the above-mentioned integral equation, then

The proof is complete.

Lemma 7. Let M > 0, W ≥ 0, N ≥ 0, L ≥ 0, L_k ≥ 0, 0 ≤ σ_k ≤ τ_k ≤ t_k − t_k−1, θ ∈ (J, J), I_k ∈ C(J, R), k = 1,2, …, m, g ∈ PC(J, R), and η ∈ PC¹(J, R) and assume that

Then problem (35) has a unique solution in PC(J, R).

Proof. We define the mapping F : PC(J, R) → PC(J, R) by

for any x ∈ PC(J, R) and G is given by Lemma 6. Then The above result and condition (40) imply that F is a contractive mapping, which completes the proof.

Corollary 8. Let M > 0, W ≥ 0, N ≥ 0, L ≥ 0, L_k ≥ 0, 0 ≤ σ_k ≤ τ_k ≤ t_k − t_k−1, θ ∈ (J, J), I_k ∈ C(J, R), k = 1,2, …, m, g ∈ PC(J, R), and η ∈ PC¹(J, R) and assume that

Then problem (35) has a unique solution in PC¹(J, R).

Theorem 9. Assume the existence of lower and upper solutions for PBVP (4) and also suppose that the following conditions hold.

(H₁) The function f ∈ C(J × R⁴, R) satisfies

where β(t) ≤ x₁ ≤ y₁ ≤ z₁ ≤ α(t), β(θ(t)) ≤ x₂ ≤ y₂ ≤ z₂ ≤ α(θ(t)), (Kβ)(t) ≤ x₃ ≤ y₃ ≤ z₃ ≤ (Kα)(t), and (Sβ)(t) ≤ x₄ ≤ y₄ ≤ z₄ ≤ (Sα)(t), t ∈ J, where M > 0, W ≥ 0, N ≥ 0, L ≥ 0, and θ ∈ C(J, J).

(H₂) The functions I_k ∈ C(R, R) satisfy

where $$, k = 1,2, …, m, where L_k ≥ 0 and 0 ≤ σ_k ≤ τ_k ≤ t_k − t_k−1, k = 1,2, …, m.

If inequalities (12) and (40) hold, then there exists a solution x of PBVP (4) such that β(t) ≤ x(t) ≤ α(t), for t ∈ J.

Proof. We consider the following modified problem relative to PBVP (4):

where g_q(t) = f(t, q(t), q(θ(t)), (Kq)(t), (Sq)(t)) − Mq(t) − Wq(θ(t)) − N(Kq)(t) − L(Sq)(t) and If x ∈ E is such that β ≤ x ≤ α on J, then x is a solution of PBVP (4) if and only if x is a solution of (46). We will show that (46) is solvable and that every solution of (46) satisfies β ≤ x ≤ α on J. Suppose that x ∈ E is a solution of (46). We will show that β ≤ x. Let p = β − x. Then, we have p(0) − p(T) = β(0) − β(T) since x(0) = x(T) and By Lemma 3, we get p(t) ≤ 0 on J; that is, β ≤ x. Similar arguments show that x ≤ α.

It remains to prove that (46) possesses at least one solution. By Lemma 6, PBVP (46) is equivalent to the following impulsive integral equation:

where $$. For any x ∈ E, define a continuous compact operator F by Let δ > 0, such that |α(t)| < δ, |β(t)| < δ, t ∈ J, and take the compact setsB = {(t, y₁, y₂, y₃, y₄) ∈ R⁵ : t ∈ J, β(t) ≤ y₁ ≤ α(t), β(θ(t)) ≤ y₂ ≤ α(θ(t)), (Kβ)(t) ≤ y₃ ≤ (Kα)(t), and (Sβ)(t) ≤ y₄ ≤ (Sα)(t)}. Since f is continuous, then we can choose a constant ρ > 0, such that |f(t, y₁, y₂, y₃, y₄)| ≤ ρ, (t, y₁, y₂, y₃, y₄) ∈ B. For λ ∈ (0,1), we see that any solution of satisfies From the continuity of I_k, k = 1,2, …, m, and β(t) ≤ q(t, x(t)) ≤ α(t) on J, we can choose some ω > 0 such that |e_k | < ω, k = 1,2, …, m. Then we have where Hence, by Schaefer’s theorem [21], we get that F has at least a fixed point x ∈ E, which is a solution of (46). Such a solution lies between β and α and, consequently, is a solution of (4). Thus, the proof is complete.

Theorem 10. Assume that there exist lower and upper solutions for PBVP (4) and assume the following.

(H₃) The function f ∈ C(J × R⁴, R) satisfies

where $$, $$, $$, and $$, t ∈ J, where M > 0, W ≥ 0, N ≥ 0, L ≥ 0, and θ ∈ C(J, J).

(H₄) The functions I_k ∈ C(R, R) satisfy

where $$, k = 1,2, …, m, where L_k ≥ 0 and 0 ≤ σ_k ≤ τ_k ≤ t_k − t_k−1, k = 1,2, …, m.

Suppose that inequalities (12) and (40) hold. Then there exist monotone sequences {α_n}, {β_n} with α₀ = α, β₀ = β, which converge uniformly on J to the extremal solutions of the periodic boundary value problem (4) in [β, α].

Proof. For any η ∈ [β, α], consider PBVP (35) with

By Lemmas 6 and 7, PBVP (35) possesses a unique solution x ∈ E. We define an operator A by u = Aη; then the operator A has the following properties:

• (i)

  β ≤ Aβ, Aα ≤ α;

• (ii)

  Aη₁ ≤ Aη₂ for any η₁, η₂   ∈ [β, α] with η₁ ≤ η₂.

First we prove (i). Let p = β₀ − β₁, where β₁ = Aβ₀. Then, we have p(0) − p(T) = β₀(0) − β₀(T) since β₁(0) = β₁(T) and By Lemma 3, we get p(t) ≤ 0 on J; that is, β ≤ Aβ. Analogously, we have Aα ≤ α.

Now, we claim (ii). Set u₁ = Aη₁, u₂ = Aη₂, where η₁, η₂ ∈ [β, α] with η₁ ≤ η₂. Let p = u₁ − u₂; by (H₃)-(H₄), we have

By Lemma 3, we have p(t) ≤ 0 on J and so u₁ ≤ u₂. Thus we may define the sequences {α_n}, {β_n} by α_n+1 = Aα_n, β_n+1 = Aβ_n, α₀ = α, and β₀ = β. From (i), (ii), we obtain and each α_n, β_n ∈ E  (∀n ∈ N) satisfies where Hence, there exist ξ, ψ such that lim⁡_n→∞α_n(t) = ψ(t) and lim⁡_n→∞β_n(t) = ξ(t) uniformly on J. Clearly, ξ, ψ satisfy PBVP (4). We will prove that ξ, ψ are extreme solutions of PBVP (4). Let x(t) be any solution of PBVP (4), which satisfies β(t) ≤ x(t) ≤ α(t), t ∈ J. Also suppose there exists a positive integer n such that for t ∈ J, β_n(t) ≤ x(t) ≤ α_n(t). Setting p(t) = β_n+1(t) − x(t), then for t ∈ J, According to Lemma 3, we get that p(t) ≤ 0 on J. Similarly, we obtain x(t) ≤ α_n+1(t) on J. Since β₀ ≤ x(t) ≤ α₀(t) on J, by induction, we have β_n(t) ≤ x(t) ≤ α_n(t) on J for all n. Therefore, ξ(t) ≤ x(t) ≤ ψ(t) on J by taking limit as n → ∞. The proof is complete.

Example 11. Consider the following periodic boundary value problem:

where k(t, s) = ts³, h(t, s) = t³s, m = 1, t₁ = 1/2, σ₁ = 1/4, τ₁ = 1/3, and T = 1.

Obviously, α₀ = 0, β₀ = −5 are lower and upper solutions for (64), respectively, and β₀ ≤ α₀.

Let

We have where $$, $$, $$, and $$, t ∈ J. It is easy to see that where $$, k = 1.

Taking L₁ = 1/2, M = 1/6, W = 1/16, N = 1/3, and L = 1/4, it follows that

Thus, Therefore, (64) satisfies all conditions of Theorem 10. So PBVP (64) has minimal and maximal solutions in the segment [β₀, α₀].