Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 40, pp. 1–16. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2024.40w NONEXISTENCE RESULTS FOR FRACTIONAL DIFFERENTIAL INEQUALITIES JEFFREY R. L. WEBB Abstract. We prove nonexistence of global solution of fractional differential inequalities of the form Dαu(t) ≥ λtβ |u(t)|p when p > 1 for each of the Riemann-Liouville and Caputo fractional derivatives. This is motivated by work of Laskri and Tatar (Comput. Math. Appl. (2010)) and Shan and Lv (Filomat (2024)). The result of Laskri-Tatar was claimed to be false by Zhang, Liu, Wu and Cui (J. Funct. Spaces (2017)) with a correction and counter- example. We show that the counter-example and the claims are not accurate. We use a different method to that of Laskri and Tatar, our result supports the one of Laskri and Tatar. We also improve on the result in Shan and Lv paper by considering a more general problem and giving a more precise conclusion. 1. Introduction Some years ago Laskri and Tatar [15] proved that global solutions of an inequality for the Riemann-Liouville (R-L) fractional derivative do not exist. They study the inequality Dαu(t) ≥ tβ |u(t)|p, t > 0, where p > 1, 0 < α < 1 and β > −α, (1.1) with initial condition (IC) I1−αu(0) = b ≥ 0, where I1−αu is the R-L fractional integral and Dαu = D(I1−αu) denotes the R-L fractional derivative; detailed defi- nitions are given later in the paper. The authors considered solutions belonging to a space they denoted Lα := {u ∈ L1 : Dαu ∈ L1}. Their result is the following. Theorem 1.1 ([15, Theorem 1]). Assume that β+α > 0 and 1 < p < 1+β 1−α . Then, problem (1.1) does not admit global nontrivial solutions when b ≥ 0. They give an example where for p ≥ 1+β 1−α non-zero solutions exist for all t, so p0 := 1+β 1−α is a critical exponent. We will show that for p ≥ p0 and b > 0 there is no solution in the space we use for solutions, while for p < 1+β 1−α any solution must fail to exist at or before some finite value T1 explicitly determined by the data and parameters. The reason for these differences is that we have b > 0, the example has b = 0. We expect that for 1 < p < 1+β 1−α solutions will become unbounded at some point T2 ≤ T1 (blow-up) but this requires knowledge of what occurs at the 2020 Mathematics Subject Classification. 34A08, 34A40, 45D05. Key words and phrases. Fractional differential equations; non-existence; Volterra integral equation. ©2024. This work is licensed under a CC BY 4.0 license. Submitted March 25, 2024. Published July 30, 2024. 1 2 J. R. L. WEBB EJDE-2024/40 endpoint of a maximal interval of existence and we do not know of any such result for R-L differential equations. Zhang, Liu, Wu, and Cui [25] claim that the proof in [15] has a flaw and that the result is not correct. They give a ‘counter-example’ and a ‘correct version’ of Theorem 1.1. Unfortunately the correction is wrong, and the counter-example has an error, which invalidates their claims. We explain this fully in Section 4. One of the problems studied by Shan and Lv [18] is the initial value problem (IVP) for the Caputo fractional derivative Dα ∗ u(t) = up(t), t > 0, with IC u(0) = u0 > 0. (1.2) Shan-Lv assert that when 1 < p ≤ 1/(1 − α) every solution of (1.2) blows-up (becomes unbounded) at some T ∗ and the same holds when p > 1/(1− α) and u0 has a sufficiently large positive lower bound. They do not fully prove this, they prove nonexistence but do not prove blow-up nor do they mention any continuation result that would give blow-up. For this Caputo derivative equation there is a continuation result of Eloe and Masthay [6, Theorem 2.4] and of Wu and Liu [24, Theorem 4.1] which asserts that when f is continuous, a solution u of Dα ∗ u(t) = f(t, u(t)), t > 0, with IC u(0) = u0, exists on a maximal interval of existence [0, T0) and, if T0 is finite, u(t) → ∞ as t → T− 0 . This could be used to justify the blow-up claims of Shan-Lv. We improve the Shan-Lv result by considering the more general inequality as in (1.1) with the possibly singular term tβ . We also prove, by a simple method, an improved result, namely that, for any initial value u0 > 0, when p > 1 global solutions do not exist. Since we have a more general case of an inequality and a singular term, we do not know how to prove that this is blow-up. There are a number of generalizations of the results in [15], for example the papers [7, 8, 9] but they use different methods and have little relevance to this paper so we do not discuss them. Some comments on existence theorems. There are existence results for Ca- puto fractional differential equations of the form Dα ∗ u(t) = t−ηf(t, u(t)), t > 0, u(0) = u0, where 0 < η < α and also for a somewhat more general non-negative function f . For local existence, existence on some possibly short interval, continuity of f is sufficient, while for global existence it is usually supposed that |f(t, u)| ≤ C1+C2|u|p where p ≤ 1, and a Gronwall or Bihari type inequality is used to get suitable a priori bounds, see for example [14, 20, 21, 22]. Our results show that for global existence p ≤ 1 cannot be improved to have p > 1. For the R-L fractional derivative case Dαu(t) = f(t, u(t)), t > 0, lim t→0+ t1−αu(t) = u0, (1.3) when |f(t, u)| ≤ k(t) + l(t)|u|, under a variety of conditions on k, l, which allow singularities in the t variable, Zhu [26, 27, 28] has proved various global existence theorems. In the papers [1, 2, 3], Becker-Burton-Purnaras give interesting results concerning existence theory for R-L fractional differential equations. With a sign condition, opposite to the sign we have, global L1 solutions are possible for p > 1. For EJDE-2024/40 FRACTIONAL DIFFERENTIAL INEQUALITIES 3 example, in [1, Example 4.12], D1/2u(t) = − √ π 2 t3/4u3/2 with the initial condition limt→0+ t1/2u(t) = 1 is shown to have an explicit global solution u(t) = 1√ t(1+t) , t > 0. A result closely related to the problem we study gives local existence. Theorem 1.2 ([2, Theorem 3.1]). Let 0 < α < 1, β > −1, and p ≥ 0 satisfy β−p+α(1+p) > 0. Suppose that f : (0,∞)×R → R is continuous. Suppose there are nonnegative constants K1,K2 such that |f(t, u)| ≤ K1+K2t β |u|p for u ∈ R and 0 < t < T0, where T0 ∈ (0,∞]. Then, for u0 ̸= 0, the problem Dαu(t) = f(t, u(t)), t > 0, lim t→0+ t1−αu(t) = u0, has a solution in Cα−1[0, T ] for some T ∈ (0, T0). Here u ∈ Cα−1 means that u is continuous on (0, T ] and limt→0+ t1−αu(t) exists. Note that β − p + α(1 + p) > 0 is equivalent to p < α+β 1−α so it is implicit that α+ β > 0. Thus local existence is possible in this case. Our Theorem 4.2 has a similar but larger critical value of p. We show that when 1 < p < 1+β 1−α solutions can exist only on a finite interval [0, T ), where T ≤ T1 for some explicitly determined T1, while for p ≥ 1+β 1−α existence in the space Cα−1[0, T ] is impossible for any T > 0. In the paper [2], for the problem Dαu(t) = un(t), t > 0, lim t→0+ t1−αu(t) = u0, (1.4) with n ∈ N, the authors showed that whether or not a solution of the initial value problem (1.4) exists on an interval (0, T ], for some T > 0, depends on the value of α. One of their results is the following. Theorem 1.3 ([2, Theorem 3.11]). Let 0 < α < 1, n ∈ N and u0 ̸= 0. The initial value problem (1.4) has a solution if and only if α > n−1 n . Moreover, the solution is unique. For the problem (1.4) with n replaced by p, a special case of our Theorem 4.2 shows that for u0 > 0 there does not exist a nontrivial solution in the space Cα−1[0, T ] if p ≥ 1 1−α , that is α ≤ p−1 p , which partially extends Theorem 1.3. Lan [11, 12, 13] has studied more general fractional problems that include the Ca- puto and R-L fractional equations as special cases, and he has proved equivalences between fractional differential and integral equations. 2. Preliminaries We consider real valued functions defined on an arbitrary finite interval [0, T ], which is, by a simple change of variable, equivalent to any finite interval. In this paper all functions are supposed to be measurable, all integrals are Lebesgue integrals and L1[0, T ] denotes the usual space of Lebesgue integrable functions; we will often simply write L1. The space of functions that are continuous on [0, T ] is denoted by C[0, T ], or simply C, and is endowed with the supremum norm ∥u∥∞ := maxt∈[0,T ] |u(t)|, C1 = C1[0, T ] will denote the space of continuously differentiable functions. When studying fractional integrals and derivatives, functions such as tα−1 arise where typically 0 < α < 1. This leads to consideration of a weighted space of 4 J. R. L. WEBB EJDE-2024/40 functions that are continuous except at t = 0 and have an integrable singularity at t = 0. For γ > −1 we define the space denoted Cγ = Cγ [0, T ] by Cγ [0, T ] := {u ∈ C(0, T ] such that lim t→0+ t−γu(t) exists}. (2.1) Then u ∈ Cγ if and only if u(t) = tγU(t) for some function U ∈ C[0, T ] and we define ∥u∥γ := ∥U∥∞. The spaces of functions with a singularity at t = 0 are C−γ where γ > 0. For 0 < γ < 1 we have C−γ ⊂ L1. We also use the space of absolutely continuous functions which is denoted AC. The space AC is the appropriate space for the fundamental theorem of the calculus for Lebesgue integrals. In fact, we have the following equivalence. u ∈ AC[0, T ] if and only if u′ ∈ L1[0, T ], u′(t) exists for almost every (a.e.) t ∈ [0, T ] and u(t)− u(0) = ∫ t 0 u′(s) ds for all t ∈ [0, T ]. (2.2) The Gamma and Beta functions frequently occur in fractional problems. The Gamma function is, for α > 0, given by Γ(α) := ∫ ∞ 0 sα−1 exp(−s) ds. (2.3) The Gamma function has the property Γ(α + 1) = αΓ(α) for α > 0. The Beta function is defined for α > 0, β > 0 by B(α, β) := ∫ 1 0 (1− s)α−1sβ−1 ds. (2.4) These are well defined Lebesgue integrals and it is well known that B(α, β) = Γ(α)Γ(β) Γ(α+ β) . We will also use the following property which is proved by the simple substitution s = tσ. Lemma 2.1. Let t > 0 and α > 0, β > 0. Then we have∫ t 0 (t− s)α−1sβ−1 ds = tα+β−1B(α, β). (2.5) These properties will be used without further mention. We will use the so-called Riemann-Liouville (R-L) fractional integral. Using this we will consider the two most often used fractional derivatives, the R-L and the Caputo versions. The R-L fractional integral is defined for L1 functions as follows. Definition 2.2. The Riemann-Liouville (R-L) fractional integral of order α > 0 of a function u ∈ L1[0, T ] is defined for a.e. t by Iαu(t) := 1 Γ(α) ∫ t 0 (t− s)α−1u(s) ds. (2.6) The integral Iαu is the convolution of the L1 functions h, u where h(t) = tα−1/Γ(α), so by the well known results on convolutions Iαu is defined as an L1 function, in particular Iαu(t) is defined and finite for a.e. t. If α = 1 this is the usual integration operator which we denote I. We define Iαu(0) := limt→0+ Iαu(t) if this limit exists, otherwise it is not defined. Detailed discussion of these operators can be found in the texts [5, 10, 17], a survey of some important results is given in the free to access paper [21]. One useful result is the semigroup property as follows, see for example [5, Theorem 2.2], [17, (2.21)], [21, Lemma 2.4]. EJDE-2024/40 FRACTIONAL DIFFERENTIAL INEQUALITIES 5 Lemma 2.3 (Semigroup property). Let α, β > 0 and u ∈ L1[0, T ]. Then IαIβ(u) = Iα+β(u) as L1 functions, thus, IαIβ(u)(t) = Iα+β(u)(t) for a.e. t ∈ [0, T ], in fact for every t for which Iα+β(|u|)(t) exists. If u is continuous this holds for all t ∈ [0, T ]. If u ∈ L1 and α+ β ≥ 1 equality again holds for all t ∈ [0, T ]. In this paper we only consider fractional derivatives of order 0 < α < 1. Let D denote the usual differentiation operator, Du = u′. The Riemann-Liouville (R-L) fractional derivative of order α ∈ (0, 1) is informally defined by Dαu(t) = D(I1−αu)(t). For D I1−αu(t) to be defined at a point t, it is necessary that I1−αu should be differentiable at t which requires some extra condition, which we now discuss. It is useful to know when the fractional derivative and fractional integral are inverse operations, that is when a fractional differential equation (FDE) with an initial condition is equivalent to an integral equation. One frequently used, but imprecise statement, is as follows. If 0 < α < 1, then u satisfies Dαu = f and I1−α(0) = c/Γ(α) if and only if u(t) = Iαf(t) + ctα−1. If u ∈ L1 then I1−αu ∈ L1 but need not be differentiable. Assuming additionally that I1−αu is differentiable almost everywhere then Dαu(t) = f(t) can be satisfied for a.e. t, but it is not equivalent to an integral equation, it is necessary to always have I1−αu ∈ AC. This was noted long ago in the monograph [17], see [17, Defi- nition 2.4] and the related comments in the ‘Notes to §2.6’. It was recalled in [21]. Therefore a suitable precise definition is as follows. Definition 2.4. For α ∈ (0, 1) and u ∈ L1 the R-L fractional derivative Dαu is defined when I1−αu ∈ AC as an L1 function by Dαu(t) := D I1−αu(t), a.e. t ∈ [0, T ]. (2.7) Then we do have an equivalence which is stated below in Proposition 3.1. The Caputo differential operator, or Caputo fractional derivative, is usually used for continuous functions u and is defined via the R-L derivative, as in the texts [5, Definition 3.2], [10, (2.4.1)]. Definition 2.5. For α ∈ (0, 1), u ∈ C and I1−α(u − u(0)) ∈ AC the Caputo derivative Dα ∗ u is defined by Dα ∗ u := Dα(u− u(0)). (2.8) This defines Dα ∗ u as an L1 function, so Dα ∗ u(t) is defined and finite for a.e. t. The Caputo derivative of a constant is 0 but the R-L derivative is not, Dαc = c Γ(1−α) t −α for t > 0. There is another commonly used definition of Caputo derivative namely: Definition 2.6. For 0 < α < 1, the Caputo derivative Dα Cu is defined for u ∈ AC as an L1 function by Dα Cu(t) := I1−αu′(t), for a.e. t. (2.9) We will not use the definition Dα Cu because it has the severe disadvantage that for u continuous the ‘equivalence’ between the fractional initial value problem (IVP) Dαu(t) = f(t), u(0) = u0 and the Volterra integral equation u(t) = u0 + (Iαf)(t) is not valid. Iα maps C[0, T ] into C[0, T ] but not (all of) C[0, T ] into AC[0, T ]; examples are in Cichon-Salem [4, Counter-Example 1], and Webb [21, addendum]. A detailed discussion is given in [14]. When u ∈ AC, Dα ∗ u = Dα Cu, so often there is no reason to use Dα C . 6 J. R. L. WEBB EJDE-2024/40 3. Riemann-Liouville equivalences For 0 < α < 1 an initial value problem for the R-L fractional differential equation (FDE) Dαu = f , I1−αu(0) = limt→0+ Iαu(t) = c0, with f ∈ L1 can be studied in the space L1. An equivalence with an integral equation is given by the following result, for example [10, Lemma 2.5(b)], [17, Theorem 2.4] and [21, Proposition 6.1]. Proposition 3.1. Let f ∈ L1[0, T ] and c0 ∈ R. Then a function u ∈ L1 such that I1−αu ∈ AC satisfies Dαu(t) = f(t) a.e. and I1−αu(0) = c0 if and only u ∈ L1 satisfies the Volterra integral equation u(t) = c0 tα−1 Γ(α) + 1 Γ(α) ∫ t 0 (t− s)α−1f(s) ds, a.e. t ∈ [0, T ]. (3.1) We illustrate part of the argument needed for the proof of Proposition 3.1 by showing a positivity result. Lemma 3.2. For 0 < α < 1 suppose that u ∈ L1 and I1−αu ∈ AC[0, T ] and that Dαu = f where f ∈ L1 and f(t) ≥ 0 a.e. on [0, T ]. Then I1−αu(0) = c0 ≥ 0 implies that u(t) ≥ c0 tα−1 Γ(α) for a.e. t ∈ (0, T ]. If, in addition, u ∈ Cα−1 then t1−αu(t) ≥ c0 Γ(α) for all t ∈ (0, T ]. Proof. Dαu = f means thatD(I1−αu) = f . Since I1−αu ∈ AC[0, T ] and I1−αu(0) = c0, this can be integrated to give I1−αu(t) = c0 + If(t), for all t ∈ [0, T ]. Applying Iα and using the semigroup property gives Iu = c0 tα Γ(α+1) +I(Iαf). Since all terms are AC, the derivatives exist a.e., which gives u(t) = c0 tα−1 Γ(α) + (Iαf)(t) for a.e. t, and proves the result since (Iαf)(t) ≥ 0 for a.e. t. The last part follows since then both sides are continuous functions of t ∈ (0, T ]. □ The problem (3.1) can also be studied in the smaller space Cα−1 = C−(1−α) when the initial condition limt→0+ I1−αu(t) = c0 is replaced by limt→0+ t1−αu(t) = c0/Γ(α). The two limits are related as follows. Lemma 3.3. Let 0 < α < 1 and suppose that u ∈ L1. Then lim t→0+ u(t)t1−α = u0 implies that I1−αu(0) = lim t→0+ I1−αu(t) = u0Γ(α). The result is proved for example in [21, Lemma 6.3], a longer proof is given in [1, Theorem 6.1], also the more general case when α ∈ C with 0 < Re(α) < 1 is proved in [10, Lemma 3.2, page 151]. The converse of this result is false. Example 3.4. Let 0 < γ ≤ α < 1 and let Q denote the rational numbers. Let u(t) := { tγ−1, for t ∈ (0, T ] ∩Q, 0, otherwise. Then u ∈ L1[0, T ], limt→0+ I1−αu(t) = 0 but limt→0+ t1−αu(t) does not exist. Proof. Since u(t) = 0 a.e. on [0, T ], (I1−αu)(t) = 0 for every t > 0 so limt→0+ I1−αu(t) = 0. Also we have t1−αu(t) := { tγ−α, for t ∈ (0, T ] ∩Q, 0, otherwise, and for α ≥ γ, limt→0+ t1−αu(t) does not exist. □ EJDE-2024/40 FRACTIONAL DIFFERENTIAL INEQUALITIES 7 The following Proposition [1, Theorem 6.2] proves the equivalence between the FDE and the Volterra integral equation in the space Cα−1. Proposition 3.5. Let 0 < α < 1, and let f be continuous on (0, T ] × J where J ⊂ R is an unbounded interval. If u : (0, T ] → J is continuous, u ∈ L1[0, T ] and t 7→ f(t, u(t)) belongs to L1[0, T ], then u satisfies the initial value problem, Dαu(t) = f(t, u(t)), t ∈ (0, T ], lim t→0+ t1−αu(t) = u0, (3.2) if and only if it satisfies the Volterra integral equation u(t) = u0tα−1 + 1 Γ(α) ∫ t 0 (t− s)α−1f(s, u(s)) ds, t ∈ (0, T ]. (3.3) The difference between these equivalence results is that Proposition 3.5 assumes thatDαu(t) exists for every t ∈ (0, T ] and functions that are continuous on (0, T ] are considered, as opposed to supposing that functions are in L1 and Dαu(t) exists only a.e. in Proposition 3.1. The conditions in Proposition 3.5 imply that I1−αu ∈ AC as shown in [21, Remark 6.6]. 4. Non-existence for R-L inequalities Henceforth for non-existence results we consider p > 1 since there are theorems of existence on an arbitrary interval [0, T ] when p ≤ 1, for example Zhu [26, 27] for the R-L case, and [14, 23] for the Caputo case. In the paper [15] Laskri and Tatar study the inequality Dαu(t) ≥ tβ |u(t)|p, t > 0, where p > 1, with IC I1−αu(0) = b. (4.1) They consider u ∈ Lα where Lα := {u ∈ L1 : Dαu ∈ L1}. Their result is as follows. Theorem 4.1 ([15, Theorem 1]). Assume that β+α > 0 and 1 < p < 1+β 1−α . Then, problem (4.1) does not admit global nontrivial solutions when b ≥ 0. From Lemma 3.2 we see that for b > 0 any solution must be positive a.e.. When b = 0 there is the trivial solution u = 0 but nontrivial solutions are possible, see Remark 4.5 below. Laskri and Tatar use a test function method due to Mitidieri and Pokhozhaev [16]. Zhang, Liu, Wu, and Cui [25] claim that the proof in [15] has some flaws and that the result is not correct. They give a ‘counter-example’ and a ‘correct version’ of Theorem 4.1. Unfortunately both the correction and the counter-example are wrong. We now explain these points. Laskri-Tatar [15] consider a nonincreasing C1[0,∞) test function φ ≥ 0 such that, for some τ > 0, φ(t) = { 1, if t ≤ τ/2, 0, if t ≥ τ. Then for a supposed positive solution u of (4.1), the integral ∫ τ 0 φ′(t)I1−αu(t) dt is estimated from above. The integrand in this integral is non-positive and is negative on an interval. The mistake in [15] is that absolute value signs have been omitted but, in fact, it is the absolute value of the integral that is estimated. When the absolute value signs are added in [15] the flaw claimed in [25] disappears. The paper [25] claims, that for 1 < p < 1+β 1−α and b ≥ 0, based on their ‘counter- examples’, the problem has infinitely many global nontrivial positive solutions. Also 8 J. R. L. WEBB EJDE-2024/40 it is claimed that the problem does not have any global nontrivial negative solutions. The last point actually follows immediately from Lemma 3.2, not from a correction of the proof of [15]. The ‘counter-example’ claimed in [25] is for the case α = 1/2, β = −1/6, p = 3/2. It is stated that, for c = Γ(3/2), u1(t) = { c2t1/2, if t ≤ 1, c2t−4/5, if t > 1, is a global solution of (4.1) with b = 0. For t ≤ 1, D1/2u1(t) = D(I1/2u1)(t) = c3 and the inequality (4.1) holds. For t > 1 the authors give D1/2u1(t) = c3t−13/10. This is not correct. It seems that they apply a known formula, but that formula is for a function equal to a single power of t for all t > 0. For h(t) = c2t−4/5 for t > 0, the formula gives D1/2h(t) = c2c1t −13/10 where the constant c1 is negative, so it is not equal to c3t−13/10. The correct calculation of D1/2u1(t) = D(I1/2u1)(t) for t > 1 starts as follows. I1/2u1(t) = 1 Γ(1/2) ∫ t 0 (t− s)−1/2u1(s) ds = c2 1 Γ(1/2) (∫ 1 0 (t− s)−1/2s1/2 ds+ ∫ t 1 (t− s)−1/2s−4/5 ds ) . The first integral is a decreasing function of t so it contributes a negative amount to the fractional derivative, the other term is not known in terms of elementary functions and its monotonicity properties are not easy to prove, but a Maple cal- culation and graph suggest that it first increases then decreases, so it would not be a counter-example. However, the authors of [25] claim that, by “the proof of the above example”, a more general case also holds, namely for u2(t) = { ktα, if t ≤ 1, kt−µ−(α+β)/(p−1), if t > 1, where p > 1 and k = Γ(α)1/(p−1), µ ≥ 1, α+β > 0, it is claimed that u2 is a global solution of the inequality (4.1). For this case we can prove this is false by giving a counter-example that can be readily checked. Let α = 1/2, β = 1, µ = 1 and p = 2. The corresponding fractional integral is as follows. For t ≤ 1, I1/2u2(t) = 1 Γ(1/2) ∫ t 0 (t− s)−1/2ks1/2 ds = ktΓ(3/2). The fractional derivative D1/2u2(t) = D(I1/2u2)(t) = kΓ(3/2) is constant for t ≤ 1. For t > 1 we have I1/2u2(t) = 1 Γ(1/2) ∫ t 0 (t− s)−1/2u2(s) ds = 1 Γ(1/2) (∫ 1 0 (t− s)−1/2ks1/2 ds+ ∫ t 1 (t− s)−1/2ks−5/2 ds ) . The second integral can be evaluated and we have, for t > 1,∫ t 1 (t− s)−1/2ks−5/2 ds = k 2 √ t− 1(t− 2) 3t2 . EJDE-2024/40 FRACTIONAL DIFFERENTIAL INEQUALITIES 9 This function of t first decreases for t ∈ (1, t1] where t1 = 4− 2 √ 2 ≈ 1.17157, then increases to a maximum at t2 = 4+2 √ 2 ≈ 6.82843 and then decreases again. Since the first part of I1/2u2(t) is a decreasing function of t we see that the fractional derivative is certainly negative except possibly for t in part of the interval (t1, t2), so can never be a global solution of (4.1) since the right side is always non-negative. For the case b = 0 it is possible to have explicit solutions of the corresponding R- L equation Dαu(t) = tβ |u(t)|p when p > 1 for some values of α, β, see the details in Example 4.6 below. The values used in the above discussion do not fit the example. We will prove non-existence results by a different method to that used by Laskri- Tatar [15] which supports their conclusion. We do not believe the space Lα is an adequate space to consider the inequality (4.1). For, in the proof of [15, Theorem 1], the integration by parts requires I1−α ∈ AC, thus a better definition of Lα is Lα := {u ∈ L1 : I1−αu ∈ AC}. We will use a smaller space and have u ∈ Cα−1 so that u ∈ L1 and is continuous on (0, T ]. For 0 < α < 1, α+ β > 0 and p > 1 we will study the following problem. Dαu(t) ≥ tβ |u(t)|p, t > 0, with IC lim t→0+ t1−αu(t) = u0. (4.2) By a solution u of (4.2) on an interval [0, T ] we will mean that u ∈ Cα−1[0, T ], I1−αu ∈ AC[0, T ], Dαu is continuous on (0, T ], tβ |u(t)|p ∈ L1[0, T ], the inequality is satisfied for all t ∈ (0, T ] and the IC is satisfied. A global solution would be a solution u(t) which exists for all finite t > 0. We will prove that no global solution exists, a precise statement is given in the following Theorem. Firstly we prove a result for an equation. Theorem 4.2. For p > 1 and u0 > 0 let f ∈ L1 with f continuous on (0, T ] and f(t) ≥ 0 for t > 0. Then for λ > 0 and β + α > 0, the equation Dαu(t) = λtβ |u(t)|p + f(t), t > 0 with limt→0 t 1−αu(t) = u0 does not have a global solution u ∈ Cα−1. In fact, for p ≥ 1+β 1−α there is no solution in the space Cα−1[0, T ] for any T > 0, while for 1 < p < 1+β 1−α there exists T1, explicitly determined by the given data and parameters, such that a solution can exist on an interval [0, T ] only if T < T1. Note that β can be negative. The hypothesis β + α > 0 ensures that 1+β 1−α > 1. Proof. From Lemma 3.2 we note that, if a solution u exists, then t1−αu(t) ≥ u0 > 0 for all t in its interval of existence, so u(t) > 0 for t > 0 in its interval of existence. Since Dαu ∈ L1, it is necessary that tβup(t) ∈ L1, otherwise no solution can exist. However, when t1−αu(t) = U(t) ≥ u0 > 0, with U continuous on [0, T ], tβup(t) = tβ−p(1−α)Up(t) is in L1[0, T ] for some T > 0 if and only if β−p(1−α) > −1, that is p < 1+β 1−α , so no solution can exist when p ≥ 1+β 1−α . We recover this again below but for now we suppose a solution exists on an interval [0, T ] with T > 0. Let v = cu where cp−1 = λ, then v ∈ Cα−1 satisfies Dαv(t) = tβvp(t) + cf(t), t ∈ (0, T ], with IC lim t→0 t1−αv(t) = c u0. By Proposition 3.5, v satisfies v(t) = c u0tα−1 + Iα(sβvp(s))(t) + cIαf(t), t ∈ (0, T ]. 10 J. R. L. WEBB EJDE-2024/40 Then w(t) = t1−αv(t) is continuous for t ∈ [0, T ]. Discarding the last non-negative term, we have w satisfies the inequality w(t) ≥ c u0 + t1−α 1 Γ(α) ∫ t 0 (t− s)α−1sβvp(s) ds = c u0 + 1 Γ(α) ∫ t 0 (1− s/t)α−1sβ−p(1−α)wp(s) ds ≥ c u0 + 1 Γ(α) ∫ t 0 sβ−p(1−α)wp(s) ds, where we used (1 − s/t)α−1 ≥ 1. Note that w(t) ≥ c u0 for all t ∈ [0, T ]. If β − p(1− α) ≤ −1, that is, p ≥ 1+β 1−α , then∫ t 0 sβ−p(1−α)wp(s) ds ≥ ∫ t 0 sβ−p(1−α)(c u0)p ds, and the last integral does not exist for any t > 0, so there is no solution in the space Cα−1[0, T ] for any T > 0. This can be thought of as instantaneous blow-up, or blow-up at 0. For β − p(1 − α) > −1, that is, for 1 < p < 1+β 1−α , let γ = p(1 − α) − β, then γ < 1. By Theorem 1.2, local solutions can certainly exist in some cases. We suppose that u and hence also w exist on an interval (0, T ] with T > 0. Thus we have w(t) ≥ c u0 + 1 Γ(α) ∫ t 0 s−γwp(s) ds for all t ∈ [0, T ] since terms are continuous. Let g(t) := c u0 + 1 Γ(α) ∫ t 0 s−γwp(s) ds, then g ∈ AC, g is strictly increasing, g(t) ≥ g(0) = c u0 > 0, and g′(t) = t−γwp(t) ≥ 1 Γ(α) t −γgp(t). Since gp ∈ AC is positive, integrating g′ gp ≥ 1 Γ(α) t −γ gives g1−p(t)− g1−p(0) 1− p ≥ 1 Γ(α) t1−γ 1− γ , that is g1−p(t) ≤ (cu0)1−p − (p− 1) 1 Γ(α) t1−γ 1− γ , for all t ∈ [0, T ]. Let T1 = [ (1−γ)Γ(α) (p−1)(cu0)p−1 ] 1 1−γ . Then we have (cu0)1−p − (p − 1) 1 Γ(α) T 1−γ 1 1−γ ≤ 0, but since g(t) is positive on its interval of existence, g1−p(t) cannot exist at t = T1. Thus w, and hence u, can exist on an interval [0, T ] only for T < T1. □ Remark 4.3. We expect that the solution u exists on an interval (0, T2) for some T2 ≤ T1 and blows up at T2 but we do not know a proof of this. If by some means we knew that T2 = T1 then we would have blow-up. Laskri-Tatar [15] prove non-existence of a global solution but do not have any estimate of the interval of existence (0, T2). Theorem 4.4. Let p > 1 and λ > 0. The fractional inequality Dαu(t) ≥ λtβup(t) for t > 0, together with limt→0 t 1−αu(t) = u0 > 0, does not have a global solution u ∈ Cα−1. In fact, for p ≥ 1+β 1−α there is no solution in Cα−1[0, T ] for any T > 0, while for p < 1+β 1−α any solution can only exist on an interval [0, T ] for T < T1, with T1 as given in Theorem 4.2. EJDE-2024/40 FRACTIONAL DIFFERENTIAL INEQUALITIES 11 Proof. Let f(t) = Dαu(t)−λtβup(t). For p ≥ 1+β 1−α and u ∈ Cα−1 the term λtβup(t) is not integrable and no solution exists. Otherwise we can apply Theorem 4.2 noting that T1 does not depend on f . □ Remark 4.5. Laskri-Tatar [15] state that p0 = 1+β 1−α is a critical exponent, arguing that no global solution exists for 1 < p < p0, while for p ≥ p0 a global solution exists by quoting an explicit example given in [10, Example 3.3]. This appears to contradict our result in Theorem 4.2, but the real reason is that the example has u0 = 0 while we have u0 > 0. The case p ≥ p0 is not discussed in [15] when u0 > 0. We now give extra details of this example since it is only stated in [15]. They cite the monograph [10, Example 3.3], but there is a missing minus sign in the constant term in the formula in [10, page 177]) which misprint is copied in [15]. Example 4.6. Consider the equation Dαu = λtβ |u(t)|p, t > 0, with lim t→0 t1−αu(t) = 0 and λ > 0, p > 0. (4.3) The zero function is a solution. We show that a positive solution of the form u(t) = ctr, c > 0 can exist. Let p0 = 1+β 1−α , where 0 < α < 1, but β can be of either sign. We have by direct calculation, I1−αu = 1 Γ(1− α) ∫ t 0 (t− s)−αcsrds = ct1−α+r Γ(1 + r) Γ(2 + r − α) , where we impose 1 + r−α > 0 for this to be AC and to have a nonzero derivative. Therefore Dαu(t) = c Γ(1+r) Γ(1+r−α) t r−α for 1 + r − α > 0. Hence, for u = ctr to be a solution, we require that r − α = β + pr and cp−1 = Γ(1 + r) λΓ(1 + r − α) . (4.4) The value p = 1 is possible only if α+ β = 0 and λ, r satisfy Γ(1+r) Γ(1+r−α) = λ. If α + β = 0 and p ̸= 1 then r = 0 which gives a constant solution with cp−1 = 1/(λΓ(1− α)). We now suppose that α + β ̸= 0. For p > 1, we have r = −(α+β) p−1 . It is readily verified that 1− α+ r > 0 if p(1− α) > 1 + β, that is p > p0. Also it then follows that u ∈ Cα−1[0, T ] and λtβup ∈ L1[0, T ] for every T > 0, thus u is a solution when p > 1 and p > p0. Note that p0 > 1 if and only if α + β > 0. Therefore if α + β > 0 we must have p > p0 > 1, whereas if α + β < 0 any p > 1 is allowed. When α + β < 0, we have r > 0 and u is continuous; this also gives an explicit solution for the Caputo derivative case when u(0) = 0. For 0 < p < 1, we have r = α+β 1−p and 1 + r − α > 0 requires p < p0. If p0 ≥ 1, that is α + β > 0, any p < 1 is allowed, while if p0 ≤ 1, that is α + β ≤ 0, it must be 0 < p < p0 < 1. Remark 4.7. Our Theorem 4.2 shows that, when α + β > 0, p0 is critical for u0 > 0 because when 1 < p < p0 solutions can exist only for 0 < t ≤ T < T1 for an explicit T1, but solutions do not exist in the space Cα−1[0, T ] for any T > 0 when p ≥ p0 > 1. Thus u0 = 0 is an exceptional case. 12 J. R. L. WEBB EJDE-2024/40 5. Blow-up for Caputo derivative inequalities For 0 < α < 1, α + β ≥ 0, λ > 0 and p > 1, we will investigate continuous functions u that satisfy the inequality Dα ∗ u(t) ≥ λtβ |u(t)|p with u(0) = u0 > 0. (5.1) Our aim is to prove nonexistence of nontrivial global solutions. Of course, for u0 = 0 the trivial solution u = 0 exists for all t. Since the Caputo derivative is defined in terms of the R-L derivative it should be no surprise that a similar result to Theorem 4.2 holds. However there are some differences. In the Caputo case it is supposed that u is continuous but Dα ∗ u need not be continuous. By a solution u of the problem (5.1) on an interval [0, T ] we will mean that u ∈ C[0, T ], I1−α ∈ AC[0, T ], and tβ |u(t)|p ∈ L1[0, T ], the inequality is satisfied for t ∈ (0, T ] and the IC is satisfied. By a global solution we mean u(t) is a solution for all t > 0. We first give a result which will prove useful. It can be deduced from more general known results, for example [11, Theorem 3.2], [14, Lemma 4], [21, Theorem 5.1]. For completeness we give the simple proof. Lemma 5.1. Let 0 < α < 1, and for f ∈ L1 suppose that u ∈ C[0, T ] and I1−α(u− u0) ∈ AC[0, T ] satisfies Dα ∗ u(t) = f(t), for a.e. t > 0, with IC u(0) = u0, (5.2) Then u satisfies the Volterra integral equation u(t) = u0 + 1 Γ(α) ∫ t 0 (t− s)α−1f(s) ds, a.e. t ∈ [0, T ]. (5.3) Proof. Since u is continuous, for M > 0 there exists δ > 0 such that |u(s)−u0| < M for 0 ≤ s < δ. Then we have, for 0 < t < δ, |I1−α(u− u0)(t)| ≤ 1 Γ(1− α) ∫ t 0 (t− s)−α|u(s)− u0| ds ≤ M Γ(2− α) t1−α, thus I1−α(u−u0)(0) = 0. By definition, Dα ∗ u = f means that D(I1−α(u−u0)) = f . Since I1−α(u−u0)) ∈ AC[0, T ], by integration and the above calculation we obtain I1−α(u − u0))(t) = If(t) for all t. Then, applying Iα and using the semigroup property gives I(u−u0)(t) = I(Iαf)(t). The functions on both sides of this equation are absolutely continuous, so are differentiable almost everywhere, and we get u(t)− u0 = Iαf(t) for a.e. t. □ Remark 5.2. The converse needs more condition since, for f ∈ L1, we only have Iαf ∈ L1, and u(t) − u0 = Iαf(t) for a.e. t does not imply u(0) = u0. There is an equivalence when f is continuous as is proved in Diethelm [5, Lemma 6.2], and there are equivalences under some conditions on f weaker than continuity, for example [11, Theorem 3.2], [14, Lemma 4] and [21, Theorem 4.6]. Shan-Lv [18] studied the Caputo IVP Dα ∗ u(t) = up(t), t > 0, with IC u(0) = u0. (5.4) This is a special case of the problem we study with β = 0. They asserted that if u0 > 0 and 1 < p ≤ 1/(1 − α) then every solution of (5.4) blows-up in finite time, [18, Theorem 2.2]. They also asserted that when p > 1/(1 − α) and u0 has EJDE-2024/40 FRACTIONAL DIFFERENTIAL INEQUALITIES 13 an explicit positive lower bound, then solutions blow-up in finite time. In fact they proved non-existence of global solutions but did not prove that this is blow-up. We improve their result by considering the more general inequality with the possibly singular term tβ . We prove that, for for every p > 1 and any initial value u0 > 0, solutions can only exist on a finite interval [0, T ] with an explicit upper bound T < T1. We start with the IVP, for α+ β > 0, f ∈ L1, f(t) ≥ 0 for t > 0. Dα ∗ u(t) = λtβup(t) + f(t) with u(0) = u0 > 0. (5.5) A solution u ∈ C[0, T ] of (5.5) satisfies u(t) = u0 + λ Γ(α) ∫ t 0 (t− s)α−1(sβup(s) + f(s)) ds, a.e. t ∈ [0, T ]. (5.6) Since p is a real number, it is implicit that solutions of (5.5) are positive. In fact, u0 > 0 and continuity imply that a solution u(t) will be positive for small t > 0 and then from (5.6) it follows that u(t) > u0 for t > 0 on its interval of existence. We first consider the case 1 < p ≤ 1+β 1−α . Theorem 5.3. Let f ∈ L1 be non-negative and let λ > 0 and α+ β > 0. Consider the problem Dα ∗ u(t) = λtβup(t) + f(t), a.e. t > 0, with IC u(0) = u0. (5.7) For any u0 > 0 and for 1 < p ≤ 1+β 1−α there does not exist a global solution u. More precisely, there exists T1 > 0, explicitly determined by the parameters of the problem, such that a solution u can only exist on an interval [0, T ] where T < T1. Proof. Suppose a solution u exists on an interval [0, T ] where T > 1; otherwise we can take any T1 > 1. Define c > 0 by cp−1 = λ. Then v = cu is a solution of Dα ∗ v(t) = tβvp(t) + cf(t), a.e. t ∈ (0, T ),with IC v(0) = v0 = cu0 > 0, By Lemma 5.1, for a.e. t ∈ (0, T ], v satisfies the equation v(t) = v0 + 1 Γ(α) ∫ t 0 (t− s)α−1sβvp(s) ds+ 1 Γ(α) ∫ t 0 (t− s)α−1cf(s) ds, = v0 + 1 Γ(α) tα−1 ∫ t 0 (1− s/t)α−1sβvp(s) ds + 1 Γ(α) ∫ t 0 (t− s)α−1cf(s) ds. (5.8) The last term can be discarded and we obtain t1−αv(t) ≥ t1−αv0 + 1 Γ(α) ∫ t 0 sβ−p(1−α)(s1−αv(s))p ds, a.e. t. Let w(t) := t1−αv(t). Then w is continuous and satisfies w(t) ≥ t1−αv0 + 1 Γ(α) ∫ t 0 sβ−p(1−α)wp(s) ds. (5.9) For t ∈ [1, T ] we have w(t) ≥ v0 + 1 Γ(α) ∫ t 1 sβ−p(1−α)wp(s) ds. (5.10) 14 J. R. L. WEBB EJDE-2024/40 Let γ = p(1 − α) − β, then 1 < p ≤ 1+β 1−α implies that γ ≤ 1 and (5.10) can be written w(t) ≥ v0 + 1 Γ(α) ∫ t 1 s−γwp(s) ds. Let g(t) = v0 + 1 Γ(α) ∫ t 1 s−γwp(s) ds for t ∈ [1, T ]. Now terms are continuous so g ∈ AC, g(1) = v0, g(t) ≥ v0 > 0 for all t ≥ 1 and g′(t) = 1 Γ(α) t−γwp(t) ≥ 1 Γ(α) t−γgp(t), so g′ gp ≥ t−γ Γ(α) , t ∈ [1, T ]. We can integrate from 1 to t ≤ T to obtain g1−p(t) ≤ v1−p 0 − (p− 1) Γ(α) (t1−γ − 1) 1− γ , for γ < 1, g1−p(t) ≤ v1−p 0 − (p− 1) Γ(α) ln t, for γ = 1. (5.11) It is clear that there exists T1 > 1 (it can be written explicitly) such that the terms on the right become zero, hence g1−p(T1) does not exist, thus u can only exist on an interval [0, T ] with T < T1. □ Corollary 5.4. Let 1 < p ≤ 1+β 1−α , λ > 0 and α + β > 0. The problem Dα ∗ u(t) ≥ λtβup(t), t > 0, with u(0) = u0 > 0, does not have a global solution. Proof. The proof of Theorem 5.3 applies since for f(t) = Dα ∗ u(t)−λtβup(t), f ∈ L1 and f(t) ≥ 0 for t > 0. □ We now can deal with the case p > 1+β 1−α very simply and it gives the following result. Theorem 5.5. Let λ > 0 and α+β > 0. For any p > 1 and any u0 > 0 there does not exist a global solution u of the problem Dα ∗ u(t) ≥ λtβup(t), t > 0,with u(0) = u0 > 0. (5.12) Proof. A solution u of (5.12) will satisfy u(t) ≥ u0 > 0 on its interval of existence. For 1 < p ≤ 1+β 1−α the result is shown in Corollary 5.4, so suppose that p > 1+β 1−α . Write p = p0 + p1 where p0 = 1+β 1−α . Then up = up1u0 ≥ up1 0 up0 and from (5.12) we obtain Dα ∗ u(t) ≥ (up1 0 λ)tβup0(t), with IC u(0) = u0 > 0. By Corollary 5.4, u can only exist on some interval [0, T ] where T < T̂1 and T̂1 is determined by the parameters of the problem. □ Remark 5.6. When α + β < 0, the opposite sign to the one we have considered, and p > 1, there exists a continuous solution of the form ctr with c > 0 for the Caputo problem with initial data 0 Dα ∗ u(t) = λtβup(t), u(0) = u0 = 0. The calculation is given above in Example 4.6. The solution is ctr for r = −α−β p−1 . This nontrivial continuous solution exists if α+ β < 0 for any p > 1. EJDE-2024/40 FRACTIONAL DIFFERENTIAL INEQUALITIES 15 Remark 5.7. In the paper by Shan and Lv [18], who have the special case β = 0, at the corresponding stage (5.8) of our proof of Theorem 5.3, the authors use the inequality (t − s)α−1 ≥ (t + 1)α−1 together with comparison principles. We discuss the more general result using a different inequality at that point and simple comparisons. The special case of β = 0 in Theorem 5.3 gives a similar conclusion to [18, Theorem 2.2 (1)]. Theorem 5.5 improves [18, Theorem 2.2 (2)], which uses an inequality that requires the initial condition u0 to be bounded below by a sufficiently large explicit constant. They write that the solution blows-up but do not mention any continuation theorem that would give the proof of this. By using Wu and Liu [24, Theorem 4.1], or Eloe-Masthay [6, Theorem 2.4], the blow-up at some T2 < T ∗ can be justified for the case of an equation. Systems of Caputo inequalities are studied in [19]. A nonexistence result for the inequality Dα Cu(t) ≥ λtβup for β ≥ 0 and p > 1 is given in [19, Proposition 3.1], using a test function and capacity method. The given condition is β + 1 ≥ βp′ where 1/p + 1/p′ = 1. This is equivalent to β < p − 1 so it cannot be a sharp estimate; perhaps there is a typo. Blow-up is claimed but it seems to need further explanation for fractional inequalities. References [1] L. C. Becker, T. A. Burton, I. K. 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Differ. Equ. No. 52, 1-12 (2022). Jeffrey R. L. Webb School of Mathematics and Statistics, University of Glasgow, Glasgow G12 8SQ, UK Email address: jeffrey.webb@glasgow.ac.uk 1. Introduction Some comments on existence theorems 2. Preliminaries 3. Riemann-Liouville equivalences 4. Non-existence for R-L inequalities 5. Blow-up for Caputo derivative inequalities References