Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 166, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF LOCAL AND GLOBAL SOLUTIONS FOR
HADAMARD FRACTIONAL DIFFERENTIAL EQUATIONS
MENGMENG LI, JINRONG WANG
Abstract. In this article, we study a class of Hadamard fractional differential
equations and give sufficient conditions on the existence of local and global of
solutions.
1. Introduction
Let 0 ≤ γ < 1, 1 < a < T and G be an open set in R. Denote a Banach
space by Cγ,ln [a, T ] := {µ(x) : (ln xa )γ µ(x) ∈ C[a, T ]} endowed with the norm
kµkCγ,ln = k(ln xa )γ µ(x)kC . In this article, we study the existence of local and
global solutions to the Hadamard type fractional differential equation
α
H Da,x y(x)
= f (x, y(x)),
α−1
+
H Da,x y(a )
= c,
0 < α < 1, x ∈ J,
c ∈ R,
(1.1)
α
where J = [a, a + h], h > 0 or [a + ∞) and the symbol H Da,x
y(x) is defined by
Z x
1
d x
dτ
α
x
(ln )−α y(τ ) .
H Da,x y(x) =
Γ(1 − α) dx a
τ
τ
We use the notation
α−1
y(x) and
= limx→a+ Ja,x
Z x
1
x
dt
α−1
Ja,x
y(x) =
(ln )α−1 y(t)) .
Γ(α) 1
t
t
α−1
+
H Da,x y(a )
Following [1, Theorem 3.28], the solution y ∈ C1−γ,ln [a, a + h] of (1.1) satisfies
Z x
1
x
dτ
y(x) = y0 (x) +
(ln )α−1 f (τ, y(τ )) , x ∈ (a, a + h]
(1.2)
Γ(α) a
τ
τ
c
where y0 (x) = Γ(α)
(ln xa )α−1 , if f : (a, a + h] × G → R and f (x, y) ∈ Cγ,ln [a, a + h]
for any y ∈ G.
Inspired by the work in [1, 2, 3], we examine other explicit sufficient conditions
on the nonlinear term f to guarantee the local existence of solutions in Cγ,ln [a, a+h]
and global existence of solutions in Cγ,ln [a, +∞).
2010 Mathematics Subject Classification. 26A33, 34A12.
Key words and phrases. Hadamard fractional differential equation; local solution;
global solution.
c
2015
Texas State University - San Marcos.
Submitted February 2, 2015. Published June 17, 2015.
1
2
M. LI, J. WANG
EJDE-2015/166
2. Main results
The following equality will be used in the sequel.
Lemma 2.1 ([4, p.296]). Let α, β, γ, p > 0, then
Z x
xθ p(γ − 1) + 1
(xα − sα )p(β−1) sp(γ−1) ds =
B
, p(β − 1) + 1 ,
α
α
0
R1
where θ = p[α(β − 1) + γ − 1] + 1 and B[ξ, η] = 0 sξ−1 (1 − s)η−1 ds.
x > 0,
Let B = {y ∈ R : ky − y0 (x)kCγ,ln ≤ b} where b will be chosen latter. Define
D = {(x, y) ∈ R × R : x ∈ J, y ∈ B}. We assume that f : D → R satisfies the
following conditions:
(H1) f (x, y) is Lebesgue measurable with regard to x on J and f (x, y) is continuous with respect to y on B.
(H2) there exists m(·) ∈ Lq (J), q > 1 such that |f (x, y)| ≤ m(x), for arbitrary
x ∈ J, y ∈ B.
Now we use Picard iterative approach to derive the existence of a local solutions to
(1.1).
Theorem 2.2. Assume that (H1)–(H2) hold for J = [a, a + h] and p, q, α satisfy
p(α − 1) + 1 > 0, p1 + 1q = 1. Then (1.1) has a solution in Cγ,ln [a, a + h] for some
h > 0.
Proof. To achieve our aim, we divide our proof into three steps.
Step 1. Linking our assumptions and using Hölder inequality via p(α − 1) + 1 > 0
and p1 + 1q = 1, one can obtain
s
Z x
dt
a1−p hp(α−1)+1
1/p
|(ln x − ln τ )α−1 f (τ, y(τ ))| ≤
km(·)kLq [a,a+h] ,
(2.1)
τ
p(α − 1) + 1
a
where we use basic inequalities: ln u − ln v ≤ u − v for u ≥ v > 1 and [5, Lemma
2.2],
Z x
a1−p (ln x − ln a)p(α−1)+1
.
(2.2)
(ln x − ln τ )p(α−1) τ −p dτ ≤
p(α − 1) + 1
a
This proves that (ln x − ln τ )f (τ, y(τ )) is Lebesgue integrable with respect to τ ∈
[a, x] for arbitrary x on J, provided that y(τ ) is Lebesgue measurable on the interval
[a, a + h].
Step 2. For a given M > 0, there exists a h0 > 0 satisfying
Z a+h0
mq (τ )dτ ≤ M q ,
(2.3)
a
0
1
p(α−1)+1 }.
, T, [ MbΓ(α)(p(α−1)+1)
ap−1 (ln T −ln a)γ ]
whenever h = min{h
For δ to be chosen latter,
define


0, if a ≤ x < a + δ, 0 < δ < nh ,




 c (ln x )α−1 , if a + δ ≤ x < a + h ,
a
n
yn (x) = Γ(α)
R x− nh

c
x α−1
1
x α−1

+ Γ(α) a
(ln τ )
f (τ, yn (τ )) dτ

Γ(α) (ln a )
τ ,



h
if a + n ≤ x ≤ a + h.
EJDE-2015/166
EXISTENCE OF LOCAL AND GLOBAL SOLUTIONS
We show that yn (x) is continuous on [a, a +
Case 1. For a < a + δ ≤ x1 < a +
h
n
2h
n ]
3
for all n.
< x2 ≤ a + h,
|c| x2
x1
(ln )α−1 − (ln )α−1 Γ(α)
a
a
Z x2 − nh
1
x2
dτ
+
(ln )α−1 |f (τ, yn (τ ))|
Γ(α) a
τ
τ
:= I1 + I2 .
|yn (x2 ) − yn (x1 )| ≤
2−α
For each ε > 0, there exists 0 < δ1 < [ aεΓ(α)δ
2|c|(1−α) ] such that for all x2 − x1 ≤ δ1
and for all n, we derive that
x2
x1
|c|
x2
x1
x1 x2
|c|
|χ( ) − χ( )| ≤
|χ0 (ξ)|| − |, ξ ∈ ( , )
Γ(α)
a
a
Γ(α)
a
a
a a
|c|(1 − α)(x2 − x1 )
≤
Γ(α)(a + δ)(ln(a + δ) − ln a)2−α
|c|(1 − α)(x2 − x1 )
<
< ε/2,
Γ(α)a(ln(a + δ) − ln a)2−α
I1 =
where χ(x) = (ln x − ln a)α−1 and χ0 (x) = (α − 1)(ln x − ln a)α−2 xa .
p
p
1
] p(α−1)+1 such that for all
For any ε > 0, there exists 0 < δ2 < [ ε Γ 2(α)(p(α−1)+1)
p a1−p M p
h
x2 − n − a ≤ x2 − x1 ≤ δ2 and for all n, we use (2.2) and (2.3) to obtain
1
Γ(α)
Z
1
≤
Γ(α)
Z
I2 =
h
x2 − n
(ln
x2 α−1
dτ
)
|f (τ, yn (τ ))|
τ
τ
(ln
x2 α−1
dτ
)
m(τ )
τ
τ
a
h
x2 − n
a
M a1−p (ln(x2 − nh ) − ln a)p(α−1)+1 1/p
≤
Γ(α)
p(α − 1) + 1
1−p
M a (x2 − nh − a)p(α−1)+1 1/p
≤
< ε/2,
Γ(α)
p(α − 1) + 1
where we use that ln u − ln v ≤ u − v, u > v > 1 again in the last inequality.
From above, we can choose δ̄ =min{δ1 , δ2 , h/n} such that for all x2 − x1 ≤ δ̄ and
for all n, such that |yn (x2 ) − yn (x1 )| ≤ I1 + I2 < ε.
Case 2. For a +
h
n
≤ x1 < x2 ≤ a +
2h
n .
One has
|yn (x2 ) − yn (x1 )|
Z x1 − nh 1
x2
x1
dτ
≤ S1 +
(ln )α−1 − (ln )α−1 |f (τ, yn (τ ))|
Γ(α) a
τ
τ
τ
Z x2 − nh
1
x2
dτ
+
(ln )α−1 |f (τ, yn (τ ))|
Γ(α) x1 − nh
τ
τ
:= S1 + S2 + S3 .
4
M. LI, J. WANG
EJDE-2015/166
h
i
h 2−α
aεΓ(α)( n
)
such that for all x2 − x1 ≤ δ¯1 ,
For each ε > 0, there exists 0 < δ¯1 <
3|c|(1−α)
we have
|c|(1 − α)(x2 − x1 )
S1 ≤
< ε/3.
Γ(α)a(ln(a + nh ) − ln a)2−α
1
i
h p
p
(p(α−1)+1) p(α−1)+1
For each ε > 0, there exists 0 < δ¯2 < ε Γ(α)
such that for
p
p
1−p
3 M a
¯
all x2 − x1 ≤ δ2 , by using the similar estimation methods of I2 we have
Z
h
x1 − n
dτ
x1
x2 α−1
)
− (ln )α−1 |f (τ, yn (τ ))|
τ
τ
τ
a
Z x1 − nh 1
x2
dτ
x1
≤
(ln )α−1 − (ln )α−1 m(τ )
Γ(α) a
τ
τ
τ
Z x1 − nh
Z x1 − nh
1/p i
h
1/p
M
x2
x1
≤
−
(ln )p(α−1) τ −p dτ
(ln )p(α−1) τ −p dτ
Γ(α)
τ
τ
a
a
h
p(α−1)+1 i1/p
M (ln x2 − ln x1 )
≤
Γ(α) ap−1 (p(α − 1) + 1)
M h (x2 − x1 )p(α−1)+1 i1/p
≤
< ε/3.
Γ(α) ap−1 (p(α − 1) + 1)
S2 =
1
Γ(α)
(ln
p
p
1
(p(α−1)+1) p(α−1)+1
, such that for
For each ε > 0, there exists 0 < δ¯3 < [ ε Γ(α)
]
3p M p a1−p
all x2 − x1 ≤ δ¯3 , by using the similar estimation methods of I2 we have
Z x2 − nh
x2
1
dτ
(ln )α−1 |f (τ, yn (τ ))|
S3 =
Γ(α) x1 − nh
τ
τ
Z x2 − nh
x2
dτ
1
(ln )α−1 m(τ )
≤
Γ(α) x1 − nh
τ
τ
≤
M a1−p (x2 − x1 )p(α−1)+1 1/p
< ε/3.
Γ(α)
p(α − 1) + 1
From the above, we choose δ̄¯ = min{δ¯1 , δ¯2 , δ¯3 , h/n} such that x2 − x1 ≤ δ̄¯, then
|yn (x2 ) − yn (x1 )| ≤ S1 + S2 + S3 < ε. Therefore, we choose δ = min{δ̄, δ̄¯} will lead
to yn (x) is continuous with regard to x on [a, a+ 2h
n ] for all positive integers n. Note
that (ln x − ln a)γ is continuous function, so yn (x)(ln x − ln a)γ is also continuous.
c
Nevertheless, for all x ∈ [a + δ, a + nh ], one has |yn (x) − Γ(α)
(ln xa )α−1 | = 0, and
for all x ∈ [a + nh , a + h], using Hölder inequality again,
x
c
x
(ln )γ |yn (x) −
(ln )α−1 |
a
Γ(α)
a
Z x− nh
1
x
x
dτ
≤
(ln )γ (ln )α−1 |f (τ, yn (τ ))|
Γ(α) a
a
τ
τ
≤
(2.4)
(ln(a + h) − ln a)γ a1−p M hp(α−1)+1
≤ b,
Γ(α)(p(α − 1) + 1)
which implies that (x, yn (x)) ∈ D for all n. Therefore, {yn (x)}∞
n=1 defined on
[a, a + h] is equicontinuous and uniformly bounded.
EJDE-2015/166
EXISTENCE OF LOCAL AND GLOBAL SOLUTIONS
5
Step 3. By using Arzelò-Ascoli lemma and Step 2, there must exist {ynk (x)}∞
k=1 :=
∞
∞
{yk (x)}∞
contained
in
{y
(x)}
,
such
that
{y
(x)}
is
uniformly
convergent
n
k
n=1
k=1
k=1
to y(x) which is continuous with regard to x on [a, a + h]. Now we only need to
prove that this limit function y(x) is a solution of (1.2).
For each ε > 0, there exists K1 > 0, such that for all k > K1 , and x ∈ [a, a + h],
we have
Γ(α + 1)ε
.
(2.5)
|f (x, yk (x)) − f (x, y(x))| <
2(ln(a + h) − ln a)γ hα
Note that
(ln x − ln a)γ |yk (x) − y(x)|
Z x
1
x
x
dτ
(ln )γ (ln )α−1 |f (τ, yk (τ )) − f (τ, y(τ ))|
≤
Γ(α) a
a
τ
τ
Z x
1
x γ
x α−1
dτ
+
(ln ) (ln )
|f (τ, yk (τ ))|
Γ(α) x− hk
a
τ
τ
:= S4 + S5 .
Using (2.5) one obtains,
Z x
x
dτ
1
x
S4 =
(ln )γ (ln )α−1 |f (τ, yk (τ )) − f (τ, y(τ ))|
Γ(α) a
a
τ
τ
γ Z a
(ln(a + h) − ln a)
x
≤
(ln )α−1 |f (τ, yk (τ )) − f (τ, y(τ ))|d(ln x − ln τ )
Γ(α)
τ
x
γ α
(ln(a + h) − ln a) h
Γ(α + 1)ε
≤
< ε/2.
aΓ(α)
2(ln(a + h) − ln a)γ hα
Also there exists
−1
ap−1 εp Γ(α)p (p(α − 1) + 1) p(α−1)+1
,
0 < K2 = h p p
pγ
2 M (ln(a + h) − ln a)
such that for all k > K2 ,
Z x
1
x
dτ
x
S5 =
(ln )γ (ln )α−1 |f (τ, yk (τ ))|
Γ(α) x− hk
a
τ
τ
Z x
1/p
γ
(ln(a + h) − ln a) M
x
≤
(ln )p(α−1) τ −p dt
Γ(α)
τ
x− h
k
h p(α−1)+1
1/p
γ
(k)
(ln(a + h) − ln a) M
≤
< ε/2.
Γ(α)
ap−1 (p(α − 1) + 1)
Hence, taking K = max{K1 , K2 } and for all k > K, one arrives at kyk (x) −
y(x)kCγ,ln < ε. Consequently, y(x) satisfies (1.2) which means that there at least
exists a solution of (1.1).
Next, we give an existence and uniqueness theorem, using the assumption
(H3) there exists a µ(·) ∈ Lq (J), 1q = 1 − p1 , p > 1 such that |f (x, y) − f (x, z)| ≤
µ(x)|y − z| for x ∈ J and y, z ∈ B,
1
Theorem 2.3. Let 0 ≤ γ ≤ min{α − 1 + p1 , 2p
}, p > 1. Assume that (H1)–(H3)
are satisfied for J = [a, a + h]. Then (1.1) has a unique solution in Cγ,ln [a, a + h]
for some h > 0.
6
M. LI, J. WANG
EJDE-2015/166
Proof. There exits h∗ > 0, such that for all x ∈ [a, a + h∗ ],
Z a+h∗
Z x
µq (τ )dτ ≤
µq (τ )dτ < g q .
a
(2.6)
a
Let Ψh = {y ∈ Cγ,ln [a, a + h] : ky(x) − y0 (x)kCγ,ln ≤ b, x ∈ [a, a + h]}, where h
is the smaller one between h∗ and h obtained in Theorem 2.2. Note Ψh endowed
with k · kCr ln is a Banach space. Define
Z x
1
x
dτ
ψ(y) = y0 (x) +
(ln )α−1 f (τ, y(τ )) , ∀x ∈ [a, a + h].
Γ(α) a
τ
τ
Firstly, assume that a ≤ x1 < x2 ≤ a+h, then according to the proof in Step 2 of
Theorem 2.2, we obtain that ψ(y) is continuous with regard to x on [a, a + h]. Note
that (ln x − ln a)γ is continuous function, so ψ(y)(ln x − ln a)γ is also continuous.
Secondly, it following (2.4) that ψ(y) ∈ Ψh with y ∈ Ψh . Thirdly, the condition
1
1
}, p > 1 implies that p(α − 1 − γ) + 1 > 0, 0 ≤ γ < 2p
.
0 ≤ γ ≤ min{α − 1 + p1 , 2p
For any y1 , y2 ∈ Ψh , then using Hölder inequality, (2.6) and Lemma 2.1, we have
(ln x − ln a)γ |ψ(y2 ) − ψ(y1 )|
Z
(ln x − ln a)γ x
dτ
x
≤
(ln )α−1 µ(τ )|y1 (τ ) − y2 (τ )|
Γ(α)
τ
τ
a
Z
(ln(a + h) − ln a)γ x
τ −γ
dτ
x α−1
≤
(ln ) µ(τ ) ky1 − y2 kCr ln
(ln )
Γ(α)
τ
a
τ
a
Z
1/p
x
(ln(a + h) − ln a)γ ga1−p ln x−ln a
ky1 − y2 kCγ ln
(ln − t)p(α−1) t−pγ dt
≤
Γ(α)
a
0
1/p
(ln(a + h) − ln a)γ ga1−p x
≤
(ln )p(α−1−γ)+1 B[1 − pγ, p(α − 1) + 1]
Γ(α)
a
× ky1 − y2 kCγ ln
hγ g p(α−1−γ)+1 1
(h
B[ , p(α − 1) + 1])1/p ky1 − y2 kCγ ln
Γ(α)
2
1
1/p i
h hα−1+ p g 1
ky1 − y2 kCγ ln ,
= a1−p
B[ , p(α − 1) + 1]
Γ(α)
2
≤ a1−p
where we use that
1
1
1
⇒ 1 − pγ > ⇒ t1−pr ≤ t 2 (0 ≤ t ≤ 1),
2p
2
Z 1
B[1 − pγ, p(α − 1) + 1] =
t−pγ (1 − t)p(α−1) dt
γ<
0
Z
≤
0
Obviously, one can choose
h≤
1
1
1
t− 2 (1 − t)p(α−1) dt = B[ , p(α − 1) + 1].
2
p
p(α−1)+1
Γ(α)
,
g(B[ 21 , p(α − 1) + 1])1/p
then
1
1/p
hα−1+ p g 1
≤ 1.
B[ , p(α − 1) + 1]
Γ(α)
2
EJDE-2015/166
EXISTENCE OF LOCAL AND GLOBAL SOLUTIONS
7
Therefore,
kψ(y2 ) − ψ(y1 )kCγ ln ≤ a1−p ky1 − y2 kCγ ln .
Obviously, a1−p < 1 due to a, p > 1, applying the Banach Contractive Mapping
Principle, one concludes that there exists a unique y ∗ (x) ∈ Ψh , such that (1.2).
The proof is compete.
Next, we give the existence of a global solution, using the assumption
(H2’) there exist ω, ν > 0 such that |f (x, y)| ≤ ω + ν|y| for x ∈ (a, +∞) and
y ∈ R.
Theorem 2.4. Assume that (H1), (H2’), (H3) hold for J = (a, +∞). Further,
1
}, p > 1. Then (1.1) has a unique solution
choose γ = 1 − α ≤ min{α − 1 + p1 , 2p
in Cγ,ln [a, +∞).
Proof. It follows (H2’) that f is locally bounded in the domain D. By Theorem 2.3,
(1.1) has a unique solution in Cγ,ln [a, a+h]. Next, we present proof by contradiction.
Assume that the solution y(x) admits a maximal existence interval, denoted by
(a, T ) ⊂ (a, +∞). To achieve our aim, it is sufficient to verify that kykCγ,ln is
bounded. In fact,
(ln x − ln a)γ |y(x)|
Z x
|c|
1
x
x
dτ
≤
+
(ln )γ (ln )α−1 |f (τ, y(τ ))|
Γ(α) Γ(α) a
a
τ
τ
Z x
1
x
dτ
|c|
x
+
≤
(ln )γ (ln )α−1 (ω + ν|y(τ )|)
Γ(α) Γ(α) a
a
τ
τ
1−α
α
|c|
ω(ln T − ln a)
(T − a)
≤
+
Γ(α)
Γ(a + 1)
Z x
x
dτ
ν
(ln )α−1 (ln x − ln a)γ |y(τ )|
+
Γ(α) a
τ
τ
Z x
ω(T − a)
ν
dτ
|c|
x
+
+
≤
(ln )α−1 (ln x − ln a)γ |y(τ )| .
Γ(α)
Γ(α + 1)
Γ(α) a
τ
τ
By applying the generalized Gronwall inequality from [3, Corollary 3.4], one
can conclude that there exists l := Eα (ν(ln T )α ) > 0 (Eα denotes Mittag-Leffler
function) such that
(ln x − ln a)γ |y(x)| ≤ l(
|c|
ω(T − a)
+
) := ρ < +∞.
Γ(α)
Γ(α + 1)
This implies that kykCγ,ln < b on [a, T ) when b is chosen as a larger number than
b=ρ+
|c|
.
Γ(α)
(2.7)
This contradicts the assumption that (a, T ) is the maximal existence interval. The
proof is complete.
To finish this article, we give an example that illustrates our theoretical results.
Consider
4|y|
3/4
2
sin x, x ∈ J = (e, e2 ] or (e, +∞),
H De,x y(x) = x +
1 + |y|
(2.8)
−1/4
+
H De,x y(e ) = 1,
8
M. LI, J. WANG
EJDE-2015/166
where α = 3/4, T = e2 , γ = 1/4, a = e, c = 1, and p = q = 2.
4|y|
Define f (x, y) = x2 + 1+|y|
sin x, µ(x) = 4 and ω = e2 + 4 and ν = 0. Thus
|f (x, y) − f (x, z)| ≤ µ(x)|y − z| and |f (x, y)| ≤ ω. Then l := Eα (0) = 1 (see [6,
2
2
−e)
2
Lemma 2]) and b = Γ(3/4)
+ (e +4)(e
(see (2.7)).
Γ(7/4)
R
2e
Let h0 = h∗ = e. Set M 2 = e (e4 + 4)2 dx = e(e4 + 4)2 (see (2.3)) and g 2 =
R 2e
1
16dx = 16e (see (2.6)). Moreover, one can find γ = 1 − α = min{α − 1 + p1 , 2p
}.
e
• According to Theorem 2.3, (2.8) admits a unique solution y ∈ C 14 ,ln (e, e + h]
where
1
n
h bΓ(α)(p(α − 1) + 1) i p(α−1)+1
h = min h0 , T,
,
M ap−1
p
o
Γ(α)
p(α−1)+1
1
1/p
g(B[ 2 , p(α − 1) + 1])
2
2
n
h ( 2 + (e +4)(e −e) )Γ(3/4) i2 h
i4 o
Γ(3/4)
Γ(3/4)
Γ(7/4)
√ 4
= e, e2 ,
.
, √ q
2 e(e + e)e
4 e B[ 1 , 1 ]
2 2
• According to Theorem 2.4, (2.8) has a unique solution y ∈ C 14 ,ln (e, +∞).
Acknowledgments. This work is supported by the National Natural Science
Foundation of China (11201091) and by the Outstanding Scientific and Technological Innovation Talent Award of Education Department of Guizhou Province
([2014]240).
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[3] J. Wang, Y. Zhou, M. Medveď; Existence and stability of fractional differential equations with
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Mengmeng Li
Department of Mathematics, Guizhou University, Guiyang, Guizhou 550025, China
E-mail address: Lmm0424@126.com
Jinrong Wang (corresponding author)
Department of Mathematics, Guizhou University, Guiyang, Guizhou 550025, China
E-mail address: sci.jrwang@gzu.edu.cn