Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 128, pp. 1–12.

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

CONTINUABILITY OF SOLUTIONS TO FRACTIONAL

DIFFERENTIAL EQUATIONS

MIROSLAV BARTUŠEK

Abstract. This article concerns the Caputo fractional differential equation

cDαa x
[n−1](t) = f(t, x(t)) + e(t), n ≥ 2

where x[n−1] is the quasiderivative of x of order (n−1) and cDαa is the Caputo
derivative of the order α ∈ (0, 1). We study the continuability and noncon-

tinuability of solutions.

1. Introduction

We consider the fractional differential equation
cDα

ax
[n−1](t) = f

(
t, x(t)

)
+ e(t) (1.1)

where a > 1, α ∈ (0, 1), n ≥ 2 is an integer, cDα
au(t) is the Caputo derivative of

order α, defined as

cDα
au(t) :=

1

Γ(1− α)

∫ t

a

(t− s)−αu′(s) ds ,

Γ(x) =

∫ ∞
0

sx−1e−sds , x > 0

(1.2)

is the Gamma function and u[i], i = 0, . . . , n − 1 are quasiderivatives of u defined
as

u[0](t) = u(t) , u[i](t) = ai(t)
(
u[i−1](t)

)′
, i = 1, . . . , n− 1 . (1.3)

Let [a, b] ⊂ [a,∞), and AC[a, b] the set of all functions defined on [a, b] that are
absolutely continuous on [a, b].

Let [a, b) ⊂ [a,∞). Then we denote by ACloc[a, b) the set of all functions defined
on [a, b) that are absolutely continuous on every compact subinterval of [a, b).

In the reminder of this article we assume the following:

(H1) ai : [a,∞)→ (0,∞) are continuous functions for i = 1, . . . , n− 1;
(H2) e : [a,∞)→ R = (∞,∞);
(H3) f : [a,∞)× R→ R is continuous.

Note that
x[n−1](t) = an−1(t)

(
an−2(t)(. . . (a1x

′(t))′ . . .
)′
.

In some places, the following assumptions will be used:

2010 Mathematics Subject Classification. 26A33, 34A08.
Key words and phrases. Caputo fractional equations; continuability; noncontinuability;
quasiderivatives.
c©2020 Texas State University.

Submitted September 15, 2019. Published December 22, 2020.

1



2 M. BARTUŠEK EJDE-2020/128

(H4) There exist continuous functions r : [a,∞) → R+ = [0,∞) and ω : R+ →
R+ such that ω(x) > 0 for x > 0, ω is nondecreasing and∣∣f(t, x)

∣∣ ≤ r(t)ω(|x|) , ∀t ∈ [a,∞) , x ∈ R ;

(H5) e ∈ ACloc[a,∞), f(t, u) ∈ ACloc[a,∞) for any fixed u ∈ R,
f(t, u) ∈ ACloc(R) for any fixed t ∈ [a,∞).

The Caputo derivative given by (1.2) is the special case of Caputo derivative of
order α > 0, defined as

cDα
au(t) :=

1

Γ(m− α)

∫ t

a

(t− s)m−α−1u(m)(s) ds ,

where m is the smallest integer greater than or equal to α, see e.g. [4, 5, 7]. Frac-
tional differential equations have attract eda great attention in the last two decades
because of their importance in applications in areas of physics, chemistry, aerody-
namics, etc., see e.g. monographs [4, 5, 9] and the references therein.

There are a lot of papers devoted to the study of asymptotic behavior of solutions
of fractal differential equations, see e.g. [6, 7, 8, 9, 10, 12]. But results of forced
fractional differential equations are relatively scarece. Equation (1.1) is studied in
[7] (when n = 2 or n = 3 and a2 ≡ 1) where sufficient conditions for boundedness
of all non-oscillatory solutions are given.

A function x : [a, b) → R, b ≤ ∞ is said to be the solution of (1.1) if x[n−1] ∈
ACloc[a, b) and (1.1) is valid on [a, b). We will suppose that x is nonextendable to
the right, i.e., if b < ∞, then x cannot be defined at t = b. Solution x is said to
be continuable if b = ∞, otherwise it is said to be noncontinuable. A continuable
solution x is said to be proper if it is nontrivial in any neighbourhood of ∞.

In this article we study problem (1.1) with

x[i](a) = di , i = 0, . . . , n− 1, (1.4)

where di ∈ R, i = 0, . . . , n− 1.
Let (1.1), (1.4) have a solution x. We investigate whether or not, x is continuable.
When α = 1, then (1.1) is the ordinary differential equation (t ≥ a)

x[n](t) = f
(
t, x(t)

)
+ e(t) (1.5)

with x[n](t) =
(
x[n−1](t)

)′
. It is known that (1.5) can have noncontinuable solutions,

see [2, 8]. A special case of (1.5) is the equation

x′′(t) = r(t)h(x) (1.6)

where λ1 > 1, λ2 ∈ (0, 1), M > 0, r ∈ C0[a,∞), h ∈ C0(R), r(t) ≥ M
t2 for large t,

h(x)x > 0 for x 6= 0, ∣∣h(x)
∣∣ ≥ |x|λ1 for |x| ≥ 1 ,∣∣h(x)
∣∣ ≤ |x|λ2 for |x| < 1 .

Then, by [1, Lemma 4], equation (1.6) has no proper solution.
Some papers only study proper solutions of (1.1) because of their great impor-

tance. In this article, we study only the part corresponding to the continuability of
solutions to (1.1). However, the methods used here can be applied for other types
of Caputo differential equations.



EJDE-2020/128 CONTINUABILITY OF SOLUTIONS 3

Notation. We denote

r̄(t) = max
a≤s≤t

|r(s)| , ē(t) = max
a≤s≤t

|e(s)| , t ≥ a .

If x is a solution of (1.1) defined on [a, b) with b ≤ ∞, we put

x̄(t) = max
a≤s≤t

|x(s)| , t ∈ [a, b) .

Let 1 ≤ j ≤ i ≤ n− 1 be integers and t ∈ [a,∞). Then we put

Ji,j(t) =

∫ t

a

a−1
j (sj+1)

∫ sj+1

a

a−1
j+1(sj+2)

∫ sj+2

a

· · ·
∫ si

a

a−1
i (σ) dσ dsi . . . dsj+1 ,

Jj,i(t) ≡ 1 if j > i .

If i, j ∈ {0, 1, . . . }, i < j and ck ∈ R for i ≤ k ≤ j, then we put
∑i
k=j ck = 0.

2. Preliminaries

The following lemmas state some properties of Caputo fractional differential
equations . For this, we define the Riemann-Liouville fractional integral operator
of order α on L1[a, b), b ≤ ∞ by

Jαa g(t) :=
1

Γ(α)

∫ t

a

(t− s)α−1g(s) ds .

Let Dg(t) = d
dtg(t).

Lemma 2.1. Let a < b ≤ ∞. Then

(i) Jαa maps ACloc[a, b) to ACloc[a, b).
(ii) If g ∈ ACloc[a, b), then J1−α

a Jαa g = J1
ag, and

cDα
a g(t) = DJ1−α

a

[
g(t)− g(a)

]
, t ∈ [a, b) .

(iii) If g ∈ ACloc[a, b), then

Jαa
cDα

a g(t) = g(t)− g(a) , t ∈ [a, b) .

For the proof of (i), see [10, Lemma 2.3]. For (ii), see [5, Theorem 2.2, Definition
3.2 and Lemma 2.11]. For (iii), see [5, Theorem 3.8].

Lemma 2.2. (i) Let x be a solution of (1.1). Then it is the solution of the nonlinear
Volterra type integral equation (t ≥ a)

x[n−1](t) = x[n−1](a) +
1

Γ(α)

∫ t

a

(t− s)α−1
[
f(s, x(s)) + e(s)

]
ds . (2.1)

Let (H5) be valid. Then equation (1.1) is equivalent to (2.1), i.e. every function x,
defined on [a, b), b ≤ ∞ such that x[n−1] ∈ AC1

loc[a, b) is the solution of (1.1) if,
and only if it is the solution of (2.1).

(ii) Let a solution x of (1.1) be defined on [a, b), b <∞. If

lim sup
t→b−

n−1∑
i=0

|x[i](t)| =∞ (2.2)

then it is noncontinuable. If (H5) holds and x is noncontinuable then (2.2) holds
and

lim
t→b−

x̄(t) =∞ . (2.3)



4 M. BARTUŠEK EJDE-2020/128

Proof. (i) Let x be a solution of (1.1) on [a, b), b ≤ ∞. Then x[n−1] ∈ ACloc[a, b)
and according to Lemma 2.1(iii) (with g = x[n−1])

x[n−1](t)− x[n−1](a) = Jαa
cDα

ax
[n−1](t) = Jαa

(
f
(
t, x(t)

)
+ e(t)

)
;

hence, (2.1) is valid.
Let (H5) hold and x be a solution of (2.1). Then x ∈ C1[a, b) and according to

(H5), f(t, x(t)) + e(t) ∈ ACloc[a, b). Using Lemma 2.1(i), Jαa
(
f(t, x(t)) + e(t)

)
∈

ACloc[a, b). From this and (2.1), we have x[n−1] ∈ ACloc[a, b). Applying Lemma
2.1(ii) and (2.1), we have

cDα
ax

[n−1] = DJ1−α
a

(
x[n−1](t)− x[n−1](a)

)
= DJ1−α

a

( 1

Γ(α)

∫ t

a

(t− s)α−1
[
f(s, x(s)) + e(s)

]
ds
)

= DJ1−α
a Jαa

(
f(t, x(t)) + e(t)

)
= DJ1

a

(
f(t, x(t)) + e(t)

)
= f(t, x(t)) + e(t)

for t ∈ [a, b). Hence (1.1) holds.
(ii) If (2.2) holds then x is clearly noncontinuable. Let (H5) hold and let x

be a noncontinuable solution of (1.1) defined on [a, b), b < ∞. We prove (2.2).

So, suppose, on the contrary, that
∑n−1
i=0 |x[i](t)| is bounded on [a, b). From this

and from b < ∞, limt→b− x
[i](t) exist for i = 0, 1, . . . , n − 2. The existence of

limt→b− x
[n−1](t) follows from (2.1). So, the solution x of (2.1) can be extended

to t = b, x[i](b) := limt→b− x
[i](t), i = 0, 1, . . . , n − 1. Moreover, as x ∈ C1[a, b],(

f(t, x(t)+e(t)
)
∈ AC[a, b], according to part (i), x is the solution of (1.1) on [a, b].

This contradicts the noncontinuability of x proves statement (2.2).
If (2.3) does not hold then (2.1) implies x[n−1] is bounded on [a, b) and, hence,

x[i], i = 0, 1, . . . , n − 2 are bounded on [a, b) that contradicts (2.2). Thus, (2.3) is
valid. �

Because of Lemma 2.2(i), we will investigate (2.1) instead of (1.1) without men-
tion it. The proofs of the main results are based on the following lemmas.

Lemma 2.3. Let u : [a, b) → R, a < b ≤ ∞ be a function such that u[n−1] exists
on [a, b) and let ∣∣u[n−1](t)

∣∣ ≤ K(t) , t ∈ [a, b) (2.4)

where K is a nondecreasing, continuous function. Then

∣∣u[1](t)
∣∣ ≤ n−2∑

i=1

J2,i(t)
∣∣u[i](a)

∣∣+ J2,n−1(t)K(t) for t ∈ [a, b). (2.5)

Proof. If n = 2, then (2.5) follows from (2.4). Hence, suppose n ≥ 3. We prove
that ∣∣u[j](t)

∣∣ ≤ n−2∑
i=j

Jj+1,i(t)
∣∣u[i](a)

∣∣+K(t)Jj+1,n−1(t) (2.6)

for j = 1, 2, . . . , n− 2. Using (1.3) we have(
u[n−2](t)

)′
=

1

an−1(t)
u[n−1](t) .



EJDE-2020/128 CONTINUABILITY OF SOLUTIONS 5

From this and from (2.4), the integration implies∣∣u[n−2](t)− u[n−2](a)
∣∣ ≤ ∫ t

a

K(σ)

an−1(σ)
dσ ≤ K(t)Jn−1,n−1(t)

and (2.6) holds for j = n − 2. We apply mathematical induction. Suppose, that
(2.6) holds for j = n− 2, n− 3, . . . , k. Then, by (1.3),(

u[k](t)
)′

=
1

ak+1(t)
u[k+1](t)

and the integration on [a, t] implies∣∣u[k](t)− u[k](a)
∣∣ ≤ ∫ t

a

a−1
k+1(σ)|u[k+1](σ)| dσ

≤
∫ t

a

a−1
k+1(σ)

[ n−2∑
i=k+1

Jk+2,i(σ)|u[i](a)|+K(σ)Jk+2,n−1(σ)
]
dσ

≤
n−2∑
i=k+1

Jk+1,i(t)|u[i](a)|+K(t)Jk+1,n−1(t) .

Hence, (2.6) is valid for j = k. Now, (2.5) is given by (2.6) for j = 1. �

Lemma 2.4. Let (H4) hold and let x be a solution of (1.1) defined on [a, b), b ≤ ∞.
Then

x̄(t) ≤M1(t) +

∫ t

a

M2(s)ω
(
x̄(s)

)
ds (2.7)

for t ∈ [a, b), where

M1(t) =
∣∣x[0](a)

∣∣+

∫ t

a

a−1
1 (s)

[ n−2∑
i=1

J2,i(s)
∣∣x[i](a)

∣∣
+
(
|x[n−1](a)|+ ē(s)

αΓ(α)
(s− a)α

)
J2,n−1(s)

]
ds ,

M2(t) =
r̄(t)

αΓ(α)
a−1

1 (t)(t− a)αJ2,n−1(t) .

(2.8)

Proof. By (2.1) and (H4), we hve∣∣x[n−1](t)
∣∣ ≤ ∣∣x[n−1](a)

∣∣+
ē(t)

αΓ(α)
(t− a)α

+
1

Γ(α)

∫ t

a

(t− s)α−1 r(s)ω
(
|x(s)|

)
ds

≤
∣∣x[n−1](a)

∣∣+
ē(t)

αΓ(α)
(t− a)α +

r̄(t)

αΓ(α)
(t− a)αω

(
x̄(t)

)
.

(2.9)

Applying Lemma 2.3 for u = x, b = t and

K(t) =
∣∣x[n−1](a)

∣∣+
ē(t)

αΓ(α)
(t− a)α +

r̄(t)

αΓ(α)
(t− a)αω

(
x̄(t)

)
,

from (2.9) we obtain∣∣(x[0](t))′
∣∣ =
|x[1](t)|
a1(t)

≤ M̄1(t) +M2(t)ω
(
x̄(t)

)
(2.10)



6 M. BARTUŠEK EJDE-2020/128

with

M̄1(t) = a−1
1 (t)

{ n−2∑
i=1

J2,i(t)
∣∣x[i](a)

∣∣+
(∣∣x[n−1](a)

∣∣+
ē(t)

αΓ(α)
(t− a)α

)
J2,n−1(t)

}
.

Hence, using the first equality in (1.3), the integration of (2.10) on [a, τ), a < τ ≤ t
implies

|x(τ)| ≤M1(t) +

∫ t

a

M2(s)ω
(
x̄(s)

)
ds ,

or

x̄(t) ≤M1(t) +

∫ t

a

M2(s)ω
(
x̄(s)

)
ds .

Hence, (2.7) is valid. �

The following two lemmas are well known.

Lemma 2.5 ([11, Lemma 2.1]). Let k > 0, λ > 1, t0 ≥ 0 be constants, F be a con-
tinuous, nonnegative function on R+ and v be a continuous, nonnegative function
on R+ satisfying the inequality

v(t) ≤ k +

∫ t

t0

F (s)vλ(s) ds , t ≥ t0 . (2.11)

If

(λ− 1)kλ−1

∫ ∞
t0

F (s) ds < 1 (2.12)

then

v(t) ≤ k
(

1− (λ− 1)kλ−1f tt0F (s) ds
)− 1

λ−1

for t ≥ t0.

Lemma 2.6 ([8, Lemma 9.2]). Let k > 0, g > 0 be a continuous function on [t0, b),
b ≤ ∞ and ω(t) > 0 for t ≥ k be a continuous function such that

∫∞
k

ds
ω(s) = ∞.

Then for any continuous function x : [t0, b)→ R+ fulfilling

x(t) ≤ k +

∫ t

t0

g(s)ω
(
x(s)

)
ds , t ∈ [t0, b)

the estimation

x(t) ≤ Ω−1
(∫ t

t0

g(s) ds
)
, t ∈ [t0, b)

holds where Ω−1 is the inverse function to Ω(s) =
∫ s
k

dτ
ω(τ) .

Consider the auxilliary system of differential equations

y′i = bi(t)yi+1 , i = 1, . . . , n− 1 , y′n = F (t, y1) , (2.13)

where bi ∈ C0[a,∞), bi > 0 on [a,∞), i = 1, . . . , n− 1 and F ∈ C0([a,∞),R).
Furthermore, suppose y0 > 0, bn ∈ C0[a,∞), bn > 0, λ > 1, β ∈ {−1, 1} exist

such that

βF (t, u) ≥ bn(t)|u|λ for t ≥ a, βu > y0 . (2.14)



EJDE-2020/128 CONTINUABILITY OF SOLUTIONS 7

A solution {yi}u1 of (2.13), defined on [a, b) with b <∞, is called noncontinuable
if it can not be extended to t = b. In this case

lim sup
t→b−

u∑
i=1

|yi(t)| =∞ .

The following lemma states sufficient conditions for the existence of noncontinuable
solutions of (2.13) with (2.14).

Lemma 2.7. Suppose (2.14) holds.

(i) If t1 ∈ (a,∞), then (2.13) possesses a noncontinuable solution {yi}ui=1 that
is defined on a subinterval [a, b) ⊂ [a, t1) and

βyi(t) ≥ y0 for t ∈ [a, b) , i = 1, . . . , n .

(ii) Let δ > 0, µi ∈ R for i = 1, . . . , n,

bi(t) ≥ δtµi , i = 1, . . . , n

and let

µn + λ

n−1∑
i=1

(1 + µi) + 1 > 0 . (2.15)

Then any solution {yi}n1 of (2.13), satisfying the initial conditions

βyi(a) > y0 , i = 1, . . . , n ,

is noncontinuable.
(iii) Let

∫∞
a
bi(t) dt =∞ for i = 1, . . . , n. Then the statement in (ii) is valid.

The above lemma follows [2, Theorems 3, 4 (for l = n)] or [3, Theorems 1, 2, 3].

3. Continuable solutions

The first theorem gives a sufficient condition for all solutions of (1.1) be con-
tinuable. It is a generalization of well known theorem by Winter and Osgood [8]
for differential equations.

Theorem 3.1. Suppose (H4) and∫ ∞
1

dx

ω(x)
=∞ . (3.1)

Then every solution of (1.1) is continuable.

Proof. Suppose, on the contrary, that x is a noncontinuable solution of (1.1) defined
on [a, b). Then according to Lemma 2.2(ii), b <∞ and

lim
t→b−

x̄(t) =∞ . (3.2)

Lemma 2.4 implies

x̄(t) ≤M1(t) +

∫ t

a

M2(s)ω
(
x̄(s)

)
ds ≤M1(b) +M

∫ t

a

ω
(
x̄(s)

)
ds

on [a, b) where M1 and M2 are given by (2.8) and M = maxa≤s≤bM2(s). From
this, (3.2) and Lemma 2.6 (with t0 = a, k = M1(b), g(t) ≡ M , x(t) = x̄(t)) we
obtain ∫ ∞

a

dτ

ω(τ)
= lim
t→b−

∫ x̄(t)

a

dτ

ω(τ)
≤ lim
t→b−

∫ t

a

M ds = M(b− a) <∞ .



8 M. BARTUŠEK EJDE-2020/128

This contradicts (3.1) and proves that x is continuable. �

If (3.1) does not hold, then noncontinuable solutions may exist (see Theorem 3.3
below). The following theorem gives us a set of initial conditions under which
solutions are continuable.

Theorem 3.2. Let λ > 1, (H4) and (H5) hold with ω(x) = xλ for x ∈ R+ and let
x be a solution of (1.1) satisfying the initial conditions dj ∈ R,

x[j](a) = dj , j = 0, . . . , n− 1 . (3.3)

If

k :=|d0|+
∫ ∞
a

a−1
1 (s)

{ n−2∑
i=1

J2,i(s)|di|

+
(
|dn−1|+

ē(s)

αΓ(α)
(s− a)α

)
J2,n−1(s)

}
ds <∞ ,

(3.4)

and
(λ− 1)kλ−1

αΓ(α)

∫ ∞
a

r̄(s)(t− a)α

a1(t)
J2,n−1(t) dt < 1 , (3.5)

then x is continuable.

Proof. Let x be a solution of (1.1) with (3.3), (3.4) and (3.5). Suppose, on the
contrary, that x is noncontinuable and it is defined on [a, b), b <∞. Then according
to Lemma 2.2(ii)

lim
t→b−

x̄(t) =∞ . (3.6)

Lemma 2.4 implies

x̄(t) ≤M1(t) +

∫ t

a

M2(s) x̄λ(s) ds (3.7)

for t ∈ [a, b) where M1 and M2 are given by (2.8). As M1 is nondecreasing, (3.4)
implies k = M1(∞) is finite.

Let T ∈ [a, b) be fixed. We define

v(t) =

{
x̄(t) if t ∈ [a, T )

x̄(T ) if t > T .
(3.8)

Then with respect to (3.7),

v(t) ≤ k +

∫ t

a

M2(s)vλ(s) ds , t ∈ [a,∞) .

Now, according to Lemma 2.5 (with t0 = a, F = M2, condition (2.12) follows from
(3.5)) we have

v(t) ≤ k
(

1− (λ− 1)kλ−1

∫ ∞
a

M2(s) ds
)− 1

λ−1

=: k1 <∞

for t ≥ a. Hence, by (3.8),

x̄(t) ≤ k1 , t ∈ [a, T ] .

As T ∈ [a, b) is arbitrary, x̄(t) ≤ k1 for t ∈ [a, b). The contradiction with (3.6)
proves that x is continuable. �

The following two theorems give us sets of initial conditions for which the solu-
tions are noncontinuable.



EJDE-2020/128 CONTINUABILITY OF SOLUTIONS 9

Theorem 3.3. Let λ > 1, x0 > 0, β ∈ {−1, 1}, t1 > a and a continuous function
r : [a, t1]→ (0,∞) exist such that

βf(t, x) ≥ r(t)|x|λ for t ∈ [a, t1] , βx ≥ x0 ,

βe(t) ≥ −x
λ
0

2
r(t) for t ∈ [a, t1] .

(3.9)

Then there exists D > 0 such that any solution of (1.1) satisfying βx[i](a) ≥ D,
i = 0, . . . , n− 1 is noncontinuable.

Proof. Let β = 1. Consider the auxiliary differential equations

y[n] =
tα−1
1

2Γ(α)
r0|y(t)|λ sgn y(t) (3.10)

for t ∈ [a, t1), y[n](t) =
(
y[n−1](t)

)′
, r0 = mina≤t≤t1 r(t) > 0. This equation can be

transformed into

y′i =
1

ai(t)
yi+1 , i = 1, 2, . . . , n− 1 ,

y′n =
tα−1
1

2Γ(α)
r0|y1(t)|λ sgn y(t)

(3.11)

with yi = y[i−1], i = 1, 2, . . . , n. Then, according to Lemma 2.7(i) (with t1 =

t1, y0 = x0, bi(t) =
(
ai(t)

)−1
for i = 1, . . . , n − 1, bn =

tα−1
1

2Γ(α)r0), (3.11) has

a noncontinuable solution y defined on [a, b) ⊂ [a, t1) such that yi(t) ≥ x0 for
t ∈ [a, b). Denote by di = yi+1(a), i = 0, . . . , n− 1. Hence, (3.10) has the solution
y with the initial conditions

y[i](a) = di , i = 0, . . . , n− 1 (3.12)

and (3.11) implies all quasiderivatives are increasing. At the same time

lim sup
t→b

n−1∑
i=0

y[i](t) =∞ . (3.13)

Let x be a solution of (1.1) with the initial conditions

x[i](a) > di , i = 0, . . . , n− 1 . (3.14)

We denote by I the intervals where both functions y and x are defined. We prove
that

x[i](t) > y[i](t) , t ∈ I , i = 0, . . . , n− 1 . (3.15)

Because of the initial conditions (3.12) and (3.13), equation (3.15) is valid in a right
neigbourhood of a. Suppose, that it is not valid on the whole interval I. Then there
is a t2 ∈ I and an index j ∈ {0, . . . , n− 1} exist such that

x[j](t2) = y[j](t2) , x[i](t) > y[i](t) for t ∈ [a, t2) , (3.16)

i = 0, . . . , n−1. First, we prove that j 6= n−1. Using (3.9) and (3.16), for t ∈ [a, t2)
we have

x[n−1](t) > dn−1 +
1

Γ(α)

∫ t

a

(t− s)α−1
[
e(t) + f

(
s, x(s)

)]
ds

≥ dn−1 +
tα−1
1

Γ(α)

∫ t

a

[
− r(s)

2
xλ0 + r(s)xλ(s)

]
ds



10 M. BARTUŠEK EJDE-2020/128

≥ dn−1 +
tα−1
1

Γ(α)

∫ t

a

r(s)

2
xλ(s) ds

≥ dn−1 +
tα−1
1 r0

2Γ(α)

∫ t

a

yλ(s) ds = y[n−1](t) .

Hence, j ∈ {0, . . . , n− 2}. If w(t) = x[j](t)− y[j](t), then w(a) > 0, w(t2) = 0 and
there exists t3 ∈ (a, t2) such that w′(t3) < 0, i.e.,(

x[j](t3)− y[j](t3)
)′

=
1

aj+1(t3)

[
x[j+1](t3)− y[j+1](t3)

]
< 0 .

This contradicts (3.16) and implies (3.15) is valid. Now, according to (3.13), (3.15)
and Lemma 2.2(ii), x is noncontinuable. So the statement of the theorem holds
with D = max(d0, . . . , dn−1) + 1.

When β = −1, the proof is similar. �

Theorem 3.4. Let λ > 1, β ∈ {−1, 1}, x0 > 0 and let a continuous function
r : [a,∞)→ (0,∞) be such that

βf(t, x) ≥ r(t)|x|λ for t ∈ [a,∞), βx ≥ x0 ,

βe(t) ≥ −x
λ
0

2
r(t) for t ∈ [a,∞) .

Let one of the following two assumptions hold:

(i) Let Cj ∈ R+, λj ∈ R, j = 1, . . . , n be such that

ai(t) ≤ Citλi , i = 1, . . . , n− 1, r(t) ≥ Cntλu (3.17)

for t ≥ a and

λn > −1 + λ
[
1− α−

n−1∑
i=1

(1− λi)
]
. (3.18)

(ii) Let
∫∞
a
a−1
i (t) dt = ∞ for i = 1, . . . , n − 2,

∫∞
a
tα−1a−1

n−1(t) dt = ∞ and∫∞
a
r(t) dt =∞.

Then any solution x of (1.1) satisfying the initial conditions

βx[i](a) > x0a
1−α , i = 0, . . . , n− 2, βx[n−1](a) > x0

is noncontinuable.

Proof. (i) Let β = 1. Consider the auxilliary integro-differential equation

y[n−1](t) = y[n−1](a) +
tα−1

2Γ(α)

∫ t

a

r(s)|y(s)|λ sgn y(s) ds (3.19)

and its solution with the initial conditions

y[j](a) = dj > 0 , j = 0, . . . , n− 1 . (3.20)

This equation is equivalent to the system

y′j =
1

aj(t)
yj+1 , j = 1, . . . , n− 2 ,

y′n−1 =
1

an−1(t)
tα−1yn ,

y′n =
(1− α)dn−1

tα
+

1

2Γ(α)
r(t)|y1|λ sgn y1 >

1

2Γ(α)
r(t)|y1|λ sgn y1

(3.21)



EJDE-2020/128 CONTINUABILITY OF SOLUTIONS 11

with
yi = y[i−1] , i = 1, . . . , n− 1 , yn = t1−αy[n−1] . (3.22)

The solution y of (3.19) and (3.20), and the solution {yi}ni=1 of (3.21) with the
initial conditions

yi(a) = di−1 , i = 1, . . . , n− 1 , yn(a) = a1−αdn−1 (3.23)

satisfy (3.22). We apply Lemma 2.7(ii) to (3.21) and (3.23) with

y0 = x0a
1−α , bi(t) = a−1

i (t) , i = 1, . . . , n− 2 ,

bn−1(t) = tα−1a−1
n−1 , bn(t) =

1

2Γ(α)
r(t) , µi = −λi, i = 1, . . . , n− 2 ,

µn−1 = −λn−1 − 1 + α , µn = λn , δ = min
(
C−1

1 , . . . , C−1
n−1,

Cn
2Γ(α)

)
.

Note, by (3.17) and (3.18), condition (2.15) is valid. Now, Lemma 2.7(ii) implies
the solutions of (3.21) and (3.23) and of (3.19) and (3.20) are noncontinuable. The
rest of the proof is similar as the one of Theorem 3.3; only (3.17) has to be replaced
by

x[n−1](t) ≥ · · · ≥ dn−1 +
tα−1

2Γ(α)

∫ t

a

r(s)yλ(s) ds = y[n−1](t) .

If β = −1, the proof is similar.
(ii) The proof is similar, we use Lemma 2.7(iii) instead of Lemma 2.7(ii). �

4. Special case

Consider the special case of (1.1), (1.4) (for n = 2)

cDα
a

(
a1(t)x′

)
= r(t)|x|λ sgnx ,

x(a) = d0 , x[1](a) = d1 ,
(4.1)

where λ > 0, d0 ∈ R, d1 ∈ R, r ∈ C[a,∞), a1 ∈ C[a,∞) and a1(t) > 0 for t ≥ a.

Corollary 4.1.

(i) If λ ≤ 1, then any solution of (4.1) is continuable.
(ii) Let λ > 1 and r > 0 on [a,∞). Then there exists D > 0 such that any solu-

tion of (4.1) satisfying |d0| ≥ D, |d1| ≥ D and d0d1 > 0 is noncontinuable.
(iii) Let λ > 1, C1 > 0, C2 > 0, λ1 ∈ R, λ2 ∈ R, either λ2 > −1 + λ(λ1 − α) or

λ1 ≤ α, λ2 ≥ −1, and let

a1(t) ≤ C1t
λ1 , r(t) ≥ C2t

λ2 for t ≥ a . (4.2)

If d0d1 > 0, then any solution of (4.1) is noncontinuable.
(iv) Let λ > 1, r ∈ ACloc[a,∞), and d0, d1 be such that

k = |d0|+ |d1|
∫ ∞
a

a−1
1 (s) ds <∞

and
(λ− 1)kλ−1

αΓ(α)

∫ ∞
a

r̄(s)

a1(s)
(s− a)α ds < 1 . (4.3)

Then any solution x of (4.1) is continuable.

Proof. In cases (i), (ii), (iii) and (iv), the proofs follow from Theorems 3.1, 3.3, 3.4
and 3.2, respectively. In Theorem 3.4 we put x0 = 1

2 min(|d0|aα−1, |d1|). �



12 M. BARTUŠEK EJDE-2020/128

Note that cases (iii) and (iv) of Corollary 4.1 are not in a contradiction. Let
(4.2) be valid. If (iii) holds, then λ2 > −1 + λ(λ1 − α) is supposed. If (iv) is valid,
then according to (4.3) we have λ2 < −1 + λ1 − α. So, the relationships between
λ1 and λ2 are different in these two cases.

Acknowledgement. This research was supported by grant GA 17-03224S.

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Miroslav Bartušek

Department of Mathematics and Statistics, Masaryk University, 611 37 Brno, Czech
Republic

Email address: bartusek@math.muni.cz


	1. Introduction
	Notation

	2. Preliminaries
	3. Continuable solutions
	4. Special case
	Acknowledgement

	References