Multiple Positive Solutions for First-Order Impulsive Integrodifferential Equations on the Half Line in Banach Spaces

Remark 2.1. Obviously, condition (H₄) is satisfied automatically when E is finite dimensional.

Remark 2.2. It is clear that if condition (H₁) is satisfied, then the operators T and S defined by (1.2) are bounded linear operators from BPC[J, E] into BPC[J, E] and | | T|| ≤ k^*,   | | S|| ≤ h^*; moreover, we have T(BPC[J, P]) ⊂ BPC[J, P] and S(BPC[J, P]) ⊂ BPC[J, P].

Lemma 2.3. If conditions (H₁)–(H₄) are satisfied, then operator A defined by (2.5) is a completely continuous (i.e., continuous and compact) operator from BPC[J, P] into Q.

Proof. Let r > 0 be given. Let

For u ∈ BPC[J, P],   | | u | |_B ≤ r, we see that by virtue of condition (H₂) and (2.6), which implies the convergence of the infinite integral On the other hand, condition (H₃) and (2.7) imply the convergence of the infinite series It follows from (2.5), (2.10), and (2.12) that which implies that Au ∈ BPC[J, P] and Moreover, by (2.5), we have It is clear that so, (2.16) and (2.17) imply It follows from (2.15) and (2.18) that Hence, Au ∈ Q. That is, A maps BPC[J, P] into Q.

Now, we are going to show that A is continuous. Let $$ (n → ∞). Then r = sup _n| | u_n | |_B < ∞ and $$. Similar to (2.14), it is easy to get

It is clear that Moreover, we see from (2.8) that It follows from (2.21), (2.22) and the dominated convergence theorem that On the other hand, for any ϵ > 0, we can choose a positive integer j such that And then, choose a positive integer n₀ such that From (2.24) and (2.25), we get hence It follows from (2.20), (2.23), and (2.51) that $$, and the continuity of A is proved.

Finally, we prove that A is compact. Let V = {u_n} ⊂ BPC[J, P] be bounded and | | u_n | |_B ≤ r (n = 1,2, 3, …). Consider J_i = (t_i−1, t_i] for any fixed i. By (2.5) and (2.8), we have

which implies that the functions {w_n(t)} (n = 1,2, 3, …) defined by ($$ denotes the right limit of (Au_n)(t) at t = t_i−1) are equicontinuous on $$. On the other hand, for any ϵ > 0, choose a sufficiently large τ > η and a sufficiently large positive integer j > m such that We have, by (2.29), (2.5), (2.8), (2.30), and condition (H₃), It follows from (2.31), (2.32), (2.33), (2.8), and [4, Theorem 1.2.3] that where W(t) = {w_n(t) : n = 1,2, 3, …}, V(s) = {u_n(s) : n = 1,2, 3, …}, (TV)(s) = {(Tu_n)(s) : n = 1,2, 3, …}, (SV)(s) = {(Su_n)(s) : n = 1,2, 3, …} and α(U) denotes the Kuratowski measure of noncompactness of bounded set U ⊂ E (see [4, Section 1.2]). Since $$ for s ∈ J, where r^* = max {r, k^*r, h^*r}, we see that, by condition (H₄), It follows from (2.34)–(2.36) that which implies by virtue of the arbitrariness of ϵ that α(W(t)) = 0 for $$.

By Ascoli-Arzela theorem (see [4, Theorem 1.2.5]), we conclude that W = {w_n : n = 1,2, 3, …} is relatively compact in $$; hence, {w_n(t)} has a subsequence which is convergent uniformly on $$, so, {(Au_n(t)} has a subsequence which is convergent uniformly on J_i. Since i may be any positive integer, so, by diagonal method, we can choose a subsequence $$ of {(Au_n)(t)} such that $$ is convergent uniformly on each J_k (k = 1,2, 3, …). Let

It is clear that v ∈ PC[J, P]. By (2.14), we have which implies that v ∈ BPC[J, P] and Let ϵ > 0 be arbitrarily given and choose a sufficiently large positive number τ such that For any τ < t < ∞, we have, by (2.5), which implies by virtue of (2.8), condition (H₃) and (2.41) that Letting i → ∞ in (2.43), we get On the other hand, since $$ converges uniformly to v(t) on [0, τ] as i → ∞, there exists a positive integer i₀ such that It follows from (2.43)–(2.45) that By (2.45) and (2.46), we have hence $$ as i → ∞, and the compactness of A is proved.

Lemma 2.4. Let conditions (H₁)–(H₄) be satisfied. Then u ∈ BPC[J, P]∩C¹[J′, E] is a solution of IBVP(1.5) if and only if u ∈ Q is a solution of the following impulsive integral equation:

that is, u is a fixed point of operator A defined by (2.5) in Q.

Proof. For u ∈ PC[J, E]∩C¹[J′, E], it is easy to get the following formula:

Let u ∈ BPC[J, P]∩C¹[J′, E] be a solution of IBVP(1.5). By (1.5) and (2.49), we have

We have shown in the proof of Lemma 2.3 that the infinite integral (2.9) and the infinite series (2.11) are convergent, so, by taking limits as t → ∞ in both sides of (2.50), we get On the other hand, by (1.5) and (2.50), we have It follows from (2.51)–(2.53) that and, substituting it into (2.50), we see that u(t) satisfies (2.48), that is, u = Au. Since Au ∈ Q by virtue of Lemma 2.3, we conclude that u ∈ Q.

Conversely, assume that u ∈ Q is a solution of (2.48). We have, by (2.48),

Moreover, by taking limits as t → ∞ in (2.33), we see that u(∞) exists and It follows from (2.55)–(2.57) that On the other hand, direct differentiation of (2.48) gives and, it is clear, by (2.48), Hence, u ∈ C¹[J′, E] and u(t) satisfies (1.5).

Corollary 2.5. Let cone P be normal. If u is a fixed point of operator A defined by (1.5) in Q and | | u | |_B > 0, then u(t) > θ for t ∈ J, so, u is a positive solution of IBVP(1.5).

Proof. For u ∈ Q, we have

so, where N denotes the normal constant of P. Since | | u | |_B > 0, (2.61) and (2.62) imply that u(t) > θ for t ∈ J.

Lemma 2.6 (Fixed point theorem of cone expansion and compression with norm type, see [3, Theorem  3] or [1, Theorem  2.3.4]). Let P be a cone in real Banach space E and Ω₁, Ω₂ two bounded open sets in E such that θ ∈ Ω₁, $$, where θ denotes the zero element of E and $$ denotes the closure of Ω₂. Let operator $$ be completely continuous. Suppose that one of the following two conditions is satisfied:

• (a)
  $$ (2.63)
  where ∂Ω_i denotes the boundary of Ω_i (i = 1,2).

• (b)
  $$ (2.64)

Then A has at least one fixed point in $$.

Remark 3.1. Condition (H₅) means that f(t, u, v, w) is superlinear with respect to u.

•  

  H₆ There exist u₁ ∈ P∖{θ},   c ∈ C[J, R₊], and σ ∈ C[P, R₊] such that

  $$ (3.2)

Theorem 3.2. Let cone P be normal and conditions (H₁)–(H₆) satisfied. Assume that there exists a ξ > 0 such that

where (for g(x, y, z), F(x), a^* and γ^*, see conditions (H₂) and (H₃)). Then IBVP(1.5) has at least two positive solutions u^*, u^** ∈ Q∩C¹[J′, E] such that 0<| | u^* | |_B < ξ<| | u^** | |_B.

Proof. By Lemmas 2.3, 2.4, and Corollary 2.5, operator A defined by (2.5) is completely continuous from Q into Q, and we need to prove that A has two fixed points u^* and u^** in Q such that 0<| | u^* | |_B < ξ<| | u^** | |_B.

By condition (H₅), there exists an r₁ > 0 such that

where N denotes the normal constant of P, so, Choose For u ∈ Q,   | | u | |_B = r₂; we have by (2.62) and (3.7), so, (2.5), (3.8), (3.6), and (2.62) imply and consequently, Similarly, by condition (H₆), there exists r₃ > 0 such that so, Choose For u ∈ Q,   | | u | |_B = r₄, we have by (3.13) and (2.62), so, similar to (3.9), we get by (2.5), (3.12), and (3.14) hence On the other hand, for u ∈ Q, | | u | |_B = ξ, by condition (H₂), condition (H₃), (3.4), we have It is clear that It follows from (3.17)–(3.19) that Thus, (3.20) and (3.3) imply From (3.7) and (3.13), we know 0 < r₄ < ξ < r₂; hence, (3.10), (3.16), (3.21), and Lemma 2.6 imply that A has two fixed points u^*, u^** ∈ Q such that r₄<| | u^* | |_B < ξ<| | u^** | |_B < r₂. The proof is complete.

Theorem 3.3. Let cone P be normal and conditions (H₁)–(H₅) satisfied. Assume that

(for g(x, y, z) and F(x), see conditions (H₂) and (H₃)). Then IBVP(1.5) has at least one positive solution u^* ∈ Q∩C¹[J′, E].

Proof. As in the proof of Theorem 3.2, we can choose r₂ > 0 such that (3.10) holds (in this case, we only choose r₂ > Nβ(1 − γ) ⁻¹r₁ instead of (3.7)). On the other hand, by (3.22), there exists r₅ > 0 such that

where Choose For u ∈ Q,   | | u | |_B = r₆, we have by (2.62) and (3.25), so, (3.23) imply It follows from (3.19), condition (H₂), condition (H₃), (3.27), and (3.24) that and consequently, Since 0 < r₆ < r₂ by virtue of (3.25), we conclude from (3.10), (3.29), and Lemma 2.6 that A has a fixed point u^* ∈ Q such that r₆≤| | u^* | |_B ≤ r₂. The theorem is proved.

Example 3.4. Consider the infinite system of scalar first-order impulsive integrodifferential equations of mixed type on the half line:

Evidently, u_n(t) ≡ 0    (n = 1,2, 3, …) is the trivial solution of infinite system (3.30).

Conclusion. Infinite system (3.30) has at least two positive solutions $$ (n = 1,2, 3, …) and $$ (n = 1,2, 3, …) such that

Proof. Let E = l¹ = {u = (u₁, …, u_n, …) : ∑_n=1∞ | u_n | < ∞} with norm $$ and P = (u₁, …, u_n, …) : u_n ≥ 0,     n = 1,2, 3, …}. Then P is a normal cone in E with normal constant N = 1, and infinite system (3.30) can be regarded as an infinite three-point boundary value problem of form (1.5). In this situation, u = (u₁, …, u_n, …), v = (v₁, …, v_n, …), w = (w₁, …, w_n, …), t_k = k  (k = 1,2, 3, …), K(t, s) = e^−(t+1)s, H(t, s) = (1 + t + s) ⁻², η = 9/2, γ = 1/2, β = 6, f = (f₁, …, f_n, …), and I_k = (I_k1, …, I_kn…), in which

It is easy to see that f ∈ C[J × P × P × P, P], I_k ∈ C[P, P] (k = 1,2, 3, …), and condition (H₁) is satisfied and k^* ≤ 1,     h^* ≤ 1. We have, by (3.32), so, observing the inequality $$, we get which implies that condition (H₂) is satisfied for a(t) = e^−5t( ^* = 1/5) and By (3.33), we have so, condition (H₃) is satisfied for γ_k = 3^−k−1(γ^* = 1/6) and On the other hand, (3.32) implies so, we see that condition (H₅) is satisfied for b(t) = (1/8)e^−5t(b^* = 1/40), τ(u) = | | u | |² and u₀ = (1, …, 1/n², …), and condition (H₆) is satisfied for c(t) = (1/8)e^−5t(c^* = 1/40), $$, and let u₁ = (1, …, 1/n², …). Now, we check that condition (H₄) is satisfied. Let t ∈ J and r > 0 be fixed, and {z^(m)} be any sequence in f(t, P_r, P_r, P_r), where $$. Then, we have, by (3.34), So, $$ is bounded, and, by diagonal method, we can choose a subsequence {m_i}⊂{m} such that which implies by virtue of (3.40) that Consequently, $$. Let ϵ > 0 be given. Choose a positive integer n₀ such that By (3.41), we see that there exists a positive integer i₀ such that It follows from (3.40)–(3.44) that Thus, we have proved that f(t, P_r, P_r, P_r) is relatively compact in E. Similarly, by using (3.37), we can prove that I_k(P_r) is relatively compact in E. Hence, condition (H₄) is satisfied. Finally, it is easy to check that inequality (3.3) is satisfied for ξ = 1 (in this case, M_ξ ≤ 17/36 and N_ξ = 1). Hence, our conclusion follows from Theorem 3.2.

Example 3.5. Consider the infinite system of scalar first-order impulsive integrodifferential equations of mixed type on the half line:

Evidently, u_n(t) ≡ 0    (n = 1,2, 3, …) is the trivial solution of infinite system (3.46).

Conclusion. Infinite system (3.46) has at least one positive solution $$ (n = 1,2, 3, …) such that

Proof. Let $$ with norm $$ and P = {u = (u₁, …, u_n, …) ∈ l¹ : u_n ≥ 0,   n = 1,2, 3, …}. Then P is a normal cone in E with normal constant N = 1, and infinite system (3.46) can be regarded as an infinite three-point boundary value problem of form (1.5) in E. In this situation, u = (u₁, …, u_n, …), v = (v₁, …, v_n, …), w = (w₁, …, w_n, …), t_k = 2k (k = 1,2, 3, …), K(t, s) = (1 + ts + s²) ⁻¹, H(t, s) = e^−ssin²(t − s), η = 7, γ = 3/4, β = 1/2, f = (f₁, …, f_n, …), and I_k = (I_k1, …, I_kn, …), in which

It is clear that f ∈ C[J × P × P × P, P],   I_k ∈ C[P, P] (k = 1,2, 3, …), and condition (H₁) is satisfied and k^* ≤ π/2, h^* ≤ 1. We have so, condition (H₂) is satisfied for a(t) = (1 + t) ⁻³(a^* = (1/2)) and and (H₃) is satisfied for γ_k = 2^−k(γ^* = 1) and From we see that condition (H₅) is satisfied for b(t) = (1 + t) ⁻³    (b^* = 1/2),     τ(u) = | | u | |³, and u₀ = (1, …, 1/n³, …). Moreover, it is clear that (3.22) are satisfied. Similar to the discussion in Example 3.4, we can prove that f(t, P_r, P_r, P_r) and I_k(P_r) (for fixed t ∈ J and r > 0;     k = 1,2, 3, …) are relatively compact in E = l¹; so, condition (H₄) is satisfied. Hence, our conclusion follows from Theorem 3.3.