Based on the work in [[15], Lemma 3.1 and Definition 3.1], in this paper, we adopt the following definition of mild solution of Cauchy problem (1.1).
Definition 3.1. By a mild solution of Cauchy problem (1.1), we mean a function u ∈ C([0, T]; X
α
) satisfying
for t ∈ [0, T].
Let us first introduce our basic assumptions.
(H
0) κ
1, κ
2 ∈ C([0, T]; [0, T]) and K ∈ C([0, T]; ℝ+).
(H
1) F, G : [0, T] × X
α
× X
α
→ X are continuous and for each positive number k ∈ ℕ, there exist a constant γ ∈ [0, β(1 - α)) and functions φ
k
(·) ∈ L
1/γ(0, T; ℝ+), ϕ
k
(·) ∈ L
∞(0, T; ℝ+) such that
(H
2) F, G : [0, T] × X
α
× X
α
→ X are continuous and there exist constants L
F
, L
K
such that
for all (t, u
1, v
1), (t, u
2, v
2) ∈ [0, T] × X
α
× X
α
.
(H
3) H : C([0, T]; X
α
) → X
α
is Lipschitz continuous with Lipschitz constant L
H
.
(H
4) H : C([0, T]; X
α
) → X
α
is continuous and there is a η ∈ (0, T) such that for any u, w ∈ C([0, T]; X
α
) satisfying u(t) = w(t)(t ∈[η, T]), H(u) = H(w).
(H
5) There exists a nondecreasing continuous function Φ : ℝ+ → ℝ+ such that for all u ∈ Θ
k
,
Remark 3.1. Let us note that (H
4) is the case when the values of the solution u(t) for t near zero do not affect H(u). We refer to[19]for a case in point.
In the sequel, we set . We are now ready to state our main results in this section.
Theorem 3.1. Let the assumptions (H
0), (H
1) and (H
3) be satisfied. Then, for u
0 ∈ X
α
, the fractional Cauchy problem (1.1) has at least one mild solution provided that
(3.1)
Proof. Let v ∈ C([0, T]; X
α
) be fixed with |v|
α
≡ 0. From (3.1) and (H
1), it is easy to see that there exists a k
0 > 0 such that
Consider a mapping Γ defined on by
It is easy to verify that (Γu)(·) ∈ C([0, T]; X
α
) for every . Moreover, for every pair and t ∈ [0, T], by (H
1) a direct calculation yields
That is, for every pair . Therefore, the fractional Cauchy problem (1.1) has a mild solution if and only if the operator equation Γ1
u + Γ2
u = u has a solution in .
In what follows, we will show that Γ1 and Γ2 satisfy the conditions of Lemma 2.5. From (H
3) and (3.1), we infer that Γ1 is a contraction. Next, we show that Γ2 is completely continuous on .
We first prove that Γ2 is continuous on . Let be a sequence such that u
n
→ u as n → ∞ in C([0, T]; X
α
). Therefore, it follows from the continuity of F, G, κ
1 and κ
2 that for each t ∈ [0, T],
Also, by (H
1), we see
and
Hence, as
we conclude, using the Lebesgue dominated convergence theorem, that for all t ∈ [0, T],
which implies that
This proves that Γ2 is continuous on .
It suffice to prove that Γ2 is compact on . For the sake of brevity, we write
Let t ∈ [0, T] be fixed and ε, ε
1 > 0 be small enough. For , we define the map by
Therefore, from Lemma 2.1 we see that for each t ∈ (0, T], the set is relatively compact in X
α
. Then, as
in view of (2.1), we conclude, using the total boundedness, that for each t ∈ [0, T], the set is relatively compact in X
α
.
On the other hand, for 0 < t
1 < t
2 ≤ T and ε' > 0 small enough, we have
where
Therefore, it follows from (H
1), Lemma 2.2, and Lemma 2.3 that
from which it is easy to see that A
i
(i = 1, 2, 3, 4) tends to zero independently of as t
2 - t
1 → 0 and ε' → 0. Hence, we can conclude that
and the limit is independently of .
For the case when 0 = t
1 < t
2 ≤ T, since
||(Γ2
u)(t
1) - (Γ2
u)(t
2)||
α
can be made small when t
2 is small independently of . Consequently, the set is equicontinuous. Now applying the Arzela-Ascoli theorem, it follows that Γ2 is compact on .
Therefore, applying Lemma 2.5, we conclude that Γ has a fixed point, which gives rise to a mild solution of Cauchy problem (1.1). This completes the proof. ■
The second result of this paper is the following theorem.
Theorem 3.2. Let the assumptions (H
0), (H
2), (H
4) and (H
5) be satisfied. Then, for u
0 ∈ X
α
, the fractional Cauchy problem (1.1) has at least one mild solution provided that
(3.2)
Proof. The proof is divided into the following two steps.
Step 1. Assume that w ∈ C([η, T]; X
α
) is fixed and set
It is clear that w ∈ C([0, T]; X
α
). We define a mapping Γ
w
on C([0, T]; X
α
) by
Clearly, (Γ
w
u)(·) ∈ C([0, T]; X
α
) for every u ∈ C([0, T]; X
α
). Moreover, for u ∈ Θ
k
, from (H
2), it follows that
where v ∈ C([0, T]; X
α
) is fixed with |v|
α
≡ 0, which implies that there exists a integer k
0 > 0 such that Γ
w
maps into itself. In fact, if this is not the case, then for each k > 0, there would exist u
k
∈ Θ
k
and t
k
∈ [0, T] such that ||(Γ
w
u
k
)(t
k
)||
α
> k. Thus, we have
Dividing on both sides by k and taking the lower limit as k → +∞, we get
this contradicts (3.2). Also, for , a direct calculation yields
which together with (3.2) implies that Γ
w
is a contraction mapping on . Thus, by the Banach contraction mapping principle, Γ
w
has a unique fixed point , i.e.,
for t ∈ [0, T].
Step 2. Write
It is clear that is a bounded closed convex subset of C([η, T]; X
α
).
Based on the argument in Step 1, we consider a mapping on defined by
It follows from (H
5) and (3.2) that maps into itself. Moreover, for , from Step 1, we have
that is,
which yields that is continuous. Next, we prove that has a fixed point in . It will suffice to prove that is a compact operator. Then, the result follows from Lemma 2.6.
Let's decompose the mapping as
Since assumption (H
5) implies that the set is bounded in X
α
, it follows from Lemma 2.4 that for each t ∈ [η, T], is relatively compact in X
α
. Also, for η ≤ t
1 ≤ t
2 ≤ T,
independently of . This proves that the set is equicontinuous. Thus, an application of Arzela-Ascoli's theorem yields that is compact.
Observe that the set
is bounded in X. Therefore, using Lemma 2.1, Lemma 2.2 and Lemma 2.3, it is not difficult to prove, similar to the argument with Γ2 in Theorem 3.1, that is compact. Hence, making use of Lemma 2.6 we conclude that has a fixed point . Put . Then,
Since and hence q is a mild solution of the fractional Cauchy problem (1.1). This completes the proof. ■