RACSAM DOI 10.1007/s13398-017-0389-4 ORIGINAL PAPER

The uniqueness result of solutions to initial value problems of differential equations of variable-order Shuqin Zhang1

Received: 18 October 2015 / Accepted: 9 March 2017 © Springer-Verlag Italia 2017

Abstract This paper is concerned with the existence and uniqueness of solution to an initial value problem for a differential equation of variable-order. The results are obtained by means of fixed point theorem. The obtained results are illustrated with the aid of examples. Keywords Derivatives and integrals of variable-order · Initial value problem · Piecewise constant functions · Uniqueness of solution Mathematics Subject Classification 26A33 · 34A08

1 Introduction The study of initial or boundary value problem for fractional differential equations has become an object of extensive study in view of their extensive applications in various scientific disciplines, such as fluid mechanics, biomathematics, ecology, visco-elastodynamics, aerodynamics, control theory, electro-dynamics of complex medium, etc., see [21]. The researchers have gained many beautiful results for existence of solutions to differential equations of fractional order by the nonlinear functional analysis methods, such as some fixed point theorems, monotone iterative method, etc., see [1–4]. In [1], the author studied the following initial value problem of fractional differential equation  q D0+ x(t) = f (t, x), 0 < t ≤ T < +∞, x(0) = x0 ,

This research is supported by the Natural Science Foundation of China (11371364).

B 1

Shuqin Zhang [email protected] Department of Mathematics, China University of Mining and Technology, Beijing 100083, People’s Republic of China

S. Zhang q

where 0 < q < 1, f ∈ C([0, T ], R), D0+ is the Riemann–Liouville fractional derivative. The author obtained the basic theory of the above fractional differential equations by the classical approach. On the other hand, the operators of variable-order, which are the derivatives and integrals whose order is a function of certain variables, attract attention due to their applied significance in various research areas, such as engineering, physics, chemistry, control of dynamical systems etc. In recent years, the operator and differential equations of variable-order have been applied in engineering more and more frequently. For the examples and details, see [5–20] and the references therein. Although the existing literature on solutions of fractional differential equations is quite wide, few papers deal with the existence of solutions to differential equations with variableorder. Further progress needs to be achieved to extend the available theorems in ordinary or fractional derivatives to variable-order derivatives. In this paper, we study the following initial value problem for differential equation of variable-order  q(t) D0+ x(t) = f (t, x), 0 < t ≤ T, (1.1) x(0) = 0, q(t)

where 0 < T < +∞, D0+ denotes derivative of variable-order defined by  t (t − s)−q(t) d q(t) D0+ x(t) = x(s)ds, t > 0, dt 0 Γ (1 − q(t)) and 1−q(t)

I0+

 x(t) = 0

t

(t − s)−q(t) x(s)ds, t > 0 Γ (1 − q(t))

denotes integral of variable-order 0 < q(t) < 1, 0 ≤ t ≤ T . Remark 1.1 Let p : [a, b] → (0, +∞)(−∞ < a < b < +∞), the fractional integral of variable-order p(t) for function x(t) is defined as following  t (t − s) p(t)−1 p(t) x(s)ds, t > a. (1.2) Ia+ x(t) = Γ ( p(t)) a Remark 1.2 Let p : [a, b] → (n −1, n] (−∞ < a < b < +∞, n is a given natural number), the fractional derivative of variable-order p(t) for function x(t) is defined as following  n  t (t − s)n− p(t)−1 d p(t) x(s)ds, t > a, (1.3) Da+ x(t) = dt Γ (n − p(t)) a The variable-order fractional derivative is an extension of constant order fractional derivative. According to (1.2) and (1.3), it is clear that when q(t) is a constant function, i.e. q(t) ≡ q q(t) q(t) (q is a finite positive constant), then Ia+ , Da+ are the usual Riemann–Liouville fractional integral and derivative [21]. Obviously, (1.1) is a generalization of the initial value problem in [1]. Ones can transform the fractional differential equations into an equivalent integral equations by means of some fundamental properties of differential and integral calculus of q q fractional order. The following properties of fractional calculus operators D0+ , I0+ play an important part in discussing the existence of solutions of fractional differential equations.

The uniqueness result of solutions to initial... γ

γ +δ

δ f (t) = I Proposition 1.1 [21] The equality I0+ I0+ 0+ f (t), γ > 0, δ > 0 holds for f ∈ L(0, b), 0 < b < +∞. γ

γ

Proposition 1.2 [21] The equality D0+ I0+ f (t) = f (t), γ > 0 holds for f ∈ L(0, b), 0 < b < +∞. Proposition 1.3 [21] Let 0 < α ≤ 1. Then the differential equation α D0+ u=0

has solution u(t) = ct α−1 , c ∈ R. α u ∈ L(0, b). Then the following equality holds Proposition 1.4 [21] Let 0 < α ≤ 1, D0+ α α D0+ u(t) = u(t) + ct α−1 , c ∈ R. I0+

We are interested in whether the above properties of fractional operators remain true for the operators of variable-order. Let’s take Proposition 1.1 for example. In the special cases: p(t) q(t) p(t)+q(t) f (t) ≡ 1, t ∈ [0, T ], we calculate I0+ I0+ f (t) and I0+ f (t). First, according to (1.2), we have   t (t − s) p(t)−1 s (s − τ )q(s)−1 p(t) q(t) f (τ )dτ ds I0+ I0+ f (t) = Γ ( p(t)) Γ (q(s)) 0 0  t  (t − s) p(t)−1 s (s − τ )q(s)−1 = dτ ds Γ ( p(t)) Γ (q(s)) 0 0  t (t − s) p(t)−1 s q(s) = ds 0 Γ ( p(t))Γ (1 + q(s))  1 t q(tr ) t p(t) = (1 − r ) p(t)−1 r q(tr ) dr. Γ ( p(t)) 0 Γ (1 + q(tr )) Second, we have p(t)+q(t)

I0+

 f (t) = 0

t

(t − s) p(t)+q(t)−1 t p(t)+q(t) f (s)ds = . Γ ( p(t) + q(t)) Γ (1 + p(t) + q(t))

Thus, for f (t) ≡ 1, we can obtain p(t) q(t)

p(t)+q(t)

I0+ I0+ f (t) = I0+

f (t)

only if  0

1

t q(tr ) β ( p(t), q(t)) q(t) (1 − r ) p(t)−1 r q(tr ) dr = t . Γ (1 + q(tr )) Γ (q(t))

However, we can’t assert that the above equality is true. We can only get  1 1 β ( p(t), q(t)) q(t) t q(t) r q(t)−1 (1 − r ) p(t)−1 dr = t . Γ (q(t)) 0 Γ (q(t)) Therefore, from the above arguments, we could obtain the following result.

S. Zhang

Lemma 1.1 Let x(t), p(t), q(t) be real functions on finite interval [0, T ], assume that p(t) p(t)+q(t) p(t) q(t) variable-order fractional integrals I0+ x(t), I0+ x(t) and I0+ I0+ x(t) defined by (1.2) exist. In general case, we could claim that p(t) q(t)

p(t)+q(t)

I0+ I0+ x(t)  = I0+

x(t), for some points in [0, T ].

(1.4)

In particular, for general functions 0 < p(t) < 1 and x(t), we have p(t) 1− p(t)

I0+ I0+

p(t)+1− p(t)

x(t)  = I0+

1 x(t) = I0+ x(t), for some points in [0, T ].

The following example illustrates that (1.4) is valid.  t, 0 ≤ t ≤ 2, Example 1 Let p(t) = q(t) = t, 0 ≤ t ≤ 6, and f (t) = 1, 0 ≤ t ≤ 6. 2, 2 < t ≤ 6, p(t) q(t)

p(t)+q(t)

We’ll verify I0+ I0+ f (t)|t=2 is not equal to I0+ 

p(t) q(t)

I0+ I0+ f (t) =

t

(t − s) p(t)−1 Γ ( p(t))

2

(t − s) p(t)−1 Γ ( p(t))

0

 =

0





f (t)|t=2 .

s

(s − τ )q(s)−1 f (τ )dτ ds Γ (q(s))

s

(s − τ )q(s)−1 dτ ds Γ (q(s))

0



0

 (t − s) p(t)−1 s (s − τ )q(s)−1 + dτ ds Γ ( p(t)) Γ (q(s)) 2 0  2  (t − s) p(t)−1 s (s − τ )s−1 = dτ ds Γ ( p(t)) Γ (s) 0 0  t  (t − s) p(t)−1 s (s − τ )s−1 + dτ ds Γ ( p(t)) Γ (s) 2 0  2  t (t − s) p(t)−1 s s (t − s) p(t)−1 s s = ds + ds. 0 Γ ( p(t))Γ (1 + s) 2 Γ ( p(t))Γ (1 + s) t

1 1 |, M2 = max2≤s≤6 | Γ (1+s) |. Then for 2 ≤ t ≤ 6, it follows Let M1 = max2≤t≤6 | Γ ( p(t)) from   t   t      p(t)−1   t − s p(t)−1  ss (t − s) p(t)−1 s s    ·   6 · ds  ≤ ds   6 Γ ( p(t))Γ (1 + s) 2 Γ ( p(t))Γ (1 + s) 2    t t − s 2−1 6 62−1 s ds ≤ M1 M2 6 2  t = M1 M2 (t − s)s 6 ds 2  t (t − s)ds ≤ M1 M2 t 6 2

M1 M2 t 6 (t − 2)2 = 2 that 

t 2

(t − s) p(t)−1 s s ds Γ ( p(t))Γ (1 + s)

 = 0. t=2

The uniqueness result of solutions to initial...

Then, we have 

p(t) q(t)

I0+ I0+ f (t)|t=2 =

0

2

(2 − s)2−1 s s ds ≈ 1.888. Γ (2)Γ (1 + s)

Moreover, p(t)+q(t) I0+



t

f (t) = 0

(t − s) p(t)+q(t)−1 t p(t)+q(t) ds = . Γ ( p(t) + q(t)) Γ (1 + p(t) + q(t))

So, we get p(t)+q(t)

I0+

f (t)|t=2 =

2 p(2)+q(2) 24 2 = = . Γ (1 + p(2) + q(2)) Γ (1 + 2 + 2) 3

Therefore, we obtain p(t) q(t)

p(t)+q(t)

I0+ I0+ f (t)|t=2  = I0+

f (t)|t=2 .

Remark 1.3 According to Lemma 1.1, we could claim that variable-order calculus of non-constant functions p(t) for x(t) defined by 1.2, 1.3 don’t have the properties like Propositions 1.2–1.4. Therefore, Propositions 1.1–1.4 are not true for the operators of variable-order defined by (1.2) and (1.3). Thus, one can not transform a differential equation of variable-order into an equivalent integral equation without these propositions. It is a difficulty for us in dealing with the initial value or boundary value problems of differential equations of variable-order. It is necessary and significant for us to conquer the difficulty and obtain the solution to a differential equations of variable-order. The paper is organized as following. In Sect. 2, we provide some necessary definitions associated with problem (1.1). In Sect. 3, we establish the unique existence of solutions for (1.1) by using Banach contraction mapping principle. In Sect. 3, some examples are given to demonstrate the theoretical results.

2 Preliminaries In this section, we introduce some definitions which are used throughout this paper. Let −∞ < a < b < +∞. Definition 2.1 A generalized interval is a subset I of R which is either an interval (i.e. a set of the form [a, b], (a, b), [a, b) or (a, b]); a point {a}; or the empty set ∅. Definition 2.2 If I is a generalized interval. A partition of I is a finite set P of generalized intervals contained in I , such that every x in I lies in exactly one of the generalized intervals J in P.

Example 2 The set P = {1}, (1, 6), [6, 7), {7}, (7, 8] of generalized intervals is a partition of [1, 8]. Definition 2.3 Let I be a generalized interval, let f : I → R be a function, and let be P a partition of I . f is said to be piecewise constant with respect to P if for every J ∈ P, f is constant on J .

S. Zhang

Example 3 The function f : [1, 6] → R defined by ⎧ 3, 1 ≤ x < 3, ⎪ ⎪ ⎪ ⎨1, x = 3, f (x) = ⎪ 5, 3 < x < 6, ⎪ ⎪ ⎩ 2, x = 6,

is piecewise constant with respect to the partition [1, 3), {3}, (3, 6), {6} of [1, 6]. Definition 2.4 Let I be a generalized interval. The function f : I → R is called piecewise constant on I , if there exists a partition P of I such that f is piecewise constant with respect to P. The following example illustrates that (1.4) is valid for piecewise constant functions p(t) and q(t) defined in the same partition of finite interval [a, b].   2, 0 ≤ t ≤ 1, 1, 0 ≤ t ≤ 1, Example 4 Let p(t) = q(t) = and f (t) = 1, 0 ≤ t ≤ 3, 1 < t ≤ 3, 2, 1 < t ≤ 3, p(t) q(t)

p(t)+q(t)

3. We’ll verify I0+ I0+ f (t)|t=2 is not equal to I0+ 

p(t) q(t)

I0+ I0+ f (t) =

t

(t − s) p(t)−1 Γ ( p(t))

1

(t − s) p(t)−1 Γ ( p(t))

0

 =

0



f (t)|t=2 .

s

(s − τ )q(s)−1 f (τ )dτ ds Γ (q(s))

s

(s − τ )q(s)−1 dτ ds Γ (q(s))

0



0

 (t − s) p(t)−1 s (s − τ )q(s)−1 dτ ds + Γ ( p(t)) Γ (q(s)) 1 0  1  (t − s) p(t)−1 s (s − τ )1−1 = dτ ds Γ ( p(t)) Γ (1) 0 0  t  (t − s) p(t)−1 s (s − τ )2−1 + dτ ds Γ ( p(t)) Γ (2) 1 0  1  t (t − s) p(t)−1 s (t − s) p(t)−1 s 2 = ds + ds, Γ ( p(t)) 0 1 2Γ (2)Γ ( p(t)) 

t

thus, we have 

p(t) q(t)

I0+ I0+ f (t)|t=2 = = = = = =

 2 (2 − s)2 s (2 − s)2 s 2 ds + ds Γ (3) 0 1 2Γ (2)Γ (3)     1 2 2 1 1 4s − 4s 2 + s 3 ds + 4s − 4s 3 + s 4 ds 2 0 4 1     4 1 1 4 1 1 2− + + (8 − 1) − (16 − 1) + (32 − 1) 2 3 4 4 3 5   31 11 1 28 + − 15 + 24 4 3 5 11 140 − 225 + 93 + 24 4 × 15 11 2 55 + 16 71 + = = . 24 15 120 120 1

The uniqueness result of solutions to initial...

Moreover, p(t)+q(t)

I0+



t

f (t) = 0

(t − s) p(t)+q(t)−1 t p(t)+q(t) ds = . Γ ( p(t) + q(t)) Γ (1 + p(t) + q(t))

So, we get p(t)+q(t)

I0+

f (t)|t=2 =

2 p(2)+q(2) 4 32 23+2 = = . = Γ (1 + p(2) + q(2)) Γ (1 + 3 + 2) 15 120

Therefore, we obtain p(t) q(t)

p(t)+q(t)

I0+ I0+ f (t)|t=2  = I0+

f (t)|t=2 ,

which implies that (1.4) is valid for piecewise constant functions p(t) and q(t) defined in the same partition [0, 1], (1, 3] of finite interval [0, 3].

3 Existence and uniqueness result In this section, we present our main results. Now we make the following assumptions: (H1 ) Let P = {[0, T1 ], (T1 , T2 ], (T2 , T3 ], . . . , (TN ∗ −1 , T ]} (N ∗ is a given natural number) be a partition of the finite interval [0, T ], and let q(t) : [0, T ] → (0, 1] be a piecewise constant function with respect to P, i.e. ∗

q(t) =

N 

qk Ik (t), t ∈ [0, T ],

(3.1)

k=1

where 0 < qk ≤ 1, k = 1, 2, . . . , N ∗ are constants, and Ik is the indicator of the interval [Tk−1 , Tk ], k = 1, 2, . . . , N ∗ (here T0 = 0, TN ∗ = T ), that is Ik = 1 for t ∈ [Tk−1 , Tk ], Ik = 0 for elsewhere. (H2 ) For 0 ≤ r ≤ qi , i = 1, 2, . . . , N ∗ , t r f : [0, T ] × R → R be a continuous function, and there exists a positive constant L satisfying q −r

L Ti i Γ (1 − r ) <1 Γ (1 + qi − r ) such that t r | f (t, x) − f (t, y)| ≤ L|x − y|, 0 ≤ t ≤ T, x, y ∈ R. (H3 ) For 0 ≤ r ≤ qi , i = 1, 2, . . . , N ∗ , t r f : [0, T ] × R → R is a continuous function, and that there exist positive constants M, μ > 1 satisfying 

qi −r

4 max0≤t≤Ti t r | f (t, 0)|Γ (1 − r )Ti Γ (1 + qi − r )

<

Γ (1 + qi − r )

qi −r

2MΓ (1 − r )Ti



1 μ−1

, (TN ∗ = T )

such that t r | f (t, x) − f (t, y)| ≤ M|x − y|μ , 0 ≤ t ≤ T, x, y ∈ R. In order to obtain our main results, we firstly carry on essential analysis to equation of (1.1).

S. Zhang

According to (H1 ), we have ∗

q(t) =

N 

qk Ik (t), t ∈ [0, T ].

k=1

Hence, we get 

N∗

 (t − s)−q(t) x(s)ds = Ik (t) Γ (1 − q(t))



(t − s)−qk x(s)ds. Γ (1 − qk )

(3.2)

 t N d  (t − s)−qk x(s)ds = f (t, x), 0 < t ≤ T. Ik (t) dt 0 Γ (1 − qk )

(3.3)

0

t

k=1

t 0

So, equation of (1.1) can be written by ∗

k=1

Then, Eq. (3.3) in the interval [0, T1 ] can be written by  t (t − s)−q1 d q1 x(t) = f (t, x), 0 < t ≤ T1 . x(s)ds = D0+ dt 0 Γ (1 − q1 ) Again, Eq. (3.3) in the interval (T1 , T2 ] can be written by  t (t − s)−q2 d x(s)ds = f (t, x), T1 < t ≤ T2 . dt 0 Γ (1 − q2 ) As well, Eq. (3.3) in the interval (T2 , T3 ] can be written by  t d (t − s)−q3 x(s)ds = f (t, x), T2 < t ≤ T3 . dt 0 Γ (1 − q3 )

(3.4)

(3.5)

(3.6)

In the same way, Eq. (3.3) in the interval (Ti−1 , Ti ], i = 4, 5, . . . , N ∗ (TN ∗ = T ) can be written by  t d (t − s)−qi x(s)ds = f (t, x), Ti−1 < t ≤ Ti . (3.7) dt 0 Γ (1 − qi ) Now, we present definition of solution to problem (1.1), which is fundamental in our work. Definition 3.1 We say problem (1.1) has a solution, if there exist functions u i (t), 1, 2, . . . , N ∗ , such that u 1 ∈ C[0, T1 ] satisfying Eq. (3.4) and u 1 (0) = 0; u 2 ∈ C[0, T2 ] satisfying Eq. (3.5) and u 2 (0) = 0; u 3 ∈ C[0, T3 ] satisfying Eq. (3.6) and u 3 (0) = 0; u i ∈ C[0, Ti ] satisfying Eq. (3.7) and u i (0) = 0 (i = 4, 5, . . . , N ∗ )(TN ∗ = T ). Remark 3.1 We say problem (1.1) has one unique solution, if function u i (t) of Definition 3.1 are unique, i = 1, 2, . . . , N ∗ . Based on the previous arguments, we have the following results. Theorem 3.1 Assume that conditions (H1 ) and (H2 ) hold, then problem (1.1) has one unique solution.

The uniqueness result of solutions to initial...

Proof According the above analysis, equation of problem (1.1) can be written as equation (3.3). Equation (3.3) in the interval [0, T1 ] can be written as (3.4). Applying operator q I0+1 to both sides of (3.4), by Proposition 1.4, we have  t 1 x(t) = ct q1 −1 + (t − s)q1 −1 f (s, x(s))ds, 0 ≤ t ≤ T1 . Γ (q1 ) 0 By x(0) = 0 and the assumption of function f , we could get c = 0. Define operator T : C[0, T1 ] → C[0, T1 ] by  t 1 T x(t) = (t − s)q1 −1 f (s, x(s))ds, 0 ≤ t ≤ T1 . (3.8) Γ (q1 ) 0 It follows from the continuity of function t r f (t, x(t)) that operator T : C[0, T1 ] → C[0, T1 ] is well defined. In fact, let g(t, x(t)) = t r f (t, x(t)), by (H2 ), one has g : [0, T ] × R → R is continuous. For x(t) ∈ C[0, T1 ], t0 ∈ [0, T1 ], we have |T x(t) − T x(t0 )|    t  t0  1  1 q1 −1 q1 −1  = (t − s) f (s, x(s))ds − (t0 − s) f (s, x(s))ds  Γ (q1 ) 0 Γ (q1 ) 0  q −r  1 t 1 =  (1 − τ )q1 −1 τ −r g(tτ, x(tτ ))dτ Γ (q1 ) 0  q −r  1  t0 1 q1 −1 −r − (1 − τ ) τ g(t0 τ, x(t0 τ ))dτ  Γ (q1 ) 0 q −r  |t q1 −r − t0 1 | 1 ≤ (1 − τ )q1 −1 τ −r |g(tτ, x(tτ ))|dτ Γ (q1 ) 0 q −r  1 t 1 (1 − τ )q1 −1 τ −r |g(tτ, x(tτ )) − g(t0 τ, x(t0 τ ))|dτ. + 0 Γ (q1 ) 0 Together with continuity of functions g and t q1 −r , we could easily obtain T x(t) ∈ C[0, T1 ]. For x(t), y(t) ∈ C[0, T1 ], we obtain that  t 1 (t − s)q1 −1 | f (s, x(s)) − f (s, y(s))|ds |T x(t) − T y(t)| ≤ Γ (q1 ) 0  t L (t − s)q1 −1 s −r |x(s) − y(s)|ds ≤ Γ (q1 ) 0  t L (t − s)q1 −1 s −r x − yds ≤ Γ (q1 ) 0 LΓ (1 − r )t q1 −r x − y = Γ (1 − r + q1 ) q −r

≤ LT

q1 −r

Γ (1−r )

LΓ (1 − r )T1 1 x − y. Γ (1 − r + q1 )

According to Γ1(1−r +q1 ) < 1, the Banach contraction mapping principle assures that operator T has one unique fixed point x1 (t) ∈ C[0, T1 ]. Obviously, we could get x1 (0) = 0. So, x1 (t) is one unique solution of Eq. (3.4) with initial value condition x(0) = 0.

S. Zhang

Also, we have obtained that Eq. (3.3) in the interval (T1 , T2 ] can be written by (3.5). In order to consider the existence result of solution to (3.5), we may discuss the following equation defined on interval (0, T2 ]  t d (t − s)−q2 q2 x(t) = f (t, x), 0 < t ≤ T2 . (3.9) x(s)ds = D0+ dt 0 Γ (1 − q2 ) . It is clear that if function x ∈ C[0, T2 ] satisfies Eq. (3.9), then x(t) must satisfy Eq. (3.5). In fact, if x ∗ ∈ C[0, T2 ] with x ∗ (0) = 0 is a solution of Eq. (3.9) with initial value condition x(0) = 0, that is  t (t − s)−q2 ∗ d q2 ∗ x (s)ds = f (t, x ∗ ), 0 < t ≤ T2 ; x ∗ (0) = 0. x (t) = D0+ dt 0 Γ (1 − q2 ) Hence, from the above equality, it holds that  t (t − s)−q2 ∗ d x (s)ds = f (t, x ∗ ), T1 ≤ t ≤ T2 . dt 0 Γ (1 − q2 ) As a result, we have that x ∗ ∈ C[0, T2 ] with x ∗ (0) = 0 satisfies equation  t (t − s)−q2 d x(s)ds = f (t, x), T1 ≤ t ≤ T2 , dt 0 Γ (1 − q2 ) which means the function x ∗ ∈ C[0, T2 ] with x ∗ (0) = 0 is a solution of Eq. (3.5). Based on this fact, we will consider existence of solution to Eq. (3.9) with initial value condition x(0) = 0. q Now, applying operator I0+2 on both sides of (3.9), by Proposition 1.4 , we have that x(t) = ct q2 −1 +

1 Γ (q2 )



t

(t − s)q2 −1 f (s, x(s))ds, 0 ≤ t ≤ T2 .

0

By initial value condition x(0) = 0, we have c = 0. Define operator T : C[0, T2 ] → C[0, T2 ] by  t 1 T x(t) = (t − s)q2 −1 f (s, x(s))ds, 0 ≤ t ≤ T2 . Γ (q2 ) 0 From the previous arguments, it follows from the continuity of function t r f (t, x(t)) that operator T : C[0, T2 ] → C[0, T2 ] is well defined. For u(t), v(t) ∈ C[0, T2 ], we obtain that  t 1 (t − s)q2 −1 | f (s, u(s)) − f (s, v(s))|ds |T u(t) − T v(t)| ≤ Γ (q2 ) 0  t L (t − s)q2 −1 s −r |u(s) − v(s)|ds ≤ Γ (q2 ) 0 LΓ (1 − r )t q2 −r u − v ≤ Γ (1 + q2 − r ) q −r

≤

LΓ (1 − r )T2 2 u − v. Γ (1 + q2 − r )

The uniqueness result of solutions to initial...

By condition (H2 ), the Banach contraction mapping principle assures that operator T has one unique fixed point x2 ∈ C[0, T2 ], that is, 1 Γ (q2 )

x2 (t) =



t

(t − s)q2 −1 f (s, x2 (s))ds, 0 ≤ t ≤ T2 .

(3.10)

0 q

2 on both sides of (3.10), by From (3.10), we know x2 (0) = 0. Applying operator D0+ Proposition 1.2, we can obtain that

q

2 D0+ x2 (t) = f (t, x2 ), 0 < t ≤ T2 ,

that is, x2 (t) satisfies equation as following 1 d dt Γ (1 − q2 )



t

(t − s)−q2 x2 (s)ds = f (t, x2 ), 0 < t ≤ T2 ; x2 (0) = 0.

0

From the previous arguments, we obtain x2 ∈ C[0, T2 ] with x2 (0) = 0 satisfies Eq. (3.5). By the similar way, for i = 3, . . . , N ∗ , we could get that Eq. (3.3) defined on (Ti−1 , Ti ] has one unique solutions xi (t) ∈ C[0, Ti ] with xi (0) = 0 (TN ∗ = T ). As a result, we obtain that problem (1.1) has one unique solution. The proof is completed.

Theorem 3.2 Assume that conditions (H1 ) and (H3 ) hold, then problem (1.1) has one unique solution. Proof The proof is similar to Theorem 3.1. By (H1 ), we have obtained that equation of problem (1.1) can be written by (3.3). And, (3.3) in the interval [0, T1 ] can be written by q (3.4). Applying the operator I0+1 to both sides of (3.4), by Proposition 1.4, we have x(t) = ct q1 −1 +

1 Γ (q1 )



t

(t − s)q1 −1 f (s, x(s))ds, 0 ≤ t ≤ T1 .

0

By initial value condition x(0) = 0 and assumption of function f , we could get that c = 0. Define operator T : C[0, T1 ] → C[0, T1 ] by T x(t) =

1 Γ (q1 )



t

(t − s)q1 −1 f (s, x(s))ds, 0 ≤ t ≤ T1 .

0

From the previous arguments, it follows from the continuity of function t r f (t, x(t)) that operator T : C[0, T1 ] → C[0, T1 ] is well defined. Let Ω1 = {x ∈ C[0, T1 ] : x ≤ R1 } be a closed subset of C[0, T1 ], where R1 is a positive constant satisfying 

q −r

4 max0≤t≤T1 t r | f (t, 0)|Γ (1 − r )T1 1 Γ (1 + q1 − r )

< 2R1 <

Γ (1 + q1 − r )

q −r

2MΓ (1 − r )T1 1

Obviously, Ω1 is a Banach space with the metric in C[0, T1 ].



1 μ−1

.

S. Zhang

For x(t) ∈ Ω1 , by (H3 ), we have that  t 1 (t Γ (q1 ) 0  t 1 (t ≤ Γ (q1 ) 0  t M ≤ (t Γ (q1 ) 0  t M (t ≤ Γ (q1 ) 0

|T x(t)| ≤

− s)q1 −1 | f (s, x(s))|ds − s)q1 −1 (| f (s, x(s)) − f (s, 0)| + | f (s, 0)|) ds  t 1 q1 −1 −r μ − s) s |x(s)| ds + (t − s)q1 −1 s −r | f (s, 0)|ds Γ (q1 ) 0 max0≤t≤T1 t r | f (t, 0)|Γ (1 − r ) q1 −r t − s)q1 −1 s −r xμ ds + Γ (1 + q1 − r ) q −r

≤

MΓ (1 − r )T1 1 max0≤t≤T1 t r | f (t, 0)|Γ (1 − r ) q1 −r μ−1 R1 R1 + T1 Γ (1 + q1 − r ) Γ (1 + q1 − r )

=

MΓ (1 − r )T1 1 max0≤t≤T1 t r | f (t, 0)|Γ (1 − r ) q1 −r 21−μ R1 (2R1 )μ−1 + T1 Γ (1 + q1 − r ) Γ (1 + q1 − r )

q −r

R1 R1 + 2 2 R1 R1 ≤ + = R1 , 2 2 ≤ 21−μ

which implies T : Ω1 → Ω1 is well defined. For x(t), y(t) ∈ Ω1 , by (H3 ), we get  t 1 (t − s)q1 −1 | f (s, x(s)) − f (t, y(s))|ds Γ (q1 ) 0  t M (t − s)q1 −1 s −r |x(s) − y(s)|μ ds ≤ Γ (q1 ) 0  t M (t − s)q1 −1 s −r x − yμ ds ≤ Γ (q1 ) 0

|T x(t) − T y(t)| ≤

q −r

≤

MΓ (1 − r )T1 1 x − y · x − yμ−1 Γ (1 + q1 − r )

≤

MΓ (1 − r )T1 1 (x + y)μ−1 x − y Γ (1 + q1 − r )

q −r

q −r

MΓ (1 − r )T1 1 (2R1 )μ−1 x − y Γ (1 + q1 − r ) 1 < x − y. 2 ≤

Hence T is contraction operator. Therefore, the Banach contraction mapping principle assures that T has one unique fixed point x1 ∈ C[0, T1 ]. Obviously, we could get x1 (0) = 0. So, x1 (t) is unique solution of Eq. (3.4) with initial value condition x(0) = 0. Equation (3.3) in the interval (T1 , T2 ] can be written by (3.5) defined by d dt

 0

t

(t − s)−q2 x(s)ds = f (t, x), T1 < t ≤ T2 . Γ (1 − q2 )

The uniqueness result of solutions to initial...

In order to consider the existence result of solution to Eq. (3.5), we may consider the following equation defined on interval (0, T2 ] d dt

 0

t

(t − s)−q2 q2 x(t) = f (t, x(t)), 0 < t ≤ T2 . x(s)ds = D0+ Γ (1 − q2 )

(3.11)

By the same arguments as done in Theorem 3.1, we see that, if function x ∈ C[0, T2 ] satisfies Eq. (3.11), then x(t) must satisfy Eq. (3.5). So, we will consider the existence of solution to Eq. (3.11) with initial value condition x(0) = 0. q Now, applying operator I0+2 on both sides of (3.11), by Proposition 1.4 , we have x(t) = ct q2 −1 +

1 Γ (q2 )



t

(t − s)q2 −1 f (s, x(s))ds, 0 ≤ t ≤ T2 .

0

By x(0) = 0, we have c = 0. Define operator T : C[0, T2 ] → C[0, T2 ] by 1 T x(t) = Γ (q2 )



t

(t − s)q2 −1 f (s, x(s))ds, 0 ≤ t ≤ T2 .

0

Also, it follows from the continuity of function t r f (t, x) that operator T : C[0, T2 ] → C[0, T2 ] is well defined. Let Ω2 = {x ∈ C[0, T2 ] : x ≤ R2 } be a closed subset of C[0, T2 ], where R2 is a positive constant satisfying 

q −r

4 max0≤t≤T2 t r | f (t, 0)|Γ (1 − r )T2 2 Γ (1 + q2 − r )

< 2R2 <

Γ (1 + q2 − r )

q −r

2MΓ (1 − r )T2 2



1 μ−1

.

It is clear that Ω2 is a Banach space with the metric in C[0, T2 ]. For x(t) ∈ Ω2 , by (H3 ), we have  t 1 (t − s)q2 −1 | f (s, x(s))|ds |T x(t)| ≤ Γ (q2 ) 0  t  t 1 M (t − s)q2 −1 s −r |x(s)|μ ds + (t − s)q2 −1 s −r | f (s, 0)|ds ≤ Γ (q2 ) 0 Γ (q2 ) 0  t 4 max0≤t≤T2 t r | f (t, 0)|Γ (1 − r ) q2 −r M (t − s)q2 −1 s −r xμ ds + ≤ t Γ (q2 ) 0 Γ (1 + q2 − r ) q −r

MΓ (1 − r )T2 2 4 max0≤t≤T2 t r | f (t, 0)|Γ (1 − r ) q2 −r 21−μ R2 (2R2 )μ−1 + T2 Γ (1 + q2 − r ) Γ (1 + q2 − r ) R2 R2 ≤ + = R2 , 2 2 ≤

which implies that T : Ω2 → Ω2 is well defined.

S. Zhang

For x, y ∈ Ω2 , by (H3 ), we have  t 1 (t − s)q2 −1 | f (s, x(s)) − f (t, y(s))|ds Γ (q2 ) 0  t M (t − s)q2 −1 s −r |x(s) − y(s)|μ ds ≤ Γ (q2 ) 0  t M (t − s)q2 −1 s −r x − yμ ds ≤ Γ (q2 ) 0

|T x(t) − T y(y)| ≤

q −r

≤

MΓ (1 − r )T2 2 x − y · x − yμ−1 Γ (1 + q2 − r )

≤

MΓ (1 − r )T2 2 (x + y)μ−1 x − y Γ (1 + q2 − r )

q −r

q −r

MΓ (1 − r )T2 2 (2R2 )μ−1 x − y Γ (1 + q2 − r ) 1 < x − y. 2 ≤

Hence, T is contraction operator. Thus, the Banach contraction mapping principle assures that operator T has one unique fixed point x2 ∈ Ω2 , that is, x2 (t) =

1 Γ (q2 )



t

(t − s)q2 −1 f (s, x2 (s))ds, 0 ≤ t ≤ T2 .

(3.12)

0 q

2 on both sides of (3.12), by Proposition By (3.12), we get x2 (0) = 0. Applying operator D0+ 1.2, we obtain

q

2 D0+ x2 (t) = f (t, x2 ), 0 < t ≤ T2 ,

that is, x2 satisfies the following equation q

2 D0+ x(t) =

1 d dt Γ (1 − q2 )



t

(t − s)−q2 x2 (s)ds = f (t, x2 ), 0 < t ≤ T2 .

0

From the previous arguments, we have x2 ∈ Ω2 satisfies Eq. (3.5). By the similar way, we get that (3.3) defined on (Ti−1 , Ti ] has the solution xi ∈ Ωi = {x ∈ C[0, Ti ] : x ≤ Ri } with xi (0) = 0, where Ri is a positive constant satisfying 

qi −r

4 max0≤t≤Ti t r | f (t, 0)|Γ (1 − r )Ti Γ (1 + qi − r )

< 2Ri <

Γ (1 + qi − r )

qi −r

2MΓ (1 − r )Ti



1 μ−1

,

i = 3, . . . , N ∗ , (TN ∗ = T ). As a result, we obtain that problem (1.1) has one unique solution. Thus we complete the proof.



The uniqueness result of solutions to initial...

4 Example Example 5 Consider the following initial value problem for differential equation of variableorder  1 q(t) D0+ x(t) = t − 4 (2 + ax), 0 < t ≤ 4, (4.1) x(0) = 0, where

 0 < a < min

q : [0, 4] → R defined by

Γ ( 45 )

17 Γ ( 12 )

13 ) Γ ( 12

, , 1 Γ ( 43 ) 2 12 Γ ( 3 ) 4 125 Γ ( 3 ) 4 4

 ,

⎧1 ⎪ 2 , 0 ≤ t ≤ 1, ⎪ ⎪ ⎪ ⎨ q(t) = 13 , 1 < x ≤ 2, ⎪ ⎪ ⎪ ⎪ ⎩2 3 , 2 < t ≤ 4.

It is clear that q(t) is piecewise constant with respect to the partition {[0, 1], (1, 2], (2, 4]} of [0, 4]. 1 Let r = 41 , it is easy to see that t 4 f (t, x) = 2 + ax ∈ C([0, 4] × R, R) satisfying 1

t 4 | f (t, x) − f (t, y)| = a|x − y|, 0 ≤ t ≤ 4, x, y ∈ R. Also, aΓ ( 43 ) Γ ( 45 )

1

< 1,

a2 12 Γ ( 43 ) 13 Γ ( 12 )

5

< 1,

a4 12 Γ ( 43 ) 17 Γ ( 12 )

< 1,

Hence, Theorem 3.1 assures that (4.1) has one unique solution. Example 6 Consider the following initial value problem for differential equation of variableorder  1 q(t) D0+ x(t) = t − 3 (2 + ax 2 ), 0 < t ≤ 4, (4.2) x(0) = 0, where

 0 < a < min

q : [0, 4] → R defined by

Γ 2 ( 76 )

,

1

,

Γ 2 ( 43 )

16Γ 2 ( 23 ) 16Γ 2 ( 23 ) 4 3 Γ 2 ( 2 ) 3 8

 ,

⎧1 ⎪ 2 , 0 ≤ t ≤ 1, ⎪ ⎪ ⎪ ⎨ q(t) = 13 , 1 < x ≤ 2, ⎪ ⎪ ⎪ ⎪ ⎩2 3 , 2 < t ≤ 4.

Obviously, q(t) is piecewise constant with respect to the partition {[0, 1], (1, 2], (2, 4]} of [0, 4].

S. Zhang 1

Let r = 13 , it is easy to see that t 3 f (t, x) = 2 + ax 2 ∈ C([0, 4] × R, xR) satisfies 1

t 4 | f (t, x) − f (t, y)| = a|x − y|2 , 0 ≤ t ≤ 4, x, y ∈ R, max t r | f (t, 0)| = max t r | f (t, 0)| = max t r | f (t, 0)| = 2,

0≤t≤1

0≤t≤2

q −r

4 max0≤t≤T1 t r | f (t, 0)|Γ (1 − r )T1 1 Γ (1 + q1 − r )

= =

q −r

4 max0≤t≤T2 t r | f (t, 0)|Γ (1 − r )T2 2 Γ (1 + q2 − r )

q −r

 3 < Γ 76   Γ 76  , 2aΓ 23

0≤t≤4



Γ (1 + q1 − r )

1  μ−1

q −r

2MΓ (1 − r )T1 1

1  μ−1    Γ (1 + q2 − r ) 2 < = 8Γ q −r 3 2MΓ (1 − r )T2 2

1

 , 2aΓ 23   1 1   μ−1 8Γ 23 4 3 Γ (1 + q3 − r ) =   < q −r Γ 43 2MΓ (1 − r )T3 3 4 Γ = 5 3  , 4 6 aΓ 23 =

4 max0≤t≤T3 t r | f (t, 0)|Γ (1 − r )T3 3 Γ (1 + q3 − r )

8Γ

2

We can get that (H3 ) holds with M = a, μ = 2. Hence, Theorem 3.2 assures that problem (4.2) has one unique solution.

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