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(DR0(C([−q, a],X))) ⊂ DR0(C([−q, a],X)). For every u ∈ C([−q, a],X), by condition (P1), we have
(7)
Choose a positive constant
(8)
For every u ∈ DR0(C([−q, a],X)), then ||ut||[−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 DR0(C([−q, a],X)). Let un ⊂ DR0(C([−q, a],X)) with un → u (n → ∞) in DR0(C([−q, a],X)). Then unt → ut as n → ∞ for all t ∈ [0, a], where unt = (un)t . Applying the condition (P3) we get that for each t ∈ [−q, 0],
By the facts unt → ut 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 DR0(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 DR0(C([−q, a],X)). Taking u and υ in DR0(C([−q, a],X)). Then u|[0, a] and υ|[0, a] in DR0(C([0, a],X)), by the hypothesis (P3) we get that
and
From the above discussion we get that
for all u, υ ∈ DR0(C([−q, a],X)), which means that Q1 : DR0(C([−q, a],X)) → DR0(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 ⊂ DR0(C([−q, a],X)),
(9)
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 ∈ DR0(C([−q, a],X))} is a family of equicontinuous functions. Then for every bounded subset D ⊂ DR0(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* = {un} ⊂ D such that
(10)
For each t ∈ [0, a], we denote by
Then, for every u, υ ∈ D* we have ut, υt ∈ Dt*, and
(11)
By (11) and Lemma 2.5 we know that
(12)
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
(13)
Since Q2(D*) is equicontinuous, by Lemma 2.1, we know that
Hence by (10) and (13), we have
(14)
For every subset D ⊂ DR0(C([−q, a],X)), by (5), (9) and (14), we have
Therefore, Q : DR0(C([−q, a],X)) → DR0(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 DR0(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:
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 ∈ DR(C([−q, a],X)), by the conditions (P4) and (P5), we know that there exist constants C¯3 and N¯2 such that
(15)
for every t ∈ [0, a], and
(16)
Taking (15) and (16) into account, we obtain that
and
Hence,
It means that Q(DR(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,
(17)
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. □
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