# Nonexistence Results for Some Classes of Nonlinear Fractional Differential Inequalities

Authors: Mohamed Jleli; Bessem Samet
Journal: Journal of Function Spaces
 (2020)

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## Abstract

We study the nonexistence of global solutions for new classes of nonlinear fractional differential inequalities. Namely, sufficient conditions are provided so that the considered problems admit no global solutions. The proofs of our results are based on the test function method and some integral estimates.


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## Body

## 1. Introduction

We first consider the problem(1)DC01+αut+DC01+βut≥DC0γutp,t>0,u0,u′0=u0,u1,where p>1, α,β,γ∈0,1, CD0κ, κ∈1+α,1+β,γ is the Caputo fractional derivative of order κ, u0∈ℝ, and u1≥0. Namely, we are interested in providing sufficient conditions for which problem (1) admits no global solution. Next, we study the same question for the inhomogeneous problem
(2)DC01∣+αut+DC01∣+βut≥DC0γutp+ft,t>0,u0,u′0=u0,u1,where p,q>1, α,β,γ∈0,1, u0∈ℝ, u1≥0, f∈Lloc1([0,∞)), f≥0, and f≠0.Due to the importance of fractional calculus in applications (see e.g. [1–5]), in the past few decades, there has been a growing interest in the study of fractional differential equations. In particular, from the theoretical point of view, the existence of solutions for different classes of fractional differential equations was investigated in many contributions (see e.g. [6–12] and the references therein).For the issue of nonexistence of solutions for fractional differential equations and inequalities, we refer to [13–22] and the references therein. In particular, in [17], Laskri and Tatar studied the problem
(3)D0αut≥tγytp,t>0,I01−αutt=0=b,where p>1, 0<α<1, γ>−α, and b≥0, D0α is the Riemann-Liouville fractional derivative of order α, and I01−α is the left-sided Riemann-Liouville fractional integral of order 1−α. It was shown that, if p≤γ+1/1−α, then problem (3) does not admit nontrivial global solution. In [16], Kassim et al. studied the problem
(4)DC0αut+DC0βut≥tγytp,t>0,ui0=bi,i=0,1,⋯,n−1,where p>1, n≥1 is an integer, n−1<β≤α<n, and bi≥0. It was shown that, if
(5)p1−β−1<γ<p−1,then problem (4) does not admit nontrivial global solution. In [15], Furati and Kirane investigated the system of nonlinear fractional differential equations
(6)u′t+CD0αut=vtq,t>0,v′t+CD0βvt=utp,t>0,subject to the initial conditions
(7)u0,v0=u0,v0,where 0<α,β<1, p,q>1, and u0,v0>0. It was shown that, if
(8)1−1pq≤maxα+βp,β+αq,then solutions to system (6) subject to (7) blow up in a finite time.For the issue of nonexistence of global solutions for fractional in time evolution equations, we refer to [6, 23–25] and the references therein.On the other hand, to the best of our knowledge, the nonexistence of global solutions for problems of types (1) and (2) was not yet investigated.Before stating our main results, let us mention what we mean by global solutions to problems (1) and (2).Definition 1.
A functionu∈AC2([0,∞)) is said to be a global solution to problem (1), if u satisfies
(9)CD01+αut+CD01+βut≥CD0γutp,for almost every where t>0, and
(10)u0,u′0=u0,u1.Definition 2.
A functionu∈AC2([0,∞)) is said to be a global solution to problem (2), if u satisfies
(11)CD01+αut+CD01+βut≥CD0γutp+ft,for almost every where t>0, and
(12)u0,u′0=u0,u1.We first consider problem (1). We discuss separately the cases u1>0 and u1=0.Theorem 3.
Letα,β,γ∈0,1, and u0∈ℝ. Ifu1>0, then for all p>1, problem (1) admits no global solution.Theorem 4.
Letα,β,γ∈0,1, α≤β, u0∈ℝ, and u1=0.
(i)
Ifγ≤α, then for all p>1, the only global solution to problem (1) is u≡u0(ii)
Ifγ>α, then for all(13)1<p<1γ−α,the only global solution to problem (1) is u≡u0.Next, we consider problem (2).Theorem 5.
Letα,β,γ∈0,1, u0∈ℝ, f∈Lloc1([0,∞)), f≥0, and f≡0. If u1>0, then for all p>1, problem (2) admits no global solution.Theorem 6.
Letp>1, α,β,γ∈0,1, α≤β, u0∈ℝ, u1=0, f∈Lloc1([0,∞)), f≥0, and f≡0. If
(14)limsupT→+∞Tα−γp+1/p−1∫0T/2ftdt=+∞,then problem (2) admits no global solution.We discuss below some special cases of Theorem6.Corollary 7.
Letα,β,γ∈0,1, α≤β, u0∈ℝ, and u1=0. Let
(15)ft=eat,t>0,where a∈ℝ and a≠0.
(i)
Ifa>0, then for all p>1, problem (2) admits no global solution(ii)
Ifa<0 and γ≤α, then for all p>1, problem (2) admits no global solution(iii)
Ifa<0 and γ>α, then for all(16)1<p<1γ−α,problem (2) admits no global solution.Corollary 8.
Letα,β,γ∈0,1, α≤β, u0∈ℝ, and u1=0. Let
(17)ft=tσ,t>0,where σ>−1.
(i)
Ifσ≥0, then for all p>1, problem (2) admits no global solution(ii)
Let−1<σ<0(a)
Ifγ≤α, then for all p>1, problem (2) admits no global solution(b)
Ifγ>α and γ−α−1≤σ<0, then for all p>1, problem (2) admits no global solution(c)
Ifγ>α and −1<σ<γ−α−1, then for all(18)1<p<σσ−γ+α+1,problem (2) admits no global solution.The rest of the paper is organized as follows. In Section2, we recall briefly some standard notions on fractional calculus and prove some properties. Section 3 is devoted to the Proofs of Theorems 3, 4, 5, and 6 and Corollaries 7 and 8.


## 2. Some Preliminaries

We denote byAC([0,∞)) the space of absolutely continuous functions on 0,∞. Given an integer n≥2, we denote by ACn([0,∞)) the space of functions f which have continuous derivatives up to order n−1 on 0,∞ such that fn−1∈AC([0,∞)). Here, fn−1 denotes the derivative of order n−1 of f.LetT>0 be fixed. Given ρ>0 and f∈L10,T, the left-sided Riemann-Liouville fractional integral of order ρ of f is defined by
(19)I0ρft=1Γρ∫0tt−sρ−1fsds,for almost everywhere 0≤t≤T. Here, Γ denotes the Gamma function. The right-sided Riemann-Liouville fractional integral of order ρ of f is defined by
(20)ITρft=1Γρ∫tTs−tρ−1fsds,for almost everywhere 0≤t≤T. Notice that, if f∈C(0,T), then I0ρf is defined for all 0<t≤T. Moreover, one has limt→0+I0ρft=0. Similarly, if f∈C(0,T), then ITρf is defined for all 0≤t<T. Moreover, one has limt→T−ITρft=0.Lemma 9 (see [5]).
Letρ,κ>0 and f∈Lτ0,T, where 1≤τ≤∞. Then
(21)I0ρI0κft=I0κI0ρft=I0ρ+κft,for almsot everywhere 0≤t≤T.Lemma 10 (see [5]).
Letρ>0, τ,μ≥1, and 1/τ+1/μ≤1+ρ (τ≠1 and μ≠1 in the case 1/τ+1/μ=1+ρ). If f∈Lτ0,T and g∈Lμ0,T, then
(22)∫0TI0ρftgtdt=∫0TITρgtftdt.Letn−1<ρ<n and f∈ACn([0,∞)), where n≥1 is an integer. The (left-sided) Caputo fractional derivative of order ρ of f is defined by
(23)CD0ρft=I0n−ρfnt,for almost everywhere t>0. Here, for n=1, AC1([0,∞))=AC([0,∞)).Forλ≫1 (λ is large enough), we define the function
(24)ξt=T−λT−tλ,0≤t≤T.Lemma 11.
Letρ>0 and 0<κ<1. Then
(25)ITρξt=Γλ+1Γρ+λ+1T−λT−tλ+ρ,0≤t<T,(26)ITρξ′t=−Γλ+1Γρ+λT−λT−tλ+ρ−1,0≤t<T,(27)ITρξ′′t=Γλ+1Γρ+λ−1T−λT−tλ+ρ−2,0≤t<T,(28)ITρIT1−κξ′′t=Γλ+1Γλ+ρ−κT−λT−tλ+ρ−1−κ,0≤t<T.Proof.
We prove only (25). Namely, differentiating (25), (26) follows. Similarly, differentiating (26), (27) follows. Moreover, taking ρ=1−κ in (27) and using a similar calculation as in the proof of (25), (28) follows.
Fort∈0,T, one has
(29)ITρξt=T−λΓρ∫tTs−tρ−1T−sλds=T−λΓρ∫tTs−tρ−1T−t−s−tλds=T−λT−tλΓρ∫tTs−tρ−11−s−tT−tλds.
Using the change of variablez=s−t/T−t, one obtains
(30)ITρξt=T−λT−tλ+ρΓρ∫01zρ−11−zλ+1−1dz=T−λT−tλ+ρΓρBρ,λ+1,where B·,· is the beta function. Using the property
(31)Ba,b=ΓaΓbΓa+b,a,b>0,one obtains (25).


## 3. Proofs

The proofs of our results are based on the test function method (see e.g. [26])and some integral estimates.Proof of Theorem 3.
Let us suppose thatu∈AC2([0,∞)) is a global solution to (1). For T>0, multiplying the differential inequality in (1) by ξ, where ξ is the function defined by (24), and integrating over 0,T, one obtains
(32)∫0TCD0γutpξtdt≤∫0TCD01+αutξtdt+∫0TCD01+βutξtdt.
Without restriction of the generality, we may suppose that(33)0<α≤β<1.
On the other hand, using Lemma10, one obtains
(34)∫0TCD01+αutξtdt=∫0TI01−αu′′tξtdt=∫0Tu′′tIT1−αξtdt.
Using an integration by parts, the initial conditions and (25), it holds that
(35)∫0TCD01+αutξtdt=u′tIT1−αξt0T−∫0Tu′tIT1−αξ′tdt=−u1IT1−αξ0−∫0Tu′tIT1−αξ′tdt.
On the other hand, by Lemma9 and using the initial conditions, one obtains
(36)u′t=ut−u0′=∫0tu′sds=I01u′′t=I0γI01−γu′′t.
Therefore, by (35), one obtains
(37)∫0TCD01+αutξtdt=−u1IT1−αξ0−∫0TI0γI01−γu′tIT1−αξ′tdt.
Using an integration by parts, the initial conditions, (26) and Lemma 10, it holds that
(38)∫0TCD01+αutξtdt=−u1IT1−αξ0−ut−u0IT1−αξ′t0T−∫0TI0γI01−γu′tIT1−αξ′′tdt=−u1IT1−αξ0−∫0TI01−γu′tITγIT1−αξ′′tdt=−u1IT1−αξ0−∫0TCD0γutITγIT1−αξ′′tdt.
Similarly, one has(39)∫0TCD01+βutξtdt=−u1IT1−βξ0−∫0TCD0γutITγIT1−βξ′′tdt.
Next, using (32), (38), and (39), one obtains
(40)∫0TCD0γutpξtdt+u1IT1−αξ0+IT1−βξ0≤∫0TCD0γutITγIT1−αξ′′tdt+∫0TCD0γutITγIT1−βξ′′tdt.
On the other hand, usingε-Young inequality with 0<ε<1/2, one obtains
(41)∫0TCD0γutITγIT1−αξ′′tdt≤ε∫0TCD0γutpξtdt+Cε,p∫0Tξt−1/p−1ITγIT1−αξ′′tp/p−1dt,where Cε,p is a positive real number that depends only on ε and p. Similarly, one has
(42)∫0TCD0γutITγIT1−βξ′′tdt≤ε∫0TCD0γutpξtdt+Cε,p∫0Tξt−1/p−1ITγIT1−βξ′′tp/p−1dt.
Hence, it follows from (40), (41), and (42) that
(43)1−2ε∫0TCD0γutpξtdt+u1IT1−αξ0+IT1−βξ0≤Cε,p∫0Tξt−1/p−1ITγIT1−αξ′′tp/p−1dt+∫0Tξt−1/p−1ITγIT1−βξ′′tp/p−1dt.
Since0<ε<1/2, one deduces from (43) that
(44)u1IT1−αξ0+IT1−βξ0≤Cε,p∫0Tξt−1/p−1ITγIT1−αξ′′tp/p−1dt+∫0Tξt−1/p−1ITγIT1−βξ′′tp/p−1dt.
On the other hand, by (25), one has
(45)IT1−αξ0=Γλ+1Γ2+λ−αT1−α,and
(46)IT1−βξ0=Γλ+1Γ2+λ−βT1−β,which yield
(47)u1IT1−αξ0+IT1−βξ0=Γλ+1u1T1−α1Γ2+λ−α+Tα−βΓ2+λ−β.
Sinceu1>0, one deduces that
(48)u1IT1−αξ0+IT1−βξ0≥C1u1T1−α,where C1=Γλ+1/Γ2+λ−α>0. Next, using (28) with ρ=γ and κ=α, one obtains
(49)∫0Tξt−1/p−1ITγIT1−αξ′′tp/p−1dt=Γλ+1Γλ+γ−αp/p−1T−λ∫0TT−tλ+γ−1−αp/p−1dt=Γλ+1Γλ+γ−αp/p−1Tγ−1−αp/p−1∫0T1−tTλ+γ−1−αp/p−1dt=Γλ+1Γλ+γ−αp/p−1Tγ−1−αp/p−1+1∫011−sλ+γ−1−αp/p−1ds,which yields
(50)∫0Tξt−1/p−1ITγIT1−αξ′′tp/p−1dt=C2Tγ−1−αp/p−1+1,where
(51)C2=Γλ+1Γλ+γ−αp/p−1∫011−sλ+γ−1−αp/p−1ds>0.
Similarly, one has(52)∫0Tξt−1/p−1ITγIT1−βξ′′tp/p−1dt=C3Tγ−1−βp/p−1+1,where
(53)C3=Γλ+1Γλ+γ−βp/p−1∫011−sλ+γ−1−βp/p−1ds>0.
Therefore, it follows from (44), (48), (50), and (52) that
(54)C1u1T1−α≤Cε,pC2Tγ−1−αp/p−1+1+C3Tγ−1−βp/p−1+1,which yields
(55)u1≤Cε,pC1Tα+γ−1−αp/p−1C2+C3Tα−βp/p−1,T>0.
Notice that for allp>1, one has
(56)α+γ−1−αpp−1<0,α−βpp−1≤0from33.
Hence, using (56) and passing to the limit as T⟶+∞ in (55), one obtains u1≤0, which contradicts the fact that u1>0. Therefore, one deduces that for all p>1, problem (1) admits no global solution.Proof of Theorem 4.
Letu1=0. First, one observes that in this case u≡u1 is a global solution to (1). Suppose now that u∈AC2([0,∞)) is a global solution to (1). Taking u1=0 in (43), one obtains
(57)∫0TCD0γutpξtdt≤C′ε,p∫0Tξt−1/p−1ITγIT1−αξ′′tp/p−1dt+∫0Tξt−1/p−1ITγIT1−βξ′′tp/p−1dt,for all T>0, where C′ε,p=Cε,p/1−2ε and 0<ε<1/2. Next, using (24) and the estimates (50) and (52), one obtains
(58)∫0TCD0γutp1−tTλdt≤C′ε,pTγ−1−αp/p−1+1C2+C3Tα−βp/p−1.
Notice that sinceα≤β, one has
(59)α−βpp−1≤0.
Moreover, ifγ≤α, then
(60)γ−1−αpp−1+1<0.
Hence, passing to the infimum limit asT⟶+∞ in (58) and using Fatou’s lemma, one obtains
(61)∫0∞CD0γutpdt=0,which yields
(62)CD0γut=0,for almost everywhere t>0. Then, using the initial conditions and Lemma 9, one deduces that
(63)I0γCD0γut=I0γI01−γu′t=I01u′t=ut−u0=0,for almost everywhere t>0. Since u is continuous (u∈AC2([0,∞))), it holds that ut=u0 for all t≥0. This proves part (i) of Theorem 4.
Suppose now thatγ>α. In this case, if 1<p<1/γ−α, then (60) holds. Hence, proceeding as above, one obtains ut=u0 for all t≥0, which proves part (ii) of Theorem 4.Proof of Theorem 5.
It is sufficient to observe that any global solution to problem (2) is a global solution to problem (1). Hence, using Theorem 3, one deduces that problem (2) admits no global solution.Proof of Theorem 6.
Let us suppose thatu∈AC2([0,∞)) is a global solution to problem (2). Proceeding as in the Proof of Theorem 4 and using that u1=0, one obtains
(64)∫0Tftξtdt≤C∫0Tξt−1/p−1ITγIT1−αξ′′tp/p−1dt+∫0Tξt−1/p−1ITγIT1−βξ′′tp/p−1dt,for all T>0, where C>0 is a constant (independent on T). On the other hand, by (24), one has
(65)∫0Tftξtdt=∫0TT−λT−tλftdt≥2−λ∫0T/2ftdt.
Moreover, by (50) and (52), one has
(66)∫0Tξt−1/p−1ITγIT1−αξ′′tp/p−1dt+∫0Tξt−1/p−1ITγIT1−βξ′′tp/p−1dt≤C2Tγ−1−αp/p−1+1+C3Tγ−1−βp/p−1+1=Tγ−αp−1/p−1C2+C3Tα−βp/p−1.
Next, it follows from (64), (65), and (66) that
(67)2−λ∫0T/2ftdt≤CTγ−αp−1/p−1C2+C3Tα−βp/p−1,which yields
(68)T1+α−γp/p−1∫0T/2ftdt≤2λCC2+C3Tα−βp/p−1.Finally, passing to the supremum limit asT⟶+∞ in (68), using (14) and the fact α≤β, a contradiction follows.Proof of Corollary 7.
For alla≠0, one has
(69)∫0T/2ftdt=1aeaT/2−1.
Ifa>0, then
(70)∫0T/2ftdt~1aeaT/2,asT⟶+∞,which yields
(71)limT→+∞Tα−γp+1/p−1∫0T/2ftdt=limT→+∞1aTα−γp+1/p−1eaT/2=+∞.
Hence, by Theorem6, one deduces that for all p>1, problem (2) admits no global solution, which proves part (i).
Ifa<0, then
(72)∫0T/2ftdt~−1a,asT⟶+∞,which yields
(73)limT→+∞Tα−γp+1/p−1∫0T/2ftdt=limT→+∞−1aTα−γp+1/p−1.
Therefore, ifα≥γ, one has
(74)α−γp+1>0,p>1,which yields
(75)limT→+∞Tα−γp+1/p−1∫0T/2ftdt=+∞,p>1.
Hence, by Theorem6, one deduces that for all p>1, problem (2) admits no global solution, which proves part (ii). On the other hand, if γ>α, one has
(76)α−γp+1>0,1<p<1γ−α,which yields
(77)limT→+∞Tα−γp+1/p−1∫0T/2ftdt=+∞,1<p<1γ−α.
Hence, by Theorem6, one deduces that for all 1<p<1/γ−α, problem (2) admits no global solution, which proves part (iii).Proof of Corollary 8.
For allσ>−1, one has
(78)∫0T/2ftdt=12σ+1σ+1Tσ+1−2σ+1,which yields
(79)∫0T/2ftdt~Tσ+12σ+1σ+1,asT⟶+∞.
Hence,(80)limT→+∞Tα−γp+1/p−1∫0T/2ftdt=limT→+∞12σ+1σ+1Tα−γp+1/p−1+σ+1.
Notice that(81)α−γp+1p−1+σ+1>0⇔α+1−γ+σp>σ.
Hence, by Theorem6, one deduces that for all p>1 satisfying
(82)α+1−γ+σp>σ,problem (2) admits no global solution.
Consider the caseσ≥0. In this case, for all p>1, one has
(83)α+1−γ+σp>σp≥σ.
Then (82) is satisfied for all p>1. Hence, one deduces that for all p>1, problem (2) admits no global solution, which proves part (i).
Suppose now that−1<σ<0. If γ≤α, then for all p>1, one has
(84)α+1−γ+σp≥σ+1p>0>σ.
Then, (82) is satisfied for all p>1. Hence, one deduces that for all p>1, problem (2) admits no global solution, which proves part (ii)(a). On the other hand, if γ>α and −1<γ−α−1≤σ<0, then
(85)α+1−γ+σp≥0>σ.
Hence, (82) is satisfied for all p>1. Therefore, for all p>1, problem (2) admits no global solution, which proves part (ii)(b). Finally, if γ>α and −1<σ<γ−α−1, then for all p>1, (82) is equivalent to
(86)1<p<σα+1−γ+σ.
Hence, for allp satisfying the above condition, problem (2) admits no global solution, which proves part (ii)(c).



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