Nonexistence of Global Solutions to Time-Fractional Damped Wave Inequalities in Bounded Domains with a Singular Potential on the Boundary

We first consider the damped wave inequality ∂2u∂t2−∂2u∂x2+∂u∂t≥xσ|u|p,t>0,x∈(0,L), where L>0, σ∈R, and p>1, under the Dirichlet boundary conditions (u(t,0),u(t,L))=(f(t),g(t)),t>0. We establish sufficient conditions depending on σ, p, the initial conditions, and the boundary conditions, under which the considered problem admits no global solution. Two cases of boundary conditions are investigated: g≡0 and g(t)=tγ, γ>−1. Next, we extend our study to the time-fractional analogue of the above problem, namely, the time-fractional damped wave inequality ∂αu∂tα−∂2u∂x2+∂βu∂tβ≥xσ|u|p,t>0,x∈(0,L), where α∈(1,2), β∈(0,1), and ∂τ∂tτ is the time-Caputo fractional derivative of order τ, τ∈{α,β}. Our approach is based on the test function method. Namely, a judicious choice of test functions is made, taking in consideration the boundedness of the domain and the boundary conditions. Comparing with previous existing results in the literature, our results hold without assuming that the initial values are large with respect to a certain norm.

The investigation of the question of blow-up of solutions to initial boundary value problems for semilinear wave equations started in the 1970s. For example, Tsutsumi [1] considered the nonlinear damped wave equation under homogeneous Dirichlet boundary conditions, where b ≥ 0 and for some κ > 0 and ρ > 0. By means of the energy method, the author established sufficient conditions for the blow-up of solutions. In [2], using a concavity argument, Levine established sufficient conditions for the blow-up of solutions to an abstract Cauchy problem in a Hilbert space, of the form where P and A are positive symmetric operators and F is a nonlinear operator satisfying certain conditions. Later, the concavity method was used and developed by many authors in order to study more general problems. For further blow-up results for nonlinear wave equations, obtained by means of the energy/concavity method, see e.g., [3][4][5][6][7][8][9][10][11] and the references therein. Fractional operators arise in various applications, such as chemistry, biology, continuum mechanics, anomalous diffusion, and materials science, see for instance [12][13][14][15][16]. Consequently, many mathematicians dealt with the study of fractional differential equations in both theoretical and numerical aspects, see e.g., [17][18][19][20][21].
In [22], Kirane and Tatar considered the time-fractional damped wave equation where p > 1, α ∈ (−1, 1), and Ω is a bounded domain of R N . Using some arguments based on Fourier transforms and the Hardy-Littlewood inequality, it was shown that the energy grows exponentially for sufficiently large initial data. By combining an argument due to Georgiev and Todorova [23] with the techniques used in [22], Tatar [24] proved that the solutions to (3) blow up in finite-time for sufficiently large initial data.
In all the above cited references, the blow-up results were obtained for sufficiently large initial data. In this paper, we use a different approach than those used in the above mentioned references. Namely, our approach is based on the test function method introduced by Mitidieri and Pohozaev [25]. Taking into consideration the boundedness of the  domain as well as the boundary conditions, adequate test functions are used to obtain  sufficient conditions for the nonexistence of global weak solutions to problems (1) and (2). Notice that our results hold without assuming that the initial values are large with respect to a certain norm.
Let us mention also that recently, methods for the numerical diagnostics of the solution's blow-up have been actively developing (see e.g., [26][27][28]), which make it possible to refine the theoretical estimates.
The rest of the paper is organized as follows: In Section 2, we provide some preliminaries on fractional calculus, and some useful lemmas. We state our main results in Section 3. The proofs are presented in Section 4.

Preliminaries on Fractional Calculus
For the reader's convenience, we recall below some notions from fractional calculus, see e.g., [17,20].
Let T > 0 be fixed. Given ρ > 0 and v ∈ L 1 ([0, T]), the left-sided and right-sided Riemann-Liouville fractional integrals of order ρ of v, are defined, respectively, by In this case, we may consider I Given a positive integer n, τ ∈ (n − 1, n), and v ∈ C n ([0, T]), the (left-sided) Caputo fractional derivative of order τ of v, is defined by for all t ∈ [0, L].
We have the following integration by parts rule.
The following inequality will be useful later.

Statement of the Main Results
We first consider problem (1). Let We introduce the test function space Remark 2. The weak formulation (9) is obtained by multiplying the differential inequality in (1) by ϕ, integrating over Q, and using the initial conditions in (1). So, clearly, any global solution to (1) is a global weak solution to (1) in the sense of Definition 1.
We first consider the case g ≡ 0. If then (1) admits no global weak solution.

Remark 3.
Comparing with the existing results in the literature, in Theorem 1, it is not required that the initial data are sufficiently large with respect to a certain norm. The same remark holds for the next theorems.
Then, all the assumptions of Theorem 1 are satisfied. Consequently, we deduce that (1) admits no global weak solution.
Next, we consider the case when where C g > 0 is a constant.
, and g be the function defined by (12). If one of the following conditions is satisfied: Then, by the statement (ii) of Theorem 2, we deduce that (1) admits no global weak solution.
Consider now problem (2). For all T > 0, let We introduce the test function space (13) is obtained by multiplying the differential inequality in (2) by ϕ, integrating over Q T , using the initial conditions in (2), and using the fractional integration by parts rule provided by Lemma 1. So, clearly, any global solution to (2) is a global weak solution to (2) in the sense of Definition 2.

Remark 4. The weak formulation
As for problem (1), we first consider the case g ≡ 0.
and one of the following conditions is satisfied: then (2) admits no global weak solution.
Next, we consider the inhomogeneous case, where the function g is given by (12).
, and g be the function defined by (12). If and one of the following conditions is satisfied: Then (17) is satisfied, σ ≥ −(p + 1), and γ > 0. Then, by Theorem 4, we deduce that (2) admits no global weak solution.

Proof of the Main Results
Throughout this section, any positive constant independent on T and R, is denoted by C. Namely, in the proofs, we use several asymptotic estimates as T → ∞ and R → ∞; therefore, the value of any positive constant independent of T and R has no influence in our analysis.

Proof of Theorem 1
Proof. Suppose that u is a global weak solution to (1). Then, by (9), for every ϕ ∈ Φ, there holds On the other hand, using Lemma 3 with ε = 1 3 and adequate choices of a and b, we obtain Using (18)-(21), we obtain where Consider now two cut-off functions ξ, µ ∈ C ∞ ([0, ∞)) satisfying the following properties: For sufficiently large and R, let where θ > 0 is a constant that will be determined later. Consider the function By the properties of the cut-off functions ξ and µ, it can be easily seen that the function ϕ defined by (24), belongs to Φ. Thus, the estimate (22) holds for this function. Now, let us estimate the terms I j (ϕ), j = 1, 2, 3. For j = 1, by (24), we obtain On the other hand, by the definitions of the function ϕ 1 and the cut-off function ξ, there holds By the definitions of the function ϕ 2 and the cut-off function µ, we obtain is the indicator function of the interval 1 2 R −1 , R −1 . Then, there holds Thus, it follows from (25)- (27) that For j = 2, I j (ϕ) can be written as By the definitions of the function ϕ 1 and the cut-off function ξ, we obtain Moreover, we have On the other hand, by (11), we have σ < p − 1, thus we deduce that Combining (29)-(31), there holds Now, let us estimate I 3 (ϕ). This term can be written as 3 (ϕ 1 )I 3 (ϕ 2 ).
A similar calculation as above yields Observe that I 3 (ϕ 2 ) = I 2 (ϕ 2 ). Thus, by (31), (33), and (34), we obtain Next, combining (28), (32), and (35), we obtain Let θ be such that Notice that by (11), we have θ > 0. Then, (36) reduces to Next, let us estimate the terms from the right side of (22). Observe that by the definition of the function ϕ, and the properties of the cut-off function µ, we have By the properties of the cut-off function ξ, we have Thus, we obtain Then, taking into consideration that u 0 , u 1 ∈ L 1 ([0, L]), by the dominated convergence theorem, we obtain Hence, by (10), for sufficiently large R, there holds Next, combining (22), (37), (38), and (40), we obtain Passing to the limit as R → ∞ in the above inequality, we obtain which contradicts (10). Consequently, (1) admits no global weak solution. The proof is completed.

Proof of Theorem 2
Proof. As was performed previously, suppose that u is a global weak solution to (1). From the proof of Theorem 1, for sufficiently large R, there holds where θ > 0 and ϕ is the function defined by (24). On the other hand, by the definition of the function ϕ, for sufficiently large R, there holds Then, by (41), we deduce that Let σ < −(p + 1). In this case, (42) reduces to Taking θ > 0 so that passing to the limit as R → ∞ in (43), and using (39), we obtain a contradiction with C > 0. This proves part (i) of Theorem 2. Let σ ≥ −(p + 1) and γ > 0. If −(p + 1) ≤ σ < p − 1, then (43) holds. Since γ > 0, there exists θ > 0 such that (44) holds. Thus, passing to the limit as R → ∞ in (43), we obtain a contradiction. If σ = p − 1, then (42) yields As in the previous case, since γ > 0, there exists θ > 0 such that (44) holds. Thus, passing to the limit as R → ∞ in the above inequality, we obtain a contradiction. If σ > p − 1, then (42) yields Taking θ such that (44) is satisfied, and passing to the limit as R → ∞ in the above inequality, a contradiction follows. Thus, part (ii) of Theorem 2 is proved.

Proof of Theorem 3
Proof. Suppose that u is a global weak solution to (2). Then, by (13), for every T > 0 and ϕ ∈ Φ T , there holds On the other hand, using Lemma 3 with ε = 1 3 and adequate choices of a and b, we obtain and where For sufficiently large T, λ, , and R, let where η is the function defined by (4), and ϕ 2 is the function given by (23). Using Lemma 2 and the properties of the cut-off function µ, it can be easily seen that the function ϕ defined by (50), belongs to Φ T . Thus, (49) holds for this function. Let us estimate the terms J j (ϕ), j = 1, 2, 3. For j = 1, by (50), we have An elementary calculation shows that Hence, using (27), (51), and (52), we obtain For j = 2, we have Moreover, by Lemma 2, we obtain Integrating over (0, T), there holds Next, taking into consideration that σ < −(p + 1) (so σ < p − 1), it follows from (31), (54), and (55) that Proceeding as above, we obtain Hence, by (53), (56), and (57), we obtain Consider now the terms from the right side of (49). By (50) and the properties of the cut-off function µ, since g ≡ 0, there holds On the other hand, using (50) and Lemma 2, for all x ∈ [0, L], we obtain Consequently, we obtain Thus, combining (49), (58)-(60), we obtain Next, taking T = R θ , where θ > 0 is a constant that will be determined later, the above inequality reduces to Suppose that (14) holds. In this case, we obtain Hence, for sufficiently large R, Combining (61) with (62), we obtain Observe that, since α < β + 1, we have Hence, taking into consideration that σ < −(p + 1), picking θ > 0 so that and passing to the limit as R → ∞ in (63), we obtain a contradiction with C > 0. Suppose that (15) holds. Then, Thus, (61) reduces to Moreover, we have for sufficiently large R. Hence, using (64), and following the same argument as above, a contradiction follows. Finally, suppose that (16) holds. In this case, we obtain Hence, for sufficiently large R, Combining (61) with (65), we obtain Taking θ > 0 such that and passing to the limit as R → ∞ in (66), a contradiction follows. This completes the proof of Theorem 3.

Proof of Theorem 4
Proof. Suppose that u is a global weak solution to (2). From the proof of Theorem 3, for sufficiently large T and R, there holds where ϕ is the function defined by (50). On the other hand, by (50) and the properties of the cut-off function µ, we have where B denotes the Beta function. Thus, by (67), we obtain Taking T = R θ , where θ > 0 is a constant that will be determined later, the above inequality reduces to Let σ < −(p + 1). In this case, for sufficiently large R, there holds Since by (17), β + γ > 0, there holds Thus, taking θ > 0 so that using (17), and passing to the limit as R → ∞ in (69), we obtain a contradiction with C > 0. This proves part (i) of Theorem 4. Let σ ≥ −(p + 1) and γ > 0. If −(p + 1) ≤ σ < p − 1, then (69) holds. Since γ > 0, there exists θ > 0 satisfying (70). Thus, passing to the limit as R → ∞ in (69), a contradiction follows. If σ = p − 1, then (68) yields As in the previous case, since γ > 0, there exists θ > 0 satisfying (70). Thus, passing to the limit as R → ∞ in (71), a contradiction follows. If σ > p − 1, then (68) yields So, taking θ > 0 satisfying (70) and and passing to the limit as R → ∞ in (72), a contradiction follows. This proves part (ii) of Theorem 4.

Conclusions
Using the test function method, sufficient conditions for the nonexistence of global weak solutions to problems (1) and (2) are obtained. For each problem, an adequate choice of a test function is made, taking into consideration the boundedness of the domain and the boundary conditions. Comparing with previous existing results in the literature, our results hold without assuming that the initial values are large with respect to a certain norm.
In this paper, we treated only the one dimensional case. It will be interesting to study problems (1) and (2)