arXiv:1710.03407v1 [math.CA] 10 Oct 2017
Fractional Differential Equations Involving Caputo Fractional Derivative with Mittag-Leffler Non-Singular Kernel: Comparison Principles and Applications Mohammed Al-Refai Department of Mathematical Sciences, UAE University, P.O.Box 15551, Al Ain, UAE. m−
[email protected]
October 11, 2017 Abstract In this paper we study linear and nonlinear fractional differential equations involving the Caputo fractional derivative with Mittag-Leffler non-singular kernel of order 0 < α < 1. We first obtain a new estimate of the fractional derivative of a function at its extreme points and derive a necessary condition for the existence of a solution to the linear fractional equation. The obtained sufficient condition determine the initial condition of the associated fractional initial value problem. We then derive comparison principles to the linear fractional equations. We apply these principles to obtain a norm estimate of solutions to the linear equation and to obtain a uniqueness result to the nonlinear equation. We also derive a lower and upper bound of solutions to the nonlinear equation. The applicability of the new results is illustrated through several examples. Key words and phrases: Fractional differential equations, Maximum principle, Fractional derivatives.
1
Introduction
Fractional models have been implemented to model various problems in several fields, [16, 20, 21, 22]. The non-locality of the fractional derivative makes fractional models more practical than the usual ones, especially for systems which involve memory. In recent years there are great interests to develop new types of non-local fractional derivative with non-singular kernel, see [1, 14]. The idea is to have more types of non-local fractional derivative, and it is the role of application that will determine which fractional model is more proper. The theory of fractional models is effected by the type of the fractional derivative. Therefore, several papers have been devoted recently to study the new types of fractional derivatives and their applications, see [4, 5] for the Caputo-Fabrizio fractional derivative and [2, 6, 7, 8, 15, 17] for the Abdon-Baleanu fractional derivative. In this paper, we analyze the solutions of a class of fractional differential equations involving the Caputo fractional derivative with Mittag-Leffler non-singular kernel of order 1
0 < α < 1. To the best of our knowledge this is the first theoretical study of fractional differential equations with fractional derivative of non-singular kernel. We start with the definition and main properties of the nonlocal fractional derivative with Mittag-leffler nonsingular kernel. For more details the reader is referred to [1, 2, 3]. Definition 1.1. Let f ∈ H 1 (a, b), a < b, α ∈ (0, 1), the left Caputo fractional derivative with Mittag-Leffler non-singular kernel is defined by Z α B(α) t α ABC α Eα − (1.1) (t − s) f ′ (s)ds. ( a D f )(t) = 1−α a 1−α where B(α) > 0 is a normalization function satisfying B(0) = B(1) = 1, and Eα (s) is the well known Mittag-Leffler function.. Definition 1.2. Let f ∈ H 1 (a, b), a < b, α ∈ (0, 1), the left Riemann-Liouville fractional derivative with Mittag-Leffler non-singular kernel is defined by Z α B(α) d t α ABR α (1.2) Eα − (t − s) f (s)ds. ( a D f )(t) = 1 − α dt a 1−α The associated fractional integral is defined by (AB a I α f )(t) =
α 1−α f (t) + (a I α f )(t), B(α) B(α)
(1.3)
where (a I α f )(t) is the left Riemann-Liouville fractional integral of order α > 0 defined by 1 (a I f )(t) = Γ(α) α
Z
t
(t − s)α−1 f (s)ds. a
The following holds true, (ABC 0 D α f )(t) = (ABR 0 D α f )(t) − (ABR a D α AB a I α f )(t) = f (t), (AB a I α ABR a D α f )(t) = f (t).
B(α) α α f (0)Eα (− t ), 1−α 1−α
(1.4) (1.5) (1.6)
The rest of the paper is organized as follows. In Section 2, we present a new estimate of the fractional derivative of a function at its extreme points. In Section 3, we develop new comparison principles for linear fractional equations and obtain a norm bound to their solutions. We also, obtain the solution solution for a class of linear equations in a closed form, and present a necessary condition for the existence of their solutions. In Section 4, we consider nonlinear fractional equations. We obtain a uniqueness result and derive upper and lower bounds to the solution of the problem. Finally we present some examples to illustrate the applicability of the obtained results.
2
Estimates of fractional derivatives at extreme points
We start with estimating the fractional derivative of a function at its extreme points, this result is analogous to the ones obtained in [9] for the Caputo and Riemann-Liouville fractional 2
derivatives. The applicability of these results were indicated in ([10]-[13]) by establishing new comparison principles and studying various fractional diffusion models. Therefore, the current result can be used to study fractional diffusion models involving the Caputo and Riemann-Liouville fractional derivatives with Mittag-Leffler non-singular kernel, and we leave this for a future work. Lemma 2.1. Let a function f ∈ H 1(a, b) attain its maximum at a point t0 ∈ [a, b] and 0 < α < 1. Then the inequality (ABC a D α f )(t0 ) ≥
α B(α) Eα [− (t0 − a)α ](f (t0 ) − f (a)) ≥ 0. 1−α 1−α
(2.1)
holds true. Proof. We define the auxiliary function g(t) = f (t0 ) − f (t), t ∈ [a, b]. Then it follows that g(t) ≥ 0, on [a, b], g(t0 ) = g ′ (t0 ) = 0 and (ABC a D α g)(t) = −(ABC a D α f )(t). Since g ∈ H 1 (a, b), then g ′ is integrable and integrating by parts with u = Eα [−
α (t0 − s)α ], and dv = g ′ (s)ds, 1−α
yields
(
ABC
aD
α
g)(t0 ) = = = =
Z α B(α) t0 Eα [− (t0 − s)α ]g ′ (s) ds 1−α a 1−α Z t0 B(α) α α d α t0 α Eα [− (t0 − s) ]g(s)|a − Eα [− (t0 − s) ]g(s)ds 1−α 1−α 1−α a ds Z t0 α α d B(α) α α Eα [0]g(t0 ) − Eα [− (t0 − a) ]g(a) − Eα [− (t0 − s) ]g(s)ds 1−α 1−α 1−α a ds Z t0 B(α) α α d α α − Eα [− (t0 − a) ]g(a) − Eα [− (t0 − s) ]g(s)ds . (2.2) 1−α 1−α 1−α a ds
We recall that for 0 < α < 1, see [18], we have Z ∞ α Eα (−t ) = e−rt Kα (r)dr, 0
where Kα (r) =
1 r α−1 sin(απ) > 0. π r 2α + 2r α cos(απ) + 1
Thus, α α d α 1/α d α Eα [− (t0 − s) ] = Eα − ( ) (t0 − s) ds 1−α ds 1−α Z Z ∞ d −r( α )1/α (t0 −s) d ∞ −r( α )1/α (t0 −s) e 1−α e 1−α Kα (r)dr = Kα (r)dr = ds 0 ds 0 Z α 1/α ∞ −r( α )1/α (t0 −s) = ( Kα (r)dr > 0, (2.3) re 1−α ) 1−α 0 3
which together with g(t) ≥ 0 on [a, b], will lead to the integral in Eq. (2.2) is nonnegative. We recall here that Eα [t] > 0, 0 < α < 1, see [19], and thus α B(α) α ABC α − Eα [− (t0 − a) ]g(a) ( a D g)(t0 ) ≤ 1−α 1−α B(α) α =− Eα [− (t0 − a)α ](f (t0 ) − f (a)) ≤ 0. (2.4) 1−α 1−α The last inequality yields −(ABC a D α f )(t0 ) ≤ −
B(α) α Eα [− (t0 − a)α ](f (t0 ) − f (a)) ≤ 0, 1−α 1−α
which proves the result. By applying analogous steps to −f we have Lemma 2.2. Let a function f ∈ H 1 (a, b) attain its minimum at a point t0 ∈ [a, b] and 0 < α < 1. Then the inequality (ABC a D α f )(t0 ) ≤
B(α) α Eα [− t0 ](f (t0 ) − f (a)) ≤ 0. 1−α 1−α
(2.5)
holds true. Lemma 2.3. Let a function f ∈ H 1 (a, b) then it holds that (ABC a D α f )(a) = 0, 0 < α < 1.
(2.6)
α (t − s)] is continuous on [a, b], then it is in L2 [a, b]. Applying the Proof. Because Eα [− 1−α Cauchy-Schwartz inequality we have 2 2 Z t Z t B 2 (α) α ABC α 2 α ′ |( ≤ Eα [− (t − s) ] ds f (s) ds. (2.7) a D f )(t)| (1 − α)2 a 1−α a
Since f ∈ H 1 (a, b) then f ′ is square integrable and it holds that result is obtained as the first integral in Eq. (2.7) is bounded.
3
Ra a
f ′ (s)
2
ds = 0. The
The linear equation
We implement the results in Section 1 to obtain new comparison principles of the linear fractional differential equations of order 0 < α < 1, and to derive a necessary condition for the existence of their solutions. We then use these principles to obtain a norm bound of the solution. We also present the solution of certain linear equation by the Laplace transform. Lemma 3.1. (Comparison Principle-1) Let a function u ∈ H 1 (a, b) ∩ C[a, b] satisfies the fractional inequality Pα (u) = (ABC a D α u)(t) + p(t)u(t) ≤ 0, t > a, 0 < α < 1, where p(t) ≥ 0 is continuous on [a, b] and p(a) 6= 0. Then u(t) ≤ 0, t ≥ a. 4
(3.1)
Proof. Since u ∈ H 1 (a, b) then by Lemma 2.3 we have (ABC a D α u)(a) = 0. By the continuity of the solution, the fractional inequality (3.1) yields p(a)u(a) ≤ 0, and hence u(a) ≤ 0. Assume by contradiction that the result is not true, because u is continuous on [a, b] then u attains absolute maximum at t0 ≥ a with u(t0 ) > 0. Since u(a) ≤ 0, then t0 > a. Applying the result of Lemma 2.1 we have α B(α) Eα [− (t0 − a)α ](u(t0 ) − u(a)) > 0. (ABC a D α u)(t0 ) ≥ 1−α 1−α We have (ABC a D α u)(t0 ) + p(t0 )u(t0 ) ≥ (ABC a D α u)(t0 ) > 0, which contradicts the fractional inequality (3.1), and completes the proof. Corollary 3.1. (Comparison Principle-2) Let u1 , u2 ∈ H 1 (a, b) ∩ C[a, b] be the solutions of (ABC a D α u1 )(t) + p(t)u1 (t) = g1 (t), t > a, 0 < α < 1, (ABC a D α u2 )(t) + p(t)u2 (t) = g2 (t), t > a, 0 < α < 1, where p(t) ≥ 0, g1(t), g2 (t) are continuous on [a, b] and p(a) 6= 0. If g1 (t) ≤ g2 (t), then it holds that u1 (t) ≤ u2 (t), t ≥ a. Proof. Let z = u1 − u2 , then it holds that Pα (z) = (ABC a D α z)(t) + p(t)z(t) = g1 (t) − g2 (t) ≤ 0, t > a, 0 < α < 1.
(3.2)
Applying virtue of Lemma 3.1 we have z(t) ≤ 0, and hence the result. Lemma 3.2. Let u ∈ H 1 (a, b) be the solution of (ABC a D α u)(t) + p(t)u(t) = g(t), t > a, 0 < α < 1, where p(t) > 0 is continuous on [a, b]. Then it holds that ||u||[a,b] = max |u(t)| ≤ M = max {| t∈[a,b]
t∈[a,b]
g(t) |}. p(t)
g(t) Proof. We have M ≥ | p(t) |, or Mp(t) ≥ |g(t)|, t ∈ [a, b]. Let v1 = u − M, then it holds that
Pα (v1 ) = (ABC a D α v1 )(t) + p(t)v1 (t) = (ABC a D α u)(t) + p(t)u(t) − p(t)M = g(t) − p(t)M ≤ |g(t)| − p(t)M ≤ 0. Thus by virtue of Lemma 3.1 we have v1 = u − M ≤ 0, or u ≤ M.
(3.3)
Analogously, let v2 = −M − u, then it holds that Pα (v2 ) = (ABC a D α v2 )(t) + p(t)v2 (t) = −(ABC a D α u)(t) − p(t)u(t) − p(t)M = −g(t) − p(t)M ≤ −g(t) − |g(t)| ≤ 0. Thus by the result of Lemma 3.1 we have v2 = −u − M ≤ 0, or u ≥ −M.
(3.4)
By combining the results of Eq’s (3.3) and (3.4) we have |u(t)| ≤ M, t ∈ [a, b] and hence the result. 5
Lemma 3.3. The fractional initial value problem (ABC a D α u)(t) = λu + f (t), t > 0, 0 < α < 1, u(0) = u0 .
(3.5) (3.6)
has the unique solution 1 α ′ u(t) = B(α)u0Eα [ωt ] + (1 − α)(g(t) ∗ f (t) + f (0)g(t)) , B(α) − λ(1 − α) in the functional space H 1 (0, b)∩C[0, b], if and only if, λu0 +f (0) = 0, where ω = and α tα−1 ∗ Eα [wtα ]. g(t) = Eα [ωtα ] + 1 − α Γ(α)
(3.7)
λα , B(α)−λ(1−α)
Proof. Since u ∈ H 1 (0, b) we have (ABC a D α u)(0) = 0. Thus, a necessary condition for the existence of a solution to Eq. (3.5) is that λu0 + f (0) = 0.
(3.8)
Applying the Laplace transform to Eq. (3.5) and using the fact that (ABC 0 D α u)(t) =
B(α) α α Eα [− t ] ∗ u′ (t), 1−α 1−α
we have α α B(α) ′ L Eα [− t ] ∗ u (t) . λL(u) + L(f (t)) = 1−α 1−α Applying the convolution result of the Laplace transform together with L(Eα [−
α α sα−1 t ]) = α α , 1−α s + 1−α
|
α 1 | < 1, 1 − α sα
will lead to λL(u) + L(f (t)) =
B(α) sα−1 α (sL(u) − u(0)). 1 − α sα + 1−α
(3.9)
Direction calculations will lead to α
sα + 1−α B(α)u0 1−α sα−1 L(u) = + L(f (t)), B(α) − λ(1 − α) sα − ω B(α) − λ(1 − α) sα − ω where ω =
λα . B(α)−λ(1−α)
(3.10)
Thus,
α−1 α α s + 1−α B(α)u0 1−α s −1 −1 u(t) = + L L L(f (t)) , B(α) − λ(1 − α) sα − ω B(α) − λ(1 − α) sα − ω α α s + 1−α 1−α B(α)u0 α −1 Eα [ωt ] + L L(f (t)) . (3.11) = B(α) − λ(1 − α) B(α) − λ(1 − α) sα − ω 6
Let G(s) =
α sα−1 α 1 sα−1 1 sα + 1−α = + , s sα − ω sα − ω 1 − α sα sα − ω
then g(t) = L−1 G(s) = Eα (ωt) +
α tα−1 ∗ Eα (ωtα ). 1 − α Γ(α)
Applying the convolution result we have α α s + 1−α −1 −1 L G(s)sL(f (t)) L(f (t)) = L sα − ω ′ −1 −1 G(s)[L(f ) + f (0)] G(s)[sL(f ) − f (0) + f (0)] = L = L ′ −1 G(s)L(f ) + f (0)G(s) = g(t) ∗ f ′ (t) + f (0)g(t). (3.12) = L The result follows by substituting Eq. (3.12) in Eq. (3.11). Corollary 3.2. The fractional differential equation (ABC 0 D α u)(t) = λu, t > 0, 0 < α < 1,
(3.13)
has only the trivial solution u = 0, in the functional space H 1 (0, b) ∩ C[0, b]. Proof. Applying the result of Lemma 3.3 with f (t) = 0, yields u(t) =
1 B(α)u0 Eα [ωtα ]. B(α) − λ(1 − α)
The necessary condition for the existence of solution yields that u0 = 0, and hence the result.
4
The nonlinear equation
In this section we apply the obtained comparison principles to establish a uniqueness result for a nonlinear fractional differential equation and to estimate its solution. We have Lemma 4.1. Consider the nonlinear fractional differential equation (ABC a D α u)(t) = f (t, u), t > a, 0 < α < 1,
(4.1)
where f (t, u) is a smooth function. If f (t, u) is non-increasing with respect to u then the above equation has at most one solution u ∈ H 1 (a, b). Proof. Assume that u1 , u2 ∈ H 1 (a, b) be two solutions of the above equation and let z = u1 − u2 . Then it holds that (ABC a D α z)(t) = f (t, u1) − f (t, u2). Applying the mean value theorem we have f (t, u1) − f (t, u2) = 7
∂f ∗ (u )(u1 − u2 ), ∂u
for some u∗ between u1 and u2 . Thus, (ABC a D α z)(t) −
∂f ∗ (u )z = 0. ∂u
(4.2)
∂f Since − ∂u (u∗ ) > 0, then z(t) ≤ 0, by virtue of Lemma 3.1. Also, Eq. (4.2) holds true for −z and thus −z ≤ 0, by virtue of Lemma 3.1. Thus, z = 0 which proves that u1 = u2 .
Lemma 4.2. Consider the nonlinear fractional differential equation (ABC a D α u)(t) = f (t, u), t > a, 0 < α < 1,
(4.3)
where f (t, u) is a smooth function. Assume that λ2 u + h2 (t) ≤ f (t, u) ≤ λ1 u + h1 (t), for all t ∈ (a, b), u ∈ H 1 (a, b), where λ1 , λ2 < 0. Let v1 and v2 be the solutions of (ABC a D α v1 )(t) = λ1 v1 + h1 (t), t > 0, 0 < α < 1,
(4.4)
(ABC a D α v2 )(t) = λ2 v2 + h2 (t), t > 0, 0 < α < 1,
(4.5)
and then it holds that v2 (t) ≤ u(t) ≤ v1 (t), t ≥ a. Proof. We shall prove that u(t) ≤ v1 (t) and by applying analogous steps one can show that v2 (t) ≤ u(t). By subtracting Eq. (4.4) from Eq. (4.3) we have ABC α a D (u − v1 ) (t) = f (t, u) − λ1 v1 − h1 (t) ≤ λ1 u + h1 (t) − λ1 v1 − h1 (t) = λ1 (u − v1 ). Let z = u − v1 , then it holds that (ABC a D α z)(t) − λ1 z(t) ≤ 0. Since λ1 > 0, then z ≤ 0, by virtue of Lemma 3.1, which proves the result. We now present some examples to illustrate the efficiency of the obtained results. Example 4.1. Consider the nonlinear fractional initial value problem (ABC 0 D α u)(t) = e−u − 2, t > 0, 0 < α < 1.
(4.6)
u(0) = − ln(2). Since e−u − 2 ≥ −u − 1, let v be the solution of (ABC 0 D α v)(t) = −v − 1, t > 0, 0 < α < 1,
(4.7)
then v(t) ≤ u(t) by virtue of Lemma 4.2. The solution of Eq. (4.7) is given by Eq. (3.7) with λ = −1, and f (t) = −1. Thus, u(t) ≥ v(t) = −
α tα−1 1 B(α)Eα [wtα ] + (1 − α)(Eα [wtα ] + ∗ Eα [wtα ] , B(α) + 1 − α 1 − α Γ(α)
α . We recall that Eq. (4.7) has a solution only if v(0) = −1. where ω = − B(α)+1−α
8
Example 4.2. Consider the nonlinear fractional initial value problem 1 (ABC 0 D α u)(t) = e−u − u2 , t > 0, 0 < α < 1. 2
(4.8)
u(0) = u0 , where u0 is the unique solution of e−u0 = 21 u20 . By the Taylor series expansion of f (u) = e−u , one can easily show that e−u − 12 u2 ≤ 1 − u. Let v be the solution of (ABC 0 D α v)(t) = −v + 1, t > 0, 0 < α < 1,
(4.9)
then v(t) ≥ u(t) by virtue of Lemma 4.2. The solution of Eq. (4.9) is given by Eq. (3.7) with λ = −1, and f (t) = 1. Thus, u(t) ≤ v(t) =
1 α tα−1 B(α)Eα [wtα ] + (1 − α)(Eα [wtα ] + ∗ Eα [wtα ] , B(α) + 1 − α 1 − α Γ(α)
α . We recall that Eq. (4.9) has a solution only if v(0) = 1. Moreover, where ω = − B(α)+1−α applying the result of Lemma 3.2 we have ||v|| ≤ 1, and hence ||u|| ≤ 1.
Example 4.3. Consider the nonlinear fractional initial value problem (ABC 0 D α u)(t) = −eu (3 + cos(u)) + 4e−t , t > 0, 0 < α < 1.
(4.10)
u(0) = 0. Let h(u) = −eu (3 + cos(u)), since h′′ (u) = eu (−3 + 2 sin(u)) ≤ 0, by the Taylor series expansion method one can easily show that h(u) ≤ h(0) + h′ (0)u = −4 − 4u. Let v be the solution of (ABC 0 D α v)(t) = −4v − 4 + 4e−t , t > 0, 0 < α < 1, v(0) = 0,
(4.11)
then v(t) ≥ u(t) by virtue of Lemma 4.2. The solution of Eq. (4.11) is given by Eq. (3.7) where λ = −4, and f (t) = −4 + 4e−t . Applying the result of Lemma 3.2 we have ||u|| ≤ ||v|| ≤ |
−4 + 4e−t | = 1 − e−t , t > 0. 4
Acknowledgment: The author acknowledges the support of the United Arab emirates University under the Fund No. 31S239-UPAR(1) 2016.
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