Smoothness and stability of the solutions for nonlinear fractional differential equations

Nonlinear Analysis 72 (2010) 1768–1777

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Nonlinear Analysis journal homepage: www.elsevier.com/locate/na

Smoothness and stability of the solutions for nonlinear fractional differential equations Weihua Deng School of Mathematics and Statistics, Lanzhou University, Lanzhou 730000, PR China

article

info

Article history: Received 12 July 2008 Accepted 1 September 2009 MSC: 26A33 34A34 58B10 58K25

abstract This paper first obtains the differentiability properties of the solutions for nonlinear fractional differential equations, and then the sufficient conditions for the local asymptotical stability of nonlinear fractional differential equations are also derived. © 2009 Elsevier Ltd. All rights reserved.

Keywords: Fractional differential equation Differentiability Stability

1. Introduction Fractional calculus is an area having a long history whose infancy dates back to three hundred years, the beginnings of classical calculus. It had always attracted the interest of many famous ancient mathematicians, including L’Hospital, Leibniz, Liouville, Riemann, Grünward, and Letnikov [1,2]. As the ancient mathematicians expected, in recent decades, fractional differential equations have been found to be a powerful tool in more and more fields, such as materials, physics, mechanics, and engineering [3–9,1,2,10]. Fractional differential operators are one kind of pseudo-differential operators. Since they are nonlocal and have weakly singular kernels, the study of fractional differential equations seems to be more difficult and less theories have been established than for classical differential equations. First, this paper studies the smoothness properties of the solutions for nonlinear fractional differential equations (FDEs). Not only the analytical properties of FDEs, but also, more important, the knowledge of the smoothness properties is indispensable for the construction of good numerical schemes [11–15]. This proves that, in general, the differentiability properties of the solutions for FDEs at initial points are different from other points. Thanks to these results, it is surprisingly found that it is hard for autonomous/nonautonomous FDEs to have periodic solutions besides fixed points, which are quite distinct from classical ODEs. On the other hand, it appears to be natural if one understands this phenomena from a physical point of view; the possible periodicity of the solutions for FDEs are destroyed by the memory effects of fractional differential operators [4]. It is well known that stability issue is a key topic for application sciences. The necessary and sufficient stability conditions for linear FDEs and linear time-delayed FDEs have already been obtained in [16,17,8]. To the best of our knowledge, the stability of nonlinear FDEs is still ‘‘open’’, although this is a very hot and urgent topic for engineers, physicists, controllers, and pure mathematicians. Many of the scientists are expecting the fractional version of the classical Hartman–Grobman theorem for hyperbolic dynamical systems of order 1 [5]. This paper provides the sufficient stability conditions for nonlinear FDEs.

E-mail addresses: [email protected], [email protected]. 0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.09.018

W. Deng / Nonlinear Analysis 72 (2010) 1768–1777

1769

The definition of fractional integral is described by −β

a Dt

x(t ) =

t

Z

1

Γ (β)

(t − τ )β−1 x(τ )dτ ,

β > 0.

a

There are three kinds of widely used fractional derivatives, namely the Grünwald–Letnikov derivative, the Riemann–Liouville derivative, and the Caputo derivative. The Grünwald–Letnikov derivative and the Riemann–Liouville derivative are equivalent if the functions they act on are sufficiently smooth, and the Riemann–Liouville derivative is meaningful under weaker smoothness requirements. So just the definitions of the Riemann–Liouville derivative and the Caputo derivative are introduced as follows: the Riemann–Liouville derivative q a Dt x

(t ) = Dm a Dqt −m x(t ),

q ∈ [m − 1, m),

and the Caputo derivative

(t ) = a Dqt −m Dm x(t ), q ∈ (m − 1, m), where m ∈ Z+ , Dm is the classical m-order derivative. C q a Dt x

For the Riemann–Liouville derivative, we have lim

q→(m−1)+

q a Dt x

(t ) =

and lim

q→m−

q a Dt x

(t ) =

dm−1 x(t ) dt m−1

dm x ( t ) dt m

.

But for the Caputo derivative, we have C q a Dt x q→(m−1)+

(t ) =

lim

and C q a Dt x q→m−

lim

(t ) =

dm−1 x(t ) dt m

dm x ( t ) dt m

− D(m−1) x(a)

.

q

q

Obviously, a Dt (q ∈ (−∞, +∞)) varies continuously with q, i.e., a Dt (q ∈ (−∞, +∞)) bridges all the gaps among the integer derivatives and the integer integrals, but the Caputo derivative cannot do this [18]. However, the Caputo derivative is extensively used in real applications because the initial conditions of FDEs with Caputo derivative have a clear physical meaning [6,13]. This introduction is closed by outlining the rest of the paper. In the next section, some preliminary lemmas are first provided and then the smoothness of the solutions for nonlinear FDEs with the Caputo derivative and with the Riemann–Liouville derivative are discussed. The theorem on sufficient stability conditions of nonlinear FDEs with the Caputo derivative is proved in Section 3. 2. Smoothness of nonlinear FDEs The FDE with the Caputo derivative is given as q a Dt x

(C

(t ) = f (t , x(t )),

Dk x(a) = xka ,

m − 1 < q < m ∈ Z +, k = 0, 1, . . . , m − 1,

(1)

and the FDE with Riemann–Liouville derivative is provided by

 q a Dt x(t ) = f (t , x(t )), i h  a Dtq−k x(t ) = xka , t =a

m − 1 < q < m ∈ Z +, k = 1, 2, . . . , m.

(2)

2.1. Preliminary lemmas The FDEs (1) and (2) can be converted to their equivalent Volterra integral equations of the second kind under some natural conditions. Lemma 2.1 ([19]). If the function f (t , x) is continuous, then the initial value problem (1) is equivalent to the following nonlinear Volterra integral equation of the second kind, m−1

x(t ) =

X xk

a

k! k=0

(t − a)k +

and its solutions are continuous.

1

Γ (q)

t

Z

(t − τ )q−1 f (τ , x(τ ))dτ , a

(3)

1770

W. Deng / Nonlinear Analysis 72 (2010) 1768–1777

Lemma 2.2. If the function f (t , x) and all the solutions of (2) are continuous, then the initial value problem (2) is equivalent to the following nonlinear Volterra integral equation of the second kind, m−1

x(t ) =

X

xka

k=1

Γ (q − k + 1)

(t − a)q−k +

t

Z

1

Γ (q)

(t − τ )q−1 f (τ , x(τ ))dτ .

(4)

a

−q

Proof. ‘‘Necessity’’. Since both f (t , x) and x(t ) are continuous, letting the operator a Dt of (2), we have x(t ) −

m h X

q −k x a Dt

(t )

i t =a

k =1

1 (t − a)q−k = Γ (q − k + 1) Γ (q)

Z

act on both sides of the first equation

t

(t − τ )q−1 f (τ , x(τ ))dτ ,

a

namely, x(t ) =

m X

xka

t

Z

1

(t − τ )q−1 f (τ , x(τ ))dτ . Γ ( q − k + 1 ) Γ ( q ) a k=1  q −m  It can be verified that xm x(t ) t =a = 0 because of the continuity of x(t ), in fact, taking M = maxt ∈[a,T ] |x(t )| (T > a), a = a Dt (t − a)q−k +

then

xm a

|

q −m x a Dt

| 6 lim | t →a

6 lim

t →a

Z t 1 q −m (t )| = lim (t − τ ) x(τ )dτ t →a Γ ( q − m + 1 ) a

M

Γ (q − m + 2)

[−(t − τ )

q−m+1

t a

]| = lim

t →a

M

Γ (q − m + 2)

(t − a)q−m+1 = 0.

So, m−1

x(t ) =

X

xka

k=1

Γ (q − k + 1)

(t − a)q−k +

1

Γ (q)

t

Z

(t − τ )q−1 f (τ , x(τ ))dτ . a

‘‘Sufficiency’’. Assume m−1

Fm [x](t ) :=

X

xka

k=1

Γ (q − k + 1)

(t − a)q−k ,

obviously, Fm [x](t ) is continuous, and suppose

χ1 := (α Γ (q + 1)/kf k∞ )1/q . Introduce the set U := {x ∈ C [a, a + χ1 ] : kx − Fm [x](t )k∞ 6 α}, clearly, U is a closed and convex subset of the Banach space of all continuous functions on [a, a + χ1 ], equipped with the Chebyshev norm. On U define the operator Y by

(Yx)(t ) := Fm [x](t ) +

t

Z

1

Γ (q)

(t − τ )q−1 f (τ , x(τ ))dτ . a

Using this operator, (4) can be rewritten as x = Yx, and now it just needs to show that Y has fixed points on U. Therefore first investigate the properties of the operator Y . Taking a 6 a + t1 6 a + t2 6 a + χ1 , 1

Z

a + t2

|(Yx)(a + t2 ) − (Yx)(a + t1 )| = Fm [x](a + t2 ) − Fm [x](a + t1 ) + (a + t2 − τ )q−1 f (τ , x(τ ))dτ Γ (q) a Z a+t1 1 − (a + t1 − τ )q−1 f (τ , x(τ ))dτ Γ (q) a m −1

6

X k =1

|xka | kf k∞ ((t2 )q−k − (t1 )q−k ) + (2(t2 − t1 )q + (t2 )q − (t1 )q ), (5) Γ (q − k + 1) Γ (q + 1)

proving that Yx is a continuous function. Moreover, for x ∈ U and t ∈ [a, a + χ1 ], it can be found that

|(Yx)(t ) − Fm [x](t )| =

1

Γ (q)

t

Z

(t − τ )q−1 f (τ , x(τ ))dτ 6 a

1

Γ (q + 1)

kf k∞ (t − a)q

W. Deng / Nonlinear Analysis 72 (2010) 1768–1777

6

1

Γ (q + 1)

kf k∞ (χ1 )q 6

1

Γ (q + 1)

kf k∞ ·

1771

α Γ (q + 1) = α. kf k∞

Thus, if x ∈ U, then Yx ∈ U, i.e., Y maps the set U to itself. Next, prove that Y is a continuous operator. Since f is continuous, it is uniformly continuous on the compact set [a, a + χ1 ] × U. Consequently, given an arbitrary  > 0, there can be found a δ > 0 such that

|f (t , x) − f (t , y)| <

ε Γ (q + 1) whenever kx − yk∞ < δ. (χ1 )q

(6)

Now, let x, x˜ ∈ U such that kx − x˜ k∞ < δ . Then, in view of (6),

|f (t , x(t )) − f (t , x˜ (t ))| <

ε Γ (q + 1) for all t ∈ [a, a + χ1 ]. (χ1 )q

Hence,

Z t Z ε Γ (q + 1) t q−1 |(Yx)(t ) − (Y x˜ )(t )| = (t − τ )q−1 dτ (t − τ ) (f (τ , x(τ )) − f (τ , x˜ (τ )))dτ 6 Γ (q) a (χ1 )q Γ (q) a ε(t − a)q = ≤ ε, (χ1 )q 1

proving the continuity of the operator Y . Then, look at the set of functions Y (U ) := {Yx : x ∈ U }. For y ∈ Y (U ), it can be found that, for all t ∈ [a, a + χ1 ],

|y(t )| = |(Yx)(t )| 6 |Fm [x](t )| + 6 |Fm [x](t )| +

1

Γ (q)

t

Z

1

Γ (q)

(t − τ )q−1 · |f (τ , x(τ ))|dτ a

kf k∞ (χ1 )q ,

which means that Y (U ) is bounded in a pointwise sense. Noting that the expression on the right-hand side of (5) is independent of x, it can be seen that the set Y (U ) is equicontinuous. Then the Arzelà–Ascoli theorem yields that every sequence of functions from Y (U ) has got a uniformly convergent subsequence, and therefore Y (U ) is relatively compact. So, the Schauder’s fixed point theorem [20] asserts that Y has fixed points, and these fixed points are the continuous solutions of (4) in the interval [a, a + χ1 ]. For t ∈ [a + χ1 , ∞), (4) can be recast as x(t ) = Fm [x](t ) +

= Fm [x](t ) + =e Rm [x](t ) +

1

Γ (q) 1

Γ (q) 1

Γ (q)

t

Z

(t − τ )q−1 f (τ , x(τ ))dτ a a+χ1

Z

(t − τ )

q −1

f (τ , x(τ ))dτ +

a t −χ1

Z

1

Γ (q)

t −χ1

Z

(t − χ1 − τ )q−1 f (χ1 + τ , x(χ1 + τ ))dτ

a

(t − χ1 − τ )q−1 f (χ1 + τ , x(χ1 + τ ))dτ ,

a

where e Rm [x](t ) is continuous, using the same analysis to above, it can be proved there exists χ2 > 0 such that the solutions of (4) are continuous in the interval [a + χ1 , a + χ1 + χ2 ]. By induction, the interval [a, a + χ1 ] can be extended to [a, ∞), so all the solutions of (4) are continuous on the interval [a, ∞). q Now, letting the operator a Dt act on both sides of (4), it can be obtained (2), this is because q a Dt

1

Γ (q) m −1 X q

a Dt

k=1

Z

t

q (t − τ )q−1 f (τ , x(τ ))dτ = a Dqt a D− t f (t , x(t )) = f (t , x(t )), a ! m −1 X
xka xka q q−k q −k ( t − a) = a D t ( t − a) Γ (q − k + 1) Γ (q − k + 1) k=1 m −1 m −1 k X X xka Γ (q − k + 1) xa = (t − a)−k = (t − a)−k = 0, Γ ( q − k + 1 ) Γ ( 1 − k ) ∞ k=1 k=1

and m−1

lim

t →a

q−j a Dt

X

xka

k=1

Γ (q − k + 1)

! ( t − a)

q−k

m−1

= lim

t →a

X

xka

k=1

Γ (q − k + 1)

m−1

= lim

t →a

X

xka

k=1

Γ (j − k + 1)



q −j a Dt

(t − a)q−k

(t − a)j−k



1772

W. Deng / Nonlinear Analysis 72 (2010) 1768–1777

= lim

xka

k=1

Γ (j − k + 1)

t →a

m −1

+ lim

t →a

=

!

j−1 X

( t − a)

xka

k=j+1

Γ (j − k + 1) !

xka

k=1

Γ (j − k + 1)

( t − a)

j−k

· 0 + xja

m −1

X xk a + lim (t − a)j−k t →a ∞ k=j+1 The proof is complete.

+ xja !

X

j −1 X

j−k

! = xja ,

j = 1, . . . , m − 1.



In what follows, we first consider a linear integral equation of the form t

Z

1

X (t ) = F (t ) +

Γ (q)

(t − τ )q−1 h(τ )X (τ )dτ .

(7)

a

Lemma 2.3. For some T > a assume F ∈ C (a, T ] ∩ L1 (a, T ) and h ∈ L∞ (a, T ). Then (7) has a unique solution X ∈ C (a, T ] ∩ L1 (a, T ). Proof. First, we prove (7) has a unique solution X ∈ L1 (a, T ). Let h0 be the essential supremum of |h(t )| on t ∈ [a, T ]. Pick S=(

Γ (q+1) 2h0

1

) q , then Z a+S

h0

Γ (q + 1)

(t − τ )q−1 dτ =

a

1 2

< 1.

The existence of a unique X (t ) ∈ L1 (a, s) on a 6 t 6 a + S follows immediately by the principle of contraction mappings on L1 (a, a + S ). Replacing t by t + S in (7), we have X1 ( t ) = F 1 ( t ) +

t

Z

1

Γ (q)

(t − τ )q−1 h(τ + S )X1 (τ )dτ , a

where X1 (t ) = X (t + S ) and F1 (t ) = F (t + S ) +

R a+S

(t + S − τ )q−1 h(τ )X (τ )dτ . Thanks to X , F1 ∈ L1 (a, a + S ), and the contraction mapping principle, it can be shown X1 (t ) ∈ L1 (a, a + S ). Continue by induction, then X ∈ L1 (a, T ) is proved. Secondly, confirm X ∈ C (a, T ]. Since X ∈ L1 (a, T ) and h ∈ L∞ (a, T ), then the function (a + δ − τ )q−1 h(τ )X (τ ) ∈ L1 (a, a + δ) for almost any δ ∈ (0, T − a). Choose such δ , then   Z a+δ Z t 1 1 q −1 X (t + δ) = F (t + δ) + (t + δ − τ ) h(τ )X (τ )dτ + (t − τ )q−1 h(τ + δ)X (τ + δ)dτ . (8) Γ (q) a Γ (q) a 1

Γ (q)

a

Note that the function F (t + δ) is continuous on [0, T − δ]. The function E, defined by the relation E (t ) =

Z

1

Γ (q)

a+δ

(t + δ − τ )q−1 h(τ )X (τ )dτ ,

a

is easily seen to be continuous on a < t 6 T . To see that E (t ) is continuous at t = a, it must be shown that for any sequence tn tending monotonically to a one has 1

Γ (q)

a+δ

Z

(tn + δ − τ )q−1 h(τ )X (τ )dτ → a

1

Γ (q)

a+δ

Z

(a + δ − τ )q−1 h(τ )X (τ )dτ . a

Since (tn + δ − τ )q−1 h(τ )X (τ ) → (a + δ − τ )q−1 h(τ )X (τ ) monotonically. Using the dominated convergence theorem, it can be shown that E (t ) ∈ C [a, T ]. So, one can recast (8) as y(t ) = e F (t ) +

1

Γ (q)

t

Z

(t − τ )q−1e h(τ )y(τ )dτ , a

where y(t ) = X (t + δ) and e F (t ) ∈ C [a, T − δ]. Applying an argument similar to the proof of Lemma 2.2, it follows that y(t ) = x(t + δ) ∈ C [a, T − δ]. Since δ > 0 can be arbitrarily small, the confirmation is finished. 

W. Deng / Nonlinear Analysis 72 (2010) 1768–1777

1773

Lemma 2.4 ([21]). The solution of a linear integral equation X (t ) = g (t ) +

t

Z

1

Γ (q)

(t − τ )q−1 X (τ )dτ , a

can be represented as X (t ) = g (t ) +

t

Z

R(t + a − τ )g (τ )dτ , a

where R(t ) (> 0), the resolvent associated with a given kernel function t q−1 /Γ (q), is defined by the unique L1 solution of the linear integral equation R(t ) =

1

Γ (q)

t

Z

1

(t − a)q−1 +

Γ (q)

(t − τ )q−1 R(τ )dτ . a

Lemma 2.5. For a pair of nonlinear integral equations Xj (t ) = Gj (t ) +

t

Z

1

Γ (q)

(t − τ )q−1 fj (τ , Xj (τ ))dτ ,

j = 1, 2,

a

assume: (i) G1 and G2 ∈ L1 (a, T ),(ii) f1 (t , x) and f2 (t , x) are continuous in (t , x) for a 6 t 6 T and all x,(iii) f1 (t , x) is Lipschitz continuous in x with Lipschitz constant L (independent of t and x), and (iv) X1 and X2 exist on 0 6 t 6 T and are L1 . Let r (t ) be the resolvent of the kernel (Lt q−1 )/Γ (q) and define Q (t ) = G1 (t ) − G2 (t ) +

t

Z

1

Γ (q)

(t − τ )q−1 {f1 (τ , X2 (τ )) − f2 (τ , X2 (τ ))}dτ . a

Then on a < t < T , one has

|X1 (t ) − X2 (t )| 6 |Q (t )| +

t

Z

r (t + a − τ )|Q (τ )|dτ . a

Proof. Define z (t ) = X1 (t ) − X2 (t ), G(t ) = G1 (t ) − G2 (t ), and F (t ) = {f1 (t , X1 (t )) − f1 (t , X2 (t ))}/z (t ) when z (t ) 6= 0 and F (t ) = 0 when z (t ) = 0. Obviously z and G ∈ L1 (a, T ), F ∈ L∞ (a, T ), and |F (t )| 6 L. According to the definition above it follows that

Z

1

t

(t − τ )q−1 {f1 (τ , X2 (τ )) − f2 (τ , X2 (τ ))}dτ Γ (q) a Z t 1 + (t − τ )q−1 {f1 (τ , X1 (τ )) − f1 (τ , X2 (τ ))}dτ Γ (q) a Z t 1 = Q (t ) + (t − τ )q−1 F (τ )z (τ )dτ , Γ (q) a

z (t ) = G(t ) +

so that

|z (t )| 6 |Q (t )| +

L

t

Z

Γ (q)

(t − τ )q−1 · |z (τ )|dτ . a

Let p(t ) be a nonnegative function such that

|z (t )| = |Q (t )| − p(t ) +

t

Z

L

Γ (q)

(t − τ )q−1 · |z (τ )|dτ . a

Since r (t ) is the resolvent of (Lt q−1 )/Γ (q), then it implies that

|z (t )| = |Q (t )| − p(t ) +

t

Z

r (t + a − τ ){|Q (τ )| − p(τ )}dτ . a

The lemma follows because r and p are nonnegative.



1774

W. Deng / Nonlinear Analysis 72 (2010) 1768–1777

Now we consider the nonlinear integral equation x(t ) = g (t ) +

t

Z

1

Γ (q)

(t − τ )q−1 f (τ , x(τ ))dτ ,

(9)

a

which can be expressed as x(t ) = g (t ) +

t

Z

1

Γ (q)

(τ − a)q−1 f (t + a − τ , x(t + a − τ ))dτ , a

and its formal derivative x (t ) = g (t ) + 0

0

f (a, x(a))

( t − a)

Γ (q) ∂ f (t ,x)

q−1

+

t

Z

1

Γ (q)

(t − τ )q−1 {f1 (τ , x(τ )) + f2 (τ , x(τ ))x0 (τ )}dτ ,

(10)

a

∂ f (t ,x)

where we denote f1 = ∂ t and f2 = ∂ x . The last equation may be written in the form X (t ) = G(t ) +

t

Z

1

Γ (q)

(t − τ )q−1 f2 (τ , x(τ ))X (τ )dτ ,

(11)

a

where X (t ) = x0 (t ) and f (a, x(a)) 1 G(t ) = g 0 (t ) + (t − a)q−1 + Γ (q) Γ (q)

t

Z

(t − τ )q−1 f1 (τ , x(τ ))dτ . a

Lemma 2.6. Suppose f (t , x) in (9) is locally Lipschitz continuous with respect to x and g (t ) ∈ W 1,∞ (a, T ), let X (t ) be the solution of (11), then the solution x(t ) of (9) is of class C [a, T ] ∩ C 1 (a, T ] and x0 (t ) = X (t ) on the interval a < t 6 T . Proof. Applying the principle of contraction mappings and the similar idea in the proof of Lemma 2.2, it can be proved that (9) has a unique continuous solution x(t ) on the interval a 6 t 6 T , here is omitted. By Lemma 2.3 it follows that X ∈ C (a, T ] ∩ L1 (a, T ). Let M = max |x(t )| on t ∈ [a, T ] and P (x) be a C ∞ -function such that P (x) = 1 if |x| 6 M + 1 and P (x) = 0 if |x| > M + 2. If the function f (t , x) in (9) is replaced by f (t , x)P (x), then nothing is changed in the range of interest. Therefore it can be assumed that f has compact support and in particular f , f1 , and f2 are bounded and f2 (t , x) is globally Lipschitz continuous in x. Fix a number δ in the range 0 < δ < (T − a)/2. Define Z (t , h) = {x(t + h) − x(t )}/h for 0 < h 6 δ and a < t 6 T − δ . Since x(t ) satisfies (9), then Z satisfies an equation of the form Z (t , h) = R(t , h) +

t

Z

1

Γ (q)

(t − τ )q−1 f2 (τ , x∗ (τ ))Z (τ , h)dτ , a

where x∗ (τ ) is between x(τ ) and x(τ + h), θ (h) ∈ (0, h), and R(t , h) = (g (t + h) − g (t ))/h +

+

Γ (q)

t +h

Z

Γ (q)

(τ − a)q−1 f (t + h + a − τ , x(t + h + a − τ ))dτ t

t

Z

1

h−1

(τ − a)q−1 f1 (t + θ (h) + a − τ , x(t + h + a − τ ))dτ . a

Let r (t ) be the resolvent of (Lt q−1 )/Γ (q). By Lemma 2.5 above,

|Z (t , h) − X (t )| 6 Q (t , h) +

t

Z

r (t + a − τ )Q (τ , h)dτ a

on a < t 6 T − δ , where Q (t , h) = |R(t , h) − G(t )| +

1

Γ (q)

t

Z

(t − τ )q−1 |f2 (τ , x∗ (τ )) − f2 (τ , x(τ ))|dτ . a

Let K > 0 be a bound for all of the functions |g 0 (t )|, |f (t , x)|, f1 (t , x), and f2 (t , x). Then, when a < τ < t, the definitions of Q , R, and G may be used to obtain the bound

|Q (τ , h)| 6 K +

Γ (q)h

6 2K +

K

Γ (q)

 6 2K

1+

τ +h

Z

K

τ

( u − a)

q −1

du +

3K

Γ (q)

T

Z

(u − a)q−1 du + |G(τ )| a

max{(τ − a)q−1 , (τ + δ − a)q−1 } + 2

Γ (q + 1)

( T − a) q +

1

Γ (q)

4K

Z

T

K

(u − a)q−1 du + (τ − a)q−1 Γ ( q ) a  q−1 max{(τ − a) , (τ + δ − a)q−1 } . Γ (q)

W. Deng / Nonlinear Analysis 72 (2010) 1768–1777

1775

Denote this bound in the form |Q (t , h)| 6 K0 + K1 β(t ), where

β(t ) = max{(t − a)q−1 , (t + δ − a)q−1 }. Given ε > 0 let K2 be a bound for r (t ) over a + δ 6 t 6 T − δ . Pick η in the range 0 < η 6 δ and so small that a+η

Z

K2 {K0 + K1 β(t )}dt < ε. a

Now pick h0 small enough such that whenever 0 < h 6 h0 , we have

|Q (t , h)| 6 ε

T

Z

r (τ )dτ + 1

−1

a

uniformly over a + η 6 t 6 T − η. Then for h and t in the range 0 < h 6 h0 , a + δ 6 t 6 T − δ one has

|Z (t , h) − X (t )| 6 Q (t , h) +

a+η

Z

r (t + a − τ )Q (τ , h)dτ +

Z

Z

r (t + a − τ )Q (τ , h)dτ a+η

a

6 ε+

t

a+η

K2 {K0 + K1 β(τ )}dτ +

Z

t

 Z

r (t + a − τ ) ε a+η

a

T



r (u)du dτ

a

6 3ε.

Since ε > 0 is arbitrary this shows that Z (t , h) → X (t ) as h → 0+ uniformly in t ∈ [a + δ, T − δ]. But δ > 0 is also arbitrary so that X (t ) is the continuous right derivative of x(t ) on the interval a < t < T . By virtue of the uniform convergence to X (t ), it follows that on any interval I = {t : a + δ 6 t 6 T − δ} the set {Z (·, h) : 0 < h < δ} is equicontinuous. Hence, lim Z (t , h) = lim Z (t − h, h) = X (t )

h→0+

h→0+

uniformly on I. Then X (t ) is also the continuous left derivative of x(t ) on the interval a < t < T . For t = T a similar argument shows that X (T ) is the left derivative of x(t ) at t = T . Now, this completes the proof.  Remark 2.7. The results of Lemma 2.6 can be extended by weakening the hypothesis to g ∈ C [a, T ] ∩ C 1 (a, T ] and a+η

Z

(g (t + h) − g (t ))/h − g 0 (t ) dt → 0 a

as η → 0 uniformly in h. The proof is similar to Lemma 2.6. 2.2. Smoothness of FDEs Based on the lemmas of above subsection, the smoothness properties of the solutions for FDEs are presented. Theorem 2.8. Assume f (t , x) is sufficiently smooth with (t , x) respectively for t on the interval [a, T ] and for all x. Then the solution of (1) satisfies the following: ∗ ∗ (A) if {Dk f (u, x(u))}|u=a = 0, k = 0, 1, . . . , k∗ − 1, and {Dk f (u, x(u))}|u=a 6= 0, then (i) x(t ) ∈ C m+k −1 [a, T ] ∩ ∗ ∗ ∗ C m+k (a, T ],(ii) x(m+k ) (t ) ∈ L1 (a, T ), x(m+k ) (t ) = O((t − a)(q−m) ) as t → a+ , and (iii) x(n) (t ) =

m−1

X

xka

n−k∗ −1

(t − a)k−n +

X

1

(t − a)q−k−1 {Dn−k−1 f (u, x(u))}|u=a ( k − n )! Γ ( q − k ) k=n k=0 Z t 1 + (τ − a)q−1 {Dn f (u, x(u))}|u=t −τ +a dτ Γ (q) a

for 1 6 n 6 m + k∗ − 1, a 6 t 6 T , and n = m + k∗ , a < t 6 T .

(B) if {Dk f (u, x(u))}|u=a = 0, k = 0, 1, . . . , +∞, or f (t ,

Pm−1 k=0

xka k!

(12)

(t − a)k ) ≡ 0 for t ∈ [a, T ], then

m−1

x(t ) =

X xk

a

k! k=0

(t − a)k .

Proof. If m + k∗ = 1, namely, m = 1 and k∗ = 0, then according to Lemma 2.6, x0 (t ) ∈ C (a, T ] ∩ L1 (a, T ) and x(t ) satisfies (12) on a < t 6 T .

1776

W. Deng / Nonlinear Analysis 72 (2010) 1768–1777

If m + k∗ ∈ (1, ∞), then the hypotheses of Lemma 2.6 are still trivially satisfied. So x0 (t ) ∈ C (a, T ] ∩ L1 (a, T ) and x(t ) satisfies (12) on 0 < t 6 T for n = 1, but it can be easily verified that x0 (a) exists and satisfies (12). Continue by induction, (12) can be established for n = 1, 2, . . . , m + k∗ − 1 by using Lemma 2.6. Applying Remark 2.7 to (12) with n = m + k∗ − 1, ∗ it can be obtained (12) for n = m + k∗ and x(m+k ) (t ) = O((t − a)(q−m) ). For the last case, if {Dk f (u, x(u))}|u=a = 0, k = 0, 1, . . . , +∞, then f (t , x(t )) is C ∞ with t on [a, T ] thanks to above analysis. It can be concluded that f (t , x(t )) ≡ 0 on [a, T ], and then x(t ) =

Pm−1 k=0

xka k!

(t − a)k because of Lemma 2.1.



Exactly the same proof, the following theorem can be achieved for (2). Theorem 2.9. Assume f (t , x) is sufficiently smooth with (t , x) respectively for t on the interval [a, T ] and for all x, and the solution of (2) is continuous. Then the solution of (2) satisfies the following: ∗ (A) if all the initial values of (2) are zero and {Dk f (u, x(u))}|u=a = 0, k = 0, 1, . . . , k∗ − 1, and {Dk f (u, x(u))}|u=a 6= 0, ∗ ∗ ∗ ∗ then (i) x(t ) ∈ C m+k −1 [a, T ] ∩ C m+k (a, T ],(ii) x(m+k ) (t ) ∈ L1 (a, T ), x(m+k ) (t ) = O((t − a)(q−m) ) as t → a+ , and (iii) x

(n)

n−k∗ −1

(t ) =

X

1

k=0

Γ (q − k )

(t − a)

q−k−1

{D

n−k−1

f (u, x(u))}|u=a +

1

Γ (q)

t

Z

(τ − a)q−1 {Dn f (u, x(u))}|u=t −τ +a dτ

(13)

a

for 1 6 n 6 m + k∗ − 1, a 6 t 6 T , and n = m + k∗ , a < t 6 T . (B) if all the initial values of (2) are zero, and {Dk f (u, x(u))}|u=a = 0, k = 0, 1, . . . , +∞ or f (t , 0) ≡ 0 for t ∈ [a, T ], then x(t ) ≡ 0 for t ∈ [a, T ]. (C) if not all of the initial values xka (k = 1, . . . , m − 1) of (2) are zero, denote the smallest k such that xka 6= 0 by l, then (i) x(t ) ∈ C m−l−1 [a, T ] ∩ C m−l (a, T ],(ii) x(m−l) (t ) ∈ L1 (a, T ), x(m−l) (t ) = O((t − a)(q−m) ) as t → a+ , and (iii) x(n) (t ) =

l X

xka

(t − a)q−k−n +

n −1 X

1

Γ (q − k − n + 1) Γ (q − k) k=1 k=0 Z t 1 (τ − a)q−1 {Dn f (u, x(u))}|u=t −τ +a dτ + Γ (q) a

(t − a)q−k−1 {Dn−k−1 f (u, x(u))}|u=a (14)

for 1 6 n 6 m − l − 1, a 6 t 6 T , and n = m − l, a < t 6 T . Remark 2.10. From Theorems 2.8 and 2.9, we note that the smoothness of the solutions for (1) and (2) at the initial point is different from other points, so it is very hard for them to have periodic solutions besides fixed points. 3. Stability of nonlinear FDEs with the Caputo derivative The sufficient conditions for the local asymptotical stability of nonlinear FDE (1) are discussed in this section. Definition 3.1. The equilibrium x∗ = 0 of nonlinear FDE (1) is said to be locally asymptotically stable if ∃δ > 0 such that ∀xa ∈ B, one has lim kx(t ; xa )k = 0,

t →+∞

where B = {y : kyk < δ} and x(t ; xa ) denotes the solution of (1) with initial value x(a) = xa . Theorem 3.2. The equilibrium x∗ = 0 of autonomous nonlinear FDE (1) with f 0 (x) ∈ C [a, +∞) and q ∈ (0, 1) is locally asymptotically stable if λ = f 0 (0) < 0. Proof. Obviously, there exists δ, λ1 , λ2 such that if |x(t )| < δ then

λ2 < f 0 (x(t )) < λ1 < 0. Note the equivalent Eq. (3) of nonlinear FDE (1) is x(t ) = xa +

t

Z

1

Γ (q)

(t − τ )q−1 f (x(τ ))dτ . a

Pick S such that 1

Γ (q)

a+S

Z

(t − τ )q−1 · (−λ2 )dτ < 1, a

and define the operator Y (x(t )) = xa +

1

Γ (q)

t

Z

(t − τ )q−1 f (x(τ ))dτ , a

with a 6 t 6 a + S .

W. Deng / Nonlinear Analysis 72 (2010) 1768–1777

1777

Clearly, Y is a continuous operator. Assume xa > 0. By using the principle of contraction mappings defined on {φ(t ) ∈ C [a, a + S ] : 0 < φ(t ) 6 xa }, it follows that x(t ) ∈ C [a, a + S ] and 0 < x(t ) 6 xa when t ∈ [a, a + S ]. Replacing t by t + S in the above equation one obtains x˜ (t ) = x˜ a (t ) +

1

Γ (q)

t

Z

(t − τ )q−1 f (x(τ + S ))dτ , a

R a+S

(t + S − τ )q−1 f (x(τ ))dτ . It can be easily seen x˜ a (t ) ∈ C [a, a + S ] and 0 < x˜ a (t ) < xa , then the contraction mappings argument yields x˜ (t ) ∈ C [a, a + S ] and 0 < x˜ (t ) 6 x˜ a (a) < xa . Continue by induction on the intervals jS 6 t 6 (j + 1)S. So one draws the conclusion that x(t ) is positive and monotonically decreases. Next we prove limt →+∞ x(t ) = 0. Suppose, on the contrary, limt →+∞ x(t ) = δ ∗ > 0; then ∃ T s.t. when t > T + a, R a +S we have 0 < x(t ) − δ ∗ < ε , where ε < 2Γ1(q) a (a + S − τ )q−1 · (−λ1 )δ ∗ dτ . But at the same time, it can be deduced x(a + T + S ) − δ ∗ < −ε < 0, and this contradiction shows that limt →+∞ x(t ) = 0. The same proof can be used for the case xa < 0.  where x˜ (t ) = x(t + S ) and x˜ a (t ) = xa +

1

Γ (q)

a

Acknowledgements This work was supported by the National Natural Science Foundation of China under Grant No. 10801067 and the Fundamental Research Fund for Physics and Mathematics of Lanzhou University under Grant No. Lzu07001. The author thanks the reviewer for his/her valuable comments and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21]

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