Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 111, pp. 1–14.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

CONVERGENCE OF SOLUTIONS OF FRACTIONAL

DIFFERENTIAL EQUATIONS TO POWER-TYPE FUNCTIONS

MOHAMMED DAHAN KASSIM, NASSER EDDINE TATAR

Abstract. In this article we study the asymptotic behavior of solutions of
some fractional differential equations. We prove convergence to power type

functions under some assumptions on the nonlinearities. Our results extend

and generalize some existing well-known results on solutions of ordinary dif-
ferential equations. Appropriate estimations and lemmas such as a fractional

version of L’Hopital’s rule are used.

1. Introduction

We consider the initial value problems

(CDα
0x)′(τ) = f(τ, x(τ),C Dβ

0x(τ)), 0 < β < α < 1, τ > 0

CDα
0x(τ)

∣∣
τ=0

= b2, = x(τ)|τ=0 = b1, b1, b2 ∈ R,
(1.1)

and
CDα

0x(τ) = f(τ, x(τ),CDβ
0x(τ)), 0 ≤ β < α < 1, τ > 0

x(τ)|τ=0 = b,
(1.2)

where CDα
0 is the Caputo fractional derivative. The definition of the Caputo frac-

tional derivative is given in the next section. We prove that the solutions of (1.1)
approach power type functions and the solutions of (1.2) are bounded. To this
end, the fractional differential problems (1.1) and (1.2) are first transformed into
equivalent integral equations in appropriate underlying spaces. Various appropriate
estimates, comparison theorems and lemmas are used. Moreover, we prove a Ca-
puto fractional version of L’Hopital’s rule. Our arguments here are quite different
from those used so far in the literature.

The behavior of solutions of various classes of ODEs (ordinary differential equa-
tions) has been discussed in fairly a large number of papers in the literature. For
example the equation

x′′(τ) + f(τ, x(τ)) = 0, (1.3)

has been studied in [17, 18, 7, 30, 32, 9, 31] and other papers. The authors proved
that, under various conditions, all solutions of (1.3) are asymptotic to cτ + b as
τ →∞, c, b ∈ R. For the equation

x′′(τ) + f(τ, x(τ), x′(τ)) = 0, (1.4)

2010 Mathematics Subject Classification. 34E10, 26A33, 34A08.
Key words and phrases. Asymptotic behavior; boundedness; fractional differential equation;
Caputo fractional derivative; Riemann-Liouville fractional derivative.
c©2020 Texas State University.

Submitted September 15, 2020. Published November 4, 2020.

1



2 M. D. KASSIM, N. E. TATAR EJDE-2020/111

see, for instance [8, 10, 19, 22, 23, 25, 27, 28]. It is proved that all solutions of (1.4)
can be expressed asymptotically as cτ + b as τ →∞, c, b ∈ R.

Medveď and Pekárková [23], studied the one-dimensional p-Laplacian equation

(|x′|p−1x′)′ = f(τ, x, x′), p > 1. (1.5)

They demonstrated that any solution of (1.5) behaves asymptotically as b+ cτ as
τ →∞ for some real numbers b, c.

In [22], the initial value problem

(Φp(x
′)Ψ(τ))′ + f(τ, x, x′) = 0, 1 < p < 2,

x(τ0) = x0, x′(τ0) = x1, τ0 ≥ 1,
(1.6)

was studied, where Φp(u) = |u|p−2u and Ψ(τ) is a continuous positive function. Suf-
ficient conditions under which all solutions of (1.6) obey the asymptotic expansion
x(τ) = b+ cτ are established.

In contrast, the fractional case of equations (1.3) and (1.4) have been studied by
comparatively a only few researchers; see, for instance [1, 2, 3, 4, 5, 6, 11, 12, 13, 14,
15, 20, 21, 24]. In 2009, Băleanu and Mustafa [1] studied the nonlinear fractional
differential equation

CDα
0x(τ) = f(τ, x(τ)), 0 < α < 1, τ > 0. (1.7)

They showed that the solutions of (1.7) are asymptotic to o(τ cα) as τ → ∞, for
some c.

In 2012, Medveď [20] studied the problem

CDα+1
a x(τ) = f(τ, x(τ)), 0 < α < 1, τ ≥ a > 1

x(a) = c1, x′(a) = c2.
(1.8)

He demonstrated that any solution of (1.8) has the asymptotic property x(τ) =
b+ cτ as τ →∞, for some c, b ∈ R.

Also, in 2013, Medveď [21] discussed the equation

CDα+1
a x(τ) = f(τ, x(τ), x′(τ)), 0 < α < 1, τ ≥ a > 1 (1.9)

and proved that every solution of (1.9) can be expressed asymptotically as b + cτ
as τ →∞, for some c, b ∈ R.

Brestovanská and Medveď [6] studied the problem

x′′(τ) + f(τ, x(τ), x′(τ)) +

m∑
i=1

ri(τ)

∫ τ

0

(τ − s)αi−1fi(s, x(s), x′(s))ds = 0

x(1) = b1, x′(1) = b2, 0 < αi < 1, i = 1, 2, . . . ,m.

(1.10)

They showed that any solution enjoys the asymptotic expansion x(τ) = b + cτ as
τ →∞, for some c, b ∈ R.

In 2015, Medveď and Posṕı̌sil [24] considered the equation

CDα
ax(τ) = f(τ, x(τ),CDβ

0x(τ)), τ > a. (1.11)

They proved that any solution x(τ) with 0 < β < α < 1, has the asymptotic
property x(τ) = cτβ + o(τβ) as τ → ∞, for some c ∈ R. Also they proved that
any solution x(τ) of (1.11), 0 < β < 1 < α < 2, has the asymptotic property
x(τ) = cτ + o(τ) as τ → ∞, for some c ∈ R. Moreover, they proved that there



EJDE-2020/111 FRACTIONAL DIFFERENTIAL EQUATIONS 3

exists a constant c ∈ R such that any global solution x(τ) of the initial value
problem

CDα
ax(τ) = f

(
τ, x(τ), x′(τ), . . . , x(n−1)(τ),CDβ1

0 x(τ), . . . ,CDβm
0 x(τ)

)
,

τ > a.x(n−1)(a) = ci, i = 0, 1, . . . , n− 1,
(1.12)

has the asymptotic property x(τ) = cτk + o(τk) as τ → ∞, where k ∈ max{n −
1, βm}, 0 < β1 < . . . . < βm < α < n, and n, m ∈ N.

In Sections 2 and 3, we prepare some material which will be needed later in our
proofs. Sections 4, 5 and 6 are devoted to the main results on the asymptotic be-
havior results and boundedness of solutions for non-fractional and fractional source
terms, respectively.

2. Preliminaries

In this section, we introduce some basic definitions, notation, properties and lem-
mas to be used in our results. We refer the reader to citeKilbas,Podl-1999,Samko
for more details.

Definition 2.1 ([16]). We introduce the space

Cη[a, b] = {h : (a, b]→ R : h(τ)(τ − a)η ∈ C[a, b]}, 1 > η ≥ 0.

Definition 2.2 ([16]). The left-sided Riemann-Liouville fractional integral of order
α > 0 is defined by

Iαaf(τ) :=
1

Γ(α)

∫ τ

a

f(s)

(τ − s)1−α ds, τ > a,

provided that the right hand side exists.

Definition 2.3 ([16]). The left-sided Riemann-Liouville fractional derivative of
order α ≥ 0, n− 1 ≤ α < n, n = −[−α], is defined by

Dα
af(τ) = DnIn−αa f(τ) = (

d

dτ
)nIn−αa f(τ)

=
1

Γ(n− α)

( d
dτ

)n ∫ τ

a

f(s)

(τ − s)α−n+1
ds, τ > a.

In particular, when α = n we have Dα
af = Dnf , and when α = 0, D0

af = f .

Definition 2.4 ([16]). The left-sided Caputo fractional derivative of order α ≥ 0,
n− 1 ≤ α < n, n = −[−α], is defined by

CDα
af(τ) = In−αa f (n)(τ), τ > a.

The fractional integral and fractional derivative of power functions have the same
effect as the integer-order integral and derivative. Namely,

Lemma 2.5 ([16]). If β > 0 and α ≥ 0, then

Iαa (τ − a)β−1 =
Γ(β)

Γ(β + α)
(τ − a)α+β+1, α > 0, τ > a,

CDα
a (τ − a)β−1 =

Γ(β)

Γ(β − α)
(τ − a)β−α−1, α ≥ 0, τ > a.

If β = 1, then (CDα
a1)(τ) = 0, τ > a.



4 M. D. KASSIM, N. E. TATAR EJDE-2020/111

The Riemann-Liouville fractional integral (Definition 2.2) satisfies the following
semigroup property.

Lemma 2.6. [16]] Let 0 ≤ η < 1, α > 0 and β > 0. If h ∈ Cη[a, b], then

IβaI
α
ah(τ) = Iβ+α

a h(τ), τ > a.

The following result provides another composition of the fractional integration
operator Iαa with the fractional differentiation operator Dα

a .

Lemma 2.7 ([16]). Let α > 0, 0 ≤ η < 1, n = −[−α]. If h ∈ Cη[a, b] and
In−αa h ∈ Cnη [a, b], then

IαaD
α
ah(τ) = h(τ)−

n∑
i=1

(Dn−iIn−αa h)(a)

Γ(α− i+ 1)
(τ − a)α−i, τ > a.

Lemma 2.8 ([16]). Let α > 0, n = −[−α]. If h ∈ Cn[a, b] or h ∈ ACn[a, b], then

Iαa
CDα

ah(τ) = h(τ)−
n−1∑
k=0

h(k)(a)

k!
(τ − a)k, τ > a.

Lemma 2.9. Let 0 < β ≤ α < 1. If h ∈ AC[a, b], then

CDβ
0h = Iα−β0

CDα
0h.

Proof. From Definition 2.4 and Lemma 2.6, we have

CDβ
0h = I1−β

0 h′ = Iα−β0 I1−α
0 h′ = Iα−β0

CDα
0h.

�

Lemma 2.10 ([12]). Let f ∈ L1(0,∞). Then

lim
τ→∞

1

τα
Iα+1

0 f(τ) =
1

Γ(α+ 1)

∫ ∞
0

f(s)ds =
1

Γ(α+ 1)
I1

0f(∞), α > 0.

Lemma 2.11. Let 0 < α < 1 and 0 ≤ η < 1. Assume that x ∈ AC[0,∞) and
I1−α

0 x′ ∈ C1
η [0,∞). Then

lim
τ→∞

x(τ)

τα
= lim
τ→∞

CDα
0x(τ)

Γ(1 + α)
. (2.1)

Proof. Since x ∈ AC[0,∞) ⊂ Cη[0,∞) and I1−α
0 x′ ∈ C1

η [0,∞), we can use Lemma
2.7 to obtain

Iα0D
α
0x
′(τ) = x′(τ)− I1−α

0 x′(0)

Γ(α)
τα−1, τ > 0. (2.2)

Applying I1
0 to both sides of (2.2), using Lemma 2.5 (with β = α) and Lemma 2.6,

we obtain

x(τ) = x(0) +
I1−α

0 x′(0)

Γ(α+ 1)
τα + I1+α

0 Dα
0x
′(τ), τ > 0. (2.3)

Dividing both sides of (2.3) by τα, we obtain

x(τ)

τα
=
x(0)

τα
+

I1−α
0 x′(0)

Γ(α+ 1)
+

1

τα
I1+α

0 Dα
0x
′(τ), τ > 0.

Next, we take the limit as τ →∞, we arrive at

lim
τ→∞

x(τ)

τα
=

I1−α
0 x′(0)

Γ(1 + α)
+

1

Γ(1 + α)
lim
τ→∞

I1
0D

α
0x
′(τ), (2.4)



EJDE-2020/111 FRACTIONAL DIFFERENTIAL EQUATIONS 5

where we used Lemma 2.10. Moreover, we conclude that

I1
0D

α
0x
′(τ) = I1

0DI1−α
0 x′(τ) = I1−α

0 x′(τ)−I1−α
0 x′(0) = CDα

0x(τ)−CDα
0x(0), τ > 0,

(2.5)
and (2.1) follows directly from (2.4) and (2.5). �

3. Some useful inequalities

First, we define the following special classes of functions

Hk = {h ∈ L1(0,∞) : h is positive and skh ∈ L1(1,∞), k > −1}, (3.1)

M =
{
F : (0,∞)× R+ → R+ where 0 ≤ F (τ, s)− F (τ, r) ≤ H(τ)(s− r),
for some continuous function H on R+, s ≥ r ≥ 0 and τ > 0

}
.

(3.2)

Φ =
{
ϕ ∈ C(0,∞) : ϕ is nondecreasing and positive on (0,∞),

1

v
ϕ(w) ≤ ϕ(

w

v
), w > 0, v ≥ 1

}
.

(3.3)

The proofs of the following lemmas are based on an application of the Bihari in-
equality which is a generalization of the Gronwall inequality.

Lemma 3.1 ([12]). Let g(τ) and z(τ) be nonnegative continuous functions defined
for τ ≥ 0, ϕ ∈ Φ and ci ∈ R, i = 1, 2, 3. Then

z(τ) ≤ c1 + c2τ
γ + c3τ

γ

∫ τ

0

g(s)ϕ(z(s))ds, τ, γ ≥ 0, (3.4)

implies

z(τ) ≤

{
E−1(E(|c1|+ |c2|) + |c3|

∫ τ
0
g(s)ds), 0 ≤ τ < 1

τγE−1(E(A) + |c3|
∫ τ

1
sγg(s)ds), τ ≥ 1,

(3.5)

where

A = |c1|+ |c2|+ |c3|ϕ(E−1(C))

∫ 1

0

g(s)ds,

C = E(|c1|+ |c2|) + |c3|
∫ 1

0

g(s)ds <∞,

and E−1 is the inverse function of

E(ξ) =

∫ ξ

ξ0

ds

ϕ(s)
.

Lemma 3.2 ([12]). Let z(τ) satisfy

z(τ) ≤ c1τγ + c2τ
γ

∫ τ

0

[F1(s, z(s) + c3) + F2(s, z(s) + c4) + h(s)]ds, τ ≥ 0, (3.6)

where h : C[R+,R+], Fj ∈M , j = 1, 2 and γ, ci > 0, i = 1, 2, 3, 4. Then

z(τ) ≤ τγf(τ), τ > 0, (3.7)

where

f(τ) =
(
c1 + c2

∫ τ

0

[F1(s, c3) + F2(s, c4) + h(s)]ds
)

× exp
(
c2

∫ τ

0

sγ [N1(s) +N2(s)]ds
)
, τ > 0,

(3.8)



6 M. D. KASSIM, N. E. TATAR EJDE-2020/111

with N1 and N2 are as in the definition of M corresponding to F1 and F2, respec-
tively.

Remark 3.3. If p, q > 1 and 1
p+ 1

q = 1, then for α > 0, p(α−1)+1 > 0⇐⇒ qα > 1.

Lemma 3.4 ([13]). If υ, λ + 1 > 1/r, for some r > 1, and g is a nonnegative
continuous function defined on [0,∞), then∫ τ

0

(τ − s)υ−1sλg(s)ds ≤ Cτυ+λ−1/r
(∫ τ

0

gr(s)ds
)1/r

, τ > 0, (3.9)

where

C = Kp(υ−1)+1,pλ =
Γ(pλ+ 1)Γ(p(υ − 1) + 1)

Γ(pλ+ p(υ − 1) + 2)
,

1

p
+

1

r
= 1.

Lemma 3.5 ([13]). Let h and z be nonnegative continuous functions defined on
[0,∞). Let ϕi(z) > 0 on (0,∞), i = 1, 2, and ϕi(z) are continuous nondecreasing
functions defined on [0,∞). If

z(τ) ≤ K1 +K2(

∫ τ

0

hq(s)ϕq1(z(s))ϕq2(z(s))ds)1/q, q > 1, τ > 0, (3.10)

where Ki ∈ R+, i = 1, 2, then

z(τ) ≤
[
E−1

(
E(2q−1K1) + 2q−1K2

∫ τ

0

hq(s)ds
)]1/q

, τ > 0,

where E−1 is the inverse of

E(ξ) =

∫ ξ

ξ0

ds

ϕq1(s1/q)ϕq2(s1/q)
, ξ > ξ0 > 0.

4. Source without fractional derivatives

We consider the asymptotic behavior of solutions of the equation

(CDα
0x)′(τ) = f(τ, x(τ)), 0 < α < 1, τ ≥ 0, (4.1)

subject to
x(0) = b1,

CDα
0x(τ)|τ=0 = b2, (4.2)

in the space

Cα,11−α[0,∞) = {x ∈ AC[0,∞), (CDα
0x)′ ∈ C1−α[0,∞)}. (4.3)

We assume the following conditions:

(C1) f(τ, x) ∈ C[[0,∞) × R,R] is such that f(·, x(·)) ∈ C1−α[0,∞) for any
x ∈ AC[0,∞).

(C2) There are continuous functions P,ϕ : [0,∞)→ [0,∞) such that

|f(τ, x(τ))| ≤ ϕ(|x(τ)|)P (τ), τ ≥ 0, (4.4)

where ∫ ∞
1

sαP (s)ds <∞, (4.5)

and ϕ ∈ Φ.

Theorem 4.1. Suppose f satisfies (C1), (C2) and x ∈ AC[0,∞) is a solution of
(4.1)-(4.2). Then

lim
τ→∞

x(τ)

τα
= a ∈ R, as τ →∞.



EJDE-2020/111 FRACTIONAL DIFFERENTIAL EQUATIONS 7

Proof. Integrating both sides of (4.1), we find

CDα
0x(τ) = b2 +

∫ τ

0

f(s, x(s))ds = b2 + I1
0f(τ, x(τ)). (4.6)

Applying Iα0 to both sides of (4.6), and using Lemmas 2.8 and 2.5, we arrive at

x(τ) = b1 +
b2

Γ(α+ 1)
τα +

1

Γ(α+ 1)

∫ τ

0

(τ − s)αf(s, x(s))ds, τ ≥ 0. (4.7)

By using (4.4) we obtain

|x(τ)| ≤ |b1|+
|b2|

Γ(α+ 1)
τα +

1

Γ(α+ 1)
τα
∫ τ

0

P (s)ϕ(|x(s)|)ds, τ ≥ 0. (4.8)

Applying Lemma 3.1 to (4.8) we obtain

|x(τ)| ≤

{
E−1

(
E(|b1|+ |b2|

Γ(α+1) ) + 1
Γ(α+1)

∫ τ
0
P (s)ds

)
, 0 ≤ τ < 1

ταE−1
(
E(A) + 1

Γ(α+1)

∫ τ
1
sαP (s)ds

)
, τ ≥ 1,

where

A = |b1|+
|b2|

Γ(α+ 1)
+

1

Γ(α+ 1)
ϕ(E−1(K))

∫ 1

0

P (s)ds,

K = E(|b1|+
|b2|

Γ(α+ 1)
) +

1

Γ(α+ 1)

∫ 1

0

P (s)ds <∞.

From (4.5) and the continuity of P on R+, we see that

|x(τ)| ≤

{
C1, 0 ≤ τ < 1,

ταC2, τ ≥ 1,
(4.9)

with

C1 = E−1
(
E
(
|b1|+

|b2|
Γ(α+ 1)

)
+

1

Γ(α+ 1)

∫ 1

0

P (s)ds
)
<∞,

C2 = E−1
(
E(A) +

1

Γ(α+ 1)

∫ ∞
1

sαP (s)ds
)
<∞.

Next, it is clear that∫ τ

0

|f(s, x(s))|ds ≤
∫ τ

0

P (s)ϕ(|x(s)|)ds

≤
∫ 1

0

P (s)ϕ(|x(s)|)ds+

∫ τ

1

P (s)ϕ(|x(s)|)ds

≤
∫ 1

0

P (s)ϕ(|x(s)|)ds+

∫ τ

1

sαP (s)ϕ
( |x(s)|
sα

)
ds, τ > 0.

(4.10)

By (4.9) and (4.10), we have

lim
τ→∞

∫ τ

0

f(s, x(s))ds <∞. (4.11)

On the other hand, integrating (4.1) yields

CDα
0x(τ) = b2 +

∫ τ

0

f(s, x(s))ds, τ > 0. (4.12)



8 M. D. KASSIM, N. E. TATAR EJDE-2020/111

From (4.11) and (4.12), we conclude

lim
τ→∞

CDα
0x(τ) = c, c ∈ R.

Further, by Lemma 2.11, we can write

lim
τ→∞

x(τ)

τα
= lim
τ→∞

CDα
0x(τ)

Γ(α+ 1)
= a,

for some real number a. �

Example 4.2. All solutions of

(CDα
0x)′(τ) = e−τxr(τ), 0 < α, r ≤ 1, τ > 0. (4.13)

satisfy limτ→∞
x(τ)
τα = a, as τ →∞, for some real number a.

To prove this claim, let ϕ(τ) = τ r and P (τ) = e−τ . Then∫ ∞
1

sαP (s)ds ≤
∫ ∞

0

sαe−sds = Γ(α+ 1) <∞.

Obviously ϕ is a nondecreasing and positive function with

uϕ(v) = uvr ≤ (vu)r = ϕ(vu), u ≥ 1, v > 0,

and ∫ ∞
0

ds

ϕ(s)
=

∫ ∞
0

ds

sr
=∞.

Then ϕ ∈ Φ. All the conditions of Theorem 4.1 are satisfied, therefore every solution

x of (4.13) has satisfy limτ→∞
x(τ)
τα = a, , a ∈ R, as τ →∞.

5. Equations with fractional source terms

We study problem (1.1) in the space Cα,11−α[0,∞) defined in (4.3) with the fol-
lowing assumptions:

(C3) f(τ, v, w) : [0,∞)×R2 → R is so that f(·, v(·), w(·)) ∈ C1−α[0,∞) for every
v, w ∈ AC[0,∞).

(C4)

|f(τ, u(τ), v(τ))| ≤ F1(τ, |u(τ)|) + F2(τ, τβ |v(τ)|), τ ≥ 0, (5.1)

where Fi ∈M , i = 1, 2.

Lemma 5.1. Suppose that f satisfies (C3), (4) and x ∈ AC[0,∞) is a solution of
(1.1). Then

max
{
|x(τ)|, τβ |CDβ

0x(τ)|
}
≤ |b1|+ z(τ), τ > 0, (5.2)

where

z(τ) = C2τ
α + C3τ

α

∫ τ

0

[F1(s, |x(s)|) + F2(s, τβ |CDβ
0x(s)|)]ds, τ > 0,

C3 = max
{ 1

Γ(α+ 1)
,

1

Γ(α− β + 1)

}
, C2 = |b2|C3.

(5.3)



EJDE-2020/111 FRACTIONAL DIFFERENTIAL EQUATIONS 9

Proof. Applying I1
0 to (1.1), we obtain

CDα
0x(τ) = b2 + I1

0f
(
τ, x(τ),CDβ

0x(τ)
)

= b2 +

∫ τ

0

f
(
s, x(s),CDβ

0x(s)
)
ds, τ > 0.

(5.4)

Next, we apply Iα0 to both sides of (5.4), using Lemmas 2.6, 2.8 and 2.5, we find

x(τ) = b1 +
b2

Γ(α+ 1)
τα + I1+α

0 f
(
τ, x(τ),CDβ

0x(τ)
)

= b1 +
b2

Γ(α+ 1)
τα +

1

Γ(α+ 1)

∫ τ

0

(τ − s)αf(s, x(s),CDβ
0x(s))ds,

(5.5)

for τ > 0. Thus, from (5.5) and (5.1) we have

|x(τ)| ≤ |b1|+
|b2|

Γ(α+ 1)
τα +

τα

Γ(α+ 1)

∫ τ

0

∣∣f(s, x(s),CDβ
0x(s)

)∣∣ds
≤ |b1|+ C2τ

α + C3τ
α

∫ τ

0

(
F1(s, |x(s)|) + F2

(
s, sβ |CDβ

0x(s)|
))
ds,

(5.6)

for τ > 0. By Lemma 2.9, we see that

CDβ
0x(τ) = Iα−β0 (CDα

0x(τ)). (5.7)

Let us insert the expression (5.4) into (5.7), using Lemmas 2.6 and 2.5, we have

CDβ
0x(τ) = Iα−β0

(
b2 + I1

0f
(
s, x(s),CDβ

0x(s)
))

(τ)

=
b2

Γ(α− β + 1)
τα−β + Iα−β+1

0 f
(
τ, x(τ),CDβ

0x(τ)
)

=
b2

Γ(α− β + 1)
τα−β

+
1

Γ(α− β + 1)

∫ τ

0

(τ − s)α−βf(s, x(s),CDβ
0+x(s))ds, τ > 0.

Then from this and (5.1) we obtain the bound

τβ |CDβ
0x(τ)|

≤ C3|b2|τα + C3τ
α

∫ τ

0

|f(s, x(s),CDβ
0+x(s))|ds

≤ C2τ
α + C3τ

α

∫ τ

0

(
F1(s, |x(s)|) + F2(s, sβ |CDβ

0x(s)|)
)
ds, τ > 0.

(5.8)

Relation (5.2) follows directly from (5.3), (5.6) and (5.8). �

Theorem 5.2. Suppose that f satisfies (C3)-(C4) and∫ ∞
0

sαNi(s)ds <∞,
∫ ∞

0

Fi(s, |b1|)ds <∞, i = 1, 2. (5.9)

Then, every solution x(τ) of problem (1.1) has the following property

lim
τ→∞

x(τ)

τα
= a, a ∈ R.



10 M. D. KASSIM, N. E. TATAR EJDE-2020/111

Proof. By using Lemma 5.1 we have

F1(τ, |x(τ)|) ≤ F1(τ, |b1|+ z(τ)), τ > 0, (5.10)

F2

(
τ, τβ |CDβ

0x(τ)|
)
≤ F2(τ, z(τ) + |b1|), τ > 0. (5.11)

Taking into account (5.3), (5.10) and (5.11) we arrive at

z(τ) ≤ C2τ
α + C3τ

α

∫ τ

0

[F1(s, |b1|+ z(s)) + F2(s, z(τ) + |b1|)]ds, τ > 0.

Then, by Lemma 3.2, we find that

z(τ) ≤ Cτα, τ > 0 (5.12)

where

C =
(
C2 + C3

∫ ∞
0

[F1(s, |b1|) + F2(s, |b1|)]ds
)

× exp(C3

∫ ∞
0

sγ [N1(s) +N2(s)]ds) <∞.

It follows from Lemma 5.1 and (5.12) that

|x(τ)| ≤ |b1|+ Cτα, τβ |CDβ
0x(τ)| ≤ |b1|+ Cτα, τ > 0. (5.13)

Again by hypothesis (5.1) we have

|
∫ τ

0

f(s, x(s),CDβ
0x(s))ds| ≤

∫ τ

0

|f(s, x(s),CDβ
0x(s))|ds

≤
∫ τ

0

[F1(s, |x(s)|) + F2(s, sβ |CDβ
0x(s)|)]ds,

for τ > 0. From this inequality and (5.13), we obtain

|
∫ τ

0

f(s, x(s),CDβ
0x(s))ds|

≤
∫ τ

0

[F1(s, |b1|+ Csα) + F2(s, |b1|+ Csα)]ds

=

∫ τ

0

{
F1(s, |b1|+ Csα)− F1(s, |b1|) + F1(s, |b1|)

+ F2(s, |b1|+ Csα)− F2(s, |b1|) + F2(s, |b1|)
}
ds, τ > 0.

As the functions Fi, i = 1, 2 are in M , we obtain∣∣ ∫ τ

0

f(s, x(s),CDβ
0x(s))ds

∣∣
≤ C

∫ τ

0

sα[N1(s) +Ns(s)]ds+

∫ τ

0

[F1(s, |b1|) + F2(s, |b1|)]ds <∞,
(5.14)

where we have used (5.9). Then

lim
τ→∞

∫ τ

0

f(s, x(s),CDβ
0x(s))ds <∞.

By (5.4) we conclude that there is b ∈ R such that limτ→∞
CDα

0x(τ) = b. Further,
by Lemma 2.11, we deduce that

lim
τ→∞

x(τ)

τα
= lim
τ→∞

CDα
0x(τ)

Γ(α+ 1)
= a,



EJDE-2020/111 FRACTIONAL DIFFERENTIAL EQUATIONS 11

and the proof is now complete. �

6. Boundedness

We consider the fractional differential problem (1.2) in the space

Cα[0,∞) =
{
x ∈ AC[0,∞) : CDα

0x ∈ C[0,∞)
}
. (6.1)

We assume the following conditions:

(C5) f : [0,∞) × R2 → R is so that f(·, v(·), w(·)) ∈ C[0,∞) for every v, w in
C[0,∞).

(C6)

|f(τ, u, v)| ≤ τγh(τ)ϕ1(|u(τ)|)ϕ2(|v(τ)|), τ > 0, (6.2)

where h, ϕ1, ϕ2 : R+ → R+ are continuous functions with ϕi, i = 1, 2, are
nondecreasing functions and h ∈ Lq(0,∞) for some q > 1

α−β , γ = 1
q − α.

Lemma 6.1. : Suppose that f satisfies (C5), (C6) and x ∈ AC[0,∞) is a solution
of (1.2). Then

max
{
|x(τ)|, |CDβ

0x(τ)|
}
≤ z(τ), τ ≥ τ0 > 0, (6.3)

where

z(τ) = |b|+K1

(∫ τ

0

hq(s)ϕq1(|x(s)|)ϕq2(|CDβ
0x(s)|)ds

)1/q

, τ > 0, (6.4)

and

K1 = max
{K1/p

1+p(α−1),pγ

Γ(α)
,
K

1/p
1+p(α−β−1),pγ

Γ(α− β)τβ0

}
,

Kα,β =
Γ(β + 1)Γ(α)

Γ(α+ β + 1)
,

1

p
+

1

q
= 1.

Proof. Applying Iα0 to (1.2) and taking into account Lemma 2.8, we have

x(τ) = b+
1

Γ(α)

∫ τ

0

(τ − s)α−1f(s, x(s),CDβ
0x(s))ds, τ > 0. (6.5)

Using the inequality (6.2), we obtain

|x(τ)| ≤ |b|+ 1

Γ(α)

∫ τ

0

(τ − s)α−1sγh(s)ϕ1(|x(s)|)ϕ2(|CDβ
0x(s)|)ds, (6.6)

for τ > 0. It follows from the assumptions β < α, q > 1
α−β and γ = 1

q − α that

p(α − 1) + 1 ≥ p(α − β − 1) + 1 > 0 and pγ + 1 = p( 1
q − α) + 1 = p(1 − α) > 0.

Then, we apply Lemma 3.4, to obtain

|x(τ)| ≤ |b|+ 1

Γ(α)
K

1/p
p(α−1)+1,pγτ

α+γ−1/q
(∫ τ

0

hq(s)ϕq1(|x(s)|)ϕq2(|CDβ
0x(s)|)ds

)1/q

≤ |b|+K1

(∫ τ

0

hq(s)ϕq1(|x(s)|)ϕq2(|CDβ
0x(s)|)ds

)1/q

, τ > 0.

(6.7)



12 M. D. KASSIM, N. E. TATAR EJDE-2020/111

Also, by Lemma 2.9, we conclude that

CDβ
0x(τ) = Iα−β0

CDαx(τ) =
1

Γ(α− β)

∫ τ

0

(τ − s)α−β−1 CDα
0x(s)ds

=
1

Γ(α− β)

∫ τ

0

(τ − s)α−β−1f(s, x(s),CDβ
0x(s))ds, τ > 0.

(6.8)

In view of (6.2), we have

|CDβ
0x(τ)| ≤ 1

Γ(α− β)

∫ τ

0

(τ − s)α−β−1sγh(s)ϕ1(|x(s)|)ϕ2(|CDβ
0x(s)|)ds,

for τ > 0. Again, from Lemma 3.4, we find

|CDβ
0x(τ)|

≤
K

1/p
p(α−β−1)+1,pγ

Γ(α− β)
τα−β+γ−1/q

(∫ τ

0

hq(s)ϕq1(|x(s)|)ϕq2(|CDβ
0x(s)|)ds

)1/q

≤
K

1/p
p(α−β−1)+1,pγ

Γ(α− β)
τ−β

( ∫ τ

0

hq(s)ϕq1(|x(s)|)ϕq2(|CDβ
0x(s)|)ds

)1/q

≤ K1(

∫ τ

0

hq(s)ϕq1(|x(s)|)ϕq2(|CDβ
0x(s)|)ds)1/q, τ ≥ τ0 > 0.

(6.9)

Therefore (6.3) follows from (6.4), (6.7) and (6.9). �

Theorem 6.2. Assume that f satisfies (C5), (C6). Then, any solution x of (1.2)
satisfies

|x(τ)| ≤ C, |CDβ
0x(τ)| < C,

for some positive constant C, τ > 0, provided that∫ ∞
ξ0

ds

ϕq1(s1/q)ϕq2(s1/q)
=∞, ξ0 > 0.

Proof. In view of Lemma 6.1 we have

ϕ1(|x(τ)|) ≤ ϕ1(z(τ)), 2(|CDβ
0x(τ)|) ≤ ϕ2(z(τ)), τ > 0. (6.10)

From this inequality and (6.4), we obtain

z(τ) ≤ |b|+K1

(∫ τ

0

hq(s)ϕq1(z(s))ϕq2(z(s))ds
)1/q

, τ > 0. (6.11)

Therefore, Lemma 3.5 implies

z(τ) ≤
[
E−1

(
E(2q−1K1) + 2q−1K2

∫ τ

0

hq(s)ds
)]1/q

<∞,

because h ∈ Lq(0,∞). This completes the proof. �

Example 6.3. Consider the problem

CD
2/3
0 x(τ) = τ1/q−2/3e−λτ (x(τ))3/5

(C
D

1/3
0 x(τ)

)1/3(
cos(CD

1/3
0 x)

)
τ > 0,

x(0) = b, q > 3, λ > 0.

(6.12)



EJDE-2020/111 FRACTIONAL DIFFERENTIAL EQUATIONS 13

Let ϕ1(τ) = τ3/5, ϕ2(τ) = τ1/3 and h(τ) = e−λτ , γ = 1/q − 2/3. Then h ∈
Lq(0,∞) and∫ ∞

ξ0

ds

ϕq1(s
1
q )ϕq2(s1/q)

=

∫ ∞
ξ0

ds

s3/5s1/3
=

∫ ∞
ξ0

ds

s14/15
=∞.

Then, by Theorem 6.2, we deduce that any solution x of (6.12) satisfies

|x(τ)| ≤ C, |CDβ
0x(τ)| < C,

for α = 2/3, β = 1/3, and τ > 0.

Acknowledgements. M. D. Kassim wants to thank Imam Abdulrahman Bin
Faisal University for its support and facilities. N. E. Tatar is grateful for the
financial support and the facilities provided by King Fahd University of Petroleum
and Minerals through project number IN181008.

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Mohammed Dahan Kassim
Department of Basic Engineering Sciences, College of Engineering, Imam Abdulrahman

Bin Faisal University, P.O. Box 1982, Dammam 31441, Saudi Arabia

Email address: mdkassim@iau.edu.sa

Nasser Eddine Tatar
King Fahd University of Petroleum and Minerals, Department of Mathematics and
Statistics, Dhahran, 31261, Saudi Arabia

Email address: tatarn@kfupm.edu.sa


	1. Introduction
	2. Preliminaries
	3. Some useful inequalities
	4. Source without fractional derivatives
	5. Equations with fractional source terms
	6. Boundedness
	Acknowledgements

	References