Well-posedness of the solution of the fractional semilinear pseudo-parabolic equation

Abstract

This article concerns the Cauchy problem for the fractional semilinear pseudo-parabolic equation. Through the Green’s function method, we prove the pointwise convergence rate of the solution. Furthermore, using this precise pointwise structure, we introduce a Sobolev space condition with negative index on the initial data and give the nonlinear critical index for blowing up.

1 Introduction

We consider the following Cauchy problem for the fractional semilinear pseudo-parabolic equation:

$$ \textstyle\begin{cases} u_{t}+k (-\triangle )^{a}u_{t}+ (-\triangle )^{a}u=u^{p}, \quad x\in R^{n}, t>0, \\ u\vert _{t=0}=u_{0}, \end{cases} $$

(1.1)

where $p>0$, $k>0$, $u_{0} (x )$ is sufficiently smooth and nonnegative. If $k=0$, (1.1) is the classical heat equation, see [1]. If $k> 0$, (1.1) is a pseudo-parabolic equation, see [2].

The pseudo-parabolic equation is used in diverse fields such as seepage theory of homogeneous liquid through cracked rock [3] (the coefficient of the third-order term represents the degree of cracks in the rock, and its decrease corresponds to the increase in the degree of cracking), the unidirectional propagation of nonlinear dispersive long waves [4, 5] (where u is amplitude or curl), and the description of racial migration [6] (where u is the population density). Because of the wide range of applications of pseudo-parabolic equations, they attract great attention of mathematicians.

Ting, Showalter, and Gopala Rao proved the existence and uniqueness of the solution on the initial boundary value problem and the Cauchy problem of linear pseudo-parabolic equations, see [2, 7, 8]. Since then, many scholars have paid great attention to the study of nonlinear pseudo-parabolic equations, including about existence, asymptotic behavior, decay of regularity and solutions, etc., see [9–12]. Recently, Yang Cao et al. proved the existence and blowing up of the solution of equation (1.1) at $a=1$, but the pointwise estimation of the solution was not discussed, see [13]. Later, Wang Weike et al. used the Green’s function to improve [13] and p only needs to satisfy $p>1+{ \frac{2}{n+s}}$ instead of $p>1+{ \frac{2}{n}}$. More specifically, they proved the pointwise estimation of the solution of equation (1.1) at $a=1$, also obtained the nonlinear critical index of the blowing up at $p>1+{ \frac{2}{n+s}}$ by limiting the initial condition, see [14].

In the above research, they focused on integer order equations. The fractional dissipation operator $(-\triangle )^{a}$ can be regarded as the infinitesimal generators of Levy stable diffusion process. Compared with the integral differential equation, it can describe some physical phenomena more accurately, see [15–17]. Therefore, more and more scientists are devoted to the research of fractional differential equations, see [16–18].

Motivated by the above works, we study the pointwise estimate and exponential decay of the solution for problem (1.1) in the fractional order case. At present, there is little research on the pointwise estimate and exponential decay of the solution of this fractional equation, and the main difficulty stems from its fractional dissipation operator term. The structure of this article is organized as follows: In Sect. 2, we recall some preliminary results and show the main results of this paper. In Sect. 3, by Green’s function method, we use the Green’s function to express the solution of fractional equation (1.1) and get the pointwise estimate result of the Green’s function. In Sect. 4, we obtain the pointwise estimate of fractional equation (1.1) with appropriate conditions p, $u_{0}$. In Sect. 5, we prove that the exponential decay of equation (1.1) still exists without $a=1$.

2 Preliminaries and main results

Let C represent a generic positive constant, which may change from line to line. The norm of $L^{p} (\varOmega )$ is written as ${ \Vert \cdot \Vert }_{L^{p} (\varOmega )}$ ($1\leq p \leq \infty$). The notation X is a Banach space with a norm ${ \Vert \cdot \Vert }_{X}$.

Definition 2.1

Suppose $f (x,t )\in L^{1} (R^{n} )$. Then the Fourier transform is as follows:

$$ \widehat{f} (\xi ,t )= \int _{R^{n}}f (x,t )e^{- \sqrt{-1}x\cdot \xi } \,\mathrm{d}x, $$

(2.1)

its inverse Fourier transform is

$$ \bigl(\mathcal{F}^{-1}\widehat{f} \bigr) (x,t )= (2 \pi )^{-n} \int _{R^{n}}\widehat{f} (\xi ,t )e^{\sqrt{-1}x \cdot \xi } \,\mathrm{d}\xi . $$

(2.2)

According to [19], we have the following two lemmas.

Lemma 2.1

If $\widehat{f} (\xi ,t )$has a compact support for ξ satisfying

$$ \bigl\vert D_{\xi }^{\beta } \bigl(\xi ^{\alpha } \widehat{f} (\xi ,t ) \bigr) \bigr\vert \leq C \bigl( \vert \xi \vert ^{{ ( \vert \alpha \vert +k- \vert \beta \vert )}_{+}}+ \vert \xi \vert ^{ ( \vert \alpha \vert +k )}t^{\frac{ \vert \beta \vert }{2}} \bigr) \bigl(1+t \vert \xi \vert ^{2} \bigr)^{m}e^{-b \vert \xi \vert ^{2}t}, $$

(2.3)

where $b>0$, α, β are any multi-indexes and $\vert \beta \vert \leq 2N$, then

$$ \bigl\vert D_{x}^{\alpha }f (x,t ) \bigr\vert \leq C_{N}t^{-n+ \vert \alpha \vert +\frac{k}{2}}B_{N} \bigl( \vert x \vert ,t \bigr), $$

(2.4)

where k and m are any positive integers, ${ (a )}_{+}=\max (0,a )$, and

$$ B_{N} \bigl( \vert x \vert ,t \bigr)= \biggl(1+ \frac{ \vert x \vert ^{2}}{1+t} \biggr)^{-N}. $$

Lemma 2.2

Let $\operatorname{supp}f (\xi )\subset O_{R}=: \{ \xi , \vert \xi \vert >R \}$and

$$ \bigl\vert D_{\xi }^{\beta }\widehat{f} (\xi ) \bigr\vert \leq C \vert \xi \vert ^{-1- \vert \beta \vert }, $$

(2.5)

then there exist distributions $f_{1} (x )$and $f_{2} (x )$satisfying

$$ f (x )=f_{1} (x )+f_{2} (x )+C_{0} \delta (x ), $$

(2.6)

where $C_{0}$is a constant and $\delta (x )$is the Dirac function. Furthermore, choosing $\varepsilon _{0}$small enough, we have the estimate

$$ \begin{gathered} \bigl\vert D_{x}^{\alpha }f_{1} (x ) \bigr\vert \leq C \bigl(1+ \vert x \vert ^{2} \bigr)^{-N}, \\ { \Vert f_{2} \Vert }_{L_{1}}\leq C,\qquad \operatorname{supp}f_{2} (x )\subset \bigl\{ x; \vert x \vert < 2 \varepsilon _{0} \bigr\} ,\end{gathered} $$

for positive integer $2N>n+ \vert \alpha \vert$.

We make the following assumptions:

$(H1 )$:

$u_{0}\in C^{\alpha +2} (R^{n} )$ for sufficiently small $u_{0}>0$;

$(H2 )$:

$u_{0}\in W^{-s,2} (R^{n} )\cap W^{-s, \infty } (R^{n} )\cap L^{\infty } (R^{n} )\cap L^{2} (R^{n} )$, $0\leq s< n$, for sufficiently small $u_{0}>0$.

Based on the above assumptions, we draw the following conclusions.

Theorem 1

Let $p>p_{c}=1+{ \frac{2a}{n}}$, $(H1 )$be satisfied. Then Cauchy problem (1.1) has the pointwise estimate of the solution u, satisfying

$$ \begin{gathered} \vert u \vert \leq 2C (1+t )^{-\frac{n}{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N}, \\ \bigl\vert D_{x}^{\alpha }u \bigr\vert \leq 2C (1+t )^{-\frac{n}{2a}- \frac{ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N},\end{gathered} $$

(2.7)

where $N>0$is an arbitrary constant, C depends on the initial value $u_{0}$and the parameter p.

Theorem 2

Suppose $p>p_{s}=1+{ \frac{2a}{n+s}}$, $(H2 )$hold. Then problem (1.1) has solution u satisfying

$$ { \Vert u \Vert }_{L^{q} (R^{n} )}\leq C (1+t )^{-\frac{n}{2a} (1-\frac{1}{q} )-\frac{s}{2a}}, \quad q\in {[} 2, +\infty ), $$

(2.8)

where C depends on the initial value data and p.

3 Pointwise estimate of the Green’s function

In this section, we will consider the pointwise estimation of the solution to linear form of problem (1.1). We study the Green’s function of Cauchy problem (1.1) and obtain the following:

$$ \textstyle\begin{cases} \partial _{t}G+k (-\triangle )^{a} G_{t}+ (- \triangle )^{a} G=0,\quad x\in R^{n}, t>0, \\ G\vert _{t=0}=\delta (x ),\end{cases} $$

(3.1)

where $\delta (x )=\delta (x_{1} )\otimes \delta (x_{2} )\otimes \cdots \otimes \delta (x_{n} )$ is the Dirac function and ⊗ represents the tensor product. Considering the Fourier transform of equation (3.1) with respect to x, we get

$$ \textstyle\begin{cases} \partial _{t}\widehat{G}+k \vert \xi \vert ^{2a}{\widehat{G}}_{t}+ \vert \xi \vert ^{2a}\widehat{G}=0,\quad \xi \in R^{n}, \\ \widehat{G}\vert _{t=0}=1.\end{cases} $$

(3.2)

By solving the above equation directly, we know that

$$ \widehat{G} (\xi ,t )=e^{\mu (\xi )t}, $$

(3.3)

where $\mu (\xi )=- \frac{ \vert \xi \vert ^{2a}}{1+k \vert \xi \vert ^{2a}}$. Now we use frequency decomposition to obtain an estimate of the Green’s function G. Let

$$\begin{aligned}& \chi _{1} (\xi )= \textstyle\begin{cases} 1,& \vert \xi \vert \leq \varepsilon , \\ 0,& \vert \xi \vert >2\varepsilon ,\end{cases}\displaystyle \quad \in C^{\infty }, \end{aligned}$$

(3.4)

$$\begin{aligned}& \chi _{3} (\xi )= \textstyle\begin{cases} 1,& \vert \xi \vert \geq R, \\ 0,& \vert \xi \vert < R-1,\end{cases}\displaystyle \quad \in C^{\infty }, \end{aligned}$$

(3.5)

$$\begin{aligned}& \chi _{2} (\xi )=1-\chi _{1} (\xi )-\chi _{3} (\xi ),\quad \in C^{\infty }, \end{aligned}$$

(3.6)

where $\chi _{1} (\xi )$ and $\chi _{3} (\xi )$ are the smooth cut-off functions, ε, R are any positive constants satisfying $2\varepsilon < R-1$. Define $\widehat{G_{i}} (t,\xi )=\chi _{i}\widehat{G} (t, \xi )$, $i=1,2,3$. From the literature [20], we know that the attenuation of the solution of the linear problem is mainly related to the low frequency part of $\widehat{G} (t,\xi )$. We use cut-off functions to divide the solution into three parts: low frequency, intermediate frequency, and high frequency.

Proposition 3.1

Let ε be a sufficiently small constant. Then there exists a constant $C>0$satisfying

$$ \bigl\vert D_{x}^{\alpha }G_{1} (x,t ) \bigr\vert \leq C_{N} (1+t )^{-\frac{n+ \vert \alpha \vert }{2a}}B_{N} \bigl( \vert x \vert ^{a},t \bigr). $$

(3.7)

Proof

In the case of low frequency, let $0< \vert \xi \vert <2R$, then Ĝ has compact support. Taking into account (3.3) and Lemma 2.1, there is

$$ \bigl\vert D_{\xi }^{\beta }\xi ^{\alpha } \chi _{1} (\xi )\widehat{G} (\xi ,t ) \bigr\vert \leq C \bigl( \vert \xi \vert ^{{ ( \vert \alpha \vert - \vert \beta \vert )}_{+}}+ \vert \xi \vert ^{ \vert \alpha \vert }t^{\frac{ \vert \beta \vert }{2a}} \bigr) \bigl(1+t \vert \xi \vert ^{2a} \bigr)^{\beta }e^{-b \vert \xi \vert ^{2a}t} $$

(3.8)

for $\forall \vert \beta \vert \leq 2N$. Then is (3.7) established. □

Actually, we can discuss obtaining (3.7) in two cases.

(1) If $\vert \beta \vert \leq \vert \alpha \vert$, we find

$$ \begin{aligned}[b] \bigl\vert x^{\beta }D_{x}^{\alpha } \chi _{1} (D )G \bigr\vert &=C \biggl\vert \int e^{ix\xi }D_{\xi }^{\beta } \bigl(\xi ^{\alpha }\chi _{1} (\xi )\widehat{G} (\xi ,t ) \bigr) \,\mathrm{d} \xi \biggr\vert \\ & \leq C \int \bigl( \vert \xi \vert ^{{ ( \vert \alpha \vert - \vert \beta \vert )}_{+}}+ \vert \xi \vert ^{ \vert \alpha \vert }t^{\frac{ \vert \beta \vert }{2a}} \bigr) \bigl(1+t \vert \xi \vert ^{2a} \bigr)^{\beta } \bigl\vert e^{ix\xi } \bigr\vert e^{-b \vert \xi \vert ^{2a}t} \,\mathrm{d}\xi \\ & \leq C \int \vert \xi \vert ^{ \vert \alpha \vert } \bigl( \vert \xi \vert ^{-{ \vert \beta \vert }_{+}}+t^{\frac{ \vert \beta \vert }{2a}} \bigr) \bigl(1+t \vert \xi \vert ^{2a} \bigr)^{\beta }e^{-b \vert \xi \vert ^{2a}t} \,\mathrm{d}\xi \\ & \leq Ct^{- \frac{n+ \vert \alpha \vert - \vert \beta \vert }{2a}}.\end{aligned} $$

(3.9)

(2) If $\vert \beta \vert > \vert \alpha \vert$, we obtain

$$ \begin{aligned}[b] \bigl\vert x^{\beta }D_{x}^{\alpha } \chi _{1} (D )G \bigr\vert &\leq C \bigl(1+t^{-\frac{ \vert \alpha \vert - \vert \beta \vert }{2a}} \bigr)t^{-\frac{n}{2a}} \\ & \leq C (1+t )^{-\frac{ \vert \alpha \vert - \vert \beta \vert }{2a}}t^{- \frac{n}{2a}} \\ & \leq C (1+t )^{\frac{ \vert \beta \vert }{2a}}t^{- \frac{n+ \vert \alpha \vert }{2a}}.\end{aligned} $$

(3.10)

On the other hand, if $\vert x \vert ^{2a}\leq 1+t$, let $\vert \beta \vert =0$, we have

$$ \bigl\vert D_{x}^{\alpha }\chi _{1} (D )G \bigr\vert \leq Ct^{- \frac{n+ \vert \alpha \vert }{2a}}. $$

(3.11)

If $\vert x \vert ^{2a}>1+t$, let $\vert \beta \vert =2aN$, we see that

$$ \bigl\vert D_{x}^{\alpha }\chi _{1} (D )G \bigr\vert \leq Ct^{- \frac{n+ \vert \alpha \vert }{2a}}\min \biggl(1, \frac{ (1+t )^{N}}{ \vert x \vert ^{2aN}} \biggr). $$

(3.12)

Because

$$ 1+\frac{ \vert x \vert ^{2a}}{1+t}\leq \textstyle\begin{cases} 2,& \vert x \vert ^{2a}\leq 1+t, \\ 2\frac{ \vert x \vert ^{2a}}{1+t},& \vert x \vert ^{2a}>1+t,\end{cases} $$

(3.13)

we know that

$$ \min \biggl(1, \frac{ (1+t )^{N}}{ \vert x \vert ^{2aN}} \biggr)\leq 2^{N}B_{N} \bigl( \vert x \vert ^{a},t \bigr). $$

(3.14)

Then, with the help of (3.11)–(3.12) and (3.14), we infer (3.7).

Next, we estimate $G_{2} (x,t )$.

Proposition 3.2

Suppose that ε and R are fixed constants. Then

$$ \bigl\vert D_{x}^{\alpha }G_{2} (x,t ) \bigr\vert \leq Ce^{- \frac{t}{2m_{0}}}B_{N} \bigl( \vert x \vert ^{a},t \bigr), $$

(3.15)

where $m_{0}$is a positive constant.

Proof

Choosing m sufficiently large and $m>\frac{1}{2} (\frac{1}{ \vert \varepsilon \vert ^{2a}}+k )$. If $\varepsilon \leq \vert \xi \vert \leq R$, it is easy to see that $\mu (\xi )\leq -\frac{1}{2m}$. This analysis reflects that

$$ \vert \widehat{G_{2}} \vert = \bigl\vert \chi _{2} (\xi ) \widehat{G} \bigr\vert \leq Ce^{-\frac{t}{2m}}. $$

(3.16)

From (3.16), there holds

$$ \begin{aligned}[b] \bigl\vert D_{x}^{\alpha }G_{2} (x,t ) \bigr\vert &\leq C \biggl\vert \int _{ \varepsilon \leq \xi \leq R}e^{ix\xi } \bigl(\xi ^{\alpha } \widehat{G_{2}} (\xi ,t ) \bigr) \,\mathrm{d}\xi \biggr\vert \\ & \leq Ce^{-\frac{t}{2m}} \int _{ \varepsilon \leq \xi \leq R} \vert \xi \vert ^{\alpha } \,\mathrm{d}\xi \\ & \leq Ce^{-\frac{t}{2m}}.\end{aligned} $$

(3.17)

Now, we apply mathematical induction to prove the following inequality:

$$ \bigl\vert D_{\xi }^{\beta } \widehat{G_{2}} (\xi ,t ) \bigr\vert \leq C (1+t )^{ \vert \beta \vert }e^{-\frac{t}{2m}}. $$

(3.18)

Obviously, the above formula holds when $\vert \beta \vert =0$. Suppose that $\vert \beta \vert \leq l-1$, the above formula still holds. Then we will prove that the formula of (3.18) also holds for $\vert \beta \vert \leq l$. Taking the Fourier transform of (3.1) with respect to x and multiplying it with $\chi _{2} (\xi )$, we get

$$ \textstyle\begin{cases} \partial _{t}\widehat{G_{2}} (\xi ,t )-\mu (\xi )\widehat{G_{2}} (\xi ,t )=0, \\ \widehat{G_{2}} (\xi ,0 )=\chi _{2} (\xi ).\end{cases} $$

(3.19)

Using $D_{\xi }^{\beta }$ to equation (3.19), we can obtain that

$$ \textstyle\begin{cases} \partial _{t}D_{\xi }^{\beta }\widehat{G_{2}} (\xi ,t )-\mu (\xi )D_{\xi }^{\beta }\widehat{G_{2}} (\xi ,t )=F (\xi ,t ), \\ \widehat{G} (\xi ,0 )=a_{0},\end{cases} $$

(3.20)

where $F (\xi ,t )=\sum_{\beta _{1}+\beta _{2}=\beta , \vert \beta _{1} \vert \neq 0}\frac{\beta !}{\beta _{1}!\beta _{2}!} (D_{\xi }^{\beta _{1}}\mu (\xi )D_{\xi }^{\beta _{2}}\widehat{G_{2}} (\xi ,t ) )$ and $a_{0}$ is a polynomial of $\vert \xi \vert$. Considering $\vert \beta \vert =l$, we see that

$$ D_{\xi }^{\beta }\widehat{G_{2}} (\xi ,t )=a_{0}\widehat{G_{2}} (\xi ,t )+ \int _{0}^{t}\widehat{G_{2}} (\xi ,t-s )F (\xi ) \,\mathrm{d}s. $$

(3.21)

By induction we have

$$ \begin{aligned}[b] \bigl\vert D_{\xi }^{\beta } \widehat{G_{2}} (\xi ,t ) \bigr\vert &\leq Ce^{- \frac{t}{2m}}+C \int _{0}^{t}e^{-\frac{t-s}{2m}} (1+t )^{ \vert \beta \vert -1}e^{-\frac{s}{2m}} \,\mathrm{d}s \\ & \leq C (1+t )^{ \vert \beta \vert }e^{-\frac{t}{2m}}.\end{aligned} $$

(3.22)

Therefore, for each $1\leq \vert \beta \vert \leq l$, we find

$$ \begin{aligned}[b] \bigl\vert x^{\beta }D_{x}^{\alpha }G_{2} (x,t ) \bigr\vert &\leq C \biggl\vert \int _{R^{n}}e^{ix\xi }D_{\xi }^{\beta } \bigl(\xi ^{\alpha }\widehat{G_{2}} (\xi ,t ) \bigr) \,\mathrm{d}s \biggr\vert \\ & \leq Ce^{- \frac{t}{2m}} (1+t )^{ \vert \beta \vert } \int _{ \varepsilon \leq \vert \beta \vert \leq R} \vert \xi \vert ^{ \vert \alpha \vert }+ \vert \xi \vert ^{ \vert \vert \alpha \vert - \vert \beta \vert \vert } \,\mathrm{d}\xi \\ & \leq Ce^{- \frac{t}{2m}} (1+t )^{ \vert \beta \vert } \\ & \leq Ce^{- \frac{t}{2m_{0}}} (1+t )^{\frac{ \vert \beta \vert }{2a}},\end{aligned} $$

(3.23)

where $m_{0}\in (0, m )$. Taking into account (3.17) and (3.23) (let $\vert \beta \vert =2N$), we deduce

$$ \bigl\vert D_{x}^{\alpha }G_{2} (x,t ) \bigr\vert \leq Ce^{- \frac{t}{2m_{0}}}\min \biggl(1, \frac{{(1+t)}^{2N}}{{\vert x\vert }^{2N}} \biggr). $$

(3.24)

Considering (3.14) and (3.24), we obtain the desired result. □

Now considering the high frequency part $G_{3} (x,t )$.

Proposition 3.3

Let R be a sufficiently small constant. Then

$$ \bigl\vert D_{x}^{\alpha } \bigl(G_{3} (x,t )-F_{l} \bigr) \bigr\vert \leq C (1+t )^{- \frac{n+ \vert \alpha \vert }{2a}}B_{N} \bigl( \vert x \vert ^{a},t \bigr), $$

(3.25)

where b is a positive constant and

$$ F_{l}=\chi _{3} (D ) \Biggl[e^{-\frac{t}{k}} \Biggl(\delta (x )+\sum_{j=1}^{l}p_{j} (t ) (- \triangle )^{-aj} \Biggr) \Biggr],\quad l= \frac{ \vert \alpha \vert +n}{2a}, $$

(3.26)

is the distribution.

Proof

Denote

$$ \rho =\frac{1}{ \vert \xi \vert ^{2a}},\qquad h (\rho )=e^{- \frac{t}{ \vert \rho \vert +k}}. $$

(3.27)

Expanded in Taylor’s series at $\vert \rho \vert \rightarrow 0$, we infer

$$ h (\rho )=e^{-\frac{t}{k}} \biggl(1+\frac{t\rho }{k^{2}}+ \frac{t^{2}\rho ^{2}}{2k^{4}}-\frac{t\rho ^{2}}{2k^{3}}+\cdots \biggr). $$

(3.28)

Then

$$ e^{-\frac{t \vert \xi \vert ^{2a}}{1+k \vert \xi \vert ^{2a}}}=e^{- \frac{t}{k}} \Biggl(1+\sum _{j=1}^{l}p_{j} (t ) \bigl( \vert \xi \vert ^{2a} \bigr)^{-j} \Biggr)\chi _{3} (\xi )+R (\xi ,t ), $$

(3.29)

where $p_{j} (t )$ is a polynomial of degree j.

Let

$$ F_{l}=\chi _{3} (D ) \Biggl[e^{-\frac{t}{k}} \Biggl(\delta (x )+\sum_{j=1}^{l}p_{j} (t ) (- \triangle )^{-aj} \Biggr) \Biggr],\quad l= \frac{ \vert \alpha \vert +n}{2a}. $$

(3.30)

With the help of Lemma 2.2 and choosing R big enough, it is easy to see that

$$\begin{aligned} \bigl\vert D_{x}^{\alpha } \bigl(G_{3} (x,t )-F_{l} \bigr) \bigr\vert \leq C (1+t )^{- \frac{n+ \vert \alpha \vert }{2a}}B_{N} \bigl( \vert x \vert ^{a},t \bigr). \end{aligned}$$

(3.31)

 □

In conclusion, we use the following lemma to explain the estimate of the regular part of G.

Lemma 3.1

Let G be the solution of the linear form of Cauchy problem (1.1). Then

$$ \bigl\vert D_{x}^{\alpha } \bigl(G (x,t )-F_{l} \bigr) \bigr\vert \leq C (1+t )^{-\frac{n+ \vert \alpha \vert }{2a}}B_{N} \bigl( \vert x \vert ^{a},t \bigr), $$

(3.32)

where $F_{l}$is the distribution and

$$ F_{l}=\chi _{3} (D ) \Biggl[e^{-\frac{t}{k}} \Biggl(\delta (x )+\sum_{j=1}^{l}p_{j} (t ) (- \triangle )^{-aj} \Biggr) \Biggr],\quad l= \frac{ \vert \alpha \vert +n}{2a}.$$

(3.33)

Proof

Note that

$$ \begin{aligned}[b] \bigl\vert D_{x}^{\alpha } \bigl(G (x,t )-F_{l} \bigr) (x,t ) \bigr\vert &\leq \bigl\vert D_{x}^{\alpha }G_{1} (x,t ) \bigr\vert + \bigl\vert D_{x}^{\alpha }G_{2} (x,t ) \bigr\vert \\ &\quad{} + \bigl\vert D_{x}^{\alpha } \bigl(G_{3} (x,t )-F_{l} \bigr) (x,t ) \bigr\vert .\end{aligned} $$

(3.34)

Considering Proposition 3.1, Proposition 3.2, Proposition 3.3, and (3.34), we have (3.32). Then Lemma 3.1 is proved. □

Lemma 3.2

Assume that $t>0$, $m>\frac{n}{2a}$, then

$$ \int _{R^{n}} \biggl(1+\frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-m} \,\mathrm{d}x\leq C (1+t )^{\frac{n}{2a}}. $$

(3.35)

Proof

Let $w= \vert x \vert$, we infer

$$\begin{aligned}& \begin{aligned}[b] \int _{R^{n}} \biggl(1+\frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-m} \,\mathrm{d}x&\leq C \int _{0}^{\infty } \biggl(1+\frac{w^{2a}}{1+t} \biggr)^{-m}w^{n-1} \,\mathrm{d}w \\ & \leq C (1+t )^{\frac{n}{2a}} \int _{0}^{\infty } \biggl(1+ \frac{w}{ (1+t )^{\frac{1}{2a}}} \biggr)^{-2am+n-1} \,\mathrm{d}\frac{w}{ (1+t )^{\frac{1}{2a}}} \\ & \leq C (1+t )^{\frac{n}{2a}}. \end{aligned} \end{aligned}$$

(3.36)

 □

Using Hausdorff–Young’s inequality, the following lemma of Green’s function is easily obtained.

Lemma 3.3

If $p\in {[} 1,\infty ]$, then

$$ { \bigl\Vert D_{x}^{\alpha } (G-F_{l} ) (t,\cdot ) \bigr\Vert }_{L^{p} (R^{n} )}\leq C (1+t )^{- \frac{n}{2a} (1-\frac{1}{p} )-\frac{ \vert \alpha \vert }{2a}} $$

(3.37)

for any multi-indexes α.

Proof

If $p\in {[} 1,\infty )$, it follows from (3.25) that

$$ \begin{aligned}[b] { \bigl\Vert D_{x}^{\alpha } (G-F_{l} ) (t,\cdot ) \bigr\Vert }_{L^{p} (R^{n} )}&= \biggl\vert \int _{R^{n}} \bigl\vert D_{x}^{\alpha } (G-F_{l} ) (t,\cdot ) \bigr\vert ^{p} \,\mathrm{d}x \biggr\vert ^{\frac{1}{p}} \\ & \leq C (1+t )^{- \frac{n+ \vert \alpha \vert }{2a}} \biggl( \int _{R^{n}} \bigl\vert B_{N} \bigl( \vert x \vert ^{a},t \bigr) \bigr\vert ^{p} \,\mathrm{d}x \biggr)^{\frac{1}{p}} \\ & \leq C (1+t )^{-\frac{n}{2a} (1-\frac{1}{p} )-\frac{ \vert \alpha \vert }{2a}}.\end{aligned} $$

(3.38)

If $p=\infty$, by (3.25), we conclude that

$$\begin{aligned}& \begin{aligned}[b] { \bigl\Vert D_{x}^{\alpha } (G-F_{l} ) (t,\cdot ) \bigr\Vert }_{L^{p} (R^{n} )}&= \operatorname{ess}\sup \bigl\vert D_{x}^{\alpha } (G-F_{l} ) (t,\cdot ) \bigr\vert \\ & \leq \operatorname{ess} \sup \bigl\{ C (1+t )^{- \frac{n+ \vert \alpha \vert }{2a}}B_{N} \bigl( \vert x \vert ^{a},t \bigr) \bigr\} \\ & \leq C (1+t )^{-\frac{n}{2a} (1-\frac{1}{p} )-\frac{ \vert \alpha \vert }{2a}}.\end{aligned} \end{aligned}$$

(3.39)

 □

4 Pointwise estimation of the solution

In this section, we get the pointwise estimate of the solution under appropriate conditions of $u_{0}$, p.

Lemma 4.1

Assume that $\vert y \vert \leq M$, $t\geq 4M^{2}$, $N>0$, then

$$ \biggl(1+\frac{ \vert y-x \vert ^{2a}}{1+t} \biggr)^{-N}\leq C_{N} \biggl(1+\frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N}. $$

(4.1)

Proof

(1) If $\vert x \vert ^{2a}\leq 1+t$, then

$$ \frac{ \vert x \vert ^{2a}}{1+t}\leq 1 $$

(4.2)

and

$$ \biggl(1+\frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N}\geq 2^{-N}. $$

(4.3)

On the other hand,

$$ \biggl(1+\frac{ \vert x-y \vert ^{2a}}{1+t} \biggr)^{-N}\leq 1. $$

(4.4)

It follows from (4.3) and (4.4) that

$$ \biggl(1+\frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N}\leq 2^{N} \biggl(1+\frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N}. $$

(4.5)

(2) If $\vert x \vert ^{2a}>1+t$, then

$$ \vert x-y \vert \geq \vert x \vert - \vert y \vert \geq \sqrt{1-t}- \vert y \vert \geq 0 $$

(4.6)

and

$$ \begin{aligned}[b] \vert x-y \vert ^{2a}&\geq \bigl( \vert x \vert - \vert y \vert \bigr)^{2a}= \bigl( \vert x \vert ^{2}-2 \vert x \vert \vert y \vert + \vert y \vert ^{2} \bigr)^{a} \\ & = \biggl(\frac{ \vert x \vert ^{2}}{2}+ \frac{ ( \vert x \vert -2 \vert y \vert )^{2}}{2}- \vert y \vert ^{2} \biggr)^{a}\geq \biggl(\frac{ \vert x \vert ^{2}}{4}+ \biggl( \frac{t}{4}-M^{2} \biggr) \biggr)^{a}.\end{aligned} $$

(4.7)

From $\frac{t}{4}-M^{2}\geq 0$, the proof is obtained. □

Proof of Theorem 1

With the help of Lemma 3.3 and considering (1.1), from Green’s function, we have

$$ \begin{aligned}[b] u&=\widehat{T}u=G (t )\ast u_{0}+ \int _{0}^{t}H (t-s, \cdot )\ast F \bigl(u (s, \cdot ) \bigr) \,\mathrm{d}s \\ & =G (t )\ast u_{0}+ \int _{0}^{t} \int _{R^{n}}H (x-y,t-s )\ast F \bigl(u (y,s ) \bigr) \,\mathrm{d}y \,\mathrm{d}s,\end{aligned} $$

(4.8)

where the symbol ∗ represents convolution, $F (u )=u^{p}$, and H satisfies

$$ \widehat{H}=\frac{1}{1+k \vert \xi \vert ^{2a}}\widehat{G}. $$

(4.9)

Applying the inverse Fourier transform, we deduce

$$ H=K_{k}\ast G,\qquad K_{k} (x )= (4\pi )^{- \frac{n}{2a}} \int _{0}^{\infty }e^{-\gamma _{1}s- \frac{ \vert x \vert ^{2a}}{4s}}s^{-\frac{n}{2a}} \,\mathrm{d}s. $$

(4.10)

Obviously, the estimated value on Ĝ is also correct on Ĥ.

By using $D_{x}^{\alpha }$ to equation (4.8), we get

$$ \begin{aligned}[b] D_{x}^{\alpha }u (x,t )&=D_{x}^{\alpha }G\ast u_{0}+ \int _{0}^{t} \int _{R^{n}}H (x-y,t-s )D_{x}^{\alpha }F \bigl(u (y,s ) \bigr) \,\mathrm{d}y \,\mathrm{d}s \\ & =D_{x}^{\alpha }G\ast u_{0}+ \int _{0}^{t} \int _{R^{n}}D_{x}^{\alpha }H (x-y,t-s )F \bigl(u (y,s ) \bigr) \,\mathrm{d}y \,\mathrm{d}s.\end{aligned} $$

(4.11)

Define

$$ \begin{gathered} \phi = (1+t )^{-\frac{n}{2a}}B_{N} \bigl( \vert x \vert ^{a},t \bigr), \\ \phi _{\alpha }= (1+t )^{- \frac{n+ \vert \alpha \vert }{2a}}B_{N} \bigl( \vert x \vert ^{a},t \bigr), \\ M (t )=\underset{0\leq s\leq t,x\in R}{\sup } \bigl\vert u (x,s ) \bigr\vert \phi ^{-1} (x,s ).\end{gathered} $$

(4.12)

Then

$$ \begin{gathered} u\leq M\cdot \phi , \\ u (x,t )\leq G\ast u_{0}+M^{p} \int _{0}^{t} \int _{R^{n}}H (x-y,t-s )\phi ^{p} (y,s ) \,\mathrm{d}y \,\mathrm{d}s, \\ D_{x}^{\alpha }u (x,t )\leq D_{x}^{\alpha }G \ast u_{0}+M^{p} \int _{0}^{t} \int _{R^{n}}D_{x}^{\alpha }H (x-y,t-s )\phi ^{p} (y,s ) \,\mathrm{d}y \,\mathrm{d}s.\end{gathered} $$

(4.13)

First of all, we discuss the singular part. By $(H2 )$, following [20], we have

$$ \begin{aligned}[b] \bigl\vert D_{x}^{\alpha }F_{l} \ast u_{0} \bigr\vert &\leq C\varepsilon e^{- \frac{t}{2ak}} \bigl(1+ \vert x \vert ^{2a} \bigr)^{-N} \\ & \leq C\varepsilon (1+t )^{-\frac{n+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N} \\ & \leq C\varepsilon \phi _{\alpha }.\end{aligned} $$

(4.14)

Secondly, we consider the nonsingular part. Since $\vert u_{0} \vert \leq (1+ \vert y \vert ^{2a} )^{-N}$, $\operatorname{supp} u_{0}\subset \{ \vert y \vert \leq M \}$, according to the definition of tight support, we can know $u_{0}$ has a compact support. If t is large enough, we find

$$ \begin{aligned}[b] \bigl\vert D_{x}^{\alpha } (G-F_{l} )\ast u_{0} \bigr\vert &\leq C (1+t )^{-\frac{n+ \vert \alpha \vert }{2a}} \int _{R^{n}} \biggl(1+\frac{ \vert y-x \vert ^{2a}}{1+t} \biggr)^{-N} \biggl(1+ \frac{ \vert y \vert ^{2a}}{1+t} \biggr)^{-N} \,\mathrm{d}y \\ & \leq C (1+t )^{-\frac{n+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N} \int _{R^{n}} \biggl(1+ \frac{ \vert y \vert ^{2a}}{1+t} \biggr)^{-N} \,\mathrm{d}y \\ & \leq C \varepsilon \phi _{\alpha }.\end{aligned} $$

(4.15)

Then

$$ \bigl\vert D_{x}^{\alpha }G\ast u_{0} \bigr\vert \leq C\varepsilon \phi _{\alpha } . $$

(4.16)

Here we still divide the nonlinear term into a singular part and a nonsingular part. Define

$$ \begin{aligned}[b] \int _{0}^{t} \int _{R^{n}}D_{x}^{\alpha }H (x-y,t-s ) \phi ^{p} (y,s ) \,\mathrm{d}y \,\mathrm{d}s&\leq \int _{0}^{t} \int _{R^{n}}D_{x}^{\alpha } (G-F_{l} )\phi ^{p} (y,s ) \,\mathrm{d}y \,\mathrm{d}s \\ &\quad{} + \int _{0}^{t} \int _{R^{n}}D_{x}^{\alpha }F_{l} \phi ^{p} (y,s ) \,\mathrm{d}y \,\mathrm{d}s \\ & =\psi _{1,1}+ \psi _{1,2},\end{aligned} $$

(4.17)

where $\psi _{1,1}=\int _{0}^{t}\int _{R^{n}}D_{x}^{\alpha } (G-F_{l} )\phi ^{p} (y,s ) \,\mathrm{d}y \,\mathrm{d}s$, $\psi _{1,2}=\int _{0}^{t}\int _{R^{n}}D_{x}^{\alpha }F_{l}\phi ^{p} (y,s ) \,\mathrm{d}y \,\mathrm{d}s$.

Estimating the nonsingular part. Recalling Lemma 3.1, it shows that

$$ \begin{aligned}[b] \psi _{1,1}&\leq C \int _{0}^{t} \int _{R^{n}} (1+t-s )^{- \frac{n+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x-y \vert ^{2a}}{1+t-s} \biggr)^{-N} \\ &\quad {}\times (1+s )^{-\frac{np}{2a}} \biggl(1+ \frac{ \vert y \vert ^{2a}}{1+s} \biggr)^{-Np} \,\mathrm{d}y \,\mathrm{d}s.\end{aligned} $$

(4.18)

Denote $\varOmega = [0,t ]\times R^{n}$, $\varOmega ^{1}=\varOmega \cap \{ s\geq \frac{t}{2} \}$, $\varOmega ^{2}=\varOmega \cap \{ s \leq \frac{t}{2} \}$. We will discuss it in two cases:

(1) If $\vert x \vert ^{2a}\leq 1+t$, then

$$\begin{aligned} \psi _{1,1}&\leq C \int _{\varOmega ^{1}} (1+t-\tau )^{- \frac{n+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x-y \vert ^{2a}}{1+t-\tau } \biggr)^{-N} (1+\tau )^{-\frac{np}{2a}} \\ &\quad {}\times \biggl(1+\frac{ \vert y \vert ^{2a}}{1+\tau } \biggr)^{-Np} \,\mathrm{d}y \,\mathrm{d}\tau +C \int _{\varOmega ^{2}} (1+t-\tau )^{-\frac{n}{2a}} \biggl(1+ \frac{ \vert x-y \vert ^{2a}}{1+t-\tau } \biggr)^{-N} \\ &\quad {}\times (1+\tau )^{- \frac{np+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert y \vert ^{2a}}{1+\tau } \biggr)^{-Np} \,\mathrm{d}y \,\mathrm{d}\tau \\ & \leq C \int _{\frac{t}{2}}^{t} (1+t-\tau )^{- \frac{n+ \vert \alpha \vert }{2a}} (1+\tau )^{- \frac{np}{2a}} (1+\tau )^{\frac{n}{2a}} \,\mathrm{d}\tau \\ &\quad {} +C \int _{0}^{\frac{t}{2}} (1+t-\tau )^{- \frac{n}{2a}} (1+\tau )^{- \frac{np+ \vert \alpha \vert }{2a}} (1+\tau )^{\frac{n}{2a}} \,\mathrm{d}\tau \\ & \leq C (1+t )^{-\frac{n+np-n}{2a}+1- \frac{ \vert \alpha \vert }{2a}} \\ & \leq C (1+t )^{-\frac{np}{2a}+1- \frac{ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-n}. \end{aligned}$$

(4.19)

(2) If $\vert x \vert ^{2a}>1+t$, then

$$ \begin{aligned}[b] \psi _{1,1}&\leq C \int _{\varOmega ^{1}} (1+t-\tau )^{- \frac{n+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x-y \vert ^{2a}}{1+t-\tau } \biggr)^{-N} (1+\tau )^{-\frac{np}{2a}} \\ &\quad {}\times \biggl(1+\frac{ \vert y \vert ^{2a}}{1+\tau } \biggr)^{-Np} \,\mathrm{d}y \,\mathrm{d}\tau +C \int _{\varOmega ^{2}} (1+t-\tau )^{-\frac{n}{2a}} \biggl(1+ \frac{ \vert x-y \vert ^{2a}}{1+t-\tau } \biggr)^{-N} \\ &\quad {}\times (1+\tau )^{- \frac{np+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert y \vert ^{2a}}{1+\tau } \biggr)^{-Np} \,\mathrm{d}y \,\mathrm{d}\tau \\ & \leq C \int _{\varOmega ^{1}} (1+t-\tau )^{- \frac{n+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t-\tau } \biggr)^{-N} (1+\tau )^{-\frac{np}{2a}} \\ &\quad {}\times \biggl(1+\frac{ \vert y \vert ^{2a}}{1+\tau } \biggr)^{-Np} \,\mathrm{d}y \,\mathrm{d}\tau +C \int _{\varOmega ^{2}} (1+t-\tau )^{-\frac{n}{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t-\tau } \biggr)^{-N} \\ &\quad {}\times (1+\tau )^{- \frac{np+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+\tau } \biggr)^{-Np} \,\mathrm{d}y \,\mathrm{d}\tau \\ & \leq C \biggl(1+\frac{ \vert x \vert ^{2a}}{1+\tau } \biggr)^{-n} \int _{\frac{t}{2}}^{t} (1+t-\tau )^{- \frac{n+ \vert \alpha \vert }{2a}} (1+\tau )^{- \frac{np}{2a}} (1+\tau )^{\frac{n}{2a}} \,\mathrm{d}\tau \\ &\quad{} +C \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+\tau } \biggr)^{-n} \int _{0}^{\frac{t}{2}} (1+t-\tau )^{-\frac{n}{2a}} (1+\tau )^{- \frac{np+ \vert \alpha \vert }{2a}} (1+\tau )^{\frac{n}{2a}} \,\mathrm{d}\tau \\ & \leq C (1+t )^{-\frac{np}{2a}+1- \frac{ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-n}.\end{aligned} $$

(4.20)

From (4.19)–(4.20), we can get

$$ \psi _{1,1}\leq C (1+t )^{-\frac{np}{2a}+1- \frac{ \vert \alpha \vert }{2a}}B_{N} \bigl( \vert x \vert ^{a},t \bigr). $$

(4.21)

Due to $p>1+\frac{2a}{n}$, we deduce

$$ (1+t )^{-\frac{np}{2a}+1- \frac{ \vert \alpha \vert }{2a}}\leq (1+t )^{- \frac{n}{2a}-\frac{ \vert \alpha \vert }{2a}}. $$

(4.22)

Estimating the singular part. By [19], we know that

$$\begin{aligned} \psi _{1,2}&= \int _{0}^{t} \int _{R^{n}}D_{x}^{\alpha }F_{l} \phi ^{p} (y,s ) \,\mathrm{d}y \,\mathrm{d}s \\ & \leq \int _{0}^{t}e^{-\frac{t}{2ka}}\phi _{\alpha }^{p} \,\mathrm{d}s \\ & \leq C\phi _{\alpha }. \end{aligned}$$

(4.23)

Coming back to the whole solution, we find

$$\begin{aligned}& \begin{aligned}[b] \bigl\vert u (x,t ) \bigr\vert &\leq \vert G \ast u_{0} \vert + \biggl\vert \int _{0}^{t} \int _{R^{n}}H (x-y,t-s )F \bigl(u (y,s ) \bigr) \,\mathrm{d}y \,\mathrm{d}s \biggr\vert \\ & \leq C \bigl(\varepsilon +M^{p} \bigr) \phi , \end{aligned} \\ & \\ & \bigl\vert D_{x}^{\alpha }u (x,t ) \bigr\vert \leq C \bigl( \varepsilon +M^{p} \bigr)\phi _{\alpha }. \end{aligned}$$

(4.24)

Therefore,

$$ M\leq C \bigl(\varepsilon +M^{p} \bigr). $$

(4.25)

Since $M (0 )\leq C\varepsilon$, applying the continuity method, we get that

$$ M (t )\leq 2C\varepsilon ,\quad \forall t\in {[} 0,+ \infty ). $$

(4.26)

By the above inequality, we have

$$ \begin{gathered} \bigl\vert D_{x}^{\alpha }u \bigr\vert \leq M\phi _{\alpha }\leq 2C\varepsilon \phi _{\alpha }, \\ \vert u \vert \leq 2C\varepsilon (1+t )^{-\frac{n}{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N},\\ \bigl\vert D_{x}^{\alpha }u \bigr\vert \leq 2C\varepsilon (1+t )^{- \frac{n+ \vert \alpha \vert }{2a}} \biggl(1+ \frac{ \vert x \vert ^{2a}}{1+t} \biggr)^{-N}.\end{gathered} $$

(4.27)

 □

5 Improvement of the initial data

In this section we consider the Cauchy problem of (1.1). It shows that the limit of the parameter p can be weaker when the initial conditions become stronger. Since we have known the proof of the existence and uniqueness of the solution, we will not discuss it. Only for attenuation of the decay estimate.

Definition 5.1

Suppose that operator $\varLambda ^{s}$, $s\in R^{n}$, satisfies

$$ \varLambda ^{s}f (x )= (2\pi )^{-n} \int _{R^{n}} \vert \xi \vert ^{s}\widehat{f} (\xi )e^{\sqrt{-1}x \cdot \xi } \,\mathrm{d}\xi , $$

(5.1)

and Sobolev space $W^{-s,p}$ indicates

$$ W^{-s,p}= \bigl\{ f|{ \bigl\Vert \varLambda ^{-s}f \bigr\Vert }_{L^{p} (R^{n} )}\leq C \bigr\} . $$

(5.2)

Lemma 5.1

Let $0< s< n$, $1< p< q<\infty$, $\frac{1}{q}+\frac{s}{n}=\frac{1}{p}$, then

$$ { \bigl\Vert \varLambda ^{-s}f \bigr\Vert }_{L^{q} (R^{n} )}\leq C_{p,q}{ \Vert f \Vert }_{L^{q} (R^{n} )}, $$

(5.3)

where $C_{p,q}$is a constant depending on p, q.

Through [21, p. 99, Theorem 1], Lemma 5.1 can be proved.

Proof of Theorem 2

We divide it into a singular part and a nonsingular part:

$$ u=G\ast u_{0}+ \int _{0}^{t}H\ast u^{p} \,\mathrm{d}\tau =u_{N}+u_{S}, $$

(5.4)

where

$$ \begin{gathered} u_{N}= (G-F_{l} )\ast u_{0}+ \int _{0}^{t} \frac{G-F_{l}}{1+k (-\triangle )^{a}}\ast u^{p} \,\mathrm{d}\tau , \\ u_{S}=F_{l}\ast u_{0}+ \int _{0}^{t} \frac{F_{l}}{1+k (-\triangle )^{a}}\ast u^{p} \,\mathrm{d}\tau .\end{gathered} $$

(5.5)

Now, let us start from the nonsingular part, we obtain

$$ \begin{aligned}[b] u_{N}&= (G-F_{l} )\ast u_{0}+ \int _{0}^{t} \frac{G-F_{l}}{1+k (-\triangle )^{a}}\ast u^{p} \,\mathrm{d}\tau \\ & =\varLambda ^{s} (G-F_{l} )\ast \varLambda ^{-s}u_{0}+ \int _{0}^{t}\varLambda ^{s} \frac{G-F_{l}}{1+k (-\triangle )^{a}}\ast \varLambda ^{-s}u^{p} \,\mathrm{d}\tau .\end{aligned} $$

(5.6)

Using Young’s inequality, we have

$$ \begin{aligned}[b] { \Vert u_{N} \Vert }_{L^{q} (R^{n} )}&\leq { \bigl\Vert \varLambda ^{s} (G-F_{l} )\ast \varLambda ^{-s}u_{0} \bigr\Vert }_{L^{q} (R^{n} )} \\ &\quad{} +{ \biggl\Vert \int _{0}^{t} \varLambda ^{s} (G-F_{l} )\ast \varLambda ^{-s}u^{p} \,\mathrm{d}\tau \biggr\Vert }_{L^{q} (R^{n} )} \\ & \leq { \bigl\Vert \varLambda ^{s} (G-F_{l} )\ast \varLambda ^{-s}u_{0} \bigr\Vert }_{L^{q} (R^{n} )} \\ &\quad{} + \int _{0}^{t}{ \bigl\Vert \varLambda ^{s} (G-F_{l} ) \bigr\Vert }_{L^{\frac{nq}{n+q}} (R^{n} )}{ \bigl\Vert \varLambda ^{-s}u^{p} \bigr\Vert }_{L^{\frac{n}{n-1}} (R^{n} )} \,\mathrm{d}\tau .\end{aligned} $$

(5.7)

Based on the initial condition and estimate of the linear part, we see that

$$ { \bigl\Vert \varLambda ^{s} (G-F_{l} ) \bigr\Vert }_{L^{q} (R^{n} )}\leq C (1+t )^{-\frac{n}{2a} (1-\frac{1}{q} )-\frac{s}{2a}} $$

(5.8)

and

$$ { \bigl\Vert \varLambda ^{s} (G-F_{l} )\ast \varLambda ^{-s}u_{0} \bigr\Vert }_{L^{q} (R^{n} )}\leq C (1+t )^{- \frac{n}{2a} (1-\frac{1}{q} )-\frac{s}{2a}}. $$

(5.9)

Recalling Lemma 5.1, there holds

$$ { \bigl\Vert \varLambda ^{-s}u^{p} \bigr\Vert }_{L^{\frac{n}{n-1}} (R^{n} )}\leq C \Vert u \Vert _{L^{\frac{np}{n+s-1}} (R^{n} )}^{p} \leq C \bigl[ (1+t )^{-\frac{n}{2a} (1- \frac{n+s-1}{np} )-\frac{s}{2a}} \bigr]^{p}. $$

(5.10)

Taking into account (5.8) and (5.10), we get

$$ \begin{aligned}[b] & \int _{0}^{t}{ \bigl\Vert \varLambda ^{s} (G-F_{l} ) \bigr\Vert }_{L^{\frac{nq}{n+q}} (R^{n} )}{ \bigl\Vert \varLambda ^{-s}u^{p} \bigr\Vert }_{L^{\frac{n}{n-1}} (R^{n} )} \,\mathrm{d}s \\ &\quad \leq C \int _{0}^{t} (1+t-s )^{-\frac{n}{2a} (1- \frac{n+q}{nq} )-\frac{s}{2a}} \bigl[ (1+t )^{- \frac{n}{2a} (1-\frac{n+s-1}{np} )-\frac{s}{2a}} \bigr]^{p} \,\mathrm{d}s \\ &\quad \leq C (1+t )^{-\frac{n}{2a} (1-\frac{1}{q} )- \frac{s}{2a}}.\end{aligned} $$

(5.11)

Considering the singular part, we infer

$$ { \Vert F_{l}\ast u_{0} \Vert }_{L^{q} (R^{n} )}\leq Ce^{- \frac{t}{2ak}}{ \Vert u_{0} \Vert }_{L^{q} (R^{n} )} \leq C\varepsilon e^{-\frac{t}{2ak}}$$

(5.12)

and

$$ \begin{aligned}[b] \int _{0}^{t}{ \biggl\Vert \frac{F_{l}}{1+k (-\triangle )^{a}} \ast u^{p} \biggr\Vert }_{L^{q} (R^{n} )} \,\mathrm{d}\tau &\leq C \int _{0}^{t}e^{-\frac{t-\tau }{2ak}} \Vert u \Vert _{{L^{p,q} (R^{n} )}}^{p} \,\mathrm{d}\tau \\ & \leq C \int _{0}^{t}e^{-\frac{t-\tau }{2ak}} (1+\tau )^{-\frac{n}{2a} (1-\frac{1}{pq} )p} \,\mathrm{d}\tau \\ & \leq C (1+t )^{-\frac{n}{2a} (1- \frac{1}{q} )-\frac{s}{2a}}.\end{aligned} $$

(5.13)

With the help of (5.9) and (5.11)–(5.13), we obtain

$$ { \Vert u \Vert }_{L^{q} (R^{n} )}\leq C (1+t )^{-\frac{n}{2a} (1-\frac{1}{q} )-\frac{s}{2a}}. $$

(5.14)

 □