Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 58, 1-17; http://www.math.u-szeged.hu/ejqtde/

ON THE EXISTENCE OF MILD SOLUTIONS TO SOME SEMILINEAR FRACTIONAL INTEGRO-DIFFERENTIAL EQUATIONS

T. DIAGANA, G. M. MOPHOU, AND G. M. N’GUÉRÉKATA

Abstract. This paper deals with the existence of a mild solution for some fractional semilinear differential equations with non local conditions. Using a more appropriate definition of a mild solution than the one given in [12], we prove the existence and uniqueness of such solutions, assuming that the linear part is the infinitesimal generator of an analytic semigroup that is compact for t > 0 and the nonlinear part is a Lipschitz continuous function with respect to the norm of a certain interpolation space. An example is provided.

1. Introduction

Let 𝕏 be a Banach space and let T > 0. This paper is aimed at discussing about the existence and the uniqueness of a mild solution for the fractional semilinear integro-differential equation with nonlocal conditions in the form:

\{ \begin{matrix}D^{β} x(t) = −Ax(t) + f(t, x(t)) + ∫_{0}^{t}a(t − s)h(s, x(s)) ds, & t ∈ [0, T], \\ x(0) + g(x) = x_{0}, \end{matrix} (1)

where the fractional derivative D^{β} (0 < β < 1) is understood in the Caputo sense, the linear operator −A is the infinitesimal generator of an analytic semigroup (R(t))_{t≥0} that is uniformly bounded on 𝕏 and compact for t > 0, the function a(·) is real-valued such that

(2)  a_{T} = ∫_{0}^{T} a(s) ds < ∞,

the functions f, g and h are continuous, and the non local condition

g(x) = ∑_{k=1}^{p} c_{k}x(t_{k}),

with c_{k}, k = 1, 2, ...p, are given constants and 0 < t_{1} < t_{2} < ... < t_{p} ≤ T.

Let us recall that those nonlocal conditions were first utilized by K. Deng [4]. In his paper, K. Deng indicated that using the nonlocal condition x(0) + g(x) = x_{0}

EJQTDE, 2010 No. 58, p. 1 to describe for instance, the diffusion phenomenon of a small amount of gas in a transparent tube can give better result than using the usual local Cauchy Problem x(0) = x_{0}. Let us observe also that since Deng’s paper, such problem has attracted several authors including A. Aizicovici, L. Byszewski, K. Ezzinbi, Z. Fan, J. Liu, J. Liang, Y. Lin, T.-J. Xiao, H. Lee, etc. (see for instance [1, 2, 3, 4, 9, 8, 7, 14, 11, 13] and the references therein).

1991 Mathematics Subject Classification. 34K05; 34A12; 34A40.

Key words and phrases. fractional abstract differential equation, sectorial operator.

This problem has been studied in Mophou and N’Guérékata [12]. In this paper, we revisit that work and use a more appropriate definition for mild solutions. Namely, we investigate the existence and the uniqueness of a mild solution for the fractional semilinear differential equation (1), assuming that f is defined on [0, T] × 𝕏_{α} × 𝕏_{α} where 𝕏_{α} = D(A^{α}) (0 < α < 1), the domain of the fractional powers of A.

The rest of this paper is organized as follows. In Section 2 we give some known preliminary results on the fractional powers of the generator of an analytic compact semigroup. In Section 3, we study the existence and the uniqueness of a mild solution for the fractional semilinear differential equation (1). We give an example to illustrate our abstract results.

2. Preliminaries

Let I = [0, T] for T > 0 and let 𝕏 be a Banach space with norm ∥ · ∥. Let (𝔹(𝕏), ∥ · ∥_{𝔹(𝕏)}) be the Banach space of all linear bounded operators on 𝕏 and A : D(A) → 𝕏 be a linear operator such that −A is the infinitesimal generator of an analytic semigroup of uniformly bounded linear operators (R(t))_{t≥0}, which is compact for t > 0. In particular, this means that there exists M > 1 such that

(3)  sup_{t≥0}∥R(t)∥_{𝔹(𝕏)} ≤ M.

Moreover, we assume without loss of generality that 0 ∈ ρ(A). This allows us to define the fractional power A^{α} for 0 < α < 1, as a closed linear operator on its domain D(A^{α}) with inverse A^{−α}(see [8]). We have the following basic properties for fractional powers A^{α} of A:

Theorem 2.1. ([15], pp. 69 -75). Under previous assumptions, then:

(i) 𝕏_{α} = D(A^{α}) is a Banach space with the norm ∥x∥_{α} := ∥A^{α} x∥ for x ∈ D(A^{α});

(ii) R(t) : 𝕏 → 𝕏_{α} for each t > 0; (iii) A^{α}R(t)x = R(t)A^{α}x for each x ∈ D(A^{α}) and t ≥ 0;

EJQTDE, 2010 No. 58, p. 2 (iv) For every t > 0, A^{α}R(t) is bounded on 𝕏 and there exist M_{α} > 0 and

δ > 0 such that

(4)  ∥A^{α}R(t)∥_{𝔹(𝕏)} ≤ \frac{M_{α}}{t^{α}} e^{−δt};
(v) A^{−α} is a bounded linear operator in 𝕏 with D(A^{α}) = Im(A^{−α}); and
(vi) If 0 < α ≤ ν, then D(A^{ν}) ↪ D(A^{α}).

Remark 2.2. Observe as in [9] that by Theorem 2.1 (ii) and (iii), the restriction R_{α}(t) of R(t) to 𝕏_{α} is exactly the part of R(t) in 𝕏_{α}. Let x ∈ 𝕏_{α}. Since

∥R(t)x∥_{α} = ∥A^{α}R(t)x∥ = ∥R(t)A^{α}x∥ ≤ ∥R(t)∥_{𝔹(𝕏)}∥A^{α}x∥ = ∥R(t)∥_{𝔹(𝕏)}∥x∥_{α},

and as t decreases to 0

∥R(t)x − x∥_{α} = ∥A^{α}R(t)x − A^{α}x∥ = ∥R(t)A^{α}x − A^{α}x∥ → 0,

for all x ∈ 𝕏_{α}, it follows that (R(t))_{t≥0} is a family of strongly continuous semigroup on 𝕏_{α} and ∥R_{α}(t)∥_{𝔹(𝕏)} ≤ ∥R(t)∥_{𝔹(𝕏)} for all t ≥ 0.

Lemma 2.3. [9] The restriction R_{α}(t) of R(t) to 𝕏_{α} is an immediately compact semigroup in 𝕏_{α}, and hence it is immediately norm-continuous.

Now, let Φ_{β} be the Mainardi function:

Φ_{β}(z) = ∑_{n=0}^{+∞} \frac{(−z)^{n}}{n!Γ(−βn + 1 − β)}.

Then

(5a) Φ_{β}(t) ≥ 0 for all t > 0;

∫_{0}^{∞}Φ_{β}(t)dt = 1; (5b)

∫_{0}^{∞}t^{η}Φ_{β}(t)dt = \frac{Γ(1 + η)}{Γ(1 + βη)}, ∀η ∈ [0, 1]. (5c)

For more details we refer to [10].

We set

(6)  𝕊_{β}(t) = ∫_{0}^{∞}Φ_{β}(θ)R(θt^{β}) dθ,
(7)  ℙ_{β}(t) = ∫_{0}^{∞}βθΦ_{β}(θ)R(t^{β}θ)dθ

Then we have the following results EJQTDE, 2010 No. 58, p. 3 Lemma 2.4. [16] Let 𝕊_{β} and ℙ_{β} be the operators defined respectively by (6) and (7). Then

(i) ∥𝕊_{β}(t)x∥ ≤ M∥x∥; ∥ℙ_{β}(t)x∥ ≤ M \frac{β}{Γ(β + 1)}∥x∥ for all x ∈ 𝕏 and t ≥ 0.

(ii) The operators (𝕊_{β}(t))_{t≥0} and (ℙ_{β}(t))_{t≥0} are strongly continuous. (iii) The operators (𝕊_{β}(t))_{t>0} and (ℙ_{β}(t))_{t>0} are compact.

Lemma 2.5. Let 𝕊_{β} and ℙ_{β} be the operators defined respectively by (6) and (7). Then

∥𝕊_{β}(t)x∥_{α} ≤ M ∥x∥_{α} , ∀x ∈ 𝕏_{α}, t ≥ 0,

∥ℙ_{β}(t)x∥_{α} ≤ \{ \begin{matrix}\frac{βM_{α}t^{−βα}Γ(2 − α)}{Γ(1 + β(1 − α))} ∥x∥ & if & x ∈ 𝕏, t > 0, \\ \frac{β}{Γ(1 + β)}∥x∥_{α} & if & x ∈ 𝕏_{α}, t > 0. \end{matrix}

Proof. Using (3) and (5b) we have for any x ∈ 𝕏_{α} and t ≥ 0,

∥𝕊_{β}(t)x∥_{α} = ∥∫_{0}^{∞}Φ_{β}(θ)R(θt^{β})x dθx∥_{α}
≤ ∫_{0}^{∞}Φ_{β}(θ) ∥A^{α}R(θt^{β})x∥ dθx
≤ M∫_{0}^{∞}Φ_{β}(θ) ∥A^{α}x∥ dθ
= M ∥x∥_{α} , ∀x ∈ 𝕏_{α}.

In view of (4) and (5c), we can write for any t > 0,

∥ℙ_{β}(t)x∥_{α} = ∥∫_{0}^{∞}βθΦ_{β}(θ)R(θt^{β})x dθ∥_{α}
≤ ∫_{0}^{∞}βθΦ_{β}(θ) ∥A^{α}R(θt^{β})x∥ dθ
≤ ∫_{0}^{∞}βθΦ_{β}(θ)∥A^{α}R(θt^{β})∥_{𝔹(𝕏)}∥x∥ dθ
≤ βM_{α}t^{−αβ}∥x∥∫_{0}^{∞}θ^{1−α}Φ_{β}(θ) dθ
≤ \frac{βM_{α}t^{−βα}Γ(2 − α)}{Γ(1 + β(1 − α))} ∥x∥ , ∀x ∈ 𝕏

and

∥ℙ_{β}(t)x∥_{α} = ∥∫_{0}^{∞}βθΦ_{β}(θ)R(θt^{β})x dθ∥_{α}
≤ ∫_{0}^{∞}βθΦ_{β}(θ) ∥A^{α}R(θt^{β})x∥ dθ
≤ M∥x∥_{α} ∫_{0}^{∞} βθΦ(θ)dθ
= M∥x∥_{α} \frac{β}{Γ(1 + β)}, ∀x ∈ 𝕏_{α}.
□

EJQTDE, 2010 No. 58, p. 4 Definition 2.6. ([5, 6]) Let 𝕊_{β} and ℙ_{β} be operators defined respectively by (6) and (7). Then a continuous function x : I → 𝕏 satisfying for any t ∈ [0, T] the equation

(8) \begin{aligned}x(t) & = & 𝕊_{β}(t)(x_{0} − g(x)) \\ & + & ∫_{0}^{t}(t − s)^{β−1}ℙ_{β}(t − s) [(f(s, x(s)) − ∫_{0}^{t}a(t − s)h(s, x(s))] ds, \end{aligned}

is called a mild solution of the equation (1).

In the sequel, we set

(9)  Kx(t) := ∫_{0}^{t}a(t − s)h(s, x(s)) ds.

We set α ∈ (0, 1) and we will denote by C_{α}, the Banach space C([0, T], 𝕏_{α}) endowed with the supnorm given by

∥x∥_{∞} := sup_{t∈I} ∥x∥_{α}, for x ∈ C.

3. Main Results

In addition to the previous assumptions, we assume that the following hold. (H_{1}) The function f : I × 𝕏_{α} → 𝕏 is continuous and satisfies the following

condition: there exists a function μ_{1}(t) ∈ L^{∞} (I, ℝ^{+}) such that

∥f(t, x,) − f(t, y)∥ ≤ μ_{1}(t)∥x − y∥_{α}
for all t ∈ I, x, y ∈ 𝕏_{α}.

(H_{2}) The function h : I × 𝕏_{α} → 𝕏 is continuous and satisfies the following condition: there exists a function μ_{2}(t) ∈ L^{∞} (I, ℝ^{+}) such that

∥h(t, x,) − h(t, y)∥ ≤ μ_{2}(t)∥x − y∥_{α}
for all t ∈ I, x, y ∈ 𝕏_{α}.

(H_{3}) The function g : C_{α} → 𝕏_{α} is continuous and there exists a constant b such

that

∥g(x) − g(y)∥_{α} ≤ b∥x − y∥_{∞}

for all x, y ∈ C_{α}.

Theorem 3.1. Suppose assumptions (H_{1})-(H_{3}) hold and that Ω_{α,β,T} < 1 where

Ω_{α,β,T} = [Mb + \frac{βM_{α}Γ(2 − α)T^{β(1−α)}}{Γ(1 + β(1 − α))(β(1 − α))}(∥μ_{1}∥_{L^{∞}(I,ℝ_{+})} + a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})})].

If x_{0} ∈ 𝕏_{α}, then (1) has a unique mild solution x ∈ C_{α}.

EJQTDE, 2010 No. 58, p. 5 Proof. Define the nonlinear integral operator F : C_{α} → C_{α} by

\begin{aligned}(F x)(t) & = & 𝕊_{β}(t) (x_{0} − g(x)) , \\ & + & ∫_{0}^{t}(t − s)^{β−1} ℙ_{β}(t − s) [f(s, x(s)) + Kx(s)] ds. \end{aligned}

where K is given by (9).

In view of Lemma 2.4- (ii), the integral operator F is well defined. Now take t ∈ I and x, y ∈ C_{α}. We have

∥(F x)(t) − (F y)(t)∥_{α} ≤ ∥𝕊_{β}(t) (g(x) − g(y)) ∥_{α}
+ ∫_{0}^{t}(t − s)^{β−1} ∥ℙ_{β}(t − s) (f(s, x(s)) − f(s, y(s)))∥_{α} ds
+ ∫_{0}^{t}(t − s)^{β−1} ∥ℙ_{β}(t − s) (K x(s) − K y(s))∥_{α} ds

which according to Lemma 2.5 and (H_{3}) gives

∥(F x)(t) − (F y)(t)∥_{α} ≤ Mb∥x − y∥_{∞}
+ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t}(t − s)^{β(1−α)−1} ∥(f(s, x(s)) − f(s, y(s)))∥ ds
+ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t}(t − s)^{β(1−α)−1} ∥(K x(s) − K y(s))∥ ds

Since (H_{2}) and (2) hold, we can write

∥Kx(s) − Ky(s)∥ = ∫_{0}^{s}a(s − τ) ∥h(τ, x(τ)) − h(τ, y(τ))∥ dτ
≤ ∫_{0}^{s}a(s − τ)μ_{2}(τ) ∥x(τ) − y(τ)∥ dτ
≤ a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})}∥x − y∥_{∞}.

Thus, using (H_{1}) we obtain

∥(F x)(t) − (F y)(t)∥_{α}≤Mb∥x − y∥_{∞}
+\frac{βM_{α}Γ(2 − α)∥x − y∥_{∞}}{Γ(1 + β(1 − α))} ∫_{0}^{t}(t − s)^{β(1−α)−1}μ_{1}(s) ds
+\frac{βM_{α}Γ(2 − α)T^{β(1−α)}}{Γ(1 + β(1 − α))(β(1 − α)}a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})}∥x − y∥_{∞}
≤Mb∥x − y∥_{∞}
+\frac{βM_{α}Γ(2 − α)T^{β(1−α)}}{Γ(1 + β(1 − α))(β(1 − α))}∥x − y∥_{∞}∥μ_{1}∥_{L^{∞}(I,ℝ_{+})}
+\frac{βM_{α}Γ(2 − α)T^{β(1−α)}a_{T}}{Γ(1 + β(1 − α))(β(1 − α))}∥x − y∥_{∞}∥μ_{2}∥_{L^{∞}(I,ℝ_{+})}
≤Ω_{α,β,T} ∥x − y∥_{∞}

EJQTDE, 2010 No. 58, p. 6 So we get

∥(F x)(t) − (F y)(t)∥_{∞} ≤ Ω_{α,β,T} (t)∥x − y∥_{∞}.

Since Ω_{α,β,T} < 1, the contraction mapping principle enables us to say that, F has a unique fixed point in C_{α},

x(t) = 𝕊_{β}(t) (x_{0} − g(x)) + ∫_{0}^{t}(t − s)^{β−1} ℙ_{β}(t − s) [f(s, x(s)) + K x(s)] ds

which is the mild solution of (1). □

Now we assume that

(H_{4}) The function f : I × 𝕏_{α} → 𝕏 is continuous and satisfies the following condition: there exists a positive function μ_{1} ∈ L^{∞}(I, ℝ^{+}) such that

∥f(t, x)∥ ≤ μ_{1}(t),

(H_{5}) The function h : I × 𝕏_{α} → 𝕏 is continuous and satisfies the following condition: there exists a positive function μ_{2} ∈ L^{∞}(I, ℝ^{+}) such that

∥h(t, x)∥ ≤ μ_{2}(t),

(H_{6}) The function g ∈ C(C_{α}, 𝕏_{α}) is completely continuous and there exist λ, γ >

0 such that

∥g(x)∥_{α} ≤ λ∥x∥_{∞} + γ.

Theorem 3.2. Suppose that assumptions (H_{4})-(H_{6}) hold. If x_{0} ∈ 𝕏_{α} and

(10)  Mλ < \frac{1}{2}

then (1.1) has a mild solution on [0, T].

Proof. Define the integral operator F : C_{α} → C_{α} by

\begin{aligned}(F x)(t) & = & 𝕊_{β}(t) (x_{0} − g(x)) , \\ & + & ∫_{0}^{t}(t − s)^{β−1}ℙ_{β}(t − s) [f(s, x(s)) + Kx(s)] ds, \end{aligned}

and choose r such that

\begin{aligned}r & ≥ & 2\frac{T^{β(1−α)}βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))(β(1 − α))}(∥μ_{1}∥_{L^{∞}(I,ℝ_{+})} + a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})}) \\ & + & 2M(∥x_{0}∥_{α} + γ). \end{aligned}

Let B_{r} = {x ∈ C_{α} : ∥x∥_{∞} ≤ r}. We proceed in three main steps. EJQTDE, 2010 No. 58, p. 7

Step 1. We show that F (B_{r}) ⊂ B_{r}. For that, let x ∈ B_{r}. Then for t ∈ I, we have

∥(F x)(t)∥_{α} ≤ ∥𝕊_{α}(t) (x_{0} − g(x))∥_{α}
+ ∫_{0}^{t}(t − s)^{β−1} ∥ℙ_{α}(t − s)f(s, x(s))∥_{α} ds
+ ∫_{0}^{t}(t − s)^{β−1} ∥ℙ_{α}(t − s)K x(s)∥_{α} ds

which according to (H_{4})-(H_{6}) and Lemma 2.5 gives

∥(F x)(t)∥_{α} ≤ M (∥x_{0}∥_{α} + λ∥x∥_{∞} + γ)
+ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t}(t − s)^{β(1−α)−1} ∥f(s, x(s))∥ ds
+ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t}(t − s)^{β(1−α)−1} ∥Kx(s)∥ ds
≤ M (∥x_{0}∥_{α} + λ∥x∥_{∞} + γ)
+ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t}(t − s)^{β(1−α)−1}μ_{1}(s) ds
+ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t}(t − s)^{β(1−α)−1}∫_{0}^{s}a(s − τ)μ_{2}(τ)dτ ds.

Consequently, using the inequality Mλ < \frac{1}{2}, which yields Mλ∥x∥_{∞} < \frac{r}{2} and the choice of r above, we get

∥(F x)(t)∥_{α} ≤ M (∥x_{0}∥_{α} + λ∥x∥_{∞} + γ)
+ \frac{∥μ_{1}∥_{L^{∞}(I,ℝ_{+})}T^{β(1−α)}}{(β(1 − α)} \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}
+ \frac{∥μ_{2}∥_{L^{∞}(I,ℝ_{+})}T^{β(1−α)}}{(β(1 − α)} \frac{βM_{α}Γ(2 − α)a_{T}}{Γ(1 + β(1 − α))} .

In view of (10) and the choice of r, we obtain

∥(F x)∥_{∞} ≤ r.

Step 2. We prove that F is continuous. For that, let (x_{n}) be a sequence of B_{r} such that x_{n} → x in B_{r}. Then

\begin{aligned}f(s, x_{n}(s)) & → & f(s, x(s)), & n → ∞, \\ h(t, x_{n}(s)) & → & h(t, x(s)), & n → ∞ \end{aligned}

as both f and h are jointly continuous on I × 𝕏_{α}.

Now, for all t ∈ I, we have

∥F x_{n} − F x∥_{α} ≤ ∥𝕊_{β}(t)(g(x_{n}) − g(x))∥_{α}
+ ∥∫_{0}^{t}(t − s)^{β−1}ℙ_{β}(t − s) (Kx_{n}(s) − Kx(s)) ds∥_{α}
+ ∥∫_{0}^{t}(t − s)^{β−1}S(t − s) (f(s, x_{n}(s)) − f(s, x(s))) ds∥_{α} ,

EJQTDE, 2010 No. 58, p. 8 which in view of Lemma 2.5 gives

∥F x_{n} − F x∥_{α}≤ M∥g(x_{n}) − g(x)∥_{α}
+ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t}(t − s)^{β(1−α)−1} ∥f(s, x_{n}(s)) − f(s, x(s))∥ ds
+ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t}(t − s)^{β(1−α)−1} ∥Kx_{n}(s) − Kx(s)∥ ds

for all t ∈ I. Therefore, on the one hand using (2), (H_{4}) and (H_{5}), we get for each t ∈ I

∥f(s, x_{n}(s)) − f(s, x(s))∥ ≤ 2μ_{1}(s) for s ∈ I,
∥Kx_{n}(s) − Kx(s)∥  ≤ ∫_{0}^{s}a(s − τ)∥h(τ, x_{n}(τ)) − h(τ, x(τ))∥dτ,
≤ 2∫_{0}^{s}a(s − τ)μ_{2}(τ)dτ
≤ 2a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})} for s ∈ I;

and on the other hand using the fact that the functions s ↦ 2μ_{1}(s)(t − s)^{β(1−α)−1} and s ↦ (t − s)^{β(1−α)−1} are integrable on I, by means of the Lebesgue Dominated Convergence Theorem yields

\begin{matrix}∫_{0}^{t}(t − s)^{β(1−α)−1} ∥f(s, x_{n}(s)) − f(s, x(s))∥ ds → 0, \\ ∫_{0}^{t}(t − s)^{β(1−α)−1} ∥Kx_{n}(s) − Kx(s)∥ ds → 0. \end{matrix}

Hence, since g(x_{n}) → g(x) as n → ∞ because g is completely continuous on C_{α}, it can easily be shown that

lim_{n→∞} ∥(F x_{n}) − (F x)∥_{∞} = 0,

as n → ∞.

In other words, F is continuous.

Step 3. We show that F is compact. To this end, we use the Ascoli-Arzela’s theorem. For that, we first prove that {(F x)(t) : x ∈ B_{r}} is relatively compact in 𝕏_{α}, for all t ∈ I. Obviously,{(F x)(0) : x ∈ B_{r}} is compact.

EJQTDE, 2010 No. 58, p. 9 Let t ∈ (0, T]. For each h ∈ (0, t), ϵ > 0 and x ∈ B_{r}, we define the operator F_{h,ϵ} by

(F_{h,ϵ}x)(t) = 𝕊_{β}(t) (x_{0} − g(x))
+ ∫_{0}^{t−h}(t − s)^{β−1}∫_{ϵ}^{∞}βθΦ_{β}(θ)R((t − s)^{β}θ)f(s, x(s))dθ ds
+ ∫_{0}^{t−h}(t − s)^{β−1}∫_{ϵ}^{∞}βθΦ_{β}(θ)R((t − s)^{β}θ)Kx(s)dθ ds
= 𝕊_{β}(t) (x_{0} − g(x))
+ R(h^{β}ϵ)∫_{0}^{t−h}(t − s)^{β−1}∫_{ϵ}^{∞}βθΦ_{β}(θ)R((t − s)^{β}θ − h^{β}ϵ)f(s, x(s))dθ ds
+ R(h^{β}ϵ)∫_{0}^{t−h}(t − s)^{β−1}∫_{ϵ}^{∞}βθΦ_{β}(θ)R((t − s)^{β}θ − h^{β}ϵ)Kx(s)dθ ds.

Then the sets {(F_{h,ϵ}x)(t) : x ∈ B_{r}} are relatively compact in 𝕏_{α} since by Lemma 2.3, the operators R_{α}(t), t > 0 are compact on 𝕏_{α}. Moreover, using (H_{1}) and (4) we have

∥(F x)(t) − (F_{h,ϵ}x)(t)∥_{α} ≤
∫_{0}^{t}(t − s)^{β−1}∫_{0}^{ϵ}βθΦ_{β}(θ) ∥R((t − s)^{β}θ)f(s, x(s))∥_{α} dθ ds+
∫_{t−h}^{t}(t − s)^{β−1}∫_{ϵ}^{∞}βθΦ_{β}(θ) ∥R((t − s)^{β}θ)f(s, x(s))∥_{α} dθ ds+
∫_{0}^{t}(t − s)^{β−1}∫_{0}^{ϵ}βθΦ_{β}(θ) ∥R((t − s)^{β}θ)Kx(s)∥_{α} dθ ds+
∫_{t−h}^{t}(t − s)^{β−1}∫_{ϵ}^{∞}βθΦ_{β}(θ) ∥R((t − s)^{β}θ)Kx(s)∥_{α} dθ ds.

Then using (4) and (H_{4}), we obtain

∥(F x)(t) − (F_{h,ϵ}x)(t)∥_{α} ≤ βM_{α}∫_{0}^{t}(t − s)^{β(1−α)−1}μ_{1}(s)∫_{0}^{ϵ}θ^{1−α}Φ_{β}(θ)dθ ds
+ βM_{α}∫_{t−h}^{t}(t − s)^{β(1−α)−1}μ_{1}(s)∫_{ϵ}^{∞}βθ^{1−α}Φ_{β}(θ)dθ ds
+ βM_{α}∫_{0}^{t}(t − s)^{β(1−α)−1}∫_{0}^{ϵ}βθ^{1−α}Φ_{β}(θ) ∥Kx(s)∥ dθ ds
+ βM_{α}∫_{t−h}^{t}(t − s)^{β(1−α)−1}∫_{ϵ}^{∞}βθ^{1−α}Φ_{β}(θ) ∥Kx(s)∥ dθ ds.

Since by (H_{5}) and (2),

∥Kx(s)∥ ≤ ∫_{0}^{s}a(s − τ)∥h(τ, x(τ))∥dτ
≤ ∫_{0}^{s}a(s − τ)μ_{2}(τ)dτ
≤ a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})},

EJQTDE, 2010 No. 58, p. 10 using (5c), we deduce for all ϵ > 0 that

∥(F x)(t) − (F_{h,ϵ}x)(t)∥_{α} ≤ \frac{t^{β(1−α)}βM_{α}∥μ_{1}∥_{L^{∞}(I,ℝ_{+})}}{β(1 − α)} ∫_{0}^{ϵ}θ^{1−α}Φ_{β}(θ)dθ
+ \frac{h^{β(1−α)}βM_{α}Γ(2 − α)∥μ_{1}∥_{L^{∞}(I,ℝ_{+})}}{β(1 − α)Γ(1 + β(1 − α))}
+ \frac{t^{β(1−α)}βM_{α}∥μ_{2}∥_{L^{∞}(I,ℝ_{+})}a_{T}}{β(1 − α)} ∫_{0}^{ϵ}θ^{1−α}Φ_{β}(θ)dθ
+ \frac{h^{β(1−α)}βM_{α}Γ(2 − α)a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})}}{β(1 − α)Γ(1 + β(1 − α))} .

In other words

\begin{aligned}∥(F x)(t) − (F_{h,ϵ}x)(t)∥_{α} & ≤ & \frac{h^{β(1−α)}βM_{α}Γ(2 − α)∥μ_{1}∥_{L^{∞}(I,ℝ_{+})}}{β(1 − α)Γ(1 + β(1 − α))} \\ & + & \frac{h^{β(1−α)}βM_{α}Γ(2 − α)a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})}}{β(1 − α)Γ(1 + β(1 − α))} . \end{aligned}

Therefore, the set {(F x)(t) : x ∈ B_{r}} is relatively compact in 𝕏_{α} for all t ∈ (0, T] and since it is compact at t = 0 we have the relatively compactness in 𝕏_{α} for all t ∈ I. Now, let us prove that F (B_{r}) is equicontinuous. By the compactness of the set g(B_{r}), we can prove that the functions F x, x ∈ B_{r} are equicontinuous a t = 0. For 0 < t_{2} < t_{1} ≤ T, we have

∥(F x)(t_{1}) − (F x)(t_{2})∥_{α} ≤ ∥(𝕊_{β}(t_{1}) − 𝕊_{β}(t_{2})) (x_{0} − g(x))∥_{α}
+ ∥∫_{0}^{t_{2}}(t_{1} − s)^{β−1} (ℙ_{β}(t_{1} − s) − ℙ_{β}(t_{2} − s)) (f(s, x(s)) + Kx(s)) ds∥_{α}
+ ∥∫_{0}^{t_{2}} ((t_{1} − s)^{β−1} − (t_{2} − s)^{β−1}) ℙ_{β}(t_{2} − s)(f(s, x(s)) + Kx(s)) ds∥_{α}
+ ∥∫_{t_{2}}^{t_{1}}(t_{1} − s)^{β−1}ℙ_{β}(t_{1} − s)(f(s, x(s)) + Kx(s)) ds∥_{α}
≤ I_{1} + I_{2} + I_{3} + I_{4},

where

\begin{aligned}I_{1} & = & ∥(𝕊_{β}(t_{1}) − 𝕊_{β}(t_{2})) (x_{0} − g(x))∥_{α} \\ I_{2} & = & ∥∫_{0}^{t_{2}}(t_{1} − s)^{β−1} (ℙ_{β}(t_{1} − s) − ℙ_{β}(t_{2} − s)) (f(s, x(s)) + Kx(s)) ds∥_{α} \\ I_{3} & = & ∥∫_{0}^{t_{2}} ((t_{1} − s)^{β−1} − (t_{2} − s)^{β−1}) ℙ_{β}(t_{2} − s)(f(s, x(s)) + Kx(s)) ds∥_{α} \\ I_{4} & = & ∥∫_{t_{2}}^{t_{1}}(t_{1} − s)^{β−1} ℙ_{β}(t_{1} − s)(f(s, x(s)) + Kx(s)) ds∥_{α} \end{aligned}

Actually, I_{1}, I_{2}, I_{3} and I_{4} tend to 0 independently of x ∈ B_{r} when t_{2} → t_{1}. Indeed, let x ∈ B_{r} and G = sup_{x∈Cα} ∥g(x)∥_{α}. In view of Lemma 2.5, we have

I_{1} = ∥(𝕊_{β}(t_{1}) − 𝕊_{β}(t_{2})) (x_{0} − g(x))∥_{α}
≤ ∫_{0}^{∞}Φ_{β}(θ) ∥R(θt_{1}^{β}) − R(θt_{2}^{β})∥_{𝔹(𝕏)} ∥x_{0} − g(x)∥_{α} dθ
≤ ∫_{0}^{∞}Φ_{β}(θ) ∥R(θt_{1}^{β}) − R(θt_{2}^{β})∥_{𝔹(𝕏)} (∥x_{0}∥_{α} + G)dθ

from which we deduce that lim_{t_{2}→t_{1}}I_{1} = 0 since by Lemma 2.3 the function t ↦ ∥R_{α}(t)∥_{α} is continuous for t ≥ 0

I_{2} ≤ ∫_{0}^{t_{2}} ∥(t_{1} − s)^{β−1} (ℙ_{β}(t_{1} − s) − ℙ_{β}(t_{2} − s)) (f(s, x(s)) + Kx(s))∥_{α} ds.

Therefore using the continuity of ℙ_{β}(t) (Lemma 2.4) and the fact that both f and K are bounded we conclude that lim_{t_{2}→t_{1}}I_{2} = 0

I_{3} ≤ ∫_{0}^{t_{2}} ((t_{2} − s)^{β−1} − (t_{1} − s)^{β−1}) ∥ℙ_{β}(t_{2} − s)(f(s, x(s)) + Kx(s))∥_{α} ds
≤ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t_{2}} ((t_{2} − s)^{β−1} − (t_{1} − s)^{β−1}) (t_{2} − s)^{−αβ} ∥f(s, x(s))∥ ds
+ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{0}^{t_{2}} ((t_{2} − s)^{β−1} − (t_{1} − s)^{β−1}) (t_{2} − s)^{−αβ} ∥Kx(s)∥ ds.

Since −(t_{2}−s)^{−αβ}(t_{1}−s)^{β−1} ≤ −(t_{1}−s)^{β(1−α)−1} because (t_{1}−s)^{−αβ} ≤ (t_{2}−s)^{−αβ}, we deduce that

I_{3} ≤ \frac{βM_{α}Γ(2 − α)∥μ_{1}∥_{L^{∞}(I,ℝ_{+})}}{Γ(1 + β(1 − α))} ∫_{0}^{t_{2}} ((t_{2} − s)^{β(1−α)−1} − (t_{1} − s)^{β(1−α)−1}) ds
+ \frac{a_{T} βM_{α}Γ(2 − α)∥μ_{1}∥_{L^{∞}(I,ℝ_{+})}}{Γ(1 + β(1 − α))} ∫_{0}^{t_{2}} ((t_{2} − s)^{β(1−α)−1} − (t_{1} − s)^{β(1−α)−1}) ds
≤ \frac{βM_{α}Γ(2 − α)∥μ_{1}∥_{L^{∞}(I,ℝ_{+})}}{β(1 − α)Γ(1 + β(1 − α))} (t_{1} − t_{2})^{β(1−α)}
+ \frac{a_{T} βM_{α}Γ(2 − α)∥μ_{1}∥_{L^{∞}(I,ℝ_{+})}}{β(1 − α)Γ(1 + β(1 − α))} (t_{1} − t_{2})^{β(1−α)}.

Hence lim_{t_{2}→t_{1}}I_{3} = 0 since β(1 − α) > 0.

EJQTDE, 2010 No. 58, p. 12

I_{4} ≤ ∫_{t_{2}}^{t_{1}}(t_{1} − s)^{β−1} ∥ℙ_{β}(t_{1} − s)(f(s, x(s) + Kx(s))∥_{α} ds
≤ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{t_{2}}^{t_{1}}(t_{1} − s)^{β(1−α)−1} ∥f(s, x(s)) + Bx(s))∥ ds
≤ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}∫_{t_{2}}^{t_{1}}(t_{1} − s)^{β(1−α)−1}(μ_{1}(s) + ∫_{0}^{s}a(s − τ)∥h(τ, x(τ))∥dτ)ds
≤ \frac{βM_{α}Γ(2 − α)}{Γ(1 + β(1 − α))}(∥μ_{1}∥_{L^{∞}(I,ℝ_{+})} + a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})})∫_{t_{2}}^{t_{1}}(t_{1} − s)^{β(1−α)−1}ds
≤ \frac{(t_{1} − t_{2})^{β(1−α)}βM_{α}Γ(2 − α)}{β(1 − α)Γ(1 + β(1 − α))}(∥μ_{1}∥_{L^{∞}(I,ℝ_{+})} + a_{T} ∥μ_{2}∥_{L^{∞}(I,ℝ_{+})})

Since β(1 − α) > 0, we deduce that lim_{t_{2}→t_{1}}I_{4} = 0.

In short, we have shown that F (B_{r}) is relatively compact, for t ∈ I, {F x : x ∈ B_{r}} is a family of equicontinuous functions. Hence by the Arzela-Ascoli Theorem, F is compact. By Schauder fixed point theorem F has a fixed point x ∈ B_{r}, which obviously is a mild solution to (1). □

4. Example

Let 𝕏 = L^{2}[0, π] equipped with its natural norm and inner product defined respectively for all u, v ∈ L^{2}[0, π] by

∥u∥_{L^{2}[0,π]} = ( ∫_{0}^{π} |u(x)|^{2}dx)^{1/2} and ⟨u, v⟩ = ∫_{0}^{π} u(x)\overline{v(x)}dx.

Consider the following integro-partial differential equation

(E) \{ \begin{matrix}\frac{∂^{β}u}{∂t^{β}}(t, x) = \frac{∂^{2}u}{∂x^{2}}(t, x) + \frac{cos(tx)}{1 + u^{2}(t, x)} + ∫_{0}^{t}e^{−|t−s|} cos(u(s, x)) ds, \\ u(t, 0) = u(t, π) = 0, t ∈ [0, 1] \\ u(0, x) + δ_{0} ∑_{k=0}^{N} ∫_{0}^{π} cos(x − y)u(t_{k}, y)dy = u_{0}(x), x ∈ [0, π] \end{matrix}

where t ∈ [0, 1], x ∈ [0, π], 0 < t_{1} < t_{2} < ... < t_{N} ≤ 1, and δ_{0} > 0.

First of all, note that f, h, a are given by

f(t, u(t, x)) = \frac{cos(tx)}{1 + u^{2}(t, x)}, a(t) = e^{−|t|}, and h(t, u(t, x)) = cos(u(s, x)),

and hence in (H_{4}) and (H_{5}) we take μ_{1}(t) = μ_{2}(t) = π. Moreover, a_{1} = ∫_{0}^{1} e^{−|t|}dt = 1 − e^{−1}.

EJQTDE, 2010 No. 58, p. 13 Let A be the operator given by Au = −u′′ with domain

D(A) := {u ∈ L^{2}([0, π]) : u′′ ∈ L^{2}([0, π]), u(0) = u(π) = 0}.

It is well known that A has a discrete spectrum with eigenvalues of the form n^{2}, n ∈ ℕ, and corresponding normalized eigenfunctions given by

z_{n}(ξ) := \sqrt{\frac{2}{π}} sin(nξ).

In addition to the above, the following properties hold:

(a) {z_{n} : n ∈ ℕ} is an orthonormal basis for L^{2}[0, π];

(b) The operator −A is the infinitesimal generator of an analytic semigroup R(t) which is compact for t > 0. The semigroup R(t) is defined for u ∈ L^{2}[0, π] by

R(t)u = ∑_{n=1}^{∞} e^{−n^{2}t}⟨u, z_{n}⟩z_{n}.

(c) The operator A can be rewritten as

Au = ∑_{n=1}^{∞} n^{2}⟨u, z_{n}⟩z_{n}

for every u ∈ D(A).

Moreover, it is possible to define fractional powers of A. In particular,

(d) For u ∈ L^{2}[0, π] and α ∈ (0, 1),

A^{−α}u = ∑_{n=1}^{∞} \frac{1}{n^{2α}} ⟨u, z_{n}⟩z_{n};

(e) The operator A^{α} : D(A^{α}) ⊆ L^{2}[0, π] ↦ L^{2}[0, π] given by

A^{α}u = ∑_{n=1}^{∞} n^{2α}⟨u, z_{n}⟩z_{n}, ∀u ∈ D(A^{α}),

where D(A^{α}) = {u ∈ L^{2}[0, π] : ∑_{n=1}^{∞} n^{2α}⟨u, z_{n}⟩z_{n} ∈ L^{2}[0, π]}.

EJQTDE, 2010 No. 58, p. 14 Clearly for all t ≥ 0 and 0 ≠ u ∈ L^{2}[0, π],

|R(t)u| = | ∑_{n=1}^{∞} e^{−n^{2}t}⟨u, z_{n}⟩z_{n}|
≤ ∑_{n=1}^{∞} e^{−t}|⟨u, z_{n}⟩z_{n}|
= e^{−t} ∑_{n=1}^{∞} |⟨u, z_{n}⟩z_{n}|
≤ e^{−t}|u|

and hence ∥R(t)∥_{B(L^{2}[0,π])} ≤ 1 for all t ≥ 0. Here we take M = 1. Set

g(u)(ξ) := δ_{0} ∑_{k=0}^{N} ∫_{0}^{π} cos(ξ − y)u(t_{k}, y)dy.

Suppose α ∈ (0, \frac{1}{2}) and

δ_{0} < \frac{\sqrt{6}}{2π^{2}N} . (11)

Now

∥A^{α}g(u)(ξ)∥_{L^{2}[0,π]}^{2} = ∑_{n≥1} n^{4α}∥z_{n}∥_{L^{2}[0,π]}^{2} |⟨g(u)(ξ), z_{n}⟩|^{2}
≤ ∑_{n≥1} n^{2}|⟨g(u)(ξ), z_{n}⟩|^{2}
= \frac{2}{π} ∑_{n≥1} | ∫_{0}^{π} g(u)(ξ) n sin(nξ)dξ|^{2}
= ∑_{n≥1} \frac{1}{n^{2}} | ∫_{0}^{π} \frac{∂^{2}}{∂ξ^{2}} g(u)(ξ) z_{n}(ξ)dξ|^{2}
≤ \frac{π^{2}}{6} ∥\frac{∂^{2}}{∂ξ^{2}} g(u)(ξ)∥_{L^{2}[0,π]}^{2}
≤ \frac{π^{2}}{6} ∥g(u)(ξ)∥_{L^{2}[0,π]}^{2}
≤ δ_{0}^{2} \frac{π^{2}}{6} N^{2}π^{2}∥u∥_{∞}^{2}

and hence ∥g(u)∥_{α} ≤ λ∥u∥_{∞} + μ where λ = \frac{δ_{0}π^{2}N}{\sqrt{6}} and μ = 0. Therefore, the

condition Mλ < \frac{1}{2} holds under assumption (11).

Using Theorem 3.2 and inequality Eq. (11) it follows that the system (E) at least one mild solution.

Acknowledgements: This work was completed when the second author was visiting Morgan State University in Baltimore, MD, USA in May 2010. She likes to thank Prof. N’Guérékata for the invitation.

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(Received May 19, 2010)

Toka Diagana, Department of Mathematics, Howard University, 2441 6th Street NW, Washington, DC, 20009, USA

E-mail address: tdiagana@howard.edu

Gisèle M. Mophou,Université des Antilles et de la Guadeloupe, Département de Mathématiques et Informatique, Université des Antilles et de La Guyane, Campus Fouillole 97159 Pointe-à-Pitre Guadeloupe (FWI)

E-mail address: gmophou@univ-ag.fr

Gaston M. N’Guérékata, Department of Mathematics, Morgan State University, 1700 E. Cold Spring Lane, Baltimore, M.D. 21251, USA

E-mail address: Gaston.N'Guerekata@morgan.edu