# Nonexistence of global weak solutions to semilinear wave equations involving time-dependent structural damping terms

Mokhtar Kirane, Ahmad Fino, Sebti Kerbal

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Mokhtar Kirane, Ahmad Fino, Sebti Kerbal. Nonexistence of global weak solutions to semilinear wave equations involving time-dependent structural damping terms. 2023. hal-04091015

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# Nonexistence of global weak solutions to semilinear wave equations involving time-dependent structural damping terms

Mokhtar Kirane<sup>a,1,1</sup>, Ahmad Z. Fino<sup>b</sup>, Sebti Kerbal<sup>c</sup>, Aymen Laadhari<sup>a</sup>

<sup>a</sup>Department of Mathematics, College of Arts and Sciences, Khalifa University of Science and Technology, Abu Dhabi, United Arab Emirates

<sup>b</sup>College of Engineering and Technology, American University of the Middle East, Egaila 54200, Kuwait

<sup>c</sup>Department of Mathematics, Sultan Qaboos University, FracDiff Research Group (DR/RG/03), Al-Khoud 123, Muscat, P.O. Box 36, Oman

## Abstract

We consider a semilinear wave equation involving a time-dependent structural damping term of the form $\frac{1}{(1+t)^\beta}(-\Delta)^{\sigma/2}u_t$.

Our results show the influence of the parameters $\beta, \sigma$ on the nonexistence of global weak solutions under assumptions on the given system data.

**Keywords:** Blow-up, damped wave equations, variable coefficients, Fractional Laplacian

**2020 MSC:** Primary: 35B44, 35A01, Secondary: 35L20, 35L71

## 1. Introduction

The aim of the paper is to establish a nonexistence of global weak solutions to the Cauchy problem for the following semilinear structurally damped wave equation

$$ \left\{ \begin{array}{ll} u_{tt} - \Delta u + \frac{b_0}{(1+t)^\beta} (-\Delta)^{\sigma/2} u_t = |u|^p & x \in \mathbb{R}^n, t > 0, \\ u(0,x) = u_0(x), \quad u_t(0,x) = u_1(x), & x \in \mathbb{R}^n, \end{array} \right. \qquad (1) $$

where $0 < \sigma < 2, n \ge 1, p > 1, b_0$ is a positive constant, and $\beta \in \mathbb{R}$. Without loss of generality, we assume that $b_0 = 1$.

Prior to our main blow-up results, we would like to emphasize that the special case of Eq. (1) for $b_0 = 0$ is governed by the classical semilinear wave equation, where the Strauss conjecture for this case states that: if $p \le p_S$ then there is no global solution for (1) under suitable sign assumptions on the initial data, and for $p > p_S$ a global solution exists for small initial data; see e.g. [10, 14, 16, 18] among many other references, where $p_S$ is the positive solution of the following quadratic equation $(n-1)p^2 - (n+1)p - 2 = 0$, and is given by

$$ p_S = p_S(n) := \frac{n + 1 + \sqrt{n^2 + 10n - 7}}{2(n - 1)}. $$

In order to have more general initial data, but still with compact support, Kato [11] obtained with a slightly less sharp blow-up result an exponent of the form $(n+1)/(n-1)$ which is less than the Strauss exponent $p_S$, for $n \ge 2$.

We would like to emphasize here that the test function method, introduced by [17] and used by [8, 9, 12], plays a similar role as of Kato's method in the proof of blow-up results. In fact, the test function is effective in the case of parabolic equations as it provide exactly the critical exponent $p_c$, but in the case of hyperbolic equations (cf. [12]) we

<footnote>*Corresponding author</footnote>

<footnote>Email addresses: mokhtar.kirane@ku.ac.ae (Mokhtar Kirane), ahmad.fino@aum.edu.kw (Ahmad Z. Fino), skerbal@squ.edu.om (Sebti Kerbal), aymen.laadhari@ku.ac.ae (Aymen Laadhari)</footnote>
get the so-called Kato's exponent $p^*$ which is less than $p_c$. This is one of the weakness of the test function method but in general it can be applied to a more general equation and system with no positive condition on solutions.

When $\beta = 0, \sigma \to 2$, and $b_0 = 1$, problem (1) is reduced to

$$u_{tt} - \Delta u - \Delta u_t = |u|^p, \quad x \in \mathbb{R}^n, t > 0, \qquad (2)$$

which is called the viscoelastic damping case. D'Ambrosio and Lucente [3, Theorem 4.2] proved that the solution of (2) blows-up in finite time when $1 < p \le (n+1)/(n-1)_+$, where $(\cdot)_+ := \max\{0, \cdot\}$, by applying the test function method. Similar result has been obtained recently by Fino [6] in the case of an exterior domain. On the other hand, D'Abbicco-Reissig [2] proved that there exists a global solution for (2) when $p > 1 + \frac{3}{n-1} (n \ge 2)$ for sufficiently small initial data. Therefore, the exact value of the critical exponent is still an open question.

Recently, problem (1) with $\sigma \to 2$,

$$u_{tt} - \Delta u - \frac{b_0}{(1+t)^\beta} \Delta u_t = |u|^p, \quad x \in \mathbb{R}^n, t > 0, \qquad (3)$$

has been studied by Fino & Hamza [7]. They proved that if $\int_{\mathbb{R}^n} u_1(x)dx > 0$,

$$\begin{cases} p \in (1, \infty) & \text{for } n = 1, 2, \\ p \in (1, \frac{n}{n-2}) & \text{for } n \ge 3, \end{cases}$$

and

$$\begin{cases} 1 < p \le \frac{n+1}{(n-1)_+} & \text{if } \beta \ge -1, \\ 1 < p \le \frac{n(1-\beta)+2}{(n(1-\beta)-2)_+} & \text{if } \beta \le -1, \end{cases}$$

then the mild solution of (3) blows-up in finite time.

The goal of this paper is to prove the nonexistence of global weak solutions of (1) which is a generalization of the results of [7] to the case of fractional Laplacian damping term by using a recent version of the test function method that relies on Lemma 1 below. The novelty of this paper is the choice of this test function which is consistent with the suppression of positivity usually assumed by many authors.

In the case of $\beta \ne 0$ and $b_0 = 1$, we give an intuitive observation for understanding the influence of the damping term $(\frac{1}{(1+t)^\beta}(-\Delta)^{\sigma/2}u_t)$ by scaling argument. Let $u(t, x)$ be a solution of the linear strong damped wave equation

$$u_{tt}(t, x) - \Delta u(t, x) + \frac{1}{(1+t)^\beta} (-\Delta)^{\sigma/2} u_t(t, x) = 0. \qquad (4)$$

When $\sigma \ge 1 - \beta$, we put

$$u(t, x) = v(\lambda(1 + t), \lambda x), \quad \lambda(1 + t) = s, \lambda x = y, \qquad (5)$$

with a parameter $\lambda > 0$, we have

$$v_{ss}(s, y) - \Delta v(s, y) + \frac{\lambda^{\sigma+\beta-1}}{s^\beta} (-\Delta)^{\sigma/2} v_s(s, y) = 0.$$

Thus, when $\sigma = 1 - \beta$ we notice that Eq. (4) is invariant, while when $\sigma > 1 - \beta$, letting $\lambda \to 0$, we obtain the wave equation without damping

$$v_{ss}(s, y) - \Delta v(s, y) = 0.$$
We note that $\lambda \to 0$ is corresponding to $t \to +\infty$.

On the other hand, when $\sigma < 1 - \beta$, we put

$$u(t, x) = v(\lambda^{\frac{\sigma}{1-\beta}}(1+t), \lambda x), \quad \lambda^{\frac{2}{1-\beta}}(1+t) = s, \lambda x = y,$$

with a parameter $\lambda > 0$, we have

$$v_{ss}(s, y) + \frac{1}{s^{\beta}}(-\Delta)^{\sigma/2}v_s(s, y) - \lambda^{\frac{2(1-\beta-\sigma)}{1-\beta}}\Delta v(s, y) = 0.$$

In this case, letting $\lambda \to 0$, we obtain the pseudo-parabolic equation

$$v_{ss}(s, y) + \frac{1}{s^{\beta}}(-\Delta)^{\sigma/2}v_s(s, y) = 0.$$

This means that the asymptotic behaviour of solutions depends on the behaviour of the coefficient of the damping term.

This paper is organized as follows. We start in Sec. 2 by stating the main theorem of our work and we prove it in Sec. 4. Sec. 3 is to collect some preliminaries.

## 2. Main results

This section is aimed to state our main results.

**Theorem 1 (Blow-up).** We assume that

$$(u_0, u_1) \in (L^1(\mathbb{R}^n) \cap H^1(\mathbb{R}^n)) \times (L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n))$$

satisfying the following condition:

$$\int_{\mathbb{R}^n} u_1(x)dx > 0. \qquad (6)$$

If

$$\left\{ \begin{array}{ll} 1 < p < \min \left\{ \frac{n+1}{(n-1)_+}; \frac{n}{(n-\sigma)_+} \right\} & \text{if } \sigma \ge 1-\beta, \\ \\ 1 < p < \min \left\{ \frac{n(1-\beta)+\sigma}{(n(1-\beta)-\sigma)_+}; \frac{n}{(n-\sigma)_+} \right\} & \text{if } \sigma \le 1-\beta, \end{array} \right. \qquad (7)$$

or

$$\left\{ \begin{array}{ll} p = \frac{n+1}{n-1}, & \text{if } \sigma \ge 1-\beta, \sigma > \frac{2n}{n+1}, n \ge 2, \\ \\ p = \frac{n(1-\beta)+\sigma}{n(1-\beta)-\sigma}, & \text{if } n(1+\beta) < \sigma \le 1-\beta, n \ge 1, \end{array} \right.$$

where $(\cdot)_+ := \max\{0, \cdot\}$, then problem (1) has no global weak solutions.

**Remark 1.** We note that, by taking the limit case $\sigma \to 2$ we recover the same results of [7].

**Remark 2.** We stress that the exponent $\frac{n+1}{(n-1)_+}$ appearing in (7) was introduced first in [11] to prove the nonexistence of global solutions to the semilinear wave equation with the nonlinearity $|u|^p$, subject to small initial data with compact support.
### 3. Preliminaries

**Definition 1.** [13]

Let $S$ be the Schwartz space of rapidly decaying $C^\infty$ functions in $\mathbb{R}^n$ and $s \in (0, 1)$. The fractional Laplacian $(-\Delta)^s$ in $\mathbb{R}^n$ is a non-local operator defined on $S$ by

$$(-\Delta)^s v(x) := C_{n,s} \text{ p.v.} \int_{\mathbb{R}^n} \frac{v(x) - v(y)}{|x - y|^{n+2s}} dy \\ = \begin{cases} C_{n,s} \int_{\mathbb{R}^n} \frac{v(x) - v(y)}{|x - y|^{n+2s}} dy, & \text{if } 0 < s < 1/2, \\ C_{n,s} \int_{\mathbb{R}^n} \frac{v(x) - v(y) - \nabla v(x) \cdot (x - y) \chi_{|x-y|<\delta}(y)}{|x - y|^{n+2s}} dy, & \text{if } 1/2 \le s < 1, \end{cases}$$

where p.v. stands for Cauchy's principal value, and $C_{n,s} := \frac{s 4^s \Gamma(\frac{n}{2} + s)}{\pi^{\frac{n}{2}} \Gamma(1 - s)}$.

In fact, we are rarely going to use the fractional Laplacian operator in the Schwartz space; it can be extended to less regular functions as follows: for $s \in (0, 1)$, $\varepsilon > 0$, let

$$\mathcal{L}_{s,\varepsilon}(\Omega) := \begin{cases} L_s(\mathbb{R}^n) \cap C^{0,2s+\varepsilon}(\Omega) & \text{if } 0 < s < 1/2, \\ L_s(\mathbb{R}^n) \cap C^{1,2s+\varepsilon-1}(\Omega), & \text{if } 1/2 \le s < 1, \end{cases}$$

where $\Omega$ be an open subset of $\mathbb{R}^n$, $C^{0,2s+\varepsilon}(\Omega)$ is the space of $2s + \varepsilon$- Hölder continuous functions on $\Omega$, $C^{1,2s+\varepsilon-1}(\Omega)$ the space of functions of $C^1(\Omega)$ whose first partial derivatives are Hölder continuous with exponent $2s + \varepsilon - 1$, and

$$L_s(\mathbb{R}^n) = \left\{ u : \mathbb{R}^n \to \mathbb{R} \quad \text{such that} \quad \int_{\mathbb{R}^n} \frac{u(x)}{1 + |x|^{n+2s}} dx < \infty \right\}.$$

**Proposition 1.** [13, Proposition 2.4]

Let $\Omega$ be an open subset of $\mathbb{R}^n$, $s \in (0, 1)$, and $f \in \mathcal{L}_{s,\varepsilon}(\Omega)$ for some $\varepsilon > 0$. Then $(-\Delta)^s f$ is a continuous function in $\Omega$ and $(-\Delta)^s f(x)$ is given by the pointwise formulas of Definition 1 for every $x \in \Omega$.

**Remark:** A simple sufficient condition for function $f$ to satisfy the conditions in Proposition 1 is that $f \in L_{\text{loc}}^1(\mathbb{R}^n) \cap C^2(\Omega)$.

Using [4, Lemma 2.11] and its proof, we have the following.

**Lemma 1.** Let $\langle x \rangle := (1 + (|x| - 1)^4)^{1/4}$ for all $x \in \mathbb{R}^n$, $n \ge 1$. Let $s \in (0, 1]$ and $\phi : \mathbb{R}^n \to \mathbb{R}$ be a function defined by

$$\phi(x) = \begin{cases} 1 & \text{if } |x| \le 1, \\ \langle x \rangle^{-n-2s} & \text{if } |x| \ge 1. \end{cases} \qquad (8)$$

Then

$$\phi \in C^2(\mathbb{R}^n) \cap L^\infty(\mathbb{R}^n) \cap H^2(\mathbb{R}^n), \quad \partial_x^2 \phi \in L^\infty(\mathbb{R}^n),$$

and the following estimate holds:

$$\max \{|\Delta\phi(x)|, |(-\Delta)^s\phi(x)|\} \le C\phi(x) \quad \text{for all } x \in \mathbb{R}^n.$$

**Lemma 2.** Let $h$ be a smooth function satisfying $\partial_x^2 h \in L^\infty(\mathbb{R}^n)$. For any $R > 0$, let $h_R$ be a function defined by

$$h_R(x) := h(R^{-1}x) \quad \text{for all } x \in \mathbb{R}^n.$$

Then, $(-\Delta)^s(h_R)$, $s \in (0, 1]$, satisfies the following scaling property

$$(-\Delta)^s h_R(x) = R^{-2s} (-\Delta)^s h(R^{-1}x), \quad \text{for all } x \in \mathbb{R}^n.$$
**Lemma 3.** [4, Lemma 2.14]

Let $s, \bar{s} \in (0, 1]$, $R > 0$ and $p > 1$. Then, the following estimate holds

$$ \int_{\mathbb{R}^n} (\phi_R(x))^{-\frac{1}{p-1}} |(-\Delta)^{\bar{s}} \phi_R(x)|^{\frac{p}{p-1}} dx \le C R^{-\frac{2\bar{s}p}{p-1}+n}, $$

where $\phi_R(x) := \phi(x/R)$ and $\phi$ as defined in (8).

**4. Proof of Theorem 1**

The proof of Theorem 1 relies mainly on the concept of weak solution of the Cauchy problem (1) and the use of the test function method. Let

$$ X_{\delta,T} = \{\varphi \in C([0, \infty), H^2(\mathbb{R}^n)) \cap C^1([0, \infty), H^\delta(\mathbb{R}^n)) \cap C^2([0, \infty), L^2(\mathbb{R}^n)), \text{ such that } \text{supp}\varphi \subset Q_T \}, $$

where $Q_T := [0, T] \times \mathbb{R}^n$, and the homogeneous fractional Sobolev space $H^\delta(\mathbb{R}^n)$, $\delta \in (0, 2)$, is defined by

$$ H^\delta(\mathbb{R}^n) = \begin{cases} \{u \in L^2(\mathbb{R}^n); (-\Delta)^{\delta/2} u \in L^2(\mathbb{R}^n)\}, & \text{if } \delta \in (0, 1), \\ H^1(\mathbb{R}^n), & \text{if } \delta = 1, \\ \{u \in H^1(\mathbb{R}^n); (-\Delta)^{\delta/2} u \in L^2(\mathbb{R}^n)\}, & \text{if } \delta \in (1, 2), \end{cases} $$

endowed with the norm

$$ \|u\|_{H^\delta(\mathbb{R}^n)} = \begin{cases} \|u\|_{L^2(\mathbb{R}^n)} + \|(-\Delta)^{\delta/2} u\|_{L^2(\mathbb{R}^n)}, & \text{if } \delta \in (0, 1), \\ \|u\|_{L^2(\mathbb{R}^n)} + \|\nabla u\|_{L^2(\mathbb{R}^n)}, & \text{if } \delta = 1, \\ \|u\|_{L^2(\mathbb{R}^n)} + \|\nabla u\|_{L^2(\mathbb{R}^n)} + \|(-\Delta)^{\delta/2} u\|_{L^2(\mathbb{R}^n)}, & \text{if } \delta \in (1, 2). \end{cases} $$

The weak formulation associated with (1) reads as follows:

**Definition 2.** (Weak solution)

Let $T > 0$, and $u_0, u_1 \in L^2(\mathbb{R}^n)$. A function

$$ u \in L^1((0, T), L^2(\mathbb{R}^n)) \cap L^p((0, T); L^{2p}(\mathbb{R}^n)), $$

is said to be a weak solution of (1) on $[0, T) \times \mathbb{R}^n$ if

$$ \begin{aligned} & \int_{Q_T} |u|^p \varphi dt dx + \int_{\mathbb{R}^n} u_1(x) \varphi(0, x) dx \\ & + \int_{\mathbb{R}^n} u_0(x) (-\Delta)^{\sigma/2} \varphi(0, x) dx - \int_{\mathbb{R}^n} u_0(x) \varphi_t(0, x) dx \\ & = \int_{Q_T} u \varphi_{tt} dt dx - \int_0^T \frac{1}{(1+t)^\beta} \int_{\mathbb{R}^n} u (-\Delta)^{\sigma/2} \varphi_t dx dt \\ & - \int_{Q_T} u \Delta \varphi dt dx + \int_0^T \frac{\beta}{(1+t)^{\beta+1}} \int_{\mathbb{R}^n} u (-\Delta)^{\sigma/2} \varphi dx dt, \end{aligned} $$

holds for all $\varphi \in X_{\sigma,T}$. We denote the lifespan for the weak solution by

$$ T_w(u_0, u_1) := \sup\{T \in (0, \infty] \text{ for which there exists a unique weak solution } u \text{ to (1)}\}. $$

Moreover, if $T > 0$ can be arbitrary chosen, i.e. $T_w(u_0, u_1) = \infty$, then $u$ is called a global weak solution of (1).
**Proof of Theorem 1.** Let $u$ be a global weak solution of (1), that is

$$ 
\begin{aligned}
& \int_{Q_T} |u|^p \varphi dt dx + \int_{\mathbb{R}^n} u_1(x) \varphi(0, x) dx \\
& + \int_{\mathbb{R}^n} u_0(x) (-\Delta)^{\sigma/2} \varphi(0, x) dx - \int_{\mathbb{R}^n} u_0(x) \varphi_t(0, x) dx \\
& = \int_{Q_T} u \varphi_{tt} dt dx - \int_0^T \frac{1}{(1+t)^\beta} \int_{\mathbb{R}^n} u (-\Delta)^{\sigma/2} \varphi_t dx dt \\
& - \int_{Q_T} u \Delta \varphi dt dx + \int_0^T \frac{\beta}{(1+t)^{\beta+1}} \int_{\mathbb{R}^n} u (-\Delta)^{\sigma/2} \varphi dx dt,
\end{aligned} 
\quad (9) $$

for all $T > 0$, and all $\varphi \in X_{\sigma,T}$.

Let $T > 0$. Now, we introduce the following test function

$$ \varphi(t, x) = \psi^{\eta}(t) \phi_{T^d}(x) = \psi^{\eta}(t) \phi\left(\frac{x}{T^d}\right) \quad (10) $$

where $\phi$ is defined in (8), $\psi(t) = \Psi(\frac{t}{T})$, $\eta \gg 1$, $d > 0$ are constants that will be determined later, and $\Psi \in C^\infty(\mathbb{R}_+)$ is a cut-off non-increasing function such that

$$ \Psi(r) = \begin{cases} 1 & \text{if } 0 \le r \le 1/2, \\ \searrow & \text{if } 1/2 \le r \le 1, \\ 0 & \text{if } r \ge 1. \end{cases} $$

From the formulation (9), we get the following inequality

$$ 
\begin{aligned}
& \int_{Q_T} |u|^p \varphi dt dx + \int_{\mathbb{R}^n} u_1(x) \varphi(0, x) dx \\
& \le \int_{\frac{T}{2}}^T \int_{\mathbb{R}^n} |u| |\varphi_{tt}| dx dt + \int_{Q_T} |u| |\Delta \varphi| dt dx \\
& + \int_{\frac{T}{2}}^T \frac{1}{(1+t)^\beta} \int_{\mathbb{R}^n} |u| |(-\Delta)^{\sigma/2} \varphi_t| dx dt + \int_0^T \frac{\beta}{(1+t)^{\beta+1}} \int_{\mathbb{R}^n} |u| |(-\Delta)^{\sigma/2} \varphi| dx dt \\
& + \int_{\mathbb{R}^n} |u_0| \left( |(-\Delta)^{\sigma/2} \varphi(0, x)| + |\varphi_t(0, x)| \right) dx \\
& =: I_1 + I_2 + I_3 + I_4 + I_5.
\end{aligned} 
\quad (11) $$

Let $\varepsilon > 0$. By applying $\varepsilon$-Young's inequality

$$ AB \le \varepsilon A^p + C(\varepsilon, p) B^{p'}, \quad A \ge 0, B \ge 0, p + p' = pp', C(\varepsilon, p) = (p-1)(\varepsilon p^p)^{-1/(p-1)}, $$

we obtain the estimation for the first integral

$$ 
\begin{aligned}
I_1 & \le \int_{Q_T} |u| \varphi^{1/p} \varphi^{-1/p} \phi_{T^d} |(\psi^\eta)_{tt}| dt dx \\
& \le \varepsilon \int_{Q_T} |u|^p \varphi dt dx + C \int_{Q_T} \phi_{T^d} \psi^{-\eta/(p-1)} |(\psi^\eta)_{tt}|^{p'} dt dx.
\end{aligned} 
\quad (12) $$

As $(\psi^\eta)_{tt} = \eta \psi^{\eta-1} \psi_{tt} + \eta(\eta-1) \psi^{\eta-2} |\psi_t|^2$, the inequality (12) becomes

$$ I_1 \le \varepsilon \int_{Q_T} |u|^p \psi dt dx + C \int_{Q_T} \phi_{T^d} \psi^{\eta-p'} |\psi_{tt}|^{p'} dt dx + C \int_{Q_T} \phi_{T^d} \psi^{\eta-2p'} |\psi_t|^{2p'} dt dx. \quad (13) $$
Proceeding similarly as for (13), we get

$$I_2 \le \varepsilon \int_{Q_T} |u|^p \varphi dt \, dx + C \int_{Q_T} \psi^\eta \phi_{T^d}^{-1/(p-1)} |\Delta \phi_{T^d}|^{p'} dt \, dx. \quad (14)$$

In the same way, we write

$$\begin{aligned} I_3 &= \int_{\frac{T}{2}}^{T} \frac{1}{(1+t)^\beta} \int_{\mathbb{R}^n} |u| \varphi^{1/p} \varphi^{-1/p} |(-\Delta)^{\sigma/2} \phi_{T^d}| |(\psi^\eta)_t| dx dt \\ &\le \varepsilon \int_{Q_T} |u|^p \varphi dt \, dx \\ &\quad + C \int_{\frac{T}{2}}^{T} \frac{1}{(1+t)^{\beta p'}} \int_{\mathbb{R}^n} \phi_{T^d}^{-1/(p-1)} |(-\Delta)^{\sigma/2} \phi_{T^d}|^{p'} |\psi^{\eta-p'}| |\psi_t|^{p'} dx dt. \end{aligned} \quad (15)$$

Clearly,

$$\frac{1}{(1+t)^{\beta p'}} \le C T^{-\beta p'}, \quad \forall t \in (\frac{T}{2}, T),$$

therefore,

$$I_3 \le \varepsilon \int_{Q_T} |u|^p \varphi dt \, dx + C T^{-\beta p'} \int_{Q_T} \phi_{T^d}^{-1/(p-1)} |(-\Delta)^{\sigma/2} \phi_{T^d}|^{p'} |\psi^{\eta-p'}| |\psi_t|^{p'} dx dt. \quad (16)$$

In the same manner,

$$\begin{aligned} I_4 &\le C \int_0^T \frac{1}{(1+t)^{\beta+1}} \int_{\mathbb{R}^n} |u| \varphi^{1/p} \varphi^{-1/p} \psi^\eta |(-\Delta)^{\sigma/2} \phi_{T^d}| dx dt \\ &\le \varepsilon \int_{Q_T} |u|^p \varphi dt \, dx + C \int_0^T \frac{1}{(1+t)^{(\beta+1)p'}} \int_{\mathbb{R}^n} \phi_{T^d}^{-1/(p-1)} |(-\Delta)^{\sigma/2} \phi_{T^d}|^{p'} |\psi^\eta| dx dt. \end{aligned} \quad (17)$$

Finally, it remains only to control the term $I_5$. By exploiting the identity $\varphi_t(0, x) = \eta\psi_t(0)\phi_{T^d}(x)$, we infer

$$I_5 \le C \int_{\mathbb{R}^n} |u_0| \left( |(-\Delta)^{\sigma/2} \phi_{T^d}| + |\psi_t(0)| \phi_{T^d} \right) dx. \quad (18)$$

Plugging (11) together with (13)-(18) and choosing $\varepsilon$ small enough, we deduce that

$$\begin{aligned} & \int_{Q_T} |u|^p \varphi \, dx \, dt + \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) \, dx \\ &\le C \int_{Q_T} \phi_{T^d} \psi^{\eta-p'} |\psi_t|^{p'} \, dt \, dx + C \int_{Q_T} \phi_{T^d} \psi^{\eta-2p'} |\psi_t|^{2p'} \, dt \, dx \\ &\quad + C \int_{Q_T} \psi^\eta \phi_{T^d}^{-1/(p-1)} |\Delta \phi_{T^d}|^{p'} \, dt \, dx \\ &\quad + C T^{-\beta p'} \int_{Q_T} \phi_{T^d}^{-1/(p-1)} |(-\Delta)^{\sigma/2} \phi_{T^d}|^{p'} |\psi^{\eta-p'}| |\psi_t|^{p'} \, dt \, dx \\ &\quad + C \int_0^T \frac{1}{(1+t)^{(\beta+1)p'}} \int_{\mathbb{R}^n} \phi_{T^d}^{-1/(p-1)} |(-\Delta)^{\sigma/2} \phi_{T^d}|^{p'} |\psi^\eta| \, dx \, dt \\ &\quad + C \int_{\mathbb{R}^n} |u_0| \left( |(-\Delta)^{\sigma/2} \phi_{T^d}| + |\psi_t(0)| \phi_{T^d} \right) \, dx. \end{aligned}$$

Taking account of the expression of $\varphi$ given by (10) and Lemmas 1-3, we infer that

$$\begin{aligned} & \int_{Q_T} |u|^p \varphi \, dx \, dt + \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) \, dx \\ &\le C T^{-2p'+1+nd} + C T^{-2dp'+1+nd} + C T^{-\beta p'-p'-\sigma dp'+1+nd} \\ &\quad + C T^{-\sigma dp'+nd} \int_0^T (1+t)^{-\frac{(\beta+1)p}{p-1}} dt + C (T^{-\sigma d} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| \, dx. \end{aligned} \quad (19)$$
Since $0 < \sigma < 2$, we notice that the cases of $\sigma \ge 1 - \beta$ and $\sigma \le 1 - \beta$ are equivalent to

$$ \sigma \ge 1 - \beta, \text{ for all } \beta > -1, \quad \text{and} \quad \sigma < 1 - \beta, \text{ for all } \beta < 1, $$

therefore, we distinguish two cases:

**I. Case:** $\sigma \ge 1 - \beta$, for all $\beta > -1$.

In this case, we choose $d = 1$.

$$ \text{Subcritical case } p < \min\left\{\frac{n+1}{(n-1)_+}; \frac{n}{(n-\sigma)_+}\right\}. $$

Note that,

$$ \int_0^T (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C \begin{cases} 1 - \frac{(\beta+1)p}{p-1} & \text{if } \beta p < -1, \\ \ln T & \text{if } \beta p = -1, \\ 1 & \text{if } \beta p > -1. \end{cases} $$

We have two cases to distinguish.

a) If $\beta \ge 0$. In this case, we have $\beta p > -1$ and so

$$ \int_0^T (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C. $$

Therefore, (19) implies

$$ \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_T(x) dx \\ \le C T^{-2p'+1+n} + C T^{-(\beta+1)p'-\sigma p'+1+n} + C T^{-\sigma p'+n} + C(T^{-\sigma} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx. $$

Using the fact $\beta p > -1 \implies -(\beta+1)p' - \sigma p' + 1 + n < -\sigma p' + n$, we conclude that

$$ \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_T(x) dx \\ \le C T^{-2p'+1+n} + C T^{-\sigma p'+n} + C(T^{-\sigma} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx. \quad (20) $$

Note that, we can easily see that

$$ -2p' + 1 + n < 0 \iff p < \frac{n+1}{(n-1)_+} \quad \text{and} \quad -\sigma p' + n < 0 \iff p < \frac{n}{(n-\sigma)_+}. $$

Letting $T \to \infty$, and using the Lebesgue dominated convergence theorem together with $u_1 \in L^1(\mathbb{R}^n)$, we conclude that

$$ \int_{\mathbb{R}^n} u_1(x) dx \le 0. $$

This contradicts our assumption (6).

b) If $-1 < \beta < 0$. We have three cases for $n$.

<figure><img src="image_2.png" /></figure>
Note that when $n = 1$, we have $n = 1 < \frac{1-\beta}{1+\beta}$.

i) If $n \ge \frac{\sigma}{1+\beta}$, then $n \ge \frac{1-\beta}{1+\beta}$ and therefore

$$p < \min \left\{ \frac{n+1}{n-1} ; \frac{n}{n-\sigma} \right\} \le -\frac{1}{\beta}, \quad \text{for all } n \ge 2,$$

i.e. $\beta p > -1$, which implies a contradiction by following the same calculations as in part a).

ii) If $\frac{1-\beta}{1+\beta} < n < \frac{\sigma}{1+\beta}$ when $\sigma > 1-\beta$, then we have

$$n < \frac{\sigma}{1+\beta} < \frac{\sigma}{2-\sigma},$$

and therefore

$$p < \min \left\{ \frac{n+1}{n-1} ; \frac{n}{n-\sigma} \right\} = \frac{n+1}{n-1} < -\frac{1}{\beta}, \quad \text{for all } n \ge 2,$$

i.e. $\beta p > -1$, which implies a contradiction by following the same calculations as in part a).

iii) If $n \le \frac{1-\beta}{1+\beta}$, then $n \le \frac{\sigma}{1+\beta}$, i.e.

<figure><img src="image_3.png" /></figure>

• If $p \le -1/\beta$, i.e. $p\beta \ge -1$, then

$$\int_{0}^{T} (1+t) \frac{-(\beta+1)p}{p-1} dt \le C \ln T, \quad \text{for all } T \gg 1.$$

Then, (19) implies

$$\int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_T(x) dx \\ \le C T^{-2p'+1+n} + C T^{-(\beta+1)p'-\sigma p'+1+n} + C T^{-\sigma p'+n} \ln T + C(T^{-\sigma} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx,$$

for all $T \gg 1$. As $p\beta \ge -1 \implies (\beta+1)p' \ge 1$, we conclude that

$$\int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_T(x) dx \\ \le C T^{-2p'+1+n} + C T^{-\sigma p'+n} \ln T + C(T^{-\sigma} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx, \quad \forall T \gg 1. \quad (21)$$

Note that, we can easily see that

$$-2p' + 1 + n < 0 \iff p < \frac{n+1}{(n-1)_+} \quad \text{and} \quad -\sigma p' + n < 0 \iff p < \frac{n}{(n-\sigma)_+}.$$

Letting $T \to \infty$, using the fact that $\ln T \le T^{\frac{\sigma p'-n}{2}}$ (because $p < \frac{n}{(n-\sigma)_+}$) and the Lebesgue dominated convergence theorem, we conclude that

$$\int_{\mathbb{R}^n} u_1(x) dx \le 0.$$
This contradicts our assumption (6).

• If $p > -1/\beta$, i.e. $p\beta < -1$, then

$$ \int_0^T (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C T^{1-\frac{(\beta+1)p}{p-1}}, \quad \text{for all } T > 0. $$

Then, (19) implies

$$ \begin{aligned} & \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_T(x) dx \\ & \le C T^{-2p'+1+n} + C T^{-(\beta+1)p'-\sigma p'+1+n} + C (T^{-\sigma} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx, \end{aligned} $$

for all $T \gg 1$. As $\sigma \ge 1 - \beta \implies n \le \frac{1-\beta}{1+\beta} \le \frac{\sigma}{2-\sigma}$, we conclude that

$$ p < \min \left\{ \frac{n+1}{(n-1)_+}; \frac{n}{(n-\sigma)_+} \right\} = \frac{n+1}{(n-1)_+} \le \frac{n+1}{(n-\beta-\sigma)_+} \implies -(\beta+1)p' - \sigma p' + 1 + n < 0, $$

and

$$ p < \min \left\{ \frac{n+1}{(n-1)_+}; \frac{n}{(n-\sigma)_+} \right\} = \frac{n+1}{(n-1)_+} \iff -2p' + 1 + n < 0. $$

Note that $\sigma \ge 1 - \beta > 1$, and $2 > \sigma > \sigma + \beta$, therefore

$$ \begin{cases} (n - \beta - \sigma)_+ = (n - 1)_+ = (n - \sigma)_+ = 0, & \text{when } n = 1, \\ (n - \beta - \sigma)_+ = n - \beta - \sigma, \quad (n - 1)_+ = n - 1, \quad (n - \sigma)_+ = n - \sigma, & \text{when } n \ge 2, \end{cases} $$

Letting $T \to \infty$, using the Lebesgue dominated convergence theorem, we conclude that

$$ \int_{\mathbb{R}^n} u_1(x) dx \le 0. $$

This contradicts our assumption (6).

Critical case: $p < \infty$ and $\sigma \ge 1$ when $n = 1$, or $p = \frac{n+1}{n-1}$ and $\sigma > \frac{2n}{n+1}$ when $n \ge 2$.

Note that when $2 \le n < \frac{\sigma}{2-\sigma}$, we have

$$ p = \frac{n+1}{n-1} < \frac{n}{n-\sigma} \quad \text{i.e.} \quad -\sigma p' + n < 0, $$

and when $\sigma \ge 1$ and $n = 1$, we also have $-\sigma p' + n < 0$.

We have two cases to distinguish.

a) If $\beta \ge 0$. In this case, we have $\beta p > -1$ and so

$$ \int_0^T (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C. $$

From the subcritical case, we can see that we have

$$ u \in L^p((0, \infty); L^p(\mathbb{R}^n)). \qquad (22) $$
On the other hand, by applying Hölder's inequality instead of Young's inequality, we get

$$ \int_{\mathbb{R}^n} u_1(x)\phi_T(x)dx \le C \left( \int_{\frac{T}{2}}^T \int_{\mathbb{R}^n} |u|^p \varphi dx dt \right)^{1/p} + C \left( \int_0^T \int_{|x|\ge T} |u|^p \varphi dx dt \right)^{1/p} \\ + C T^{-\sigma p'+n} + C (T^{-\sigma} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)|dx. $$

Letting $T \to \infty$ and taking into consideration (22), we get a contradiction.

b) If $-1 < \beta < 0$. We have three cases for $n$.

<figure><img src="image_4.png" /></figure>

Note that when $n = 1$, we have $n = 1 < \frac{1-\beta}{1+\beta}$.

i) If $n > \frac{1-\beta}{1+\beta}$, then

$$ p = \frac{n+1}{n-1} < -\frac{1}{\beta}, $$

i.e. $\beta p > -1$, which implies a contradiction by following the same calculations as in part a).

ii) If $n = \frac{1-\beta}{1+\beta}$, then

$$ p = \frac{n+1}{n-1} = -\frac{1}{\beta}, $$

i.e. $\beta p = -1$, and therefore

$$ \int_{0}^{T} (1+t) \frac{(\beta+1)p}{p-1} dt \le C \ln T, \quad \text{for all } T > 0. $$

From the subcritical case (20), we can see easily that we have

$$ u \in L^p((0, \infty); L^p(\mathbb{R}^n)). \qquad (23) $$

On the other hand, by applying Hölder's inequality instead of Young's inequality and using $p = \frac{n+1}{n-1}$, we get

$$ \int_{\mathbb{R}^n} u_1(x)\phi_T(x)dx \le C \left( \int_{\frac{T}{2}}^T \int_{\mathbb{R}^n} |u|^p \varphi dx dt \right)^{1/p} + C \left( \int_0^T \int_{|x|\ge T} |u|^p \varphi dx dt \right)^{1/p} \\ + C T^{-\sigma p'+n} \ln T + C (T^{-\sigma} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)|dx. $$

Letting $T \to \infty$ and taking into consideration (23) and the fact that $\ln T \le T \frac{\sigma p' - n}{2}$ (because $p < \frac{n}{n-\sigma}$), we get

$$ \int_{\mathbb{R}^n} u_1(x) dx \le 0. $$

This contradicts our assumption (6).
iii) If $n < \frac{1-\beta}{1+\beta}$, then

$$p = \frac{n+1}{(n-1)_+} > -\frac{1}{\beta},$$

i.e. $p\beta < -1$. In this case, we change the test function $\psi$ by $\psi(t) = \Psi(\frac{t}{K^{-1}T})$ where $K \ge 1$ is independent of $T$. Then

$$\int_{0}^{K^{-1}T} (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C K^{-1+(\beta+1)p'} T^{1-(\beta+1)p'}, \quad \text{for all } T > 0.$$

From the subcritical case (20), we can see easily that we have

$$u \in L^p((0, \infty); L^p(\mathbb{R}^n)). \qquad (24)$$

On the other hand, by applying Hölder's inequality in (19) instead of Young's inequality, we get

$$\int_{\mathbb{R}^n} u_1(x)\phi_T(x)dx \le C K^{2p'-1} \left( \int_{\frac{K-1}{2}}^{K-1} \int_{\mathbb{R}^n} |u|^p \varphi dx dt \right)^{1/p} + C K^{-1} \left( \int_0^{K-1} \int_{|x|\ge T} |u|^p \varphi dx dt \right)^{1/p} \\ + C K^{-1+(\beta+1)p'} T^{-(\beta+1)p'-\sigma p'+1+n} + C(K^\sigma T^{-\sigma} + KT^{-1}) \int_{\mathbb{R}^n} |u_0(x)|dx.$$

As $\sigma \ge 1 - \beta$ and $2 > \sigma > \sigma + \beta$, we conclude that

$$p = \frac{n+1}{n-1} \le \frac{n+1}{n-\beta-\sigma} \implies -(\beta+1)p' - \sigma p' + 1 + n \le 0, \quad \text{when } n \ge 2,$$

and

$$-(\beta+1)p' - \sigma p' + 1 + n < 0, \quad \text{when } n=1,$$

so

$$\int_{\mathbb{R}^n} u_1(x)\phi_T(x)dx \le C K^{2p'-1} \left( \int_{\frac{K-1}{2}}^{K-1} \int_{\mathbb{R}^n} |u|^p \varphi dx dt \right)^{1/p} + C K^{-1} \left( \int_0^{K-1} \int_{|x|\ge T} |u|^p \varphi dx dt \right)^{1/p} \\ + C K^{-1+(\beta+1)p'} + C(K^\sigma T^{-\sigma} + KT^{-1}) \int_{\mathbb{R}^n} |u_0(x)|dx.$$

Letting $T \to \infty$ and taking into consideration (24), we get

$$\int_{\mathbb{R}^n} u_1(x) dx \le C K^{-1+(\beta+1)p'}.$$

Letting $K \to \infty$ and using $p\beta < -1 \implies -1 + (\beta+1)p' < 0$, we infer that

$$\int_{\mathbb{R}^n} u_1(x) dx \le 0.$$

This contradicts our assumption (6).

II. Case: $\sigma < 1 - \beta$, for all $\beta < 1$.

In this case we take $d = \frac{1-\beta}{\sigma} > 1$.

Subcritical case $p < \min\left\{\frac{n(1-\beta)+\sigma}{n(1-\beta)-\sigma}; \frac{n}{n-\sigma}\right\}$.

Note that,

$$\int_0^T (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C \begin{cases} 1-\frac{(\beta+1)p}{p-1} & \text{if } \beta p < -1, \\ \ln T & \text{if } \beta p = -1, \\ 1 & \text{if } \beta p > -1. \end{cases}$$

We have two cases to distinguish.
a) If $0 \le \beta < 1$. In this case, we have $\beta p > -1$ and so

$$ \int_{0}^{T} (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C. $$

Therefore, (19) implies

$$ \begin{aligned} & \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) dx \\ & \le C T^{-2p'+1+\frac{n(1-\beta)}{\sigma}} + C T^{-\frac{2(1-\beta)}{\sigma} p'+1+\frac{n(1-\beta)}{\sigma}} \\ & \quad + C T^{-(1-\beta)p'+\frac{n(1-\beta)}{\sigma}} + C (T^{-(1-\beta)} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx. \end{aligned} $$

As

$$ \sigma < 1 - \beta \implies -\frac{2(1-\beta)}{\sigma} p' + 1 + \frac{n(1-\beta)}{\sigma} < -2p' + 1 + \frac{n(1-\beta)}{\sigma}, $$

we conclude that

$$ \begin{aligned} & \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) dx \\ & \le C T^{-2p'+1+\frac{n(1-\beta)}{\sigma}} + C T^{(1-\beta)}[-p'+\frac{n}{\sigma}] + C (T^{-(1-\beta)} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx, \end{aligned} $$

Note that, we can easily see that

$$ (1-\beta)\left[-p' + \frac{n}{\sigma}\right] < 0 \iff p < \frac{n}{(n-\sigma)_+} \quad \text{and} \quad -2p' + 1 + \frac{n(1-\beta)}{\sigma} < 0 \iff p < \frac{n(1-\beta) + \sigma}{n(1-\beta) - \sigma}. $$

Note that $n(1-\beta) - \sigma \ge 1 - \beta - \sigma > 0$, for all $n \ge 1$. Letting $T \to \infty$, and using the Lebesgue dominated convergence theorem together with $u_1 \in L^1(\mathbb{R}^n)$, we conclude that

$$ \int_{\mathbb{R}^n} u_1(x) dx \le 0. $$

This contradicts our assumption (6).

b) If $\beta < 0$. We have two cases for $n$.

$$ \frac{\bullet}{\sigma} \xrightarrow[n]{(1+\beta)_+} n $$

Note that $n(1-\beta) - \sigma \ge 1 - \beta - \sigma > 0$ and $\sigma < \beta + 1 < 1 \le n \implies n - \sigma > 0$, for all $n \ge 1$.

i) If $n \ge \frac{n(1-\beta) + \sigma}{(1+\beta)_+}$, we have

$$ p < \min \left\{ \frac{n(1-\beta) + \sigma}{n(1-\beta) - \sigma}; \frac{n}{n-\sigma} \right\} = \frac{n}{n-\sigma} \le -\frac{1}{\beta}, $$

i.e. $\beta p > -1$, which implies a contradiction by following the same calculations as in part a).

ii) If $n < \frac{n(1-\beta) + \sigma}{(1+\beta)_+}$, we get

$$ p < \min \left\{ \frac{n(1-\beta) + \sigma}{n(1-\beta) - \sigma}; \frac{n}{n-\sigma} \right\} = \frac{n(1-\beta) + \sigma}{n(1-\beta) - \sigma}, $$

and

$$ -\frac{1}{\beta} < \frac{n(1-\beta) + \sigma}{n(1-\beta) - \sigma}, $$

i.e.
<figure><img src="image_5.png" /></figure>

• If $p \le -1/\beta$, i.e. $p\beta \ge -1$, then

$$ \int_0^T (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C \ln T, \quad \text{for all } T \gg 1. $$

Then, (19) implies

$$ \begin{aligned} & \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) dx \\ & \le C T^{-2p'+1+\frac{n(1-\beta)}{\sigma}} + C T^{-\frac{2(1-\beta)}{\sigma}p'+1+\frac{n(1-\beta)}{\sigma}} \\ & \quad + C T^{-(1-\beta)p'+\frac{n(1-\beta)}{\sigma}} \ln T + C(T^{-(1-\beta)} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx, \end{aligned} $$

for all $T \gg 1$. As

$$ \sigma < 1 - \beta \implies -\frac{2(1-\beta)}{\sigma}p' + 1 + \frac{n(1-\beta)}{\sigma} < -2p' + 1 + \frac{n(1-\beta)}{\sigma}, $$

we conclude that

$$ \begin{aligned} & \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) dx \\ & \le C T^{-2p'+1+\frac{n(1-\beta)}{\sigma}} + C T^{(1-\beta)}[-p'+\frac{n}{\sigma}] \ln T + C(T^{-(1-\beta)} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx, \end{aligned} $$

Note that, we can easily see that

$$ (1-\beta)[-p' + \frac{n}{\sigma}] < 0 \iff p < \frac{n}{n-\sigma} \quad \text{and} \quad -2p' + 1 + \frac{n(1-\beta)}{\sigma} < 0 \iff p < \frac{n(1-\beta)+\sigma}{n(1-\beta)-\sigma}. $$

Letting $T \to \infty$, using the fact that $\ln T \le T^{\frac{(1-\beta)(p'-\frac{n}{\sigma})}{2}}$ (because $p < \frac{n}{n-\sigma}$) and the Lebesgue dominated convergence theorem, we conclude that

$$ \int_{\mathbb{R}^n} u_1(x) dx \le 0. $$

This contradicts our assumption (6).

• If $p > -1/\beta$, i.e. $p\beta < -1$, then

$$ \int_0^T (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C T^{1-\frac{(\beta+1)p}{p-1}}, \quad \text{for all } T > 0. $$

Then, (19) implies

$$ \begin{aligned} & \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) dx \\ & \le C T^{-2p'+1+\frac{n(1-\beta)}{\sigma}} + C T^{-\frac{2(1-\beta)}{\sigma}p'+1+\frac{n(1-\beta)}{\sigma}} + C(T^{-(1-\beta)} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx. \end{aligned} $$
As

$$ \sigma < 1 - \beta \implies -\frac{2(1-\beta)}{\sigma}p' + 1 + \frac{n(1-\beta)}{\sigma} < -2p' + 1 + \frac{n(1-\beta)}{\sigma}, $$

we conclude that

$$ \begin{aligned} & \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) dx \\ & \le C T^{-2p'+1+\frac{n(1-\beta)}{\sigma}} + C (T^{-(1-\beta)} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx, \end{aligned} $$

Note that, we can easily see that

$$ -2p' + 1 + \frac{n(1-\beta)}{\sigma} < 0 \iff p < \frac{n(1-\beta) + \sigma}{n(1-\beta) - \sigma}. $$

Letting $T \to \infty$, using the Lebesgue dominated convergence theorem, we conclude that

$$ \int_{\mathbb{R}^n} u_1(x) dx \le 0. $$

This contradicts our assumption (6).

Critical case: $p = \frac{n(1-\beta) + \sigma}{n(1-\beta) - \sigma}$ and $\sigma > n(1+\beta)$.

Note that when $n(1+\beta) < \sigma$, we have

$$ p = \frac{n(1-\beta) + \sigma}{n(1-\beta) - \sigma} < \frac{n}{n-\sigma} \quad \text{i.e.} \quad -\sigma p' + n < 0, $$

We have two cases to distinguish.

a) If $0 \le \beta < 1$. In this case, we have $\beta p > -1$ and so

$$ \int_{0}^{T} (1+t)^{-\frac{(\beta+1)p}{p-1}} dt \le C. $$

From the subcritical case (20), we can see easily that we have

$$ u \in L^p((0, \infty); L^p(\mathbb{R}^n)). \qquad (25) $$

On the other hand, by applying Hölder's inequality instead of Young's inequality, (19) implies

$$ \begin{aligned} \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) dx \le & C \left( \int_{\frac{T}{2}}^{T} \int_{\mathbb{R}^n} |u|^p \varphi dx dt \right)^{1/p} + C \left( \int_{0}^{T} \int_{|x| \ge T^d} |u|^p \varphi dx dt \right)^{1/p} \\ & + C T^{(1-\beta)} [-p' + \frac{n}{\sigma}] + C (T^{-(1-\beta)} + T^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx. \end{aligned} $$

Letting $T \to \infty$ and taking into consideration (25) we get

$$ \int_{\mathbb{R}^n} u_1(x) dx \le 0. $$

This contradicts our assumption (6).
b) If $\beta < 0$. As $n(1 + \beta) < \sigma \implies n < \frac{\sigma}{(1 + \beta)_+}$, so

$$p = \frac{n(1 - \beta) + \sigma}{n(1 - \beta) - \sigma} > -\frac{1}{\beta},$$

i.e. $p\beta < -1$. In this case, we change the test function $\psi$ by $\psi(t) = \Psi(\frac{t}{K^{-1}T})$ where $K \ge 1$ is independent of $T$. Then

$$\int_{0}^{K^{-1}T} (1 + t)^{-\frac{(\beta+1)p}{p-1}} dt \le C K^{-1+(\beta+1)p'} T^{1-(\beta+1)p'}, \quad \text{for all } T > 0.$$

From the subcritical case (20), we can see easily that we have

$$u \in L^p((0, \infty); L^p(\mathbb{R}^n)). \qquad (26)$$

On the other hand, by applying Hölder's inequality in (19) instead of Young's inequality and using $p = \frac{n(1 - \beta) + \sigma}{n(1 - \beta) - \sigma}$, (19) implies

$$\begin{aligned} & \int_{Q_T} |u|^p \varphi dx dt + \int_{\mathbb{R}^n} u_1(x) \phi_{T^d}(x) dx \\ & \le C K^{2p'-1} \left( \int_{\frac{K^{-1}T}{2}}^{K^{-1}T} \int_{\mathbb{R}^n} |u|^p \varphi dx dt \right)^{1/p} + C K^{-1} \left( \int_{0}^{K^{-1}T} \int_{|x| \ge T^d} |u|^p \varphi dx dt \right)^{1/p} \\ & + C K^{-1+(\beta+1)p'} + C (K^{1-\beta}T^{-(1-\beta)} + KT^{-1}) \int_{\mathbb{R}^n} |u_0(x)| dx. \end{aligned}$$

Letting $T \to \infty$ and taking into consideration (26), we get

$$\int_{\mathbb{R}^n} u_1(x) dx \le C K^{-1+(\beta+1)p'}.$$

Letting $K \to \infty$ and using $p\beta < -1 \implies -1 + (\beta + 1)p' < 0$, we infer that

$$\int_{\mathbb{R}^n} u_1(x) dx \le 0.$$

This contradicts our assumption (6).

<figure><img src="image_6.png" /></figure>

## Acknowledgments

M.K. gratefully acknowledges the financial support by KUST through the grant FSU-2021-015. A.L. gratefully acknowledges the financial support by KUST through the grant FSU-2021-027.

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