J. Appl. Math. Comput. DOI 10.1007/s12190-016-1049-0 ORIGINAL RESEARCH
Existence and stability for Hadamard p-type fractional functional differential equations Haihua Wang1 · Yanqiong Liu1 · Huanran Zhu2
Received: 3 July 2016 © Korean Society for Computational and Applied Mathematics 2016
Abstract In this paper, we study the existence and Hyers–Ulam stability of solutions for a Hadamard p-type fractional order functional and neutral functional differential equations involving initial value problems. Sufficient conditions which guarantee the existence and Hyers–Ulam stability are new and obtained. Keywords Hadamard fractional derivative · p-function · Existence · Hyers–Ulam stability Mathematics Subject Classification 26A33 · 34K10
1 Introduction Let J ⊂ R. Denote C(J, R) as the Banach space of all continuous functions from J into R with the norm x = supt∈J |x(t)|. Given any r > 0, let C = C([−r, 0], R) denote the space of continuous functions on [−r, 0]. For any element φ ∈ C, define the norm φ∗ = supθ∈[−r,0] |φ(θ )|. This paper is concerned with the existence of solutions and stability for initial value problems of Hadamard p-type fractional order functional differential equations. In Sect. 3, we are interested in the existence and stability of the following initial value problems:
B
Haihua Wang
[email protected]
1
Department of Mathematics, Guangxi Science and Technology Normal University, Laibin 546100, Guangxi, People’s Republic of China
2
Department of Mathematics, Hunan University of Science and Technology, Xiangtan 411201, Hunan, People’s Republic of China
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q
Da+ x(t) = f (t, xt ), xa = ϕ, (a, ϕ) ∈ ,
(1.1)
q
where Da+ is the Hadamard fractional derivative of order 0 < q < 1 and a > 0, is an open subset of (0, +∞) × C. f : → R is a given function satisfies some assumptions that will be specified later, ϕ ∈ C and ϕ(0) = 0. xt ∈ C is defined by xt (θ ) = x( p(t, θ )), where −r ≤ θ ≤ 0, p(t, θ ) is a p-function. Section 4 is devoted to Hadamard fractional neutral functional differential equations: q Da+ [x(t) − g(t, xt )] = f (t, xt ), (1.2) xa = ϕ, (a, ϕ) ∈ , where a, f , ϕ and are as in Problem (1.1), and g : C → R is a given function such that g(a, ϕ) = 0. Fractional differential equations have gained considerable importance due to their applications in various science, such as physics, chemistry, engineering, polymer rheology, etc. In recent years, there has been a significant development in ordinary and partial differential equations involving fractional derivatives, see [4–6,9,14,16,19]. Fractional derivatives provide an excellent tool for the description of memory and heredity properties of various materials and processes. These characteristics of the fractional derivatives make the fractional order models more realistic and practical than the classical integer order models. In consequence, the subject of fractional differential equations is gaining much importance and attention. For detail, see the monographs of Kilbas et al. [12], Lakshmikantham et al. [13], Miller and Ross [15], Podlubny [18], Samko et al. [20] and the references therein. Let us notice that most of the work on the topic is based on Riemann–Liouville or Caputo type fractional differential equations up to now. We also note that Hadamard [8] introduce another fractional derivative, which differs from the preceding ones in the sense that the kernel of the integral (in the definition of Hadamard derivative) contains logarithmic function of arbitrary exponent. It is remarkable that some properties has been investigated extensively and some interesting results are reported in [1–3,10,11]. Recently, in [21], authors investigated a p-type fractional differential equations of the initial value problem:
Dq x(t) = f (t, xt ), t > t0 xt0 = ϕ,
(1.3)
where Dq is the Caputo’s fractional derivative of order 0 < q < 1. xt ∈ C is defined by xt (θ ) = x( p(t, θ )). Existence and uniqueness are obtained by various criteria. Some processes, especially in areas of population dynamics, ecology, and pharmacokinetics, involve hereditary issues. The theory and applications addressing such problems have heavily involved functional differential equations. From the following Lemma 2.2 and Definition 2.4, it is easy to see that p-type fractional functional differential equation is a very general type of fractional functional differential equations which includes fractional functional differential equations with bounded as well as unbounded delays. Why? In fact, we are very familiar with the frequently used
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symbol xt , xt (θ ) = x(t + θ ) in the theory of functional differential equations with bounded delay. For example, choose p(t, θ ) = t + θ , −r ≤ θ ≤ 0, we know that xt is a partial case of x( p(t, θ )). On the other hand, we choose p(t, θ ) = t + θ 3 , or p(t, θ ) = t + θ eθ , −r ≤ θ ≤ 0, we see that x( p(t, θ )) is not the usual symbol xt in this cases. Moreover, up to now, we can not find the relevant application of p-function with respect to Hadamard fractional functional differential equations. Inspired by the above works, we investigate the existence and Hyers–Ulam stability of Problems (1.1) and (1.2). Our approach is based on the Banach fixed point theorem and Schauder’s fixed point theorem. Our results are new and can be considered as a contribution to this emerging field.
2 Preliminaries and Lemmas Definition 2.1 [12] Let a ≥ 0, the Hadamard fractional integral of order q > 0 for a function g : [a, b] → R is defined as q Ja+ g (t) =
1 (q)
t t q−1 g(s) log ds. s s a
Definition 2.2 [12] Let a ≥ 0, the Hadamard derivative of fractional order q > 0 for a function g is defined by q n−q Da+ g (t) := δ n Ja+ g (t) =
t n−q−1 g(s) d n t 1 log ds, t (n − q) dt s s a
d where δ = t dt is the δ-derivative, n = [q] + 1 and log(·) = loge (·).
Definition 2.3 [17] Let J be an interval in R, (B, | · |) a (real or complex) Banach space, A : C(J, B) → C(J, B) an operator. The fixed point equation x = A(x)
(2.1)
has the Hyers–Ulam stability if there exists c > 0 such that for each solution x ∈ C(J, B) of |x(t) − A(x)(t)| ≤ ε, ∀t ∈ J, there exists a unique solution x ∗ of (2.1) such that |x(t) − x ∗ (t)| ≤ cε, ∀t ∈ J. Lemma 2.1 [12] Let q > 0, n = [q] + 1 and 0 < a < b < ∞. If g(t) ∈ L(a, b) n−q and (Ja+ g)(t) ∈ ACδn [a, b], then n−q n δ n−k Ja+ g (a)
q q t q−k log Ja+ Da+ g (t) = g(t) − . (q − k + 1) a
(2.2)
k=1
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Lemma 2.2 [13] Let p ∈ C(R × [−r, 0], R), where r > 0 is a constant. Then p(t, θ ) is called a p-function if it has the following properties: (i) p(t, 0) = t, (ii) p(t, −r ) is a nondecreasing function of t, (iii) there exists a σ ≥ −∞ such that p(t, θ ) is an increasing function for θ for each t ∈ (σ, ∞), (iv) p(t, 0) − p(t, −r ) ≥ λ > 0 for t ∈ (σ, ∞), where λ is a constant. Throughout the following text we suppose t ∈ (σ, ∞). Definition 2.4 [13] Let t0 ≥ 0, A > 0 and x ∈ C([ p(t0 , −r ), t0 + A], R). For any t ∈ [t0 , t0 + A], we define xt by xt (θ ) = x( p(t, θ )), −r ≤ θ ≤ 0, so that xt ∈ C. Lemma 2.3 Suppose that p(t, θ ) is a p-function. Then for any a > 0, p (t, θ ) = 1 p(at, θ ) is also a p-function. a Proof (i) p (t, 0) = a1 p(at, 0) = a1 at = t for every t ∈ R. (ii) p (t, −r ) = a1 p(at, −r ) is nondecreasing in t ∈ R. (iii) For any fixed t ∈ (σ, ∞), p(t, θ ) is increasing in θ , and so is a1 p(at, θ ) for any p (t, θ ) is increasing with respect to θ for fixed t > σa . fixed t > σa , that is, (iv) p (t, 0) − p (t, −r ) = a1 ( p(at, 0) − p(at, −r )). Hence p(at, 0) − p(at, −r ) > 0 for at > σ implies that p (t, 0) − p (t, −r ) > 0 for t > σa . Let α > 0, ⊂ [a, a + α] × C be an open set, assume that the function f : → R satisfies the following conditions. (H1) f (t, φ) is Lebesgue measurable with respect to t for any (t, φ) ∈ , (H2) f (t, φ) is continuous with respect to φ for any (t, φ) ∈ . For each (a, ϕ) ∈ , let p (t, θ ) = C([ p (1, −r ), +∞), R) by
1 a
p(at, θ ). Define the function η ∈
η(t) = 0, 1 ≤ t ≤ +∞, η( p (1, θ )) = ϕ(θ ), θ ∈ [−r, 0].
Let x ∈ C([ p(a, −r ), a + α], R), α < A and x(at) = η(t) + z(t) for p (1, −r ) ≤ t ≤ 1 + αa . Lemma 2.4 [7] Let U be a closed, convex and nonempty subset of a Banach space E. Let F : U → U be a continuous mapping such that FU is a relatively compact subset of E. Then F has at least a fixed point in U .
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Lemma 2.5 x is a solution of Problem (1.1) on [ p(a, −r ), a + α] if and only if z(t) satisfies ⎧ ⎪ ⎨ z(t) =
1 (q) ⎪ ⎩ z 1 = 0,
t t q−1 f (as, ηs + zs ) α , log ds, t ∈ 1, 1 + s s a 1
(2.3)
where ηt = η( p (t, θ )), z t = z( p (t, θ )), −r ≤ θ ≤ 0. Proof Since xt is continuous in t, xt is a measurable function, therefore according to conditions (H1), f (t, xt ) is Lebesgue measurable on [a, a + α]. For t ∈ [a, a + α] and β ∈ (0, q), we see that t a
t log s
q−1
1
1−β
s
1 1−β
ds = t =t
β β−1
β β−1
t a
q−1
(log s) 1−β s
≤
β
a 1−β
ds
1 t a
q−1
β
(log s) 1−β s −1 s 1−β ds
1
1
2β−1 1−β
1−β q −β
(2.4)
q−β t 1−β log , a
1 q−1 1 which gives that log st ∈ L 1−β [a, t]. In light of Hölder inequality, we obtain s q−1 f (s,xs ) that log st is Lebesgue integrable with respect to s ∈ [a, t] for all t ∈ s [a, a + α] and
t t q−1 t q−1 f (s, xs ) log ds ≤ log s s s a
1 s
1
L 1−β [a,t]
m
1
L β [a,t]
.
Hence, from Lemma 2.1, we know that x is a solution of problem (1.1) if and only if it satisfies the relation ⎧ ⎪ ⎨ x(t) =
1 (q) ⎪ ⎩ xa = ϕ,
t t q−1 f (τ, xτ ) log dτ, t ∈ [a, a + α], τ τ a
or setting τ = as, ⎧ ⎪ ⎨ x(at) = ⎪ ⎩
1 (q) xa = ϕ.
t t q−1 f (as, xas ) α log , ds, t ∈ 1, 1 + s s a 1
(2.5)
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By the relation x(at) = η(t) + z(t) for p (1, −r ) ≤ t ≤ 1 + αa , we have xat (θ ) = x ( p(at, θ )) = x (a p (t, θ )) = η( p (t, θ )) + z( p (t, θ )) α . = ηt (θ ) + z t (θ ), t ∈ 1, 1 + a In particular, xa (θ ) = η1 (θ ) + z 1 (θ ). Hence xa = ϕ if and only if z 1 = 0 according to η1 (θ ) = ϕ. It is clear that x(t) satisfies (2.5) if and only z(t) satisfies (2.3), the proof is completed.
3 FDEs of fractional order Denote by
E(δ, γ ) = z ∈ C
δ , R : z 1 = 0, z t ∗ ≤ γ for t ∈ [0, δ] , p (1, −r ), 1 + a
where δ, γ are positive constants. Obviously, E(δ, γ ) is a bounded closed convex subset of the Banach space C p (1, −r ), 1 + aδ , R endowed with the norm · . Let us define the operator T on E(δ, γ ) by ⎧ ⎪ ⎨ (T z)(t) =
t t q−1 f (as, δ ηs + zs ) 1 log ds, t ∈ 1, 1 + , (q) 1 s s a ⎪ ⎩ (T z)(t) = 0, t ∈ [ p (1, −r ), 1].
(3.1)
Theorem 3.1 Assume that f : → R satisfies (H1), (H2) and 1
(H3) There exist constant β ∈ (0, q), γ > 0 and a L β -integrable function m such that | f (t, z)| ≤ m(t) for almost every t ∈ [a, a + δ], z ∈ E(δ, γ ),where ⎧ ⎨
β 1+ aη 1 η q−β δ = max η : η ≤ α, log 1 + (m(s)) β ds ⎩ a 1 q − β 1−β ≤ γ (q) . 1−β
Then Problem (1.1) has at least one solution on [ p(a, −r ), a + δ].
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Proof For any z ∈ E(δ, γ ), by the Hölder inequality and (H3), we get t t q−1 | f (as, ηs + z s )| log ds s s 1 t t q−1 m(as) 1 log ds ≤ (q) 1 s s ⎤1−β ⎡ t q−1 β t 1−β 1 t 1 ⎣ 1 β ds ⎦ log ≤ ds (m(as)) 1 (q) s 1 1 s 1−β β 1+ δ 1 a 1 − β 1−β 1 q−β ≤ (m(as)) β ds (log t) (q) q − β 1 β 1+ aδ 1 1 − β 1−β δ q−β 1 log 1 + ≤ (m(as)) β ds (q) q − β a 1 δ , ≤ γ , for t ∈ 1, 1 + a
1 |(T z)(t)| ≤ (q)
which implies that T is a mapping from E(δ, γ ) into itself. For z ∈ E(δ, γ ), we get T z ≤ γ , this means that {T z : z ∈ E(δ, γ ) is uniformly bounded. In addition, T is continuous from the condition (H2). Now, Let t1 , t2 ∈ 1, 1 + aδ with t1 < t2 . Then we have |(T x)(t2 ) − (T x)(t1 )| ηs + t2 q−1 zs ) 1 t1 t1 q−1 f (as, log ds = − log (q) s s s 1
t2 q−1 f (as, ηs + z s ) log ds + s s t1 q−1 t1 1 ηs + t1 z s )| t2 q−1 | f (as, ≤ log ds − log (q) 1 s s s t2 t2 q−1 | f (as, 1 ηs + z s )| log ds + (q) t1 s s t1 1 t1 q−1 t2 q−1 m(as) ≤ log ds − log (q) 1 s s s t2 t2 q−1 m(as) 1 log ds + (q) t1 s s ⎫1−β ⎧ β 1 ⎬ t1 1 1 ⎨ t1 t2 q−1 1−β 1 t1 q−1 β ≤ − log ds (m(as)) ds log 1 (q) ⎩ 1 s s 1 s 1−β ⎭
t2
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⎤ 1 ⎡ 1−β q−1 β t2 1 t2 1−β 1 1 ⎣ t2 β ⎦ log + ds (m(as)) ds 1 (q) s t1 t1 s 1−β ⎫1−β ⎧ ⎡ ⎤ β q−1 q−1 1+ δ 1 a t1 1−β 1 ⎨ t1 ⎣ t2 1−β ⎦ 1 ⎬ log ds ≤ − log (m(as)) β ds (q) ⎩ 1 s s s ⎭ 1 ⎤ 1 ⎡ β 1−β q−1 1+ δ 1 a t2 1−β 1 ⎦ 1 ⎣ t2 log ds + (m(as)) β ds (q) s s t1 1 β 1+ aδ 1 1 β = (m(as)) ds (q) 1 1 − β 1−β t2 q−β q−β q−β × −log(t2 ) + 2 log log(t1 ) . q −β t1 Obviously, the right hand side of the above inequality tends to zero. The equiconp (1, −r ) ≤ t1 ≤ 1 ≤ t2 is obvious. tinuous for the case p (1, −r ) ≤ t1 < t2 ≤ 1 and Therefore, it follows by the Arzela-Ascoli theorem, T : E(δ, γ ) → E(δ, γ ) is completely continuous and hence T E(δ, γ ) is relatively compact. By Lemma 2.4, T has at least a fixed point z in E(δ, γ ). This completes the proof. Theorem 3.2 Assume that f : → R satisfies (H1), (H2) and (H3)’ there exist constant γ > 0 and a L ∞ -integrable function m such that | f (t, z)| ≤ m(t) for almost every t ∈ [a, a + δ], z ∈ E(δ, γ ), i.e., there exists a M > 0 such that | f (t, z)| ≤ M for almost every t ∈ [a, a + δ], z ∈ E(δ, γ ), where
$
δ = min α, a e
γ (1+q) M
1 q
% −1
.
Then Problem (1.1) has at least one solution on [ p(a, −r ), a + δ]. Proof The proof is similar to Theorem 3.1, we need only notice that t M t q−1 1 log ds (q) 1 s s M (log t)q = (1 + q) δ q M δ log 1 + . = ≤ γ , for t ∈ 1, 1 + (1 + q) a a
|(T z)(t)| ≤
The rest of the proof is similar to that of Theorem 3.1 and it is omitted.
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Theorem 3.3 Assume that f : → R satisfies (H1) and (H4) there exists a constant γ > 0 such that f satisfies the Lipschitz condition with respect to the second variable, i.e. there exists a constant k > 0 such that | f (t, x) − f (t, y)| ≤ kx − y∗ for all t ∈ [a, a + δ] and x, y ∈ E(δ, γ ), where
$
δ = min α, a e
(1+q) k
%
1 q
−1
,
(3.2)
then Problem (1.1) has a unique solution on [ p(a, −r ), a + δ]. Proof For any z, w ∈ E(δ, γ ), z1 − w 1 = 0, note that sup 1≤t≤1+ aδ
zt − w t ∗ = = =
sup
sup |z( p (t, θ )) − w( p (t, θ ))|
1≤t≤1+ aδ −r ≤θ≤0
sup
sup
p (t,−r )≤s≤t 1≤t≤1+ aδ
sup p (1,−r )≤s≤1+ aδ
|z(s) − w(s)|
|z(s) − w(s)| = z − w.
Then for t ∈ 1, 1 + aδ , we have T z − T w = =
sup p (1,−r )≤t≤1+ aδ
sup 1≤t≤1+ aδ
|(T z)(t) − (T w)(t)|
|(T z)(t) − (T w)(t)|
t t q−1 | f (as, ηs + z s ) − f (as, ηs + w s )| log ds s s 1 a t t q−1 zs − w s ∗ k log sup ds ≤ (q) 1≤t≤1+ δ 1 s s a t t q−1 1 kz − w log sup ds ≤ (q) 1≤t≤1+ δ 1 s s a δ q kz − w log 1 + = . (1 + q) a (3.3) It follows from (3.2) that T is a contraction mapping. Therefore T has a unique fixed point by Banach’s contraction principle. The proof is completed. ≤
1 sup (q) 1≤t≤1+ δ
Theorem 3.4 Assumption of Theorem 3.3 are satisfied. Then the fixed point equation z = T z is Hyers–Ulam stability.
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Proof We denote that z ∗ be the unique solution of z = T z. It follows that |z(t) − z ∗ (t)| = |z(t) − (T z ∗ )(t)| ≤ |z(t) − (T z)(t)| + |(T z)(t) − (T z ∗ )(t)| ≤ |z(t) − (T z)(t)| t 1 t q−1 | f (as, ηs + z s ) − f (as, ηs + z s∗ )| + log ds (q) 1 s s q δ k log 1 + ≤ε+ z − z ∗ , (1 + q) a hence, |z(t) − z ∗ (t)| ≤ z − z ∗ ≤
1−
k (1+q)
1 q ε := cε. log 1 + aδ
Together with Definition 2.3, z = T z is Hyers–Ulam stability. The proof is completed.
4 NFDEs of fractional order Solution of (1.2) is equivalent to the solution of ⎧ t ⎪ t q−1 f (τ, xτ ) ⎨ x(t) = g(t, x ) + 1 log dτ, t ∈ [a, b], t (q) a τ τ ⎪ ⎩ xa = ϕ, in analogy to Lemma 2.5, we consider the operator T1 : E(δ, γ ) → E(δ, γ ) defined by ⎧ t ⎪ t q−1 f (as, δ ηs + zs ) 1 ⎨ +(T z)(t) = g(at, log ds, t ∈ 1, 1 + , η + z ) + t t 1 (q) 1 s s a ⎪ ⎩ (T1 z)(t) = 0, t ∈ [ p(1, −r ), 1].
(4.1) Theorem 4.1 Assume that f : → R satisfies (H1) and (H2), furthermore there exist constants δ1 ∈ (0, α) and γ > 0 such that (H5) g is continuous and completely continuous, and for any bounded set B in E(δ1 , γ ), the set {t → g(t, xt ) : xt ∈ B} is equicontinuous in C([a, a +δ1 ], R). 1
(H6) there exist constant β ∈ (0, q) and a L β -integrable function m such that | f (t, z)| ≤ m(t) for almost every t ∈ [a, a + δ1 ], z ∈ E(δ1 , γ ). Then problem (1.2) has at least one solution on [ p(a, −r ), a + δ] for some positive number δ.
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Proof Since g is continuous and xt is continuous in t, there exists 0 < δ ≤ δ1 such
that for 1 < t < 1 + δa , |g(at, ηt + z t )| = |g(at, ηt + z t ) − g(a, ϕ)| <
γ , 2
(4.2)
Let ⎧ ⎨
β 1+ aη 1 η q−β δ = max η : η ≤ δ , log 1 + (m(s)) β ds ⎩ a 1 q − β 1−β ≤ 2γ (q) . 1−β
For any z ∈ E(δ, γ ), in light of (4.2) and Hölder inequality, we have t ηs + t q−1 | f (as, z s )| 1 log ds (q) 1 s s t γ t q−1 m(as) 1 ≤ + log ds 2 (q) 1 s s β 1+ aδ 1 δ q−β 1 − β 1−β 1 γ log 1 + ≤ + (m(as)) β ds 2 (q) q − β a 1 δ , ≤ γ , for t ∈ 1, 1 + a
|(T1 z)(t)| ≤ |g(at, ηt + z t )| +
which implies that T1 z ∈ E(δ, γ ). For z ∈ E(δ, γ ), we get T1 z ≤ γ , this means that {T1 z : z ∈ E(δ, γ ) is uniformly bounded. In addition, T1 is continuous from the condition (H2) and (H5). Similar to Theorem 3.1, together with (H2) and (H5), it suffices to show that T1 is continuous and completely continuous and it is thus omitted. By Lemma 2.4, T1 has at least a fixed point z in E(δ, γ ). This completes the proof. Theorem 4.2 Assume that f : → R satisfies (H1) and (H2), furthermore there exist constants δ1 ∈ (0, α) and γ > 0 such that (H5) is satisfied and (H6) there exist constant β ∈ (0, q) and a L ∞ -integrable function m such that | f (t, z)| ≤ m(t) for almost every t ∈ [a, a + δ1 ], z ∈ E(δ1 , γ ), i.e., there exists a M > 0 such that | f (t, z)| ≤ M for almost every t ∈ [a, a +δ1 ], z ∈ E(δ1 , γ ). Then problem (1.2) has at least one solution on [ p(a, −r ), a + δ] for some positive number δ. The proof is similar to that of Theorems 3.2 and 4.1, we omit it. Theorem 4.3 Assume that f : → R satisfies (H1), in addition, there exist constants δ1 ∈ (0, α) and γ > 0 such that
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(H7) there exists a nonnegative constant 0 < k < 1 such that |g(t, x) − g(t, y)| ≤ kx − y∗ , | f (t, x) − f (t, y)| ≤ kx − y∗ , for all t ∈ [a, a + δ1 ] and x, y ∈ E(δ1 , γ ). then problem (1.2) has a unique solution on [ p(a, −r ), a + δ] for some positive number δ. Proof Let
$
δ = min δ1 , a e
(1−k)(1+q) k
1 q
% −1
,
for z, wδ ∈ E(δ, γ ), then following the steps of Theorem 3.3, we have for t ∈ 1, 1 + a , T1 z − T1 w = = ≤
sup p (1,−r )≤t≤1+ aδ
sup 1≤t≤1+ aδ
sup 1≤t≤1+ aδ
|(T1 z)(t) − (T1 w)(t)|
|(T1 z)(t) − (T1 w)(t)| |g(at, ηt + z t ) − g(at, ηt + w t )|
t t q−1 | f (as, ηs + z s )− f (as, ηs + w s )| log ds s s 1 a t t q−1 zs − w s ∗ k sup ds log ≤ k zt − w t ∗ + (q) 1≤t≤1+ δ 1 s s a ⎡ ⎤ q−1 t 1 t 1 ≤ kz − w ⎣1 + log sup ds ⎦ (q) 1≤t≤1+ δ 1 s s a 1 δ q = kz − w 1 + , log 1 + (1 + q) a (4.3) which implies that T1 is a contraction. Therefore T1 has a unique fixed point by Banach’s contraction principle. The proof is completed. +
1 sup (q) 1≤t≤1+ δ
Similar to Theorem 3.4, we have the following Hyers–Ulam stability result. Theorem 4.4 Assumption of Theorem 4.3 are satisfied. Then the fixed point equation z = T1 z is Hyers–Ulam stability.
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5 Examples Example 5.1 Let r > 0, γ = 2.1, consider the problem ⎧ ⎨
1
2 D0.5+ x(t) = f (t, xt ), t > 0.5,
(5.1)
⎩ x(0.5 + θ ) = ϕ(θ ), θ ∈ [−r, 0]. Here, q = 21 , a = 0.5, p(t, θ ) = t + θ , θ ∈ [−r, 0], ⎧ ⎪ ⎪ ⎨t
1 , 0.5 < t ≤ 1, 1 + |x| f (t, x) = 1 ⎪ ⎪ ⎩ , t > 1, t (1 + |x|) and ϕ(0) = 0. Moreover, | f (t, x)| ≤ t := m(t), t ∈ [0.5, 1]. Let β = 13 , it is easy to calculate that for δ = 0.5, all conditions of Theorem 3.1 are satisfied, hence, by the Problem (5.1) has at least one solution in [0.5 − r, 1]. Example 5.2 Let γ > 0 and δ1 = 5, we consider the problem ⎧ 1 |xt | ⎪ ⎨ D 3 x(t) − t − 2 xt = t + , t > 2, 2+ 1+t 10(1 + |xt |) ⎪ ⎩ x(2 + θ 3 ) = ϕ(θ ), θ ∈ [−r, 0].
(5.2)
Here, q = 13 , a = 2, p(t, θ ) = t + θ 3 , θ ∈ [−r, 0], r > 0, f (t, x) = t + g(t, x) =
t−2 1+t x,
|x| 10(1+|x|) ,
t ∈ [2, 5], x ∈ E(δ1 , γ ) and ϕ(0) = 0. |y| 1 |x| − 10 (1 + |x|) (1 + |y|) 1 |x − y| ≤ 10 (1 + |x|)(1 + |y|) 1 ≤ x − y∗ , 10
| f (t, x) − f (t, y)| =
and
t −2 |x − y| 1+t 1 ≤ x − y∗ . 2
|g(t, x) − g(t, y)| ≤
Taking k = 21 , by Theorem 4.3, the problem (5.2) has a unique solution in [2 − + δ] for δ = min{5, 2.0765} = 2.0765. As a result, the Hyers–Ulam stability result is obtained with the help of Theorems 4.4 to the fixed point equation z = T1 z.
r 3, 2
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H. Wang et al. Acknowledgments The authors are grateful to the editor and the anonymous reviewers for their constructive comments and suggestions which improved the quality of the paper. The first author’s work is supported by Research Fund of Hunan Provincial Education Department (15C0538) and partially supported by Hunan Provincial Natural Science Foundation of China (2015JJ2068).
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