Electronic Journal of Differential Equations, Vol. 2009(2009), No. 26, pp. 1–15.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SOLUTIONS TO BOUNDARY-VALUE PROBLEMS FOR
NONLINEAR DIFFERENTIAL EQUATIONS OF
FRACTIONAL ORDER
XINWEI SU, SHUQIN ZHANG
Abstract. we discuss the existence, uniqueness and continuous dependence
of solutions for a boundary value problem of nonlinear fractional differential
equation.
1. Introduction
Fractional differential equations have gained considerable popularity and importance during the past three decades or so, due mainly to their varied applications in
many fields of science and engineering. Analysis of fractional differential equations
has been carried out by various authors. As for the research in solutions and also
many real applications for factional differential equations, we refer to the book by
Kilbas, Srivastava and Trujillo [4] and references therein. Boundary-value problems
for fractional differential equations have been discussed in [1, 2, 3, 5, 7, 8, 9]. Bai
and Lü [2] used fixed-point theorems on a cone to obtain the existence and multiplicity of positive solutions for a Dirichlet-type problem of the nonlinear fractional
differential equation
D0α+ u(t) + f (t, u(t)) = 0,
0 < t < 1, 1 < α ≤ 2,
u(0) = u(1) = 0,
where f : [0, 1] × [0, ∞) → [0, ∞) is continuous and D0α+ is the fractional derivative
in the sense of Riemann-Liouville. However, as mentioned in [8], the RiemannLiouville fractional derivative is not suitable for nonzero boundary values. Therefore, Zhang [8] investigated the existence and multiplicity of positive solutions for
the problem
Dα
0+ u(t) = f (t, u(t)),
0 < t < 1, 1 < α ≤ 2,
0
u(1) + u0 (1) = 0,
u(0) + u (0) = 0,
with the Caputo’s fractional derivative Dα
0+ and a nonnegative continuous function
f on [0, 1]×[0, ∞). The existence of solutions for the nonlinear fractional differential
2000 Mathematics Subject Classification. 34B05, 26A33.
Key words and phrases. Boundary value problem; fractional derivative; fixed-point theorem;
Green’s function; existence and uniqueness; continuous dependence.
c
2009
Texas State University - San Marcos.
Submitted June 25, 2008. Published February 3, 2009.
1
2
X. SU, S. ZHANG
EJDE-2009/26
equation
C δ
0 Dt u(t)
= g(t, u(t)),
u(0) = α 6= 0,
0 < t < 1, 1 < δ < 2,
u(1) = β 6= 0
δ
has been discussed using the Laplace transform method in [9], where C
0 Dt denotes
the Caputo’s fractional derivative and g : [0, 1] × R → R is a given continuous
function. By means of Schauder fixed-point theorem, Su [7] proved an existence
result for the problem
Dα u(t) = f (t, v(t), Dµ v(t)),
0 < t < 1,
Dβ v(t) = g(t, u(t), Dν u(t)),
0 < t < 1,
u(0) = u(1) = v(0) = v(1) = 0,
where 1 < α, β < 2, µ, ν > 0, α − ν ≥ 1, β − µ ≥ 1, f, g : [0, 1] × R × R → R are
given functions and D is the standard Riemann-Liouville differentiation.
Motivated by the previous results, we present in this paper analysis of a boundary value problem for the fractional differential equation involving more general
boundary conditions and a nonlinear term dependent on the fractional derivative
of the unknown function
C
D0α+ u(t) = f (t, u(t), C D0β+ u(t)), 0 < t < 1,
a1 u(0) − a2 u0 (0) = A, b1 u(1) + b2 u0 (1) = B,
(1.1)
where 1 < α ≤ 2, 0 < β ≤ 1, ai , bi ≥ 0, i = 1, 2, a1 b1 + a1 b2 + a2 b1 > 0, C D0α+
and C D0β+ are the Caputo’s fractional derivatives and f : [0, 1] × R × R → R is
continuous. We impose a growth condition on the function f to prove an existence
result for (1.1). For f Lipschitz in the second and third variables, the uniqueness of
solution and the solution’s dependence on the order α of the differential operator,
the boundary values A and B, and the nonlinear term f are also discussed.
Throughout this work, we denote by I0α+ and D0α+ the Riemann-Liouville fractional integral and derivative respectively. The definitions and some properties
of fractional integrals and fractional derivatives of different types can be found in
[4, 6]. In order to proceed, we recall some fundamental facts of fractional calculus
theory.
Remark 1.1. If α = n is an integer, the Riemann-Liouville fractional derivative
of order α is the usual derivative of order n. The following properties are well
known: I0α+ I0β+ f (t) = I0α+β
f (t), D0α+ I0α+ f (t) = f (t), α > 0, β > 0, f ∈ L1 (0, 1);
+
α
I0+ : C[0, 1] → C[0, 1], α > 0.
Remark 1.2. For α = n, the Caputo’s fractional derivative of order α becomes
a conventional n-th derivative. The Caputo’s fractional derivative is defined in [4]
Pn−1 (k) +
as follows: C D0α+ f (t) = D0α+ (f (t) − k=0 f k!(0 ) tk ), provided that the right-side
derivative exists. In particular, C D0α+ C = 0 for any constant C ∈ R, α > 0.
Moreover, we can derive the following useful properties from [4, Lemmas 2.21 and
2.22]: C D0α+ I0α+ f (t) = f (t), α > 0, f (t) ∈ C[0, 1]; I0α+ C D0α+ f (t) = f (t) − f (0), 0 <
α ≤ 1, f (t) ∈ C[0, 1].
Similar composition relation below between I0α+ and C D0α+ can be found in [8,
Lemma 2.3], but the author did not point out the space to which u(t) belongs.
Besides, the subscript n of the coefficient cn is wrong.
EJDE-2009/26
SOLUTIONS TO BOUNDARY-VALUE PROBLEMS
3
Lemma 1.3. Assume that u(t) ∈ C(0, 1) ∩ L1 (0, 1) with a derivative of order n
that belongs to C(0, 1) ∩ L1 (0, 1). Then
I0α+ C D0α+ u(t) = u(t) + c0 + c1 t + c2 t2 + · · · + cn−1 tn−1
for some ci ∈ R, i = 0, 1, 2, . . . , n − 1, where n is the smallest integer greater than
or equal to α.
We can also prove this lemma using [2, Lemma 2.2] and Remark 1.1. This proof
is obvious and we omit it here.
2. Existence and uniqueness results
In this section, we first impose a growth condition on f which allows us to
establish an existence result of solution, and then utilize the Lipschitz condition on
f to prove a uniqueness theorem for the problem (1.1). Our approaches are based
on the fixed-point theorems due to Schauder and Banach.
Let I = [0, 1] and C(I) be the space of all continuous real functions defined on
I. Define the space X = {u(t) | u(t) ∈ C(I) and C D0β+ u(t) ∈ C(I), 0 < β ≤ 1}
endowed with the norm kuk = maxt∈I |u(t)| + maxt∈I |C D0β+ u(t)|. Then by the
method in [7, Lemma 3.2] and Remark 1.2 we can know that (X, k · k) is a Banach
space.
Now we present the Green’s function for boundary value problem of fractional
differential equation.
Lemma 2.1. Let 1 < α ≤ 2. Assume that g : [0, 1] → R is a continuous function.
Then the unique solution of
C
D0α+ u(t) = g(t), 0 < t < 1,
a1 u(0) − a2 u0 (0) = 0, b1 u(1) + b2 u0 (1) = 0,
is u(t) =
R1
G(t, s)g(s)ds, where
 1
[(t − s)α−1 − a2lb1 (1 − s)α−1 − a1lb1 (1 − s)α−1 t]


 Γ(α) 1

a2 b2
α−2
+
− a1lb2 (1 − s)α−2 t],
Γ(α−1) [− l (1 − s)
G(t, s) =
a2 b1
1
α−1

− a1lb1 (1 − s)α−1 t]


Γ(α) [− l (1 − s)


1
+ Γ(α−1)
[− a2lb2 (1 − s)α−2 − a1lb2 (1 − s)α−2 t],
0
s ≤ t,
t ≤ s,
here l = a1 b1 + a1 b2 + a2 b1 .
This lemma can be proved using Lemma 1.3 and Remark 1.1. For details, we
refer the reader to [8, Lemma 3.1].
Similarly, we can obtain the solution for the boundary-value problem with homogeneous equation and nonhomogeneous boundary conditions.
Lemma 2.2. Let 1 < α ≤ 2. Then the unique solution of
C
D0α+ u(t) = 0,
0
a1 u(0) − a2 u (0) = A,
is u(t) =
(b1 +b2 )A+a2 B
l
+
a1 B−b1 A
t.
l
0 < t < 1,
b1 u(1) + b2 u0 (1) = B,
4
X. SU, S. ZHANG
EJDE-2009/26
In the following discussion, we denote
ϕ(t) :=
(b1 + b2 )A + a2 B
a1 B − b1 A
+
t,
l
l
and use the assumption
(H) 1 < α ≤ 2, 0 < β ≤ 1, ai , bi ≥ 0, i = 1, 2, a1 b1 + a1 b2 + a2 b1 > 0,
f : [0, 1] × R × R → R is a given continuous function.
Lemma 2.3. Assume that (H) holds. Then (1.1) is equivalent to the nonlinear
integral equation
Z 1
G(t, s)f (s, u(s), C D0β+ u(s))ds + ϕ(t).
(2.1)
u(t) =
0
In other words, every solution of (1.1) is also a solution of (2.1) and vice versa.
Proof. Let u ∈ X be a solution of (1.1), applying the method used to prove Lemma
2.1, we can obtain that u is a solution of (2.1).
Conversely, let u ∈ X be a solution of (2.1). We denote the right-hand side of
the equation (2.1) by w(t); i.e.,
w(t) = I0α+ f (t, u(t), C D0β+ u(t)) +
−a2 b1 − a1 b1 t α
I0+ f (1, u(1), C D0β+ u(1))
l
−a2 b2 − a1 b2 t α−1
I0+ f (1, u(1), C D0β+ u(1)) + ϕ(t).
l
Using Remarks 1.1 and 1.2, we have
+
a1 b1 α
I + f (1, u(1), C D0β+ u(1))
l 0
a1 b2 α−1
a1 B − b1 A
−
I + f (1, u(1), C D0β+ u(1)) +
l 0
l
a
b
1
1
= I0α−1
f (t, u(t), C D0β+ u(t)) −
I α+ f (1, u(1), C D0β+ u(1))
+
l 0
a1 b2 α−1
a1 B − b1 A
−
I0+ f (1, u(1), C D0β+ u(1)) +
,
l
l
w0 (t) = D01+ I01+ I0α−1
f (t, u(t), C D0β+ u(t)) −
+
C
D0α+ w(t) = D0α+ (w(t) − w(0+ ) − w0 (0+ )t) = D0α+ I0α+ f (t, u(t), C D0β+ u(t))
= f (t, u(t), C D0β+ u(t)),
namely, C D0α+ u(t) = f (t, u(t), C D0β+ u(t)). One can verify easily that a1 u(0) −
a2 u0 (0) = A, b1 u(1) + b2 u0 (1) = B. Therefore, u is a solution of (1.1), which
completes the proof.
Lemma 2.3 indicates that the solution of the problem (1.1) coincides with the
fixed point of the operator T defined as
Z 1
T u(t) =
G(t, s)f (s, u(s), C D0β+ u(s))ds + ϕ(t).
0
Now, we give the main results of this section.
Theorem 2.4. Let the assumption (H) be satisfied. Suppose further that
EJDE-2009/26
SOLUTIONS TO BOUNDARY-VALUE PROBLEMS
5
(H1)
lim
(|x|+|y|)→∞
maxt∈I |f (t, x, y)|
lΓ(α)Γ(2 − β)
<
=: K.
|x| + |y|
(2l + a2 b2 )Γ(2 − β) + 2l
Then there exists at least one solution u(t) to the boundary-value problem (1.1).
Proof. For any t ∈ I, we find
Z 1
|G(t, s)|ds
0
≤
1 Γ(α)
Z
0
t
(t − s)α−1 ds +
a2 b1
l
Z
1
(1 − s)α−1 ds +
0
a1 b1 t
l
Z
1
(1 − s)α−1 ds
0
Z
a b Z 1
a1 b2 t 1
1
2 2
α−2
(1 − s)
ds +
(1 − s)α−2 ds
+
Γ(α − 1)
l
l
0
0
1
a2 b1 + a1 b1 t
1 a2 b2 + a1 b2 t
α
=
t +
+
Γ(α + 1)
l
Γ(α)
l
1 a2 b1 + a1 b1
a2 b2 + a1 b2 ≤
1+
+
Γ(α)
l
l
2l + a2 b2
=
lΓ(α)
and
Z 1
|G0t (t, s)|ds
0
Z t
Z 1
Z 1
1
a1 b1
a1 b2
α−2
α−1
≤
(t − s)
ds +
(1 − s)
ds +
(1 − s)α−2 ds
Γ(α − 1) 0
lΓ(α) 0
lΓ(α − 1) 0
tα−1
a1 b1
a1 b2
=
+
+
Γ(α) lΓ(α + 1) lΓ(α)
1 a1 b1
a1 b2 2
≤
1+
+
≤
.
Γ(α)
l
l
Γ(α)
Therefore, |G(t, ·)| and |G0t (t, ·)| are integrable for any t ∈ I.
h(x,y)
Denote h(x, y) = maxt∈I |f (t, x, y)| and Choose ε = 12 (K−lim(|x|+|y|)→∞ |x|+|y|
).
It follows from the condition (H1) that there exists a constant d1 > 0 such that
h(x, y) ≤ (K − ε)(|x| + |y|) for |x| + |y| ≥ d1 . Let M = max{h(x, y) : |x| + |y| ≤ d1 }
and choose d2 > d1 such that M/d2 ≤ K −ε. Then we get h(x, y) ≤ (K −ε)d2 , |x|+
|y| ≤ d2 . Therefore, h(x, y) ≤ (K − ε)c, |x| + |y| ≤ c for any c ≥ d2 .
Let k1 = maxt∈I |ϕ(t)|, k2 = maxt∈I |ϕ0 (t)|, k = max{k1 , k2 /Γ(2 − β)}, d3 =
2Kk/ε and d = max{d2 , d3 }. Define
U = {u(t) : u(t) ∈ X, ku(t)k ≤ d, t ∈ I}.
Then U is a convex, closed and bounded subset of X. Moreover, for any u ∈ U ,
h(u(t), C D0β+ u(t)) ≤ (K − ε)d.
Now we prove that the operator T maps U to itself. For any u ∈ U , we can get
Z 1
|T u(t)| ≤ |ϕ(t)| +
|G(t, s)h(u(s), C D0β+ u(s))|ds
0
Z
≤ k1 + d(K − ε)
1
|G(t, s)|ds
0
6
X. SU, S. ZHANG
≤ k1 + d(K − ε)
EJDE-2009/26
2l + a2 b2
,
lΓ(α)
|C D0β+ (T u)(t)|
Z t
1
(t − s)−β (T u)0 (s)ds
=
Γ(1 − β) 0
Z t
Z 1
1
|G0s (s, τ )f (τ, u(τ ), C D0β+ u(τ ))|dτ + |ϕ0 (s)| ds
≤
(t − s)−β
Γ(1 − β) 0
0
Z t
Z 1
1
−β
0
≤
|ϕ (s)| +
(t − s)
|G0s (s, τ )h(u(τ ), C D0β+ u(τ ))|dτ ds
Γ(1 − β) 0
0
Z t
Z t
Z 1
1
k2
−β
−β
|G0s (s, τ )|dτ ds
(t − s) ds + d(K − ε)
(t − s)
≤
Γ(1 − β) 0
Γ(1 − β) 0
0
k2
2
≤
+ d(K − ε)
Γ(2 − β)
Γ(2 − β)Γ(α)
for 0 < β < 1, and
0
|(T u) (t)| = Z
0
1
G0t (t, s)f (s, u(s), C D0β+ u(s))ds + ϕ0 (t)
0
Z
≤ |ϕ (t)| +
0
1
|G0t (t, s)h(u(s), C D0β+ u(s))|ds
Z
1
≤ k2 + d(K − ε)
|G0t (t, s)|ds ≤ k2 + d(K − ε)
0
2
Γ(α)
for β = 1. Hence,
kT uk ≤ 2k + d(K − ε)
(2l + a2 b2 )Γ(2 − β) + 2l
ε
1
≤ d + d(K − ε) = d.
lΓ(α)Γ(2 − β)
K
K
Note also that
T u(t) = I0α+ f (t, u(t), C D0β+ u(t)) +
+
−a2 b1 − a1 b1 t α
I0+ f (1, u(1), C D0β+ u(1))
l
−a2 b2 − a1 b2 t α−1
I0+ f (1, u(1), C D0β+ u(1)) + ϕ(t),
l
a1 b1 α
I + f (1, u(1), C D0β+ u(1))
l 0
a1 B − b1 A
a1 b2 α−1
I0+ f (1, u(1), C D0β+ u(1)) +
,
−
l
l
(T u)0 (t) = I0α−1
f (t, u(t), C D0β+ u(t)) −
+
C
0
D0β+ (T u)(t) = I01−β
+ (T u) (t)
a1 b1 α−β+1
I+
f (1, u(1), C D0β+ u(1))
l 0
a1 b2 α−β
a1 B − b1 A
−
I + f (1, u(1), C D0β+ u(1)) + I01−β
.
+
l 0
l
= I0α−β
f (t, u(t), C D0β+ u(t)) −
+
It is easy to see T u(t), C D0β+ (T u)(t) ∈ C(I). Therefore, T : U → U .
Claim: T is a continuous operator. In fact, for un , n = 0, 1, 2, . . . and u ∈ U
such that limn→∞ kun − uk → 0, we have
|T un (t) − T u(t)|
EJDE-2009/26
=
SOLUTIONS TO BOUNDARY-VALUE PROBLEMS
Z
1
0
G(t, s) f (s, un (s), C D0β+ un (s)) − f (s, u(s), C D0β+ u(s)) ds
≤ max |f (t, un (t),
t∈I
≤
7
C
D0β+ un (t))
− f (t, u(t),
C
D0β+ u(t))|
1
Z
|G(t, s)|ds
0
2l + a2 b2
max |f (t, un (t), C D0β+ un (t)) − f (t, u(t), C D0β+ u(t))|,
lΓ(α) t∈I
|C D0β+ (T un )(t) − C D0β+ (T u)(t)|
Z t
1
(t − s)−β ((T un )0 (s) − (T u)0 (s))ds
=
Γ(1 − β) 0
Z t
Z 1 1
−β
G0s (s, τ ) f (τ, un (τ ), C Dβ+ un (τ ))
≤
(t − s)
0
Γ(1 − β) 0
0 − f (τ, u(τ ), C D0β+ u(τ )) dτ ds
≤ max |f (t, un (t), C D0β+ un (t)) − f (t, u(t), C D0β+ u(t))|
t∈I
Z t
Z 1
1
−β
×
(t − s)
|G0s (s, τ )|dτ ds
Γ(1 − β) 0
0
≤ max |f (t, un (t),
t∈I
≤
C
D0β+ un (t))
− f (t, u(t),
C
D0β+ u(t))|
Γ(1
2
− β)Γ(α)
Z
t
(t − s)−β ds
0
2
max |f (t, un (t), C D0β+ un (t)) − f (t, u(t), C D0β+ u(t))|
Γ(2 − β)Γ(α) t∈I
for 0 < β < 1, and
|(T un )0 (t) − (T u)0 (t)|
Z
1 0
=
Gt (t, s) f (s, un (s), C D0β+ un (s)) − f (s, u(s), C D0β+ u(s)) ds
0
≤ max |f (t, un (t), C D0β+ un (t)) − f (t, u(t), C D0β+ u(t))|
t∈I
Z
1
|G0t (t, s)|ds
0
2
max |f (t, un (t), C D0β+ un (t)) − f (t, u(t), C D0β+ u(t))|
Γ(α) t∈I
≤
for β = 1. Then in view of the uniform continuity of the function f on I × [−d, d] ×
[−d, d], we obtain that T is continuous.
The last step is to prove that T is a completely continuous operator. Let t, τ ∈ I
be such that t < τ and N = maxt∈I,u∈U |f (t, u(t), C D0β+ u(t))| + 1. Then we have
|T u(t) − T u(τ )|
Z 1
=
(G(t, s) − G(τ, s))f (s, u(s), C D0β+ u(s))ds + ϕ(t) − ϕ(τ )
0
Z τ
Z t
≤N
|G(t, s) − G(τ, s)|ds +
|G(t, s) − G(τ, s)|ds
Z
+
τ
0
1
t
|G(t, s) − G(τ, s)|ds + |ϕ(t) − ϕ(τ )|
8
X. SU, S. ZHANG
EJDE-2009/26
h Z t (τ − s)α−1 − (t − s)α−1
a b (1 − s)α−1
1 1
+ (τ − t)
Γ(α)
l
Γ(α)
0
α−2 a1 b2 (1 − s)
ds
+
l Γ(α − 1)
Z τ
a b (1 − s)α−1
(τ − s)α−1
a1 b2 (1 − s)α−2 1 1
+
ds
+ (τ − t)
+
Γ(α)
l
Γ(α)
l Γ(α − 1)
t
Z 1
a b (1 − s)α−1
a1 b2 (1 − s)α−2 i
1 1
+
(τ − t)
ds + |ϕ(t) − ϕ(τ )|
+
l
Γ(α)
l Γ(α − 1)
τ
Z
Z
t
h τ (τ − s)α−1
a b Z 1 (1 − s)α−1
(t − s)α−1
1 1
=N
ds −
ds + (τ − t)
ds
Γ(α)
Γ(α)
l
Γ(α)
0
0
0
Z
a1 b2 1 (1 − s)α−2 i
|a1 B − b1 A|
+
ds + (τ − t)
l
Γ(α − 1)
l
0
h τ α − tα
a b
i
a
b
|a1 B − b1 A|
1 1
1 2
=N
+ (τ − t)
+
,
+ (τ − t)
Γ(α + 1)
lΓ(α + 1) lΓ(α)
l
≤N
|C D0β+ (T u)(t) − C D0β+ (T u)(τ )|
Z
Z 1
t
1
−β
(t − s)
G0s (s, θ)f (θ, u(θ), C D0β+ u(θ))dθ + ϕ0 (s) ds
=
Γ(1 − β) 0
0
Z τ
Z 1
−
(τ − s)−β
G0s (s, θ)f (θ, u(θ), C D0β+ u(θ))dθ + ϕ0 (s) ds
0
0
Z
Z 1
t
1
(t − s)−β
G0s (s, θ)f (θ, u(θ), C D0β+ u(θ))dθ ds
≤
Γ(1 − β) 0
0
Z t
Z 1
−
(τ − s)−β
G0s (s, θ)f (θ, u(θ), C D0β+ u(θ))dθ ds
0
0
Z 1
1
(τ − s)−β
G0s (s, θ)f (θ, u(θ), C D0β+ u(θ))dθ ds
Γ(1 − β) 0
0
Z τ
Z 1
−β
0
−
(τ − s)
Gs (s, θ)f (θ, u(θ), C D0β+ u(θ))dθ ds
0
0
Z
Z t
t
1
−β 0
+
(t − s) ϕ (s)ds −
(τ − s)−β ϕ0 (s)ds
Γ(1 − β) 0
0
Z
Z τ
t
1
+
(τ − s)−β ϕ0 (s)ds −
(τ − s)−β ϕ0 (s)ds
Γ(1 − β) 0
0
Z t
Z τ
2N
2N
≤
((t − s)−β − (τ − s)−β )ds +
(τ − s)−β ds
Γ(1 − β)Γ(α) 0
Γ(1 − β)Γ(α) t
Z
Z
|a1 B − b1 A| t
|a1 B − b1 A| τ
+
((t − s)−β − (τ − s)−β )ds +
(τ − s)−β ds
lΓ(1 − β) 0
lΓ(1 − β) t
2N
|a1 B − b1 A| 1−β
≤
+
(τ
− t1−β + 2(τ − t)1−β )
Γ(2 − β)Γ(α)
lΓ(2 − β)
+
Z
t
for 0 < β < 1, and
|(T u)0 (t) − (T u)0 (τ )|
EJDE-2009/26
=
SOLUTIONS TO BOUNDARY-VALUE PROBLEMS
9
1
Z
0
Z
G0t (t, s)f (s, u(s), C D0β+ u(s))ds + ϕ0 (t)
1
G0τ (τ, s)f (s, u(s), C D0β+ u(s))ds − ϕ0 (τ )
0
Z τ
Z t
N
α−2
α−2
≤
((t − s)
− (τ − s)
)ds +
(τ − s)α−2 ds
Γ(α − 1) 0
t
N
α−1
α−1
α−1
≤
(τ
−t
+ 2(τ − t)
)
Γ(α)
−
for β = 1.
Now, using the fact that the functions τ α − tα , τ α−1 − tα−1 and τ 1−β − t1−β are
uniformly continuous on the interval I, we conclude that T U is an equicontinuous
set. Obviously it is uniformly bounded since T U ⊆ U . Thus, T is completely
continuous. The Schauder fixed-point theorem asserts the existence of solution in
U for the problem (1.1) and the theorem is proved.
The following corollary is obvious.
Corollary 2.5. Let the assumption (H) be satisfied. Suppose further that there
exist two nonnegative functions a(t), b(t) ∈ C[0, 1] such that |f (t, x, y)| ≤ a(t)|x|ρ +
b(t)|y|θ , where 0 < ρ, θ < 1. Then there exists at least one solution for the boundary
value problem (1.1).
Example 2.6. Consider the problem
1
1/2
3/2
D0+ u = (t − )3 (u(t) + C D0+ u(t)), 0 < t < 1,
2
a1 u(0) − a2 u0 (0) = A, b1 u(1) + b2 u0 (1) = B.
√
√
Using Γ(1/2) = π, a simple computation shows K = (lπ)/(2(2l + a2 b2 ) π + 8l).
Since |f (t, x, y)| = |t − 12 |3 |x + y| ≤ |x|+|y|
,
8
C
lim
(|x|+|y|)→∞
|x|+|y|
8
|x| + |y|
=
1
,
8
√
then, if 1/8 < (lπ)/(2(2l + a2 b2 ) π + 8l) (for example, we choose a1 = b2 = 1, a2 =
b1 = 0, then K ≈ 0.2082 > 1/8), Theorem 2.4 ensures the existence of solution for
this problem.
Theorem 2.7. Let the assumption (H) be satisfied. Furthermore, let the function
f fulfill a Lipschitz condition with respect to the second and third variables; i.e.,
|f (t, x, y) − f (t, u, v)| ≤ L(|x − u| + |v − y|) with a Lipschitz constant L such that
0 < L < K, where K is as the same as that in Theorem 2.4. Then the boundary
value problem (1.1) has a unique solution u(t) ∈ X.
Proof. We have shown in Theorem 2.4 that T u(t), C D0β+ (T u)(t) ∈ C(I); i.e., T :
X → X. To apply the Banach fixed-point theorem, we need to verify that T is a
contraction mapping. For any u, v ∈ X, we have
|T u(t) − T v(t)|
Z
1
=
G(t, s) f (s, u(s), C D0β+ u(s)) − f (s, v(s), C D0β+ v(s)) ds
0
10
X. SU, S. ZHANG
EJDE-2009/26
1
Z
≤ Lku − vk
|G(t, s)|ds
0
≤
(2l + a2 b2 )L
ku − vk,
lΓ(α)
|C D0β+ (T u)(t) − C D0β+ (T v)(t)|
Z t
1
(t − s)−β ((T u)0 (s) − (T v)0 (s))ds
=
Γ(1 − β) 0
Z t
Z 1 1
G0s (s, τ ) f (τ, u(τ ), C Dβ+ u(τ ))
≤
(t − s)−β
0
Γ(1 − β) 0
0
− f (τ, v(τ ), C D0β+ v(τ )) dτ ds
Z t
Z 1
1
−β
≤ Lku − vk
(t − s)
|G0s (s, τ )|dτ ds
Γ(1 − β) 0
0
2L
ku − vk
≤
Γ(2 − β)Γ(α)
for 0 < β < 1, and
|(T u)0 (t) − (T v)0 (t)|
Z
1 0
=
Gt (t, s) f (s, u(s), C D0β+ u(s)) − f (s, v(s), C D0β+ v(s)) ds
0
Z
≤ Lku − vk
1
|G0t (t, s)|ds ≤
0
2L
ku − vk
Γ(α)
2
L
2 b2
for β = 1. Thus, kT u − T vk ≤ L( 2l+a
lΓ(α) + Γ(2−β)Γ(α) )ku − vk = K ku − vk. Hence,
the Banach fixed-point theorem yields that T has a unique fixed point which is the
unique solution of the problem (1.1). The proof is therefore complete.
3. Dependence on the parameters
The present section is devoted to the study of the dependence of solution on the
parameters α, A and B, and f for the problem (1.1), provided that the function
f (t, x, y) is Lipschitz with respect to x and y.
Theorem 3.1. Suppose that the conditions of Theorem 2.7 hold. Let u1 (t), u2 (t)
be the solutions, respectively, of the problems (1.1) and
C
C β
D0α−ε
D0+ u(t)),
+ u(t) = f (t, u(t),
0
a1 u(0) − a2 u (0) = A,
0 < t < 1,
b1 u(1) + b2 u0 (1) = B,
(3.1)
where 1 < α − ε < α ≤ 2. Then ku1 − u2 k = O(ε).
Proof. Let G1 (t, s) = G(t, s) and
 1
α−ε−1
− a2lb1 (1 − s)α−ε−1 − a1lb1 (1 − s)α−ε−1 t]

Γ(α−ε) [(t − s)



a2 b2
1
α−ε−2
+
− a1lb2 (1 − s)α−ε−2 t],
s ≤ t,
Γ(α−ε−1) [− l (1 − s)
G2 (t, s) =
a2 b1
1
α−ε−1

− a1lb1 (1 − s)α−ε−1 t]

 Γ(α−ε) [− l (1 − s)


1
+ Γ(α−ε−1)
[− a2lb2 (1 − s)α−ε−2 − a1lb2 (1 − s)α−ε−2 t],
t ≤ s,
EJDE-2009/26
SOLUTIONS TO BOUNDARY-VALUE PROBLEMS
11
be the Green’s function of (3.1). Then
Z 1
G1 (t, s)f (t, u1 (s), C D0β+ u1 (s))ds + ϕ(t),
u1 (t) =
0
Z
u2 (t) =
0
1
G2 (t, s)f (t, u2 (s), C D0β+ u2 (s))ds + ϕ(t).
First we show that
Z 1
|G1 (t, s) − G2 (t, s)|ds = O(ε),
Z
1
|G01t (t, s) − G02t (t, s)|ds = O(ε).
(3.2)
0
0
Observing that
Z t
Z 1
(t − s)α−1
(t − s)α−ε−1 |G1 (t, s) − G2 (t, s)|ds ≤
ds
−
Γ(α)
Γ(α − ε)
0
0
Z 1
(1 − s)α−1
a2 b1
a1 b1 t
(1 − s)α−ε−1 ds
+(
+
)
−
l
l
Γ(α)
Γ(α − ε)
0
Z 1
(1 − s)α−2
a2 b2
a1 b2 t
(1 − s)α−ε−2 +(
+
)
−
ds
l
l
Γ(α − 1)
Γ(α − ε − 1)
0
and
Z 1
|G01t (t, s) − G02t (t, s)|ds
0
Z
(t − s)α−2
(t − s)α−ε−2 a1 b1 1 (1 − s)α−1
(1 − s)α−ε−1 −
ds +
−
ds
≤
Γ(α − 1)
Γ(α − ε − 1)
l
Γ(α)
Γ(α − ε)
0
0
Z
(1 − s)α−ε−2 a1 b2 1 (1 − s)α−2
−
ds,
+
l
Γ(α − 1)
Γ(α − ε − 1)
0
t
Z
without loss of generality, we only estimate one of the integrals in the right-hand
side of the two inequalities above.
Z t
(t − s)α−1
(t − s)α−ε−1 −
ds
Γ(α)
Γ(α − ε)
0
Z t
Z t
(t − s)α−ε−1
(t − s)α−1
(t − s)α−ε−1 (t − s)α−ε−1 ds +
ds
≤
−
−
Γ(α)
Γ(α)
Γ(α)
Γ(α − ε)
0
0
Z t
Z
t
α−1
1
1
1
x
=
− xα−ε−1 dx + −
(t − s)α−ε−1 ds
Γ(α) 0
Γ(α) Γ(α − ε) 0
1
1
1
1 1
1
≤
−
+
−
Γ(α) α − ε α
Γ(α) Γ(α − ε) α − ε
1
|Γ0 (α − ε + θε)| =ε
+
,
α(α − ε)Γ(α) (α − ε)Γ(α)Γ(α − ε)
for some θ such that 0 < θ < 1. So we arrive at the relations in (3.2). Furthermore,
|u1 (t) − u2 (t)|
Z
Z
1
C β
=
G1 (t, s)f (t, u1 (s), D0+ u1 (s))ds −
Z
≤
0
0
1
G1 (t, s)(f (t, u1 (s), C Dβ+ u1 (s))ds
0
0
1
G2 (t, s)f (t, u2 (s), C D0β+ u2 (s))ds
− f (t, u2 (s), C D0β+ u2 (s))ds
12
X. SU, S. ZHANG
Z
+
1
G1 (t, s)f (t, u2 (s), C Dβ+ u2 (s))ds − G2 (t, s)f (t, u2 (s), C Dβ+ u2 (s))ds
0
0
Z
0
1
≤ Lku1 − u2 k
Z
|G1 (t, s)|ds + |kf k|
0
≤
EJDE-2009/26
1
|G1 (t, s) − G2 (t, s)|ds
0
(2l + a2 b2 )L
ku1 − u2 k + |kf k|
lΓ(α)
Z
1
|G1 (t, s) − G2 (t, s)|ds,
0
where |kf k| = sup0<ε<α−1 {maxt∈I |f (t, u2 (t), C D0β+ u2 (t))|}.
|C D0β+ u1 (t) − C D0β+ u2 (t)|
Z t
Z 1 1
G01s (s, τ )f (τ, u1 (τ ), C Dβ+ u1 (τ ))
(t − s)−β
≤
0
Γ(1 − β) 0
0
− G02s (s, τ )f (τ, u2 (τ ), C D0β+ u2 (τ ))dτ ds
Z t
Z 1 1
G01s (s, τ )f (τ, u1 (τ ), C Dβ+ u1 (τ ))
≤
(t − s)−β
0
Γ(1 − β) 0
0
− G01s (s, τ )f (τ, u2 (τ ), C D0β+ u2 (τ ))dτ
Z 1
0
G1s (s, τ )f (τ, u2 (τ ), C Dβ+ u2 (τ )) − G02s (s, τ )f (τ, u2 (τ ), C Dβ+ u2 (τ ))dτ ds
+
0
0
0
Z t
Z 1
1
(t − s)−β
|G01s (s, τ )|dτ ds
Γ(1 − β) 0
0
Z t
Z 1
1
(t − s)−β
+ |kf k|
|G01s (s, τ ) − G02s (s, τ )|dτ ds
Γ(1 − β) 0
0
2L
ku1 − v2 k
≤
Γ(2 − β)Γ(α)
Z t
Z 1
1
+ |kf k|
(t − s)−β
|G01s (s, τ ) − G02s (s, τ )|dτ ds
Γ(1 − β) 0
0
≤ Lku1 − u2 k
for 0 < β < 1, and
|u01 (t) − u02 (t)| ≤ Lku1 − u2 k
Z
1
|G01t (t, s)|ds + |kf k|
1
Z
0
|G01t (t, s) − G02t (t, s)|ds
0
2L
ku1 − u2 k + |kf k|
≤
Γ(α)
Z
1
|G01t (t, s)
− G02t (t, s)|ds
0
for β = 1. It follows that
ku1 − u2 k
≤
Z t
h
Z 1
1
1
|kf k|
(t − s)−β
|G01s (s, τ ) − G02s (s, τ )|dτ ds
1 − L/K
Γ(1 − β) 0
0
Z 1
i
+ |kf k|
|G1 (t, s) − G2 (t, s)|ds
0
for 0 < β < 1 and
ku1 − u2 k
1
≤
|kf k|
1 − L/K
Z
0
1
|G01t (t, s)
−
G02t (t, s)|ds
Z
+ |kf k|
0
1
|G1 (t, s) − G2 (t, s)|ds
EJDE-2009/26
SOLUTIONS TO BOUNDARY-VALUE PROBLEMS
13
for β = 1. Thus, in accordance with (3.2), we obtain ku1 − u2 k = O(ε), which
completes the proof.
Theorem 3.2. Assume the conditions of Theorem 2.7 are valid. Let u1 (t), u2 (t)
be the solutions, respectively, of the problems
C
D0α+ u(t) = f (t, u(t), C D0β+ u(t)),
0
0 < t < 1,
0
a1 u(0) − a2 u (0) = A, b1 u(1) + b2 u (1) = B,
and
C
D0α+ u(t) = f (t, u(t), C D0β+ u(t)),
0
0 < t < 1,
b1 u(1) + b2 u0 (1) = B + ε2 ,
a1 u(0) − a2 u (0) = A + ε1 ,
Then ku1 − u2 k = O(max{ε1 , ε2 }).
Proof. Let
(b1 + b2 )A + a2 B
a1 B − b1 A
+
t,
l
l
(b1 + b2 )(A + ε1 ) + a2 (B + ε2 ) a1 (B + ε2 ) − b1 (A + ε1 )
ϕ2 (t) =
+
t.
l
l
ϕ1 (t) =
Then
1
Z
G(t, s)f (s, u1 (s), C D0β+ u1 (s))ds + ϕ1 (t),
u1 (t) =
0
1
Z
u2 (t) =
0
G(t, s)f (s, u2 (s), C D0β+ u2 (s))ds + ϕ2 (t).
So we obtain
Z
|u1 (t) − u2 (t)| ≤ Lku1 − u2 k
1
|G(t, s)|ds + |ϕ1 (t) − ϕ2 (t)|
0
≤
(b1 + b2 )ε1 + a2 ε2
|a1 ε2 − b1 ε1 |
L(2l + a2 b2 )
ku1 − u2 k +
+
,
lΓ(α)
l
l
|C D0β+ u1 (t) − C D0β+ u2 (t)|
Z t
Z 1
1
−β
≤ Lku1 − u2 k
(t − s)
|G0s (s, τ )|dτ ds
Γ(1 − β) 0
0
Z t
1
(t − s)−β |ϕ01 (s) − ϕ02 (s)|ds
+
Γ(1 − β) 0
2L
1
|a1 ε2 − b1 ε1 |
≤
ku1 − u2 k +
Γ(2 − β)Γ(α)
Γ(2 − β)
l
for 0 < β < 1, and
|u01 (t) − u02 (t)| ≤ Lku1 − u2 k
Z
1
|G0t (t, s)|ds + |ϕ01 (t) − ϕ02 (t)|
0
2L
|a1 ε2 − b1 ε1 |
≤
ku1 − u2 k +
Γ(α)
l
for β = 1. Thus,
ku1 − u2 k
14
X. SU, S. ZHANG
EJDE-2009/26
(b + b )ε + a ε
1
|a1 ε2 − b1 ε1 |
1
|a1 ε2 − b1 ε1 | 1
2 1
2 2
+
+
1 − L/K
l
l
Γ(2 − β)
l
max{ε1 , ε2 } b1 + b2 + a2
a1 + b1
1
a1 + b1 .
≤
+
+
1 − L/K
l
l
Γ(2 − β)
l
≤
Therefore, the conclusion of the theorem follows.
Theorem 3.3. Suppose the conditions of Theorem 2.7 are satisfied. Let u1 (t), u2 (t)
be the solutions, respectively, of the problems
C
D0α+ u(t) = f (t, u(t), C D0β+ u(t)), 0 < t < 1,
a1 u(0) − a2 u0 (0) = A, b1 u(1) + b2 u0 (1) = B,
and
C
D0α+ u(t) = f (t, u(t), C D0β+ u(t)) + ε, 0 < t < 1,
a1 u(0) − a2 u0 (0) = A, b1 u(1) + b2 u0 (1) = B.
Then ku1 − u2 k = O(ε).
R1
Proof. Note that u1 (t) = 0 G(t, s)f (s, u1 (s), C D0β+ u1 (s))ds + ϕ(t),
R1
u2 (t) = 0 G(t, s)(f (s, u2 (s), C D0β+ u2 (s)) + ε)ds + ϕ(t). Thus,
Z 1
Z 1
|u1 (t) − u2 (t)| ≤ Lku1 − u2 k
|G(t, s)|ds + ε
|G(t, s)|ds
0
0
L(2l + a2 b2 )
ε(2l + a2 b2 )
≤
ku1 − u2 k +
,
lΓ(α)
lΓ(α)
|C D0β+ u1 (t) − C D0β+ u2 (t)|
Z t
Z 1
1
−β
≤ Lku1 − u2 k
(t − s)
|G0s (s, τ )|dτ ds
Γ(1 − β) 0
0
Z t
Z 1
ε
+
(t − s)−β
|G0s (s, τ )|dτ ds
Γ(1 − β) 0
0
2ε
2L
ku1 − u2 k +
≤
Γ(2 − β)Γ(α)
Γ(2 − β)Γ(α)
for 0 < β < 1, and
|u01 (t)
−
u02 (t)|
Z
≤ Lku1 − u2 k
1
|G0t (t, s)|ds
0
≤
Z
+ε
1
|G0t (t, s)|ds
0
2L
2ε
ku1 − u2 k +
Γ(α)
Γ(α)
for β = 1. Then,
ku1 − u2 k ≤
2l + a b
ε
2
2 2
+
,
1 − L/K
lΓ(α)
Γ(2 − β)Γ(α)
and we get the desired result.
EJDE-2009/26
SOLUTIONS TO BOUNDARY-VALUE PROBLEMS
15
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Department of Mathematics, China University of Mining and Technology, Beijing,
100083, China
E-mail address, Xinwei Su: kuangdasuxinwei@163.com
E-mail address, Shuqin Zhang: zhangshuqin@tsinghua.org.cn