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Laplace Transform and Fractional Differential Equations View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Applied Mathematics Letters 24 (2011) 2019–2023 Contents lists available at ScienceDirect Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml Laplace transform and fractional differential equations Li Kexue ∗, Peng Jigen Department of Mathematics, Xi'an Jiaotong University, Xi'an 710049, PR China article info a b s t r a c t Article history: In this paper, we give a sufficient condition to guarantee the rationality of solving constant Received 13 May 2010 coefficient fractional differential equations by the Laplace transform method. Received in revised form 25 May 2011 ' 2011 Elsevier Ltd. All rights reserved. Accepted 31 May 2011 Keywords: Laplace transform Riemann–Liouville fractional derivative Caputo fractional derivative Fractional differential equations 1. Introduction This paper deals with the rationality of Laplace transform for solving the following fractional differential equation C Dαx.t/ D Ax.t/ C f .t/; 0 < α < 1; t ≥ 0; 0 t (1) x.0/ D η; C α· × where 0 Dt is the Caputo fractional derivative operator, A is a n n constant matrix, f .t/ is a n-dimensional continuous vector-valued function, which is usually called the forcing term. Fractional derivatives describe the property of memory and heredity of many materials, and it is the major advantage over integer-order derivatives. Fractional differential equations are differential equations of arbitrary (noninteger) order. Fractional differential equations have attracted considerable attention recently, due to its extensive applications in engineering and science; see the books of Podlubny [1], Miller and Boss [2], Oldham and Spanier [3] and the papers by Friedrich [4], Chen and Moore [5], Ahmad and Sivasundaram [6], Odibat [7]. An effective and convenient method for solving fractional differential equations is needed. Methods in [2,3] for rational order fractional differential equations is not applicable to the case of arbitrary order. Some authors used the series method [3,4], which allows solution of arbitrary order fractional differential equations, but it works only for relatively simple equations. Podlubny [1] introduced a method based on the Laplace transform technique, it is suitable for a large class of initial value problems for fractional differential equations. However, we find that the existence of Laplace transform is taken for granted in some papers to solve fractional differential equations (see, e.g., [8,9]). In fact, not every function has its Laplace transform, for example, f .t/ D 1=t2; f .t/ D 2 et , do not have the Laplace transform. In this paper, to guarantee the rationality of solving fractional differential equations by the Laplace transform method, we give a sufficient condition, i.e., Theorem 3.1. ∗ Corresponding author. E-mail addresses: [email protected], [email protected] (L. Kexue), [email protected] (P. Jigen). 0893-9659/$ – see front matter ' 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2011.05.035 2020 L. Kexue, P. Jigen / Applied Mathematics Letters 24 (2011) 2019–2023 The paper has been organized as follows. In Section 2, basic definitions and lemmas related to fractional derivatives and fractional integrals have been summarized. In Section 3, we give the main results. 2. Preliminaries In this section, we give some definitions and lemmas which are used further in this paper. Definition 2.1. The Laplace transform is defined by Z 1 − F.s/ D LTf .t/UD e st f .t/dt; (2) 0 where f .t/ is n-dimensional vector-valued function. Definition 2.2 ([1]). The Mittag-Leffler function in two parameters is defined as 1 X zk Eα,β .z/ D ; z 2 C; (3) Γ k C kD0 (α β/ where α > 0; β > 0, C denotes the complex plane. Definition 2.3 ([10]). The Riemann–Liouville fractional integral of order 0 < α < 1 of a function h V .0; 1/ ! Rn is defined by 1 Z t α D − α−1 0It h.t/ .t s/ h.s/ds: (4) Γ (α/ 0 Definition 2.4 ([10]). The Riemann–Liouville fractional derivative of order 0 < α < 1 of a function h V .0; 1/ ! Rn is defined by 1 d Z t α D − −α 0Dt h.t/ .t s/ h.s/ds: (5) Γ .1 − α/ dt 0 Definition 2.5 ([10]). The Caputo fractional derivative of order 0 < α < 1 of a function h V .0; 1/ ! Rn is defined by C α D αT − U 0 Dt h.t/ 0 Dt h.t/ h.0/ : (6) Definition 2.6. A function f on 0 ≤ t < 1 is said to be exponentially bounded if it satisfies an inequality of the form kf .t/k ≤ Mect (7) for some real constants M > 0 and c, for all sufficiently large t. Lemma 2.1. Let C be complex plane, for any α > 0; β > 0 and A 2 C n×n, β−1 α α−β α −1 Lft Eα,β .At /g D s .s − A/ (8) holds for <s > kAk1/α, where <s represents the real part of the complex number s. < k k1/α P1 k −.kC1)α D α − −1 Proof. For s > A , we have kD0 A s .s A/ . Then " 1 α k # β−1 α β−1 X .At / LTt Eα,β .At /UD L t Γ k C kD0 (α β/ 1 AkLTtαkCβ−1U D X Γ k C kD0 (α β/ 1 − X − C D sα β Aks .k 1)α kD0 α−β α −1 D s .s − A/ : L. Kexue, P. Jigen / Applied Mathematics Letters 24 (2011) 2019–2023 2021 Lemma 2.2 ([11]). Suppose b ≥ 0; β > 0 and a.t/ is a nonnegative function locally integrable on 0 ≤ t < T (some T ≤ C1), and suppose u.t/ is nonnegative and locally integrable on 0 ≤ t < T with Z t − u.t/ ≤ a.t/ C b .t − s/α 1u.s/ds (9) 0 on this interval. Then Z t 0 u.t/ ≤ a.t/ C θ Eβ (θ.t − s//a.s/ds; 0 ≤ t < T ; (10) 0 where 1 1/β X nβ 0 d θ D .bΓ (β// ; Eβ .z/ D z =Γ .nβ C 1/; E .z/ D Eβ .z/; β dz nD0 0 − 0 1 E .z/ ' zβ 1=Γ (β/ as z ! 0C; E .z/ ' ez as z ! C1; β β β ' 1 z ! C1 ≡ ≤ (and Eβ .z/ β e as z ). If a.t/ a, constant, then u.t/ aEβ (θt/. 3. Main results In this section, we discuss the reliability of the Laplace transform method for solving (1). Theorem 3.1. Assume (1) has a unique continuous solution x.t/, if f .t/ is continuous on T0; 1/ and exponentially bounded, C α then x.t/ and its Caputo derivative 0 Dt x.t/ are both exponentially bounded, thus their Laplace transforms exist. Proof. Since f .t/ is exponentially bounded, there exist positive constants M; σ and enough large T such that kf .t/k ≤ Meσ t for all t ≥ T . It is easy to see that Eq. (1) is equivalent to the following integral equation Z t 1 − x.t/ D η C .t − τ/α 1TAx(τ/ C f (τ/Udτ; 0 ≤ t < 1: (11) Γ (α/ 0 For t ≥ T , (11) can be rewritten as Z T Z t 1 − 1 − x.t/ D η C .t − τ/α 1TAx(τ/ C f (τ/Udτ C .t − τ/α 1TAx(τ/ C f (τ/Udτ Γ (α/ 0 Γ (α/ T In view of assumptions of Theorem 3.1, the solution x.t/.x.0/ D η/ is unique and continuous on T0; 1/, then Ax.t/ C f .t/ is bounded on T0; T U, i.e., there exists a constant K > 0 such that kAx.t/ C f .t/k ≤ K. We have Z T Z t Z t K − 1 − 1 − kx.t/k ≤ kηk C .t − τ/α 1dτ C .t − τ/α 1kAk kx(τ/kdτ C .t − τ/α 1kf (τ/kdτ Γ (α/ 0 Γ (α/ T Γ (α/ T Multiply this inequality by e−σ t and note that e−σ t ≤ e−σ T , e−σ t ≤ e−σ τ , kf .t/k ≤ Meσ t .t ≥ T / to obtain −σ t Z T −σ t Z t − − Ke − e − kx.t/ke σ t ≤ kηke σ t C .t − τ/α 1dτ C .t − τ/α 1kAk kx(τ/kdτ Γ (α/ 0 Γ (α/ T −σ t Z t e − C .t − τ/α 1kf (τ/kdτ Γ (α/ T −σ T Z t − Ke kAk − − ≤ kηke σ T C .tα − .t − T /α/ C .t − τ/α 1kx(τ/ke σ τ dτ αΓ (α/ Γ (α/ T −σ t Z t e − C .t − τ/α 1kf (τ/kdτ Γ (α/ T −σ T Z t − Ke kAk − − ≤ kηke σ T C .tα − .t − T /α/ C .t − τ/α 1kx(τ/ke σ τ dτ αΓ (α/ Γ (α/ 0 Z t M − − C .t − τ/α 1eσ (τ t/dτ Γ (α/ 0 α −σ T Z t Z t − KT e kAk − − M − − ≤ kηke σ T C C .t − τ/α 1kx(τ/ke σ τ dτ C sα 1e σ sds αΓ (α/ Γ (α/ 0 Γ (α/ 0 2022 L. Kexue, P. Jigen / Applied Mathematics Letters 24 (2011) 2019–2023 α −σ T Z t Z 1 − KT e kAk − − M − − ≤ kηke σ T C C .t − τ/α 1kx(τ/ke σ τ dτ C sα 1e σ sds αΓ (α/ Γ (α/ 0 Γ (α/ 0 α −σ T Z t − KT e M kAk − − ≤ kηke σ T C C C .t − τ/α 1kx(τ/ke σ τ dτ; t ≥ T : α αΓ (α/ σ Γ (α/ 0 Denote α −σ T − KT e M kAk − a D kηke σ T C C ; b D ; r.t/ D kx.t/ke σ t ; αΓ (α/ σ α Γ (α/ we get Z t − r.t/ ≤ a C b .t − τ/α 1r(τ/dτ; t ≥ T : (12) 0 By Lemma 2.2, 1 X .bΓ (α//ntnα r.t/ ≤ aEα(θt/ D a ; t ≥ T : Γ n C 1 nD0 .
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