Oscillation theorems for three classes of conformable fractional differential equations

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R E S E A R C H

Open Access

Oscillation theorems for three classes of

conformable fractional differential equations

Limei Feng

1

_{and Shurong Sun}

1*

*_{Correspondence:} [email protected]

1_{School of Mathematical Sciences,}

University of Jinan, Jinan, P.R. China

Abstract

In this paper, we consider the oscillation theory for fractional diﬀerential equations. We obtain oscillation criteria for three classes of fractional diﬀerential equations of the forms

Tt0 αx(t) +

m

i=1

pi(t)x

(τ

i(t))= 0, t≥t0,

Tt0

α

(

r(t)(Tαt0

(

x(t) +p(t)x

(τ

(t)))) β

)

+q(t)xβ

(σ

(t))= 0, t≥t0, and

Tt0

α

(

r2Tαt0

(

r1

(

Tαt0y

)

β

))(

t) +p(t)(Tt0 αy(t))

β

+q(t)f

(

y

(

g(t)))= 0, t≥t0, whereTαdenotes the conformable diﬀerential operator of order

α, 0 <

α

≤1.

MSC: 34C10; 26A33; 65Q10

Keywords: Oscillation; Conformable fractional calculus; Diﬀerential equation

1 Introduction

Fractional differential equations have been of great interest recently. Apart from diverse areas of mathematics, fractional differential equations arise in rheology, dynamical pro-cesses in self-similar and porous structures, fluid flows, electrical networks, chemical physics, and many other branches of science.

The oscillation of fractional differential equations as a new research field has received significant attention, and some interesting results have already been obtained. We refer to [1–11] and the references therein. The definition of the fractional-order derivative used is either the Caputo or the Riemann–Liouville fractional-order derivative involving an integral expression and the gamma function. Because of the definition, the oscillation of these types of fractional equations cannot be studied by regular methods, for example, by the Riccati transformation. It can only be studied by transforming it into an integer-order equation. In 2012, Chen et al. [4] studied the oscillation behavior of the following fractional differential equation:

r(t)Dα_–η(t)–q(t)f

∞

t

(v–t)–αy(v)dv = 0 fort> 0,

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whereDα

–ydenotes the Liouville right-sided fractional derivative of orderα,

Dα_–y(t) := – 1

Γ(1 –α)

d dt

∞

t

(v–t)–α_y(v)dv _for

t∈R+:= (0,∞).

By the Riccati transformation the authors obtained some suﬃcient conditions.

Recently, Khalil et al. [12] introduced a new well-behaved definition of local fractional derivative, called the conformable fractional derivative, depending just on the basic limit definition of the derivative. This new theory is improved by Abdeljawad [13]. For recent results on conformable fractional derivatives, we refer the reader to [14–23]. This new def-inition satisfies formulas of the derivatives of the product and quotient of two functions and has a simpler chain rule. In addition to the definition of conformable fractional deriva-tive, a definition of conformable fractional integral, the Rolle theorem, and the mean value theorem for conformable fractional differentiable functions were given. These properties are more conducive to the study of the oscillation of fractional-order equations.

In fact, some works in this ﬁeld have shown the signiﬁcance of conformable fractional derivative. For example, [24] discusses the potential conformable quantum mechanics, [25] discusses the conformable Maxwell equations, and [26,27] show that the conformable fractional derivative models present good agreements with experimental data, but there are less oscillation results.

In the paper, we study oscillation criteria of conformable fractional diﬀerential equa-tions. Our main goal is to generalize the oscillatory criteria in [28–37] to the conformable fractional derivative. The three equations represent three classes of equations of diﬀerent orders. For example, in 2016, Akca et al. [33] studied the equation

x(t) +

m

i=1 pi(t)x

τi(t)

= 0, t≥0,

and obtained the following:

Theorem 1.1 Assume that0 <ς:=lim inft→∞τt(t) m

i=1pi(s)ds≤1_e and for some r∈N,

we have

lim sup

t→∞ t

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ>1 +lnλ0

λ0 ,

where h(t) = max1≤i≤mhi(t), hi(t) = sup0≤s≤tτi(s), a1(t,s) := exp{ t

s

m

i=1pi(ζ)dζ},

ar+1(t,s) :=exp{ t

s

m

i=1pi(ζ)ar(ζ,τi(ζ))dζ}, andλ0 is the smaller root of the equation eς λ₌_λ_._{Then the above equation oscillates.}

From this we can unify the oscillation theory of integral-order and fractional-order dif-ferential equations. Through the inequality principle, iterative sequences, and the Riccati transformation method this can be extended to the conformable fractional derivatives by Lemma2.2.

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Sect.2, we introduce some notation and deﬁnitions on conformable fractional integrals. In Sect.3, we present the main theorems onα-order equations. Section4is devoted to the oscillatory results on 2α-order equation. In Sect.5, we demonstrate the oscillatory results for 3α-order equations. In each section, we give examples to illustrate the signiﬁcance of the results.

2 Conformable fractional calculus

For the convenience of the reader, we give some background from fractional calculus the-ory. These materials can be found in the recent literature, see [12,13,23].

Deﬁnition 2.1 ([13]) The (left) fractional derivative of a functionf : [a,∞)→Rof order

α∈(0, 1] starting fromais deﬁned by

T_αaf(t) =lim ε→0

f(t+ε(t–a)1–α_{) –}_f_(t)

ε .

Whena= 0, we writeTα.

Note that iff is diﬀerentiable, then (Ta

αf)(t) = (t–a)1–αf(t).

Deﬁnition 2.2 ([13]) The left fractional integral of orderα∈(0, 1] starting atais deﬁned by

I_αaf(t) = t

a

f(x)dα(x,a) = t

a

(x–a)α–1_f_(x)_dx.

Deﬁnition 2.3 ([13]) Letf: [a,∞)→Rbe a continuous function, and letα∈(0, 1]. Then, for allt>a, we have

TαaIαaf(t) =f(t).

Deﬁnition 2.4 ([13]) Letf : (a,b)→Rbe let a diﬀerentiable function, and letα∈(0, 1]. Then, for allt>a, we have

IαaTαa(f)(t) =f(t) –f(a).

Proposition 2.1 ([13])Let f : (a,∞)→ ∞ →Rbe a twice diﬀerentiable function,and let 0 <α,β≤1be such that1 <α+β≤2.Then

T_αaT_βa(t) =T_αa₊_βf(t) + (1 –β)(t–a)–β_Ta

αf(t).

Proposition 2.2 ([23])Letα∈(0, 1],and let f and g beα-diﬀerentiable at a point t> 0on [a,∞).Then

(1) Ta

α(af +bg) =aTαa(f) +bTαa(g)for alla,b∈R, (2) Tαa(λ) = 0for all constant functionsf(t) =λ, (3) Ta

α(fg) =fTαa(g) +gTαa(f), (4) Ta

α(

f g) =

gTa α(f)–fTαa(g)

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(5) Ta

α(tn) =ntn–αfor alln∈R,and (6) Ta

α(f ◦g)(t) =f(g(t))Tαa(g)(t)forf diﬀerentiable atg(t).

Lemma 2.1 ([13])Let f,g: [a,b]→Rbe two functions such that fg is diﬀerentiable,and letα∈(0, 1].Then

b

a

f(x)T_αa(g)(x)dα(x,a) =fg

b

a

– b

a

g(x)T_αa(f)(x)dα(x,a).

Lemma 2.2 Let f : (t0,∞)→Rbe diﬀerentiable,and letα∈(0, 1].If Tαt0f(t) =M(t),then

for all t>s>t0,we have

f(t) –f(s) =It0 αM(t).

Proof We can conclude fromTt0

αf(t) =M(t) that

t–t0

t–s 1–α

Tαsf(t) =M(t),

that is,

Tαsf(t) =

t–t0

t–s

α–1 M(t).

Then applyingIαto the latter fromstot, we have

IαsTαsf(t) =Iαs

t–t0

t–s

α–1 M(t)

,

that is,

f(t) –f(s) =It0 αM(t).

The proof of Lemma2.2is complete.

3

α

-Order conformable fractional differential equations with ﬁnite nonmonotone delay arguments

In this section, we deal with the diﬀerential equations of the form

Tt0 αx(t) +

m

i=1 pi(t)x

τi(t)

= 0, t≥t0, (3.1)

whereTαdenotes the conformable diﬀerential operator of orderα∈(0, 1],pi(t), 1≤i≤m,

are nonnegative functions,τi(t), 1≤i≤m, are nonmonotone functions of positive real

numbers such that

τi(t)≤t, t≥t0, lim

t→∞τi(t) =∞, 1≤i≤m.

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Lemma 3.1 Assume that x(t)is an eventually positive solution of(3.1)and ar(t,s),r∈N+,

is deﬁned as

a1(t,s) =exp t

s

(ζ–t0)α–1

m

i=1

pi(ζ)dζ

,

ar+1(t,s) =exp t

s

(ζ–t0)α–1

m

i=1 pi(ζ)ar

ζ,τi(ζ)

.

(3.2)

Then

x(t)ar(t,s)≤x(s), 0≤s≤t,r∈N+. (3.3)

Proof Letx(t) be an eventually positive solution of equation (3.1). Then there existst1>t0 such thatx(t) > 0 andx(τi(t)) > 0, 1≤i≤m, for allt≥t1, so

Tt0 αx(t) = –

m

i=1 pi(t)x

τi(t)

≤0, t≥t1.

This means thatx(t) is monotonically decreasing, that is,x(τi(t))≥x(t), 1≤i≤m, and it

is easy to put it into the original equation:

Tt0

αx(t) +x(t)

m

i=1

pi(t)≤0, t≥t1.

Dividing this equation byx(t), we get

Tt0 αx(t)

x(t) ≤–

m

i=1

pi(t), t≥t1,

that is,

(t–t0)1–α x(t)

x(t) ≤–

m

i=1

pi(t), t≥t1.

Integrating the last inequality fromstot, 0≤s≤t, we get

lnx(ζ)

t

s

≤ t

s

(ζ–t0)α–1

–

m

i=1 pi(ζ)

dζ,

that is,

lnx(t)≤lnx(s) – t

s

(ζ–t0)α–1

m

i=1

pi(ζ)dζ.

So

x(s)≥x(t)exp

t

s

(ζ–t0)α–1

m

i=1

pi(ζ)dζ

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that is, estimate (3.3) is valid for r= 1. Supposing that (3.3) is established forr=n, we obtain

x(t)an(t,s)≤x(s),

so

Tt0 αx(t) +

m

i=1

pi(t)x(t)an

t,τi(t)

≤0.

Repeating these steps can, we obtain

x(s)≥x(t)exp

t

s

(ζ–t0)α–1

m

i=1 pi(ζ)an

ζ,τi(ζ)

dζ

,

that is,x(t)an+1(t,s)≤x(s). So Lemma3.1is proved by mathematical induction.

Lemma 3.2 Assume that x(t)is an eventually positive solution of(3.1)and

0 <β:=lim inf

t→∞

t

h(t)

(ζ–t0)α–1

m

i=1

pi(ζ)dζ≤

1

e, (3.4)

where

h(t) = max

1≤i≤mhi(t), hi(t) =0max≤s≤tτi(s), t≥0. (3.5)

Then

γ =lim inf

t→∞

x(h(t))

x(t) ≥λ0, (3.6)

whereλ0is the smaller root of the equationλ=eβλ.

Proof Letx(t) be an eventually positive solution of equation (3.1). Then there existst1>t0 such thatx(t) > 0 andx(τi(t)) > 0, 1≤i≤m, for allt≥t1. Thus we can conclude from (3.1) that

Tt0 αx(t) = –

m

i=1 pi(t)x

τi(t)

≤0, t≥t1.

This means thatx(t) is monotonically decreasing and positive. By (3.4), for anyε∈(0,β), there istεsuch that

t

h(t)

(ζ–t0)α–1

m

i=1

pi(ζ)dζ≥β–ε, t≥tε≥t1.

We will show that

lim inf

t→∞ x(h(t))

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whereλ1is the smaller root of the equation

e(β–ε)λ=λ.

For contradiction, we assume that

γ =lim inf

t→∞ x(h(t))

x(t) <λ1.

Therefore

e(β–ε)γ>γ. (3.8)

Then for anyδ∈(0,γ), there existstδsuch thatx(_xh₍(_tt₎))≥γ–δfort≥tδ. Dividing both sides

of (3.1) byx(t), we have

–T

t0 αx(t)

x(t) =

m

i=1 pi(t)

x(τi(t))

x(t) ≥

m

i=1 pi(t)

x(h(t))

x(t) ≥(γ–δ)

m

i=1 pi(t).

Integrating the latter fromh(t) tot, we obtain

– t

h(t) x(s)

x(s) ds≥ t

h(t)

(s–t0)α–1

(γ–δ)

m

i=1 pi(s)

ds,

or

– t

h(t) x(s)

x(s) ds≥(γ–δ)(β–ε),

so

x(h(t)) x(t) ≥e

(γ–δ)(β–ε)_.

Therefore

γ =lim inf

t→∞ x(h(t))

x(t) ≥e

(γ–δ)(β–ε)_,

which implies

γ ≥e(β–ε)γ,

which is a contradiction to hypothesis (3.8). So (3.7) is true. Since (3.7) implies (3.6), the

proof of Lemma3.2is complete.

Theorem 3.1 Assume that(3.4)holds and for some r,we have

lim sup

t→∞ t

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ>1 +lnλ0

λ0

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where h(t)is deﬁned by(3.5),ar(t,s)is deﬁned by(3.2),andλ0is the smaller root of the equation eβλ₌_λ_._{Then equation}₍_3.1₎_oscillates.

Proof If equation (3.1) has a solutionx(t), then –x(t) is also a solution of equation (3.1), so we only consider the situation where a solution of (3.1) is eventually positive, that is, there is an integert1≥t0such thatx(t) > 0 andx(τi(t)) > 0, 1≤i≤m, for allt≥t1. By (3.1) we have

Tt0 αx(t) = –

m

i=1 pi(t)x

τi(t)

≤0, t≥t1.

It is shown thatx(t) is an eventually decreasing function.

By Lemma 3.2inequality (3.6) holds. It can be easily seen thatλ0> 1, so for any real number 0 <ε≤λ0– 1, we have

x(h(t))

x(t) ≥λ0–ε, t≥t2≥t1.

Then there ist∗∈(h(t),t) satisfying

x(h(t))

x(t∗) =λ0–ε, t≥t2. (3.10)

Then integrating fromt∗totequation (3.1) and substituting into (3.3), we have

x(t) –xt∗+xh(t)

t

t∗

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ≤0.

Combining this with (3.10), we have

t

t∗

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ≤ x(t

∗₎

x(h(t))= 1

λ0–ε

. (3.11)

Dividing (3.1) byx(t), substituting into (3.3), and then integrating fromh(t) tot∗, we have

– t∗

h(t) x(ζ)

x(ζ) dζ≥ t∗

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)

x(h(t)) x(ζ) ar

h(t),τi(ζ)

dζ,

and because ofTt0

αx(t) < 0, we have

(λ0–ε) t∗

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ

≤ t∗

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)

x(h(t)) x(t) ar

h(t),τi(ζ)

dζ

≤– t∗

h(t) x(ζ)

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that is,

t∗

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ≤ 1

λ0–ε

lnx(h(t))

x(t∗) . (3.12)

Adding (3.12) to (3.11), we get

t

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ≤1 +ln(λ0–ε)

λ0–ε

.

This inequality holds for all 0 <ε≤λ0– 1, so asε→0, we obtain

lim sup

t→∞

t

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ≤1 +lnλ0 λ0

.

This is a contradiction to (3.9). The proof of Theorem3.1is complete.

Lemma 3.3 Assume that x(t)is an eventually positive solution of(3.1)and thatβand h(t) are deﬁned by(3.4)and(3.5).Then

lim inf

t→∞ x(t) x(h(t))≥

1 2

1 –β–1 – 2β–β2_:=_A(_β_). _(3.13)

Proof Assume thatx(t) > 0 fort>T1≥t0. Then there existsT2≥T1such thatx(τi(t)) > 0,

i= 1, 2, . . . ,m. In view of (3.1),Tt0

αx(t)≤0 on [T2,∞). Clearly, (3.13) holds forβ= 0. If 0 <β≤1_e, then for anyε∈(0,β), there existsNεsuch that

t

h(t)

(ζ–t0)α–1

m

i=1

pi(ζ)dζ >β–ε, t>Nε. (3.14)

For ﬁxedε, we will show that for eacht>Nε, there existsλtsuch thath(λt) <t<λtand

λt

t

(ζ–t0)α–1

m

i=1

pi(ζ)dζ=β–ε. (3.15)

In fact, for a given t>Nε,f(λ) :=

λ

t (ζ –t0)

α–1m

i=1pi(ζ)dζ is continuous. Because of

limt→∞h(t) =∞and (3.14), we havelimλ→∞f(λ) >β–ε> 0. Hence there existsλt >t

such thatf(λ) =β–ε, that is, (3.15) holds. From (3.14) we have

λt

h(λt)

(ζ–t0)α–1

m

i=1

pi(ζ)dζ>β–ε=

λt

t

(ζ–t0)α–1

m

i=1

pi(ζ)dζ,

and thereforeh(λt) <t.

Integrating (3.1) fromt(>T3=max{T2,Nε}) toλt, we have

x(t) –x(λt)≥

λt

t

(ζ–t0)α–1

m

i=1 pi(ζ)x

τi(ζ)

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We see thath(t)≤h(ζ)≤h(λt) <tfort≤y≤λt. Integrating (3.1) fromτi(ζ) tot, we have

Noting the known formula

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we have

λt

t

(ζ–t0)α–1

m

i=1 pi(ζ)

ζ

t

(u–t0)α–1

m

i=1

pi(u)du dζ

=1 2

λt

t

(ζ–t0)α–1

m

i=1 pi(ζ)

λt

t

(u–t0)α–1

m

i=1

pi(u)du dζ

=1 2

λt

t

(ζ–t0)α–1

m

i=1

pi(ζ)dζ

2

=1 2(β–ε)

2_.

Substituting this into (3.18), we have

x(t) >x(λt) +x(t)(β–ε) +

1 2(β–ε)

2_x_h(t)_. _(3.19)

Hence

x(t) x(h(t))>

(β–ε)2 2(1 –β+ε):=d1,

and then

x(λt) >

(β–ε)2 2(1 –β+ε)x

h(λt)

=d1x

h(λt)

≥d1x(t).

Substituting this into (3.19), we obtain

x(t) >x(t)(m+d1–ε) + 1 2(m–ε)

2_x_h(t)_,

and hence

x(t) x(h(t))>

(β–ε)2 2(1 –β–d1+ε)

:=d2.

In general, we have

x(t) x(h(t))>

(β–ε)2 2(1 –β–dn+ε)

:=dn+1, n= 1, 2, . . . .

It is not diﬃcult to see that ifεis small enough, then 1≥dn>dn–1,n= 2, 3, . . . . Hence

limn→∞dn=dexists and satisﬁes

–2d2+ 2d(1 –β+ε) = (β–ε)2,

that is,

d=1 –β+ε±

1 – 2(β–ε) – (β–ε)2

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Because ofTt0

α ≤0, we haved< 1. Therefore, for all larget,

x(t) x(h(t))>

1 –β+ε–1 – 2(β–ε) – (β–ε)2

2 .

Lettingε→0, we obtain that

x(t) x(h(t))>

1 –β–1 – 2β–β2

2 =A(β).

This shows that (3.13) holds.

Theorem 3.2 Assume(3.4)holds and that for some r,we have

lim sup

t→∞ t

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ

> 1 –1 2

1 –β–1 – 2β–β2_, _(3.20)

where h(t)is deﬁned by(3.5),ar(t,s)is deﬁned by(3.2),andλ0is the smaller root of the equation eβλ₌_λ_._{Then equation}₍_3.1₎_oscillates.

Proof If equation (3.1) has a solutionx(t), then –x(t) is also a solution of equation (3.1), so we only consider the situation where a solution of (3.1) is eventually positive, that is, x(t) > 0 andx(τi(t)) > 0, 1≤i≤m, for allt≥T3. By (3.1) we have

x(t) –xh(t)(t–t0)α–1

m

i=1

pi(t)≤0, t≥T3.

Integrating fromh(t) totthe latter and substituting into (3.3), we have

x(t) –xh(t)+xh(t)

t

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ≤0.

Consequently,

t

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ≤1 – x(t) x(h(t)),

which gives

lim sup

t→∞

t

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ≤1 –lim inf

t→∞

x(t) x(h(t)),

and by (3.13) the last inequality leads to

lim sup

t→∞

t

h(t)

(ζ–t0)α–1

m

i=1 pi(ζ)ar

h(t),τi(ζ)

dζ≤1 –1 2

1 –β–1 – 2β–β2_,

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Example3.1 We consider the delay diﬀerential equation

r(t) attains its maximum. In particular,

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+1 4

3k+2.6

3k+2

exp

1 4

h(t) –τ1(ζ)

+exp

1 4

h(t) –τ2(ζ) dζ

≈1.5052,

and therefore

lim sup

t→∞ F1(t)≥1.5052.

Now we see that

β=lim inf

t→∞ t

h(t)

ζ–12

m

i=1

pi(ζ)dζ=

1 4

t–h(t)=1 4≤

1 e.

The solution ofλ=eβλ_is_λ

0= 1.435, so we get

1.5052 >1 +lnλ0

λ0 ≈

0.9485,

1.5052 > 1 >A(β).

Therefore equation (3.21) satisﬁes the conditions of Theorems3.1and3.2, and thus equa-tion (3.21) oscillates.

4 Oscillation of2

α

-order neutral conformable fractional differential equation

In this section, we deal with diﬀerential equations of the form

Tt0 α

r(t)Tt0

α

x(t) +p(t)xτ(t)β+q(t)xβσ(t)= 0, t≥t0, (4.1)

whereTαdenotes the conformable diﬀerential operator of orderα∈(0, 1],β≥1 is a

quo-tient of odd positive integers, and the functions r,p,q,τ, σ are such thatr,p,q,τ,σ ∈

C1_([t

0,∞), (0,∞)). We also assume that, for all t≥t0, τ(t)≤t, σ(t)≤t, Tαt0σ(t) > 0, limt→∞τ(t) =limt→∞σ(t) =∞, 0≤p(t) < 1,q(t)≥0, andqdoes not vanish eventually.

We further use the following notation:

ε:=β/(β+ 1)β+1, Q(t) :=q(t)1 –pσ(t)β,

z(t) =x(t) +p(t)xτ(t)<∞, π(t) := ∞

t

(s–t0)α–1r(s)–1/βds.

Lemma 4.1 Letβ≥1be a ratio of two odd numbers.Then

A(β+1)/β– (A–B)(β+1)/β≤4 2B

1/β

β(1 +β)A–B, AB≥0.

–Cv(β+1)/β₊_Dv_≤ β β

(β+ 1)β+1 Dβ+1

Cβ , C> 0.

(4.2)

Theorem 4.1 Assume thatπ(t) =_t∞(s–t0)α–1r(s)–1/βds<∞and there exists a function

ρ∈C1([t0,∞), (0,∞))such that

lim sup

t→∞ I

t0 α

ρ(t)Q(t) –σ(t) –t0

(1–α)β (Tαt0ρ+(t))β+1r(σ(t)) (β+ 1)β+1_ρβ(t)_(Tt0

ασ(t))β

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Suppose that there exists a functionδ∈C1_([t

0,∞), (0,∞))such that

lim sup

t→∞ I

t0 α

ψ(t) –δ(t)r(t)((ϕ(t))+)

β+1

(β+ 1)β+1

=∞, (4.4)

where

ψ(t) :=δ(t)

q(t)

1 –pσ(t)π(τ(σ(t)))

π(σ(t)

β

+ 1 –β

r1/β_(t)_πβ+1_(t)

,

p(t) <π(t)/πτ(t), ϕ(t) :=T

t0 αδ(t)

δ(t) +

1 +β

r1/β_(t)_π_(t),

and(ϕ(t))+:=max{0,ϕ(t)}.Then equation(4.1)oscillates.

Proof Letx(t) be a nonoscillating solution of (4.1) on [t0,∞). Without loss of generality, we may assume that there existst1≥t0such thatx(t) > 0,x(τ(t)) > 0, andx(σ(t)) > 0 for allt≥t1. Thenz(t)≥x(t) > 0, and since

Tt0 α

r(t)Tt0

α

z(t)β= –q(t)xβσ(t)≤0, (4.5)

the function [r(t)Tt0

αz(t)]βis nonincreasing for allt≥t1. ThereforeTαt0z(t) does not change

sign eventually, that is, there existst2≥t1such that eitherTαt0z(t) > 0 orTαt0z(t) < 0 for all

t≥t2.

Case I. Assume ﬁrst thatTt0

αz(t) > 0 for allt≥t2. Note thatTαt0z(t)|t=σ(t)=Tαt0(z(σ(t))).

Then

r(t)Tt0 α

z(t)β≤rσ(t)Tt0 αz

σ(t)β,

from which it follows that

Tt0 α

zσ(t)≥Tt0 α

z(t) r(t) r(σ(t))

1/β

. (4.6)

Sincex(t)≤z(t), we see that

x(t)≥1 –p(t)z(t), t≥t2. (4.7)

In view of (4.7) and (4.1),

Tt0 α

r(t)Tt0

α

x(t) +p(t)xτ(t)β+Q(t)zβσ(t)≤0, t≥t2. (4.8)

Put

w(t) =ρ(t)r(t)(T

t0 αz(t))β

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Clearly,w(t) > 0. ApplyingTt0

α to (4.9) and using (4.6) and (4.8), we obtain

Tt0 α

w(t)

=T

t0 αρ+(t)

ρ(t) w(t) +ρ(t) Tt0

α(r(t)(Tαt0z(t))β)

zβ₍_σ_(t)) –ρ(t)

r(t)(Tt0

αz(t))ββz(σ(t))Tαt0σ(t)

zβ+1₍_σ_(t))

≤Tαt0ρ+(t)

ρ(t) w(t) –ρ(t)Q(t) –

βTt0 ασ(t)

(ρ(t)r(σ(t)))1β₍_σ_{(t) –}_t

0)1–α

wβ+1β _(t),

whereTt0

αρ+(t) =max{Tαt0ρ(t), 0}. Set

F(v) =T

t0 αρ+(t)

ρ(t) v–

βTt0 ασ(t)

(ρ(t)r(σ(t)))β1₍_σ_{(t) –}_t

0)1–α vβ

+1

β _, _v_{> 0.}

By calculation lettingv0= (σ(t) –t0)(1–α)β₍_β₊₁₎1 β

(Tαt0ρ+(t))β

ρβ–1₍_t₎

r(σ(t)) (Tαt0σ(t))β

, we have that when

v=v0,

the functionF(v) attains its maximumF(v0). So

F(v)≤F(v0) =

σ(t) –t0

(1–α)β (Tαt0ρ+(t))β+1r(σ(t)) (β+ 1)β+1_ρβ_(t)(Tt0

ασ(t))β

.

Therefore

Tt0 α

w(t)≤–ρ(t)Q(t) +σ(t) –t0

(1–α)β (Tαt0ρ+(t))β+1r(σ(t)) (β+ 1)β+1_ρβ_(t)(Tt2

ασ(t))β

.

ApplyingIαto the last inequality fromt0tot, we have

0 <w(t)≤w(t0) –Iαt0

ρ(t)Q(t) –σ(t) –t0

(1–α)β (Tαt0ρ+(t))β+1r(σ(t)) (β+ 1)β+1_ρβ(t)_(Tt0

ασ(t))β

.

Lettingt→ ∞in this inequality, we get a contradiction to (4.3). Case II. Assume now that Tt0

αz(t) < 0 for all t ≥ t0. It follows from (4.1) that Tt0

α(r(Tαt0z)β) < 0 for alls≥t≥t2, and thus

Tt0 αz(s)≤

r(t) r(s)

1/β

Tt0

αz(t). (4.10)

Dividing (4.10) by (s–t0)1–αand then integrating fromttol,l≥t≥t2, we have

z(l) –z(t)≤ l

t

(s–t0)α–1

r(t) r(s)

1/β

Tt0 α

z(t)ds

=r(t)1/βTt0 α

z(t)

l

t

(s–t0)α–1r(s)–1/βds.

Lettingl→ ∞, we get

z(t)≥–π(t)r1/β(t)Tt0 α

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which implies that

Tt0 α

z(t)

π(t) =

π(t)Tt0

αz(t) –z(t)Tαt0π(t)

π2_(t) =

π(t)Tt0

αz(t) +z(t)r–1/β(t))

π2_(t)

≥π(t)Tαt0z(t) –π(t)Tαt0z(t)

π2_(t) = 0.

Hence we conclude that

x(t) =z(t) –p(t)xτ(t)≥z(t) –p(t)zτ(t)≥

1 –p(t)π(τ(t))

π(t) z(t). (4.12)

Using (4.12) in (4.5), we have

Tt0 α

r(t)Tt0

α

z(t)β≤–q(t)

1 –pσ(t)π(τ(σ(t)))

π(σ(t))

β

zβσ(t)≤0. (4.13)

Deﬁne a generalized Riccati substitution by

w(t) :=δ(t)

r(t)(Tt0 αz(t))β

zβ_(t) +

1

πβ_(t)

. (4.14)

By (4.11),w(t)≥0 for allt≥t2. ApplyingTαt0 to (4.14), we have

Tt0 αw(t) =

Tt0 αδ(t)

δ(t) w(t)

+δ(t)

Tt0

zβ –

βr(t)(Tt0 αz(t))β+1

zβ+1_(t) –βπ

–(β+1)_Tt0 απ(t)

=T

t0 αδ(t)

δ(t) w(t)

+δ(t)T

t0

zβ –βδ(t)r(t)

w(t)

δ(t)r(t)– 1 r(t)πβ_(t)

(β+1)/β

+ βδ(t)

r1/β_(t)_π1+β_(t). (4.15)

LetA:=w(t)/(δ(t)r(t)) andB= 1/(r(t)πβ_{(t)). Using Lemma}_4.1_{, we conclude that}

w(t)

δ(t)r(t)– 1 r(t)πβ_(t)

β+1 β

≥

w(t)

δ(t)r(t)

β+1 β

– 1

βr1/β_(t)_π_(t)

(1 +β) w(t)

δ(t)r(t)– 1 r(t)πβ_(t)

.

On the other hand, we get by (4.13) thatTt0

αz< 0 and fromσ(t)≤tthat

Tt0

α{r(t)[Tαt0(z(t))]β}

zβ_(t) ≤–q(t)

1 –pσ(t)π(τ(σ(t)))

π(σ(t))

β

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Thus (4.15) yields

ApplyingIαto the latter inequality fromt0tot, we have

It0

which contradicts (4.4). Therefore (4.1) oscillates.

Example4.1 We consider the equation

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and it is obvious that (4.3) holds. Because ofϕ(t) = 2/√t,ψ(t) = (q0(1 – 2 √

2p0))/t=34₂₅. So

It0 α

ψ(t) –δ(t)r(t)((ϕ(t))+)

β+1

(β+ 1)β+1

=It0 α

34 25–

1

tt

2₍_√2

t)

2

22

=It0 α

9 25,

and we can conclude that condition (4.4) is satisﬁed. Hence by Theorem4.1we deduce that (4.18) oscillates.

5 Oscillation of3

α

-order damped conformable fractional differential equation

This section deals with oscillatory behavior of all solutions of the 3α-order nonlinear delay damped equation of the form

Tt0 α

r2Tαt0

r1

Tt0

αy

β

(t) +p(t)Tt0 αy(t)

β

+q(t)fyg(t)= 0, t≥t0, (5.1)

where 0 <α≤1, andβ≥1 is the ratio of positive odd integers. We further assume that the following conditions are satisﬁed:

(H1) r1,r2,p,q∈C(I,R+), whereI= [t0,∞),R+= (0,∞);

(H2) g∈C1_(I,_R₎_,_Tt0

αg(t)≥0andg(t)→ ∞ast→ ∞;

(H3) f ∈C(R,R)is such thatxf(x) > 0forx= 0, andf(x)/xγ_≥_k_{> 0}_{, where}_γ _{is the ratio} of positive odd integers.

We deﬁne

R1(t,t0) =Iαt0

1

r1/₁β(t), R2(t,t0) =I

t0 α

1 r2(t)

, and R∗(t,t0) =Iαt0

R2(t,t0)

r1(t) 1/β

fort0≤t1≤t≤ ∞and assume that

R1(t,t0)→ ∞, t→ ∞, (5.2)

and

R2(t,t0)→ ∞, t→ ∞. (5.3)

A functionyis called a solution of (5.1) ify,r1(Tαt0y) β_,_r

2(r1(Tαt0y)

β₎_∈_C1_([t

y,∞),R) and

ysatisﬁes (5.1) for [ty,∞) for somety≥t0. For brevity, we deﬁne

L0y(t) =y(t), L1y(t) =r1(t)

Tt0 α(L0y)

β

(t),

L2y(t) =r2(t)Tαt0(L1y)(t), L3y(t) =Tαt0(L2y)(t)

onI. Then (5.1) can be written as

L3y(t) + p(t) r1(t)

L1y(t) +q(t)f

yg(t)= 0.

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de-lay

Next, we state and prove the following lemmas.

Lemma 5.1 Let y be a nonoscillatory solution of(5.1)on I.Suppose(5.4)is nonoscillatory.

Then there exists t2∈[t1,∞)such that y(t)L1y(t) > 0or y(t)L1y(t) < 0,t≥t2.

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Lemma 5.2 If y is a nonoscillatory solution of(5.1)and y(t)L1y(t) > 0,t≥t1≥t0,then

L1y(t)≥R2(t,t0)L2y(t) for all t≥t1 (5.5)

and

y(t)≥R∗(t,t0)(L2y)1/β(t) for all t≥t1. (5.6)

Proof Ifyis a nonoscillatory solution of (5.1), theny(t) > 0,y(g(t)) > 0, andL1y(t) > 0 for t≥t1≥t0. It is easy to see that

L3y(t) = – p(t) r1(t)

L1y(t) –q(t)f

yg(t)≤0,

which implies thatL2y(t) is nonincreasing on [t1,∞). ApplyingIαtoTαt0L1y(t) =L_r2₂y₍(_tt₎)from t1totand Lemma2.2, we get

L1y(t) =L1y(t1) +Iαt0

L2y(t)

r2(t)

≥L2y(t)Iαt0

1 r2(t)

=L2y(t)R2(t,t0) for anyt≥t1.

Then

Tt0 αy(t)≥

R2(t,t0)

r1(t) 1/β

(L2y)1/β(t).

Now, applyingIαto the last inequality fromt1tot, we can obtain from Lemma2.2that

y(t)≥y(t1) +Iαt0

R2(t,t0)

r1(t) 1/β

(L2y)1/β(t)

≥(L2y)1/β(t)Iαt0

R2(t,t0)

r1(t) 1/β

=R∗(t,t0)(L2y)1/β(t) fort≥t1.

This completes the proof.

In the following two lemmas, we consider the second-order delay diﬀerential inequality

Tt0 α

r2Tαt0x(t)

≥Q(t)xh(t), t>t0, (5.7)

where the functionr2is as in (5.1),Q(t)∈C(I,R+), andh(t)∈C1(I,R) is such thath(t)≤t, Tt0

αh(t)≥0 fort≥t0, andh(t)→ ∞ast→ ∞.

Lemma 5.3 If

lim sup

t→∞ R2

h(t),t0

It0

αQ(t) > 1, (5.8)

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Proof Letx(t) be a bounded nonoscillatory solution of (5.7), sayx(t) > 0 andx(h(t)) > 0 for t≥t1for somet1≥t0. By (5.7),r2Tαt0x(t) is strictly increasing on [t1,∞). Hence, for any t2≥t1, applyingIαfromt2totinTαt0x(t) =

r2(t)Tαt0x(t)

r2(t) and Lemma2.2yield

x(t) =x(t2) +Iαt0

r2(t)Tαt0x(t)

r2(t)

>x(t2) +r2(t2)Tαt0x(t2)Iαt0

1 r2(t)

=x(t2) +r2(t2)Tαt0x(t2)R2(t,t0),

soTt0

αx(t2) < 0, as otherwise (5.3) would implyx(t)→ ∞ast→ ∞, a contradiction to the boundedness ofx. Altogether,

x> 0,Tt0

αx< 0, and T

t0 α

r2Tαt0x

> 0 on [t1,∞).

Now, forv≥u≥t1, repeating the previous steps, we have

x(u) >x(u) –x(v) = –It0 α

r2(v)Tαt0x(v)

r2(v)

≥–r2(v)Tαt0x(v)Iαt0

1 r2(v)

= –r2(v)Tαt0x(v)R2(v,t0). (5.9)

Fort≥s≥t1, settingu=h(s) andv=h(t) in (5.9), we get

xh(s)> –r2

h(t)Tt0 αx

h(t)R2

h(t),t0

.

ApplyingIαto (5.7) fromh(t)≥t1tot, we obtain from Lemma2.2that

–r2

h(t)Tt0 αx

h(t)>r2(t)Tαt0x(t) –r2

h(t)Tt0 αx

h(t)

≥It0 α

Q(t)xh(t)

> –r2

h(t)Tt0 αx

h(t)R2

h(t),t0

It0

αQ(t),

that is,

1 >R2

h(t),t0

It0 αQ(t).

Takinglim supast→ ∞on both sides of this inequality yields a contradiction to (5.8).

Lemma 5.4 If

lim sup

t→∞,u→∞R2(u,t0)I

t0

αQ(t) > 1, (5.10)

then all bounded solutions of (5.7)are oscillatory.

Proof Letxbe a bounded nonoscillatory solution of (5.7), sayx(t) > 0 andx(h(t)) > 0 for t≥t1for somet1≥t0. As in Lemma5.1, we obtain

x> 0, Tt0

αx< 0, and Tαt0

r2Tαt0x

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ApplyingIαto(5.7) fromu≥t1tot, we obtain from the previous forms that

–r2(u)Tαt0x(u) >r2(t)Tαt0x(t) –r2(u)Tαt0x(u)≥Iαt0

Q(t)xh(t)≥xh(t)It0 αQ(t),

so

–Tt0 αx(u) >

1 r2(u)

It0 αQ(t) x

h(t). (5.11)

We obtain from (5.11) that

xh(t)>xh(t)–x(u)≥xh(t)It0 α

1 r2(u)

It0 αQ(t) ,

that is,

1 >R2(u,t0)Iαt0Q(t).

Takinglim supasu,t→ ∞on both sides of this inequality yields a contradiction to (5.10).

Theorem 5.1 Assume that(5.2)and(5.3)hold andβ≥γ.Suppose that there exist two functions m,h∈C1_(I,_R₎_{such that}

g(t)≤h(t)≤t, Tt0

αh(t)≥0, and m(t) > 0, t∈I,

satisfying

lim sup

t→∞

It0 α

km(t)q(t) –A 2_(t)

4B(t)

=∞, (5.12)

and for t≥t1, ⎧ ⎨ ⎩

A(t) =T

t0 αm(t)

m(t) –

p(t)

r1(t)R2(t,t0),

B(t) =c∗m–1_(t)Tt0

αg(t)(R∗(g(t),t0))

γ–1₍R2(g(t),t0)

r1(g(t)) )1/

β_(t_–_t

0)α–1,

(5.13)

and that(5.8)or(5.10)holds with

Q(t) =ckq(t)R1

h(t),t0 γ

– p(t) r1(t)≥

0, t≥t1,

with c,c∗> 0.Then every solution y of (5.1)and L2y(t)are oscillatory.

Proof Letybe a nonoscillatory solution of (5.1) on [t1,∞),t1≥t0. We assume thaty(t) > 0 andy(g(t)) > 0 fort≥t1. From Lemma5.1we haveL1y(t) < 0 orL1y(t) > 0 fort≥t1.

Step 1. We assume thatL1y(t) > 0 on [t1,∞). By (5.1)L2yis strictly decreasing. Hence, for anyt2≥t1, we have from Lemma2.2that

L1y(t) =L1y(t2) +Iαt0

L2y(t)

r2(t)

≤L1y(t2) +L2y(t2)Iαt0

1 r2(t)

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SoL2y(t2) > 0 as otherwise (5.3) would implyL1y(t)→–∞ast→ ∞, a contradiction to the positivity ofL1y. Altogether,L2y> 0 on [t1,∞).

Deﬁne the following generalized Riccati transformation:

w(t) =m(t) L2y(t)

yγ_(g(t)), t∈[t1,∞). (5.14)

By the product and quotient rules,α-diﬀerentiatingw, we obtain

Tt0

αw(t) =Tαt0

m(t) L2y(t) yγ_(g(t))

=Tt0 αm(t)

L2y(t) yγ_(g(t))

+m(t)T

t0

α(L2y(t))yγ(g(t)) –γ[yγ–1(g(t))]y(g(t))Tαt0g(t)L2y(t) y2γ_(g(t))

=T

t0 αm(t)

m(t) w(t) +m(t) Tt0

α(L2y(t))

yγ_(g(t)) –m(t)

γy(g(t))Tt0

αg(t)L2y(t)

yγ+1_(g(t))

=T

t0 αm(t)

m(t) w(t) + Tt0

α(L2y)(t) L2y(t)

w(t) –γTt0 αg(t)

y(g(t)) y(g(t))w(t).

Using (5.1), (5.5), and assumption (H3) onf, we obtain

Tt0 αm(t)

m(t) w(t) + Tt0

α(L2y)(t) L2y(t)

w(t)

=T

t0 αm(t)

m(t) w(t) –

p(t)

r1(t)L1y(t) +q(t)f(y(g(t)))

L2y(t)

w(t)

=T

t0 αm(t)

m(t) w(t) –

p(t)

r1(t)L1y(t)

L2y(t)

w(t) –q(t)f(y(g(t))) L2y(t)

w(t)

≤Tαt0m(t)

m(t) w(t) – p(t) r1(t)

R2(t,t0)w(t) –km(t)q(t)

=

Tt0 αm(t)

m(t) – p(t) r1(t)

R2(t,t0)

w(t) –km(t)q(t)

=A(t)w(t) –km(t)q(t).

By the deﬁnition ofL1y(t) and (5.5) we obtain

(t–t0)1–α

yg(t)=Tt0 αy

g(t)=

1 r1(g(t))

L1y

g(t) 1/β

≥

R2(g(t),t0) r1(g(t))

1/β

L2y

g(t)1/β

≥

R2(g(t),t0) r1(g(t))

1/β

L2y(t) 1/β

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Then

y(g(t))

y(g(t)) ≥(t–t0)

α–1

R2(g(t),t0) m(t)r1(g(t))

1/β

m1/β_(t)(L

2y)1/β(t) yγ/β_(g(t)) y

γ/β–1_g(t)

(5.14)

= (t–t0)α–1

R2(g(t),t0) m(t)r1(g(t))

1/β

w1/β_(t)yγ/β–1_g(t)_,

and we obtain

Tt0

αw(t)≤A(t)w(t) –km(t)q(t)

–γTt0

αg(t)(t–t0)α–1

R2(g(t),t0) m(t)r1(g(t))

1/β

w1/β(t)yγ/β–1g(t)w(t)

≤A(t)w(t) –km(t)q(t)

–γTt0

αg(t)(t–t0)α–1w1/β+1(t)yγ/β–1

g(t)R2(g(t),t0) m(t)r1(g(t))

1/β

. (5.15)

SinceL3y(t) < 0, we have 0 <L2y(t)≤L2y(t1),L2y(t1) =c1fort≥t1. Then

r2(t)Tαt0(L1y)(t) =L2y(t)≤c1, t≥t1,

and thus we get from Lemma2.2that

r1(t)

Tt0 αy

β

(t) =L1y(t) =L1y(t1) +Iαt0

r2(t)Tαt0(L1y(t)) r2(t)

≤L1y(t1) +c1Iαt0

1 r2(t)

=L1y(t1) +c1R2(t,t0) =

L1y(t1) R2(t,t0)

+c1

R2(t,t0)

≤

L1y(t1) R2(t2,t0)

+c1

R2(t,t0) =c˜1R2(t,t0)

(note thatL1y(t1) > 0), where

˜ c1=c1+

L1y(t1) R2(t2,t0)

.

Therefore, we get for allt≥t2that

y(t) =y(t2) +Iαt0

Tt0

αy(t)

≤y(t2) +Iαt0

_˜ c1R2(t,t0)

r1(t) 1/β

=y(t2) +c˜11/βR∗(t,t0) =

y(t2) R∗(t,t0)

+c˜11/β

R∗(t,t0)

≤

y(t2) R∗(t2,t0)

+c˜11/β

R∗(t,t0)

=c2R∗(t,t0)

(note thaty(t2) > 0), where

c2= y(t2) R∗(t2,t0)

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Then we get

yγ/β–1_g(t)_≥_cγ/β–1 2

R∗g(t),t0 γ/β–1

, t≥t2. (5.16)

By (5.14) and (5.6) we have

w(t) =m(t) L2y(t)

yγ_(g(t))≤m(t)

L2y(g(t)) yγ_(g(t))

≤m(t)R∗g(t),t0 –β

yβ–γg(t), t≥t2. (5.17)

Using (5.16) in (5.17), we get

w(t)≤cβ₂–γm(t)R∗g(t),t0 –γ

, t≥t2.

Then

w1/β–1(t)≥c₂(1/β–1)(β–γ)m1/β–1(t)R∗g(t),t0

–γ(1/β–1)

, t≥t2. (5.18)

Using (5.16) and (5.18) in (5.15), we get

Tt0 αw(t)

≤A(t)w(t) –km(t)q(t)

–γc–₂β+γm–1Tt0 αg(t)

R∗g(t),t0

γ–1R2(g(t),t0) r1(g(t))

1/β

(t–t0)α–1w2(t)

=A(t)w(t) –km(t)q(t) –B(t)w2(t)

= –km(t)q(t) –B(t)w(t) – A(t) 2√B(t)

2 +A

2_(t)

4B(t)

≤–km(t)q(t) +A 2_(t)

4B(t), t≥t2, (5.19)

wherec∗=γcγ₂–β, andAandBare as in (5.13). ApplyingIαto (5.19) fromt0tot, we get

It0 α

km(t)q(t) –A 2_(t)

4B(t)

≤w(t0) –w(t)≤w(t0),

which contradicts (5.12).

Step 2. LetL1y(t) < 0 on [t1,∞). We consider the functionL2y(t). The caseL2y(t)≤0 cannot hold for all larget, sayt≥t2≥t1, since by double integration of

Tt0 αy(t) =

L1y(t)

r1(t) 1/β

≤

L1y(t2) r1(t)

1/β

, t≥t2,

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we have

Proceeding exactly as in the proofs of Lemmas5.3and5.4, we arrive at the desired

con-clusion, thus completing the proof.

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It0

and we obtain that

It0

is nonoscillatory, and one nonoscillatory solution of (5.21) isz(t) = 18t13_{. Then we get that}

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Example5.2

T1 2

t–12_T1

2

t–12_T1

2y(t)

+ 2t–12_T1

2y(t) + 3y(t) = 0, t≥0, (5.22)

wherer1(t) =r2(t) =t–

1

2,p(t) = 2t–12,q(t) = 3,k= 1,g(t) =t,_α=1

2,β=γ = 1,c∗=c= 1. Lettingm(t) = 1 andh(t) =t, we can obtain

R2(t,t0) =t, A(t) = –2t, B(t) =t, Q(t) = 3t– 2,

so all conditions except (5.12) are satisﬁed. Equation (5.22) can be rewritten as

y(t) + 2y(t) + 3y(t) = 0.

It is obvious that the equation is nonoscillatory. It has a nonoscillatory solution x= e12tcos

√ 2

2 t. We can obtain that condition (5.12) indispensable.

6 Conclusion

In this paper, we study three kinds of diﬀerent order conformable fractional equations and obtain oscillatory results of three equations. Those results unify the oscillation theory of the integral-order and fractional-order diﬀerential equations.

Funding

This research is supported by the Natural Science Foundation of China (61703180, 61803176), and supported by Shandong Provincial Natural Science Foundation (ZR2016AM17, ZR2017MA043).

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

The authors declare that the study was realized in collaboration with the same responsibility. Both authors read and approved the ﬁnal manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 11 April 2019 Accepted: 17 July 2019 References

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