Necessary Conditions for the Existence of Global Solutions to Nonlinear Fractional Differential Inequalities and Systems
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.
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).
(A1) a′(t) > 0 for all t ≥ 0
(A2) a(0) = 0
(A3) a(t) ≥ t for all t ≥ 0
Our aim is to study the influence of a(t) on the large time behavior of solutions. By a global solution to (3), we mean a function u ∈ AC([0, ∞)) (an absolutely continuous function) satisfying the fractional differential inequality in (3) for almost everywhere t > 0, and the initial condition u(0) = u0. Our aim is to derive sufficient conditions for which (3) admits no global solution. Namely, the following result is obtained.
We provide below some examples where (4) is satisfied.
Example 2. Consider problem (3) with a(t) = t, ρ ≡ 1, μ ≡ 0, and u0 > 0. Then,
Hence, by Theorem 1, we deduce that for all p > 1, (3) admits no global solution.
Example 3. Consider problem (3) with a(t) = t, ρ ≡ 1, μ ≡ 1, and u0 ∈ ℝ. Then
Hence, by Theorem 1, we deduce that for all p > 1, (3) admits no global solution.
By a global solution to (13), we mean a pair of functions (u1, u2) ∈ AC([0, ∞)) × AC([0, ∞)) satisfying the fractional differential inequalities in (13) for almost everywhere t > 0, and the initial condition .
We have the following result.
Theorem 6.
- (i)
Let and . If
- (ii)
Let and . If
We provide below some examples for which (14) or (15) are satisfied.
Example 7. Consider (1), that is, System (13) with ai(t) = t, ρi ≡ 1, μi ≡ 0, and , for all i ∈ {1, 2}. Then
Hence, by Theorem 6, 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 ≤ max{α1 + α2/p1, α2 + α1/p2}.
Example 8. Consider System (13) with
Hence, by Theorem 6, 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 Section 2, 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]).
Lemma 9 (see [25], Lemma 2.7.)Let σ > 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) ∈ Lr(I) × Ls(I), then
For θ ≫ 1 (θ is sufficiently large), let
The following results can be found in [19].
Lemma 10. Let σ > 0. Then
for all t ∈ 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 Theorem 1. We use the contradiction argument. Namely, suppose that u ∈ AC([0, ∞)) 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
By the initial condition u(0) = u0, we have
Proof of Theorem 6. Suppose that (u1, u2) ∈ AC([0, ∞)) × AC([0, ∞)) 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
4. Conclusion
then (13) admits no global solution.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Authors’ Contributions
All authors made equal contributions and read and supported the last original copy.
Acknowledgments
The authors extend their appreciation to the Deanship of Scientific Research at Al Imam Mohammad Ibn Saud Islamic University for funding this work through Research Group no. RG-21-09-01.
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Data Availability
No data were used to support this study.