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Article

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

by
Areej Bin Sultan
,
Mohamed Jleli
and
Bessem Samet
*,†
Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Fractal Fract. 2021, 5(4), 258; https://doi.org/10.3390/fractalfract5040258
Submission received: 1 November 2021 / Revised: 26 November 2021 / Accepted: 29 November 2021 / Published: 6 December 2021

## Abstract

:
We first consider the damped wave inequality $∂ 2 u ∂ t 2 − ∂ 2 u ∂ x 2 + ∂ u ∂ t ≥ x σ | u | p , t > 0 , x ∈ ( 0 , L ) ,$ 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 $g ( t ) = t γ$, $γ > − 1$. Next, we extend our study to the time-fractional analogue of the above problem, namely, the time-fractional damped wave inequality $∂ α u ∂ t α − ∂ 2 u ∂ x 2 + ∂ β u ∂ t β ≥ x σ | u | p , t > 0 , x ∈ ( 0 , L ) ,$ where $α ∈ ( 1 , 2 )$, $β ∈ ( 0 , 1 )$, and $∂ τ ∂ t τ$ 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.
MSC:
35B44; 35B33; 26A33

## 1. Introduction

In this paper, we first consider the damped wave inequality
$∂ 2 u ∂ t 2 − ∂ 2 u ∂ x 2 + ∂ u ∂ t ≥ x σ | u | p , t > 0 , x ∈ ( 0 , L ) , ( u ( t , 0 ) , u ( t , L ) ) = ( f ( t ) , g ( t ) ) , t > 0 , u ( 0 , x ) , ∂ u ∂ t ( 0 , x ) = ( u 0 ( x ) , u 1 ( x ) ) , x ∈ ( 0 , L ) ,$
where $L > 0$, $σ ∈ R$, and $p > 1$. It is supposed that $u 0 , u 1 ∈ L 1 ( [ 0 , L ] )$, $f ∈ L l o c 1 ( [ 0 , ∞ ) )$, and $g ( t ) = C g t γ$, where $C g ≥ 0$ and $γ > − 1$, are constants. Namely, we establish sufficient conditions depending on the initial values, the boundary conditions, p, and $σ$, under which (1) admits no global weak solution, in a sense that will be specified later.
Next, we study the time-fractional analogue of (1), namely the time-fractional damped wave inequality
$∂ α u ∂ t α − ∂ 2 u ∂ x 2 + ∂ β u ∂ t β ≥ x σ | u | p , t > 0 , x ∈ ( 0 , L ) , ( u ( t , 0 ) , u ( t , L ) ) = ( f ( t ) , g ( t ) ) , t > 0 , u ( 0 , x ) , ∂ u ∂ t ( 0 , x ) = ( u 0 ( x ) , u 1 ( x ) ) , x ∈ ( 0 , L ) ,$
where $α ∈ ( 1 , 2 )$, $β ∈ ( 0 , 1 )$, and $∂ τ ∂ t τ$, $τ ∈ { α , β }$, is the time-Caputo fractional derivative of order $τ$.
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
$∂ 2 u ∂ t 2 − Δ u + b ∂ u ∂ t = F ( u ) ,$
under homogeneous Dirichlet boundary conditions, where $b ≥ 0$ and
$F ( s ) s − 2 ( 2 κ + 1 ) ∫ 0 s F ( τ ) d τ ≥ d 0 | s | ρ + 2 , s ∈ R ,$
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
$P ∂ 2 u ∂ t 2 + A u + Q ∂ u ∂ t = F ( u ) ,$
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
$∂ 2 u ∂ t 2 − Δ u + ∂ 1 + α u ∂ t 1 + α = a | u | p − 1 u , t > 0 , x ∈ Ω , u ( t , x ) = 0 , t > 0 , x ∈ ∂ Ω , u ( 0 , x ) , ∂ u ∂ t ( 0 , x ) = ( u 0 ( x ) , u 1 ( x ) ) , x ∈ Ω ,$
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.

## 2. 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
$( I 0 ρ v ) ( t ) = 1 Γ ( ρ ) ∫ 0 t ( t − s ) ρ − 1 v ( s ) d s a n d ( I T ρ v ) ( t ) = 1 Γ ( ρ ) ∫ t T ( s − t ) ρ − 1 v ( s ) d s ,$
for almost everywhere $t ∈ [ 0 , T ]$, where $Γ$ denotes the Gamma function. It can be easily seen that, if $v ∈ C ( [ 0 , T ] )$, then
$lim t → 0 + ( I 0 ρ v ) ( t ) = lim t → T − ( I T ρ v ) ( t ) = 0 .$
In this case, we may consider $I 0 ρ v$ and $I T ρ v$ as continuous functions in $[ 0 , T ]$, by taking
$( I 0 ρ v ) ( 0 ) = ( I T ρ v ) ( T ) = 0 .$
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
$d τ v d t τ ( t ) = I 0 n − τ d n v d t n ( t ) = 1 Γ ( n − τ ) ∫ 0 t ( t − s ) n − τ − 1 d n v d t n ( s ) d s ,$
for all $t ∈ [ 0 , L ]$.
We have the following integration by parts rule.
Lemma 1
(see the Corollary in [17], p. 67). Let $ρ > 0$, $q , r ≥ 1$, and $1 q + 1 r ≤ 1 + ρ$ ($q ≠ 1$, $r ≠ 1$, in the case $1 q + 1 r = 1 + ρ$). If $( v , w ) ∈ L q ( [ 0 , T ] ) × L r ( [ 0 , T ] )$, then
$∫ 0 T ( I 0 ρ v ) ( t ) w ( t ) d t = ∫ 0 T v ( t ) ( I T ρ w ) ( t ) d t .$
Lemma 2.
For sufficiently large λ, let
$η ( t ) = T − λ ( T − t ) λ , 0 ≤ t ≤ T .$
Let $ρ ∈ ( 0 , 1 )$. Then
$( I T ρ η ) ( t ) = Γ ( λ + 1 ) Γ ( ρ + λ + 1 ) T − λ ( T − t ) ρ + λ ,$
$( I T ρ η ) ′ ( t ) = − Γ ( λ + 1 ) Γ ( ρ + λ ) T − λ ( T − t ) ρ + λ − 1 ,$
$( I T ρ η ) ′ ′ ( t ) = Γ ( λ + 1 ) Γ ( ρ + λ − 1 ) T − λ ( T − t ) ρ + λ − 2 .$
Proof.
We have
$( I T ρ η ) ( t ) = 1 Γ ( ρ ) ∫ t T ( s − t ) ρ − 1 η ( s ) d s = T − λ Γ ( ρ ) ∫ t T ( s − t ) ρ − 1 ( T − s ) λ d s = T − λ Γ ( ρ ) ∫ t T ( s − t ) ρ − 1 ( ( T − t ) − ( s − t ) ) λ d s = T − λ ( T − t ) λ Γ ( ρ ) ∫ t T ( s − t ) ρ − 1 1 − s − t T − t λ d s .$
Using the change of variable $z = s − t T − t$, we obtain
$( I T ρ η ) ( t ) = T − λ ( T − t ) λ + ρ Γ ( ρ ) ∫ 0 1 z ρ − 1 ( 1 − z ) λ d z = T − λ ( T − t ) λ + ρ Γ ( ρ ) B ( ρ , λ + 1 ) ,$
where B denotes the Beta function. Using the property (see e.g., [20])
$B ( a , b ) = Γ ( a ) Γ ( b ) Γ ( a + b ) , a , b > 0 ,$
we obtain
$( I T ρ η ) ( t ) = T − λ ( T − t ) λ + ρ Γ ( ρ ) Γ ( ρ ) Γ ( λ + 1 ) Γ ( ρ + λ + 1 ) = Γ ( λ + 1 ) Γ ( ρ + λ + 1 ) T − λ ( T − t ) ρ + λ ,$
which proves (5).
Next, calculating the derivative of $I T ρ η$, we obtain
$( I T ρ η ) ′ ( t ) = − ( ρ + λ ) Γ ( λ + 1 ) Γ ( ρ + λ + 1 ) T − λ ( T − t ) ρ + λ − 1 .$
On the other hand, by the property (see e.g., [20])
$Γ ( a + 1 ) = a Γ ( a ) , a > 0 ,$
we obtain
$Γ ( ρ + λ + 1 ) = ( ρ + λ ) Γ ( ρ + λ ) .$
Hence, we deduce that
$( I T ρ η ) ′ ( t ) = − Γ ( λ + 1 ) Γ ( ρ + λ ) T − λ ( T − t ) ρ + λ − 1 ,$
which proves (6).
Differentiating $( I T ρ η ) ′$ and using (8), we obtain
$( I T ρ η ) ′ ′ ( t ) = ( ρ + λ − 1 ) Γ ( λ + 1 ) Γ ( ρ + λ ) T − λ ( T − t ) ρ + λ − 2 = ( ρ + λ − 1 ) Γ ( λ + 1 ) ( ρ + λ − 1 ) Γ ( ρ + λ − 1 ) T − λ ( T − t ) ρ + λ − 2 = Γ ( λ + 1 ) Γ ( ρ + λ − 1 ) T − λ ( T − t ) ρ + λ − 2 ,$
which proves (7). □
The following inequality will be useful later.
Lemma 3
(Young’s Inequality with Epsilon, see [29], p. 36). Let $ε > 0$ and $p > 1$. Then, for all $a , b ≥ 0$, there holds
$a b ≤ ε a p + C ε , p b p p − 1 ,$
where $C ε , p = ( p − 1 ) p − 1 ( ε p ) − 1 p − 1$.
Remark 1.
For a function $u : ( 0 , ∞ ) × ( 0 , L ) → R$, the notation $∂ α u ∂ t α$ used in (2), where $1 < α < 2$, means the following:
$∂ α u ∂ t α ( t , x ) = I 0 2 − α ∂ 2 u ∂ t 2 ( · , x ) ( t ) , t > 0 , 0 < x < L ,$
i.e.,
$∂ α u ∂ t α ( t , x ) = 1 Γ ( 2 − α ) ∫ a t ( t − s ) 1 − α ∂ 2 u ∂ t 2 ( s , x ) d s .$
Similarly, the notation $∂ β u ∂ t β$ used in (2), where $0 < β < 1$, means the following:
$∂ β u ∂ t β ( t , x ) = I 0 1 − β ∂ u ∂ t ( · , x ) ( t ) , t > 0 , 0 < x < L ,$
i.e.,
$∂ β u ∂ t β ( t , x ) = 1 Γ ( 1 − β ) ∫ a t ( t − s ) − β ∂ u ∂ t ( s , x ) d s .$

## 3. Statement of the Main Results

We first consider problem (1). Let
$Q = [ 0 , ∞ ) × [ 0 , L ] .$
We introduce the test function space
Definition 1.
Let $u 0 , u 1 ∈ L 1 ( [ 0 , L ] )$ and $f , g ∈ L l o c 1 ( [ 0 , ∞ ) )$. We say that u is a global weak solution to (1), if
(i)
$x σ | u | p ∈ L l o c 1 ( Q )$, $u ∈ L l o c 1 ( Q )$;
(ii)
for every $φ ∈ Φ$,
$∫ Q x σ | u | p φ d x d t + ∫ 0 ∞ f ( t ) ∂ φ ∂ x ( t , 0 ) − g ( t ) ∂ φ ∂ x ( t , L ) d t + ∫ 0 L u 1 ( x ) φ ( 0 , x ) − u 0 ( x ) ∂ φ ∂ t ( 0 , x ) + u 0 ( x ) φ ( 0 , x ) d x ≤ − ∫ Q u ∂ 2 φ ∂ x 2 d x d t + ∫ Q u ∂ 2 φ ∂ t 2 d x d t − ∫ Q u ∂ φ ∂ t d x d t .$
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 $g ≡ 0$.
Theorem 1.
Let $u 0 , u 1 ∈ L 1 ( [ 0 , L ] )$, $f ∈ L l o c 1 ( [ 0 , ∞ ) )$, and $g ≡ 0$. Suppose that
$∫ 0 L u 0 ( x ) + u 1 ( x ) ( L − x ) d x > 0 .$
If
$σ < − ( p + 1 ) ,$
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.
Example 1.
Consider problem (1) with
$f ( t ) = 1 t , t > 0 , g ≡ 0 , u 0 ( x ) = − ( L − x ) , u 1 ( x ) = 2 ( L − x ) , σ = − 4 , p = 2 .$
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
$g ( t ) = C g t γ , γ > − 1 , t > 0 ,$
where $C g > 0$ is a constant.
Theorem 2.
Let $u 0 , u 1 ∈ L 1 ( [ 0 , L ] )$, $f ∈ L l o c 1 ( [ 0 , ∞ ) )$, and g be the function defined by (12). If one of the following conditions is satisfied:
(i)
$σ < − ( p + 1 )$;
(ii)
$σ ≥ − ( p + 1 )$, $γ > 0$,
then (1) admits no global weak solution.
Example 2.
Consider problem (1) with
$f ( t ) = e t t , t > 0 , u 0 ( x ) = x , u 1 ( x ) = x 2 , g ( t ) = t , t > 0 , σ = − 2 , p = 2 .$
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
$Q T = [ 0 , T ] × [ 0 , L ] .$
We introduce the test function space
$Φ T = φ ∈ C 2 ( Q T ) : φ ≥ 0 , φ ( · , 0 ) = φ ( · , L ) ≡ 0 , ∂ ( I T 2 − α φ ) ∂ t ( T , · ) ≡ 0 .$
Definition 2.
Let $u 0 , u 1 ∈ L 1 ( [ 0 , L ] )$ and $f , g ∈ L l o c 1 ( [ 0 , ∞ ) )$. We say that u is a global weak solution to (2), if
(i)
$x σ | u | p ∈ L l o c 1 ( Q )$, $u ∈ L l o c 1 ( Q )$;
(ii)
for all $T > 0$ and $φ ∈ Φ T$,
$∫ Q T x σ | u | p φ d x d t + ∫ 0 T f ( t ) ∂ φ ∂ x ( t , 0 ) − g ( t ) ∂ φ ∂ x ( t , L ) d t + ∫ 0 L u 1 ( x ) ( I T 2 − α φ ) ( 0 , x ) − u 0 ( x ) ∂ ( I T 2 − α φ ) ∂ t ( 0 , x ) + u 0 ( x ) ( I T 1 − β φ ) ( 0 , x ) d x ≤ − ∫ Q T u ∂ 2 φ ∂ x 2 d x d t + ∫ Q T u ∂ 2 ( I T 2 − α φ ) ∂ t 2 d x d t − ∫ Q T u ∂ ( I T 1 − β φ ) ∂ t d x d t .$
Remark 4.
The weak formulation (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.
As for problem (1), we first consider the case $g ≡ 0$.
Theorem 3.
Let $u 0 , u 1 ∈ L 1 ( [ 0 , L ] )$, $f ∈ L l o c 1 ( [ 0 , ∞ ) )$, and $g ≡ 0$. If
$σ < − ( p + 1 ) ,$
and one of the following conditions is satisfied:
$α < β + 1 , ∫ 0 L u 1 ( x ) ( L − x ) d x > 0 ;$
$α = β + 1 , ∫ 0 L u 0 ( x ) + u 1 ( x ) ( L − x ) d x > 0 ;$
$α > β + 1 , ∫ 0 L u 0 ( x ) ( L − x ) d x > 0 ,$
then (2) admits no global weak solution.
Example 3.
Consider problem (2) with
$f ( t ) = 1 t , t > 0 , u 0 ≡ 0 , u 1 ( x ) = 2 ( L − x ) , α = 3 2 , β = 3 4 , σ = − 4 , p = 2 .$
Since (14) is satisfied and $σ < − ( p + 1 )$, by Theorem 3, we deduce that (2) admits no global weak solution.
Next, we consider the inhomogeneous case, where the function g is given by (12).
Theorem 4.
Let $u 0 , u 1 ∈ L 1 ( [ 0 , L ] )$, $f ∈ L l o c 1 ( [ 0 , ∞ ) )$, and g be the function defined by (12). If
$α > max 1 − γ , 1 , β > max { − γ , 0 } ,$
and one of the following conditions is satisfied:
(i)
$σ < − ( p + 1 )$;
(ii)
$σ ≥ − ( p + 1 )$, $γ > 0$,
then (2) admits no global weak solution.
Example 4.
Consider problem (2) with
$f ( t ) = 1 t , t > 0 , u 0 ( x ) = − x , u 1 ( x ) = x 2 , g ( t ) = t 2 3 , t > 0 , α = 3 2 , β = 1 2 ,$
and
$σ = − 3 , p = 3 .$
Then (17) is satisfied, $σ ≥ − ( p + 1 )$, and $γ > 0$. Then, by Theorem 4, we deduce that (2) admits no global weak solution.

## 4. 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.

#### 4.1. Proof of Theorem 1

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

#### 4.2. 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
$− ∫ 0 ∞ g ( t ) ∂ φ ∂ x ( t , L ) d t + ∫ 0 L u 1 ( x ) φ ( 0 , x ) − u 0 ( x ) ∂ φ ∂ t ( 0 , x ) + u 0 ( x ) φ ( 0 , x ) d x ≤ C R θ + p + 1 + σ p − 1 + R θ 1 − p p − 1 ∫ 1 2 R − 1 L x − σ p − 1 d x ,$
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
$∂ φ ∂ x ( t , L ) = − φ 1 ( t ) , t > 0 ,$
which yields
$− ∫ 0 ∞ g ( t ) ∂ φ ∂ x ( t , L ) d t = ∫ 0 ∞ g ( t ) φ 1 ( t ) d t = C ∫ 0 ∞ t γ ξ ℓ ( R − θ t ) d t ≥ C ∫ 0 1 2 R θ t γ d t = C R θ ( γ + 1 ) .$
Then, by (41), we deduce that
$C + R − θ ( γ + 1 ) ∫ 0 L u 1 ( x ) φ ( 0 , x ) − u 0 ( x ) ∂ φ ∂ t ( 0 , x ) + u 0 ( x ) φ ( 0 , x ) d x ≤ C R − θ γ + p + 1 + σ p − 1 + R − θ γ + p p − 1 ∫ 1 2 R − 1 L x − σ p − 1 d x .$
Let $σ < − ( p + 1 )$. In this case, (42) reduces to
$C + R − θ ( γ + 1 ) ∫ 0 L u 1 ( x ) φ ( 0 , x ) − u 0 ( x ) ∂ φ ∂ t ( 0 , x ) + u 0 ( x ) φ ( 0 , x ) d x ≤ C R − θ γ + p + 1 + σ p − 1 + R − θ γ + p p − 1 .$
Taking $θ > 0$ so that
$θ γ > p + 1 + σ p − 1 ,$
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
$C + R − θ ( γ + 1 ) ∫ 0 L u 1 ( x ) φ ( 0 , x ) − u 0 ( x ) ∂ φ ∂ t ( 0 , x ) + u 0 ( x ) φ ( 0 , x ) d x ≤ C R − θ γ + p + 1 + σ p − 1 + R − θ γ + p p − 1 ln R .$
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
$C + R − θ ( γ + 1 ) ∫ 0 L u 1 ( x ) φ ( 0 , x ) − u 0 ( x ) ∂ φ ∂ t ( 0 , x ) + u 0 ( x ) φ ( 0 , x ) d x ≤ C R − θ γ + p + 1 + σ p − 1 + R − θ γ + p p − 1 + σ p − 1 − 1 .$
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. □

#### 4.3. 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
$∫ Q T x σ | u | p φ d x d t + ∫ 0 T f ( t ) ∂ φ ∂ x ( t , 0 ) − g ( t ) ∂ φ ∂ x ( t , L ) d t + ∫ 0 L u 1 ( x ) ( I T 2 − α φ ) ( 0 , x ) − u 0 ( x ) ∂ ( I T 2 − α φ ) ∂ t ( 0 , x ) + u 0 ( x ) ( I T 1 − β φ ) ( 0 , x ) d x ≤ ∫ Q T | u | ∂ 2 φ ∂ x 2 d x d t + ∫ Q T | u | ∂ 2 ( I T 2 − α φ ) ∂ t 2 d x d t + ∫ Q T | u | ∂ ( I T 1 − β φ ) ∂ t d x d t .$
On the other hand, using Lemma 3 with $ε = 1 3$ and adequate choices of a and b, we obtain
$∫ Q T | u | ∂ 2 φ ∂ x 2 d x d t ≤ 1 3 ∫ Q T x σ | u | p φ d x d t + C ∫ Q T x − σ p − 1 φ − 1 p − 1 ∂ 2 φ ∂ x 2 p p − 1 d x d t ,$
$∫ Q T | u | ∂ 2 ( I T 2 − α φ ) ∂ t 2 d x d t ≤ 1 3 ∫ Q T x σ | u | p φ d x d t + C ∫ Q T x − σ p − 1 φ − 1 p − 1 ∂ 2 ( I T 2 − α φ ) ∂ t 2 p p − 1 d x d t ,$
and
$∫ Q T | u | ∂ ( I T 1 − β φ ) ∂ t d x d t ≤ 1 3 ∫ Q T x σ | u | p φ d x d t + C ∫ Q T x − σ p − 1 φ − 1 p − 1 ∂ ( I T 1 − β φ ) ∂ t p p − 1 d x d t .$
Using (45)–(48), we obtain
$∫ 0 T f ( t ) ∂ φ ∂ x ( t , 0 ) − g ( t ) ∂ φ ∂ x ( t , L ) d t + ∫ 0 L u 1 ( x ) ( I T 2 − α φ ) ( 0 , x ) − u 0 ( x ) ∂ ( I T 2 − α φ ) ∂ t ( 0 , x ) + u 0 ( x ) ( I T 1 − β φ ) ( 0 , x ) d x ≤ ∑ j = 1 3 J j ( φ ) ,$
where
$J 1 ( φ ) = ∫ Q T x − σ p − 1 φ − 1 p − 1 ∂ 2 φ ∂ x 2 p p − 1 d x d t , J 2 ( φ ) = ∫ Q T x − σ p − 1 φ − 1 p − 1 ∂ 2 ( I T 2 − α φ ) ∂ t 2 p p − 1 d x d t , J 3 ( φ ) = ∫ Q T x − σ p − 1 φ − 1 p − 1 ∂ ( I T 1 − β φ ) ∂ t p p − 1 d x d t .$
For sufficiently large T, $λ$, , and R, let
$φ ( t , x ) = η ( t ) φ 2 ( x ) , t ≥ 0 , x ∈ [ 0 , L ] ,$
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
$J 1 ( φ ) = ∫ 0 T η ( t ) d t ∫ 0 L x − σ p − 1 φ 2 − 1 p − 1 ( x ) | φ 2 ′ ′ ( x ) | p p − 1 d x .$
An elementary calculation shows that
$∫ 0 T η ( t ) d t = T λ + 1 .$
Hence, using (28), (51), and (52), we obtain
$J 1 ( φ ) ≤ C T R σ + 2 p p − 1 − 1 .$
For $j = 2$, we have
$J 2 ( φ ) = ∫ 0 T η − 1 p − 1 ( t ) | ( I T 2 − α η ) ′ ′ ( t ) | p p − 1 d t ∫ 0 L x − σ p − 1 φ 2 ( x ) d x .$
Moreover, by Lemma 2, we obtain
$η − 1 p − 1 ( t ) | ( I T 2 − α η ) ′ ′ ( t ) | p p − 1 = Γ ( λ + 1 ) Γ ( 1 − α + λ ) p p − 1 T − λ ( T − t ) λ − α p p − 1 .$
Integrating over $( 0 , T )$, there holds
$∫ 0 T η − 1 p − 1 ( t ) | ( I T 2 − α η ) ′ ′ ( t ) | p p − 1 d t = C T − α p p − 1 + 1 .$
Next, taking into consideration that $σ < − ( p + 1 )$ (so $σ < p − 1$), it follows from (31), (54), and (55) that
$J 2 ( φ ) ≤ C T 1 − α p p − 1 .$
Proceeding as above, we obtain
$J 3 ( φ ) ≤ C T 1 − β p p − 1 .$
Hence, by (53), (56), and (57), we obtain
$∑ j = 1 3 J j ( φ ) ≤ C T R σ + 2 p p − 1 − 1 + T 1 − β p p − 1 .$
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
$∫ 0 T f ( t ) ∂ φ ∂ x ( t , 0 ) − g ( t ) ∂ φ ∂ x ( t , L ) d t = 0 .$
On the other hand, using (50) and Lemma 2, for all $x ∈ [ 0 , L ]$, we obtain
$( I T 2 − α φ ) ( 0 , x ) = Γ ( λ + 1 ) Γ ( 3 − α + λ ) T 2 − α φ 2 ( x ) : = C 1 T 2 − α φ 2 ( x ) , ∂ ( I T 2 − α φ ) ∂ t ( 0 , x ) = − Γ ( λ + 1 ) Γ ( 2 − α + λ ) T 1 − α φ 2 ( x ) : = − C 2 T 1 − α φ 2 ( x ) , ( I T 1 − β φ ) ( 0 , x ) = Γ ( λ + 1 ) Γ ( 2 − β + λ ) T 1 − β φ 2 ( x ) : = C 3 T 1 − β φ 2 ( x ) .$
Consequently, we obtain
$∫ 0 L u 1 ( x ) ( I T 2 − α φ ) ( 0 , x ) − u 0 ( x ) ∂ ( I T 2 − α φ ) ∂ t ( 0 , x ) + u 0 ( x ) ( I T 1 − β φ ) ( 0 , x ) d x = ∫ 0 L C 1 T 2 − α u 1 ( x ) + C 2 T 1 − α u 0 ( x ) + C 3 T 1 − β u 0 ( x ) φ 2 ( x ) d x = ∫ 0 L C 1 T 2 − α u 1 ( x ) + C 2 T 1 − α u 0 ( x ) + C 3 T 1 − β u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x .$
Thus, combining (49), (58)–(60), we obtain
$∫ 0 L C 1 T 2 − α u 1 ( x ) + C 2 T 1 − α u 0 ( x ) + C 3 T 1 − β u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≤ C T R σ + 2 p p − 1 − 1 + T 1 − β p p − 1 .$
Next, taking $T = R θ$, where $θ > 0$ is a constant that will be determined later, the above inequality reduces to
$∫ 0 L C 1 R θ ( 2 − α ) u 1 ( x ) + C 2 R θ ( 1 − α ) u 0 ( x ) + C 3 R θ ( 1 − β ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≤ C R θ + σ + 2 p p − 1 − 1 + R θ 1 − β p p − 1 .$
Suppose that (14) holds. In this case, we obtain
$lim R → ∞ R − θ ( 2 − α ) ∫ 0 L C 1 R θ ( 2 − α ) u 1 ( x ) + C 2 R θ ( 1 − α ) u 0 ( x ) + C 3 R θ ( 1 − β ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x = C 1 ∫ 0 L u 1 ( x ) ( L − x ) d x > 0 .$
Hence, for sufficiently large R,
$∫ 0 L C 1 R θ ( 2 − α ) u 1 ( x ) + C 2 R θ ( 1 − α ) u 0 ( x ) + C 3 R θ ( 1 − β ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≥ C R θ ( 2 − α ) .$
Combining (61) with (62), we obtain
$C ≤ R θ ( α − 1 ) + σ + 2 p p − 1 − 1 + R θ α − β p p − 1 − 1 .$
Observe that, since $α < β + 1$, we have
$α − β p p − 1 − 1 < 0 .$
Hence, taking into consideration that $σ < − ( p + 1 )$, picking $θ > 0$ so that
$θ < − ( p + 1 ) − σ ( p − 1 ) ( α − 1 ) ,$
and passing to the limit as $R → ∞$ in (63), we obtain a contradiction with $C > 0$.
Suppose that (15) holds. Then,
$( I T 2 − α φ ) ( 0 , x ) = ( I T 1 − β φ ) ( 0 , x ) .$
Thus, (61) reduces to
$∫ 0 L C 1 R θ ( 2 − α ) u 0 ( x ) + u 1 ( x ) + C 2 R θ ( 1 − α ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≤ C R θ + σ + 2 p p − 1 − 1 + R θ 1 − β p p − 1 .$
Moreover, we have
$lim R → ∞ R − θ ( 2 − α ) ∫ 0 L C 1 R θ ( 2 − α ) u 0 ( x ) + u 1 ( x ) + C 2 R θ ( 1 − α ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x = C 1 ∫ 0 L u 0 ( x ) + u 1 ( x ) ( L − x ) d x > 0 ,$
which yields
$∫ 0 L C 1 R θ ( 2 − α ) u 0 ( x ) + u 1 ( x ) + C 2 R θ ( 1 − α ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≥ C R θ ( 2 − α ) ,$
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
$lim R → ∞ R − θ ( 1 − β ) ∫ 0 L C 1 R θ ( 2 − α ) u 1 ( x ) + C 2 R θ ( 1 − α ) u 0 ( x ) + C 3 R θ ( 1 − β ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x = C 3 ∫ 0 L u 0 ( x ) ( L − x ) d x > 0 .$
Hence, for sufficiently large R,
$∫ 0 L C 1 R θ ( 2 − α ) u 1 ( x ) + C 2 R θ ( 1 − α ) u 0 ( x ) + C 3 R θ ( 1 − β ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≥ C R θ ( 1 − β ) .$
Combining (61) with (65), we obtain
$C ≤ R θ β + σ + 2 p p − 1 − 1 + R − θ β p − 1 .$
Taking $θ > 0$ such that
$θ < − σ − ( p + 1 ) β ( p − 1 ) ,$
and passing to the limit as $R → ∞$ in (66), a contradiction follows. This completes the proof of Theorem 3. □

#### 4.4. 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
$− ∫ 0 T g ( t ) ∂ φ ∂ x ( t , L ) d t + ∫ 0 L C 1 T 2 − α u 1 ( x ) + C 2 T 1 − α u 0 ( x ) + C 3 T 1 − β u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≤ C T R σ + 2 p p − 1 − 1 + T 1 − β p p − 1 ∫ 1 2 R − 1 L x − σ p − 1 d x ,$
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
$− ∫ 0 T g ( t ) ∂ φ ∂ x ( t , L ) d t = ∫ 0 T g ( t ) η ( t ) d t = T − λ ∫ 0 T t γ ( T − t ) λ d t = B ( γ + 1 , λ + 1 ) T γ + 1 : = C T γ + 1 ,$
where B denotes the Beta function. Thus, by (67), we obtain
$C + ∫ 0 L C 1 T 1 − α − γ u 1 ( x ) + C 2 T − γ − α u 0 ( x ) + C 3 T − β − γ u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≤ C T − γ R σ + 2 p p − 1 − 1 + T − β p p − 1 − γ ∫ 1 2 R − 1 L x − σ p − 1 d x .$
Taking $T = R θ$, where $θ > 0$ is a constant that will be determined later, the above inequality reduces to
$C + ∫ 0 L C 1 R θ ( 1 − α − γ ) u 1 ( x ) + C 2 R − θ ( γ + α ) u 0 ( x ) + C 3 R − θ ( β + γ ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≤ C R − θ γ + σ + 2 p p − 1 − 1 + R − θ β p p − 1 + γ ∫ 1 2 R − 1 L x − σ p − 1 d x .$
Let $σ < − ( p + 1 )$. In this case, for sufficiently large R, there holds
$∫ 1 2 R − 1 L x − σ p − 1 d x ≤ C .$
Hence, (68) yields
$C + ∫ 0 L C 1 R θ ( 1 − α − γ ) u 1 ( x ) + C 2 R − θ ( γ + α ) u 0 ( x ) + C 3 R − θ ( β + γ ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≤ C R − θ γ + σ + 2 p p − 1 − 1 + R − θ β p p − 1 + γ .$
Since by (17), $β + γ > 0$, there holds
$β p p − 1 + γ > 0 .$
Thus, taking $θ > 0$ so that
$θ γ > σ + p + 1 p − 1 ,$
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
$C + ∫ 0 L C 1 R θ ( 1 − α − γ ) u 1 ( x ) + C 2 R − θ ( γ + α ) u 0 ( x ) + C 3 R − θ ( β + γ ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≤ C R − θ γ + σ + 2 p p − 1 − 1 + R − θ β p p − 1 + γ ln R .$
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
$C + ∫ 0 L C 1 R θ ( 1 − α − γ ) u 1 ( x ) + C 2 R − θ ( γ + α ) u 0 ( x ) + C 3 R − θ ( β + γ ) u 0 ( x ) ( L − x ) μ ℓ ( R x ) d x ≤ C R − θ γ + σ + 2 p p − 1 − 1 + R − θ β p p − 1 + γ + σ p − 1 − 1 .$
So, taking $θ > 0$ satisfying (70) and
$θ β p p − 1 + γ > σ p − 1 − 1 ,$
and passing to the limit as $R → ∞$ in (72), a contradiction follows. This proves part (ii) of Theorem 4. □

## 5. 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) in a bounded domain $Ω ⊂ R N$ under different types of boundary conditions, such as Dirichlet boundary conditions, Neumann boundary conditions, and Robin boundary conditions.

## Author Contributions

Investigation, A.B.S.; Supervision M.J. and B.S. All authors have read and agreed to the published version of the manuscript.

## Funding

The second author is supported by the Researchers Supporting Project number (RSP-2021/57), King Saud University, Riyadh, Saudi Arabia.

Not applicable.

Not applicable.

Not applicable.

## Conflicts of Interest

The authors declare no conflict of interest.

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Bin Sultan, A.; Jleli, M.; Samet, B. Nonexistence of Global Solutions to Time-Fractional Damped Wave Inequalities in Bounded Domains with a Singular Potential on the Boundary. Fractal Fract. 2021, 5, 258. https://doi.org/10.3390/fractalfract5040258

AMA Style

Bin Sultan A, Jleli M, Samet B. Nonexistence of Global Solutions to Time-Fractional Damped Wave Inequalities in Bounded Domains with a Singular Potential on the Boundary. Fractal and Fractional. 2021; 5(4):258. https://doi.org/10.3390/fractalfract5040258

Chicago/Turabian Style

Bin Sultan, Areej, Mohamed Jleli, and Bessem Samet. 2021. "Nonexistence of Global Solutions to Time-Fractional Damped Wave Inequalities in Bounded Domains with a Singular Potential on the Boundary" Fractal and Fractional 5, no. 4: 258. https://doi.org/10.3390/fractalfract5040258