Global existence and uniqueness for implicit differential ... - Ele-Math

F ractional Differential Calculus Volume 5, Number 2 (2015), 199–208

doi:10.7153/fdc-05-17

GLOBAL EXISTENCE AND UNIQUENESS FOR IMPLICIT DIFFERENTIAL EQUATION OF ARBITRARY ORDER S AGAR T. S UTAR AND K ISHOR D. K UCCHE Abstract. The aim of this paper is to establish the existence result for implicit differential equation of fractional (arbitrary) order via topological transversality theorem known as Leray-Schauder alternative. Further we prove the uniqueness results. The Grownwall’s lemma for singular kernels play an important role to prove our results. We verify our results by providing an example.

1. Introduction It is well known that behavior of many physical systems can be properly described by using the various forms of fractional differential and hence the theory of fractional differential has a great importance. Detail basic theory of fractional calculus, fractional differential equations and its applications, can found in the monographs [5, 9, 12, 15]. We mention few papers [1, 2, 6, 10, 11] in which the interesting theory results of fractional differential equations has been investigated. Recently, in papers [4, 13] the existence, uniqueness and other properties of solutions for implicit fractional differential equations with initial conditions have been established by using the technique of fixed point theorems and approximated methods. Inspired by the works of [4, 13] here we study the existence and uniqueness of solution of initial value problem for implicit differential equation of fractional order with Caputo fractional derivative given in the form: c

Dα x(t) = f (t, x(t),c Dα x(t)), k

n

x (0) = xk ∈ R ,

t ∈ [0, b], b > 0, n − 1 < α n,

k = 0, 1, · · · , n − 1,

(1.1) (1.2)

where f : [0, b] × Rn × Rn → Rn is a nonlinear continuous function, x : [0, b] → Rn and c Dα denotes the Caputo fractional derivative of order α . The initial value problem (1.1)–(1.2) and their special forms have been studied [3, 4, 13] by using different approaches. The method which we have used here gives existence as well as the maximal interval of existence. Further, the uniqueness result is obtained with the same assumption of existence result. We organize this paper as follows: Preliminaries, the basic definitions of fractional calculus, and the theorems which are required in this paper are given in Section 2. In Section 3, we establish existence and uniqueness theorems for problem (1.1)–(1.2). We present an example to illustrate the theorem in Section 3. Mathematics subject classification (2010): 26A33, 34A08. Keywords and phrases: Implicit fractional differential equation, Leray-Schauder alternative, existence and uniqueness, integral inequality. c , Zagreb Paper FDC-05-17

199

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S. T. S UTAR AND K. D. K UCCHE

2. Preliminaries Let Rn is an Euclidean n -space with the norm · and denote by B = C([0, b], Rn ) – the Banach space of all continuous functions from [0, b] into Rn with the suprimum norm xB = sup{x(t) : t ∈ [0, b]} . Let C[a, b] denotes the space of all continuous functions defined over [a,b] and Cn [a, b] be the set of all real valued functions on [a,b] having nth order continuous derivatives. Here we give some basic definitions and the results [5, 9, 12, 15] which are required throughout this paper. D EFINITION 2.1. Let f ∈ C[0, b] and α > 0 then Riemann-Liouville fractional integral of order α of a function f is defined as I α f (t) =

1 Γ(α )

t 0

(t − s)α −1 f (s)ds,

provided the integral exists. Note that, I 0 f (t) = f (t). D EFINITION 2.2. For a function f ∈ Cn [0, b], the Caputo derivative of order α is defined as c

Dα f (t) =

1 Γ(n − α )

t 0

(t − s)n−α −1 f (n) (s)ds,

n − 1 < α n.

L EMMA 2.3. Let f ∈ Cn [0, b], and α > 0 , then n−1

f k (0) k t , k=0 Γ(k + 1)

I α [c Dα f (t)] = f (t) − ∑

n − 1 < α n.

L EMMA 2.4. Let f (t) = t β , where β 0 and let n − 1 < α n , n ∈ N, then 0 if β ∈ {0, 1, · · ·, n − 1} c α D f (t) = Γ(1+β ) β −α if β ∈ N, β n or β ∈ / N, β > n − 1. Γ(1+β −α ) t L EMMA 2.5. [8] Suppose δ 0 , α > 0 and a(t) is a nonnegative function locally integrable on 0 t < T (some T ∞), and suppose u(t) is nonnegative and locally integrable on 0 t < T with u(t) a(t) + δ

t 0

(t − s)α −1 u(s)ds

on this interval; then u(t) a(t) + θ

t 0

Eα (θ (t − s))a(s)ds, 0 t < T,

E XISTENCE AND UNIQUENESS OF SOLUTIONS

201

d zα −1 Eα (z), Eα (z) Γ( α ) as dz z → 0+ , Eα (z) α1 ez as z → +∞ (and Eα (z) α1 ez as z → ∞+ ). If a(t) ≡ a , constant, then u(t) aEα (θ t). nα

1

z where θ = (δ Γ(α )) α , Eα (z) = ∑∞ n=0 Γ(nα +1) , Eα (z) =

D EFINITION 2.6. If f ∈ C([0, b] × Rn × Rn , Rn ), then x ∈ B given by x(t) =

n−1

xk 1 ∑ Γ(k + 1) t k + Γ(α ) k=0

t 0

(t − s)α −1 f (s, x(s),c Dα x(s))ds, t ∈ [0, b],

is the solution of implicit fractional differential equation (1.1)–(1.2). To prove the existence result, we use the topological transversality theorem known as Leray-Schauder alternative given below. L EMMA 2.7. ([7]) Let S be a convex subset of a normed linear space E and assume that 0 ∈ S . Let F : S → S be a completely continuous operator, and let

ε (F) = {x ∈ S : x = λ Fx, 0 < λ < 1}. Then either ε (F) is unbounded or F has a fixed point in S . Study of integer order differential equations via topological transversality can be found in [14] and the references therein. 3. Existence and uniqueness of solution T HEOREM 3.1. (Existence) Let f : [0, b] × Rn × Rn → Rn be a continuous function that satisfies the conditions: (H1) f (t, x, y) p(t)x + Ly , t ∈ [0, b] and x, y ∈ Rn , where p(t) : [0, b] → (0, ∞) is continuous function and 0 < L < 1 . (H2) There exist constants M > 0 , 0 < N < 1 such that f (t, x, y) − f (t, x, y) Mx − x + Ny − y, t ∈ [0, b], x, x, y, y ∈ Rn . Then the initial-value problem for implicit fractional differential equation (1.1)–(1.2) has at least one solution on [0, b]. Proof. Firstly we establish the priori bounds on the solutions of implicit fractional differential equation of the form: c

Dα x(t) = λ f (t, x(t),c Dα x(t)), k

n

x (0) = xk ∈ R , for λ ∈ (0, 1).

t ∈ [0, b], b > 0, n − 1 < α n,

k = 0, 1, · · · , n − 1,

(3.1) (3.2)

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Let x(t) be the solution of (3.1)–(3.2) then it is equivalent to integral equation x(t) =

n−1

xk λ ∑ Γ(k + 1) t k + Γ(α ) k=0

t 0

(t − s)α −1 f (s, x(s),c Dα x(s))ds.

Then for any t ∈ [0, b] we have x(t)

n−1

x 1 ∑ Γ(k +k 1) t k + Γ(α ) k=0

n−1

xk 1 ∑ Γ(k + 1) t k + Γ(α ) k=0

t 0

t 0

(t − s)α −1 f (s, x(s),c Dα x(s))ds (t − s)α −1 (p(s)x(s) + Lc Dα x(s)) ds

(3.3)

From (3.1) and the hypothesis (H1) for any t ∈ [0, b] we have, c Dα x(t) p(t)x(t) + LcDα x(t), and hence c Dα x(t)

p(t) 1−L x(t).

Thus for any t ∈ [0, b], (3.3) gives

n−1

xk k 1 x(t) ∑ t + Γ(k + 1) Γ( α) k=0

n−1

xk

t

1

∑ Γ(k + 1) bk + Γ(α )

k=0

0

(t − s)

t 0

α −1

(t − s)α −1

Lp(s) x(s) ds p(s)x(s) + 1−L

p(s) x(s)ds 1−L

As p(t) : [0, b] → (0, ∞) is continuous on compact set [0, b] there exist constant P such that p(t) P < ∞, t ∈ [0, b]. Therefore, x(t)

n−1

xk P ∑ Γ(k + 1) bk + (1 − L)Γ(α ) k=0

t 0

(t − s)α −1 x(s)ds. t ∈ [0, b].

xk k By an application of Lemma 2.5, with a(t) = ∑n−1 k=0 Γ(k+1) b , a constant, and δ = P , we obtain (1−L)Γ(α )

1 α P Γ(α ) t (1 − L)Γ(α ) 1 n−1 α P xk k b Eα =∑ t 1−L k=0 Γ(k + 1) n−1

xk k b Eα x(t) ∑ k=0 Γ(k + 1)

ξ , t ∈ [0, b], where ξ is some nonnegative constant. Therefore xB ξ . This proves the solution of (3.1)–(3.2) is bounded. Now let y ∈ B and define x(t) =

k=n−1

∑

k=0

xk t k + y(t), t ∈ [0, b]. Γ(k + 1)


203

Then y(t) satisfies n−1 xk xk k c α k s + y(s), D s + y(s) (t − s) f s, ∑ ds ∑ 0 k=0 Γ(k + 1) k=0 Γ(k + 1) t n−1 1 xk α −1 k c α = (t − s) f s, ∑ s + y(s), D y(s) ds Γ(α ) 0 k=0 Γ(k + 1)

1 y(t) = Γ(α )

t

α −1

n−1

if and only if x(t) satisfies x(t) =

n−1

xk 1 ∑ Γ(k + 1) t k + Γ(α ) k=0

t 0

(t − s)α −1 f (s, x(s),c Dα x(s))ds, t ∈ [0, b]

Denote B0 = {y ∈ B : y(k) (0) = 0, k = 0, 1, ..., n − 1} then clearly B0 is a convex subset of the space B and 0 ∈ B0 . Define an operator F : B0 → B0 by: 1 (Fy)(t) = Γ(α )

t 0

α −1

(t − s)

f

xk k c α s + y(s), D y(s) ds. s, ∑ k=0 Γ(k + 1) n−1

Let yn → y in B0 . Then using (H2) for any t ∈ [0, b] we have

t

n−1 1 xk α −1

k c α (Fyn )(t) − (Fy)(t) s + yn(s), D yn (s) (t − s)

f s, ∑

Γ(α ) 0 k=0 Γ(k + 1)

n−1 xk

k c α s + y(s), D y(s) ds − f s, ∑

Γ(k + 1) k=0

1 Γ(α )

t 0

(t − s)α −1 (M(yn − y)(s) + Nc Dα (yn − y)(s)) ds.

But (H1) gives c Dα (yn − y)(t) Therefore

p(t) (yn − y)(t), t ∈ [0, b]. 1−L

t 1 N p(s) (yn − y)(s) ds (t − s)α −1 M(yn − y)(s) + Γ(α ) 0 1−L t M(1 − L) + NP (t − s)α −1 (yn − y)(s)ds (1 − L)Γ(α ) 0 (M(1 − L) + NP)bα yn − yB, t ∈ [0, b]. (1 − L)Γ(α + 1)

(Fyn )(t) − (Fy)(t)

It follows that Fyn − FyB

(M(1 − L) + NP)bα yn − yB, (1 − L)Γ(α + 1)

204


Thus Fyn → Fy in B0 whenever yn → y in B0 . This proves F is continuous on B0 . Next, we show that F is completely continuous. Let {un } be a bounded sequence in B0 .Then there exists a constant K > 0 such that un K ∀n . Using the definition of F and the hypothesis (H1), for any t ∈ [0, b], we obtain

n−1 xk

k c α (t − s) s + un (s), D un (s) ds

f s, ∑

Γ(k + 1) 0 k=0 t n−1 xk k 1 α −1 c α (t − s) b + un (s) + L D un (s)) ds p(s) ∑ Γ(α ) 0 k=0 Γ(k + 1) t n−1 x Lp(s) 1 k α −1 k (t − s) p(s) ∑ b + un (s) + un (s) ds Γ(α ) 0 1−L k=0 Γ(k + 1) t n−1 x 1 PK k (t − s)α −1 P ∑ bk + ds Γ(α ) 0 1−L k=0 Γ(k + 1) n−1 PK xk k bα P∑ b + . Γ(α + 1) 1−L k=0 Γ(k + 1)

1 (Fun )(t) Γ(α )

t

α −1

This proves the set {(Fun )(t) : un B K, 0 t b} is uniformly bounded in Rn . Next aim is to show that the sequence {Fun } is equicontinuous. Let any t1 ,t2 ∈ [0, b]. By definition of F , assumption (H1) and letting 0 t1 t2 b , we have (Fun )(t1 ) − (Fun )(t2 )

n−1 xk 1

t1 α −1 k c α f s, ∑ s + un (s), D un (s) ds

(t1 − s) Γ(α ) 0 k=0 Γ(k + 1)

t2

n−1 xk

− (t2 − s)α −1 f s, ∑ sk + un (s), c Dα un (s) ds

Γ(k + 1) 0 k=0

n−1 xk 1

t1 α −1 α −1 k c α − (t2 − s) ] f s, ∑ s + un (s), D un (s) ds

[(t1 − s) Γ(α ) 0 k=0 Γ(k + 1)

t2

n−1 xk

α −1 k c α s + un (s), D un (s) ds

(t2 − s) f s, ∑ −

t1 k=0 Γ(k + 1)

t1

n−1 1 xk

α −1 α −1

k c α s + un (s), D un (s) ds [(t1 − s) − (t2 − s) ] f s, ∑

Γ(α ) 0 Γ(k + 1) k=0

t2

n−1 1 xk

+ sk + un (s), c Dα un (s) ds. (t2 − s)α −1 f s, ∑ (3.4)

Γ(α ) t1 Γ(k + 1) k=0

Noting that,

n−1 n−1 PM xk xk k

k c α s + un (s), D un (s) P ∑ b + ,

f s, ∑

1−L k=0 Γ(k + 1) k=0 Γ(k + 1)

from the inequality (3.4) we obtain 1 (Fun )(t1 ) − (Fun )(t2 ) Γ(α + 1)

n−1

xk k PM P∑ b + Γ(k + 1) 1 −L k=0

[t1α − t2α + 2(t2 − t1 )α ]

205


2 Γ(α + 1)

n−1

PM xk k b + P∑ Γ(k + 1) 1 −L k=0

(t2 − t1 )α .

(3.5)

From the inequality 3.5 we conclude that {Fun } is equicontinuous family and hence by Arzela-Ascoli argument the operator F is completely continuous. Finally we see that the set ε (F) = {y ∈ B0 : y = λ Fy, 0 < λ < 1} is bounded in B, since for any y ∈ ε (F), the function x(t) =

n−1

tk

∑ Γ(k + 1) xk + y(t),

t ∈ [0, b],

k=0

is a solution of (3.1)–(3.2) for which we have shown that xB ξ and hence yB bk ξ + ∑n−1 k=0 Γ(k+1) xk . Therefore by Lemma 2.7, the operator F has a fixed point y ∈ B0 . Then x (t) =

n−1

tk

∑ Γ(k + 1) xk + y (t),

t ∈ [0, b],

k=0

is the solution of (1.1)–(1.2). This completes the proof.

R EMARK 3.1. Note that an application of the topological transversality theorem (Leray-Schauder alternative) does not guarantee an uniqueness of the solution. With the same assumption of Theorem 3.1 we have following uniqueness theorem. T HEOREM 3.2. (Uniqueness) Let f : [0, b] × Rn × Rn → Rn satisfies the assumptions (H1) and (H2). Then the initial-value problem for implicit fractional differential equation (1.1)–(1.2) has a unique solution. Proof. Let x1 (t) and x2 (t) be any two solution of (1.1)–(1.2) then by using (H2) for any t ∈ [0, b] we have x1 (t) − x2(t)

t 0

(t − s)α −1 ( f (s, x1 (s),c Dα x1 (s)) − f (s, x2 (s),c Dα x2 (s))) ds

1 Γ(α )

t 0

(t − s)α −1 (M(x1 − x2 )(s) + Nc Dα (x1 − x2)(s)) ds.

By (H1) for any t ∈ [0, b], c Dα (x1 − x2 )(t)

p(t) (x1 − x2)(t). 1−L

Hence for any t ∈ [0, b] we have t 1 N p(s) (x1 − x2 )(s) ds (t − s)α −1 M(x1 − x2)(s) + Γ(α ) 0 1−L M(1 − L) + NP t (t − s)α −1 (x1 − x2 )(s)ds. Γ(α )(1 − L) 0

(x1 − x2 )(t)

206


Using Lemma 2.5 and noting that a(t) = 0 for above inequality , we obtain (x1 − x2 )(t) = 0, ∀t ∈ [0, b]. This proves uniqueness of solutions of (1.1)–(1.2).

4. Example In this section, we give an example to illustrate the results. E XAMPLE . Consider the implicit fractional differential equation c

5

D 2 x(t) =

et

5 1 et x(t) + sin(c D 2 x(t)), t ∈ [0, b], 0 < b < ∞. +5 2

(4.1)

Define f : [0, b] × Rn × Rn → Rn is given by

5 f t, x(t),c D 2 x(t) =

et

5 1 et x(t) + sin(c D 2 x(t)), t ∈ [0, b] +5 2

Then 5 1 et x(t) + sin(c D 2 x(t)) +5 2 5 et 1 t x(t) + c D 2 x(t). e +5 2

5

f t, x(t),c D 2 x(t)

et

t

We see that p(t) = ete+5 : [0, b] → (0, ∞) is a continuous function and L = 12 . u−v sin 2 |u − v|, for any u, v ∈ Further, noting that | sin u − sin v| = 2 cos u+v 2 R, we have

5 5

f t, x(t),c D 2 x(t) − f t, x(t),c D 2 x(t)

5 5 et 1 t x(t) − x(t) + sin(c D 2 x(t) − sin(c D 2 x(t)) e +5 2 5 5 et 1 t x(t) − x(t) + c D 2 x(t) − c D 2 x(t) e +5 2

5 1

5 Mx(t) − x(t) + c D 2 x(t) − c D 2 x(t) , 2 t where M > 0 is a constant such that ete+5 M , t ∈ [0, b]. Since all the assumptions of Theorems 3.1 and 3.2 are satisfied the initial value problem (4.1) has unique solution on [0, b] for any 0 < b < ∞. Acknowledgements. We are very grateful to the anonymous referees for their valuable comments and suggestions to improve our paper. The work of K. D. Kucche has been supported by UGC, New Delhi, through the Minor Research Project F. No. 42997/2013(SR).


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(Received September 25, 2015)

Sagar T. Sutar Department of Mathematics Raje Ramrao Mahavidyalaya, Jath, Sangli-416404 Maharashatra, India e-mail: [email protected] Kishor D. Kucche Department Of Mathematics, Shivaji University Kolhapur-416 004 Maharashatra, India e-mail: [email protected]

Fractional Differential Calculus

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