Applied Mathematics Letters 24 (211) 219 223 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 7149, PR China a r t i c l e i n f o a b s t r a c t Article history: Received 13 May 21 Received in revised form 25 May 211 Accepted 31 May 211 In this paper, we give a sufficient condition to guarantee the rationality of solving constant coefficient fractional differential equations by the Laplace transform method. 211 Elsevier Ltd. All rights reserved. Keywords: Laplace transform Riemann Liouville fractional derivative Caputo fractional derivative Fractional differential equations 1. Introduction his paper deals with the rationality of Laplace transform for solving the following fractional differential equation C Dα t x(t) = Ax(t) + f (t), < α < 1, t, x() = η, (1) where C Dα t 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) = 1/t 2, f (t) = e t2, 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., heorem 3.1. Corresponding author. E-mail addresses: kexueli@gmail.com, likexue@yahoo.com (L. Kexue), jgpeng@mail.xjtu.edu.cn (P. Jigen). 893-9659/$ see front matter 211 Elsevier Ltd. All rights reserved. doi:1.116/j.aml.211.5.35
22 L. Kexue, P. Jigen / Applied Mathematics Letters 24 (211) 219 223 he 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. he Laplace transform is defined by F(s) = L[f (t)] = e st f (t)dt, where f (t) is n-dimensional vector-valued function. (2) Definition 2.2 ([1]). he Mittag-Leffler function in two parameters is defined as E α,β (z) = k= z k Γ (αk + β), z C, where α >, β >, C denotes the complex plane. (3) Definition 2.3 ([1]). he Riemann Liouville fractional integral of order < α < 1 of a function h : (, ) R n is defined by I α t h(t) = 1 (t s) α 1 h(s)ds. (4) Definition 2.4 ([1]). he Riemann Liouville fractional derivative of order < α < 1 of a function h : (, ) R n is defined by D α h(t) = 1 d t Γ (1 α) dt (t s) α h(s)ds. (5) Definition 2.5 ([1]). he Caputo fractional derivative of order < α < 1 of a function h : (, ) R n is defined by C Dαh(t) = t D α t [h(t) h()]. (6) Definition 2.6. A function f on t < is said to be exponentially bounded if it satisfies an inequality of the form f (t) Me ct (7) for some real constants M > and c, for all sufficiently large t. Lemma 2.1. Let C be complex plane, for any α >, β > and A C n n, L{t β 1 E α,β (At α )} = s α β (s α A) 1 (8) holds for Rs > A 1/α, where Rs represents the real part of the complex number s. Proof. For Rs > A 1/α, we have k= Ak s (k+1)α = (s α A) 1. hen L[t β 1 E α,β (At α )] = L t β 1 (At α ) k Γ (αk + β) k= A k L[t αk+β 1 ] = Γ (αk + β) k= = s α β A k s (k+1)α k= = s α β (s α A) 1.
L. Kexue, P. Jigen / Applied Mathematics Letters 24 (211) 219 223 221 Lemma 2.2 ([11]). Suppose b, β > and a(t) is a nonnegative function locally integrable on t < (some + ), and suppose u(t) is nonnegative and locally integrable on t < with u(t) a(t) + b on this interval. hen where u(t) a(t) + θ (t s) α 1 u(s)ds θ = (bγ (β)) 1/β, E β (z) = E β (θ(t s))a(s)ds, t <, (1) z nβ /Γ (nβ + 1), n= E β (z) = d dz E β(z), E β (z) zβ 1 /Γ (β) as z +, E β (z) 1 β ez as z +, (and E β (z) 1 β ez as z + ). If a(t) a, constant, then u(t) ae β (θt). (9) 3. Main results In this section, we discuss the reliability of the Laplace transform method for solving (1). heorem 3.1. Assume (1) has a unique continuous solution x(t), if f (t) is continuous on [, ) and exponentially bounded, then x(t) and its Caputo derivative C Dα t 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 such that f (t) Me σ t for all t. It is easy to see that Eq. (1) is equivalent to the following integral equation x(t) = η + 1 For t, (11) can be rewritten as x(t) = η + 1 (t τ) α 1 [Ax(τ) + f (τ)]dτ, t <. (11) (t τ) α 1 [Ax(τ) + f (τ)]dτ + 1 (t τ) α 1 [Ax(τ) + f (τ)]dτ In view of assumptions of heorem 3.1, the solution x(t) (x() = η) is unique and continuous on [, ), then Ax(t) + f (t) is bounded on [, ], i.e., there exists a constant K > such that Ax(t) + f (t) K. We have x(t) η + K (t τ) α 1 dτ + 1 (t τ) α 1 A x(τ) dτ + 1 Multiply this inequality by e σ t and note that e σ t e σ, e σ t e σ τ, f (t) Me σ t (t ) to obtain x(t) e σ t η e σ t + K e σ t + e σ t η e σ + K e σ + e σ t η e σ + K e σ + M (t τ) α 1 dτ + e σ t α (tα (t ) α ) + A α (tα (t ) α ) + A (t τ) α 1 e σ (τ t) dτ η e σ + K α e σ α + A (t τ) α 1 A x(τ) dτ (t τ) α 1 x(τ) e σ τ dτ (t τ) α 1 x(τ) e σ τ dτ (t τ) α 1 x(τ) e σ τ dτ + M s α 1 e σ s ds
222 L. Kexue, P. Jigen / Applied Mathematics Letters 24 (211) 219 223 Denote we get η e σ + K α e σ α + A η e σ + K α e σ α + M σ α + A a = η e σ + K α e σ r(t) a + b By Lemma 2.2, r(t) ae α (θt) = a α + M σ α, (t τ) α 1 x(τ) e σ τ dτ + M (t τ) α 1 x(τ) e σ τ dτ, t. b = A, r(t) = x(t) e σ t, s α 1 e σ s ds (t τ) α 1 r(τ)dτ, t. (12) (b) n t nα Γ (nα + 1), t. n= Recall that the Mittag-Leffler function is defined as then F α (t) = n= t n Γ (αn + 1), α >, r(t) af α (bt α ), t. (13) It is known that (see [1], p. 21) the Mittag-Leffler type function F α (ωt α ) satisfies the following inequality F α (ωt α ) Ce ω1/α t, t, ω, < α < 2, (14) where C is a positive constant. With (13) and (14), we have then r(t) ace (b)1/α t, t. x(t) ace [(b)1/α +σ ]t, t. From Eq. (1), we obtain C Dα t x(t) A x(t) + f (t) a A Ce [(b)1/α +σ ]t + Me σ t (a A C + M)e [(b)1/α +σ ]t, t. aking Laplace transform with respect to t in both sides of (1), we obtain ˆx(s) = s α 1 (s α A) 1 η + (s α A) 1ˆf (s), (15) where ˆx(s) and ˆf (s) denotes the Laplace transform of x(t) and f (t), respectively. he inverse Laplace transform using (8) yields x(t) = E α,1 (t)η + (t τ) α 1 E α,α (A(t τ) α )f (τ)dτ. (16) Remark 3.1. he Laplace transform method is suitable for constant coefficient fractional differential equations, but it demands for forcing terms, so not every constant coefficient fractional differential equation can be solved by the Laplace transform method.
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