Laplace Transform of Fractional Order Differential Equations

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Electronic Journal of Diﬀerential Equations, Vol. 2015 (2015), No. 139, pp. 1–15. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

LAPLACE TRANSFORM OF FRACTIONAL ORDER DIFFERENTIAL EQUATIONS

SONG LIANG, RANCHAO WU, LIPING CHEN

Abstract. In this article, we show that Laplace transform can be applied to fractional system. To this end, solutions of linear fractional-order equations are first derived by a direct method, without using Laplace transform. Then the solutions of fractional-order differential equations are estimated by employing Gronwall and Hölderinequalities. They are showed be to of exponential order, which are necessary to apply the Laplace transform. Based on the estimates of solutions, the fractional-order and the integer-order derivatives of solutions are all estimated to be exponential order. As a result, the Laplace transform is proved to be valid in fractional equations.

1. Introduction Fractional calculus is generally believed to have stemmed from a question raised in the year 1695 by L’Hopital and Leibniz. It is the generalization of integer-order calculus to arbitrary order one. Frequently, it is called fractional-order calculus, including fractional-order derivatives and fractional-order integrals. Reviewing its history of three centuries, we could find that fractional calculus were mainly in- teresting to mathematicians for a long time, due to its lack of application back- ground. However, in the previous decades more and more researchers have paid their attentions to fractional calculus, since they found that the fractional-order derivatives and fractional-order integrals were more suitable for the description of the phenomena in the real world, such as viscoelastic systems, dielectric polariza- tion, electromagnetic waves, heat conduction, robotics, biological systems, finance and so on; see, for example, [1, 2, 8, 9, 10, 16, 17, 19]. Owing to great efforts of researchers, there have been rapid developments on the theory of fractional calculus and its applications, including well-posedness, stability, bifurcation and chaos in fractional differential equations and their control. Several useful tools for solving fractional-order equations have been discovered, of which Laplace transform is frequently applied. Furthermore, it is showed to be most efficient and helpful in analysis and applications of fractional-order systems, from which some results could be derived immediately. For instance, in [11, 12], the authors investigated stability of fractional-order nonlinear dynamical systems

2010 Mathematics Subject Classification. 26A33, 34A08, 34K37, 44A10. Key words and phrases. Fractional-order differential equation; Laplace transform; exponential order. c 2015 Texas State University - San Marcos. Submitted October 10, 2014. Published May 20, 2015. 1 2 S. LIANG, R. WU, L. CHEN EJDE-2015/139 using Laplace transform method and Lyapunov direct method, with the introduction of Mittag-Leffler stability and generalized Mittag-Leffler stability concepts. In [5], Deng et al studied the stability of n-dimensional linear fractional differential equation with time delays by Laplace transform method. In [18], Jocelyn Sabatier et al obtained the stability conditions in the form of linear matrix inequality (LMI) for fractional-order systems by using Laplace transform. The Laplace transform was also used in [6, 7, 13, 14, 15, 21]. Although it is often used in analyzing fractional-order systems, the validity of Laplace transform to fractional systems is seldom touched upon when it is applied to fractional systems. In this paper, its validity to fractional systems will be justified. It is showed that Laplace transform could be applied to fractional systems under certain conditions. To this end, solutions of linear fractional-order equations are first derived by direct method, without using the Laplace transform. The obtained results match those obtained by the Laplace transform very well. The method provides an alternative way of solution, different from the Laplace transform. Then solutions of fractional-order differential equations are estimated. They are showed to be of exponential order, which is necessary to apply the Laplace transform. Finally, the Laplace transform is proved to be feasible in fractional equations. The article is organized as follows. In Section 2, some preliminaries about fractional calculus are presented. In Section 3, solutions of linear fractional-order equations are expressed by the direct method, without using Laplace transform. Section 4 is devoted to the estimates of solutions of fractional-order equations. The Laplace transform is proved to be valid in fractional-order equations in Section 5. Finally, some conclusions are drawn in Section 6.

2. Preliminaries In fractional calculus, the traditional integer-order integrals and derivatives of functions are generalized to fractional-order ones, which are commonly defined by Laplace convolution operation as follows. Definition 2.1 ([9, p. 92]). Caputo fractional derivative with order α for a function x(t) is defined as 1 Z t C Dα x(t) = (t − τ)m−α−1x(m)(τ)dτ, t0 Γ(m − α) t0 where 0 ≤ m − 1 ≤ α < m, m ∈ Z+, and t = t0 is the initial time and Γ(·) is the Gamma function. Definition 2.2 ([9, p. 69]). Riemann-Liouville fractional integral of order α > 0 for a function x : R+R is defined as 1 Z t Iα f(t) = (t − τ)α−1f(τ)dτ, t0 F (α) t0 where t = t0 is the initial time and Γ(.) is the Gamma function. Definition 2.3 ([9, p. 70]). Riemann-Liouville fractional derivative with order α for a function x : R+R is defined as 1 dm Z t RLDα x(t) = (t − τ)m−α−1x(τ)dτ, t0 m Γ(m − α) dt t0 EJDE-2015/139 LAPLACE TRANSFORM 3 where 0 ≤ m − 1 ≤ α < m, m ∈ Z+, and t = t0 is the initial time and Γ(.) is the Gamma function. Definition 2.4 ([17, pp. 16-17]). The Mittag-Leffler function is defined as ∞ X zk E (z) = , α Γ(ka + 1) k=0 where α > 0, z ∈ C. The two-parameter Mittag-Leffler function is defined as ∞ X zk E (z) = , α,β Γ(ka + β) k=0 where α > 0, β > 0, z ∈ C. There are some properties between fractional-order derivatives and fractional- order integrals, which are expressed as follows. n−α Lemma 2.5 ([9, pp. 75-76, 96]). Let α > 0, n = [α]+1 and fn−α(t) = (Ia f)(t). Then fractional integrals and fractional derivatives have the following properties. 1 n (1) If f(t) ∈ L (a, b) and fn−α(t) ∈ AC [a, b], then n (n−j) X f (a) (IαRLDαf)(t) = f(t) − n−α (t − a)α−j, a a Γ(a − j + 1) j=1 holds almost everywhere in [a, b]. (2) If f(t) ∈ ACn[a, b] or f(t) ∈ Cn[a, b], then n−1 X f (k)(a) (IαC Dαf)(t) = f(t) − (t − a)k. a a k! k=0 Note that the Laplace transform is a useful tool for analyzing and solving or- dinary and partial differential equations. The definition of Laplace transform and some applications to integer-order systems are recalled from [20]. They will be useful for later analysis. Definition 2.6 ([20, pp. 1-2]). The Laplace transform of f is defined as Z ∞ Z τ F (s) = L(f(t))(s) = e−stf(t)dt = lim e−stf(t)dt, 0 τ→∞ 0 whenever the limit exists (as a finite number). Definition 2.7 ([20, p. 10]). A function f is piecewise continuous on the interval + [0, ∞) if (i) limt→0+ f(t) = f(0 ) exists and (ii) f is continuous on every finite interval (0, b) except possibly at a finite number of points τ1, τ2, . . . , τn in (0, b) at which f has a jump discontinuity. Definition 2.8 ([20, p. 12]). A function f is of exponential order γ if there exist γt constants M > 0 and γ such that for some t0 > 0 such that |f(t)| ≤ Me for t ≥ t0. Some existence results of Laplace transform for functions and their derivatives are listed as follows. Theorem 2.9 ([20, p. 13]). If f is piecewise continuous on [0, ∞) and of exponential order γ, then the Laplace transform L(f(t)) exists for Re(s) > γ and converges absolutely. 4 S. LIANG, R. WU, L. CHEN EJDE-2015/139

Theorem 2.10 ([20, p. 56]). If we assume that f 0 is continuous [0, ∞) and also of exponential order, then it follows that the same is true of f. Theorem 2.11 ([20, p. 57]). Suppose that f(t), f 0(t), . . . , f (n−1)(t) are continuous on (0, ∞) and of exponential order, while f (n)(t) is piecewise continuous on [0, ∞). Then L(f (n)(t))(s) = snL(f(t)) − sn−1f(0+) − sn−2f 0(0+) − · · · − f (n−1)(0+). Although the Laplace operator can be applied to many functions, there are some functions, to which it could not be applied, see for example [20, p.6]. The following inequalities will also be helpful for later analysis. Lemma 2.12 ([21]). Suppose β > 0, a(t) is a nonnegative function locally integrable on 0 ≤ t < T (some T ≤ +∞) and g(t) is a nonnegative, nondecreasing continuous function deﬁned on 0 ≤ t < T , g(t) ≤ M (constant), and suppose u(t) is nonnegative and locally integrable on 0 ≤ t < T with Z t u(t) ≤ a(t) + g(t) (t − s)β−1u(s) ds 0 on this interval. Then ∞ Z t h X (g(t)Γ(β))n i u(t) ≤ a(t) + (t − s)nβ−1a(s) ds, 0 ≤ t < T. Γ(nβ) 0 n=1

Lemma 2.13 (Cauchy inequality [22]). Let n ∈ N, and let x1, x2, . . . , xn be nonnegative real numbers. Then for ϑ,

n ϑ n X ϑ−1 X ϑ xi ≤ n xi . i=1 i=1 Lemma 2.14 (Gronwall integral inequality [4]). If Z t x(t) ≤ h(t) + k(s)x(s) ds, t ∈ [t0,T ), t0 where all the functions involved are continuous on [t0,T ), T ≤ +∞, and k(s) ≥ 0, then x(t) satisﬁes Z t R t k(u)du x(t) ≤ h(t) + h(s)k(s)e s ds, t ∈ [t0,T ). t0 If, in addition, h(t) is nondecreasing, then

R t k(s) ds x(t) ≤ h(t)e t0 , t ∈ [t0,T ).

3. Solutions of linear fractional-order equations by a direct method Consider the one-dimensional linear fractional-order equation α D0 x(t) = λx(t), (3.1) where D denotes RLD orC D, l − 1 < α ≤ l, l ∈ N, λ ∈ R. Take Laplace transform on both sides of (3.1), then the solutions of (3.1) could be ﬁgured out, see [9, pp.284, 313]. The solutions are presented as follows. EJDE-2015/139 LAPLACE TRANSFORM 5

(a) When D denotes RLD, the solution is represented as l X x(t) = djxj(t), (3.2) j=1

RL α−j (l−j) α−j α where dj = ( D x)(0+) = xl−α (0+), xj(t) = t Eα,α+1−j(λt ), and j = 1, 2, . . . , l. (b) When D denotes C D, the solution is represented as l−1 X x(t) = bjxej(t), (3.3) j=0 (j) j α where bj = x (0), xej(t) = t Eα,j+1(λt ), and j = 1, 2, . . . , l − 1. Now we employ the direct method to derive the solutions of (3.1). The whole process will be formulated after the following theorem is introduced. Theorem 3.1. Suppose that α > 0, u(t) and a(t) are locally integrable on 0 ≤ t < T (some T ≤ +∞), and |a(t)| ≤ M(constant). Suppose x(t) is locally integrable on 0 ≤ t < T with 1 Z t x(t) = u(t) + (t − τ)α−1a(τ)x(τ)dτ Γ(α) 0 on this interval. Then ∞ Z t h X 1 i x(t) = u(t) + (t − τ)nα−1an(τ)u(τ) dτ. Γ(nα) 0 n=1

1 R t α−1 Proof. Let Bφ(t) = Γ(α) 0 (t − τ) a(τ)φ(τ)dτ, t ≥ 0, where φ is the locally integrable function. Then x(t) = u(t) + Bx(t) implies n−1 X x(t) = Bku(t) + Bnx(t). k=0 Let us prove by mathematical induction that 1 Z t Bnx(t) = (t − τ)nα−1an(τ)x(τ)dτ, (3.4) Γ(nα) 0 and Bnx(t) → 0 as n → +∞ for each t in 0 ≤ t < T . We know that the relation (3.4) is true for n = 1. Assume that it is true for n = k. If n = k + 1, then the induction hypothesis implies Bk+1x(t) = B(Bkx(t)) 1 Z t h 1 Z s i = (t − s)α−1a(s) (s − τ)kα−1ak(τ)x(τ)dτ ds. Γ(α) 0 Γ(kα) 0 By interchanging the order of integration, we have 1 Z t Z t Bk+1x(t) = [ (t − s)α−1(s − τ)kα−1ds]ak+1(τ)x(τ)dτ, Γ(α)Γ(kα) 0 τ where the integral Z t Z 1 (t − s)α−1(s − τ)kα−1ds = (t − τ)kα+α−1 (1 − z)α−1zkα−1dz τ 0 6 S. LIANG, R. WU, L. CHEN EJDE-2015/139

= (t − τ)(k+1)α−1B(kα, α) Γ(α)Γ(kα) = (t − τ)(k+1)α−1 Γ((k + 1)α) is evaluated with the help of the substitution s = τ + z(t − τ) and the deﬁnition of the beta function. The relation (3.4) is proved. Since 1 Z t Bnx(t) = (t − τ)nα−1a(τ)x(τ)dτ Γ(nα) 0 1 Z t ≤ |(t − τ)nα−1an(τ)x(τ)|dτ Γ(nα) 0 1 Z t ≤ (t − τ)nα−1|an(τ)||x(τ)|dτ Γ(nα) 0 M n Z t ≤ (t − τ)nα−1|x(τ)|dτ Γ(nα) 0 M nT nα−1 Z t ≤ |x(τ)|dτ → 0, Γ(nα) 0 as n → +∞ for t ∈ [0,T ). Then the proof is complete. The way to prove the theorem can also be found in [21] and [23]. The theorem provides a direct method to solve the linear fractional-order equation (3.1). RL α (A) When D denotes D, take the operator I0 on both sides of (3.1), then from Lemma 2.5 we have l (l−j) X x (0+) 1 Z t x(t) = l−α tα−j + (t − τ)α−1λx(τ)dτ. Γ(α − j + 1) Γ(α) j=1 0 Let l (l−j) l X x (0+) X dj l−α tα−j = tα−j = u(t), Γ(α − j + 1) Γ(α − j + 1) j=1 j=1 from Theorem 3.1, one obtains ∞ Z t h X 1 i x(t) = u(t) + (t − τ)nα−1λnu(τ) dτ. (3.5) Γ(nα) 0 n=1 Assume that Z t ∞ dj h X 1 dj i A = tα−j + (t − τ)nα−1λn τ α−j dτ j Γ(α − j + 1) Γ(nα) Γ(α − j + 1) 0 n=1 d = j tα−j Γ(α − j + 1) ∞ n (n+1)α−j Z 1 X h djλ t τ τ α−j τ i + (1 − )nα−1 d Γ(nα)Γ(α − j + 1) t t t n=1 0 ∞ n (n+1)α−j dj X djλ t = tα−j + B(nα, α − j + 1) Γ(α − j + 1) Γ(nα)Γ(α − j + 1) n=1 EJDE-2015/139 LAPLACE TRANSFORM 7

∞ n (n+1)α−j dj X djλ t = tα−j + Γ(α − j + 1) Γ(nα + α − j + 1) n=1 α−j α = djt Eα,α+1−j(λt ) = djxj(t). Then l l X X x(t) = Aj = djxj(t). (3.6) j=1 j=1 It means that the solution of linear fractional-order diﬀerential equations with Rieman-Liouville derivative could be solved by the direct method above. C α (B) When D denotes D, take the operator I0 on both sides of (3.1), then from Lemma 2.5 we have l−1 X x(j)(0) 1 Z t x(t) = tj + (t − τ)α−1λx(τ)dτ. j! Γ(α) j=0 0 Let l−1 X x(j)(0) tj = u(t), j! j=0 from Theorem 3.1 one obtains ∞ Z t h X 1 i x(t) = u(t) + (t − τ)nα−1λnu(τ) dτ. (3.7) Γ(nα) 0 n=1 Assume that ∞ x(j)(t) Z t h X 1 x(j)(0) i B = tj + (t − τ)nα−1λn tj dτ j j! Γ(nα) j! 0 n=1 (j) ∞ (j) n nα+j Z 1 x (t) X hx (0)λ t τ τ j τ i = tj + (1 − )nα−1 d j! Γ(nα)Γ(j + 1) t t t n=1 0 ∞ x(j)(t) X x(j)(0)λntnα+j = tj + B(nα, j + 1) j! Γ(nα)Γ(j + 1) n=1 ∞ x(j)(0) X x(j)(0)λntnα+j = tj + j! Γ(nα + j + 1) n=1 (j) j α = x (0)t Eα,j+1(λt ) = bjxej(t). Then l−1 l−1 X X x(t) = Bj = bjxej(t). (3.8) j=0 j=0 That is, the solution of linear fractional-order diﬀerential equations with Caputo derivative could also be solved by the direct method.

4. Estimates of solutions to fractional-order differential equations Consider the nonlinear fractional-order differential equation α D0 x(t) = Ax(t) + f(x) + d(t), (4.1) 8 S. LIANG, R. WU, L. CHEN EJDE-2015/139 where D denotes RLD or C D, l − 1 < α ≤ l, l ∈ N, λ ∈ R, x ∈ Rn, f(x) is the nonlinear part and continuous in x ∈ Rn, f(0) = 0, d(t) means the input of the equation. To obtain the main results, make the following assumptions. (i) f(x) satisfies the Lipschitz condition, that is, there exists a constant L > 0 such that kf(x)k ≤ Lkxk; (ii) d(t) is bounded, that is, there exists a constant M > 0 such that kd(t)k ≤ M. Then we have the following result. Theorem 4.1. When t > 1, D denotes C D or RLD and (4.1) satisfies assumptions (i) and (ii), then the solution of (4.1) satisfies

p1t kx(t)k ≤ Mf1e , (4.2) where h−1 h h 2 (kAk + L) Γ v (va − v + 1) p1 = + α + 1, αh−h+ h hv v Γh(α) 1 1 h = 1 + , v = 1 + α, Mf1 = (l + 1)M, α 2 (l−α) ((l−α)) n kx (0)k kx (0)k kx(l−α)(0)k M = max l−1 , l−2 ,..., 0 , Γ(α − 1 + 1) Γ(α − 2 + 1) Γ(α − l + 1) M kx(0)k kx0(0)k kx(l−1)(0)ko , ,..., . Γ(α + 1) 0! 1! (l − 1)! α Proof. Applying the operator I0 on both sides of (4.1), we have 1 Z t x(t) = u(t) + (t − τ)α−1(Ax(τ) + f(x(τ)) + d(τ))dτ, (4.3) Γ(α) 0 where  x(l−j)(0) Pl l−α a−j RL j=1 Γ(a−j+1) t ,D = D, u(t) = (j) Pl−1 x (0) j C  j=0 j! t ,D = D. Taking norms of both sides of (4.3), one obtains 1 Z t kx(t)k ≤ ku(t)k + (t − τ)α−1(kAkkx(τ)k + kf(x(τ))k + kd(τ)k)dτ, (4.4) Γ(α) 0 where  kx(l−j)(0)k Pl l−α α−j RL j=1 Γ(aα−j+1) t ,D = D, ku(t)k ≤ (j) Pl−1 kx (0)k j C  j=0 j! t ,D = D. From assumptions (i) and (ii), one has 1 Z t kx(t)k ≤ ku(t)k + (t − τ)α−1[(kAk + L)kx(τ)k + M]dτ Γ(α) 0 (4.5) Mtα 1 Z t = ku(t)k + + (t − τ)α−1(kAk + L)kx(τ)kdτ. Γ(α + 1) Γ(α) 0 Let t > 1, Mˆ = (l + 1)M and (l−α) ((l−α)) n kx (0)k kx (0)k kx(l−α)(0)k M = max l−1 , l−2 ,..., 0 , Γ(α − 1 + 1) Γ(α − 2 + 1) Γ(α − l + 1) EJDE-2015/139 LAPLACE TRANSFORM 9

M kx(0)k kx0(0)k kx(l−1)(0)ko , ,..., . Γ(α + 1) 0! 1! (l − 1)! Then we have 1 Z t kx(t)k ≤ Mtˆ α + (t − τ)α−1(kAk + L)kx(τ)kdτ Γ(α) 0 (4.6) 1 Z t = Mtˆ α + (t − τ)α−1eτ−tet−τ (kAk + L)kx(τ)kdτ. Γ(α) 0 1 Let h = 1 + α , v = 1 + α, from (4.6) and H¨olderinequality, we have kx(t)k t t kAk + Lh Z v i1/vh Z h i1/h ≤ Mtˆ α + (t − τ)α−1eτ−t dτ et−τ kx(τ)k dτ Γ(α) 0 0 kAk + Lh Z t i1/vh Z t i1/h = Mtˆ α + (t − τ)vα−vevτ−vtdτ eht−hτ kx(τ)khdτ . Γ(α) 0 0 (4.7) Note that Z t Z t (t − τ)vα−ve−(vt−vτ)dτ = svα−ve−svds 0 0 1 Z tv = uvα−ve−udu v 0 (4.8) Z +∞ 1 vα−v −u ≤ vα−v+1 u e du v 0 1 = Γ(va − v + 1), vvα−v+1 where s = t − τ,u = sv. Submitting (4.8) into (4.7), one has (kAk + L)Γ1/v(va − v + 1)h Z t i1/h kx(t)k ≤ Mtˆ α + eht−hτ kx(τ)khdτ . (4.9) α−1+ 1 v v Γ(α) 0 From Lemma 2.13 and (4.9) it follows that kx(t)kh h h ht t (kAk + L) Γ v (va − v + 1)e Z (4.10) ≤ 2h−1Mˆ hthα + 2h−1 e−hτ kx(τ)khdτ, αh−h+ h v v Γh(α) 0 then we can obtain kx(t)khe−ht h−1 h h t 2 (kAk + L) Γ v (va − v + 1) Z ≤ 2h−1Mˆ hthαe−ht + e−hτ kx(τ)khdτ αh−h+ h v v Γh(α) 0 (4.11) h−1 h h t 2 (kAk + L) Γ v (va − v + 1) Z ≤ 2h−1Mˆ hthα + e−hτ kx(τ)khdτ. αh−h+ h v v Γh(α) 0 From Lemma 2.14, we have

kx(t)khe−ht ≤ 2h−1MˆhthαeKt˜ ≤ 2h−1Mˆhe(K˜ +hα)t, (4.12) 10 S. LIANG, R. WU, L. CHEN EJDE-2015/139 where h−1 h h 2 (kAk + L) Γ v (va − v + 1) K˜ = . αh−h+ h v v Γh(α) Then one obtains 1 ( K˜ +α+1)t kx(t)k ≤ Meˆ h . (4.13) 2 ˜ 1 ˆ K p1t Let Mf1 = 2 M, p1 = h + α + 1, then ||x(t)|| ≤ Mf1e . The proof is complete. Theorem 4.2. (1) When 0 < t ≤ 1, D denotes C D and the equation (4.1) satisfy assumptions (i) and (ii), then the solution of (4.1) satisﬁes

p2t kx(t)k ≤ Mf2e , (4.14) 1 where Mf2 = 2 (l + 1)M, h−1 h h 2 (kAk + L) Γ v (va − v + 1) p2 = + 1. αh−h+ h hv v Γh(α) (2) When 0 < t ≤ 1, D denotes RLD and (4.1) satisﬁes assumptions (i) and (ii) for any b > 0, then the solution of (4.1) satisﬁes

p3t kx(t)k ≤ Mf3e (t > b > 0), (4.15)

h−1 h h where M = 1 (l + 1)Mbα−l, p = 2 (kAk+L) Γ v (va−v+1) + 1. The expressions M, f3 2 3 αh−h+ h hv v Γh(α) h and v are the same as in Theorem 4.1. Proof. (1) When 0 < t ≤ 1, D denotes C D, we can write (4.5) as 1 Z t kx(t)k ≤ Mˆ + (t − τ)α−1(kAk + L)kx(τ)kdτ Γ(α) 0 (4.16) 1 Z t = Mˆ + (t − τ)α−1eτ−tet−τ (kAk + L)kx(τ)kdτ. Γ(α) 0 Following the same process as in Theorem 4.1, we have

p2t kx(t)k ≤ Mf2e , (4.17) 1 where Mf2 = 2 (l + 1)M, h−1 h h 2 (kAk + L) Γ v (va − v + 1) p2 = + 1. αh−h+ h hv v Γh(α) (2) When 0 < t ≤ 1, D denotes RLD, and t > b, we can write (4.5) as 1 Z t kx(t)k ≤ Mbˆ α−l + (t − τ)α−1(kAk + L)kx(τ)kdτ Γ(α) 0 (4.18) 1 Z t = Mbˆ α−l + (t − τ)α−1eτ−tet−τ (kAk + L)kx(τ)kdτ. Γ(α) 0 Following the same process as in Theorem 4.1, we have

p3t kx(t)k ≤ Mf3e , (4.19) 1 α−l where Mf3 = 2 (l + 1)Mb , h−1 h h 2 (kAk + L) Γ v (va − v + 1) p3 = + 1. αh−h+ h hv v Γh(α) EJDE-2015/139 LAPLACE TRANSFORM 11

The proof is complete. The methods to prove Theorems 4.1 and 4.2 are similar to those in [24] and [3]. From Theorems 4.1 and 4.2, we can have the following theorem. Theorem 4.3. (1) When D denotes C D, and (4.1) satisﬁes (i) and (ii), there exist constants Mf > 0 and p > 0 such that kx(t)k ≤ Mef pt, (4.20) for all t > 0. (2) When D denotes RLD, and (4.1) satisﬁes (i) and (ii), there exist constants Mf > 0 and p > 0 for any b > 0 such that kx(t)k ≤ Mef pt, (4.21) for all t > b > 0.

5. Validity of Laplace transform for fractional-order equations Consider the one-dimensional fractional-order diﬀerential equation α D0 x(t) = ax(t) + f1(x(t)) + d1(t), (5.1) RL C where D denotes D or D, l − 1 < α ≤ l, l ∈ N, λ ∈ R, x ∈ R, f1(x) is the nonlinear part and continuous in x ∈ R, d1(t) is the input of the equation. And f1 and d1 also satisfy the assumptions (i) and (ii). Before the validity of Laplace transform method is justiﬁed, some lemmas and theorems are needed. Lemma 5.1 ([9, p. 84]). Let Re(α) > 0 and f ∈ L1(0, b) for any b > 0. Also let the estimate |f(t)| ≤ Aep0t (t > b > 0) α hold for some constants A > 0 and p0 > 0. Then the relation L(I0 f(t)) = −α s L(f(t)) is valid for Re{s} > p0.

n−α d n−α dn−1 Theorem 5.2. If α > 0, n = [α] + 1, and x(t),I0 x(t), dt I0 x(t),..., dtn−1 n−α RL α I0 x(t) are continuous in (0, ∞) and of exponential order, while D0 x(t) is piecewise continuous on [0, ∞). Then n−1 X d(k−1) L(RLDαx(t)) = sαL(x(t)) − sn−k−1 In−αx(0+). 0 dt(k−1) 0 k=0

RL α d(n) n−α n−α RL α Proof. Since D0 x(t) = dt(n) I0 x(t), let f(t) = I0 x(t), then D0 x(t) = d(n) dt(n) f(t). Under the assumptions, from theorem 2.11 we have n−1 RL α (n) n X n−k−1 (k) L( D0 x(t)) = L(f (t)) = s L(f(t)) − s f (0+) k=0 n−1 X d(k−1) = sαL(x(t)) − sn−k−1 In−αx(0+). dt(k−1) 0 k=0 This completes the proof. Theorem 5.3. When D denotes RLD, the Laplace transform can be taken on both sides of (5.1), if assumptions (i) and (ii) are satisﬁed and x(t) is continuous. 12 S. LIANG, R. WU, L. CHEN EJDE-2015/139

Proof. From (5.1) and Theorem 4.3, there exist constants M1 > 0 and P1 > 0 such that RL α | D0 x(t)| ≤ |a||x(t)| + |f1(x(t))| + |d1(t)| ≤ (|a| + L)x(t) + M

p1t ≤ M1e .

RL α dn n−α This implies D0 x(t) = dtn I0 x(t) begin of exponential order. Then from The- n−α d n−α orems 2.10 and 4.3 and Lemma 5.1, we have that x(t), I0 x(t), and dt I0 x(t), dn−1 n−α ..., dtn−1 I0 x(t) are of exponential order. From theorem 5.2, the Laplace transform can be taken on both sides of (5.1). Theorem 5.4. If α > 0, n = [a]+1, and x(t), x0(t), x00(t), x(n−1)(t) are continuous C α on [0, +∞) and of exponential order, while D0 x(t) is piecewise continuous on [0, ∞). Then n−1 C α α X α−k−1 (k) L( D0 x(t)) = s L(x(t)) − s x (0). k=0 Proof. Since Z t C α 1 m−α−1 (n) D0 x(t) = (t − τ) x (τ)dτ Γ(n − α) 0 1 = tm−α−1 ∗ x(n)(t). Γ(n − α) Under the assumptions in Theorem 2.11, one has 1 L(C Dαx(t)) = L(tn−α−1) ·L(x(n)(t)) 0 Γ(n − α) n−1 X = sαL(x(t)) − sα−k−1x(k)(0). k=0 C α (j) Lemma 5.5. If D0 x(t) is of exponential order and n = [α] + 1, then x (t)(j = 1, . . . , n − 1) is also of exponential order.

C α Proof. Since D0 x(t) is of exponential order, then there exist constants M2 > 0 and P2 > 0 such that

C α C α−j (j) p2t | D0 x(t)| = | D0 x (t)| ≤ M2e ; that is,

p2t C α−j (j) p2t −M2e ≤ D0 x (t) ≤ M2e .

Then there exist functions M1(t) ≤ 0,M2(t) ≤ 0 such that

p2t C α−j (j) p2t −M2e + M1(t) = D0 x (t) = M2e − M2(t). This is equivalent to n−i−1 X x(i+j)(0) ti − Iα−j(M ep2t) + Iα−jM (t) i! 0 2 0 1 i=0 EJDE-2015/139 LAPLACE TRANSFORM 13

n−i−1 X x(i+j)(0) = x(j)(t) = ti + Iα−j(M ep2t) − Iα−jM (t). i! 0 2 0 2 i=0

α−j α−j In view of I0 M1(t) ≥ 0 and I0 M2(t) ≥ 0, we have

n−i−1 n−i−1 X x(i+j)(0) X x(i+j)(0) ti − Iα−j(M ep2t) ≤ x(j)(t) ≤ ti + Iα−j(M ep2t). i! 0 2 i! 0 2 i=0 i=0 Note that 1 Z t α−j p2t α−j−1 p2τ I0 e = (t − τ) e dτ Γ(α − j) 0 1 Z t = ep2t (t − τ)α−j−1ep2(τ−t)dτ Γ(α − j) 0 1 Z t = ep2t sα−j−1e−sp2 ds Γ(α − j) 0 1 1 Z p2t p2t α−j−1 −u = e α u e du Γ(α − j) p2 0 1 Z +∞ −α+j p2t α−j−1 −u ≤ p2 e u e du Γ(α − j) 0

−α+j p2t ≤ p2 e , where s = t − τ, u = p2s. It is not diﬃcult to get that there exist M3 > 0 and p3 > 0 such that

(j) p3t |x (t)| ≤ M3e , where j = 0, . . . , n − 1.

C α Theorem 5.6. When D denotes D0 , the Laplace transform can be taken on both sides of (5.1), if assumptions (i) and (ii) are satisﬁed and x(t) is continuous.

Proof. From (5.1) and Theorem 4.3, then there exist constants M4 > 0 and P4 > 0 such that

C α | D0 x(t)| ≤ |a||x(t)| + |f1(x(t))| + |d1(t)| ≤ (|a| + L)x(t) + M

p4t ≤ M4e .

C α It means that D0 x(t) is of exponential order. Then from Lemma 5.5 and Theorem 4.3 we have x(j)(t)(j = 0, . . . , n − 1) is of exponential order. From Theorem 5.4, the Laplace transform can be taken on both sides of (5.1).

Conclusions. By Gronwall and H¨older inequalities, solutions of fractional-order equations are showed to be of exponential order. Based on that, the fractional-order and integer-order derivatives are all estimated to be of exponential order. Conse- quently, the Laplace transform is proved to be valid for fractional-order equations under general conditions. So the validity of Laplace transform of fractional-order equations is justiﬁed. 14 S. LIANG, R. WU, L. CHEN EJDE-2015/139

Acknowledgments. This work was supported by the National Science Founda- tion of China (No. 61403115), by the Specialized Research Fund for the Doctoral Program of Higher Education of China (No.20093401120001), by the Natural Sci- ence Foundation of Anhui Province (No. 11040606M12, 1508085QF120), and by the 211 project of Anhui University (No.KJJQ1102, KJTD002B).

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Song Liang School of Mathematics, Anhui University, Hefei 230601, China E-mail address: [email protected]

Ranchao Wu School of Mathematics, Anhui University, Hefei 230601, China E-mail address: [email protected]

Liping Chen School of Electrical Engineering and Automation, Hefei University of Technology, Hefei 230009, China E-mail address: lip [email protected]