Existence of solutions for delay evolution equations with nonlocal conditions

Lemma 2.1

If the semigroup T (t) (t ≥ 0) is equicontinuous and m ∈ L([0, a], ℝ⁺), then the set

is equicontinuous for t є [0, a].

For each u ∈ C([−q, a],X), the restriction of u on [0, a] denoted by u|_{[0, a]} is an element of C([0, a],X). For simplicity, we also write g(u|_{[0, a]}) as g(u). For any subset D of the space C([−q, a],X), denote D|_{[0, a]} = {u|_{[0, a]} : u ∈ D}. Similarly, we also write g(D|_{[0, a]}) as g(D).

Next, we recall some properties about the Kuratowski measure of noncompactness, which will be used in the sequel. In this article, we denote by α(.) and α_{[a, b]}(.) the Kuratowski measure of noncompactness on the bounded subsets of X and C([a, b],X), respectively. For any t ∈ [a, b] and D ⊂ C([a, b],X), set D(t) = {u(t) | u ∈ D} ⊂ X. If D ⊂ C([a, b],X) is bounded, then D(t) is bounded in X and α(D(t)) ≤ α[a, b](D). For the details of the definition and properties for the Kuratowski measure of noncompactness, we refer to the monographs [17] and [18].

To discuss the existence of mild solutions for the problem (1) we also need the following lemmas.

Lemma 2.2 ([17])

Let X be a Banach space, and let D ⊂ C([a, b],X) bounded and equicontinuous be a bounded and equicontinuous set. Then α (D(t)) is continuous on [a, b], and

Lemma 2.3 ([19])

Let X be a Banach space, and let D = {u_n} ⊂ C([a, b], X) be a bounded and countable set. Then α (D(t)) is Lebesgue integrable on [a, b], and

Lemma 2.4 ([4, 20])

Let X be a Banach space, and let D ⊂ X be bounded. Then there exists a countable subset D* ⊂ D, such that α(D) ≤ 2α(D*).

Lemma 2.5 ([21])

Let A with norm || • ||_A and C with norm || • ||_C be bounded sets in Banach space X. If there is surjective map Q : C → A such that for any c, d ∈ C one has ||Q(c) − Q(d)||_A ≤ ||c − d ||_C, then α(A) ≤ α(C).

In order to introduce the concept of mild solution for problem (1), by comparison with the linear initial value problem

whose properties are well known [22], we associate (1) to the integral equation

Definition 2.6

A continuous function u : [−q, a] → X is called a mild solution of the problem (1) if u₀ = ɸ + g(u) and the integral equation (4) is satisfied.

It is supposed that for each positive number R:

then D_R is clearly a bounded closed convex set in C([−q, a],X)). Obviously, if u ∈ D_R, then u|_{[0, a]} is an element of D_R(C([0,a],X)).

Theorem 3.1

Assume that the conditions (P₁), (P₂) and (P₃) are satisfied. Then for every ϕ ∈ C([−q, 0],X), problem (1) has at least one mild solution provided that

and

Proof

We consider the operator Q : C([−q, a],X) → C([−q, a],X) defined by

where

With the help of Definition 2.6, we know that the mild solution of problem (1) is equivalent to the fixed point of operator Q. Next, we shall show that Q has at least one fixed point by using the famous Sadovskii’s fixed point theorem, which can be found in [23]. To do this, we first prove that there exists a positive constant R₀ big enough such that Q(D_R0(C([−q, a],X))) ⊂ D_R0(C([−q, a],X)). For every u ∈ C([−q, a],X), by condition (P₁), we have

Choose a positive constant

For every u ∈ D_R0(C([−q, a],X)), then ||u_t||_{[−q, 0]} ≤ ||u||_{[−q, a]} ≤ R₀ for all t ∈ [0, a]. By the condition (P₃) we get that for t ∈ [−q, 0],

Hence,

On the other hand, by the condition (P₃) and (7) we know that for every t ∈ [0, a],

which means that

Notice that M ≤ 1 yields that K₁ ≤ K₂. Therefore, by (6) and (8) we obtain

Hence, we have proved that

Next, we will prove that Q is continuous on D_R0(C([−q, a],X)). Let u_n ⊂ D_R0(C([−q, a],X)) with u_n → u (n → ∞) in D_R0(C([−q, a],X)). Then u_nt → u_t as n → ∞ for all t ∈ [0, a], where u_nt = (u_n)_t . Applying the condition (P₃) we get that for each t ∈ [−q, 0],

By the facts u_nt → u_t when n → ∞ for every t ∈ [0, a] and the function f is continuous, we have

Combined this fact with Lebesgue dominated convergence theorem, we know that

Hence

which means that the operator Q is continuous on D_R0(C([−q, a],X)).

In what follows, we will prove that Q is a condensing operator. For this purpose, we firstly prove that Q₁ is Lipschitz continuous on D_R0(C([−q, a],X)). Taking u and υ in D_R0(C([−q, a],X)). Then u|_{[0, a]} and υ|_{[0, a]} in D_R0(C([0, a],X)), by the hypothesis (P₃) we get that

and

From the above discussion we get that

for all u, υ ∈ D_R0(C([−q, a],X)), which means that Q₁ : D_R0(C([−q, a],X)) → D_R0(C([−q, a],X)) is Lipschitz continuous with Lipschitz constant ML. Therefore, combining this fact with Lemma 2.5 we know that for any bounded subset D ⊂ D_R0(C([−q, a],X)),

On the other hand, by the fact that the semigroup T(t) (t ≥ 0) generated by −A is equicontinuous and Lemma 2.1, it is easily to see that {Q₂u : u ∈ D_R0(C([−q, a],X))} is a family of equicontinuous functions. Then for every bounded subset D ⊂ D_R0(C([−q, a],X)), we know that Q₂(D) is bounded and equicontinuous. It follows from Lemma 2.4 that there exists a countable set D* = {u_n} ⊂ D such that

For each t ∈ [0, a], we denote by

Then, for every u, υ ∈ D* we have u_t, υ_t ∈ D_t*, and

By (11) and Lemma 2.5 we know that

For every t ∈ [−q, 0], (Q₂D*)(t) = 0, and therefore α((Q₂D*)(t)) = 0. For every t ∈ [0, a], taking the assumption (P₂), Lemma 2.3 and (12) into account, we get that

Therefore, for every t ∈ [−q, a], we get that

Since Q₂(D*) is equicontinuous, by Lemma 2.1, we know that

Hence by (10) and (13), we have

For every subset D ⊂ D_R0(C([−q, a],X)), by (5), (9) and (14), we have

Therefore, Q : D_R0(C([−q, a],X)) → D_R0(C([−q, a],X)) is a condensing operator. By Sadovskii’s fixed point theorem, we know that the operator Q has at least one fixed point u in D_R0(C([−q, a],X)) which means that the problem (1) has at least one mild solution u ∈ (C([−q, a],X).This completes the proof of Theorem 3.1.   □

If the nonlinear term f and the nonlocal term g satisfy the following growth conditions:

(P₄)The function f: [0, a] × X × C([−q, 0],X) → X is continuous and there exist positive constants C¯₁, C¯₂, C¯₀ and γ ∈ [0,1) such that

(P₅)There exist positive constants N¯₁, N¯₀ and μ ∈[0,1) such that

we have the following existence result:

Theorem 3.2

Assume that the conditions (P₂),(P₃),(P₄) and (P₅) are satisfied. Then for every ɸ ∈ C([−q, 0],X) nonlocal problem (1) has at least one mild solution provided that

Proof

Let Q be the operator defined in the proof of Theorem 3.1. By conditions (P₃) and (P₄), one can use the same argument as in the proof of Theorem 3.1 to deduce that Q is continuous on C([−q, a],X) and the mild solution of problem (1) is equivalent to the fixed point of the operator Q.

Next, we will demonstrate that Q maps bounded sets of C([−q, a],X) into bounded sets. For any R > 0 and u ∈ D_R(C([−q, a],X)), by the conditions (P₄) and (P₅), we know that there exist constants C¯₃ and N¯₂ such that

for every t ∈ [0, a], and

Taking (15) and (16) into account, we obtain that

and

Hence,

It means that Q(D_R(C([−q, a],X))) is a bounded set in C([−q, a],X).

Using similar argument as getting the proof of Theorem 3.1 deduce that Q : C([−q, a],X) → C([−q, a],X) is a condensing operator.

Next, we show that the set S = {u ∈ C([−q, a],X) | u = λQu, 0 < λ < 1} is bounded. Let u ∈ S. Then u = λQu for every 0 < λ < 1. Applying the conditions (P₄) and (P₅) we get

for t ∈ [−q, 0], and

for t ∈ [0, a]. Hence,

By (17) and the fact that Ɣ, μ ∈ [0,1), we can prove that S is a bounded set. If this is not true, then for any u ∈ S we have that ||u||_{[−q, a]} → +∞. Dividing both sides of (17) by ||u||_{[−q, a]} and taking the limits as ||u||_{[−q, a]} → + ∞, we get

This is a contradiction. It means that the set S is bounded. By condensing mapping fixed point theorem of topological degree, which can be found in [24], there is a fixed point u of Q on C([−q, a],X), which is just the mild solution of problem (1). The proof is completed.   □

Theorem 4.1

Theorem 4.1. Assume that the conditions(P₂), (P₆) and (P₇) are satisfied. Then for every ɸ ∈ C([−q, 0],X), nonlocal problem (1) has at least one mild solution provided that

and

where c = || ɸ ||_{[−q, 0]} + N.

Proof

Consider the map Q : C([−q, a],X) → C([−q, a],X) defined in the proof of Theorem 3.1. It is easily seen that the fixed point of Q is equivalent to the mild solution of the problem (1). Similar to the proof of Theorem 3.1, we can show that Q is continuous by usual technique.

Firstly, we see that if u ∈ C([−q, a],X) with ||u||_{[−q, a]} ≤ k for some positive constant k, then ||u_t ||_{[−q, 0]} ≤ k for all t ∈ [0, a]. Therefore, by the conditions (P₆) and (P₇) we know that

and

This together with (20) imply that Q maps bounded subsets of C([−q, a],X) into bounded subsets.

By lemma 2.1 and the compactness of g involving Ascoli-Arzela’s theorem, it is easily seen that the operator Q maps bounded subsets of C([−q, a],X) into equicontinuous sets.

Next, we shall demonstrate that Q ia a condensing operator. Let D be abounded subset of C([−q, a],X). From the proof of Theorem 3.1, we know that

On the other hand, the compactness of g implies that

Therefore, for every subset D ⊂ D_R(C([−q, a],X)), due to (18), (21) and (22), we get that

which means that Q : C([−q, a],X) → C([−q, a],X) is a condensing operator.

Now, we show that the set

is bounded. Let u ∈ S. Then u= λQu for every 0 < λ < 1. Thus

for t ∈ [0, a]. It follows from the conditions (P₆) and (P₇) that

for t ∈ [0, a]. We consider the function ρ defined by

Then ||u(t)|| ≤ ρ(t) and ||u_t|| [−q, 0] ≤ ρ(t) for t ∈ [0, a]. By the previous inequality (23), we have

Let us take the right-hand side of the above inequality as υ(t). Then we have

and

Using the nondecreasing character of φ and (24) we get that

By (19) and (25), we have

It follows that there exists a constant L* such that υ(t) ≤ L*, and therefore ρ(t) ≤ υ(t) ≤ L* for t ∈ [0, a]. Since for every t ∈ [0, a], ||u_t||_{[−q, 0]} ≤ ρ(t), we have

where L* depends only on a and the functions p and φ. Noticing that ρ(t) ≤ ||ɸ||_{[−q, 0]} + N for t ∈[−q, 0]. Therefore, the set S is bounded. By condensing mapping fixed point theorem of topological degree, there exists a fixed point u of Q on C([−q, a],X), which is just the mild solution of problem (1). The proof is completed.   □