Implicit Fractional Differential Equations via the ... - Semantic Scholar
Mathematics 2015, 3, 398-411; doi:10.3390/math3020398
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mathematics ISSN 2227-7390 www.mdpi.com/journal/mathematics Article
Implicit Fractional Differential Equations via the Liouville–Caputo Derivative Juan J. Nieto 1,2, *, Abelghani Ouahab 3 and Venktesh Venktesh 1,4 1
Department of Mathematical Analysis, University of Santiago de Compostela, Santiago de Compostela 15782, Spain; E-Mail: [email protected] 2 Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia 3 Laboratory of Mathematics, Sidi-Bel-Abbès University, P.O. Box 89, Sidi-Bel-Abbès 22000, Algeria; E-Mail: [email protected] 4 Birla Institute of Technology & Science, Pilani 333031, India * Author to whom correspondence should be addressed; E-Mail: [email protected]; Tel.: +34-881-813-177; Fax: +34-981-597-054. Academic Editor: Hari M. Srivastava Received: 16 March 2015 / Accepted: 13 May 2015 / Published: 25 May 2015
Abstract: We study an initial value problem for an implicit fractional differential equation with the Liouville–Caputo fractional derivative. By using fixed point theory and an approximation method, we obtain some existence and uniqueness results. Keywords: fractional differential equations; fractional integral; fractional derivative; Liouville–Caputo derivative; implicit
1. Introduction Differential equations of fractional order have recently been proven to be valuable tools in the modeling of many physical phenomena [1–3]. There has also been a significant theoretical development in fractional differential equations in recent years; see the monographs of Abbas et al. [4], Kilbas et al. [5], Miller and Ross [6], Podlubny [7] and Samko et al. [8]. The basic theory for initial value problems for fractional differential equations involving the Riemann–Liouville and Liouville–Caputo differential operator was discussed by Diethelm [9].
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Recently, fractional functional differential equations, fractional differential inclusions and impulsive fractional differential equations with different conditions were studied, for example, by Aghajani et al. [10], Ahmad and Nieto [11], Benchohra et al. [12], Chalishajar and Karthikeyan [13,14], Henderson and Ouahab [15,16], Ouahab [17,18] and in the references therein. In this paper, we study the existence of solutions of the following implicit fractional differential equation with initial condition: c α c α D y(t) = f (t, y(t), D y(t)), t ∈ J (1) y(0) = y0 where J = [0, b], 0 < α < 1 and f : J × R × R → R is a continuous function. This problem is motivated by the importance of implicit ordinary differential equations of the form: f (t, x(t), x0 (t), . . . , x(n−1) (t)) = 0.
(2)
under various initial and boundary conditions. Implicit equations have been considered by many authors [19–26]. Furthermore, our intention is to extend the results to implicit differential equations of fractional order. Very recently, some existence results for an implicit fractional differential equation on compact intervals were investigated [27–29]. Our goal in this work is to give some existence and uniqueness results for implicit fractional differential equations. 2. Fractional Calculus According to the Riemann–Liouville approach to fractional calculus, the notation of the fractional integral of order α (α > 0) is a natural consequence of the well-known formula (usually attributed to Cauchy) that reduces the calculation of the n−fold primitive of a function f to a single integral of the convolution type. The Cauchy formula reads: Z t 1 n (t − s)n−1 f (s) ds, t > 0 , n ∈ N. [I f ](t) = (n − 1)! 0 Definition 1. The fractional integral of order α > 0 of a function f ∈ L1 (a, b) is defined by: Z t (t − s)α−1 α Ia+ f (t) := f (s)ds, Γ(α) a where Γ is the classical gamma function. When a = 0, we write I α f (t) = f (t) ∗ φα (t), where φα (t) = tα−1 for t > 0, φα (t) = 0 for t ≤ 0 and φα → δ as α → 0+ , where δ is the Dirac delta function. For Γ(α) consistency, I 0 = Id (Identity operator), i.e., I 0 f (t) = f (t). Furthermore, by I α f (0+ ), we mean the limit (if it exists) of I α f (t) for t → 0+ ; this limit may be infinite. After the notion of the fractional integral, that of the fractional derivative of order α (α > 0) becomes a natural requirement, and one is attempted to substitute α with −α in the above formulas. However,
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this generalization needs some care in order to guarantee the convergence of the integral and to preserve the well-known properties of the ordinary derivative of integer order. Denoting by Dn , with n ∈ N, the operator of the derivative of order n, we first note that: Dn I n = Id,
I n Dn 6= Id,
n ∈ N,
i.e., Dn is the left inverse (and not the right inverse) to the corresponding integral operator I n . We can easily prove that: n−1 X tk n n I D f (t) = f (t) − f (k) (0+ ) , t > 0. k! k=0 As a consequence, we expect that Dα is defined as the left inverse to I α . For the fractional derivative of order α > 0 with integer n, such that n − 1 < α ≤ n, we have: Definition 2. For a function f given on interval [a, b], the Riemann–Liouville fractional-order derivative of order α of f is defined by: n Z t d 1 α (t − s)−α+n−1 f (s)ds, Da f (t) = Γ(n − α) dt a provided the right-hand side is defined. Defining for consistency, D0 = I 0 = Id, then we easily recognize that: Dα I α = Id,
α ≥ 0,
(3)
and D α tγ =
Γ(γ + 1) γ−α t , α > 0, γ > −1, t > 0. Γ(γ + 1 − α)
(4)
Of course, Properties (3) and (4) are a natural generalization of those known when the order is a positive integer. Note the remarkable fact that the fractional derivative Dα f is not zero for the constant function f (t) = 1, if α 6∈ N. In fact, Equation (4) with γ = 0 illustrates that: Dα 1 =
t−α , α > 0, t > 0. Γ(1 − α)
(5)
It is clear that Dα 1 = 0, for α ∈ N, due to the poles of the gamma function at the points 0, −1, −2, . . .. We now observe an alternative definition of fractional derivative, introduced by Caputo [30,31] in the late 1960s and adopted by Caputo and Mainardi [32] in the framework of the theory of linear viscoelasticity (see a review in [2]). Definition 3. Let f ∈ AC n ([a, b]). The Liouville–Caputo fractional-order derivative of f is defined by: Z t 1 c α ( Da f )(t) := (t − s)n−α−1 f (n) (s)ds. Γ(n − α) a
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This definition is of course more restrictive than the Riemann–Liouville definition, in that it requires the absolute integrability of the derivative of order n. Whenever we use the operator c Dα , we (tacitly) assume that this condition is met. We easily recognize that in general: Dα f (t) := Dn I n−α f (t) 6= I n−α Dn f (t) := c Dα f (t),
(6)
unless the function f (t), along with its first n − 1 derivatives, vanishes at t = 0+ . In fact, assuming that the passage of the n−derivative under the integral is legitimate, we recognize that, for n − 1 < α < n and t > 0, n−1 X tk−α α c α f (k) (0+ ), (7) D f (t) = D f (t) + Γ(k − α + 1) k=0 and therefore, recalling the fractional derivative of the power Function (4): Dα
f (t) −
n−1 X k=0
!
k
t f (k) (0+ ) Γ(k + 1)
= c Dα f (t).
(8)
The alternative definition, that is Definition 3, for the fractional derivative thus incorporates the initial values of the function and of lower order. The subtraction of the Taylor polynomial of degree n − 1 at t = 0+ from f (t) means a sort of regularization of the fractional derivative. In particular, according to this definition, the relevant property for which the fractional derivative of a constant is still zero is satisfied, i.e., c α D 1 = 0, α > 0. (9) At the end of this section, we present some properties of a special function. Denote Eα,β the generalized Mittag–Leffler special function defined by: Eα,β (z) :=
∞ X k=0
zk . Γ(kα + β)
Also, 1 Eα,β (z) = 2πi
Z Υ
λα−β eλ dλ λα − z 1
where Υ is a contour, which starts and ends at −∞ and encircles the disc |λ| ≤ |z| α counterclockwise. Lemma 1. [9] Let α > 0, n = dαe and λ ∈ R. The solution of the initial value problem: c α D y(t) = λy(t) + q(t)
y
(k)
(0)
= yk , k = 0, 1, . . . , n − 1,
where q ∈ C[0, b] is a given function, can be expressed in the form: y(t) =
n−1 X k=0
yk uk (t) + y∗ (t)
(10)
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with: y∗ (t) =
α I0 q(t)
1 λ
Rt 0
if λ = 0
q(t − s)u00 (s)ds
if λ 6= 0,
where uk (t) = I0k eα (t), k = 0, 1, . . . , n − 1 and eα (t) := Eα (λtα ). Remark 1. In the case 0 < α < 1, we can rewrite the solution of Problem (10) in the form: Z t α y(t) = y(0)Eα (λt ) + α q(t − s)sα−1 Eα0 (λsα )ds 0
Lemma 2. Let v : [0, b] → [0, ∞) be a real function, and w is a nonnegative, locally-integrable function on [0, b]. Assume that there are constants a > 0 and 0 < β < 1, such that: Z t v(s) v(t) ≤ w(t) + a ds, β 0 (t − s) then there exists a constant K = K(β), such that: t
Z v(t) ≤ w(t) + Ka 0
w(s) ds, (t − s)β
for every t ∈ [0, b]. Proof. From Gronwall’s lemma for singular kernels, whose proof can be found in Lemma 7.1.1 on page 188 of [33], we know that: Z t 0 v(t) ≤ w(t) + c E1−β (c(t − s))w(s)ds 0 0 where c ∈ R is a constant dependent on β. Now, E1−β (s) is bounded for s ∈ [0, b] and for s → t− : 0 E1−β (c(t − s)) ≈
a − s)β
c−β (t
where a > 0 is a constant. This implies that: Z v(t) ≤ w(t) + Ka 0
t
w(s) ds (t − s)β
with constant K = K(β). Lemma 3. Let (X, d) be a non-empty complete metric space with a contraction mapping T : X → X. Then, T admits a unique fixed-point x∗ in X(i.e., T (x∗ ) = x∗ ). Lemma 4. If K is a convex subset of a topological vector space V and T is a continuous mapping of K into itself, so that T (K) is contained in a compact subset of K, then T has a fixed point. Lemma 5. Consider a sequence of real-valued continuous functions {fn }n∈N defined on a closed and bounded interval [0, b]. If this sequence is uniformly bounded and equicontinuous, then there exists a subsequence (fnk ) that converges uniformly.
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For further readings and details on fractional calculus, we refer to the books and papers by Kilbas et al. [5], Podlubny [7] and Samko et al. [8]. 3. Existence and Uniqueness In this section, we prove some existence results and describe the structure of the solution set. We first note that if y ∈ C[0, b] is an absolutely continuous function on [0, b] satisfying Equation (1), then: Z t 1 (t − s)α−1 f (s, y(s), c Dα y(s))ds. y(t) = y0 + Γ(α) 0 Theorem 1. Assume that there exist K1 , K2 > 0, such that: |f (t, x, y) − f (t, x¯, y¯)| ≤ K1 |x − x¯| + K2 |y − y¯|, for each x, x¯, y, y¯ ∈ R. If
bα K1 α
+ K2 < 1, then there exists unique y ∈ C(J, R), which satisfy Equation (1).
Proof. Consider N : C(J, R) → C(J, R) defined by: Z t 1 α−1 (t − s) z(s)ds, z(t) , N (z(t)) = f t, y0 + Γ(α) 0
t ∈ J.
Let z1 , z2 ∈ C(J, R), then: Z t K1 |N (z1 (t)) − N (z2 (t))| ≤ (t − s)α−1 |z1 (s) − z2 (s)|ds + K2 |z1 (t) − z2 (t)| Γ(α) 0 K1 tα ≤ ||z1 − z2 ||∞ + K2 |z1 (t) − z2 (t)| Γ(α) Hence:
kN (z1 ) − N (z2 )k∞ ≤
K 1 bα + K2 kz1 − z2 k∞ , for each z1 , z2 ∈ C(J, R). Γ(α)
From the Banach fixed point theorem, Lemma 3, there exists a unique z ∈ C(J, R), such that z = N (z). Therefore: Z t 1 α−1 z(t) = f t, y0 + (t − s) z(s)ds, z(t) , t ∈ J. Γ(α) 0 Set 1 y(t) = y0 + Γ(α) c
Z
t
(t − s)α−1 z(s)ds.
0
α
This implies that D y(t) = z(t), and hence: c
Dα y(t) = f (t, y(t),c Dα y(t)),
for every t ∈ J.
The goal of the second result of this section is to apply the Schauder fixed point theorem, Lemma 4. For the study of this problem, we present some auxiliary lemmas.
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Lemma 6. Let , 0 ∈ (0, 1) and L : C 1 (J, R) → C(J, R) be an operator defined by: L(z) = c Dα z + 0 z. Then, L is a linear, continuous and invertible operator. Proof. It is clear that L is a linear operator. For every z ∈ C 1 (J, R), we have: −α+1 b 0 kL(z)k∞ ≤ + (kz 0 k∞ + kzk∞ ) for every z ∈ C 1 (J, R). Γ(α) Hence, L is continuous. Now, we show that if L(z) = 0, then z = 0, that is L is injective. Indeed, let z ∈ C(J, R), such that L(z) = 0, then: c Dα z(t) + 0 z(t) = 0, t ∈ J ⇒ z(0) = 0. From Lemma 1, we have: z = 0. Let h ∈ C(J, R); we consider the fractional Cauchy problem: c
Dα z +
0 h(t) z= ,
z(0) = h(0).
Again from Lemma 1, we obtain that: 0 0 Z t α α α−1 0 z(t) = h(0)Eα − t + α (t − s) Eα − (t − s) h(s)ds. 0 This implies that L is bijective, and from the Banach isomorphism theorem, L−1 is a continuous operator. Lemma 7. Let F : C 1 (J, R) → C(J, R) be an operator defined by: F (z) = c Dα z + 0 (y0 + I α f (., z, c Dα z)). We note 1 [L(z)](t) := y0 + Γ(α)
Z
t
(t − s)α−1 z(s)ds,
t ∈ J.
0
Assume: (H1 ) there exist M1 , M3 > 0, M2 ∈ (0, 1), such that: |f (t, x, y)| ≤ M1 |x| + M2 |y| + M3 , for each x, y, ∈ R, t ∈ J. Then, F is continuous and compact. Proof. The proof will be given in three steps. Step 1: F is continuous. Let (zn )n∈N ⊂ C 1 (J, R) be a sequence, such that (zn )n∈N → z ∈ C 1 (J, R). Then:
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Z t α (t − s)α−1 (zn (s) − z(s))ds ≤ t kzn − zk∞ . α 0 Thus: kL(zn ) − L(z)k∞ ≤
bα kzn − zk∞ → 0, n → ∞. Γ(α + 1)
Furthermore: Z t 1−α (t − s)−α (zn0 (s) − z 0 (s))ds ≤ t kzn0 − z 0 k∞ . 1 − α 0 and hence: b1−α kz 0 − z 0 k∞ → 0, n → ∞, Γ(2 − α) n
¯ n ) − L(z)k ¯ kL(z ∞ ≤ where: ¯ L(z) := c Dα z(t) =
1 Γ(1 − α)
t
Z
(t − s)−α z 0 (s),
t ∈ J.
0
Hence: kF (zn ) − F (z)k∞ ≤
0 bα ¯ n )) − f (·, z, L(z))k ¯ kf (·, zn , L(z ∞ Γ(α + 1) ¯ n ) − L(z)k ¯ +kL(z ∞ → 0, as n → ∞.
Step 2: F sends bounded sets in C 1 (J, R) into bounded sets in C(J, R). For each t ∈ J and z ∈ B(0, r), we have: Rt 1 b1−α 0 α−1 ¯ k(F z)(t)k ≤ Γ(2−α) r + |y0 | + 0 Γ(α) (t − s) |f (s, z(s), (Lz)(s))|ds Rt 1 b1−α ¯ r + 0 |y0 | + 0 Γ(α) (t − s)α−1 (M1 |z(s)| + M2 |L(z)(s)| + M3 )ds ≤ Γ(2−α) Rt b1−α 0 α−1 M2 rb1−α ≤ Γ(2−α) r + 0 |y0 | + Γ(α) (t − s) + M r + M 1 3 ds. Γ(2−α) 0 Then: kF (z)k∞
b1−α r 0 bα 0 ≤ + |y0 | + Γ(2 − α) Γ(α + 1)
M2 rb1−α + M1 r + M3 . Γ(2 − α)
Step 3: F maps bounded sets into equicontinuous sets. Let t1 , t2 ∈ J, such that t1 < t2 and z ∈ B(0, r); we have: Rt Rt r [ t12 (t2 − s)−α ds + 0 1 ((t2 − s)−α − (t1 − s)−α )ds] |(F z)(t2 ) − (F z)(t1 )| ≤ Γ(1−α) R t2 0 ¯ ((t2 − s)α−1 − (t1 − s)α−1 ))|f (s, z(s), (Lz)(s))|ds + Γ(α) 0 R 0 t2 ¯ + (t2 − s)α−1 )|f (s, z(s), (Lz)(s))|ds ≤
Γ(α) t1 r (t Γ(2−α) 2 0 l (t Γ(α+1) 2
0 l )+ (t (tα−1 − tα−1 1 Γ(α+1) 1 2 −α −α r (t (t − t1 )) Γ(2−α) 1 2
− t1 )1−α + − t1 )α +
where:
M2 rb1−α + M3 . Γ(2 − α) Then, the right-hand side tends to zero as t2 − t1 → 0. By the Arzelá-Ascoli theorem, Lemma 5, we conclude that F : C 1 (J, R) → C(J, R) is a compact continuous operator. l = M1 r +
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Now, we present a result without Lipschitz condition. Theorem 2. Suppose that there is a constant M > 0, such that for each λ ∈ [0, 1) and z ∈ C 1 (J, R) with kzk1 = M , we have: (H2 ) c Dα z + 0 z 6= λ[ c Dα z + 0 (y0 + I α f (t, z(t), c Dα z))]. Then, the implicit Problem (1) has at least one solution. Moreover, if 0 < M2 < 1, then F ix(L−1 F ) is compact. Proof. From Lemmas 6 and 7, the operators L, F : C 1 (J, R) → C(J, R) are linear, continuous, invertible and continuous and compact, respectively. Hence, L−1 F : C 1 (J, R) → C 1 (J, R) is a compact continuous operator. Set: U = {z ∈ C 1 (J, R) : kzk1 < M }. Assume that there exists λ ∈ [0, 1) and z ∈ ∂U , such that z = λL−1 F (z) ⇒ L(z) = λF (z). Hence: c Dα z(t) + 0 z(t) = λ[ c Dα z + 0 (y0 + I α f (t, z(t), c Dα z)], which contradicts with (H2 ). As a consequence of the nonlinear alternative of Leray–Schauder, we deduce that L−1 F has a fixed point z in U , which is a solution to Problem (1). Now, we show that F ix(L−1 F ) is compact. Let z ∈ F ix(L−1 F ), then: z = (L−1 F )(z) ⇒ z = y0 + I α f (·, z(·), c Dα z) and c Dα z = f (·, z(·), c Dα z). Thus, 1 M3 bα + |z(t)| ≤ |y0 | + Γ(α + 1) Γ(α)
t
Z
(t − s)α−1 (M2 |c Dα z(s)| + M1 |z(s)|) ds
0
and: |c Dα z(t)| ≤
1 (M1 |z(t)| + M3 ) 1 − M2
Let: µ(t) = sup |z(τ )| τ ∈[0,t]
Thus: Z µ(t) ≤ C1 + C2
t
(t − s)α−1 µ(s)ds.
0
where: C1 = |y0 | + and: C2 =
M3 bα M3 M2 bα + Γ(α + 1) Γ(α + 1)(1 − M2 ) M2 M1 M1 + . Γ(α)(1 − M2 ) Γ(α)
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By Lemma 2, there exists K3 > 0, such that: kzk∞ ≤ K3 . Therefore, kzk1 = k(L−1 (F (z))k1 ≤ kL−1 kk(F z)k∞ . Additionally, since kzk∞ ≤ K3 : 1 (M1 K3 + M3 ) 1 − M2
⇒ kc Dα zk∞ ≤ and:
bα M2 (M1 K3 + M3 ) + M1 K3 + M3 Γ(α + 1) 1 − M2
¯ kI f (., z(.), L(z))k ∞ ≤ α
This implies that there exists K4 > 0, such that: kzk1 < K4 . Therefore, F ix(L−1 F ) is bounded. Hence, L−1 F (F ix(L−1 F )) is compact. Now, since L−1 F is a continuous operator, the set F ix(L−1 F ) is a closed set. It is clear that F ix(L−1 F ) ⊂ L−1 F (F ix(L−1 F )), and thus, we conclude that F ix(L−1 F ) is compact. By the approximation method, we present the existence of the solution for Problem (1): Theorem 3. Assume that: (H3 ) for each > 0, there exists δ > 0 and k = k(), with: lim→0+ k() = 0 such that, if |t1 − t2 | < δ, |x − x¯| < δ, |y − y¯| < k(), then we have: |f (t1 , x, y) − f (t2 , x¯, y¯)| < k(). If the approximation sequence (zn )n∈N defined by: z0 (t) = y0 , t ∈ J, zn (t) = f
1 t, y0 + Γ(α)
Z
t α−1
(t − s)
zn−1 (s)ds, zn−1 (t) ,
0
is bounded, then there exist at least one solution of Problem (1).
t ∈ J, n ∈ N.
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Proof. Since (zn )n∈N is bounded, we only need to show that (zn )n∈N is equicontinuous. We show this fact by induction. We consider n = 1; let t1 , t2 ∈ J. Then, by (H3 ) we have, for > 0; we consider r >0, such that k(r) ≤ there exists δ > 0, such that for |tα1 − tα2 | < η, |t1 − t2 | < η, η ≤ δΓ(α+1) min 2δ , 2(1+|y , δΓ(α+1) , we have: 0 |) 2M (1+|y0 |) |z1 (t1 ) − z1 (t2 )| = tα1 tα2 = f t1 , y0 − y0 , y0 − f t2 , y0 − y0 , y0 ≤ Γ(α + 1) Γ(α + 1) ≤ k(r) ≤ . We suppose that p = n + 1:
|tα1
−
tα2 |
< η, |t1 − t2 | < η, η ≤ min
δ δΓ(α+1) , , δΓ(α+1) 2 2(1+|y0 |) 2M (1+|y0 |)
for each p ≤ n. Let
Z t1 1 α−1 |zn+1 (t1 ) − zn+1 (t2 )| = f t1 , y0 + (t1 − s) zn (s)ds, zn (t1 ) Γ(α) 0 Z t2 1 α−1 −f t2 , y0 + (t2 − s) zn (s)ds, zn (t2 ) . Γ(α) 0 Set: 1 F∗ (t) = Γ(α)
Z
t
(t − s)α−1 zn (s)ds.
0
Without loss of generality, we can assume that t1 ≥ t2 : Z t1 Z t2 M α−1 α−1 α−1 ((t1 − s) − (t2 − s) ))ds + | (t2 − s) ds | |F∗ (t1 ) − F∗ (t2 )| ≤ Γ(α) 0 t1 M ≤ (tα2 − tα1 ), Γ(α + 1) where M = kzk∞ . By (H3 ), we have: Z t1 1 α−1 |zn+1 (t1 ) − zn+1 (t2 )| = f t1 , y0 + (t1 − s) zn (s)ds, zn (t1 ) Γ(α) 0 Z t2 1 α−1 −f t2 , y0 + (t2 − s) zn (s)ds, zn (t2 ) Γ(α) 0 ≤ k(r) ≤ . Hence, by the Arzelá-Ascoli theorem, Lemma 5, we conclude that (zn )n∈N is relatively compact in C(J, R). Then, there exists a subsequence {znk }, which converges to some limit z. Since f is continuous, this implies that: Z t 1 α−1 z = f t, y0 + (t − s) z(s)ds, z(s) . Γ(α) 0 Example 1. Consider the Cauchy problem: c
Dα y(t) = f (t, y) + h(c Dα y(t)), t ∈ J := [0, b],
y(0) = y0 ∈ R,
(11)
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where f : J × R → R, h : R → R are continuous functions. Assume that there exist K1 , K2 > 0, such that: |f (t, x¯) − f (t, x)| ≤ K1 |¯ x − x|, for all, x¯, x ∈ R and t ∈ J; |h(¯ y ) − h(y)| ≤ K2 |¯ y − y|, for all, y¯, y ∈ R. bα K1 α
+ K2 < 1, then, from Theorem 1, Problem (11) has a unique solution. For instance, take J = [0, 1] and α = 12 with f (t, y) = t + 16 cos y and h(c Dα y(t)) = 21 sin(c Dα y(t)), t ∈ J and y ∈ R. Then, Problem (11) becomes:
If
c
Dα y(t) = t +
1 1 cos y + sin(c Dα y(t)), t ∈ J, 6 2
y(0) = y0 ∈ R,
(12)
It is clear that the functions f and h in Problem (12) are continuous and: 1 |f (t, y¯) − f (t, y)| ≤ |¯ y − y|, 6
1 |h(c Dα y¯(t)) − h(c Dα y(t))| ≤ |c Dα y¯(t) −c Dα y(t)|, 2 for all, y, y¯ ∈ R, t ∈ J.
We have:
bα K 1 1 1 5 + K2 = + = < 1 α 3 2 6 This implies that Problem (12) has a unique solution.
4. Conclusions We have proven an existence result for implicit fractional differential equations. In the future, we will extend the results to other fractional derivatives and boundary value problems. Acknowledgments The research has been partially supported by the Ministerio de Economía y Competitividad of Spain under Grants MTM2010-15314 and MTM2013–43014–P, XUNTA de Galicia, the local government, under Grant R2014/002 and co-financed by the European Community fund FEDER. Author Contributions Each of the authors, Juan J. Nieto, Abelghani Ouahab and V. Venktesh, contributed to each part of this study equally and read and approved the final version of this manuscript. Conflicts of Interest The authors declare no conflict of interest. References 1. Gaul, L.; Klein, P.; Kempfle, S. Damping description involving fractional operators. Mech. Syst. Signal Process. 1991, 5, 81–88.
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2. Mainardi, F. Fractional calculus: Some basic problems in continuum and statistical mechanis. In Fractals and Fractional Calculus in Continuum Mechanics; Carpinteri, A., Mainardi, F., Eds.; Springer-Verlag: Wien, Austria, 1997; pp. 291–348. 3. Metzler, F.; Schick, W.; Kilian, H.G.; Nonnenmacher, T.F. Relaxation in filled polymers: A fractional calculus approach. J. Chem. Phys. 1995, 103, 7180–7186. 4. Abbas, S.; Benchohra, M.; N’Guérékata, G.M. Topics in Fractional Differential Equations; Springer: New York, NY, USA, 2012. 5. Kilbas, A.A.; Srivastava, H.M.; Trujillo, J.J. Theory and Applications of Fractional Differential Equations; Elsevier: Amsterdam, The Netherlands, 2006; Volume 204, North-Holland Mathematics Studies. 6. Miller, K. S.; Ross, B. An Introduction to the Fractional Calculus and Differential Equations; John Wiley: New York, NY, USA, 1993. 7. Podlubny, I. Fractional Differential Equations; Academic Press: San Diego, CA, USA, 1999. 8. Samko, S.G.; Kilbas, A.A.; Marichev, O.I. Fractional Integrals and Derivatives, Theory and Applications; Gordon and Breach: Yverdon, Switzerland, 1993. 9. Diethelm, K. The Analysis of Fractional Differential Equations. An Application-Oriented Exposition Using Differential Operators of Caputo Type; Springer-Verlag: Berlin, Germany, 2010; Volume 2004, Lecture Notes in Mathematics. 10. Aghajani, A.; Bana´s, J.; Jalilian, Y. Existence of solutions for a class of nonlinear Volterra singular integral equations. Comput. Math. Appl. 2011, 62, 1215–1227. 11. Ahmad, B.; Nieto, J.J. Anti-periodic fractional boundary value problems with nonlinear term depending on lower order derivative. Fract. Calc. Appl. Anal. 2012, 15, 451–462. 12. Benchohra, M.; Henderson, J.; Ntouyas, S.K.; Ouahab, A. Existence results for fractional order functional differential equations with infinite delay. J. Math. Anal. Appl. 2008, 338, 1340–1350. 13. Chalishajar, D.N.; Karthikeyan, K. Existence and uniqueness results for boundary value problems of higher order fractional integro-differential equations involving Gronwall’s inequality in Banach spaces. Acta Math. Sci. 2013, 33, 758–772. 14. Chalishajar, D.N.; Karthikeyan, K. Boundary value problems for impulsive fractional evolution integro-differential equations with Gronwall’s inequality in Banach spaces. Discontin. Nonlinearity Complex. 2014, 3, 33–48. 15. Henderson, J.; Ouahab, A. Fractional functional differential inclusions with finite delay. Nonlinear Anal. 2009, 70, 2091–2105. 16. Henderson, J.; Ouahab, A. Impulsive differential inclusions with fractional order. Comput. Math. Appl. 2010, 59, 1191–1226. 17. Ouahab, A. Some results for fractional boundary value problem of differential inclusions. Nonlinear Anal. 2008, 69, 3877–3896. 18. Ouahab, A. Semilinear fractional differential inclsions. Comput. Math. Appl. 2012, 64, 3235–3252. 19. Benavides, T.D. An existence theorem for implicit differential equations in a Banach space. Ann. Mat. Pura Appl. 1978, 4, 119–130.
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20. Emmanuele, G. Convergence of successive approximations for implicit ordinary differential equations in Banach spaces. Funkc. Ekvac. 1981, 24, 325–330. 21. Emmanuele, G.; Ricceri, B. On the existence of solutions of ordinary differential equations in implicit form in Banach spaces. Ann. Mat. Pura Appl. 1981, 129, 367–382. (In Italian) 22. Hokkanen, V.M. Continuous dependence for an implicit nonlinear equation. J. Differ. Equ. 1994, 110, 67–85. 23. Hokkanen, V.M. Existence of a periodic solution for implicit nonlinear equations. Differ. Integral Equ. 1996, 9, 745–760. 24. Li, D. Peano’s theorem for implicit differential equations. J. Math. Anal. Appl. 2001, 258, 591–616. 25. Ricceri, B. Lipschitz solutions of the implicit Cauchy problem g(x0 ) = f (t, x), x(0) = 0, with f discontinuous in x. Rend. Circ. Mat. Palermo 1985, 34, 127–135. 26. Seikkala, S.; Heikkilä, S. Uniqueness, comparison and existence results for discontinuous implicit differential equations. Nonlinear Anal. 1997, 30, 1771–1780. 27. Benchohra, M.; Lazreg, J.E. Nonlinear fractional implicit differential equations. Commun. Appl. Anal. 2013, 17, 471–482. 28. Benchohra, M.; Lazreg, J.E. Existence and uniqueness results for nonlinear implicit fractional differential equations with boundary conditions. Rom. J. Math. Comput. Sci. 201 4, 4, 60–72. 29. Vityuk, A.N.; Mikhailenko, A.V. The Darboux problem for an implicit differential equation of fractional order. Ukr. Mat. Visn. 2010, 7, 439–452, 585, translation in J. Math. Sci. (N. Y.) 2011, 175, 391–401. (In Russian) 30. Caputo, M. Elasticità e Dissipazione; Zanichelli: Bologna, Italy, 1969. 31. Caputo, M. Linear models of dissipation whose Q is almost frequency independent, Part II. Geophys. J. R. Astr. Soc. 1967, 13, 529–529. 32. Caputo, M.; Mainardi, F. Linear models of dissipation in anelastic solid. Riv. Nuovo Cimento (Ser. II) 1971, 1, 161–198. 33. Henry, D. Geometric Theory of Semilinear Parabolic Partial differential Equations; Springer-Verlag: Berlin, Gremany; New York, NY, USA, 1989. c 2015 by the authors; licensee MDPI, Basel, Switzerland. This article is an open access article
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Implicit Fractional Differential Equations via the Liouville–Caputo Derivative Juan J. Nieto 1,2, *, Abelghani Ouahab 3 and Venktesh Venktesh 1,4 1
Department of Mathematical Analysis, University of Santiago de Compostela, Santiago de Compostela 15782, Spain; E-Mail: [email protected] 2 Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia 3 Laboratory of Mathematics, Sidi-Bel-Abbès University, P.O. Box 89, Sidi-Bel-Abbès 22000, Algeria; E-Mail: [email protected] 4 Birla Institute of Technology & Science, Pilani 333031, India * Author to whom correspondence should be addressed; E-Mail: [email protected]; Tel.: +34-881-813-177; Fax: +34-981-597-054. Academic Editor: Hari M. Srivastava Received: 16 March 2015 / Accepted: 13 May 2015 / Published: 25 May 2015
Abstract: We study an initial value problem for an implicit fractional differential equation with the Liouville–Caputo fractional derivative. By using fixed point theory and an approximation method, we obtain some existence and uniqueness results. Keywords: fractional differential equations; fractional integral; fractional derivative; Liouville–Caputo derivative; implicit
1. Introduction Differential equations of fractional order have recently been proven to be valuable tools in the modeling of many physical phenomena [1–3]. There has also been a significant theoretical development in fractional differential equations in recent years; see the monographs of Abbas et al. [4], Kilbas et al. [5], Miller and Ross [6], Podlubny [7] and Samko et al. [8]. The basic theory for initial value problems for fractional differential equations involving the Riemann–Liouville and Liouville–Caputo differential operator was discussed by Diethelm [9].
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Recently, fractional functional differential equations, fractional differential inclusions and impulsive fractional differential equations with different conditions were studied, for example, by Aghajani et al. [10], Ahmad and Nieto [11], Benchohra et al. [12], Chalishajar and Karthikeyan [13,14], Henderson and Ouahab [15,16], Ouahab [17,18] and in the references therein. In this paper, we study the existence of solutions of the following implicit fractional differential equation with initial condition: c α c α D y(t) = f (t, y(t), D y(t)), t ∈ J (1) y(0) = y0 where J = [0, b], 0 < α < 1 and f : J × R × R → R is a continuous function. This problem is motivated by the importance of implicit ordinary differential equations of the form: f (t, x(t), x0 (t), . . . , x(n−1) (t)) = 0.
(2)
under various initial and boundary conditions. Implicit equations have been considered by many authors [19–26]. Furthermore, our intention is to extend the results to implicit differential equations of fractional order. Very recently, some existence results for an implicit fractional differential equation on compact intervals were investigated [27–29]. Our goal in this work is to give some existence and uniqueness results for implicit fractional differential equations. 2. Fractional Calculus According to the Riemann–Liouville approach to fractional calculus, the notation of the fractional integral of order α (α > 0) is a natural consequence of the well-known formula (usually attributed to Cauchy) that reduces the calculation of the n−fold primitive of a function f to a single integral of the convolution type. The Cauchy formula reads: Z t 1 n (t − s)n−1 f (s) ds, t > 0 , n ∈ N. [I f ](t) = (n − 1)! 0 Definition 1. The fractional integral of order α > 0 of a function f ∈ L1 (a, b) is defined by: Z t (t − s)α−1 α Ia+ f (t) := f (s)ds, Γ(α) a where Γ is the classical gamma function. When a = 0, we write I α f (t) = f (t) ∗ φα (t), where φα (t) = tα−1 for t > 0, φα (t) = 0 for t ≤ 0 and φα → δ as α → 0+ , where δ is the Dirac delta function. For Γ(α) consistency, I 0 = Id (Identity operator), i.e., I 0 f (t) = f (t). Furthermore, by I α f (0+ ), we mean the limit (if it exists) of I α f (t) for t → 0+ ; this limit may be infinite. After the notion of the fractional integral, that of the fractional derivative of order α (α > 0) becomes a natural requirement, and one is attempted to substitute α with −α in the above formulas. However,
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this generalization needs some care in order to guarantee the convergence of the integral and to preserve the well-known properties of the ordinary derivative of integer order. Denoting by Dn , with n ∈ N, the operator of the derivative of order n, we first note that: Dn I n = Id,
I n Dn 6= Id,
n ∈ N,
i.e., Dn is the left inverse (and not the right inverse) to the corresponding integral operator I n . We can easily prove that: n−1 X tk n n I D f (t) = f (t) − f (k) (0+ ) , t > 0. k! k=0 As a consequence, we expect that Dα is defined as the left inverse to I α . For the fractional derivative of order α > 0 with integer n, such that n − 1 < α ≤ n, we have: Definition 2. For a function f given on interval [a, b], the Riemann–Liouville fractional-order derivative of order α of f is defined by: n Z t d 1 α (t − s)−α+n−1 f (s)ds, Da f (t) = Γ(n − α) dt a provided the right-hand side is defined. Defining for consistency, D0 = I 0 = Id, then we easily recognize that: Dα I α = Id,
α ≥ 0,
(3)
and D α tγ =
Γ(γ + 1) γ−α t , α > 0, γ > −1, t > 0. Γ(γ + 1 − α)
(4)
Of course, Properties (3) and (4) are a natural generalization of those known when the order is a positive integer. Note the remarkable fact that the fractional derivative Dα f is not zero for the constant function f (t) = 1, if α 6∈ N. In fact, Equation (4) with γ = 0 illustrates that: Dα 1 =
t−α , α > 0, t > 0. Γ(1 − α)
(5)
It is clear that Dα 1 = 0, for α ∈ N, due to the poles of the gamma function at the points 0, −1, −2, . . .. We now observe an alternative definition of fractional derivative, introduced by Caputo [30,31] in the late 1960s and adopted by Caputo and Mainardi [32] in the framework of the theory of linear viscoelasticity (see a review in [2]). Definition 3. Let f ∈ AC n ([a, b]). The Liouville–Caputo fractional-order derivative of f is defined by: Z t 1 c α ( Da f )(t) := (t − s)n−α−1 f (n) (s)ds. Γ(n − α) a
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This definition is of course more restrictive than the Riemann–Liouville definition, in that it requires the absolute integrability of the derivative of order n. Whenever we use the operator c Dα , we (tacitly) assume that this condition is met. We easily recognize that in general: Dα f (t) := Dn I n−α f (t) 6= I n−α Dn f (t) := c Dα f (t),
(6)
unless the function f (t), along with its first n − 1 derivatives, vanishes at t = 0+ . In fact, assuming that the passage of the n−derivative under the integral is legitimate, we recognize that, for n − 1 < α < n and t > 0, n−1 X tk−α α c α f (k) (0+ ), (7) D f (t) = D f (t) + Γ(k − α + 1) k=0 and therefore, recalling the fractional derivative of the power Function (4): Dα
f (t) −
n−1 X k=0
!
k
t f (k) (0+ ) Γ(k + 1)
= c Dα f (t).
(8)
The alternative definition, that is Definition 3, for the fractional derivative thus incorporates the initial values of the function and of lower order. The subtraction of the Taylor polynomial of degree n − 1 at t = 0+ from f (t) means a sort of regularization of the fractional derivative. In particular, according to this definition, the relevant property for which the fractional derivative of a constant is still zero is satisfied, i.e., c α D 1 = 0, α > 0. (9) At the end of this section, we present some properties of a special function. Denote Eα,β the generalized Mittag–Leffler special function defined by: Eα,β (z) :=
∞ X k=0
zk . Γ(kα + β)
Also, 1 Eα,β (z) = 2πi
Z Υ
λα−β eλ dλ λα − z 1
where Υ is a contour, which starts and ends at −∞ and encircles the disc |λ| ≤ |z| α counterclockwise. Lemma 1. [9] Let α > 0, n = dαe and λ ∈ R. The solution of the initial value problem: c α D y(t) = λy(t) + q(t)
y
(k)
(0)
= yk , k = 0, 1, . . . , n − 1,
where q ∈ C[0, b] is a given function, can be expressed in the form: y(t) =
n−1 X k=0
yk uk (t) + y∗ (t)
(10)
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with: y∗ (t) =
α I0 q(t)
1 λ
Rt 0
if λ = 0
q(t − s)u00 (s)ds
if λ 6= 0,
where uk (t) = I0k eα (t), k = 0, 1, . . . , n − 1 and eα (t) := Eα (λtα ). Remark 1. In the case 0 < α < 1, we can rewrite the solution of Problem (10) in the form: Z t α y(t) = y(0)Eα (λt ) + α q(t − s)sα−1 Eα0 (λsα )ds 0
Lemma 2. Let v : [0, b] → [0, ∞) be a real function, and w is a nonnegative, locally-integrable function on [0, b]. Assume that there are constants a > 0 and 0 < β < 1, such that: Z t v(s) v(t) ≤ w(t) + a ds, β 0 (t − s) then there exists a constant K = K(β), such that: t
Z v(t) ≤ w(t) + Ka 0
w(s) ds, (t − s)β
for every t ∈ [0, b]. Proof. From Gronwall’s lemma for singular kernels, whose proof can be found in Lemma 7.1.1 on page 188 of [33], we know that: Z t 0 v(t) ≤ w(t) + c E1−β (c(t − s))w(s)ds 0 0 where c ∈ R is a constant dependent on β. Now, E1−β (s) is bounded for s ∈ [0, b] and for s → t− : 0 E1−β (c(t − s)) ≈
a − s)β
c−β (t
where a > 0 is a constant. This implies that: Z v(t) ≤ w(t) + Ka 0
t
w(s) ds (t − s)β
with constant K = K(β). Lemma 3. Let (X, d) be a non-empty complete metric space with a contraction mapping T : X → X. Then, T admits a unique fixed-point x∗ in X(i.e., T (x∗ ) = x∗ ). Lemma 4. If K is a convex subset of a topological vector space V and T is a continuous mapping of K into itself, so that T (K) is contained in a compact subset of K, then T has a fixed point. Lemma 5. Consider a sequence of real-valued continuous functions {fn }n∈N defined on a closed and bounded interval [0, b]. If this sequence is uniformly bounded and equicontinuous, then there exists a subsequence (fnk ) that converges uniformly.
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For further readings and details on fractional calculus, we refer to the books and papers by Kilbas et al. [5], Podlubny [7] and Samko et al. [8]. 3. Existence and Uniqueness In this section, we prove some existence results and describe the structure of the solution set. We first note that if y ∈ C[0, b] is an absolutely continuous function on [0, b] satisfying Equation (1), then: Z t 1 (t − s)α−1 f (s, y(s), c Dα y(s))ds. y(t) = y0 + Γ(α) 0 Theorem 1. Assume that there exist K1 , K2 > 0, such that: |f (t, x, y) − f (t, x¯, y¯)| ≤ K1 |x − x¯| + K2 |y − y¯|, for each x, x¯, y, y¯ ∈ R. If
bα K1 α
+ K2 < 1, then there exists unique y ∈ C(J, R), which satisfy Equation (1).
Proof. Consider N : C(J, R) → C(J, R) defined by: Z t 1 α−1 (t − s) z(s)ds, z(t) , N (z(t)) = f t, y0 + Γ(α) 0
t ∈ J.
Let z1 , z2 ∈ C(J, R), then: Z t K1 |N (z1 (t)) − N (z2 (t))| ≤ (t − s)α−1 |z1 (s) − z2 (s)|ds + K2 |z1 (t) − z2 (t)| Γ(α) 0 K1 tα ≤ ||z1 − z2 ||∞ + K2 |z1 (t) − z2 (t)| Γ(α) Hence:
kN (z1 ) − N (z2 )k∞ ≤
K 1 bα + K2 kz1 − z2 k∞ , for each z1 , z2 ∈ C(J, R). Γ(α)
From the Banach fixed point theorem, Lemma 3, there exists a unique z ∈ C(J, R), such that z = N (z). Therefore: Z t 1 α−1 z(t) = f t, y0 + (t − s) z(s)ds, z(t) , t ∈ J. Γ(α) 0 Set 1 y(t) = y0 + Γ(α) c
Z
t
(t − s)α−1 z(s)ds.
0
α
This implies that D y(t) = z(t), and hence: c
Dα y(t) = f (t, y(t),c Dα y(t)),
for every t ∈ J.
The goal of the second result of this section is to apply the Schauder fixed point theorem, Lemma 4. For the study of this problem, we present some auxiliary lemmas.
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Lemma 6. Let , 0 ∈ (0, 1) and L : C 1 (J, R) → C(J, R) be an operator defined by: L(z) = c Dα z + 0 z. Then, L is a linear, continuous and invertible operator. Proof. It is clear that L is a linear operator. For every z ∈ C 1 (J, R), we have: −α+1 b 0 kL(z)k∞ ≤ + (kz 0 k∞ + kzk∞ ) for every z ∈ C 1 (J, R). Γ(α) Hence, L is continuous. Now, we show that if L(z) = 0, then z = 0, that is L is injective. Indeed, let z ∈ C(J, R), such that L(z) = 0, then: c Dα z(t) + 0 z(t) = 0, t ∈ J ⇒ z(0) = 0. From Lemma 1, we have: z = 0. Let h ∈ C(J, R); we consider the fractional Cauchy problem: c
Dα z +
0 h(t) z= ,
z(0) = h(0).
Again from Lemma 1, we obtain that: 0 0 Z t α α α−1 0 z(t) = h(0)Eα − t + α (t − s) Eα − (t − s) h(s)ds. 0 This implies that L is bijective, and from the Banach isomorphism theorem, L−1 is a continuous operator. Lemma 7. Let F : C 1 (J, R) → C(J, R) be an operator defined by: F (z) = c Dα z + 0 (y0 + I α f (., z, c Dα z)). We note 1 [L(z)](t) := y0 + Γ(α)
Z
t
(t − s)α−1 z(s)ds,
t ∈ J.
0
Assume: (H1 ) there exist M1 , M3 > 0, M2 ∈ (0, 1), such that: |f (t, x, y)| ≤ M1 |x| + M2 |y| + M3 , for each x, y, ∈ R, t ∈ J. Then, F is continuous and compact. Proof. The proof will be given in three steps. Step 1: F is continuous. Let (zn )n∈N ⊂ C 1 (J, R) be a sequence, such that (zn )n∈N → z ∈ C 1 (J, R). Then:
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Z t α (t − s)α−1 (zn (s) − z(s))ds ≤ t kzn − zk∞ . α 0 Thus: kL(zn ) − L(z)k∞ ≤
bα kzn − zk∞ → 0, n → ∞. Γ(α + 1)
Furthermore: Z t 1−α (t − s)−α (zn0 (s) − z 0 (s))ds ≤ t kzn0 − z 0 k∞ . 1 − α 0 and hence: b1−α kz 0 − z 0 k∞ → 0, n → ∞, Γ(2 − α) n
¯ n ) − L(z)k ¯ kL(z ∞ ≤ where: ¯ L(z) := c Dα z(t) =
1 Γ(1 − α)
t
Z
(t − s)−α z 0 (s),
t ∈ J.
0
Hence: kF (zn ) − F (z)k∞ ≤
0 bα ¯ n )) − f (·, z, L(z))k ¯ kf (·, zn , L(z ∞ Γ(α + 1) ¯ n ) − L(z)k ¯ +kL(z ∞ → 0, as n → ∞.
Step 2: F sends bounded sets in C 1 (J, R) into bounded sets in C(J, R). For each t ∈ J and z ∈ B(0, r), we have: Rt 1 b1−α 0 α−1 ¯ k(F z)(t)k ≤ Γ(2−α) r + |y0 | + 0 Γ(α) (t − s) |f (s, z(s), (Lz)(s))|ds Rt 1 b1−α ¯ r + 0 |y0 | + 0 Γ(α) (t − s)α−1 (M1 |z(s)| + M2 |L(z)(s)| + M3 )ds ≤ Γ(2−α) Rt b1−α 0 α−1 M2 rb1−α ≤ Γ(2−α) r + 0 |y0 | + Γ(α) (t − s) + M r + M 1 3 ds. Γ(2−α) 0 Then: kF (z)k∞
b1−α r 0 bα 0 ≤ + |y0 | + Γ(2 − α) Γ(α + 1)
M2 rb1−α + M1 r + M3 . Γ(2 − α)
Step 3: F maps bounded sets into equicontinuous sets. Let t1 , t2 ∈ J, such that t1 < t2 and z ∈ B(0, r); we have: Rt Rt r [ t12 (t2 − s)−α ds + 0 1 ((t2 − s)−α − (t1 − s)−α )ds] |(F z)(t2 ) − (F z)(t1 )| ≤ Γ(1−α) R t2 0 ¯ ((t2 − s)α−1 − (t1 − s)α−1 ))|f (s, z(s), (Lz)(s))|ds + Γ(α) 0 R 0 t2 ¯ + (t2 − s)α−1 )|f (s, z(s), (Lz)(s))|ds ≤
Γ(α) t1 r (t Γ(2−α) 2 0 l (t Γ(α+1) 2
0 l )+ (t (tα−1 − tα−1 1 Γ(α+1) 1 2 −α −α r (t (t − t1 )) Γ(2−α) 1 2
− t1 )1−α + − t1 )α +
where:
M2 rb1−α + M3 . Γ(2 − α) Then, the right-hand side tends to zero as t2 − t1 → 0. By the Arzelá-Ascoli theorem, Lemma 5, we conclude that F : C 1 (J, R) → C(J, R) is a compact continuous operator. l = M1 r +
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Now, we present a result without Lipschitz condition. Theorem 2. Suppose that there is a constant M > 0, such that for each λ ∈ [0, 1) and z ∈ C 1 (J, R) with kzk1 = M , we have: (H2 ) c Dα z + 0 z 6= λ[ c Dα z + 0 (y0 + I α f (t, z(t), c Dα z))]. Then, the implicit Problem (1) has at least one solution. Moreover, if 0 < M2 < 1, then F ix(L−1 F ) is compact. Proof. From Lemmas 6 and 7, the operators L, F : C 1 (J, R) → C(J, R) are linear, continuous, invertible and continuous and compact, respectively. Hence, L−1 F : C 1 (J, R) → C 1 (J, R) is a compact continuous operator. Set: U = {z ∈ C 1 (J, R) : kzk1 < M }. Assume that there exists λ ∈ [0, 1) and z ∈ ∂U , such that z = λL−1 F (z) ⇒ L(z) = λF (z). Hence: c Dα z(t) + 0 z(t) = λ[ c Dα z + 0 (y0 + I α f (t, z(t), c Dα z)], which contradicts with (H2 ). As a consequence of the nonlinear alternative of Leray–Schauder, we deduce that L−1 F has a fixed point z in U , which is a solution to Problem (1). Now, we show that F ix(L−1 F ) is compact. Let z ∈ F ix(L−1 F ), then: z = (L−1 F )(z) ⇒ z = y0 + I α f (·, z(·), c Dα z) and c Dα z = f (·, z(·), c Dα z). Thus, 1 M3 bα + |z(t)| ≤ |y0 | + Γ(α + 1) Γ(α)
t
Z
(t − s)α−1 (M2 |c Dα z(s)| + M1 |z(s)|) ds
0
and: |c Dα z(t)| ≤
1 (M1 |z(t)| + M3 ) 1 − M2
Let: µ(t) = sup |z(τ )| τ ∈[0,t]
Thus: Z µ(t) ≤ C1 + C2
t
(t − s)α−1 µ(s)ds.
0
where: C1 = |y0 | + and: C2 =
M3 bα M3 M2 bα + Γ(α + 1) Γ(α + 1)(1 − M2 ) M2 M1 M1 + . Γ(α)(1 − M2 ) Γ(α)
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By Lemma 2, there exists K3 > 0, such that: kzk∞ ≤ K3 . Therefore, kzk1 = k(L−1 (F (z))k1 ≤ kL−1 kk(F z)k∞ . Additionally, since kzk∞ ≤ K3 : 1 (M1 K3 + M3 ) 1 − M2
⇒ kc Dα zk∞ ≤ and:
bα M2 (M1 K3 + M3 ) + M1 K3 + M3 Γ(α + 1) 1 − M2
¯ kI f (., z(.), L(z))k ∞ ≤ α
This implies that there exists K4 > 0, such that: kzk1 < K4 . Therefore, F ix(L−1 F ) is bounded. Hence, L−1 F (F ix(L−1 F )) is compact. Now, since L−1 F is a continuous operator, the set F ix(L−1 F ) is a closed set. It is clear that F ix(L−1 F ) ⊂ L−1 F (F ix(L−1 F )), and thus, we conclude that F ix(L−1 F ) is compact. By the approximation method, we present the existence of the solution for Problem (1): Theorem 3. Assume that: (H3 ) for each > 0, there exists δ > 0 and k = k(), with: lim→0+ k() = 0 such that, if |t1 − t2 | < δ, |x − x¯| < δ, |y − y¯| < k(), then we have: |f (t1 , x, y) − f (t2 , x¯, y¯)| < k(). If the approximation sequence (zn )n∈N defined by: z0 (t) = y0 , t ∈ J, zn (t) = f
1 t, y0 + Γ(α)
Z
t α−1
(t − s)
zn−1 (s)ds, zn−1 (t) ,
0
is bounded, then there exist at least one solution of Problem (1).
t ∈ J, n ∈ N.
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Proof. Since (zn )n∈N is bounded, we only need to show that (zn )n∈N is equicontinuous. We show this fact by induction. We consider n = 1; let t1 , t2 ∈ J. Then, by (H3 ) we have, for > 0; we consider r >0, such that k(r) ≤ there exists δ > 0, such that for |tα1 − tα2 | < η, |t1 − t2 | < η, η ≤ δΓ(α+1) min 2δ , 2(1+|y , δΓ(α+1) , we have: 0 |) 2M (1+|y0 |) |z1 (t1 ) − z1 (t2 )| = tα1 tα2 = f t1 , y0 − y0 , y0 − f t2 , y0 − y0 , y0 ≤ Γ(α + 1) Γ(α + 1) ≤ k(r) ≤ . We suppose that p = n + 1:
|tα1
−
tα2 |
< η, |t1 − t2 | < η, η ≤ min
δ δΓ(α+1) , , δΓ(α+1) 2 2(1+|y0 |) 2M (1+|y0 |)
for each p ≤ n. Let
Z t1 1 α−1 |zn+1 (t1 ) − zn+1 (t2 )| = f t1 , y0 + (t1 − s) zn (s)ds, zn (t1 ) Γ(α) 0 Z t2 1 α−1 −f t2 , y0 + (t2 − s) zn (s)ds, zn (t2 ) . Γ(α) 0 Set: 1 F∗ (t) = Γ(α)
Z
t
(t − s)α−1 zn (s)ds.
0
Without loss of generality, we can assume that t1 ≥ t2 : Z t1 Z t2 M α−1 α−1 α−1 ((t1 − s) − (t2 − s) ))ds + | (t2 − s) ds | |F∗ (t1 ) − F∗ (t2 )| ≤ Γ(α) 0 t1 M ≤ (tα2 − tα1 ), Γ(α + 1) where M = kzk∞ . By (H3 ), we have: Z t1 1 α−1 |zn+1 (t1 ) − zn+1 (t2 )| = f t1 , y0 + (t1 − s) zn (s)ds, zn (t1 ) Γ(α) 0 Z t2 1 α−1 −f t2 , y0 + (t2 − s) zn (s)ds, zn (t2 ) Γ(α) 0 ≤ k(r) ≤ . Hence, by the Arzelá-Ascoli theorem, Lemma 5, we conclude that (zn )n∈N is relatively compact in C(J, R). Then, there exists a subsequence {znk }, which converges to some limit z. Since f is continuous, this implies that: Z t 1 α−1 z = f t, y0 + (t − s) z(s)ds, z(s) . Γ(α) 0 Example 1. Consider the Cauchy problem: c
Dα y(t) = f (t, y) + h(c Dα y(t)), t ∈ J := [0, b],
y(0) = y0 ∈ R,
(11)
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where f : J × R → R, h : R → R are continuous functions. Assume that there exist K1 , K2 > 0, such that: |f (t, x¯) − f (t, x)| ≤ K1 |¯ x − x|, for all, x¯, x ∈ R and t ∈ J; |h(¯ y ) − h(y)| ≤ K2 |¯ y − y|, for all, y¯, y ∈ R. bα K1 α
+ K2 < 1, then, from Theorem 1, Problem (11) has a unique solution. For instance, take J = [0, 1] and α = 12 with f (t, y) = t + 16 cos y and h(c Dα y(t)) = 21 sin(c Dα y(t)), t ∈ J and y ∈ R. Then, Problem (11) becomes:
If
c
Dα y(t) = t +
1 1 cos y + sin(c Dα y(t)), t ∈ J, 6 2
y(0) = y0 ∈ R,
(12)
It is clear that the functions f and h in Problem (12) are continuous and: 1 |f (t, y¯) − f (t, y)| ≤ |¯ y − y|, 6
1 |h(c Dα y¯(t)) − h(c Dα y(t))| ≤ |c Dα y¯(t) −c Dα y(t)|, 2 for all, y, y¯ ∈ R, t ∈ J.
We have:
bα K 1 1 1 5 + K2 = + = < 1 α 3 2 6 This implies that Problem (12) has a unique solution.
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