# Necessary Conditions for the Existence of Global Solutions to Nonlinear Fractional Differential Inequalities and Systems

Authors: Ibtisam Aldawish; Bessem Samet
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, fractional differential inequalities and systems of fractional differential inequalities involving fractional derivatives in the sense of Caputo are investigated. Namely, necessary conditions for the existence of global solutions are obtained. Our approach is based on the test function method and some integral inequalities.


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## Body

## 1. Introduction and Main Results

In several studies, the usefulness of fractional derivatives in the mathematical modeling of various phenomena from physics and engineering has been demonstrated (see, e.g., [1–7], and the references therein). Due to this fact, the study of fractional differential equations has received a great deal of attention from many researchers. The existence of solutions is one of the most important topics of fractional differential equations. The study of sufficient conditions for the existence of solutions has been investigated by many authors using different approaches from functional analysis (see, e.g., [8–18], and the references therein). The study of necessary conditions for the existence of global solutions in the context of fractional differential equations has been initiated by Kirane and his collaborators (see, e.g., [19–24], and the references therein).In [19], Furati and Kirane investigated the system of fractional differential equations:
(1)u1′t+DC0α1u1t≥u2tp2,t>0,u2′t+DC0α2u2t≥u1tp1,t>0,u10,u20=u10,u20,where 0<α1,α2<1, u10,u20>0, and p1,p2>1. Here, DC0αi, i∈1,2, denotes the Caputo fractional derivative of order αi. Namely, it was shown that if
(2)1−1p1p2≤maxα1+α2p1,α2+α1p2,then (1) admits no global solution.Motivated by Furati and Kirane [19], in this paper, we first consider the fractional differential inequality
(3)u′t+CD0αut≥ρtuatp+μt,t>0,u0=u0,where 0<α<1, p>1, u0∈ℝ, ρ∈C0,∞, ρ>0, μ∈Lloc10,∞, and μ≥0. It is supposed that a:0,∞⟶0,∞ is a C1 function and satisfies(A1)a′t>0 for all t≥0(A2)a0=0(A3)at≥t for all t≥0Our aim is to study the influence ofat on the large time behavior of solutions. By a global solution to (3), we mean a function u∈AC0,∞ (an absolutely continuous function) satisfying the fractional differential inequality in (3) for almost everywhere t>0, and the initial condition u0=u0. Our aim is to derive sufficient conditions for which (3) admits no global solution. Namely, the following result is obtained.Theorem 1.
Suppose that(4)limsupT⟶+∞Tαp/p−1u0T1−α+∫0T/2μtdt∫0Tρa−1ta−1′t−1/p−1dt=+∞.
Then (3) admits no global solution.We provide below some examples where (4) is satisfied.Example 2.
Consider problem (3) with at=t, ρ≡1, μ≡0, and u0>0. Then,
(5)Tαp/p−1u0T1−α+∫0T/2μtdt∫0Tρa−1ta−1′t−1/p−1dt=u0Tα/p−1⟶+∞,asT⟶+∞.Hence, by Theorem1, we deduce that for all p>1, (3) admits no global solution.Example 3.
Consider problem (3) with at=t, ρ≡1, μ≡1, and u0∈ℝ. Then
(6)Tαp/p−1u0T1−α+∫0T/2μtdt∫0Tρa−1ta−1′t−1/p−1dt=u0Tαp/p−1−1+12Tαp/p−1⟶+∞,asT⟶+∞.Hence, by Theorem1, we deduce that for all p>1, (3) admits no global solution.Example 4.
Consider problem (3) with u0>0 and
(7)at=t+1γ−1,ρt=t+1β,μt=tδ,where
(8)γ≥1,−1<δ<−α,γ1−α<1+β,p>max1+βγ,1.In this case, after elementary calculations, we obtain(9)Tαp/p−1u0T1−α+∫0T/2μtdt∫0Tρa−1ta−1′t−1/p−1dt~Cu0Tβ+1−γ1−α/γp−1⟶+∞,asT⟶+∞,where C>0 is a constant independent of T. Hence, by Theorem 1, we deduce that for all p>max1+β/γ,1, (3) admits no global solution.Example 5.
Consider problem (3) with u0∈ℝ and
(10)at=t+1γ−1,ρt=t+1β,μt=tδ,where
(11)γ≥1,δ>−α,γ1−α<1+β,p>max1+βγ,δ+1−1+β/γα+δ,1.In this case, after elementary calculations, we obtain(12)Tαp/p−1u0T1−α+∫0T/2μtdt∫0Tρa−1ta−1′t−1/p−1dt~CTpα+δ−δ+1−1+β/γ/p−1⟶+∞,asT⟶+∞,where C>0 is a constant independent of T. Hence, by Theorem 1, we deduce that for all p>max1+β/γ,δ+1−1+β/γ/α+δ,1, (3) admits no global solution.In the second part of this paper, we extend the previous study to the system of fractional differential inequalities:(13)u1′t+CD0α1u1t≥ρ2tu2a2tp2+μ2t,t>0,u2′t+CD0α2u2t≥ρ1tu1a1tp1+μ1t,t>0,u10,u20=u10,u20,where for i∈1,2, 0<αi<1, pi>1, ui0∈ℝ, ρi∈C0,∞, ρi>0, μi∈Lloc10,∞, and μi≥0. Moreover, it is supposed that ai:0,∞⟶0,∞ is a C1 function and satisfies (A1)–(A3). Notice that in the special case ρi≡1, ait=t, and μi≡0, (13) reduces to (1).By a global solution to (13), we mean a pair of functions u1,u2∈AC0,∞×AC0,∞ satisfying the fractional differential inequalities in (13) for almost everywhere t>0, and the initial condition u10,u20=u10,u20.We have the following result.Theorem 6.
(i)
Letu10≥0 and u20∈ℝ. If(14)limsupT⟶+∞Tα1p1+α2p1p2/p1p2−1∫0T/2μ1tdt+T1−α2u20∫0Tρ1a1−1ta1−1′t−1/p1−1dtp1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp1p2−1/p1p2−1=+∞,then (13) admits no global solution
(ii)
Letu10∈ℝ and u20≥0. If(15)limsupT⟶+∞Tα2p2+α1p1p2/p1p2−1∫0T/2μ2tdt+T1−α1u10∫0Tρ1a1−1ta1−1′t−1/p1−1dtp2p1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp2−1/p1p2−1=+∞,then (13) admits no global solutionWe provide below some examples for which (14) or (15) are satisfied.Example 7.
Consider (1), that is, System (13) with ait=t, ρi≡1, μi≡0, and ui0>0, for all i∈1,2. Then
(16)Tα1p1+α2p1p2/p1p2−1∫0T/2μ1tdt+T1−α2u20∫0Tρ1a1−1ta1−1′t−1/p1−1dtp1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp1p2−1/p1p2−1=Tα1p1+α2/p1p2−1u20⟶+∞,asT⟶+∞.Hence, by Theorem6, we deduce that for all p1,p2>1, (1) admits no global solution. This improves [19] (Theorem 1), where the nonexistence of a global solution was obtained only when 1−1/p1p2≤maxα1+α2/p1,α2+α1/p2.Example 8.
Consider System (13) with
(17)a1t=tt+2,a2t=t,ρ1t=ρ2t=t+13,μ1t=μ2t=tδ,α1=α2=α,−1<δ<−α,u10,u20>0,p1>2,p2>4.In this case, after elementary calculations, we obtain(18)Tα1p1+α2p1p2/p1p2−1∫0T/2μ1tdt+T1−α2u20∫0Tρ1a1−1ta1−1′t−1/p1−1dtp1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp1p2−1/p1p2−1~CTα+3p1+α+1/p1p2−1⟶+∞,asT⟶+∞,(19)Tα2p2+α1p1p2/p1p2−1∫0T/2μ2tdt+T1−α1u10∫0Tρ1a1−1ta1−1′t−1/p1−1dtp2p1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp2−1/p1p2−1~CTαp1+1+p2/p1p2−1⟶+∞,asT⟶+∞.Hence, by Theorem6, we deduce that for all p1>2 and p2>4, (13) admits no global solution.The rest of this paper is organized as follows. In Section2, we recall briefly some notions related to fractional calculus and provide some lemmas that will be used in the proofs of our main results. In Section 3, we prove Theorems 1 and 6.


## 2. Preliminaries

We first recall some basic notions and properties related to fractional calculus (see, e.g., [25, 26]).LetT>0 and I=0,T. The left-sided Riemann-Liouville fractional integral of order σ>0 of a function f∈L1I is defined by
(20)I0σft=1Γσ∫0tt−sσ−1fsds,for almost everywhere t∈I, where Γ is the Gamma function. The right-sided Riemann-Liouville fractional integral of order σ>0 of a function f∈L1I is defined by
(21)ITσft=1Γσ∫tTs−tσ−1fsds,for almost everywhere t∈I.The Caputo fractional derivative of orderσ∈0,1 of a function f∈ACI is defined by
(22)DC0σft=I01−σf′t=1Γ1−σ∫0tt−s−σf′sds,for almost everywhere t∈I.Lemma 9 (see [25], Lemma 2.7).
Letσ>0, r,s≥1, and 1/r+1/s≤1+σ (r≠1, s≠1, in the case 1/r+1/s=1+σ). If f,g∈LrI×LsI, then
(23)∫0TI0σftgtdt=∫0TftITσgtdt.
Forθ≫1 (θ is sufficiently large), let
(24)ϕt=T−θT−tθ,t∈I.The following results can be found in [19].Lemma 10.
Letσ>0. Then
(25)ITσϕt=Γθ+1Γσ+θ+1T−θT−tθ+σ,ITσϕ′t=−Γθ+1Γσ+θT−θT−tθ+σ−1
for allt∈I.


## 3. Proofs of the Main Results

The proofs of our main results are based on the test function method developed by Mitidieri and Pohozaev [27].Proof of Theorem1.
We use the contradiction argument. Namely, suppose thatu∈AC0,∞ is a global solution to (3). Multiplying the fractional differential inequality in (3) by the function ϕ defined by (24) with θ≫1, and integrating over 0,T, T≫1, we obtain
(26)∫0Tρtuatpϕtdt+∫0Tμtϕtdt≤∫0Tu′tϕtdt+∫0TDC0αutϕtdt.
By the initial conditionu0=u0, we have
(27)∫0Tu′tϕtdt=uTϕT−u0ϕ0−∫0Tutϕ′tdt.Notice that by the definition ofϕ, we have ϕT=0 and ϕ0=1. Therefore, it holds that
(28)∫0Tu′tϕtdt=−u0−∫0Tutϕ′tdt.Next, using Lemma9, we obtain
(29)∫0TDC0αutϕtdt=∫0TI01−αu′tϕtdt=∫0Tu′tIT1−αϕtdt.Integrating by parts and using the initial conditionu0=u0, we obtain
(30)∫0Tu′tIT1−αϕtdt=uTIT1−αϕT−u0IT1−αϕ0−∫0TutIT1−αϕ′tdt.On the other hand, by Lemma10, we have IT1−αϕT=0 and IT1−αϕ0=Γθ+1/Γ2−α+θT1−α. Therefore,
(31)∫0Tu′tIT1−αϕtdt=−Γθ+1Γ2−α+θu0T1−α−∫0TutIT1−αϕ′tdt.Combining (29) with (31), we deduce that
(32)∫0TDC0αutϕtdt=−Γθ+1Γ2−α+θu0T1−α−∫0TutIT1−αϕ′tdt.Hence, it follows from (28) and (29) that
(33)∫0Tu′tϕtdt+∫0TDC0αutϕtdt=−1+Γθ+1Γ2−α+θT1−αu0−∫0Tutϕ′tdt−∫0TutIT1−αϕ′tdt.Consider now the terms from the left side of (26). By a change of variable, using the properties (A1)–(A3), and the decay property of ϕ, we obtain
(34)∫0Tρtuatpϕtdt=∫0aTρa−1tutpϕa−1ta−1′tdt≥∫0Tρa−1tutpϕta−1′tdt.Next, by (24), and using that μ≥0 (we have also μ∈Lloc10,∞), we obtain
(35)∫0Tμtϕtdt=T−θ∫0TμtT−tθdt≥T−θ∫0T/2μtT−tθdt≥2−θ∫0T/2μtdt.Combining (34) with (35), we deduce that
(36)∫0Tρtuatpϕtdt+∫0Tμtϕtdt≥∫0Tρa−1tutpϕta−1′tdt+2−θ∫0T/2μtdt.Hence, it follows from (26), (33), and (36) that
(37)∫0Tutpψtdt+2−θ∫0T/2μtdt+1+Γθ+1Γ2−α+θT1−αu0≤∫0Tutϕ′tdt+∫0TutIT1−αϕ′tdt,where
(38)ψt=ρa−1ta−1′tϕt,t∈0,T.Next, using Young’s inequality, we obtain(39)∫0Tutϕ′tdt≤12∫0Tutpψtdt+C∫0Tψt−1/p−1ϕ′tp/p−1dt,(40)∫0TutIT1−αϕ′tdt≤12∫0Tutpψtdt+C∫0Tψt−1/p−1IT1−αϕ′tp/p−1dt,where C denotes a positive constant (independent of T) whose value may change from line to line. Then, it follows from (37), (39), and (40) that
(41)2−θ∫0T/2μtdt+1+Γθ+1Γ2−α+θT1−αu0≤CI1T+I2T,where
(42)I1T=∫0Tψt−1/p−1ϕ′tp/p−1dt,I2T=∫0Tψt−1/p−1IT1−αϕ′tp/p−1dt.Let us estimate the termsIiT, i=1,2. For all t∈0,T, by (24) and (38), we have
(43)ψt−1/p−1ϕ′tp/p−1=ρa−1ta−1′tϕt−1/p−1ϕ′tp/p−1=Cρa−1ta−1′t−1/p−1Tθ/p−1T−t−θ/p−1T−θp/p−1T−tθ−1p/p−1≤Cρa−1ta−1′t−1/p−1T−θTθ−p/p−1=Cρa−1ta−1′t−1/p−1T−p/p−1.Integrating over0,T, we obtain
(44)I1T≤T−p/p−1∫0Tρa−1ta−1′t−1/p−1dt.Next, by (38) and Lemma 10, for all t∈0,T, we have
(45)ψt−1/p−1IT1−αϕ′tp/p−1=Cρa−1ta−1′tT−θT−tθ−1/p−1T−θT−tθ−αp/p−1=Cρa−1ta−1′t−1/p−1T−θT−tθ−αp/p−1≤Cρa−1ta−1′t−1/p−1T−αp/p−1.Integrating over0,T, we obtain
(46)I2T≤T−αp/p−1∫0Tρa−1ta−1′t−1/p−1dt.Therefore, it follows from (41), (44), and (46) that
(47)2−θ∫0T/2μtdt+1+Γθ+1Γ2−α+θT1−αu0≤CT−p/p−1∫0Tρa−1ta−1′t−1/p−1dt+T−αp/p−1∫0Tρa−1ta−1′t−1/p−1dt≤CT−αp/p−1∫0Tρa−1ta−1′t−1/p−1dt,which yields
(48)∫0T/2μtdt+T1−αu0≤CT−αp/p−1∫0Tρa−1ta−1′t−1/p−1dt.Hence, we deduce that(49)Tαp/p−1∫0T/2μtdt+T1−αu0∫0Tρa−1ta−1′t−1/p−1dt≤C,which contradicts (4). The proof is completed.Proof of Theorem6.
Suppose thatu1,u2∈AC0,∞×AC0,∞ is a global solution to (13). Multiplying the first fractional differential inequality in (13) by the function ϕ defined by (24) with θ≫1, and integrating over 0,T, T≫1, we obtain
(50)∫0Tρ2tu2a2tp2ϕtdt+∫0Tμ2tϕtdt≤∫0Tu1′tϕtdt+∫0TDC0α1u1tϕtdt.Following the same steps of the Proof of Theorem1, we obtain
(51)∫0Tu1′tϕtdt+∫0TDC0α1u1tϕtdt=−1+Γθ+1Γ2−α1+θT1−α1u10−∫0Tu1tϕ′tdt−∫0Tu1tIT1−α1ϕ′tdt,(52)∫0Tρ2tu2a2tp2ϕtdt+∫0Tμ2tϕtdt≥∫0Tρ2a2−1tu2tp2ϕta2−1′tdt+2−θ∫0T/2μ2tdt.Hence, it follows from (50), (51), and (52) that
(53)∫0Tu2tp2ψ2tdt+2−θ∫0T/2μ2tdt+1+Γθ+1Γ2−α1+θT1−α1u10≤∫0Tu1tϕ′tdt+∫0Tu1tIT1−α1ϕ′tdt,where
(54)ψ2t=ρ2a2−1ta2−1′tϕt,t∈0,T.Arguing similarly with the second inequality in (13), we obtain
(55)∫0Tu1tp1ψ1tdt+2−θ∫0T/2μ1tdt+1+Γθ+1Γ2−α2+θT1−α2u20≤∫0Tu2tϕ′tdt+∫0Tu2tIT1−α2ϕ′tdt,where
(56)ψ1t=ρ1a1−1ta1−1′tϕt,t∈0,T.Next, by Hölder’s inequality, we have(57)∫0Tu1tϕ′tdt≤∫0Tu1tp1ψ1tdt1/p1∫0Tψ1t−1/p1−1ϕ′tp1/p1−1dtp1−1/p1,(58)∫0Tu1tIT1−α1ϕ′tdt≤∫0Tu1tp1ψ1tdt1/p1∫0Tψ1t−1/p1−1IT1−α1ϕ′tp1/p1−1dtp1−1/p1.Similarly, we have(59)∫0Tu2tϕ′tdt≤∫0Tu2tp2ψ2tdt1/p2∫0Tψ2t−1/p2−1ϕ′tp2/p2−1dtp2−1/p2,(60)∫0Tu2tIT1−α2ϕ′tdt≤∫0Tu2tp2ψ2tdt1/p2∫0Tψ2t−1/p2−1IT1−α2ϕ′tp2/p2−1dtp2−1/p2.Fori∈1,2, let
(61)Ji=∫0Tuitpiψitdt,KψiT=∫0Tψit−1/pi−1ϕ′tpi/pi−1dtpi−1/pi,LψiT=∫0Tψit−1/pi−1IT1−αiϕ′tpi/pi−1dtpi−1/pi.It follows from (53), (57), and (58) that
(62)J2+2−θ∫0T/2μ2tdt+1+Γθ+1Γ2−α1+θT1−α1u10≤J11/p1Kψ1T+Lψ1T.Similarly, by (55), (59), and (60), we obtain
(63)J1+2−θ∫0T/2μ1tdt+1+Γθ+1Γ2−α2+θT1−α2u20≤J21/p2Kψ2T+Lψ2T.Now, we consider the caseu10≥0 and u20∈ℝ. In this case, using (62) and (63), we obtain
(64)J1+2−θ∫0T/2μ1tdt+1+Γθ+1Γ2−α2+θT1−α2u20≤J11/p1p2Kψ1T+Lψ1T1/p2Kψ2T+Lψ2T.On the other hand, using Young’s inequality, we obtain(65)J11/p1p2Kψ1T+Lψ1T1/p2Kψ2T+Lψ2T≤1p1p2J1+p1p2−1p1p2Kψ1T+Lψ1Tp1/p1p2−1Kψ2T+Lψ2Tp1p2/p1p2−1.Therefore, combining (64) with (65), it holds that
(66)2−θ∫0T/2μ1tdt+1+Γθ+1Γ2−α2+θT1−α2u20≤CKψ1T+Lψ1Tp1/p1p2−1Kψ2T+Lψ2Tp1p2/p1p2−1,which yields
(67)∫0T/2μ1tdt+T1−α2u20≤CKψ1T+Lψ1Tp1/p1p2−1Kψ2T+Lψ2Tp1p2/p1p2−1.Next, using the estimates (44) and (46), we obtain (after elementary calculations)
(68)Kψ1T+Lψ1Tp1/p1p2−1Kψ2T+Lψ2Tp1p2/p1p2−1≤CT−α1p1−α2p1p2/p1p2−1∫0Tρ1a1−1ta1−1′t−1/p1−1dtp1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp1p2−1/p1p2−1.Combining (67) with (68), we obtain
(69)Tα1p1+α2p1p2/p1p2−1∫0T/2μ1tdt+T1−α2u20∫0Tρ1a1−1ta1−1′t−1/p1−1dtp1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp1p2−1/p1p2−1≤C,which contradicts (14).In the second case, whenu10∈ℝ and u20≥0, following the same steps as used in the previous case, we obtain
(70)Tα2p2+α1p1p2/p1p2−1∫0T/2μ2tdt+T1−α1u10∫0Tρ1a1−1ta1−1′t−1/p1−1dtp2p1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp2−1/p1p2−1≤C,which contradicts (15). The proof is completed.


## 4. Conclusion

In this paper, problems (3) and (13) are investigated. Namely, using the test function method and some integral inequalities, sufficient conditions for the nonexistence of global solutions (or equivalently, necessary conditions for the existence of global solutions) to the considered problems are obtained. For problem (3), we proved that (see Theorem 1) under assumptions (A1), (A2), and (A3), if
(71)limsupT⟶+∞Tαp/p−1u0T1−α+∫0T/2μtdt∫0Tρa−1ta−1′t−1/p−1dt=+∞,then (3) admits no global solution. For the system of fractional differential inequalities (13), always under assumptions (A1), (A2), and (A3), we proved that (see Theorem 6), if u10≥0 and
(72)limsupT⟶+∞Tα1p1+α2p1p2/p1p2−1∫0T/2μ1tdt+T1−α2u20∫0Tρ1a1−1ta1−1′t−1/p1−1dtp1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp1p2−1/p1p2−1=+∞,oru20≥0 and
(73)limsupT⟶+∞Tα2p2+α1p1p2/p1p2−1∫0T/2μ2tdt+T1−α1u10∫0Tρ1a1−1ta1−1′t−1/p1−1dtp2p1−1/p1p2−1∫0Tρ2a2−1ta2−1′t−1/p2−1dtp2−1/p1p2−1=+∞,then (13) admits no global solution.



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